text:
What happens when you add a weak acid to a solution buffered at a pH of 8.0?
There are a lot of textbook examples showing what happens to the pH of a buffered solution when you add strong acids and strong bases. Searching for an example calculation where you add weak acids or bases instead, I did not find any information.
The purpose of this question is to establish a model answer for this type of problem. To give a specific example, we start with a phosphate buffer (total phosphate concentration of 100 mmol/L) with a pH adjusted to 8. What will happen when you add weak acid to a concentration of 10 mmol/L? Because the strength of the weak acid might matter, give the answer for a p$K_\mathrm{a}$ of 6.0, 8.0 and 10.0. Does it matter if the weak acid is one of the substances already present in the buffer?
This question is related, but not a duplicate of questions about titrating weak acids.

Such cases are well solved as 2 pH buffer systems, linked by mutual neutralization, with conjugate pair ratios fitting the same pH.

I would just say pretty much nothing happens, and call it a deal. Calculation would be difficult and resulting pH very close to original. Questions are about strong acids, because they have a serious effect. Also, now I'm tempted to find a duplicate :)

related https://chemistry.stackexchange.com/questions/139062/is-this-buffer-possible https://chemistry.stackexchange.com/questions/26303/can-i-make-a-buffer-from-two-different-weak-acids-or-bases

The buffer pH depends on the ratio of weak acid to its conjugate base. Adding any weak acid will increase the acid content lowering the pH of the buffer solution and possibly changing the buffer capacity

@Poutnik Do you know of a good reference that explains broad range buffer  mixtures? I do see a connection between them and this question.

Related: https://chemistry.stackexchange.com/a/26305/72973

Have a look at this answer https://chemistry.stackexchange.com/questions/100346/calculate-ph-of-a-mixture-of-a-strong-base-and-acid-knowing-only-the-ph-wt-v/100355#100355 and this one  https://chemistry.stackexchange.com/questions/60068/how-to-set-up-equation-for-buffer-reaction/95566#95566 for general method to solve these kinds of problems.

Weak acids at a concentration of 10 mmol/L and the given range of p$K_\mathrm{a}$ values dissociate very little in pure water. How much it dissociates in a buffered solution depends on the pH of the solution and the p$K_\mathrm{a}$ of the weak acid. Here, the pH is close to 8. With the information given, it is impossible to calculate the exact pH change. It is possible, however, to quantify how much strong acid you would have to add to get the same effect.

p$K_\mathrm{a} = 6$: The weak acid is fully (or to 99%) dissociated. It's effect on the buffer is similar to that of adding strong acid as long as you don't exceed the buffer capacity.

p$K_\mathrm{a} = 10$: The weak acid hardly dissociates (99% protonated). It's effect on the buffer is negligible.

p$K_\mathrm{a} = 8$: If the weak acid you are adding is identical to the conjugate acid of the buffer system, you are increasing its concentration by 10 mmol/L. If it is a different acid, about 50% of it will dissociate. This is the same as adding half the amount of strong acid. However, the weak acid will also increase the buffer strength.

So it really depends on the p$K_\mathrm{a}$ of the weak acid added. It can either have an effect comparable to that of a strong acid, or hardly any effect at all, or something in between.


text:
Brown sediment in shower, nothing works
I have an ensuite shower with white floor (or call it pad) which is from some kind of composite plastic material.
Over time, brown discoloration and sediments started to acumulate in areas where the foam from shower session is not able to dissipate immediately, and stic kthre for extended time periods.
This, contrary to calcium sediments on the glass door, which can be easily cleaned by applying cytric acid solution, is resisting all my efforts to remove.
I tried mechanical removal (to the extent possible to not damage the white material), bleach, citric acid, baking soda, all the £5 cleaning solution from my local supermarket and DIY shops. Nothing works.
Can you please tell me what it is and recommend a chemical solution that is safe to humans, but can dissolve this thing?
See two photos of the issue here ->
https://postimg.cc/gallery/xCFzw2G

These brown deposits are probably containing iron(III) hydroxides, oxides  or carbonates. They should be easily eliminated by concentrated $\ce{HCl}$, but this acid may also attack the floor. You might try using wet Rochelle salt, which is the common name for potassium sodium tartrate. Action of this compound should clean the spot, but not immediately. Try !

source: Stack Exchange

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

sort: votes


text:
Does germanium(IV) chloride have to be handled under inert atmosphere?
I am finding conflicting information on the proper handling of germanium(IV) chloride. Is the liquid air/moisture sensitive? Does $\ce{GeCl4}$ have to be handled in a glove box?

Probably only moisture sensitive afaict.

One look at Wikipedia gives plenty of hints that this is moisture sensitive. You can use this compound in DRY air.

At times like this, you always refer to the material safety data sheet (MSDS). Have a look:

Upon contact with water or moist air this material reacts to release corrosive and toxic hydrogen chloride gas and dense white fumes of germanium dioxide

Please also check the "HANDLING AND STORAGE" section.

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


text:
Major product formed when HBr is added to 1-phenyl but-2-ene
Addition of $\ce{HBr}$ to $\ce{Ph-CH2-CH=CH-CH3}$ possibly yields two products:

$\ce{Ph-CH2-CH2-CHBr-CH3}$ or
$\ce{Ph-CH2-CHBr-CH2-CH3}$

Which one of them should possibly be the major product? Markovnikov's rule (incorrectly) predicts that both of them should be the major product because the double-bonded carbon atoms in the substrate contains equal number of hydrogen atoms.

That would depend on reaction parameters. Also you're missing a rearrangement product, quite possibly even more important than the two.

@Mithoron Could you please elaborate about which rearrangement product you're talking of?

Normal carbocation rearrangement when you got aromatic ring in neighbourhood. If you still don't get it check out your org. chem. book.

Phenyl groups are bulky and cause steric hindrance.
This makes the $\ce{Br^-}$ nucleophile have an easier time targeting the carbon atom in the double bond that's closer from the right side of the reactant molecule.
On the other hand, $\ce{H^+}$ is small enough to bypass said steric hindrance.
Therefore, the first product is more likely to form.


text:
Is there really no solvent (or any other really) glue for polyethylene terephthalate?
PVC can be glued well with solvent cement wherein it melts somewhat and then sticks together really well as if it was welded, which is really convenient. But I can't find anything similar online for PET. Some sources say there just isn't.
Is there really no room temperature solvent for PET?
The only thing in my experience that holds PET together is this glue which still does it quite poorly and doesn't seem to melt it, only adheres to it somewhat.

PET isn't polyethylene. It is polyethylene terephthalate, a completely different class of polymer. And one with very different "stickiness". Which one was the question about: polyethylene or PET?

my best choice for that could be a polychloroprene-based glue, a SMP Polymer glue or polyacrylate mix glue.

A polychloroprene-based glue showed no resistance whatsoever when I disconnected the PET pieces. Can you suggest a specific brand that worked for you perhaps?


text:
How to add any polymer to azobenzene?
I want to make a polymer by adding two polymeric substituent groups to azobenzene (I would like only N-azobenzene-N where N is a polymer and repeats). I have been reading about how they make the polymer. The synthetic methods are quite difficult and expensive.
I thought that azobenzene could be obtained as the result of the oxidation of aniline with potassium permanganate, and then see how to polymerize the azobenzene group as a substituent in a polymer already having the monomer.
The simplest synthesis I can find is using
isocyanate oligomer
4-aminoazobenzene=azobenzene
(1,4-diazabicyclo[2.2.2]octane) =DABCO (catalyst)
I do not understand almost anything about the reaction mechanism, and I'm looking for simpler ways to link a polymer chain to azobenzene.
My references:
https://pubs.rsc.org/en/content/articlelanding/2009/jm/b918130j
https://pubs.rsc.org/en/content/articlelanding/2008/SM/b805434g ,
https://www.pnas.org/doi/full/10.1073/pnas.1817082116

And why is inorganic chemistry tag to this question?

@NilayGhosh I don't know if it's wrong but I think azobenzene polymers use a lot of solid state stuff, like for example inorganic catalysts and things like clousters are needed for some syntheses.

Interesting that azobenzene may be conjugated and strictly hindered enough not to polymerize without adding in another monomer.


text:
What is the composition of air (oxygen and nitrogen) as it is displaced by propane?
We have purchased an oxygen analyzer that is calibrated with pure nitrogen and pure oxygen. It measures 20.91% when it samples air in the atmosphere. The analyzer is reading in %volume.
We have built a container that allows gas to escape. It is initially filled with air. The container is about 6 cu.ft and we have about 4 inch diameter hole with a mechanism to close and open the hole. The oxygen analyzer is on the other side of the room which the gas mixture unit is isolated in the explosive room designed to withstand explosion. We then introduce propane into the system. As more propane we have introduced into the container, the oxygen decreases to 20.14%.
We want to use the measurement of oxygen to calculate the air-fuel ratio. The combustion reaction is:
$$\ce{C3H8 + 5O2-> 3CO2 + 4 H2O},$$
so a stoichiometric air-fuel mixture should have 1 part propane and 5 parts pure oxygen, or 1 part propane and 25 parts air, approximately (by volume).
Does the nitrogen-to-oxygen ratio remain constant (79/21=3.76) when propane is introduced in the container and partially displaces the air?

Nitrogen / oxygen ratio remains constant until propane is ignited or mixture explodes.

@Poutnik that's what I suspect as well. I do not think it's possible to have only oxygen decreased but not the nitrogen.

If we assume that the Temperature and Pressure conditions under which this is done are such that all gases behave like ideal gases, then we have:
$$X_{air}+X_{fuel}=1$$
Solving for $X_{air}$:
$$X_{air}=1-X_{fuel}$$
Dividing by $X_{fuel}$:
$$\frac{X_{air}}{X_{fuel}}=\frac{1-X_{fuel}}{X_{fuel}}$$
We also have:
$$X_{O_2}+X_{N_2}+X_{fuel}=1$$
As you noted, the ratio of $N_2$ : $O_2$ in air is:
$$\frac{X_{N_2}}{X_{O_2}}=3.76$$
So we have:
$$X_{N_2}=3.76\;X_{O_2}$$
Substituting and solving for $X_{fuel}$:
$$X_{O_2}+3.76\;X_{O_2}+X_{fuel}=1$$
$$X_{fuel}=1-4.76\;X_{O_2}$$
Substituting into the air/fuel ratio equation:
$$\frac{X_{air}}{X_{fuel}}=\frac{1-1+4.76\;X_{O_2}}{1-4.76\;X_{O_2}}$$
Simplifying:
$$\frac{X_{air}}{X_{fuel}}=\frac{4.76\;X_{O_2}}{1-4.76\;X_{O_2}}$$
This is the Air-Fuel ratio in terms of molar fractions or volume fractions, since we're assuming all species are ideal gases.
The volume or molar fraction of oxygen needed to satisfy stoichiometric relationship with propane, while using a different $N_2 : O_2$ ratio $\alpha$ can be calculated by:
$$X_{O_2}+X_{N_2}+X_{fuel}=1$$
$$X_{O_2}+\alpha\;X_{O_2}+0.2X_{O_2}=1$$
Finally:
$$X_{O_2}=\frac{1}{1.2+\alpha}$$
$$\alpha=\frac{X_{N_2}}{X_{O_2}}$$

Thank you for your detail explanation. This is still based on the fact that N2/O2 ratio stays constant when fuel is introduced into the system. Thank you for confirming that. Since we want to determine the Stoichiometric Mixture of propane from the oxygen reading, I am using 78.08% as N2 and 20.95% as O2 for the calculation which gives the N2/O2 ratio of 3.73. The stoichiometric mixture for propane is 15.67. If using N2/O2 ratio of 3.76, the oxygen level at stoichiometric mixture will be approximately 19.75%. With N2/O2 ratio of 3.73, it will be approximately 19.89%.

You're welcome. The stoichiometric oxygen percentages I calculated with 3.73 and 3.76 ratios are slightly different (20.28% and 20.16%, respectively), but they are close to yours. I edited my response above to include my calculation.

My apology. The stoichiometric mixture of propane-air that I have stated (15.67) was incorrect. That is ratio by mass. The Stoichiometric mixture for propane in volume should be 23.9. After recalculation, I have got 20.165% of O2 with the ratio of 3.76 and 20.306% of O2 with the ratio of 3.73. This leads the percentage of fuel to be 4.01% which is closer to Propane's LEL of 2.1%. I expect the percentage of propane's stoichiometric mixture tp be somewhat in the middle of 6.1%.

Is your goal to determine %$O_2$ required for %fuel to be lower than propane's LEL, or what condition in terms of LEL are you trying to satisfy?

The goal is to determine the %O2 and % fuel at stoichiometric mixture since there is no easy way to measure the composition of air-fuel mixture. We thought it is easier to use an oxygen analyzer to monitor the % fuel in the container.

In terms of molar percentages, the stoichiometric %fuel would be equal to the product of %$O_2$ you're measuring, times 0.2. This would be equal in terms of volume percentages if all gases behave like ideal gases. If they do not behave like ideal gases, a correction factor must be included. What Pressure and Temperature conditions are you using for this operation? (this is important to know since it will determine if the gases behave ideally or not).

The air is in standard atmosphere pressure and room temperature. The fuel (propane) has bee injected into a container with small leak, which is close to atmosphere pressure at 70°F. I have also added a link of the excel spreadsheet for calculating mixture of N2, O2, and propane. See [Link] (https://docs.google.com/spreadsheets/d/1muH6MUjvY-8YFfMFDvxZzJSWSw1vJTDF/edit?usp=sharing&ouid=105155522786863337336&rtpof=true&sd=true)

Under those conditions, the three gases behave reasonably ideally. In that case, what i would do is: (1) Measure the %$O_2$ with the instrument you have. (2) calculate %fuel with: $X_{fuel}=1-X_{O_2}(1+\alpha)$ (3) Calculate stoichiometric %$O_2$ with: $X_{O_2}=5;X_{fuel}$

When %$O_2$ measured is equal to 20.1613%, you will have reached the point of stoichiometric composition, assuming $\alpha$ = 3.76

Thank you for confirming with that and the oxygen level of 20.1613% is what I found to be the point of stoichiometric composition. If my calculation is correct, this % fuel at the oxygen level of 20.1613% would be 4.03%.

I am just surprised that the point of stoichiometric composition is not somewhere in the middle of UEL and LEL. I thought that the point of stoichiometric would be somewhat 6.1% of fuel.


text:
Energy Conservation
So as we start Valence Bond Theory in Chemical Bonding from Inorganic Chemistry.
Then  we are introduced to reasoning behind bond formation as follows:
As it is a well known fact that $$Potential\;Energy\;\alpha\;\frac{1}{Stability}$$
As in the atoms of molecule (here $H_2$) attractive forces tend to bring molecule closer whereas repulsive forces tend to push them apart.
Magnitude of new attractive forces is more than new repulsive forces and ultimately they reach a critical distance $r_o$ where the attractive forces balance the repulsive forces. The effect of distance is visible in the graph attached. Since maximum lowering is at distance $r_o$ there is a bond formed and this is the bond length (74 pm in case of $H_2$).

Now my problem is that how is the principle of conservation of energy being satisfied?
Like how come total energy is constant since it is clear that potential energy of the two atom system varies but its kinetic energy does not compensates for the same.
Even if the energy is radiated then how can it be verified?

Born–Oppenheimer approximation $\;$In which it is supposed that the nuclei, being so much heavier than an electron, move relatively slowly and may be treated as stationary while the electrons move in their field.

It is called a potential energy curve because the kinetic energy of the stationary nuclei is zero.

It is evident that nucleus does move to alter kinetic energy.
Please confirm the energy conservation here.
Thank You

Who says the kinetic energy does not compensate? Both protons move. If the excited H2 molecule does not
pass energy elsewhere while protons bounce together and away again in 1 vibration period, both atoms separate again, having just enough energy for that.

If the hydrogen atoms travel straight toward each other, they accelerate as they come nearer and potential energy transforms into kinetic. This is not shown in the diagram because you don't plot kinetic energy. It would go up as the atoms get closer, up to the potential energy minimum at 74 pm, then decrease again as the repulsive interactions increase faster than attractive ones.
While it is not explicit, there are implicit assumptions within the drawn potential. For instance, it implies the existence of attractive forces that become substantial at approximately twice the bond length (shown as 74 pm) which would set the nuclei into motion toward each other. Therefore, stating that the nuclei can be stationary isn't quite right unless what is shown is not the total potential (which includes forces that inhibit motion of the nuclei toward each other).
Another piece of information missing from the diagram is the orientation of the individual paths of the nuclei (which presumably are in the gas phase and not being retained by some unknown external potential). The drawn potential is a schematic of what it would look like if you could bring the nuclei toward each other along a line connecting them. In reality the nuclei might take glancing paths past each other in all sorts of random directions rather than carom into each other (sampling the repulsive part of the potential) and instead just deflect each other. In either case energy would be conserved as it is converted into kinetic and then back into potential energy again.

Phet has a great simulation of this where you can see the potential energy turning into kinetic.  Thus, we can see that energy is conserved. @Buck

@Avogadro very neat and super relevant, thanks for sharing

Hydrogen atoms in a vacuum have low attractive forces for each other certainly not enough to generate the potential energy curve. What they do have is intersecting velocity vectors with the associated kinetic energy and momentum. In the collision the kinetic energy is converted into the potential energy of the distorted 1s orbitals. If the KE is high enough repulsion of the protons is paramount. The energy conservation is the original KE of the atoms and the incipient PE in the collision Their sum will be constant. I do not know if this transfer of energy involves excited energy levels or if it is similar to a Raman effect. It seems more of the latter because the removal of energy to form the molecular orbital is by a third body collision not by radiation. It is possible the H2 molecules form in an excited molecular state.
What I am saying is that low temperature hydrogen atoms will not react to form H2 because they will not collide and any collisions will not distort the orbitals enough for rearrangement into molecular orbitals.  Imagine a solid block of H atoms at absolute zero as close as possible and unable to react. The epitome of frustration.

You asked several questions, I will focus on this one:

how is the principle of conservation of energy being satisfied?

Depending on the conditions, the energy is used to heat up the surrounding or dissipates otherwise. You can do experiments to verify that hydrogen molecules have lower potential energy than atoms, or do calculations from first principles.
The potential energy curve makes no assumption about kinetic energy of the nuclei (of course, the electrons always have kinetic energy). The nuclei can be fairly stationary (in a solid at low temperature), or they can move around (translate, rotate, atom pairs vibrate). All these movements are compatible with the potential energy function. As the OP states, the Born-Oppenheimer approximation allows us to calculate the potential energy for a given snapshot with a certain bond distance.


text:
Finding a Heterogeneous Catalyst for Hydrogen Peroxide
I'm designing an experiment to test the effect of surface area of a catalyst on the activation energy of hydrogen peroxide decomposition. However, I don't think I can use a powdered catalyst because determining its surface area would be too complex (specific surface area is beyond the scope of my course). So, I proposed using a potato as a catalyst because I can get a chunk of it and split it into smaller and smaller pieces every time I want to increase surface area but keep concentration constant. However, my chemistry teacher isn't a fan of the potato because it deals with biology as well. So I'm wondering if there is a heterogeneous catalyst for hydrogen peroxide that I can get in chunks and reasonably split it into smaller pieces? Preferably, it should be at least somewhat common to find in laboratories.

Maybe titanium oxide (skittles, sun screen, road paint), see https://doi.org/10.1021/jp300255h

Potato is bad in that the catalyst is the juice, not the surface. Titanium oxide is powdered. Come to think of it, I can hardly imagine finding anything which is not powdered and still active enough to be measured.

Try manganese metal pieces. It works great. The bottle of it that I had was in the form of thin pieces with lumpy surfaces on one side and flat smooth surfaces on the other side. If I was doing it, I would melt some wax and place the manganese pieces lumpy side down on the wax. Then the smooth surfaces would be the relevant ones.

If you search on ebay for “manganese metal”, you can see lots of the flake (electrolytic) manganese for sale. Pretty cheap, actually. Best of success with your project!

I think your question needs refinement.
Activation Energy is not affected by the surface area of the catalyst.  The catalyst is providing a different mechanism to reach the final product than without it.  It does this when particles of the reactants interact with the particles on the surface of the catalyst (since this is a heterogeneous catalyst).
The reason the reaction would go much faster if the heterogenous catalyst had more surface area is because of more particle-particle interactions able to take place between the catalyst and the reaction.  This would be similar to why reacting a piece of iron in hydrochloric acid would go much slower than if the same sized iron were cut into 1000 pieces and placed into the same solution.  The higher chance of particles to interact with each other, more likely they'll have the activation energy to form products.
A possible answer and improvement to your experiment:
If your goal is to determine how activation energy is affected by surface area, then performing the hydrogen peroxide reaction within a coffee cup calorimeter would be great.  You could measure out the same amount of water for the calorimeter, use the same mass of reactants, and measure the temperature changed caused for each surface area tested.  If activation energy is affected, the temperature change in the water of the coffee cup would be different.
Manganese metal suggested by @EdV is a good choice.  A pricier option would be silver metal.  It is very good at decomposing hydrogen peroxide, but at a higher price.

(+1) For the experiment improvement suggestions! I was going to suggest silver metal strip, since I have some and it works fine, but manganese is cheap and works well for this purpose. It is also very good for hydrogen generation from common mineral acids. Welcome to Chemistry Stack Exchange!

Thanks both of you for these pieces of advice, suggestions, and improvements for my investigation. My only question is how I would go about manipulating the surface area of the catalyst. For the metal strips you suggested, for example, would cutting the strips create a big enough change in surface area to be noticeable? I also have to actually calculate the surface area so I can't use any weird shapes or anything. I originally wanted a solid chunk of something so that it would have some depth to it, allowing me to increase surface area easily by separating it into smaller pieces.

I have posted two relevant photos at our temporary location: https://chemistry.meta.stackexchange.com/a/5143/79678. I tried the 99.9% pure (aka ‘fine’ or 3N) silver strip and a 20 g piece of silver of the same purity. It fails to do anything. I will check the hydrogen peroxide: maybe it has already decomposed. I know silver powder works, but then the surface area is unknown and there is no simple way, that I know of, to measure accurately the surface area of fine powder. I would go with the flake manganese. As for the silver strip failure, I will try roughening the surface with sandpaper.

@EdV Thank you for testing this.  I know I've done the powder as well.  Silver does oxidize quite easily, so I am 'hoping' that the issue is with that and not somehow the metal in this form doesn't work.  Which would at least give the original question asker something to discuss.  Also thanks for the welcome!  I <3 chemistry!

Sanding the silver strip works: https://chemistry.meta.stackexchange.com/a/5143/79678. Of course, now the surface area is not that of the formerly flat surface, but it is proportional, at least approximately, so the OP can use different size pieces. Anyway, good to know that roughened bulk silver metal works!

If you have any use for any of my photos I linked, feel free to use them however you please. I will delete them from the chemistry meta Sandbox III in two days or so. I frequently use that location to post temporary items, mostly photos or information, that might be helpful to others, but it is only temporary and I don’t want to run afoul of any site expectations.


text:
Why is 10 times phosphate-buffered saline more acidic (~6.9) than 1 time phosphate-buffered saline (~7.4)?
Recently, I've been told that similar pH shift (of about 0.5 units) was observed with 20 times NPS solution as well.
In case of NPS the authors provide the following recipe for the 20 times solution:
H2O 900 ml,
(NH4)2SO4 66 g,
KH2PO4 136 g,
Na2HPO4 142 g.
with expected pH to be about 6.75.
Why is that?

Asking for answers and not providing relevant details, purpose or background of questions are contradictory decisions.
Also, questions should be self containing, making sense even without link target availablity.

What is the meaning of the words "10xPBS stock" ? Is it ten times something ? Ten times what ? What is PBS ? What is 1xsolution ?

Thanks! Edited for better understanding

Activity coefficients of ions with bigger charge ($\ce{HPO4^2-}$ versus $\ce{H2PO4-}$)  drop with growing ionic strength faster, therefore the phosphate buffer gets more acidic with higher concentrations, and vice versa.

@Poutnik, thanks. Maybe you can post your comment as an answer?

Activity coefficients of ions with bigger charge ($\ce{HPO4^2−}$ versus $\ce{H2PO4−}$) drop with growing ionic strength faster, therefore the phosphate buffer gets more acidic with higher concentrations, and vice versa.


text:
Relative atomic mass calculation
In my textbook it states that relative atomic mass of an element Y is equal to the average mass off one atom of element Y multiplied by 12 divided by the mass of one atom of carbon 12.
What is the intuition behind his? I can't understand where this formula comes from. Additionally isn't the mass of carbon 12 defined to be 12 units hence the 12s cancel out anyways?

Carbon-12 is a name, much like Ivan is my name. You can't cancel anything out of it.

$$\frac {12x}{y}=  \frac {x}{(\frac {y}{12})}$$
Relative atomic masses refer to 1/12 of the mass of 1 atom of carbon-12 ( 126C ), which is conventionally chosen as the reference unit.

You could rewrite the definition of your textbook like this :
The relative atomic mass of an element Y is equal to the average mass of one atom of element Y divided by 1/12 of the mass of one atom of carbon 12. And I think you will accept this new definition.

By "average mass of atom Y" is this referring to the SI mass?

Atomic masses are awkwardly small in SI units: for convenience they are measured relative to a standard
We could measure atomic mass in SI units which would result in multiples of about $1.66 \times 10^{-27} kg$ but this would be pretty inconvenient when doing any calculations or comparing nuclear masses.
When chemists want to talk about relative mass of atoms it is far more convenient to use a different unit that makes the relative numbers small. This is sometimes called the dalton (Da) or atomic mass unit (AMU). But these numbers are not exact multiples of the number of nuclei as the energy involved in nuclear forces is so large that it impacts the resulting mass (nuclear masses are lower than the sum of the masses of the protons and neutrons involved). So chemists have to choose a specific nucleus as the reference against which other nuclei can be weighed. That nucleus is, historically, $\ce{^12C}$ which has 12 nucleons. So other nuclei are "weighed" relative to that nucleus (or, strictly 1/12 of it so the results are approximately relative to a single nucleon).
This explains the math.


text:
What is the number of Gaussian chains of network per unit volume of rubber?
In this chapter, page 70, I encountered the term `Gaussian chains of network'. In equation 3.34 (page 73) of this chapter, the phrase 'n is the number of moles of Gaussian chains' occurs in the description of a formula which is also referred to in a laboratory guide document:

An advanced three-dimensional molecular model of rubber elasticity (affine network model or freely jointed chain model, yields the equation of state
$$f = n Ak_B(α − \frac{1}{α^2})T $$
with n the number of polymer chains per unit volume, $A$ the cross-sectional area of the unstretched rubber band, and $α = L/L_0$ the ratio of the length to the unstretched length $L_0$.

Are there standard values on the internet for the number of Gaussian chains of network per unit volume of a rubber? Or maybe estimates or a rough indication? Can this value be expressed in other numbers which are easier to work with?

Note that "number of moles" is a layman synonymum for molar amount. We do not say "number of kilograms" either, but mass.

@BuckThorn I have intended my comment as the general note, not targetting anybody in particular. The document repeatedly writes the number of moles of near to the Eq. 3.34, but somewhere "of moles" is omitted.

@Poutnik Never mind. The document does use "moles of xxx". However since the OP is quoting verbatim a pass is allowed. Being stringent on the detail will not help much.

@BuckThorn I agree for number of moles vs molar amount. But "of moles" is kind of essential.

For readers, Gaussian chains are referred as "ideal gaussian chains (or freely-jointed chains)" in Polymer Scattering Wikipedia page , leading to Ideal Chain


text:
How can I find references related to pH for water solutions of salts
I have searched the web pretty much all day because I want to work out what the pH will be of various salts when they are in water solution. I have searched for references related to pKa and pH, but I find nothing. Is it really true that there doesn't exist a database anywhere in the world were these numbers have been collected? The salts I'm looking for in my case are polyphosphates; dipotassium dihydrogenpyrophosphate and tetrapotassium pyrophosphate for example.
If you can provide a reference for literature/database/table that is sufficiently large that it contain common salts at the same time as these compounds, I will accept it as the answer. If you can provide any other reference that in some way may be used to calculate the pH for the same compounds, that is helpful too.

You will have to calculate it: Pyrophosphoric acid  H4P2O7
Ka pKa
1.2 × 10−1  0.91 //  7.9 × 10−3 2.10 //
2.0 × 10−7 6.70 // 4.8 × 10−10 9.32
https://chem.libretexts.org/Ancillary_Materials/Reference/Reference_Tables/Equilibrium_Constants/E1%3A_Acid_Dissociation_Constants_at_25C

Pyrophosphoric acid is unstable in water, hydrolyzing at the $\ce{P-O-P}$ bond, but only slowly without catalysis or acidity.
$$\ce{K2H2P2O7 + H2O  ->  2H2KPO4}$$
$$\ce{K4P2O7  +  H2O  ->  2 HK2PO4}$$
pH would be based on the concentration of the hydrolyzed forms, buffering at their respective pKa's.
A 1% solution of tetrasodium pyrophosphate has a pH of 10.3 as compared to pH 11.9 of a $\ce{Na3PO4}$ 1% solution.
Apparently, the $\ce{P-O-P}$ linkage reduces the basicity of the $\ce{O- Na+}$ in the pyrophosphate compared with the phosphate.

It is better to write $\ce{KH2PO4}$ since it is ionic $\ce{K+H2PO4-}$. I left for you to edit that.


text:
Why does population difference matter in NMR
I know that population difference matters in NMR for sensitivity, that the splitting is dependent upon the gyromagnetic ratio of the nucleus, and that the population of the spin states is almost identical with a slight bias towards the lower energy spin state; but I don't really understand why both the magnitude of the splitting and the population difference matter in terms of sensitivity. For instance, DNP causes a massive splitting of energy levels, but why does that matter? I can't seem to find any good sources on the rationale that goes beyond a comparison of an electron's gyromagnetic ratio to that of a proton. Thanks!

The magnitude of the splitting (let's call it $\Delta E$) leads to a larger equilibrium population difference, via the Boltzmann distribution. To be precise, we have that
$$\frac{p_\text{upper}}{p_\text{lower}} = \exp\left(-\frac{\Delta E}{kT}\right),$$
where $p_\text{upper}$ ($p_\text{lower}$) is the population of the upper (lower) state. If you further impose the condition that the populations must sum to $1$:
$$p_\text{upper} + p_\text{lower} = 1,$$
and use the approximation that $\exp(x) = 1 + x$ (which is valid when $|x| < 1$, and is definitely true of $\Delta E/kT$), then you can show that the population difference, $p_\text{lower} - p_\text{upper}$, is proportional to $\Delta E$. (See also: https://chemistry.stackexchange.com/a/133818/16683)
So, the remaining question is why a larger population difference means a larger signal.
To start off, let's look at what the $z$-angular momentum of this sample is. For a $\ce{^1H}$ nucleus (a proton), the 'lower' state is $\alpha$ (spin-up), and the 'upper' state is $\beta$ (spin-down). The $\alpha$ state has $z$-angular momentum of $+\hbar/2$, and the $\beta$ state $-\hbar/2$. So, the net angular momentum is
$$I_z = \frac{\hbar}{2}Np_\alpha - \frac{\hbar}{2}Np_\beta \propto \frac{\hbar}{2}\Delta E,$$
where $N$ is the total number of molecules, and $p_\alpha$ and $p_\beta$ are the populations of the $\alpha$ and $\beta$ states.
In an NMR experiment, you don't measure this angular momentum directly: what you measure is the voltage induced in a coil by the net magnetisation of the chemical sample. The larger this magnetisation, the stronger the signal. The magnetisation, in turn, is directly proportional to the angular momentum:*
$$M = \gamma I,$$
where $\gamma$ is a physical constant (the gyromagnetic ratio of the nucleus being studied). And the angular momentum is proportional to the population difference, which is proportional to the energy difference.
Much of this is covered in standard magnetic resonance textbooks such as Keeler's Understanding NMR Spectroscopy and Levitt's Spin Dynamics, which I'd strongly suggest reading.

DNP causes a massive splitting of energy levels

That's not true; DNP doesn't change the energy levels. Hyperpolarisation techniques (including DNP) seek to establish a non-equilibrium population difference which is much larger than the Boltzmann population difference.

* The magnetisation that is detected must lie along the $xy$-plane. The NMR experiment must rotate this magnetisation from the $z$-axis into the $xy$-plane. However, this doesn't affect the conclusions above, as the rotation cannot change the amount of magnetisation available. So the initial $z$-magnetisation still determines how much signal is observed. (For the very nitpicky, I'm ignoring non-unitary transformations.)

In addition to your clear answer I would add that that in absorbing electromagnetic radiation of any type a difference in population between two levels is essential, otherwise no net absorption occurs, and the bigger the population difference the bigger the chance of absorption.  In NMR the large static magnetic field splits the nuclear energy into two levels (in the case of protons) with a small population difference as described above. The bigger the mag field the bigger $\Delta E$ and the bigger the nmr signal.


text:
NAD+ and NADH /H+
The $\ce{NAD^+}$ molecule has an electron-poor carbon susceptible to a nucleophilic attack by a hydride ion: It's the C4 of the pyridine ring, who loses its aromaticity in this way.
Usually, it is written that $\text{NAD}$ + is reduced with the acquisition of two hydrogen atoms, so the reduced form of $\text{NAD}^+$ would be $\text{NADH} + \ce{H}^+ $. But the $\ce{H}^+$ ion, where does it come from?

It comes from one of the two hydrogen atoms that you just mentioned.

One of them is the hydride ion who binds to C4 in the pyridine ring. And the other?

You just said it all yourself. We are given two (2) neutral hydrogen atoms. One becomes hydride ion and goes to NADH. Another becomes H+.

Here is the oxidation half reaction:
$$\ce{NADH -> NAD+ + H+ + 2e- }$$
When you balance the half reaction, you put a hydrogen ion on the product side, and this would go into aqueous solution.
And here is the full redox reaction with formaldehyde reacting to methanol (small molecules for simplicity; you could also use pyruvate and lactate):
$$\ce{NADH + HCHO + H+ -> NAD+ + CH3OH}$$
Notice that the $\ce{H+}$ appears on the opposite side. When NADH reduces a carbonyl, it provides the hydride anion, and the hydrogen ion (i.e. cation) comes from the solution.
The reaction with molecule hydrogen would be written like this (but I am not sure if it is observed in nature - it is certainly not part of the primary animal or plant metabolism):
$$\ce{NADH +  H+ -> NAD+ + H2}$$
Again, NADH provides the hydride, and the aqueous solution provides the hydrogen ion.

See also: http://web.mit.edu/7.01x/7.014/documents/Redox.pdf


text:
Terms used to describe the Pauli exclusion principle
This is a very simple question. Today, someone explained to me that the Pauli exclusion principle says that "no two electrons can occupy the same energy state". However, I believe that this language is not precise as two electrons can occupy the same energy state, allowing they have different spin. Rather it is the quantum state that cannot be the same.
So, my question is, is it wrong to say energy state?

I’d go one step further and argue that the term ‘energy state’ is not well-defined. If they meant that two electrons can’t have the same energy, then they should just say that (even though it’s obviously wrong).

The principle means that electrons cannot have the same set of quantum numbers and be in the same state. You realise this by saying that they must have different spin. (Electrons have two spin quantum numbers. Spin quantum number in your question means magnetic/projection/azimuthal spin $m_s=\pm1/2$, whereas the other electron spin quantum number is always the same and is $S=1/2$)

Depending on the level at which it is taught, you will find different statements, some of which are just wrong. Related: https://chemistry.stackexchange.com/q/111062/72973


text:
Is "pentel" a real chemistry term?
Allegedly "pentel" is a synonym for "pnictogen," that is, an element of group 15, the nitrogen group. However, I have not been able to find any sources to support the existence of this sense of the word except Wikipedia, Wiktionary, Wikidoc, and these two pages at thoughtco.com by Anne Marie Helmenstine, Ph.D. No other dictionary I've consulted has it, not even the OED. They only mention "pentel" as a certain kind of pen. Helmenstine's article cites

Modern Chemistry (South Carolina). Holt, Rinehart and Winston. Harcourt Education (2009)

but this citation is for the whole article, not specifically for "pentel."
So is this real, or is it a case of citogensis?

Use Google Scholar (scholar.google.com/) when needed. All I searched was: pentels pnictogen, and the third search result leads to an article in Angewandte Chemie which in turn leads to the original source of your requested term. There is a footnote in the Angew. Chem. article "A Deltahedral Zintl Ion Now Made in the Solid-state" by Virginie Queneau and Slavi C. Sevov.
Footnote 1 reads

The use of triels, tetrels, and pentels or pnictogens has been
suggested by the Commission of Nomenclature of Inorganic Chemistry in
the Inorganic Chemistry Division of IUPAC: B. P. Block, W. H. Powell,
W. C. Fernelius Inorganic Chemical Nomenclature. Principles and
Practice, American Chemical Society, Washington, DC, 1990, p. 24.

I respect the Oxford English Dictionary, but sometimes they are very stubborn in not including some chemical terms, such as banana bonds, and anation (reaction). I suggested several times but it fell on deaf ears.
BTW, the 22-volume OED does have Pentel but not in the chemical sense; it is "  A proprietary name for: a type of ballpoint pen typically having a non-metallic tip and a free flow of ink resembling that of a felt-tip pen."

But they do include words like exptl and muzjik (?).

Group 13 and group 14 elements are trivially called triels and tetrels respectively (see previous posts- this and this). By same analogy, group 15 elements are called "pentels".

pentel (pnicogen, pnictogen)
N,P,As,Sb,Bi derived (with contraction) from pent(a)-and element-referring to the fifth main group of the periodic table
Elsevier's Dictionary of Chemoetymology: The Whys and Whences of Chemical Nomenclature and Terminology by Alexander Senning, Elsevier, 2006

However, this usage is so rare and uncommon in chemistry that if you search for "pentel" on Google, it will lead to certain ball point pen.

well... i don't think any amount of usage in chemistry would make it competitive in searches with said ballpoint pen company - not least because ballpoint pens are a lot less niche

@somebody And group 15 probably has a smaller SEO budget than an international stationary manufacturer.


text:
Does ionization energy play a role in absorption spectrometry of non metals?
Absorption spectrometry is useful for analysing metals in the samples. But it's not recomended for non metals though. I know that it's difficult to analyse non metals because most of them absorbs in the same wavelenght as atmosphere (oxygen). Anyway, I would like to know if ionization energy plays a role in this analytical technique. I can't find any strong source.

Your intuitive argument is not that far off. I am sure you are talking about atomic absorption spectrometry (AAS). Analyzing metals is routine but analyzing non-metals is indeed near impossible because one has to work in the vacuum UV region, but some non-metals like phosphorous can be determined with very low sensitivity (=large detection limits). Indirect methods for non-metals exist though.
The main point to remember in AAS is that you want free atoms not ions. So nobody is interested in ionizing metals or non-metals (hence their ionization energy), rather it is desired that ionization be suppressed in AAS using ionization suppression buffers. What one is interested in AAS in the electronic transition from the ground state to the first excited state. Ionization is far far away.
Look at this data from Perkin Elmer's AAS Cookbook for P, and its corresponding harsh conditions to even atomize phosphorous. A nitrous oxide is flame is required.

So the argument of ionization energy is invalid. It all makes sense now.

I think so. There is Saha equation that talks about ionization but that is for stars but some people did connect to flame spectroscopy as well. I studied this decades ago.


text:
What balances charges that may build up as reduction and oxidation occur in a voltaic cell?
There's a disagreement between me and my high school chemistry teacher regarding the answer to this question. I'm not convinced by his argument and neither is he convinced by my argument.
The problem goes as following

What balances charges that may build up as reduction and oxidation
occur in a voltaic cell?
a. the salt bridge.
b. the electrolyte solutions
c. one of the half-cells
d. the moving electrons

We both agree that the word "charges" in the question refers to electrons. I argue that the answer is "a. the salt bridge". The salt bridge allows ions to pass to between the half-cells which prevents the build-up of charge on one of the half-cells due to the electrons, allowing the reaction to continue. So the salt bridge counter-acts or balances the charges. My teacher claims that the answer is "d. the moving electrons". He argues that moving the electrons balances the charges (I can't really phrase his argument).
My understanding of "balances the charges" is "counter-acts the effect of the charges". Perhaps this is where I just don't understand the question or my teacher. How can the movement of the electrons balance the charges or, in my understanding, counter-act the effect of the charges when it, in my understanding, is what creates the effect of the charges.
What is the correct answer and why?

A bridge is optional, need not to be present if half cells share the electrolyte. Electrons are not present in electrolytes but ions. Free electrons reduce water in less then millisecond.

Also, electrons, coming and leaving, disbalance charges.

What a terribly asked exercise! You're both kind of correct and kind of wrong at the same time. The electrons from the oxidation half cell just move to the reduction half cell, answer d. At the same time, ions must travel the other way. This could be via a salt bridge, but that doesn't make it the correct answer.

Let us clarify a couple of fundamental aspects. In a functional galvanic/voltaic cell there is always a very small charge imbalance at the electrode solution interface, this is why voltage difference develops between the two electrodes. However the bulk of solution is electrically neutral. This bulk neutrality is indeed achieved by a salt bridge. If you go deeper, the salt bridge can also have a slight charge imbalance due to different speeds of the anion and the cation in the gel. So your argument is correct about salt bridge in a general sense.
The conceptual problem with the question is the fact that none of you are specifying the part of the galvanic cell. Are we talking about the bulk solution, are we talking about the electrode-electrolyte interface or are we talking about the external circuit?
It just reminds me of the famous but really old school poem, The Blind Men and the Elephant, when six keen scholars went on to explore an elephant. One holds the legs and thinks that the elephant is a trunk. The second one holds the broad side and calls it a wall, the third touches the tusks and thinks that the elephant is nothing but a spear ...and the fight continues...
And so these men of Indostan, disputed loud and long,
each in his own opinion, exceeding stiff and strong,
Though each was partly in the right, and all were in the wrong!


text:
Does charging Li-Ion batteries at very low C-rates reduce their lifespan?
Rapid charging of “energy cell” (e.g. lithium-cobalt) type Li-Ion batteries, say at 2C, 3C or higher, is well known to reduce battery lifespan.
On the other extreme, does charging at very low rates, say C/10 or C/20, have any effect on their lifespan?
For instance, I charge my portable devices such as phones and tablets using a 2.5 W charger. My thinking is that any extra time spent charging is time not spent discharging, which would contribute to extra cycles on the battery, as well as slightly reducing temperatures, since high temperatures are also known to be detrimental to battery lifespan. I also use limited-wattage-output power banks to (very slowly) charge my laptop battery while away from a power outlet.
While the effect on these variables may be minor, it should be beneficial or at most irrelevant. What I’m looking for are any references to detrimental effects from slow charging.

Why people keep thinking battery lifespan is a chemistry issue?

@Mithoron It is about matter, right? And there are ongoing chemical reactions, right? Then it must be the dark magic of chemists, the mages in white coats, gloves and googles, who understand all the mysteries of the matter, right?  :-) // It seems to me physicists are more successful than chemists in delegation of  part of their science out of their science. (chemistry is physics of interaction of atomic electrons) //
In fact, it is a chemistry issue, but of applied chemistry within the battery technology, design and manufacture. Li-Ion/pol degradation has very interesting chemical aspect.

A very low charge could reduce the coulombic efficiency of the battery, specially the NiCd and NiMH. While a fast charge drop the CE to 90%, a very low charge could drop the CE to 70%

I just came across this paper. Table 3 shows a series of effects that contribute to degradation, and indicates whether they depend on low/high T, V, SoC, I. Many depend on high current, none on low current. Despite some noise here about CE without supporting references, all signs point to low current not being an issue.

Low charging rates are issue only in cases if it affects proper detections of charge stopping thresholds. Typical case are NiMH cells, where signs saying to stop charging at C/10 or lower are not significant enough. Li-ion/pol cells have strictly given max voltage threshold which is easy to control.
Where it is applicable, like for Li-ion/pol laptops batteries, more important than the charging rate is the charge level management. Because, fully charged lithium cells have several times faster static aging than cells charged to 40-50%. And, near fully charged cells are much more sensitive to high temperature.
The typical scenario to be avoided is

high ambient temperature
device heated by high power load
battery being charged, with current charge > 70(80)% of its capacity.

Similar effect has the deepness of charging cycle. Shallow charging can provide within the cell lifetime 2-4 times larger total cumulative capacity, compared to full charge/discharge cycle.
E.g. in case the laptop is attached to AC most the time, and only occasionally running on battery, it is good to have charging start/stop thresholds near 50% of the charge capacity. E.g. 50% - 60%.
My experiance with Lenovo Thinkpads (T400/T430/T470/T14) is both Linux and Windows offer ways to manage it. ( Lenovo (commercial) Vantage for Windows, tpacpi-bat/acpi_call utils for Linux ).

Thank you for this answer. Indeed I also watch out for heat and avoid high SoCs -- in a Mac this can be done automatically using e.g. the AlDente app. I concur that low C-rate charging is a tertiary concern.


text:
Does Sodium carbonate and magnesium hydroxide react?
I understand Sodium carbonate and calcium hydroxide reacts, but I am surprised to see there is not any description regarding Sodium carbonate and magnesium hydroxide reaction. Is there a reason? do these two compounds react? Looking at solubility rule they should react and form precipitate.
Double Displacement Reactions on libretexts.org

magnesium hydroxide is practically insoluble in water (solubility is 0.0064 g/L at room temperature) compared to calcium hydroxide (solubility of 1.6g/L).

The key is taking solubility product of magnesium hydroxide and magnesium carbonate and evaluate the equilibrium.

As I understand, sodium carbonate is used as a water softener. It supposedly reacts with magnesium hydroxide to form 'magnesium carbonate' (https://www.tatachemicalseurope.com/applications/water-purification).
The 'magnesium carbonate' might actually be magnesium carbonate hydrates (https://www.nature.com/articles/s41598-021-02261-8).
Presumably, since sodium carbonate can raise the pH to values > 11, one might have to work with diluted solutions of sodium carbonate and magnesium hydroxide, such that the pH is not much lower than 9 (magnesium carbonate dissolution) and not much higher than 10 (magnesium hydroxide precipitation).


text:
Does norcamphor have enantiomers?
The following two molecular structures have two chiral centers, so in total four chiral centers:

I did the absolute configuration multiple times and I keep getting R,R and S,S, which should be enantiomers. My book says these two are “identical molecules that are superimposable”. What am I doing wrong?

These two are enantiomers.

Just try superimposing these two structures! BTW your CIP assignments are incorrect. Assignments to bicyclic compounds are not straightforward. Reading left to right, they are (1S,4R)-bicyclo[2.2.1]heptan-2-one and (1R,4S)-bicyclo[2.2.1]heptan-2-one. (Take a look at the case of camphor here. (Don’t be confused by the R,R assignment to camphor.)


text:
sp mixing in $N_2$ vs sp mixing in CO
We were first taught that in B, C and N the bonding $\sigma_{2s}$ orbital repels with the bonding $\sigma_{2p}$ orbital which cause the energy level of $\sigma_{2p}$ orbital to increase. This increase in energy level causes the $\sigma_{2p}$ to switch positions with $\pi_{2p}$orbitals. We were told that this phenomenon is known as sp mixing.
But then we were taught the MOT diagram of CO. Over here, it looks like the bonding $\sigma_{2s}$ and the bonding $\sigma_{2p}$ orbital are not repelling each other but rather mixing (or hybridising?) with each other.  So what does exactly happen during sp mixing? Do the orbitals repel or mix?

My main doubt is that why are the 2s orbitals in N2 only contributing to the 2s sigma bonds, while the 2s orbitals in CO are contributing to both non bonding orbitals and sigma sp?
Why isn't a sigma sp being formed in N2?

Can you cite a source for the concept that the orbitals repel each other?

@Andrew I don't particularly have a resource since my teacher gives us notes ( We dont use textbooks, only workbooks). But I think this site is also saying the same thing? https://courses.lumenlearning.com/suny-albany-chemistry/chapter/molecular-orbital-theory/   correct me if i am wrong.

I don't see anything in the linked article about MOs repelling each other. It says that the change in energy is due to s-p mixing, and that mixing is less in O,F and Ne because of the larger starting energy difference between the s and p orbitals which they attribute to the pairing energy of filled p orbitals (I would add to that the higher effective nuclear charge). Nothing about repulsion between MOs

@Andrew maybe repel is the wrong word, but I can't find sources that say that the s and p orbitals mix (hybridize) either. Most of the stuff online say that the orbitals interact and influence their energies. What exactly does interact mean here?

@Andrew For example- https://www.google.com/url?sa=t&source=web&rct=j&url=https://en.m.wikipedia.org/wiki/Molecular_orbital_diagram%23s-p_mixing&ved=2ahUKEwjK_Oi57vL6AhWdgGMGHSfABKkQ0gIoAXoECBcQAg&usg=AOvVaw3JkXQSOUP8kAi9xHW-Jl9h

Although it's called "s-p mixing", it isn't the s and p atomic orbitals mixing, but rather the molecular orbitals derived from them, specifically the $\sigma_s$ and $\sigma_p$. This is different from hybridization, as it results in a change in energy of the eigenfunction orbitals of the energy operator, whereas hybridization creates an alternative orbital basis set that are not eigenfunctions.

@Andrew Why does this happen in CO but not in N2?

@Pumpkin_Star the site you linked (https://courses.lumenlearning.com/suny-albany-chemistry/chapter/molecular-orbital-theory/) has the answer to your question: "...the 2p orbitals for O, F, and Ne are higher in energy than the 2p orbitals for Li, Be, B, C, and N. Because of this, O2, F2, and N2 only have negligible s-p mixing (not sufficient to change the energy ordering), and their MO diagrams follow the normal pattern..."


text:
Cyclic voltammetry graph of graphite electrode in NaOH
I'm a student who is a beginner to cyclic voltammetry and university electrochemistry in general. For my lab experiment, I performed cyclic voltammetry on a graphite electrode in 1 M NaOH with Ag|AgCl reference electrode.
I was wondering why I got this kind of graph, instead of a leaf shape indicative of the formation of an electrochemical double layer. I tried searching but I can't find any source writing about such a simple setup.
I am hoping if anyone can explain the shape of this graph, and why I did not get a leaf shape CV graph.

Did you clean the electrode? It seems something Is getting reduced, maybe some impurities?
Also, what is the diameter of the electrode you are using? and the scan rate, maybe use a faster one? If you are a beginner, I recommend you also read this ARTICLE :


text:
What determines the higher priority in CIP rules, if both groups have the same atoms?
Which of the following groups has the highest priority in the Cahn-Ingold-Prelog sequence rules?
a) CH2CH3
b) CH=CH2
c) C≡CH
d) C(CH3)3
I narrowed it down to either C or D. The carbon in answer C is triple bonded to another carbon. The carbon on answer D is also bonded to three carbons. How do I decide between the two? My book gives C as the answer.

Type of bond (Triple > Double > Single) has higher priority than total number of single bonds.

I don't think the type of bond part has been formalised. The rule are in the blue book to which I currently have no access. An answer should include this reference.

In lieu of the IUPAC Blue Book, you might visit this link.

@Sam202 Based on what I remember, that statement is not correct. A triple bond is treated as a a carbon plus two "phantom" carbons. If you compare this to an ethyl group, the triple bond will have higher priority because in the ethyl group, the connected carbon only has one carbon attachment. But against a t-butyl group, the t-butyl group should have higher priority than an alkyne. Cf. https://en.wikipedia.org/wiki/Cahn%E2%80%93Ingold%E2%80%93Prelog_priority_rules#Double_and_triple_bonds

@Zhe Organic Chemistry by L.G. Wade puts isopropyl's CIP priority under vinyl group. Situation should be analogous between t-butyl and ethynyl.

@Sam202 I don't think that's right...

@Zhe Well, it seems 2 books support that (Wade and OP's book). Maybe you could cite one that states otherwise?

Looks like I gave away my copy of Elial, but I found IUPAC blue book P-92.1.4.2. You need to expand the multiple bond, duplicating the atom on the end, but the duplicates each have phantoms attached with 0 priority. So, a t-Bu is a carbon with three carbons attached, while an ethynyl is also a carbon with three carbons attached. But in the next sphere out, the t-Bu has more real hydrogens where the ethynyl has only one real hydrogen and the rest are phantoms.

Comparing ethynyl vs t-butyl in terms of shells we have, respectively: [C,C,C][C,C,H] vs [CCC][HHH]. Ethynyl should have higher priority.

I don't understand what you're saying there. Ethynyl has [C][CCC][H] and tBu has [C][CCC][HHHHHHHHH]

Let X* represent a duplicate atom and X a real atom. In ethynyl, the first C is bonded to C*, C*, and C. The second C is bonded to C*, C*, and H. On the other hand, in t-butyl, the first C is bonded to C, C, and C. Then, whichever C you choose as the second layer will be bonded to H,H, and H. [CCH] beats [HHH].

@Sam202 You're totally right. I forgot about the phantoms on the other end...

@Zhe just a final point for clarification. Your source describes two different types of imaginary atoms: "duplicate" and "phantom". While phantoms have 0 priority, duplicates keep the priority of the real atom they're representing. Distinction between the two is important if other sources name these imaginary atoms differently.

@Sam202 That is correct. I also erred on my original statement.

As the old adage goes, "A Picture is Worth a Thousand Words." Perhaps a picture is required. I apologize in advance for any redundancy with explanations provided in the Comments.
There are four functional groups in question. I have arranged them as substituents about a single carbon atom with arbitrarily assigned stereochemistry and atom numbers. A digraph is the best way to assign priorities to the four substituents. Duplicate atoms (ghost, phantom, etc.) are shown in parentheses and they have the same atomic number as their non-duplicate (original) atom. However, duplicate atoms have three atoms of atomic number zero attached to them and they are not displayed in the digraph. Their designation is {0,0,0} and, by default, have the lowest possible ranking.
In sphere 2 (red numbers) all of the carbons are equivalent. No priorities may be assigned. Progressing to sphere 3, one has $\ce{C2}${3,H,H}, $\ce{C4}${5,(4),(4)}, $\ce{C6}${7,8,9} and $\ce{C10}${11,(10),H}. With the atoms in braces ranked in descending order of priority and looking for a first difference, it is clear that $\ce{C2}${3,H,H} is the lowest priority atom with $\ce{C10}${11,(10),H} the next to lowest priority. To distinguish between $\ce{C4}$ and $\ce{C6}$, one must proceed to sphere 4. Here $\ce{C4}$ displays {(5),(5),H} which is higher in priority than that which is offered by $\ce{C7}$, $\ce{C8}$ or $\ce{C9}$--namely--{H,H,H}. Therefore, the priorities of the carbon atoms in sphere 2, and the groups of which they are comprised, have the ranking $\ce{C4}$>$\ce{C6}$>$\ce{C10}$>$\ce{C2}$. In other words, yne>t-but>ene>ane. The compound has the (S)-configuration.


text:
Isotopes \ monoisotopic elements
I am confused about the definition of isotopes
I know that isotopes are atoms of an element with different numbers of neutrons. What I don't understand is that we have monoisotopic elements which are elements that have only a single stable isotope.
Why do we say isotope in this sentence, we are not comparing the atoms of that element with each other (for example we can not say the number of neutrons is different cause there is only one type of atom)?

Unknow what you know. Isotopes are atoms of an element with the same numbers of neutrons.

Always try to find answers on your own in offline and online resources, before asking. It gives you more than when just receiving the answer.

I have read my book and all my notes and PDF from different sites. But I couldn't understand this is why I am here

There are many available resources for the topic of isotopes. Multiple Wikipedia pages as the comprehensible summary and source of primary references.

This post could use a more descriptive title.

You were taught a sloppy definition of "isotope. ;)  An isotope is a type of atom that has a defined number of protons AND neutrons.

An isotope of the given element is an atom (or element consisting of just such atoms) with the particular neutron or nucleon number.
Note that every element has more than one isotope, often dozens of isotopes, but many, most or all are unstable. Search for Wikipedia articles with names "Isotopes of <element>"
In this context, elements with odd proton number have maximally 2 stable isotopes, sometimes just one like fluorine or none like technetium.
E.g. fluorine has 18 known isotopes.
See also

26 monoisotopic elements which have only a single stable isotope (nuclide).
21 mononuclidic elements, found naturally on Earth essentially as a single nuclide (which may, or may not, be a stable nuclide).

You may notice also the term nuclide with slightly different meaning than an isotope, and used in a different context. Generally, the term isotope is used in context of chemical or physical atomic properties, while nuclides in context of nuclear properties. Differences arise if a nucleus is known to exist in different states ("nuclear isomers") with  different energy, spin, stability and possibly decay mode schema.


text:
NaOt-Bu + chloroethane

What is the major product formed by reaction of sodium tert-butoxide and chloroethane?

I thought NaOt-Bu is a bulky base and the major product will be ethene due to E2. But the answer says t-butyl ethyl ether (due to SN2). Can you please help?

I think the answer is wrong. I'm on my phone so I can't give references, but if you search "t-butoxide reaction with primary haloalkanes" you will find references that support this.

Here is a link to what @Waylander mentioned. NaO-t-But is rarely used as a base but KO-t-But is the preferred base. This has nothing to do with their inherent properties but rather their ease of preparation. Sodium (mp 98) does not melt in boiling HO-t-But (bp 82) where as potassium (mp 63) does. Potassium reacts faster than sodium in the solid state with alcohols. Potassium has the added advantage in this instance of being in the liquid state with a larger surface area.

It seems this is an edge case where steric hindrance of base and alkene stability are opposing factors and perhaps both the products will be have significant yields. It'd be good if the reaction could've been performed for real.

Firstly, it's important to identify which species is the nucleophile/base, and which species is the substrate.
In this case, $\ce{(CH3)3CO^-}$ is the nucleophile/base, while $\ce{CH3CH2Cl}$ is the substrate.
This substrate is a primary (1°) alkyl halide because the $\ce{C}$ atom the $\ce{Cl}$ atom is bonded to, is only bonded to one $\ce{C}$ atom.
For $SN2$ reactions, substrate reactivity in terms of number of carbon substituents around the $\ce{C-X}$ bond is:
$$1°>2°>3°$$
For $E2$ reactions, substrate reactivity is the opposite:
$$3°>2°>1°$$
This is explained by the following:
Methyl groups ($\ce{CH3 -}$) act as electron donating groups, which help stabilize the positive partial charge that is formed in the $\ce{C}$ atom from which an $\ce{H}$ is extracted by a base (in an E2 reaction).
In other words, a double bond that is formed with a higher number of neighboring methyl groups will be more stable than one formed with a lower number.
If you try to perform $E2$ between a 1° alkyl halide (substrate) with a base (bulky or not), the resulting transition state will be very unstable because there are no substituents providing stabilization to the double bond that would be later formed.
This lower stability, in terms of kinetics, translates to higher activation energy and therefore, lower reaction rate.
According to S. Bailey and A. Bailey:

When alkyl halides are treated with a nucleophile, they can undergo either substitution or elimination. The most important factor in determining which will occur is the stability of the alkene that could be formed by elimination should it predominate.

The steric hindrance you're referring to (bulky reactant) as a way of discriminating between $E2$ vs $SN2$ dominance, is more impactful when the substrate is what's bulky, not the nucleophile/base itself.
For example, if we switched things around and had a small nucleophile/base like $\ce{CH3O-}$ react with a bulky substrate like $\ce{(CH3)3C-Cl}$,$\;$ $E2$ would dominate over $SN2$, which is consistent with the conclusion you reached.
I could not find an explicit source with porcentual yield estimates for the competing products in this reaction by $SN2$ and $E2$ (ETBE and ethene, respectively), but I will try to make the case for why I believe, based on what I've researched so far, that $SN2$ would slightly dominate in this case.
First, I will use a similar reaction with known product yields as a reference point.
According to Yurkanis Bruice, the reaction between t-butoxide and 1-bromopentane yields 15% of the $SN2$ product (1-tert-butoxypentane), and 85% of the $E2$ product (1-pentene):

While the reaction we're dealing with here is:

There's 3 important differences that would promote $SN2$ and not $E2$ when comparing both reactions:
(1) $R$ group size in substrate:
The R group in the reference reaction is pentyl, while in our reaction it's ethyl, which is a difference of 3 carbon groups.
According to Yurkanis Bruice, in similar cases, a difference of just 1 carbon group implies more than a two-fold increase in $SN2$ rate:

The rate of an SN2 reaction depends not only on the number of alkyl groups attached to the
carbon that is undergoing nucleophilic attack but also on their size. For example, bromoethane and
1-bromopropane are both primary alkyl halides, but bromoethane is more than twice as reactive in
an SN2 reaction, because the bulkier alkyl group on the carbon undergoing nucleophilic attack in
1-bromopropane provides greater steric hindrance to back-side attack.

So, we can expect that the absence of 3 carbon groups in the substrate of our reaction will boost the $SN2$ rate by a significant factor.
(2) Polarity of C-X bond in substrate:
$X$ is bromine in the reference reaction, while it is chlorine in our reaction.
The dipole moment $\mu$ of the C-Cl bond is higher than C-Br and even C-F. Their value according to L.G. Wade is:

This means that the carbon atom in C-Cl is more electrophilic (has a higher positive partial charge) than the carbon atom in C-Br.
In other words, the attraction between the nucleophile (t-butoxide) and such carbon atom is higher in our reaction, which also promotes $SN2$.
(3) Alkene product stability:
The $E2$ product in the reference reaction is a mono-substituted alkene (1-pentene), while it is a non-subsituted alkene (ethene) in our reaction. As mentioned before, substituted alkenes are more stable than non-substituted ones.
As a secondary reference point to estimate how important alkene stability is for $E2$, Marc Loudon claims the yield of ethene when bromoethane reacts with ethoxide is only 1%:

Furthermore, the addition of just 1 carbon group to the substrate R chain results in a 10-fold increase in yield for the resulting mono-substituted alkene:

I think it's reasonable to believe that the combination of these 3 differentiating factors (in this very particular case) offset the factor of steric hindrance in t-butoxide and lead to the $SN2$ product being the major product (although probably by a small margin), as claimed by OP's book and according to S. Bailey's alkene stability trend.
As a final note, Yurkanis Bruice claims Williamson ether synthesis (i.e. the particular $SN2$ reaction taking place in our case) is significantly more efficient when bulky alkoxides are used compared to when bulky substrates are used:

Consequently, a Williamson ether synthesis should be designed in such a way that the
less hindered alkyl group is provided by the alkyl halide and the more hindered alkyl
group comes from the alkoxide ion.

Do you have any citations to support your claim that the substrate steric bulk is significantly more impactful than the nucleophile bulk?

@Waylander "When alkyl halides are treated with a nucleophile, they can undergo either substitution or elimination. The most important factor in determining which will occur is the stability of the alkene that could be formed by elimination should it predominate". Organic Chemistry by Bailey Jr. (page 223). Book can be consulted here: https://archive.org/details/organicchemistry0000bail/page/n551/mode/2up

that quote says nothing about the effect of nucleophile sterics. See the reference I quote in my answer.

In the context the quote is made, it says that when weighing in all the factors previously discussed in terms of S vs E selectivity (i.e. solvent, steric hindrance, alkene stability, etc.), alkene stability is the most important one.

According to this book chapter by Robert J. Ouellette and J. David Rawn quoted here, original  here

If a primary haloalkane is treated with tert-butoxide ion instead of ethoxide, the amount of elimination product increases significantly. The tert-butoxide ion is not only more basic than the ethoxide ion, it is also much more sterically hindered. The combination of these two factors favors elimination by an E2 process over substitution by an SN2 process.

It goes on to offer the example of 1-bromobutane reacting with t-butoxide to give 10% of the t-butoxybutyl ether (SN2 product) and 90% but-1-ene (E2 product).
I see no reason why chloroethane should behave any differently. and

According to this source, the preparation of an unsymmetrical ether (like ETBE, i.e. the product in this reaction assuming SN2 dominance) claims a primary alkyl halide and a tertiary alkoxide ion are the best reagents.

As the data I referred to shows, it is possible to make tBuOR ethers using Williamson. It is just not a very good way of doing it as you get more elimination. All the industrial preparations are done using tBuOH under acid catalysis e.g. https://www.sciencedirect.com/science/article/abs/pii/S1381514899000929

That's not necessarily true. If R=methyl, preparation of MTBE with t-butoxide and chloromethane won't produce any E2 product since there's no second carbon atom in the substrate to form a double bond with. In the case we're dealing with here, R=ethyl, and the formation of ethene (assuming E2 domination) implies forming an alkene (ethene) with 0 substituents to stabilize it. Conversely, in the example you cited, 1-bromobutane produces a mono-substituted alkene (1-butene), which is more stable than a non-substituted one.

Your example involves 3 different variables with respect to this problem: (1) Electronegativity (Cl > Br). (2) Etyl-X vs Butyl-X. (3) Alkene product stability (0 substituents  vs 1 substituent). I think it's not an analogous example to draw a valid conclusion from.

@Sam202  I disagree. In my opinion it is the best example available to us as the actual reaction is not reported in the literature (unless someone knows better). The principle quoted by Ouellette & Rawn accords with my own experience as a practising synthetic chemist.

Well, not to discredit your experience and expertise in the subject, but it seems OP's book result is consistent with what my book claims about alkene stability being more important than steric hindrance. As I said above, steric hindrance in t-butoxide does not impede $CH_3-Cl$ to exclusively react by SN2 to form MTBE, and $CH_3CH_2-Cl$ is much more similar to it than $CH_3CH_2CH_2CH_2-Br$ is.

Again I disagree. SN2 is the only reaction mode availalble to Me-Cl. Elimination is key to the discussion and it cannot occur in this case. There is no definitive answer to this unless you have the facilities to run the reaction - I do not.


text:
Compare the covalent character of ZnCl2, CdCl2,HgCl2
I want to know how to arrange the covalent character of ZnCl2, CdCl2, and HgCl2.
By Fajans' rules, We know that a smaller cation equals higher covalent character.
We also know that atomic size increases as we go down a group. So, covalent character should decrease as we go down the group. Meaning covalent character should be in the order ZnCl2 > CdCl2 > HgCl2.
However, ZnCl2 and CdCl2 are ionic whereas HgCl2 is covalent. What is the explanation behind this?

I think you misinterpreted that rule. When atoms have a similar size (when big) in a compound they will behave as covalent bond character. Also, is good to remember that they usually have similar electronegativity values. So for example check the diatomic molecules as Cl2 or O2. They obviously are not compounds like you mentioned but you can take them as an example.
When you have smaller atoms in a compound such as NaF, you sure have similar size but their outermost shell are different from HgCl2. Sodium has low electronegativity and fluorine has a large energy gain when receive an electron. Also is good to remember that in NaF you have ions, differently from compounds with covalent character.
Electronegativity for the cited atoms:

chlorine: 3,16
mercury: 2
fluorine: 4
zinc: 1,65
sodium: 0,93
oxygen: 3,44
cadmium: 1,68

EDIT: also you have to check the polarizability of certain atoms. The bigger the atoms the higher polarizability. So in compounds with different in size between the atoms you have to consider the polarizability and when you have compounds such as NaF and HgCl2 you have to check the electronegativity.


text:
Equilibrium constant vs Reaction rate constant
For a reaction, e.g.,
$$a X + b Y → c Z$$
Its reaction rate constant is
$${\displaystyle r=k_f(T)[\mathrm {X} ]^{m}[\mathrm {Y} ]^{n}}$$
where the exponents m and n are called partial orders of reaction and are not generally equal to the stoichiometric coefficients a and b.
However, for the equilibrium constant, the exponents must be stoichiometric, e.g.,
$$K_c=\frac{[Z]^c}{[X]^a [Y]^b}$$
However, for reversible chemical reaction, $K_c=k_f/k_b$ which means that stoichiometric coefficients are required for equilibrium. Does this mean that when the exponents of the rate reaction equation are similar to that of stoichiometric coefficients, the reaction is in equilibrium?

If there are more complex reactions, reaction rate equations do not follow the numerator and denominator of the equilibrium constant equation.

They may not match stoichiometric coefficients.
The may have non-integer exponents.
They may contain additive and/or ratio terms.

Such reactions are not elementary and have a reaction schema.
Therefore, such forward/backward reaction rates, involving initial reactants and final products, do not belong to the opposite directions of the same elementary reaction.
The ratio of their rate constants then need not to match the equation for the equilibrium constant.
As example, see surprisingly complex forward reaction rate equation(small PDF) for the reaction
$$\ce{H2(g) + Br2(g) <=> 2 HBr(g)}$$
$$\frac{\mathrm{d}[\ce{HBr}]}{\mathrm{d}t}=\frac{k_1[\ce{H2}]{[\ce{Br2}]}^{\frac 12}}{1 + k_2\frac{[\ce{HBr}]}{[\ce{Br2}]}}$$
(simplified)

There is no contradiction if the partial order of reaction differ from the stochiometric parameters of the reaction. This will depend on the reaction mechanism.
If you consider a reaction that is kinetically reversible, for example
$$ \ce{A + B <=> C + D} $$
both the forward and backward steps of the reaction are elementary.Since the reactions are elementary in both the reactions, the partial orders of the reactions coincide with the stoichiometric coefficients.
$$
\begin{align}
\ce{A + B ->  C + C} & \quad v_f = k_f |A||B|\\
\ce{C + D ->  A + B} & \quad v_b = k_b |C||D|
\end{align}
$$
And the expressión of $K_c$ is consistent with the one you expected.
However, if the reaction mechanism does not correspond to a secuence of reversible steps you can not use the stoichiometric coefficients to predict the form of the rate equations. An example of this case would be ammonium equilibrium
$$\ce{3 H_2(g) + N_2(g) <=> 2NH_3(g)}$$


text:
What happens when you mix a salt with a strong base?
This doesn't come off from a question but it came to my mind. Take an example of $NH_4Cl$ as the salt and $NaOH$ as the strong base. What would happen to the pH for example? I'm guessing that the salt would dissociate $$NH_4 Cl\rightarrow NH_4^+ + Cl^-$$ $$c_0 \ \ \ \ \ c_0 \ \ \ \ c_0$$ and then the $NH_4$ would hidrolise like this: $$NH_4^+ +H_2O\rightarrow NH_3+H_3O^+$$ and then the hydroxide sodium would dissociate into hidroxide to find the pH? Not sure what would happen

You can format chemical substances and reactions using "mhchem" notation. For instance  write $\ce{NH+4 +H2O->NH3 + H3O+}$ as $\ce{NH+4 +H2O->NH3 + H3O+}$.

Also, please browse the site for similar questions. What you describe (as the answer correctly points out) is the reaction of a weak acid with a strong base. Use the search bar to start a search: https://chemistry.stackexchange.com/search?q=weak+acid+NH4%2B+strong+base

Convenient reference for text/formula formatting:
Notation basics  /  Formatting of math/chem expressions  /
upright vs italic  // For more:  Math SE MathJax tutorial.  // Not to be applied in CH SE titles.

Sodium hydroxide would dissociate into sodium cation and hydroxide anion.
Hydroxide anion would react with both $\ce{NH4+}$ cation and $\ce{H+}$ cation because of their strength ($\ce{OH-}$ is a strong base).
So as you can see $\ce{NH4Cl}$ is an acid salt because the solution that remains after its addition into water contains $\ce{H+}$ produced from the hydrolysis of ammonium cation. Therefore if you add NaOH that is a strong base it will end up as a basic solution. In spite of the final solution you can calculate both pOH and pH when information are provided.

Please check spelling, correct use of chemical nomenclature, and use mhchem formatting as need be. For instance write $\ce{NH4+}$ as $\ce{NH4+}$. For more on formatting see: https://chemistry.meta.stackexchange.com/questions/86/faq-how-can-i-format-math-chemistry-expressions-on-chemistry-stack-exchange

Thank you. English is not my first language and I am newbie here.

Your English is not the issue (in fact, it's pretty good), chemical notations and overall formatting are (which are international). Chemical names are common nouns, not proper ones. Symbols for chemical elements follow strict capitalization rules not allowing arbitrary use of capital letters. Charges and indices must be properly formatted in order to avoid ambiguity, hence recommendation to use mhchem.


text:
I have liquids A,B, and C. B is soluble in both A and C, but A and C are insoluble in each other. If I add B to A and C, what happens?
My AP Chem class was doing a solubility lab, and I thought of something but did not have substances that fit the parameters to try it. I have three liquids, A,B, and C. A and C are insoluble in each other, but B is soluble in both. If I put A and C in a test tube, they should separate. If I add B to the tube, which will it dissolve in? Both? If only one, why?

If you knew the partition coefficient of this combination, you'd know. Or vice versa. ;) There will always be some B in both A and C. Law of mass action.

If we knew what A, B and C were, then we could do some educated guessing.

Theoretically, A and B and C will react to get new compounds -- like D and E, but we don’t know if they react to each other. We don’t have enough data to draw a conclusion:)

Particular result will depend on liquids and their ratios.
Few principles to consider and apply:

Let assume very low mutual solubility A-C and unlimited mutual solubility A-B and C-B.
Initially, there is very little of A in C and vice versa.
Small addition of B leads to distribution of B between A and C, keeping the ratio of concentrations of B in A, resp. C, according to the distribution constant.
Simultaneously, mutual content of A in C and vice versa increases, as solubility of A in (C + B) and C in (A + B) increases.
At some point, with progressive B addition, one of the above solubilities is able to absorb all the A or C liquid in the other phase, forming just the single phase, containing A + B + C.
If B is not fully miscible with either of A or C, or neither of them, there is possible to have initially 2 to 3 phases, that transform to two or just one phase with huge B excess.


text:
Can we replace the Cu electrode with Ag in a Daniell Cell?
In a Daniell cell, the reaction at the anode involves the oxidation of zinc metal atoms to zinc(2+) ions, and the reaction at the cathode involves the reduction of copper(2+) ions to copper metal atoms. The copper metal atoms get deposited onto the copper electrode. As far as I can see, the copper electrode itself is not involved in the reaction, except as a conductor and a place for the newly formed copper atoms to deposit. Is the electrode involved in any other way? If not, I should be able to replace the copper electrode with a silver electrode as long as it is still in CuSO4 solution. Can I do this?

The silver electrode would become a copper electrode, with copper depositing on it in the galvanic cell mode.
The copper electrode is involved.
The open circuit state:
$$\ce{Cu(s) <=> Cu^2+(aq) + 2 e-}$$
The galvanic/voltaic state;
$$\ce{Cu(s) <<=> Cu^2+(aq) + 2 e-}$$
The electrolytic)charging state;
$$\ce{Cu(s) <=>> Cu^2+(aq) + 2 e-}$$
It is analogical to thermal energy exchange between cold and warm objects at molecular level. It happens always in both directions, with the net value positive, zero, or negative.


text:
What's the stoichiometric ratio of Kerosene to a mixture of 50% water and 50% hydrogen peroxide
So I'm wanting to do some testing with combustion of certain liquids, and I'd like to know what the right ratio is for these liquids. What I'm going to be using is regular kerosene (NOT RP-1) and 50% pure hydrogen peroxide. If you could go over how you got your results, that would be great, but it's not necessary.

Working with 50 % hydrogen peroxide and not being able to calculate it seems to me a bit controversial.
Kerosene contains hydrocarbons typically with 9-16 carbon atoms. If we take $\ce{C12H26}$ as an representative:
$\ce{C12H26 + 37 H2O2 -> 12 CO2 + 50 H2O}$
It leads to approximately (assuming integer molar masses give enough accuracy due unknown variables):

$12 \times \pu{12 g} + \pu{26 g} = \pu{170 g}$ of kerosene
$37 \times \pu{34 g} \approx \pu{1260 g}$ of 100 % hydrogen peroxide, i.e. $\pu{2520 g}$ of 50 % one
This leads to mass ratio roughly  1 : 15. The ratio will somewhat vary with the particular kerosene composition.
As kerosene density is $\pu{0.78–0.81 g/mL}$ and for
50% hydrogen peroxide it is $\pu{1.2 g/mL}$, the typical volume ratio would be 1 : 9.9.

I assume the practical ratio would be higher due possible hydrogen peroxide decomposition without oxygen being used for burning.
Considering all the water to evaporate, no wonder they used 100 % hydrogen peroxide as the oxidant, with all the danger it brought. It is unclear how intense the oxidation can be with 50 %.


text:
Is calcium carbonate a Brønsted–Lowry base, a Lewis base, or both?
What type of base would $\ce{CaCO3}$ be in the following example:
$$\ce{CaCO3 + H2SO4 -> CaSO4 + CO2 + H2O}?$$
My understanding is that calcium carbonate cannot be a Brønsted–Lowry base because it does not gain a proton. Can someone please help me out?

If $\ce{NaOH}$ cannot gain a proton, but $\ce{OH^-}$ can, does that mean $\ce{NaOH}$ is not a base?

$\ce{CO3^2- ->[H+] HCO3- ->[H+] H2CO3 -> H2O + CO2}$

It might be worth browsing earlier posts describing the behavior of BL acids/bases, eg https://chemistry.stackexchange.com/questions/97341/is-sodium-hydroxide-a-bronsted-lowry-base?rq=1 or https://chemistry.stackexchange.com/search?q=lowry

Most bases are both.

With such wording this q. is ripe for closing. It has hardly anything to do with base definitions.

The reaction you write, in aqueous solution, would happen with the ions produced by dissociation of the salts, not with the parent compounds, and would involve not one but two acid-base reactions (with carbonate ion, and then bicarbonate) and finally decomposition of carbonic acid to $\ce{CO2}$.
So while the complete reaction is
$$ \ce{CaCO3 + H2SO4 -> CaSO4 + CO2 + H2O}$$
there are at least the following steps involved (all species aqueous except solid $ \ce{CaCO3}$ and $\ce{CaSO4}$)$^1$:
$$ \ce{CaCO3 -> Ca^{2+} + CO3^{2-}} \\  \ce{H2SO4 -> 2H+  + SO4^{2-}}\\ \ce{Ca^{2+}   +   SO4^{2-} -> CaSO4 }\\\ce{ CO3^{2-} +  H+   -> HCO3- }\\\ce{ HCO3- +  H+   -> H2CO3 }\\ \ce{H2CO3 -> CO2 + H2O}$$
You can then see that bicarbonate $\ce{ HCO3-}$ can act as either acid or base (in the Brønsted-Lowry sense) whereas $\ce{ CO3^{2-}}$ acts as a base (proton acceptor).
Note also, regarding the Lewis definition, that Brønsted-Lowry acid/bases form a subset of Lewis acids/bases.

It might be worth noting that $\ce{CaSO4}$ is slightly soluble in water (on the order of g per L at RT), and that carbonic acid is not stable in the presence of water at RT.

I think this answer could be improved.  The dissociation of $\ce{CaCO3}$ is more properly: $\ce{CaCO3 (s) <=> Ca^2+ (aq) + CO3^2- (aq)}$.  Calcium sulfate is also soluble in water and would not precipitate in this process (depending on concentrations).  It also might be noted that $\ce{HCO3^- (aq) + H^+ (aq) -> H2CO3 }$ should probably be shortened to $\ce{HCO3^- (aq) + H^+ (aq) -> H2O (l) + CO2 (g) }$ because the presence of $\ce{H2CO3}$ doesn't occur at normal temperatures.


text:
Calculating rate constant for reaction between ethyl acetate and sodium hydroxide
My research question is investigating how temperature affects the rate constant for the reaction between ethyl ethanoate (acetate) and sodium hydroxide. I understand that I must titrate sodium ethanoate with an acid e.g. $\ce{HCl}$, but that will just yield me the concentration of sodium ethanoate in the sample. How would I go about calculating the rate constant?
Possibilities:

Would I use the integrated rate law, with a concentration against time graph?

Would I use the rate law expression? For this one I am unsure what I would use as the "rate".

@Poutnik Would there be an issue in titrating sodium ethanoate with hydrochloric acid, using bromocresol green as indicator (equivalence point about 5.2)

@Poutnik google tells me equivalence point for a reaction between weak base and strong acid is between 5 - 5.5. Is that wrong or am I misinterpreting the situation?

@Poutnik That makes sense, thanks. What would you recommend I do instead? I need to titrate the sodium acetate to find the concentration change.

@Poutnik Could you please rephrase your initial comment? I don't understand it.

There are unknown concentrations of NaOH and CH3COONa, equal in sum to initial NaOH concentration. Titration by HCl until discolouring phenolphthaleinum at pH 8.2 titrates NaOH, while all (>99.97%) CH3COONa still remains. Concentration of CH3COONa is then the difference between the initial and current NaOH concentrations.

@Poutnik So if I understand your comment correctly, do these statements follow?: $[NaOH]_t+[CH_3COONa]_t=[NaOH]_0$ and $[CH_3COONa]_t=[NaOH]_0-[NaOH]_t$.

What could be the alternative explanation?

Convenient reference for text/formula formatting:
Notation basics  /  Formatting of math/chem expressions  /
upright vs italic  // For more:  Math SE MathJax tutorial.  // Not to be applied in CH SE titles.

The general process of trying to solve this problem:

Determine starting molarities of each reactant. Let's say 0.1 M
Have a set of sample collection tubes (test tubes) pre-measured with 0.1 M of standardized HCl that would yield a slight excess of moles of NaOH in the starting mixture.  This will be used to STOP the reaction (more on this).  I would choose at least 5.
Setup your reaction batch and mix the reactants with enough volume.  Probably 100mL of each would suffice. As well as make sure the reactants and mixture maintain a temperature desired.
During set intervals, withdrawal 10mL of the reaction mixture at periodic intervals and place them into one of the sample tubes.  Placing it in the tube will immediately quench the NaOH in the reaction sample and convert any remaining acetate ions to acetic acid.
After each sample tube has been used, you can titrate the remaining HCl + acetic acid with a standardized NaOH solution and an appropriate indicator such as bromthymol blue to obtain the amount of acid present.

You can now determine the amount of NaOH extracted from the reaction by using the number of moles of HCl present in the tube to start subtracted by the number of moles of acid that remain (step 5) to determine the amount of NaOH.
Since we now know the amount of NaOH extracted in a 10 mL sample of the reaction batch, you can determine the concentration of the NaOH at time of extraction within the reaction batch AND the time it took to reach this concentration.
Repeating this for each tube, we now have the change of the molarity of NaOH over time.  This is also allows us to plot the change of the molarity of ethyl acetate since their reaction is a 1 mol : 1 mol  ratio.
Using the integrated rate laws, a plot of these concentrations over time, you can determine the slope of the plot which creates a linear graph and thus find the reaction constant $(k)$.
Thanks to @Poutnik for the help on refining the answer.

A very nice post. But it seems you have not mentioned acetic acid being titrated together with HCl. There will remain the constant amount of HCl, that would neutralize either NaOH either sodium acetate. And the variable part will be acetic acid on constant HCl baseline, titrated to first colour on phenolphthalein. With the equivalent amount of HCl Vs NaOH, it  would be solely titration of acetic acid.

Without considering it,, one could be tempted to use e.g. methyl orange or other indicators for acidic pH ranges, what would be a big mistake.

@Poutnik Since the molar relationship between acetic acid and $\ce{HCl}$ are equal in the test tube, you could titrate using phenolphthalein to find the amount of both substances and halve it.  But a better indicator probably could be chosen, either methyl violet or thymol blue.  I can update the answer with the indicator option as that is probably the more appropriate experimental answer.

2 mmol NaOH + 3 mmol HCl -> 1 mmol HCl // 1 mmol NaOH + 1 mmol AcONa + 3 mmol HCl -> 1 mmol AcOH + 1 mmol HCl // With just 2 mmol HCl, just 0 or 1 mmol AcOH // What dividing by 2?

@Poutnik My apologies, it took me a little to understand what was being communicated, but I do get it now after sitting with an ICE chart.  I updated the answer to mention that you will be back titrating both excess HCl and acetic acid.

Thymol blue (red 1.2- yellow 2.8, 8.0 yellow-9.6 blue, not to be confused by the first one) or phenolphthalein (clear 8.2-10.0 magenta) are OK, but bromthymol blue (yellow 6.0-7.6 blue) has too low pH transition. At pH 6.0 is there about 6% of acetic acid not neutralised yet. 99.9% of acid is neutralized at pH 7.75 and the pH of e.g. 0.1 M AcONa is near 8.9.


text:
Electron count of complexes
For counting the electrons of the complex (RO)3W≡CMe, the book considers three RO- groups that donate a total of 10 electrons and it regards ≡CMe as [carbyne]- group in the ionic model and has a total of 16 electrons.
What confuses me is that:

The RO- group used to consider a 4 electron donor because they can form a sigma bond with metal and they can also donate 2 lone pair electrons to the metal (as a pi-donor), so they can donate overall 4 electrons. However, this seems not to be the case. I wonder whether it's because that 10=4+4+2, which means that only 2 RO- act as pi-donor and the other is sigma-donor only?
I used to consider M≡CR as M(III) and CR(-3) in the ionic model considering the electronegativity. I don't understand why carbyne can be regarded as the group with only 1 negative charge. Nevertheless, when I consider M(III) and CR(-3) and count the total electrons, it also turns out to be 16e. Does that mean the two ways are both available?


text:
Which one has higher priority in CIP rules -- O or OH?
I know that if the atom is the same, I have to look at the next atom. In the first choice, Oxygen is not connected to any other atom (besides the chiral carbon). It is an anion. Whereas in the second choice, OH has Hydrogen attached to Oxygen. So how do I decide which one gets higher priority?

"oxygen is not connected to any other atom" — How is that possible? Oxygen needs two bonds. Unless you're talking about an anion, or a radical, perhaps? Could you clarify?

Sorry, Oxygen is an anion, so it is bonded to only the chiral carbon.

A lone pair on an anion is considered to be a substituent that has an atomic number of 0

@SafdarFaisal I see, so in this case, OH will get the higher priority, as H has an atomic number of 1?