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How can we define a non-spontaneous reaction?
Consider a reaction:$$\ce{$aA + bB$ <=> $cC + dD$}$$
The value of reaction quotient at a certain time $t$, $${Q_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}}$$
where the concentrations $[A], [B], [C]$ and $[D]$ are at time $t$.
Let the reaction start initially at $t = 0$, with only reactants $i.e.$ $[A]$ and $[B]$ equal to say $1$ mol and $[C]$ and $[D]$ equal to $0$ mol. Hence, $$Q_c = 0$$
We know that the value of equilibrium constant $K_c$ must be such that, $$K_c > 0$$
Thus, $$Q_c < K_c$$
Which is the only condition for the advancement of reaction in forward direction. This condition does not consider the value of change in Gibbs energy $\Delta G$.
Now, considering the relation $$\Delta G = \Delta G^o + RT~\mathrm{ln}~Q_c$$
When $Q_c = 0+$ then $\mathrm{ln}~Q_c \to -\infty$, which means $\Delta G << 0$ and reaction is spontaneous in forward direction.
Hence, can it be concluded that every reaction is spontaneous in forward direction if it starts with only reactants?
If so, how can we define a non-spontaneous reaction?

No reaction is at equilibrium when one of the species (reactant or product) is zero. So you can make any reaction that goes to equilibrium go forward (start with no products) or in reverse (start with no reactants).

This spontaneous / non-spontaneous thing seems to be a big bee in some teachers´ bonnets. I never found it useful after I had understood what chemical equillibrium means.

@KarstenTheis It seems very logical that "if a reaction is going to be in equilibrium in future than that has to be spontaneous in forward direction at start with only reactants." This means that any reversible reaction has to be spontaneous in forward direction at start if we start with only reactants. Thanks for your approach!

Providing knowledge at basic level will be like Free energy of products should be more than Free energy of reactants
for a rxn to be non spontaneous. ∆G > 0

It sounds like your confusion arises from not making a distinction between $\Delta G$ and $\Delta G^\circ$ when describing a reaction as spontaneous or not.
The $\Delta G^\circ$ is the free energy change for the reaction at the defined "standard" conditions of 1 M solute concentrations and/or 1 bar gas partial pressures of both the reactants and products. When we make the general statement that a reaction is spontaneous or not, we are usually referring to whether this $\Delta G^\circ$ is greater or less than 0.
If $\Delta G^\circ < 0$, then the reaction is described as spontaneous, and it will proceed in the forward direction when starting from a mix of 1 M reactants and 1 M products until equilibrium is reached.
In contrast, $\Delta G$ is the free energy change for the reaction in whatever state is of interest. In your case, you have chosen a state with no product present at all. When we say that a reaction is spontaneous based on $\Delta G$, we mean whether it will go in the forward or reverse direction from this specific state of interest, not the standard state.
Obviously, if there is no product present, the reaction cannot go in the reverse direction (forming reactant from product), so all reactions will go in the forward direction if started in a state of $0$ product. That is what you have shown quantitatively using $Q=0$ so that $\Delta G <<0$.
So you are absolutely correct that every reaction in a state in which no product is present is spontaneous. That does not mean, however, that the reaction will necessarily happen anytime soon. Very often, the activation energy barrier is high enough that the reaction rate is extremely slow. Or it is possible that the equilibrium constant may favor reactants so much that even one molecule of product is sufficient to reach equilibrium and the reaction stops very quickly.

Or the equilibrium constant may favor reactants so much that one molecule of product is already way past equilibrium, and you'll be back to zero in short order even if any product is ever produced. Or some other reaction might consume everything and leave you with no product or reactant for the reaction in question.

(1) That's an insight that for comparing $\Delta G^o$ starting point should be  a mix of 1 M reactants and 1 M products (standard states). (2) Both $\Delta G$ and $\Delta G^o$ can be compared for spontaneity, depending on the conditions. (3) $\Delta G^o = -RT~\mathrm{ln}~K$ only at equilibrium when $\Delta G = 0$. Can we equate $\Delta G^o = \Delta H^o - T\Delta S^o = -RT~\mathrm{ln}~K$ at any (constant) temperature? (4) Only an irreversible reaction can be non-spontaneous at start because a reversible reaction (at start) will go in forward direction to attain equilibrium.

@Andrew Can you please verify the above four statements?

(1) yes (2) yes (3) No. $\Delta G^\circ = -RT \ln K$ is always true. It does not matter what $\Delta G$ is, since that isn't in the equation. For the second part, true technically only if $\Delta H^\circ$ and $\Delta S^\circ$ have been determined for your T of interest. They don't vary a lot with T, though, so we treat them as invariant if the temp change isn't too great from standard. (4) Not sure what you're saying here. Irreversible usually means forward is so favored that reverse is not observed, so it would definitely be spontaneous.

@Andrew So the question remains "What is a non-spontaneous reaction?" Can we say that any reaction can be spontaneous or non-spontaneous depending upon temperature and pressure conditions?

A non-spontaneous reaction is one in which net conversion of product to reactant will occur in order to attain thermodynamic equilibrium. Assuming that we require some amount of reactant to be present in order to consider a state to be a reaction condition, there will be some reactions for which it is not possible to add enough product to make the reaction non-spontaneous.

Or no path leading to a decrease of Free energy exist. But this point is may be too trivial.

In general it is necessary to consider any entropy changes in determining whether a system is at equilibrium or if a spontaneous change will occur.
As there must be an increase in entropy in actual  processes then $dS_{system}+dS_{surr}=dS_{irrev} \ge 0$. By using the first law  with the last expression and after several steps, we find that  in an isothermal  reaction (surrounding temperature $T$) in which no work is done other than $pV$ (gas expansion) work the condition becomes
$$dG + TdS_{irrev}=0$$
thus a process can take place spontaneously with entropy production only if the free energy decreases. If there is no change in free energy, no change can take place and the system is at equilibrium.
This is what you have determined in your question but with  the implicit assumption that reaction will occur.
A non-spontaneous process is one that is already at equilibrium or that will require input of some work to make it change.

Thanks! I mentioned another form of this relation in this question. For spontaneity, $\Delta S_{total} > 0$ but $\Delta S_{sys}$ can be positive or negative.

I think  that my $S_{irrev}$ is your $S_{total}$ so we agree. For a reversible process $dS_{irrev}=0$.

I agree with Andrew that it depends on whether you define spontaneity based on  $\Delta G^\circ <0$ or  $\Delta G < 0$.  Typically, freshman chemistry books use the former.
However, I've never liked equating spontaneity with the sign of $\Delta G^\circ$, prefering to instead use the sign of $\Delta G^\circ$ as an indicator of whether reactants or products are "favored".
The reason I don't like using the sign of $\Delta G^\circ$  (as opposed to that of $\Delta G$) as an indicator of spontaneity is it causes the very confusion you are experiencing.  It also obscures the concept of chemical equilibrium.
I thus favor equating spontaneity with the sign of $\Delta G$ (which in turn depends not only on $K_{eq}$, but also upon where the reaction mixture is relative to it), which is the view you are articulating.  For clarifying this latter view, I think the following diagram, which I've borrowed and modified from one of my earlier answers (Are all exothermic reactions spontaneous?), is helpful:

I deleted my comment. Your approach makes sense, and I think everyone learning about Gibbs energy of reaction should be exposed to the graph you posted. I'd be happy if the term "spontaneous" disappears from textbooks because it always sounds like it is talking about kinetics, which is not the intent. For relating the direction of the reaction to the Gibbs energy, I think textbooks should include the following relationship: $$\Delta_r G = RT \ln{\frac{Q}{K}}$$ which relates the Gibbs energy to the ratio of the reaction quotient and the equilibrium constant.

... If you start with that relationship, you can plug in $Q=1$ (i.e. standard state) to get the relationship between standard Gibbs energy of reaction $\Delta_r G^\circ$ and equilibrium constant. The relationship as is shows the sign of Gibbs energy of reaction when $Q < K$ and $Q > K$, something students are already familiar with at that point, so it is a quantitative extension of that concept. For biochemistry, you can use the relationship directly to get the Gibbs energy available from a proton gradient or any other gradient (by plugging in K = 1).

@KarstenTheis In this expression, why does not it matter whether the equilibrium constant is $K_p$ or $K_c$, considering $K_p = K_c(RT)^{\Delta n}$?

@Apurvium When you switch from $K_c$ to $K_p$, you also switch from $Q_c$ to $Q_p$, so the factor $(RT)^{\Delta n}$ cancels out.

@KarstenTheis $\Delta G^o = -RT\ln K=ΔH^o−TΔS^o$; in this expression, there can be $K_c$ or $K_p$ but it is equated to $ΔH^o−TΔS^o$ also?

Your standard state will be different (either 1 atm or 1 mol/L), so the standard change of enthalpy and the standard change of entropy will be different, too.

@KarstenTheis $K_p = K_c(RT)^{\Delta n}$; based on different values of $R$ in different units, we can get different values of $K_p$ in pascal, bar, atm, etc. When we put these values in $\Delta G^{\circ}=-RT\ln K_p$, we will get different values of change in standard Gibbs energy, as $\ln K_p$ is now dimension less.

This question shows that you have probably not really understood what the free enthalpy (or Gibbs energy, or free energy) is. I will try to explain it qualitatively without too much thermodynamics. Let's go !
The origin of the Gibbs energy is coming from Gibbs' reflexions on the spontaneity of chemical reactions. He was trying to find a potential energy similar to the mechanical energy for predicting a spontaneous reaction. In mechanics, if an object is situated at a high level, let's say on a table, and that it has the possibility of going to a lower level, on the ground, it will spontaneously do it and fall, without any effort greater than a flip from the observer. Energy is produced in the fall. The contrary is never spontaneous. You may heat it, it will never jump back up higher onto the table. You always have to pick up the object, give it some amount of mechanical work to put it back on the table.
So Gibbs was trying to find some sort of potential energy that atoms and molecules may have that will always decrease in spontaneous reactions. Enthalpy, or heat content, is not this energy, because endothermic reactions may be spontaneous. For example, the mixture of NaHCO3 and HCl solution reacts spontaneously, but the reaction is endothermic. The temperature of the system decreases. To go back to the original temperature, you have to heat the final system.
As heat content is not the energy Gibbs was looking for, he tried to discover the energy which is always exported in a chemical reaction, namely both in endothermic and exothermic reaction. Is there such an energy somewhere ? Yes, there is such an energy, and this is electricity. A cell works out by a chemical reaction. This reaction may be exo- or endothermic. But the galvanic cell will always deliver electric energy. A cell cannot absorb electric energy. It would be a non-sense. So Gibbs decided that in all spontaneous chemical reaction, "his energy", called G, must decrease, whatever the thermic effect. And $\ce\Delta$${G}$, the variation of $\ce{G}$, is related to the emf  $\ce{E}$ of the cell by the relation : $\ce{\Delta G = - zEF}$
As a consequence, he established the free energy of formation $\ce{\Delta G_{form}}$ for all substances consumed and produced in galvanic cells. And  $\ce\Delta$${G}$ of any chemical reaction can be worked out by substracting the $\ce{\Delta G_{form}}$ of the products minus those of the reactants. A difficulty happened when Gibbs realized that the emf $\ce{E}$ of a cell depends upon the concentrations of the substances, via the Nernst equation. So $\ce{\Delta G_{form}}$ must also be changing with the concentrations. As a consequence, the tabulated values of $\ce{\Delta G_{form}}$ are always reported for pure substances. In this case, $\ce{\Delta G_{form}}$ is called $\ce{\Delta G°_{form}}$, and it is this $\ce{\Delta G°_{form}}$ which is equal to $\ce{ -zEF}$.
Later on, but it is too long to explain it here, Gibbs was also able to calculate $\ce{\Delta G°_{r}}$ for reactions happening out of galvanic cell. He found that $\ce{\Delta G°_{r}}$ is related to the equilibrium constant of the reaction, by the expression $\ce{\Delta G°_{r}}$ = - RT ln K.
Surprisingly, $\ce{\Delta G°_{form}}$ and $\ce{\Delta H°_{form}}$ are not very different from one another. More surpringly, the emf and of course $\ce{\Delta G°_{r}}$ changes with the temperature T, although $\ce{\Delta H°_{r}}$ is nearly independant on T. And if $\ce{\Delta G°_r}$ and $\ce{\Delta H°_{r}}$ are reported on the same graph versus T, you will see that both values vary linearity with T, $\ce{\Delta H°_r}$ being nearly horizontal. Most surprisingly, both lines have a crossing point exactly at O K.
Apparently the difference $\ce{\Delta G°_{form}}$ - $\ce{\Delta H°_{form}}$ is proportional to T. And the ratio ($\ce{\Delta G°_{form}}$ - $\ce{\Delta H°_{form}}$)/T is constant. This ratio is called $\ce\Delta S$ and it is the entropy difference of the reaction. But this is another story.
Have you followed my explanation ?

Yes, most of it, except "$\Delta G^o_{form}$ and $\Delta H^o_{form}$ are not very different from one another." Nice way to relate $\Delta G^o_r$ with mechanics. You have mentioned that $\Delta H^o_r$ is nearly independent on temperature. But what about Kirchhoff's law related to it?

One more thing. The standard state of a substance is its pure form at 1 bar pressure at a specified temperature. Is temperature not defined for standard state?

Can you please check the above statements?

I think it is best to start with a real-world example of a spontaneous reaction as cited in electrochemistry, namely, as occurring in an electrochemical (or battery) cell. In particular, comments from Wikipedia:

A spontaneous electrochemical reaction (change in Gibbs free energy less than zero) can be used to generate an electric current in electrochemical cells. This is the basis of all batteries and fuel cells.

However, for those with experience with the startup of a so-called electrochemical cell, there can be an obvious inception (or waiting) period for these 'spontaneous electrochemical reactions' to proceed!
To explain the lag effect, I further note this comment from an educational source on entropy:

Changes in internal energy, that are not accompanied by a temperature change, might reflect changes in the entropy of the system in so called 'spontaneous' reactions.

Apparently, an increase in entropy corresponds to an increase in disorder. And, in the case of the battery cell with a salt bridge, the removal of electrons via a wire from the cell also results in a corresponding movement of cations and anions (to maintain charge balance in the respective half-cells). The latter occurs more slowly, but apparently is required for the continued efficient flow of electrons.
So, as illustrated by the so called 'spontaneous' electrochemical cells, there is also an increase in entropy component to balance the decrease in Gibbs free energy.
Interestingly, my cited example/argument seems to suggest that the word 'spontaneous' as coupled to 'spontaneous reactions' may implied, in particular cases, a more relative measure in terms of actual passage of time.


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How do the two half cells of Daniel cell know when they are connected?
I'm currently studying electrochemistry in school, and there is something I don't can't quite understand about Daniel cell.
When you put $\ce{Zn}$ solid into $\ce{ZnSO4}$ solution, does the electrode become negatively charged even before connecting it with the $\ce{Cu}$ half cell?
I searched but I wasn't able to form a clear understanding, I read that the answer to the previous question is no, the the two cells need to be connected for the redox reaction to happen.
If that was the case, how do the half cells communicate, how do they know when they are connected even when there is distance between them?
If the electrodes were charged before connecting them that would make perfect sense for me, the electrons repel each other and when connecting the two half cells with a wire, the electrons move because of the negatively charged electrode, but from what I've read that is not the case?
I'm really sorry if that is a stupid question but I really can't find an answer.

No, electrodes do not know about each other.
When an electrode is inserted to an electrolyte, the electrochemical reaction is ongoing in both directions. If reduction direction overruns oxidation, the potential of the electrode is increasing ( or vice versa ) until the rate of both reaction gets equal, the net reaction rate is zero and the electrode reaches the equilibrium potential.
That may be quick or slow, depending on the electrode and respective reaction kinetics.
When both electrodes are galvanically connected, the electron exchange due the potential difference disbalances the equilibrium potentials of both electrodes and the net electrochemical reactions get the steady non zero rate.
Note that the electrode potential is conventionally referred to the Standard hydrogen electrode (SHE), which itself has potential $\pu{+4.44 \pm 0.02 V}$ ( Wikipedia ) wrt the potential of a free electron in vacuum. So saying $\ce{Zn}$ electrode potential is negative rather means it is less positive.

You may want to try searching various terms in wikipedia and following relevant internal/external links. The great source of compactly explaned topics is hyperphysics, but with limited scope. It is a kind of hyperlinked quick cards/cheat sheets, very handy, if you want to understand the basics.  This link is particularly for Electrochemistry

Does that mean that each half cell have a potential even before connecting the two cells? And when the two are connected with a wire electrons move because of the potential difference? Also is it possible that you direct me to where I can read more about this?

Yes, I have tried to say exactly that.  See the answer update for more.

Another way of describing the electric behavior of the electrode is to state that when dipped in water, one Zn atom quickly looses two electrons and gets dissolved in the solution. The electron remains in the metal and prevents any second Zn atom from bringing more electrons in the metal. Simultaneously the solution gets positively charged and the first $\ce{Zn^{2+}}$ ion  repels any second $\ce{Zn^{2±}}$ ion from trying to enter the solution.

1 atom does not make much difference. Imagine the electrode has capacity just 1 pF, ( in fact much more) so 1 electron would make potential difference 0.16 microV.( in fact much less ). 1 Extra electron just very very slightly speeds up reduction and slows down oxidation.

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Thermal reduction of other metal oxides beside Fe2O3
In the firing of ceramic glazes you can make an effect called oil spot.
This is done by making two layers of glaze. A bottom layer rich in iron oxide (+15%) and a top white layer that's fairly permeable. Once your reach 1225° C, the iron loses it's bond with the oxygen and goes into reduction, despite being in an oxygen rich electric kiln. So the oxygen bubbles up to the surface, taking along a bit of the iron. It then turns into FeO, and it also has something to do with magnetite which I don't understand.
Anyway, the result is that you get a white glaze with dark spots surrounded with grey/green parts, or some yellow.This effect is very nice but it only works with iron oxide as far as I know.
So my question is: are there other oxides that can have the same thermal reduction effect, and if so at what temperature. I know somebody found a way to make it in blue (it took him 4 years) so maybe he used cobalt oxide, though that looks to me to be to obvious to take 4 years. Anyway, he's not a chemist, he's a potter.
For a potter this is hard to find out but maybe for a chemist this is not that difficult to resolve, so I thought why not go to the people who know this stuff much better than us potters.

As far as I remember, Fe2O3 is transformed into magnetite Fe3O4 at high temperature. But here, it may be different because of the presence of the glaze, whose composition is probably determinant for understanding your observations. But you don't give us the composition of this glaze.

Developing colors in glazes is complicated; It is done on the basis of experience and "cut and try".The matrix composition affects the color of additives, not just the atmosphere. One variable is the composition of gas in the kiln , such as ratio of CO to CO2.  I have often noticed on a crossection cut in glazed tiles : Near the surface is red ( Fe2 O3) and below is dark grey ( Fe3 O4 or Fe O).  So it absorbed oxygen at the kiln temperature.

One candidate is manganese dioxide, $\ce{MnO2}$.  This decomposes at 535°C, releasing oxygen.  As the phase diagram from Ref. [1] shows below, the reduced manganese fuses with excess silica between 1250 and 1300°C, forming rhodonite ($\ce{MnSiO3}$).  You would see this as a pink color.  But, of course, I'm a chemist rather than a potter, so I make no guarantees on whether or how it works in pottery!

Reference:
1.
Y.-B. Kang, I.-H. Jung, H.-G. Lee, "Critical thermodynamic evaluation and optimization of the MnO–SiO2–TiO2–Ti2O3" system, Calphad
Volume 30, Issue 3, September 2006, Pages 226-234.  https://doi.org/10.1016/j.calphad.2006.05.002

If I were not a chemist, and wanted blue, I would suspect cobalt first, but after 4 years, I might suspect copper, but not just Cu or CuO or Cu2O. Copper chlorides can be blue, but maybe not in a hot (or cold) glaze.
I'm not a potter, but I wonder if you can use the idea behind the borax bead test (https://en.wikipedia.org/wiki/Bead_test) to get your different colors. Two approaches: 1) combine the metal (oxide) and borax and melt them together, then grind them to the desired size before glazing.   2) Just combine them and dump them into the glaze. Borax has water of hydration which will bubble up and do things like the oxygen from Fe2O3.
There are glazes for sale on the internet which give colors, which could be combined with the Fe2O3 to give different results, but as blacksmith said, it's cut and try! One site I found (https://ceramicartsnetwork.org/daily/ceramic-glaze-recipes/glaze-chemistry/recipes-cool-ceramic-glaze-colors/) showed these colors:

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Why is average bond enthalpy measured by homolytic fission?
Average bond enthalpy: the energy required for 1 mole of a specific gaseous bonds to undergo homolytic fission
Why specifically homolytic fission? Is it to allow comparison, but if that were the case, couldn't heterolytic also be used?

I would think homolytic because homolytic fission returns the two atoms as they would be in neutral state, prior to forming the covalent bond. Heterolytic fission could result in ionic species. Ionic species didn't originally form the bond (you assume) so you use homolytic fission.

@arevmelikyan This makes sense. So a greater energy would be required for heterloytic fission?

Not necessarily, its just that using homolytic fission would provide a better indicator of the average bond enthalpy, as I have said. But yes, usually it would require more energy to separate two oppositely charged ions.

Try to think the other way round. If you would like to determine the strength of a bond like H-H, would you start form one H+ and one H- ion, or from 2 H· atoms ?

I would guess that it's because that's how the the bond dissociation energy is defined: http://goldbook.iupac.org/terms/view/B00699


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Redox flow battery
What is meant by decoupling of power and energy in context of redox flow battery? I understand that the electrolyte containing the active ions can be stored in tanks and that adds flexibility in scale up. But what does decoupling of power and energy mean in the context of redox flow battery.

Please explain ! Power and energy cannot be coupled, because they are different concepts. You cannot couple length and temperature, weight and time, concentration and height.

In my understanding, by energy/power  decoupling is here meant the possibility of independent scaling up of the available sustained battery power and the possibly stored energy.
For normal batteries of electrochemical cells, regardless of their chemistry, scaling up increases both battery power and energy in more or less comparable rate.
But for redox flow batteries, like $\ce{V^2+/V^3+ + VO2+/VO^2+}$, there is possibility to scale up both power and energy independently.
Scaling up energy can be done by increasing the total volume of working solutions.
Scaling up power can be done by increasing the total active surface of electrodes.

+1 This seems to be the reasonable interpretation of what the OP wanted, so I will delete my comment in a couple of hours.


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How does adding HCl convert AgCl into a sol?
According to the text I follow, in peptization the ions of the added electrolyte (peptizing agent) get adsorbed by the particles of the precipitate and turn the precipitate into a sol. An example of this is silver chloride, which can be converted into a sol by adding hydrochloric acid (chloride being the common ion).
What is the mechanism behind this?
And why does having a common ion matter?

What is the mechanism behind this?
From the NCERT textbook:

During peptization, the precipitate adsorbs one of the ions of the electrolyte on its surface. This causes the development of positive or negative charge on precipitates, which ultimately break up into smaller particles of the size of a colloid. Note that the presence of equal and similar charges on colloidal particles is largely responsible in providing stability to the colloidal solution, because the repulsive forces between similarly charged particles prevent them from coalescing or aggregating when they come closer to one another. Thus, they remain in dispersion medium as stable colloidal sols.

And why does having a common ion matter?

The sol particles acquire positive or negative charge by preferential adsorption of positive or negative ions. When two or more ions are present in the dispersion medium, preferential adsorption of the ion common to the colloidal particle usually takes place.

RE: "This causes the development of positive or negative charge on precipitates, which ultimately break up into smaller particles of the size of a colloid." // Think this sentence needs to be tweaked. It implies that a large single crystal would breakup which really isn't true. Rather the sol size particles of $\ce{AgCl}$ coalesced, but excess $\ce{Cl-}$  disperses the particles. // Ostwald ripening actually favors formation of large particles.

Please cite references when quoting.

Extra chloride ions adsorb on surface of solid $\ce{AgCl}$ particles.
The charged particles get then more easily hydrated, what  stabilizes the colloid.


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How to evaluate the basicity of molecules? (Ethanamide and Ethanamine, Example)
Hi there i recently came in a position where was asked to evaluate the basicity of molecules, and managed to understand the heirachy of most molecules, however i couldn't figure out who was more basic between ethanamide and ethanamine, i thought if i look at which ever was more acidic it would mean it would be less basic. So i came to the conclusion that Ethanamide would be more acidic since it has ressonance and the negative charge would be distributed through the Nitrogen and Oxygen, so by definition ethanamine would be the most basic, if we look at the Lewis perpective of things since it has no ressonance it would donate a a pair of electrons more willingly.
However when i googled the pKa of both compounds i found that in reality Ethamine was more acidic, and I was mad, since it was more acidic it couldn't be the most basic out of the two. I found the pKa of Ethanamine to be 10,8 and Ethanamide to be 16,75.
How do i correctly evaluate the basicity of molecules, what are the variables to look out for? and what am i missing?
I further apologize for any errors, english is not my native language.
I also apologize for the long post. I get mad when i can't understand the logic behind something and I can't do anything else before I understand it :)

Probably you found the pKa of the ion Ethanammonium, and the pKa of the Ethanamide., which is not an ammonium ion. In my tables, I get pKa = -1.1 for the ammonium ion obtained with ethanamide plus H+.

@Maurice So, does that mean my first assessment is correct?

@ Miguel. Yes. Your first assessment is correct. Ethanamide is more acidic than Ethanamine.

@Maurice And therefore the latter is more basic?

Not therefore - these properties are only tangentially related when comparing different compounds. In this case you had a lucky guess, though.

Mithoron and Maurice, thanks for the help!! I will revise the subject. On another note perhaps this question might be misleading to other, is it worth removing?

So, there are many factors that affect acidity or basicity of a compound, the main factors being bond electronegativities and resonance structures. However, I believe you are correct in assuming that those pKa values seem a bit off. Upon further investigation, I found that 10.8 is the pKa of the conjugate acid of ethanamine, so the pKa of the actual compound is much higher, making it the more basic of the two.

Thanks! I feel stupid, i simply saw the same information, but didn't pay attention to it. I understand this subject alot better now :) Thanks once more


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Can you determine when the orbtials of an atom will hybridize?
We know that when it comes to C its orbitals hybridize (they become sp3 , sp2 or sp) and the energy levels of an hybridized orbital of this kind is lower than that of the p orbital that formed it but higher than the s orbital so for example energetically speaking is more convenient for an atom to have its electrons (even if they are non-bonding and they don't do too much) in an hybridized orbital if possible.And I know that O and N also can hybridize so the energy barrier between their orbitals (the s and the p orbitals) is not that high to be unable to do this so why don't they just hybridize first in O2 or N2 and then form the molecular orbitals like is the case with C-C bonds , why in the case of O-O bonds we have no hybridization? I mean ok I get the perspective of being able to form a double bond but O2 can have a double bond even if it was sp2 hybridized and then all the non-bonding electrons would be in lower energy orbitals (the other sp2).

Hybridization occurs in our heads, not in atoms.

Ok , but hybridization is a concept based on a theoretical framework , so what I ask is how would this problem/question be solved using this concept.

The molecule $O_2$ does not have a double bond O=O. It is a biradical which has a single bond O-O, and two independant electrons.
Anyway in diatomic molecules, it is not necessary to consider hybridization. It does not help. Hybridization is an approach that has been invented by Pauling to explain the geometry of polyatomic molecules. Hybridization is not really necessary. It allows to easily describe the shape of the molecules, but the same result may be obtained without hybridization, although it is much more complicated to get to the same result.


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Which double bond will react with m-CPBA to form an epoxide?

The solution to this reaction sequence given was,

What I don't understand is the last step. How do we know which double bond will react to form an epoxide?
My guess was that the double bond adjacent to the methyl group should react as it has a more double bond character than the other one which is in resonance with the lone pair of oxygen.
Any help would be appreciated.

mCPBA is electrophilic, so electron-rich alkenes react first -- related: https://chemistry.stackexchange.com/q/116760/16683

The epoxide shown may well be unstable in the presence of the by-product m-chlorobenzoic acid and rearrange to an alpha-methoxycyclohexenone.

@Tony: how do you explain the first step?

When two or more double bond are present then the more electron rich double bond will epoxidise. In above example the double bond containing the methyl group  has  electron donating  inductive  effect while the double bond containing the methoxy group  will have an electron donating resonance effect. The resonance effect is dominant over the inductive effect. Therefore the double bond containing the methoxy group is more electron rich. Therefore epoxidation occur on it.


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Will phosphorous Ylide participate in carbene generation and insertion?
The following question was sent to me . The answer for this question was given to me as  option b

I attempted it as following , recognizing the first step as a ylide reaction with a carbonyl compound.

In order to get option B , I have to assume ylide generating a carbene and this may insert to double bond.
I have not come across phosphorous ylide as carbene generating reagent.
My question is as follows:
Will phosphorous Ylide participate in carbene generation and insertion in the presence of double bond ?
An answer in this regard would be quite help full.

I wasn't sure about phosphorous ylides, but cyclopropanation would be taken place with sulfur ylides. For example, the reaction of the Corey-Chaykovsky reagent with enones give corresponding cyclopropanes by undergoing a 1,4-addition that is followed by ring closure: Tetrahedron Lett. 2003, 44(18), 3629-3630 (https://doi.org/10.1016/S0040-4039(03)00732-9).

For review on above reaction: https://doi.org/10.1021/cr960411r

I agree. This is not a carbene reaction. I think this takes place because of the high stability of the carbanion between the rings with elimination of Ph3P


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Effect of concentration on rate

How do I solve this question?
My thinking is this: since O2 is not a reactant in any step, and since the reaction rate depends only on the reactants, increasing the concentration of O2 does not affect the rate of the reaction.
Is my thinking correct?

Hint: What defines a reactant ?

A reactant is a species that undergoes effective collisions to give product species. But O2 does not undergo collision to form product species. So it is not a reactant.. right?

Hint No 2: What do the bidirectional arrows mean ?

That it is a reversible reaction. So increasing concentration of O2 would increase concentration of O3. O3 is a reactant in the rate determining step. So the rate of reaction increases??

Hint No 3 (last): Enumerate equations of reaction rates, determine conditions for steady state of one of the intermittent species and express decomposition rate as function of O2 concentration. Be aware the site is not a free online service for solving homework or alike tasks.

Can you elaborate on what you mean by determining the conditions for steady state? I'm not sure I know how to do that. I don't want just the answer, I want to know how to solve these kinds of problems. I would appreciate if you have any links related to this problem which I can read, and understand how to solve these type of problems. Thanks for replying.

steady state for [X] means d[X]/dt=0, rate of production X equals the rate of its consumption.

Let us continue this discussion in chat.

Question is solved. Poutnik taught me to do it using steady state approximation :)

Just adding the image for somebody who might find this in future.


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How is presupposition of the order of reaction justified in Michaelis-Menten model?
from what I have understood, to derive Michaelis-Menten equation, we first suppose that
$\ce{E + S <=>[k_1][k_{-1}] ES->[k_2]E + P }$
and make the 'steady-state assumption', that $k_1[E][S]=(k_{-1}+k_2)[ES]$.
But from what I've learned, the order of reaction cannot be derived from its equation, but can only be decided by experiment. How can we 'presuppose' the order of these, especially when both $E$ and $S$ are macromolecules? The left side might as well be $k_1[E][S]^2$, for example.
What am I missing out?

These are elementary steps.

This is a basic model for the most simple scenario. There are many much more complicated enzyme kinetic models for reactions in which there are multiple substrates or inhibitors binding in random order or specific order, etc.

What you suggest about stoichiometry and what the reaction steps actually are is true in general, but in this particular case this model has been confirmed by experiment and that is the only reason it is still used.

Oh, so it was just because of the simplicity of the model. Thank you all!


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What is the characterstics of additives that it must posses to make any mixture hypergolic?
This is about hypergolicity of rocket propellant.Turpentine gave hard-start with RFNA and WFNA but excellent with mixed acid(mixture of HNO3 & H2SO4).Similarly,Furfuryl alcohol & WFNA(also RFNA) become hypergolic mixture.So,what properties must additives posses to ignite particular mixture at any time without requiring oxygen?

In a mixture $HNO_3 + H_2SO_4$, there are more $NO_2^+$ ions than in pure $HNO_3$.

@Maurice So,NO2+ ion makes the mixture hypergolic? & RFNA also contain NO2.That idea seems contradicting

@ Auberron. Please don't mix up $NO_2$ which is a brown and neutral molecule, and $NO_2^+$ which a colorless ion.

@Maurice Yeah you are right.

Say i have a propellant & I want to light it up in space.So how can i choose additives such that it reacts with the propellant and the mixture ignites in space?

Do you really want to carry out this operation in space ? In space ? Are you working in Cape Canaveral ? Or is it a theoretical problem given by a teacher to his students ?

@Maurice No,I just want to illustrate my question clear to all viewers.

Just use chlorine trifluoride: it is hypergolic with everything including most containers. seriously, though, there are whole books on the topic so it is unlikely you will get a good general answer. Try John D Clarke's famous "Ignition: an informal history of liquid rocket propellants" for a good overview.

The ideal formulation of rocket fuel is a balanced mix of  oxidizer and fuel. A strong oxidizer and a highly flammable compound or strong reducer, which is termed the "fuel", Is the Ideal combination for a fuel that will combust hypergolically when mixed. The compounds in hypergolic fuels are ideally liquids, because fluidity allows for easier mixing.
As oxidizers, Nitrates work well because the large quantity of oxygen atoms attached to the single nitrogen atom makes for a strong oxdizer. Other compounds, such as Sulfates, Phosphates, Chromates & Dichromates, Chlorates & Perchlorates, Peroxides, and Permanganates also serve as fairly good oxidizers for this reason. However, strong oxidizers do not always require high oxygen content. For example, Chlorine Trifluoride is a very strong oxidizer and is commonly used in many different hypergolic rocket fuel combinations
As far as the rest of the mixture, the fuel, most flammable liquids work well. Aniline, Liquid H2, Hydrazine and it's analogues, Kerosene, Gasoline, and most short-chain alkenes work well with the right oxidizer. An ideal choice when it comes to Hypergolic fuels is fairly small alcohols and ethers, like ethanol, diethyl ether, furfuryl alcohol, and especially Glycerol. It seems that the occasional oxygen atom in the fuel makes it easier and faster to oxidize when the time comes.
In conclusion, Adding a small amount of a stronger oxidizer in solution with your main oxidizer should help spark the reaction if it has trouble getting started. Also, while adding a fuel booster doesn't always work, sometimes adding a little ethanol or glycerol can help get the reaction in motion as well as help it run more smoothly. This is similar to how ethanol is often added to gasoline to increase the efficiency of the burn.
I kind of have an interest in the chemistry of rocket fuels, so if you ever find yourself looking for another answer, feel free to mention me directly.


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Organic chemistry
Could someone tell me what happens when a compound having a benzene ring reacts with a mixture of NaOH/HCHO?
What reaction and mechanism does this follow?

Could you please choose a more specific title? Also exactly what compound is this? What other functional groups does it have, or is it just benzene (doesn't seem to be based on your description)? A bit more detail will go a long way in getting a better answer...

Sounds to me as if you need some other reactants or substituents, and it would help if you gave us more detail on the reaction conditions.

@orthocresol yes ok sorry, I'll attach an image of the compound

As your question stands, there is no reaction.

One example of a reaction like this is the reaction of Benzaldehyde and Formaldehyde in the presence of a basic hydroxide (OH-).
This reaction proceeds as follows:

Thus, using Sodium Hydroxide (NaOH), The reaction would yield Phenylmethanol (C6H5CH2OH) and Sodium Formate (HCOONa, sodium salt of formic acid)
If you want a better answer, or if this was not quite what you were looking for, you can try to better define your "Benzene Compound"

Thanks a ton! This helps!! My compound was phenol with a OMe (Methoxy) group in the ortho position

@user762391 "My compound was phenol with a OMe (Methoxy) group in the ortho position" – Then why don't you tell us in the question? And this compound has a name, so use it!


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Topical hand sanitizer production question from tutor; Concentration and dilution of chemicals of known concentrations
The question is as follows:
You are in charge of making hand sanitiser based on the WHO Guide to handrub formulations,
FORMULATION 1
Final concentrations:
• Ethanol 80% (v/v),
• Glycerol 1.45% (v/v),
• Hydrogen peroxide 0.125% (v/v)
You are given;
Ethanol @94.45% ABV
H2O2 @6% concentration
Glycerin @100% concentration
Water
Using 10,000L of ethanol @94.45%, what volume of (a)glycerin and (b)H2O2 are needed to achieve the final concentration in FORMULA 1? What would the volume of H2O2 be if given a 3% concentration solution?
(c)What volume of water is needed to be added?
The following is my attempt;
Batch production
For 100ml
100*80 = 8000 / 94.45 = 84.7008999ml
For 1L Batch
847.008999ml = 0.847008999L
Have 10,000L @94.45% ABV
10000L / 0.847008999L = 11806.25L Total hand san made at 80% v/v ethanol
If 10000L ethanol @ 94.45% then
1806.25L used for dilution between H2O2, glycerine, and water
(a) glycerine @100% conc 1.45% v/v
Glycerine
11806.25L / 100 = 118.0625 X 1.45 = 171.19L
(b) H2O2 @3%
11806.25L / 100 = 118.0625 X 0.125 = 14.78L
14.78L / 3 = 4.919 X 100 = 491.93L
If have 6% conc H2O2
14.78 / 6 = 2.459 X 100 = 245.96L
(c)
1806.25 – 171.19 – 491.93 = 1143.13L water
Am a bit unsure if correct, especially with part (b) H2O2, Thanks in advance!

The formulation is given by amount / volumes to be mixed. Forget final % concentration. To get exactly the same result (which might be overshooting but I am not here for medical advice) just makes conversion on the ingredients side. Proportions also. This is because volume is not addictive. H2O2 in particular could be omitted. It serves to destroy spores eventually in water. A health organisation cannot give non perfect recipes. That is a multi purpose sanitizer to be deployed in every scenario.

Yes, I agree in the technical proportions, and here is another web reference with a topical solution prep starting with 96% Ethanol (which may be of interest to you and others).
As a side comment, however, there is only economic loss involved by substituting a higher % H2O2 (if you are interested in a potentially more powerful mix) for your personal use. Supporting comments from Wikipedia:

Hydrogen peroxide may be used for the sterilization of various surfaces,[64] including surgical tools,[65] and may be deployed as a vapour (VHP) for room sterilization.[66] H2O2 demonstrates broad-spectrum efficacy against viruses, bacteria, yeasts, and bacterial spores.[67] In general, greater activity is seen against Gram-positive than Gram-negative bacteria; however, the presence of catalase or other peroxidases in these organisms may increase tolerance in the presence of lower concentrations.[68] Higher concentrations of H2O2 (10 to 30%) and longer contact times are required for sporicidal activity.[69]

where the key takeaway for me is that even with 10% to 30% H2O2, longer contact times can be required for select conditions.
Also, there may be possibly a harsher effect on one's skin, but from a marketing perspective, the hydrogen peroxide boost may be a plus.
If for personal use only, a less convenient option to consider, in my opinion, is not adding the 'distilled water component' until the time of point-of-use in quantities needed. I would consider substituting for distilled water inexpensive raw 'tap water'. This will likely reduce shelf-life and the stability of the H2O2, which could result in the problematic liberation oxygen in the dispensing container. Possibly offsetting, in my view, is the introduction of favorable transition metals salts.
Possibly chemistry with ferrous ions that may be present, for example, which could engage in a Fenton reaction  (web link to 'A convenient lecture demonstration of the Fenton reaction is presented with references to applications in medicine and environmental engineering') associated with the creation of powerful hydroxyl radicals (capable of breaking down organic compounds). Unfortunately, this usually requires more acidic pH (in the current context, from say further adding citric acid). Tap water may also contain manganese, cuprous,..., ions, which could also engage in related Fenton-type reactions (with radical products). Interestingly, with copper ions, for example, this usually occurs closer to more neutral pH conditions.


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Grouping of symmetry classes in chemistry character tables
In the C3v point group, it lists three sigma Vs together in one column (class), and it doesn't list the  cVs separately. Consequently there are three irreducible representations.
In the C2v point group, however, it lists 2 separate cVs in separate columns. Consequently there are four irreducible representations.
My question is, why are the cVs grouped together in C3v, but not grouped together in C2v?

The $\sigma_\mathrm v$ mirror planes in the $C_\mathrm{3v}$ point group are themselves related by symmetry: note that they can be interchanged via rotation by 120 degrees about the preexisting $C_3$ axis. To be technically precise, they belong to the same conjugacy class, in the sense that applying $C_3$, then one mirror plane, and then the inverse of $C_3$ is the same as applying a different mirror plane:
$$C_3\sigma_{\mathrm v, i}C_3^{-1} = \sigma_{\mathrm v, j}$$
As a result of this, they have to be treated together and not separately, in the sense that the characters in the character table represent how each irrep transforms under all the three mirror planes collectively. Unfortunately, this is hard to explain properly without delving into the actual matrices. But it should be possible to see that the two mirror planes in $C_\mathrm{2v}$ are not related to each other in the same way.

@KameronShrum, if you think it addressed your queries sufficiently, can I encourage you to accept the answer (green tick near the voting buttons) -- same goes for your other questions, too. See: https://chemistry.stackexchange.com/help/someone-answers Of course if you don't think it's enough (like if you want a more rigorous explanation), then feel free to not accept it, by all means. I do not desperately need reputation. :-)


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Nucleophilic Aromatic substitution in 3-Bromoanisole
I came across an example in my textbook where $\ce{3-Bromoanisole}$ is reacted with $\ce{NaNH2}$ in liq. $\ce{NH3}$. My understanding is that the reaction would proceed by a Benzyne intermediate, with the triple bond between $\ce{C2}$ and $\ce{C3}$. Now, I know that the carbanion must be formed where it is most stable. Since $\ce{C3}$ is at meta position with respect to the $\ce{-OCH3}$ group, the electron density there must be lower, and hence that is where the carbanion should be formed, leaving us with the $\ce{-NH2}$ group at $\ce{C3}$, forming a product that will be more stable due to hydrogen bonding between the $\ce{-OCH3}$ and $\ce{-NH2}$ groups. However, in my textbook it says that a meta-substituted product is formed with no explanation. To me, it seems like that would be very unstable as the $\ce{-OCH3}$ group would be increasing the electron density by both resonance and the inductive effect. Can someone please explain what I'm missing?

The methoxy group is electron-withdrawing by induction, so the nucleophile will attack C3 to place the carbanion as close as possible to the methoxy group.


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Can ring inversion in Cis-Decalin allow the E2 reaction?

This is a tutorial question I need to do, I initially thought the question was testing whether students properly understand (1) chair conformations of decalin, and (2) that the leaving group and the breaking bond should be 'anti-periplanar'.
However, the third molecule also turned out to have undergone the elimination reaction when the first didn't, so I am very confounded. I have drawn all possible conformations of cis-Decalins, but no single conformation seems to allow $\ce{C-OTs}$ bond and $\ce{C-OH}$ bond to be anti-periplanar. Am I missing out something?

The issue in this Grob fragmentation is whether or not the $\ce{C-C}$ and $\ce{C-OTs}$ bonds are anti-periplanar. The $\ce{C-OH}$ bond is free to rotate and the unpaired electrons of the negative oxygen can align with the other two bonds. In the first example a), the red bonds are gauche to one another. Standard E2 elimination is likely the source of the mixture. Examples b), c) and d) all have red bonds anti-periplanar. The blue bonds predict the geometry of the double bonds in the product. Examples b) and c) give E-alkenes while d) affords a Z-alkene.


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Breaking of carbon chains due to annihilation of electrons using positron beams
I and my friends were taking part in a competition(physics and chem expertise needed) and we were having a few doubts about something. I was hoping if anyone of you can help in the clarification of the doubt.
We use a positron beam(from the DESY II particle accelerator) directed towards a polymer(eg polythene), which will annihilate a few valence electrons of the atoms in the chain. We’ve not found any source to support our theory that bonds will break although they should as electrons are being annihilated which support bonding. We searched and found nothing because no one else has had a similar experiment before.
We also wanted to know if we annihilate an inner core electron then will it lead the de-excitation of the valence electrons further leading to destabilization of bonds.
We can’t confirm how the electron donors will react with carbocations/radicals formed, also since there are many possibilities and we are unable to consider all of them.
The momentum of the positrons can be set to any value between 0.5 GeV/c and 6 GeV/c.
We needed some confirmation to the feasibility of the experiment.
All we need is to break the carbon chain, any other ways to do that would be appreciated.

That's just using ionising radiation, specific kind is largely irrelevant. It does some damage. If you irradiate it enough, you might find some crosslinking, depolymerization and general weakening of material.

This a problem of radiation chemistry. It has been thoroughly studied in the years 1950 - 1960 - 1970 with gamma rays. When intense source of gamma rays strikes water, electrons are extracted from the water molecule. And some of these electrons are coming from the valence bond O-H. The bond is then weakened, and may even be broken. Usually the electron soon comes back to its original atom. This produces individual H atoms and OH radicals. Usually H and OH soon recombine and produce again H2O a few nanoseconds later. But if the irradiation is intense, they may recombine with similar neighbors. And this produce dissolved H2 and H2O2 molecules. So the water sample contains more and more dissolved H2 and H2O2. Titration of H2O2 in solution is possible, and may be used to determine the doses and the intensity of the radiation absorbed by the sample.
The same experiments have been done with hydrocarbons. After irradiation, hexane, for example, is partly converted into dodecane, with a lot of different isomers. The distribution of these isomers allows to determine where the initial breaking in hexane happened, in other words which C-H or C-C bond has been initially broken, and in which proportion. Why don't you read the publications by Ausloos or Gäumann or others in this old literature. Today nearly nobody is interested any more in this chemistry, which is considered to be "finished".

In our case, we are working on electron-positron annihilation and not breaking bonds using gamma rays. As the bonds consist of electrons when we annihilate electrons, will it lead to the instability of the carbon chain? And can we then use electron donor groups to break the bonds and form new small carbon chains?

Your beam is mostly energy, and only incidentally positrons. The scattering on high energy e-, e+ and gamma rays are all pretty much the same until the energy drops below 1MeV

Right. Since the energy of the positron beam generated in DESY II is about 0.5 GeV , we want to reduce its energy to about 500 keV. We need a suitable material to reduce the kinetic energy of the positron beam , when we pass the beam through it. By online sources we found Al suitable for the task but still we aren’t sure and want to know if it will work. Is there any other way you can think of?

@MihirGupta I'm not sure there's a need to complicate it. You can just check how high energy particles are gonna damage it. If you already have access to high energy source, why waste it?


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Why are Friedel Crafts reaction not possible for molecules less activated than benzene?
I know that Friedel Crafts alkylation reaction are not possible for aniline and phenol since they form complexes with Lewis Acids. But our teacher told us that Friedel Crafts reactions aren't possible for any benzene derivative which is less activated than benzene. Why does that happen?

Strong electron-withdrawing groups deactivate the ring by reducing the electron denisity available to the attacking cationic species and to stabilise the intermediate cation.

Also, aniline and phenol would be more activated than benzene since the electron density on the rind would be higher otherwise.

You should ask your teacher to explain how the FC alkylation actually works. Recognising dis/activating substituents is elementary, and not by learning them by heart! Don´t you have a proper textbook?

@Waylander If strong electron withdrawing groups deactivate the ring for EAS(electrophilic aromatic substitution) then why does compounds like nitrobenzene give normal EAS reactions giving meta as major product, the ring has been deactivated there also.

All the sites are relatively deactivated, the meta sites are less deactivated than o and p. You need forcing conditions to nitrate nitrobenzene because of this.

Well I got the answer in an another question it is because the withdrawing group takes the electron density from the ring and then becomes electron rich and then finally forms complex with lewis acids like in case of aniline and phenols.

Technically, the teacher's claim might be slightly modified.  Deactivated benzenoid aromatics do fail to undergo Friedel-Crafts reactions, but according to a summary description from the University of Calgary:

Deactivated benzenes are not reactive to Friedel-Crafts conditions, the benzene needs to be as or more reactive than a mono-halobenzene.

So, a monohalogenated ring which is only weakly deactivated, can still react.  However, stronger deactivation, for example with an acyl group, would kill the F-C reaction.
This property can actually be quite useful when we need to do the F-C reaction selectively.   Using acyl as the deactivating group, we can acylate an aromatic ring and be sure the reaction occurs only once.  The ring becomes too deactivated to react further.  In principle this can then lead to a monoalkylated product with good yield by subsequently reducing the acyl group, whereas direct alkykation which activates the ring could give polyalkylated products.

OP quote: I know that Friedel Crafts alkylation reaction are not possible for aniline and phenol since they form complexes with Lewis Acids.

This is true partially for Lewis acids (Note: I said partially because Lewis acids work with anisole and other phenol ethers; Ref.1). However, Brønsted Acid has been used in Friedel-Crafts alkylation of phenol and phenol ethers before. For example, anisole and other $\ce{O}$-alkyl derivatives have been used in Tandem-Friedel-Crafts alkylation reactions to synthesize hitherto unknown perylene derivatives (Ref.1):

When phenol is subjected to Friedel-Crafts alkylation with a Brønsted Acid such as p-toluenesulfonic acid (TsOH), the reaction would produce multiple products as a result. For example, the reaction of phenol and cyclokexene (4:1 mole ratio), the products are mixture of 2-cyclohexylphenol, 4-cyclohexylphenol, 2,4-dicyclohexylphenol, and 2,6-dicyclohexylphenol (major product) with traces of other unidentified products (Ref.2). Surprisingly, no 2,4,6-tricyclohexylphenol was found among the product mixture.
Keep in mind that when chlorobenzene was used as the solvent, the reaction gave traces of alkylated chlorobenzene products in refluxing conditions, even though these reactions are believed to be worked only with activated phenyl nucleus.
A detailed study of Friedel-Crafts alkylation of ctivated phenyl nucleus using TsOH as a Brønsted Acid catalyst is given in Ref.3.
References:

Mark A. Penick, Mathew P. D. Mahindaratne, Robert D. Gutierrez, Terrill D. Smith, Edward R. T. Tiekink, George R. Negrete, "Tandem Friedel–Crafts Annulation to Novel Perylene Analogues," J. Org. Chem. 2008, 73(16), 6378–6381 (https://doi.org/10.1021/jo800558c)(PDF).
Mathew P. D. Mahindaratne, Ph.D. Dissertation 2000, Wichita State University, Wichita, KS.
Mathew P. D. Mahindaratne, Kandatege Wimalasena, "Detailed Characterization of p-Toluenesulfonic Acid Monohydrate as a Convenient, Recoverable, Safe, and Selective Catalyst for Alkylation of the Aromatic Nucleus," J. Org. Chem. 1998, 63(9), 2858–2866 (https://doi.org/10.1021/jo971832r).


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Is the carbon atom in the carbon dioxide molecule partially positive?
I had a question about nonpolar molecules that have symmetrical dipole vectors.
Let's take $\ce{CO2}$ as an example. Each of the $\ce{C=O}$ bonds are pulling in the opposite way. My teacher says that this causes all atoms in $\ce{CO2}$ to be equally charged as there mustn't be any net 'force'.
However, I disagree. Intuitively, it seems that the oxygen atoms would pull electron density away from the central carbon atom and make the carbon atom slightly positive and the oxygen atoms slightly negative, like so:
$$\large\ce{\overset{\small\delta-}{O}=\overset{\small\delta+}{C}=\overset{\small\delta-}{O}}$$
This process should make the carbon atom slightly positive and the oxygen atoms slightly negative. However, if I was right, then why don't we say $\ce{CO2}$ has a dipole (there is a separation of charge)? Perhaps I may have the wrong definition of a dipole.

It might help you to look up the definition of “quadrupole”

See Quadrupole of a molecule

Non-polar is arguably a misnomer. It specifically means "non-dipolar". It doesn't mean the charge distribution is actually constant.

There's a difference between a molecule's overall dipole and local dipoles/bond dipoles within a molecule. A molecule with completely nonpolar bonds cannot have an overall molecular dipole. However, this does not imply that molecules with polar bonds must have an overall molecular dipole - having bond dipoles is a necessary but not sufficient condition. $\ce{CO2}$ is a case of a molecule with bond dipoles which cancel out exactly, and leave no overall molecule dipole. However, as has been stated, $\ce{CO2}$ does have a quadrupole moment.

You're correct in assuming that the carbon atom in $\ce{CO2}$ has a partial positive charge. This is because the oxygen atoms are much more electronegative, so they pull the electrons away from the carbon atom. However, this molecule is still nonpolar. This is because, when you draw a dipole moment, you have to take all bonds into account. Take water for example:

In this molecule, there are two bonds, each with their own dipole. But these cancel out like any other vectors, leaving you with a vertical net dipole. The dipoles in carbon dioxide cancel out in a similar way; however, they cancel each other completely, because the bond is linear, not bent like in water:

This produces a zero net dipole, rendering the molecule nonpolar.

This is correct, and we can actually test it. Replace one of the O with an S to break the symmetry,  Now the direction of the O=C dipole is exactly opposite to that of the C=S dipole, but the magnitudes are not equal, so we get a net dipole moment (of 0.65 D, according to https://en.wikipedia.org/wiki/Carbonyl_sulfide).

but if the carbon is positive then surely the partially negative oxygen of another CO2 molecule can form a dipole dipole bond with the C?

It does still somewhat affect the properties of the compound, because it encourages the molecules to stack in a staggered pattern, but technically it is still nonpolar, since there is no way for another molecule to align it's own dipole on the same axis as the original molecule's dipole. It's nonpolar because you can't say that one entire side of the molecule is more positive/negative than the entire other side

@mpprogram6771 But if you use that as the yardstick then I feel like there's some  carbocation or something that you could say is also non-polar... Maybe I'm just confused on what a "side" is?

One side meaning like draw a line straight through the center and take one half. CO2 is electrochemically symmetrical, so one side won't have a charge any different from the other side And yes, you could probably find something like that, but that doesn't mean it's not an electrophile just because the molecule is nonpolar. The carbon atom would still have all the regular implications of a partial positive charge, just the molecule as a whole has no clear positive end and negative end, because it's symmetrical.

Also, about the carbocation, nonpolar doesn't necessarily mean non-ionic. Take the carbonate anion for example: it is trigonal planar, and so is symmetrical (radially). Each C-O bond is 1 and 1/3 of a bond, (4 carbon bonds divided by 3 oxygen atoms) so each oxygen atom has a formal charge of -2/3, for a total charge of -2. the carbon atom is partially positive.

Think about it this way; If you drew a perfect circle around the entire molecule, and then drew a line from the edge of the circle, through the center of the molecule, to the opposite side of the circle, then if the molecule is nonpolar, then no matter how you draw the line, one endpoint of the line won't have a different charge than the other endpoint. The line may cross some different charges on it's way, but that doesn't matter. This is how we know that there's no net dipole. If you try that with water, the strongest difference in the endpoint charges is along the dipole.

@mpprogram6771 so I didn't want to just throw out "some carbocation" and leave it at that so I googled around some. I did find something called tropylium but I think what I actually meant was Lewis acids instead of ions, BF3 for example.

@mpprogram6771 Ooh okay I didn't see that last comment until my browser refreshed but that makes more sense. So basically point symmetrical. Still I feel weird calling an acid, or even an ion if I understand you correctly, non-polar. Like if you had liquid BF3 (or maybe not BF3 but BCl3?) It wouldn't be a non polar solvent would it?

The boiling point of BCl3 is a little low, but technically, it could be considered a nonpolar solvent. However, it reacts or forms adducts with many organic compounds, and hydrolyses in water to form two very acidic compounds, so it might not be your first choice. I guess it is kind of weird to think about it that way, isn't it?

This answer is a great starting point because it focuses on the idealistic model of $\mathrm{C}\mathrm{O}_2 ,$ where it's time-averaged, in a vacuum, both Oxygen atoms are of the same isotope, there're no significant fields, etc..  As a simple, informative idealization, it's great information to present first -- but, it should still be noted that this idealization is a starting place rather than a complete description.
It may be worth noting this limitation then pointing toward some of the other answers which build from this starting point.

@mpprogram6771 I saw that you committed to joining Materials Modeling Stack Exchange. Did you notice that it was launched now? Please take a second to join: https://materials.stackexchange.com

The other answers have done a great job explaining why, even though its bonds are polar, $\ce{CO2}$ lacks a permanent dipole: the molecule's symmetry cancels out the polarity of its bonds.
But that's not the whole story. I'd like to add to this a very interesting and environmentally important charateristic of $\ce{CO2}$ — namely that, while it lacks a permanent dipole, it does exibit transient (dynamic) dipoles.
Specifically, $\ce{CO2}$ lacks a dipole only when the two oxygens are both equidistant from, and in alignment with, the carbon.  In $\ce{CO2}$'s symmetric vibrational mode, that symmetry is maintained.  But $\ce{CO2}$ has three other vibrational modes: a linear asymmetric vibrational mode, and two bending vibrational modes (the collection is nicely pictured here:  Is carbon dioxide IR inactive?).
Why is this important environmentally?  In order for $\ce{CO2}$ to absorb IR light (i.e., in order for it to be a greenhouse gas), it needs to have a dipole.  And it does, transiently, because of these asymmetric vibrational modes.
This animation, added by Karsten Theis, shows the dipoles dynamically created by one of $\ce{CO2}$'s bending modes (aka "The Floss"):

[According to Karsten, the "GIF is via jsmol from molcalc.org, with the arrow added using Camtasia".]

re: your useful comment about greenhouse effect. The argument can be extended to oxygen and nitrogen which having no dipole do not absorb or radiate ir radiation. This means that when they gain energy from the CO2 by collision, they can only lose it by collisions and so heat the atmosphere.

You are correct, the carbon does have a positive charge. We cannot measure a dipole, but that doesn't prove anything. However, $\ce{CO2}$ does have a quadrupole moment. Imagine a $\ce{CO2}$ molecule oriented along the $x$-axis, and a bit further along the $x$-axis there's also a $\ce{H2O}$ molecule with its dipole oriented along the $x$-axis. Its dipole moment interacts with both dipole moments of $\ce{CO2}$, but one of the two dipoles in $\ce{CO2}$ is closer to the water dipole. So, schematically you get
        H
O=C=O    O
        H

If there was no charge distribution on the $\ce{CO2}$, we wouldn't see this.
Mathematically, this happens because space is 3D. Forces between two charges drop of with the square of their distance.

The previous answers by mpprogram6771 and MSalters nailed it. I'd like to add that, as $\ce{CO2}$ is a very small molecule, you can, with a bit of effort, set up a little numeric experiment to answer your own question, and even get approximate partial charges in each atom, and dipole moment of the whole molecule, using just free/open source software.
First, you need to install molecular modelling software in your machine. The one I like most is Avogadro. It has wonderful usability and many features to design and visualize your compounds. Ghemical was also good, but it seems to be unmaintained for years now. I could not getting it working properly anymore.
In my machine I use Ubuntu MATE 18.04 (a GNU/Linux variant) as the operating system. There I'm able to install Avogadro with a simple command in the terminal:
sudo apt-get install avogadro

With Avogadro you can assemble the $\ce{CO2}$, joining the carbon atom and both oxygen atoms with double bonds. Beyond the molecular editor, you will need another piece of software, able to take the data about the molecule you assembled and do a series of quantum mechanical calculations over it, to give you an approximate answer to your questions.
There is a great variety of quantum mechanical software, as this page on Wikipedia shows. Unfortunately, IMHO the landscape of free/open source tools in this field is fragmented, and most lag far behind Avogadro in terms of usability, stuck in the average level of user-friendliness of the 1980s (sometimes at the compile-it-yourself level), and the proprietary alternatives have restrictive licenses and/or are eye-watering expensive, out of reach of people with no institutional affiliation. Academia treats its voluntary toolmakers badly, as some great people in mathematics can tell you, firsthand. Sooner or later we must fix that. We need a William Stein in computational chemistry. I just hope he/she gets better treatment after stepping up to the task.
Yet, among the several packages supported by the Avogadro input generator, my recommendation is Psi4, for a beginner. It's as easy to install as Avogadro, if you're under Ubuntu or any Debian-based distribution.
sudo apt-get install psi4

They have a well documented site, with a section dedicated to education with simple projects and friendly message boards. The version available in the Ubuntu repository is functional, but quite outdated, 1.1.5, as of Mar 2020. If one is serious about learning it, my advice is to download it directly from their site. The latest stable version as Mar 2020 is 1.3.2. But for the sake of this answer, the repository default is enough.
After assembling your molecule and doing some preliminary geometry optimization inside Avogadro, you can generate a preliminary input text file with its Psi4 plugin under menu Extras → PSI4. My preliminary version started like this:
set basis  aug-cc-pVDZ
molecule {
0 1
   C       -3.47367        0.73246        0.22361
   O       -2.43476        1.12414       -0.22175
   O       -4.51237        0.34053        0.66926
}
optimize('B3LYP-D')

The Avogadro plugin for Psi4 is very basic, so we will need to tune the template by hand. A set of good templates you can change to fit your needs is a great thing to have when learning to use a new package. We should have more of these. But first things first, let's see what we have on our proto-input. It has three sections. The first section specify a basis set, aug-cc-pVDZ (computational chemists love to feast on alphabet soup). To be short, a basis set is a jury-rigged set of easy-to-calculate mathematical functions, used emulate the real, hard-to-calculate atomic and molecular orbitals, kind of like this:

The second section has the x,y,z coordinates of every atom in the molecule, and also its overall charge (in this case 0) and multiplicity (in this case 1, as all electrons are paired). The third section says what kind of information we want to calculate from our initial information, in this case, the optimum geometry of the molecule (optimize), and the algorithmic machinery chosen to process it, in this case, B3LYP-D (another serving of alphabet soup), a variant of density functional theory (DFT).
I changed the Avogadro generated template as follows:
memory 4 Gb

set basis  aug-cc-pVTZ
molecule {
0 1
   C       -3.47367        0.73246        0.22361
   O       -2.43476        1.12414       -0.22175
   O       -4.51237        0.34053        0.66926
}
optimize('B3LYP-D')
E, wfn = energy('B3LYP-D', return_wfn=True)
oeprop(wfn, "MULLIKEN_CHARGES", "DIPOLE", title = "CO2 B3LYP-D")

I optionally raised the limit on system memory to 4 GB, from the system default, as my machine has a good amount of memory. As the molecule is small and the impact on runtime will be probably acceptable, I also changed the previous basis set, aug-cc-pVDZ, to one more detailed, aug-cc-pVTZ. Also added a section asking Psi4 to return a wavefunction (wfn) object for the system, besides its energy (E). Finally, following the guidance on the Psi4 manual here, I added a section asking for our information of interest, the estimated partial charges on each atom, given by Mulliken analysis, and the estimated dipole moment on the $\ce{CO2}$ molecule.
Now we can save the text file with our input data and run Psi4 in the terminal:
psi4 carbon_dioxide.in

After some time, Psi4 will finish the run, and return its results to an output file named carbon_dioxide.out that has a huge amount of information. But the section of more interest to your question is right at the end:
Properties computed using the CO2 B3LYP-D density matrix

  Nuclear Dipole Moment: (a.u.)
     X:    -0.0000      Y:     0.0000      Z:     0.0000

  Electronic Dipole Moment: (a.u.)
     X:     0.0000      Y:     0.0000      Z:    -0.0000

  Dipole Moment: (a.u.)
     X:     0.0000      Y:     0.0000      Z:    -0.0000     Total:     0.0000

  Dipole Moment: (Debye)
     X:     0.0000      Y:     0.0000      Z:    -0.0001     Total:     0.0001

  Mulliken Charges: (a.u.)
   Center  Symbol    Alpha    Beta     Spin     Total
       1     C     2.80993  2.80993  0.00000  0.38015
       2     O     4.09503  4.09503  0.00000 -0.19006
       3     O     4.09504  4.09504  0.00000 -0.19008

   Total alpha = 11.00000, Total beta = 11.00000, Total charge =  0.00000

*** Psi4 exiting successfully. Buy a developer a beer!

The results indicate exactly the situation you intuitively predicted, with both oxygen atoms pulling electron density away from the central carbon atom and the carbon atom becoming slightly positive and the oxygen atoms slightly negative. In fact, we were able to use the computer as a sort of power armor for the mind.
At first, your intuition could only provide vague guidance in the direction of electron density transfer, from oxygen to carbon. Now we can corroborate that, and augment our intuition with numeric estimates, an average loss of 0.38 electrons in the carbon atom and an average gain of 0.19 electrons in each oxygen atom. Wonderful.
Despite the charge separation, the results of our little numeric experiment also point to near zero dipole moment, as we see. It doesn't tell us explicitly why. But our geometric intuition suggest a way out. As there are two oxygen atoms, the effect of charge separation on both may cancel out. The output of Psi4 corroborate that, as the partial charge on each oxygen atom is the same within four decimal places, and both take opposite positions in a linear geometry.
There's a similar molecule, but without the possibility of charge separation cancelling out, $\ce{CO}$, carbon monoxide, with a single oxygen. To do a comparison, I created the equivalent input file for it.
memory 4 Gb

set basis  aug-cc-pVTZ
molecule {
0 1
   C       -3.99710        1.44942        0.00000
   O       -2.86898        1.44942        0.00000
}
optimize('B3LYP-D')
E, wfn = energy('B3LYP-D', return_wfn=True)
oeprop(wfn, "MULLIKEN_CHARGES", "DIPOLE", title = "CO B3LYP-D")

And ran it.
psi4 carbon_monoxide.in

Again the results point to some measure of charge separation.
Properties computed using the CO B3LYP-D density matrix

  Nuclear Dipole Moment: (a.u.)
     X:     0.0000      Y:     0.0000      Z:     0.0023

  Electronic Dipole Moment: (a.u.)
     X:     0.0000      Y:     0.0000      Z:     0.0348

  Dipole Moment: (a.u.)
     X:     0.0000      Y:     0.0000      Z:     0.0371     Total:     0.0371

  Dipole Moment: (Debye)
     X:     0.0000      Y:     0.0000      Z:     0.0944     Total:     0.0944

  Mulliken Charges: (a.u.)
   Center  Symbol    Alpha    Beta     Spin     Total
       1     C     2.95397  2.95397  0.00000  0.09206
       2     O     4.04603  4.04603  0.00000 -0.09206

   Total alpha =  7.00000, Total beta =  7.00000, Total charge =  0.00000

*** Psi4 exiting successfully. Buy a developer a beer!

But this time the dipole was nonzero, with an estimated value around 0.094 debye. The Wikipedia article on carbon monoxide gives us a measured value of 0.122 debye. So we got an estimate around 23% lower than the real value. The difference may arise either as a intrinsic limitation of our model (the science vs. engineering thing), or because I fumbled somewhere either in the input I gave to Psi4 or in my assumptions to treat the problem (always very likely).
It would be interesting to check the literature in the subject, if one wants to go deeper. Anyway, the contrast in the results between $\ce{CO2}$ and $\ce{CO}$ point clearly to mutual cancellation to explain the lack of a dipole in $\ce{CO2}$.

Wow! you put a ton of effort into this! That's a definite upvote!

I will go through this carefully this weekend. Five years ago I asked How can I calculate the charge distribution of a water molecule? and started to try to figure out how to run PyQuante but then realized I'd have to do much more reading before I'd understand what I was doing.

Wow this is really impressive. I want to give it a try. Thank you so much for your effort!

My teacher says that this causes all atoms in CO2 to be equally charged as there must be no net 'force'.

I don't think the other answers have explained why this is wrong. If you have a set of three point charges arranged like $Q$ ... $q$ ... $Q$, then it's easy to show that the forces all cancel when $q/Q=-1/4$. However, this can't be the physical situation, for two reasons. (1) The net charge $2Q+q$ is nonzero unless $q=Q=0$. (2) The equilibrium is unstable.
So based on this argument using Coulomb's law and Newtonian mechanics, your teacher would actually be right that the charges can't be nonzero. However, even in the case of $q=Q=0$, the equilibrium isn't stable. In this case there is no binding force at all, so the atoms would just drift off. In reality, CO2 is bound.
In general, we simply don't expect to be able to explain the stability of matter using classical physics and electrostatic forces. There is a theorem called Earnshaw's theorem that shows this is impossible. Quantum physics is required in order to explain the stability of matter.


text:
What is the best way to go about matching proton-NMR spectra to molecules?
I've attatched an example question. I have already solved this one, I'm wondering about what, in your experience, is the most time-efficient way of approaching a question like this one. I can solve these questions backwards pretty decently (ie If I know the solutions, then I can justify them in pretty decent detail) but when it comes to deciding on which spectrum is assigned to which molecule, I find that it takes me ages. Is there any 'standard approach' that I should be aware of, or is it simply a matter of practice?

You said you have already solved this one. So, show your work and we direct you to better way if your work is longer than needed. This is not hard as you think it is.

The compounds with vinylic and methyl protons are a good place to start (B,G,E,3), that should be relatively easy. Also, look for symmetry. C should also be simple.

Where are you studying chemistry, do they teach there, or just expect you to somehow turn into accomplished chemist? ;))


text:
How to compute change in Gibbs energy given change in entropy and enthalpy found over a temperature range?
I was given the following question by my tutor:

The variation of enthalpy ($\Delta H$) and entropy ($\Delta S$) in a reaction carried out at constant pressure in the temperature range between  $\pu{300^\circ C}$ and $\pu{350^\circ C}$, are $\pu{1700 kcal}$ and $\pu{15 kcal/K}$, respectively. Calculate the range of temperatures in which this reaction is spontaneous.

But my book on thermodynamics says that the change in Gibbs' energy for a process can only be calculated if we start and end at the same temperature (the temperature can vary during the process as heat is added or lost from the system to the surroundings, but the end points must have constant temperature and pressure.) In this question the changes in entropy and enthalpy are found over a range of temperatures where the final temperature is not the same as the initial temperature, but the pressure is always constant.
Is my tutor's question invalid, is my book wrong in saying that the temperature has to be the same at the initial and final values, or am I thinking about this wrong?

$\ce{\Delta G}$ can be calculated by the expression ${\Delta G =  \Delta H - T \Delta S}$. Why don't you calculate the change of H and S, then put ${\Delta G = 0}$. This an equation with one unknown T. A simple substraction plus a division gives you the minimum temperature for the reaction to become spontaneous.

@Maurice Yes, this is the "solution" I was given, but my question is not regarding answering the question, it's whether the question is valid since the enthalpy and entropy were taken over a range of temperatures rather than at constant end points of temperature and pressure (Gibb's function is defined so that a spontaneous process with starting pressure $P_0$ and temperature $T_0$ and the same pressure $P_0$ and temperature $T_0$ at the end point yields a $\Delta G < 0$, according to my book!)

@ Son1. You are right in stating that in order to calculate $\Delta G$, the temperature of the system must be the same in the beginning and at the end of the reaction. But it does not mean that this temperature must stay at a given value. The reaction may occur at any temperature, provided it is the same at the beginning and at the end of the reaction. $\Delta G$ can be calculated at any temperature, provided you know the change in enthalpy and entropy, and provided these changes in H and S  are not changing with the temperature.

I think what is (too loosely) being referred to in the problem statement is the standard changes in enthalpy and entropy for the reaction.  I think the intent is for you to assume that they are nearly constant over the temperature range from 300 C to 350 C.  And I think you are supposed to use the rule of thumb that a reaction is spontaneous at a given temperature if the change in Gibbs free energy is less than zero.

This should address your confusion about constant T:
$$G=H-TS\Rightarrow\mathrm dG=\mathrm dH-T\,\mathrm dS-S\,\mathrm dT$$
$$\Rightarrow \text{ (at constant $T$) } \mathrm dG=\mathrm dH-T\,\mathrm dS\Rightarrow\Delta G=\Delta H-T\,\Delta S$$
So, yes, you need $\Delta T=0$ for $\Delta G=\Delta H-T\Delta S$ to hold.
However, this holds at any $T$.
E.g., suppose you do the reaction at $300\ \mathrm K$. $\Delta T=0$, so $\Delta G=\Delta H-T\,\Delta S$ holds.
But suppose you repeat reaction at $350\ \mathrm K$. It's still the case that $\Delta T=0$, so $\Delta G=\Delta H-T\,\Delta S$ still holds.
As far as the question itself goes, I agree with Chet Miller that it appears to be badly worded.  It seems what your tutor meant to say was:
"Between $300\ \mathrm{^\circ C}$ and $350\ \mathrm{^\circ C}$, assume that $\Delta H$ and $\Delta S$ are constant, with values of $1700\ \mathrm{kcal}$ and $15\ \mathrm{kcal/K}$, respectively. Calculate the range of temperatures over which $\Delta G\le0$, i.e., the range the textbook labels as 'spontaneous'".


text:
Why does Sodium Bicarb. Reduce Combustion Rate?
In many common solid-propellant model rocket engines, Sodium Bicarbonate, $\ce{NaHCO3}$, is used to predictably regulate/limit the rate of combustion of the fuel. My questions are as follows:

What is the mechanism for this (reaction?)?
How would I know how much $\ce{NaHCO3}$ to add to the fuel for a certain reaction delay time?
Is $\ce{NaHCO3}$ somewhat unique in it's ability to do this, or is this something that many common chemicals can do?

The use of dry NaHCO3 to extinguish a fire works in two ways:
First, the smothering of the flame with CO2 and water per the decomposition reaction:
$\ce{2NaHCO3 ⟶ Na2CO3 + CO2 + H2O}$
Second, the decomposition reaction is endothermic, so it lowers the temperature of the fire. To quote a source:

In an endothermic reaction heat must be continuously supplied and the reaction is not self-sustaining. Stop heating NaHCO3 and the decomposition will slow down and stop.

When heated, $\ce{NaHCO_3}$ is easily decomposed and produces much $\ce{CO_2}$ in the chemical reaction : $$\ce{2 NaHCO_3 -> Na_2CO_3 + CO_2  +  H_2O}$$ and $\ce{CO_2}$ is a gas that tends to prevent Oxygen from maintaining the fire. As to your question 2, I have no idea how to solve it. And as to your question 3, $\ce{NaHCO_3}$ is by far the best compound and the cheapest. The other possibilities are the other alcaline bicarbonates, and apart from potassium bicarbonate, they are rather expensive.

I guess that rules out francium bicarbonate! ;-)

Which would vaporize from its own radiation anyway.  I've read in Wikipedia that magnesium hydroxide works too, by endothermically evolving water vapor.


text:
Difference in a array of probability density plots for the electron in its lowest-energy states of Hydrogen atom

Dynamic Periodic Table
If you visit Dynamic Peridic Table, You will notice that for n=4 l=3,2 m=0,0, and n=2,l=1, m=0 there is a difference in  the probability density plots for electron in its stationary energy state of Hydrogen atom  and image of arrays of probability density plots attached herewith titled'Hydrogen Wave Function'.
Why do we found such discrepancies? or are these plots  same?
If any member knows the answer to this question, may reply.

What difference? In not inclined to check random site just you say there's some "difference". One could colour such pics in whatever way for any particular electron density.


text:
what is the relationship of Salt, White Vinegar and Water to create 1 liter of Hypochlorous Acid
What is the relationship of Salt, White Vinegar, and Water to create 1 liter of Hypochlorous Acid (for disinfecting and cleaning purposes), and how much time do I have to put it under a 9v battery induced electrolyzing process?

Don´t do that..

Vinegar and water do not make any chemical reaction. So they will not produce Hypochlorous acid. If you want to produce hypochlorous acid, you have to electrolyze salt alone,  without vinegar. The main point is not the tension (9 V), but it is the currant. And the currant depends upon the geometry : surface of electrodes, distance between electrodes and resistance of the solution. The duration of the electrolysis may be many hours. But it depends on the currant. Vinegar may be added at the end of the electrolysis.

The vinegar is added to the salt solution to raise the pH to shift the product equilibrium from sodium hypochlorite to hypochloric acid

Vinegar would lower the pH.

The action of a weak acid (even H2CO3, acetic acid, Vitamin C,...) are generally all stronger than HOCl and will displaced it from NaOCl. Best preparation reaction is CaCl2 + NaOCl  (Chlorine Bleach) + NaHCO3 (Baking Soda) creating HOCl, a precipitate of CaCO3 and Na+, Cl+ ions. Decant to separate out the CaCO3. Best stored cold and used within a day or two. Unstable, Cl2, Cl2O fumes...

This path is based on an old patent citing the action of NaHCO3 on Calcium hypochlorite (or the action of CO2 on aqueous hypochlorite). My version is used when Ca(OCl)2 is not available. You should dilute the solution with distilled water to increase its stability and do store cold. Expect chlorine fumes and will react with transition metal to further produce powerful bleaching (and disinfecting) hydroxyl radicals, *OH, capable of breaking down organics by extracting a H atom creating H2O.


text:
Why doesn't water wash off the coronavirus alone?
So the coronavirus has a lipid bilayer where the hydrophilic heads are facing outward. So why can't water just stick to these ends and flush out the virus particles from our hands? Is it because our hands stick better to the virus? If so, why and how does the virus stick better to our hands than water?

Because your hands are covered in a thick crust of dirt, dust and fats, in which the virus can hide.
But anyway still nobody has been able to show that this virus is regularily (or even rarely) transmitted via smear infection. (If someone knows about any evidence to the contrary, please come forward. And yes, some other diseases, especially bacterial infections, do even mainly take this route. That´s how you usually get the "shits".)
It can be detected on surfaces after hours via PCR, however PCR is as sensitive to broken down pieces of a virus particle as it is to the fresh, contagious specimen. Obviously the RNA pieces also of a destroyed virus would not evaporate, so the result of these investigations proves nothing but the extremely obvious. Newspapers are spreading it as a piece of clickbait and tinder, causing people to buy lots of "sanitizers".
Anyway, use soap, leave the disinfectants to the health care people who absolutely, direly need them, and be done.

We shouldn't be commenting on any questions about Covid-19 due to the medical exclusion.

I think the non-polar fatty part  of the soap damage the  lipid bilayer of virus causing it dysfunction and the polar part  of the soap attracted to the water.

It might be somehow similiar to "why not to use just water while washing hands". Washing would remove the virus. Soap even more by 1) disrupting it and 2) by removing other dirt in which it can be. I feel free to comment as for the message is WASH YOUR HANDS WITH WATER AND SOAP frequently. I also feel to add STAY AT HOME. Many places are still behind Italy so get the message. Here we did a lot even at beginning, simply wasn't enough.

It has now been tried to find "viable" viruses, i.e. ones that can be replicated in a cell culture, on all kinds of surfaces in households where multiple currently infected and contagious people live. Everything is full of virus RNA (detected via PCR), but no success. Which is hardly a surprise.


text:
How much sodium hydroxide would it take to dissolve one tonne of bauxite?
Generally speaking what proportion of NaOH would be typically required to dissolve one tonne of crushed bauxite of average purity?
Also are varying quantities required for gibbsite, böhmite and diaspore respectively and how do they differ?

Going to depend on the composition of the Bauxite. Is it Gibbsite or Bohmite? How pure is it? Here is the basic process  http://bauxite.world-aluminium.org/refining/process/

Thanks @Waylander that's my source. I was just curious what the ratio would be on average.

Bauxite contains usually 45 - 60 % Al2O3. To get an order of magnitude, one ton bauxite contains about 500 kg Al2O3, or about 5000 mole Al2O3. If 2 NaOH are needed to dissolve 1 Al2O3, it means that a minimum of 10'000 mol NaOH is needed to dissolve Al2O3, which weighs 400 kg. But more NaOH is necessary, because the bauxite contains also some silica, which is also dissolved in NaOH. So the amount of NaOH must be higher than 400 kg.

Thanks @Maurice, that's a great answer and just what I was looking for.

Maurice, you provided a good starting basis (hence my upgrade). Alas, working in bulk is likely more daunting, hence my follow-up advice.

My advice is start with the best theoretical amount (per above) and apply it with varying deviations of concentration to some test samples. Heat or stir as you may be able to perform when working with larger quantities in bulk.
My understanding, even after your sample testing, you will find that working in larger quantities is more likely difficult to get the best results than you did experience in your small test sampling efforts. Here is a source noting an issue concerning surface charging properties of metal oxides (including Al2O3) in bulk.
Perhaps, applying some sonolysis (which could introduce powerful radicals from the water itself with sound frequencies) to your bulk preparations, which are harder to evenly heat, may prove fruitful.
There are also an mechanochemical option.


text:
Chemoselectively dihydroxylate a terminal alkene in the presence of an internal alkene?

The goal is to dihydroxylate the exo alkene, and then convert to a ketone.
Any suggestions would be apprecitated to either chemoselectively convert the terminal alkene, or directly convert the terminal alkene to a ketone.
Thank you!

Maybe complete gradual oxidation of methyl having allylic hydrogens would be better idea.

The Lemieux-Johnson oxidation (details here) using catalytic $\ce{OsO4}$ or $\ce{RuO4}$ turned over with $\ce{NaIO4}$ will do the transformation you want in one step. These conditions should be selective for the exo double bond.

Thank you! I was hitting a roadblock.

@Ellie If you are happy with this, please accept the answer

To confirm, you want to make (R)-4-acetyl-1-methylcyclohex-1-ene? Do you have to start from (S)-limonene?
If not, consider the Diels-Alder reaction of isoprene and methyl vinyl ketone (e.g. https://doi.org/10.1039/B701333G , though this is not an enantioselective synthesis I imagine you can add chiral auxiliaries somewhere to increase that selectivity)

Thank you. I was considering that as well.


text:
Solution of a first- and second-order consecutive reaction scheme
I have a compartmental model system X, a simplified version of which, looks like this:
$$
\text{System }x: \enspace \text{STO} \xrightleftharpoons[k_2]{k_1}  \text{blood} \xrightarrow[]{k_3} \text{urine}
\equiv
x_1 \xrightleftharpoons[k_2]{k_1} x_2 \xrightarrow[]{k_3} x_3
$$
The system of differential equations is then:
$$
\frac{dx_1}{dt} = -k_1 x_1 + k_2 x_2 \\
\frac{dx_2}{dt} = k_1 x_1 - k_2 x_2 - k_3 x_2\\
\frac{dx_3}{dt} = k_3 x_2 \\
$$
I can solve this system of equations using the following matrix approach.
First, I write the rate matrix [R]. From [R] one can obtain a new matrix [A] by first replacing each diagonal element of [R] by the negative of the sum of each of row elements, and then transposing it:
$$
[R] =
\begin{bmatrix}
    0            & k_2    & 0 \\
    k_1          & 0      & k_3 \\
    0            & 0      & 0
\end{bmatrix} \\
[A'] =
\begin{bmatrix}
    -k_2            & k_2    & 0 \\
    k_1          & -(k_1 + k_3)      & k_3 \\
    0            & 0      & 0
\end{bmatrix} \\
[A] =
\begin{bmatrix}
    -k_2            & k_1    & 0 \\
    k_1          & -(k_1 + k_3)      & 0 \\
    0            & k_3      & 0
\end{bmatrix} \\
$$
I can calculate the amount in each compartment by doing the following:
$$
x(t) = e^{[A]t}x(0) \enspace \text{where} \enspace x(0) = \begin{bmatrix} 0 \\ 1 \\0
 \end{bmatrix}
$$
In python:
RMatrix = model_matrix.as_matrix()
row, col = np.diag_indices_from(RMatrix)
RMatrix[row, col] = -(RMatrix.sum(axis=1)-RMatrix[row,col])
AMatrix = RMatrix.T

def content(t):
    cont = np.dot(linalg.expm(t*AMatrix), x0))

This method is working well for me.
Now, I have a little more complicated model where reactants in compartments 1 and 2 of Systems X and Y combine to get product in System Z.
$X + Y \rightarrow Z$, with a reaction constant of $k_R$.
$$
\begin{align}
&x_1 & \xrightleftharpoons[k_2]{k_1} \enspace & x_2 &\xrightarrow[]{k_3} & x_3 \\
&+  & \ &  + \\
&y_1 & \xrightleftharpoons[k_5]{k_4} \enspace & y_2 &\xrightarrow[]{k_6} & y_3 \\
&\downarrow^{k_R} & \  & \downarrow^{k_R} \\
&z_1 & \xrightleftharpoons[k_8]{k_7} \enspace  & z_2 &\xrightarrow[]{k_9} & z_3 \\
\end{align}
$$
, and the corresponding system of differential equations would be:
$$
\begin{align}
\frac{dx_1}{dt} &= -k_1 x_1 + k_2 x_2 - k_R x_1 y_1 \\
\frac{dx_2}{dt} &= k_1 x_1 - k_2 x_2 - k_3 x_2 -  k_R x_2 y_2\\
\frac{dx_3}{dt} &= k_3 x_2 \\
\frac{dy_1}{dt} &= -k_4 y_1 + k_5 y_2 - k_R x_1 y_1 \\
\vdots& \\
\frac{dz_z}{dt} &= -k_7 z_1 + k_8 z_2 + k_R x_1 y_1 \\
\end{align}
$$
I am struggling with a method to solve this system of differential equations (1st and 2nd order) to calculate the amount in each compartment at a certain time t, given the initial conditions, $k_R$, and the transfer rates $k_1$, $k_2$, $k_3$, etc...
Can I solve it using the matrix method like the one above for a system of first order differential equations? What other options in Python do I have?
Thanks in advance!

I don't know python well enough to direct you to the right code, but a common approach to these is to run simulations and minimize the difference between the simulation and actual data points by least squares. That's easier than finding the explicit solution.

Look for equation solvers in Numpy/ Scipy and load at the top of your script scipy.integrate import odeint and import numpy as np . There are examples of how to numerically solve differential equations in the examples and on line. The Master equation approach does not work for second order steps. There is also the Gillespie method which is a sort of Monte-Carlo method see this answer which has an example code Probability of the single molecule reaction to happen