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Wave function of a perturbed system
In perturbation theory, wave function of a perturbed system can be expressed in a power series of some $\lambda$ as $\psi^{(0)} + \lambda\psi^{(1)}+\lambda^2\psi^{(2)}......$.
Then we express Hamiltonian operator as $H^{(0)} + \lambda H^{(1)}$.
Introduction of $\lambda$ in $\psi$ can be understood but why we introduce $\lambda$ in Hamiltonian operator and what is the significance of this $\lambda$ factor.

You use it but it does not appear in the final result. It is used to sort out zero, first, second etc powers of perturbation. In chemical terms very loosely sort of like a catalyst.

For distinguishment purpose, we can introduce $\lambda$ in $\psi$ but why in Hamiltonian operator. Why we can't write $H$ as just $H^{(0)}+H^{(1)}$. As it also gives first order correction as $\lambda$ is present in $\psi$ while solving Schrödinger's equation?

If you write $H$ as just $H^{(0)}+H^{(1)}$, then you don't know which of these is big, and which is small, and which is to be neglected.

Also when I solve Schrödinger's equation using $H$ as $H^{(0)}+H^{(1)}$, I get ($H^{(0)}+H^{(1)}$)× $\psi$ = $E^{(0)}$*$\psi$, which is a contradiction as it implies $H^{(1)}=0 but system is perturbed.

I don't understand your argument, i.e, to compare $H^{(1)}$ and $H^{(0)}$. Please explain a bit more.

The method is only outlined below, the full details should be in your textbook or lecture notes.
Your $H^{(1)}$ is just the (small) perturbing potential energy added to the original hamiltonian so you could write $H^{(0)}+\lambda V$.  ($\lambda$ is dimensionless and varies from 0, no perturbation, to 1, full perturbation ).
The Schroedinger equation is now $(H^{(0)}+\lambda V)\varphi_n=E_n\varphi_n$  where $\varphi_n,\; E_n$ are the new wavefunctions and energy. These are expanded as a Taylor series in $\lambda$, e.g.  $\displaystyle \varphi_n \approx \varphi_n^0 + \lambda \frac{d\varphi_n}{d\lambda}+\frac{\lambda^2}{2!} \frac{d\varphi_n^2}{d\lambda^2}+ ..\equiv \varphi_n^{(0)}+\lambda\varphi^{(1)}_n +\lambda^2\varphi^{(2)}_n+..$ and similarly for $E_n$.
Next this result and that for  $E_n$ are substituted back into the Schroedinger eqn and powers of $\lambda$ factored out. These remaining equations are solved but none now contain $\lambda$; for example the first order term is $\lambda\left((V-E^{(1)}_n)\varphi^{(0)}_n+(H^{(0)}-E_n^{(0)})\varphi^{(1)}\right)$ where in this new notation  $\varphi^{(0)}$ is the original unperturbed wavefunction and similarly for the energy.
The equation to solve is now $(V-E^{(1)}_n)\varphi^{(0)}_n+(H^{(0)}-E_n^{(0)})\varphi^{(1)}=0$.
After (formally) integrating the equation some fiddling around you find that $\varphi^{(1)}$ is needed. The important step now is to expand this wavefunction in terms of the original basis set of unperturbed wavefunctions $\phi$. This can be done because these form a complete orthonormal set, thus $\varphi^{(1)}= \sum_k a_k\phi_k$ where $a_k$ are the expansion coefficients and these have to be found.  Again after some more fiddling with equations and using the orthonormality of the $\phi_n$ the perturbation energy $E^{(1)}_n = \int \varphi^{(0)*}_nV\varphi^{(0)}_nd\tau = \int\phi^{0*}V\phi^0 d\tau$.
Notice how only the unperturbed wavefunctions contribute to the energy.  The new wavefunction is $$ \varphi^{(1)}_n=\sum_{n\ne k} \frac{E_n^{(1)}}{E_n^{(0)}-E_k^{(0)}}\varphi^{(0)}$$
which is a linear combination of the unperturbed wavefunctions with the $a$ coefficients as ratio of energies.

As we use $d\psi_n/d\lambda$, it can happen that the wave function may not be continuous with respect to $\lambda$ or is not differentiable at a particular $\lambda$, because the postulates of quantum mechanics specify that wave function is continuous with respect to x,y and z, not with respect to some other quantity like extent of perturbation($\lambda$). Do we assume that $\psi$ is also continuous for $\lambda$?

It has to be assumed that the derivatives behave normally, but as it happens its only a formal differentiation, and then integration; the wavefunction is only used symbolically. Only the unperturbed wavefunctions are used to calculate first order and the function for these (say sin, cos , exp etc. ) are known by solving the unperturbed equation, and as the $V$ is known and also the $E_n$ the wavefunction for $\varphi^{(1)}$ can be calculated.


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How does the pKa of a neutral molecule inform the protonation state of its positively charged molecule?
Let's say we have some neutral molecule A and its protonated form, A+. The pKa of molecule A is known, but we do not know the pKa of molecule A+. Can the pKa of molecule A+ be determined only knowing the pka of molecule A? For example, is there an expression that relates the pKa values of arginine together? (For reference, arginine's pKa values are are 2, 9, 12.5.)
The context of this question lies in trying to understand how the protonation state of molecule A+ will change in an inhomogeneous environment (for example, a lipid bilayer). The pKa of molecule A+ is estimated to change +2 pK units in the bilayer relative to in aqueous solution. The pKa of molecule A is known to be 7.09, but the pKa of molecule A+ is not known. How can any change in protonation state of molecule A+ be predicted given only this information?
Note that this question is NOT asking the protonation state of a molecule if we know THAT molecule's pKa and if we know how ITS pKa changes (for example, as was asked previously).

No, these values are quite independent of each other.

@IvanNeretin, I see. Does that mean the second question also becomes moot? Is there any information that can be drawn from A+ and its predicted change in pKa?

Yes, pretty much so. These are independent species. Sure, the pKa of second protonation must be higher than the first (otherwise why would it be second?), but that's all about it.

For clarification: Do you mean A and A+ are at the same position of the molecule (like acetate vs. acetic acid vs H3CCOOH2+ (the 2 oxygens protonated) or the situation at several functional groups within a molecule, like in your arginine example above?

@imalipusram, My molecule has just one expected protonation state, so one position and not like the arginine example.

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How can I purify Ferric or Ferrous acetate with a very basic lab?
I’ve been trying to make Ferric acetate by mixing vinegar with steel wool and allowing it to react. The result is successful in dying leather or wood black and acting as a mordant. Theoretically there are ways of crystallizing, precipitating, filtering, or separating it with solvents.
There is only one problem: I can’t remove it from the vinegar solution without it oxidizing. Adding Hydrogen peroxide causes it to create iron oxide, heating it causes it to hydrolyze and create iron oxide. Leaving it in solution turns some of it into iron oxide.
One idea I had was using glacial acetic acid instead of vinegar but I’m hoping I don’t have to because that’s fairly expensive. I could also do the entire thing under vacuum but it still wouldn’t fix the hydrolysis reaction. Could I maybe crystallize it? I’m open to suggestions with an emphasis on stuff that doesn’t require expensive reagents or obscenely overpriced labware.

Both the Handbook of Chemistry and the Merck Index state that $\ce{Ferric Acetate}$ exists only in solution. If this solution is evaporated, a basic ferric acetate is obtained that is insoluble in water, with a formula $\ce{Fe(OH)(CH_3COO)_2}$. This basic acetate is soluble into acetic acid.
Nevertheless there has been recent reports stating that solid Trinuclear Ferric Acetate may be obtained in rather concentrated solutions with the following structure : $\ce{[Fe_3O(OAc)_6(H_2O)_3]OAc}$. Its synthesis has been described in A. Laurikenes et al., Lithuan. J. Phys. 56, 1. p.35 (2016).
They mix $11.2$ g Iron powder ($0.200$ mol) and $12$ mL $\ce{CH_3COOH}$ $50$% ($0.100$ mol) at $75°C$. Why this choice ? The contrary would have been a much better choice (interchange $0.1$ and $0.2$)... They obtain a concentrated green solution, which they filtrate, and they add $6$ mL $\ce{H_2O_2}$ $30$%, and filtrate the obtained red mixture. The obtained filtrate is very concentrated. It is then evaporated under vacuum. And they obtain $2.95$ g of the trinuclear compound. The yield is miserable.  But they say that they have been able to recrystallize this compound in ethanol.
It is interesting to know that this substance has three equivalent $\ce{Fe}$ atoms in the apexes of a triangle, with an Oxygen atom in the center of the triangle, and that each $\ce{Fe}$ atom is included in the center of an octahedron, where all apexes are occupied by Oxygen atoms.
See also : K. I. Turte et al. Synthesis and Structure of Trinuclear Iron Acetate, J. Struct. Chem. 43, 108-117 (2002)

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In the reaction mentioned what percentage of reaction proceeds via SN1 mechanism?
2-iodo butane (having radioactive iodine) reacts with KI (having non radioactive iodine). Rate of loss of optical activity was 1.96 times the rate of loss of radioactivity. What percentage of reaction proceeds via SN1 mechanism?
Assume that 100% inversion takes place in SN2 reaction while in SN1 there is 50% retention and 50% inversion.
I proceeded with KI substituting the iodide ion with the radioactive iodide ion but  couldn't understand how would 2-iodo butane lose its optical activity(it still has a chiral C). So I couldn't go any further.

It has to do with an (equimolar) Racemic mixture being formed! Note that a racemic mixture still has chiral atoms but cannot contribute to any net rotation of the plane polarized light making it optically inactive as a combined product

https://chemistry.stackexchange.com/questions/132885/in-the-reaction-mentioned-what-percentage-of-reaction-proceeds-via-sn1-mechanism

This question does not make sense. At the beginning of the reaction, the rate of loss of radioactivity is the rate of both types of substitution. But the rate of loss of optical activity is tempered by the fact that only one of the two mechanisms produces more of the other enantiomer to progress towards racemization. By that logic, how could the rate of loss of optical activity be greater than the loss of radioactivity?

@Zhe As I have written in my answer, rate of loss of optical activity is proportional to extent of racemisation. If I take a substrate with optical activity having only one enantiomer, then initially, there is no racemisation. Suppose 10% of substrate reacts and undergoes inversion, then the solution will be 20% racemic since a racemic mixture consists of equimolar amounts of two enantiomers. So rate of loss of optical activity is twice of rate of reaction for SN2 mechanism.

"Suppose 10% of substrate reacts and undergoes inversion, then the solution will be 20% racemic" That statement is unfortunately not correct. As soon as any of the initial starting material reacts, you generate the other enantiomer which reacts at the same rate as the initial enantiomer.

@Indian135 I'm going to sleep, but I will write something substantial tomorrow.

First, for simplicity, I am assuming that the initial substrate (2-iodobutane with radioactive iodine) consists of only one enantiomer.

How would 2-iodobutane lose its optical activity (it still has a chiral $\ce{C}$)?

In $\mathrm{S_N1}$ mechanism, there is 50% inversion and 50% retention, so a racemic mixture starts to form which decreases the optical activity of reaction mixture due to external compensation.
In easy terms: In a racemic mixture, although the two enantiomers rotate plane-polarized light in opposite directions, the rotations cancel because they are present in equal amounts. And in a partially racemic mixture or still undergoing reaction mixture with racemisation, the optical activity is decreased (rotation of plane polarised light is less).
So, 2-iodobutane doesn’t lose its optical activity, it’s optical activity is only countered by its inverted enantiomer.

What percentage of reaction proceeds via $\mathrm{S_N1}$ mechanism?

The rate of loss of radioactivity is proportional to amount of substrate that has reacted (2-iodobutane with radioactive iodine), that is rate of reaction.
And, the rate of loss of optical activity is proportional to rate formation of racemic mixture.
Now suppose,
$x =$ rate of reaction of substrate at an instant in moles per unit time.
Then, $1.96x$ is the rate of racemisation.
So, during a small time period $t$, $xt$ moles of substrate has reacted and $1.96xt$ moles of racemic mixture is obtained.
The racemic mixture consists of 50% inverted product and 50% with original configuration. Note that the reaction mixture is not purely racemic. It has substrate and retention product (both of which have original configuration) and inversion product. So, by racemic mixture I mean all the inverted product and equal amount of molecules with original configuration.
Since no inverted molecule was present before the reaction (assumption at start of answer), $0.98xt$ moles of substrate proceeded via inversion and the rest $0.02xt$ moles via retention ($x$ moles had reacted)
So, only $\mathrm{S_N1}$ can cause retention and that also only 50% of reaction via $\mathrm{S_N1}$ gives retention product, $0.04xt$ moles of substrate must have proceeded via $\mathrm{S_N1}$ mechanism.
Thus, percentage of $\mathrm{S_N1}$ mechanism $= \dfrac{0.04x}{x} \times 100\% =4\%$.

See my comment to the question on why I don't believe you should be able to solve this problem...

This really doesn't answer the question, but the discussion is too long for comments.
The question only makes sense if you start with a mixture that has optical activity.
Let's call the concentration of the starting enantiomer $x$ and the other enantiomer $y$.
Optical activity is unfortunately, not a concentration, so we need a definition to make the two rates comparable.
$$\alpha = \frac{\phi l}{10}(x - y)$$
$$\frac{d\alpha}{dt} = \frac{\phi l}{10}\frac{d(x - y)}{dt}$$
For the rate of loss of optical activity, we can strip off the leading constant to convert to a concentration based rate. But we could just as well replace it with any constant, and I'll address this later. For now, we have:
$$\mathrm{rate}_{\mathrm{optical}}=-\frac{d(x-y)}{dt}$$
This mostly makes sense, since $\ce{[x]}-\ce{[y]}$ represents the concentration of the excess of the major enantiomer. Thus, the rate of change in this value, which should be negative, represents the rate of loss of optical activity. As a sanity check, it has the maximum value at the beginning of the reaction and is equal to zero when the system is racemic.
We also should make clear that the respective rates of $S_{\mathrm{N}}1$ and $S_{\mathrm{N}}2$ are:
$$\mathrm{rate}_{1} = k_{1}\ce{[X]}$$
$$\mathrm{rate}_{2} = k_{2}\ce{[X]}$$
where I've absorbed the assumed constant concentration of salt into the rate constant of the $S_{\mathrm{N}}2$ reaction.
The loss of radioactivity is equal to the rate of types of substitution because both types of substitution replace the labelled iodine with an unlabelled iodine.
$$\mathrm{rate}_{\mathrm{label}} = \left(k_{1} + k_{2}\right)[\mathrm{electrophile}]$$
Half the products of the $S_{\mathrm{N}}1$ reaction result in retention while the other half result in inversion. All of the $S_{\mathrm{N}}2$ product results in inversion.
Therefore:
$$\frac{d(\ce{[x]}-\ce{[y]})}{dt}=\frac{k_{1}}{2}[\mathrm{electrophile}] - k_{2}[\mathrm{electrophile}] - \frac{k_{1}}{2}[\mathrm{electrophile}] = - k_{2}[\mathrm{electrophile}]$$
But this means that, at the start of the reaction, the magnitude of the rate of loss of optical activity is smaller than the magnitude of the rate of loss of radioactivity.
The real issue here is that you cannot compare apples and oranges. The rate of loss of radioactivity is a concentration based rate, but the rate of loss of optical activity is an a change in angle over change in time. Under the simplest assumption (using a conversion of unity), this problem cannot be solved.


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Net production of CO2 in plants
I am trying to understand in a qualitative way photosynthesis. I am not a chemist so please have some patience.
I read on the Wikipedia page for photosynthesis that the way plants use $\ce{CO2}$ is through the reaction:
$\ce{energy + 6 CO2 + 6 H2O -> C6H12O6 + 6O2}$
but then for cellular aerobic respiration, I find exactly the opposite reaction:
$\ce{ C6H12O6 + 6O2 -> 6 CO2 + 6 H2O + energy}$
So, from this simplified view where only net reactions are written it looks that this mechanism can be used to transfer and store energy. But from a $\ce{CO2}$ point of view it looks that plants do not "consume" carbon dioxide since what they use via photosynthesis they give back through respiration. Clearly this cannot be true. So where is the catch ?

One must be very careful of 'balanced' equations such as your Eqn 1 when considering photosynthesis.  In PS, oxygen (all of it) is produced from the splitting of water, and it takes  $12 H_2O$ to produce $6 O_2$.  A better equation, maybe, is (discussed here): $\ce{6 CO2 + 12 H2O -> C6H12O6 + 6O2 + 6H2O \tag{1}}$

Plants are able to store energy as carbohydrates, i.e. they can make more carbohydrates than they need for their metabolism  when there is no sunlight, and store it for later use. So the amount of energy in your equations is not equal.
Some plants are offering this stored energy as sugars as nectar or fruits in exchange for pollination and spreading their seeds or use it as energy storage for times when they can't make enough carbohydrates for quick use, e.g. in winter or as storage for high demand, e.g. quick growth to outsmart competitors for space.
In addition, quite much of this fixed CO2 goes into structures like stems, roots and leaves and thus is sequestered for some time from the atmosphere.

Thanks this is useful (+1). So, just considering carbon, I understand that not all CH2O is reconverted via respiration. Some will be reconverted at a later point ( which nevertheless over time gives a zero balance ), but some is also sequestred into stems, root and leaves, or given to insects and will follow a different "chemical" path. I guess that than insects and other microorganism will produce again some CO2 when using these sugars so the calculation of the net effect of CO2 storage is a bit difficult to evaluate. Or am I missing some important mechanism that "stores" this "unused" sugar?

Thank you, Thomas. I think you got it :)
Of course, most of the CO2 sequestration is temporally, but nevertheless an important sink that removes some of it from the atmosphere. In the long run, a little of this fixed carbon dioxide will be buried in the ground and finally form coal and oil again.

Thanks again. It is a bit sad that you that "a little". That would mean that only a little of CO2 is sequestred into a long-term carbon storage by plants. Anyway I guess we should see this is as a cycle. Carbon does not get destroyed anyway ( without nuclear reactions ) and cycles into its different forms. The game would be to have not large quantities of it in its most dangerous form for a long time. So temporary sequestrations help to limit this, even if the longer is this "temporary" the better :)

( PS: FYI I will wait for a couple of days to see if there are other contributions and than also accept your answer )

That's fine, there should be more possible and probably better answers.
The concentration of CO2 in the atmosphere has increased substantially by burning fossil fuels in the past centuries which have needed millions of years to form. It won't be an easy task to get the level down in the long run. Marine algae which will sink to the seabed can make the carbon disappear for some substantial time and in significant amounts, but also that will take time.

Biomass
The net production of $\ce{CO2}$ in all living organisms is close to zero. The argument is that if the total biomass on earth does not change over a period of time, and the net transfer from biomass to other reservoirs of carbon does not change much, the production and consumption should balance out. Wikipedia's article on biomass states:

The total live biomass on Earth is about 550–560 billion tonnes C, and the total annual primary production of biomass is just over 100 billion tonnes C/yr.

Plants only

[OP]But from a $\ce{CO2}$ point of view it looks that plants do not "consume" carbon dioxide since what they use via photosynthesis they give back through respiration.

Plants do produce some $\ce{CO2}$ through respiration, offsetting the $\ce{CO2}$ consumed through photosynthesis, but this is not a net zero balance. For an individual plant that grows, the carbon in its biomass comes from carbon dioxide, so it is a net consumer. When a plant dies and gets decomposed by bacteria, or when it gets eaten by an animal, or when it burns in a forest fire, the carbon eventually turns into carbon dioxide again. Depending on your perspective, you could count that as $\ce{CO2}$ production by the plant, or by bacteria, fire and animals.
Overall carbon distribution
For some context, here is a figure showing the carbon distribution on earth, together with the amount that exchanges from one reservoir to the next (found at http://acmg.seas.harvard.edu/people/faculty/djj/book/bookchap6.html, from cElroy, M.B., The Atmosphere: an Essential Component of the Global Life Support System, Princeton University Press):

This is the pre-industrial situation. The numbers in the boxes ("inventories") show the mass of carbon in $\pu{e15 g}$. The numbers outside of the boxes show the flow per year. So the vast majority of carbon is in the ocean sediments. This is shown as steady state. In the current era, the carbon in the atmosphere increases at the expense of carbon in the crust (fossil fuel).

Your second equation (respiration) happens less than your first equation (photosynthesis) while the plant is growing.
About a third of the mass of a typical plant is cellulose, which is created by linking the glucose that is product of the first reaction. So the fact that you have a plant in front of you at all means that the plant has performed the first reaction more times than the second.


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Peculiar deprotonation of an alkenyl proton
In this synthesis of Punctatin A, there is this peculiar deprotonation step. When subjected to the base $\ce {NaH}$, we would expect deprotonation at the $\alpha$-position, at the side opposite to the double bond. However, what is observed here is, in fact, the deprotonation at the double bond. Not only is the $\ce {C(sp^2)-H}$ bond stronger, there is also no delocalisation of the negative charge in the conjugate base that is produced. Note that the lone pair in the carbanion conjugate base is held in an $\ce {sp^2}$ hybrid orbital that is perpendicular to the $\ce {p}$ orbitals of the enone $\pi$ system. Does anyone have any idea what is going on here?

Decided to convert this to an answer anyway. Disclaimer: I'm not an organic chemist (anymore), and corrections/suggestions are more than welcome.
One distinct mechanistic possibility here is that you deprotonate γ to the carbonyl, then the extended enolate reacts at the position you see, and then after that there's some tautomerisation which brings the double bond back into conjugation with the carbonyl (maybe during workup).

These are specific conditions under which you might expect deprotonation at the γ-position over the other α-position. In particular, the extended enolate resulting from this is more stable than the cross-conjugated enolate resulting from α-deprotonation, so this is called the "thermodynamic enolate". Paquette et al. use this term in the paper (J. Am. Chem. Soc. 1987, 109 (10), 3017–3024):

Toward this end, the thermodynamic enolate of 6b was exposed to [...] 1-bromo-2-methylpropane.

I won't comment on the specifics of this reaction, but will mention that it can be very dangerous if scaled up.
If you wanted to deprotonate the other α-position, you would probably have more success with something like LDA. That deprotonation relies on chelation of the carbonyl oxygen to lithium (via a cyclic transition state*), so goes for the proton nearer to the carbonyl. That would be the "kinetic enolate".

* The six-membered chair-like transition state is widely taught, but is actually of questionable accuracy in this specific case. I think the transition state is cyclic, but is eight-membered or something like that. See e.g. slides 50 onwards of this very informative presentation from David Macmillan's group: "Lies My Sophomore Organic
Professor Told Me".

What can we say about the proton acidity at the gamma position compared to the alpha position at the other side? While in theory, resonance structures can be drawn for the conjugate base resulting from gamma deprotonation, illustrating extensive delocalisation of the lone pair, I am slightly skeptical of the significance of these resonance forms as the oxygen atom and gamma carbon are 3 atoms apart.

Resonance doesn't really care that much whether the oxygen is one or three bonds away, as long as there's an intervening π system. I think the easiest way of looking at it is to not overcomplicate it - pKa reflects the HA/A- equilibrium, and since HA in this case is the same compound, pKa is a direct reflection of the conjugate base stability. The system with linear conjugation is more stable than the one with cross conjugation, unless there's a significant difference in e.g. ring strain which I don't think is the case here.

... But I'm not sure exactly how different the pKa values are. Might be worth a computation...

That's a relatively low yield for an alkylation, it would be interesting to know what the other products are - whether there is a significant amount of the alpha alkylation product

@Waylander, apparently with an allylic iodide they get 40% desired product + 20% dialkylation of the extended enolate at the alpha position (hence the double bond can't move back into conjugation) - but no comment on the reaction with the plain alkyl iodide (they say "exclusive formation of 15", 15 being the desired product)

While Scheme III shows alkylation with 1-bromo-2-methylpropane, the experimental section employs the iodide. Alkylation occurs (54%) with recovery of a lesser amount of recovered starting enone. With the non-allylic electrophile, reconjugation is faster than double alkylation.


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If the gaseous molecules didn't have any kinetic energy would the surface of a filled balloon still be firm?
The most of the ideal gas law equation is based on the fact that the gaseous molecules are moving around and cause collisions and thus have some pressure. If these molecules were to not have any kinetic energy, and thus consequently would not suffer any collisions would the surface of let's say a balloon, a filled balloon, which is firm only due to the pressure offered by the gas inside, still be firm? would the volume of the gad contribute to its firmness and give it shape? won't the volume be affected too

What about the gas around the balloon? What do you think it does?

If ideal gas molecules had had no kinetic energy, there would have been no gas pressure. Inside nor outside the balloon. The outer gas would have been laying on the ground, the inner gas on the bottom of the balloon.
In such a case, the only remaining force, aside of gravity, would have been the elastic forces of the balloon, pulling the balloon rubber together to its original shape and volume.
For real gases compressed to volume about equal to tight arrangement of molecules, a sudden repulsive force and therefore pressure would have progressively appeared, due repulsion of electrons.

Thanks a lot for answering! I still had a question though, given that the gas molecules aren't showing any kinetic energy, they'd probably have a fixed position over time, giving them a fixed shape, and as you said they'd be lying at the bottom. So what would cause to bring about an elastic force in the rubber balloon? Just this matter at the bottom?

Elastic rubber force is gas independent. If you stretch a piece of rubber then after  stretching is released, it contracts black again, regardless of if there is a gas around, or not.

You would not be able to blow up the balloon in the first place, at least now in the way how you are used to. You would be able to fill it by a densely packed gas, but the balloon would be heavy and hard.

That's right. So, the balloon obviously wouldn't hold structure as it would before when filled with a gas, and would instead behave like a deflated rubber balloon filled with a less fluid substance, almost like a solid in its physical properties due to the negligible kinetic energy of the gas molecules

Yes, like solid. And everything around such a gas would freeze solid as well. Rubber would break when you would touch it. Note that even at absolute zero 0 K, matter does have residual kinetic energy, related to zero point oscillations.


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Difference between homogeneous and heterogeneous mixture
When i mix two equal but large number of stones it is possible (though extremely unlikely) that stone a and stone b perfectly segregate themselves into two halves or that they are uniformly distributed but we still call it a heterogeneous mixture because almost always this doesn't happen this is discontinuity in the pattern always given a large enough sample size is taken. However ink mixed in water is homogeneous yes it is distributed uniformly but the ink particles may seperate themselves and lose order and shift to one side of the container (very low chance but still a possibility at a very small time) So my question is how can we truly say if a mixture is homogeneous or heterogeneous when probability and approximations must be taken into account.

I’m voting to close this question because it's incoherent.

Homogenity/heterogenity is a conditional parameter.
Homogenity means that at given scale, resolution or sensitivity, you cannot tell the difference in properties in different places within the mixture.
Sand dunes are homogenous mixture, seen from far enough. Any matter is heterogenous, if observed at atomic level, because of its non-continuous, atom based nature.


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What is the hybridisation of [Co(H₂O)₆]³⁺?
There's a lot of ambiguity in the hybridisation of $\ce{Co^3+}$ in its hexaaqua complex; is it a high spin or low spin complex? As theoretically the CFSE is less than P so it should be low spin, but in a lot of places, it's assumed to be high spin.

Similar question was asked earlier..

Holleman/Wiberg states (translation by me)

$\ce{Co^{3+}}$ usually forms octahedral low-spin complexes, because it is the only way to achieve a high ligand-field stabilization energy. Octahedral high-spin complexes are typically an exception; they are formed with the "weakest" ligands (fluoride): $\ce{[CoF6]^{3-}}$, $\ce{[CoF3(H2O)3]}$ (the ion $\ce{[Co(H2O)6]^{3+}}$ is in the low-spin state).

and further

That octahedral $\ce{Co^{3+}}$ complexes from a magnetic standpoint are in the low-spin state most of the time, while complexes of the isoelectronic $\ce{Fe^{2+}}$ prefer the high-spin state most of the time, is due to the higher amount of ligand field splitting, following the increased charge on $\ce{Co^{3+}}$.

Source: Holleman, Wiberg: Lehrbuch der Anorganischen Chemie, 101. edition, de Gruyter, 1995.


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Calculation of the cell potential of a lemon battery experiment
I have the following problem that I would like to get checked.
I’m told to calculate the cell potential of a simple lemon battery experiment.
After Finding out the oxidizing agents and the reducing agents I find that the cell reaction is as follows:
$$\ce{2H+(aq) + Zn(s) —> H2(g) + Zn^2+(aq)}$$
with a cell potential of $E_\mathrm{cell} = +\pu{0.76 V}$.
Are the reactions and the cell potential correct?

Check the charge balance on the reaction equation. Something's amiss.

Hello, thanks for answering. I have made changes to the reaction. Is it expressed properly?

I am afraid that nobody can predict the true potential of Zn/ Cu lemon battery. It is not a standard system. The electrode potentials you see in the published tables are measured or calculated under unknown or controlled conditions and one of the basic requirement is that the metal must be in equilibrium with its own ions.
The lemon battery, although looks very simple, it is too complicated to deal with.

When you insert the electrodes into the lemon, there are no metallic ions to begin with, i.e., the concentration of Zn$^{2+}$ and Cu$^{2+}$ is essentially zero. Just using plug & chug approach for the Nernst equation will give you infinite potential difference. This is not what is experimentally observed.

1b. Even if we assume that some metal dissolves around in a few seconds, still the metal concentration is not high enough to allow a reliable use of Nernst equation.

Citric acid is not only an acid, but it has complexing/chelating property as well.
Any voltage which comes close the calculated potential from redox tables is a mere coincidence.

I think the proper way to determine the potentials generated by contact potentials. This is the realm of physics. Two dissimilar metals, when brought in contact, will generate a potential difference. See my previous answer Electrochemical reactions in a salt water battery

See https://en.m.wikipedia.org/wiki/Lemon_battery for experimental data.

I upvoted the answer because it clearly points out significant issues with this user-friendly, though misunderstood, voltaic cell. Not only does the “theory” of the lemon cell take unsupportable liberties with the Nernst equation, but the inert cathode composition matters, even though hydrogen gas is evolved. So copper, silver, gold and platinum cathodes do not all give the same cell potential, other things equal.


text:
How does methyl vinyl ketone dimerise with heating?

(A) to (B) is simply reduction, tosylation followed by intramolecular reaction and removal of tosylate. However, I am not able to guess how dimerization will take place, seems similar to Rauhut–Currier reaction, but can't guess. Any hint?

TsOH does not form a tosylate. That's likely just the workup for the borohydride reduction.

My hint for you is that you can determine the which of A through D is mostly likely to be the product (B) based on several factors. Then working backwards, you can figure out what (A) is.

@Zhe Don't TsOH tosylate like TsCl? I didn't know that, if that's true. Thanks, any other way to get the answer? Any proposed mechanism?

That's kind of like how you can (sort of) alkylate stuff using alkyl chlorides RCl, but you can't alkylate stuff using alcohols ROH.

No tosic acid is not like tosyl chloride, in the same way methyl alcohol is not like methyl chloride. I have a putative mechanism, but I thought I'd let you try to figure it out for yourself first.

... I think I'll stop jumping in here....

@orthocresol Wow. The timing is impeccable.

The dimer of methyl vinyl ketone (1) is not shown in the choices in the question. The dimer is A shown below. It is a "tail-to-tail" dimer. Product B is an intramolecular ketal as a probable mixture of epimers. Compare with this paper.

B. P. Mundy, et al., J. Org. Chem., 1978, 43, 2347.

Don't have access to the paper... And I still can't get the mechanism for dimerization....

You can usually predict Diels-Alder reactions by using ionic charges. This one is probably radical in nature. Whoever made up the structure (B) fell into this trap.


text:
Do radioisotopes emitted particles influence their chemistry?
Supposedly, a isotopes have the same chemistry, i.e. how they bond to other atoms. Is this true for radioisotopes?  I can imagine a nucleus emitting particles that interact with existing bonds in its outer shell, i.e. it may break an existing bonds, or reconfigure them. Does this happen at all?

The recoil from a nuclear reaction is frequently enough to break a bond.

The radiated particle can break a thousand bonds and still keep running. Of course it does influence chemistry a great deal. Radiation chemistry is a thing, after all.

An atom that transmutes into another element .... is extremely unlikely to have the same chemistry afterwards.

https://chemistry.stackexchange.com/questions/65809/what-happens-to-a-radioactive-carbon-dioxide-molecule-when-its-carbon-14-atom-de

You asked about nucleus emitting particles. As Karl pointed out that would tend to transmute the atom to another element. There are two forms of radioactivity that would allow an element to stay the same element - neutron emission, and a metastable nucleus which emits a gamma ray.


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What is the easiest way to concentrate a solution without causing it to hydrolyze?
I’ve been working with iron acetate solutions in dilute acetic acid (vinegar) and I’ve been running into a problem. Every time I try to boil off the solvent the solution hydrolyzes to form iron oxides. I’ve given up trying to dry it completely because supposedly iron acetates can’t exist outside of solution, but is there a way of concentrating the solution?

Is it iron (II) or iron (III)? BTW, both exist in the solid state, though not without certain reservations. Anyway, I'd try working with less dilute vinegar, and use a generous excess of it, too.

You can evaporate any solution without heat by vacuum evaporation. But heating iron(III) acetate solution does not produce an oxide. It produces a basic acetate which is insoluble, like iron oxide (and the same color).

See (basic) ferric acetate and ferrous acetate

"If you can't synthesize it, buy it"

Could we introduce acetic anhydrous, converting all the water to acid, and then boil/evaporate? Ulterior question is, is that how the priduct you buy is made?

Or vacuum, room temperature evaporation. Or lyofilization (freeze-drying, but that is seldom available technique for private persons.

Try adding Na2SO4 to the mix and freeze forming Na2SO4.7H2O. Slow, but at least, no oxygen-related chemistry, discussed below.
As a sidebar, you likely have formed basic ferric acetate from boiling, which adds oxygen, to your acidic (vinegar) and aqueous ferrous acetate mix.
Likely reaction mechanics are complex, see, for example, Table S1 here . Per the source my take of a possible reaction sequence:
$\ce{ O2 (aq) + Fe^{2+} → Fe^{3+} + .O2− }$ ( moderately fast) )
$\ce{ . O2− + Fe^{2+} + 2 H+ → Fe^{3+} + H2O2}$  (fast )
$\ce{ H2O2 + Fe^{2+} → Fe^{3+}  + .OH + OH− }$ ( slow)
$\ce{  .OH + Fe^{2+} → Fe^{3+} + + OH− }$ ( fast )
Net radical reaction:
$\ce{ O2 + 4 Fe^{2+} + 2 H+ → 4 Fe^{3+} + 2 OH-}$
The creation of FeOOH is also possible in less acidic conditions per his source: 'Air Oxidation of Ferrous Iron in Water'.

Some good news here as the redox potential of iron (III) to iron (II) is very favorable:
Fe +++ + 1 e -   ---->  Fe ++   + 0.77 V
Almost as strong as silver 1 to metallic silver
Ag +  + 1 e -    ----> Ag.  + 0.80 V

You may try to extract iron (II) acetate into ether, which can be much more easily removed by vacuum distillation (under nitrogen for safety)$^1$.

It is very difficult to "boil off" water from vinegar$^3$ (acetic acid) because their boiling points are very close (100 C and 118C). With vinegar, you are probably boiling off all of the acetic acid$^4$.  Iron (II) hydroxide is almost insoluble in water, and could be formed during attempts at distillation as:
2 CH3COO- + Fe ++  + 2 H2O  ---->  CH3COOH (volatile) + Fe(OH)2 (solid)
And finally, unfortunately, O2 is strong enough to oxidize even Iron (II) to Iron (III) under acidic conditions:
O2 + 4 H+ + 4 e-  ----> 2 H2O. + 1.23 V
or even:  H2O2 + H+ + 2 e-   ---->  2 H2O. + 1.76 V. $^2$
But in basic solution:
O2 + 2 H2O + 4 e-  ----> 4 OH-   + 0.39 V. (no reaction)
So, one really would not want to acidify iron (II) and heat it in the presence of oxygen$^5$.
$^1$ this may also reduce unwanted oxidation of Fe (II) to Fe (III)
$^2$ what is this colorless ferric solution?
$^3$ vinegar, being dilute acetic acid, boils much closer to 100 C.
$^4$ acetic acid would have significant vapor pressure at 100 C.
$^5$ If a reducing agent such as iron wool is used, conceivably, fractional distillation could be employed, or, one could use more concentrated acetic acid and iron carbonate in the synthesis process.


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(a) Determine the symmetries (irreducible components) of the MOs responsible for the d-d metal-metal quadruple bonding in this molecule
The Re2Cl8 2- ion is composed of two square planar ReCl4 units joined by four Re- Re bonds (quadruple bonding) and has D4h symmetry.
(a) Determine the symmetries (irreducible components) of the MOs responsible for the d-d metal-metal quadruple bonding in this molecule.
This is the entire question. However, I only need help genereating the irreducible representations. There are four sets of irreps that need to be generated that are based on the dz^2 orbital, dxy orbital, dyz orbital, and dxz orbital. I have only been able to determine the irrep for the dz^2 orbital. Here is what I did:

I transformed the dz^2 orbital under every symmetry operation from the D4h point group. (E, 2C4, C2, 2C2', 2C2'', i, 2S4, sigma h, 2 sigma v, and 2 sigma d. )
I got a reducible representation of E=2, 2C4=2, C2=2, 2C2'=0, 2C2''=0, i=0, 2S4=0, sigma h=0, 2 sigma v=2, 2 sigma d=2
Lastly, I used tabular reduction to get that dz^2 = A1g +A2u

For the other three d orbitals I can not seem to get the correct irreps.

I'm confused how you are getting your reducible representation. Since dz^2 is a single orbital, it should only correspond to a irrep. I'm not sure how you are getting 2 for the character of the identity operator. The symmetry of the Cartesian functions can be found in the last two columns of most character tables if you want to check your work.

I can add a picture of the work for the dz^2 irreps. The dz^2 irreps were solved by my instructor.  I know how to look at the character table to get each irreps corresponding to the d-orbital. But I am asked to find the irreps by hand. This is also very confusing to me as far as how to do it. I know that you treat the d orbital with each operation and see how the orbital changes. If it stays the same then the value=1, if it moves the value is =0, if it inverts the value is =-1

The procedure you describe sounds correct. So when you apply E to the dz^2 orbital, what happens to it? As to the character table, you can just use that to check that your by hand result is correct.

Under the E operation nothing happens so I would have a value of +1, or in this case +2 because I am looking at both d z^2 oprbitals on the Re2

Okay sorry I see what you are doing now. Your method will work fine. Could you should what you are getting for the reducible representation of one of the other orbitals.

I just added the photo of my work. I honestly got frustrated with trying to visualize the d orbitals and how they transform with each symmetry operation so i'm quite sure my work is not right. If you could help explain, I would really appreciate it. I have been working on this problem for a few days now.

I'll try to walk through how it works for $d_{xy}$. You will need to consider $d_{yz}$ and $d_{xz}$ as a pair later when you attempt those.
Obviously $\chi(E)=2$. Now for the $C_4$ rotation, just think about rotating one of the orbitals in the $xy$ plane (the same thing will happen to the other orbital, so we will then double the character). When we do this, the orbital remains in place, but it's positive and negative lobes invert. So it will contribute -1 and for both orbitals $\chi(C_4)=-2$. For $C_2$, the rotation again leaves the orbital in place and leaves the lobes in each location with the same sign so $\chi(C_2)=2$.
$\chi(C_2')=\chi(C_2'')=\chi(i)=\chi(S_4)=\chi(\sigma_h)=0$ since all of these operations involve a rotation/reflection around a parallel line/plane so they will move the orbitals.
$\sigma_v$ runs along the $\ce{Re-Cl}$ bonds. Along and on either side of this plane, the sign of the lobes is the same and so $\chi(\sigma_v)=2$. $\sigma_d$ runs between the $\ce{Re-Cl}$ bonds and so this plane inverts the sign and $\chi(\sigma_d)=-2$.
Plugging these into a calculator I get a $b_{1g}$ and a $b_{2u}$ irrep. I will leave it to you to figure out which of these is the bonding combination and which is antibonding.

Thank you, this makes a lot more sense. One question, what do you mean by consider dyz and dxz as a pair? I know on the character table they are a degenerate set, but how will that affect how I go about finding their irreps?

So when I apply the C4 operation to dxz I get dyz, so what does that mean as far as the character? (is it +2, +4?) Also when I do the C2 operation for the dxz orbital, I get the inverted dxz orbital so is the character -2?

Basically, I just mean that the irreps you eventually get will be degenerate and if you just try to break the reducible representation of one orbital or the other into irreps it won't work.


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Can reactions produce microwave or radio wave radiation?
I know that many reactions produce visible light,  heat and sound,  but is there a reaction (preferably one example for an inorganic and organic one)  that could produce electromagnetic waves in the microwave and radio band?

Interestingly, the reaction *H + *OH -> H2O + Photon (of Light) but no radiation.

Explosions produce high velocities and dust and that combination can produce moving charge and that can produce radio noise. Radio noise from dust storms is a known phenomenon on Mars, but this is probably too much of a stretch for this question. Also I think pulling tape can make RF noise for similar reasons, electric charge flow can even make X-rays! https://www.nytimes.com/2008/10/28/science/28xray.html

The production of microwave radiation probably accompanies most any exothermic reaction in which a molecule is the product, because microwave radiation results from rotational motion (and occasionally barrier tunnelling as in NH3) and exothermic reactions initially have a non-boltzmann distribution of energies. Only in magnetic fields as in an NMR machine would you expect to emit radio frequency radiation as the energy gap here is far smaller than any molecular process such as rotation or vibration.

An ammonia maser qualifies as an inorganic RF-emitting reaction. As the nitrogen changes position in the $\ce{NH3}$ molecule, ~24 GHz radiation is emitted.
Similarly, organic methanol, $\ce{CH3OH}$ makes a maser found in nature, emitting at ~36.2 GHz and also at ~6.7 GHz.
BTW, if the reactants are  moving away rapidly (e.g. interstellar gases), visible radiation is red-shifted to microwaves... or lower, so wavelength depends on your frame of reference. So, as Brumley and Stevens state, listen in to radio astronomy.
Though these might be due to physical excitation, and not an exact answer for your question, an interesting experiment would be to form $\ce{NH3}$, e.g. with the Haber process, and try to detect the 24 GHz radiation, perhaps due to initial formation of the molecule in a metastable state, or due to rearrangement. A similar experiment could be done with methanol, forming it from $\ce{CO}$ and $\ce{H2}$, looking for 36.2 GHz or 6.7GHz signals. I'd be very interested to know if you or anyone else tries that!

you've piqued my curiosity, so I've just asked Does the nitrogen atom move with in the molecule in an ammonia maser, or does the molecule flip?

What about ring current in aromatic compounds like benzene if benezene undergoes reversible hydrogenation in an equilibrium won't that produce RF

@DrMoishe Pippik , I meant a direct chemical   reaction not a physical action on a molecule

@mohamed, I've edited the answer.


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Assumption behind Gibbs energy and maximum work

I'm somewhat confused by this derivation (in Schroeder's Book of Thermal Physics) of the fact that in a closed system, the change in the Gibbs Energy is always less than the Non-Expansion work done on the system.
If we take the change in the entropy of the universe (system + ideal surroundings maintained at constant temperature and pressure also assuming that reactions are taking place in system only i.e. the composition of various species present in surroundings remain constant), then it turns out (in this derivation) that it equals to negative of change in Gibbs Energy divided by Temp of system (which is same as that of surroundings).
Now since entropy of Universe always increases it turns out that Gibbs Energy of system will always decrease. But this is only true in absence of Non-Expansion work.
However during this derivation nowhere it's mentioned that 'no non expansion work assumed' yet the result only holds true in absence of non expansion work.
I think that maybe it was assumed somewhere and I'm not able to spot where. Please help me to clear my doubt.

The whole point of Gibbs free energy is to quantify a bound on possible non-expansion work at constant p and T. That non-expansion work is implicit as part of the term "dU" which in the reversible case can be written as
$$dU = -pdV + TdS + dw_{\text{non-exp,rev}}$$
The $-pdV$ term is the expansion work if the process is carried out reversibly. You remove that contribution by adding a  $pdV$  term:
$$dU + pdV= -pdV + TdS + dw_{\text{non-exp,rev}} + pdV = TdS + dw_{\text{non-exp,rev}}$$
Finally if you subtract the entropy term you are left with the maximum possible non-expansion work (obtained when the process is performed in a reversible fashion):
$$dU + pdV - TdS= dw_{\text{non-exp,rev}}$$

I think you did not understand my question.  What I mean to say that in general the change in Gibbs energy is less than the non expansion work done on system, and in the absence of non expansion work the Gibbs energy of system falls. However during this derivation the final result is the latter one. That's the point which confused me. Because to get the latter result one has to explicitly assume the absence of non expansion work, which I dont think is assumed in this proof. Can you please point out where in the derivation they assumed absence of non expansion work?

@ShivanshJ please see my edit. Hopefully clearer now that I understand your question better.

Buck Thorn: Thanks! Now it's much clearer!

A small question (to make sure I understand correctly). In equation 5.27 (In the photo I attached with my question). This thermodynamic identity assumes no non expansion work done on system. Am I right?

@ShivanshJ Yes, and in hindsight you are right, that derivation is not clear.

Buck Thorn: Can you please recommend an article or a book where I can read more about this. The most general form of thermodynamic identity. I'm asking this because in the book I mentioned (in my question) explains P/T as the rate of change of entropy w.r.t volume in a beautiful and intuitive way. However now I know that this is not complete there are other terms (apart from 3 terms mentioned in book) which have to be included in the rate of change of entropy equation (thermodynamic identity).

At an introductory level Atkins, more advanced maybe McQuarrie.

Okay Thanks! Btw I've been pondering over this since last night and came to this conclusion only that something is missing in equation 5.27.

Yes it's not general, you can include more extensive coordinates as needed. For instance, if there is area expansion work (as in a bubble), you might include the area A as an extensive coordinate. The derivative of the energy wrt area is the surface tension (a type of generalized force).

BTW writing "free energy as a force toward equilibrium" is already bound to be potentially confusing. A force is a derivative of energy wrt some coordinate.


text:
Distinguishing between potassium chloride and sodium chloride
I have these very large blocks of salt intended for a water softener in my house, but I'm not sure if they're potassium chloride or sodium chloride, according to what I've read they're both widely used for water softeners. I've thought about distinguishing them by weight but I don't have any sodium chloride with the same granularity as the salt blocks.

If the blocks are pure salts, flame coloring may be possible.

They taste different.

@Mithoron I used to distinguish various xylenes by their smell when I worked closely with them, but now I won't be able to do so, unless I have a standard to rely upon. Same thing here.

@TAR86 Flame coloring worked great, turned out it was sodium chloride in case anyone was wondering

Wonder how pure the KCl for such a use would be. Sodium gives a very strong flame test, potassium much weaker. So for even like 5% sodium (or less..) you'd see a yellow Na flame. It would help to have a piece of cobalt blue glass to look though.

@MaxW According to this site they're going to be upwards of 99% purity http://blog.watertech.com/what-type-of-salt-should-i-use-with-my-water-softener/

@Faissaloo - Super fantastic! That is much lower than I guessed.

Just a spectroscopic suggestion. Even if there is 1% NaCl, it will mask potassium emission. This is why potassium emission is usually seen with the help of cobalt blue glass.

Flame test is the best test for distinguishing alkali metal ion but it can also be distinguished chemically. In basic qualitative analysis, there is a method of distinguishing $\ce{Na+}$ and $\ce{K+}$. For $\ce{Na+}$, potassium pyroantimonate is used and for $\ce{K+}$ sodium cobaltinitrate/picric acid is used. See: https://chemistry.stackexchange.com/questions/51653/how-to-separate-and-analyse-a-sample-of-cations-of-alkaline-and-earth-alkaline-c/52051#52051

@ToddMinehardt It's sold simply as 'block salt'and supplied by the water softener company, they don't provide any details beyond that and it's not added to the drinking water. I also don't see why that would be a reason to close the question.

Related: How can I identify whether a substance is KOH or NaOH?

Per Wikipedia:

..., MgSO4·H2O, can be combined with a solution of potassium chloride to produce potassium sulfate.

So, cooling an aqueous solution KCl in the presence of Epsom's Salt:
$\ce{2 KCl + MgSO4 <=> MgCl2 + K2SO4}$
creates potassium sulfate, which falls out of the solution forming distinct orthorhombic crystals.
If Na2SO4.7H2O is created instead from aqueous NaCl in place of KCl, you will get something that resembles a gluey paste.

Nice. Pure KCl might be hard to find at home, but a grocery store will have plenty of NaCl. Wonder what would happen with Morton Lite salt Na:K::1.4:1

Cream of tartar is potassium bitartrate. It would be low in Na. Not sure if there would be so little as to not see any yellow in the flame.


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Copper acetate solution turning clear
I made a solution of copper acetate using 5% Acetic acid 3%hydrogen peroxide and copper wire and copper pipe and it had produced a blue solution of copper acetate. I left it over night to continue reacting, but when I came back In the morning the solution turned clear. I didn’t do any Stoichiometry so I don’t know if they’re was a excess of H2O2 or a excess of acetic acid. I’m wondering if you know what happened

Was a precipitate created?

Was anything precipitated?  Corrosion on the pipe?

I am asking again whether any precipitate was created and whether any testing was done on samples.  Disagreement has emerged in the answers and we need more experimental evidence to resolve it.

No there is no precipitate and after sitting for a few days nothing has changed, and I may continue testing.

Wikipedia states that upon heating, copper(II) acetate reacts with metallic copper to make copper(I) acetate, which would be colorless.  While this is no guarantee of what happened with your solution, you can mix a sample of the solution with saltwater (chloride ions).  White precipitate = copper(I).
In light of a comment, I am compelled to expand on the possibility of copper(I) in solution.  According to the Wiley Online Library, this salt is soluble in water with hydrolysis to cuprous oxide.  At the same time we have the copper Pourbaix diagram reported from Ref. 1, indicating that the oxide is stable only above a pH of 4.

This means any significant leftover acetic acid could prevent the hydrolysis and keep the copper(I) in solution, apparently stabilized by complexing with the acetate ion given that disproportionation is not reported.  (Such a stabilization is not included in the above diagram but would contract the Cu2O region further.)
Since this is only indirect evidence, again testing for a copper(I) chloride precipitate by mixing a sample with saltwater is recommended.
Reference
1.
Oh, Youn-Jin & Park, Gyung-Soon & Chung, Chan-Hwa. (2006). Planarization of Copper Layer for Damascene Interconnection by Electrochemical Polishing in Alkali-Based Solution. Journal of The Electrochemical Society - J ELECTROCHEM SOC. 153. 10.1149/1.2200288.

In the presence of H2O2, Cu(I) cannot survive, assuming there is some residual H2O2.

Is there a way to reverse this

@mfarooq right, but we don't know if h2o2 would be present.  It may not survive itself with the copper pipe.

@15 I would test for the reduction to copper(i) before pursuing any sort of recovery.  Let us know what happened if you run the suggested test.

The reaction is Cu + Cu(II) <--> 2 Cu(I) and is an equilibrium reaction, so this should not be an explanation as to how the solution becomes colorless.

Usually solutions of Copper(I) are colorless. But they are very sensitive to oxygen; They only stay colorless in closed containers. In presence of air, they soon become colored.

A reaction equilibrium is established between Copper metal and Cuprous in acetic acid and H2O2:
$\ce{Cu + Cu(II) (aq) <=> Cu(I) (complex)}$
where I assume that initially, the cuprous acetate is soluble in excess acetic acid, forming a so-called acid salt. As such, the coloration of the solution is maintained due to reaction the equilibrium.
However, in time, with excess H2O2 and acetic acid acting on Copper metal, the acetic acid is likely consumed. This now results in the reaction:
$\ce{Cu + Cu(II) (aq) -> Cu(I) (s)}$
where now an insoluble cuprous acetate is formed, resulting in a clear solution also containing a clear cuprous acetate crystal precipitate.

This also may not really work.  Wiley Online Library lists cuprous acetate as soluble in water with hydrolysis to cuprous (not cupric) oxide, so either you get the precipitated oxide or, with remaining acetic acid to suppress the hydrolysis, it remains in solution.  If you read carefully you will see that I had recommended testing for copper(I), not said definitely it was there.


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How to create .CIF file from DRX data in a paper, for compound not found in crystallographic database
I tried to recover a .CIF file with the crystal structure for the compound discussed in this 2006 paper, strontium hydroxide octahydrate - $\ce{Sr(OH)2·8H2O}$ - from Cambridge Crystallographic Data Centre (CCDC), but couldn't find it. The paper itself has tables with information recovered from x-ray diffraction data, as shown below.

Is there any free software or open source tool where I can just type the data from these tables to assemble a crystal structure like the ones I'm used to download from the database?

If you cannot find any software, as the system is tetragonal and you know $a=b \ne c$  and $\alpha=\beta=\gamma=90$ and the fractional positions given the table, you should be able to reproduce the structure quite easily.

Recently, this question was asked on materials.SE

You can fairly easily create your own CIF with VESTA using the space group and unit cell parameters from Table 1 and atomic coordinates from Table 2.

You can't find this entry in CCDC's CSD, because the scope of this database are small organic, or organometallic molecules' crystal structures.  (For large organic molecules, analogue studies' data may be found, e.g., on wwpdb.org.)  Because your compound is considered as inorganic*, you'll find it in the ICSD, the inorganic crystal structure database.
Note, the paper includes the statement:

«Further details of the crystal structure investigation may  be  obtained  from  Fachinformationszentrum Karlsruhe, 76344 Eggenstein-Leopoldshafen, Ger-many (fax: +49-7247-808-666; e-mail: crysdata@fiz-karlsruhe.de,   http://www.fiz-informationsdienste.de/en/DB/icsd/depotanforderung.html) on quoting the depository number CSD-417100.»

Because CCDC and ISCD established in 2018 a search interface in common, and entries' labels in both databases differ from each other, it is easy to find the record here.  It is not as comfortable as their own GUI and API interfaces (example ICSD), but it is possible to download the record's .cif.
Note that complementary to these two databases, there are others like e.g., the American Mineralogist Crystal Structure Database and the COD.  Entries 2100155, 2100156, and 2100157 of the later for example point to the .cif about $\ce{Sr(OH)2 ·8H2O}$ of a study by Ricci et al. at 20, 100 and 200 K, published 2005 in Acta Cryst. B.

*) A simplified definition of «organic compound» is that it is molecular and contains at least one carbon atom excluding carbon oxides ($\ce{CO, CO2}$, and others), their hydrates, and simple metal carbonates ($\ce{H2CO3, HCO^-_3, CO^{2-}_3}$).


text:
Preferred IUPAC name of Protoporphyrin IX
I was looking up protoporphyrin IX and was surprised to find that no two websites gave the same IUPAC name for the compound. Here are some examples:

3,3'-(7,12-Diethenyl-3,8,13,17-tetramethylporphyrin-2,18-diyl)dipropanoic acid (EPA)
3,3'-(3,7,12,17-Tetramethyl-8,13-divinyl-2,18-porphyrindiyl)dipropanoic acid
(CHEMSPIDER)
3-[18-(2-carboxyethyl)-8,13-bis(ethenyl)-3,7,12,17-tetramethyl-22,23-dihydroporphyrin-2-yl]propanoic acid (PUBCHEM)

Anyone know if an IUPAC name is agreed upon among scientists or if there is a preferred IUPAC name?

I'd assume that the name from pubchem is right. I've not kept up all the naming rules, but they must have changed a number of times since organic chemists are clever fellows and create all sorts of weird molecules that no one considered that would have to be named. Would you have thought that a  buckyball could actually be created?

At a first glance, the Chemspider name with "vinyl" and "2,18-porphyrindiyl" cannot be the preferred IUPAC name.

I think both EPA and ChemSpider names are acceptable. However, I don't agree with dihydroporphyrin part in PubChem name. This is because the Gold Book gives following framework for porphyrin nucleus:

According to the numbering the IUPAC name of protoporphyrin IX is 3,3'-(7,12-Diethenyl-3,8,13,17-tetramethylporphyrin-2,18-diyl)dipropanoic acid. This name based on least locant numbers for ethynyl group, which have higher priority before methyl groups (3,3'- is based on positions of propanoic acid). The same name is given to protoporphyrin IX in Ref.2, without 3,3'- locants.
References:

IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific Publications, Oxford (1997). Online version (2019-) created by S. J. Chalk. ISBN 0-9678550-9-8. https://doi.org/10.1351/goldbook.
IUPAC-IUB Joint Commission on Biochemical Nomenclature (JCBN): Nomenclature of tetrapyrroles (Recommendations 1986); Prepared for publication by Gerard P. Moss, Queen Mary College, London, Eur. J. Biochem. 1988, 178(2), 277-328 (https://doi.org/10.1111/j.1432-1033.1988.tb14453.x) (PDF).


text:
Does the nitrogen atom move with in the molecule in an ammonia maser, or does the molecule flip?
This intriguing answer to Can reactions produce microwave or radio wave radiation? says (in part):

An ammonia maser qualifies as an inorganic RF-emitting reaction. As the nitrogen changes position in the $\ce{NH3}$ molecule, ~24 GHz radiation is emitted.
Similarly, organic methanol, $\ce{CH3OH}$ makes a maser found in nature, emitting at ~36.2 GHz and also at ~6.7 GHz

The first link is to The Ammonia Maser in Feynman's Lectures on Physics and it includes a picture of a triangular pyramid with the nitrogen pointing "up" in one case and "down" in the other as an illustration of the transition.
But how does it get there? Does it "tunnel through" the nitrogen triangle, or does the whole molecule flip around? Can we even ask that question? Could we, if for example we labeled the corners with $\ce{{}^{1}H}$ $\ce{{}^{2}H}$ and $\ce{{}^{3}H}$ as a gedankenexperiment to make the two distinguishable?

The nitrogen atom pierces the triangle of the hydrogen atoms to go on the other side. If you wanted to label something then it should be the hydrogen atoms. There is only one nitrogen atom. But if you arranged $\ce{^1H}$, $\ce{^2H}$, and $\ce{^3H}$ clockwise then you could tell if the nitrogen atom was above or below the triangle of hydrogen atoms.

@MaxW yikes $\ce{HN3}$! sorry about that. Because it's such a glaring error I'm going to fix the question accordingly. But how does one know that it pierces the triangle? Hopefully a posted answer will support it's postion. Thanks!

Consider the ammonia molecule stationary in its own frame of reference. With the emission or adsorption of a photon the center of gravity would have to recoil appropriately from the photon's momentum.

@MaxW That's always true for any system of atoms that emits a photon. A molecule will recoil after radiating a photon no matter what the nature of the transition nor of what if any internal rearrangement happens. If you know how it is known that the nitrogen pierces the triangle then pease consider writing that up as an answer where it can be voted on. Thanks!

Tunnelling occurs from one structure to the other and as a result the levels below the barrier are doubled. The double well potential and some energy levels is shown in this answer, https://chemistry.stackexchange.com/questions/132449/particle-in-a-box-with-different-barriers-between-two-molecules/132499#132499. The spectroscopic data shows that in ammonia the barrier is approx 2000 cm$^{-1}$. The N atom is 38 pm out of the plane of the 3 H atoms and the lowest energy transition is 0.97 cm${-1}$. see Swalen, J. D.; Ibers, J. A. J. Chem. Phys. 1966, 36, 1914 - 1918.

Thanks! I'm on a bus right now, I see it in 1962, when I get to a library I'll see if I can access it. I wonder if in the following 60 years there's been any independent way to physically verify that this is really quantum mechanical tunneling proper without question, and all other possibilities have been ruled out, or if it's still "the numbers match so it must be right".

I think that even in 1966 the idea of tunnelling was at least 30 years old.  I think that spectroscopy is /was the way to test this. There are many other examples in molecules, also radioactivity and scanning tunnelling microscopes.

@porphyrin I'm familiar with electron tunneling in STM but when it's a complex system like a whole atom I guess I'm just curious how it's known that that's definitely what's happening in this case.

If there was not tunnelling in ammonia the energy levels in each of the two wells would be independent of one another, i.e. be single in each well, i.e. there would be no interaction between one half and the other, the wavefunction would not be able to pass from one well to the other, which is the signature of tunnelling. As there is interaction, there is the possibility of tunnelling,  the levels in each well are doubled and split in energy and this is confirmed by spectroscopic experiment. It is the only reasonable explanation.

continued. The theory is the same whether it's electrons or whatever tunnelling, it's just the extend of this that changes, being far less probable as the mass increases.

@porphyrin There are several possible ways think about going from initial to final state and trying to cover all of them in comments is probably "over-constrained" but until all possible explanations are addressed it's hard to say "it's the only..." I think an answer post would probably provide more space. Just for example though, what rules out the much lighter hydrogens from tunneling?

if I remember correctly from my undergraduate lectures, all saturated N centres (ammonia & amines) flip continually back and forth at a very high rate. This is the reason that a tertiary amine with 3 differing R groups is not chiral. This was always explained by the sp3 orbital (that the lone pair on the Nitrogen atom occupies) has a tiny probability of being on the opposite side of the N atom. This allows the N centre to invert with a low energetic barrier.

@uhoh You should consider the reduced mass of the system as the mass that tunnels. In a simplified picture, the hydrogen atoms are considered to move simultaneously with respect to the nitrogen atom and the reduced mass is given by $\mu_\text{red}=3m_\text{N}m_\text{H}/(m_\text{N}+3m_\text{H})\approx 2.5$ u, which is a relatively light and can easily tunnel. Another proof that the ammonia really tunnels is by comparing the observed splitting of NH$_3$ with that of ND$_3$ and NT$_3$ which follows exactly what you expect for tunneling.

The reason why QM tunneling often plays a role in masers based on molecules (in methanol, the hindered internal rotation of the OH group with respect to the CH$_3$ group also represents a tunneling motion), is that the tunneling splitting provides nearby energy levels of opposite parity which can be used to generate a population inversion in the upper component of the tunneling level which can decay to the lower level via dipole transitions providing the maser radiation.

@Paul I think you have everything ready to post an answer. It's really hard being on the receiving end of great insight compressed into SE comments. I wrote this a few years ago: What is the overall positive result of leaving compressed answers as comments?. Probably "another proof" should really be "additional strong evidence". Anyway, all that needs to be answered is the question posted and based on your comments I think you have it!

Let me expand a little on my earlier comment. If we only consider the inversion of the ammonia molecule (we ignore rotation and vibrations other than the inversion mode), the Schrödinger equation is given by
$$
-\frac{\hbar^2}{2\mu_\text{red}}\frac{\text{d}^2\psi{z}}{\text{d}z^2}+V(z)\psi(z)=E\psi(z),
$$
with $\mu_\text{red}$ the reduced mass and $z=d_\text{NH}\sin{\alpha}$, where $d_\text{NH}$ is the NH bond distance and $\alpha$ the so-called umbrella angle. The potential V(z) is shown in the link given by porphyrin in one of his comments. Although the reduced mass does, in principle, depend on $\alpha$ (see Eq. 4 of this paper), the dependance is weak and we may consider that the hydrogen atoms move simultaneously with respect to the nitrogen atom. In this case
$$
\mu_\text{red}=\frac{3m_\text{H}m_\text{N}}{3m_\text{H}+m_\text{N}}\approx2.5\text{ u}.
$$
The tunneling does not change the center of mass (c.o.m.) of the molecule, so while the nitrogen moves up, the plane of hydrogen atoms moves down to ensure the stationary c.o.m. (and vice versa). I made the following animated gif at some point during my PhD (sorry for the watermark, but I've  had some negative experiences in the past):

The first paper - to my knowledge - that treats the tunneling of ammonia appeared already in 1932. The first application of QM tunneling was done by Gamow appeared 4 years prior to the ammonia work and solved the mystery of the huge range of observed lifetimes of alpha emitters (20 orders of magnitude).
In the 1932 paper, Dennison and Uhlenbeck also derive the magnitude of the tunneling splitting as a function of the reduced mass
$$
E_\text{inv}=\frac{h\nu_0}{\pi}\exp\left \{-\frac{2\pi}{h}\int_\limits{-z_0}^{z_0}\left [2\mu_\text{red}\left (V(z)-E \right ) \right ]^{1/2} \right \},
$$
where $\pm z_0$ are the equilibrium positions corresponding to the potential minima, $\nu_0$ is the vibration frequency in one of the two wells and $E$ is the energy of the system with respect to the potential minimum. This expression is derived from the WKB approximation and it basically is the product of the tunneling probability with the frequency with which the reduced mass 'hits' the barrier. As can be seen from the formula, the tunneling frequency scales exponentially with the square root of the reduced mass, which is also observed experimentally. The following picture is taken from my thesis and basically a copy of Fig. 2 of this paper.

If the molecule would simple rotate instead of tunnel, the change in the energy splitting would be inversely proportional to the reduced mass (via the moment of inertia) and this is clearly not observed.

This is of course an excellent answer! I'm off to the library today to get these...

hmm... I'm not sure what happened but I don't seem to have followed through on this, So I'm off to the library today again @@ to get these...  ;-)


text:
IUPAC nomenclature for tricarboxylic acids

What is the rule for naming compounds with three carboxyl groups?
I have the two following coumpunds

4-(Carboxymethyl)heptanedioic acid
Pentane-1,3,5-tricarboxylic acid

They both have 7 carbon atoms in their parent chain, yet the second coumpunds is called a pentane. Also both have three carboxyl groups yet the first is a dioc acid. Also in the second one, its name is written as tricarboxylic acid. Why not trioic acid.
All these anomalies are confusing me. Can anybody explain why both are named so?

Related: IUPAC name for citric acid

Systematic names of carboxylic acids are generally formed using the suffix ‘oic acid’ $\ce{{}-(C)OOH}$, or ‘carboxylic acid’ $\ce{{}-COOH}$, or the prefix ‘carboxy’ $\ce{{}-COOH}$. Note that the $\ce{{}-(C)OOH}$ group of ‘oic acid’ formally replaces a terminal $\ce{{}-CH3}$ group whereas the $\ce{{}-COOH}$ group of ‘carboxylic acid’ or ‘carboxy’ is a usual substituent that replaces just a $\ce{{}-H}$.
The relevant rules in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) read as follows.

P-65.1.2.1 Carboxylic acid groups, $\ce{{}-COOH}$, that conceptually replace a $\ce{{}-CH3}$ group of methane or terminate an unbranched hydrocarbon chain are named by replacing the final ‘e’ of the name of the corresponding hydrocarbon by the suffix ‘oic acid’. No locants are necessary to denote the positions of the carboxylic acid groups in a hydrocarbon chain; locants are used when hydrocarbon chains are modified by skeletal replacement, as shown in P-15.4.3.2.3. Except for formic acid, acetic acid, oxalic acid (see P-65.1.1.1), and oxamic acid (see P-65.1.1.1), systematically formed names are preferred IUPAC names; the names given in P-65 .1.1.2 are retained names for use in general nomenclature.

That’s why the name of the unsubstituted parent structure of the first compound that is given in the question is a heptanedioic acid.

P-65.1.2.2.1 If an unbranched chain is linked to more than two carboxy groups, all carboxy groups are named from the parent hydride by substitutive use of the suffix ‘carboxylic acid’, preceded by the appropriate numerical prefix ‘tri’, ‘tetra’ etc. and appropriate locants.

That’s why the name of the second compound is pentane-1,3,5-tricarboxylic acid.

P-65.1.2.2.3 The prefixes ‘carboxy’ and ‘oxalo’
  When another group is present that has priority for citation as suffix, for example, a free valence, or when all carboxylic acid groups cannot be described by a suffix, carboxylic acid groups, $\ce{{}-COOH}$, are indicated by the preferred prefix ‘carboxy’ (also used in general nomenclature). The prefix ‘oxalo’ is recommended as the preferred prefix for $\ce{{}-CO-COOR}$, but cannot be used to lengthen a carbon chain. In general nomenclature, the compound prefix ‘carboxycarbonyl’ may be used, but the compound prefix ‘carboxyformyl’ is not recommended.

That’s why the complete name of the first compound is 4-(carboxymethyl)heptanedioic acid.


text:
Formation of Acetal from Hemiacetal?
In the reaction mechanism of formation of Acetal from Hemiacetal, why doesn't H+ attack the Oxygen of OR Group instead of the OH?
Since O in OR group has +I effect from two alkyl groups, shouldn't it donate its lp more easily?

Of course. The point is, you can protonate on OR. But after you protonate on OR, what is the next step? Either ROH is kicked out (which brings you back to a carbonyl), or the proton falls back off and you got back to square one. That's what we sometimes call an unproductive step: it most certainly can happen, but it doesn't get you to the product which you want to get. Only by protonating on OH can you eventually get to an acetal. If the equilibrium is indeed biased towards acetal, then the system will get there eventually, and it has to get there by protonating OH, not OR. @user93057

The difference in protonatability between the two Oxygens due to +I effect is not enough that the OH is never protonated. Some is, and as that is the only pathway by which the reaction can proceed, that is what appears in the reaction mechanism diagram.

The acetal reaction is performed in $\ce{H+/ROH}$ condition (e.g., $\ce{H+/CH3OH}$). The general mechanism can be found in any Organic Chemistry textbook. The reaction is reversible. The first step is protonation of aldehyde (or ketone) carbonyl oxygen by an acid ($\ce{H+}$):

The protonation makes the carbonyl carbon more susceptible to nucleophilic attack by available alcohol (e.g., methanol in the scheme). This gives the tetrahedral protonated hemiactal intermediate, which undergoes deprotonation to give stable hemiacetal. All of these steps are reversible, but forward reaction is favored due to highly available alcohol molecules (Le Chatelier's Principle).
The hemiketal has two oxygen molecules, $\ce{OH}$ and $\ce{OR}$, both of which are susceptible to protonate. However, if $\ce{OR}$ is protonate, reaction would go backward. Of cause, it protonates because the reaction is reversible (see the scheme). Yet, more favored is $\ce{OH}$, which would subsequently eliminate as a $\ce{H2O}$ molecule. The resultant positively charged carbonyl compound undergoes second nucleophilic attack by another $\ce{CH3OH}$ molecule to give protonated acetal, which proceeds to give expected acetal by deprotonation. Since we are using the alcohol as the reaction solvent, forward reaction is favored by multiple times.
OP's suggestion of protonation of $\ce{OH}$ group is legitimate, but reaction goes backward for a moment, but forward reaction favorably proceed to the end (Recall Fischer esterification). If you want back your aldehyde (or ketone), you can make backward reaction favored by using water as your solvent ($\ce{H+/H2O}$).

Thank you. You explained that protonation of OH is more favourable, is it due to steric hindrance involving OR?

Steric hindrence applied to both ways. For example, acetylation of ketones is slower than that of aldehydes.


text:
Can an element have a third energy shell without completely filling the second energy shell?
I came across an electronic configuration diagram, it had three shells. The first shell had only one electron, the second had six and the third energy shell had five. An electron was also present outside all the shells.
Is this configuration possible?

Share a picture.

Yes, it is possible. When an element is excited, some electrons from lower shell jumps to higher shells. For eg, in-ground state suppose an element has electronic configuration
$\ce{(1s)^2 (2s)^2 (2p)^5}$ . If we energize it, then some of the electrons jump up for a short period of time(till the energy is available) . The config becomes $\ce{(1s)^2 (2s)^1(2p)^6 }$.

Jdeep's information is not an answer to the question. Hazika would like to discuss an excited state whose configuration does not differentiate between $s$ and $p$ shells. This configuration  could be described by the unusual electronic configuration $\pu{1^1}$  $\pu{2^6}$  $\pu{3^5}$ This would be a highly excited state, because $1$ electron is missing on the level $\pu{n = 1}$, and $2$ electrons are missing on the level $\pu{n = 2}$.


text:
Stability of carbocation directly attached to a halogen atom
I have two compounds..
How is the first one more stable?
As I have been taught that -I effect of Chlorine dominates its +R effect but to explain this case, My teacher is saying that because of the achievement of the complete octet, chlorine donates its electrons to the carbocation(showing +R effect). If that's the case then the nucleophilic addition of alkanoyl chloride should be the slowest but experimentally proven, they show the fastest nucleophilic addition reaction by increasing the electrophilicity of the carbon atom(showing -I effect). Moreover, on treatment of alkynes with HCl, we get gem dihalides(which also shows the +R effect of chlorine).
Please help me with this because I am really confused between the -I and the +R effect of halogens.

Don't relate stability with reactivity. Usually more stable means less reactive, but 'usually'. Also, Octet stabilization dominates other factors

https://chemistry.stackexchange.com/questions/43751/why-does-fluorine-stabilise-a-carbocation

some people say that its because of octet completion of the carbocation
and the NS reactivity that you mentioned is due to the -I effect of chlorine which makes it the most reactive and not the least reactive


text:
Synthesis of Schiff Bases in Aqueous Media
The paper https://www.hindawi.com/journals/jchem/2013/909217/ describes a possibility to synthesize salicylaldehyde-based Schiff Bases using water as a solvent. Yields shown for the microwave-assisted procedure are high, exceeding 90%, though the procedure requires no water removal, Dean-Stark equipment etc.
Is it the exception that proves the rule or the Schiff bases synthesis, in general, is always possible in aqueous media? Or maybe the microwave does the magic?

There's a passing reference to this at https://chemistry.stackexchange.com/a/89718/16683, although to stop anybody getting ideas, I'll say that this isn't a duplicate.

I wonder if the reactants are truly in solution, or are they esentially dissolving in each other as a microemulsion where reaction occurs.

The reference post given by @orthocresol describes "conventional' i.e., not microwave-assisted reactions. The cited reference concludes that vacuum simply
helps in the formation of imine by removing the water formed in the imine condensation. The case described in this post is different: solid product was obtained in reaction mixture. As far as I understand it means that water is present after the completion of the reaction.

I have two opinions here: (1) Microwave sometimes does weird things that make them superior to regular reactions, so that may just be it. (2) If this were a general reaction, I would assume that the paper would not refer to salicyladehydes and be in a better journal.

Marcin Stec: That;s what exactly @orthocresol was saying. Your post is not a duplicate, but strangely similar.

The experimental section of the paper you cite says the product was filtered, not extracted. Thus, the equilibrium was shifted by precipitation of the product in the same way that a Dean-Stark trap shifts an equilibrium. I suspect the procedure may not be effective with aliphatic primary amines.


text:
What are the organic reactions that are taking place in this reaction?
I know that A is the product of a Knoevenagel condensation between the aldehyde and the diethyl malonate, due to the acidic protons of the latter, but I don't know what the reaction between that product and potassium cyanide would be.

The are only 2 possible sites of nucleophilic attack on A for CN-. One attack is completely reversible and represents a dead end, so it must be the other.

I'm guessing then that you're referring to a nucleophilic attack on the double bond, because the attack on the ester would be pointless. That was one of my first thoughts, but I discarded that option because I didn't know where to get the proton that was missing, maybe from the solvent? Thanks for your answer!

The 1,4 attack on the double bond is the pointless one. Think through attack on one of the ester carbonyls

@Zenix  not as I see it. C is the cinnamic acid and D is the mixed anhydride.

@Waylander correct me, C is p-cyclohexyl cinnamic acid (after decarboxylation of geminal di-carboxylic acid). Thus, D will be mixed anhydride, what will AlCl3/CH2Cl2 do?

I would suggest OP to edit the title of question. As it stands, the title looks vague. Be specific with the title. A title should always reflect the body of question. Check this meta post to see what to write in a title.

Here is what I think is going on:

$Step$ $1$: As the OP correctly identified is a Knoevenagel condensation to give A diethyl 4-cyclohexyl-benzalmalonate.
$Step$ $2$ is an example of little-used ester hydrolysis using cyanide ion to give initially the acyl cyanide which is unstable in aq EtOH and gives the diacid B
$Step$ $3$ is loss of one carboxy group by acid catalysed decarboxylation to give the cinnamic acid C
$Step$ $4$ formation of the mixed anhydride D
$Step$ $5$ cyclisation through formation of an aluminium-acyl complex literature example here to give the final product 6-cyclohexyl-Inden-1-one E

Mechanism of step 5?

Formation of an acyl-aluminium complex similar to acyl halide/AlCl3 cyclisation. I've edited to include a reference. Not the way I would choose to do it (PPA would be my choice), but I think this question is trying to show alternative ways this could be done

@Waylander has provided a well-reasoned solution to the post. I offer a different interpretation. Knoevenagel product 1a has been converted in high yield to cyano ester 2a and subsequently hydrolyzed to phenylsuccinic acid (3a) 1.  In the addition of cyanide to the double bond of 1a and protonation, one of the labile ester groups is cleaved by cyanide. (e.g., Diethyl malonate is readily monosaponified by base in the cold.) Cyclization of anhydride 4a has been accomplished by standard Friedel-Crafts methods but, like Waylander, I am a big fan of polyphosphoric acid (PPA). The cyclohexyl series b (R = c-C6H11) is likely to lead to ketoacid 5b.

1) C. F. H. Allen and H. B. Johnson, Organic Synthesis, Coll. Vol. 4, 804, 1963.; Vol. 30, 83, 1950.


text:
Why does a strong acid displace salt of weak acid?
This thing has been bugging me for a few days. I don't understand why a strong acid displaces a weak acid from it's salt solution.

And what do you mean by "displacing"? What would happen according to you if some HCl was added to solution of sodium acetate (for example)?

Maybe you prefer an image. Bases behave with $\ce{H+}$ like magnets with nails. Suppose a $\ce{H+}$ ion (a proton) is like a piece of iron (nail) attracted by a weak magnet. It forms a couple which may more or less look like an acid molecule. If now a big and strong magnet (a strong base) approaches, it will pick up the nail, and the weak magnet will stay alone. If the nail is fixed on a small magnet  it makes a weak acid. And if you approach a strong magnet (a strong base like OH-), the proton will leave the weak base, and be fixed against the strong base. For example, if the weak base is the ion acetate, the initial couple weak base + proton is the acetic acid. If a strong base like OH- approaches, there will be a proton transfer, from the weak base to the strong base $\ce{OH^-}$. Water will be produced, leaving the weak base (acetate) alone. The proton transfer is like a nail transfer between magnets.
Strong bases are like strong magnets. Strong acid is a strong donor. So the corresponding magnet is extremely small and weak. A tiny magnet + nail is able to donate its nail to any other base. A strong acid is able to donate its proton to any medium or strong base.


text:
Are there any fluorophores with lifetimes in the micro to milli second range?
I'm looking for a fluorophore with exceptionally long lifetime to use in an oxygen optode (the longer the lifetime the less timing precision needed in the electronics).
So far the longest I have encountered in my searches is 600nS for Ru(bpy)3[PF6]2 found in this table but that resource appears less than exhaustive; Is there a more comprehensive data sources available where I could compare lifetimes and relevant wavelengths for all known fluorophors?
.. also just to clarrify the fluorophore needs to undergo quenching by oxygen ...

Then you're rather looking for phosphorescence.

Uranyl ion has a fluorescence lifetime of about 1.5 microseconds: Catalysis of ionic reactions by polyelectrolytes. III. Quenching of uranyl ion fluorescence by iron(II) ions in poly(vinylsulfonic acid) solution, Herbert MorawetzIssam A. I. Taha
Cite this: J. Am. Chem. Soc. 1971, 93, 4, 829-833
Publication Date:February 1, 1971
https://doi.org/10.1021/ja00733a005

Thanks, while heartening to find the microseconds lifetime is not out of the realm of possibilities - Uranyl is a bit to exotic for my purposes (the end product needs to be compatible with international and domestic postal services)

You might consider Europium or Terbium compounds. Eu3+ and Tb3+ complexes can have long lifetimes in the range of hundreds of microseconds to milliseconds (https://www.tcichemicals.com/US/en/c/10226).

According to the article Indicators for optical oxygen sensors; Michela Quaranta, Sergey M. Borisov, Ingo Klimant

Among all the indicators that have been used for optical oxygen sensors, Pt(II) and Pd(II) porphyrins are the mostpopular luminophores since they possess strong phosphorescence at room temperature, moderate to high molar absorption coefficients and large Stokes’shifts. Additionally, phosphorescence lifetimes are rather long (microsecond to millisecond) and can be tuned by varying the nature of the central atom.

They then go on to include a table (Table 3 on page 131) of 36 dyes based on that chemistry, 35 with lifetimes in the microseconds and 1 (PdTFPP) with a lifetime of 1.65 milliseconds.
In relation to my quest for a data source to compare fluorophors I managed to stumble upon the website fluorophores.org that currently includes 952 substances and can be filtered by excitation, emission, quantum yield, and lifetime parameters.

What is your analytical application? You mentioned oxygen optode but where do you plan to use it.

Inside permeable product packaging to monitor effectiveness of oxygen scavenger / absorber materials in real time.

Is it something like this commercial product https://www.oxysense.com/how-oxysense-works.html?

On first glance that appears to be EXACTLY what I need! Do you know of any commercial products that use this in there packaging so I could test it out? I will be contacting the company directly but who knows how long that will take.

I am afraid not. But I am sure they have something safe. These metal containing fluorophores will be expensive and who knows how safe they are. Do a heavy and deep literature search again.

Thanks, according to OxySense's patent (US20070243618A1) they use a ruthenium based fluorophore: "The present invention provides a ruthenium-based luminescence indicator composition that includes a tris-4,7-diphenyl-1,10-phenanthroline ruthenium(II) compound and one or more dioctylphthalate dispersed within a gas permeable polyacrylate matrix. The tris-4,7-diphenyl-1,10-phenanthroline ruthenium(II) compound has a fluorescence lifetime that is affected by exposure to one or more gases and exposure to the one or more gases is monitored"

Great. So you have clues now to see the new directions.


text:
The pH of a drop of water containing ammonia
Presented with the following equations: What would the $\mathrm{pH}$ of a drop of water be, while in equilibrium with $\ce{NH3(g)}$ at a concentration of $\pu{150 ppb}$?
$$\ce{NH3_{(g)} + H2O_{(l)} <=> NH3_{(aq)} + H2O} \qquad K_\mathrm{H} = \pu{1.76 \cdot 10^{-2} M/atm}$$
$$\ce{NH4+_{(aq)} <=> NH3_{(aq)} + H+}  \qquad K_\mathrm{a} = 5.62 \cdot 10^{-10}$$
So my issue is that I don't know how to express the concentration of $\ce{[NH4+]}$.
Calculating the concentration of $\ce{NH3_{(aq)}}$ is simple: $K_\mathrm{H} \cdot P_\ce{NH3}$, while $P_\ce{NH3} = C \cdot P_\mathrm{air} = 150 \cdot 10^{-9} \cdot \pu{1atm}$, yields $\ce{NH3_{(aq)}} = \pu{2.64 \cdot 10^{-9} M}$.
If we are to tally up all contributions of $\ce{H+}$ then we get:
$\ce{[H+]} = K_\mathrm{w}/\ce{[H+]}$ (from water dissociation) + $K_\mathrm{a} \ce{[NH4+]}/\ce{[H+]}$.
Then $\ce{[H+]} = (K_\mathrm{w} + K_\mathrm{a}\ce{[NH4]})^{0.5}$ and we'd get the $\mathrm{pH}$.
I'm stuck unsure how to express $\ce{[NH4]}$.

The provided Kh must be errorneous, considering the huge solubility of ammonia in water.

HINT You don't know $\ce{[NH4+], [H+], [OH-]}$ -- So you need three independent equations.

There is a general way to solve these type of problems, see the answers here https://chemistry.stackexchange.com/questions/60068/how-to-set-up-equation-for-buffer-reaction/95566

Well, I decided to look at the dissociation of ammonia instead: NH3+H2O <->NH4(+)+OH(-), using Kb (Kw/Ka). We get 1 mol each of OH- and ammonium ion, so with my given data of NH3 conc (2.64*10^-9 M), I arrived at pOH=6.66 or pH=7.33.

@porphyrin isn't this way applicable only when you have the concentrations of both acid+conj base or base+conj acid? in my case we only know the conc. of the conj. base of ammonium ion.

It still applies, you have $\ce{NH3, NH4+,  H+, OH-}$. As the $\ce{NH3}$ concentration is so small so that approximately $\ce{[H+]=[OH-]= 10^{-7} }$  you know that the pH is going to be approximately 7.


text:
Groups of the periodic table
The groups go from 1 to 18 in the periodic table. When looking at the transition metals in the periodic table, You see roman numerals labelled with 'B,' in the order of III, IV, V, VI, VII, VIII, VIII, VIII, I, II.
Why does III - VIII come before I and II? And why does VIII get repeated three times?
Also, what is the difference in the groups labelled A and B, besides the fact that the B groups are the transition metals?

If you don't get old numeration than just uses new Table.

It doesn't. I see that there are 18 groups but why are they separated into A and B? Is there more significance to that than "transition vs non-transition?" And, within the B transition metals, why are the numerals ordered the way that they are? why does the numeral VIII appear 3 times whereas the other numerals appear only once? And why does III - VIII comes before I and II? These are my questions.

The numbering with Roman numerals is outdated. Modern tables just label them 1-18. For a bit about why there are three columns of VIII, see the comments here

Even though your question get downvoted, I think it is an legitimate question for people to understand the trend. I believe I-VIII Roman numeral nomenclature belong to CAS system while 1-18 Arabic numeral groups are recommended by IUPAC (Ref.1). The IUPAC recommended Periodic Table is given below (ignore the Roman numerals, which I put in there to explain):

According to Ref.1, the confusion of Periodic Table of Elements has been addressed as follows:

H.G. Deming used the long periodic table in his textbook General Chemistry (See following diagram with Roman numerals only; Ref.2), which appeared in the USA for the first time in 1923, and designated the first two and the last five Main Groups with the notation "A", and the intervening Transition Groups with the notation "B". The numeration was chosen so that the characteristic oxides of the B groups would correspond to those of the A groups. The iron, cobalt, and nickel groups were designated neither A nor B. The Noble Gas Group was originally attached by Ueming to the left side of the periodic table. The group was later switched to the right side and usually labeled as Group VlllA. This version of the periodic table was distributed for many years by the Sargent-Welch Scientific Company, Skokie, Illinois, USA. $[\cdot\cdot\cdot]$
The designations A and B have been extensively and rather arbitrarily used in the meantime in textbooks and in publications. An investigation of the application of the subgroup designations A and B in all articles, which appeared between 1972 and 1981 and covered by Chemical Abstracts, revealed a completely arbitrary use of the designation. Moreover, more than 10% of the articles it was nearly impossible, from the wording of the text, to recognize which elements were being discussed without further information (Ref.3).

The IUPAC Commission on the Nomenclature of Inorganic Chemistry (CNIC) proposed the designations of groups by Arabic numerals in 1984 (Ref.4), which was approved by the Nomenclature Commission of the American Chemical Society (ACS) and rest is history.
Thus, it is safe to say that this nomenclature is now outdated and not worth discussing.

References:

E. Fluck, “New notations in the periodic table,” Pure and Applied Chemistry 1988, 60(3), 431-436 (https://dx.doi.org/10.1351/pac198860030431).
Horace Grove Deming, General Chemistry; J. Wiley & Sons, Inc.: New York, NY, 1923.
W. C. Fernelius, W. H. Powell, “Confusion in the periodic table of the elements,” J. Chem. Educ. 1982, 59(6), 504-504 (https://doi.org/10.1021/ed059p504).
K. L. Loening, “Recommended Format for the Periodic Table of the Elements,” J. Chem. Educ. 1984, 61(2), 136-136 (https://doi.org/10.1021/ed061p136).


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Relativistic contraction of orbitals
I am confused about what seems to me two very different explanations for the relativistic contraction of orbitals for heavy elements.
My teacher told me that since 1s electrons in heavy elements travel very fast their momenta is much larger than predicted by classical physics. Hence the 1s orbital contracts lead to the contraction of all ns orbitals so that their overlap continues to be zero. D and f orbitals on the other hand do not contract. Because of their symmetry, they do not need to contract for overlap with the 1s orbital to remain 0. They in fact expand because of increased shielding from s electrons.
My book however says the reason for the contraction is that since ns orbitals are non-zero at the nucleus they experience the effects of nuclear charge more and hence relativistic contraction is most significant in them (it as if implies that the ns orbitals are all traveling at relativistic speeds). Which one is correct? Or are both of them oversimplifications?

I would rather close this then put a bounty...

Both effects are happening.  For orbitals to contract two conditions must be met, 1) the electrons in the orbital must travel at relativistic speeds and 2) the orbitals must have a non-zero probability density at the nucleus (s, p).  See this earlier answer for more details.

s orbitals always overlap. More generally, there is no requirement for zero orbital overlapping and they generally do overlap each other.