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Lowering the freezing point of a lipid by adding another lipid
I'm trying to stop bacon grease from congealing in the fridge by adding another lipid with a lower freeze point to it, but I'm not sure how much to add, as I'd like to keep the ratio as bacon-grease heavy as possible.
Of the oils I have available, corn oil has the lowest freeze point at -11C (http://www.veganbaking.net/articles/tools/fat-and-oil-melt-point-temperatures), and bacon grease has a freezing point of 26C (https://www.doesitgobad.com/does-bacon-grease-go-bad/). I need to bring the freezing point down to 5C or lower.
Can I calculate the melting point of the mixture simply by finding the average melting temp between the two according to their ratios? So for corn oil:
T = (26b + -11c) / (b + c)
Where T=Freezing temp. of resulting mixture, b=# parts bacon grease, and c=# parts corn oil

If not, how should I calculate the ratio of corn oil to bacon grease to bring the freezing point to 5C? Can I calculate it the same way if I use cottonseed oil (-48C freezing point) instead of corn oil?

I doubt that will work, but chemistry is an experimental science so try it with a small quantity of bacon grease and oil. -- I'd guess that you'll get something that is part solid and part liquid.

I doubt it would be the average. That kind of stuff happens in fairy tales alone.

Have a look at this: https://en.wikipedia.org/wiki/Freezing-point_depression

From observing virgin olive oil, there is not one single melting point. Which makes sense if there is any variation in chain length. So you may get a broader range of melting, but that isn't what you seem to be looking for.


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Are the C1 and C3 carbons of cis 1-bromo-3-chloro-cyclobutane chiral?
I have tried to approach this from 2 directions:
The ‘official’ method my textbook gives, which gives the correct answer - They are achiral
It consists of going ‘priority-wise’ from the two carbons to see if they are unsymmetrical. Which, both carbons connected to C1 (and C3 in the other case) ultimately meeting the same atom C3 (and C1 in the other case), gives no as an answer.
I tried to approach this by interchanging the positions of H and Br on C1 and seeing if I could superimpose this and the original structure, which I should be able to because C1 is symmetrical.
Are they superimposable? Am I doing the superimposing wrong? I have been trying to superimpose them in my mind for quite some time and have been unsuccessful (I’ll get the ‘balls and sticks’ kit after the lockdown ends to get a more intuitive understanding).
Or are they not superimposable and I’m missing something?
Edit: Adding a picture of the compound (It is meant to be in cis form) :

C1 and C3 are stereogenic but achirotopic. A change of stereochemistry at either carbon creates cis/trans isomers (stereogenic) but the cis/trans isomers will always be achiral.

A drawing(s) on some sort would go along way for people to better interpret the question.

@user55119 So the 2 structures can be superimposed?

I know you’re waiting to get a kit, but perhaps look into a 3D drawing program like Avogadro. Sorry for not mentioning this earlier.

Gist - There is a mirror plane through the Br, Cl, and the two carbons to which the halogens are attached.

Imagine the cis and trans isomers as ghosts. Slide one into the other. Do all atoms match?

They are achiral. If you take the mirror image (first picture) and rotate 180 degrees (second picture), you can overlay them. Images made with Avogadro

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Van 't Hoff equation
In thermodynamics many definition has been made from partial derivatives at constant conditions for instance:
Let $U$ be $U:f(S,V,m_)$ then:
$$T = \left( \frac{\partial U}{\partial S} \right)_{V, m_{k}};  p = \left( \frac{\partial U}{\partial V} \right)_{S, m_{k}}; \mu_k = \left( \frac{\partial U}{\partial m_k} \right)_{V, S}$$
In a particular case, heat of reaction and van 't Hoff relation
Being $H:H(T,p,\xi)$
$$\mathrm dH= \left( \frac{\partial H}{\partial T} \right)_{p, \xi}\,\mathrm dT + \left( \frac{\partial H}{\partial p} \right)_{T, \xi}\,\mathrm dp + \left( \frac{\partial H}{\partial \xi} \right)_{T, p}\,\mathrm d\xi$$
Where $C_p$ is the specific heat at constant pressure, $h_T$ is the heat compressibility and $r_{T,p}$ is the heat of reaction at constant temperature and pressure:
$$C_p = \left( \frac{\partial H}{\partial T} \right)_{p, \xi} ; h_T = \left( \frac{\partial H}{\partial p} \right)_{T, \xi}; r_{T,p} = \left( \frac{\partial H}{\partial \xi} \right)_{T, p}$$
In terms of variation of affinity with temperature, the heat of reaction is given by (where $K$ is the equilibrium constant of a chemical reaction):
$$r_{T,p} = -RT^{2}\frac{\partial }{\partial T} \ln K(T,p)$$
So the question is: How can I define an equation where the left side needs constant temperature, but the right side implies the derivative of temperature?

The T that is held constant on the LHS of your equation has a different meaning from the T that is allowed to vary on the RHS.
The van 't Hoff equation answers the following:
Suppose I have a reaction that takes place at const T, p at T = T1.  Now I change to a new temperature, T2, and allow that reaction to again take place at const T, p, except now T = T2.  How does the equilibrium constant change?
Now that you understand this, you can see there is no inconsistency between "constant T" and "changing T", since they refer to two different things.
I.e., the fact that the reaction is constrained to take place at constant T in no way prevents me from testing the reaction at a different, but still constant, T.
The analogy I'd give is this.  Suppose we want to determine the effect of altitude on 100 m sprint times.  The 100 m is always run on a level surface (no change in altitude during the race).  Yet this does not in any way prevent us from comparing 100 m run times at different altitudes (e.g,  sea level vs. 10,000 feet).
So your question is like asking:  "How can we determine the effect of changing altitude on 100 m run times, given that the 100 m run is always done at constant altitude?"  Well, we can, of course.

Good point I was thinking about the same think, but I was getting trouble, for instance the temperature definition implies that V and chemical composition ($m_k$) must be constant hence Does it implie that chemical reaction with a change in volume has no temperature to be measurable?

@JuanSebastianGarciaPerez  Not sure I understood your question -- there may be a language barrier -- but it seems you are saying that there will be no measurable temperature change because V and chem. comp. are constant.   That's not the case.  The reason there is no T change is because we've designed the experimental conditions that way: the reaction is taking place in a constant-T environment.

Furthermore, it's not the case that V and chem. composition are constant.  The van 't Hoff equation applies for reactions at constant T and p.  This means that, typically, as the reaction takes place, V will not be constant (unless the molar volumes of the reactants and products are identical).  And since a reaction is taking place, the composition will not be constant either.

What I am trying to say is I understand your answer about Van't Hoff equation but now What happen with temperature definition? Does temperature only exist when V is constant?

No, temperature is always defined for any thermodynamic system.  Your first equation is simply saying that the rate at which U changes with S at constant V and chemical composition happens to be equal to T. There are other relationships that also equal T.  For instance, using your notation, we also have:  $T = \left( \frac{\partial H}{\partial S} \right){p, m{k}}.$  If you want to understand the thermodynamic concept of temperature, you should read about the zeroth law of thermodynamics.

But the zeroth law of thermodynamics is not actually a law, it implies the concept of equilibrium, temperature should be a transversal concept, exist in both equilibrium and non-equilibrium processes, how are we able to explain the heat flux by a temperature gradient? (which is a non-equilibrium process).

If you want an answer to that, you should post it as a separate question.

There are a few different ways to derive the van't Hoff equation. The question here however is to justify the mathematical form of the equation. A good (if not entirely satisfying) way to do so is to be (a little) more mathematically rigorous. You then find that the form of the equation follows from the fact that the Gibbs' free energy G is a thermodynamic potential with continuous second derivatives and so observes Schwartz's theorem, which in the context of thermodynamics is used to derive Maxwell's relations. Thermodynamic properties like the Gibbs free energy, pressure and temperature that obey Schwartz's theorem are called state functions. Algebraic combinations of these are also state functions.
In the case of the free energy divided by the temperature, the total differential can be written as
$$d(G/T) = \left(\frac{\partial (G/T)}{\partial P} \right)_{T,\xi}dP + \left(\frac{\partial (G/T)}{\partial T} \right)_{\xi,P}dT + \left(\frac{\partial (G/T)}{\partial \xi}\right)_{P,T}d\xi \tag{1}$$
Since G/T is a state function it follows that the order of differentiation of the first derivatives of the potential does not matter and that the following relation between second derivatives holds:
$$\left(\frac{\partial ^2 (G/T)}{\partial T \partial \xi} \right)_{P} = \left(\frac{\partial ^2 (G/T)}{\partial \xi \partial T}\right)_{P} \tag{2}$$
If you evaluate the first partial derivatives of (G/T) in this equation and perform appropriate manipulations you can arrive at the van't Hoff equation. The question then about how the odd relation arises in the van't Hoff equation, in which a derivative is taken wrt a variable that was previously held fixed, can be asked already here, because Equation 2 could be written as follows:
$$\left(\frac{\partial }{\partial T }\left(\frac{\partial  (G/T)}{\partial \xi} \right)_{T,P}\right)_{P} = \left(\frac{\partial }{\partial \xi }\left(\frac{\partial (G/T)}{ \partial T}\right)_{\xi,P}\right)_{P} \tag{2'}$$
You see on the left-hand-side that you first hold T constant and later take the derivative wrt T.

How do you relate the heat of reaction $_{,}$ with an enthalpy accord to Maxwell's relations?

You can rewrite the mixed derivative on the rhs in equation 2' in terms of $r_{T,p}$

I edited my answer. As written earlier it was more difficult to perform the manipulations that lead to the van't Hoff equation.

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Doping of a solid with impurities
This question is a continuation of the previous question of mine:press me if you wish
And my question is how can I dope a solid with impurities and if the process can be understood with basic knowledge of solid state physics.

I give a very general answer: it depends on the solid and the dopant.
One of the main issues is best viewed from a thermodynamic point of view: you want to create something similar to a solution. The dopant should randomly occupy some positions in the solid, and this should be comparably stable or more stable than the pure phases themselves. The dopant can substitute an atom (then it is often chosen to be with similar size and chemical properties), but it can go originally unoccupied places, too.
Some examples:

You can introduce dopants with co-crystallization. You have a method to prepare the host solid through crystallization from its melt or from a solution, then you can mix the dopant during the synthesis, and if you are lucky it works. With molecular materials or simple inorganic salt, it can be a relatively easy process.
Chemical vapor deposition techniques are often used to create (doped) thin films.
Some dopants can be introduced by simple immersion. Hydrogen can diffuse inside of the host material, and act as a dopant in metals and other materials. The preparation can be just letting the sample stay in hydrogen gas for a longer period. Similarly, there are many oxides where oxygen acts as a dopant, and its level can be controlled by heat treatment in oxygen (or in a vacuum, to remove the extra oxygen). My understanding is that the boron doping in silica using diborane gas is also based on similar principles.


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What do 'sigma' and 'pi' in front of ligands in the formula of a coordination compound mean?
The following coordination compound was given in the book Concise Inorganic Chemistry by J.D. Lee (Adapted by Sudarsan Guha) under the topic 'Effective Atomic Number (EAN)':
$\ce{[Ti(\sigma -C5H5)2 (\pi  -C5H5)2]^0}$
What is meant by the quantifiers $\sigma$ and $\pi$ in front of the ligand?  I have not seen this format with other ligands? If this is to signify $\sigma$ and $\pi$ bonds, why is it not used in the formulas of metal carbonyls which have synergic bonds (a combination of both $\sigma$ and $\pi$ bonds)?
In all the 67 occurrences of 'cyclopentadienyl' in the IUPAC Red Book, I was unable to see $\ce{C5H5-}$ having a $\sigma$ or $\pi$ in front of it. Instead I only saw $\eta$ being used in the name. I came to know that it is meant to denote 'hapticity'. Is this related to the presence of $\sigma$, $\pi$?

This question is similar to yours: https://chemistry.stackexchange.com/questions/66911/%CE%B71-or-sigma-complex-involving-cyclopentadienyl-anion

Concise Inorganic Chemistry1 has the following to say about this tetra(cyclopentadienyl) titanium complex:

Tetra(Cyclopentadienyl) compounds such as $\ce{Ti(C5H5)4}$ may be made from $\ce{TiCl4}$ and $\ce{Na(C5H5)}$. The formula may be written as $\ce{[Ti(\eta_5-C5H5)2(\eta_1-C5H5)2]}$ where two cyclopentadienyl rings are attached by 5 C-atoms ($\pi$ - bonded) and two rings are attached by one C-atom ($\sigma$ - bonded).
NMR studies on these tetra(cyclopentadienyl) compounds suggest that in the $\eta_1$ rings the C bonded to Ti continually changes, and the $\eta_1$ and $\eta_5$ rings change their roles.

The first paragraph explains quite well what the author intends by the quantifiers $\sigma$ and $\pi$. Two rings are bonded by $\pi$ donation with a hapticity of 5 and two rings are $\sigma$ bonded with a hapticity of 1. This picture should clear things up

References:

Lee, J. D. Concise Inorganic Chemistry. 5th ed, Wiley publishers & Blackwell Science, pp. 696, 1996


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E-Z system without a double bond
Can we apply E-Z nomenclature to geometrical isomers without a double bond. For example, can we use it for cis and trans isomers of 1,2-dimethylcyclohexane?

No, those are (R,R)/(R,S)/(S,S).

@orthocresol: I didn't get your point. Shouldn't we use E-Z system as an alternate to cis-trans system of nomenclature when we go by the order of priority using CIP rule over R-S system which is meant for the spatial orientation about a chirality centre?

@GuruVishnu, if you think about the point you're making, and think about whether the carbons in 1,2-dimethylcyclohexane are chiral... You're smart enough to figure it out. Furthermore, the trans diastereomer has two enantiomers, so you're restricting yourself to two letters to describe three different things. Let's be very clear here: the descriptors cis and trans have basically nothing to do with the IUPAC descriptors Z and E. They just so happen to be the same thing in the special case of a 1,2-disubstituted alkene.


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What is the value of isothermal compressibility of a van der Waals gas at critical point?
For a van der Waals gas, isothermal compressibility is $$\kappa=\frac{V^2(V-nb)^2}{nRTV^3-2an^2(V-nb)^2}$$ If one substitutes the values of critical temperature and volume in this formula, isothermal compressibility goes to infinity. What is the physical meaning of this? Is it right to use this equation at critical point?

Of course it goes to infinity. What else did you expect? At critical point we have ${\partial{\mathfrak p}\over\partial V}=0$; that's the definition.

Generally, the vdW equation is better to be avoided near critical point. Particularly,   some partial derivatives of state variables go to infinity at the critical point.

Indeed, the vdW equation becomes less and less reliable near critical point. This particular detail, however, is reliable.

I agree that a critical point is a point of inflection. If one has to compress at critical point at constant temperature, one has to move along the critical isotherm. How does this shed a physical meaning to $$\kappa\rightarrow\infty$$?


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The true shape of p orbitals
I'm trying to model  $\mathrm{p}$ orbitals using a 3D program. What confuses me is that there are 2 shapes of $\mathrm{p}$ orbitals. I'm not sure which one is the accurate model. And what is the reason to have 2 types of shape?
Shape 1

Shape 2

The second one is more accurate. The first one is easier to comprehend.

Out of curiosity, what 3D program are you using for modelling orbitals? The renders you've posted look pretty good.

@andselisk those are not modelled by me they are 2d images i found on internet.not sure what program they have  used. but i guess any popular 3d program can do that.i'm using blender by the way

The Wikipedia Page on atomic orbitals (https://en.wikipedia.org/wiki/Atomic_orbital) is actually quite good. One should keep in mind that the electron density of an atom is spherical, so the orbitals also need to reflect that.

These are not the shape of p orbital. These shapes are the place around the subshell which tells you the possibility of electron around that subshell.( Probability of finding electron in some area is greater than 60% is mostly considered as a shape of that subshell.
P orbital whose l=1 has dumbbell shape which is represented by both the figures

There is a misconception here. A p orbital is a 3D-function, and these functions don't have shapes, they have values at any point in space. If you describe an electron distribution with one of these functions, you can plot contours at a chosen value, and these contours have shapes. For example, you could choose a contour level such that an electron has a 90% chance of being inside the shape defined by the contour level.

[OP...] What confuses me is that there are 2 shapes of p orbitals.

There are actually four common graphs depicting p orbitals, and a lot of conceptual drawings that don't have an exact mathematical shape. The graphs differ in whether they show the wave function or the square of it (the latter is proportional to the probability of finding the electron in that location). They also differ in whether they use a cartesian coordinate system, or plot the angular dependence in a polar plot.
The one you probably want is a contour plot of the probability density (you could also say the electron density). Here is an example from an article in the Journal of Chemical Education (center panel):

The shape of the contour lines of the wavefunction are the same, but you can have positive and negative values (and because of the squaring involved, the contour lines also have different values). Your "shape 2" is a contour plot of the wavefunction, with two contours of equal magnitude and opposite sign.
Polar plots
The other type of depiction is a polar plot, where the distance from the origin is the value of a function. This is different from a Cartesian coordinate system, where the nucleus is at the origin, and the distance from the origin corresponds to the distance from the nucleus.
If you plot the angular component of the wavefunction (you can think of this as keeping the distance from the nucleus constant, and sampling different angles), you get what looks like two spheres. A polar plot has no good way of dealing with negative values, so often one part of the curve is shown in a different color.

If you plot the square of the angular component, you get a shape closer to your "shape 1". For technical reasons, the plots are rotated by 90 degrees (the angle zero is defined along the x-axis in Wolfram Alpha, which I used to plot these).

So what does your "shape 1" represent? The left-most panel looks like it wants to show an electron density, but it is very inaccurate. There are other artists' interpretations floating around, some that look like they are actual graphs of functions, such as this one:

It is not clear which function, if any, this shows a graph of.

[OP...] The true shape of p orbitals

There is none. You have a choice of representations, and it depends on the context which one, if any, makes sense.

Which representation you would use depends on what kind of information you would want to represent. The p-orbitals (and all orbitals with a finite angular momentum quantum number $l$, for that matter) have a spatially dependent phase: The nodal planes in your bottom representation separate the parts of the p-orbitals that have opposite phases. In other words, there is a phase difference of $\pi$.
Whether you would want to display the phase difference depends on the property of the p-orbitals that you would like to represent. If you just want to display the part of the space that captures e.g. 90% of the probability density to find an electron, then the top representation will suffice. However, once you start to consider interactions with other orbitals, the phase will be important, as it will determine whether the interaction is bonding or antibonding. In that case, the colour of the p-orbital and other involved orbitals can be included to emphasise the type of interaction.


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Acidic cleavage of an ether
In the following question;

Apparently, an SN2 reaction ensues, where the iodine atom's supposed to cleave off with the less sterically hindered alkyl branch from both the oxygens. This is the solution:

That's what happens with the lower oxygen, but what's going on with the top one? In the second step, why doesn't the methyl attached to it cleave off with the iodine atom?

For the second step, the SN1 mechanism can take place due to the stability of the carbocation formed. (The positive charge can be delocalized over benzene ring). This means that MeOH can be a leaving group, and you get the product shown in the reaction scheme.

Okay, thanks. Can you put it down as an answer so I can mark it as one?

The Wikipedia page on ether cleavage says:

Depending on the specific ether, cleavage can follow either $\mathrm{S_N1}$ or $\mathrm{S_N2}$ mechanisms. Distinguishing between both mechanisms requires consideration of inductive and mesomeric effects that could stabilize or destabilize a potential carbocation in the $\mathrm{S_N1}$ pathway. Usage of hydrohalic acids takes advantage of the fact that these agents are able to protonate the ether oxygen atom and also provide a halide anion as a suitable nucleophile.

So in essence, the mechanisms for such reactions are usually a trade-off between $\mathrm{S_N1}$ and $\mathrm{S_N2}$ pathways, with adequate carbocation stability playing a major role in the determination of the pathway (as intermediate cases like secondary cations often lead to a "confusing" case where it is difficult to say qualitatively whether the path is majorly $\mathrm{S_N1}$ or $\mathrm{S_N2}$).
In particular, the Wikipedia page cited above has pretty much answered your question under the $\mathrm{S_N2}$ ether cleavage tab:

If the potential carbocation can not be stabilized, ether cleavage follows a bimolecular, concerted $\mathrm{S_N2}$ mechanism. In the example, the ether oxygen is reversibly protonated. The halide ion (here bromide) then nucleophilically attacks the less sterically hindered carbon atom, thereby forming methyl bromide and 1-propanol.

So for your case, the upper oxygen after protonation can cleave the $\ce{C-O}$ via a $\mathrm{S_N1}$ pathway to generate a reasonably stable potential carbocation which is delocalized over the benzene ring.
Hence,it can simply remove a molecule of methanol(which is a good leaving group by itself) on it's own without needing an iodide to assist in the removal.
The strongly acidic solution of $\ce{HI}$ will anyway ensure it that the methanol undergoes $\mathrm{S_N2}$ to form methyl iodide later. Also, from the point of view of a nucleophile,it would be much easier to perform an $\mathrm{S_N2}$ reaction on a separated methanol molecule than on the entire substrate molecule initially given in the question due to steric considerations,so that also becomes a contributing factor.
But as for the lower oxygen,the poor fellow does not have such a choice. The potential carbocation that can be generated by cleaving the lower $\ce{C-O}$ bond is a phenyl cation,which is pretty unstable as explained here. So, like the Wikipedia page said and illustrated in the example above, the pathway has to be a concerted $\mathrm{S_N2}$ mechanism now so as to take the reaction forward.
Note: Even if the said $\mathrm{S_N2}$ starts out slow, Le Chatlier's principle will ensure that ultimately, most of  the reactant goes into the product as equilibrium is attained.

Correct representation is $\mathrm{S_N2}$, instead of $\ce{SN2}$, you can use \mathrm{S_N2} for it.


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Alkali metals chemistry
Ok my understanding of the electronic structure of atoms could be wrong but this is why I don't understand:Why alkali metals are so much electropositive.
Some high school professor in chemistry would say because if they lose an electron they have an electronic configuration of noble gas etc.
If alkali metals gained 1 electron then the wavefunction of the each electron in the s orbital would shield some of the charge of the nucleus for the other electron in the s orbital->the overall net charge on each electron would be decreased but because now we have 2 electrons instead of one potential energy would be decreased.
That is clearly not the case.Why?

Dimers of Sodium or Potassium atoms do exist in the gas phase and in the absence of air. But the proportion of dimer to the total of atomes is weak.

Maurice notes that alkali metal diatomic molecules are at best a very minor constituent.  But alkali metals do fill their $s$ valence shells in other ways.
Most alkali metals are known to form alkalides, in which an electron is transferred from one alkali metal to another (or from barium to an alkali metal, in a more recent example).  To effect this electron transfer, a complexing agent is used to stabilize the cation, such as 2.2.2 cryptand in the compound $\ce{[Na}\text{(2.2.2-crpytand)}\ce{]^+Na^-}$.  A combination of low ionization energy, positive electron affinity and the stabilization offered by complexing the cation enable this outcome.
Lithium apparently fails to form such alkalides, but not because it refuses to adopt an $s^2$ valence electron configuration.  Quite the opposite: it tends to form polar covalent bonds with that configuration.  Organolithium compounds are a widely used example of this electron structure, and the covalent bonding component in lithium salts may contribute to their solubility in some organic solvents.


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Bond strength and length of HF and H2O
I know that acidity of an acid increases across a period (with electronegativity increase of the atom bonded to hydrogen) and hence $\ce{HF}$ is more acidic than $\ce{H2O}$ or $\ce{NH3}$. But the reason I learnt it is this way, is that difference in electronegativities of the atoms increases and hence bond becomes more polar and weaker (hence easier to donate a proton).
But shouldn't bond strength increase with difference in electronegativities? Also bond length of $\ce{HF}$ is shorter than that of $\ce{H2O}$ ($\pu{92 pm}$ vs $\pu{96 pm}$) and bond enthalpy of $\ce{HF}$ is $\pu{565kJ/mol}$ as opposed to $\pu{490kJ/mol}$ of water. Isn't this contradictory? Why is $\ce{HF}$ still more acidic than $\ce{H2O}$?
Is it because  the conjugate base of $\ce{HF}$ is highly solvated and more stable in an aqueous solution?  Is there not a direct correlation between ease of bond dissociation and acidity?
P.S.: I understand the acidic trend down the group (stability of conjugate base increases and since the halogen atom becomes bigger, resulting in weaker bonds).

see this link . Did you get your answer?

Your welcome :-)


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Do quantum particles have mass?
Quantum particles are never objects but are always waves. But do they have mass or can they carry mass?

I don't think all the downvotes without suggestions are particualrly helpful

@LigninPauling agreed. While perhaps OP may have found the answer by doing more digging before asking (not saying they didn’t), this type of question could help someone in the future.

I'm only suggesting that those who downvote should explain why they dislike the question rather than simply downvoting. I thought your answer below was very useful.

@LigninPauling, There are persistent pests which automatically down vote without having the courtesy to say why? Most likely they have nothing positive to say.

You are made of quantum particles, i.e atoms. Wave particle duality means that in some experiments designed to show  wave behaviour, this is observed (i.e. double slit experiment on atoms and molecules), in others, particle behaviour is observed in experiments designed to measure this, i.e. mass spectrometer. But not both behaviours observed at the same time.

welcome to the chemistry StackExchange!
I think it would be helpful for you to review the wikipedia page on particle-wave duality. If you have access to a textbook on the origins of quantum mechanics, that would help clear things up,too.
The quantum mechanical model for particles states they have both particle-like and wave-like behavior, so it is incorrect to say they are never particles, which is what I assume you mean by "objects".
At the most fundamental level, quantum mechanics is just another tool with which to view and understand the world around us. It's an especially useful tool when we are looking at things in very fine detail (let's say, usually < $10^{-7}$ m). So, we can look at a given particle from the framework of classical mechanics or from the framework of quantum mechanics. This is an important point, because it is not quantum or classical mechanics that attributes mass to the particle. Rather, the particle had that mass to begin with.
So, to answer your question: A quantum mechanical treatment neither prescribes nor removes mass from a particle or an object. A photon is massless, a proton has mass, and that's that!

A photon has zero rest mass, but has momentum / mass due to energy mass equality.

Momentum can exist without mass. And photons do not have mass, they can only have relativistic mass.


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What is the difference between enthalpy change and energy change?
This question seems really dumb but I rather be safe than sorry. I read about enthalpy change in my chemistry textbook which defines enthalpy change as the energy exchange between a chemical reaction and the surroundings at constant pressure. My confusion came from when I read this BBC Bitesize page about energy change, can someone explain to me how they're different as they use the same units?

Gibbs free energy and enthalpy generally share the same units, which might be a source of your confusion. Enthalpy is a component of Gibbs free energy.

From 1st law of thermodynamics we get dU=DQ+DW(I wrote 'D' in place of 'd' because U is a total differential and a function of state only but heat or work are no such property of system, they are transient form of energy and so it's 'D')
Any way if Pop is the external pressure on system then the work term becomes DW= - Pop.dV
dU=DQ-Pop.dV=DQ-PdV(if Pop=pressure of gas=P,i.e. reversible)
=>dU=DQ if there is no change in volume of the system,i.e. (DQ)v=dU
Whereas,
dH=d(U+PV)=dU+pdV+VdP=DQ+vdP
=>dH=DQp(if the whole process is isobaric i.e. constant pressure and reversible)
So you get internal energy change of a system as the heat exchange in constant volume process and enthalpy change of a system as the heat exchnge in a constant pressure process
Actually U and H both are measures of  some amount of energy. Try to understand their definations. Definition of  H is H=U+PV. Don't consider only chemical reactions. It's valid any where.
Actually most of the chemical reactions are done in the lab at constant pressure and so the heat exchange of chemical reactions is more convenient to express in terms of ∆H.

It seems that you do not understand the difference between the internal energy $\ce{U}$ and the enthalpy $\ce{H}$. I will try to present these fundamental notions in a different way.
The internal energy U is the sum of all the energies stored between the atoms in the chemical bonds. If a chemical reaction happens, the bonds are losing or absorbing energy. Heat is getting in or out of the system, and this heat $\Delta$Q can be measured and correspond to the change in internal energy. $\ce\Delta$U = $\Delta$Q, if the volume is constant.  But if the transformation is carried out in the atmosphere, at constant pressure, a part of this energy is converted into work $\Delta$w to repel the atmosphere, because the volume of the system may change. Usually this volume change is small (except if a gas is produced or consumed). And this volume change may sometimes be difficult to know with precision. But it must be taken into account to calculate $\Delta$U. So if this small amount of work is neglected, a correction must be added to the internal energy U, that depends on pressure and volume. Neglecting this volume correction gives a sort of “apparent internal energy”, which is called enthalpy H, with H = U + PV. And then, in all transformations made at ordinary pressure, the following expression is valid : $\Delta$H = $\Delta$Q.
The difference between $\Delta$U and $\Delta$H is similar to the difference between the weight of an object and its weight in the vacuum. After Archimedes, all objects dipped in a fluid lose in weight the amount of fluid (air or water) they displace. When an object is weighed in air, the obtained value is the apparent weight, smaller than the weight in a vacuum. It should be increased by the Archimedes’ principle. Nobody does it. It is about the same for the enthalpy. Enthalpy is a sort of apparent internal energy.


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Effect of clemmensen reagent on 4-oxo-4-phenylbutanoic acid
When I first solved the question, I had reduced the ketonic group. Then as the medium is acidic the -OH in carboxylic acid, will get protonated and there will be formation of stable carbocation and it will attach to the benzene ring. I hope you get it. But in the answer, the carboxylic acid retained and only ketonic group is reduced. Please help.

Clemmensen is done in cHCl, but this is not the dehydrating conditions necessary to create the carbonyl cation necessary to give cyclisation as there is still a lot of water present. You need something like polyphosphoric acid.

Then for the compound  3-hydroxycyclopentanone on applying same reagent, the answer given is  chlorocyclopentane. Then how is that it possible?

Secondary alcohol in cHCl which is stable. You're protonating an alcohol not an acid and you have a better nucleophile present in chloride. If you produced an acyl chloride under Clemmensen conditions it would just hydrolyse back to the acid.

How can u say H2O is a better nucleophile than cl-. As charged molecules are always better nucleophile than neutral. Also on second attempt is think after formation of ring the acidic group is now converted into a keto group so I think it would also get reduced. So I think it is a special case in which  carboxylic acids will get reduced. Please help where am I wrong.


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Can ZrO2 be made fluoride resistant?
$\ce{ZrO2}$ has great properties, biocompatibility, chemical resistance, decent scratch resistance, great thermal resistance and refraction, but it could be better. It, like glass, will slowly react with $\ce{HF}$ and other fluorine substances, specifically oxidizing agents with fluorine.
Normally to make stuff resistant to fluorine the only thing to do is react it with fluorine in such a way that the reacted surface stays where it is. Steel flasks, for example, can be polished in a low oxygen environment and treated with fluorine to make the surface of the steel an iron fluoride.
Unlike iron oxides, fluorine won’t react with it because fluorine is a more effective oxidizing agent and essentially nothing else can break apart the intramolecular bonds keeping the iron fluoride together under normal conditions. It’s like, you can’t burn something twice, you can’t fluorinate something twice. That might not work with $\ce{ZrO2}$, I was going to say cause it doesn’t burn or oxidize but it’s already an oxide so that’s a bad reason.
Anyways PTFE doesn’t react with fluorine cause it’s just fluorine and carbon already.
So here’s my question: Can you do that with $\ce{ZrO2}$? Would Zirconium form a stable molecule with fluorine? If so, would it have similar properties, or at least be solid at STP? If not, is there another way of making $\ce{ZrO2}$ fluorine resistant? Or something else like $\ce{ZrO2}$ that’s more fluorine resistant?

Zirconium fluoride ZrF4 is the equivalent of ZrO2 with fluorine, see https://en.wikipedia.org/wiki/Zirconium_tetrafluoride .

Zirconium tetrafluoride melts at 910 c compared to ZrO2’s 2715 c. It’s used in really expensive and underwhelming optical equipment, and some nuclear reactors. It’s soluble in water, less dense than ZrO2, and much more reactive. I’m trying to figure out if you can HF-proof ZrO2, not make a fairly reactive precursor to more interesting chemicals or make overhyped lenses.

Sorry if that sounded mean, it was meant to be funny and interesting and provoke further discussion and I couldn’t figure out how to do that well.


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Electrophilic addition to benzene
A question deals with electrophilic addition to the following compound;

Apparently, the benzylic carbon gets a positive charge during electromerism, invoking resonance stabilization.
Benzene is an electron-withdrawing group, and therefore a negative charge on this carbon atom should be stable, as it has resonance stabilisation to boot.
Why isn't this so?

Is this 1‐phenyl‐1‐propene? If so, does the question specifies whether it's a cis or trans isomer? Please note that adding proper name for the compounds helps future visitors a lot. Also, please avoid using a [tag:homework] tag: it's deprecated.

This is an electrophillic reaction. These reactions make cationic intermediates. There's never been a question about stabilizing intermediate carbanions in electrophillic reactions. Thus, you are talking about apples and oranges. Please correct that part in your question.
Now, let's talk about relavent chemistry on electrophillic reactions. The alkene (electron donor) reacts with the electrophilic (electron seeking) proton to make the stable carbocation following Markovnikov's rule. In this case the more stable carbocation will be the carbocation adjacent to the benzene ring, a benzylic carbocation. The benzylic carbocation is more stable than secondary carbocation, because the adjacent benzene ring can stabilize the $\ce{C+}$ center through resonance as depicted in following scheme:


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Is the Beckmann Rearrangement technically an "SN2" reaction at an "sp2" center?
Below is the mechanism from this link of the Beckmann Rearrangement.

In the alkyl migration step, the alkyl group migrates simultaneously as the $\ce{H2O}$ leaving group is expelled. This seems as though it is an $\mathrm{S_N2}$ reaction at the $\mathrm{sp^2}$-hybridized nitrogen atom, but I thought that it was geometrically impossible to have an $\mathrm{S_N2}$ reaction at an $\mathrm{sp^2}$ center, per this link.
What am I missing in my reasoning?

Please do not use \ce{…} expression in formatting other than chemical equations and formulas.

For the record, the link you provide in "per this link", contains an answer of mine where I argue that $\mathrm{S_N2}$ reactions at $\mathrm{sp^2}$ centers are quite feasible, not impossible. For example, the Burgi-Dunitz angle is in accord with such reactions at carbonyl carbons.  Sometimes the Pauling "bent-bond" heuristic makes things more understandable than the equivalent Huckel "sigma-pi" approach.

When you were taught that we don't see $S_{\mathrm{N}}2$ reactions at $sp^{2}$ centers, what evidence or rationale was given? In organic chemistry, very few things are definite. When they are, it's usually because of being forbidden by quantum mechanics (e.g., symmetry in pericyclic reactions). Otherwise, they are just very unlikely, unless you somehow magically make the conditions right.
Greg Fu used to give this example: J. Am. Chem. Soc. 1998, 120, 2275-2282.
In it, a trans-substituted alkene undergoes nucleophilic substitution to become a cis-substituted alkene. Using that fact and other data, the authors conclude that the mechanism most consistent with the observed reaction data is a vinylic $S_{\mathrm{N}}2$. If you have a better mechanism, consider writing a paper and submitting it to JACS.

My textbook, Clayden et al., says: "To attack from the back, the nucleophile would have to appear inside the benzene ring and invert the carbon atom in an absurd way. This reaction of course is not possible." It then generalizes this rule by saying: "SN2 at sp2 C does not occur".

That's for a benzene ring. What about an alkene?


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Which salts precipitate from a potassium/sodium chloride/sulfate solution?
This is not a homework question. I am not in any course of study. The following question arises from my personal curiosity. If 1 mole of sodium sulfate and 2 moles of potassium chloride are dissolved in water, and the solution is gradually evaporated, which salts precipitate, and in which order?
There are four possible single salts that could be precipitated, namely $\ce {Na2SO4, K2SO4, NaCl, KCl}$, as well as mixed salts with the general formula $\ce{((Na,K)Cl)_x.((Na,K)2SO4)_{1-x}}$, but I don't know how to work out which ones, or what the sequence would be, since I am a complete ignoramus regarding the chemical thermodynamics of these salts, their solutions, and their precipitation from solution. I do understand the basic concept of Gibbs free energy, which I guess would be involved here.

There is no simple answer to this problem. The result of the evaporation of concentrated solutions of two cations and two anions cannot be predicted by calculation, and are not dependent on the solubilities of the pure salts. It has to be established by experience. Sometimes double salts are obtained, sometimes not. it is entirely empirical. You have to establish the so-called phase diagrams. It is a rather special science.

@Maurice :Thank you for your helpful answer, which at least establishes that this could not plausibly be a “homework” question.

Actually @Maurice is again overestimating difficulty of problem, by talking about general situation. In this particular case you'd, most probably, first crystallise Na2SO4 and then KCl. While interactions between solute can complicate things, in this case I think the difference in solubilities to be enough.

related https://chemistry.stackexchange.com/questions/9406/evaporation-of-a-mixture-of-salts https://chemistry.stackexchange.com/questions/132927/distinguishing-between-potassium-chloride-and-sodium-chloride/ https://chemistry.stackexchange.com/questions/7514/how-to-separate-kcl-from-nacl-and-mgcl%e2%82%82

@ Mithoron. I agree that your opinion is the most probable. But it is only probable. To be sure, the phase diagram K-Na-SO4-Cl should be studied. Don't forget that Sodium Sulfate may produce $\ce{Na_2SO_4, Na_2SO_4·7H_2O, Na_2SO_4·10H_2O, 3K2SO4·Na_2SO_4}$ depending upon the temperature and the concentrations.


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Does CO2 dissolve in water?
First of all not a homework question, but one day it suddenly popped into my head while opening a bottle of soda and accidentally leaving a glass out for a while.
I get that $\ce{CO2}$ in water is not the same as Carbonic acid, however this also raises the question if $\ce{CO2}$ can combine with water (It does so in acid rain in the atmosphere(I would guess normal temperature and pressure)), then shouldn't dissolved $\ce{CO2}$ just form carbonic acid and hence become unusable for aquatic plants? And also if Carbonic acid does form in the atmosphere ( Again,I would guess normal temperature and pressure (feel free to correct me!)), why does the fizzing happen in soda bottles when opened in the first place, should Carbonic acid be stable at normal conditions?
Please do note that I talk about $\ce{CO2}$ dissolving under normal temperature and pressure
I feel like I'm missing something super basic and obvious here and I just can't put a finger on it.
 Thanks in advance!

"it can't be dissolved in H20 on normal temperature and pressure" — says who? BTW, there is carbon dioxide dissolved in your blood right now.

Do you drink soda?

The amount of CO2 dissolved in water is proportional to the outer pressure.  At 20°C, 1 liter water dissolves about 1.7 g CO2 at normal pressure (1 atm). If the pressure is twice as large, the amount of dissolved CO2 is twice as much, 3.4 g.

When you open that bottle, does all of the carbon dioxide escape? Observation 1 is quite good.

But the soda will lose almost all of it's carboration if left in an open glass in a relatively short amount of time so I thought CO2 wasn't as likely to dissolve in H20.. but thanks I understand now

The reason why carbonated beverages go flat when left in the open is due to equilibria and gas exchange. Dissolved carbon dioxide is in equilibrium with gas outside the solution but the latter will diffuse away unless the container is closed. This constantly draws carbon dioxide out of solution to reestablish the equilibrium.

I want to extend Maurice's comment:

The amount of $\ce{CO2}$ dissolved in water is proportional to the outer pressure. At $\pu{20 °C}$, 1 liter water dissolves about $\pu{1.7 g}$ $\ce{CO2}$ at normal pressure (1 atm). If the pressure is twice as large, the amount of dissolved $\ce{CO2}$ is twice as much, $\pu{3.4 g}$.

To talk about solubility of gases in liquids, we take the help of Henry's Law which states that:

The amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid.
Mathematically,
$$\pu{S_g=k_HP_g^\circ}$$
where $\pu{S_g}$ is the solubility of the gas, $\pu{k_H}$ is the Henry's law
constant which is different for different gases and $\pu{P_g^\circ}$ is the
partial pressure of the gas.

For $\ce{CO2}$ the value of $\pu{k_H}$ in water is $\pu{1.67}$ when the pressure is taken in atm at $\pu{20^\circ C}$. Also note that generally the solubility of gases decrease with increase in temperature and increase with increase in pressure.
There is another reason why $\ce{CO2}$ can be more soluble in water:
$$\ce{H2O + CO2 <=> H2CO3}$$
$\ce{CO2}$ reacts with water to form carbonic acid which then further dissociates to form $\ce{H+}$ ions and hence the overall solubility increases.
For your first observation, the amount of $\ce{CO2}$ released by fish(es) is so small that it is almost impossible to observe the formation of bubbles.
For your second observation, the pressure while fizzing the soda bottles is very high which leads to higher solubility of $\ce{CO2}$ and when the bottle cap is opened, the pressure inside the bottle is decreased and hence again following Henry's law, excess $\ce{CO2}$ comes out in the form of bubbles since now its amount is not so less.
Hope this answer cleared some of your doubts :).

When I used to make beer, I would use CO2 and put pressure in the keg of beer that needed to be carbonated. Just putting pressure in the keg would cause the CO2 gas regulator to hiss as the pressure in the keg came up. In order to get more gas into the keg, one could either put the keg and CO2 bottle in the refrigerator with pressure in the keg OR put a nominal low pressure from the CO2 bottle and agitate the keg and the CO2 regulator could be heard injecting gas as long as the agitation was continuous.
For our atmosphere, during rain events, the atmosphere is active (roiling, updrafts, etc) which is the proper environment for dissolving CO2 gas into the water that is precipitating out of the atmosphere. This means that CO2 gets scrubbed from the atmosphere during rainstorms and is carried into the ground by the water it is dissolved in. So, the earth indeed has a built-in mechanism for scrubbing CO2 from the atmosphere.


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Studying chemical engineering
My question is pretty simple. Does studying chemical engineering require memorizing manufacturing processes and other complex chemical mechanisms, or does it rely more on the practical side of things focusing on how stuff actually works.

Every subject in the world will require some level of factual retention. You cannot say every time, "Hey, let me check the reference before I can answer you." Secondly, you will need a solid background in mathematics to succeed (differential equations, some programming knowledge etc.)

Your question will probably be closed because there is no yes-no answer to your query. Opinion based questions are discouraged here.

As a chemical engineer, my answer to your question is that it involves applying the basics of thermodynamics and physical chemistry to quantitatively design and analyze the operation of processing equipment in chemical plants, including chemical reactors, absorption columns, cooling towers, heat exchanges, etc.  So it's going to involve a lot of physical and mathematical modeling calculations, and relatively less memorization.

@M.Farooq well when it comes to meths and fundamental understanding I'm all about that just because I'm pretty bad at memorizing so u have to rely on understanding how something works. Also mathematics aren't really the problem here. I actually hope that studying chemical engineering will involve more maths and less memorizing. Thanks for your answer.

@user254610, I would say talk to students who are actually studying chem engineering in your university of choice. Never ever listen to a negative advice by someone is is actually studying the subject. Those are immature students who will continue to change their mind and their current vision is narrow. Look at the profile of professors in the chem. engineering dept. If you are from India, just keep in mind, the highest number of unemployed PhDs are from pure chemistry.

I can speak based on my experience studying chemical engineering as an undergraduate at an ABET accredited university in the United States and working in industry.
A typical curriculum would include a focus on fundamentals which can then be applied in specific industries or in different cases rather than an in depth study of specific manufacturing processes or specific chemical reactions. This process is intended to develop a core of knowledge to approach specific processes or reactions and also as a preparation for learning more
So for example an undergraduate might start off taking calculus, and general chemistry and proceed to additional chemistry like organic and possibly phsyical chemistry. Starting in chemical engineering a student will encounter fundamental fields of chemical engineering which are fundamental tools such as mass and energy balance, thermodynamics, fluid dynamics, heat transfer, mass transfer, and chemical kinetics. These are all tools which can be used as a chemical engineer to evaluate processes. Also a review and study of types of what's called unit operations would be included.
Unit Operations are like the building blocks of many processes. The include operations like heat exchangers, distillation or absorbion columns, cooling towers, reactors, tanks, and more.
An understanding of chemical engineering fields such as kinetics or thermodynamics or heat transfer combined with an understanding of unit operations can be combined to learn more about a specific process.
I am sure there are some folks out there with more insight so I'll be very interested to hear from them as well!


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What is the effect of pressure upon the melting and boiling points of a substance?
I want to know about the inter-relation between the effect of pressure upon the melting and boiling points of a substance. Can anyone help?

How does pressure enter in the Gibbs free energy?

Does this answer your question? The effect of pressure on boiling point?

When we apply pressure on a substance, its melting point increases or decreases depending upon the change of volume which occurs during the phase change of that substance. To be specific, if the volume of the substance's liquid phase is greater than the volume of the solid phase, its melting point will increase upon the increase of volume. Again, if the volume of the substance's liquid phase is less than the volume of the solid phase, its melting point will decrease upon the increase of volume.
The boiling point of liquids always increases when pressure is applied on that liquid. This is because the molecules of the liquid will require comparatively more energy to turn into gaseous state when pressure is applied on that substance.

That is direct application of Le Chatelier's principle

Will the Le Chatelier's principle predict that a liquid will become a solid at very high pressure?


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Little to no reaction in my hot process soap
I tried to make hot process soap and it was a failure. I carefully weighed the ingredients, but when I added the KOH to the melted oil, nothing happened. When I watch videos on hot process soap making, the oil turns cloudy immediately when the alkali is added. In my case there was no color change and the oil remained clear.
I heated the mixture for over 2 hours at around 180F and got very little reaction. There was some slight coagulation of small white beads on the surface of the oil, but that was it. In videos it shows the oil turning into a mashed potato consistency in as little as 20 minutes. That did not happen in my case at all.
It was though my alkali component was grossly deficient. However, I followed the measures carefully. The formula I used was coconut oil, 5.6 ounces, palm oil 8 ounces, moringa oil 2.4 ounces, water 6.08 ounces, KOH 3.49 ounces. The KOH was in flake form. The package reads caustic potash, 93% purity, KOH.
Could the 93% purity have been the problem?
Another issue is possible water absorption by the KOH. I put some of it in the microwave and it quickly formed quite a puddle, which I mistakenly thought was molten KOH, but it must have been water on second thought, so it is possible the KOH had a high water content.

KOH certainly absorbs water. Some labs use it as a drying agent

Did you stir ? Your mixture must be stirred as much as possible, specially in the beginning, because oil and KOH solution are not miscible. The reaction starts at the contact surfaces of the two phases. After a while an emulsion is produced and it is less important to stir. The proportion of watering KOH is not the main parameter.

I agree, stirring is crucial

How old is the KOH? If not cared for appropriately or if there's a loose lid, over time you get potassium hydrogencarbonate.  It's not clear how you mixed the KOH and the oil.


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m-CPBA Selectivity between Baeyer Villiger Oxidation and Epoxidation?
If both a ketone and an alkene are present in a molecule and you add one equivalent of m-CPBA, what are the factors that determine which group will be oxidized, the ketone through Baeyer-Villiger oxidation or the alkene through concerted epoxidation? Is one of the reactions particularly faster and better than the other?
EDIT: Since the literature seems to be conflicted and there is not a clear general answer, let us take the below as an example because I am interested in the competing factors here. For this example, I am leaning towards epoxidation just due to electron-rich alkene, and there is no base to make m-CPBA a better nucleophile for BV.

For your particular compound, epoxidation occurs, see: Danishefsky, S.; Kitahara, T.; Schuda, P. F.; Etheredge, S. J. A remarkable epoxide opening. An expeditious synthesis of vernolepin and vernomenin. J. Am. Chem. Soc. 1976, 98 (10), 3028-3030. DOI: 10.1021/ja00426a066 (compound 8 → 9) where it's claimed that "strong acceleration by the neighbouring hydroxyl group" helps.

Indeed, the presence of the adjacent hydroxyl group facilitates epoxidation by forming hydrogen bonds with one of the oxygen atoms of m-CPBA in the transition state.

@orthocresol thanks, you are truly a wonder at finding things. How in the world did you find this paper?

I just happen to remember almost all of the organic chemistry literature. More seriously, I don't, but there are databases that do. https://www.reaxys.com/

The answer appears to be "it depends" (perhaps this should not surprise an experienced organic chemist).
According to the abstract of the paper by Mendelovici and Glotter [1] in steroid chemistry:

Treatment of 3β- and 3α-hydroxy-(acetoxy-)cholest-4-en-6-one and of 6β-and 6α-hydroxy-(acetoxy-)cholest-4-en-3-one with MCPBA gives two types of product, depending on the initial site of the peroxy acid attack. Attack at the carbonyl group gives a Baeyer–Villiger rearrangement leading first to enol lactones and then by epoxidation of the latter to epoxy lactones. Alternatively, attack at the double bond gives epoxy ketones which can subsequently undergo a Baeyer–Villiger rearrangement leading to epoxy lactones. With one exception (3α-hydroxycholest-4-en-6-one), the Baeyer–Villiger oxidation of the enone is the dominant process. Epoxidation of the double bond is suppressed in the presence of an axial 3α- or 6β-acetoxy group.

However, Kazmeier and Horn [2] working with unsaturated aldehydes find

Purified m CPBA is a useful reagent for the oxidation of several classes of aldehyde. Although linear unbranched aliphatic aldehydes are oxidized to the corresponding carboxylic acids, α‐branched ones undergo Baeyer–Villiger oxidation to formates. α‐Branched α,β‐unsaturated aldehydes provide enolformates and/or epoxides, which can be saponified to α‐hydroxy ketones with shortening of the carbon chain by 1 carbon. Unbranched α,β‐unsaturated aldehydes undergo an interesting Baeyer–Villiger oxidation/epoxidation/formate migration/BV oxidation cascade, which results in formyl‐protected hydrates with an overall loss of two carbon atoms.

Overall it appears that both processes can occur so Murphy's Law applies. It would be a good idea to explore alternative epoxidation conditions for an unsaturated ketone.
References

Mendelovici, M.; Glotter, E. Epoxidation and Baeyer–Villiger Oxidation of γ-Hydroxy-Αβ-Unsaturated Ketones on Exposure to m-Chloroperbenzoic Acid. J. Chem. Soc., Perkin Trans. 1 1992, No. 13, 1735–1740. DOI: 10.1039/P19920001735.
Horn, A.; Kazmaier, U. Purified MCPBA, a Useful Reagent for the Oxidation of Aldehydes. European Journal of Organic Chemistry 2018, 2018 (20–21), 2531–2536. DOI: 10.1002/ejoc.201701645.

thanks for your answer. It is interesting how there is no clear winner here. Care to check out my edit? That example should have a selective formation (its a problem) but I'm trying to figure out why

The selective Baeyer-Villiger Oxidation of unsaturated ketones is definitely depends on the conditions used. I do not want to eloborate this fact any further because it was well documented in Waylander's answer. Howevr, so far into the literature suggest that use of peracids in presence of unsaturation in the substrate ketone is not a good idea. The quote "Attack at the carbonyl group gives a Baeyer–Villiger rearrangement leading first to enol lactones and then by epoxidation of the latter to epoxy lactones" clearly suggests Baeyer-Villiger Oxidation predominates under the condition, but the following epoxidation cannot be avoided. How ever, recent literature suggest that use of suitable catalyst may stop the reaction at lactome formation, avoiding the forthcoming epoxidation.
$\ce{Sn}$-doped zeolite beta can be used in stopping side reactions after Baeyer-Villiger Oxidation (Ref.1 and Ref.2). The use of solid acids such as zeolites as selective catalyst has benefited to avoid use of peracids as the oxidant (the traditionally used oxidant), and replacing it with cheaper and less polluting hydrogen peroxide ($\ce{H2O2}$). The major problem with the traditional conditions of Baeyer-Villiger Oxidation is their lack of sufficient selectivity for the desired product if the starting material contains functional groups other than the ketone group. This shortcoming is well met with use of  zeolite beta after incorporation of 1.6 weight per cent tin into its framework as the catalyst of choice (Ref.1). This $\ce{Sn}$-doped zeolite beta acts as an efficient and stable heterogeneous catalyst for the Baeyer-Villiger oxidation of saturated as well as unsaturated ketones by $\ce{H2O2}$, with the desired lactones (e.g., vinyl ester) forming more than 98% of the reaction products (Ref.1).
When $\ce{Sn}$‐Beta is used as a heterogeneous catalyst for the Baeyer–Villiger reaction with $\ce{H2O2}$, cyclic ketones are transformed into the corresponding lactones, while unsaturated ketones are oxidized to the corresponding unsaturated lactones with very high chemoselectivity (Ref.2). The catalyst is also selective for the oxidation of aromatic aldehydes ($\ce{Ar-CHO}$) with $\ce{H2O2}$, producing the formate ester ($\ce{HCOO-Ar}$) or the corresponding hydrolyzed product, that is the corresponding phenol ($\ce{Ar-OH}$). Shape‐selective oxidations are observed for isomeric reactants with different molecular shapes (Ref.2).
Seemingly, $\ce{H2O2}$ is the choice of oxidant to avoid further oxidation of double bond. Zhang et al. have used organoselenium catalyst with $\ce{H2O2}$ to perform Baeyer-Villiger Oxidation on $\alpha,\beta$-unsaturated ketones (Ref.3), abstract of which states that:

By carefully screening the organoselenium pre‐catalysts and optimizing the reaction conditions, simple dibenzyl diselenide was found to be the best pre‐catalyst for Baeyer–Villiger oxidation of (E)‐$\alpha,\beta$-unsaturated ketones with the green oxidant hydrogen peroxide at room temperature. The organoselenium catalyst used in this reaction could be recycled and reused several times. This new method was suitable not only for methyl unsaturated ketones, but also for alkyl and aryl unsaturated ketones. Therefore, it provided a direct, mild, practical, highly functional group‐tolerant process for the chemoselective preparation of the versatile (E)‐vinyl esters from the readily available (E)‐$\alpha,\beta$-unsaturated ketones. A possible mechanism was also proposed to rationalize the activity of the organoselenium catalyst in the presence of hydrogen peroxide in this Baeyer–Villiger oxidation reaction.

Further, Ref.4 describes a metal-Free Baeyer–Villiger synthesis of vinyl acetates using Oxone as the oxidant:

Oxone, a cheap, stable, and nonhazardous oxidizing reagent, transforms $\alpha,\beta$-unsaturated ketones of defined stereochemistry into their corresponding vinyl acetates through a Baeyer–Villiger reaction. This process is general and straightforward, tolerating a wide range of functional groups.

Both graphical abstracts of Ref.3 and Ref.4 (Scheme A and Scheme B, respectively) are depicted in following diagram for your convenience:

Note that Oxone (potassium peroxomonosulphate) has been used in epoxidation of alkenes in diphasic conditions in presence of a ketone (usually acetone) and phase transfer catalyst (e.g., Ref.5 and Ref.6). However, in these conditions, epoxidation has been failed when the substrate alkene is $\alpha,\beta$-unsaturated ketone (Ref.5).

References:

A. Corma, L. T. Nemeth, M. Renz, S. Valencia, “$\ce{Sn}$-zeolite beta as a heterogeneous chemoselective catalyst for Baeyer–Villiger oxidations,”  Nature 2001, 412, 423–425 (https://doi.org/10.1038/35086546).
Michael Renz, Teresa Blasco, Avelino Corma, Vicente Fornés, Robert Jensen, Laszlo Nemeth, “Selective and Shape-Selective Baeyer-Villiger Oxidations of Aromatic Aldehydes and Cyclic Ketones With $\ce{Sn}$-beta Zeolites and $\ce{H2O2}$,” Chemistry - A European Journal 2002, 8(20), 4708-4717 (https://doi.org/10.1002/1521-3765(20021018)8:20<4708::AID-CHEM4708>3.0.CO;2-U).
Xu Zhang,  Jianqing Ye,  Lei Yu,  Xinkang Shi,  Ming Zhang, Qing Xu,  Mark Lautens, “Organoselenium‐Catalyzed Baeyer–Villiger Oxidation of $\alpha,\beta$-Unsaturated Ketones by Hydrogen Peroxide to Access Vinyl Esters,” Advanced Synthesis and Catalysis 2015, 357(5), 955–960 (https://doi.org/10.1002/adsc.201400957).
Belén Poladura, Ángel Martínez-Castañeda, Humberto Rodríguez-Solla, Ricardo Llavona, Carmen Concellón, Vicente del Amo, “General Metal-Free Baeyer–Villiger-Type Synthesis of Vinyl Acetates,” Org. Lett. 2013, 15(11), 2810–2813 (https://doi.org/10.1021/ol401143q).
Norio Hashimoto, Atsushi Kanda, “Practical and Environmentally Friendly Epoxidation of Olefins Using Oxone,” Org. Proc. Res. Dev. 2002, 6(4), 405–406 (https://doi.org/10.1021/op025511f).
Scott E. Denmark, David C. Forbes, David S. Hays, Jeffrey S. DePue, Richard G. Wilde, “Catalytic Epoxidation of Alkenes with Oxone,” J. Org. Chem. 1995, 60(5), 1391–1407 (https://doi.org/10.1021/jo00110a049).


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Measuring vapour pressure in a closed system
My experiment requires the determination of vapour pressure from which I will calculate the number density (or number of moles of the molecule) later, for a closed system.
Here's my setup :

Liquid is placed in a steel container with the steel tube connecting it to a t-junction. One branch goes to the liq N2 immersed catch flask, followed by a pump.
Other branch goes to a pressure transducer (Si sensor, which measures absolute pressure, linear from 0.2 atm to 4 atm).
Procedure :
Liquid, in my case n-hexane, is frozen in the container using liq. N2. The gases in the container and the tubes are removed using pump. While the valves are closed, the liquid is brought to room temperature. The freeze-pump-thaw cycle is repeated twice again, to remove dissolved gases from the  liquid.
Later the liquid is brought to room temperature and the system is left for few hours for equilibrium while the container is immersed in a water bath.  Valve for pump is closed during this time, while the valve towards the pressure sensor is open. Temperature measurement of the water bath is done using thermocouple.
Issues :
The pressure sensor reading is always more than the saturated vapour pressure of the liquid for a given temperature. This maybe due to a leak. From Raman measurements of gases in the tube it was found that some N2 is present in the system.

What would be a good way to determine the vapour pressure of the liquid in this closed system.
Would it be a good approximation to assume liquid-vapour equilibrium and use the tables of vapour pressure vs temperature to determine the corresponding value for my case.

Importantly, maximal error allowed in the determined vapour pressure value is up to 2%.
Thanks.

Updates :
Pressure reading while melting n-hexane at room temperature.

Pump used is an oil pump.

What is the residual vacuum pressure, compared to vapour pressure ?

a) fix the leak, b) test with empty sample container, c) calibrate your thermocouple, d) how was your pressure sensor calibrated, e) why does it take a few hours to equillibrate?

Upon repeated freeze-pump-thaw the obtained pressure is below 0.0068 atm (~0.1psi or 0.68 kPa). Vapour pressure measurement for n-hexane gave ~5 psi (0.34 atm or 34.47 kPa). Temperature is  25.5 C / 296.6 K.
Thermocouple was calibrated by an standardization institute upto 1000K. Its error is below 0.2K from 285-1000K.
Pressure sensor is a NIST calibrated absolute sensor, calibrated between 0 to 3.4atm (https://sea.omega.com/tw/pptst/PX409-USBH.html)

What about dynamic measurement, watching how boiling point decreases with decreasing of pressure ?

Seven millibars residual pressure? Are you using a membrane pump?

@Poutnik I have not checked dynamic measurement against boiling point.

@Karl Probably,  a membrane pump. I will check and report here. Its a very common type found in organic chem labs.

@Karl Pressure sensor reading show, initially fast and then slow but gradual increase in the reading  when melting to the liquid I will add the relevant plot in the post.

Schlenck lines are run with oil pumps, which should go down to 0.01 mbar, 0.001 with fresh oil.

By "calibration", I mean "when did you last check the calibration?" If it was more than two days or two months ago (depending on who handled the instrument in the meantime), check it again. An OC lab full of solvent vapour is not a nice environment for delicate equipment.

@Karl Pressure calibration was not checked for ~6 months. Thanks for the advice. :-)   We work in  a clean room, fixed temperature and humidity. Pump is located in a different room and the tube runs through to my instrument. We are doing spectroscopy of the vapour and we need vap pressure for the analysis.

@ankit7540 I have not said you did. It was alternative arrangement suggestion, that could or could not be applied to your case.


text:
How does pH depend on temperature, when taking into account volume expansion?
Since autodissociation of water is endothermic, increasing the temperature will drive the following equilibrium to the right, which has the effect of increasing $[\ce{H+}]$:
$$\ce{H2O <=> H+ + OH-}$$
However, increasing the temperature of liquid water will also increase its volume, which has the effect of reducing $[\ce{H+}]$ (cf. this table on Wikipedia showing how density of water varies with temperature).
So as we increase the temperature, which of the two effects dominates? Does $[\ce{H+}]$ experience a net increase or decrease?

pH is NOT inversely proportional to [H+](it would be pH = k/[H+]), as pH = -log([H+]).

The concentration of [H+] is $c\alpha$. The alpha is related to equilibrium constant of dissociation of acid($K_a$). We know that $K_a$ is related to temperature by van't Hoff equation. So $K_a$ increases with temperature if reaction is endothermic , thus increases [H+] and decreasing pH.

I edited your question for clarity.

Meta post regarding closure.

Your qualitative understanding is correct.  As you increase the temperature, you have two effects:

The dissociation reaction, which is endothermic, shifts to the right.  This effect increases the concentration of hydrogen ions.
The concentration of hydrogen ions is expressed as molarity, which is number per unit volume.  As you increase $T$, water expands, resulting (everything else being equal) in all concentraitons decreasing.

And you're asking which one wins.
The answer is the first, overwhelmingly.  For instance, as you go from $\pu{25 ^\circ C}$ to $\pu{50 ^\circ C}$, $K_\mathrm w$ increases by five times.  By contrast, the increase in volume over this temperature range is only $1\%$.

For pure water, $\mathrm{pH}=- \frac 12 \cdot \log{K_\mathrm{w}}$. As $K_\mathrm{w}$ grows with temperature, $\mathrm{pH}$ decreases.
The influence of water thermal dilation is much smaller than the temperature dependency of water autodissociation constant.
For solutions with acids, bases, acidic or alkaline salts, buffering systems  it is much more complicated. It depends how acidity/alkality constants depend on temperature.


text:
How does electrolytic conductance depend on solvation?
My textbook says: If the ions are strongly solvated their effective size will increase and hence their conductance will decrease
P.s after reading my textbook, I don't think I'm sure what solvation means. If you could explain that, my understanding would be better.

Have you tried wikipedia? https://en.m.wikipedia.org/wiki/Solvation

@AniruddhaDeb I don't feel as though I've got the entirety of what solvation means. Yeah sure, I can define what solvation is but I don't really understand it.

Suppose you are in a field filled with a crowd of children and some of them are holding your hand, some are holding your shirt but you have an urge to reach from point A ( one electrode) to point B (another electrode)? In that case you effective size has increased. How easy would it be to travel from A to B, if you were all alone?
Change in conductance due to solvation is somewhat similar. When an ion wishes to travel under the influence of an electric field between two electrodes, solvent molecules like to travel with the ion, making it "bulkier", hence the ion moves with a retarded velocity.
P.S. Solvation comes from the word solvate which means "A more or less loosely bonded complex formed between a dissolved species and the solvent." It is a general term used in physical chemistry.
Even more formally, http://goldbook.iupac.org/terms/view/S05747
"Any stabilizing interaction of a solute (or solute moiety) and the solvent or a similar interaction of solvent with groups of an insoluble material (i.e. the ionic groups of an ion-exchange resin). Such interactions generally involve electrostatic forces and van der Waals forces, as well as chemically more specific effects such as hydrogen bond formation."

Thanks! This makes a lot of sense. The analogy of small children really helped, hehe.


text:
Comparing axial and equatorial bond lengths in square pyramidal structures
I found this image on wikipedia:

It got me wondering why the axial bond angles were shorter than the equatorial ones. I found this extract elsewhere on the internet -
Extract: (ww.nrcresearchpress.com)

For $AX_{5}E$ molecules which have a square pyramidal shape the extra repulsion exerted by the lone pair would be expected to cause the neighboring four bonds, i.e., those at the base of the square
pyramid to be slightly longer than the fifth bond to the apex. Experimental evidence on
molecules of this type is scanty (5). The four square-planar bonds are longer than the
single bond to the apex in $SbCl_{5}^{2-}$ but they appear to be shorter in $SbF_{5}^{2-}$.

The extract is from a research paper from 1961. It states that the rule isn't followed in $SbF_{5}^{2-}$, and that there's "scanty" evidence anyway.
Now, given that the paper is pretty old, I was wondering if there has been any new evidence collected on this subject. I essentially would like to understand why the bond length of BrF5 follows this order, and why it isn't followed by $SbF_{5}^{2-}$. I'd appreciate some references as well for evidence and perhaps other researched examples.
Note: This isn't a duplicate of this, since that question doesn't address the deviation in $SbF_{5}^{2-}$.

If the lone pair repels other electrons in the Br-F bonds by causing the bond angle to decrease  to approx 85 degrees, would it not follow that the bond length would also be increased and so reduce repulsion. This lengthening would also reduce repulsion to the axial F atom. With luck you may get an expert answer :)

@porphyrin Thanks, that was helpful. Do you know how to explain the anomaly with SbF5 2−?


text:
Electrolytic Refining of Silver Nitrate
I read that the electrolytic refining of Silver is done using dilute silver nitrate as an electrolyte and a small percentage of free $\ce{HNO3}$. The following reactions are possible at the anode:
$$\ce{Ag ->Ag+ + e-}$$
$$\ce{ OH- -> 1/4O2 + 1/2H2O + e-}$$
But the second reaction has $E^\circ = \pu{+0.4V}$ while the first reaction has
$E^\circ = \pu{+0.8V}$. So why does $\ce{Ag}$ of the anode get oxidised? Is it because of a low concentration of $\ce{OH-}$ ions due to poor dissociation? Or is there another reason?

The second reaction has E° = + 0.4 V, if the concentration of OH- is 1 M. Please apply Nernst equation to this electrode.

Oxidation always occurs at anode. I think you meant the reverse reaction for Ag

@Maurice The value changes with concentration of the electrolyte..so that explains why Ag is oxidised .

You have mistaken reactions. Note that: $$\ce{Ag+ + e- <=> Ag} \quad E^\circ = \pu{0.7996 V}$$.

And also, $$\ce{O2 + 2H2O + 4e- <=> 4OH-} \quad E^\circ = \pu{0.401 V}$$.

@ Chopin. I don't understand what you mean. Ag+ is not oxidized. It is reduced into Ag metal at the cathode. And there is no OH- in the solution. On the contrary there are H+ ions.

@Maurice But doesn't Ag of the silver anode get oxidised and move into the solution? Like it happens in electrolytic refining of copper? If not what happens at the anode? About the OH- ions, I guess the dissociation of water is very less to have a considerable OH- concentration.

@Mathew Mahindaratne I wrote down the equations like they would happen at the anode but specified the values of the reduction potential.(reverse reaction)

Electrolysis or electrolytic refining is a technique used for extraction as well as purification of metals obtained by other refining methods. In the electrolytic refining process, a block or a strip of impure metal is used as the anode, and a thin sheet of pure metal is used as the cathode. Both the cathode and anode are dipped in an electrolytic cell containing an aqueous solution of the metal salt. Upon the application of direct electric current of a suitable voltage, pure metal is deposited at the cathode (reduction) by the dissolution of impure metal at the anode (oxidation).
Since electrolytic refining id done in aqueous medium, the key to successful refining is the oxidation of the metal should be more facile than the oxidation of water (see the standard oxidation potentials below). Hence the solid metal would dissolve into the solution as cataions, leaving behind many of the impurities, which are inactive to electrolysis. The possible anodic oxidation reactions would be:
$$\ce{Ag <=> Ag+ + e-} \quad E_\mathrm{ox}^\circ = \pu{-0.7996 V} \tag1$$
$$\ce{4H2O <=> O2 + 4H+ + 4e- } \quad E_\mathrm{ox}^\circ = \pu{-1.229 V} \tag2$$
When value of $E_\mathrm{ox}^\circ \ (\ce{Ag})$ is compared with that of $E_\mathrm{ox}^\circ \ (\ce{H2O})$, it is safe to say that $\ce{Ag}$ oxidation at anode is more spontaneous (less negative value) than that of $\ce{H2O}$ (greater negative value) at anode. As a result, impure $\ce{Ag}$ anode will be slowly dissolved into solution as $\ce{Ag+}$ ins as direct current applied.
Similarly, the possible cathodic reduction reactions would be:
$$\ce{Ag+ + e- <=> Ag}  \quad E_\mathrm{red}^\circ = \pu{0.7996 V} \tag3$$
$$\ce{ 2H2O + 2e- <=> H2 + 2OH-} \quad E_\mathrm{red}^\circ = \pu{-0.8277 V} \tag4$$
Since the refining solution is dilute $\ce{HNO3}$ to provide the counter-ion for $\ce{Ag+}$ cation, we should consider possibility of  another reduction reaction at cathode:
$$\ce{NO3- + 3H+ + 2e- <=> HNO2 + H2O} \quad E_\mathrm{red}^\circ = \pu{0.934 V} \tag5$$
$$\text{And, } \ce{NO3- + 4H+ + 3e- <=> NO + 2H2O} \quad E_\mathrm{red}^\circ = \pu{0.957 V} \tag6$$
When the value of $E_\mathrm{red}^\circ \ (\ce{Ag})$ is compared with that of $E_\mathrm{red}^\circ \ (\ce{H2O})$, it is clear that $\ce{Ag}$ reduction at cathode is much more prominent (high positive value) than that of $\ce{H2O}$ (greater negative value) at cathode. However, the competition reduction would happen as well when comparing the value of $E_\mathrm{red}^\circ \ (\ce{Ag})$ with those of $E_\mathrm{red}^\circ \ (\ce{NO3-})$, both of which  are slightly higher positive values. However, if you use appropriately dilute nitric acid to begin with (c.f., Nernst equation), you will be able to avoid nitrate reduction at cathode.

Why is the reaction involving OH- ions at the anode not a possibility?

The solution is acidic, therefore not enough $\ce{OH-}$ ions floating around the solution. But there are plenty of $\ce{H2O}$ there. If you have used $\ce{NaOH}$ instead of $\ce{HNO3}$, that is a possibility.

@ Mathew Mahindaratne . If NaOH is added to a silver nitrate solution, it would produce a precipitate of silver oxide Ag2O, and silver can not be deposited at the cathode any more.

The process of purifying silver by anodic oxidation is working if the impurities are made of metals with redox potentials higher than 0.800 V. If the impurity is made of metals with redox potentials lower than 0.800 V, like copper, this impurity will be oxidized before silver, and silver will be precipitated at the bottom of the container. This is what happens when refining copper mixed with noble metals by anodic oxidation. At the anode, copper is dissolved and impurities like silver are deposited at the bottom of the solution. At the cathode, pure copper is obtained.


text:
Obtaining energy and wave function for excited states
Suppose I have done a HF calculation and obtain the energy and wave function of ground state. If I want to produce the spectrum of the specie (say Hydrogen molecule), should I use CI (configuration interaction) method? In general how can I obtain energy and wave function of excited states?
Sorry if my question is awkward, because I'm new

You might find this article helpful: http://www.ccl.net/cca/documents/dyoung/topics-orig/excited.html The question itself is quite broad, as the method you'll be using is dependent on system size and the parameters you need. It's also a hot topic in development, so the more details you can add to your query, the better it will be received, and the sooner you may get an answer.

Related: Why is time-dependent density-functional theory (TD-DFT) used to describe excited states? (It's less broad then your question and still got closed.) More: Excited states and electron addition/removal energies within Hartree-Fock, CI, etc There are other examples, so please browse our site for even more.

@Martin - マーチン Thanks a lot for useful links. If I do a CI calculation after the HF method and obtain ground state energy, then how can I obtain the energy and wave function of first excited state using CI?


text:
Order of activating ability in electrophilic aromatic substitutions
From A Guidebook to mechanism in organic chemistry by P. Sykes [1, p. 155]:

The activating effect of $\ce{Y}$ on the nucleus is found to increase, i.e.
the overall rate of substitution increases, in the approximate order:
$$\ce{OCOR} < \ce{NHCOR} < \ce{OR} < \ce{OH} < \ce{NH2} < \ce{NR2}$$
$\ce{NR2}$ is more powerfully activating than $\ce{NH2}$ because of the electron-donating effect of the $\ce{R}$ groups.
It should not be forgotten, however, that in acid solution, e.g. in nitration, these two groups will be converted into $\ce{^\oplus NHR2}$ and $\ce{^\oplus NH3},$ respectively; the nucleus will then be deactivated and substitution will be predominantly m- (cf. $\ce{^\oplus NR3},$ p. 151).
The $\ce{OH}$ group is sufficiently activating to cause phenol to be brominated instantaneously to the 2,4,6-tribromo derivative (the p- and both o-positions all attacked) by bromine water at room temperature.
The groups $\ce{OCOR}$ and $\ce{NHCOR}$ are less powerfully activating than $\ce{OH}$ and $\ce{NH2},$ respectively, because of the reduction in electron-availability on $\ce{O}$ and $\ce{N}$ by delocalisation over the adjacent, electron-withdrawing carbonyl group:

Why is $\ce{OR}$ a less activating group than $\ce{OH},$ but $\ce{NR2}$ is a more activating group than $\ce{NH2}?$
One would expect that that alkyl groups increase the electron density via their inductive effect and thus make the group more activating towards electrophilic aromatic substitutions (EAS) reactions, both in the case of $\ce{OR}$ as well as $\ce{NR2}.$ However, $\ce{OH}$ is a better activating group despite such an effect. Why is this so? Why would the same effect not be applicable for $\ce{NH2}?$
Reference

Sykes, P. A Guidebook to Mechanism in Organic Chemistry, 6th ed.; Longman; Wiley: Harlow, Essex, England; New York: New York, NY, 1986. ISBN 978-0-582-44695-3.

The most likely reason for this is that an $\ce{OH}$ group can act as an acid and exists in an equilibrium with the $\ce{O-}$ anion, which is significantly more activating.
$\ce{NH2},$ on the other hand, is not really that acidic. It does not benefit enough in this manner to be significant.
The loss of the anionic form is making $\ce{OR}$ less activating despite the inductive effect. $\ce{NR2},$ however, does not have as much to lose and only gains in activating ability.


text:
Baking Soda and Vinegar: Are there any secondary reactions?
I want to run a Lego Pneumatic Engine on the CO2 that is produced, but before I do, I want to be sure that that's the ONLY gaseous product from a real-world reaction.  Everyone seems to assume that it is, just by omission, but no one actually says it.
I'm sure it would run, but I'd rather not damage the ABS plastic or probably-steel parts (shiny metal piston shaft), or a rubber seal or the factory lubricant or something else that I can't see.  CO2 seems inert enough by itself, but is that really ALL I'll get?
(assuming of course, that I keep the foam out)

Vinegar contains about 50 - 60 g acetic acid per liter. The other acids are tantric acids and lactic acid, which are present at 1 to 3 g/L. All acids react with baking soda with production of $\ce{CO_2}$ No other substances can produce a gas, with reaction with baking soda.

You won't get pure $\ce{CO2}$ out of an aqueous solution. There will be water vapor in the $\ce{CO2}$. Over a period of time that will almost certainly cause any steel Lego parts to corrode.

@MaxW That makes sense.  If there's water involved, then inevitably some will evaporate.  So how does that compare to normal humidity?  These parts are about 20 years old, with no special care except possibly by accident (open air and regular use for the first 10 years or so, then in sealed plastic bags for maybe another 10 years, all climate-controlled), and no discernible damage so far.

Acetic acid and dissolved salts will promote corrosion, unless it is stainless steel. And even in that case, the shiny polished surface may got not so shiny over time.

So far, the biggest concern seems to be the metal shaft.  Plastic, rubber, and a bit of grease are all okay by omission then?

As another data point, I had one of them in water about 20 years ago (the bathroom sink, just playing with the possibility), and it rusted shortly afterwards.  But you couldn't tell that now.  I guess the regular use after that scraped off the corrosion?  I don't know what that means, but maybe someone else does.

Re: Humidity. Because of the bubbling, microscopic droplets not only of water, but also the dissolved reactants and products, are released. You can reduce this by putting a filter, e.g. some fiberglass from a HVAC filter, in the neck of the reaction vessel.

It seems to me, according to the comments on the question, that there are NO additional reactions or products to worry about.
However, because of the foam, some of the original reactants can be tossed up, and find their way into the later parts of the apparatus as a haze/fog/cloud that is not completely inert.
Is that more-or-less accurate?  I'm trying to get an answer here, and not just a string of comments.

Your answer is an accurate summary and is fine. So afterwards, rinse the LPE with water, dry it and maybe put a small amount of oil on the steel shaft.

@EdV I guess the easiest way to do that is to run it on baking soda and vinegar as planned, then run it on water to rinse it out, then run it on air to dry it.  Then the question becomes, "How to hook up a garden hose to an LPE?", and is probably better answered on bricks.stackexchange.  A sports inflator needle fits wonderfully, so I can use that to hook up a shop air compressor.  (and dial the regulator down a bit!)  I've been using a bicycle pump for testing, but I don't think I want to pump long enough to dry it!

I have to admit iI have no idea how big the LPE is, but getting all the reaction residue (and whatever might be is excess), washed away, is good. Just reasonably dry is OK: the rusting of iron and steel requires both oxygen and moisture, so dryer is better. Best of success! BTW, have an upvote: upvotes here are easier than at the EE stack exchange, where they are tossed around like manhole covers. ;-)


text:
Inductive effect of -NHMe
Why does $\ce{-NHMe}$ show -I effect. I think it should be opposite. I know that $\ce{-Me}$ groups are electron donating in nature. So the nitrogen atom will have a small $\delta^-$ charge. So then why would it still be electron withdrawing?

You're looking at the result and trying to use that as the starting point. The end result tells you what happened, not what is going to happen. Where did the nitrogen atom get that partial negative? By withdrawing electron density from neighboring atoms.

@Zhe ok I think I got it. The nitrogen atom is electronegative, that is why it is withdrawing electrons from all three bonds. My reasoning was like saying that we have a sink hole, and since water is pouring in from the left, water can't pour in from the right. Thanks

The inductive effect is reflective of the comparison of electronegativity between an -R group and carbon. There is an -I effect because the nitrogen is more electronegative than carbon, so it will inductively withdraw electrons. I will also note that the inductive effect is measured relative to the -R group being hydrogen.


text:
Can the titrant and the unknown material be flipped?
We know the typical experiment of determining the concentration of $\ce{Fe^2+}$, where we titrate $\ce{MnO4^-}$ which has dark purple color and watch the color of the bottle.
Can we also determine the $\ce{MnO4^-}$ concentration using $\ce{Fe^2+}$ as titrant and how?

You can see that the iron is inside the beaker and we titrate with  $\ce{MnO4^-}$. Can we do it backward, i.e. having $\ce{MnO4^-}$ inside the beaker with purple color and we titrate $\ce{Fe^2+}$.

Yes and no. The titrant is picked to give a good end point detection. It is easier to see a color appearing than to see a color disappearing.

For human eyes, it is much easier to observe colourless -> colour, than the other way around - so in this case, I'd say it won't work, but there might be other cases where it could by switching to a different indicator.

Also, the equivalence point is light pink when using $\ce{MnO4-}$. Observing pink to light pink is way harder than observing colourless to light pink.

Refer primary standard solution

Additional complication: the unknown substance has an unknown concentration. If you add it to the titrant, what if you don't have enough?

In principle yes. In practice is more complicated due to facts already mentioned. One is relative to detection. The most important one, as user600016 said, is that not everithing is a good standard.

The titration reaction does not know or care whether one reagent is in the buret and one is in the conical flask (Erlenmeyer flask, it is not a beaker in your figure). And certainly you can standardize/titrate potassium permanganate with ferrous ammonium sulfate solution to determine the concentration of permanganate ion. Ferrous ammonium sulfate hexahydrate is a primary standard in analytical chemistry.
If you are extraordinary careful, you can fill the iron (II) solution in the buret and have permanganate in the conical flask. The end-point will be slightly vague (deep purple to colorless). You really have to be extra-cautious. The question is why bother with this painful eye-stressing procedure? Always take the permanganate in the buret.
A dark solution can be used in the conical flask. For example, in iodometric titrations, we start with a deep brown solution, which fades, to nearly a colorless solution upon titration with thiosulfate. In order to confirm the end-point, you would put a drop of starch before your expected equivalence point. Again, the color change is from deep blue to colorless.

I disagree that there is an equivalence. If you add all of the unknown via the burret, but it's not enough, what do you do then?

Well, a trained analytical chemist will never start randomly. There is something called as "pilot reading" when you have no clue of the sample concentration, which roughly tells us what should be the volume of the titrant. You always have some idea of the sample analyte concentration and they would adjust the concentrations so that they are within the buret volumes.


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Two electron integrals between 1s STO
I am trying to understand SCF cycle by trying to code up solved example from Quantum Chemistry by Levine (page 443, 5th edition). The problem is shown below:

One electron integrals are straight forward and I was able to get the correct answer, however I still can get correct values for two election integral, lets say (11|11). Given below is my attempt in octave
clear all;
N=2000;
zeta1 = 1.45;
zeta2 = 2.91;

r = linspace(0.000001,10,N)';
dr = r(2)-r(1);
chi = @(zetad,x) (2*zetad.^(3/2))*exp(-zetad*x).*x;
chichi = 0;
for i =1:N
    chichi = chichi + dr*(chi(zeta1,r(i))*chi(zeta1,r(i))*chi(zeta1,r')*(chi(zeta1,r)./((r(i)-r) + 0.000001)));
end
chichi*dr

However my values are way off in this case. Can anyone please shed a light on it? Value of (11|11) = 5/8 zeta1 = 0.9062.
Two electron integrals are defined in the book as:

You replaced the 6-fold integration by one in spherical coordinates. Not sure if that can work as easily as you wrote it and I'm not an expert on that transformation, but I think that some multiplicative constants from the angle integrations are missing.

I am closing this question as it has been asked and answered on [mattermodeling.se]: One-center two-electron integrals between 1s STO

@TAR86 was correct, I mistook the volume integral as simple 1d integral. Also integrating $1/|r_2 - r_1|$ would result in a singularity at $r_1 == r_2$ which would yield really wrong results.
As per @user1271772 suggestions, materials modelling stack exchange was quite useful (see question here). There as per Susi Lehota's suggestion I was able to get correct result by integrating Legendre Expansion.
My octave code for two electron integral (11|11) follows:
zeta1=1.45;
chi1111 = @(r1,r2) 16*zeta1^3*zeta1^3*exp(-2*zeta1.*(r1)).*exp(-2*zeta1.*(r2)).*r1.*r1.*r2.*r2./max(r1,r2);
dblquad(chi1111,0.000001,10,0.0000001,10,1E-3)
ans =  0.90623

Similarly other integrals can be encoded, and I get results exactly as Levine example.
Thank you everyone


text:
What would be the best combination to clean polyurethane
I am struggling to find the best composition to clean old golf balls and make them white again.
The balls are made out of different types of coating, e.g. polyurethane or an isomer.
I have found hydrogen peroxide to be so far a good candidate.
I cover the balls in hydrogen peroxide and expose them to UV light.
What I'm trying to figure out is how to find a good process or accelerator to the compound.
For example, should the balls only be soaked for 10 seconds or do they have to be in for 5 hours due to the material?
Would any other item (e.g. bleach or oxi washing powder) add to the cleaning power of hydrogen peroxide?
Any insights on a chemical level of what should be the best process forward or suggestion on how to evaluate my methods?

Sounds like you are on the right path. As for the exact procedure, I suspect one might need to play around with different conditions to find the optimal one. Still, a good question regarding underlying chemistry.


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Can the Belousov-Zhabotinsky reaction be simulated with the Brusselator?
I'm trying to simulate the Belousov-Zhabotinsky reaction mathematically, with a system of ODEs. I know how to do this with the Oregonator and now I'm looking for another model to simulate the BZ-reaction.
The Brusselator model is often mentioned as a model to simulate an oscillating reaction and now I'm looking for a paper/article that uses the Brusselator to simulate the BZ-reaction. But till now, I have not found much, so I'm wondering if it even is possible to simulate th BZ-reaction with the Brusselator?
If it is possible, can you point me to a paper that uses the Brusselator to simulate the Oregonator?
If not, why can't the BZ-reaction be simulated with the Brusselator? And are there alternative models I can use, besides the Oregonator?
Thanks a lot in advance!

reference 1 at the end of the Wikipedia page on the Brusselator compares different models so this might be a good place to start.

There are alternative models based on real chemical species that can be used, I mentioned a few in my answer here: https://chemistry.stackexchange.com/questions/79106/how-does-an-oscillating-reaction-work/81924#81924