text:
Medroxyprogesterone

Hi, I have a question regarding my university assignment. In this case; it’s medroxyprogesterone. We all know that it’s a hydrophobic drug, hence, we are assigned to create a sterile injection in aqueous formulation. Aqueous means it has to be solubilized in water, but, the problem here is, medroxyprogesterone is poorly-water soluble but is really soluble in chloroform. So, I just wanna discuss about it here. How do we make it to be soluble in water (is it possible to do it so?) Thank you. And the questions below (the handwritten ones) I hope you can explain that also, theyre the things we have to present.. Help me, thank you!

That's a broad question.  But under https://www.merckmanuals.com/professional/appendixes/brand-names-of-some-commonly-used-drugs you find an entry about this drug in general, and a subsection about dosage forms which mentions some of the carriers teaming up for such a more lipophilic than hydrophilic molecule: methylparaben, polyethylene glycol, propylparaben.  If your API tends to form easily crystals with acids or bases, then formulation as co-crystals may alter the both the cinetics as well as thermodynamic solubility parameters quite a lot.

Medroxyprogesterone is a steroid. However, active drug is its acetate form, medroxyprogesterone acetate, which is sold under the brand name Depo-Provera among others (Wikipedia). Nonetheless, either is hydrophobic, and hence, should have prepared suitable way for administration to have better bio-availability. One of these methods are liposomal preparation (Ref.1). An easy loposormal preparation of a steroidal drug is summarized as follows (Ref.2):

The standard liposome formulation was prepared under aseptic conditions: Dipalmitoyl phosphatidylcholine ($\pu{6 mg}$), egg yolk phosphatidic acid ($\pu{1 mg}$), and cortisol palmitate ($\pu{3 mg}$) were dissolved in chloroform ($\pu{1.25 mL}$). This solution was rotary evaporated to dryness in vacuo at $\pu{55 ^\circ C}$, and the dried lipid film was hydrated at the same temperature by vortexing with $\pu{3 mL}$ sterile $\pu{5 mM}$ sodium phosphate buffered saline (PBS), $\mathrm{pH\: 7.4}$. After $\pu{1 h}$ at $\pu{21 ^\circ C}$ the liposome suspension was centrifuged at $\pu{50000 G}$ for $\pu{10 min}$, and the lipid pellet was resuspended in PBS ($\pu{2 mL}$). This procedure was repeated twice, and the cortisol concentration in the final suspension ($\pu{\approx 0 45 mg/mL}$) was determined by the fluorometric method of Zenker and Bernstein.

Further review for water insoluble drug delivery strategies: Read Refs.3-5.
References:

Chee Wai Wong, Bertrand Czarny, Josbert M. Metselaar, Candice Ho, Si Rui Ng, Amutha Veluchamy Barathi, Gert Storm, Tina T. Wong, "Evaluation of subconjunctival liposomal steroids for the treatment of experimental uveitis," Scientific Reports 2018, 8(6604), 11 pages (https://doi.org/10.1038/s41598-018-24545-2).
Nigel C. Phillips, D. Peter Page Thomas, C. Graham Knight, John T. Dingle, "Liposome-incorporated corticosteroids. II. Therapeutic activity in experimental arthritis," Annals of the Rheumatic Diseases 1979, 38(6), 553-557 (http://dx.doi.org/10.1136/ard.38.6.553).
Huabing Chen, Chalermchai Khemtong, Xiangliang Yang, Xueling Chang, Jinming Gao, "Nanonization strategies for poorly water-soluble drugs," Drug Discovery Today 2011, 16(7-8), 354-360 (https://doi.org/10.1016/j.drudis.2010.02.009).
Sandeep Kalepu, Vijaykumar Nekkanti, "Insoluble drug delivery strategies: review of recent advances and business prospects," Acta Pharmaceutica Sinica B 2015, 5(5), 442–453 (https://doi.org/10.1016/j.apsb.2015.07.003).
Jolanda M. van den Hoven, Sophie R. Van Tomme, Josbert M. Metselaar, Bastiaan Nuijen, Jos H. Beijnen, Gert Storm, "Liposomal Drug Formulations in the Treatment of Rheumatoid Arthritis," Molecular Pharmaceutics 2011, 8(4), 1002–1015 (https://doi.org/10.1021/mp2000742).


text:
Why do statistical effects impact stepwise stability constants?
I have a question regarding the stepwise stability constants of metal complex formation/substitution.
My lecture notes say that if we have a metal complex, say $\ce{[MX_{6}]}$, and a substitution reaction $\ce{[MX_{6}] + L -> [MX_{5}L]+X}$ with equilibrium constant $K_{1}$, and 5 other equilibrium constants $K_{2}$, $K_{3}$, ..., $K_{6}$ defined similarly for the successive substitutions of $\ce{L}$, then we have
$$K_{1} < K_{2} < K_{3} < \dots < K_{6} $$
One of the reasons given for this is that there are fewer positions available for complexation of the $L$ ligand after some substitutions have been made, and thus it is less likely for the successive collisions to occur. This is described as a "statistical effect". As far as I understand it though, the proability of a successful colision for a given reaction affects its rate, but not its thermodynamic properties, like the equilibrium constant. Am I wrong in this?
If not, why does this "statistical effect" affect the values of $K_{i}$?

Related: https://chemistry.stackexchange.com/q/112749 The equilibrium constant can be defined using the ratio of the forward and reverse rate constants for the reaction.

Important to note that ligand exchange in octahedral complexes is typically dissociative; i.e. an X leaves before an L can join. But I can’t get enough thoughts lined out to give an answer atm.

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text:
The reaction of an oxyacid with water
I read in my book this:
$$\ce{H2SO3 + H2O -> HSO3- +H3O+}$$
while I rather would write (not balanced):
$$\ce{H2SO3 + H2O -> 2 H+ + SO3^2- + 2 H2O -> SO3^2- + 2H3O+}$$
I mean I don't understand why the book writes only a partial deprotonated acid. Someone could explain this?

What is justification to ever write not balanced chemical reaction equation ?

The middle equation would presumably show intermediate forms. There is no free H+
ion floating freely in the water as an intermediate.

$\ce{H2A}$ and $\ce{A^2-}$ forms of a biprotic acid seldom coexist as major components, expecially if the acid is weak, unless $\mathrm{p}K_\mathrm{a1}$ and $\mathrm{p}K_\mathrm{a2}$ are similar. Instead, coexistence of major components $\ce{H2A / HA-}$ or $\ce{HA- / A^2-}$ occurs.
$\ce{SO3^2-}$ starts to form in about neutral or alkalic solutions, while $\ce{SO2}$ solution is acidic.
As $\mathrm{p}K_\mathrm{a2}=7.18$  - see polyprotic_acids
The wikipedia  article about Sulfurous_acid says:

Raman spectra of solutions of sulfur dioxide in water show only signals due to the $\ce{SO2}$ molecule and the bisulfite ion, $\ce{HSO3−}$. The intensities of the signals are consistent with the following equilibrium:
$\ce{SO2 + H2O <=> HSO3- + H+}$
$K_\mathrm{a} = 1.54 \cdot 10^{−2}$
$\mathrm{p}K_\mathrm{a} = 1.81$

Note that $\ce{H+}$ is just convenience for $\ce{H3O+}$, so it is rather  $\ce{SO2 + 2 H2O  <=> HSO3- + H3O+}$
See hydronium for complex chemistry of proton hydration.
Similarly, the strong acid $\ce{H2SO4}$ in not too diluted solution disociates just to $\ce{HSO4-}$. $\ce{SO4^2-}$ is a minor ion.

Kudos for pointing out the "real" equilibrium between SO2 and bisulfite. Problems ought to reflect reality so that the student is learning chemistry at the same time as how the mathematics of chemistry works.

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Should we expect Jahn–Teller distortion in bis(ethylenediamine)nickel(II)?

(ii) The stepwise stability constants of the following complexes in aqueous solution at $\pu{25 °C}$ are given below:
$$
\begin{array}{cll}
\hline
\ce{M} & \ce{[M(en)2(H2O)2]^2+} & \ce{[M(en)3]^2+} \\
\hline
\ce{Cu} & \pu{1E9} & 0.1~\text{(estimated)} \\
\ce{Ni} & \pu{1E6} & \pu{1E4} \\
\hline
\end{array}
$$

I got this data from an assignment problem. In the case of copper it makes complete sense for the stability constant of formation of $\ce{[Cu(en)_3]^{2+}}$ to be that low given that we have $\mathrm{d^9}$ configuration and strong Jahn–Teller (JT) distortion leading to strain in the rings formed in the complex.
But I can't figure out if there'll be any distortion in the $\ce{[Ni(en)_3]}$ ($\mathrm{d^8}$ configuration). Initially, I assumed there'll be no distortion and the complex would be satisfied with the chelation alone (compromising on the stability provided by JT distortion). But then some of my friends pointed out that $K$ reduces from $10^6$ to $10^4$ so there must be some kind of distortion. After some googling, the only article with some data I could find is by Sohail et al. [1].
In the linked article, there are no comments on JT distortion. Instead, packing effects have been held responsible for the natural distortion of the complex (the last line under Comment section). But I am not entirely convinced; so my questions are:

Would the $\mathrm{e_g}$ orbitals be symmetrically or asymmetrically filled?

If asymmetrically, then why is the JT distortion so profound in the copper(II) complex but weaker in the nickel(II) complex? And if it's symmetrically filled, then what is responsible for the decrease in the stepwise stability constant?

References

Sohail, M.; Molloy, K. C.; Mazhar, M.; Kociok-Köhn, G.; Khosa, M. K. Tris(ethylenediamine)nickel(II) tetraiodocadmate(II). Acta Cryst E Struct Rep Online 2006, 62 (2), m394–m396. https://doi.org/10/b82dng. (Open Access)

In the $d^9$ complex the $e_g$ orbitals are equally split in energy (one up , one down by same amount) by the Jahn Teller effect causing an overall lowering in energy, i..e. 2 electrons lowered and one increased in energy. In $d^8$ one electron is in each $e_g$ orbital so if split there is would be no change in overall energy so this will not happen. (The $t_{2g}$ are also split but as they are full there is no overall energy change).

@porphyrin what do you mean by "if split there is would be...." ?

The $eg$ is doubly degenerate and splits by a very small amount (far, far less than $e_g \to t_{2g}$ split) into two levels (one up and one down by same amount) with a perturbation such as Jahn Teller. (Splitting is far less than spin pairing energy). One electron in each level has the same energy as if no perturbation is present.

There is a key difference between the $\mathrm d^8$ state and the $\mathrm d^9$ state which lies in exactly that extra electron. The result of the extra electron is that instead of nickel(II)’s $\mathrm{(t_{2g})^6\ (e_g^*)^2}$ state copper(II) has a $\mathrm{(t_{2g})^6\ (e_g^*)^3}$ state. There are two $\mathrm{e_g^*}$ orbitals, thus occupying each with a single electron with parallel spins leads to no degeneracy problems. An additional electron will mean that one of the two $\mathrm{e_g^*}$ orbitals is doubly occupied while the other is singly occupied—but this is a degenerate state as the energy is the same whether it is $\mathrm d_{z^2}$ or $\mathrm d_{x^2-y^2}$ that is doubly occupied.
This degeneracy of possible states in copper(II)—or any other $\mathrm d^9$ system—is the underlying reason for Jahn-Teller distortions. By lowering symmetry from $O_\mathrm h$ to $D_\mathrm{4h}$, these two orbitals are no longer degenerate and they can (and must) diverge in energy. Thus, you have a clear ‘winner’—the lower-energy orbital which receives an electron pair—and a ‘loser’—the higher-energy orbital which receives the single electron. Looking only at $\mathrm d_{z^2}$ and $\mathrm d_{x^2-y^2}$ it doesn’t matter whether the symmetry reduction is performed by moving the $z$-ligands further away or closer but it does make a difference for the other three d orbitals ($\mathrm d_{xy}, \mathrm d_{xz}, \mathrm d_{yz}$) for which lengthening the $\ce{M-L}$ bonds along the $z$ axis proves favourable overall.
A $\mathrm d^8$ system at first glance has nothing to gain by this distortion. Rather than having a stabilisation that benefits two electrons at the expense of one, it can only stabilise one electron at the expense of one. This is energetically not favourable as there is nothing to be gained (but probably something to be lost) so a ‘simple’ Jahn-Teller distortion does not occur; nickel(II) prefers to remain octahedral in $\ce{[ML6]}$ systems.
Note however that there is one possibility when such a stabilisation is beneficial: if the ‘distortion’ is of such a large magnitude that the energy difference between $\mathrm d_{x^2-y^2}$ and the next-lowest orbital is larger than the spin-pairing energy. In this case, we could say that nickel(II) drops from a high-spin state to a low-spin state, although these terms are not usually used in this context. Geometrically, this ‘distortion’ is so significant that we term the resulting complex square planar rather than distorted octahedron; in general there is no longer any evidence of coordination bonding along the $z$ axis. A number of nickel(II) complexes with strong mid-field or strong-field ligands are indeed square planar for this reason such as $\ce{[Ni(dmg)2]}$, the dimethylglyoxime (or diacetyldioxime) complex.


text:
When disolving a iron pill for titration, why do I use sulfuric acid instead of nitric or hydrochloric acid?
In a titration to decide the iron content of a pill we dissolved the iron in sulfuric acid and then titrated it with cerium(IV) sulfate. But I just wondered what's wrong with using nitric and hydrochloric acid in this scenario?

Hint: Ceric sulfate is a strong oxidizing agent, and so is nitric acid. HCl can be oxidized. Can you think of interferences in redox titrations?

@ M. Farooq Not really sure what to look for. If i used nitric acid it would already oxidise my iron to Fe3+ ions i suppose?. And if i were to guess when using  HCL the Ceric sulfate would react with the CL- ions.? not really sure

Yes this sounds plausible.

In this case, cerium(IV) sulfate is a strong oxidising agent, hence, using $\ce{HCl}$ would likely be oxidised to form $\ce{Cl}$ of higher oxidation state. Therefore, the titre would be vastly higher than expected.
$\ce{HNO3},$ on the other hand, is an oxidising agent and would oxidise $\ce{Fe}$ to a higher oxidation state than expected (of $\ce{Fe^3+}$). Hence, not as much cerium(IV) sulfate would be needed to reach the endpoint, resulting in a titre lower than expected.
Therefore, $\ce{SO4^2-}$, being a relatively unreactive species, is the best reactant in this case.


text:
Is an ion or a polar molecule more attracted to water in general?
I realize that there are exceptions but as a general rule of thumb, is an ion or a polar molecule attracted to water? I think that it would be ions since the charge is much larger than the net dipole of polar molecules but I can't find anything definitive regarding this.

An ion or dipole is not 'attracted to water' in the sense that a piece of iron is attracted to a magnet. Possibly what you ask is whether a polar molecule or an ion is energetically more favourable to dissolve in  a  polar solvent (water, acetonitrile etc) vs. a non-polar solvent, such as hexane or benzene.

@porphyrin Hm, If we consider force being proportional to the gradient of potential energy=f{distance),, ions and water molecules are mutually attracted. It is similar as for atoms forming bonds. But this attraction is short range and weakens rapidly with distance.

@Poutnik,  I think that I had my pedants hat on;  of course you are right but I meant attracted to water molecules not water per se. In a liquid there always has to be multiple attractions otherwise its a gas/vapour, so I thought that thinking about energy was more helpful.

In general, ions are more attracted to water ( in sense of favourable energy state), due bigger and more localized charge, compared to polar molecules.
But, ions are often even more than to water attracted to other molecules or ions, forming charged or neutral complexes, or precipitates.


text:
Comparing ionisation energy between an element and an ion

Choose one that has the biggest ionization energy:

$\ce{Br-}$ and $\ce{Kr}$
$\ce{Cl-}$ and $\ce{Ar}$
$\ce{Cl}$ and $\ce{Cl-}$

I am not sure how I can compare the ionization energy between an ion and an element.

With few execptions (e.g. $\ce{H+}$) where removing an electron obviously doesn't make sense, think of the atoms or ions like box which does contain electrons which after removal of an electron has one additional positive charge.  This approach does not care if you start with a neuter atom like Kr, an anion like $\ce{Cl-}$ or even a cation which already donated an electron earlier (e.g., $\ce{K+}$ or $\ce{Ca^{2+}}$).
You then compare initial state and final state, e.g. for $\ce{Br-}$ vs. $\ce{Br}$ yielding a difference in energy A, and the difference B while comparing $\ce{Kr}$ vs. $\ce{Kr+}$.  Because eventually you either compare difference A against difference B based on experimental data, or you balance your argument with which of the two differences is greater, requiring more energy to extract this one electron.  Possible criteria may be: reaching / departing a neuter state, breaking / disobeying octet rule, radii of the atoms / ions vs. how many charge the reactant / product ion / atom has to accomodate.  Sometimes, a single criterion dominates the argumentation; sometimes you need multiple criteria to consider.
Many ionization energies are tabulated (e.g., here), so you may use these values as a test bed for your argumentation.


text:
What is the electrochemical reaction of this battery?
I've watched a YouTube video "Soda Can Battery - Power From Trash - DIY Project" by Source Port Services. I understand that the anode can be oxidised, but what is the reduction reaction at the cathode? They are using a salt water solution $\ce{Na+}$ $\ce{Cl-}$ and a copper cathode.
$$
\begin{array}{cc}
\hline
\text{Half-reaction} & E^\circ/\pu{V} \\
\hline
\ce{Cu^2+ + 2 e− <=> Cu(s)} & 0.337 \\
\ce{Na+ + e- <=> Na(s)} & -2.71 \\
\ce{Al^3+ + 3 e− <=> Al(s)} & -1.662 \\
\hline
\end{array}
$$
Thus we can see that the aluminium is oxidised, but there are no $\ce{Cu^2+}$ ions to be reduced also $\ce{Na+}$ reduction would lead to a negative $E_\mathrm{cell}$ standard and a positive Gibbs energy. Is it an air cathode where $\ce{O2}$ is being reduced?

I took a liberty to convert a list of reduction potentials for elements into a table. Standard potentials determined for half-reactions, not elements.

Thus we can see that the Al is oxidised but there are no Cu2+ ions to
  be reduced also Na+ reduction would lead to a negative Ecell standard
  and a positive Gibbs. Is it an air cathode where O2 is being reduced?

Well, there is no free lunch in this world. This is not a conventional cell in which each electrode is sitting in its own ions and one can easily calculate their Nernst potential. What we are probably seeing here is the very concept of contact potential. When two dissimilar metals are brought into contact via some electrolyte, a potential difference appears between the two. Recall the first experiment of Volta. He made a long series of
-silver[wet cardboard]zinc-silver[wet cardboard]zinc-silver[wet cardboard]zinc- cells, where the cardboard was wetted with salt-water. He was able to generate large potential differences this way by making very long series of Ag-Zn couple. Is this set-up good for drawing current-probably not.
These experiments were done in the 17th century by Volta. Atoms and ions did not exist then. These are contact potentials and you do not need a redox system to begin the experiment (metal dipping in its salt). Let us say you start with Al, Al will start to dissolve in a corrosive medium (salt is very corrosive) to form Al ions. You have a valid point, there are no copper ions to reduce? What else do we have in the system? Water, it has some ionized $\ce{H+}$, they start to reduce, one should see hydrogen bubbles at the copper electrode. Certainly you can draw small currents to power a LED from such batteries but not for long.
I do have some doubt about the design of the experiment. All soda cans are lined with a very very thin polymer layer. So the contact of Al is not that good with the copper coil. If one scrubs the can very well, only then that polymer layer would come off. The claim of seven LEDs is a stretch.

But this setup is being used to power 7 LEDs and a potato battery can be used to power a motor but it too has no ion to reduce. I am still wondering what is the reduction reaction?

@ChemEng, I edited the answer to clarify your query.

I actually made a mistake he said 9 LEDs and it lasted for a week plus in the vid

M. Farooq pointed to the cathodic reduction: H2O + e-  --> OH- + H. ; then 2H.  --> H2. In a dry cell, the H2 bubbles (or H.) would be oxidized to H2O by MnO2 to prevent pressure build-up. In these soda can cells, if the sand is not saturated, some air might get to the Cu wire and be reduced to water: 4H. + O2 --> 2H2O.


text:
Why is the bottom white in a lucas test?
Consider this positive result from the Lucas test (adding $\ce{ZnCl2/HCl}$ to a tertiary alcohol):

My question is quite straightforward but I can't find the answer. What is in the top layer (thin and clear) and what is in the bottom milky layer? Also, what is making it milky?

Do you know your $3^\circ$-alcohol? How much of it you used for the test? Do you used any organic solvent to dissolve it? Your answers might be within answers to those questions.

The solution might be tertiary alkyl chloride.

This was not from my experiment - but say 2-methylpropan-2-ol and 3mL?

Your picture is from Wikipedia. Hence, I 'm strongly suspect this as a homework question.

Yes it is from wikipedia - why does this mean it's a homework question? I am simply curious what is it that is making it white/cloudy. The picture is for illustration purposes. Even if it WAS a homework question - it still asks for conceptual reasoning which is allowed by the site's policy

I think the bottom layer is the haloalkane, because we're switching the $\ce{OH}$ group of mass $\pu{17 u}$, with chlorine which is $\pu{35.45 u}$. This makes the chloride more dense than the water, which is also formed in the process
$$\ce{ROH + HCl ->[ZnCl2 catalyst] RCl + H2O}$$


text:
Is m-cresol or p-cresol more reactive towards electrophilic substitution?

Which is more reactive towards electrophilic substitution, meta-cresol or para-cresol?

Both oxygen and methyl are activating groups. I don't know on what basis do I compare their reactivity. There's just a change in the position of one of the group.
My book says meta-cresol is more reactive and the reason given was

If two groups are working for the same effect at the same time, then their effect decreases.

I don't understand the reasoning and hence the answer. Any help would be appreciated.

Mono- and double-substitutied benzenes react in pattern of preference during electrophilic substitutions.  Assuming ortho-cresol, you still have four different sites where the reaction may happen.  So we speak about which position(s) here?

@Buttonwood Well, nothing was mentioned about the positions in the question. It just asked which is more reactive towards electrophilic substitution. Maybe it considers all the positions to be equivalent(At least for answering this question). By the way, we are talking about meta-cresol and not ortho-cresol.

@Tony:  Agreed about my err on o, m, p-cresol.  In terms of positions and their relative ease to react, I had in mind a scheme like «rates on nitration» on https://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/benzrx1.htm.  Unable to localize a source now, I recall there are similar ones for already double-substituted benzene (like the cresols) undergoing a third electrophilic substitution since sometimes the influence of the two may be synergistic.

Can anybody explain why is my post on "hold"?

@Tony Probably these close-votes accumulated during figuring out which cresols to compare. Since now it's been sorted out, I just brushed up formatting a little; could you please add a reference for "your book"?

@andselisk I fear it's not available online. So I can't provide a link. It's my school textbook though.

@Tony A reference/source is always required when something is quoted, otherwise this may be considered a plagiarism. It doesn't matter whether the textbook is available online or not. Every textbook should at least have an author, title, publisher, year and page numbers.

To start, it is important that you understand that the methyl and hydroxy groups are activating only in the ortho and para positions. For example, if I were to substitute plain phenol, which contains an activating hydroxy group, which position would my electrophile (in this a bromine electrophile) add to? You can see that the intermediary sigma complex involves a positive charge. When the bromine attacks at the ortho or para (I did not include a diagram for the para attack, but it is similar to the ortho attack) positions, it is stabilized by an additional resonance structure provided by the oxygen lone pair. In the case of the meta attack, as you can see in the diagram below, this does not happen, and the intermediate is left with one less resonance structure.

Most activating groups will have the same directing effects as shown with the activating hydroxyl group above (para-ortho directors), so a methyl group, which contains no lone pair (and thus no additional resonance structures) will still stabilize the intermediate by donating electron density in much the same way as if an additional resonance structure was present.
Now getting back the para and meta cresol, the meta cresol contains two activating groups

Both of these activating groups activate the same positions, so this combination of activating effect will make the meta cresol more reactive than para-cresol. In para-cresol, the two groups do not work to activate the same positions, so, no matter which carbon is substituted, it will not be activated by both groups simultaneously like in meta-cresol.


text:
Why when deriving the Clapeyron equation can you assume that dH = dq, as surely the pressure is not constant?
Part of the Clapeyron equation involves replacing $\Delta S$ with $\Delta H/T$. I can see the step from $\Delta S$ to $\Delta q/T$ but I'm not sure why you can replace $\Delta q$ with $\Delta H$. From my understanding $\mathrm dH = \mathrm dq_V +V\mathrm dp$ when assuming no work other than $pV$ work is done. I don't understand how you can substitute $\Delta H$ for $\Delta q$ without assuming that the pressure is constant?

Heat absorbed at constant volume is change in internal energy i.e dU=nCvdT =qv this relation is true even when volume is not constant similarly change in enthalpy is heat absorbed at constant pressure .im yet to find the reason how Above 2 facts are true , but im quite sure that they are true. I was thinking of asking this question myself.

It is also interesting to note that mayers relationship is derived by stating the above mentioned facts to be true from dH = dU + Vdp -> Qp =Qv + nRdT -> nCpdT = nCvdT + nRdT -> Cp =Cv -R

Thank you, I understand that for a fixed volume ∆u = q assuming no non PV work is done, but I don't think ∆H is always equal to ∆q, or if it is I am not sure why. All of the textbooks I've read state that for a phase transition ∆H is ∆q but I am not sure if this is always the case, or if it is the case I am not sure why

At any given temperature, the change from a liquid to a vapor occurs at constant pressure.  What we are trying to find with the Clapeyron equation is how this constant pressure changes as the given temperature changes.
At equilibrium at a given temperature and constant pressure, we have $$G_{vapor}=G_{liquid}$$ and $$S_{vapor}-S_{liquid}=\frac{(H_{vapor}-H_{liquid})}{T}$$
Now, if we want to find out how temperature and pressure change in tandem along the equilibrium line, we can write:$$dG_{vapor}=-S_{vapor}dT+V_{vapor}dP$$and
$$dG_{liquid}=-S_{liquid}dT+V_{liquid}dP$$Since, along the equilibrium line, we must have $dG_{vapor}=dG_{liquid}$, we can write:$$-S_{vapor}dT+V_{vapor}dP=-S_{liquid}dT+V_{liquid}dP$$or equivalently:$$\frac{dP}{dT}=\frac{(S_{vapor}-S_{liquid})}{(V_{vapor}-V_{liquid})}=\frac{(H_{vapor}-H_{liquid})}{T(V_{vapor}-V_{liquid})}$$This is the Clapeyron equation.

From the definition of the Gibbs free energy $G=H-TS$,
$$\Delta G = \Delta H - T \Delta S \tag{constant T}$$
For a pure substance, equilibrium between two phases (such as for a point on a coexistence line) reduces the degrees of freedom to one, such that p is constrained by the choice of T, or vice-versa.
Since the condition for equilibrium at constant T and p is
$$G_{m,\textrm{phase 2}}-G_{m,\textrm{phase 1}}  = \Delta G_m =0 \tag{constant T and p}$$
where the molar Gibbs free energy $G_{m,i}$ is equal to the chemical potential $\mu_i$, it follows that on any point on a two-phase coexistence line $$\Delta H_m = T \Delta S_m $$
Note also that only for a reversible process $\Delta S = q/T$, and the reversible phase transitions in question occur at constant T and p, such that $\Delta H = q$.

The derivation of the Clapeyron equation starts from the premise that two points on a two-phase coexistence line share the same difference in chemical potential ($\Delta \mu =0$ for all points on the line), therefore if you move between two neighboring points on the line you must satisfy $d\mu_1=d\mu_2$. From this equality you can derive the equation by noting that $d\mu=V_mdp -S_mdT$.

Hi, thank you for answering my question, you explained it very clearly. Would you mind showing me how that fits in with the rest of the derivation of the Clapeyron equation as I am still unsure as to how the assumptions fit in when you change the pressure and temperature to find the gradient of the phase diagram?


text:
Acid-base titration: Unknown base in burette or flask?
While finding the concentration of $\ce{NaOH}$ via titration with $\pu{0.05M}$ Oxalic acid solution (in our lab) we put NaOH in the burette whereas added phenolphthalein to oxalic acid in titration flask. But in this video the demonstrator does just opposite ( i.e., $\ce{NaOH}$ in flak with phenolphthalein). I am confused in deciding which one might be correct. My view is that either would do the job but I need some clarification.

Both are perfectly correct methods as long as the volume measurements are accurate.

@M.Farooq - I disagree. It is easier to detect the change from colorless to colored. So, if possible, the titrant should be a base when phenolphthalein is the indicator.

No matter what you do students will always come to you holding the flask asking if this is the right end-point :-) For a skilled analytical chemist, both techniques should be fine and of course colorless to color is easier to detect. One is less painful than the other.

The base should be in the burette, and the acid in the flask. With this technique, the point of equivalence is obtained with a rather high degree of accuracy.

This is not a universal practice in analytical chemistry. For example, if we wish to determine NaOH contaminated with some sodium carbonate, one has to use HCl in the buret, and two use two indicators, NaOH solution is in the flask.


text:
Preparation of Ca/PO4/acetate solution
EDIT: This is not a homework problem and we need this for an exploratory study on teeth demineralization. There is further work planned around this activity, which may not be relevant to the question.
For this purpose, few publications have quoted the use of Ca/PO4/acetate solution containing 2mmol/L calcium, 2 mmol/L phosphate and 0.075 mol/L acetate at pH 4.5. This composition will supposedly emulate tooth demineralization environment closely.
We are a team of computer and electrical engineers with rusty basic (school level) chemistry skills. We tried to prepare few equations considering commercially easily available compounds e.g. calcium phosphate, calcium acetate, calcium oxide etc. But, we could not derive any balanced combination that could result in mentioned desired concentrations.
We avoided the use of elemental calcium, primarily because we wanted to avoid handling of such reactive element as it may further necessitate special lab equipment. But, if this is the only way to go, then please let us know.
So, can anybody help us with the method to prepare such a solution, if possible using compounds/mixtures available commonly in the market?
Acknowledging that this can very well be an artefact of our rusty and limited chemistry knowledge, so we are welcoming hints/pointers/keywords/references as well.
Appreciate your time and attention to our request.

So you want to prepare a solution containing 2 mmol/L calcium, 2 mmol/L phosphate and 0.075 mol/L acetate at pH 4.5. It is not an easy problem. You may dissolve 2 mmol CaCl2 (0.111 g), 2 mol NaH2PO4 (=0.283 g), and 0.075 mol acetic acid ( 4.5 g). If the pH is not 4.5, it may be adapted by adding some drops of NaOH 0.1 M ou HCl 0.1 M.  But take care of the fact that instead of getting anhydrous compounds CaCl2, NaH2PO4, you may get hydrated powders, like CaCl2·6H2O, or NaH2PO4·2H2O. In that case the masses have to be increased to take care of the number of water molecules..

So you want to prepare a solution containing 2 mmol/L calcium, 2 mmol/L phosphate and 0.075 mol/L acetate at pH 4.5. It is not an easy problem. You may dissolve 2 mmol CaCl2 (0.111 g), 2 mmol NaH2PO4 (=0.283 g), and 0.075 mol acetic acid ( 4.5 g). If the pH is not 4.5, it may be adapted by adding one or two drops of NaOH 0.1 M ou HCl 0.1 M.
But take care of the fact that instead of getting anhydrous compounds CaCl2, NaH2PO4, you may only obtain bottles of hydrated powders, like CaCl2·6H2O, or NaH2PO4·2H2O. Anhydrous substances are not always easy to get. In that case the masses have to be increased to take care of the number of water molecules. The molar masses of the different hydrates are : 178 for Na2HPO4·2 H2O, 268 for   Na2HPO4·7H2O, 358 for Na2HPO4·12H2O, 147 for CaCl2·2H2O, and 219 for CaCl2·6H2O.

Thanks Maurice for your prompt, detailed and very helpful response. Your response definitely provides lot of clarity on the problem and a possible solution.
Fortunately, I can see that mentioned anhydrous compounds are available for sale on websites of local suppliers.
We will give this a try and update how it goes.


text:
Gibbs energy derivative vs equilibrium for aqueous reaction
I am trying to find the Gibbs-energy equilibrium for the reaction $$\mathrm{A}+\mathrm{B}\rightleftharpoons\mathrm{C}.$$ According to most sources the equilibrium concentrations in an ideal solution should form a ratio $$\Delta G^0=-RT\ln\frac{[\mathrm{C}]}{[\mathrm{A}][\mathrm{B}]}.$$
I am having trouble seeing this when the stoichiometric coefficients on either side of the reaction are unequal, as in this example above.
The Gibbs energy for the solution should be $$G=n_A(\mu_A^0+RT\ln[\mathrm{A}])+n_B(\mu_B^0+RT\ln[\mathrm{B}])+n_C(\mu_C^0+RT\ln[\mathrm{C}]).$$
Assuming we started with a 1 L solution containing 1 mol of A and 1 mol of B, $\xi$ mol of each react to form $\xi$ mol of C, this equation should then be $$G=(1-\xi)(\mu_A^0+RT\ln(1-\xi))+(1-\xi)(\mu_B^0+RT\ln(1-\xi))+\xi(\mu_C^0+RT\ln\xi).$$
My understanding is that if the Gibbs free energy is at its minimum the reaction above will not be spontaneous in either direction and so the solution will be at equilibrium. The derivative of the above equation gives $$0=-(\mu_A^0+RT\ln(1-\xi))-RT-(\mu_B^0+RT\ln(1-\xi))-RT+(\mu_C^0+RT\ln\xi)+RT.$$
Some rearrangements result in $$\Delta G^0=-RT\ln\frac{[\mathrm{C}]}{[\mathrm{A}][\mathrm{B}]}+RT.$$
In other words, my expression is off by $\Delta nRT$. Having been unable to find any help on this matter on the internet, I feel I have either made some silly logical/mathematical mistake or have some fundamental misunderstanding. Please point out the error.

Maybe the free energy of the solvent changed because the concentration of solute increased, see https://chemistry.stackexchange.com/questions/112430/what-is-the-gibbs-energy-of-dissolving-a-solid-in-pure-water

To derive the Gibbs' free energy of an ideal solution you would invoke Raoult's law, which says that
$$\mu_i(l) = \mu_i^*(l)+RT\ln\left(\frac{p_i}{p*}\right)= \mu_i^*(l)+RT\ln\chi_i$$
Here the asterisk refers to the pure substance. Using the notation in the OP, the total number of moles of reagent and product in solution is
$$n_{tot}=2(n_{ini}-\xi)+\xi+n_{solv}=2n_{ini}-\xi+n_{solv}$$
where $n_{ini}$ is the initial amount of A or B (assumed equal), so that one can write for reagents A and B:
$$\mu_i(l) = \mu_i^*(l)+RT\ln\left(\frac{n_{ini}-\xi}{2n_{ini}-\xi+n_{solv}}\right)$$
and for product C:
$$\mu_C(l) = \mu_C^*(l)+RT\ln\left(\frac{\xi}{2n_{ini}-\xi+n_{solv}}\right)$$
There is also a term for the solvent:
$$\mu_{solv}(l) = \mu_{solv}^*(l)+RT\ln\left(\frac{n_{solv}}{2n_{ini}-\xi+n_{solv}}\right)$$
which I ignored in earlier edits.
Adding terms to compute the total Gibbs energy and taking the derivative wrt the extent of reaction $\xi$ then results in
$$\begin{align} \Delta G' &= \Delta \mu^* + RT \ln\left(\frac{\xi(2n_{ini}-\xi+n_{solv})}{(n_{ini}-\xi)^2}\right)  \\  &= \Delta \mu^* + RT \ln\left(\frac{n_Cn_{tot}}{n_A n_B}\right)  \\&= \Delta \mu^* + RT \ln \left(\frac{\chi_C}{\chi_A \chi_B}\right) \end{align} $$
($^\dagger$)
Note that here the standard state for each substance refers to the pure liquid in equilibrium with its vapor at the same temperature. To obtain an expression relating a change in standard free energy to the concentration equilibrium constant requires expressing mole fractions in terms of concentrations. This is generally done for dilute solutions where $n_{solv} \approx n_{tot}$ by writing $\chi_i\approx c_iM_{solv}/\rho_{solv}$, which leads to
$$\Delta G' = \Delta \mu^* - RT \ln \left( \frac{M_{solv}}{\rho_{solv}} \right) + RT \ln \left(\frac{c_C}{c_A c_B}\right)  $$
Then by redefining the standard states as
$$\mu_i^\circ = \mu_i^* + RT \ln \left( \frac{M_{solv}}{\rho_{solv}} \right) $$ leads to $$\Delta G' = \Delta \mu^\circ + RT \ln \left(\frac{c_C}{c_A c_B}\right)  $$
If you think the standard state looks strange, consider that if $\chi_i = c_iM_{solv}/\rho_{solv}$ then $$\chi_i(\pu{c^\circ_i= 1 M}) = c_i^\circ \frac{M_{solv}} {\rho_{solv}}$$ and so we can write
$$\mu_i^\circ = \mu_i^* + RT \ln \left( \chi_i(\pu{c^\circ_i= 1 M}) \right) $$ which is more in line with canonical definitions of standard states. The new standard state is defined as a 1 M ideal solution with chemical potential related to that of the pure solute by the above equation.

$^\dagger$/TL;DR :  The mysterious extra $RT$ term in the OP is gone. It's important to keep a term for the solvent. And a lesson from these repeated edits: careful doing math when tired!

That would explain it. It seems odd to use the ideal gas law for a liquid; I suppose it is unfair to assume that the volume is constant in a liquid chemical reaction? If I had an open beaker of water and simply stirred in some A and B to make an ideal solution, what would be a more valid way to derive the equilibrium relation? A gas phase reaction having constant pressure is easier for me to visualize, but I am having trouble conceptualizing a solution.

@ErwinDerek I will alter my answer a little later to show how to directly obtain the expected equation.


text:
Alkene reaction with acid via SN1 reaction
I want to ask a question about alkenes and their reactivity with acids.
From high school studies, I recall that if you had an acid $\ce{H3O+}$ i.e. $H^+$ that it would react with an alkene as shown below:

However, when I was given a lecture today on Allylic alcohols in a $\ce{S_{N}1}$ reaction, the acid does not react with the alkene at all:

As you may understand, attempting to google or find an equivalent mechanism to explain this observation didn't yield a successful result.
The two mechanisms in my opinion appear to contradict each other, but I couldn't work out how the second mechanism for the Allylic Alcohol functioned whereby the alkene did not react with the acid. My assumption was that it was easier to protonate the alcohol group than react with the acid (with maybe a reasoning along the lines of pKa) but I couldn't find anything sufficient to justify my idea.
Why does the acid not react with the alkene group in the allylic alcohol shown above?

My suspicion is that the resonance structures accessible to the allylic carbocation provide a lower energy state than the formation of the C--H single bond and the isolated carbocation that your expected pathway would require.
In order of reactivities, you are right to think that the alcohol is the controlling factor because of its pKa. Under acidic conditions it will behave more as a base than the alkene (It is much more basic than an alkene, or think of the Oxygen in an alcohol as being generally a stronger nucleophile than the alkene). This was one main consideration my O-Chem professor stressed was for us to be aware of how other functionalities in our molecules would behave under the reaction conditions of other steps (Eg, if going into a strongly acidic/basic environment after installing a reactive group). I am not an organic chemist by training, but I suspect that if you wanted to access the alkene reactivity, you would want some kind of alcohol protecting group.
This page has a brief discussion about three classes of nucleophile (and the next page describes what makes a good nucleophile): https://www.masterorganicchemistry.com/2011/03/04/the-three-classes-of-nucleophiles/

Here the issue in question is about rate. Of course, in a real life scenario, both reactions probably take place, the question is which reaction is faster?
Well, looking at the reactants here, you are having a strong acid. The hydroxy group is way more basic/nucleophilic than the double bond simply due to electronic effects. Hence, the Sn1 substitution is expected to be faster and more kinetically favourable. Therefore, though alkene might be hydrated, the predominant reaction is expected to be the Sn1 reaction.


text:
Electrolysis of calcium hydroxide in glycerol
Calcium hydroxide is soluble in glycerol. Assuming anhydrous conditions, could such a solution be electrolysed to yield calcium metal, or would glycerol OH groups interefere with the process?

There would be even equilibrium with some water created!

Calcium hydroxide is probably slightly soluble in glycerine due to tight chelation of Ca++ ion by the three hydroxyls of (probably 2) glycerine(s). But the hydroxyls will be present too, and on hydrolysis, will yield O2 plus H2O.
But beyond that, imagine you produce some calcium metal: it will be in glycerine, which has hydroxyl groups that will react with the Ca to produce H2. It could be a competition where some conditions allow production of measurable amounts of Ca metal, but it doesn't look promising. PEG 400 (polyethylene glycol with a mw of 400) has been used as a solubility enhancer for Ca(OH)2 (similar to glycerine), but has many fewer hydroxyls, and might do the job better. https://www.mdpi.com › pdf yields a pdf with more information.
Glycerine and PEG seem to be used to increase solubility of Ca(OH)2 in water rather than in the other solvent alone. I suspect increased viscosity is an issue.
Another way to get some calcium would be to use mercury as the cathode; then calcium will deposit and amalgamate (I'm pretty sure. I've done it with sodium and ammonium, but not with calcium.).


text:
Does Amoxicillin (as trihydrate) degrade in stomach acid?
I crushed my antibiotics (amoxicillin + potassium Clavulanate). Does it deform in the stomach acid and therefore become useless? Or does the acid have no effect and the amoxicillin goes directly into the intestines and blood stream subsequently?

Because amoxicillin is available as oral tablets, it can be ascertained that it is not significantly degraded to a useless form in the stomach.  In fact, it has an oral bioavailability of 95% (meaning 95% of the orally ingested material goes into the blood stream rather than being destroyed in the stomach). The term for an oral medication being destroyed before it enters the blood stream is "first pass metabolism". So oral amoxicillin can be said to undergo minimal first pass metabolism. Oral potassium clavulanate also has a high oral bioavailability.

So, crushing a amoxicillin tablet which has an extended release profile will just mean that we are overdosing amoxicillin instead of deactivating the API ?

I'd say that's a proper interpretation.

The accepted answer is correct in some respects, though the assertion that an oral dose can be assumed to be safely crushed is only generally true. The Institute of Safe Medication Practices maintains a list of medications that can't be crushed (available for "free" on some sites, but with overlap in lists like this one). These medications are usually formulated in a way that they will slowly release a large dose over time or are coated to protect the drug from stomach acid ("enteric coated").
Importantly, the amoxicillin/clavulanate you crushed (Augmentin is a common brand name of the medication) is often given in extended release form containing 875 mg amox and 125 mg clav, and should not be crushed *. If that is the case, I would discuss the specific changes to dosing with your doctor or pharmacist that this would entail; generally it's not a big deal, but this site is not the proper forum to advise you.
* From your comment on the accepted response, it seems you surmised this yourself. However, because of some important info left out of the accepted response, I leave this answer as is.

Indications that crushing is OK are if the info sheet says: "you may dissolve / prepare a suspension", "chewing tablet". Info sheet saying that you may halve the tablet is so-so in that respect: in most cases it's OK to cut it into smaller parts but not in all. Indications that crushing is strictly not OK are "stomach acid resistant coating", "do not divide", "sustained release tablet" or "retard" in the name.

@cbeleitessupportsMonica True. If it doesn't say you can/may crush it, it probably shouldn't be crushed. To elaborate on the splitting being a so-so indication, I'd say it's a poor-to-bad indication; if it is scored (marked with lines to indicate breaking points), splitting there is encouraged, but one should presume splitting outside that area (e.g., by further splitting or crushing) is not quality tested or recommended. In general, just take your oral doses with a bit of water and don't do anything extra the info sheet doesn't say you can do; or chat with your pharmacist :)


text:
Predicting a precipitate based on Ksp

$\pu{105 mL}$ of $\pu{0.10 M}$ $\ce{AgNO3}$ is added to $\pu{125 mL}$ of $\pu{0.35 M}$ $\ce{K2CrO4}.$ Will a precipitate form? $(K_\mathrm{sp}$ for $\ce{Ag2CrO4}$ is $\pu{1.12E-12}.)$

Answer:

 Yes, $Q = \pu{4.0E-4}$ which is greater than $K_\mathrm{sp}.$

Can someone please help me out with this question? I know that I have to find $Q_\mathrm{sp}$ and compare this to the $K_\mathrm{sp}$ given, and if $Q_\mathrm{sp}$ is bigger, then a precipitate will form. However, I do not know how to find $Q_\mathrm{sp}$ when two substances are mixed together.

Assume the volumes of the solution are additive. This isn't absolutely true, but close enough given the relatively poor exactness of the Ksp's.

To obtain $Q_\mathrm{sp}$ for
$$\ce{Ag2CrO4 <=> 2 Ag+ + CrO4^2-}$$
$$K_\mathrm{sp} = [\ce{Ag+}]^2[\ce{CrO4^2-}]$$
$$Q_\mathrm{sp} = c(\ce{Ag+})^2\cdot c(\ce{CrO4^2-})\label{eqn:1}\tag{1}$$
you need to use new concentrations of ions in solution $c(\ce{Ag+})$ and $c(\ce{CrO4^2-})$, which can be found from the corresponding values for initial concentration $c_0$ and volume $V_0$:
$$
\begin{align}
c(\ce{Ag+}) &= \frac{c_0(\ce{AgNO3})\cdot V_0(\ce{AgNO3})}{V_0(\ce{AgNO3}) + V_0(\ce{K2CrO4})} \\
  &= \frac{\pu{0.10 mol L-1} × \pu{105 mL}}{\pu{105 mL} + \pu{125 mL}}\\
  &= \pu{4.57E-2 mol L-1} \label{eqn:2}\tag{2}
\end{align}
$$
$$
\begin{align}
c(\ce{CrO4^2-}) &= \frac{c_0(\ce{K2CrO4})\cdot V_0(\ce{K2CrO4})}{V_0(\ce{AgNO3}) + V_0(\ce{K2CrO4})} \\
  &= \frac{\pu{0.35 mol L-1} × \pu{125 mL}}{\pu{105 mL} + \pu{125 mL}} \\
  &= \pu{1.90E-1 mol L-1} \label{eqn:3}\tag{3}
\end{align}
$$
Finally, plug in the concentrations of ions from \eqref{eqn:2} and \eqref{eqn:3} in \eqref{eqn:1} and you find a value for $Q_\mathrm{sp},$ which indeed exceeds $K_\mathrm{sp}$:
$$Q_\mathrm{sp} = (\pu{4.57E-2 mol L-1})^2 × \pu{1.90E-1 mol L-1} = \pu{4.00E-4 mol3 L-3}$$

You are in right track on this problem. But you admitted that you don't know how to find the concentrations, when two substances (of solutions) are mixed together. First, no other information is given, you have to assume that these two solutions are additive. Then, you can treat it as a dilution.
For example, if $v_1$ volume of a solution with $c_1$ concentration of one substance is added together with $v_2$ volume of a solution with $c_2$ concentration of another substance, we can assume final volume is $v_1+v_2$ (only if volumes are additive). Now, you can treat this as you are diluting $v_1$ volume of a solution with $c_1$ concentration of one substance to  $v_1+v_2$ volume. If the final concentration of the substance is $c_f$ after the dilution, you can use $c_iv_i=c_fv_f$ for this calculation.
Now, let's look at the problem in hand: $\pu{105 mL}$ $\left\{v_i(\ce{AgNO3})\right\}$ of $\pu{0.10 M} \; \ce{AgNO3}$ $\left\{c_i(\ce{AgNO3})\right\}$ is added to $\pu{125 mL}$ $\left\{v_i(\ce{K2CrO4})\right\}$ of of $\pu{0.35 M} \; \ce{K2CrO4}$ $\left\{c_i(\ce{K2CrO4})\right\}$. The final volume for both dilution is:
$$ v_f = \left\{v_i(\ce{AgNO3})\right\} + \left\{v_i(\ce{K2CrO4})\right\} = \pu{105 mL} + \pu{125 mL} = \pu{230 mL}$$
Thus, we can calculate the concentration of $\ce{AgNO3}$ after dilution using $c_iv_i=c_fv_f$:
$$c_f\left\{(\ce{AgNO3})\right\} = \frac{c_i\left\{(\ce{AgNO3})\right\} \times v_i\left\{(\ce{AgNO3})\right\}}{\pu{230 mL}}= \frac{\pu{0.10 M} \times \pu{105 mL}}{\pu{230 mL}} = \pu{0.0457 M}$$
Similarly, we can also calculate the concentration of $\ce{K2CrO4}$ after dilution using same equation:
$$c_f\left\{(\ce{K2CrO4})\right\} = \frac{c_i\left\{(\ce{K2CrO4})\right\} \times v_i\left\{(\ce{K2CrO4})\right\}}{\pu{230 mL}}= \frac{\pu{0.35 M} \times \pu{125 mL}}{\pu{230 mL}} = \pu{0.190 M}$$
Since $\ce{AgNO3 (aq) -> Ag+ (aq) + NO3- (aq)}$, $[\ce{Ag+}] = c_f(\ce{AgNO3}) = \pu{0.0457 M}$.
And also, since $\ce{K2CrO4 (aq) -> 2K+ (aq) + CrO4^2- (aq)}$, $[\ce{CrO4^2-}] = c_f(\ce{K2CrO4}) = \pu{0.190 M}$.
Now, using these final concentrations, you can find $Q_\mathrm{sp}$ (well defined in andselisk's answer elsewhere) that is $\approx 4.0\times 10^{-4}$. The numerical value of $Q_\mathrm{sp}$ is greater than $K_\mathrm{sp}$, and hence $\ce{Ag2CrO4}$ would precipitate.

Hmm, I'm trying to understand how this answer is different from mine. It looks like you've calculated $Q$ as $c(\ce{Ag+})\cdot c(\ce{CrO4^2-})$ (which doesn't seem correct) instead of $c(\ce{Ag+})^\color{red}{2}\cdot c(\ce{CrO4^2-})$. Also, some notations leave me confused. Volume is usually $V,$ not $v$: this distinction is important in applications of kinetic gas theory to distinguish it from velocity $v$. Also, curly brackets are normally used to denote activity; I'm not sure what purpose do they serve here.

@andselisk: Good catch. It was just a mental mistake. I have corrected the answer. Thanks for the warning.


text:
Why doesn't this reaction happen?
In the synthesis of DMAD, we eliminate two bromides from dibromosuccinic acid in basic environment to get acetylenedicarboxlyic acid (Step 2).

However, this reaction could also happen:

The reaction in Organic Syntheses (http://www.orgsyn.org/demo.aspx?prep=cv2p0010) went with a yield of 72-88 %, which means that these pathways probably do happen, but only in a minority of cases.
Why is that so, given that the elimination of CO2 is entropically favoured?

iI there is any propargylic acid or acetylene formed, the yield is not an indication. Most OrgSyn yields are less than 100%. Bromination of methacrylic acid and heating in pyridine gives loss of CO2, pyridine hydrobromide and 2-bromopropene by the mechanism you suggest.

@user55119 Any references for the methacrylic acid?

E. A Braude and E. A. Evans, J. Chem. Soc., 3333 (1956). I ran the reaction about 50 years ago. In your example the rate of the observed elimination is faster than the one you question.

Is it possible that the solvent is the determining factor?

Something like that just doesn't happen - you need stabilised carbanion as intermediate. Having some leaving group nearby simply isn't enough for carboxylate instability.

This elimination is unfavorable because C--C bonds are generally quite stable. Decarboxylation would involve the formation of an unstable carbanion before proceeding to the elimination as you have outlined. (Note, this is more favorable under different functional group conditions, namely when a neighboring carbonyl can stabilize the carbanion through the formation of an enol  https://www.masterorganicchemistry.com/tips/decarboxylation/ ) So, the formation of the alkyne is favored here over the elimination of CO2.

But C-H bonds are stronger than C-C bonds and when eliminating CO2, the newly formed C=O double bond is also very strong.

You not only need to look at the strength of the bonds broken and formed but the stability/energetics of the intermediates the reaction would have to pass through. Since this would necessitate the formation of an unstable (and unstablized) carbanion intermediate, this process is unfavorable energetically.

The elimination could be concerted, avoiding the carbanion intermediate.

And also, in the comments above, it was mentioned that methacrylic acid gives the CO2 elimination product, even though the carbanion intermediate would be less stable than in this case.

This isn’t entirely true; the eliminations can be concerted if the atoms are aligned properly (think of the mechanism of the Swern oxidation, especially the breakdown of oxalyl chloride). I suspect that alignment is the problem here.


text:
Shouldn't mineral NPK fertilizer ingredients add up to 100%?
I use a mineral based  fertilizer in powder form which - according to the label - should not contain any fiber and is fully soluble in water (I dissolved 2 times the recommended dosage and found no solids remaining in the solution):

The NPK + Mg + trace elements add up to about 60%.
What constitutes the rest?

Note that the fertilizer does not actually contain chemical compounds like $\ce{P2O5}$, $\ce{K2O}$, and $\ce{MgO}$. These are just the reference compounds for the calculation of results of the elemental analysis.

I'm not sure if I can follow. From what I have read so far if chemical compounds are listed after the percentage this does indeed indicate that the fertilizer contains these compounds in the indicated mass percentage. To calculate the percentage of one element one has to calculate the the percentage of that element in the compound using atomic weight. For example with K2O. K = 39u, O = 16u. Total atomic weight = 239 + 16. Percentage of potassium is 239 / (2*39 + 16) ~ 0.83 ~ 83% of K2O is potassium

You got it wrong. The listed compounds are simply not there, as blunt as that.

If the compounds are not there why do we have to calculate the elemental percentage in the compounds as described in the wikipedia article?

E.g. K2O is very probably present as K2SO4. So 10% of K2O is equivalent to 30 . 172/94 = 54.9% K2SO4. Etc for the other components. By other words, there is so much potassium sulphate equivalent to 30% of potassium oxide.

I called the company a while ago and asked why there is no sulphur in their fertilizer. They told me that sulphur is not necessary for plants, although there is clearly evidence to the contrary. So I doubt that in this specific case K2SO4 is present..

@CuriousIndeed You still do not get it. The listed fertilizer composition is NOT it's chemical composition. It is list of selected content, expressed in equivalent amount of respective, historically chosen reference compounds, mostly oxides. It does not mean these oxides are there chemically present.

It means if nitrogen/phosphorus/potassium/magnesium compounds were  chemically converted to N2/P2O5/K2O/MgO, the mass of N2/P2O5/K2O/MgO would be 10/20/30/2.6%  of the mass of the fertilizer.

About K2SO4, sulphur is not mentioned, as it is not the major nutrition element, but is present in nearly any combined fertilizer.

@Poutnik How could I determine if sulfur is indeed present without sophisticated analysis? I'm really fond of this fertilizer but have discouraged / limited its use because of the supposed lack of sulfur..

You cannot. You can use a crystal ball or ask provider. How much sulphur is to be present to be the needed amount ? Sulphur is seldom a deficient element in plant nutrition. Plants like garlic and onions, containing explicitly large amount of sulphur compounds, often use specialized fertilizers. One of common fertilizers is ammonium sulphate.

At least in Europe sulfur deficiency seems to be a problem. "In particular, incidence of S deficiency has increasingly been reported in Brassica and cereal crops in Western Europe over the last decade, mainly as a consequence of a massive decrease of atmospheric S inputs." (https://www.sciencedirect.com/science/article/pii/S0733521098902417?via%3Dihub) and this was 20 years ago. However thanks for your information. I will probably order a lab analysis for this specific fertilizer.

Also related is https://chemistry.stackexchange.com/questions/122230/what-is-the-mass-of-ozone-present-in-10-6-g-of-na2co3  in particular look at the end part of my answer.

There is a specific convention how the nitrogen, potassium and phosphate concentration is given, and it might be surprising because it mentions substances that are not in the fertilizer, namely nitrogen atoms, $\ce{P2O5}$, $\ce{K2O}$ and $\ce{MgO}$. You have to imagine that all of the nitrogen, phosphorous, potassium and magnesium in your sample reacts to form these species, and then you express the masses of those fictitious products as percentage of the original mass.
$$\text{fertilizer}\ce{-> x N + y P2O5 + z K2O + q MgO}$$
For an example, look at this answer: https://chemistry.stackexchange.com/a/64282 which links to this reference https://en.m.wikipedia.org/wiki/Labeling_of_fertilizer

According to the link on wikipedia, phosphorus and potassium percentages have to be adjusted according to their portion of mass in the chemical compound. However this is not true for nitrogen. What is the rationale behind this convention?

@CuriousIndeed It’s elemental, my dear Watson ... well, atomic, actually.

But it is listed as nitrate or ammonium..(Nitratstickstoff, Ammoniumstickstoff)

@curiousindeed yes, of the nitrogen, some is in the form of nitrate and some as ammonium, but percentages are based on mass of nitrogen (adds up correctly by just adding percentages).

Is there a specific reason for this convention? I find it pretty confusing...

Historical reasons. More consistent would to express all as element content. But it would not give 100% either, as less relevant elements would be omitted as well as they are now.

@Poutnik So is it correct to say that because P2O5 and K2O are not present in the fertilizer but their mass percentages are only indicative of how much P or K are present, the remaining 40% of the mass percentage is constituted of the remaining elements in the compounds in which P and K are part of??

Yes, they are indicative of element content, expressed for historical reason as the element oxide. The missing mass are formed by compounds the listed elements take part in, or possibly  by auxiliary content, improving handling the fertilizer.

The rest is probably Oxygen in nitrates, Hydrogen in ammonium, and Sulfur in sulfates. Let us make some calculations !
In 1000 g fertilizer, there is :
- 74 g N (or 74/14 = 5,28 mol N) as nitrate ion

26 g N (or 26/14 = 1.85 mol N) as ammonium ion
200 g P2O5, or 200/142 = 1.41 mol P2O5, or 2.82 mol P, or 2.82 mol PO4 ion.
300 g K2O, or 300/94 = 3.13 mol K2O, or 6.26 mole K
26 g MgO, or 26/24.3 = 1.07 mole MgO = 1.07 mol Mg.

There must be the same number of moles of positive ions as of negative ions. To start with negative ions, one must triple the number of phosphate and add the double of magnesium plus the number of moles of K. This total gives : 3·2.82 + 5.28 + 2·1.07 = 14.81 mol.
For positive ions, it is more difficult because some H ions are added to PO4^3- ions. Phosphate ions are too much basic to exist in the ground. In neutral mixture, it must be transformed into a mixture of H2PO4^- and HPO4^2-. So, to get the sum of positive ions, one must add 1.85 mol NH4, 6.26 mol K, 2·1.07 mol Mg, and about 1.5 times the number of P for obtaining the number of H atoms bound to PO4 ion. This makes : 1.85 + 6.26 + 2.14 + 1.5·2.82 = 14.48 mol.
The agreement between positive (14.48 mol) and negative (14.81 mol) charges is extremely good !
It is a success. And it is not necessary to add any sulfate ions !...
We may carry out the same calculations with the masses, to see whether the total is 1000 g.

5.28 mol NO3 weighs 327.4 g
1.85 mol NH4 weighs 33.3 g
2.82 mol PO4 weighs 267.6 g
6.26 mol K weighs  244 g
1.07 mol of Mg weighs 26 g
2·2.82 mol H weighs 5.64 g

The total of these masses is 906 g. It is not 1000 g. It shows that there are some non ionic substances in the fertilizer, like clay or organic substances.

I like your approach but aren't Mg and K positive ions? Could it be that you mixed positive with negative?

Mg and K are elements, occuring in fertilizers in form of ions $\ce{Mg^2+}$ and $\ce{K+}$

That's why I'm asking. He wrote "To start with negative ions, one must triple the number of phosphate and add the double of magnesium plus the number of moles of K"

Note that there are hydrogen or dihydrogen phosphate. One has to calculate charge balance.

I thought we have to add the positive and negative charges. So why does he add phosphate (PO4 3-) to the positive ions Mg2+ and K+?

@CuriousIndeed I do not know, but there is no significant amount of $\ce{PO4^3-}$, otherwise the fertilizer solution would smell by ammonia and magnesium would precipitate as $\ce{ NH4MgPO4}$ or $\ce{Mg(OH)2}$

My Simple check shows there is about 1.77 of charge per 1 P, so there is ratio HPO4^2-/H2PO4- about 3:1.

About the mass deficit, there can be some crystal water.


text:
Freezing Point Depression and Temperature Decrease
I don't have a problem understanding the chemical mechanisms underlying freezing point depression or colligative properties in general, but what I cannot seem to grasp is how if you add salt to ice you end up with a salt-ice-water solution that is actually colder than the ice you started with.
Two other places on this site go into detail about this matter but I feel like it is either not totally correct or just not fully explained. I assume I must not fully understand in the first place, also. The two links are posted immediately below pertaining to the subject.
Where does the energy come from to lower the temperature of a brine solution?
Why does ice water get colder when salt is added?
How can the ice be cooling as it melts, as these links seem to claim?  I understand that it would be cooling its surroundings by needing to absorb heat to melt, but it's a phase transition so the temperature of the ice should remain constant. Furthermore, if the temperature remains constant, and the ice continues to absorb heat from the water surrounding it in order to melt since it's above its depressed freezing point, wouldn't you have the impossible feat of heat flowing from cold to hot as the water gets continuously colder, as alleged?
Energy is being used in the phase change from solid ice to liquid water. Heat is being absorbed by the ice. Yes, the ice freezes at a much lower temperature and so it will be liquid at a certain range of higher temperatures than it otherwise would be without salt. As it melts it tries to approach its new
ice-water-salt equilibrium which will be at a depressed freezing point. Why does the temperature fall? Especially if phase changes occur at constant temperature?
The energy must be coming from somewhere and going to somewhere, but where? And how can a hot-to-cold temperature difference for heat transfer actually be maintained if the salt solution is getting colder but the phase change (melting ice) is constant temperature?
I'd really appreciate some sort of insight to this.

Phase change is not at constant temperature, phase equilibrium is

[OP] it's a phase transition so the temperature of the ice should remain constant

Here is a counter example: If you add ice cubes to hot water, the ice will melt, cooling down the hot water. In this system, there is no thermal equilibrium, so not all of the components are at the melting point of water. If you wait a while, you might see the following: All the ice is gone, or some ice and some water remains, or (if the ice was really cold) all the water is frozen. Only if there is some ice and some water left do you know the equilibrium temperature - equal to the melting point.
A system can gain thermal energy from other types of energy
If you stir a viscous solution, it will get measurably warmer. (This is a classic experiment of thermodynamics.) There is no heat transfer - nothing gets colder. For melting ice, it is a conversion of thermal energy into chemical energy (breaking hydrogen bonds).

How can the ice be cooling as it melts?

It can't, at least not for pure water (see below for adding water). The temperature of the ice is below the melting point (or at it), and the temperature of the water is above the melting point (or at it, or a bit lower if it is super-cooled). The thermal energy will come from the liquid water, cooling it. If everything is at the melting point, melting and freezing will happen at the same rate, and nothing changes temperature. In a real system where surrounding is warmer, thermal energy from the surrounding flows into the system, melting the ice while the system stays at the melting point (there will be small fluctuations so that the thermal energy reaches the ice).
Scenario of adding salt
Starting with an ice/water equilibrium (at melting temperature), we can add salt (at the same temperature, and let's say there is no enthalpy of dissolution). After adding the salt, the temperature is still the same, but the melting temperature is lower. With the temperature higher than the melting temperature, water molecules break away from the ice, which requires energy. This energy comes from thermal energy in the liquid and in the remaining ice. As a result, the entire system cools down.
Will the temperature change?

In your example where ice cubes were added to hot water, wouldn't the pure water ice cube remain at 32 F until it fully melted? How is this any different than having an ice cube on a hot plate and melting it? The temperature during the heating of the cube doesn't increase because the absorbed heat is being used to break the chemical bonds of the solid ice.

If the system is not at thermal equilibrium (different temperatures within the system), some parts of the system have to change temperature before equilibrium is reached. The hot water will certainly decrease in temperature. If the ice cubes are colder than the melting point at the start, they will warm up until they have reached the melting point. (Ice can have any temperature below the melting point. For example, ice from a freezer is at about $\pu{-20^\circ C}$.)
If the system is at thermal equilibrium from the start but not at phase equilibrium (or not a chemical equilibrium for a reaction that is endothermic or exothermic), the temperature will also change. In my example of adding salt to a water: ice mixture, the process of melting is endothermic, so the temperature at equilibrium will be colder (and equal to the freezing point of the salt solution). As ice melts, the temperature decreases and the freezing point increases (because the pure melt-water dilutes the salt solution). Once the two temperatures match, the system is at equilibrium (thermal equilibrium and phase equilibrium).

This is starting to make a lot more sense now.  So if the system is at phase equilibrium the phase change happens at constant temperature, but if we are not at phase equilibrium the phase change occurs with a change in temperature?

In your example where ice cubes were added to hot water, wouldn't the pure water ice cube remain at 32 F until it fully melted?  How is this any different than having an ice cube on a hot plate and melting it? The temperature during the heating of the cube doesn't increase because the absorbed heat is being used to break the chemical bonds of the solid ice.

@Mattgen They could start out colder, and warm up and melt. Or warm up until all the water is frozen, not quite reaching 32 degrees but staying colder

I'm just saying that those ice cubes, while melting in the cold water, will remain at 32 degrees F, until they are entirely melted. That has to remain the case.

First of all, it is useful to envision an adiabatic process: add ice at the normal melting temperature to water or cold brine slightly above the MP in a perfectly insulated container (zero heat transfer to the outside) at constant pressure. What do you expect to happen? If the salt concentration is nil (pure water) then ice will melt and the temperature of the liquid will decrease until it matches the MP at that pressure, then not much more: equilibrium will be established. If the solution contains brine, ice will continue to melt until the temperature has sunk below the MP of pure ice (the normal MP). That's freezing point depression. Why does it happen? Because dissolved salt reduces the chemical potential of water in the solution relative to that in the solid: at the normal MP water prefers to be in the salty solution ($\Delta G<0$ for transfer of water from ice to solution). However, melting requires breaking bonds in the solid lattice ($\Delta H>0$). Heat is therefore transferred to the melting solid from the rest of the substance. As both ice and water get colder, both of their chemical potentials decrease, but that of the liquid decreases faster than that of the solid, eventually converging to the same value ($\Delta G = 0$) and establishing equilibrium.

This makes a lot of sense and I sort of had my lightbulb moment, but what still confuses me is, if heat is being absorbed from the water by the ice to melt it, then how is the ice also getting colder? Shouldn't the ice temperature be constant since absorbed energy is used to break the bonds for melting to occur? Does the ice absorb its own thermal energy as well as the water's?

The process of melting requires a transfer of energy in the form of heat. That transfer of energy cools both the liquid and the solid (it's easier to think of the transfer being from the liquid to the melting solid but in fact both the liquid and solid provide energy to the solid undergoing the phase transition).


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Equilibrium of an electrochemical cell
I am quite new to the field. So try to figure out what is meant by the equilibrium state of an electrochemical cell?
Is it, where both half cells passing the same number of electrons to either side? if yes, how do the electrons transfer during non-equilibrium state (is it like, there is a net electron transfer due to electron change from both electrodes).
Any explanation would be much appreciated.

At equilibrium the Ecell = 0 Volts.

You should consider the Nernst equation
$$
\Delta G = \Delta G^{\ominus} + RT\ln Q=- RT\ln Keq + RT\ln Q
 $$
At equilibrium deltaG is zero and Q the reaction coefficient is Keq also note
$$
\Delta G = -nFE
 $$
so E is zero at equilibrium. Outside equilibrium Q will move towards K decreasing E.

An electrochemical cell is producing electricity. So it is out of equilibrium. Reagents which are included in the cell are slowly consumed. The chemical composition of the cell changes, and will slowly tend towards equilibrium. When the reactions are finished, the cell will not produce any electricity later on. The equilibrium is obtained, and the cell is "out of use" : its chemical composition will not change any more in the future.


text:
Conformers of a meso-compound
In a particular question in one of my reference book, I was asked if the given compound (meso-butane-2,3-diol) was a meso compound or not.  In it’s solution part, to show that the compound was actually meso, they did the following.

As I was not convinced about rotation of C2, I searched the web where I found this which says

While rotating the bonds about any atom
  (chiral) the stereochemistry of the molecule
  does not changes.

This is what I am confused. I mean if we rotate meso-butane-2,3-diol in the given way such that we have a conformer other than eclipsed or staggered, wouldn’t it be optically active, without any plane or centre of symmetry, which means rotation does change stereochemistry. Where am I going wrong?

Duplicate of:  https://chemistry.stackexchange.com/questions/48995/is-2r-3s-butane-2-3-diol-chiral/87415#87415 Any conformations of the meso cmpd that do not have a plane or center of symmetry constitute a racemic pair.

@user55119 I agree about the racemic pair. What I was asking is that different conformers of meso compounds are either optically active or inactive, so rotation of groups about C2 and C3 results in different compounds (with different stereochemistry), unlike mentioned in the question.

Of the three staggered conformations about the C2-C3 bond, one has a center of symmetry and two are a racemic pair. If you could "freeze" the trio, say on Pluto, and put each one in a polarimeter, each enantiomer would have equal and opposite rotations. The center of symmetry one would have no rotation. On Earth, the trio is in rapid equilibrium and gives no rotation. Drawing the eclipsed conformation with a plane of symmetry is merely a test for a meso compound and a minor contributor. See my link above.

@user55119 I think I am getting it. So all conformers would have same configurations (R,S), which in turn result in same stereochemistry, but since in eclipsed form, the POS is easily visible, they are rotating it as shown, correct?

Rotation doesn't affect R,S configuration. Make 3 molecular models and see for yourself!

The method shown in OP's question to find a meso-compound is the way most people used. However, I usually used the Cahn-Ingold-Prelog (CIP) assignments of chiral centers for this purpose. It is a fact that the mirror image of (R)-chiral center is corresponding (S)-chiral center. For example, bromochlorofluoromethane is a chiral compound and the morror image of its (R)-stereoisomer is its (S)-stereoisomer as depicted in the diagram below:

When (S)-isomer is rotated $180^\circ$ vertically, the resultant structure would not be superimposed on (R)-isomer, as shown in the diagram. This is to justified the above mentioned fact.
Now, let's assign the compound in hands with CIP system. I'd call it (2 R, 3 S)-butane-2,3-diol (it could also be (2 S, 3 R)-butane-2,3-diol, based on the numbering). Either way, carbon #2 has three different groups attached to it, which are identical to those on carbon #2. Based on the given assignments on carbon #2  and #3, if you'd put a mirror between these carbons, you must see the mirror image of one part (carbon #2) on other side (carbon #3) as depicted in the diagram. Therefore, the compound has plane of symmetry (the mirror) and hence, it is the meso-isomer.
To conclude this fact, I included two enanthiomers of this compound ((2 R, 3 R)-butane-2,3-diol and (2 S, 3 S)-butane-2,3-diol), none of which gives mirror images when a mirror placed in between two chiral carbons (see the diagram).

The staggered conformation of the meso-isomer you have shown does not have a (mirror) plane of symmetry but rather a center of symmetry, which accounts for this conformations lack of optical activity. See the links in the comments above.

@user55119: I disagree. If you rotate carbon #2 $180^\circ$, you'd see that plane of symmetry.

I believe you misunderstood my point. The molecule is achiral and meso. The staggered conformation you have shown has a center of symmetry in the center of the C2-C3 bond. The other two staggered conformations form a racemate.  Yes, the rotation by 180 degrees gives the eclipsed conformation with a plane of symmetry, no rotation and low concentration. The eclipsed conformation is a test for a meso cmpd. The other 2 eclipsed conformations are also a racemate.  Your diagram should show "center of symmetry".

I just want to confirm one thing-that all conformers of a meso compound are meso compounds?


text:
Why aren't plastic bottles of coca cola full of microplastics?
My understanding is that disposable plastic water bottles should not be reused or used at all after some time, because the bottle will eventually leech microplastics into the water. Since coca cola is pretty acidic and can be used as toilet cleaner, doesn't that mean it would "dissolve" its contain bottle and get those microplastics very quickly?

PET is perhaps not as resistant as PE or PP (which are resistant even to concentrated mineral acids), but is very indifferent to a pH of around 2.

You are talking about two different things and phenomena. One is leaching-which means chemicals in the plastic dissolve into the solution with the passage of time. The other thing is literally physical pieces of microplastics- they are not dissolved but they are actual microscopic pieces of plastic just like very fine dust is present in air. It is not dissolved but it is suspended in air. Coke claims to filter before filling in the bottles. Filtration is an effective way of removing microscopic solids.
Read this report Microplastics and WHO

"Microplastic" breaks off from solid plastic material by mechanical wear, often following a degradation of the material by environmental influence like oxygen, light, ...
The inside of reusable water bottles does not experience mechanical wear.  Single use bottles even less. Any microplastic particles in there must obviously have come with whatever beverage has been filled in, OR were in there even before, most likely as dust from the plastic bottle factory, or the machine that cleans the reusable bottles, where a lot of chips from the battered outside fall off.


text:
Can someone explain my professor's answers for these alkene additions?

(a) Why would cyclohexane expand into cycloheptane? Cyclohexane is stable.
(b) Why would cyclopentane not ring expand into cyclohexane? In the process a tertiary carbocation would be formed just like with the methyl shift, but you would also get rid of ring strain.
(c) Why is there only one product? Wouldn't the double bond have hydrogen addition on both carbon sides leading to a potential cyclobutane to cyclohexane expansion?
(e) Once again like in (c), why is there only one product shown?
(f) Wouldn't there be two products once more?
POTENTIAL EXPLANATION: By examining more mechanisms in class, I have noted that potentially the methyl/alkyl shifts could be occurring in concert with the hydrobromination so perhaps that is the reason why.

Maybe ask one question at a time? For (b): there is no ring strain for cyclopentane, so your premise is incorrect. To combine a couple of the questions, you could ask what determines whether a carbocation undergoes hydride shift, alkyl shift or phenyl shift, and what happens when more then one of these is possible.

In example A, you are right that the cyclohexane ring is more stable than the cycloheptane ring. There are two choices for rearrangement in this case:
1) Ring expansion to form the cycloheptane ring
2) 1,2-hydride shift to yield the tertiary alkyl bromide
The second product will be more stable and thus be the major product
In example B, you are right, a ring expansion is possible, so there should be the ring expansion product too. You will have a mixture of products in this case (1,2-alkyl shift, ring expansion, and no carbocation rearrangement)
In examples C, E and F, it is possible for the H-A to add across the double bond either way, so again, a mixture of products would be obtained
A lot of times, only one product is shown for addition reactions such as hydrohalogenation; however, it is important to remember that there are often multiple products possible.
Chemists very rarely utilize reactions that involve a carbocation rearrangement because a mixture of products is obtained, which results in very poor yields

At least in examples (F), it is not possible for $\ce{HX}$ to add across the double bond either way, because protonation followed by hydride transfer gives very stable carbocation ($\ce{Ph2C^.R}$), which followed to give the given product.

@MathewMahindaratne But the initial addition across the double bond should have rather similar transition states for both cases and not be regioselective, right? The hydride transfer does give a very stable carbocation, but that is in subsequent steps and wouldn't be rate-determining. As the initial carbocations formed are both secondary and because the initial process is endothermic they would have similar transition states energies and thus rates.

@ebehr: Do not forget the rate determining step is the addition of halo ion in the final step. Until then, all steps are in equilibrium, so as soon as most stable carbo cation formed, it would be attack by halide ion and reaction is done.

@MathewMahindaratne link according to this it isn't though

@ebehr: Theoretically it is possible but practically would never happens. Please pay attention to relevant activation energies in the given site (msu.edu). If you pay attention, one is for formation of primary carbo cation. Practically it is impossible.


text:
Is the graph that is attached (a form of the arrhenius equation) only used when the temperature is varied?

determine the approximate value of Ea with its appropriate units.

In my lab course, I am doing an experiment on the rate of reaction, but all three of my trials were using the same temperature. That would just result in a straight vertical line, correct?
EDIT: It was brought to my attention that I would have to use varying temperatures for this form of graph. I will be using a similar form of the example shown, but I'll be using log instead of ln which can be seen in problem 6. Would this result in a positive slope? Am I able to estimate the activation energy of all 3 parts or
just the activation energy overall, and how do I go about doing it?

That would be no use.

You see that the plot varies $T$ (expressed as $1/T$) and running the experiments at one temperature only leaves you in an underdetermined situation where you access only one of the red dots. Without accessing the other red dots about the reaction run at other temperatures, you don't access the slope (even less than «a straight vertical line»).  Keep in mind that the approach assumes the reaction to proceed by the same mechanism for all the temperatures varied; a significant deviation from the straight line should ring a bell.

Thank you so much. How do I estimate the Ea(s) for the graph?


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Using Le Chatelier's Principle to Change Mechanical Energy to Chemical Energy
I was interested in an application of Le Chatelier's principle, where an increase in pressure could result in a decrease in the moles of gas.
In particular, a situation where when the pressure increased, it drove the reaction in one direction (towards fewer moles), but when the pressure was lower, then the reaction went the other direction.
What pairs of two reactive room-temperature gases can both combine to make a third room-temperature gas where the change in chemical energy can be used to store the energy from the work done on the system by compression?
So far, I've found these possible candidates:
$$
\begin{align}
\ce{2 SO3 &<=> 2 SO2 + O2}\\
\ce{2 CO2 &<=> 2 CO + O2}\\
\ce{2 NH3 &<=> N2 + 3 H2}
\end{align}
$$
I'm imagining a situation where a mixture of gasses could be compressed to a significant fraction of its original volume without too big of an increase in temperature.
Maybe a catalyst might be critical to make the activation energy less of an issue.
I don't know if this is possible, though.
Additional
I'm teaching a module on energy conversion.
We have 5 common types of energy: mechanical, electrical, thermal, chemical and light.  And we are discussing conversions between them.
The example of changing mechanical energy to chemical energy I had been using was the compression of graphite to make diamonds. I wasn't happy with this one, though.
Then I half-remembered something from my college chemistry class where the direction of a chemical reaction could be shifted by changing the pressure or the temperature, and I wanted to investigate that idea.
That's the inspiration for this question.  If this process is useful, all the better.  If it's not, then it's a way to introduce some important concepts and something they'll see later in their college chemistry class.
Second Additional
If we have a cylinder of gas, a piston compressing that gas acts a bit like a spring.  But compressed gasses heat up and lose thermal energy to the environment.
Could a chemical system like this be interesting because the energy of compression is partially stored in the chemical bonds and not completely in the motion of the molecules, and so might not lose as much energy to the environment?

It seems to me to be rather an expensive and complicated way to decrease efficiency of the already deploying standard gas compression technology. In contrary to the gas compression, there would be bigger difference between compressing and expanding pressure, and between compressing and expanding work.

I think you would want a reaction that was endothermic in the direction towards the lower-volume side.  That way, as you compress the gas, the thermal energy from compression could be better converted to chemical energy.  Unfortunately, that is not the case for any of the three reactions you've listed in your post--all of them are endothermic in the forward direction (i.e., forward as you've written them).  I'd suggest consulting tables of enthalpies of formation to see if you can find a reaction that meets this criterion (unless someone here on Chem SE can suggest such a reaction).

I'll add it could be a challenge to find a reaction that is endothermic in the lower-volume direction, since lower-volume means fewer gas molecules, and collecting the same number of atoms into fewer molecules means, on average, more bonds, and more bonds typically means exothermic.  You can see this if you compare the total no. of bonds on the LHS vs. RHS of your reactions.


text:
Energy cost of C−N rotation in hydroxamate group
I am looking for a literature reference on the energetic cost of C−N rotation in hydroxamate group such in LC5 in water. I have been googling for a while.

Reference 1 reports the isomerization barrier for N‐methylacetohydroxamic acid. From the abstract:

We have now determined the activation barrier to C–N rotation from major to minor conformation of a small secondary HA in $\ce{D2O}$ to be 67.3 kJ/mol. The HA rotational barrier scales with solvent polarity, as is observed in amides, although the HA barrier is less than that of a comparable tertiary amide in aqueous solution.

N‐methylacetohydroxamic acid (image from PubChem):

References

Stefanie P. Sippl, Paul B. White, Charles G. Fry, Sarah E. Volk, Lingxiao Ye, Heather L. Schenck. Kinetic and thermodynamic characterization of C–N bond rotation by N‐methylacetohydroxamic acid in aqueous media. Mag Res Chem 54 (2016), 46-50. DOI: 10.1002/mrc.4318.


text:
Is lithium carbide an ionic compound or covalent?
I read in my chemistry textbook that lithium carbide (Li4C) is the only carbide amongst alkali metals. As we know, carbides are ionic compounds, and calcium carbide is predominantly ionic (calcium having similar electronegativity value to lithium).
So, is the bond between lithium and carbon ionic? Or is it covalent?
Does a Li4C exist? If so, what is it's nature?

Which lithium carbide, Li2C2? You can treat it as more ionic then covalent.

Do you mean an illusive $\mbox{Li}_4 \mbox{C}$?

Yes, I mean the Li4C .

https://chemistry.stackexchange.com/questions/34099/why-do-different-elements-form-different-types-of-carbides


text:
Why can DAP be used to elute a protein in affinity chromatography?
A little background: as part of my bioinformatics degree I have to take Protein Chemistry course, but I miss chemistry basics (have CS background), so that's why I'm asking the following question.
We performed protein (Galanthus Nivalis L. agglutinin) purification in class, using affinity chromatography.
The mannose column was loaded with protein (mixed with ammonium sulphate), washed with ammonium sulfate and then the protein was eluted using 20 mM DAP. While I understand the principle of affinity chromatography, I can't really figure out the chemistry of using DAP to elute the protein.
Would appreciate any help.

It is always a good idea to define the abbreviations. DAP can be diammonium phosphate or diaminopropane.

In this report, diaminopropane is used to elute a lectin from an affinity column. They mention the pH, which is the key to the question.

The  bound  lectin was  eluted  with  20  mM diaminopropane  containing  0.15M NaCl,  pH ~11.

The diaminopropane is unbuffered, and the high pH weakens the interactions between lectin and carbohydrates.


text:
If cigarette smoke is exhaled, what happens to the accompanying exhaled breath?
I want to know what happens to that breath. To help you answer this question, consider in your mind these subquestions:

Is there a necessary principle at work, by which you know that it vanishes/solves into the air after a set amount of time or distance?
Is it attached to the smoke, and does it travel with it?
Does it get detached from it?

Please explain in clear, essential concise language suitable for a non-chemist and a humanities person (logic, philosophy, theology, etc.).
Explanation:
I am trying to help someone in need overcome a traumatic experience that has had demanding negative effects, after the person had already suffered much, and this is the most direct way I have at the moment. Your answer might just be crucial to the person's recovery.

Please make clear and not overly broad questions suitable for Chemistry SE...

Thist is a kind of empirical. An architect and regulators might have special formula. If you mean how far you must be from a smoker it can be from very close in a open place in windy day or even doesn't matter in a small and close room.

Hi Alchemista. I've changed the question, as was demanded by some. I hope it is sufficiently clear and clearer. If possible, please include and consider different yet essential situations/principles (that are applicable to many situations).

No, it doesn't make sense - that's the problem - it's an X & Y problem - question stems from false premises and fundamental misunderstanding. Breath is just air, it doesn't "detach" there's no "threshold" in mixing of gases, no specific moment when something stops being smoke.

What false premisses? They are QUESTIONS.

And while you're at it, go see a doctor or rather a psychologist. You seem to lack a heart.

You can never see single molecules with your naked eyes. If you can see the smoke, it must be something larger (much larger) than molecules and the wavelength of light. When organic materials, like tobacco, burn incompletely they generate soot which is just carbon particles. These carbon particles scatter white light and appear as bluish-grey material when someone exhales smoke. What others (poor passive smokers and the smoker) smell are literally thousands organic molecules which the human nose can sense and many of them are cancer causing chemicals including tar, benzopyrenes, and radioactive junk among others. Tobacco plant requires lot of phosphorous fertilizers those fertilizers have traces of unwanted radioactive metals. In short it is not cool to smoke and I have seen lungs of smokers which have turned black from soot in museums.

Thank you. What about other things, such as breath? How and how far does exhaled breath travel, if accompanied by exhaled cigarette smoke? Is there also maybe a point at which it---for all I know---detaches or ceases to exist?

One cannot see breath. However one can the smoker in a room which means the molecules (after burning tobacco) diffuse quite fast in the room.

I did not fully understand that last phrase in your comment starting with "However one can the smoker . . . ."

Sorry, I meant "one can smell the presence of a smoker"...


text:
Why does Pubchem show cis-1,3-cyclohexanediol with the hydroxyls in the equatorial position?
I am a beginning organic chemistry student and trying to reason about something based on information in my book, but seeing that Pubchem gives a different structure than I'd expect.
In considering a disubstituted cyclohexane with hydroxyl groups at 1 and 3, I'd think, ok, there's a competition between steric strain and hydrogen bonding.  If hydrogen bonding were the more important force, the molecule would be more stable with the hydroxyl groups in the axial position. If steric strain were more important, the molecule would be more stable with the groups in the equatorial position.
The solution manual for Klein's Organic Chemistry [1, p. 128] says in an answer to the problem 4.75 that the stabilizing effect of intramolecular hydrogen bonding between two hydroxyl groups in a gauche conformation is 20 kJ/mol:

This calculation must be modified when we take into account the effect of two $\ce{OH}$ groups that are gauche to each other, as seen in conformers A and B.  We should expect an $\ce{OH-OH}$ gauche interaction to be less than a $\ce{Me-Me}$ gauche interaction (less than 3.8 kJ/mol), because an OH group appears to be less sterically encumbering than a methyl group (compare $\ce{CH3}$ and $\ce{OH}$ in Table 4.8). Therefore, the destabilizing effect associated with an $\ce{OH-OH}$ gauche interaction (less than 3.8 kJ/mol) should be overshadowed by the stabilizing effect that results from the hydrogen bonding interactions between the two $\ce{OH}$ groups, which is expected to be approximately 20 kJ/mol (see section 1.12). As a result, we expect extra stabilization to be associated with any conformer in which two $\ce{OH}$ groups are gauche to each other.

My book also (in the same answer) says the upper bound for the steric strain in an $\ce{OH-OH}$ gauche interaction is 3.8 kJ/mol, i.e., lower than the value for an $\ce{Me-Me}$ interaction.  It further tells us that 1,3-diaxial interactions are gauche interactions. Klein's Organic Chemistry [2, p. 166] gives in a different table the value of two $\ce{OH-H}$ 1,3-diaxial interactions in a monosubstituted cycloalkane to be 4.2 kJ/mol.

It seems to me like the OH's have steric interactions with each other (3.8 kJ/mol), and they each have an interaction with an axial H (4.2 kJ/mol for both of them).  So at 8 kJ/mol steric strain, it seems like the hydrogen bonding effect should swamp the steric effects.
But Pubchem shows the hydroxyl groups in the equatorial position.  I assume it's showing me the lowest energy conformer.  I don't understand what's wrong with my reasoning above.
References

Klein, D. R. Student Study Guide and Solutions Manual, 3e, 3rd ed.; Wiley, 2017. ISBN 978-1-119-37869-3.
Klein, D. R. Organic Chemistry, 3rd ed.; John Wiley & Sons, Inc: Hoboken, 2017. ISBN 978-1-119-31615-2.

1,3-Dihydroxycyclohexane is more stable in the diequatorial conformation than in the diaxial one. Table 4.8  is for monosubstituted cyclohexanes. Do not confuse 1,3-diaxial interactions with 1,3-disubstituted. The groups in the Table have 2 gauche interactions with the ring in the axial conformation.

I know it's for monosubstituted cyclohexanes.  I'm just using it to get a value for what two OH-H gauche interactions add up to.  The disubstituted version has an OH-OH gauche interaction, and two OH-H gauche interactions.  So that 4.2 kJ/mol value should be equivalent to the two OH-H gauche interactions in the disubstituted version.  Then I add the 3.8 kJ/mol for OH-OH to get the 8 kJ/mol total.   Maybe this is wrong but I would like to know why.

It is difficult to make this calculation for the diaxial isomer. Diequatorial is taken as 0 kJ/mol. The diaxial diol has not only gauche interactions but also positive hydrogen bonding. The 1,3-trans isomer will work.

The diaxial diol has not only gauche interactions but also positive hydrogen bonding
That's what I'm saying in OP -- according to my text the stabilizing effect of H-bonding in 2 gauche OH's is 20kJ/mol.  That's why I thought diaxial would be more stable
The 1,3-trans isomer will work
I'm only thinking about cis.
It is difficult to make this calculation for the diaxial isomer
Maybe none of these back of the envelope calculations make sense or are legitimate but that's not what my book teaches.  I wish I understood why using my book's reasoning is giving a wrong answer.

In general, cis-1,3-diequatorial is more stable than diaxial. Hydrogen bonding can change the equilibrium.

I believe you.  I believe Pubchem.  I think my question may have been unclear -- it's not "which is more stable?" It's "what is wrong with this reasoning?"

"My book (Wiley, 3rd edition)" isn't a proper reference. There are two distinctive literature sources you are quoting (corrected, please take a note for the future qustions).

@andselisk thanks for the edit.  I didn't even notice that I had put the publisher instead of the author, as well.