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Concentration of solution when a battery reaches equilibrium
Problem: We want to build a battery employing as electrodes ${\text{PbSO}_4}_{(s)} | {\text{Pb}}_{(s)}\ (E^º=-0.351\ V)$ i ${\text{Cd}}_{(aq)}^{2+} | {\text{Cd}}_{(s)}\ (E^º=-0.4026\ V)$ in a solution of ${\text{CdSO}_4}$ of molar concentration M. At what concentration of ${\text{CdSO}_4}$ will the battery reach equilibrium?
Data: $T=298.15$ K; $P=1$ atm; $F=96500\ C/mole\ e^{-}$.
My attempt:
There's reduction happening on the ${\text{PbSO}_4}_{(s)} | {\text{Pb}}_{(s)}$ electrode and oxidation on ${\text{Cd}}_{(aq)}^{2+} | {\text{Cd}}_{(s)}$. Thus, the global reaction should be
${\text{Cd}}_{(s)} + {\text{PbSO}_4}_{(s)} \rightleftarrows {\text{Pb}}_{(s)} + {\text{Cd}}_{(aq)}^{2+} + {\text{SO}_4}_{(aq)}^{2-}$
I'll use Nernst's equation:
$E=E^º-\dfrac{RT}{nF}\ln{Q}$
At equilibrium:
$0=E^º-\dfrac{RT}{nF}\ln{K}$
So $K=[{\text{CdSO}_4}]=\exp\left({\dfrac{nFE^º}{RT}}\right)=\exp\left({\dfrac{2\cdot 96500\cdot ((-0.351)-(-0.4026))}{8.31\cdot 298.15}}\right)\approx 55.67$ M.
The thing is this is not the actual answer. I don't know what else can I do...

Take care ! $\ce{K = [Cd^{2+}]·[SO4^{2-}] = [Cd^{2+}]^2}$

That actually makes sense. Thanks. The actual answer should be $7.5$ M approx.

Technically  $K$ is  dimensionless, because the exponential is dimensionless. The way around this is to divide the concentrations by 1 mol/litre each and so  $\ce{K=[Cd^{2+}]^2/(1 M^{2} )}$.

Note that at ion concentrations usable for galvanic cells, concentrations are far out from the range where we can afford to consider activity coefficients to be 1 and therefore usual calculations are strongly off the track.

So the activity of $\ce{Cd^{2+}}$ is $\sqrt{55.67} = 7.461$. It is not far from the expected result $7.5$ M.


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How does red iron oxide (Fe2O3) increase the burn rate of sugar rocket fuel?
My question is similar to  this one, however that question seems to ask about how it improves combustion in general, not specifically with respect to sugar rocket fuel.
The answer to this question states:

Now molecular oxygen is by far a better oxidisation agent than Fe2O3 however in fine powdered form, these iron oxide molecules are in immediate contact with your OM; we do not need to wait for molecular oxygen to be used up and for more to diffuse in.

If I am understanding correctly, this is suggesting that the iron oxide acts as an oxidation agent when molecular oxygen is not available. However, in sugar rocket fuel, there should be potassium nitrate in direct contact with the sucrose molecules (especially because the fuel mix was dissolved in water before heating), so I feel like this point is invalidated.
This is what I think the reactions is for the oxidation using potassium nitrate:
$\ce{48KNO_3 + 5C_12H_22O_11 -> 24K_2CO_3 + 36CO_2 + 55H_2O + 24N2}$
And for iron (iii) oxide:
$\ce{24Fe_2O_3 + C_12H_22O_11 -> 48FeO + 12CO_2 + 11H_2O}$
Is the increase in burn rate due to a lower activation energy or a faster reaction rate for the reaction with the iron oxide than the reaction with the potassium nitrate? Or is it that only 9.6 molecules of potassium nitrate are consumed for every sucrose molecule, while 24 molecules of iron oxide are consumed for every sucrose molecule?

Maybe is it simply a catalytic effect of $\ce{Fe2O3}$ ? It reminds me of another experiment. If you try to burn sugar cube with a match or with a flame, it does not work. Sugar is melting and transformed into caramel but does not burn : no flame !. But if, before burning, you rub the surface of the cube with ashes, it burns properly when touched by a flame. This is a catalytic effect.

@Maurice How does the catalytic effect of iron oxide work?

I don't know how the catalytic effect of ashes or iron oxide on sugar burning works.

@Maurice do you know how I would be able to find that out?

I don't know how and where to find out this information. I don't even know if the explanation of the catalytic effect has been published somewhere.

@Maurice OK. Thank you for your help.

source: Stack Exchange

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In hydroboration oxidation reaction .. why do not we use sodium hydroxide alone as a nuclophile?
Why do not we use sodium hydroxide alone as a nuclophile instead of using hydrogen peroxide along with sodium hydroxide

It converts the borate ester into alcohol. So it's important part of the reaction.

Care to expand on how it does this? A mechanism diagram showing the key steps would be good.

The combination of hydroxide and peroxide is essential for the oxidation of boranes to alcohol, both components are needed. Firstly the hydroxide deprotonates peroxide to give peroxide anion  - a good nucleophile - which attacks the empty orbital of the R-BH2 borane giving a negatively charged boron species. Then the pair of electrons in the C–B bond migrates to oxygen, leading to breakage of C–B and formation of C–O, along with rupture of the O–O bond. Finally hydroxide ion attacks the empty p orbital of boron, the O–B bond breaks and the alkoxide anion picks up a proton from water to give the product alcohol.
Images from here

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Activation Energy in silicon etching
I'm trying to understand what does activation energy means when we're talking about etching. Here's what I understand so far:

activation energy is the energy required by the system to undergo a chemical reaction. This means we need to input this much energy into the system for the chemical bonds to break (etch).
Higher activation energy means etching species have less energy and more thermal energy is needed for etching to occur.

Here's where I'm stumped. It's well known that (111) Si planes etch slower than (100) Si planes (I'm looking at KOH specifically but it goes for most other etchants too). BUT somehow Si(100) has a higher activation energy than (111). If the second point I mentioned above is correct, it should be the other way around i.e. (111) should have higher activation energy.
I'm also confused with why would the activation energy of a particular plane change with crystallographic orientation. One of the papers I read mentioned that Si(111) has a different activation energy for (100) orientation versus (110) orientation (see image below). They did provide a possible explanation for it but I don't seem to understand it fully. I think what they mean is that for (110) surface orientation of the silicon wafer, the number of atoms of the Si(111) plane exposed to the etchant is different than the number of (111) atoms exposed when the surface orientation is (100). Am I correct?

A link to your source would be a good thing. One thing to remember is that thermodynamics and kinetics are two different things. Further, looking at the diamond cubic crystal structure, the 111 planes (the close packing stacking direction) have the fewest unsatisfied bonds on the surface, and both 100 and 110 have ledges allowing for fast etching by kinks running along them.

Here's the link to the paper I cited above but I've read similar things in other papers too..https://doi.org/10.1016/0924-4247(94)00854-B

What you said makes sense and that was my understanding of the Anisotropy too. What doesn't make sense to me is the activation energy of (111) being low. Since (111) has fewest unsatisfied bonds it makes sense that it's harder to remove since it's more stable than (100) or (110) in a way. And activation energy is the amount of energy you need to ENTER into the system to remove atoms. So shouldn't (111) have higher activation energy than the other two planes?

Well, one issue is that, should you look layer-by-layer at the 111 bonding, there are two very different unsatisfied bond densities (either 1 or 3 per atom). No consider that the observed 'activation energy' may be some average value, not directly attributable to one single step.

From my understanding of etching, (111) has 1 unsatisfied bond, since the etching molecules attack the top layer.


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Should nitric acid be classified as a strong acid? Does it completely dissociate in water?
By definition, a strong acid must completely ionize in aqueous solution. However, in order to completely dissociate in water, strong acids must be more acidic than a hydronium ion and hence have a pKa < -1.74.
I have always thought of nitric acid as a strong acid, however, nitric acid has a pKa value of -1.5, which technically should indicate that it is less acidic than the hydronium ion and hence cannot completely dissociate in water like a strong acid does, right?
Could someone please explain whether nitric acid dissociates completely in water or not? There seems to be a lot of vagueness surrounding these definitions and classifications and it seems to me like they contradict each other.

We are really splitting hairs here. As described in this answer with respect to bases, whether an acid or base is "strong" or "weak" is not cut and dried because the dissociation constant varies continuously across different materials and the apparent behavior may depend on the type of reaction.
Nitric acid does have a lower dissociation constant than hydrochloric or sulfuric acid, but that constant is still strong enough to produce about 97% dissociation at one molar concentration (90% dissociation at one molar would correspond to $pK_a\approx-1$). So in dilute solutions there is no real difference between whether you call nitric acid "strong" or "weak" because of that (to me, dubious) $-1.74$ boundary. And at higher concentrations the reactivity of nitric acid is going to be dominated by its oxidizing power instead of its acidity anyway.


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Claim that the temperature of steam is not the boiling point on a stovetop?
Came across this claim here that "The liquid water is at exactly the boiling point temperature. The water vapor (steam) can be hotter. You can have a thermometer immersed in the water which will show the boiling point, then pull it up out of the water and the temperature rises."
Why wouldn't the temperature of steam be the boiling point?

In everyday life the term boiling generally means the nucleation and growth of bubbles of steam. Typically in a saucepan or a kettle the water is heated from below and the bubbles of steam form on the bottom of pan, or on the element at the bottom of a kettle, then grow, break free and rise to the surface.
The reason for the steam being hotter than 100°C is that there is an energy barrier to nucleation and to nucleate a bubble of steam we have to superheat the steam. So if you were able to measure the temperature inside a bubble of steam as it formed at the bottom of the pan you would indeed measure a temperature above 100°C.
However I am unconvinced that the temperature would rise as you pull the thermometer upwards from the bottom of the pan. The upper surface is being rapidly cooled by evaporation and I would expect the bubbles of steam to cool as they rise and approach a temperature equal to the water surrounding them as they reach the surface i.e. 100°C. Hence I would expect a bubble of steam to start out hotter than 100°C at the bottom of the pan but to have cooled to 100°C but the time it reached the surface.
As for the temperature of the surrounding water, I would expect convection currents to provide rapid enough mixing that there was little or no temperature gradient in the water. However I have to concede that I have never done the experiment.

Hm, so the heat energy does not first go towards converting all the water to steam before further increasing the temperature of the system? What if I was to boil water in the absence of gravity, where the steam didn't need to form a bubble and rise, would then the temperature of steam be 100?

@xasthor Formation of a bubble has an energy barrier whether the bubble is forming in a gravitational field or not. Have a look at my answer to Why does soda fizz when it meets ice? for more on this. That question is about a CO₂ bubble forming in soda rather than a stem bubble forming in water, but the principle is the same.

@xasthor The issue is that heat is transferred from the bottom only. A microwave in outer space might help with your model.

@JohnRennie Okay. So why is it taught that after the temperature hits 100°, the heat goes towards phase change until the last drop, after which it goes again towards increasing the temperature of the steam. In this case, it appears that heat is simultaneously going towards superheating the steam as well as phase change. Is it a good approximation of what's happening?

@xasthor we tend to simplify things when we first teach students about phase changes and we don't mention superheating (or super cooling). In real life we almost always get some degree of heating (or cooling) past the transition temperature before the phase change occurs.

@JohnRennie Is it at least a good approximation?

@xasthor In most cases it's a reasonable approximation, but the amount of superheating can be surprisingly large. See for example Why droplets of water under oil explode when heated?

@JohnRennie Is there any way of estimating what the error would be?

@xasthor The amount of superheating or supercooling depends on the system and it's hard to make general rules. I have seen water supercooled to -40°C, but that was under very carefully controlled conditions.

Ironically that wasn't academic research. It was when I worked at Unilever as an industrial scientist and the research was being done by the part of the company that makes ice cream. You'd be surprised how much interesting research goes into apparently simple things like ice cream.

@JohnRennie Interesting, haha

There is a saying that "Paper never refused ink", the same goes for the web. At 1 atm  pressure, if you boil water, and water vapor exist at equilibrium and the vapor pressure of water is also 1 atm at 100 centigrade. The temperature of water and steam will be the same. Having a different temperature of steam than bulk water defies the concept of equilibrium temperature. The quora author appears to be confused between the concept of latent heat and temperature.
As the graph shows, from Hyperphysics, as long we have water, the temperature of bulk water cannot exceed 100 centigrade. Although heat is being added its temperature is not increasing. However, one can independently heat steam from another heat source and its temperature can be way higher than boiling water.

It is hard to be at equilibrium in the real world. When you boil water in a surrounding that is at room temperature, you are never at thermal equilibrium. There will be some superheated liquid and some supercooled steam (in need of nuclei), and probably also some liquid below the boiling point (near the sides of the pot) and some steam above boiling point (near the bottom of the pot). It would be different if the temperature of the air is near the boiling point of water, or the entire room is filled with steam (like an autoclave).

Right Karsten, I agree but chemists live in an ideal world.

Your answer appears to be saying something different from the other posted answer here. Could you have a look at it?

@xasthor, The other answer is talking about microscopic temperature. I do not have any proof or studies handy which show that a water vapor bubble inside the bulk of water is at higher temperature than bulk of water. What one can guarantee is that the bulk temperature of water cannot exceed 100 C during boiling at 1 atm. I don't think the bubble temperature is any higher than 100 C during boiling.

Now there are certain bubble (or better cavitation) phenomena where the temperature inside the bubble in water is 5000 C or more!! That is another world. Check cavitation and sonoluminescence but it has nothing to do with your query about normal boiling process.

@AChem Agreed, the water temperature won't rise above 100°, but what is being claimed is the steam temperature will be above 100 in the other answer. This is what was claimed by the original person I linked as well.

@xasthor, I do not agree with the conjecture that steam is at higher temperature in boiling water. To be specific, I mean the water vapor above the bulk surface.

What's being claimed by him too is that the bubbles cool by the time they reach the surface, but that initially they start out at a higher temperature. What do you think about this?

If they cool by the time they reach the surface, there will be no boiling i.e., the bubble will collapse. When water is boiling vigorously, you have tons, of rising bubbles, and if the local temperature of the bubbles at the bottom were higher than bulk water, the water temperature would have certainly risen above 100 C. This is never the case. I think this is close to hair splitting (this could be an interesting serious research).  The query is local variation in temperature at bubble points.


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Putting salt and ice in a can causes frost to form, while doing this again without salt will only cause condensation, why?
If this is the wrong place to ask my question, then blame my friend, and also tell me where to post this since I'm new. Anyway, we did an experiment in class where we took 2 soda cans with no top. We filled both 3/4 the way full with ice but only put salt in one of them. The one with salt formed frost while the one without salt on formed condensation. It wasn't really explained very well though. My theory is that the salt melts the ice, but the melted liquid is still below freezing. Since it is liquid it gets good contact with the can cooling it down to below freezing, causing frost. With the one with no salt, I think the solid ice can't get good contact with the can, so it won't be able to cool to below freezing. The liquid also is below freezing. This means only condensation will form. Now, I would still expect the ice to just move so that way it is making contact with the can cooling it down until it is below freezing, making frost. Why doesn't that happen? Thanks for any answers!

https://en.wikipedia.org/wiki/Freezing-point_depression

Melting ice occurs at $0$°C, and stays at this temperature. It will never be colder. But a mixture of melting ice and salt is always at a temperature lower than $0$°C. The minimum temperature obtained with a precise mixture is -$21$°C. Usually, without special care, the temperature is between $0$°C and -$21$°C. If some mixture ice + salt is in a soda can, condensation of atmospheric water vapor on the outer part of the can produces frost (or ice), because its temperature is under $0$°c. If the can contains only ice plus water (no salt), its temperature is never lower than $0$°C; the condensation of the humid air can only produce liquid water droplets, and no frost.

How does the ice remain frozen if it isn't below 0 degrees Celsius?

A mixture of pure water + ice remains at $0$°C as long as one of these two phases do coexist. This mixture is even the definition of zero degree Celsius.

Oh, that makes sense! Thanks!

Wait, so why does frost not form in the non-salt can since the ice should cool it down to below freezing?

Pure ice is and stays at zero degree if water is present. It cannot be cooler than $0$°C. If the temperature of the mixture ice + water goes down to <$0$°C, the water will be totally transformed into ice.

Oh, that makes sense, thanks!


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Destination of carbonyl compounds free toluene
I am currently in the process of trying to get toluene free of carbonyl compounds using a distillation processes with DNPH.
Unfortunately I am no distillation expert. And am running into some problems.
The method is take one liter of toluene and 4g of dnph and 0.5 ml of HCl. all technical grade. Reflux for 2 hours and distill. Discard the first 50 ml.
DNPH has a amount of water. That's I assume you need to discard the first 50ml
Apparently when distilling you need a fractional column. Don't know why but assume that compound with a boiling point that differs from the clean toluene.
Now the problem is I do not get clean toluene. I think this has something to do with the boiling point of hydrazine. This  has almost the same boiling point as toluene. How would one get better separation in a distillation process?

If you mean DNPH for 2,4-dinitrophenylhydrazine (which is solid), its boiling point is way higher than that of toluene. Also, use of that in distillation process raised safety hazards.

You are correct it is 2,4 dinitrophenylhydrazine. I know it's a safety concern. But this is how it's described in the unclear method. Dnph does dissolve at a certain temperature in toluene. Does  dissolve poorly in tolueen at room temperature. Needed to find that out as well. Any guesses on how to solve the problem? It's like destilling de last 4 procent of water out of ethanol I believe.

Ensure not to distill to the dryness when DNPH present.

Yes according to the method I need to discard 50 ml in the beginning and need to distill to about  75% of the volume added  at beginning of the distillation


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Calculating total fluorescence based on thin slices in a slab
I am modeling an acid-base reaction that includes a fluorescence indicator. My simple model is a slab of acid next to a slab of base and so on. These slabs are very thin, typically less than 10 microns and are assumed to be laminar such that diffusion is the only method of mass transport. At the interface of acid and base the reaction occurs and the pH is 10^-7 (which is shown as zero on the attached figure).
The attached figure shows an example of the concentration gradients of acid and base at some time t. Because of symmetry only a half slab of acid is shown next to a half slab of base. I want to calculate the total fluorescence emitted from both half slabs together. With regards to the concentrations on the figure, for the acid slab the concentration is the hydrogen ion concentration.  For the base slab the concentration is the hydroxyl ion concentration. I have an equation which gives me the fluorescence as a function of pH.
So the way I have been calculating the fluorescence is to get the pH for each slice shown in the figure and assume that the pH is constant for that slice (of course as the slice thickness is decreased this becomes a better assumption). Then the fluorescence is calculated for each slice based on the known pH and the equation.
Now comes my question: If I want to calculate the total fluorescence for both slabs, can I just simply add up the fluorescence from each slice in the acid slab and base slab? In other words, the total fluorescence emitted from both slabs together is the sum of the individual fluorescence for each slice. Is this the correct methodology?

Is the fluorescent emitter the same in both acidic and basic solutions ? Why do you think the neutral mixture is not fluorescent ?

Yes, I am using umbelliferone in both the acid and base and the concentrations are the same, so there is no diffusion gradient for the umbelliferone.  I did not mean to mislead regarding the pH7.  A pH of 7 is definitely fluorescent.  I was just showing the concentration gradients and that bit of information is more important for the analytical model than for my question.  Sorry to confuse and thanks for bringing up the point about possible diffusion of the umbelliferone.  Any thoughts on my question regarding adding up the slice fluorescences to get the total for both half slabs?

What is the meaning of the two curves  c(x,t) going to zero in the middle of the diagram ? Are the curves theoretical or experimental ? Why are there no numerical values on the Ox axis ? Why is the time not mentioned ?

You say you have fluorescence yield vs pH, (f(pH) )so the total will be weighted according to this, so you have f(pH)Conc(acid)+f(pH)Conc(base) = total  fluorescence so you seem not to have enough information, one measurement but two values you need. But I'm not clear about what you are actually doing

The figure is just meant to provide an idea of the concentration profiles at a specific time. Where the curves go to zero is the reaction plan of acid and base.  The mathematical model assumes that acid and base cannot coexist and sets this value as zero.  However, as mentioned above, the value is actually 10-7, which for all practical purposes is zero and does not affect any model calculation determining the pH of the curves. The curves are theoretical.  On the x axis the values can be listed as -L on the left side and +L on the right side. L is the half slab thickness.

@porphyrin:  Let me make sure I understand what you are saying.  You are saying that I just add the calculated fluorescence value for the base slab to the calculated fluorescence value for the acid slab.  If so, that is what I have been doing.   However, in each slab the pH is not constant; hence the slices for each slab. The pH/fluorescence equation that I have is based on a thickness, sm.   I am calculating the pH and then fluorescence for each slice where it is assumed that these values are constant within the thin slice.  Comment continues..........

The pH for a slice is calculated from Fick's diffusion model (attached diagram shows the concentration profiles (or alternatively, pH profiles) at a specific time, t). Then the equation is used to get the fluorescence at that pH.  The equation, however, provides a fluorescence value for a known thickness, sm. So to get the fluorescence for the slice, multiply the calculated fluorescence by the slice thickness, dx, divided by the thickness, sm.  The foregoing is accomplished for all slices in both half slabs.  Comment continues below.....

Finally, all of the slice fluorescences are summed to get the total fluorescence for the two half slabs, which is a thickness of 2L (see my previous comment on half slab thickness).  Please let me know if this is not what you mean.

Your method seems fine, as long as your excitation light source is strong enough in intensity.  You don't want the light absorption by any of your slices to significantly affect the excitation intensity in the layer below.  Also, you might want to make sure that the absorbance of your "slices" for the emitted light is negligible too.  If both of these assumptions are met, then each slice really is independent of the adjacent ones, and you can probably add up all the fluorescence to get the total

@Curt F.  The procedure I outlined above is for a specific time.  My simulation actually runs for about a thousand iterations or so, so from time, t = 0 to t = tfinal.  My simulated total fluorescence as a function of time profile is then compared to the experimental one.  I guess values of the slab thickness until the profiles match.  For about 25% of my data, no correction value is needed in order to get the profiles to match (at a specific value of L). However, for the other 75%, the simulation seems to understimate the fluorescence when compared to the experimental fluorescence.

But I find that if I multiply all of the simulated fluorescence values (values at each time step) by a single constant value, then the profiles will match reasonably well.  So a single correction factor applied to the entire simulation fluorescence curve will result in it matching the experimental curve. I have been puzzled by why this is the case.  Perhaps it is due to the cautions that you outline in your post.  Unfortunately, I don't feel that I have a full understanding of the issues that you bring up.

You appear to be interested in monitoring pH gradients across the glass slab based on diffusion of an acid and base by monitoring the variation in fluorescence intensity or emission maximum shifts. Before you proceed further on this fancy experiment you will have to establish certain fluorescence facts. Hence control experiments are extremely necessary.
A quick survey on Google Scholar will show that umbelliferone emission is pH dependent. I am sure you will have access to a good quality spectrofluorometer. Make sure you collect emission spectrum at the pH range of interest with identical concentrations of the dye in various pH solutions. By identical concentration of the dye in each pH solution, I mean prepare dye solutions as accurately as analytically possible. One could be prepare a stock solution of the dye and perhaps weigh the volume transferred just to be very accurate in each pH solution.
Once you collect emission spectra (fixed excitation), the emission maxima will/might change. The key observation here is that do emission intensities also change or not as a function of pH?
When the fluorescence intensity and emission are changing, it implies that we are not dealing with identical species of the dye, thus different "species" have different fluorescence quantum yields. It is then perhaps not kosher to add up fluorescence intensities. Alternatively, you might try even a fancier experiment and measure quantum yields as a function of pH.
Of course, the measuring instrument is oblivious to these things, it is just counting photons from the region you are collecting light from. In short, the experiment is not trivial and the deeper we explore, the more problems will be exposed. For example, is the detector uniformly sensitive to all the wavelengths of your interest or not.

Unfortunately, I wasn't clear that the medium here is liquid. To provide more information, I am impinging an acid sheet of liquid with a base sheet of liquid and watching the fluorescence change within the combined sheet that is produced.  I have already shown that within the combined sheet, the liquid flows along radial lines (i.e. there is no crossover of liquid from one radial line to the next) and thus the attached diagram is applicable to actual conditions.  However as noted above, the diagram only shows one time. I should have attached a diagram showing profiles at different times.


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Do asymmetrical HER/HOR catalysts exist?
Suppose I have an electrochemical hydrogen pump (Wikipedia), whereby I have an anode and cathode, each with supported catalyst, separated by a proton exchange membrane. When I apply an appropriate potential, H$_2$ is effectively transported across the membrane. Now assume I do not have platinum catalysts on both electrodes, but have different catalysts on anode and cathode. What I want is that the anode catalyst strongly favours the hydrogen oxidation reaction and the cathode catalyst strongly favours the hydrogen evolution reaction. Such that if I reverse bias the cell, so that the electrode that is normally the anode becomes the cathode and vice versa, I will have a much reduced flow of hydrogen through the membrane with respect to regular polarity.
Do such catalysts allowing for such an asymmetrical cell exist? Is such a thing possible?
Thanks

catalysts can't favor one reaction direction over the other

.. as if they could, you could create catalyst perpetuum mobile by inserting the catalyst in step 1 and removing it in the step 2..

Note that there is a good habit to explain in text less usual acronyms to avoid guessing or searching.

Yes such a thing is possible.
One way to achieve this, to allow oxidation to preferably happen compared to reduction (or vice versa) one could look at semiconductor electrodes. What happens at a semiconductor electrolyte interface with an applied potential? It depends on the nature of your semiconductor (if the current is carried by holes or free electrons, the location of the band edges, the band gap etc.), typically a p-type semiconductor allows for a significant anodic current but a reduced cathodic current (and vice versa for the n-type). The electrode-electrolyte interface effectively behaves like a diode.
For more information (given I am unsure what your background is on this subject) I refer you to
http://www.chm.bris.ac.uk/pt/diamond/gustavothesis/chapter4.htm
or any general text describing electrochemistry with semiconductors would suffice.


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Burning Plastic: can it be done without harmful pollution?
Can plastic be melted, incinerated or recycled without emitting harmful gases and CO2? Specifically, can this be done without releasing gases harmful to humans?
This question pertains to the melting performed during initial product production (in molds) and later during the recycling state.
I ask because I see differing positions on whether it’s more environmentally friendly to bury vs incinerate. I’m also curious how factories that mold plastic from those pellets insure there is no air pollution. Maybe some exhaust devices capture the harmful gases. Very curious because I visited Thailand once where there were factories and it smelled like burning plastic outside.
Hoping to hear a chemistry viewpoint.

CO2 does not impact ozone layer.

Many plastics that can be safely  burnt are better to be recycled.

You cannot burn something containing carbon without producing carbon dioxide.

Even many simple plastics containing C, H and O (PET, PP, PE, PS) which might be thought to be burnable to CO2 and water are not easy to burn because of the other additives in the actual plastic. Many contain plasticisers and flame retardants and other additives that will generate much more toxic emissions when burned.

Strongly depends on the kind of plastic you burn. As long as they do not contain any elements besides carbon, hydrogen, oxygen and nitrogen and given that you provide enough oxygen to the combustion process there is not real problem and the only combustion products will be water, carbon dioxide and nitrogen.
So, plastics like PE, PP, PET, polyesters, polyamides and so on are not something to  worry about. Much more challenging are plastics that contain elements like chlorine (PVC), fluorine (PTFE, PFA) or sulfur (rubber) because they can produce toxic gases when burned.

You can't exactly "burn" PTFE. Also burning polyamides  is a terrible idea unless you have something like industrial grade furnace.

So PVC, PTFE and sulfur are the plastics the other poster here states release harmful fumes (the ones I typically hear are bad for humans to breath)?

Also for PVC, PTFE and sulfur are there capture mechanisms for the harmful gases?  I still think of my experience in Thailand where I could smell burnt plastic in the air after leaving Bangkok and wonder if it was due to manufacturing (which is why I was there).

Usually burning plastics is not recommended as it can emits harmful fumes like carbon monoxide, nitrogen oxides, carbon dioxide, heavy metals like lead, mercury, arsenic and carcinogens like dioxins, furans, polychlorinated biphenyls (PCBs), brominated polyaromatic hydrocarbons (PAHS) but there are still incinerators that burn plastics and the heat generated is converted to electricity (as a form of recycling energy). You can check below articles:

https://oceana.ca/en/blog/burning-plastic-is-not-a-recycling-solution-its-more-pollution/
https://www.nationalgeographic.com/environment/article/should-we-burn-plastic-waste
https://www.bbc.com/news/science-environment-43120041
https://flo-bro.com/burn-plastics/

However, scientists and researches are figuring out a way to burn plastics in a more greener way. In one articleref, researchers at have come up with a process named pyrolytic gasification. Instead of directly burning the plastic in open air, their reactor will heats the material to 800 °C in a completely oxygen-free environment. This causes the plastic to become a gas, which is then mixed with air before it is burned as a clean fuel free of oxygen-free organics like dioxins and furan.
The patent-pending process has one more advantage: when the researchers introduced a bit of stainless steel into the reactor, they found it acted as a catalyst for growing uniform carbon nanotubes which can be used in other applications.
Ref.: https://phys.org/news/2013-12-plastic-cleanly-natural-gas.html

Any comments on the initial process of filling molds for products?  Is there some sort of capture device for the gases, especially with PVC, PTFE and sulfur?


text:
I have been able to make sulfuric acid from alum using a membrane cell. What else other than alum can I use to improve yields
Here is how I was able to make sulfuric acid.
A membrane cell was made by placing a earthen clay pot in a larger vessel.
1.Make concentrated solution of potash alum.
2.Pour alum sol into both sides of the cell.
4. Using carbon electrodes applied a voltage of 14V.
The cell was run for 4-5 hr.
Sulfuric acid was formed on the anode, and potassium and al hydroxide was formed on the cathode side.
Tested with sodium bicarbonate to give of effervescence. No other tests were done, but what other acid can form if not sulfuric acid?
I am planning to repeat it to increase the yields, like substituting alum for other sulphate salts, thinner membrane, adding hydrogen peroxide on the anode side.
What else can I do to improve the yields?

All solutions of sulfates will produce sulfuric acid in your cell. At the anode (positive pole), the electrolysis will produce $\ce{H2O ->  O2  +  4 H+  +  4 e-}$. So the solution becomes acidic, due to $\ce{H2SO4}$ appearing in solution. But it contains more of the original sulfate. At the cathode, the reaction will be $\ce{2 H2O + 2 e- -> H2 + 2 OH-}$. So the cathodic solution becomes basic. The only way of improving the yield is to increase the duration of the electrolysis. Anyway you will never obtain pure sulfuric acid. You will always obtain a mixture sulfate + sulfuric acid.

Ok so I have to distill it then?
Also I did this a long time back and I think I only kept the sulfate solution on the cathode side only, the other side had distilled water.The SO4(2-) must have traversed the membrane to the other side. Have to repeat if this was what I had done.

If one of the side is pure water, no current will flow through the cell. Pure water does not conduct electricity.

Distillation is no use for purifying sulfuric acid solution. Sulfuric acid is not volatil. Heating such a solution will simply eliminate water. Sulfate and sulfuric acid will not be separated, and will remain mixed in the flask.

ok ya right its bp is higher than water, can hydrogen peroxide improve the yields?

No ! There is no way for increasing the yield by adding something. The only thing that is necessary is removing the metallic ions from the original copper sulfate or aluminium sulfate, or sodium sulfate. And this is difficult to do.

You may want to find a gold and silver refiner who has aqueous Copper Sulfate as a waste product.  Your reactions become:
$$\ce{Cathode:  2Cu++ + 4e-  ->   2 Cu metal + 0.34 V}$$
$$\ce{Anode:    2H2O  ->  4H+ + 4e- + O2 gas    - 1.23 V}$$
Valuable copper metal is produced before:
$$\ce{Cathode: 4 H+ + 4e-  ->  2 H2 gas    + 0.00 V}$$
Observation of H2 gas at Cathode would indicate reaction completion.  Copper metal helps pay for the process.  Dilute sulfuric acid can be concentrated by removing water.
Aluminum sulfate may not be the best as a starting material:
$$\ce{Cathode: Al+++ + 3 e-  -> Al metal  - 1.66 V}$$
 Al +++ will not reduce before aqueous H+, or even water inself:
$$\ce{Cathode:  2H2O + 2 e-  -> H2 + 2 OH-  - 0.83 V}$$

Metallic aluminum is never produced by electrolysis in aqueous solution. NEVER ! It is not a question of acidity ! Pure water reacts with metallic aluminum, whatever its pH. Metallic aluminum is produced by electrolysis in an anhydrous solvent (usually cryolithe)

@Maurice edited for easier reading.


text:
Why surface tension acts tangentially if there is a net inward force?
I understand that surface tension is caused by an imbalance of forces for molecules on the surface in comparison to those that lie within. As a result, this creates a net inward force that seeks to reduce surface area. However, if this force acts inwards, why does surface tension act tangentially? I've seen explanations that state bonds on the surface become stronger when the surface area is minimised. If this is true then why is this the case? It doesn't seem clear to me why a net downward force causes the surface to act as an elastic membrane.

I’m voting to close this question because this is not a chemistry question. Please check Physics.SE

Consider that in the interior, all molecules are surrounded by equal attractive forces,  but at an interface, the attractive forces are missing (or greater, in the case of adhesion to another substance).

This question is related to soft-matter chemistry, so solid grasp of the phenomenon in physics is indeed relevant, just like how you got struggle enough to understand introduction of quantum mechanics for quantum chemistry.

@NilayGhosh Surface tension is an essential concept taught in physical chemistry.

Good point. An imbalance of forces (a net force) can arise where there is uneven curvature or three phases meet. A small volume of water suspended in air meets the requirement
until becoming a tidy sphere.

Haowen Xie's question deserves a bounty. Nobody has yet found an answer !

@Karl. I have never set up a bounty. How to do it ?

@Karl. The Help Center said that the button situated below the upvote and downvote marks should be pressed to set up a bounty. I have pressed this button. It does not give access to a bounty...

@Maurice Please see https://chemistry.stackexchange.com/help/bounty You have to wait at least 2 days from the day it was asked.

Each molecule at the surface of a drop of water is pulled towards the others around it (by cohesion, i.e. in this case hydrogen bonds), those at the surface, and those below. There is nothing above it, so the net force is pointing towards the center of the drop.
In the orthogonal surface area, the forces cancel out, except if the surface was previously disturbed/deformed, i.e. the area increased from the planar/spherical ideal.
When you increase the surface area by deforming the drop, that means you pull out water molecules from the body, against that net downwards force.
Ergo, surface tension and that perpendicular force are essentially the same thing.

It's a pity this post has been closed, because nobody has ever answered the fundamental question : How and why a radial force produces a tangential force ?


text:
Lewis structure and resonance for arginine ion
I'm working through a problem set in MIT's open courseware 'Principles of Chemical Science' course here. In question 5, it asks:

The skeletal structures of two amino acids, leucine and arginine, are drawn below. Non-zero formal charges are indicated. Provide the Lewis structure(s), including double bonds and lone pairs, for each of these molecules. If there are equivalent resonance forms (which may include moving the formal charge on N), include them

This arginine structure is given for part (b):

I've been able to produce the resonance structures given in the solution sheet (here).

But I've also produced another resonance structure which I think is valid but it was not included in the solutions. I've put the formal charge on on the non-terminal N atom. My structure seems to satisfy the octet rule for all atoms and has the same number of bonds and electron pairs as the other solutions. Is there something I'm missing here?

Your third structure is correct. In fact it is more stable than the other two structures because of inductive effect.

As it is mentioned in the question , you have to produce "equivalent" canonical structures   only. Your third structure is not equivalent to the other two but the both above are equivalent w.r.t each other.

The first given skeletal arg with formal charge is not correct.

@CHEMUMAN can you explain what is meant by 'equivalence'? Why is moving the charge between terminal N's equivalent while the third structure is not?

@AlRi The two structures are structurally equivalent as you cannot distinguish them without numbering the terminal N atoms. Equivalent canonical structures have same stabilities and same contribution to the 'Resonance Hybrid'. Your third structure is obviously not equivalent to the other two as you can easily distinguish it from the other two just by the sight , moreover if you analyze the three structures chemically , you will land on to the conclusion that the third structure has different stability. E.g.: Often you draw two "equivalent" Kekulé structures (canonical form) for Benzene.

@AlRi Moreover there are no equivalent structures for your third structure.                                       For better insights refer to :https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/04%3A_Lewis_Formulas_Structural_Isomerism_and_Resonance_Structures/4.03%3A_Resonance_Structures

@AlRi For future reference: two, or more of mesomeric formulae are put into a single pair of square brackets in common.  It doesn't make sense to enclose each form into its own pair of square brackets.  As an example, see the depiction of the nitrite anion in Wikipedia's article here.

Thanks for the responses @CHEMUMAN. It is making more sense to me now. My third structure is valid, but not structurally equivalent to the given skeletal arginine. The question is asking specifically for equivalent structures, so only two structures are shown in the answer.

@Buttonwood thanks for letting me know, that image was copied out of the MIT answer sheet.


text:
What processes generate entropy as heat flows across temperature gradient
Suppose, we have a source at high
temperature $T_\mathrm h$ and sink at temperature lower temperature $T_\mathrm l$.
If $Q_\mathrm h$ amount of heat flow from source to sink, then change in entropy of source is $\frac{-Q_\mathrm h}{T_\mathrm h}$ and change in entropy of sink is $\frac{+Q_\mathrm h}{T_\mathrm l}$. I saw a similar question here:
https://physics.stackexchange.com/questions/358142/entropy-generation-during-heat-transfer-processes.
And while I understand that entropy is being generated in the partition, my question is
how does heat conduction across a finite temperature gradient generates entropy

Not sure what you mean by part b)

I meant that source and sink separated by some distance

Therefore thermally insulated(not considering radiation)? In such a case, entropy of reservoirs and whole system is constant, as there is no heat flow.

How are they thermally insulated? Won't heat still flow from source to sink? Thanks

Describe it better then. Confusing description leads to confused responses. How b) differs from having partition allowing  heat flow between reservoirs?

I'm sorry, English is not my first language. Did you read the linked question? I think I know where it is confusing. In the linked question I think by partition they mean that source and sink are separated by a finite distance. Sorry for confusion.

For steady state heat conduction in the slab between the source and sink, we have $$\frac{d}{d x}\left(k\frac{d T}{d x}\right)=0$$where k is the thermal conductivity.  If we divide this equation by absolute temperature T, we obtain$$\frac{1}{T}\frac{d}{d x}\left(k\frac{d T}{d x}\right)=0$$Next, making use of the product rule for differentiation, we have:$$\frac{1}{T}\frac{d}{d x}\left(k\frac{d T}{d x}\right)=\frac{d}{d x}\left(\frac{k}{T}\frac{dT}{dx}\right)+\frac{k}{T^2}\left(\frac{dT}{dx}\right)^2=0$$Assuming that the source is at x = 0 and the sink is at x = L, if we integrate this equation between x =  0 and x + L, we obtain: $$-\frac{q}{T_C}+\frac{q}{T_H}+\int_0^L{\frac{q^2}{k}dx}=0\tag{1}$$ where the heat flux q is given by $$q=-k\frac{dT}{dx}$$If we rearrange Eqn. 1 slightly, we obtain:$$\frac{q}{T_C}=\frac{q}{T_H}+\int_0^L{\frac{q^2}{k}dx}\tag{2}$$The quantity $\frac{q}{T_H}$ is the rate at which entropy enters the slab per unit area, and the quantity $\frac{q}{T_C}$ is the rate at which entropy exits the slab per unit area of slab.  Eqn. 2 tells us that the rate at which entropy exits the slab is equal to the rate at which entropy exits the slab plus the rate at which entropy is generated within the slab.  According to Eqn.2, the rate of entropy generation within the slab per unit volume is $q^/k$.  This entropy generation rate is, of course, positive definite.

Thank you so much. I have one more question though, you showed above how entropy generation rate is positive definite, may I ask how is entropy being generated though? I mean, it is being generated this much is clear, but what physical processes are happening in the slab which generates entropy? I hope I'm being clear

The physical process is the irreversible transport of heat by conduction at finite rate.

Thanks. I do not know much about transport of heat by conduction at finite rate, so I'll look it up now. Also, is there a thing such as reversible transport of heat? Or perhaps heat transferred through infinitesimal difference of temperature is the same thing as reversible transport of heat?

Also, is there a book which will help me understand how heat transfer generate entropy intuitively. Thank you

See Transport Phenomena, by Bird, Stewart, and Lightfoot, Chapter 11, Homework problem 11.D.1

Thanks, I'll check it out.


text:
Find volume of methanol produced via reaction between carbon monoxide and hydrogen gases

Methanol is manufactured by the reaction of carbon monoxide gas with hydrogen gas. If a $95~\%$ yield is usually obtained, approximately how many liters of hydrogen gas at $\pu{350 °C}$ and $\pu{300 atm}$ are usually required to make $1.0$ liter of methanol measured at $\pu{20.0 °C}?$ The density of methanol at $\pu{20.0 °C}$ is $\pu{0.8 g mL^-1}.$
(a) $\pu{0.89 L}$
(b) $\pu{2.5 L}$
(c) $\pu{4.8 L}$
(d) $\pu{5.0 L}$
(e) $\pu{8.1 L}$

 The correct answer is (d) $\pu{5.0 L}.$

I got two different answers, both wrong. I have detailed them below.
Attempt 1
I found the mass of methanol:
$$m(\ce{MeOH}) = \rho V = (\pu{0.8 g mL^-1})(\pu{1000 mL}) = \pu{800 g},\tag{1.1}$$
and calculated its theoretical yield:
$$m_\mathrm{theor}(\ce{MeOH}) = 0.95\times\pu{800 g} = \pu{842 g}.\tag{1.2}$$
The amount of methanol is
$$n(\ce{MeOH}) = \frac{\pu{842 g}}{\pu{32 g mol^-1}} = \pu{26.3 mol}.\tag{1.3}$$
The amount of hydrogen from balanced equation is $n(\ce{H2}) = \pu{52.6 mol}.$ From here
$$
\begin{align}
V(\ce{H2}) &= \frac{nRT}{p} \\
  &= \frac{(\pu{52.6 mol})(\pu{8.31 J mol^-1 K^-1})(\pu{(350 + 273) K})}{\pu{3E7 Pa}} \\
  &= \pu{9.08E-3 m^3} \\
  &\approx \pu{9.1 L}. \tag{1.4}
\end{align}
$$
Attempt 2
Assuming the pressure of methanol is $\pu{1 atm} = \pu{1E5 Pa},$
$$
\begin{align}
n &= \frac{pV}{RT} \\
  &= \frac{(\pu{1E5 Pa})(\pu{1000 mL})}{(\pu{8.31 J mol^-1 K^-1})(\pu{293 K})} \\
  &= \pu{41071 mol}. \tag{2.1}
\end{align}
$$
Theoretical amount of methanol is
$$n_\mathrm{theor}(\ce{MeOH}) = \pu{41071 mol}/0.95 = \pu{43232 mol}.\tag{2.2}$$
The amount of hydrogen from the balanced equation is $n(\ce{H2}) = \pu{86464 mol}.$ From here
$$
\begin{align}
V(\ce{H2}) &= \frac{nRT}{p} \\
  &= \frac{(\pu{86464 mol})(\pu{8.31 J mol^-1 K^-1})(\pu{(350 + 273) K})}{\pu{3E7 Pa}} \\
  &= \pu{14.92 m^3}. \tag{2.3}
\end{align}
$$
Attempt 2 is definitely wrong. I don't get what I did wrong in my attempt 1. What am I missing?

It is always good to summarize and/or complement the verbal description by algebra.
Good practice is starting with symbolic algebraic expressions and keeping it this way until all is ready to plug in literal numbers. It helps in focusing on principles, mistakes are easier to spot, orientation is improved, Q/A is reusable and has bigger  permanent value.
You may find useful  formatting mathematical/chemical expressions/formulas.

If the correct answer is d) the volume of methanol is 5 L not 5mL. And that is the volume of the liquid so no calculations using the gas equation are relevant to it.

@matt_black It is volume of pressurized (300 atm) hydrogen, not methanol.

@Poutnik The wording for attempt 2 says "Assuming the pressure of methanol is 1 atm" so maybe it is careless wording, but sounds wrong.

@matt_black Take the bold text.  approximately how many liters of hydrogen gas at 350 °C and 300 atm

@Natalie  Methanol at 20 deg C is liquid, not gas, and you cannot use for it the ideal gas state equation. (attempt 2 )

@matt_black Yeah sorry it should be 5 L

@Poutnik Yeah my attempt 2 is definitely wrong, couldn't think of anything else though. I don't really get why attempt 1 is wrong. What am I missing?

I think the yield has units of mols per mol (i.e. actual moles per theoretical moles is 0.95), not grams per gram.  I don't think it makes a difference numerically but it'd be cleaner to convert to moles first, then adjust for yield.

Your first calculation was right. The volume of $\ce{H2}$ must be nearly $9$ L, and not nearly $5$ L. $5$ L is not far from the expected volume of $\ce{CO}$.

@Karl This question is from a past paper of a Chemistry Olympiad test. You’re right I don’t think they considered how practical this question is.

@CurtF. What do you mean? I used density=mass/volume to find the mass of methanol and I used that mass in the yield calculation

@Maurice I thought it might be an error in the mark scheme but 9.1 L is not even one of the MCQ options so it couldn’t have been a typo. Like I mentioned earlier it’s from a past preliminary Olympiad test. Usually the questions are fully vetted. I really want confirm that it’s an error in the question and my answer is right before getting back to the student who asked me this.

I have evaluated the attempt 1 some time ago and it does seem to me to be an error in the task formulation or answer list preparation. If we considered hydrogen as a real gas at this high pressure, the resulting volume would be even higher.

@Karl I really don't understand the criticism of the question. Surely students should learn to calculate yield via stoichiometry? It's not a fictional process: "The industrial production of synthetic methanol abroad has been known for some time and in May, 1927, an industrial production of 4500 gallons of methanol daily was announced by one American company. In this instance, carbon dioxide and hydrogen, hitherto waste gases from the production of butanol and acetone from corn, were utilized." Methanol from Hydrogen and Carbon Monoxide, 1928


text:
Why can't we use concentrated acids to test for carbonates?
According to BBC Bitesize:

Carbonate ions, $\ce{CO_3^{2-}}$, are detected using a dilute acid. Bubbles are given off when an acid, usually dilute hydrochloric acid, is added to the test compound.

Is there a requirement that we use dilute acids for this purpose? Can't this test be done using relatively concentrated acids with the likely advantage that the reaction happens more vigorously?

Dilute HCl or HNO3 are quite acidic enough for the purpose. Safety first, which means you only take the potent reagents when they're really needed! And conc. acids might oxidise some other carbon-containing compound, which you then mistake for a carbonate.

Nitric acid (dil) or (conc.) is never used for carbonate testing and the same applies to sulfuric acid.

If you would post your comments as answers, I would gladly slap the checkmark on them. ;)

BBC's story is quite an incomplete one. Many anions can form gaseous products when acidified. The reason for using a dilute acid for carbonate test is simply that it works and therefore one does not need a stronger concentration! Even lemon juice will work for carbonates. Since carbonate is a common anion, for elementary purposes, dilute HCl is the acid of choice because it does not form insoluble salts with commonly encountered carbonates such as alkali (Li, Na, K) and alkaline earth metals (Mg, Ca, Ba) etc. Sulfuric acid forms insoluble salt layer with Ca/Ba which can prevent further raction. Nitric acid (dilute or concentrated) is to avoided because it can also generate nitrogen oxide(s) bubbles in contact with certain substances depending on whatever is present along with carbonates.


text:
Endothermic peak in DSC
I have an endothermic peak, with no mass loss and no change in (organic) crystal structure (monitored by vt-pxrd). What can cause such effect? Also, excluding the options of impurities or instrumental errors.

See this document: Interpreting Unexpected Events and Transitions in DSC Results, https://www.tainstruments.com/pdf/literature/TA039.pdf

Without knowing any experimental detail, it will be a futile exercise to speculate anything.

Can you provide more detail about the material in question?  (Maybe your PI doesn't offer clearance prior to filing a patent/submitting the work to a journal.)  Yet in analogy to zeolithes (e.g., ZSM-5), there are metal organic frameworks stable enough and with sufficient porosity for reversible absorption/desorption of molecules (e.g., water).  Once «entrapped» in the cavities, it becomes difficult to monitor these molecules of water because of its light atoms of smallish diffracting power (though this depends a bit on the X-ray source's brilliance and wavelength engaged).

@Buttonwood, OP says that there is no corresponding mass loss in that DSC event. It means nothing is leaving or entering the crystal system.

@AChem There may be multiple sites (both interior, as well as exterior) on (porous) materials, each with its own adsorption/desorption mode and energy for the molecules adsorbed.  Then, the system may identify just an other energetic minimum of adsorption when the adsorbed molecule(s) shuttle around; all without the adsorbed molecule leaving substrate for good (especially with more shallow energetic minima compared to ZSM-5, lighter molecules to adsorb/interaction with organic molecules).


text:
Differentiating Ca2+ and Sr2+ ion
$\ce{Ca^{+2}}$ and $\ce{Sr^{+2}}$ both give a bit similar red color in flame test. What is the easiest way (chemical test) to distinguish $\ce{Ca^{+2}}$ and $\ce{Sr^{+2}}$?

What have you searched so far?

Use Potassium ferrocyanide solution, you will get a mixed salt of calcium (white ppt).

@AChem this is what my tb says.

@Shub, Are you concerned about similar flame colors? I agree. Since this is a state approved textbook, I gather you will have to follow that. This video is from India (NCERT) and it explains how to distinguish the three ions. https://www.youtube.com/watch?v=HzbEfWqzBhs

Simplest tests would be visual emission spectroscopy or mass spectroscopy... A single test would show all three species present and their proportions. That said, to pass your test, use the required response.

Does this answer your question? Magnesium vs calcium vs barium ions from these 3 tests

You can differentiate based on the solubility of their sulfates. Please see my answer^

@CHEMUMAN Good time plugging my comment: https://chemistry.stackexchange.com/questions/170053/can-i-use-potassium-ferricyanide-instead-of-potassium-ferrocyanide-for-the-ident#comment357032_170053 (it has one paper linked on this test)

So Ca+2 will give white ppt in potassium ferrocyanide and none in Ba+2 and Sr+2? @NilayGhosh

@NilayGhosh Actually, I hadn't seen that question before. I have read about this test in my chemistry class few days ago only. Btw you have mentioned about mixed salt when tested in ferricyanide solution.

@Shub No, only barium and magnesium interfere with the test. I have addressed in a separate answer.

@CHEMUMAN Essentially yes. Ferricyanide is a different species and if it is used, it might give a mixture of ferricyanide + ferriferrocyanide precipitate giving a false positive. Nobody has confirmed it though since potassium ferrocyanide test works so well but I assume that happening as mentioned in the second paper linked.

The video document given by AChem is not really clear. It should be replaced by the following method, used in my university.
Let's start from $5$ mL of an initial solution containing possibly any one or all of the three ions $\ce{Ca^{2+}, Sr^{2+}, Ba^{2+}}$.
§$1$. Check if this initial solution contains the ion $\ce{Ba^{2+}}$ by taking $3$ drops of the initial solution, adding $3$ drops of $0.1$ M $\ce{K2Cr2O7}$ plus $1$ mL $0.1$ M $\ce{CH3COONa}$. If a yellow precipitate appears, it means that Barium is present in solution : Go to $§2$.
If the solution remains clear, the solution does not contain $\ce{Ba^{2+}}$ : skip §$2$, and go to §$3$ with the rest of the initial solution.
§$2$. Add the following solutions to the rest of the initial solution : first $3$ mL $0.1$ M $\ce{K2Cr2O7}$ solution, plus $5$ mL $0.1$ M $\ce{CH3COONa}$. Filtrate the yellow $\ce{BaCrO4}$ precipitate. The filtrate must be yellow, and does not contain $\ce{Ba^{2+}}$ any more.
§$3$. Add enough $\ce{Na2CO3}$ to make the solution basic (pH > $8$). Heat to boiling temperature, cool, filter and wash the precipitate containing $\ce{CaCO3 + SrCO3}$. Throw away the yellow filtrate.
$§4$. Add enough $\ce{HNO3}$ to dissolve the precipitate. Usually $1$ mL $1$ M $\ce{HNO3}$ is enough. Drop the liquid in a ceramic capsule. Heat it so as to evaporate it totally, and to eliminate all crystallization waters. Then cool it down and add $1$ mL $100$% pure ethanol. Grind the stuff with a glass stick. Filtrate on a small filter paper. Wash twice with $0.5$ mL pure ethanol ($100$%). The residual insoluble part is made of $\ce{Sr(NO3)2}$. See how to use the filtrate in $§6$.
§$5$. The obtained residue of $\ce{Sr(NO3)2}$ is washed several times to eliminate traces of Calcium nitrate. Dissolve it in $3$ mL water. Add $1$ mL 1 M $\ce{H2SO4}$. A white precipitate is produced if the substance contains Strontium. Go to §6 with the filtrate.
§$6$. The filtrate contains calcium nitrate. It can be proven by first evaporating to dryness. Then the residue is dissolved in $1$ mL water. $1$ drop $1$ M $\ce{CH3COOH}$ is added to the obtained solution. The $1$ mL $0.1$ M ammonium oxalate is added. A white $\ce{CaC2O4}$ precipitate shows the presence of Calcium.

+1 Although the protocol is useful and superior, but students have no option to change texbook protocols, this is why I suggested that the OP follow his local teaching scheme for state exams.

A good test for distinguishing calcium ions and strontium ions is potassium ferrocyanide test.
$$\ce{Ca^2+ + 2K4[Fe(CN)6] -> \underset{white}{K2Ca[Fe(CN)6]} \downarrow}$$
In presence of ammonium chloride, test is more sensitive. In this case, potassium is replaced by ammonium ions in the precipitate. However, the downside of this test is barium and magnesium interfere. You can separate them as discussed in my other answer.. You can find more details on this test on a paper linked in a previous comment of mine. However, the best test is checking the solubility of the corresponding sulfates (explained in my other answer).

What happens to Strontium with ferrocyanide ? Does it precipitate ?

@Maurice No, it won't precipitate.

And how will barium and magnesium interfere ?

@Maurice barium and magnesium also give precipitation with pot. Ferrocyanide. See: https://pubs.acs.org/doi/pdf/10.1021/ja01959a010. So, they recommend separating barium and magnesium first and using this test to confirm the presence of calcium.

So, there is just and only Strontium which does not make a precipitate with ferrocyanide ions. It seems strange !


text:
Diffusion, Migration and Einstein Equation
In the textbook: Electrochemical Systems by Newman and Alyea, 3rd edition, chapter 11.9: Moderately Dilute Solutions, equation for the mole flux of the component $i$ is given by: $$ N_i = - \frac {u_i c_i} {z_i F} \nabla \bar\mu_i\ + c_i v \tag {1}$$
where $u_i$ is the ionic mobility, $c_i$ is the concentration, $\bar \mu_i$ is the electrochemical potential and $v$ is the velocity of the streaming fluid (assuming a low concentration of the component $i$ as situation becomes more complicated at higher concentrations).
One thing to note is that I wrote ionic mobility $u_i$ a bit differently compared to the textbook. I defined it as a terminal velocity of the ion in the unitary electric field, so its measuring unit is $[\frac {m^2}{Vs}]$ in my case. In the textbook they defined it also as a terminal velocity in the unitary electric field, but divided by $z_i F$. This is something to keep in mind to avoid confusion.
Basically, equation (1) states that the mole flux of the component has three contributions: diffusion, migration and convection. Diffusion and migration are accounted for in the first term (electrochemical potential gradient) and convection is accounted for in the second term.
My problem is with the first term $-\frac {u_i c_i} {z_i F} \nabla \mu_i$ which I know how to derive and I will do this here as I don't understand something about the equation (1).
To start, we define diffusion mole flux from as the 1st Fick's law: $$N_{i,dif} = -\frac {D_i c_i} {RT} \nabla \mu_i \tag {2}$$
where $D_i$ is the diffusion coefficient and $\mu_i$ is the chemical potential of the component $i$
Migration mole flux is given by: $$N_{i,mig} = -u_i c_i \nabla \phi \tag {3}$$
where $\phi$ is the electric potential.
If there is no convection, we can write total mole flux of the component as: $$N_i = -(\frac {D_i c_i} {RT} \nabla \mu_i + u_i c_i \nabla \phi) \tag {4}$$
We recall the definition of electrochemical potential $ \bar \mu_i$: $$ \bar \mu_i = \mu_i + z_i F \phi \tag {5} $$
If we assume that system is in thermodynamic equilibrium than electrochemical potential gradient  and net mole flux vanish: $$ \nabla \bar \mu_i = 0 \tag {6} $$ $$ N_i = 0 \tag {7}$$
From equations (5) and (6), we can write: $$\frac {\nabla \mu_i}{\nabla \phi} = -z_i F \tag {8}$$
From equations (4) and (7), we can write: $$ \frac {\nabla \mu_i}{\nabla \phi} = - \frac {u_i RT}{D_i} \tag {9}$$
By equating (8) and (9), we get the Einstein equation, which relates diffusion coefficient $D_i$ and ionic mobility $u_i$ in the thermodynamic equilibrium: $$D_i = \frac {u_i RT}{z_i F} \tag {10}$$
If we know substitute the equation (10) into equation (4), we can write: $$N_i = -u_i c_i(\frac {1}{z_i F} \nabla \mu_i + \nabla \phi) = - \frac {u_i c_i} {z_i F} \nabla \bar\mu_i\ \tag {11}$$
After the derivation of the first term in the equation (1), I can get to my point. Namely, that the first term in the equation must be equal to zero since it is impossible to write this term in that form (via gradient of the electrochemical potential) without using Einstein equation. We saw earlier that Einstein equation is only valid in the thermodynamic equilibrium and therefore equation (6) applies.

I would check another book containing the same topic for the same derivation. Have you checked Bard and Faulkner?

I didn't, but I will.

@AChem Just checked Bard and Faulkner, Chapter 4.2: Migration. Basically, they explained nothing. They stated that the connection is given by the Einstein equation, but without any derivation.

Dario, We will to check the original literature now. You don't stop or rest until you find a satisfying answer. This is the true essense of science. This requires going deeper and deeper in terms of literature search. Visit a library and check other electrochemistry textbooks. Also, ask your instructor.

@AChem Yes, I'll definetly ask my supervisor. Internet provides some idea about the derivation from random walk which might give a good explanation.

Before writing, I understand your derivation, but I do not agree with Eq. (8). The gradient of a scalar field is a vector, so it is not correct to divide a vector quantity like $\underline{\nabla}\mu_i$ with $\underline{\nabla}\phi$.
Indeed the flux of species $j$, in a $1$-dimensional system, can be written as
$$ N_j = -D_j\dfrac{dC_j}{dx} - u_j C_j \dfrac{d\phi}{dx} \tag{1} $$
Here is a somewhat informal derivation of your Eq. (10). Imagine the charged species $ j $ being subjected to two forces, the drag force and the electrical force, but both being balanced. By Newton's 2nd law we have
\begin{align}
  F_D &= F_E \\
  6 \pi \mu r_j v_D &= z_j e E \\
  \dfrac{v_D}{E} &= \dfrac{z_j e}{6 \pi \mu r_j} \\
  u_j &= \dfrac{k_BT}{6 \pi \mu r_j}\dfrac{z_j e}{k_BT} \hspace{1 cm}
  \left(D_j = \dfrac{k_BT}{6 \pi \mu r_j} \hspace{0.5 cm} \text{(Einstein-Stokes law)}\right) \\
  u_j &= D_j\dfrac{z_j e N_A}{N_Ak_B T} \\
  u_j &= D_j\dfrac{z_j F}{RT} \tag{2} \\
\end{align}
This is the Einstein-Smoluchowski equation. Combining Eqs. (1) and (2) we have
$$ N_j = -D_j\dfrac{dC_j}{dx} - \dfrac{F}{RT} z_jD_jC_j \dfrac{d\phi}{dx} \tag{3} $$
If you couple Eq. (3) with the convection part, i.e. $ C_j\mathbf{v} $, the result is the famous Nernst-Planck equation.
Eq. (2) was derived by considering that the charged species $ j $ is in an equilibrium between its forces. Although I didn't, you can derive Einstein-Stokes law by doing the same, but considering: (1) the balance between the chemical force $d\mu_j/dx $ and the drag force, and (2) comparing the derived expression with Fick's law.
Thus, how come this relation, obtained by an equilibrium between forces, is combined with the expression to describe the motion of species in $N_j$?. The easiest answer is the following: so we can reduce the number of phenomenological and constant parameters, to describe the motion, from two to one.
Nernst-Planck equation is a good approximation to the molar flux, if the concentrations are sufficiently low. This means that you can neglect ion-ion interactions in the medium. This is the reason why you are able to change the activity by the concentration in the chemical part of $\tilde{\mu}_j$. It also assumes that the flux can be written as a linear combination of driving forces, and the phenomenological coefficients, are typically considered as constants. But indeed, coupling it with Eq. (2) goes much further. If you want to be more precise, Eq. (1) should be written
$$ N_j = -D_j^\infty\dfrac{dC_j}{dx} - u_j^\infty C_j \dfrac{d\phi}{dx} \tag{4} $$
So you need two phenomenological parameters to evaluate the flux. However, sometimes it is hard to obtain both coefficients, $ D_j^\infty $ and $ u_j^\infty $, so we go around with this by applying Eq. (2). I will give you an example of my own. I am modeling lithium-sulfur batteries, and there are publications where $\ce{Li+}$ (and other sulfur species) are as high as $4-5$ mol/dm$^3$, and other authors and myself still use Eq. (3).

Thank you for the detailed answer. Yes, you are correct that it isn't very rigorous to write vectors in fractions although I think it leads to correct results nevertheless. It's similar to when scientists treat differentials algebraically, it's wrong mathematically, but it still delivers correct result.

Problem with my question is that it merely proves that Einstein equation must be valid in thermodynamic equilibrium and not that it isn't valid outside of it (which I claim in the question). Basically the same derivation is carried on the wikipedia.
The thing is that to prove general validity of Einstein equation, one needs to go beyond macroscopic thermodynamics and electrochemistry. Einstein derived it using the concept of random walk which is a mathematical idea quite different from laws of thermodynamics and electrochemistry.

It is interesting that you model lithium-sulfur batteries. I research hydrogen fuel cell air electrodes (SOFC/SOEC).


text:
Why does cooking oil solidify
I am in a tropical climate. The vegetable cooking oil I buy remains liquid for about 3 or 4 months and by the time I'm finishing a container, it has solidified.
Why does this happen? Is there any way to avoid this?

It appears to be palm oil. Yes it liquefies again although in my case, it solidifes again. I warm it up every night before cooking supper.

What kind of oil exactly? That may be important. Also, does it liquefy back upon heating?

@IvanNeretin: It appears to be palm oil. Yes it liquefies again although in my case, it solidifes again. I warm it up every night before cooking supper :)

Palm oil has reported freezing point 35 deg C and minimal recommended storage temp 50 deg C ( to keep it as oil ).

Worth noting that many oils only freeze only slowly even if the temperature is lower than their freezing point. Palm oil may only freeze slowly even at 30°. Olive oil often slowly freezes at ambient temperatures in the UK.

@matt_black I assume oils have high tendency to be supercooled. I have added such a note. I also assume some oils may be oil mixtures with significantly lowered freezing point. ( Mixtures have frequently lower melting/freezing point than expected from linear interpolation for pure substances parameters)

As revealed later, first check, if it is temperature dependent temporary solidification. It is natural and reversible process, the same process as when melted fat solidifies back when cooled down. Prevention is to keep oil warm enough and the cure is warming it up until it melts.
Every fat has its typical melting point ( or range). Solid fats melt at temperature higher than room one, vegetable oils melt at temperature lower than room one. Some fats are in this aspect in a gray zone, melting or freezing according to the current room temperature.
Note in the table below palm oil has exceptionally high freezing point $\pu{35 ^{\circ}C}$. If solidification is an issue, consider using other oil with lower melting point. Oils have tendency to be supercooled (hesitating to start freezing up), so it may need lower than expected temperature to observe solidification, that may be triggered by random events.
I also assume some oils may be oil mixtures with lowered freezing point. .
See below the table  digeoris.com: Freezing Point of Vegetable Oil

Oil or fat
Freezing / melting point °C
Transportmin °C
Transportmax °C
Storagemin °C
Storagemax °C

Castor oil
-3
20
25
30
35

Coconut oil
25
27
32
40
45

Cottonseed oil
48
Ambient
Ambient
20
25

Linseed oil
-24
Ambient
Ambient
15
20

Maize (corn) oil
-11
Ambient
Ambient
15
20

Olive oil
-6
Ambient
Ambient
15
20

Palm oil
35
32
40
50
55

Canola (rapeseed)oil
-10
Ambient
Ambient
15
20

Safflower oil
-17
Ambient
Ambient
15
20

Sesame oil
-6
Ambient
Ambient
15
20

Sheanut butter
34-38
38
41
50
55

Soybean oil
-16
Ambient
Ambient
20
25

Sunflower oil
-17
Ambient
Ambient
15
20

If the above does not apply:
It probably polymerizes due present double bonds, reacting with air oxygen, being a vegetable oil. Famous for this is lineseed oil, that is used in oil paints.
I suggest keeping oil stocks in cold place, as the reaction is slower at lower temperature. If applicable, even in fridge, where it may  temporarily freeze ( and melt back when warmed up in a room).
And/or it may be more applicable to buy smaller oil packages or divide their content in several smaller vessels, kept closed, cold if possible, taken sequentially. The remaining vessels would not then be repeatedly exposed to air and if possible, would be staying in cold place.

Assuming implicit chemistry skills of the OP, I have intentionally written this rather from point of view of a cook than a chemist. So both answers may complement each other.

For some reason I can't mark this answer as accepted. There seems to be some problem with the stackexchange interface. Will contact stackexchange staff about it, please be patient!

Cooking oils are consist of esters of fatty acids. Some of them are saturated, others are unsaturated; the later refers to the presence of one, or multiple $\ce{C=C}$ double bonds.  An example for this later is oleic acid:

(image credit: English edition of wikipedia).
Over time, promoted by heat (not only cooking, ambient temperature suffice given enough time), sun light, and traces of metal salts of these fatty acids, exposure to air leads to polymerization.  Oxygen (about $\pu{21 vol\%}$ in air) is a diradical, which can attack these double bonds.  Eventually, individual molecules with $\ce{C=C}$ double bonds then cross-link with each other on expense of these double bonds.  So

the $\ce{C=C}$ double bonds are consumed
individual small molecules bind with each other on positions of former $\ce{C=C}$ double bonds
the new molecules formed are larger
compared to the starting material, the newly formed products have a higher melting and boiling temperature.

Because cooking oil is a blend of compounds, oleic acid isn't the only compound undergoing this reaction.  Then, the hardened cooking oil equally is a blend of multiple chemicals, too.  The principle is not restricted to cooking, though.  In painting, this process is triggered by intentional addition of a siccatives, compounds known to accelerate this process.


text:
Confusion about equilibrium constant in simultaneous equilibrium
Ok, so the equilibrium constant in terms of pressure is the ratio of partial pressures at equilibrium of gaseous products to reactants (all raised to the powers equal to the stoichiometric coefficients). But if I have multiple equilibriums taking place simultaneously, where, let's say, there is one (gaseous) product common to both reactions, then that means the partial pressure of that substance will actually be equal to the sum of partial pressures of the moles of that substance created from both reactions.
For example:
$$\ce{X(s)} \ce{<=>} \ce{Y(g)} + \ce{Z(g)} $$
$$\ce{R(s)} \ce{<=>} \ce{Y(g)} + \ce{S(g)}  $$
Here, let's say at equilibrium the partial pressure of Y and Z (the amount of them created from reaction 1) are $P_1$, and that of Y and S created in reaction 2 is $P_2$.

My teacher said that the partial pressure of Y in the entire reaction mixture would be $P_1 + P_2$, which makes sense. But if we write the expression for $K_1$ and $K_2$, my teacher says we will take the partial pressure of Y to be $P_1 + P_2$. I'm having a hard time understanding why we won't just take the partial pressures of Y created in those specific reactions?

Also, is it not true that the equilibrium constant of each of these reactions would change if we put them in a mixture together, since the concentration of $Y$ would change because that is present as a product in both reactions? Or would it just be that the equilibrium will move in the backward direction since more product is added?

If the answer to (2) is that $K$ will not change, then that means if I only had reaction 1 taking place, equilibrium constant for that would also take pressure of Y to be $P_1 + P_2$, which seems obviously wrong to me- it should only be $P_1$.

Overall, stuff is not adding up in my mind for simultaneous equilibriums, if anyone could clear my concepts I would appreciate it!

Chemical equilibria are dynamical. This means that even in the equilibrium, the reactions are still happening but in both directions with the same reaction rate. Now think about a molecule of $\ce{Y}$. Once it is formed, it has no memory of where it came from. When it is formed, it is part of the whole collection of $\ce{Y}$ molecules in the container, and all of them together exert a what we call the partial pressure $P(\ce{Y})$, and act equally in both reactions. Even if a molecule was formed in the second reaction, it can then be used in the first reaction. Therefore, you cannot ...

... separate the two partial pressures of $\ce{Y}$ coming from the first or the second reaction because all $\ce{Y}$ molecules are indistinguishable. I hope this helps clearing up the confusion :)

@Domen Yes, thank you! :)

The equilibrium constant $K$ is just that, i..e constant at a fixed temperature so (2) is not correct. If now the amount of Y (or any other species) is changed then the amounts of the all species change until equilibrium is re-established;  $K$ remains unchanged.

Related: https://chemistry.stackexchange.com/q/111041


text:
why does the single crystal ferroelectric have a different hysteresis curve compared to polycrystalline ferroelectric?
For the ferroelectric hysteresis curve, is there a reason why single crystal ferroelectrics have a 'hard' ferroelectric hysteresis curve compared to 'soft' ferroelectric hysteresis curve for polycrystalline ferroelectrics?

If the title length is comparable to the body length, the former is too long and the latter too short.

Crystals of polycrystallic materials do not have uniform orientation of their crystal lattices and it is well known crystallic materials have usually anisotropic properties.

@Poutnik - and being ferroelectric requires the lack of a symmetry center

Keywords to complememt «ferroelectric hysteresis»: Curie temperature, Néels temperature, Weiss domains. And conceptually, there are similarities between domains within a single crystal (note that twinning may occur) , individual crystals, or large grains.

@JonCuster Nice to see an echo of crystallography here.


text:
How to change the counterion of a nucleotide salt?
I need to take the sodium nucleotide salts I have available in my lab and modify them to instead contain some more exotic counterions. Can I do this by binding my nucleotides to a strong anion exchange resin, washing, and eluting with a salt of my preferred counter ion? Seems like it should work, but Im a biologist, not a chemist. Any other suggestions would also be appreciated! Thanks very much!

Ion-exchange chromatography is typically used to achieve that. This is how ionic liquids get their different "exotic" anions.


text:
Feasibility comparison of symmetric ether synthesis and HCl + alcohol SN2 reactions
According to Solomons & Fryhle 10$^{th}$ edition Ch 11.8,

Because the chloride ion is a weaker nucleophile than bromide or iodide ions, hydrogen chloride doesn't react with primary/secondary alcohols unless ZnCl$_2$ or a similar lewis acid is added to the mixture as well.

However, we also know that symmetric ethers can be formed by an $S_N2$ reaction between two alcohols, such as the synthesis of diethyl ether from two molecules of ethanol under dilute acidic conditions, despite the fact that alcohols are generally bad nucleophiles.
With regards to the Mayr-Patz equation, ethanol has $N = 6.68, s_N=0.85$ and chloride has $N = 13.00, s_N = 0.6$ (both in 80% Ethanol, 20% water albeit chloride's $s_N$ is estimated). Unfortunately, I don't know where to find the E value for a protonated ethanol.
Given this information, I'm now wondering if ethanol is necessarily always a worse nucleophile than chloride (considering it has a higher $s_N$ value), and if it is worse, how is it possible for symmetric ether synthesis to be feasible with just acid, while $S_N2$ happens at a negigible rate between ethanol and HCl?


text:
Is the process of volume decrease in NaOH solutions reversible during evaporation of water from the solution?
While preparing w/w 50% NaOH solution I noticed a decrease in solution volume (5-10mL) in a 2L volumetric flask, so I am wondering if there is an increase in water volume during the water evaporation or concentration of NaOH? Is it the same amount and is it measurable?
QuantumYitian question regarding the decrease of volume during the dissolution of NaOH

Try to reformulate your question to make more clear, what exactly you are asking about.

There is no water volume in solution. $\ce{Na+}$ and $\ce{OH-}$ ions and $\ce{H2O}$ molecules are packed together in solution in average tighter than is the average packing tightness for $\ce{NaOH(s)}$ and $\ce{H2O(l)}$ alone.
At water evaporation, the mass concentration of $\ce{NaOH}$ obviously increases until solution saturation. After that, extra $\ce{NaOH}$ crystalizes.
If all water evaporates and then condenses, the initial volumes are obtained, if we neglect experimental troubles of removing all water.

The difference of volumes during dissolution (assuming the same initial and final temperature) can be determined from experimental or tabulated values of volume, mass or density:

Measured mass/volume and measured/known density of solute
Measured volume/mass and measured/known density of solvent
Measured volume or measured/tabulated density of solution at give concentration

Many solutions, including $\ce{NaOH}$ have tabulated densities that can be interpolated and the volume difference calculated.
Here is the simple formula to calculate the difference:
$$\Delta V = V_\mathrm{solution} - V_\mathrm{solute} - V_\mathrm{solvent} = \\
\frac{m_\mathrm{solute} + m_\mathrm{solvent}}{\rho_\mathrm{solution}} - \frac{m_\mathrm{solute}}{\rho_\mathrm{solute}} - \frac{m_\mathrm{solvent}}{\rho_\mathrm{solvent}}=
\\
\frac{m_{\ce{NaOH}} + m_{\ce{H2O}}}{\rho_\mathrm{solution}} - \frac{m_{\ce{NaOH}}}{\rho_{\ce{NaOH(s)}}} - \frac{m_{\ce{H2O}}}{\rho_{\ce{H2O}}}=
\\
\frac{m_{\ce{NaOH}} + V_{\ce{H2O}}\cdot \rho_{\ce{H2O}}   }{\rho_\mathrm{solution}} - \frac{m_{\ce{NaOH}}}{\rho_{\ce{NaOH(s)}}} - V_{\ce{H2O}}$$

$V$ is volume
$\rho$ is density
$m$ is mass

How do you obtain the volume of a solute, e.g., NaOH in this case?

Sorry, I am still thinking. You mean the density of solid NaOH?

@Achem  Both solid and solution.  V(NaOH(s))=m(NaOH(s)) /rho(NaOH(s)) // You wieght NaOH, obtain its density and calculate its volume.

Okay. Thanks for the clarification.


text:
What is the difference between ionic bonding and the general transmission of charge?
I understand that in ionic bonding electrons are transferred from atom to atom and these ions are then attracted to each other forming a new compound. How is this different from any other transfer of charge situation? For instance, when silk and glass are rubbed together, electrons are transferred between them. These two items then attract each other but we do not say that a new chemical compound has formed, we still just have silk and glass.

Electron transmission in ionic bond is just a myth! The ionic bond is only a result of force of attraction. Rubbering silk and than detaching is result of breakage of ionic bond.

It is matter of proportion. When you receive an electron from some material, you do not notice it. When an atom or small molecule receives an electron from another one, it is a life changing event for both. If every molecule of your body received an electron, it would lead to a coulombic explosion of gigantic proportions.

When silk and glas are rubbed together, the number of electrons being transferred is probably included between $10$ and $10^6$. This is enough to show the electrostatic effect. When an ionic bond is made between  macroscopic amounts of $\ce{Na}$ and $\ce{Cl}$ to make up $\ce{NaCl}$, each atom reacts : the number of electrons so transferred is probably between $10^{15}$ to $10^{23}$

Maurice's comment gives an idea of magnitude in the glass-wool-cat fur situation. It is quite different in a lightning discharge. Every transfer of electrons, if traced back far enough, involves atoms, molecules, electronic orbitals. It is just that some orbitals are of higher energy and can lose electrons more easily, but those electrons still must have somewhere to go except possibly when in flight in a cathode ray tube. That is why when you walk along a rug in winter and touch a metal post all those electrons stored in your molecules antibonding orbitals readily transfer to empty conduction bands in the metal.  Physical forces as well as magnetic and electrical forces can transfer electrons among orbitals. It may be hard to analyze but that is what is happening. Some of that glass and wool is most likely chemically changed. Some of the molecules in the lightning bolt most certainly do get rearranged.

General transmission of change does not specify

the source of the charge
the receiver of the charge
the amount of transmitted charge
the distance of charge transmission

On the other hand ionic bond is partial electron transfer between adjacent atoms in contact, being bond by electrostatic force between the opposite charges.
No ionic bond is purely ionic, but is in fact partly ionic, partly covalent. As cations always more or less deform  electron density of anions, decreasing the value of the electrostatic dipole.
As the rule of thumb to determine 50% of bond "ionicity" is usually considered electronegativity difference 1.7.


text:
How to determine the charge of amino acid at certain pH?
Problem

What would charge would you expect on alanine when placed in a solution with a pH of 1.00?

Answer

 +1. Since alanine is nonpolar, we know that the only parts of the amino acid that can be charged are the N-terminus and the C-terminus.

 In an acidic solution, there is an excessive amount of protons available to protonate the amino acid. As a result, the carboxylic acid end and the amine end will both be fully protonated. This will result in an overall charge of +1, due to the nitrogen having three hydrogens attached.

Question
Let's say I am given a certain pH of 2.00 rather than 1.00 for the "acidic solution" and the pH in this example is of that of a non-polar acid (alanine) changes to +1.
Does the pH dictate how many protons are added to amino or carboxyl groups? What determines the number of protons being added to that amino acid on the carboxylic end? I know a carboxylic acid can take up to four protons being added on the molecule.

4 protons to carboxyl? Where did you get it?

A carboxyl group can accept four protons. bonding abilitys to the oxygens (which are apart of the group carboxyl)

No, it cannot. You cannot have $\ce{R-COH2^{2+}-OH3^2+}$. Neither you can have $\ce{H4O^2+}$

What does "neither you can have h4O2+"  is that even english? please elaborate on that last thought.

Even if it was not proper English, it could not be such a big deal to realize there is no ion $\ce{H4O^2+}$. // You cannot have A.  Neither you can have B. IMHO it is correct, but if it is not, I apologize.

P.S.: Correct would be "Neither can you have ...."

Chemistry is not English =) couldn't resist!

Behaviour of amino acids in water solution
Amino acids exist in the solution in the form of zwitterions (molecules that contain an equal number of positively and negatively charged functional groups). Thus, the following equilibrium is established in the alanine solution:
$$\ce{H2N-(CH3)CH-COOH <=> ^+H3N-(CH3)CH-COO-}$$
Alanine as an ampholyte
Since actually there are two equilibria in the solution, alanine can be considered as an ampholyte:
$$\ce{^+H3N-(CH3)CH-COO- + H2O <=> H2N-(CH3)CH-COO- + H3O+}$$
$$\ce{^+H3N-(CH3)CH-COO- + H2O <=> ^+H3N-(CH3)CH-COOH + OH-}$$
At $\textrm{pH}\ 2$ there is an excess of $\ce{H3O+}$ and a lack of $\ce{OH-}$ so the first equilibrium shifts to the left while the second equilibrium shifts to the right according to Le Chatelier's principle. Therefore, the main form in which alanine exists in water solution at $\textrm{pH}\ 2$ is $\ce{^+H3N-(CH3)CH-COOH}$. The overall charge of this molecule is $+1$. And vice versa, in an alkali media ($\textrm{pH}>7$) alanine will have the main form $\ce{H2N-(CH3)CH-COO-}$ and the net charge equal to $-1$.
It is a possible explanation from the point of view of acid-base theory. Please, check out references for further reading.
References

Skoog, Douglas A., West, Donald M., Holler, F. James, Crouch, Stanley R. (2014). Fundamentals of Analytical Chemistry (9th ed.). Singapore: Cengage Learning. p. 371-372.
Clayden, J., Greeves, N., & Warren, S. (2012). Organic Chemistry (2nd ed.). Oxford University Press. p. 167.

The change of $\mathrm{pH}$ from $+1$ to $+2$ does not change much in the main formula of alanine, which is and stays $\ce{H3\overset{+}{N}−(CH3)CH−COOH}$, up to $2.34$ which is the first $\mathrm{p}K_\mathrm{a}.$

This isn't quite correct: there is only a certain fraction of given protonated species, not unity.

@andselisk. OK for the fraction. I thought it was obvious.


text:
Can pyruvate get reduced after glycolysis?
During cellular respiration, the first step is glycolysis where the glucose is split into 2 pyruvates, 2 NADH, and 2 ATP. The pyruvate either goes through the citric acid cycle or goes through lactic acid fermentation. If you remove the citric acid cycle and fermentation, is there a specific way to reduce pyruvate? I am relatively new to chemistry and I was wondering if pyruvate can either store additional electrons or if it can get reduced to another substance and collect additional electrons. If pyruvate doesn't get reduced in any scenario, can ATP get reduced?

Role of ATP is to provide energy to reactions by P-O-P bond hydrolysis. It does not get reduced in redox sense.

ethanol is the other very common product of reduction of pyruvate (after it's been decarboxylated).  There are of course many other anaerobic fermentation schemes that are more complicated but all of them balance out the NADH.

Thank you, that is helpful but what is the chemical reaction for pyruvate to ethanol?


text:
Why is sodium hexafluoroaluminate formed instead of "tetra"?
$$\ce{AlF3 + NaF \to Na3[AlF6]}$$
My question is why don't we get $\ce{Na[AlF4]}$, since when we do the same thing with hydroxides instead of fluorides ($\ce{Al(OH)3}$), we end up with $\ce{Na[Al(OH)4]}$. I know that it is a 3rd period element so it can expand its octet, so boron being in the same group forms $\ce{NaBF4}$. I've also heard that aluminium is larger so it can accommodate more fluorine atoms around it. But why would it want to do that?

You can check these papers: https://www.semanticscholar.org/paper/On-the-physical-and-thermodynamic-stability-of-Bj%C3%B8rseth-Herstad/3950337480aa6b9bf37f5bb9ef4b592975b470c5 and https://www.semanticscholar.org/paper/The-Thermochemistry-of-Crystalline-M3AlF6%2C-NaAlF4%2C-Nikitin-Skokan/7789aecece126165c88bb74412a8373b17bf69df

In fact sodium tetrafluoroaluminate can be formed[1]. However, it requires quenching vapor from a temperature of 1000°C, definitely a non-equilibrium process.
The formation of the hexafluoroaluminate under more common conditions correlates with two features of this compound. First, it can form a relatively stable solid lattice, which formerly appeared in nature as the mineral cryolite. (Mining of this mineral for use in aluminum smelting has made it rare, and now the flux is made synthetically for industrial use.) The stability of the lattice is evident from cryolite being essentially insoluble in plain water, a rarity for sodium salts.
Second, $\ce{Al-F}$ bonding is more amenable to forming an octahedral complex than $\ce{Al-OH}$ or $\ce{B-F}$. As explained here, when an element with $s$ and $p$ valence orbitals forms an "extended octet" species such as $\ce{SF6}$ or the formally isoelectronic $\ce{AlF6^{3-}}$, there are antibonding interactions between the ligands which must be minimized. Using a larger central atom (aluminum instead of boron) and ligands that hold the electrons more tightly (fluoride instead of hydroxide) does that minimization.
Reference

E. H. Howard (1954). "Some Physical and Chemical Properties of a New Sodium Aluminum Fluoride", J. Am. Chem. Soc. 76, 8, 2041–2042.
https://doi.org/10.1021/ja01637a003


text:
How does borane target a specific double bond on an indene?

According to the text I am using (Organic Chem, Seyan Ege, 3rd ed, problem 12.57) the halide in the electrophile borane (BH3) binds at site G. But I need to make sure I understand why. First, I believe that the double bond F=G is targeted to begin with because it is the double bond furthest from the electron-withdrawing OR group. Next, why does G get the halide and not F? I believe that is because if F gets the hypothetical positive charge, that charge can be distributed, and there is greater resonance than if the positive charge was localized at G. Is that correct? (Please note I am not a student asking you to do my homework, I am an older person working through the text on my own.)

Edmund: I'm not sure where halide comes into the picture. Hydroboration is a concerted reaction but not necessarily synchronous. In the transition state the new  G-boron bond is more developed than the F-hydrogen bond. Hence, there is residual positive charge on F as you surmise. This advantage is lost in the regioisomeric transition state. Also recall that $\ce{BH3}$ that boron is more electropositive thathan hydrogen.

I should also add that the aromatic ring is too stabilized to react with more borane


text:
Ambidentate Behavior of Cyanide
Why do haloalkanes give alkyl cyanides when treated with KCN but give alkyl isocyanide with AgCN?

It is because KCN(K is an alkali metal) is majorly ionic but AgCN is majorly covalent(due to metal-carbon bond as Ag is a transition metal). So in AgCN, N acts a nucleophile.

Note the mouseover hint at down pointing grey triangle as the default reason for the question downvoting:  "This question does not show any research effort; it is unclear or not useful." Just in case you would wondering why the question got downvotes or was even closed.

Looks like homework: https://www.toppr.com/ask/en-us/question/haloalkanes-react-with-kcn-to-form-alkyl-cyanides-as-main-product-while-agcn-forms-isocyanides/

Here is a review to start researching. "Nucleophilic isocyanation was first described in the late 19th century, but the advancement and extension of this chemistry are still limited. This is probably due to the difficulty in the selective formation of isonitrile over the thermodynamically and kinetically favored isomer, nitrile. Moreover, excess amounts of toxic metal cyanides are usually required for this transformation. Currently, the relatively reliable dehydration of formamides is most often chosen for the synthesis of isonitriles."

Thanks Karsten, I had Googled the question but wasn't satisfied with the explanation. Maurice's answer is much more complete. Will keep this in mind.

The $\ce{KCN}$ substance is made of two ions : $\ce{K^+}$ and $\ce{CN^-}$ The active ion is cyanide, and the active atom in cyanide ion is the carbon atom, which is charged $-1$. This negative C atom attacks haloalkanes to produce alkylcyanide.
On the contrary, $\ce{AgCN}$ is not made of two ions. It is not soluble in water. The bond $\ce{Ag-C}$ is mainly covalent. So the active atom in $\ce{AgCN}$ is the outer $\ce{N}$ atom. When interacting with a haloalkane, the $\ce{N}$ atom has a free doublet which can attack it via a $\ce{S_N2}$ mechanism to produce an isocyanide.

I tried to find out what the structure of AgCN is in solution (and in what solvent it is soluble in) but came up short. In the crystal structure, AgCN forms linear chains with similar Ag-C and Ag-N bond lengths, so the argument does not work in the solid state. AgCN is insoluble in water, but the complex $\ce{Ag(CN)2-}$ is soluble.

How can $\ce{AgCN}$ make linear chains, if $\ce{Ag-C}$ and $\ce{Ag-N}$ have similar bond lengths ? Does it mean that the X-ray structure is made of long and parallel chains looking like $\ce{Ag-C-N···Ag-C-N···Ag-C-N··· Ag··}$, etc. This does not prevent the $\ce{N}$ atom to attack the haloalkane to produce an alkyl  isocyanide.

https://pubs.acs.org/doi/abs/10.1021/ic9714697 shows AuCN and describes AgCN as very similar, "consists of rows of AgCN chains with alternating long Ag−C = 2.15(6) and short Ag−N = 1.86(8) Å." So the bond lengths are significantly different in the solid, but both shorter than a van der Waals contact.