text:
Why is para-aminobenzoic acid more acidic than ortho-aminobenzoic acid?

Compare the acidic strength of o-, m-, p-aminobenzoic acids.

I got that meta will be the most acidic as it won't be able to show R+ effect of $\ce{NH2}.$
But among the other two, i.e. ortho and para, shouldn't ortho be more acidic as it will cause ortho effect?
But according to my book and the data I found online:
$$
\begin{array}{lc}
\hline
\text{Compound} & \mathrm{p}K_\mathrm{a}\text{(amino)} \\
\hline
\textit{o}\text{-Aminobenzoic acid} & 4.89 \\
\textit{m}\text{-Aminobenzoic acid} & 4.79 \\
\textit{p}\text{-Aminobenzoic acid} & 4.77 \\
\hline
\end{array}
$$

Do you think there may be hydrogen bonding between the amino group and the carboxylic acid?

https://chemistry.stackexchange.com/questions/20493/why-is-benzoic-acid-a-stronger-acid-than-3-aminobenzoic-acid https://chemistry.stackexchange.com/questions/4720/in-aminobenzoic-acid-do-we-have-three-different-pkas

The problem in ortho-aminobenzoic acid is that the acidic hydrogen of carboxylic group is H-bonded with the lone pair of nitrogen in amino group. As a result it is more difficult to extract it compared to that in para-aminobenzoic acid since the H-bond must also be broken during acid-base reaction. Para-aminobenzoic acid does not have a H-bond due to the distance between the two groups.

I would like to add that since the hydrogen of the $\ce{-COOH}$ group is more acidic and $\ce{N}$ has a lone pair. This is why H-bond occurs with $\ce{-NH2}$ group as electron donor and not the other way round.
Also, I presume that by ortho effect you intended to tell that due to repulsion of $\ce{-NH2}$ group $\ce{-COOH}$ group would go out of the plane. However, as we have seen, it is difficult as it is already H-bonded with the nitrogen atom and moving out of the plane would put strain into the H-bond. Moreover, the repulsion by $\ce{-NH2}$ group is not too intense so it would (in most part) avoid the strain.

So why is Salicylic acid more acidic than benzoic acid?


text:
Why is carbon dioxide a Lewis acid?
I know this has been asked before, but can anyone explain me in simpler terms? Why is $\ce{CO2}$ a Lewis acid? The carbon atom doesn't have any vacant orbitals to accept a lone pair. How can it be a Lewis acid?

Suppose one of the doublets of one C=O bond moves to one of these two atoms. It will not be to the Carbon atom. The Oxygen atom is a better choice. And now this Oxygen atom would be negatively charged. As a consequence the Carbon atom is positively charged and has a vacant orbital able to accept a lone pair, for example from an $\ce{OH-}$ ion or from water. This will produce the ion $\ce{HCO3^-}$ (from $\ce{OH-}$) or $\ce{H2CO3}$ (from $\ce{H2O}$).

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text:
What happens at the inert anode in an electrolytic cell?
Recently I did an experiment using graphite as the anode and iron as the cathode for an electrolytic cell. I tried plating different metals such as copper, nickel, and zinc with CuSO4, NiSO4, and ZnSO4 respectively. I know that normally when the anode is made from the same material as the metal being used for plating, the anode is decomposed. What I am wondering is what happens at the anode when it is made of an inert material like graphite? Is the SO4 ion oxidized there, what does it become?
During the experiment, I have also observed bubbles forming at the anode, and I hypothesized that it may be oxygen from the electrolysis of water, but I had no way to be sure.

https://chemistry.stackexchange.com/questions/102636/why-cant-sulphate-ion-be-discharged-in-the-anode-of-electrolysis

There is oxidized oxygen in water
$$\ce{2 H2O -> 4 H+(aq)  + O2(g) + 4 e-}.$$
$\ce{HSO4-}$ in sulphuric acid solution can get oxidized as well at high current density and therefore high enough anodic potential:
$$\ce{HSO4-(aq) -> S2O8^2-(aq) + 2 H+(aq) + 2 e- }$$
This is the industrial way to produce potassium peroxodisulphate.

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text:
Is there a reusable and safe exothermic reaction
I am looking for an alternative to the sodium acetate reaction used in e.g. heat packs. I need something similiar but that generates more heat, so that I could for instance warm 200ml of water to a temperature of 37 °C.
So is their any safe and reusable reaction as used in heat packs but that generates more heat?
NOTE: I am a beginner when it comes to chemistry.

I am afraid the requirements (reusable, safe, exothermic enough, closed system ) almost mutually exclude themselves. Sodium thiosulphate pentahydrate gives up to 48 deg C, but you would need a really big pack for warming 200 mL water to 37 deg C.  I am not sure about comparision of generated heat per gram of this and sodium acetate.

A rechargeable battery, connected to a heater, can do the job. But then the chemistry is hidden away inside the battery. This is a common fate of real world chemistry: it is one of the major unsung heroes that makes modern life possible.

I was hoping to not use any battery, would like a heat pack-like solution. I guess I could use several small heat packs to heat the water?

It all depends on how easy it is to re-charge the heat pack. For a short period of time, my supermarket sold self-heating cans of coffee. You activated the internal heating mechanism, which caused calcium oxide (“quick lime”) to react with water. This is a highly exothermic reaction that surrounded the separate internal compartment of coffee. An environmental disaster and market failure. But you can regenerate calcium oxide from calcium hydroxide by application of heat. Lots of heat. So it is reversible, but not so practical outside a lab.

By the way, the self-heating coffee can was not gigantic, the volume of coffee was above 200 mL, and the final temperature was well above body temperature. I bought one can for use as a demo in class. As for taste, I have had better coffee ...

Okey, yeah I would like something easy as reusable heat packs, like boil it to dissolve the crystallisation.


text:
How does precipitate affect cell potential?

A cell is set up with copper and lead electrodes in contact with $\ce{CuSO4(aq)}$ and $\ce{Pb(NO3)2(aq)},$ respectively, at $\pu{25 °C}.$ The standard reduction potentials are:
$$
\begin{align}
\ce{Pb^2+ + 2 e- &-> Pb} &\quad E^\circ &= \pu{-0.13 V} \\
\ce{Cu^2+ + 2 e- &-> Cu} &\quad E^\circ &= \pu{+0.34 V}
\end{align}
$$
If sulfuric acid is added to the $\ce{Pb(NO3)2}$ solution, forming a precipitate of $\ce{PbSO4},$ what will happen to the cell potential?

I know the lead is the anode while the copper is the cathode:
$$\ce{Pb(s) + Cu^2+ -> Pb^2+ + Cu(s)}$$
The addition of $\ce{H2SO4}$ will cause less $\ce{Pb^2+}$ to be in the solution, thus causing the equilibrium to push to the right creating more $\ce{Pb^2+}.$ With more $\ce{Pb^2+},$ if you were to put it in the Nernst equation, would the greater number in the log expression cause the cell potential to decrease?

[Pb^2+] is determined by [SO4^2-] and PbSO4 solubility product.

Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.

The solubility product of $\ce{PbSO4}$ is about $10^{-8}$. Now suppose $[\ce{Pb^{2+}}$] falls down from $\pu{1 M}$ to an arbitrary low value like $\pu{10^{-8} M},$ due to addition of $\ce{SO4^{2-}}$ ions. In this case, Nernst's law can be applied, and the potential of the lead electrode falls from $E^\circ_\ce{Pb} = \pu{-0.13 V}$ down to
$$E_\ce{Pb} = \pu{-0.13 V} + \pu{0.0296 V}\times\log{10^{-8}} = \pu{-0.37 V}\tag{1}$$
As a consequence, the $\ce{Pb}$ potential decreases. But the cell potential increases and goes from the initial value (without sulfate)
$$E^\circ_\mathrm{cell, i} = \pu{0.34 V} - (\pu{-0.13 V}) = \pu{0.47 V}\tag{2}$$
to the final value (after precipitation of lead sulfate)
$$E_\mathrm{cell, f} = \pu{0.34 V} - (\pu{-0.37 V}) = \pu{0.71 V}\tag{3}$$


text:
Does aqueous ammonium fluoride solution release hydrogen fluoride vapors?
Can $\ce{HF}$ vapors be released from $\ce{NH4F}$ solution in water $(w(\ce{NH4F}) = 40\,\%)$ at room temperature?
I assume that dissolving $\ce{NH4F}$ in water produces small amount of $\ce{HF}.$

The question is not "Can ?" but "How much ?". There will be equilibrium F-(aq)<=>HF2-(aq)<=>HF(aq)<=>HF-(g) But as the solution would be slightly alkalic, I expect the partial HF(g) pressure as negligibly low. It would be higher for NH4HF2 or an acidized solution.

Also, it sublimes when heated to give ammonia and HF. The reaction is reversible.


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Dissociation equilibria of ammonium bifluoride in water
Ammonium bifluoride or ammonium hydrogen fluoride is a salt of a weak base and a weak acid.
The weak acid is because the second equilibria of $\ce{HF}$ written as:
$$\ce{HF + F- <=> HF2-}$$
and it has constant of $\pu{3.963 M}$.
But does the bifluoride anion further dissociate?
$\ce{HF}$ in water has many species: $\ce{H+},$ $\ce{F-},$ $\ce{HF2-},$ $\ce{H2F2}$ and $\ce{(HF)_nF-}.$ The $\ce{NH4HF2}$ has a more acidic $\mathrm{pH}$ than a solution of $\ce{NH4F}$.

Perhaps rather associate than dissociate. In concentrated HF(aq) and HF(l) there is also H2F+.

You need to write out all the possible equilibria just like you do in the case of a multiprotic acid like Phosphoric acid. You will also need to have a pKa for every equation. The bifluoride ion will obviously dissociate in water

HF2- -> HF+F- (K=3.963 M), HF -> H+ + F- (K=6.85 x 10-4 M)... Like this?


text:
Why is the solution of ammonium bifluoride more acidic than aqueous ammonium fluoride?
At low concentrations of aqueous $\ce{NH4HF2}$ $(w = 1–2\,\%)$ $\mathrm{pH}\approx 3.$
But aqueous $\ce{NH4F}$ $(w = 20–40\,\%)$ has $\mathrm{pH}\approx 7.$
Yet the equilibria of
$$
\begin{align}
\ce{HF &-> H+ + F-} &\quad K &= \pu{6.85E-6 M} \tag{1} \\
\ce{HF + F- &-> HF2-} &\quad K &= \pu{3.963 M^-1} \tag{2}
\end{align}
$$
I understand that $\ce{NH4F}$ comes from a weak acid and a weak base. So, the $\mathrm{pH}$ is around neutral. I don't understand how this equilibria makes $\ce{NH4HF2}$ more acidic. Does the second equilibria make the $\ce{HF2-}$ a stronger acid? I'm having a hard time understanding this.

Does this answer your question? Dissociation equilibria of ammonium bifluoride in water

You're asking about pretty much the same thing as earlier.

HF is stronger acid than NH4+, is not it ?

This question and the one referenced by Mithoron address $\ce{NH4HF2}$ as an entity in itself which must be examined as a whole.
It is easier conceptually to rewrite the formula for the compound as $\ce{NH4F·HF}$. Consider adding $\ce{NH4F}$ to water: you get a $\mathrm{pH}$ near $7.$ In a separate container, add $\ce{HF}$ to water (approx. $\pu{0.1 M});$ you get a $\mathrm{pH}\approx 1$.
Combine the solutions and the $\mathrm{pH}$ will settle out somewhere between $1$ and $7,$ because of the common ion effect: $\ce{F-}$ from the $\ce{NH4F}$ inhibits the ionization of $\ce{HF},$ but the solution is still acidic. The approximate $\mathrm{pH}$ can be calculated from the ionization constant of $\ce{HF}$ and the concentrations of $\ce{NH4F}$ and $\ce{HF}.$
The concept of the $\ce{HF2-}$ moiety dominating the discussion for a solution is probably making the analysis murkier than it needs to be. It's just a hydrogen-bonded complex that happens to retain some semblance of existence in the solid. It is fairly stable: mp $\pu{126 °C},$ bp $\pu{240 °C}$ (with decomposition).

"126$^0$ C" reads as "one hundred twenty-six to the power of zero coulombs". "NH$_4$F.HF" has a dot that doesn't make sense and the formula breaks across lines incorrectly. Please consider using proper formatting tools (mhchem, Unicode, plain MathJax) so that your notations make sense.

@andselisk: Well, thanks, I guess. It's not clear to me what straightforward notation, according to my screen, gets retranslated to a different format. I have already been notified about certain subscripts, but not the temperature. Is there a way that I can see the wrong view before I post it?

Let's try this: I corrected the answer a bit, and once you have some spare time, feel free to review the edits, maybe also by (re)visiting this page, this page and this one on how to format your future posts better with MathJax and Markdown. There are no other tools other than the preview windows, standardized notations (which everyone is supposed to look up) and common sense, I'm afraid.


text:
Electrochemistry - reduction potentials - an IB Chemistry Question

I have re-read the Electrochemistry chapter from my textbook just now to remind my self of the exact formulae. I still don't understand how to approach this question, however - how can we work it out? I can either give you the mark scheme answers straight away and we can work backwards, or I can wait for an answer and then tell you if that is the correct solution. What ever is the norm on this forum! Thank you.

The norm on this Q&A site is for you to show what you have tried (or are stuck on) and then help may arrive. No one is just supposed to post an answer to a homework problem or the like.

@ Ed V - thank you for the clarification. As I state above, I have re-read the electrochemistry chapter in my textbook. So I know how to calculate the standard cell potential from individual reduction potentials. However, I have never come across this type of question before and so cannot offer any suggestions. I am happy to do some extra reading to find out, but didn't know what to google as, again, this wasn't something I had come across. If you could point me in the right direction, that would be very useful. Thank you.

This is a pair of puzzle questions. For the first question, you need a voltaic cell with positive potential, so the reaction would be spontaneous, with the second half reaction in the top table being the cathode reduction. Now which of the 5 half reactions in the bottom table would make a viable anode oxidation reaction? Maybe several of them. But then you narrow the choices by using the reduction half reaction in the first line in the upper table. Think about it: it is mostly a logic problem, not chemistry per se.

First put your redox potentials on a horizontal line by order of increasing standard redox potentials. You will see that it starts on the left-hand side by $\ce{Zn^{2+}/Zn}$ at $\pu{-0.76 V}$. Then, going to the right, you find $\ce{V^{III}/V^{II}}$ at $\pu{-0.26 V}$. Now you find $\ce{Pb^{2+}/Pb}$, and so on.
Now try to learn how to use this scale. The reduced species of any couple can react with any oxidized species of any couple situated to its right. For example, $\ce{Zn}$ is the reduced species of the couple $\ce{Zn^{2+}/Zn}$. So $\ce{Zn}$  can reduce all oxidized species, and for example all oxidized forms of Vanadium ions. But $\ce{Zn^{2+}}$ cannot be reduced, because there is no reduced species at the left hand side of $\pu{- 0.76 V}$.
On the other end of the scale, you find $\ce{Cl2/Cl-}$ at $\pu{+1.36 V}$. Hopefully you admit that $\ce{Cl2}$ can oxidize all reduced species, and $\ce{Cl-}$ cannot reduce anything.
Now let's consider $\ce{Pb^{2+}/Pb}$, at $\pu{-0.13 V}$. $\ce{Pb}$ can reduce all oxidized species situated on the positive side of the axis, for example $\ce{VO^{2+}}$ which is at $\pu{+0.34 V}$. But it cannot reduce $\ce{V^{3+}}$ to $\ce{V^{2+}}$, because $\ce{V^{3+}/V^{2+}}$ is on tis left-hand side. And this is exactly your first question.
Now think a little bit about how to solve the second question. It is exactly the same approach.


text:
Addition reaction of alkenes
During an addition reaction of an alkene, how does a double bond break to form two bonds with the two atoms of the reactant plus a bond between the carbon atoms? Not sure if this is a silly question or not, but if it’s a double bond, how does it end up forming three bonds?

Well, you add something, don't you?

There are two possibilities to explain this addition reaction. It could be an ionic mechanism or a radical mechanism. I only explain here the radical mechanism, because it is quickly done. You should know that doublets or bonds are made of two electrons. Imagine that for some reason the $2$nd bond of $\ce{R2C=CR2}$ is suddenly broken, then closed, then broken, and so on. When it is broken it becomes $\ce{·R2C-CR2·}$, with two free electrons, one on each carbon atom. If by chance a molecule $\ce{H2}$ or $\ce{Cl2}$ is present in the vicinity, and is also broken maybe by a ultra-violet light, the $ \ce{2 H·}$ or the two $\ce{Cl·}$ may recombine. But they may also recombine with the free electrons on the "opened" alkene molecule. This will produce an addition of $\ce{H2}$ (or $\ce{2H}$) or of $\ce{Cl2}$ (or $\ce{2 Cl}$) on the alkene. Sorry for the "serious" chemists who will read this simplified "theory". It works well with ordinary high school students, who are not scientifically inclined.


text:
Isobaric entropy relation from isothermal Gibbs free energy relation
I have been given a problem to derive a relation for change in entropy from change in Gibbs free energy:

Starting with the isothermal equation
$$G(p_2) = G(p_1) + nRT\ln\frac{p_2}{p_1}\label{eqn:1}\tag{1}$$
derive
$$S(V_2) = S(V_1) + nR\ln\frac{V_2}{V_1}\label{eqn:2}\tag{2}$$
using the relation
$$\mathrm dG = V\,\mathrm dp - S\,\mathrm dT.\label{eqn:3}\tag{3}$$

The derivation wants you to assume that pressure is constant and thus $\mathrm dp = 0,$ and \eqref{eqn:3} can be arranged to
$$\left(\frac{\partial G}{\partial T}\right)_p = -S.\label{eqn:4}\tag{4}$$
Now I understand that then differentiating \eqref{eqn:1} w.r.t $T$ and using the fact $p_1 V_1 = p_2 V_2$ (for constant $T)$ can produce the required equation \eqref{eqn:2}.
However, I don't understand quite how we can justify the derivation of an isobaric equation from an isothermal one as this method appears to. Equation \eqref{eqn:1} describes isothermal processes and is a function of pressure, $p_2.$ How is it valid then to suddenly say now pressure is constant and therefore use \eqref{eqn:4} when in equation \eqref{eqn:1} the pressure is clearly not constant?
Might it have something to do with Gibbs free energy being a state function meaning, once there is a relationship for it, constraints (for different state variables) can be changed to derive new relationships using a different constraint?

Who says you are assuming that the pressure is constant?

@ChetMiller The answer to the question says it. Is it wrong?

For an ideal gas at constant temperature $\Delta H=0$, so $\Delta G=-\Delta (TS)$, and, at constant temperature, $$\Delta G=-T\Delta S$$

Is the derivation set out in the question flawed then? I know there are many ways to derive the equation, I'm just unsure about the method behind the original post derivation.

I don't follow the derivation set out in the question.

Maybe, what they expect you to do is to write:  $$-\left(\frac{\partial S}{\partial P}\right)_T=\frac{\partial ^2 G}{\partial P\partial T}$$From Eqn. 1, this gives, for an ideal gas $$-\left(\frac{\partial S}{\partial P}\right)_T=\frac{nR}{P}$$


text:
Comparing O−O bond length in hydrogen peroxide and dioxide(2−) ion
Problem

How can we qualitatively compare the bond lengths between $\ce{H2O2}$ and $\ce{O2^2-}?$

Answer

 Bond length in $\ce{O2^2-}$ is slightly larger than in $\ce{H2O2}.$

Attempt
I tried using molecular orbital theory, but since both have peroxide $\ce{O2^2-},$ the differentiating factor must be some sort of repulsion, but I'm not getting it.

This compilation actually gives the same bond length for both species, and also does so for ozone and ionic superoxides where both oxygen-oxygen bond orders are 1.5.

The way to tackle this is to look at the lone pair repulsions between the 2 molecules. It is known that $\ce{N-N};$ $\ce{O-O};$ $\ce{F-F}$ single bonds are quite unstable due to lone pair (lp) repulsions.
An evidence of this is that $\ce{N}$ exists in molecular state as $\ce{N2}$ using multiple bonds, so that it's lone pairs do not repel each other, however $\ce{P}$ doesn't do that. It is capable of forming $\ce{P4}$ because in the third period the size of the atoms are large enough to properly distance the lone pairs.
One more evidence of this act is that among diatomic halogen molecules the bond length is in the order $\ce{Cl2} < \ce{Br2} < \ce{F2} < \ce{I2}$ the bond length of $\ce{F2}$ should have been lower (2p–2p overlapping is much stronger hence the bond should have been the shortest). However, due to lp–lp repulsions it is now much longer bond.
With this we can now see that in $\ce{H2O2}$ there is only two lone pairs on each oxygen atom while in $\ce{O2^2-}$ it has three lone pairs on each atom, hence it has increased lp–lp interaction.
As a result of this $\ce{O2^2-}$ tends to elongate (relatively more than $\ce{H2O2})$ its bond to decrease this intense lp–lp interaction, hence the longer bond.

I don't think "lone pair repulsion" stands up as an argument for pi bonding in N2 and O2 but not in S6 and P4. Much more reasonable is that the pi bonds in second row are strong enough to overcome general electron-electron repulsion because of better side-by-side p orbital overlap than in 3rd row. For the specific question, the bond length difference in $\ce{O2^2-}$ vs $\ce{H2O2}$ is easily explained by general increased charge repulsion without invoking the specific number of lone pairs

@Andrew I suppose, in a sense it is equivalent here: in H2O2 we replace one lone pair with a bond pair (albeit one that is heavily polarised towards oxygen); and the bond pairs repel each other less than the lone pairs do. I certainly prefer what you say though, as it gets at the heart of the issue, and is more general, as you mentioned.

@Andrew I do realize that it would be a less mis-directing in nature, however I think I should leave it and let people find out through the comments why it is misdirecting. But I must ask : should I edit it to reflect this correction?

Edits are totally up to you. If you feel like the answer could be improved and you want to take the time to do that, you're welcome to, but you're not obligated to. If a commenter feels strongly enough that your answer is insufficient or incorrect, they always have the option of posting a new answer and letting the community decide which is more helpful.


text:
Why does titanium have lower affinity than scandium?
Electron affinity of atoms in the same row usually increases with the atomic number until the shell or half-shell gets full, and then it drops down. Some transition metals are exceptions from the rule, but the strangest exception to me is Titanium: while scandium ($3d^1$) has an affinity of 18 kJ/mol, titanium ($3d^2$) has an affinity of 8 kJ/mol.
It does not seem to make sense: the effective charge of titanium nucleus should be larger compared to scandium (Slater's rule), and exchange energy also favors titanium.
Does anyone have a good explanation?

No scandium has lower electro affinity because electro affinity increased with in a period and decrease with in a group

@DagmawitAmanuel I'm afraid the experimental data I can find contradicts your assertion and supports the data presented by the OP. For instance https://en.wikipedia.org/wiki/Electron_affinity_(data_page) gives similar numbers to those quoted above, with references back to papers. Yes the original question would be MUCH better if the source for the numbers was given, but a bit of googling supports them.

Note that it's similar, although more complicated with Lanthanum ([Xe]5d-1 6s-2) and Hafnium ([Xe]4f-14 5d-2 6s-2). The electron affinity of the former is higher.
At the same time, the same exception does not apply to Yttrium ([Kr]4d-1 5s-2) and Zirconium ([Kr]4d-2 5s-2), but the interesting thing there is that Palladium ([Kr]4d-10) is close, and it has a different exception: a filled d sub-shell in the ground state of the atom, at the cost of a "harvested" s sub-shell.


text:
phenolsulphonic acid
Phenolsulfonic Acid( PSA) is produced after sulphonation of Phenol at around 100 degrees. They use this product in tinplating.
Once a stable PSA is formed after sufficient reaction , unreacted impurities are Phenol 2% max , Sulphate  3% max. PSA is to be 65%  minimum and rest is water.
After few days , Sulphate is increasing to 3.5 to 4.5% or even more, while others remain constant.
The question is:
Why sulphate concentration is increasing from the initial value of 3% on day 1 to 4% on day 5/6. I dont have a chemistry background so any help would be appreciated. Thank you

a crystalline PSA is formed after mixing phenol and sulphuric acid at 100 degrees, take the crystalline product and cool it down to about 50 degrees and add water.  we were getting final PSA with unreacted impurities of Phenol 2% max, Sulphate 3% max. PSA is to be 65% minimum and rest is water. After 2-3 days , Sulphate is increasing to 3.5 to 4.5% or even more, phenol remain constant. small company only makes PSA for the last 15 years and never encountered this issue. The question is: Why sulphate concentration can elevate above 3% after 2 days. Any help/suggestions is appreciated. Thanks


text:
Confusion in solution for this Question regarding mole concept
$\pu{48g}$ of graphite and $\pu{16g}$ of $\ce{O2}$ is given. The question is to count the no of molecules of $\ce{CO2}$ produced?
Now, the solution in book is this
No of $\ce{C}$ atoms = $\frac{\pu{48g}}{\pu{12g}}$ = $\pu{4mol}$. No of $\ce{O}$ atoms = $\frac{16}{16}$ =$\pu{1 mol}$ , and no of $\ce{O2}$ atoms = $\pu{0.5 mol}$.
The reaction given is
$\ce{C + 2O ->CO2}$.
$\ce{CO2}$ produced is $\pu{0.5 mol}$ here.
My question is why not to use the reaction $\ce{C + O2 ->CO2}$?
What would the difference be in the answer ?

If a chemistry textbook in this day and age uses terms and symbols such as "gm", "gms", "gmd" "no of X" and proposes reaction with atomic oxygen out of blue,  it's not a chemistry textbook. It is something that may have an appearance or a title of a chemistry textbook, but it is not one. Probably a reprint from BuzzFeed or a comic book, but definitely not a chemistry textbook.

@andselisk I am sorry for that. But it would be more helpful if you help me out here ?

@Sarabsrimt - The point is that we'd like chemistry problems to stay within the bounds of reality as much as possible to teach real chemistry as well as being an exercise. In this case you can't go to your chemical supplier and buy a tank of oxygen atoms. You must buy a tank of oxygen molecules.

@Sarabsrimt It's not your fault, just avoid textbooks like this one. And yes, given the context your chemical reaction makes much more sense: $$\ce{C(s) + O2(g) -> CO2(g)}$$ The limiting reagent is oxygen: $$n(\ce{C}) = \frac{m(\ce{C})}{M(\ce{C})} = \frac{\pu{48 g}}{\pu{12 g mol-1}} = \pu{4 mol}$$ $$n(\ce{O2}) = \frac{m(\ce{O2})}{M(\ce{O2})} = \frac{\pu{16 g}}{\pu{32 g mol-1}} = \pu{0.5 mol}$$ and the answer is $$N(\ce{CO2}) = n(\ce{CO2})\times N_\mathrm{A} = n(\ce{O2})\times N_\mathrm{A} = \pu{0.5 mol}\times\pu{6.022E23 mol-1} \approx \pu{3.0E23}$$

@Sarabsrimt Actually, feel free to check community's assembly of Resources for learning Chemistry. The resources listed there are arguably one of the best.

@andselisk Ok . Book I used was ICSE class 10 Viraj F Dalal. But I will surely check the resources. Thanks a lot.

@Sarabsrimt No prob, good luck and have a good one.

@andselisk Are both reaction correct to use ?

@Sarabsrimt No, $\ce{C + 2 O -> CO2}$ doesn't make sense unless there is extra context provided. As MaxW already told you, oxygen doesn't casually dwell in atomic form.

@andselisk Does C + 2O —> CO2 happen like this : C + O —> CO and then CO + O —> CO2

@andselisk Ok. One more thing. Someone told me that these reaction only tell us how much are ingredients produced and not specifically tells if the reaction will happen or not. Is it true ?

@andselisk Right. So reaction isn’t possible but does it mean it tells us the amount of reactant or product that would be needed ?

@Sarabsrimt Stoichiometry.

16 g of dioxygen contains 16 g of oxygen atoms. To find out how much carbon dioxide is produced, you can either use a realistic description of the reaction, or you can go atom by atom. Either path gives the same solution, but the one chosen by the textbook is not modern chemistry, and makes some of us angry or sad.

My question is why not to use the reaction $\ce{C + O2 -> CO2}$?

You should because that chemical equation describes the reactants and products much better than what the textbook has. It does not describe the mechanism (i.e. the steps to arrive at that net reaction), though.

What would the difference be in the answer.

Same numerical answer (unless you get confused because the reaction does not match the actual reactants).


text:
Removing wood stain from a ceramic surface
I have a bathtub that I accidentally poured wood stain into. Now the stain is all over the white ceramic surface of the tub. I tried using bleach but it didn't do anything. I'm curious if there is a chemist who would know the answer.
I used Varathane® Premium Fast Dry Interior Oil-Base Cabernet Wood Stain with the following properties (Technical Specifications — PDF):

Try some acetone (nail varnish remover) or white spirit on a small area.

CAREFUL - You may have plastic drain pipes and it would be  very bad to have acetone in contact with such pipes. Also acetone is very flammable so you need very good ventilation.


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Does 3‐ethyl‐1,1‐dimethylcyclohexane have a stereocenter?
I knew that we have a stereocenter if the carbon is bonded to four different groups.
According to the stereocenter rules, I should have $\ce{CH2}$ bonded to the carbon of the stereocenter (marked with asterisk), an ethyl and a hydrogen. How is it possible to have two equal groups? What did I do wrong?

I've redrawn the structure, assigned the compound's name omitting R/S notational part and slightly corrected the wording. Hopefully this is OK, but please do feel free to double-check so that the intended gist of a question is preserved.

@Amiga 500: The structure drawn by andselisk has its corresponding enantiomer. That means compound has a chiral carbon as indicated. Justifying it is your homework assignment.

@MathewMahindaratne it's not a homework, the question is whether this molecule is a stereocenter or not. As I've drawn, there are three groups: Ch2 and Ch2 molecules, an ethyl, and the hydrogen. According to the rule,it  is a stereocenter if it has 4 groups. Which is the fourth group?

@Amiga500 The stereocenter in this molecule has four different branching: all atoms counterclockwise in this cyclohexane ring; all atoms clockwise in the cyclohexane ring; the hydrogen; and the ethyl. So it's not about just the one linked $\ce{CH2}$, but all the linked atoms. The mirror image of this molecule will not overlay on the above, but will be a different molecule.

@z1273 but we have two Ch2 molecules, so is it still a stereocenter? I mean, my question is: is the C atom linked to two different groups Ch2??

There's more than single $\ce{CH2}$s. When considering the ethyl, you didn't stop at the linked $\ce{CH2}$, but you included all of the linked atoms there, i.e. an additional $\ce{CH3}$. The same is true for the cyclic $\ce{CH2}$s. One has $\ce{CH2CH2CH2C(CH3)2CH2}$ and the other has $\ce{CH2C(CH3)2CH2CH2CH2}$ - they're different. The four substituents of the chiral center are different. As a result the mirror image of the whole molecule does not overlay the original molecule. That can be attributed to that one stereocenter.

Imagine to be the C and look around. Each time you are bearing along a bond you see 4 different landscape. H- ethyl- and the the rest of the cycle but from different not equivalent sides. In other words do not stop at the CH2....


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Why is the base-centered orthorhombic crystal lattice a unique crystal system?
I'm having difficulty understanding why the base-centered orthorhombic crystal system is a unique crystal system. When I draw two base-centered orthorhombic unit cells next to each other there appears to be a primitive crystal that is contained within the union of the two lattices as shown in the image below.

This primitive crystal system appears to be monoclinic--though I wouldn't be surprised if I'm wrong on that front--if you take δ to be the typical β angle of a monoclinic crystal system (I've set δ ≠ 120° to avoid the rhombohedral scenario). Why does this not make the base-centered orthorhombic lattice redundant? Am I misguided in trying to understand these crystal systems in terms of their lattice points and unit cell lengths instead of their symmetry?

All the centered Bravais lattices are redundant in the sense that it is possible to use a primitive cell of smaller volume instead. However, this often means that crystal symmetries are no longer aligned with the cell axes. For the orthorhombic system, you either have 2-fold rotation or screw axis parallel to the cell edges, or mirror planes or glide planes perpendicular to the cell edges. If you switch to a primitive lattice for the centered cases, these symmetry elements will no longer be aligned in this simple way (they might be on diagonals, making the symmetry operations harder to figure out).
So it is a convention, and one that is useful.

Hexagonal until cells are often reduced to rhombic prisms consisting of only one-third of the hexagonal prism, which simplifies presentations.  The sixfold or threefold axis is implied to pass through the obtuse-angle vertices of the rhombic bases, whose angles measure 120°.

Monoclinic lattices do not have their two oblique axes equal; or in terms of point group symmetry, the $C_\mathrm{2h}$ symmetry characteristic of monoclinic lattices is promoted to $D_\mathrm{2h}$, therefore orthorhombic, when those two axes are equal.  Your primitive cell has what would be the two oblique axes of a monoclinic cell equal, therefore retaining the $D_\mathrm{2h}$ orthorhombic symmetry.  Compare with this answer where a specific axis ratio turns an apparently tetragonal lattice into a cubic one.


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Concept of Equivalent Weight
I'm having a hard time understanding what equivalent weight means physically.  I've looked at many articles on Google but the definition always seems nebulous to me.
For example, if the equivalent of $\ce{H2SO4}$ is $\pu{49 g/mol}$, what exactly does that mean?

Wow, equivalent weights keep popping up again and again. It is not a nebulous term, the point is you are checking recent webpages. You have to look up chemistry books from 1930 to 60. If corona were not an issue I would have suggested a library trip to read Vogel's Quantitative Chemical Analysis (oldest edition).
Equivalent weight has nothing to do with mol. It predates moles story because the unit originated centuries ago. Mixing mole with equivalents is perhaps the main source of confusion.
Equivalent is the weight of X in grams, that will furnish or combine with 1 gram H or 16 gram O atoms.
Let us take sulfuric acid, ask yourself, what weight (=mass) of sulfuric acid will provide 1 gram H. Note I am not involving any mole concept here. The answer is that if we take 49 g of sulfuric acid we will be able to "generate" 1 g hydrogen atoms.

Okay thanks.  So then I guess my next question is: why is that important?

Because it makes titration calculations extremely easy. It is still used in titration problems.

@M.Farooq How about a trip to archive.org?

Yes, it is an excellent and legal resource of classical books.

The only opportunity where the equivalents are still used today is in the measurement of pollution in used waters. As the nature of the dissolved impurities is not known, their amount cannot be precisely known or weighed. But they can be characterized by the amount of electrons consumed by a strong oxidant like $\ce{K2Cr2O7^}$ or $\ce{KMnO4}$  if this oxidizing stuff is used for oxidizing all the pollution (whatever its nature, solid, colloidal or liquid). This amount of used electrons, whatever their organic origin, does characterize the equivalent amount of pollution.
For example, if  $\pu{12 mL}$ $\pu{ 0.1 M}$ $ \ce{K2Cr2O7}$ is needed to oxidize all the organic impurities present in $\pu{1 L}$ used water, it means that $\pu{1.2E-3 mol}$ $\ce{K2Cr2O7}$ has been used in this titration. And as the ion $\ce{Cr2O7^{2-}}$ needs $6$  electrons to be reduced to $\ce{2Cr^{3+}}$, the amount of electrons used in this titration is  $6$ times the amount of $\ce{Cr2O7^{2-}}$ ions, and this is
$\pu{7.2E-3mol} $. So it makes sense to say that this amount of pollution is $\pu{7.2E-3}$ equivalent per liter.


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Is acac- a weak field ligand?
From the chemical structure of acac- compared to NH3 and en, is acac- a stronger or weaker ligand strength? How and why do we know this?

Hi, Welcome to Chemistry.SE! Could you perhaps expand on the question a little bit, and mention what exactly are you confused by, and what research you did on it? Homework type questions without showing research effort are generally closed: https://chemistry.meta.stackexchange.com/questions/29/lets-decide-on-a-homework-policy

Related : https://chemistry.stackexchange.com/questions/14575/what-makes-a-ligand-stronger-than-another?rq=1

Ligands that produce a large splitting are called strong field ligands, and those that produce a small splitting are called weak field ligands. So when crystal spittling occurs and of the 2 ligands, the one which causes greater splitting is a strong field ligand. So the strength of the ligand can be found by experimental ways . Or you to be able to decide how a ligand becomes strong field or weak field, you need to know and understand LIGAND FIELD THEORY. Whether a ligand produces a "strong" or a "weak" field was initially determined experimentally, which resulted in the so-called spectrochemical series, which lists ligands according to their field strengths. After that one can make facts and statements out of the experiments .
An easy way to identify this: The ligands in which the donor atom is Nitrogen and carbon is a strong field ligand and the ligand in which the donor atom is a halogen or oxygen is a weak field ligand. So in acac the donating atoms are Oxygen making it a weak field whereas in NH3 the donating is N making it strong field. The reason being that halogens have vacant d orbitals in their valence shell, and therefore their capacity to form a coordinate-covalent bond (by donating an electron pair) with the metal atom is relatively low, compared to ligands having pi bonds such an CN-, CO etc.


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Removing denatured Crystalbond?
I have some water-soluble crystalbond that is attached to silicon and, because of some overheating (to about 150 Celcius) during a final processing step, the crystalbond seems to have charred (white/grayish colour, see photo at the bottom) and is no longer dissolvable in water. Since this silicon is important and cannot be replaced, I really need to find a way to chemically remove this overheated crystalbond residue without using abrasives that could scratch of the thin SiO$_2$ coating on silicon.
Looking at the MSDS (see here), crystalbond is composed of the following

97% Nonylphenol Polyethylene Glycol Ether(127087-87-0) EC-No.
500-315-8
3% Polyethylene Glycol(25322-68-3)EC-No. 500-038-2

My question is: does anyone have any ideas for what crystalbond decomposes into around 150 C and how one could remove this overheated crystalbond residue?
Things I have tried (unsuccessfully)

Hot DI water (in a cup, and in a ultrasonic bath)
Acetone
2-propanol/ethanol
"Piranha" etch (asked a lab friend to do this, too scary for me!!)
Photoresist remover (lab friend also tried this)
Hexane
Dunking in LN2
Putting on a hot plate at 225C for about 24 hours
Burning under a 1100C torch for about 30 seconds

Any bright ideas, especially from organic chemists? All I can say is: damn this stuff doesn't want to come off.
Some examples of the residue below.
Lots of crystalbond residue on this Silicon wafer

Another silicon wafer with crystalbond residue. Less than the one above, but still significant.

Chloroform - 5% methanol is worth a try.

@Waylander interesting suggestion, so is that 5% chloro in 95% methanol or the reverse? Any insight forwhy  this suggestion might work?

5% MeOH in Chloroform. In my experience this is a pretty good solvent system for a lot of organic matter. The crystalbond is organic and has a lot of lipophilic groups. As for what you made by overheating it I can only speculate. I would guess that you dehydrated some of the alcohol groups perhaps creating higher density polymers

A possible option: paint remover from your local hardware or big box store. These are formulated to remove stuff that was intended to adhere well for a long time. Worth a try. And an ultrasonic bath always helps!

Thank you Ed! Do you have any specific paint removers that you know are strong. I.e. what active ingredient to look for

I have no specific suggestion for paint removers, but they typically are mixtures of solvents that work more or less synergistically. If it was me, I would try anything reasonable that claims to remove burned on gunk. Even spray on oven cleaner and treat it accordingly. And treating a small area first, not near the center, is wise.

In view of all you have tried thus far, based on your comments, the next step may be very fine sandpaper or the equivalent. I have sandpapers as fine as 5000 grit and they make sandpapers as fine as 60000 grit, which is equivalent to 0.25 $\mu $m particle size. So, basically application of elbow grease and sandpaper. Best of success, whatever you finally do!

Just a quick answer without looking too much into it.
You have been using a lot of polar solvents without any luck. The reason is that the molecule no longer has the groups that make it water-soluble because of the thermal decomposition. Only hydrocarbons chain remain. You need to use an organic solvent (toluene, benzene, hexane). Since you probably don't have access to a lab, I would try with gasoline.
Let me know if it works.
Cheers.

Thanks! Among the three non-polar solvents you mentioned, which would be best? I do know someone with access to hexane.

Hey, it seems that you do work in a lab. For the look of your Si wafer (I'm guessing is amorphous Si for the color) you were trying to modify the surface and perform a surface Science Experiment on a heated stage. I might be wrong. Either way, have you tried cleaning on an ultrasonic bath ?. You might not even need the solvent. In any case, I would try to use toluene first.

Somebody recommended a paint remover. That might work as long as it is based on an organic solvent (e.g, xylene).

Thank you! I will try to see which non-polar solvents I can find to try, especially toluene. Yes I did try ultrasonic bath in hot DI water (~75C) for about 20 minutes, but alas not much luck. If the non-polar solvent seems effective, I can try to use ultrasonic cleaning with it.

I won't have another chance to make these Si wafer devices for about 3 months, so I am desperate to find a solution. Thanks for your help :)

I gave a shot at hexane today, didn't seem to do much at sll unfortunately. I did 10min in pure hexane (no ultrasonication though). I guess James Gaidin may be right here, it's the other ingredients causing problems here.

Crystalbond™ 555-HMP is an adhesive. The SDS lists only hazardous ingredients, none of which should decompose at 150$^o$C to char. The char is probably from the non-hazardous ingredients, probably something like a hot-melt adhesive.
The adhesive char now bonded to the SiO$_2$ layer probably involves carboxyl oxygens, and if the char is not removed by the Piranha solution, that solution may have been too weak, or not immersed long enough. Another cleaner for organic residues is potassium dichromate in concentrated sulfuric acid. This may be more potent than Piranha solution.
One easy method to try (probably with a low probability of success) is to drop the wafer into a Dewar of liquid nitrogen: the thermal shock might dislodge a char layer that is distinct from the SiO$_2$ layer, but those final surface atoms may be different from what was there originally. But if any char falls off, the resulting surface may be better cleanable.
Silicon is thermally oxidized at temperatures above 800$^o$C, and any char should dissipate in oxygen (air?) at temperatures significantly below that. Whatever you did to the silicon to produce the oxide, do that at a reduced temperature (so as not to undo what you did originally) for long enough to remove the char without modifying your original treatment.

Oh, very helpful indeed, thank you James! I will look to see if there are others who can do a longer Piranha or the potassium dichromate methods. But I can certainly try the latter two methods (LN2, heating to high temperature). Would vacuum heating also dissipate the char? Or is the oxygen/air required?

Probably the char needs something more than vacuum to go away. Oxygen or air could be aggressive enough if you give it some heat.

I tried heating at 225C for about 24 hours, but no changes there. similarly, no changes when held against a torch at ~1100C for about 30 seconds. Liquid nitrogen also showed no changes. This stuff is really tenacious!

Maybe this is not a lab disaster, maybe you have invented something of great importance, like a permanent Magic Marker. Is silicon carbide black? How about blasting with a NaHCO3 sandblaster until clean?

At this point, I would just go and finish burning that char in air. You need air to complete the combustion, don't do it under vacuum. 300-400 C should do the trick. Although, if you are very concerned about the temperature, just do a TGA to the crystal bond under air and determine the temperature where it starts to combust; then apply that temperature to your wafer isothermally for an hour or so as to minimize damage to the surface layer.


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Calculating temperature of partial condenser, flash calculations
The iterative method I have learned to make flash calculations for a binary system where I know the inlet feed (F [mol/s]), total composition of the two components ($z_1$ and $z_2$), the amount of the mixture which is evaporated (q) and the total pressure (P [mmHg]) is as follows:

We seek T, $x_i$ (composition of liquid) and $y_i$ (composition of gas). We know F, $z_i$, q, P.
Guess T
Calculate $P_1°$ and $P_2°$. In our course we are using Antoine's equation to do this. Antoine's equation can be written in many ways but lets for simplicity stick with this version. That is:
$$P_i°(T)=10^{A_i-\frac{B_i}{C_i+T}}$$
where T is in °C and P in mmHg. $A_i,B_i,C_i$ are the Antoine constants and they are known.
Calculate $x_i$ and $y_i$. For simplicity we say that $\gamma_i=1$. This gives us that:
$$x_i=\frac{Fz_i}{V\frac{P_i°(T)}{P}+L}$$
and
$$y_i=\frac{P_i°(T)x_i}{P}$$
where V is the gas outlet stream ($V=q\cdot F$) and L is the liquid outlet stream ($L=(1-q)\cdot F$).
If $\sum x_i=1$ $\underline{and}$ $\sum y_i=1$ you have found the correct T, $x_i$ and $y_i$. Else go back to 2.

So my question is. Do we really need to calculate $\underline{both}$ $y_i$ and $x_i$? If I find the temperature which satisfies $\sum x_i=1$ doesn't that guarantee that also $\sum y_i=1$? I plotted this and played around with different values for the Antoine constants and values for F, q and P. Here is a link to my plot: https://www.desmos.com/calculator/m37slkdebb. For all of my cases it was true that if for a temperature T $\sum x_i=1$ then also $\sum y_i=1$ for that same temperature T. I asked my professor about this and he insisted that I would have as a requirement to stop my iteration when $\sum x_i=1$ $\underline{and}$ $\sum y_i=1$.
Am I correct that $\sum x_i=1$ guarantees $\sum y_i=1$?
Are there cases I have missed where you can have several temperatures where $\sum x_i=1$ but only one of these temperaturs satisfies both $\sum x_i=1$ and $\sum y_i=1$?
What happens when $\gamma_i\neq1$, is this simplification too much? If i use another model for $P_i°$ what will happen then?

In my judgment your professor is correct.  The sum of the mole fractions of vapor is a constraint on this problem.  There is only one T and set of x's that satisfies this constraint.


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Hydroboration, steroselectivity and anti-markovnikov product
My first question has to do with the fact that Hydroboration produces an anti-markovnikov product, why?. I don't see how that happens. A wikipedia artical says that B has a partial positive charge because it reacts slightly faster than H to form the bond but how is that possible when both H and C are more electronegative.
My second question is that why are boranes, that are highly substituted, lead to a greater yield in anti-markovnikov product?
Is it because something like 9-BBN can't be on top of a CH2 group because it's too big, and so it has to attach to the terminal C atom.

The regioselectivity of hydroboration is covered in practically every organic chemistry textbook. If Wikipedia's explanation doesn't suit you, then you should try a different source (Clayden is good). The reason is partly electronic (which is what Wikipedia was trying to explain: the C–B bond is formed slightly earlier than the C–H bond, so there's positive charge buildup on the carbon to which the hydrogen will go, and hence more substituted is better) and partly steric (which explains the second part).

The actual hydroboration step of an unsymmetrical alkene is a Markovnikov addition because boron is more electropositive than hydrogen. Alkaline peroxide oxidation of the borane exchanges boron bound to carbon for oxygen. Because electronegativity increases in the order B>H>O, it is this exchange process that makes the overall reaction anti-Markovnikov. In addition, the overall reaction is electroneutral in that the first step is a two-electron reduction while the second step is a two-electron oxidation.


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Decline in current during electrolysis
I observed a sharp decrease in the current when I carried out the electrolysis of water. Just to set the context, I used a voltage of $\pu{12V}$ and the concentration of my electrolytes were $\pu{0.15 mol dm^{-3}}$ for $\ce{Ca(NO3)2}$ and I used graphite rods which were flaking but I am not sure whether that might cause such a significant dip in the current. I did also notice some white material forming on the cathode. So I want to know why there is such a dip?

Sounds interesting. Could you add a little more detail? How much current, peak and valley? Any bubbling that covers the electrodes? How large were the electrodes and how far apart?

Did you observe any white, solid material forming at your cathode?

The current was highest as soon as electrolysis started at 0.85A. Then I recorded the current value in 10s intervals and I got 0.5A, 0.37A, 0.3A, 0.26A and so on. This is for 0.15 mol dm-3 Ca(NO3)2 solution. The graphite rods with diameter 0.62cm and a height of 5.65cm with a separation of 2.56cm. It was a pretty small setup.

I did notice some whitening of the electrode.

Why did you choose calcium nitrate as electrolyte ? Why did you choose such a high voltage as $\pu{12 V}$. Have you ever observed a similar phenomena with a more common electrolyte ? On which electrode did you see a white deposit ? Have you observed an increase of temperature ? What sort of currant generator have you got ? Is it able to stabilize its tension ? Don't you think the trouble may be due to the feeder ?

Given the whitening observed at the cathode, one potential explanation is that calcium salts were precipitating there, forming a layer blocking passage of part of the current.
I usually write the cathodic hydrogen-evolution reaction in neutral solution as:
$$
\ce{2 H2O + 2 e- -> H2 + 2OH-}
$$
Thus, the pH in the vicinity of the cathode will be considerably greater than 7 during the electrolysis process, due to the $\ce{OH-}$ formed. Since you're using a calcium-based electrolyte, it seems likely that some combination of $\ce{CaO}$/$\ce{Ca(OH)2}$ is precipitating in that alkaline surface layer. Once formed, in my experience such solids will dissolve only slowly in neutral aqueous solution.
If possible, I'd recommend trying the experiment again but in a different electrolyte, such as $\ce{KNO3}$ or $\ce{NaNO3}$; those alkali cations shouldn't exhibit any precipitation.


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Finding Kp using log Kp
I am working through some MIT lecture notes trying to understand adiabatic flame temperature. There is a section in these notes where $K_p$ is calculated for the following reaction at $\pu{1600K}$.
$$\ce{CO + \frac{1}{2}O_2 \rightleftharpoons CO_2}$$
The notes find $K_p$ by using the following values:
@$\pu{1600K}$:
$$\log K_p (\ce{CO}) = 8.234$$
$$\log K_p (\ce{O2}) = 0$$
$$\log K_p (\ce{CO2}) = 12.940$$
$$\log K_p (\text{RXN}) = 12.940-0-8.234=4.706 \implies K_p \approx 5 \times10^4$$
What is the significance of these $\log K_p$ values? I'm confused because I thought that $K_p$ was a value pertaining to a reaction as a whole, not specific reactants. Also, where can I find these values online? My notes pointed me towards https://webbook.nist.gov/chemistry/, but I am having difficulty finding these $\log K_p$ values in that database. I would appreciate any guidance/intuition you may be able to provide!

There is use of $K_p$ notation for skin permeability. In that use $\log K_p$ is handy.

You can check this yourself, but my guess is that the Kp's here are those for the standard formation reactions (i.e., the reaction used to determine the free energy and enthalpy of formation of a compound).  That's why it's zero for O2 (delta G = -RT ln Kp; if ln Kp = 0, then delta G = 0)

@theorist For CO, Atkins 9th ed. gives $\Delta_\mathrm{f}G^\circ(\ce{CO}) = \pu{-137.17 kJ/mol} \implies -\Delta_\mathrm{f}G^\circ/RT = 10.312$; and $\Delta_\mathrm{f}G^\circ(\ce{CO2}) = \pu{-394.36 kJ/mol} \implies -\Delta_\mathrm{f}G^\circ/RT = 29.646$. Granted, those thermodynamic values are for 298 K.

The $K_p$'s here are those for the standard formation reactions.
If you go to the following link, you will see $\Delta G^o_f$ values for $\ce{CO2(g)}$  and $\ce{CO(g)}$, in $100 \,\mathrm{K}$ increments, up to $1500 \,\mathrm{K}$:
https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&ved=2ahUKEwjbt56Z9IbvAhXVjp4KHbD-BS0QFjAJegQIAhAD&url=https%3A%2F%2Flink.springer.com%2Fcontent%2Fpdf%2Fbbm%253A978-3-319-41616-8%252F1.pdf&usg=AOvVaw3rJgF0nzpqXtq9ML9qvGaV
Thus, doing a simple linear extrapolation to $1600 \,\mathrm{K}$ gives us the following:
$$\Delta G^o_f,\ce{CO_2(g)} = -396.309 \mathrm{\frac{kJ}{mol}}$$
$$\Delta G^o_f, \ce{CO(g)} =  -252.3298 \mathrm{\frac{kJ}{mol}}$$
Solving for $\ln K_p$ (using $\Delta G^o = -RT \ln K_p$) gives:
$$\ln K_p,\ce{CO_2(g)} = 29.7906$$
$$\ln K_p, \ce{CO(g)} = 18.9677$$
Dividing by $\ln(10)$ to convert from the natural logarithm to the base-10 logarithm gives:
$$\log K_p,\ce{CO_2(g)} = 12.940$$
$$\log K_p, \ce{CO(g)} = 8.238$$
[$\Delta G^o_f$ for $\ce{CO(g)}$ changes more rapidly with temperature than that for $\ce{CO2(g)}$, so my extrapolation for the former may have a larger error, which could explain the small discrepancy vs. the value you were given.]
In addition, if you appropriately combine the formation reactions to give the net reaction (as you would for a Hess's Law problem), you will see that you get the supplied equation for $\log K_p$ for the net reaction (I'll leave this as an exercise for you).


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TEMPO Radical Analogue
After investigating what makes the persistent radical TEMPO stable, I was wondering if another compound which is more feasible for my needs would be comparable to it.
Would oxidizing 2,2’-dimethyl-diphenylamine result in a compound similar to TEMPO? It will have an Aminoxyl group and have the steric hindrance of the methyl group next to it. By similar I mean if it will have a stable radical in the aminoxyl group.

Electronically it is a rather different species with the phenyl groups in conjugation.

By similar I mean if it will have a persistent/stable radical on the aminoxyl group

This doesn't seem like the best question to ask here. I think a literature search would be more comprehensive.

@Evamentality - being electronically different will greatly affect the stability of the radical you propose.

I have tried to search for information regarding this compound and I can’t seem to find any on this specific configuration. The phenyl groups should make this more stable due to resonance, however organic chemistry can always surprise you

Aminoxyl (trivial name: a nitroxyl radical or a nitroxide) denotes a radical functional group with general structure $\ce{R2N–O^•}$. It is well known that sterically unhindered aminoxyls baring $\alpha$‐hydrogens are unstable and undergo rapid disproportionation to nitrones and hydroxylamines (Ref.1: A new mechanism for the decomposition of $\alpha$‐hydrogen nitroxides is proposed). For example, decomposition of one of such compound is displayed below (The scheme is from Ref.1):

Sterically hindered aminoxyls without $\alpha$‐hydrogens such as TEMPO are stable radicals. However, it is not clear that steric hindrance at carbon $\alpha$ to nitrogen atom is important in the mechanism. Yet, the lone pair of electrons on nitrogen has played a significant role in the mechanism (Ref.1). Therefore, I speculate that 2,2'-dimethyldiphenylamine, when oxidized to relevant aminoxyl radical, may not be as effective as TEMPO, mainly due to the delocalization of the lone pair of electrons on nitrogen between two 2-methylphenyl groups.
Nonetheless, there are few available aminoxyl radicals, which are more efficient than TEMPO (Both following images are from Ref.2):

The research in Ref.2 has performed catalytic oxidation using few N-oxyl radicals, namely 2,2,6,6-tetramethylpiperidine-1-oxyl (TEMPO), 2-azaadamantane N-oxyl (AZADO), 1-methyl-AZADO (1-Me-AZADO), 9-azanoradamantane N-oxyl (Nor-AZADO), and 1,5-dimethyl-9-azanoradamantane N-oxyl (DMN-AZADO), in water at $\mathrm{pH} \  10$ on wood cellulose to determine the influence of different N-oxyl radical chemical structures on the reaction kinetics and structures of oxidized celluloses. The reaction time required for complete oxidation of wood cellulose using AZADO, 1-Me-AZADO, or Nor-AZADO was less than $\pu{20 min}$, while DMN-AZADO and TEMPO required $\pu{70 min}$ and $\pu{120 min}$, respectively.
There is also some comparative work has been done in Ref.3, which also showed the superiority of other N-oxyl radicals over TEMPO (Synthetic procedure to prepare Nor-AZADO has been given in the reference; see attached PDF).
References:

Aaron Nilsen,  Rebecca Braslau, "Nitroxide decomposition: Implications toward nitroxide design for applications in living free‐radical polymerization," J. Polymer Sci., Part A: Polymer Chem. 2006, 44(2), 697-717 (DOI:https://doi.org/10.1002/pola.21207).
Hiromasa Hondo, Tsuguyuki Saito, Akira Isogai, "Preparation of oxidized celluloses in a $\ce{NaBr/NaClO}$ system using 2-azaadamantane N-oxyl (AZADO) derivatives in water at $\mathrm{pH} \  10$," Cellulose 2019, 26, 1479–1487 (DOI:10.1007/s10570-018-2177-5).
Masaki Hayashi, Yusuke Sasano, Shota Nagasawa, Masatoshi Shibuya, Yoshiharu Iwabuchi, "9-Azanoradamantane N-oxyl (Nor-AZADO): a highly active organocatalyst for alcohol oxidation," Chem. Pharm. Bull. 2011, 59(12), 1570–1573 (DOI: https://doi.org/10.1248/cpb.59.1570)(PDF).

Cool, but what about diaryl radicals? I think what OP mentions isn't nearly as good, but still...

@Mithoron: I'd agree Waylander on his comment. It won't be nearly as  good as TEMPO.  OP is looking for another compound which is more feasible for his/her needs and would be comparable to TEMPO. Diaryl compound is one of compound he/her wonder.

Thing is OP's issue is if the one he mentions (or similar, I guess) is viable replacement - perhaps he has this amine already - not what's actually better then TEMPO.

@Mithoron: See my addition to the answer.

Thanks for the contribution (Love being called OP). Attaching TEMPO to my compound is not feasible due to the synthesis route meetings things, so I was wondering if this could work in creating a stable aminoxyl radical. It doesn't have to be as efficient for the time being. As a matter of fact, the diary in my compound is actually part of a much bigger aromatic chain, but I simplified it to two benzene rings. The main idea is to see how it operates. I will improve the efficiency when I cross that bridge, now I just need to see if the proof of concept is solid.

I am actually going to perform photoelectron spectroscopy, so I need the aminoxyl radical to be as stable as possible as to not react with anything until it can regain an electron.

I just noticed some appalling mistakes in my comment. Guess that is what I get for using an old laptop configured to a different language setting in the heat of things


text:
How can someone in the lab synthesize n-allyl-4methyl-benzenesulfonamide from 4-toluenesulfonyl chloride and allylamine
I am trying to figoure out the conditions and choice of solvents and catalysts in order to run this synthesis.

Welcome to ChemSE! Your question will likely be considered as homework. You need to make some effort to answer the question yourself . Take a look here about homework policy. https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange.

The name of your starting material - 4-toluenesulfonyl is incomplete. The question cannot be answered until you name your starting material correctly. Do you mean 4-toluenesulfonyl chloride?

This is straightforward. 4-Toluenesulfonyl Chloride is readily attacked by nucleophiles and allylamine is a good nucleophile.
Dissolve 2.05 equivalents of allylamine (or 1.05 eq allylamine and 1 eq of an tertiary organic base such as triethylamine) in dichloromethane, cool in an ice/water bath (reaction is exothermic) then slowly add 1 eq of 4-toluenesulfonyl chloride dissolved in dichloromethane. Leave it half an hour then wash the reaction mixture with dil HCl soln, dried the organic phase, filter and concentrate to give your product.


text:
Is it feasible to calculate the rate law of the oxidation of ethanol using colorimetry?
My IB Chemistry class needs to have our Internal Assessment ideas finalized by next week, and I'm torn between a few experiments. I'm considering performing an oxidation of ethanol and potentially using colorimetry to measure the rate of the reaction (I believe using sodium dichromate as the oxidizing agent will create a color change, but I'm not sure if it is measurable enough to calculate kinetics). I'm also considering comparing the rate of the reaction dependent on temperatures, but am unsure about how that would work (potentially with a water bath?). I've also watched some oxidation of ethanol videos (because it's an AS Chemistry practical), but am unsure of how to move forward with it. Please just tell me if it's a feasible experiment or not, I'm quite literally just asking as to make a decision about what to do for this project.

I don't think it's slow enough to measure its rate like that.


text:
Finding rate constant from a change in pressure experiment
I measured the change in pressure for the decomposition of hydrogen peroxide in the presence of a manganese(IV) oxide catalyst at 5 different temperatures:
$$\ce{2 H2O2 (l) ->[MnO2 (s)] O2 (g) + 2 H2O (l)}$$
Is there any way that I can find concentrations and then find the rate constant, $k$? And, using them, make an Arrhenius graph to find activation energy? Any help would be much appreciated.
Edit:
Change in pressure was found using a Vernier Gas Pressure Sensor. The starting concentration of hydrogen peroxide was 3% or $\pu{0.88 mol}$. The starting pressure was always around $\pu{98kPa}$. The catalyst and hydrogen peroxide were mixed together after the stopper was put on the flask. The increasing pressure was measured and the highest it went to (at $\pu{70 ^\circ C}$) was around $\pu{145kPa}$.
Here is an example of what I have done so far to try and find the rate constant to plot on an Arrhenius graph:
At $\pu{303.15 K}$:
$PV=nRT$
$\frac{P}{RT}= \frac{n}{v}$, which is the same as concentration.
$\frac{120.3}{8.314 \times 303.15} = \pu{0.477 mol dm-3}$
$\pu{0.477 mol dm-3}$ is the final concentration of oxygen created after $\pu{180s}$ had elapsed.
$\frac{\pu{0.477 mol dm-3}}{\pu{180 s}} = \pu{0.00265 mol dm-3 s-1}$
The ratio between oxygen and hydrogen peroxide is 1:2, therefore the rate at which hydrogen peroxide was depleted at is $\pu{0.0053 mol dm-3 s-1}$.
The ending concentration must be $\pu{0.88 M} - \pu{0.0053 mol dm-3 s-1} \times \pu{180 s} = -\pu{0.074 mol dm-3}$.
And this is where I think something has gone wrong... The pressure continued to increase even after this $\pu{180s}$ so I do not think the reaction went to completion.
I would then use the equation   $k =  \frac{\text{rate}}{\text{[final concentration hydrogen peroxide]}}$ to find $k$.

There is a way you can find the activation energy in this method. But, it is not the activation energy of original reaction. It'd be the reduced activation energy due to $\ce{MnO2}$ catalyst.

Yes that is what I want to find. Thank you so so much for the clarification!! I do want to find the activation energy with the catalyst. How would I do that? How do I find the rate constant? @MathewMahindaratne I was thinking about somehow using P/(RT)=n/v which is the same as concentration but I do not know if that is even right or where to go from there

I already voted to close your question due to lack of details. If you put some details including what you have done so far (e.g., how you find the change in pressure), I can reconsider my vote and decide to help you.

Thank you I have made these edits and shown what I have done so far. @MathewMahindaratne

Thanks for the addition. I retract my vote as promised and upvoted your effort.

Are you doing this for a science fair or your laboratory experiment for grading? Ant way, your equation for $k$ is not correct. Also remember, you ate measuring pressure over water bed so saturated vapor pressure at that temperature should be included in your equation. You have also to find $k$ in different temperatures and plot those values appropriately to find activation energy. The ratio of $\ce{MnO2}$ to $\ce{H2O2}$ should be constant during all experiments.

@MathewMahindaratne Thank you for all your help. It is for assessment not grading. Would this be the correct equation? : R = K [Initial Hydrogen peroxide] ? And for the vapour pressure at 30C do I just subtract 4.24 kPa from the 120 kPa before doing all of the rest of the equations? I have also been thinking that I should use the difference in pressure (120.3-98.32=21.98kPa) instead of the final pressure.
I have done this experiment at 5 different temperatures with everything else constant.

You find the rate equation experimentally by measuring $\ce{O2}$ pressure in time intervals and plotting the values against the time (you may convert pressure values to concentration terms before plotting). By doing so, you can calculate rate for one $\ce{[H2O2]}$. Do the same for few different $\ce{[H2O2]}$. Then plot rates versus $\ce{[H2O2]}$ find $k$ for that temperature. Repeat the procedure for few different temperatures. You got the picture now.

@MathewMahindaratne I already did the experiment and I kept the concentration of hydrogen peroxide the same throughout... am I out of luck for trying to find the rate constant and activation energy? Again thank you so much for all your help.


text:
Why is there a mirror image effect in the absorbance and fluorescence spectra?
I've read some answers online but I still can't seem to understand the mirror image effect. Why is the highest energy absorption (v" = 0 to v' = 4) the lowest energy fluorescence (v' = 4 to v" = 0)? Shouldn't the molecule that absorbs the highest-energy light emit the highest-energy light?

Please look at Jablonski diagram and, in particular, the second figure shown there. Then follow the links for Stokes shift and Kasha’s rule.

See the answer here https://chemistry.stackexchange.com/questions/134675/why-do-some-molecules-show-non-symmetric-absorption-and-emission-bands/134699#134699

@porphyrin Thanks for the help, I still don't understand why the highest energy absorption would have the lowest energy fluorescence. Am I missing something here?


text:
Why do we only take into account the first product while calculating a limiting reactant?
High school chemistry student here. Currently, we are learning stoichiometry and I noticed that when we calculate the limiting reactant of a reaction with multiple products, we almost always just take into account the first product. One sample of this can be found at chem.libretext

In this case, why are we only worried about the amount of CO2 that is produced? It would make sense to me sometimes choosing a product other than the first one, maybe H2O in this case, would affect the result we would get.
Am I thinking about this right?

It doesn't affect the result, you can choose whichever one you want. The amount of CO2 produced is directly proportional to the amount of H2O produced. (What's the proportionality factor? Well, the chemical equation tells you.)

Well actually you don't need to calculate the moles or mass of any of the products. The stoichiometry of the reaction gives you the ratio of the two reactants.
For the reaction:
$$\ce{4C2H3Br3 + 11O2 -> 8CO2 + 6H2O + 6Br2}$$
we can divide all the coefficients by 4 so:
$$\ce{C2H3Br3 + 3.75O2 -> 2CO2 + 1.5H2O + 1.5Br2}$$
Now from the given answer there are 0.286 moles of  $\ce{C2H3Br3}$ and 1.53 moles of $\ce{O2}$
If:

Moles($\ce{O2}$)/Moles($\ce{C2H3Br3}$) < 3.75 then $\ce{O2}$ is the limiting reagent
Moles($\ce{O2}$)/Moles($\ce{C2H3Br3}$) = 3.75 then the reaction is perfectly balanced.
Moles($\ce{O2}$)/Moles($\ce{C2H3Br3}$) > 3.75 then $\ce{C2H3Br3}$ is the limiting reagent

$\dfrac{1.53}{0.286} = 5.35$ so $\ce{C2H3Br3}$ is the limiting reagent.


text:
Can someone help me understand the motivation behind and visualize the dihedral potential and the Urey-Bradley potential?
I am studying statistical mechanics and force-fields, and I see a lot of this formula being thrown around with no explanation:
$$U_{\mathrm{dihedral}} = \begin{cases}
       k(1+\cos (n\psi + \phi)),&  n>0 \\
      k(\psi - \phi)^2, & n=0 \\
   \end{cases} $$
The only explanation I read is: this is a 4-body torsion angle (known as the dihedral angle) potential, which embodies the potential due to the angular spring between the planes formed by the first 3 atoms, and the other 3 atoms. $\psi$ is the angle between the planes, and $\phi$ is the phase shift angle, while $k$ is a multiplicative constant.
Could anyone please provide a visualization for this? and why should there be a $\cos$ term in the potential? Why is there an $n$ in the mix? Looks an awful lot like an eigenvalue of sorts.
Same applies for the Urey-Bradley potential. But I want to understand why such a potential was introduced in the first place. The 3-body potential described angular vibrational motion:
$$U_{UB} = k(\theta - \theta_0)^2 + k_{ub}(r_{ik} - r_{ub})^2$$
where $\theta$ is the angle between $r_{ij}$ and $r_{jk}$, and the second term is for the covalent spring bond between atoms $i$ and $k$. I don't understand - if these atoms are not bonded, why cook up a new bond?

See http://cbio.bmt.tue.nl/pumma/index.php/Theory/Potentials

this is great, thank you @KarstenTheis

It turns out you cannot complete an internal coordinate system for a molecule without including parameters that include four atoms, in many cases.


text:
Why is Cu(II) more stable than Cu(I)?
Seeing Copper has an exceptional electron filling in its valence shell(4s1 3d10), the way I see it is that on +1, a d10 configuration should be more stable and hence easier to attain than a d9 which would come on +2. Or is there some reorganisation of electrons? Either way, please help explain. Thank you.

Also, I suppose your question refers to the scenario in an aqueous medium?

I suppose so. I meant to ask why the +2 oxidation state for Copper is more common than its +1. I realise there is another question on the same topic, but i didn't understand their answer.  Yours, I do. Is the greater tendency of hydration the only reason for greater stability of Cu+2 compounds? Or is there some rearrangement of energy at the electronic level?

Cu2+-2Z having higher hydrothermal stability than [Cu(OH)]+ is said to be the most contributing factor in attributing to its overall stability.                                                             This is found at : Related: https://www.osti.gov/pages/servlets/purl/1437018

Ooh that page is a little hard to understand for me, but i take your word for it. Thank you!

Yeah, no issues, I just meant that hydration is what imparts overall stability.

The stability of an ion in an aqueous medium is determined by the tendency of attraction between the ion and the water molecules termed as hydration enthalpy. It, in turn, depends on the charge on the ion and its size.
Smaller the ion greater the hydration power of the ion. It is because when the ion is small the distance between its centre and the molecule that solvates it is less hence providing a stronger attraction, thereby releasing more energy attributing to stability. The example of this fact is provided by Cu2+ which has a smaller cationic radius due to a stronger hold of the nucleus on the remaining 27 electrons after the loss of two as compared to Cu+ where only one electron is lost and the hold of the nucleus is less. The stability gained by energy loss in hydration is more stabilising than the destabilising effect of the removal of 2 electrons.
Also, as mentioned earlier, the greater the charge on the ion greater is its tendency to attract and keep the water molecules solvated to it. Cu2+ having a greater charge has a greater tendency of hydration than Cu+.


text:
Are there any ADME predictions tools available for Python/RDKit?
Is there any Pythonic way (a.k.a: a module, an rdkit package, etc) or an API I can use for ADME predictions? I need to prioritize some compounds for purchase/synthesis and some sort of ADME scoring might help me rule out most of them.
I am aware of tools like SwissADME but given the amount of compounds I want to test, it isn't a viable option. Should I train my own model?

You mean, KNIME would be just fine, too? As this 2020JChemInfModel2660, for example?

This might be better suited for matter modeling

@Buttonwood yeah, found that right after posting the question. It is my understanding that KNIME requires a license for batch jobs such as this one, though. Do yuo have any experience with it? Care to share? :)

@Leitouran The topic is too broad for ChemSE and really better suited for the sibling site (MatterModelling). Literature search (e.g. this), the RDKit UserGroupMeetings (you equally find recordings on youtube) empirical application of rules (e.g., Lilly rules, or DataWarrior to crunch many relevant data at once) may be examples of tools and resources useful for you, so ChemSE can't give a hand-on tutorial.

I don't think is necessarily better suited for a modelling SE, since I'm not asking "how to make my own model". I'm not asking for a tutorial on how to do ADME modeling. I know there's enough literature on how to do it... I'd just rather not to, and I was asking if there was any tool readily available to so.


text:
Formula for rate constant for the first order reaction
My lecturer mentioned that the formula for the rate constant $k$ for the first order reaction is
$$k = \frac{2.0303}{t_{1/2}}\,\log\frac{[\ce{A}]_0}{[\ce{A}]_{t_{1/2}} - [\ce{A}]_0},\tag{1}$$
where $t_{1/2}$ is the half-life; $[\ce{A}]_0$ is the initial concentration; $[\ce{A}]_{t_{1/2}}$ is the concentration at half-life.
In which what I know recently is that the rate constant for the first order reaction is
$$
\begin{align}
&  & \ln\frac{[\ce{A}]_t}{[\ce{A}]_0} &= -kt \tag{2}\\
&\implies & [\ce{A}]_t &= [\ce{A}]_0\,\mathrm e^{-kt}\tag{3}
\end{align}
$$
I ask my teacher, but he said I should go and do more research, but I couldn't find it.

Your first equation has some typos, the second and third are correct. You need to use  $[A]{1/2}=[A_0]/2$  and let $t\to t{1/2}$

@Muhammad Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.

Indeed, there are typos!

Can anyone help me with the derivation of the first one @porphyrin

I am attaching a written answer as typing the log and other exponents is hard :(

The two equations are pretty the same. The number 2.30 is used to convert the natural logarithm in decimal logarithm: $\ln 10 = 2.303$. Also, substitute $[A]_{1/2} = [A]_0 /2$, from the definition. The concentration at the half-life is exactly the half of the initial concentration, then you will obtain $\ln 2$ inside the logarithm term.

Except for the typo that makes the argument of the logarithm in equation (1) of the OP negative.

The half-life $t_½$ is the time it takes for the concentration of the first order reaction to decrease by a factor of two. If we plug this into equation (2) given by the OP, we get
$$\ln\frac{1}{2} = -k t_½$$
We solve this for $k$ to get:
$$k = \frac{ \ln(2) }{t_½} \approx \frac{0.693}{t_½} $$

My lecturer mentioned that the formula for the rate constant $k$ for the first order reaction is
$$k = \frac{2.303}{t_{1/2}}\,\log\frac{[\ce{A}]_0}{[\ce{A}]_{t_{1/2}} - [\ce{A}]_0},\tag{1}$$

The argument of the logarithm is two (once you swap the two parts of the difference shown in the denominator - probably a typo), and $2.303 \log(x)$ is the same as $\ln(x)$, so this is correct but probably just a step on the way to the relationship given above that is more direct and useful. If you do the math, you will find
$$ 2.303 \log(2) \approx 0.693 $$

@porphyrin - I fixed it, thanks (another "typo" in the formula reported by the OP).

I think what OP need is the derivation of the equation $(1)$ in the question (as indicated in the OP's comment, "an anyone help me with the derivation of the first one").
The first part of this derivation, the formula for the rate constant $k$ for the first order reaction, is done in the last part of Chemistry student's answer, and hence I'd not repeat it. Thus,
$$[\ce{A}]_t = [\ce{A}]_0\,\mathrm e^{-kt}\tag{3}$$
Or when the rearrangement of the equation $(3)$ gives:
$$ -kt  = \ln\frac{[\ce{A}]_t}{[\ce{A}]_0}$$
$$ \implies kt  = \ln\frac{[\ce{A}]_0 }{[\ce{A}]_t} \tag{2}$$
In equations $(2) \ \& \ (3)$, $[\ce{A}]_0$ is the initial concentration and $[\ce{A}]_{t}$ is the concentration at time $t$. If $t_{1/2}$ is the half-life and $[\ce{A}]_{t_{1/2}}$ is the concentration at half-life, we can rewrite the equation $(2)$ as:
$$ kt_{1/2}  = \ln\frac{[\ce{A}]_0 }{[\ce{A}]_{t_{1/2}}} \tag{4}$$
Actually, $[\ce{A}]_{t_{1/2}}$ is $\frac{1}{2}[\ce{A}]_0$, and therefore, $[\ce{A}]_{t_{1/2}}$ can be written as $[\ce{A}]_0 - [\ce{A}]_{t_{1/2}}$. Substituting $[\ce{A}]_{t_{1/2}} = [\ce{A}]_0 - [\ce{A}]_{t_{1/2}}$ in the equation $(4)$, you get:
$$ kt_{1/2}  = \ln\frac{[\ce{A}]_0 }{[\ce{A}]_0 - [\ce{A}]_{t_{1/2}}} \tag{5}$$
Applying $\ln x = 2.303 \log x$ in the equation $(5)$ and rearranging it to get $k$, we finally get:
$$ k  = \frac{2.303}{t_{1/2}} \log \frac{[\ce{A}]_0 }{[\ce{A}]_0 - [\ce{A}]_{t_{1/2}}} \tag{6}$$
As evident and shown by others, OP's lecturer mentioned formula has two errors: one is making the $\log$ function negative, and the other is converting $\ln x$ to $\log 2.0303 x$, which is incorrect:
$$\ln x = \log_e x = \frac{\log_{10} x}{\log_{10} e} = \frac{\log_{10} x}{\frac{\log_{e} e}{{\log_{e} 10} }} = (\log_{10} x)(\log_{e} 10) = 2.303\log_{10} x$$
Thus, the equation $(6)$ is the most accurate equation in that regards.


text:
Enthalpy, heat of solution test: Calcium chloride vs magnesium chloride in water
I did a school experiment to measure the heat of solution (enthalpy) of various salts dissolved in water.
My results are completely different to the teacher's answer.
Specifically my test showed Calcium Chloride increasing the temperature of the cup of water by 6 degrees Celcius, while the Magnesium Chloride only increased the temperature of the same volume of water by 1 degree Celcius.
The teacher's answer shows Magnesium Chloride with an enthalpy more than double (twice as negative) as that of Calcium Chloride.
I'm happy to be wrong about this but could someone point me to an "official" answer so I can conclude that the experiment was done incorrectly.

Welcome to Chemistry SE. Please take a moment to share more specifics of how the results were obtained as it may help with context. This may increase the chances of a response.

It may be caused by the real content of hydrate water in the used calcium and magnesium chlorides. As there is possible disagreement between declared and true water content, due long term reaching equilibrium with air water vapour. I have once seen a glass bottle with reportedly solid $\ce{LiCl}$, sitting on an shelf for years. The content was a clear liquid of LiCl solution.(*)
Both chlorides form multiple hydrates, stable at different humidity or temperature conditions. Generally, with increasing of crystal/hydrate water content, the dissolution enthalpy raises for both salts from negative to positive values. E.g. anhydrous $\ce{CaCl2}$ warms up water when dissolved, $\ce{CaCl2 . 6 H2O}$ cools water down significantly. Note that both anhydrous chlorides are highly hygroscopic.
I suggest to make repetition of the experiments with salts of different origin and short history. Or, heating them well in an oven to get anhydrous stuff.

(*) This hygroscopicity of $\ce{LiCl(s)}$ is used in some humidity meters, measuring equilibrium electric resistence of a soaked fabric.

This makes sense to me. @GabbyChen134 It would be worth asking your teacher about the conditions of storage of the materials. Did any other students get results similar to yours?

Thanks everyone for your help. I'll follow up with some class mates to see if I can get some other results. It has been extremely humid in Sydney.


text:
Spurious term in Thomas-Fermi-Hartree energy functional
I am quite a beginner in DFT and I am studying the theory behind the computational tools. I have some doubts about the Coulumb approximate functional term:
$$U[n] = \frac{1}{2} \int d^3 r \int  d^3 r^\prime \frac {n(\mathbf{r}) n(\mathbf{r}^\prime)}{|\mathbf{r} -\mathbf{r}^\prime|}$$
If I use the definition of the electronic density $$n[\mathbf(r)] = \sum_i^N \delta(\mathbf{r} -\mathbf{r}_i)$$ in the previous equation I obtain the square of the electronic charge over $\pu{0}$. In other words the energy diverges. Am I wrong?

No, you are not wrong. DFT in almost all forms includes an interaction of the electron with itself as the approximate exchange functional doesn't exactly cancel this contribution to the Hartree term, unlike Hartree-Fock theory. For delta function charge distributions this self-interaction is infinite - try again with Gaussians for something that will work.


text:
What is the n factor of H2O2 undergoing disproportionation reaction liberating oxygen
There is a lot of confusion about the n-factor of hydrogen peroxide in its disproportionation into water and oxygen.
$$\ce{2H2O2 ->  2H2O + O2}$$
Some sources(1,2) say n-factor of $\ce{H2O2}$ is 2
and other say n-factor of $\ce{H2O2}$ is 1 for this reaction.
Actually my main problem is "the relationship between normality and volume strength of $\ce{H2O2}$". According to my respected teacher:
$$\pu{Normality = \frac{Volume~strength}{5.6}}$$
and
$$\pu{Molarity = \frac{Volume~strength}{11.2}}$$
implying that n-factor of $\ce{H2O2}$ is 2 but I am calculating n-factor of $\ce{H2O2}$ as 1 because the definition of n-factor is "the no of electron gained  or lost per mole".
As we can see 2 electrons are participating in this reaction                            so electron exchanged per mole is 1 which is equal to n-factor. Please clarify.
Sources

https://socratic.org/questions/what-is-equivalent-weight-of-h2o2-in-decomposition-reaction#629354
https://www.meritnation.com/ask-answer/question/what-is-the-n-factor-of-h2o2-and-o2-in-the-reaction-h2o2-h2o/some-basic-concepts-of-chemistry/8526405

biggest source my teacher and some people on quora say 2   accoding to me n factor should be 1  beacause n factor  means loss or gain of electron per mole so in this reaction 2 electrons are perticipating in the reaction so 2 electron/2 mole = 1

What is the definition of that "n" factor? What do you do with it?

Have a look at this one.

yeah i saw it i used the same formula to find the n factor in h2o2 it came out to be 1 .. here is a link to the source which says n factor is 1  https://socratic.org/answers/629354

there is a problem with Indian education system people dont want to confront anyone even if they are teaching wrong concepts there are teachers on youtube you can go just type volume  strength of h202 on the search bar you will find teacher telling n factor of h 2o2 in disprotionation reaction is 2
I

I found this answer worth sharing https://www.meritnation.com/ask-answer/question/what-is-the-n-factor-of-h2o2-and-o2-in-the-reaction-h2o2-h2o/some-basic-concepts-of-chemistry/8526405  ....this site says n factor of h2o2 is 2  i am really confused

"n factors" are really old form of chemistry, nowadays stoichiometry + molarity is enough.

Yea. Tell your teacher chem.SE says hi and he's welcome to skip the 20th and come directly into the 21st century.

I want to elaborate the answer of Ashish Ahuja because he is also perplexed by the n-factors. First of all for general readers n-factors are of historical interest only. Due to my interest in historical analytical methods, the following points would clarify this situation:
A) Equivalent weights do not require any knowledge of molarity, moles, electrons because it pre-dates all of these.
B) Equivalent weight is totally dependent on the reaction, so half reaction for an equivalent weight is meaningless without a context of how it is being used.
C) In the comments, I showed that the equivalent weight of $\ce{H2O2}$ can be calculated from the very original definition of equivalent weights without invoking the useless n-factors.
Equivalent weight of X (oxidizing agent) is the weight of X that will furnish 1 gram-equivalent of oxygen = 8g. So we can write
$$\ce{H2O2-> H2O +0.5O2}$$
Which means that 8 grams of oxygen as [O] would be provided by 34.01 g of $\ce{H2O2}$. There was no need to involve moles
This means that the molarity and normality of $\ce{H2O2}$ for a disproportionation reaction must be the same.
D) The volume strength is another archaic unit, which was very practical. The OP wrote

the relationship between normality and volume strength of
$\ce{H2O2}$". According to my respected teacher:
$$\pu{Normality = \frac{Volume~strength}{5.6}}$$
$$\pu{Molarity = \frac{Volume~strength}{11.2}}$$

Please gently tell your respected teacher that is both incorrect AND correct! The reason is that volume strength was determined by decomposition of hydrogen peroxide into water and oxygen (exactly by this disproportionation reaction). So if you are quoting normality with respect to this disproportionation reaction, then normality and molarity are the same.
Now if you are using the hydrogen peroxide in a redox titration reaction, it "consumes" two hydrogen ions. Again equivalent weight is the amount that furnishes/consumes 1 gram equivalent of oxygen or 1 gram equivalent hydrogen.
$$\ce{H2O2 + 2H+ + 2e → 2H2O}$$
Here 34.01 g of $\ce{H2O2}$ consumes two gram equivalents of hydrogen as an oxidizing agent. So the equivalent weight of this compound as an oxidizer in acidic medium is 34.01/2.
The moral of the story is that normality is solely dependent on what reaction is being used! The examiner or the teacher must provide a context. It is meaningless otherwise!

We can split the reaction
$$\ce{H2O2 -> H2O + O2}$$
into the respective reduction and oxidation half-reactions.
$$\ce{H2O2 -> O2 + 2H+ + 2e-}$$
$$\ce{H2O2 + 2H+ + 2e- -> 2H2O}$$
Since the n-factor of $\ce{H2O2}$ for both these half-reactions is 2, the n-factor is:
$$\frac{1}{n_f} = \frac{1}{2} + \frac{1}{2} = 1$$
$$n_f = 1$$
Based on the comments under your post I believe you have gotten this far, but some sources claim that the n-factor for this reaction is 2. Well, how do you verify what is correct? In such a case, I'd simply write the balanced chemical reaction and verify whether the law of chemical equivalence holds or not.
$$\ce{2H2O2 -> 2H2O + O2}$$
The n-factor for $\ce{O2}$ should be $2$ as can be seen in the first half-reaction, and for $\ce{H2O}$ the n-factor should be $1$, since we have $2$ moles of $\ce{e-}$ in the second half-reaction and $2$ moles of $\ce{H2O}$, and n-factor is defined per-mole of the substance.
From the law of chemical equivalence, the number of equivalents of $\ce{H2O2}$, $\ce{H2O}$ and $\ce{O2}$ should all be equal. If the claim I make above that the n-factor of $\ce{H2O2}$ is $1$ is true, this should hold.
Plugging in the moles from the balanced chemical equation and using the n-factors described above,
Equivalents of $\ce{H2O2} = 2 \times 1$, $\ce{H2O} = 2 \times 1$, $\ce{O2} = 1 \times 2$ (moles $\times$ n-factor)
Thus the n-factor is indeed $1$. The n-factor of $\ce{H2O2}$ for the half-reactions is $2$.
(Note that all the n-factors above are with respect to the reaction $\ce{H2O2 -> H2O + O2}$ and they (most probably) will change for a different reaction)
Update (the above part answered the question before the edit):
From [1]:

The "volume strength" of hydrogen peroxide is defined as the number of times its own volume of oxygen a sample of hydrogen peroxide solution will evolve if decomposed naturally.
$$ \ce{H2O2 -> H2O + O2}$$

As you've mentioned in your question,
$$\pu{Normality = \frac{Volume~strength}{5.6}}$$
and
$$\pu{Molarity = \frac{Volume~strength}{11.2}}$$
This does not match with the reasoning that the n-factor of $\ce{H2O2}$ for the above reaction is $1$, which would imply that the normality and molarity should be equal.
From [2]:

Now, equivalent weight of $\ce{H2O2} = \frac{34}{2} = 17$

A (N) $\ce{H2O2}$ solution contains 17 gms of $\ce{H2O2}$ per litre.

So in fact when we are relating the normality of an $\ce{H2O2}$ solution to it's volume strength, we are not referring to the normality of $\ce{H2O2}$ in the decomposition reaction. You would be right in saying that the normality of $\ce{H2O2}$ in the decomposition reaction is equal to its molarity.
One way of justifying the n-factor of $2$ of an $\ce{H2O2}$ solution (not undergoing any reaction) is due to the $-2$ charge on the $\ce{O2^{2-}}$ peroxide ion.
I can't really explain why this has been done, since it would make just as much sense to consider the n-factor of $\ce{H2O2}$ in the decomposition reaction, it is just a historical fact. Since n-factor is not a constant for a particular compound, it should be really important to specify context about normality when we relate it to volume strength, which is rarely done when the relation is taught which is what leads to this confusion. Both the n-factor's are correct in different contexts, so I would just recommend sticking to the n-factor of $2$ when talking about volume strength since that is what you have been taught.
[1]: Chemical Calculations by King A archive.org
[2]: An Introduction To Chemistry Vol. 1 Ed. 3rd by Das, Ranajit archive.org

ashish ahujha  thankyou for your answer i  want to ask one more thing i am preparing for jee for engineering course in the paper a question about volume strength is asked in which they take the n factor 2  to get the formula (Normality = Volume strength/5.6) and (Molarity = volume strength /11.2)  if you can give me some clarity about this because if the n factor is 1  for this reaction molarity and normality should be equal

@gurdeepsingh That's actually an interesting question. I've learnt the same thing, and I never realized that till now. I'm afraid I don't have the answer right now, I personally feel that you should ask a separate question for that since that deviates from the original question here and then other users will also be able to answer. I'll let you know if I find the answer.

https://chemistry.stackexchange.com/q/125269/92638

the question is been asked by someone but there was no clear answer  and if i ask it again they will mark it duplicate

@gurdeepsingh hmm I'm not sure what can be done then. Right now, I don't have an answer. I probably will in some time, so you'd have to wait, and then as well I can't guarantee anything. I will update my answer whenever I do find the answer.

If I can't find the answer, then I'd probably just ask a new question myself. You can wait if you'd like to.

@AshishAhuja, I am afraid your answer is wrong. The reason is that the equivalent weight is defined with respect to a chemical reaction. You cannot calculate eq. wt the way you showed it.

@M.Farooq I have defined equivalent weight for the reaction $\ce{H2O2 -> H2O + O2}$. Could you explain?

@M.Farooq If you're talking about the way I've defined the n-factor for $\ce{O2}$ and $\ce{H2O}$ using the half-reactions, we get the same result by checking oxidation states in the original reaction. So unless the n-factor for $\ce{O2}$ and $\ce{H2O}$ is wrong, applying chemical equivalence on the original balanced reaction gives us n-factor = 1, doesn't it?

What is the original definition of equivalent weight of X? I will not invoke electrons or moles because equivalents existed way before then. I will use the original definition. Equivalent weight should be the weight that combines with the weight of one oxygen atom. So if we write H2O2= H2O + 0.5O2 (=8 grams), then yes in this decomposition reaction the equivalent weight should be 34.01 g.

@M.Farooq So then why is my answer wrong? Equivalent weight of 34 g implies an n-factor of 1 (which might not have been defined when equivalent weight originally started out) which is what my answer claims.

I am against these n-factor calculations, because not only these are confusing, but nobody uses them anymore. Actually, I have not seen n-factor being discussed in any modern chemistry book by Western authors.

@M.Farooq Ok, I do agree with that, but that doesn't make my answer wrong does it?

Ashish, you answer is numerically correct. The approach of n-factor is bad. It is not your fault, this is the way it is taught.

Let us continue this discussion in chat.

i get your point but i am really dissapointed by these entrance paper asking so much vague and conceptually wrong questions thankyou for you effort


text:
What happens if you do not allow time for an exothermic reaction to cool down?
I realise I did not give some time for the chemicals (which are releasing heat during an exothermic reaction) to cool down first and instead, I added another chemical substance to the reaction mixture immediately. Is it a red flag that I might lose some of my products and get a lower percentage yield? The experiment that I will be doing is Jacobsen's ligand asymmetric synthesis.

You burn your fingers when manipulating the flask ;-)

It depends on the reaction, some are more temperature sensitive than others. You might get a slightly reduced yield or, at the other end of the spectrum, a runaway reaction leading to a redecorated fumehood.
For the Jacobsen asymmetric epoxidation this quote from the Wikipedia article here seems relevant:

The degree of enantioselectivity depends on numerous factors, namely the structure of the alkene, the nature of the axial donor ligand on the active oxomanganese species and the reaction temperature.

So at the least you'll get reduced enantioselectivity. For other reactions (e.g. lithium-halogen exchange) getting the temperature wrong can generate completely different products than the one you were trying for.


text:
why is entropy maximum at equilibrium for a closed system?
I don't understand why entropy is maximum at equilibrium for a closed system:(
I tried understanding some of the answers her on chemistry StackExchange but they involved the use of probabilistic functions and stuff and I find them difficult to understand.
SO can anyone explain this in simpler terms as I am a high school student:)

Can you explain, in your words, what entropy is?

Do you mean "isolated system"? Whose entropy are you maximizing? That of the system or that of the universe?

https://en.wikipedia.org/wiki/Closed_system

@Zhe i keep two ideas in mind :1.measure of disorder,2:Distribution of energy

Suppose that an ideal gas is held in a flask connected at another evacuated flask with a tap and let this whole arrangement be isolated. The system is at equilibrium and no heat will enter or leave. Naturally, the gas completely fills this first flask. Now open the tap and the gas escapes into the second flask but does no work as it expands into a vacuum. Again the system comes to equilibrium but the entropy has increased because the gas occupies a bigger volume. The entropy cannot increase any more as the gas fills the whole volume instead of only part of it. We conclude therefore that entropy is at a maximum at equilibrium. Experience shows us that the gas does not spontaneously go back into the first flask, which would reduce the entropy, the chance of this happening is so incredibly small it has never been observed.
Notes: The reason that the entropy increases with volume increase is that there are more ways of arranging molecules (i.e. more positions) in a bigger volume than in a smaller one. Work is $p\Delta V$ so if the external pressure $p=0$ no work is done.

always at equilibrium.