text:
Mathematical equivalence of ionic-bond resonance structures and d-orbital participation in hypervalent molecules
It is now routine in the scientific community that "main-group hypervalent molecules, under certain conditions, tend to be better represented by resonance structures involving ionic bonds, rather than the utilisation of d-orbitals". For example, PF5 is better represented by a total of 5 "[PF4]+ [F]-" resonance structures, than sp3d hybritals on the phosphorus atom.
However, after reading the Wikipedia article for  "molecular orbital"s carefully, I began to think that they are, strictly speaking, mathematically the same-
The article said that "When the energy difference between the atomic orbitals of two atoms is quite large, one atom's orbitals contribute almost entirely to the bonding orbitals, and the other atom's orbitals contribute almost entirely to the antibonding orbitals. Thus, the situation is effectively that one or more electrons have been transferred from one atom to the other. This is called an (mostly) ionic bond.": maybe the ionic nature of a molecular orbital formed by a 3d orbital of P and a 2p orbital on F can be considered a special case of this statement, as P's 3d orbitals >>>>> F's 2p orbitals in terms of energy levels.
The question is- is my guess true?

You are essentially correct, but the distinction is the degree of bonding.
As you described, in a polarized or partially ionic bond, the atomic orbital of the more electronegative atom is lower in energy. Thus, the bonding molecular orbital is closer in energy to that atomic orbital, and the antibonding molecular orbital is closer in energy to the atomic orbital of the less electronegative atom. Because of this, the relative contribution of the lower energy AO to the bonding orbital is greater, and the contribution from the higher energy AO correspondingly lesser.
The distinction between "covalent" and "ionic" is therefore blurry. There's a continuum of degree of ionic bonding. In the extreme case of a completely ionic bond, there is no stabilization of the "bonding" MO relative to the lower energy AO. Furthermore, the higher energy AO no longer contributes at all. At that point, we no longer have bonding and antibonding MO's, but rather both are nonbonding orbitals comprised entirely of a single AO each.
But there are plenty of cases where there is just a little bit of stabilization, and we still describe the bond as "ionic". That's partly because it's something of a judgement call at what point we conclude that there is no bonding interaction. We can't solve the Schrodinger equation exactly, so we use computational approximations. Our computational methods aren't precise enough to tell the difference between a tiny bit and none at all. For the $d$ orbitals in the compounds in question, most methods predict at least a small participation of the $d$ orbitals in most compounds. But how much above 0 does it have to be before it's no longer correct to say that $d$ orbitals are not involved? We can say with certainty that the participation of the $d$ orbitals is nowhere near the $sp^3d^2$ view that used to be taught.
In your question, you described the compounds as "better represented" by ionic bonding "rather than utilization of $d$ orbitals."
You're correct that this distinction isn't either/or, but the key words are "better represented". Even if the $d$ orbitals contributed as much as 5% to the bonding MO, isn't the ionic representation "better"?


text:
Estimation of Sodium Hydroxide
Why is estimation of sodium hydroxide more accurate with oxalic acid rather than with sulfuric acid or any other strong acid?

Who told you that ?

Consider the possibility that as oxalic acid is a solid it can be weighed out with greater accuracy than an acidic solution

You are right, Waylander ! Thank you !

There are a few reasons. Some of the most important are:
-Oxalic acid dihydrate isn't hygroscopic (it doesn't absorb water from the atmosphere, whereas sulfuric acid does). If a substance can absorb water from the atmosphere, then when you weigh it out, you won't be able to accurately calculate the number of moles used because you don't know how much is water and how much is your substance. This won't be the case with oxalic acid dihydrate.

It's chemically stable when stored.
As Waylander said (in his comment elsewhere), it's a solid so is easily weighed out.
It has a moderately high molecular mass of $\pu{126 g/mol}$ (as the dihydrate) so any error in the mass weighed corresponds to a relatively small error in the calculated amount in $\pu{moles}$.

As a result, oxalic acid has become a primary standard: So you can use it to make a solution of accurately known concentration to determine the concentration of a solution of sodium hydroxide (by titration).

I'm curious: since you are using "number of moles" for the amount of substance, then why don't you use a number of kilograms in place of "mass"?:)

Well, because I was specifically referring to calculating the number of moles, which I think is clearer than referring to calculating the amount of substance, whereas "mass weighed" is unambiguous.

In my opinion, the gain in clarity is debatable, whereas terminology definitely suffers. Some consider this a nitpicking, but using the "number of moles" isn't quite literate (see Meta).

Well I edited your answer to make it better as  andselisk suggested. If you are not agreeing with the corrections, you have rights to change it since you are the author. I'd also like to include the fact that before estimating $\ce{NaOH}$ with $\ce{H2SO4}$, you need to standardize the concentration of $\ce{H2SO4}$ solution using primary standard because concentration of conc. $\ce{H2SO4}$ is uncertain.  Also, dilute $\ce{H2SO4}$ is not hygroscopic as its concentration version. :-)

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text:
How energy is released when chemical bond is formed?
In books it is commonly written that whenever a chemical bond is formed the energy of the molecules/atoms gets lowered and hence energy is released. This is generally explained by diagrams like this

But I'm really unable to understand this process, I accept that before bonding the potential energy of the molecules were more and after bonding the potential is less and hence this difference of energy must be released but how does this lowering of potential energy happens? Let's just take an example of electrostatic situation (and as far as I know its the electric potential energy only that get's lowered and hence some energy gets released), say I have an electron at the origin and a proton at $x=x_1$ and then somehow it is moved to $x=x_2$ like this
   We know from equation of electrostatics that potential energy of proton at $x=x_1$ is $$ U_1 = \frac{1}{4\pi \epsilon_0}~\frac{-e^2}{x_1} \\
\\
\textrm{and at} ~x=x_2 \\
U_2 = \frac{1}{4\pi \epsilon_0}~\frac{-e^2}{x_2}$$
Since, $x_1$ is greater than $x_2$ (from figure) therefore $U_1 \gt U_2$. So, we energy at $x_2$ is less therefore there must be some release of energy, but how? If we let the field to do the work then no energy will be released because our proton gonna get the kinetic energy, if we do the work to move it from $x_1$ to $x_2$ then according the to the laws of physics I will do the negative work and hence energy will transfer from the proton to me. So, how does energy ever get released?
How in chemistry the energy gets released by the formation of bonds? I know that we can say without intricacies that potential energy gets lowered and hence the difference of energy is the energy that we get, but how does this happens? If we have some cation and then we put some anion in the reaction mixture then, of course, cations will get attracted towards the anion and hence will gain kinetic energy and according to me that's the only possible way to lower the potential energy. Is that kinetic energy that we measure as heat and say that the reaction is exothermic?
Any help or clarification will be much appreciated.

Yes, in some cases it is just the kinetic energy that we perceive as heat. In other cases, photons are an important part of the picture.

That first picture above is a horrible supersimplification. Can you guess why? And please say wherefrom you have it. Whoever made it deserves a good public shaming.

@Karl I’m sorry but I’m unable to see What you are pointing out.

The curve looks like this should be a transition state, but the atoms above do certainly not depict one.

@Karl it might be unrealistic but I do not find it horrible. It might well correspond to a path, although passing a too high energy point.

@Alchimista No, definitely NOT, sorry. A transition state is a bound state. If you put in enough energy to break all bonds at once, the whole thing flies apart. You show such pictures if you want your students to learn facts, but never understand. Im sorry, but that is a despicable attitude for a teacher or textbook author. It comes from thinking students are idiots, which is a self fulfilling prophecy.

@Kark I agree that it is misleading. I wanted to say that the graph qualitatively corresponds to the energetic situation depicted. But I agree that the situation depicted is an extremely unlikely one. And to OP:  basically the answer is the first comment above, that by Ivan Neretin.

@Alchimista The Kinectic Energy of cations is the heat ? I mean during their acceleration towards anion they collide with air molecules or beaker and hence we feel the heat, is that it?

Yes. I just wanted to signal that you've got the answer  - in case a comment might go unnoticed.

Let us assume the simplest case to form a bond from two atoms (assumed to be in their ground state) that approach each other from an infinite distance. For convenience, we put the potential energy of the system at zero here. If the atoms come closer, they start to interact via long-range forces (such as van der Waals forces) and the potential energy is slightly lowered. If the atoms come even closer, their wave functions start to overlap and the potential is reduced rather steeply as compared to the long-range forces. Moving the atoms even closer increases the potential as the electrons will start to repel each other. The potential energy as a function of the interatomic distance can be approximately described by a Morse potential (the blue curve in the figure below (taken from Wikipedia))

In the figure you also see horizontal lines in the Morse potential which represent the energies of the vibrational states that are supported by the potential. Now that we know what the potential looks like we can take our two atoms at infinite distance and move them towards each other, that is, the total energy $E_\text{tot}(r)=E_\text{Morse}(r)+E_\text{kin}(r)>0$. As the atoms get closer, the potential energy gets lower and the velocity and kinetic energy increase, the total energy is - of course - conserved. When the atoms come so close that they start to repel each other (when they hit the repulsive "wall" on the left), they fly away in opposite direction and end up at infinity again! This is the same result that you would find with an ideal marble track (no friction) that is similarly shaped to the Morse potential: the marble will roll down, first accelerate, then climbs up the wall on the left and rolls back to where it started. In order to from a bond (or to stop the marble in the middle) we need to remove part of the energy of the system. As Ivan Neretin pointed out, this can be done by emitting a photon. When the atoms are separated at $r_e$, the release of a photon with energy $h\nu=D_0+E_\text{kin}$ will result in a loss of energy and the atoms will be trapped in the potential. (In our marble track, friction can play a similar role)
This process is called radiative association and it is believed that it played an important role in the chemistry of the early Universe. However, not all molecules can emit a photon. There are some rules - selection rules - which have to be fulfilled. The molecule needs to have a dipole moment for instance. Therefore $\text{H}_2$ molecules are unlikely to be formed via radiative association.
Another, more likely, method to remove the energy is to have a third atom (or molecule) colliding with the other two atoms. This atom can then take away the excess energy in the from of kinetic energy in a so-called three-body collision.
When you have more complex molecules that collide to form bonds, the excess energy can be distributed over the different degrees of freedom that the molecule possesses. For instance, the molecule's vibrational and/or rotational energy is increased which results in an increase in temperature.

Let us take the case where a third atom collides with the two atoms to take away the excess energy in the form of kinetic energy. Here is the question : After forming the bond between the two atoms, where does the energy lost due to reduction of potential energy go before the collision with the third particle ?

The three particles collide simultaneously, so the excess energy is instantly converted to kinetic energy. In some cases there can exists a metastable excited state of the product that may live for pico seconds or so and the energy is stored in the excited state. If no collision occurs with a third body before the excited state decays into the reactants, no products can be formed

You may simply say that energy is needed to break a bond. Breaking a bond is endothermic. So the inverse is exothermic. Energy is released when a bond is formed.
Apparently you want to just discuss what is happening when an electron approaches a proton from far away (x1 in your drawing) to a shorter distance (x2). The electron is supposed to have no kinetic energy at the beginning, in x1. When arriving at x2, the difference in potential energy has been transformed into kinetic energy. When the electron arrives at the x2 position, it has a great velocity. It will not stay here. It will continue its way until it will nearly touch the proton at the abscissa x3 < x2. At this point, all its energy is potential again (no kinetic energy) : it will be repelled back to higher x values. It will go back exactly to the initial point x1, where the kinetics energy was zero, then start again the same oscillation to x3 and back, again and again. The only way of stopping this oscillation is a collision with a third body, to which the electron gives a part of its kineticenergy. This will stabilize its movement. After a couple of collisions, the electron will get to the stablest position at x = 0.5 Å, as it is in a stable H atom.

See, the title asks "why", but the question asks "how".

I request you to please read my question.

@adeshmishra "The only way of stopping this oscillation is a collision with a third body" does answer your question, I think. The answer by Paul includes a different possibility in special cases, namely emitting a photon.

@KarstenTheis I’m reading Paul’s answer again and again for understanding it properly therefore I’m taking some time.

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text:
What is the significance of the downward arrows in energy level diagrams in organic chemistry?
The following text is from Solomons, Fryhle and Snyder Organic Chemistry Third Edition, chapter 1 "The Basics: Bonding and Molecular Structure", page 13, topic 1.5 "Resonance Theory", sub topic 1.5B "Schematic Energy Level Diagrams of Some Mesomeric Molecules/Ions":

Energy level diagrams of some resonating structures are given as follows. (Note: The downward arrow marks the conventional mesomeric energies and indicates energy level).

It is given that the downward arrows in examples 1, 2, 3, and 4 mark the conventional mesomeric energies. What is meant by "conventional mesomeric energy"? I searched for this word on Google as well as on this site, but I couldn't find any relevant information.
What is the purpose of the downward arrows in the above energy level diagrams? Is that to depict the structure with least energy or the one which is the most stable? I thought this was the reason until I saw example 5. The arrow goes from the structure which has the highest stability to the one which has the least stability. Is there any specific reason for this upward arrow or is it just a misprint? I was unable to see any other energy level diagrams with these arrows, so is this an outdated convention followed by the book?

The downward arrow has no meaning. it is just a way of showing the name of the preferred structure. And "conventional" mesomeric energy means "qualitatively determined"

@Maurice: Thank you for your comment. Could you tell, what does the upward arrow in the last example mean?

The upward arrow is drawn in the figure 5 just to explain that energy is needed to pass from the pure covalent molecule to the excited dipole. The same arrow could have been drawn in the figures 3 and 4.


text:
Computed partial charges for protonated amine: discrepencies between Gasteiger and other methods
Consider the following molecule:

SMILES: COC(=O)[C@@H](c1ccccc1)[C@@H]1CCCC[NH2+]1
I would expect the computed partial chage on the amine to be positive, due to its protonation state.
When using Gasteiger charges calculator, it is affected a positive charge of 0.2422.
However, when using more sophisticated methods (AtomicChargCalculator or Antechamber's AM1-BCC implementation) it is affected a negative charge (-0.649), in  these results for example.
It seems the charges computed for the other atoms are roughly similar, as shown in this graph (x-axis: AtomicChargCalculator charge, y-axis: Gasteiger)

Does this amine really have a negative partial charge ? Why the discrepency ? Is there something I am missing ?

Related: https://chemistry.stackexchange.com/a/16908/5017

tldr - atomic partial charges can be non-intuitive
Formal charges in a Lewis structure don't always match up with partial charge assignment methods.
Typically, one wants 'accurate' partial charge models, for some definition of 'accurate' - but what that means can vary considerably. I explained some of this in a related question.
While I haven't worked extensively with AM1-BCC, most of the common partial charge methods either attempt to divide the electron density, assign charges based on best fits to the electrostatic potential at the van der Waals surface, or approximate one or both of these concepts. The electronegativity equalization method (EEM) server you linked mentions fits to electrostatic potential models.
I think you have two parts to your question:
1) Why does Gasteiger give a different answer than AM1-BCC and EEM?
This one is easy. The Gasteiger model uses the molecular connection table (i.e., the valence bond diagram) and thus considers the initial formal charge on the nitrogen. AM1-BCC begins with a quantum calculation, and EEM is attempting to approximate a quantum calculation. You can't specify a formal charge on an atom to quantum mechanics -- the best you can do is express that the molecule is a cation and let the electrons go wherever.
2) Why do AM1-BCC and EEM methods give a negative charge to the nitrogen?
This is less clear to me, but consider that most of these models are attempting to fit point charges at an atomic center to best fit the electrostatic potential on the molecular van der Waals surface. In this molecule, the nitrogen is buried beneath four bonds - two to external hydrogen atoms that are assuredly positive partial charges.
There's ample literature that assigning the point charge to such a buried atom is very difficult because the nitrogen doesn't really have exposed VdW surface.
I haven't taken much time to look at other charge models, but I can tell you that the N+ in your structure does have a more positive charge  than the corresponding neutral N when you strip off the extra proton (i.e, B3LYP RESP gives -0.194 for NH2+ and -0.755 for NH).
That's a very quick, rough calculation, but it should indicate that relative partial charges are often better considerations. In other words, this nitrogen is more positively charged than the neutral analogue.


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Why Freundlich adsorption isotherms fails at high pressure conditions?

Experimentally it was determined that extent of gas adsorption varies directly with pressure, and then it directly varies with pressure raised to the power 1/n until saturation pressure Ps is reached. Beyond that point, the rate of adsorption saturates even after applying higher pressure. Thus, the Freundlich adsorption isotherm fails at higher pressure.

This is an extract taken from Wikipedia explaining limitation of Freundlich isotherm.But analyzing the freundlich adsorption equation we can reach the same conclusions
The Freundlich adsorption isotherm is mathematically expressed as:
$$X/M=KP^{1/n}$$
Giving whole number values to 'n' we can explain adsorption at various pressure conditions.If so why then it is said that Freundlich adsorption isotherm fails at high pressure?
Edit

It cannot be used for a straight linear isotherm part occurring either at low pressures as the
  value n =1 should be then assumed, or at high pressures as the curve increases
  unreservedly, whereas a surface has a limited value and it must be in the condition of
  saturation.

Why is it given here that it's not possible to assume a value of 1 for n?

If n is equal to 1, then as the pressure tends to infinity, the amount of gas adsorbed would also tend to infinity, which is not possible if the adsorbent has a limited surface area. As for the failure of Freundlich isotherm at high pressures, you might want to check out Langmuir's isotherm, which tends to a constant value at infinity.

Does n=1 ,when pressure equals 0 and n equals infinity when pressure becomes infinity.

no. $n$ is a constant, which depends upon the gas chosen

I knew n is temperature dependent quantity varying with nature of adsorbent and adsorbate .But my question is why this isotherm fails at high pressure

There is a saturation level for the adsorption of any adsorbate on the adsorbent due to limited surface area of the adsorbent. So once that saturation level is reached , more pressure won't increase the number of particles adsorbed .

1/n is a parameter which is constant for a fixed system at a particular temperature. n depends on nature of adsorbent and gas at a particular temperature.(1/n doesn't change with pressure). Let 1/n = 0.4. then it will be 0.4 at all pressures(high, medium or low). Now we can not change (1/n) from 0.4 to 0 at high pressure. So adsorption will continue to occur even at high pressures and there won't be any saturation. This is the limitation.
Actually, 1/n varies with Pressure for the equation to hold valid with experimental graph of adsorption. And 1/n = 0 at high pressure.

We can take the value of n as 1.
Moreover the value of n lies between 0 and 1.
So according to Freundlich  for high pressure conditions n is always taken as 1. And hence x/m=1
But experimentally it was found that saturation still increases even after increasing the pressure.
This is how Freundlich equation fails at higher pressure


text:
Why did they ‘split’ this H2+ ion in the ionic equation?
I am asked this chemistry question in my textbook:

Write an ionic equation for this reaction:
  $$\ce{Mg(s) + H2SO4(aq) -> MgSO4(aq) + H2(g)}$$

This is my thought process and answer:
$\require{cancel}$
\begin{align}
  \ce{Mg(s) + H2SO4(aq) &-> MgSO4(aq) + H2(g)}\\
  \ce{Mg(s) + H2+(aq) + SO4^{2-}(aq)
    &-> Mg^{2+}(aq) + SO4 ^{2-}(aq) + H2(g)}\\
  \ce{Mg(s) + H2+(aq) + \cancel{\ce{SO4^{2-}(aq)}}
    &-> Mg^{2+}(aq) + \cancel{\ce{SO4^2-(aq)}} + H2(g)}\\\hline
  \ce{Mg(s) + H2+(aq) &-> Mg^{2+}(aq) + H2(g)}
\end{align}
This was the answer:

$$\ce{Mg(s) + 2H+(aq) -> Mg^{2+}(aq) + H2(g)}$$

Why did they ‘split’ the $\ce{H2+(aq)}$ ion into $\ce{2H+(aq)}$?

Because there is no such thing as $\ce{H2+(aq)}$.

$\ce{H2+ (aq)}$ ions don’t exist. Furthermore, even if they did, they would not make a neutral compound $\ce{H2SO4}$ with the sulfate ion, since the sulfate ion has a $2-$ charge.
So you must split $\ce{H2SO4(aq)}$ into $\ce{2 H+}$ and $\ce{SO4^2-}$ ions.


text:
How are the pKa's of very strong acids derived mathematically?
It is often said that "for very strong acids (HCl, H2SO4, etc.), the pKa value cannot be experimentally derived from standard titration and instead needs to be theoretically computed". However, I could not find any way to compute it theoretically. More specifically - I know that the pKa is a constant multiple of the logarithm of the (magnitude of the) Gibbs energy for deprotonation, and I know that the enthalpy part of the Gibbs energy can be computed easily (just decompose both sides into its constituent elements and use Hess' law), but I can't find any mathematical way to compute the entropy part on Wikipedia.

related https://chemistry.stackexchange.com/questions/28022/how-is-the-pka-of-extremely-weak-acids-determined


text:
Are there any common (preferably somewhat complex) organic molecules that help combat ocean pollution?
I'm participating in an activity where I need to present a topic/molecule in organic chemistry, and I am pretty passionate about "reversing"/helping with cleaning the ocean. I was wondering if there are any organic molecules that help with this or with lessening the effects of pollution. If you could let me know what solutions they are used in that would be a great help also. Thank you very much.

Probably not in the sense that you might mean. What do you mean by "combat pollution"? Perhaps you can be more specific, for instance, do you mean point source pollution, such as water draining from a river or other outlet, or human waste that has already been released and that you wish to recapture. And which pollutants? It is a broad question and you need to narrow it down. Still, you already received a reasonable answer.

OH radicals are not exactly complex, but they destroy anything that floats near the surface over time.

Chemical surfactants or dispersants are useful for cleaning up oil spills. They break up the oil slick and allow the oil to become dispersed in the water, so that naturally occurring bacteria can break down and oxidize the hydrocarbons in the oil. Dispersants are typically designed to be water-soluble at one end and oily at the other end--basically soaps or detergents. See http://masgc.org/oilscience/oil-spill-science-dispersant-bkgrnd.pdf for more details.


text:
Finding the partial pressure
So I saw this problem but can't seem to figure out how to get the answer. I got a test tomorrow lol.
41) $\ce{SO2Cl2}$ decomposes in the gas phase by the following reaction:
$$\ce{SO2Cl2 -> SO2(g) + Cl2(g)}$$
The reaction is first order in $\ce{SO2Cl2}$ and the rate constant is $\ce{3.0 x 10}$^-6 at $600k$.A vessel is charged with 3.3 atm of $\ce{SO2Cl2}$ at 600k. The partial pressure of $\ce{SO2}$ at $\ce{3.0 x 10^5}$
s is ______ atm.
Answer is 2

What I tried:
Starting with PV = nRT(rearranging it to solve for m/v), and using the integrated rate law for a first order equation, I found the concentration of $\ce{Cl2}$ and $\ce{SO4}$. However, I don't know how to isolate the amount of moles to create a mole fraction.
$\ce{ln[SO2Cl2]_i = ln[SO2Cl2]_f -(3.0 * 10^{-6})(3.0 * 10^5)}$
$\ce{[SO2Cl2]_i = \frac{MP}{RT}}$
$\ce{[SO2Cl2]_i = 9.046}$
Plugging back into the first equation:
$\ce{[SO2Cl2]_f = 3.67805}$
$\ce{\frac{m_{Cl}_2 + m_{SO}_2}{V}} = 5.36849$
Not sure where to go from here...
Appreciate your help!

Why would you need mole fractions (or moles, for that matter)?

@IvanNeretin I guess you don't necessarily need it, but not sure how to find the relative partial pressure of each otherwise.

Why, the pressure of SO2Cl2 just decreases exponentially, and that of SO2 grows accordingly.

@IvanNeretin How do you know it decreases exponentially? Is there a relation between order and pressure change? or concentration and pressure change?

There is a relation between concentration and pressure; that's quite enough.


text:
Reaction between acrolein and hydrazine
When acrolein is heated with hydrazine we get dihydropyrazole. But I can't see how. First I think hydrazine should add on acrolein to form corresponding hydrazone, but what will happen next? Also is my first step correct?

If your first step is correct is the resulting hydrazone a Michael acceptor?

Count the number of atoms on the chain in the resulting hydrazine, starting from the carbonyl carbon as 1. You'll see the last nitrogen is 5. This is a setup for a nice 5-membered ring. So, the next step after hydrazine formation is dehydration to give dihydropyrazole.

Your suggestion of resulting corresponding hydrazone by the reaction between acrolein and hydrazine is correct. It is normal hydrazine-hydrozone reaction you perform in idetifocation of aldehyde and ketone carbonyl groups. Usually, E-isomer is preferred (Ref.1). E- to Z-isomerization can be occurred in presence of trace amount of acid (Ref.1). I expect that may be the case during constant heating. As a result, Z-isomer can undergo intramolecular Michael addition as depicted in following mechanism to give dihydropyrazole:

References:

Shigehisa Uchiyama, Masanori Ando, Shohei Aoyagagi, "Isomerization of aldehyde-2,4-dinitrophenylhydrazone derivativesand validation of high-performance liquid chromatographic analysis," Journal of Chromatography A 2003, 996(1-2), 95–102 (https://doi.org/10.1016/S0021-9673(03)00542-9).

The reference you cite did not deal with acrolein but rather the hydrazones of saturated, aliphatic aldehydes. While the cyclization you propose is a 5-endo-trig cyclization (Baldwin's Rules), it is not precluded since there is no competitive cyclization, i.e., 5-exo-trig. However, conjugate addition of hydrazine to the acrolein hydrazone would permit a preferred 5-exo-trig cyclization with regeneration of hydrazine by elimination.

@user55119: To my knowledge, hydrozone formation is preferred over Michael addition with hydrazene. OP is a novice to organic chemistry so I don't want to confused him. I just gave a reasonable mechanism for formation of dihydropyrazole. Also, Baldwin's rule do not apply to all cyclizations. There should be sme exceptions. Don't you think? :-)

I did account for exceptions. That is why I said " it is not precluded since there is no competitive cyclization". Baldwin's Rules apply to competitive processes. Just another possibility.


text:
Acidity order checking
Why is Methanol more acidic than Ethanol eventhough the conjugate base of Ethanol is stabilised by hyperconjugation?

What hyperconjugation is this?

Means?. I only know one kind of hyperconjugation

But shouldn't u consider -H effect for stability of anions ...but that's the not the case here, there is no -H group...how can u compare based on that?

Hyperconjugation is seen when there is a vacant p orbitals to where Alpha hydrogens can relocate their electrons.Conjugate base of ethanol does not have any vacant p orbitals. So hyperconjugation doesn't play any role in stabilization of conjugate base.
The stability of alkoxide ions in solution is determined by how much they get solvated in aqueous solutions.Smaller ions are more solvated than bigger ions,thus methoxide being a smaller ion is more solvated than ethoxide.Solvation of charged ions provide them stability.Thus methoxide have more stability in aqueous solution than ethoxide.Thus explaining why ethanol is less acidic than methanol aqueous solution.


text:
How to account for the common ion effect on osmolarity quantitatively?
In adjusting the osmolarity of a solution of Benzalknonium Cl, Disodium Editate, and Disodium borate/boric acid buffer to be isotonic (about 290 mOsm), the addition of NaCl showed unexpected behavior, possibly non-linear.
I strongly suggest that the common ion effect plays a role here, but how to account for it quantitatively? in other words how to calculate the amount of molecules of say benzalkonium Cl shifted to undissociated state after addtion of
 NaCl ?
Thanks

How would you describe the unexpected behavior of NaCl in your mixture ?

NaCl was added to adjust osmolarity, but the calculated amount of which to acheive isotonicity turned out to be too low  in first time, increseaed osmolarity by only 15 mOsm rather than the required 205 mOsm. In second time increment was too high = 260mOsm while the required was 190 mOsm. The amounts of NaCl were calculated by linear relationship with 0.9 NaCl aqueous solution for which osmolarity=300 mOsm


text:
Copper electrode in a Daniel cell
Suppose I have a beaker containing $\ce{CuSO4}$ solution and I dip $\ce{Cu}$ rod in it. What reactions are going to take place? Is the $\ce{Cu}$ rod going to acquire any positive charge?
I am asking this question to clear my doubt  regarding a Daniell cell as to why $\ce{Cu}$ rod used in the cell should be positively charged. I don't get it, though $\ce{Zn}$ rod dipped in $\ce{ZnSO4}$ solution acquires negative charge due to oxidation of $\ce{Zn}$ to $\ce{Zn^2+}$ is clear to me, but how can $\ce{Cu}$ rod dipped in $\ce{CuSO4}$ be positively charged. I suppose it to be neutral.
What is the answer to this conceptual question?

The absolute charge on the rod is not a question worth pursuing. It has no importance whatsoever. What is important is that Zn is charged more negatively (or less positively) than Cu.

The thought experiment is certainly pursuing and such questions were pursued by the giants of electrochemistry. Search electrostatic electrochemistry by Bard and single electrode reactions. We should not teach students to think along the beaten path. Let them wander and wonder scientifically.

@M.Farooq yes, also search "absolute electrode potential," which is the relative potential-difference between electrolyte and metal plate.  E.g. for the standard hydrogen electrode, it's roughly 4.44V, not zero.  These are genuine potential-differences, not "absolute." But we'd have to ask Alessandro Volta for help, employing electrostatic measurement instruments.These potential-differences require non-contact methods (we're not allowed to introduce large artifacts by touching the electrolyte with a metal probe!)  See also: Volta Potential, and Galvani Potential.

@wbeaty, Right. People have been thinking about absolute electrode potentials for ~100 years. Bockris & Reddy discuss all these concepts in great detail.

@M.Farooq  Yet when single-electrode is explained on "answer websites,", those concepts often draw hostility and denial!  (And entries angrily downvoted away.)  This might be caused by chem undergrad, where I suspect it's never mentioned, and the instructors themselves never encountered it.  "If it was real, it would have been in my textbooks!"  Riiiiiight.  Also do they disbelieve Helmholtz double-layer?   (I first encountered the topics in ancient Bockris edition, looking into nano-physics at capacitor interface. Fortunately I never had chem undergrad, didn't acquire dogmatic error.)

@wbeaty, You must read Chemistry - Fact or Fiction by Gillespie (1975). He shows why really bright students never choose to pursue chemistry. There is so much outdated information/ rote memorization in chem undergrad that sometimes I feel that gen chem should be abandoned and physical chemistry should be introduced in the second year after students have taken a course on calculus and two semesters of physics. Electrochemistry is another drama. Recently, I had a chance to see my relative's first year chemistry notes (in a very good school). It was so disappointing to see the content.

Found Gillespie, also ordering Pera "Ambiguous Frog."  . See if this sounds familiar:  Anderson "Why do they leave Physics?" https://pdfslide.net/documents/andersonwhy-do-they-leave-physics.html, and "germ-theory of education" http://amasci.com/miscon/germ.html

@wbeaty- This is current dilemma of physics too. Long time ago, one of my favorite professors and a well respected electrochemist in North America said "We have given up curiosity driven science." He was so right. When he taught, I had to undo so many electrochem misconceptions and garbage that was fed to us.

Suppose I have a beaker containing $\ce{CuSO4}$ solution and I dip $\ce{Cu}$ rod in it. Now my question is what reactions are going to take place?

On a microscopic level, copper ions can approach the rod and pick up two electrons. Or copper atoms can detach from the rod, leaving two electrons behind. On a macroscopic level there will be no net reaction because there is no source or sink for electrons.
$$\ce{Cu^2+ + 2e^-<=> Cu}$$
If the liquid is inhomogeneous, with high concentrations of copper ions touching part of the rod and low concentrations of copper ions touching another part of it, you might get a situation resembling a concentration cell, and have a current carry electrons along the rod.
Without special tricks, though, this would correspond to a half-reaction, and you need another half-reaction connected to it (via a wire and a salt-bridge, for example), to get a net reaction.

Is the $\ce{Cu}$ rod going to acquire any positive charge? I am asking this question to clear my doubt  regarding a Daniell cell as to why $\ce{Cu}$ rod used in the cell should be positively charged.

The rod is not positively charged. The (+) and (-) in diagrams indicate the potential difference or the flow of electrons. The conventions for (+) and (-) vary, so it is easier to understand the chemistry by looking for the anode and cathode labels.

I don't get it, though $\ce{Zn}$ rod dipped in $\ce{ZnSO4}$ solution acquires negative charge due to oxidation of $\ce{Zn}$ to $\ce{Zn^2+}$ is clear to me, but how can $\ce{Cu}$ rod dipped in $\ce{CuSO4}$ be positively charged.

Again, they don't acquire substantial charges, both the solutions and the electrodes are roughly neutral.

If viewing these as charged objects, we're using a physics/electrostatics viewpoint, not usually done in electrochem.  Or at best, it's a topic for postdoc courses in surface nano-chemistry (yet I think if it were discussed in undergrad classes, it would clear up much student misconception!)  The metal/electrolyte interface forms a capacitor.  For this reason we must deal with actual potentials, NOT the shifted values shown in tables of Standard Electrode Potentials.
You're right to be confused.  As with the zinc rod, the copper rod will also acquire a negative charge.  This occurs because the metal surface dissolves, and the electrolyte pulls some positive metal ions out of the metal lattice.
From a physics standpoint, the two electrodes in contact are 1) the electrolyte and 2) the metal.  A half-cell then has two conductive electrodes, not one.  A complete electrolysis cell has four.  In a half-cell, a genuine voltage appears between the metal and the electrolyte.  The voltage isn't undefined or somehow abstract or "ghostly."  It's just difficult to measure.  A simple voltmeter cannot do it.
These concepts go back over a century, to when chemistry and physics weren't fully separate.  Alessandro Volta and Michael Faraday studied electrolysis from this viewpoint.  I've seen "single" electrode potentials discussed in 1930s undergrad textbooks, but not in the modern ones.  (Well, Bockris electrochem has it, I think.)
Search: Galvani Potential, also Volta Potential, both found at surfaces and junctions between differing materials.  Perhaps also look for Helmholtz layer as well as Helmholtz Double Layer.

.
Whenever metal first contacts electrolyte, a brief oxidation reaction takes place which establishes a static potential appearing across the double-layer located at the metal surface.  Metal corrodes, and loses positive ions.  The net reaction halts once these potentials appear.  The copper is then left with a negative charge, same as the zinc, with the electrolyte having net positive charge.
However, if either a salt-bridge or a common electrolyte is connecting the half-cells, the zinc and copper rods will always acquire two different amounts of negative surface-charge, leading to a well-known potential-difference measured between them.
During work with with these physics/nanochem concepts, it's useful examine the S.H.E. reference electrode.  The relative potential-difference of the Standard Hydrogen Electrode is roughly 4.44V, not zero.  This is the Galvani potential between metal and electrolyte, also called "Absolute electrode potential," although that term comes from chemistry, and a physicist would insist that this is a genuine potential-difference, nothing "absolute."  (The two electrodes after all are the conductive electrolyte on one side, and conductive metal on the other!)  If the SHE potential-difference is 4.44V, then all the values in the table of Electrodes need to be shifted by that amount.  In that case, the polarity of every electrode is negative.  Zinc and copper will both end up with the same polarity.  Copper is not positive-charged.
However, once these concepts are grasped, there's no major need for them, since measurements in electrochemistry are always biased by the inherent Galvani potential of various Reference Electrodes (SHE, or SCE, or Ag/AgCl.)  If we ignore electrostatics considerations, and deny that the electrolyte is itself a conductive electrode, then we can take the reference electrode potential-difference to be zero.  In that case, voltage measurements become trivially easy.  No vibrating non-contact probes needed, no quadrant electrometers or Faraday-cup charge-detectors.
But in that case, surface nanochemistry questions such as @Sharad 's cannot be answered unless we first discuss the zero SHE-potential being a "convenient lie," and the table of Standard Electrode Potentials must be shifted by 4.44V to reflect electrostatic reality.  (All the potentials in that table have the same polarity.  Oxidizing of the solid metal always removes positive ions, even for the metals appearing on the table "below" the SHE's zero-voltage entry.)
See also: we already learned all this in undergrad classes, right?

Here are my comments on the associated chemistry not usually discussed in presentations of the Daniell cell (see, for example, this overly simple rendition).
The electrochemistry starts with a divided galvanic cell with the zinc displacing the copper metal from the aqueous CuSO4 (see prior reference for half-cell reactions).
Importantly, the concentrated solution of copper sulfate acts both as an electrolyte and more, per this source, where MagLab comments:

A container lined with copper (the cathode) forms the outer portion of this device, and is filled with copper sulfate, an electrolyte that reacts chemically with the solid copper.

My supposition on this comment is the following reaction:
$\ce{Cu + Cu(II) ⇌ 2 Cu(I)}$
The above often referenced reaction, per my research, suggests that it can apparently proceed more rapidly in acidic conditions.
Note, as to your question of what happens upon inserting a copper rod into the cupric sulfate solution, above is what could occur over time with respect to standard chemistry (and, not electrochemistry, so forget charge issues).
In the current context, cuprous sulfate is likely created, which is unstable decomposing on warming and, although Cu2SO4 is water-soluble, it reputedly undergoes decomposition in aqueous conditions, per Wikipedia (which is the reverse of its formation reaction above).
Operationally in the presence of air/oxygen, a possible electrochemical reaction (well known for ferrous) between the formed cuprous and oxygen in acidic conditions (which apparently take places with even other transition metals than Fe and Cu, like Mn, Co,...) could also be occurring (albeit limited depending on surface area exposure to air) with time:
$\ce{4 Cu(I) + O2 + 2 H+ -> 4 Cu(II) + 2 OH-}$
This actually introduces a second electrochemical reaction, which may importantly regenerate the CuSO4 (as could perhaps the simply application of heat). Maintaining a high copper sulfate concentration is likely beneficial to the efficient operation of the battery cell. In practice, I would advise, adding H2SO4 to the CuSO4 to maintain an acidic pH, as H+ is consumed per the above reaction. Interesting, per Wikipedia on the original Daniell Cell, to quote:

The copper vessel was filled with sulfuric acid solution saturated with copper sulfate to above the level of the perforated disc.
Over time, copper buildup will block the pores in the earthenware barrier and cut short the battery's life.

The above-cited issue from Wikipedia on the Daniell cell with respect to blockage caused by copper is a problem, but perhaps addressable, in part, by mechanically applied scrubbing.
An interesting experiment for some may be to amend the battery cell with an efficient oxygen interface (as found in metal-air batteries) and record changes in performance.
Note: there are other examples of electrochemical and other chemistry occurring together as in, for example, the action of oxygen on copper in a solution of ammonia. See this source, as an example: "Kinetics and Mechanism of Copper Dissolution In Aqueous Ammonia", (source link).

You are postulating a hypothetical mechanism. The question is about the development of charge on a $isolated$ copper rod dipped in copper sulfate solution. Sadly nobody knows the answer as yet.

Below is an equivalent circuit showing the battery at open circuit equilibrium. You can see that I modeled the two electrode-electrolyte interfaces as capacitors (each with different charges). These capacitors are labeled C_(Zn double layer) and C_(Cu double layer) for the zinc and copper double layers, respectively. You can also see the charges on the capacitor plates. Note how both electrodes have negative charges at the electrode-electrolyte interfaces! The voltage of electrode-electrolyte interfaces were taken from a table of the Standard Potentials for Selected Reduction Half-Reactions at 25°C. Something simple (but easy to gloss over) is that capacitors have equal and opposite charges on their plates. These voltages are: VZn= V0 -(-0.76) and VCu= V0 -(+0.34). The value of V0 is unimportant, the only important thing is the voltage difference. Now, the zinc and copper electrodes will have some capacitance between them, which I label C_(Z-Cu electrodes). This capacitance is likely super small. But still, it is there, and it must be there if we are to define a voltage between the electrodes. Remember that the charge on each plate of a capacitor are equal and opposite (in case you forgot, this stems from the fact that E-field that emanate from positive electrostatic charges must terminate on the same number of negative charges, which ultimately comes from Maxwell's equations). The battery voltage between the electrodes (i.e., the voltage across the equivalent capacitance between the electrodes) is calculated to be: Vbatt= VZn - VCu = 0.76 + 0.34 =1.1V. So, we have a positive voltage, even though both electrodes are negatively charged.


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What is the effect of basis set on charge transfer?
I am calculating the energy and charge transfer of a globally neutral system, composed by two opposely charged monomers (EMIM-BF4), coming closer to each other, with DFT (B3LYP) and MP2.
As a thumb rule, the larger the basis set, the better the energy result should be (given BSSE is taken in account). Is this also valid for charge transfer? What is the role of polarizability? Is there a theoretical argument to trust the results from one basis set against another one, besides the size of the basis set?
I am comparing Pople's STO (minimal bs), Pople's N-311G(p,d) (triple split bs),
Dunning-type Correlation Consistent basis (CCQ or cc-pVQZ) and APCseg-1 ( Augmented Polarization Consistent bs) and I get different results for each case when distance reaches 3 Angstrom (check the attachment).

Which one should I trust?
Thank you

By which method/measure is the ionic charge determined?

Mulliken charge is shown here

I don't see an attachment - suffice to say that Mulliken charges are known to be basis-set dependent and usually aren't regarded as useful. If you want to understand the effects of charge transfer and polarization, I strongly suggest methods like EDA or SAPT. Not sure if they're implemented in GAMESS, but Psi4 and other programs do have these decomposition methods.

Thank you @GeoffHutchison (image should be available now)

@MarcoDiGennaro: I'm not sure if the answer below was perfect for you. There's currently an effort to launch a stack exchange just for materials modeling and quantum chemistry, and many of the top questions were about correlation consistent basis sets, and Gaussian basis sets in general. Please would you consider supporting the proposal? Materials Modeling Stack Exchange

Mulliken charges are not representative of charge transfer, I would argue, in any case. Mostly because of their enormous basis set dependency, as Geoff pointed out.
In any case, several points are missing here; for instance whether the geometries were optimized with each method/basis set combination. This is not trivial.
Assuming a single static geometry (or set of geometries) for all single point calculations, if you go with an MP2 approach, the larger basis set will give you the best result. This is so by definition. In fact, you should always use a large basis set with MP2 because this method converges slowly and requires a large amount of virtual orbitals. In DFT you never know, but I would try to use modern basis sets, such as Ahlrich's (RIP) "def2tzvp" (naming conventions change, but you will find it). Such basis sets are optimized for DFT, and are generally diffuse enough (as a rule of thumb, you should always include diffuse functions if you have an anion in your system).
For your goal, several methodological problems arise. First, if you are indeed optimizing the geometry, B3LYP might be very bad. Dispersion corrections or range separation might help, but you can check the B3LYP-MP2 difference to see if anything suspicious arises.
Secondly, defining charge transfer is not easy. It is not trivial in SAPT (if you want to use SAPT i'd recommend PSI4 dearly for speed, there's a paper by A. Misquitta about charge transfer in SAPT but its not trivial to implement), which is the most rigorous energy term decomposition scheme, and in principle free from intermonomer BSSE.
Defining charge-transfer is somewhat easier in some flavours of the many available EDAs, as in the Kitauma-Morokuma scheme which I think is implemented in GAMESS. Real space EDAs, as the Interacting Quantum Atoms method, might be useful here as well. I think real space charges are far more convenient than Mulliken's in any case. You see, charge-transfer is a nice term but a very diffuse quantity.
Otherwise, you might be better off by checking some more available, well defined observable. For example, the dipole moment of the dimer will capture the charge separation.
In any case, good luck with this project!


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Interstitial defects in crystals: "high energy configurations"?
In the Wikipedia article on crystallographic defects, a type of point defect known as an interstitial defect is described as follows:

Interstitial defects are atoms that occupy a site in the crystal structure at which there is usually not an atom. They are generally high energy configurations. Small atoms (mostly impurities) in some crystals can occupy interstices without high energy, such as hydrogen in palladium.

What is meant here by "high energy configurations"? If these defects possess a "high energy configuration", then it sounds like they would be difficult to come about? If so, then how do these defects come about (where do they gain the energy from to occur)?
I would greatly appreciate it if people would please take the time to clarify this.

Yes, interstitial defects increase potential energy, as stated in Wikipedia: "In interstitial defects three or more atoms may share one lattice site, thereby increasing its total energy." See Physics on StackExchange for an explanation of potential energy in crystals.
How can they form? One example is during rapid crystallization, where there is insufficient time for atoms to segregate. Think of this as "freezing in place" a defect. If warmed and cooled again, the defect can be excluded and a more-perfect crystal forms. This is the basis for the zone refining, Czochralski process and float-zone purification, where extremely slow crystal growth allows time for impurities to be pushed out.
On the other hand, one can intentionally induce interstitial defects through repeated folding, heating, and cooling to make metals harder and tougher, such as Wootz or Damascus steel. It is easy to see how this laborious process requires energy to insert the defects.

Thanks for the answer. So "high energy configuration" is referring to a configuration that has high potential energy? If so, then what does it mean for a configuration to have "high potential energy"? What are the consequences of this?

That question has already been answered in Physics, https://physics.stackexchange.com/questions/328365/potential-energy-for-a-crystal-lattice

I understand now; thanks! But it wont allow me to award you the bounty?

Never mind; I am now able to award the bounty.


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Is the activation energy always inversely proportional to the reaction rate?
When I obtained the activation energy using the Arrhenius equation for a 10-minute glowstick (rapid, rigorous reaction), it was almost 4 times that of a 12-hour glowstick (slow, mild reaction).
However, I learned in chemistry that a reaction with a lower activation energy need less energy for successful collisions, therefore will have a higher reaction rate.
This clearly does not hold true in my case, since a shorter 10-minute glowstick obviously has a higher reaction rate compared to that of a 12-hour glowstick(and thats why it only glows for 10 minutes).
Can anyone provide an explanation to this?

Well, the activation energy is not the only variable in the Arrhenius equation.

I don't know the composition of these glow sticks. But I assume that they cannot be compared, because the chemical reactions may be absolutely different in both sticks.

Activation energy is definitely always not proportional to the reaction rate, since rates are exponentionally related to energies. "Proportional" implies a linear relationship.


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Why does water vapour deviate much more from ideal gas behaviour as it is cooled compared to nitrogen gas?
My guess is that it is something to do with the difference in boiling point between the two aforementioned molecules. But I can seem to come up with a logical explanation to why this is the case.

Well, an ideal gas has no attraction force between its molecules and does not have a boiling point, as you cannot make it liquid. Also, the sum of  molecule own volumes is supposed negligible wrt the gas volume.
Water does not form anything close to an ideal gas because of its hydrogen bonds.  You have to heat water to 100 °C to make it boiling to overcome these bonds between molecules of liquid water. Molecules of water vapour interacts with each other as well, especially at  high pressure.
If water had not had these bonds, it would have reportedly boiled at -120 °C. And if water had not been polar, what makes intermolecular attraction as well, its boiling point would  have been comparable with nitrogen.
Nitrogen is a non polar gas, with minimal molecular attraction, compared to water, due weak van Der Waals force, which leads to nitrogen boiling at -196 °C.

I have changed the second sentence a little, please check if the meaning is still what you intend.

Yeah, it is OK.

For deriving ideal gas law we assume that there exist no intermolecular forces between molecules and their size are negligible.As the size or intermolecular interactions increase there would be a increase in deviation from ideal gas law.Obviously here interactions is the major factor.(H-bonding in water)


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How to determine the formula for calculating the enthalpy change of a reaction?
$$\ce{4 HCl(g) + O2(g) -> 2 Cl2(g) + 2H2O(g)}$$
In the reaction above, knowing the enthalpy of formation of each individual compound, in order to calculate the standard enthalpy change of the reaction, how do I determine whether it is:
$\pu{[products] - [reactants]}$ or $\pu{[reactants] - [products]}$
.

There's a general principle that applies anytime you calculate a change in a property:
The change in the value of a property in going from an initial state to a final state is always:
             value(final state) - value(initial state)

For instance, what is the change in elevation (which we'll call height, h; not to be confused with enthalpy, H!) when we go from 500 m to 1000 m?  Here h(initial state) = 500 m and  h(final state) = 1000 m, so:
$$\Delta h = h(\text{final state}) – h(\text{initial state}) = \text{1000 m – 500 m = 500 m}$$
Likewise, what is the change in temperature when going from 60 C to 40 C?  Here T(initial state) = 60 C and  T(final state) = 40 C, so:
$$\Delta T = T(\text{final state}) - T(\text{initial state}) = \text{40 C - 60 C = –20 C}$$
Calculating enthalpy change is no different:
$$\Delta H = H(\text{final state}) - H(\text{initial state})$$
If we want to calculate the enthalpy change for a chemical reaction, the initial state consists of reactants, and the final state consists of products.  Thus:
$$\Delta H_{rxn} = H(\text{final state}) - H(\text{initial state})=\Delta H_f(\text{products}) - \Delta H_f(\text{reactants})$$
I.e., the fact that it's products minus reactants has nothing to do with the nature of enthalpy.  Rather, "products minus reactants" follows directly from the basic mathematical convention we use to calculate the change (in anything).

The enthalpy change is always Products minus Reactants.

I'm curious as to the rationale for giving this answer two downvotes.  I personally had given it an upvote.

@ChetMiller One downvote is mine. Even though the answer is, of course, correct, it is too brief and in my opinion should be a comment as it lacks explanation as to why it is so (or even citation of the origin). I intensely disagree with the opinion that answers using such dogmatic postulates being useful as it's not how scientific method operates and not how teaching should be done.

I agree with @andselisk. I would like to humbly request him to edit his answer accordingly, and share his valuable knowledge with us all. Expecting an informative answer soon :) Or else we could have had a stimulating discussion in the comment section as well.


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Hardy Schulze Rule
There was a question which was asked in JEE Main 2020 held on 8th January. The question was:

As per Hardy–Schulze formulation, the flocculation values of the
following for ferric hydroxide sol are in the order :
(1) $\ce{K3[Fe(CN)6] < K2CrO4 < KBr = KNO3 = AlCl3  }$
(2) $\ce{K3[Fe(CN)6] < K2CrO4 < AlCl3 < KBr < KNO3}$
(3) $\ce{K3[Fe(CN)6] > AlCl3 > K2CrO4 > KBr > KNO3}$
(4) $\ce{AlCl3 > K3[Fe(CN)6] > K2CrO4 > KBr = KNO3}$

Answer given by the offocial body was (1).
Yes, Hardy-Schulze rule does say that for a positive sol as the charge of the anion in the salt increases the flocculating value decreases(if the charges of anions in 2 salts are different then does not depend on the number of negative ions per molecule of salt and the rule still holds good). But I have never read anywhere that if the charge on the anions is the same then the flocculating values are equal.
Shouldn't the flocculating value of AlCl3 be less than flocculating value of KBr and KNO3? because it will give 3 anions per molecule whereas the other two ( KBr and KNO3) give only 1.

The Joint Entrance Examination board should update their outdated set of questions relying on rote memorization. Why do they assume the student has to memorize the charge of the ferric hydroxide particles and in what medium? Charge on the colloids is dependent on the medium and pH! They should also mention that the molar concentration of the anions is the same. With this information, the question becomes complete.
If the zeta potential of ferric hydroxide is experimentally found to be positive, then the anion should affect them. So, notice the charge on the anion in (1): (-3), (-2), and then monovalents.


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Is it more energy efficient to desalinate seawater by flash boiling or by flash freezing?
In Saudi Arabia I think there are two types of desalination using flash boiling
from electric power production and also from flash freezing using solar absorption
refrigerant freezing.  One requires fossil fuel the other solar energy.

Reverse osmosis?

Distill water with thermal solar energy is even simpler than solar adsorsption refrigeration.

and btw. why do you think so? Sources? ;)

The cheapest way is probably to use the sun, as it is done in Tunisia. In Tunisia, they dig long channels on the beach along the sea. The channels must be long and narrow, and they are filled by sea water. Now the channel is covered by an inclined glass roof. The sun heats the water through the roof, and the steam  is slowly evaporated and condensed on the roof. It makes mist that streams and flow in a drainpipe placed at the lower part of the glass, just above but nearly on a level with the sea. It produce pure water for free.

"Free" is relative. But our whole planet has a problem with recognising the value of infrastructure. Notoriously on continents starting with "A", except Antarctica. ;)


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What are Degenerate Rotational Energy Levels Physically?
I'm in introductory chem, so I don't have much of the physical background to really understand what's going on yet, but we're learning about rotational energy levels, and how some rotational energy levels for molecules are degenerate.
As I understand it, degenerate energy levels correspond to different "ways in which the molecule could rotate" such that it gains the same amount of potential energy.
I have two questions...

Why does a molecule "gain potential energy" when it rotates? Does it want to stop rotating for some reson?
What do degenerate energy levels correspond to physically in terms of the molecule's rotation?

Thanks!

When I put you on a carousel, I surely need energy to start rotating you, and your hand gets hot when you put it on the ground to decelerate. Right? OK, that was the stupid part of your question. 2. makes more sense. ;)

and if your teacher calls rotational energy potentional energy, he has rightfully spotted a likeness, but otherwise is imo an idiot. omg

That's a rather... strong way of putting it, but I agree that calling it "potential energy" is wrong. I'm fairly sure that the Hamiltonian for rotational motion involves just the kinetic energy term, $-(\hbar^2/2m) \nabla^2$. See, e.g. https://en.wikipedia.org/wiki/Rigid_rotor#Quantum_mechanical_linear_rigid_rotor Unlike, say, vibrational motion, there isn't a potential energy term. If you've never heard of a Hamiltonian, then arguably you shouldn't be taught QM (yet)... but well.

I have two questions...
Why does a molecule "gain potential energy" when it rotates? Does it
  want to stop rotating for some reason?
What do degenerate energy levels correspond to physically in terms of
  the molecule's rotation?

Do you remember the Newton's law of motion? A body in motion will always remain in motion until and unless there is a force acting on it. So you can imagine, that once a force caused the molecule to rotate, the molecule will continue to rotate forever. As a crude analogy can you recall the time when the Earth stopped spinning on its axis? It is spinning for billions of years. In the same way you can think that the isolated molecule, once excited, will rotate for a pretty long time (on a spectroscopic time scale).
You can think of degenerate rotation as a a way of physical rotation of a molecule in such a way that if you calculate the rotational energy in each case, it turns out to be the same number.
Third point, as pointed out by orthocresol, rotational energy is kinetic energy not potential energy. So no need to invoke potential energy.
Sometimes, we should not actually try to visualize what is happening physically at very small molecular levels. This is where mathematics enters. Many results originate out of pure mathematical calculations which can be hard to imagine mentally. For instance, Bohr's got correct H2 spectrum results by assuming the electron is rotating around the proton. The question is it actually rotating like a planet around a proton? A short story is worth sharing which will help you in future. Feynman (Physics Nobel) used to think a lot about of photons as a student and discussed them with his father. When he finished his PhD in physics, he realized, he still could not explain what a photon is. Sorry I cannot recall the reference. But mathematics has no problem. Yet on the web we can "see" photons in pictures as balls, which are obviously pure but meaningless imaginations.

That's a great answer @M. Farooq. Would you consider helping us launch a stack exchange just for spectroscopy and materials modeling? We would very much appreciate your support here: Materials Modeling

As pointed out in the comments the rigid rotating molecule only gains kinetic energy when it rotates not potential energy.
The degeneracy describes the fact that some levels have exactly the same energy and this depends the value of the angular momentum rotational quantum number $J$. The number of degenerate levels is given by the multiplicity $2J+1$. The $J$ quantum number has values $J=0,1, 2,3,\cdots$. A second quantum number $m_j=-J\cdots J $ describes each of the degenerate levels and 'orientation in space'. Thus if $J=2$ then $m_j$ has five values $m_j=-2,-1,0, 1,2$. Normally the $m_j$ levels cannot be distinguished by experiment because they are degenerate in energy, but in a magnetic or electric field these energy levels can be made to split (degeneracy is removed but multiplicity remains the same) and transitions between them measured. These effects are called the Zeeman and Stark effect respectively. (Historically the Zeeman effect was of great importance and showed that electrons have spin angular momentum, unknown hitherto.)
As the molecule is a quantum object some care is needed in interpreting what is meant by 'rotation'. When $J=0$ the molecule has no kinetic energy, i.e. is stationary and so has a wavefunction like that of an s orbital. To satisfy the Heisenberg Uncertainty Principle the molecule has an equal probability of  pointing in any direction in space, i.e. the wavefunction looks like a sphere. When $J >0$ the molecule has some rotational kinetic energy but it now becomes more difficult to imagine but when $J=1$ the molecule can 'rotate' in any of three directions, with wavefunctions shaped like p orbitals.  When $J=2$ the wavefunctions look like those of d orbitals and so on.
(Quantum mechanics often presents problems because we are so familiar in thinking of how every day objects move, e.g. a rotating wheel, and its easy and sometimes useful to imagine molecules do the same thing, but this is misleading as the wavefunction describes what happens and this often does not seem to make physical sense. )


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Type of equipment to measure exact amounts of liquid
I have to sample $\pu{10mL}$ of liquid hundreds of times a day at work.
We use a $\pu{10mL}$ syringe to do so, but it takes a lot of time as they often allow more than $\pu{10mL}$ to be drawn in (i.e. rather than a blind action, we have to check the measurement increments to make sure it’s exactly $\pu{10mL}$).
Is there a type of syringe of other piece of equipment designed to draw in exactly $\pu{10mL}$ (or other amounts) of liquid?

Well, duh! https://en.wikipedia.org/wiki/Pipette

In particular automated pipetting system could do it all for you.

What liquid are you sampling, what accuracy do you need, and what are you doing this for?  Also, what's your budget? The more details you can supply, the better able we will be to recommend an appropriate tool.

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

If you need to measure $\pu{10.0\pm0.1 mL}$ repititively, then you can use an any available Bottletop Dispensers. For example, check this and this out (see image below):

If you are in a educational facility, Fisher would offer attractive discount (request a quote from fisher representative in your area. For example, I got $\pu{1.0-10.0 mL}$ dispensers for less than $100/each in that way (I bought 15 of them though). Still, it is a good price.

I don't know why, but I find these so satisfying to use!

Yep, they are very convenient. I use them in undergraduate organic chemistry labs so students do not waste solvents using grad cylinders any more. Save time as well. :-)

I only remember getting to use them once when I was an undergrad - it was used for CDCl3, to measure out ca. 0.5 ml of solvent to make up an NMR sample. I guess it only really makes sense if you have a repetitive task (and a chemical that's reasonably harmless, of course).

Still, we even use them to measure nitrating reagent (just for safety issue). However, it need to be flush with at least a $\pu{1 L}$ of water every once a while to keep it from degrading. Last longer enough to be useful.

This could be a great solution if the OP needs to repeatedly dispense 10 mL of liquid; but it's possible he instead needs to repeatedly sample 10 mL of liquid (e.g., for some sort of production line QC).  That's why I asked him to explain what he needs it for.

Mathew, is that really that accurate? I thought that was the accuracy of volumetric glassware (class B) 10.0±0.1 mL?

Manufacturers have claimed it. I found they are pretty consistent (more of precision). Yet, I never attempted to check the accuracy since it wasn't my concern.

And, OP is using a syringe. Accuracy of this is compatible with syringes and a way fast.


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Reconciling electron configuration and valence electron distribution over separate energy levels in energy band theory
The electron configuration of a Silicon atom in its ground state is $\ce{1s^2 2s^2 2p^2 3s^2 3p^2}$, or equivalently, $\ce{[Ne] 3s^2 3p^2}$.
When looking at the energy-level scheme of a silicon atom, the valence electrons are distributed over two separate energy levels; specifically, the $\ce{3s}$ and $\ce{3p}$ energy levels. It is then said that, as you bring multiple Silicon atoms closer together, the two energy levels will split and overlap. And because the two energy levels are splitting and overlapping, they will produce two energy bands separated by a region of energy that is not allowed. Of these two energy bands, the lower energy band is referred to as the valence band, and the higher energy band is referred to as the conduction band.
I am familiar with the phenomena of valence and conduction bands, as well as the "region of energy that is not allowed", otherwise commonly referred to as the "forbidden energy gap".
However, I'm wondering what this phenomenon of the valence electrons being distributed over two separate energy levels is? How does one reconcile this with the electron configuration $\ce{1s^2 2s^2 2p^2 3s^2 3p^2}$, which seems to suggest that the electrons remain in discrete, well-defined energy levels (rather than overlapping)?
I would appreciate it if people would please take the time to clarify this.

I have been asking leading spectroscopists as to how they came up the assigning electron configurations of elements. Most chemists are clueless yet we are happy to teach and regurgitate electronic configurations. Better ask in a physics forum where you might get an idea as to who came up with the concept of conduction band and valence bands and why. This was all done by physicists.

@M.Farooq Oh really? So it would be better to ask physics.stackexchange?

Yes, and even better find the original papers (you can add a request in the question). By going through the original papers you will get a feeling of their line of thinking and reasoning.

Honestly I might agree with the comments above but I don't understand the question.

@Alchimista I'm essentially asking for an explanation of how valence electrons can be "distributed" over many separate energy levels, rather than being contained in a single energy level at a time. The latter is what seems to be suggested by the "electron configuration" that we are taught in basic chemistry, but the former is what we are taught in solid-state/semiconductor physics.

It is the spacing between levels that goes away. It is how an atom combines to give you a chunk of substance. Symmetry /Energy determines what combines with what and the separation between the atomic levels reflects in the separation between bands. Which can get more or less closer and even overlap.

@Alchimista Hmm, what do you mean by "it is the spacing between the levels that go away"?

If you combine N same levels you get an almost continuous bands, isn't?

https://chemistry.stackexchange.com/questions/22354/why-do-molecular-orbitals-in-solids-merge-to-bands might be what you are looking for. Plus does this pic, eg, helps? - - > http://wiki.chemprime.chemeddl.org/articles/m/o/l/File~Molecular-Orbital_Energies_Corresponding_to_Delocalization_of_Valence_Electrons_Over_Increasing_Numbers_of_Li_Atoms_.jpg_e9b7.html

Does this answer your question? Why do molecular orbitals in solids merge to bands?

@Alchimista Yes, it does! Thank you for that!


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Deprotonated form of phenolphthalein
The wikipedia page states clearly that the coloured form of phenolphthalein is the one with two protons removed. What is the evidence that phenolphthalein loses both protons in basic solution? And does the only once deprotonated form ever appear?

@OscarLanzi You probably haven't noticed, but it was me who answered your question.

Yeah I did. Didn't I accept?

I found this paper (it's in Japanese, but the relevant things are legible) https://www.jstage.jst.go.jp/article/yakushi1947/117/10-11/117_10-11_764/_pdf, which says that phenolphthalein has 2 pKas, pKa1 = 9.05 and pKa2 = 9.50. It also says that the pink form is the twice deporotonated; the 1- form exists, but is colourless.

The Wikipedia article has a nice picture of the stages of deprotonation of phenolphthalein:
https://en.wikipedia.org/wiki/Phenolphthalein

However, there is a big jump here between H2In and In2-. In fact, smaller steps, showing just one deprotonation, are very reasonable. Whoever heard of two protons coming off at the very same time?

http://www.ch.ic.ac.uk/vchemlib/course/indi/indicator.html
Of course, a fuller explanation is needed.
Thymol Blue has a structure similar to phenolphthalein in that it has two phenolic groups. They deprotonate at different pHs and thymol blue has two color changes.(https://en.wikipedia.org/wiki/Thymol_blue):

The uk reference also lists the pH for a color change of various indicators and also the range over which the color change occurs.
And litmus. Litmus is a water-soluble mixture of different dyes extracted from lichens. The litmus mixture has the CAS number 1393-92-6 and contains 10 to 15 different dyes. All of the chemical components of litmus are likely to be the same as those of the related mixture known as orcein, but in different proportions. In contrast with orcein, the principal constituent of litmus has an average molecular mass of 3300.
https://en.wikipedia.org/wiki/Litmus

In the list of indicators, all have a range of color change of 1.6 pH units, even thymol blue, which has two phenolic groups. Except phenolphthalein, which changes with a broader range of 1.8 pH units. Then there's litmus, with a color change range of 3.8!
The range for litmus can be ascribed to its being an extract of many similar dyes, a natural product, with slight differences between individual compounds. Thymol blue has a sulfonate moiety which will ionize more easily than the carboxyl in phenolphthalein. It seems that the colorless form is singly ionized, being a single quinone moiety and two aromatic rings, without extended conjugation.
In thymol blue, the second deprotonation gives a phenolate which more certainly conjugates with the quinone, establishing a larger conjugated system, thus highly colored. (Yeah, the unconjugated neutral compound is red, how do I explain that - I guess it's just the starting color of a pretty aromatic compound.)
Now in phenolphthalein, the first deprotonation yields an ion, shown in the uk reference. (Although they say it is pink, I am going to disregard that. In addition, I'm going to postulate that the mono-ion would be better described as an ester with a phenolate group, not as a quinone-carboxylate.) This first-deprotonated ion resembles the colorless form of thymol blue: a quinonoid structure with two phenyl rings attached to a central carbon. I might guess/conclude that this form could be colorless and that the second deprotonated form is the pink/fuchsia compound. The (slightly) larger range for the color transformation could well come from the lower tendency of the carboxylate (compared to the sulfonate in thymol blue)  to dissociate after the first proton is drawn off; as long as the central carbon is tetrahedral, the molecule has one phenol and one phenolate. The central carbon goes planar in phenolphthalein AFTER the second proton comes off, whereas the second proton comes off in thymol blue from an already planar structure.
To restate: in phenolphthalein, the central carbon is tetrahedral and the two phenolic groups have very similar pKa; after both phenolic protons are removed, the carboxyl detaches and the extended resonant system is formed (pink/fuchsia). In thymol blue, the first proton removal is easy because the sulfonate is a more stable ionic group than carboxyl, so this proton is more acidic than phenol, but the second proton is just part of a substituted phenol.
Just for reference, the pKa of phenol is 9.95 (https://en.wikipedia.org/wiki/Phenol).

"Whoever heard of two protons coming off at the very same time?"  Not exactly the same thing, but when nitric "acid" is placed in sulfuric acid to make nitryl ion for nitration two protons are transferred in one reaction.  One proton transfer gives $\ce{NO2^+ + H2O}$ and then the water molecule immediately grabs another proton.  The structural change associated with the first proton transfer (the nitrate function loses an oxygen atom) drives the second one.

On another part:  there may be an explanation for thymol blue being red in strong acid.  Suppose a contributing structure where the highly electronegative $\ce{-SO3}$ function takes away the bonding pair to the bridgehead carbon.  Then we have a conjugated carbocation in this contributing structure, much like that in protonated phth, actually -- and apparently with the same color.


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Silver mirror and copper mirror
There is a formation of silver mirror in the silver mirror test. why does the silver does not precipitate, it forms the mirror in the silver mirror test? Is there any test that copper can form mirror? Why silver can form silver mirror but copper cannot?

Related: Copper substituting silver in Tollen's reagent

I see the answers focus on the easiness of reduction rather than the actual forming of a mirror. The two things might be related but, though this might be unsatisfactory, the mirror formation should depend on the stiffness and adesivity of silver to glass. Not the first things that one analises in details when doing chemistry in test tubes.

A copper mirror can be formed, for example by reaction between $\ce{Cu(NH3)4^{2+}}$ and hydrazine in basic solution.
That being said, copper in general forms mirrors less easily than silver because copper is harder to reduce to the metal.  This data page gives a standard reduction potential of about 0.34 V for reducing copper(II) to the metal but about 0.80 V for reducing silver(I).

A copper mirror can be obtained by heating two or three grams of dry copper formate $\ce{Cu(HCOO)2}$ in a test tube with a Bunsen burner. For some reason, the copper atoms are deposited on the inner surface of the test tube. Try it! It is easy to do!


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how to Find $P_{tot}$ Given that $K_p = 28.5$ and the partial pressure at equilibrium
I have the equilibrium equation :
$4KO_2{_{(s)}} + 2CO_2{_{(g)}} \ce{<=>} 2K_2CO_3{_{(s)}} + 3O_2{_{(g)}}~~k_p = 28.5 ~at ~25C$.
Given that The partial Pressure of $CO_2$ is $0.191~atm$ at equilibrium Find the total pressure ($P_{tot})$.
so i constructed this :
$~~~~~~4KO_2{_{(s)}} + 2CO_2{_{(g)}} \ce{<=>} 2K_2CO_3{_{(s)}} + 3O_2{_{(g)}}$
$t_{int}~~$   ____       $~~~~~~~\frac{2191P_{tot}}{3000}~~~~~~~~~~~~~~~$____   $~~~~~~~~~~~~0$
$t_{eql}~~$   ____       $~~~~~~~\frac{191P_{tot}}{1000}~~~~~~~~~~~~~~~$____   $~~~~~~~~~~~~0.809P_{tot}$
so $K_p = \frac{[0.809P_{tot}]^3}{[0.191P_{tot}]^2} = 28.5 $
so $\fbox{$P_{tot}$ = 1.96365 atm}$
but my answer is wrong and i dont know why.
i also tried to calculate the total pressure at the initial state as an answer but that also gave wrong answer.

Bro pressure of $CO_2$ is 0.191 and pressure of $O_2$ is ($P_{tot}$-0.191)

i dont get you the partial pressure of $CO_2$ is $0.191 P_{tot}$  and so the partial pressure of $O_2$ is : $ P_{tot} - 0.191P_{tot} = 0.809P_{tot}$

The partial pressure of $CO_2$ is 0.191atm not 0.191$P_{tot}$ if you have not make any mistake in writing above question.

my bad but yes and so the pressure of $CO_2$ is what i wrote above

the result still the same its the concept

can someone solve this problem because i'v been trying to solve since yesterday

Let $P_{tot}$ - 0.191= x, then $x^3$ = 28.5(0.191)², this gives $P_{tot} approximately      1.2

it gave me a right answer , the question said that it was partial pressure but from your solution i understand that it was regular pressure of the gas . right ?

They both are same in the way you are treating regular pressure


text:
Is anode actually ever negative?
Disclaimer: This question comes from a physics student.
I was studying electrostatics and I was wondering how to charge a conductor at home without any fancy furs etc. So an idea came to me: take a battery, touch for example the negative terminal with the conductor and then remove it. Since the negative battery terminal is at a negative potential and the conductor was previously grounded ($V = 0$) then as their potentials must become equal, it means that electrons will flow from the battery terminal to the conductor for a short period of time. Afterwards, I disconnect the conductor and the electrons remain 'trapped' in it hence the conductor is now negatively charged.
At least that was the idea. Now suppose that the battery is a galvanic cell ($Zn, Cu$). The question is whether there are free electrons floating inside the $Zn$ anode when it is not connected to the cathode?
The whole system (battery, wires) is electroneutral while the current is flowing e.g. Zinc atoms at anode dissolve and electrons stay at the anode while the $Zn^{2+}$ ions go to the solution, but in the same instant other electrons leave the anode.
When the electrodes are not connected, I'm not sure would the redox reactions happen. In my understanding, $Zn$ oxidizes because $Cu$ on the other side has greater electron affinity, but $Zn$ would have to be able to 'sense' $Cu$ on the other side somehow, otherwise the oxidation reaction happens anyways for a short time until it reaches some equilibrium point. This is what I think actually happens and that's why the anode would actually be negative until connected to the cathode.
So am I on the right track with this and ultimately, will my attempt to charge the conductor a bit succeed?

See also https://chemistry.stackexchange.com/questions/16785/positive-or-negative-anode-cathode-in-electrolytic-galvanic-cell

If the anode is ever negative depends in the first place on the large scale electrostatic gradient between the ionosphere and the Earth, than on mesoscale deviation around thunderstorms and than on local electrostatic conditions. Electrode potential is conventionally related to standard hydrogen electrode, that has itself potential cca +4.4 V wrt free electron. But the potential of the free electron depends on environment potential, what brings us to the beginning.

@Poutnik That was another question that was bothering me, thank you for mentioning it. I always wondered what could be the voltage relationship between the battery terminals and a piece of electroneutral conductor and could I always or ever take it to be $0V$. But now you've cleared it up to me a lot. Also, this shouldn't be a problem in attempt to charge the conductor because even though there could be no voltage between a conductor and one battery electrode because of outside conditions, we still would have the other electrode to make the charging process possible.

See the comments underneath the post as well

It is a very good questions from a student who seems to be thinking deeply. I have been thinking to write on this topic for the last 7-8 years. In short the poles of the batteries are electrostatically charged and one can actually "sense" this charge with the help of sensitive electroscopes which were known in the time of Volta.
The label anode or cathode is not defined with respect to the electrostatic sign as explained in the post by Maurice.
Think of a large parallel capacitor. Connect one terminal of the plate to one positive terminal and the other one the negative terminal. The capacitor plates would show a electrostatic charge.
By all means, using one single cell you would not be able to detect the electrostatic charge, like a charged comb. Only specialized electroscopes can sense that.

So, just to clarify, my attempt at charging a piece of conductor by touching an electrode would succeed?

@lojle Yes you will, see my edited answer above.

Thank you for the clarification and I guess you were referring to ballistic galvanometers in the last paragraph. Now when I'm sure that a small current would rush into the conductor, I could in theory connect the conductor to a terminal via a ballistic galvanometer and observe a deflection on the galvanometer. Now the amount it deflects is a matter of conductor capacitance, battery voltage and the galvanometer sensitivity.

@lojle, no galvanometers can only detect current. Those specialized electroscopes were called quadrant electroscopes or gold-leaf electroscopes connected to capacitors, a device well known in Volta's time.

I think that the idea is that there would be a brief, small current not a steady one.  E.g. if you connect a capacitor to a battery you will get a transient current.  Whether it would be measurable in the OP's case without sophisticated equipment is a harder question.

@badjohn exactly

An anode is the place where oxidation occurs. In a galvanic cell, like the Daniell cell ($Zn - Cu$), the $Zn$ is oxidized at the Minus pole. $Zn$ is oxidized into $Zn^{2+}$ ion and the electrons are sent in the electric connexion going to the $Cu$ electrode. $Zn$ is the anode (negative sign) and $Cu$ is the cathode (positive sign)
If the same setup is used in electrolysis, the phenomena a going in the reverse sense. Electrons are obliged to flow from the copper plate to the zinc plate. $Zn^{2+}$ ions are reduced to metallic $Zn$ on the $Zn$ plate. The zinc plate is now working in the anode mode. But it is still the Minus pole, as, in electrolysis, cations (positive ions) are attracted by the negative pole. In the electrolysis, the $Cu$ plate is oxidized into $Cu^{2+}$ ions : Copper is the anode.
When comparing the same chemical equation working in a galvanic cell or in an electrolysis, $Zn$ is always the negative pole and $Cu$ is always the positive pole. The signs PLUS (+) or MINUS (-) can be printed on the plates. But Zn is an anode in a galvanic cell, and becomes cathode in an electrolysis system.
So you see that an anode is not always negative. The anode is the negative pole (Zn) in a galvanic cell. The anode is the positive pole (Cu) in electrolysis.

That is a good point. But my question wasn't Is the anode always negative, but rather Is the anode ever negative? Namely, is the anode  metal plate negatively charged even without considering the cathode. Maybe the question could be reduced to Does $Zn$ oxidize in $ZnSO_4$ solution at least a bit without cathode existing at all? But since in the battery we also have the salt brigde, I am not sure whether this simplification is possible...

Zinc placed in water will oxidize reducing the water to H2, the metal acts as both anode and cathode. [adding acid will accelerate everything] The chemistry is determined by the electrode potentials and the physical layout. The question you seem to be asking is were I to place a pure zinc rod in a large amount of water would zinc ions migrate from the rod leaving the electrons on the zinc. That is what seems to happen but as the charge builds up those electrons don't sit there idlily.


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IUPAC name of Dewar benzene
Dewar benzene according to Wikipedia has the IUPAC name bicyclo[2.2.0]hexa-2,5-diene.
Is it correct to give this name because resonance takes place and if we fix a certain numbering, as resonance takes place, a different carbon would occupy that numbering in a cyclic manner? Similarly like saying benzene's IUPAC name is cyclohexa-1,3,5-triene would be wrong.
Kindly clarify.

There's no resonance in it and otherwise it would still be no problem.

@Mithoron could you elaborate on this as a full answer?


text:
Why does sensitivity of nmr spectrum improve with increased magnetic field strength
I am struggling to find an equation that illustrates this and a qualitative explanation to go along with it. I presume that the noise remains at the same intensity and somehow the intensity of the genuine signal increases in intensity thus improving the signal to noise ratio.

Homework! ;) The increase is in first approximation quadratic. Hint: One factor has a reason that starts with "B" and ends in "mann".

The classic paper (>1000 citations): D. I. Hoult, R. E. Richards: The signal-to-noise ratio of the nuclear magnetic resonance experiment In addition to what Karl has hinted at, I think that the actual voltage in the detection coil is also proportional to the Larmor frequency $\omega_0 = -\gamma B_0$ (Faraday's law), so increasing the static magnetic field $B_0$ improves this too.

@Karl Ah I think I have it: bigger applied magnetic field leads to bigger energy difference between spin states, hence bigger populations differences, which by the Boltzmann distribution leads to more transitions, hence bigger intensity peaks. Is this right?

Jup. The other effect is that the induced voltage is proportional to the derivative of the (oscillating, nuclear) magnetic field. Higher frequency, stronger induction. Then you have to take into account the effect on the noise level etc, and it becomes more complicated. See the article by Hould orthocresol linked.

I'm voting to keep this open because of the good discussion in the comments.


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Contradictions in acidity and solubility of phenol derivatives
o-Nitrophenol is less acidic in comparison to p-nitrophenol due to intramolecular hydrogen bonding. This effect also is the reason for o-nitrophenol being less soluble in water.
At the same time, hydroquinone (para) is less acidic than catechol (ortho).
Of course there is the inductive effect, but a reason mentioned quite often as well is hydrogen bonding. Isn't it contradictional that the reason given for o-nitrophenol being less acidic than p-nitrophenol is the same reason given for catechol being more acidic than hydroquinone?

In the case of o-nitrophenol, the acidic hydrogen is hydrogen-bonded to the nitro group's oxygen atom, making it less acidic. In the case of catechol, one acidic hydrogen is hydrogen-bonded to the adjacent OH group's oxygen atom, but the other, more acidic hydrogen is not hydrogen bonded. This hydrogen is more acidic than that of hydroquinone. The resulting anion is even more strongly hydrogen-bonded than neutral catechol, stabilizing it even further. It is the second acidic hydrogen of catechol that is analogous to the acidic hydrogen of o-nitrophenol.

So the oxygen in the adjacent OH group in catechol is also involved in hydrogen bonding with the first OH group, thus weakening its own O–H bond?

That's that basic idea. But that OH bond is not necessarily weaker (relative to homolysis), but rather more polarized and hence tending more to ionize.


text:
Order of nucleophilicity of Nitrogen atoms in adenine
I recently stumbled upon a question that got me thinking. It goes something like this:

Find the major product of the following reaction:

There were 3 options given:

So it basically aims at finding out the most suitable $\ce{N}$ to attack the $\ce{CH3I}$.
i.e. The most nucleophilic $\ce{N}$ in adenine. At first, I felt that should be the $\mathrm{sp^3}$ hybridized amine group(7). But then I had to consider the delocalization of its lone pair in the ring. Then it struck me that I was forgetting about the base, which made me think about the possibility of the deprotonation of the $\ce{N}$(9) to make the 5-membered ring aromatic, which is in fact a very stabilizing phenomenon. Thus the electron density on it should be higher making it more nucleophilic. So I guessed the answer to be the first product given in this list.
I looked up in the internet for sources, but of no avail. I was just searching on this platform for further verification and I found a similar question regarding order of basicity of $\ce{N}$
in adenine.
where @DavePhD pointed out that the $\ce{N}$(1) is the most basic one followed by 3 and 7. Thus suggesting probably option 3 in the list. Though I do know that basicity and nucleophilicity aren't simply the same, but still I cant just rule out the chance of me being wrong here.
So my question is whether or not my deductions are correct and if not, then I would be very obliged to get a detailed know-how about all the possible factors influencing the formation of major product in this reaction.
Acknowledgement:
ChemDoodle 2D sketcher for helping me to draw the diagrams of organic compounds mentioned in this question.

Each answer is missing a double bond. Would you please correct it?

I go ahead and corrected the structure. And also, $\ce{N}$7 is not $\mathrm{sp^3}$-hybridized here.

Oh I'm very much sorry. Actually the double bond wasn't there in the options though , it was definitely a typo. Thanks for the edit. And yeah why don't you call the $\ce{-NH2 sp3}$ ? Is it because its delocalized? Also could you provide any insight into the reaction as such?


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What is the logic behind the terminology lyophilic and lyophobic colloids?
Its given in my text book that-

Colloidal dispersions of starch,rubber,proteins etc are lyophilic colloids whereas colloidal dispersions of metals,sulphides etc are lyophobic in nature.

But even the protein molecules have both hydrophilic as well as hydrophobic ends. And metals being ionised have strong affinity for polar dispersion medium. They why is it given that metals sols are lyphobic whereas sols generated from macromolecules are  lyophilic colloids.
Is it because we can create protein or other macromolecular sols just by simply shaking in dispersions medium and generation of metal sols require certain chemical and mechanical means.
What is the real definition for lyophilic colloids and lyophobic colloids?

The ease of preparation of such macromolecular sols over metal sols is virtue of being lyophilic. Lyophilic ('solvent loving') colloids are the colloidal systems in which the dispersed phase is strongly bonded to the dispersion medium,they are extensively hydrated due to the presence of polar groups. Unlike lyophobic ('solvent hating')colloids where the electrical charges of the solid particles of the dispersed phase and that of dispersion medium establish repulsion forces, which help to remain away from each other in the colloidal system. Thus destabilizing it and making it hard to prepare.

@SirArthur7, this might be a good answer, rather than a comment.

@DrMoishePippik Yeah I think you're right. The thing is I just knew it superficially and that too was taught to me long before, so I wasn't totally sure and was looking for a 2nd opinion. I know the comment went too long, sure would post an answer soon.