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what is volume of produced gases in water electrolysis
I decided to try and determine the combined volume of O2 and H2 that could be produced through electrolysis of water. so i looked up a bunch of stats for the relevant chemicals;

H
O
H2O

g/mol
1.00794
15.9994
18.01528

g/L
0.08988
1.429
997

L/g
11.12594571
0.699790063
0.001003009027

and then tried to do the math;
I imagined 1L of water being completely consumed with no other reactions and assumed that it would be a ratio of 2:1 H2 to O2 so i took the 997g/L of water and divided it into, 997*(2/3) and 997*(1/3) then I multiplied those numbers the L/g of H and O respectively, added them up and got 7627.60881 which deferentially seems wrong. I must have misunderstood or gone wrong somewhere because it expanding over 7000x in volume (at atmospheric pressure) seems ridiculous.

To nit pick (1) The density of water and gases of course depend on the particular temperature. (2) Given 3 significant figures for the density of water you have a ridiculous number of significant figures in the L/g row of the table.

thats because i calculated the L/g figures in excel (where i was doing the math) and it went a bit nuts with the significant figures

Excel doesn't know anything about significant figures. Significant figures are a crude way to do error propagation, but the method does keep you from inventing such wildly false precision. I remember my days as a teaching assistant. A student reported that a sample was 115% Fe. I asked Doesn't that seem a bit strange?!? The student replied - But that is what the calculator said!!

@MaxW I actually saw something like ‘90–105 %’ for the content of one element on the label of commercially purchased chemical, sooo …

You're not as far off as you think!
Here's the reaction:
$$\ce{H_2O(l)->H_2(g) + 1/2 O_2(g)}$$
It's often useful to first do a "back-of-the-envelope" calculation before proceeding with the formal calculation, since that helps give you a better intution for the problem and, in addition, gives you a rough ideal of what the answer should be (and thus serves as a check against egregious errors).  So:
As a rule of thumb, under typical lab conditions (room temp, 1 atm pressure), gases take up ~1000 x as much volume as liquids.  And since we're creating 1.5 molecules of gas for every molecule of liquid, our expected increase in volume is ~1500 x.
Now let's do the formal calculation to determine the actual increase:
Density liquid water = 1 g/mL => specific volume liquid water = 1 ml/g.
1 mole water = 18 g => volume of 1 mole of liquid water = 18 mL = 0.018 L.
Volume of 1 mole of gas at 1 atm, 23 C:
$$V = \frac{n R T}{p} = \frac{1 mole \circ  0.082 \frac{L atm}{mol K} \circ (23+273) K}{1 atm} = 24.3 L $$
Since 1 mole of liquid water decomposes in to 1.5 moles of gas:
Volume of 1.5 moles of gas at 1 atm, 23 C = 24.3 L x 1.5 = 36.5 L
Finally, the volume ratio is:
$$\frac{36.5 L}{0.018 L } = 2030 \simeq 2000$$
Thus complete electrolysis of liquid water at 1 atm and 23 C will increase its volume by 2000-fold.

Please correct the units "volume of 1 mole of liquid water = 18 mL = 0.018 mL."

@M.Farooq Thanks for catching the typo.


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Iodometric titration
When you carry out iodometric titration for determining the Peroxide value of a phospholipid mixture, the end point is the solution turning colourless. But, after some time the colour reappears. Do you stop at the first discoloration or continue titrating with sodium thiosulphate?

After how long the color re-appears?

It takes little more than about 5 mins

As per the American Oil Chemists’ Society (AOCS) Recommended Official Method Cd 8-53."The mixture is titrated with 0.1M sodium thiosulfate solution with constant stirring until the yellow colour has almost disappeared, when a starch indicator solution (0.5 mL) is added. The thiosulfate solution is added drop-wise until the blue colour just disappears." Make sure you do a blank as well, and it is clear that first disappearance marks the end point.

It has been a while, but I think that the appearance of the color again is due to oxygen in the air reacting with iodide to make iodine. If I'm wrong some one will surely nail me on that point. It has been 50 years since I did such titrations in quant. :-)

Thank you so much. Yes, it is the same AOCS Method we are following. The method says stop titration till the colour disappears. But a colleague is doing it till the reappeared colour disappears , so his values are always higher. Where can I get the reaction of oxygen (from air) reacting with the iodide?

You have to do a blank titration as well as recommended by AOCS. Iodide is quite slowly oxidized by air in water but what happens in the presence of chloroform and acetic acid and some water, I don't know. Change the batch of both reagents and see if this happens again. If the blue color reappears in the blank after 5 min as you observe, then ignore the second reappearance.

Oh Okay. Will do that . BTW, I am taking blank. Thank you

Check if the blue color appears in the blank too.

source: Stack Exchange

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

sort: votes


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What concentration is phenolphthalein indicator ? 1% or 0.1%
I read some information about phenolphthalein indicator, some websites suggested phenolphthalein indicator of 1%, others mentioned phenolphthalein indicator of 0.1%. What is the difference of both of these concentrations when we use them in titrations?
I assume that 1-2 drops of phenolphthalein indicator of 1% can be used as an indicator but for phenolphthalein indicator of 0.1%, it needs to drop more with 4-5 drops to obtain the same effeciency.
Look forward to hear from you all

In my university, all our indicators are available at concentrations 0.1%. The students know they may need 3 - 4 drops for their titrations.

I remember the same, even from labs in my analytical chemistry focused high school. 1% is an unusual high concentration.

@Maurice Thank you for your answer. It is helpful to my experiments.

During titration of small amounts of acids, the molar amount of the indicator in 1-2 drops of $\pu{1 \%}$ indicator solution may not be negligible compared to the acid molar amount, affecting the result.
So for that cases, $\pu{0.1 \%}$ solution is used, to be able to dose smaller indicator amounts.
As the phenolphalein molar mass is about $M=\pu{318 g/mol}$, 1 drop of $\pu{1 \%}$ solution is equivalent about 3 drops of $\pu{0.01 M}\ \ce{NaOH}$.

Thank you a lot for your answer, even i don't understand well your explanation. All statements from other answers support for using phenolphalein indicator 0.1%. I really appreciated your help Poutnik.

@VanThanhDo And so does mine. unless perhaps for rare exceptions, pH indicators are not used at so high concentration as 1%. By adding pH indicator, you add an acid or a base to the one you want to determine. Your interest is this addition should be negligible.

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


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How is ECR (Electron Counting Rule) done in a structure?
I've been told how to apply the ECR to GaAs{001}(2×4) in my lectures but it makes no sense to me and I don't seem to be able to find any other sources that explain this. (This is a further question on another one posted here .)
Could anyone please tell me how this is done? Or give me some sources on how to apply the ECR?
"This question has been previously asked on the Physics Stack Exchange where it is bountied, but I was told that I should try and post it here as I might be more successful in getting an answer."

This is the explanation I have been given for the following structure:

Count the number of electrons, going down to two of As atoms level with the bottom of the trench and count per unit mesh. There are 8+4 As atoms = 60 valence electrons. There are 6 Ga atoms = 18 electrons. Hence a total of 78 electrons.
Count the bonds that need to be filled :
1. 3 As dimer bonds = 6 e
2. 6 As dangling bonds (2 per dimer) = 12 e
3. Ga dangling bonds (sideways into trench) are empty = 0 e
4. 8 backbonds from top As dimers = 16 e
5. 12 backbonds from Ga to 3rd row As = 24 e
6. 16 backbonds (bulk-like) from 3rd row As to underlying Ga: each takes 5/4 e from As giving 5×16/4 = 20 e
78 electrons for both totals so structure is neutral and should be stable.
I am trying to learn how to do this as I am trying to prove that this is not applicable to the GaAS{001}(2x1) structure, which I believe to be represented like the image below from the help I was given at a previous related question, here.

You do know that cross-posting is frowned upon? BTW that bounty on Physics is probably serious waste of rep.

I was informed to do so, I didn't know that it was frowned upon, my apologies. I shall take the question. And it does not seem a waste of rep for me as I do not know how to do this and would love to learn.

Mithoron, if you could help me that would be very appreciated. Thank you for telling me about the rules.

Umm, GaAs is simply tetragonal and I don't see a reason for some electron counting.


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Porous materials: Is the expression of a mean free path $\overline{\ell}$ of the gas molecule valid for a binary mixtures?
Suppose that I have a binary gas mixture at low pressure. Can I calculate the mean free path $\overline{\ell}$ of the gas molecule in a pores of solid material by substituting a molecular weight of a single component by Molecular weight of the binary mixture $M_{m,\rm mix}$ as well as the specific gas constant of this mixture $R_{\rm mix}$, such that I get:
\begin{equation}
\overline{\ell} \:=\: \dfrac{3.2\cdot \mu_g}{p}\cdot \left(\dfrac{R_{\rm mix}\cdot T_{\rm abs}}{2\pi\cdot M_{m,\rm mix}}\right)\:\:\:\rm\left[m\right]
\end{equation}
I am calculating $\overline{\ell}$ in order to determine the main diffusion mechanism in the pores (Bulk diffusion (Fick), Knudsen or its combination).

What is the size of pores ? Is is much greater than the mean free pass in a free gas ?

Mean radius of the pore is $R_p = 8\cdot 10^{-9}:\left[\rm m\right]$. If used the equation above for the mixture, then the ratio is $\sim 0.725$, which would suggest a combination of Fick' and Knudsen diffusion. However, when I determined a Knudsen number $\rm Kn = \overline{\ell}\big/ d_p$, I obtained $\rm Kn > 10$, which would suggest Knudsen diffusion to be dominant.


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How much lipoic acid is contained within octanoic acid?
As I understand it, lipoic acid is synthesized from octanoic acid:

Lipoic acid is synthesized de novo in mitochondria from octanoic acid, an 8-carbon fatty acid (C8:0), bound to the acyl-carrier protein (ACP; see article on Pantothenic Acid) during the process of fatty acid synthesis (Figure 2).

However, I am wondering how much is actually contained within octanoic acid? Such as if you had 100 ml of octanoic acid, how much lipoic acid would you consume?

Whatever the acid chosen, it cannot "contain" another acid. It does not contain any other substance

I would be very surprised if you ever found any detectable amount of lipoic acid in any prep of octanoic acid, even if it were purified from a biological source. Lipoic acid concentrations are never high in the cell relative to other fatty acids, and any procedure for isolating octanoic from other acids would easily remove it. The typical challenge is removing other acids with slightly longer or short chains which may be fairly abundant depending on the source oil.

@Andrew Can you please explain then how is it derived from octanoic acid?

As with most biological molecules, there are enzymes that have evolved to perform this task. Look up lipoid acid biosynthesis ina biochemistry text or online for details

The phrasing is a bit confusing, perhaps. When you ask how much lipoic acid you would "consume", did you mean "produce"? Also, chemical reactions are... kind of like recipes, but they aren't recipes: firstly as z1273 rightly mentioned one would typically measure in terms of moles, not 100 ml, and secondly we don't really talk about a reactant being "contained within" a product like how a cake contains 150 g of chocolate, because when the reaction is done, the reactants cease to exist — they aren't a thing anymore. This is what Maurice's comment means, and why there's some confusion in ...

... in the answers about what you're actually asking about. The two answerers have quite different interpretations of your question. Andrew is answering the more literal question of "how much lipoic acid can you find in an extract of octanoic acid", and z1273 the implied question of "how much lipoic acid is used to make one molecule of octanoic acid". Which is your question? I don't mind helping to edit your question, but I can't really tell. Ping me if you need any help.

@orthocresol Thanks for your well thought out response. I'm not in chemistry so perhaps I just got some things confused (which seems that way). The reason why I was trying to understand how much "lipoic acid" was consumed was because I heard it was derived from octanoic acid; so I thought it was contained within it or something of that nature, much the way they use fractionation as a process to separate MCT oil from coconut oil. But seems lipoic acid isn't "available" unless a process is used to derive it from octanoic acid? Hope that makes sense.

Ah, ok. Yeah, that's not what "derived" means. Here, "derived" doesn't mean "physically extracted" like how one might extract a component of crude oil; it's more like "synthesised", as in a machine (well, a series of proteins) which you feed octanoic acid to, it does a bunch of things with it, and spits out lipoic acid. So indeed, as you say, there is no lipoic acid "available" before these proteins work their magic on octanoic acid.

Ahh understood, I get it now, thanks :)

Background
One lipoic acid molecule is enzymatically produced from one octanoic acid molecule in the mitochondrion or in the bacterium by the enzyme lipoyl synthase (LipA).

To get there, first the octanoic acid must be covalently linked to the small protein acyl carrier protein (ACP). This attachment is made at its carboxylic acid end (right end of the molecule shown above).
Then, the octanoate-AcylCarrierProtein binds to the LipA enzyme. LipA contains a [4Fe-4S] center that reacts with the lipidic end of the octanoic acid (left end of the molecule above). The iron-sulfur cluster is highly reactive. At the outset of the reaction, the enzyme leaves one sulfur atom of its iron-sulfur cluster to the third carbon atom from the left and another sulfur atom to the terminal CH3. The resulting product is released from LipA.
Separately the ACP is removed and the two sulhydryl groups chemically oxidize together to a disulfide bond, thus producing the Lipoic acid molecule as illustrated above.
I recommend examining the Scheme 2 in McLaughlin et al., 2016 for details of the catalytic mechanism in a bacterium, Mycobacterium tuberculosis leading to the Lipoic acid product, in the acyl carrier protein bound form.
OP questions
a) The whole of the Octanoic acid, except for two hydrogen atoms are included in the resulting lipoic acid. So it is one to one in molar ratio.
b) I think it may be easier to refer to molar equivalents instead of volume equivalents.

OPs question is how much lipoic acid is contained in octanoic acid, not how many atoms of octanoic acid are incorporated into lipoic acid.

@Andrew I think the OP further specified their question with 'Can you please explain then how is it [lipoic acid] derived from octanoic acid?" Hopefully the details I provided help along those lines, particularly as I tried to place it in the context of the original post. Just trying to help..

Fatty acids that are purified from biological sources are typical isolated from oils or fats, both of which are comprised of triglycerides. That is, there are three fatty acids attached to a single glycerol molecule via ester linkages.
Different oils and fats contain different amounts of each fatty acid. For example, a good source of octanoic acid is coconut oil, of which about 8% of the fatty acids are octanoic acid.
Production of pure octanoic acid requires hydrolyzing the oil and using various refining techniques to separate the fatty acids by length.
Triglycerides are formed at the end of the biosynthesis of the individual fatty acids. Although lipoic acid is derived from octanoic acid, the octanoic acid used for this purpose is diverted into the lipoic acid pathway before incorporation into triglycerides. Thus, the triglycerides used for octanoic acid isolation are very unlikely to contain any lipoic acid.
Instead the biosynthesized lipoic acid is contained in the cell attached to various proteins that use it as a reactant in the reactions they catalyze.


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Base/acid reaction -> alcohol amphoteric? why no pkb value?
I have a question concerning acid/base reactions. Is it true that an alcohol is a better base than water?
My Professor listed the basicity (because we talked about E1/E2 and Sn1/Sn2 reactions) of the following compounds the following way: RO- > OH- > ROH > H2O (with stronger base being the RO- and the weakest here H2O). I would like to understand that with respect to pka/pkb values...
Now I came across the following problem:
Bronstedt Bases/Acids are determined by their equilibrium constant in WATER (pka/pkb). Alcohols (like ethanol have a pka about 16-19), water has a pks of 14, which makes alcohol a base by definition.

if that happens -> ROH + H2O -> ROH2(+) + OH(-), wouldn't you calculate the pkb for ROH and the pks for ROH2+? The alcohol is the base here? Unfortunately I haven't found any pkb value for ROH on the internet... (I don't mean the conjugated base of ROH!) - alcohols are even considered amphoteric? how is that possible? why do I only find pka (and no pkb) values then, like for ammonia?
Ammonia is considered amphoteric and has a pka and a pkb value. But if these equilibrium constants are determined in water, how can it have a pka value, if it only acts as a base there? Shouldn't be water the only molecule that can be considered amphoteric in water?
If we would say molecules like H2O and NH3 can "generally" donate a proton or take up a proton and because of that are considered amphoteric, aren't there many more molecules than can take up and donate a proton when a strong acid/base is present? wouldn't there be many more amphoteric molecules?

So I'm really trying to understand why if alcohols can act as base AND as acid, why there are not two values for that (like for water (pka and pkb=14) and ammonia (pka=23 and pkb= 4,75), even though that would require two different reactions in alcohols too:
ROH -> ROH2(+)
ROH -> RO(-)
I hope I haven't confused you too much and explained my problem at least somewhat clearly. :)

Some pKa tables will list values for $\ce{R-OH2+}$ compounds, typically with values around -2 to -4.  The pKa of $\ce{H3O+}$ is usually given as around -1.75.  You can convert these to pKb if you prefer to work with that scale. Also remember that the values for water and its derivatives are distorted by water being the bulk solvent, so the values are not calculated using the normal 1M of everything. There are a number of questions on this site about the "actual" pKa of H2O if you need a refresher.

Thanks so much!! I guess I was just overthinking too much. Getting the pkb of the conjugated base of R-OH2+ totally makes sense. :)


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Why did my vinegar not create the traditional volcano?
At the weekend, my son and I tried the volcano experiment. You know the one: vinegar introduced to sodium bicarb (and a little red dye) within a conical "volcano".

Not actually ours; photo borrowed from here.
So, after assiduously funnelling some (slightly clumpy) bicarb into the volcano and meticulously counting drops of red food colouring... nothing happened? You can imagine our disappointment.
As luck would have it, we have two types of vinegar in the house. So I went about swapping the "non-brewed condiment" (more on this below) that we had used previously for some fancy balsamic vinegar. Lo and behold, this time we were treated to an eruption to rival Vesuvius!
I'm left to conclude that the first vinegar we tried must have gone bad somehow. It still looked, smelled and tasted like malt vinegar, but why else would it have failed us when we needed it most?
So, my question: Why would our vinegar have failed to react with sodium bicarb?

Note 1: In the UK, "non-brewed condiment" is the glamorous name for the cheap type of vinegar that is very common in fish & chip takeaway shops. It's basically identical to malt vinegar. (In fact, I only realised mine wasn't malt vinegar when preparing this question!) It's a brown, semi-transparent liquid with a sharp taste and smell. The principle ingredients are water, acetic acid, and caramel for colouring. Since acetic acid is the same acid found in real (i.e. brewed) vinegar, surely it shouldn't matter whether it's real vinegar or not?
Note 2: For completeness, it's worth mentioning that traditional (in the UK at least) vinegar bottles aren't really very air-tight. It has occurred to me that the acetic acid has been subject to some sort of chemical/physical reaction with the air. (Perhaps oxidation? Or even just evaporation?)

At a guess I would think your vinegar is very dilute. Acetic acid cannot be oxidised further by air

@Waylander, see https://www.tandfonline.com/doi/pdf/10.1080/00380768.1995.10419569, bacteria that oxidize acetic acid.

Acetic acid can be decomposed by methane-producing bacteria. Note that the citation is for soil bacteria, but these methanogens can be found almost anywhere.  Indeed, there is an entire microbial community in vinegar, including nematodes (worms). The more "natural" the product, the more organic material to nourish the microbes and to speed decomposition.
That said, check the labels for acetic acid concentration when purchased. The nominal concentration for distilled vinegar is between 5% and 10%, but some types of vinegar may be only 4% acetic acid.

This suggests next week's experiment: how much can we dilute the balsamic before it stops working? Thanks for the answer Dr Pippik!

@TomWright if the balsamic is true I would buy something untrue for experiments. I think is worth trying again with a non brewed one.


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Why doesn’t AlCl3 form the dative bond within the same molecule?
AlCl3 is said to be an electron deficient molecule as Al only has 6 electrons in its valence shell. Why doesn’t it just form a dative bond with one of the chlorine’s? I know this gives the more electronegative chlorine a positive formal charge, but then the diner Al2Cl6 shouldn’t exist either.
How does forming a dative bond with a chlorine from a different molecule instead of its own  make the situation more stable?

It does - https://www.chemguide.co.uk/atoms/bonding/dative.html

You misunderstood my question. I’m talking about the dative bond coming from the same molecule

If you wonder why dimerisation of AlCl3 is exothermic then the wording of your post isn't particularly good.

I’m not asking if the dimerisation itself is exothermic. I’m asking why Al doesn’t form a dative bond with one of its own Chlroines, when not in a diner state, to complete its octet

:/ AlCl3 has 3 partially double bonds.

They kind of do. Martin's answer here https://chemistry.stackexchange.com/a/64600/16683 shows that although the pi backbonding is not as strong as in BCl3, there is some degree of pi backbonding from Cl to Al in the monomer. You need to think in terms of "somewhere between one and two bonds". The followup question would then be why it prefers to dimerise rather than stay on its own, but that's addressed by the linked question.


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Why does lower concentration cause the metal atoms to dissipate more?
In a concentration cell, one can measure a voltage because in one of the half cells (the one with lower concentration electrolyte), the atoms tend to dissipate more into the electrolyte as ions, leaving more electrons behind (compared to the other half cell), thus creating a current.
I am not sure if I understand, why a lower concentration causes the atoms to dissipate more. Is it because the atoms of the metal have more contact to water?

Think about it as the net effect of 2 opposite processes.

It is not primarily that the process $\ce{M(s) -> M^n+(aq) + n e-}$ is faster for more diluted solutions.
The main reason is that the opposite process $\ce{M^n+(aq) + n e- -> M(s)}$ is slower for more diluted solutions.
As the net effect, there is dissolution of $\ce{M^n+(aq)}$ ions and the electrode gaining more negative potential, until the rate of both processes gets equal and the electrode reaches its equilibrium potential.
Because the rate of ion dissolution grows with increased potential,
while the rate of ion depositions grows with decreased potential.
The final effect is, the electrode in then more diluted ion solution has the more negative potential.


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Photochemical reduction of benzophenone: why inverted flask?
I've come across an interesting protocol for photochemical reduction of benzophenone to benzhydrol (diphenyl methanol) with sodium alcoholate in alcohol, the 2nd protocol in https://prepchem.com/synthesis-of-diphenylmethanol/. The reaction proceeds via ketyl radical initially to benzopinacol, which is immediately cleaved by a catalytic amount of sodium alcoholate to the benzhydrol and benzophenone. The latter repeats the reaction until it's used up. The alcohol acts as a reducing agent and solvent.
I'm particularly interested in the following part:

the flask is filled to the neck with more of isopropyl alcohol ... The flask is tightly stoppered and inverted upside down in a beaker.

I want to know why is the flask inverted?
My best guess is to avoid exposure to air, since that would interfere with the reduction. But then why isn't it enough to just stopper the full flask?
Might this have more to do with the quality of exposure to the light source (neck and stopper producing a shadow in the upright position)?
Also, is there some official term for this inverted flask technique?

Notes (extra info for the interested):

The ketyl radical (which is actually a radical anion) is brightly colored, which provides a convenient way to monitor the reaction. Upon exposure to air, the color disappears, presumably because the radical anion gets oxidized back to the ketone.
Any strong enough base should work for cleavage of the pinacol. In a related synthesis where benzopinacol is the desired product, a small amount of acid is added, because the glass itself is alkaline enough to cleave the product. The sodium alcoholate is convenient in that it's the conjugate base of the alcohol. It's formed by reaction of the alcohol with sodium metal.
Cleavage of the pinacol proceeds by deprotonation to the pinacolate anion. This rearranges to the ketone and alkoxide, which protonates to the alcohol, regenerating the base.

I think the reason for inverting the the round flask is to capture sunlight very efficiently. A round bottomed flask filled with a fluid is like a ball lens. Sun rays can be considered parallel and see how they focus. If light is coming from all directions, every part will be illuminated well as the Sun moves. See ball lenses here ball lenses

The clue comes from the instruction in the procedure given in your link "Then the flask is exposed to the best available sunlight. "

Hollow glass container filled with water

Surely if all of it is in sunlight it would not seem to matter if the flask was upside down or not. It might be best to rotate it to get even absorption. If only a part illuminated then placing the neck where the focus is in your diagram would be best.

@porphyrin I guess the round flask rest on the beaker walls and has to be pointed to the sun disk.

Wouldn't you get the same effect by simply tilting the neck out of the way of sunlight (not inverting completely)? I guess inverting is less error-prone and more reproducible, especially with a moving light source such as the Sun?

Neck is like a cylinder, but inverting the flask completely is a matter of convenience along with ball-lens effect.


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Why is H2O a weaker acid than acetylacetone? Shouldn't a hydrogen connected to an oxygen be more acidic than a hydrogen connected to a carbon atom?
I recently started studying chemistry after a long pause in my education so pardon me if it is a simple question.
In the organic chemistry course I am taking, the professor told us that the first factor defining the stability of the conjugate base (and thereby the acidity of the compound) is the atom that is connected to the acidic hydrogen. By this definition, the hydrogen in $\ce{H2O}$ should be more acidic than the hydrogen in acetylacetone as the first one is connected to oxygen but the second one is connected to carbon (as oxygen is more electronegative than carbon).
My professor mentioned other factors such as the existence of resonance structures, inductive polarity, and the effect of side chains, but she mentioned these factors come next and only if both hydrogens in our comparison are bound to the same element.

Compare comparable. Water is stronger acid than e.g. methane. And If you replace the central -CH2- by -OH, then resulting acetic acid is stronger acid than acetylacetone.

Your instructor is right, up to a certain point. There are factors that are more important (the atom to which it is bound) and factors that are less important (resonance, etc.), and it's good to be able to order these relative factors. The devil is in the details, though. If your resonance stabilisation is strong enough (and in this case it is) then it can overcome the more important factor.
Before starting, let me point out that CH's are less acidic than OH's, but this isn't actually to do with electronegativity. If it was about electronegativity, then OH's should be more acidic than SH's as O is more electronegative than S, but the opposite is true. Same applies to HF and HCl: F is more electronegative but HF is actually a weak acid. It has more to do with the X–H single bond strength than anything, as a weaker bond is easier to break (homolytically). The X–H heterolytic cleavage, relevant for an acid–base reaction, is sort of like a homolytic cleavage, plus an electron transfer from H to X. The electronegativity comes into play in the electron transfer step, but it's just not so important a consideration, compared to the bond strength which describes the homolytic cleavage.

I'm giving pKa's to the nearest 5(ish) in this discussion. The exact numbers don't matter too much, we're more interested in broader trends.
If you compare water and methane, their pKa's are ~15 and ~50 respectively, which is a difference of about 35 units. That's roughly how much the adjacent-atom factor counts for, when comparing C and O.
If you then compare acetone and methane, pKa's are ~20 and ~50 respectively, which is a difference of about 30 units. Not quite as big as the previous one, which was 35.
Acetylacetone, though, has two carbonyl groups, that's twice as many as in acetone. Now, resonance stabilisation doesn't scale linearly in that fashion and hence the additional stabilisation over acetone is only ~10 units. However, it's still enough to bring the pKa down to ~10, which is more acidic than water.

We really need to think carefully about a lot of things when thinking about acid/base reactions: charge, resonance/aromaticity, electronegativity, hybridization, and occasionally things like sterics (things like proton sponge) or reversibility (for instance where one product is emitted as a gas like in the case of NaH). In many cases these factors are interconnected.

The primary stabilization factor in the deprotonated form of acetylacetone is the delocalization of the negative charge via resonance onto the oxygen atoms (both delocalizing charge across several atoms and placing the negative charge on the more electronegative oxygen atoms). Secondarily, the resultant anion forms a conjugated system. The actual structure is a weighted linear combination of three structures, called a resonance hybrid, where two of the constituent resonance structures place the negative charge on a the electronegative oxygen atom are weighted more heavily.


text:
What starts the reaction in a voltaic cell?

In a zinc-copper voltaic cell,
What causes the initial electron movement from the zinc to the copper?
I understand that copper would pull the electrons more strongly than zinc, but only if it is electron deficient. Both half cells of the voltaic cell are electrically neutral (same amount of electrons and protons), so I don't understand what would cause the electron transfer / oxidation.
Also, I understand that $\ce{CuSO4}$ solution in the cathode half cell is needed so that the $\ce{Cu++}$ will accept the electrons, but what's the function of the $\ce{ZnSO4}$ in the anode half cell?

The fact that the chemical potential of the two separate half is not the same. I don't claim it is simple. But you can ask the same for every reaction not only for redox ones in an electrochemical apparatus.

For both electrodes, regardless of their galvanic connection, are ongoing the opposite processes
$$\ce{M(s) <=> M^n+(aq) + n e-}$$.
The forward process charges the electrode negatively, but slows down at more negative potentials.
The backward process charges the electrode positively, but slows down at more positive potentials.
As the resulting net effect, an electrode establishes such an equilibrium potential, where both processes have the equal rate and the zero net effect.
These equilibrium potentials are different for different combinations of electrode material and solution composition.

When you connect 2 electrodes ( connected also internally by a free ion passage ) by a galvanic circuit, different electrostatic potentials lead to the electric current. The current disbalances the equilibrium state on electrodes and one of the respective processes takes the upper hand.

That's the core of my question. Where is the difference in the electrostatic potential? If we count the electrons and protons in each half cell, they're the same, so what would still cause the electrons to move?

They are not exactly same and there are differences within the half cells, between electrodes and electrolyte.

We started with neutral electrodes and neutral electrolyte, where did the difference in potential came from? I'm sorry if I wasn't clear in the original question, but that is exactly what's confusing me. If I'm able to see how the electrostatic potential is created, all the other pieces will fall into place for me.

Dissolving a metal electrode charges it negatively and solution positively. The opposite process - metal depositioning - does the opposite. The metal with the greater tendency to dissolve causes the electrolyte to be more positive than the one in the other cell, cause positive ions ions to migrate there and negative here. As result, the net charge of half cells is not zero anymore and neither the charge of the electrodes.

The "spectator" ions like $\ce{SO4^2-}$ do not take part in electrode reactions. They electromigrate according to the local solution electrostatic potential gradient away from the $\ce{Cu}$ side, where they are becoming excessive and forming slightly negative solution charge. They pass the salt bridge ( or a porous membrane or a diaphragm) to the $\ce{Zn}$ side, where they are needed to counterpart being created $\ce{Zn^2+(aq)}$ to cancel slighly positive solution charge there.

When you connect 2 electrodes by a galvanic circuit, different electrostatic potentials lead to the electric current. - That's the core of my question. Where is the difference in the electrostatic potential? If we count the electrons and protons in each half cell, they're the same, so what would still cause the electrons to move?

We started with neutral electrodes and neutral electrolyte, where did the difference in potential came from? I'm sorry if I wasn't clear in the original question, but that is exactly what's confusing me. If I'm able to see how the electrostatic potential is created, all the other pieces will fall into place for me.

I think your iterpretation of electrochemical cell is incorrect.
Firstly, the electrode is not dipped
in a neutral electrolyte. You must have seen that whenever an electrolyte is represented for an electrochemical cell, it always has "(aq)" attached to it. This means that the electrolyte, say ZnSO4 is not as a neutrally bonded compound (as i think you are interpreting the electrolyte), in fact, it is in form of its ions Zn²+ and SO4²-. This is the key to the difference in charges which initiates the reaction.
Consider an electrochemical cell:
Zn|ZnSO4(aq)||CuSO4(aq)|Cu
We can see that the Zn (neutral) electrode is dipped in a sea of Zn²+ and SO4²– ions. Sometimes, the SO4²– ions attract the Zn in the neutral electrode (Note that at molecular level, "sometimes" can even mean a million times because there are just so many atoms/particles/species) which falls into the electrolyte as Zn²+ towards that attracting SO4²- ion. Since so many Zn²+ ions leave the electrode (again, sometimes at molecular level means a lot of times), the electrode acquires a negative charge. These electrons are attracted by the cathode... Voila! the cell reaction has been initiated.
Also consider that previously there were equal amount of Zn²+ and SO4²- ions in the electrolyte so that the net charge of the electrolyte is zero. But since "some" SO4²- ions attracted more Zn²+ ions from the electrode, the electrolyte becomes positively charged. To satisfy the positive charge (so that electrolyte becomes neutral again) an equal amount of (an amount equal to the amount of extra Zn²+ in the electrolyte) negatively charged ions are dropped from the Salt Bridge to the electrolyte (Say if the salt bridge is KCl, then Cl- ions will be dropped).
What keeps the cell reaction going once it has been initiated?
Notice that there are less number of Zn atoms in the electrode as some have been pulled into the electrolyte by SO4²- until there are so much extra Zn²+ ions in the electrolyte that SO4²- hardly pulls any more atoms from the electrode. This should stop the cell reaction. But the twist comes with the arrival of even more negatively charged ions (Cl- ions from the salt bridge). These extra negative ions can pull even more Zn²+ from the neutral Zn electrode.
Once again the electrons remaining in the anode due to loss of these Zn atoms flow to the cathode and even more Cl- ions are dropped into the electrolyte to again neutralise the electrolyte and the reaction goes on.
Why do we use the "(aq)" notation after the electrolyte to represent that it is in ionic form instead of just writing it as ions? (like
Zn|Zn²+SO4²-||Cu²+SO4²-|Cu)
Well in some cases the electrolyte is written just as ions but there are two good enough points (i think) as to why the (aq) is used to denote the ionisation of electrolytes.
First off, "aq" generally implies that the substance is dissolved in water. Whenever an ionic compound (electrolytes considered in cell reactions are ionic) is dissolved in water, it dissociates into separate ions. This property is so common infact that in electrochemistry, "aq" denotes ionisation more than it denotes solution in water.
The second point though is based on my opinion. In science we generally try to shorten our representations and since, in an aqueous solution, most ionic compounds dissociate into ions, and since electrolytes are ionic and need to be in ionic form for the cell reaction to happen, why not just use "aq" to denote the ionisation of electrolyte as as well!
Hope this answers your question! :D

You may prefer a "human-like" picture. You may prefer to interpret the chemistry of the cell in saying that the zinc atom "feels like" losing its electrons to become a zinc cation. On the other side, the copper electrode has not at all the same tendency. It is even ready to accept foreign electrons to discharge its own cations. I know this explanation is not very scientific, and I beg your pardon for stating it. From my experience in teaching chemistry to unmotivated high school students, I know it helps.
Now the function of the $\ce{ZnSO4}$ in the anodic solution is to help electrically speaking, because pure water is an insulator. If the anodic compartment were just pure water, the first $\ce{Zn^{2+}}$ ions created in the water from zinc metal would prevent the next ones from appearing in solution, and the cell would soon stop working. But if there are already some $\ce{Zn^{2+}}$ in the solution, the newly created $\ce{Zn^{2+}}$ ions would simply push the already dissolved cations through the solution up to the bridge where the sulfate ions are coming from the cathodic region.

the zinc atom "feels like" losing its electrons to become a zinc cation - but why that happens? If we count the electrons and protons, they're the same amount. Is it caused by something that doesn't boil down to electrostatic energy? If it is, then what is it?

@Toi Yiot This happens because redox potential is negative


text:
Is incineration of a solid fuel complete or incomplete?
If I have some solid material like biomass and incinerate it at 1000 Celsius degrees for 15 minutes in an oxidized atmosphere within an incineration oven. As an output it gives me ash. Is the incineration complete or incomplete? Like for example we have incomplete combustion (lack of oxygen) and complete combustion (enriched oxygen medium). What about this incineration?

Organic substances have a tendency to be first transformed into some form of black charcoal when being overheated. Then, if the combustion is well done, this charcoal burns producing CO2. The remaining ashes are made of mineral resides made form calcium and potassium carbonates, plus some oxides like rust, alumina and silica. A part from rust, these substances don't have any color. So if your ashes are black, the combustion is incomplete. If your ashes are grayish brown, the combustion is complete.

The comment by Maurice is perfectly good for analytical purposes.
For commercial use, in practical applications, the ash residue is not the only product of combustion that must be completely oxidized. The gaseous products should be $CO_2$, $H_2O$ and $N_2$, maybe a little $SO_2$ - but sometimes organic compounds volatilize before burning and escape. In backyard leaf-burning, or in fireplace fires, this escape is mostly innocuous (except for creosote buildup in chimneys), but when tons of garbage are pyrolyzed, the gaseous products can be obnoxious. In some cases, extra fuel is added to a pyrolytic garbage treatment to raise the temperature of burning and reduce noxious emissions, but that increases the cost.
So if you are going to ask about complete combustion, you might analyze for $CO$, nitrogen oxides, and other volatiles that may have escaped.


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How are the toxic components in a mixture identified?
I was doing research on buckwheat sprouts (as to whether I should include them in my diet) and I came across a website that said eating too much can cause fagopyrism, which is caused by compounds called fagopyrins:

I was just wondering how they came to the conclusion that it was these compounds that were dangerous? If someone reports side effects, with the number of different chemical compounds in a plant, how would they narrow it down to that one?
Do they just toxicity test all of the possible compounds? Or do they look for certain functional groups and structures to narrow down their search? Ie is there a way you can predict from the structure that a compound could be potentially toxic?
Thanks!

I changed your title as that seems to be the real basis of your question. Your train of thought is good, but by the end of it you have quite a different question, which can have quite a different answer: it goes from "how do you know which component is toxic" to "how to you predict toxicity from the structure".

Drinking too much water would kill you(like the man advertising drinking many litres daily.). Does it mean drinking water is dangerous ? It is not. Drinking/eating too much of anything would cause some problems or others. Only the dose and circumstances decides if it is a poison. E.g. drinking of infusion of Hypericum perforatum is generally very healthy. But hypericine may kill you if combined with excessive sunbathing, what was abused in ancient Rome. Fagopyrism is similar case.

Fagopyrin is located almost exclusively in the cotyledons of the buckwheat herb. When ingested, fagopyrins cause sensitivity to light.

All what you suggest is in part true. I do not think you are asking about technical procedures like analysis isolation etc but rather on how to assess which compound is toxic. Ubiquitous or largely diffused molecule can be let apart. Then one focus on the characteristic ones. Plus unfortunately comes experimenting...

You have many questions nested together. Whenever you have a mixture + identification of components in your mind, bring chromatography to your mind. You must be familiar with paper chromatography. Same idea, but applied to more sophisticated instruments.
When you have a completely unknown mixture, you start a so-called screening procedure with a high-performance liquid chromatograph and a mass spectrometer. The HPLC separates the components, and the mass spectrometer identifies the molecules from the fragmentation pattern and molecular mass.
You start to see how many components are there in the HPLC-MS chromatogram. Let us say you had natural products and you see 100 peaks (which is not unusual at all).
Later you scale up the process, go at a preparative scale and start collecting fractions of mixtures. You perform biological tests on them, and the relevant fraction is further studied by more advanced tools such as recycling the HPLC-NMR system.
The question you ask is equivalent to a decent Ph.D. project in natural product isolation and characterization followed by investigating their biological activity. In pharmaceutical companies, HPLCs run day and night with a typical HPLC may have 10-15 columns for screening as to which column under which mobile phase will produce the best separation.
I did not check how fagopyrins were discovered, but the process is general. Wiki gave a hint on the first German reference Fagopyrin, ein photodynamischer Farbstoff aus Buchweizen (Fagopyrum esculentum) = Fagopyrin, a photodynamic dye out of buckwheat, locked behind paywall. This procedure is from 1950s, when HPLCs did not exist but the technique of chromatography was very well known.

Wir haben aus einem Methanolextrakt von getrockneten Buchweizenblüten
durch fraktionierte Verteilung und chromatographische Adsorption einen
photodynamisch wirksamen Farbstoff in Form eines dunkelroten, fast
schwarzen mikrokrystallinen Pulvers isoliert, der im folgenden als
Fagopyrin bezeichnet ist.
We have isolated from a methanol extract of dried buckwheat flowers by
fractional partitioning and chromatographic adsorption a
photodynamically active dye in the form of a dark red, almost black
microcrystalline powder, hereafter referred to as fagopyrin.


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Physical significance of chemical potential and fugacity
I'm studying equilibria and thermodynamics and came across these two terms.
My problem is, unlike other thermodynamic properties that I can understand physically like volume, pressure, enthalpy etc. do these two quantities have physical significance or, is it just that we define fugacity (without any physical meaning) just to make sure the equation
$$\mu - \mu_0 = RT\ln\frac{p}{p_0}$$
looks the same for real gases? If yes, then what is the motivation for the name “fugacity” meaning literally “escaping tendency”?
Similarly, do we just define chemical potential randomly, since it's extremely useful in various calculations when the composition varies, or does it have some physical significance?
Also, why do we call it chemical potential?

I have written a very detailed reply as to why fugacity got this name in a previous Q&A post [What is fugacity?][1]. I think the concept of chemical potential can also looked from a historical point of view. Wikipedia has it https://en.wikipedia.org/wiki/Chemical_potential#History

The definition of chemical potential is not arbitrary (random).

The chemical potential is the partial derivative of the Gibbs free energy with respect to the number of moles of the specified species at constant T, P, and numbers of moles of all other species.  The equation you wrote is the what this partial derivative reduces to for a species in an ideal gas mixture, where P is the partial pressure of the species in the mixture.  For a real gas, the fugacity of the species replaces the partial pressure of the species in the same equation for the chemical potential.

Chet, Still what is the physical consequence or meaning of a partial derivative of Gibbs free energy with respect to moles.

@M.Farooq I really don't ascribe any physical significance to it, although, in combination with the chemical potentials of the other species in the mixture, it determines the driving force for interphase mass transfer of a species and for chemical reaction.

@ChetMiller thanks for the answer, but do you mean fugacity and chemical potential have no physical meaning by themselves and we just define them as mathematical terms like  $(dG)/(d{n_i})$ since they are important/handy tools??

Saying chemical potential has no physical meaning is IMHO like saying molar heat capacity or molar conductivity have no physical meaning.

Well @Poutnik, molar heat capacity is the partial derivative of either molar internal energy or molar enthalpy with respect to temperature.  And, as I said, chemical potential is the partial derivative of Gibbs free energy with respect to the number of moles of a species.  Is that the kind of physical meaning you were referring to?

@ChetMiller, I think there must be a physical interpretation of the partial derivative of Gibbs free energy with respect to moles. Do you agree with the Wiki explanation "In thermodynamics, chemical potential of a species is energy that can be absorbed or released due to a change of the particle number of the given species, e.g. in a chemical reaction or phase transition."?

@M.Farooq I guess it can be interpreted that way, although that is the same as the partial derivative I described.  And, I'm not sure I would consider it energy in the same sense as internal energy.

@ChetMiller I am aware of what they are, so I agree with you. I have written them as a close analogy, where either all of them have physical meaning, either none of them.

@M.Farooq One of intepretation is the rate of increase of work the system can do, when it is scaled up.

By layman terms, chemical potential is a measure how much "powered up" a substance is at given conditions. E.g. if we compare the same concentrations  then the same ions in well solvating solvent or the same molecules in well dissolving solvent are less "powered up" than ions "almost naked" or molecules barely dissolving. Difference in chemical potentials at the same concentration leads to different distribution constants between solvents.

In less mathematical terms, the chemical potential of a component in a mixture at a specific temperature and pressure is the amount of free energy that can be ascribed to one mole of that component under the conditions found in the mixture at that T and p. Alternately it is how much the free energy of the system, if it were scaled to an infinitely large size, would increase if you added one mole of that component.
The chemical potential is conceptually useful in that it can inform you on whether a species in a sample is content to be where it is located or in the chemical form it is in. For instance, if the chemical potential of an atom is lower in one chemical form than in others, it will prefer to be in the lower-potential form, thereby driving a reaction to that form. Minimization of the sum of the chemical potentials of all species (which is just the Gibbs free energy) drives processes such as chemical reactions (at constant T and p).
Another example is that if the chemical potential is lower in one phase of a multiphase system, then the system may spontaneously undergo a phase transformation, the driving force dissipating when equilibrium is reached and the species has the same chemical potential in all available phases. We say that at equilibrium each species in a system has the same chemical potential in all phases.
The fugacity of a species describes an effective partial pressure in the gas phase. At low pressure of the system the fugacity of a species tends to be equal to its partial pressure. However, as the partial pressures and thereby intermolecular interactions increase in importance, a simple relation between pressure and chemical potential breaks down and fugacity becomes useful when discussing the state (chemical potential).
The mathematical relation between fugacity and chemical potential was chosen to resemble that between partial pressure and chemical potential in the ideal state. This allows one formulation to describe both real scenarios and the limiting ideal case. A fugacity greater than the partial pressure would suggest that a gas has a greater chemical potential than is suggested by extrapolation from low pressures where it behaves ideally. A greater chemical potential suggests greater reactivity or a higher tendency to escape to regions of lower chemical potential (therefore escaping tendency is an appropriate term).

One more thing, you wrote, 'The fugacity of a species describes an effective partial pressure in the gas phase'. You meant to say effective partial pressure of real gas??

@NatashaJ Outside the realm of theory all gases are real, so yes :)

@ Buck Thorn, hey, sorry for bothering you again, but do you know why was there a need of fugacity when chemical potential had already been established?

The internal energy of a closed system is $dU=TdS-PdV $. In an isolated system $dU=0, dV=0$ which would imply that $dS=0$, However, the entropy can increase due to chemical reaction or diffusion mixing different substances which were initially separate. Similarly $dG=-SdT+VdP$ and if $T$ and $P$ are held constant this does not mean that $G$ cannot change as this is an extensive quantity and thus depends on the amount of material. For example in a galvanic cell more useful work can be done if more material is present, e.g. adding copper sulphate to a Daniel cell, so that $G$ must increase. This means that in a chemical system work can be done, i.e. lifting a weight etc., without there being any change in volume between the initial and final states, i.e. work done other than $PV$ work.
The difficulties with $dU=TdS-PdV $ and  $dG=-SdT+VdP$ are due to the fact that we ignore the amount of material present. This changes the energy by an amount $\sum_i \mu_idn_i$  where $\mu$ is the chemical potential. It is called this by analogy to gravitational and electrical potential. In a gravitational field a mass $m$ has energy $mgh$ where $gh$ is the potential and $m$ the quantity.  A charge $q$ has energy $q\varphi$ where $\varphi$ is the potential and $q$ the quantity. In chemical energy, $dn$ is the quantity and $\mu$ the potential so the energy is $\mu dn$ for each species.
In a homogeneous phase containing several different substances with $n_1,n_2..$ moles of species $1,2...$ the total differential of $U$ in terms of $S, T$ and $n_1,n-2..$ is
$$\displaystyle dU=TdS-PdV +\sum\mu_idn_i,\qquad \mu_i \equiv \left(\frac{\partial U}{\partial n_i} \right)_{S,V,n_j}$$
where subscript $j$ indicates all other species except $i$ are constant. Similarly for the Gibbs energy
$$\displaystyle dG=-SdT+VdP+\sum\mu_idn_i,\qquad \mu_i \equiv \left(\frac{\partial G}{\partial n_i} \right)_{T,P,n_j}$$
and this last definition provides the chemical potential shorthand label as the Gibbs energy /mole or more properly the gradient of the Gibbs energy/mole for a given species.
With reversible change in a closed system using  $dU=TdS-PdV +\sum\mu_idn_i$ the heat absorbed by the system is $TdS$, thus the remaining terms represent the total work done. Thus  $\sum\mu_idn_i$ is a form of work/energy in the absence of a change in volume.
The fugacity is used as a proxy for pressure for example when an imperfect gas is studied. It is defined as $\mu = \mu^\text{o} +RT\ln(f),\quad f/P\to 1 ; P\to 0$ so that fugacity equals the pressure when the ideal gas law is obeyed.


text:
Could shear force from blending denature alpha amylase / glucoamylase enzymes?
I am trying to break down the starches in a certain variety of oats in the most efficient way possible. I have to break the whole oats down after cooking, so they are finer particles for the amylases to do its work in the liquid.
I have been adding the amylases at the beginning of blending of the whole oats (at a temperature of about 35 centigrade) because I thought it would help the enzyme process, but could I be damaging / denaturing the enzyme from the shear forces of the blender (I am using a 600 Watt stick blender)

Some proteins can be denatured mechanically, e.g. when whisking egg whites. This denaturing may or may not be reversible. I don't know how to estimate whether these exact enzymes under your conditions will denature or not and if so whether it will be reversible. You can add them after the blending step to be on the safe side.

thanks - I edited my question to explain why I initially thought to add the enzyme at the beginning of the blending stage

are there any kinds of denaturing which are reversible?

It really depends on how automatically the protein folds itself. But addition of salts (i.e. changing the ionic strength) may be reversible. The same might be true for changes in the pH. Sometimes denaturing (loss of function) occurs even with small changes to the natural environment of the enzyme. Those are likely to be reversible.


text:
Frequency Factor of the Arrhenius Equation
The question came to my mind as to whether the effect of a catalyst on a reaction included the modification of the frequency factor (or pre-exponential factor) of a reaction, shown in Arrhenius' Equation below as "$A$":
$$ k = A\exp{\left(\frac{-E_\mathrm{a}}{RT}\right)}$$
On the one hand, we know that catalysts reduce the activation energy of a reaction, thus enabling more particles of reactants to acquire the minimum energy needed to break bonds and form products, therefore speeding up the reaction. But, does the frequency factor stay the same? My guess is that it does, and the effect of the catalysis is manifested solely on the change in activation energy. My question is if this intuition is correct and, if so, why.
Maybe it would help if someone could give a little more insight into the actual meaning of the frequency factor. Is it constant a constant specific to each reaction? What does it represent?
Thanks!

There are a number of websites and books that try to explain this (example), or you could also look in a physical chemistry textbook.

A catalyst changes the mechanism of a reaction so naturally both $E_a$ and $A$ are different. The  factor $A$ describes the rate constant when the activation energy is zero or the temperature infinite and is the fastest the reaction can proceed at and this is controlled by how quickly the species can collide with one another.

This, although I knew it already, let me realise that too often (in organic chemistry at least) we ear "catalysts only reduce the activation energy". It can be practically true but not rigorously correct.

The short answer is yes, we can imagine effects on the frequency factor. Without going into too much detail, the frequency factor combines two conceptual elements - the frequency of molecular collisions if reactants are present at 1 M or 1 bar (called z in collision theory) and the probability that such a collision occurs with the correct spatial orientation that a reaction is possible (called $\rho$ in collision theory).
Let's start with $\rho$. A catalyst that binds multiple reactants can do so in such a way that they are optimally spatially oriented to react. Examples can be found in enzymes which are often large enough to bind the reactants in very specific conformations. So in addition to stabilizing the transition state of the reaction (thus lowering the activation energy), they can also increase $\rho$.
Now let's consider z. This one is a bit more complicated. In homogeneous catalysis, we could imagine z actually decreasing, since we now need the reactants and the catalyst all coming together via sequential collisions. In heterogeneous catalysis, the situation is even more complicated, since the mathematics of colliding are more complicated when the participants are in different states.


text:
Can you get potassium by reducing KCl with lithium?
I would have said that the answer was no, because lithium is less reactive than potassium is, so you would expect the reaction to go the other way: with potassium and lithium chloride reacting to produce lithium metal and potassium chloride.
However, the YouTube video
Cody'sLab — Potassium Metal From Bananas! appears to show that the answer is yes.
Can anyone explain this result, which contradicts what I thought I had learned in high school chemistry?

Define "less reactive." Potassium reacts more violently with water, but lithium is, from a electrochemical perspective, the more reactive metal. Case in point: lithium reacts with nitrogen and potassium does not. If you're not convinced, check the reduction potentials or any activity series with both metals.

To clarify the bounty and expand upon the existing question, I am asking, why does the lithium displace the potassium, if the lithium is less reactive (electronegativity 0.98) than potassium (electronegativity 0.82)?

@TylerM It has nothing to do with any lame high school parameter, like electronegativity, whatever you mean by it - there at least four different "electronegativities".

Note that element electronegativity $\chi$ is a combined indicative quantity from covalent bond context, so its comparative values do not always apply in scenario analysis what happens or should be preferred.
Reactivity (kinetics) and tendency of reaction outcome (thermodynamics) are independent. TNT is much more reactive than coal, but released specific energy during coal burning is 10 times bigger than during TNT explosion.
Potassium reacts more violently with water than lithium. But lithium has more negative standard redox potential (which applies in water context only). That is because higher ionization energy is more than compensated by more negative hydration enthalpy of $\ce{Li+(aq)}$, compared to $\ce{K+(aq)}$, being a smaller ion with stronger electrostatic potential gradient. Hydration enthalpy contributes to the reaction Gibbs energy of $$\ce{M(s) + H+(aq) -> M+(aq) + 1/2H2(g)}$$ and therefore to the metal standard redox potential.
$\ce{Li}$ is less electropositive/more electronegative than $\ce{K}$ and has higher ionization energy than $\ce{K}$. But $\ce{LiCl}$ has higher lattice energy ( more negative lattice formation enthalpy) than $\ce{KCl}$.
$\ce{K}$ boiling point is near $\ce{KCl}$ melting point, while $\ce{Li}$ boiling point is much higher, so $\ce{K}$ boils away, what supports reaction.
There is equilibrium $$\ce{Li(\ell) + KCl(\ell) <=> K(g) + LiCl(\ell)}$$
with potassium vapors being eliminated by condensation on colder surfaces of the apparatus, used in the video.
Thermodynamic parameters are shown in the table below. higher ionization of lithium is more than compensated by the higher lattice energy and pushing the equilibrium as potassium vapor partial pressure is being decreased by vapor condensation.

Compound
Melting Point
boiling point
Ioniz. energy
Lattice energy
Cl EA

$\ce{LiCl}$
$\pu{605–614 °C}$
$\pu{1382 °C}$

$\ce{829 kJ/mol}$
$\pu{349 kJ/mol}$

$\ce{KCl}$
$\pu{770 °C }$
$\pu{1420 °C }$

$\pu{698 kJ/mol}$
$\pu{349 kJ/mol}$

$\ce{K}$
$\pu{63.5 °C}$
$\pu{759 °C}$
$\pu{419 kJ/mol}$

$\pu{349 kJ/mol}$

$\ce{Li}$
$\pu{180.50 °C}$
$\pu{1330 °C}$
$\pu{520 kJ/mol}$

$\pu{349 kJ/mol}$

In the video, you can hear the narrator talk about “potassium vapor” condensing to form the product. Based on that, I’m going to assume that he’s heating the vessel above the boiling point of potassium, and at a slightly higher temperature (above around 770 °C) , potassium chloride and lithium are both liquid.
At this point, I would think that all that needs to happen is for the molten lithium to reduce some of the $\ce{K+}$ in the liquid salt to potassium metal, some of which would vaporize, escape and condense, as Cody seems to describe.
Since the process presumably takes place above the boiling point of potassium, the equilibrium would not even need to favor the formation of $\ce{LiCl}$ for the reaction to progress, although as a commenter before me pointed out, the reduction potential of $\ce{Li+}$ is lower than that of $\ce{K+},$ indicating that lithium metal could reduce $\ce{K+}$ to potassium.


text:
Describe the formation of the [PbCl4]2- complex ion
The $\ce{[PbCl4]^{2-}}$ complex ion is formed when $\ce{PbCl2}$ is exposed to excess $Cl^-$ ions in solution as explained here and here through the following reversible reaction:
$$\ce{PbCl2(s) + 2Cl^-(aq) <=> [PbCl4]^{2-}(aq)}$$
What I wish to understand is the mechanism behind this reaction. I managed to find the mechanism to a similar complex ion formation of $\ce{[CuCl4]^{2-}}$. It is explained here that $\ce{[CuCl4]^{2-}}$ is formed by the hybridisation of the $Cu^{2+}$ 4s and 4p orbitals:
$$sp^3:\boxed{\uparrow \ }\boxed{\uparrow \ }\boxed{\uparrow \ }\boxed{\uparrow \ }$$
Would the $\ce{[PbCl4]^{2-}}$ complex ion be obtained in a similar way by $Pb^{2+}$ forming the following hypridisation orbitals?
$$sp^3:\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \ }\boxed{\uparrow \ }$$
or no need for the above hybridisation since the 6s orbital was already filled and there is space in the empty 6p orbital so the configuration for the complex ion would be as follows?
$$6p:\boxed{\uparrow \downarrow}\boxed{\uparrow \ }\boxed{\uparrow \ }$$

There's no hybridisation in there.

https://chemistry.stackexchange.com/questions/8717/what-is-the-inert-pair-effect

Thank you for your input, Mithoron. The inert pair effect would explain why PbCl2 is more stable than PbCl4 and also why no hybridisation takes place in the above (the outer s orbital acting as an inner orbital rather than a valence orbital). Does this mean that when forming the complex ion, the empty p orbital is simply filled like my second suggestion in my question?

Could the inert pair effect also explain why the total valence and shared electrons on the Pb element in the [PbCl4]2- ion add up to 10 thus overshooting the octet rule? because the 2 valence electrons in the s orbital of Pb do not really count as valence electrons?

It's more like octet rule hardly works here. Such cases can be fitted to it like SF6, though https://chemistry.stackexchange.com/a/5242/9961 https://chemistry.stackexchange.com/a/49844/9961

Mithoron, I think it would make more sense that the 6p and 6d orbitals of the Pb hybridise since co-ordinate covalent bonds will form with 4 pairs of electrons from 4 Cl- ions. what do you think?

sight best don't mention this hypervalent menace here. There was a reason I put these links specifically.


text:
Does the mechanism of AChE inhibition by Isoparathion depend on chirality?
It is known that the inhibition of acetylcholinesterase by isomalathion can proceed either with diethyl succinate as the leaving group or thiomethyl, depending on the specific stereoisomer of isomalathion. Does the mechanism of inhibition of AChE by the Parathion isomer O,S-Diethyl O-4-Nitrophenyl thiophosphate (Or its dimethyl analog) likewise depend on the chirality of the molecule, or is 4-nitrophenyl the leaving group for both stereoisomers?
Reference:
Doorn, J. A., Thompson, C. M., Christner, R. B., & Richardson, R. J. (2003). Stereoselective Inactivation ofTorpedo californicaAcetylcholinesterase by Isomalathion:  Inhibitory Reactions with (1R)- and (1S)-Isomers Proceed by Different Mechanisms. Chemical Research in Toxicology, 16(8), 958–965. doi:10.1021/tx030026e

It would appear that the mechanism of inhibition of AChE by both stereoisomers of isoparathion (Or, at least, its methyl analog) proceeds through the expulsion of the 4-nitrophenoxy leaving group. Several papers have studied the inhibition kinetics of isomalathion and compared the results to those of isoparathion. All of them seem to agree that, at least for isoparathion methyl, AChE inhibition proceeds through the expulsion of the same leaving group, but the O,S-dimethyl phosphorylated AChE would be chiral, as is expected. Furthermore, the inhibition, reactivation, and aging reaction rate constants would also differ depending on the stereoisomer of isoparathion.
References:
Berkman, C. E., Quinn, D. A., & Thompson, C. M. (1993). Interaction of acetylcholinesterase with the enantiomers of malaoxon and isomalathion. Chemical Research in Toxicology, 6(5), 724–730. doi:10.1021/tx00035a019
https://pubs.acs.org/doi/pdf/10.1021/tx00023a004
Jianmongkol, S., Marable, B. R., Berkman, C. E., Talley, T. T., Thompson, C. M., & Richardson, R. J. (1999). Kinetic Evidence for Different Mechanisms of Acetylcholinesterase Inhibition by (1R)- and (1S)-Stereoisomers of Isomalathion. Toxicology and Applied Pharmacology, 155(1), 43–53. doi:10.1006/taap.1998.8608


text:
Why is T-shaped geometry preferred to trigonal planar geometry for the bonds around atoms with 3 bonds and 2 lone pairs?
This is regarding a situation where an atom has five electron domains; two domains are lone pairs, and three are bonds to other atoms.
The course materials I am using (focusing on VSEPR theory) suggest that, as a rule of thumb, T-shaped geometry is the configuration the electron domains around an atom will take in the above situation. However, that feels unintuitive to me, because the separation between the lone pairs (which ought to repel everything else more strongly) is not maximised in that configuration - it makes more sense in my head that a trigonal planar geometry of bonds would be adopted, with a lone pair above and below the plane of the bonds, to minimize the repulsion between those two lone pairs, as well as to distance the three bonded atoms further apart.
Why, then, is T-shaped geometry preferred in most cases? Is that really the case, or a misrepresentation on the part of the course materials? Many thanks.

In each scenario, for each lone pair, count the neighbors at 90°.

You can look again at the schematic and see that the trigonal bipyramidal  minimise the global repulsion as no 90° angles are between the T arms and the lone pairs. But it is true that it is the more difficult case to visualise quickly.


text:
Why does Delta H change when the coefficients change in a reaction?

So I understand that if twice as much of the reactants are present, then twice as much energy is released. But isn’t the energy released per mole of reactant still the same? You are just scaling up both sides of the equation, I.e. the ratios are still the same, so how is it that your are producing more energy per unit of mass of reactants?

Chemistry SE prefers plain text titles. For formatting of chem/math formulas/expressions, use can use  MathJax with mhchem extension \ce{}. Using photos/screenshots of  text instead of typing text itself is  discouraged.  The image text content cannot be indexed nor searched for, cannot be reused nor referred in answers.

These values of $\Delta H$ is given in ... per mole of ..... of what is written in the equation. For your last equation, it would be "per mole of $3$ times what is written". And as it is non-sense to say "per mole of $3$ $\ce{CO2}$", it is better to say "per $3$ moles of $\ce{CO2}$ or “for $ 3$ moles of $\ce{CO2}$"

By convention, the values given for the changes in thermodynamic state functions $(\Delta G, \Delta H, \Delta S, \Delta A, \Delta U$, etc.) are for the reaction as written.  What's confusing is that the "per mole" in the units does not mean per mole of any given reactant or product.  Rather, it means per the number of moles* of each reactant and product specified in this reaction, where the number of moles is equal to the compound's stoichiometric coefficient.
Thus the way to interpret the "per mole" for the value of $\Delta H$ for the first reaction is:
This is the enthalpy change per 1 mole of methanol consumed, per 3/2 mole of oxygen consumed, per 1 mole of carbon dioxide produced, and per 2 moles of water produced.
If the reaction were instead written:
$$\ce{2CH_3OH(l) +3 O_2(g) -> 2 CO_2(g) + 4 H_2O(l)},$$
then $\Delta H$ would be 2 x –726 kJ/mol = –1452 kJ/mol, and the "per mole" would mean per 2 moles of methanol consumed, per 3 moles of oxygen consumed, per 2 moles of carbon dioxide produced, and per 4 moles of water produced.
I.e., more formally, what kJ/mol means here is, for any of the species in the reaction (whether reactant or product):
$$\frac{ \text{kJ}}{\text{mol} \circ \text{stoichiometric coefficient     }}$$
[The stoichiometric coefficient is dimensionless.]
*To those who wish to complain that I am using the phrase "number of moles" instead of "moles", I will argue that here it does make sense.  For instance, if I were were talking about liters instead of moles, I would still say the "number of liters" specified for each reactant and product.  Or if the stoichiometric coefficients told us how many dozens of atoms and/or molecules were involved, then I would say "number of dozens," etc.

Hm, when the reaction enthalpy is calculated, the result is IMHO in kJ, not in kJ/mol. $$\Delta H_\mathrm{r}[\mathrm{kJ}]=\sum {\Delta H_\mathrm{form,prod,i}[\mathrm{kJ/mol}] \cdot n_\mathrm{i}[\mathrm{mol}]} - \sum {\Delta H_\mathrm{form,react,j}[\mathrm{kJ/mol}] \cdot n_\mathrm{j}[\mathrm{mol}]},$$ and as you say, for the reaction as it is written.

@Poutnik   Yeah, when I started to write my answer I thought the use of per mole was wrong.  But then I looked in a few of my pchem texts, and noticed some do adopt this usage.  So I tried to understand why, and realized they are trying to communicate, explicitly, that the stoichiometric coefficients refer to moles rather than particles.   Thus one could argue either for or against including the per mole.

If used kJ/mol, it should be specified "per mol of ...". In this case it would be methanol, but generally it is unclear. And in such a case, both equation sides would be formally divided by molar amount of the referred substance.

@Poutnik  That's not what kJ/mol means here.  It means $\frac{ kJ}{mol \circ \text{ stoichiometric coefficient     }}$.  [I'll add this to my answer.] Thus there is no need to specify "per mol of", since it applies to all species—both reactants and products.

Yes, this is the good way. It is analogous to the reaction rates.


text:
How do oxide formations affect voltaic cells?
I asked a question previously on here but still have some confusion. To briefly summarize, I made a voltaic cell for a school project with 0.1 M copper nitrate and varying concentrations of aluminum, zinc and nickel nitrates. My results were fairly consistent with the theoretical values from the Nernst equation for zinc, however, the aluminum and nickel cells produced substantially less voltage than their theoretical values. The answer I got from my previous question explained how the low voltage in the aluminum cell was a result of its oxide layer.
However, I still had some confusion over why there were differences between aluminum and nickel, and zinc. Firstly, it's my understanding that a metal's ability to form an oxide layer is related to its reactivity. This would explain aluminum's tendency to form such a strong oxide layer however, zinc has a higher reactivity than nickel so wouldn't it also form an oxide layer and its outputted voltage would be lowered similarly.
My next explanation for the differences between aluminum and nickel, and zinc was related to the ability of the oxides to either react or dissolve in the nitrate solutions, however, I found that nickel oxide is soluble in acids and aluminum oxide reacts with acids.
I was also wondering how I even got a voltage in the aluminum cell if the oxide layer does not allow any aluminum ions to be transferred.
I would appreciate some guidance from the chemistry wizards because at this point I am stumped.

Aluminum and nickel both develop passive oxide layers, different in style. The aluminum oxide layer is thick, quite adherent, and fairly impermeable (tho not perfectly) and continues to thicken in the atmosphere, but slower and slower. A thick panel of aluminum will become quite rough and non-shiny after exposure to air and water, especially after exposure to seawater (chloride!). The entire oxide layer does not have to dissolve completely before the corrosion potential of the aluminum exerts itself in your experiment. My own results with $HCl$ and aluminum foil show rapid reaction within seconds, not minutes. (In the nuclear power industry, large amounts of $Al(NO_3)_3$ are needed, but cannot be obtained simply by dissolving aluminum in $HNO_3$; the aluminum is dissolved first in $HCl$, then $HNO_3$ is added and the $HCl$ is boiled off. The pesky $Al_2O_3$ layer is oxidized by the $HNO_3$ and strengthened, whereas the $HCl$ bites right thru it.) An interesting experiment would be to see if your voltaic cells have a large current capacity, or whether the potential is developed thru a tiny imperfection in the oxide layer, and can provide only a tiny current, as if the aluminum electrode were very small.
Nickel develops a very thin adherent oxide layer, similar to the oxide layer on stainless steel, and is used in many corrosive environments for this reason. The cell potentials may well be affected by the anions involved, for nitrate could strengthen the oxide film, resisting the corrosion needed to provide electrons, while doing the same series with the chlorides, especially where voltages (at very low current) are measured could align more perfectly with your expectations. The tables of electric potentials are not developed from actual experiments, but from a totality of data on reaction heats and energies where reactions are assumed, not stifled by passivity.
Experiments with copper and zinc work nicely, perhaps because the layer of oxide on the zinc is even less perfect than on aluminum (but still fairly resistant to water corrosion in the mid-pH range). The chemistry in books seems so straightforward and almost perfectly logical (with a few explainable quirks, of course), but in practice, labwork is can be complicated and frustrating, and you can learn as much about chemistry by analyzing your "deficient" results as from getting a perfect match with theory.

There are several questions in your text.
First, you imagine that "a metal's ability to form an oxide layer is related to its reactivity". No. It is not. Aluminum is rather reactive with water. But sodium is more. And metallic sodium is not protected by any oxide layer. On the contrary. Sodium reacts dangerously with water. Aluminum is a strange metal, as it forms an oxide that has the same density as the metal. So this oxide can perfectly cover the surface of the metal.
Then you ask about the difference of reactivity between zinc and nickel. I cannot answer this question, as I don't know enough of the nickel chemistry.
Then you say that aluminum oxide reacts with acids. It is right, but this reaction is extremely slow. If some aluminum piece or foil is put into a solution of $\ce{HCl}$ made by mixing the concentrated solution ($36$%) by an equal amount of water, you will have to wait a couple of minutes before the oxide layer is dissolved according to : $$\ce{Al2O3 + 6 HCl -> 2 AlCl3 + 3 H2}$$ And this last reaction is so slow that the expected reaction producing hydrogen gas happens only two or three minutes later according to $$\ce{2 Al  +  6 HCl  ->  AlCl3  +  3 H2}$$ In this case, the production of gas is violent and highly exothermic, but it only happens with delay.
At the end, you ask why "I even got a voltage in the aluminum cell if the oxide layer does not allow any aluminum ions to be transferred". This is a good question, for which I don't have a real answer. It may be due to some default of the alumina layer.

Thank you for your answer! In your previous answer on my last question you stated that "Aluminum ions cannot cross [the oxide] barrier" even after any attempts to remove it like the sanding process I used. The other response to this question by James Gaidis states that this layer was "fairly impermeable" but not perfectly. I was hoping for some clarification on the nature of aluminum oxide. Thanks!

On pristine aluminum, islands of oxide grow, connect, and build up by oxygen penetration thru the film already formed. https://pubs.acs.org/doi/10.1021/acsami.7b17224 Pitting is corrosion by destruction of the already-formed film and loss of aluminum. "By nature, aluminum is a reactive metal but it is also a passive metal." https://fractory.com/aluminium-corrosion/ So, the film is intact (until it isn't) over a wide range of conditions, but if it corrodes, either something gets in or aluminum gets out - not easy either way.


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Finding Gibbs energy at different temperatures just given Gibbs energy at one temperature
How would it be found the Gibbs energy at a certain temperature, if they just give you another Gibbs energy at a temperature? For example: Given $\Delta G = -230 Kcal/mol$ at $773K$ for the reaction $4/3Al + O_2 > 2/3Al_2O_3$, find the Gibbs energy at $573K$
With that data I can only think of doing $(-230*573)/773$, but as far as I know the Gibbs energy is not only related to the temperature.

You have to know the entropy change of the reaction $\Delta S$, because the variation of $\Delta G$ with temperature is equal to -$\Delta S$. So when you know $\Delta S$, you multiply it by the $\Delta T$, here $200 K$, and subtract the result from your original value of $\Delta G$. This calculation is a first approach. To be more precise, you would have to take into account the change of heat capacity of the three substances ($\ce{Al, O2, Al2O3}$) when $T$ goes from $773 K$ to $573 K$. Usually this change is small and can be neglected.

Yeah I know how the $\Delta G$ varies, the problem is the exercise, just determining the $\Delta G$ at two different temperatures. But as you say, maybe I could assume the change of the heat capacity and entropy can be neglected.

No. You should know the entropy of your substances, calculate the difference product minus reagent, multiply by $\Delta T$ and substance it to $\Delta G$. This is sufficient. And do ignore the changes of heat capacity, which is much smaller than the effect of$\Delta S$.


text:
Potential energy expectation value and virial theorem
I have been trying to solve this problem

Part a) can be proved using the formula .
For part b), the expectation value of the potential energy can be calculated from the integral
,
where

Therefore, the integral to be solved is

which gets simplified using spherical polar coordinates into

However, I have seen that the expectation value for the potential energy should be

Obtained using the same integral that I used, but using this expression for the wavefunction

which, from my understanding, should be the same expression but not in atomic units.
I am not familiar with conversions using atomic units, Bohr radius, Hartree, and so on. So I am not sure if what I have done is correct or if there's something missing.
For part c), using the virial theorem, the expectation value of the kinetic energy is
[T] = [E] - [V] = -0.5 - [V]
I am also confused by the Hamiltonian and the Laplacian which I did not use in my solution, am I missing something?
If I am not being clear please let me know and I will try to explain myself better.
Thanks very much to whoever is willing to help me understand this problem better!

I think the first step is to understand atomic units. In atomic units, the charge of the electron $e$ is set to 1, the Bohr radius is set to 1, and your energies will come out in Hartrees, without you having to do any conversions between eV and Hartrees. This will help with at least one thing: notice that when you write $e$ sometimes you're referring to the charge of the electron $\pu{1.602 \times 10^-19 C}$ and sometimes it's the base of the natural logarithm $2.718$. You can also use LaTeX-like syntax (MathJaX) to typeset equations here: $\frac{1}{2}$ $\to$ $\frac{1}{2}$

I also presume that with part (a), you're not supposed to use that formula, but rather explicitly evaluate the integral $\langle \psi | \hat{H} | \psi \rangle$.

Thanks a lot for your reply!
The elucidation about atomic units was very helpful.
For the potential energy, e is the charge of the electron; for the wavefunction, e is Euler's number. I should stick to the exp notation to avoid confusion.
Next time i'll use the LaTeX syntax, thanks.
Yeah, for part (a) it makes more sense to evaluate the integral.
The topic seems more clear now, I'll review the problem with these new notions.
Thanks again for your help!

if the potential $\sim r^n$ then $2T=nV$ where $T$ is average KE and V total potential energy and for coulomb potential  $n=-1$


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Is 1% m/m the same as 1% w/w?
Hydrocortisone cream seems to be sold as 1% w/w, but my cream is 1% m/m, which is confusing. Are they the same units or the same concentrations?

“[M]ass is the amount of "matter" in an object (though "matter" may be difficult to define), whereas weight is the force exerted on an object by gravity.” (Wikipedia quoting [de Silva, G.M.S. (2002), Basic Metrology for ISO 9000 Certification, Butterworth-Heinemann, 214p.]
Well-known relationship:
$$
W = m \times g
$$
where W is weight, m is mass and g is the acceleration of gravity.
Because g is a constant that cancels out in a ratio, mass/mass and weight/weight are identical.
Likewise, the ratios % m/m and % W/W are identical.

Indeed m/m would be more appropriate. Thumb up for the pharmacists.

Let's be thankful the % units were labeled at all!

The symbol “w/w” is more often used in chemistry, whereas its equivalent “m/m” is favored in pharmaceutical industry.
For instance, European Pharmacopoeia in section 1.2. Other provisions applying to general
chapters and monographs suggests the following methods for expressing concentration [1, pp. 4–5]:

Expression of content. In defining content, the expression ‘per cent’ is used according to circumstances with one of 2 meanings:
– per cent m/m (percentage, mass in mass) expresses the number of grams of substance in 100 g of ﬁnal product;
– per cent V/V (percentage, volume in volume) expresses the number of millilitres of substance in 100 mL of final product.
The expression ‘parts per million’ (or ppm) refers to mass in mass, unless otherwise specified.

Please note, however, that this is a somewhat flawed definition: neither of physical quantities such as “mass” or “volume” should be called or referred to as to “number
of grams” or “number of millilitres”, respectively.
Also, neither “w/w”, “V/V” nor “m/m” — despite their widespread usage — are considered correct by the BIPM [2, p. 149]:

5.4.2 Quantity symbols and unit symbols
Unit symbols must not be used to provide specific information about the quantity and should never be the sole source of information on the quantity.
Units are never qualified by further information about the nature of the quantity; any extra information on the nature of the quantity should be attached to the quantity symbol and not to the unit symbol.
For example:
The maximum electric potential difference is $U_\mathrm{max} = \pu{1000 V}$ but not $U = \pu{1000 V_\mathrm{max}}.$
The mass fraction of copper in the sample of silicon is $w(\ce{Cu}) = \pu{1.3E−6}$
but not $\pu{1.3E−6 w/w}.$

Considering the aforementioned guidelines, the hydrocortisone (HCT) content should ideally be depicted as $w(\text{HCT}) = 1\,\%$ to avoid ambiguity.
References

Council of Europe; European Directorate for the Quality of Medicines & Healthcare. European Pharmacopoeia, 10th ed.; Council of Europe: Strasbourg, 2019. ISBN 978-92-871-8912-7.
BIPM. Le Système International d’unitès / The International System of Units (“The SI Brochure”), 9th ed.; Bureau international des poids et mesures: Sèvres, 2019. (PDF)


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catalysts in catalytic hydrogenation
I am studying the very BASIC introduction of catalysis in general chemistry and encountered an example of a heterogeneous catalyst, which is catalytic hydrogenation of alkene into alkane using Pt/C or Pd/C, I have some questions:

why sometimes we need to use Pt, and sometimes we need to use Pd? Are there any differences in reactivity, or just purely depends on what is at hand?

I know the addition of activated carbon powder is to increase the surface area between reactants and catalyst. However, the carbon powder isn't added all the time when doing different heterogeneous catalysis reactions, why is it so if it is beneficial?

Bond dissociation energy of H2 = 436kJ/mol. But when H2 is adsorbed on the metal surface, the bond is broken. How come this simple step can break this strong bond? I suspect the metal-H bond formed is stronger than H-H bond, which means it is favorable, but I still have no idea where does the 436kJ/mol energy comes from.

Thanks

Q3-Hydrogenation of Alkenes is an exothermic process.

Pt is rather more reactive than Pd as a catalyst so you might prefer to use Pd when there are other potentially reducible groups present.


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How is the central carbon in 3‐bromopentane‐2,4‐diol chiral?
My teacher told us the middle carbon in 3‐bromopentane‐2,4‐diol is chiral:
$$\ce{CH3-CH(OH)-\overset{\star}{C}H(Br)-CH(OH)-CH3}$$
Since there are two identical groups around it, I would guess it should not be chiral.
It was a pre-recorded class and I was not able ask this question. Has my teacher made a mistake?

The carbons either side bearing the -OH groups are chiral so are potentially non-identical

The carbon can only be pseudochiral. Its configuration can matter but in that case, it will sit in the plane of symmetry.

This issue is addressed here: http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/isomers/RS14272/RSrs.html

Also see this; https://chemistry.stackexchange.com/questions/91539/mismatch-of-plane-of-symmetry-vs-gold-books-definition-for-a-pseudochiral-carbo/91892#91892

Carbons #2 and #4 are chiral centres themselves. There are four possible diastereoisomers:

If C2 and C4 have different configurations, which is the case in molecules A and D, then C3 is termed a pseudoasymmetric centre and labelled with a small 'r' or 's'. These diastereomers are meso compounds: although they contain chiral centres, they are not chiral, because they have an internal plane of symmetry.
On the other hand, in molecules B and C, C2 and C4 have the same configuration and C3 is not a chiral centre (and not a pseudoasymmetric centre either). These diastereomers are chiral, though: the chirality comes from C2 and C4.

So it's chiral in some cases and not in others ? Like Zhe said in the comments ?

Actually, some of the diastereoisomers are chiral molecules, i.e. without internal mirror image or rotational symmetry. I hope this helps.

Rotational symmetry isn't related to chirality. To be more precise, the lack of improper rotations $S_n$ is associated with chirality; but proper rotations $C_n$ have no relation to chirality. PS sorry about the edits. I need to double check this.

Think that should be right now..


text:
Precise estimation of air properties as a function of temperature at 1 atm
Are there any tables with empirical equations to calculate properties of air as a function of a temperature at 1 atm?
For example, NIST provides such equations for a large number of gases, but I was not able to find such solution for air, where I would specifically like to obtain:

Dynamic viscosity,
Specific heat,
Prandtl number,
Thermal conductivity,
Thermal diffusivity.

It would be best, if I could put these calculation in MS Excel, because I need to change these values with respect to temperature in multiple equations.

Properties of air are not precise.

As @Mithoron states, "air" varies, particularly as to water vapor. This was a problem for carburetors, the design of which often ignored that, leading to icing of the venturi.

Here are some links that look like they might be useful, there seems to be promising information on some engineering resource sites:

https://onlinelibrary.wiley.com/doi/pdf/10.1002/9780470516430.app2
This is the appendix from what appears to be an engineering textbook, it gives a chart of the information you're looking for at 1 atm and 15 C. It also walks you through the temperature dependent equations for some of the properties.
https://www.me.psu.edu/cimbala/me433/Links/Table_A_9_CC_Properties_of_Air.pdf
This is a table of the properties of air at 1 atm from -150C to 2000C. It doesn't provide equations but you could maybe enter the data into excel and run a regression to extract a an equation.
https://irc.wisc.edu/properties/ This is a handy little calculator that outputs a whole host of properties of fluids, including dry air.

The point that has been made about water vapor fluctuations is valid, but most calculations of air properties assume dry air with constant composition.

I have a similar table for air properties at 1 atm. I also found some other calculators, but calculating every values in a calculator is not very productive for my purposes, however, the "Appendix B" looks very promising - I will have to try the differences for values computed as a temperature function to those in the chart, if the correlation is satisfying, I'm all settled :).

A lot can be found at engineeringtoolbox , like air-prandtl-number-viscosity-heat-capacity-thermal-conductivity.
The data are ( at least mostly ) presented as combination of an online calculator, tabelated numerical values and charts, instead of specific empirical formulas. But it should not be a big deal to invent some interpolation or approximation formulas for those tabelated numerical values.
See also the Related documents section in the latter link, a chosen topics:

Air - Composition and Molecular Weight
Air - Density at varying pressure and constant temperatures
Air - Density, Specific Weight and Thermal Expansion Coefficient at Varying Temperature and Constant Pressures
Air - Diffusion Coefficients of Gases in Excess of Air
Air - Dynamic and Kinematic Viscosity
Air - Molecular Weight and Composition - Dry air
Air - Properties at Gas-Liquid Equilibrium Conditions
Air - Specific Heat at Constant Temperature and Varying Pressure
Air - Thermal Conductivity
Air - Thermal Diffusivity
Air - Thermophysical Properties

Thanks, as matter of fact, the engineeringtoolbox was the first thing I found, however, when I had to always fill out their calculator (which is great for its purpose), I realized that I need to put calculations to MS Excel, because it was very slow.

But putting a table to Excel and processing some numerical analysis.... E.g. the obvious and trivial least square method for a polynomial or rational function via Excel Solver..... At least you know this is the last resort option if you do not find anything better.

I know... it won't be horrible. In general, I was just very curious whether I missed any empirical equations, which would solve my problem most efficiently.


text:
How does an irreversible process maintain internal energy while dissipating excess heat?
From my understanding, if you were to look at a pressure-volume graph of a reversible, isothermal expansion, the area under it would represent reversible work. Similarly, if you were to graph an irreversible, isothermal expansion on top of this graph, you’d see a smaller rectangular area representing the irreversible work.

What I have been told is that the area between these areas, irreversible and reversible, is wasted energy dissipated as heat. This confuses me, however. If the work the irreversible system does is less, then it should draw less heat from the surroundings (w=-q). So, if it still releases the same quantity of energy, just more of it as heat and less as work, then how would it maintain a constant internal energy? In other words, if the system dissipates the unused work as heat, which is not replenished, then how does it stay at constant T?

I don't understand your question.  The irreversible does less work and draws less heat from the surroundings.  What's the problem?

That part I understand. What I’ve been told is that, while the irreversible process does less work, the remaining work it didn’t do (compared to an identical reversible process) is dissipated as heat. While now the irreversible process has expelled the same amount of energy (as work and heat) as the reversible one, but only been “reimbursed” for the work (w=-q)

Who told you this?

While, my professor, however the idea appears again in this thread: https://physics.stackexchange.com/questions/426853/what-happens-to-the-heat-generated-in-an-irreversible-process

The change in entropy of the system in both cases is the same, but, in the irreversible case, the entropy of the surroundings decreased less.  This is because less heat was transferred from the surroundings to the system.  In this sense, all the entropy generated within the system in the irreversible process is transferred to the surroundings.  So the interpretation would be that the "heat generated" within the system causes less heat to be transferred from the surrounding to the system and less work to be done.  In my judgment, this is a desperate attempt to provide some physical interpretation to what is happening.