text:
Why is the momentum operator imaginary?
I'm a beginner in quantum mechanics, and I'm pretty much confused with the momentum of electron in hydrogen atom.
The wave function of the electron in 1s orbital of hydrogen atom is $\Psi_{1s} = \frac{1}{{a_0}^{\frac{3}{2}}\sqrt\pi}.e^{\frac{-r}{a_0}}$, $a_0$ is the Bohr radius.
The momentum operator is $\hat p = -i \hbar \nabla $. The momentum of the electron can be found using the equation, $\hat p \Psi = \bar p \Psi$.
Which yields $$\bar p = \frac{i \hbar}{ra_0} (x \hat i + y \hat j + z \hat k)$$which is imaginary, which doesn't make any sense to me.
My questions are:
Is the momentum really imaginary? If yes, is it because of the imaginary momentum operator? If no, what is the correct way to find the momentum?

[The expectation value of the momentum (and the position) is zero - not what you have calculated, which is sort of like half the integral you need but not in spherical polar coordinates. It also has to be an observable, hence not imaginary.] Why the operator is imaginary is a result of (at least) Bohr's quantization constraint and the uncertainly relation in quantum mechanics - you can accept this as faith or work through the math. I highly recommend Part II, Chapter 9 of David Bohm's ${\it Quantum\ Theory}$, Dover (1989).

The Wikipedia page also gives a simple rationale in terms of a generalized plane wave.

Why is the momentum operator imaginary?

The simplest explanation hinges on the fact that observables are represented by Hermitian operators in quantum mechanics. Once we accept this, then we can show that the momentum operator $\hat{p} = -\mathrm{i}\hbar\nabla$ is Hermitian precisely because of the factor of $\mathrm{i}$. We need to show that for any two states $f$ and $g$, $\langle f | \hat{p} | g\rangle = \langle g | \hat{p} | f\rangle^*$. For simplicity we work in one dimension, such that $\hat{p} = -\mathrm{i}\hbar(\mathrm{d}/\mathrm{d}x)$. The proof is taken from Binney and Skinner's The Physics of Quantum Mechanics (p 36), although truthfully it is quite straightforward, the main idea being integration by parts:
$$\begin{align}
\langle f | \hat{p} | g\rangle &= -\mathrm{i}\hbar \int_{-\infty}^{\infty} f^* \frac{\mathrm{d}g}{\mathrm{d}x} \,\mathrm{d}x \\[8pt]
                               &= -\mathrm{i}\hbar \left\{ \left[f^*g\right]_{-\infty}^{\infty} - \int_{-\infty}^{\infty} g \frac{\mathrm{d}f^*}{\mathrm{d}x} \,\mathrm{d}x \right\} \\[8pt]
                               &= \mathrm{i}\hbar \int_{-\infty}^{\infty} g \frac{\mathrm{d}f^*}{\mathrm{d}x} \,\mathrm{d}x \\[8pt]
                               &= \left( -\mathrm{i}\hbar \int_{-\infty}^{\infty} g^* \frac{\mathrm{d}f}{\mathrm{d}x} \,\mathrm{d}x \right)^* \\[8pt]
                               &= \langle g | \hat{p} | f \rangle ^*
\end{align}$$
(Well-behaved functions vanish as $x \to \pm \infty$, hence the term $[f^*g]_{-\infty}^{\infty}$ goes to zero.) In a similar way, it can be shown that without the the operator $\mathrm{d}/\mathrm{d}x$ itself is not Hermitian, and thus cannot be an operator corresponding to an observable quantity in QM.

The momentum of the electron can be found using the equation, $\hat p \Psi = \bar p \Psi$

This is an eigenvalue equation for the momentum, but it does not actually give you the value of the momentum. When you apply an operator to a state, you get out another state, and there's nothing that forbids this second state from being complex / imaginary. It's only the final expectation value at the end that has to be real, and this expectation value is given by $\langle \psi | \hat{p} | \psi \rangle$.

Is the momentum really imaginary?

It turns out that the factor of $\mathrm{i}$ doesn't matter, because if you calculate this correctly, then the momentum turns out to be zero. This makes sense, because all directions in the hydrogen atom are equal, and it can't be that the electron has a net momentum in some direction.
Calculating it correctly, however, is a slightly tricker topic than expected. In this case, the momentum operator $\hat{p}$ is a vector operator ($\equiv (\hat{p}_x, \hat{p}_y, \hat{p}_z)$), and consequently when you integrate it you need to be very careful to not mix up vector and scalar quantities. If you wish to do this in spherical coordinates, I first suggest going to Physics Stack Exchange and searching for questions about "momentum of hydrogen atom". You will find lots of similar questions asking about the factor of $\mathrm{i}$ and how it doesn't make sense.
Personally, I think a more direct, and slightly easier approach, is to simply integrate one (Cartesian) component of the vector at a time. Thus, we can try to evaluate $\langle \psi | \hat{p}_x | \psi \rangle$. For simplicity, I will drop the normalisation constant and use $\psi = \exp{(-r/a_0)}$. Also, note that
$$\begin{align}
\frac{\mathrm{d}r}{\mathrm{d}x} &= \frac{\mathrm{d}}{\mathrm{d}x} (x^2 + y^2 + z^2)^{1/2} \\
&= \frac{1}{2}(x^2 + y^2 + z^2)^{-1/2} \cdot (2x) \\
&= \frac{x}{r}
\end{align}$$
Thus, we have that
$$\begin{align}
\langle \psi | \hat{p}_x | \psi \rangle &= -\mathrm{i}\hbar \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \exp\left(-\frac{r}{a_0}\right) \frac{\mathrm{d}}{\mathrm{d}x}\left[\exp\left(-\frac{r}{a_0}\right)\right] \,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z \\[8pt]
                                        &= -\mathrm{i}\hbar \iiint \exp\left(-\frac{r}{a_0}\right) \frac{\mathrm{d}}{\mathrm{d}r}\left[\exp\left(-\frac{r}{a_0}\right)\right]\frac{\mathrm{d}r}{\mathrm{d}x} \,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z \\[8pt]
                                        &= -\mathrm{i}\hbar \iiint \exp\left(-\frac{r}{a_0}\right) \left(-\frac{1}{a_0}\right)\exp\left(-\frac{r}{a_0}\right)\frac{x}{r} \,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z \\[8pt]
                                        &= \frac{\mathrm{i}\hbar}{a_0} \iiint \exp\left(-\frac{2r}{a_0}\right) \frac{x}{r} \,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z \\[8pt]
\end{align}$$
At this point, it's easiest to bring in symmetry arguments. Notice that $r = \sqrt{x^2 + y^2 + z^2}$ is an even function of $x$, $y$, and $z$. Thus, the integrand is odd with respect to $x$ (because of the extra factor of $x$), but even with respect to $y$ and $z$.* The entire integral therefore goes to zero.
The integrals for $\hat{p}_y$ and $\hat{p}_z$ proceed in exactly the same way.

* Although it also diverges at the origin. I'd be very glad if someone could tell me how to rigorously resolve this.

When you say "simplest" I'm not sure I agree!

@matt_black, the least complicated? :-)

It doesn't give any intuition about the "why" and requires heavy math (which might be the best rigorous answer but doesn't help those without the math knowledge). I might try a simpler answer with an analogy.

@matt_black, I don't really think integration by parts is all that heavy: I expect that most people studying QM should have seen it before. However, I agree it's not intuitive. I've never seen this explained without the maths before, though, so I'd be pretty keen to see what you have.

"I'd be very glad if someone could tell me how to rigorously resolve this." - Spherical coordinates?

@Antimon Thanks - yes, spherical coordinates would simplify it, but somehow we still have to exclude the point $r = 0$. My idea was to take a limit like $\lim_{a \to 0^+, b \to \infty} \int_a^b \ldots ,\mathrm{d}r$, but I'm really not all that good at maths, so have no clue whether that's correct (or even necessary)!

No, in spherical coordinates the integral is actually finite, thanks to the $r^2$ factor in the volume integral.

Because periodic waves can be represented using exponential functions using imaginary arguments
I'm going to attempt a non-rigorous explanation that might give some intuition as to why imaginary numbers appear in the mathematics of quantum stuff. Most of this doesn't require a knowledge of what a Hermitian operator is or about vector spaces or eigenvalues. You need to understand these if you want to really want to be a quantum mechanic, but the basic intuition can come from a single very famous math equation and the much simpler math used to study periodic signals in electronic circuits.
The first thing you need to understand is that electron orbitals are described by wave functions. This is an essential foundation of QM derived from de Broglie's insight that particles can be described as waves. This implies that solutions for possible orbitals of an electron orbiting a hydrogen atom must be quantised (like the possible vibrations of a violin string): waves that don't meet certain strict requirements will cancel out just as only certain waves on a vibrating string are possible. I'm simplifying a lot (not least because orbitals are 3D unlike violin strings and the "waves" are spherical harmonics (don't ask, but you will be familiar with the pictures of possible orbital shapes)).
If you accept this highly simplified picture (and given how fundamental it is to everything quantum related you really have to) the question arises what sort of math can be used to describe the wave functions?
And here we have to look at one of the most famous equations in mathematics: Euler's function:
$e^{i {\theta}} = \cos(\theta) + i \sin(\theta) $
The point here is that a periodic function can be described by an exponential function with an imaginary exponent (one containing i or the $ \sqrt {-1} $ ). And using this identity makes manipulating the math of periodic functions a great deal easier. This insight is not confined to quantum chemistry, it is essential to electrical engineering where it provides a compact way to describe the behaviour of electrical circuits involving anything more complicated than pure resistance when AC is involved. In most circuits the voltage and current vary over time in a periodic way. But the voltage and current are usually not in phase because of the inductance and capacitance of the circuit. The exponential captures the nuances of this periodic behaviour in a simple mathematical way.
Ultimately the exponential form of the math of electrical circuits captures the periodic solutions to differential equations describing the components of the circuit and the result is periodic functions describing voltage and current.
Chemically relevant wavefunctions are far more complex (not least they are 3D as I mentioned before). But they represent valid solutions to the differential equations describing the electron wave function and those solutions are periodic waves in 3D space.
So the basic leap is that, once you have accepted electrons are waves, the math describing them will involve periodic functions and the best way to describe those is exponential functions with imaginary arguments because, Euler's formula.
After that leap, QM gets complicated and you start having to grasp concepts like Hilbert spaces, Hermitian operators and Eigenvalues. But the essential leap is to realise the implications of Euler's formula: wave behaviour can be captured by exponential functions with imaginary inputs.

I'm afraid I don't really follow. You've suggested why QM as an entire field requires complex numbers; but that doesn't necessarily imply that the momentum operator $\hat{p}$ must have a factor of i in it, which is what OP's specific question is about (or at least, that's the way I understood it).

@orthocresol I'm not trying to be rigorous, just to give an intuition as to why anything with wave equations is going to involve complex numbers. Your argument is that the math doesn't work if you don't use i. My argument is a very general illustration illustrating the intuition that i is required to give wavefunction like solutions when solving differential equations. QM is a complex example of this but it all comes back to Euler: if you want waves, you need factors of i in the math.

Hmm, but you can get wavefunctions even if the operator itself is real: the Hamiltonian, for example, is real and you can solve $\hat{H}\psi = E\psi$ to get a set of wavefunctions (e.g. the orbitals). Imho, the solution to this has to somehow focus on momentum specifically, not wavefunctions in general. Perhaps, the link I'm looking for is that the eigenfunctions of momentum are plane waves, which (as your answer explains) are necessarily complex (or formally $\exp(-\mathrm{i}\vec{k}\cdot\vec{x})$), explaining the factor of i in the momentum operator. Either way, thanks for posting!

@orthocresol I'm not trying to get into the deeper quantum explanation of why those specific momentum equations have to have an i. I'm just trying to create a very simple illustration (eg the comparison with much simpler electrical signals) where the math is much simpler than QM but where i makes wave solutions work.

Yeah, I do get what you're trying to say. I just don't think that was the question OP was asking. But it's OK; I've posted what I think is an answer, so that's enough from me! I don't want to be, or come off as, confrontational here.


text:
When does an aromatic ether break its bonds?
I was reading some reactions of ethers and I came across Zeisel determination. Later that day, I was asked to determine the products of the reaction :

And I made the product:

This reaction is standard Kucherov Reaction
Now my doubt is that ether linkage can get hydrolyzed under acidic conditions, so why can't the below product be formed?

Zeisel's method follows this mechanism, so then why can't it happen in this case?

Does this need a very strongly acidic condition? like that provided by HI

Strongly acidic and a good nucleophile like $\ce{I-}$.

Aromatic methyl ethers are very robust and need specialist condtions to cleave them. 48% HBr in refluxing AcOH for example.

@Waylander ok so they are extremly non reactive this means right? Only poweful acids can do any harm to them right?

@Damstridium not just strong acid but as Mithoron says with a good nucleophile. Strong Lewis acids such as AlCl3 or BBr3  will do it

@Damstridium The Zeisel reaction as described here reads like submitting the substrate to refluxing $\ce{HI}$ with HOAc. This seems much more aggressive than the oxymercuration conditions in the first scheme, where you need a bit of $\ce{H+}$ to transfer the intermediate enol into the more stable ketone by keto-enol tautomerism. (Cf. with an OrgSyn.) The ketone likely would not survive the conditions to cleave off this methyl ether.

@Waylander Ok so ,adding a good lewis acid can also help. Is this because oxgen of ether will donate its lone pair to this good lewis base(say $AlCl_3$)  and will pull in the electrons bonded with any aliphatic hydrocarbon (as strong neucleophiles must be used here so I think $SN_2$ is likely to occure, hence primary and secondary aliphatic hydrocarbon chain is desirable here I think. ) leaving the neucleophile to attack via sn2. Am I thinking this process correctly?

@Buttonwood Ok yes I had an intutuion of this in my spine ,while I was reading the very first comments on this question. I am thankful  for clarification of this and Please submit this comment as an answer and I will accept it

@Buttonwood  the ketone will do a neucleophilic addition reaction with HI I think.UNder these conditions right??

@Damstridium  Pretty much yes. There is an explanation of the process in this paper https://chemistry-europe.onlinelibrary.wiley.com/doi/10.1002/ejoc.201501042

The Zeisel reaction as described here reads like submitting the substrate to refluxing $\ce{HI}$ with $\ce{HOAc}$. This seems much more aggressive than the oxymercuration conditions in the first scheme, where you need a bit of $\ce{H+}$ to transfer the intermediate enol into the more stable ketone by keto-enol tautomerism. Compare e.g., with an entry on Organic Synthesis:

(credit to Org. Synth. 1973, 53, 94)
which, in the case of acetylenic bonds looks like this:

(credit to Wikipedia)
Thus, in your case:

My anticipation (though speculation) is, the reaction conditions to cleave off the methyl ether easily knock-down the enolizable ketone.

Hydration of the triple bond is opposite that of the original example, which is the likely site of hydration. Mercuric ion hydration of triple bonds is catalytic, not stoichiometric. NaBH4 is not needed. Hydration of double bonds is stoichiometric and requires NaBH4 reduction. NaBH4 here would reduce any ketone present.

@user55119 Thanks for spotting that the other side was used. But based on the protocol by Org. Syn., are $\ce{C#C}$ triple bonds so much more susceptible to oxymercuration than $\ce{C=C}$ double bonds? I agree for the point of reduction of the ketone by $\ce{NaBH4}$.

The story I heard, perhaps apocryphal, was that the hydration reaction was discovered by the accidental breakage of a mercury thermometer while stirring a vessel containing an alkyne and aqueous sulfuric acid lending credence to the catalytic nature of the reaction. BTW: A gold catalyst shows 100% regioselectivity for this substrate in favor of the anisole end. [DOI:10.1039/d1cc01837j ]

Teflon, the tube that was clogged; post-it, the super-glue that did not stick; penicillin, the Petri dish left on the desk; saccharin, the sweet taint on Fahlberg's coat when performing a coal analysis; LSD and Hofman's famous bicycle ride in Basle; nitroglycerine allegedly administered to Alfred Nobel (not as explosive, but to treat an angina); there are plenty finds not anticipated.  «You can plan research, but not the results of research.» (Dieter Seebach in Angewandte Chemie, doi 10.1002/anie.199013201)

Demethylation of phenolic ethers generally requires some pretty extreme conditions. In verifying the structure of vanillin Tiemann and Haarmann (page 620) demethylated it successfully with dilute hydrochloric acid at up to 200°C. To illustrate the "usual" extreme conditions for demethylation, the authors first tried fuming hydroiodic acid at up to 130°C, but that gave an "amorphous, iodine-containing, black" decomposition product.
More recent work by Bomon et al. successfully used catalytic dilute HCl in an autoclave at up to 250°C to demethylate various phenolic ethers. The authors also give a very good review of demethylation reagents/techniques for comparison.

Tiemann and Haarmann, Ber., 7, 608 (1874)
Bomon et al., Green Chem., 23, 1995 (2021)

Funny how little has changed in those almost 150 years!

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Reduction of orthoesters and acetals with LiAlH4
As I know, orthoesters react with $\ce{LiAlH4}$. After searching on the Internet I’ve found that acetals react with $\ce{LiAlH4}$ in the presence of $\ce{AlCl3}$. Why can’t acetals be reduced only in the presence of $\ce{LiAlH4}$, like orthoesters? Why is Lewis acid required for reduction of acetals, but for orthoesters is not? Is it caused by more δ+ on orthoester or there’re other reasons?

Note, chemical information may be advantageously formatted using on ChemSE with mhchem. Take moment to familiarize with this. You are encouraged to use it in the body of questions, answers, and comments. Because it is something special not all web browsers understand well, do not use it in the title of questions or answers.

Where is the polarization of the $\ce{C=O}$ double bond in an acetal, where $\ce{LiAlH4}$, a $\sigma$-nucleophile, may act on?  What would be a good leaving group when they approach for a nucleophilic substitution on an acetal?

I’m not sure, but I supposed that leaving group, mechanism may be the same as in reduction of orthoesters. What’s the difference between reduction of acetals and orthoesters with LiAlH4, due to which orthoesters react, but acetals don’t?

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Bond frequency vs Bond Strength in Kinetic Isotope Effect
I am confused by the following: in Ansyln's Modern Physical Organic Chemistry, a higher vibrational frequency defines a stronger bond due to the deeper/sharper potential well (page 76). However, the kinetic isotope effect is explained as being due to the lower frequency of C-D bonds in comparison to C-H, resulting in a lower zero point energy and subsequently a higher bond dissociation energy (page 423). These two concepts seem at odds with one another, and I was wondering what factor I am not considering.

The first statement is false. Compare for instance the dissociation energies and vibrational frequencies of $\ce{H_2}$ and $\ce{N_2}$. The vibrational frequency is determined both by the steepness of the potential and the reduced mass of the atoms. Note that the potential for different isotopologues is the same within the Born-Oppenheimer approximation.

Since a higher vibrational frequency leads to a higher ZPE, how can I rationalize the fact that a stronger bond (C-C vs C=C for instance) has a higher ZPE, yet the ZPE of C-D is lower than C-H (and the bond is still stronger)?

Increasing the the force constant for a given pair of masses increases the frequency.  [That's what the first statement refers to.]  Increasing the mass for a given force constant decreases the frequency.  [That's the kinetic isotope effect you describe—or, more precisely, there you are significantly increasing one of the masses while leaving the force constant about the same.]  See equation here (pictured at right) relating masses, force constant, and frequency: https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/infrared/irspec1.htm

Sure, but since the energy of the vibrational level is proportional to the vibrational frequency, why can one state that the lower zero point energy of C-D vs C-H leads to greater energy needed to break that bond while at the same time a stronger bond (C=C vs C-C) has a higher zero point energy yet still needs more energy to break its bond?

The experimentally relevant dissociation energy is $D_0$, that is $D_0=D_e-\tfrac{1}{2}\omega_0\hbar$. A lower ZPE thus gives a larger dissociation energy.

So in the case of C=C vs C-C, C=C may have a higher ZPE than C-C, but De is larger as well, so that it overall has a higher dissociation energy.


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Literature recommendation - non-stationary heat transfer in porous solids
I am trying to calculate a non-stationary heat transfer to a porous spheres by conduction.
These spheres are placed in a thin metal tubes with a known heat capacity. Hot/cold air (or similar gaseous mixture) flows around the tubes to either cool or heat them.
I am particularly interested in calculating the time it takes for a defined temperature to appear in the center of these sphere. The material of these spheres has homogenous composition.
Do you have any tips on the literature that would describe this problem in a simplified way or better yet, provided concrete problems solutions?

What about a combination of «chemical engineering» with «heat transfer» and «sphere»? In one of the search engines I use, Faith Morrison, Michigan Tech appears again and again with lecture notes like this pdf; or engineeringnotes.com. I didn't assume this could be done with Excel, but this may depend on the level of detail you aim for.

@Buttonwood Thank you. Honestly, I went through library and looked to a well-known books of Bejan - Convection Heat Transfer and Thermodynamics by McGraw-Hill, which did not turn out to be an ideal sources. The presentation is especially great! I did not expect analytical solution, so this is more than fine. Actually, I need an accurate estimation, as my goal is to fasten the heat transfer by getting a dependance on the particle size, heat transfer properties including the metal tube material and the flowing medium (temperature, velocity).

With your addition of «dependance on the particle size, heat transfer properties including the metal tube material and the flowing medium (temperature, velocity)», you tap more into chemical/civil engineering (heat exchangers, flow reactors with catalyst beads, etc.) since each of them (assuming a fixed incoming flux of heat) may affect at least the temperature (incl. locally along the direction of flux). So you need to parameterize the approach and to define if the process is going to be run under isothermic/stationary or adiabatic conditions (are tubes thermally insulated, or not).

Yes, my specification was rather poor in the post. The first calculation will be done for constant conditions (size, packing density, physical parameters of the materials  - density, conductivities, heat flux (air highly concentrated with CO$_2$ at 393K) and its volumetric flow / velocity). After I get results, I would like to make sort what you have described - a parameters optimization (maintaining values of available metals / particles (shapes, porosities)) and find optimal scenario for which the heating / cooling takes the least amount of time while maintaining the process economy.


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DLVO Theory of Colloid Stability
DLVO theory gives the curve of potential energy vs distance of two colloid particles. Potential energy curve is derived for colloids being only electrostatically stabilized and not sterically.
Looking at the image which shows potential energy curve, we can see two local minima and one maximum. Between local maximum and secondary minimum is a domain of repulsive forces since potential energy increases by making distance between particles smaller.
Theory predicts that if colloid particles have enough energy (usually energy of Brownian motion), they can overcome repulsive forces and bond at primary  minimum which causes flocculation and colloid instability.
If they don't have enough energy colloid should be stable according to this theory, however this doesn't make sense since particles can still bond at secondary minimum. Bonding at secondary minimum forms much weaker bond which can be seen on the curve, but it should still cause flocculation since particles arrive at local minimum.
If this is so, how can this theory predict colloid stability?

I had read about the DLVO theory and colloids a long time ago, so in case I make a mistake, feel free to correct me.
The secondary minimum is not deep enough to hold the particles for a long enough time to cause flocculation. Basically, in room temperature, colloid particles should have enough energy to get out of that secondary minimum and drift away from each other (i.e. to the right of the curve you have drawn).  However, they should not have enough energy to cross the barrier on the left (so it cannot go to the primary minimum). When two particles collide due to thermal motion, they cannot cross the barrier on the left, so they rebound away from each other (i.e. move to the right in the graph).
Only at a very low temperature would you see enough particles arrive at the secondary minimum at the same time so as to cause some kind of coagulation. At higher temperatures, the thermal motion ensures that particles are always traversing that potential energy surface.
Also notice that at the secondary minimum, there is a considerable distance between the two particles. So even if you could get all the particles to stay in that shallow minimum at the same time, I am not sure it can be called coagulation.

Yes, that is correct, it came to me later.


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Is the activation energy higher in incomplete combustion or in complete combustion?
I am looking into high school level chemistry where it is asked if the incomplete combustion of methane into carbon monoxide has a higher activation energy then its complete combustion.
My reasoning is somewhat empirical, if there is enough oxygen, complete combustion is preferred. From which I think I can deduce that the activation energy of incomplete combustion must be higher. But is this correct? And even so does this reasoning constitute a general heuristic for activation energy in complete vs incomplete combustion?


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Aluminum Alloy with most alkaline resistance
Previously i asked this question in Engineering SE: What aluminum grade is the most resistant to corrosion in alkaline environment?
Some input from others seem indicating that no mass-produced aluminum grade suitable for alkaline aqueous environment (sodium hydroxide). So I was thinking if there is a way to produce it?
Any suggestion of what might be added to pure aluminum to make it highly resistant to corrosion in alkaline?
Other parameter (strength, workability, etc) is not significant thus can be neglected

I can't say for sure this is impossible but it is going to be hard (and fairly certainly not worth the cost). Al is very reactive to alkalis like NaOH as is the normal oxide coating that protects Al from other forms of corrosion. There are even well known reactions where Al is dissolved out of other alloys by using NaOH!

Here is an idea based on the concept of a sacrificial anode. Per Wikipedia, to quote:

A galvanic anode, or sacrificial anode, is the main component of a galvanic cathodic protection (CP) system used to protect buried or submerged metal structures from corrosion.They are made from a metal alloy with a more "active" voltage (more negative reduction potential / more positive electrochemical potential) than the metal of the structure. The difference in potential between the two metals means that the galvanic anode corrodes, so that the anode material is consumed in preference to the structure.

So perhaps, experiment with adding to the aluminum alloy, or introducing surface strips of pure Zinc, which is more anodic than Aluminum, as noted in this source. Now, Zn must all be removed before Al corrodes, and with a large surface area anode of zinc relative to say an incidental (or provided spot) of carbon (serving as the noble cathode), the galvanic reaction, at least, will be slow.
Related concept: Galvanizing Steel, which per a source, to quote:

Galvanizing, or galvanization, is a manufacturing process where a coating of zinc is applied to steel or iron to offer protection and prevent rusting. There are several galvanizing processes available, but the most commonly offered and used method is called hot-dip galvanizing.

Galvanized aluminum?

Thanks Ed V, Galvanizing aluminum helps protect metal from oxidation and corrosion. ... Cold galvanizing aluminum surfaces with zinc fusion provides effective metal protection in extreme temperatures and environments containing high concentrations of corrosive elements at a fraction of the cost for hot galvanizing.Apr 24, 2017 here https://sciencing.com/galvanize-aluminum-8664634.html

But why not just use one of the many zinc alloys that are available now? Nickel-plated aluminum is common in some applications, e.g., electrodes for excimer lasers, and nickel-plated zinc is commonly used for fasteners and such at my local hardware store. If cost is not a major factor, titanium is an option.

Nickel is at the wrong end of the anodic index (see https://www.zygology.com/cms/upload_area/pdf/Zyg-Anodic-Index.pdf).

But nickel would not be acting as sacrificial anode: it would be like the tin on a steel can, not like zinc on a galvanized trashcan.


text:
Troubleshooting an AutoDock Vina error on Chimera: “Could not find an atomic number for Hn Hn”
After minimizing a ligand (prepared with MarvinSketch) and preparing a target enzyme to dock said ligand onto (collected from RCSP PDB — 5DGN (Crystal structure of human FPPS in complex with compound 13)), I was met with an error
ValueError: Could not find atomic number for Hn Hn
cannot prepare ligand for AutoDock Vina; please look in Reply Log and/or run Chimera with --…

As a new user to the Chimera software, I neither knew what Hn Hn meant or how to troubleshoot this issue. I can also attach the mol files for the minimized ligand and prepared target enzyme if necessary to help with troubleshooting.

Not sure about Chimera specifically, but a google search for Hn turns up either 'Azanylidene' or maybe the old name for Hassium (Hs) which was "hahnium" (Hn)?

@Thunderbolt114 - It's not the PDB file, that's for sure. Can you do grep -A 5 -B 5 'Hn' minimized_file and report what it says? It's going to turn out to be something involving a missing hydrogen added to a nitrogen is my guess, but let's have a look.

Is that a command I input into the software? If so, how can I input it? I'm a bit confused.

It's a command for a Linux or UNIX shell searching for Hn and printing five lines before and after. Pretty much like the search box in a Texteditor, just not worth the unnecessary graphical interface.

I'd also assume that questions like this have a bigger audience at [mattermodeling.se]. Please be aware that cross posting is discouraged.

@Mas Again, grep is covered by one of the tutorials of software carpentry.

I have the same error for thyroxine (T44.mol2). Definitely two hydrogens on the amine nitrogen. From the mol2 file:
     31 H6'        20.4230   40.7840   26.1950 H         1 T44    0.0000
     32 HN81       21.6870   45.6570   36.2000 H         1 T44    0.0000
     33 HN82       23.1470   46.0670   35.4150 H         1 T44    0.0000
     34 HO4'       22.9030   42.4090   22.4520 H         1 T44    0.0000

I presume these are the problem Hn Hn. Another mol2 file, triiodothyronine, has the same lines, but doesn't have the error!
The problem is definitely the hydrogens on the nitrogen. However, it is with the writing of the *.ligand.pdb file. If, as above, the original .mol2 (.pdb) file has HN81 and HN82 and HO4' all three will cause errors. The fix is the clean out the ligand's .mol2 (or .pdf) file and just have the element symbol in columns 9 and 10. Problem fixed. If the element name is 4 characters long, the software writes the "H" in the wrong column, 9 not 10 of the *.ligand.pdb file, to get space.

Ahhh, what a wonderful, modern, flexible format .pdb is ...


text:
Mechanism of this addition-elimination process
In Part 1 of the synthesis of Maoecrystal V (found on synarchive), the last step subjects a $\ce {\alpha, \beta}$-unsaturated ketone to $\ce {I2}$ and pyridine, as shown below:

I am wondering what is the mechanism that takes place for the elimination process. It is unlikely to be an E2 process since that requires anti stereochemistry for the $\ce {H}$ and $\ce {I}$ that are being eliminated. After anti-addition of the $\ce {I2}$ molecule to the double bond, the $\ce {H}$ and $\ce {I}$ atoms would both be syn to each other. Since this is a ketone with acidic $\ce {\alpha}$ hydrogens, a E1cB mechanism is likely operating and thus, we observe the elimination of the $\ce {\alpha}$ hydrogen and the iodine atom on the $\ce {\beta}$ position. Is my hypothesis correct?

This is a reasonable hypothesis. It could also be proceeding by a Baylis-Hillman type halogenation via the pyridinium enolate.

Yeah, it's rather pyridine being eliminated.

@Mithoron Well... E1cB or Baylis-Hillman, both are just equally likely hypotheses


text:
Cu-Zn Ice tray battery: 2 possible cathodes?: the piece of Copper OR the liquid in which is plunged the Zinc nail?
This is the ice-tray battery:

Nails are in Zinc, wrapped with Copper wire.
The Copper wire and the Zinc nails are each plunged into white vinegar (8% acidity).
So I see 2 possible ways to consider what is the cathode here:

Electrons are ripped from Zinc. They flow until the Copper wire, where, reaching the liquid in which is plunged the Copper wire, the electrons are given to the H+ of the white vinegar and form hydrogen gas.
So the cathode is the piece of Copper, the anode is the zinc nail.

OR: like it's shown on the picture (one leg of the LED plunged into the liquid in which is plunged the Zinc nail and the other into the liquid in which is plunged the Copper wire).
The liquid in which is plunged the Copper wire loses H+ (which are forming Hydrogen gas): So it's more negative. On the other side, the liquid in which is plunged the Zinc nail gets an accumulation of Zn2+ ions which dissolved in it. So it's more positive.
So the cathode is the liquid in which the Zinc nail is plunged, and the anode the liquid in which the copper wire is plunged.

Is that true?

Liquid keeps its net charge neutrality.

I didn't know that. But there IS current when you plunge one leg of the LED into one liquid and the other leg into the other liquid. So one of the liquid (the one with Zinc nail) is the cathode, right?

No, liquid is not cathode. You need 2 conductors in contact with liquid. One is an anode, where oxidaction occurs, the other, where reduction occurs, is a cathode.

This is basically a lemon battery, with vinegar instead of lemon juice, so the zinc is the anode and copper is the cathode, exactly as @Poutnik said.

Your picture is not clear enough. It looks as if there is an electrical contact between the lower end of the nail and the copper wire coming from the next nail ? Is it so ?

As stated previously, the zinc on the galvanized nails is the anode, i.e., negative electrode. This is a very poor battery in many ways, even including having the LED leads sticking into two vinegar reservoirs instead of contacting two different metal electrodes. You would be much better off with a couple of lemons or limes. Even spuds can work OK.

Consider the LED lead in the electrolyte -- what type of metal is it?

@DrMoishePippik No idea. It's difficult to find info about what LED legs are made of. Some alloy of Aluminium or Silver maybe. But I don't think it's important for the experiment.

@Maurice The LED legs aren't touching the metal (the Copper wire or the Zinc nail). They are only immerged into the well.

@ThePhi. My point is not related to the LED, but to the other compartments, which are all similar and filled by a solution, a nail ans two copper wires, on entering, and one getting out. Is there any physical contact between them ?

@ThePhi, yes, the material is relevant. Just as you put in iron nails and copper wire as electrodes, the LED lead is another metal. Where in the electromotive series is it?

OK, some details, on this galvanic cell construction where, this reference cites the following step, to quote:

Tightly wrap a piece of copper wire five times around a galvanized nail just under the head of the nail.

So, as "galvanized" means Zinc coated, iron is not actually (at least at the start) part of this galvanic cell.
As such, without doubt the highly anodic Zinc is the anode, leaving either copper or air (oxygen) as possible cathodes (or a cathodic zone).
Another instruction:

You’ll create a circuit by placing each nail into one well of vinegar and making sure the copper wire is bent so that it goes into the vinegar in the next well. The wire is an electron bridge from one well into the next. Make sure the wire from one nail does NOT touch the nail or wire in any other well. There should be only vinegar between the metals in each well.

And finally:

Place one “leg” of the LED into the well with only a copper wire inside it and place the other “leg” into the well with only a nail in it. If the bulb lights up, you nailed it! If the bulb doesn’t light up, just lift the bulb out of the well, rotate it a half-turn to move the legs to the opposite well, and lower it back into the vinegar. It lights up!

The provided explanation on the functioning of the cell asserts:

The two metal components are electrodes. They’re the sites in a battery where moving electrons (the current) enter and then leave the battery. With this vinegar, zinc, and copper set-up, the current will flow from the wire bridge (copper), across the vinegar, to the nail (zinc), and then along the bridge to the vinegar and… well, you get the idea. The electricity passes easily through the acidic solution of vinegar in the wells.

So, copper is the anode to answer your question.

In the comment, @EdV and (Poutnik) say the opposite: " You need 2 conductors in contact with liquid. One is an anode, where oxidaction occurs, the other, where reduction occurs, is a cathode. – This is basically a lemon battery, with vinegar instead of lemon juice, so the zinc is the anode and copper is the cathode"

I generally agree, however, the copper wire here is also employed  to connect the electrodes.


text:
Don't understand how to conduct a sugar/water solution experiment
I am helping my some with simple chemistry experiments by following a book on experiments. I don't understand how to conduct one experiment thought (please see pic, experiment 16). It says

dissolve sugar in hot water till saturation point
let the water evaporate.

How will the water evaporate on its own? Will it evaporate on its own even when I am not heating the water?

If you washed the floor, would water evaporate spontaneously or would you need to heat it?

my doubt is that water in beaker would be quite more than the thin layer on the floor. How long it might take to evaporate ?

Very long time and it is adviced to minimize the depth of the solution. Generally, all crystalization procedures take intentionally very long time for crystals to grow large. You may try to pour some volume into a flat plate or  similar, to have the depth just few mm.

The answers to your questions are basically in the description you posted.

How will the water evaporate on it's own... haven't you seen or felt sweat?

just to be clear, I presumed that I need to use a beaker full of water and create a solution. That caused that confusion about the time it would take to evaporate

Evaporation is not the most efficient way of obtaining sugar crystals, because usual sugar is extraordinarily soluble into water. $100$ mL water dissolves $170$ g sugar at $0°$C. And it dissolves $450$ g sugar at $100$°C.
So if you want to get not too tiny sugar crystals, you better try to get them by cooling down the hot saturated solution. As it may take quite a long time to dissolve $450$ g sugar in $100$ mL water at 100°C, I would start to mix $400$ g sugar and $100$ mL water, heat the whole to boiling temperature, until a nice solution is obtained, and simply let it cool down slowly. If the cooling step is fast, you will obtain a lot of tiny crystals, like the usual sugar powder from the market. If the cooling step is slow, individual crystals can find the time to grow slowly.


text:
How can I roughly measure a fraction of a gram?
My son got a crystal making kit (from a relative) and it lists seed crystal weights of 0.5g for a small crystal solution and 0.75 g for a large crystal solution. The trouble is I (his father) don't have a scale that can measure a fraction of gram, so I guessed how much.
Can I roughly get the weight that a fraction of a gram without buying a scale just for this crystal making kit?

I notice that you are in Japan, so it may help to know that the mass of a 1 yen coin is 1 gram. As already suggested, you can make an equal-arm pan balance. For small mass measurements it is important to keep the balance mass small. Use a 20 cm long balsa stick for the beam and sheet balsa squares for the pans. Hang everything with fine silk sewing thread. For small weights, buy a box of tiny washers (M2 or similar) and count the number needed to balance the coin. For even smaller increments, use lengths of fine copper wire.

I might be mistaken, but isn't a 0.5 g seed crystal rather large? Do you have such big crystals already? Or do you add many small crystals that accumulate to 0.5 g? And do you have a scale with an accuracy of, let's say, 1 g?

What is the material to crystallize?  Is it copper sulfate, or alum?

The seeds are various minerals and the core solution is Alum I think.

@user2617804 Well if so, the OP might be interested in dyeing crystals.  This may lead to crystals where colour is all across the body, or just in sectors.  It isn't just a curiosity, it provides insight into mechanism of crystal growth (e.g., Bart Kahr group, elder Chem. Rev. review with many examples about different dyes, a newer CrystEngCom review, one of IUCr's public instructions).

While there are some great and creative answers here, reasonable mg scales are really cheap these days.  I can see $35 being a big expense for some just the same, but if it's not then it's actually a really handy thing to have around.  You might find you end up using it more than you expected once you have one.

How do I control for air movements- we have electronic scale at work and they move by 0.2 g easily.

Are you sure you've read the measuring instructions correctly? Most crystal growing kits for kids just measure the various dry chemicals volumetrically (e.g. "one small measuring scoop of powder A"). It's not that precise, but fortunately it needn't be.

@user2617804 Multiple approaches one may combine (not an exhaustive list): 0) Keep your balance clean, let it be checked and calibrated. 1) Place the balance on a upright vibration-less balance table e.g., with a board of granite. 2) Reduce movement of air around it, shut doors & windows to the room where you use it. Some groups use dedicated balance rooms, some only stick to the housing analytical balances offer. 3) Ensure balance and the items to weigh are at the same temperature.

You ask:

Can I roughly get the weight that a fraction of a gram without buying a scale just for this crystal making kit?

Absolutely.
You're in Japan. You have access to Japanese coins which conveniently are just what you need for this project. Because they are currency, you can also be assured that they weigh exactly as advertised.

You need one each of 5-yen, 10-yen, and 50-yen coins

For the scale, make something like this balance, and note that you might want to scale it down given the small amounts you're working with

To get 0.5 g, add enough crystal to the cup with a 50-yen coin in it such that it's balanced with the one with a 10-yen coin in it

To get 0.75 g, add enough crystal to the cup with a 5-yen coin in it such that it's balanced with the one with a 10-yen coin in it

I suggest you tape the coins to the weighing cups so you don't have to get the powder all over them. You might tape the coins on the underside of the cups and be sure to use a similar size section of tape so the weight of it is more or less equal.

Our 3 lightest coins (1, 2, 5 Czech crowns)  have $\pu{3.6 \pm 0.15 g}$, $\ce{3.7 \pm 0.15 g}$, $\pu{4.8 \pm 0.15 g}$. I guess coin mass differences may have generally  rather high relative SD.

Repeated measurements will take care of minimizing the sum total SD of weights. The Japanese Mint doesn't indicate differences, so I'm assuming they are minimal.

Hm, scientists usually consider data with explicit uncertainty or tolerance as more reliable.

Given the project in question, "rough" estimate asked for by the OP, and my general lack of knowledge of coinage and minting processes in Japan and other countries, I'm willing to let this one slide ;)

A scaled-down version of that balance could be made from Lego, mainly Technic, and fishing line.  Lego tolerances are impressively tight (though some bricks have changed mould over the years - use the same age bricks) so zeroing the empty scale may not even be needed. Cross-axle parts can be used to provide a continuously-variable adjustment if necessary.

I think the mass of the seed is not critical. But you can create your own scale - or even better, involve the son in its creation.
You can create a 2-shoulder scale, with 2 stuff holders hanging at equal distances from the central holding.
Or, you can create a 1-shoulder scale, with the stuff holder hanging on some elastic string or wire.
For a scale weight for the former, or as the calibration of the latter, you can use the fact the typical water drop has volume $\pu{0.03 mL}$, i.e. $\pu{30 mg}$. You can check the actual value by counting drops filling a known small volume.
You can find a web tips or precedures to make homemade scales like in this Kid science article.

Complementary: a grain of barley $\approx \pu{65 mg}$, grain of wheat $\approx \pu{50 mg}$ (Wikipedia). At least, the grain is «the only unit that is equal throughout the troy, avoirdupois, and apothecaries' systems of mass.» (same source). Though greater in mass (which may be compensated by the lengths of the arms of a balance), the mass of coins in your pocket are a standardized alternative, too.

If you know the density of the crystal (usually a salt), you can estimate the mass of a crystal based on its volume, i.e. by measuring the lengths of its sides.
For example, table salt (sodium chloride) has a density of $\rho=2.16\ \mathrm{g\ cm^{-3}}$.
Thus a cube of $10\ \mathrm{mm}\times10\ \mathrm{mm}\times10\ \mathrm{mm}$ has a mass of about $2.2\ \mathrm g$.
And a cube of $5\ \mathrm{mm}\times5\ \mathrm{mm}\times5\ \mathrm{mm}$ has a mass of about $0.3\ \mathrm g$.
The precision of this estimate should be enough to select a suitable seed crystal.

The  main problem with this solution is its a powder (beside knowing the substance) so it will trap a lot of air and I need to make a press to make it work.

Very roughly, you could take a hint from the old British Imperial measurements. The smallest unit of weight is the grain, which before standardisation was the weight of a single grain of wheat, which was then standardized as something which translates roughly to 64mg. (Although with plant breeding since then, I would not be surprised if modern wheat grains are much heavier! )
I suspect rice grains are lighter than wheat grains.
You could weigh some grains of a raw cereal (or peppercorns, mustard seeds ...) against a coin of known weight, count them, and divide, to get a somewhat more accurate idea of the weight of each.

Use a balance with uneven arms like in  a factor of 1:10. Test it with known weights to see if the factor is correct or recalculate to the exact factor and use it to measure with the lowest weight on the longest arm.


text:
What will the correct value of spin quantum number for last electron of Na+ ion?
Electronic configuration of $Na^+$ ion is $1s^2,2s^2,2p^6,3s^0$.
I have noticed usually , we take spin quantum number as +1/2 first & then -1/2.
So , like for the above Q:It is for first 3e=+1/2 & last 3 = -1/2.
But , according to Hunds rule in my textbook. There are 2 ways spin quantum number is given I.e for N atom as example:

Electronic configuration of N = $1s^2,2s^2,2p^3$
For the last 3 electrons , either it is +1/2 or -1/2 as starting for the  first 3 electrons.
So , I am having difficulty in understanding whether also for Na+ ions , we can start the order of spin in any way I.e it could be +1/2 or -1/2 for starting ?
I can’t share an image as I have less than 15 reputation. I can share if the image u can help me with getting 15 reputation to share the image & after some time. You can decide if you want to continue the upvote or remove it.

Yes, both values i.e. positive and negative are valid for starting to fill an orbital.

But in my textbook , when I was practicing Q. It was an MCQ based Q & only 1 option was correct form it i.e one with -1/2. Why is that ?

Can you also share the image of the question you refer to?

Image is not loading. It is getting failed again & again. I’ll share as this problem resolves. Must be internet connection. I also had difficulty in posting this Q earlier.

Do you know which rule this represent to & why is this true ? @NisargBhavsar

Maybe the question was assuming you to follow the general convention of filling the +1/2 electrons first.

Let us continue this discussion in chat.


text:
What kind of electronic configurations are possible?
Let us take $d^6$ orbital as an example:
A) ↑ | ↑ | ↑ ↓ | ↑ | ↑ |
Now , this one 1 way. My Q is that how is it possible that in the 3rd box , we have a paired electron but not in the 1st one. Like it should pair in a series right ? From 1st , 2nd & then 3rd one.
B) We know ↑ ↓ | ↑ | ↑ | ↑ | ↑ |
This one I know is correct as all the examples in my textbook are written in this order but in a Q , Option A is also correct.
I want to know why & how ?

This misconception arrises because most of the textbooks and websites starts pairing electrons from left to right in a series. But in general we can pair an electron in any of the degenerate orbitals if all degenerate orbitals are filled with an electron having same spin either up or down ( which is called Hunds rule).

Note: The orbitals which are having same energies are called degenerate orbitals.

 In an atom all five d-orbitals having same principle quantum number are degenerate orbitals. As all are of equal energies you can pair an electron in any of the five d-orbitals you like.

This type of degeneracy is lost when it is attacked by strong field ligands. This is called crytal field splitting of orbitals. (which is out scope of this question)

A further misconception is that the electrons are confined to specific orbitals at all. . . .


text:
How long would atmospheric oxygen last on Earth if plants stopped making it?
A few years ago a rather dubious paper appeared in a mathematical biology journal. The gist of it was that warming oceans could stop oceanic plankton making oxygen (according to a somewhat simplistic model). If this happened, one of the authors claimed that sea-level air could be reduced to the oxygen levels at the top of everest in 100 years. This report quotes one author saying:

By 2100, the earth at sea level could have atmospheric oxygen levels comparable to the top of Mount Everest today. And as far as I know, people cannot normally stay on Everest without oxygen masks for more than a few minutes...

This poses several problems and the authors have subsequently modified their claims.
So I posed a question about it on Skeptics.SE but that got bogged down in arguments about whether answers counted as original research.
So I thought I'd ask a simpler, but relevant, question here since the result is about atmospheric chemistry. Let's assume all oxygen production on earth has stopped. How long would the oxygen in the atmosphere last?
NB I'm not interested in the model (it has been widely rubbished). I'm interested in the basic facts of atmospheric chemistry which should be a far simpler background question to answer.

I’m voting to close this question, because it's rather about EarthScience then chemistry.

Also what's the criterion of "lasting" would be - partial pressure on Everest?

@mithoron WE have an atmospheric chemistry tag. It is as good a question here as it would be on earth science especially since we don't need any special knowledge about atmospheric processes to answer it. A simple answer would need an estimate of the total amount of oxygen and how rapidly it is used. basic calculations do the rest. But feel free to move if you disagree.

@Mithoron A simple calculation of the result would state its premises in the calculation and define the end point. enough of a loss to get to Everest-levels is a good as any other end point.

Maybe getting too much in the weeds, but for humans the ratio of oxygen to carbon dioxide may be the most important value so one should also account for decreased CO2 uptake by photosynthesizing organisms?

It's not a matter of atmospheric chemistry - lowering of O2 would be caused by respiration, decay of plant leftovers, fire and uncountable other effects - most of them close to the surface of Earth, more of an atmospheric effects, like oxidation of emitted  gases would be secondary.

Well, it's not like I can migrate it, but this question seems poorly defined. It's more of something for xkcd What If? then SE. Highly speculative scenarios need to be very precisely defined to make a specific answer possible.

@mithoron I'm not sure why it is poorly defined. Three simple observations: the atmosphere is in equilibrium; We have estimates as to how much photosynthesis occurs each year; the stock of oxygen is known. So what happens if we make no more oxygen? That's a clear scenario that should put the estimate in the controversial quote in context. More complex analysis is not required.


text:
pH of water on paper vs meter
We have both a professional (lab) pH meter, and several different version of lab grade pH paper, but are experiencing a very weird phenomenon. We have calibrated all the papers against standards, and they work fine. (And of course our meter is regularly calibrated.) However, when we use the meter to measure the pH of various plain waters (e.g., de-ionized vs. tap) we get consistent readings ranging from 4 to 10 (for different sources), but the pH papers never register anything outside of ~6 for any of these waters. Our tap water, for example, reads 10 on the meter and 6 on the paper. (dry paper reads about 6 -- it's a bit hard to tell whether it's changing at all in the water, because, of course, just wetting it makes it look slightly different, but it's definitely not changing substantially.) Whereas the same paper reads perfectly well in various standards between 2 and 12!) Any idea what might be going on here? Thanks in advance!

I'm not sure how one calibrates indicator paper against standards, but what do pH-meter and paper show for common fresh buffer standards, say, for pH 4.00 and 9.18? Can you attach the photo of the paper strips before and after the measurement, the model of your pH-meter, and the procedure for paper testing?

TL;DR: under proper conditions I'd trust the pH-meter, not the paper. Indicator itself needs to participate in acid-base reaction and this consumes dissolved alkali transferred to the strip. Considering that a very small amount of the measured solution gets on the paper, the amount of alkali might be comparable to the amount of the indicator on the paper, thus lowering the readings towards lower pH values. Also, pH paper itself possess small buffer capacity, and, since it has a mix of indicators to make it universal, it has several non-specific pH regions, but this depends on manufacturer.


text:
Does hydrogen peroxide remove chloramine from water?
I know that a small amount of hydrogen peroxide will convert chlorine in drinking water into table salt, oxygen, and water:
$$\ce{NaOCl + H2O2 → NaCl + H2O + O2}$$
What I can't find is what effect it has on chloramine. Does it safely get rid of the chloramine, or does it produce undesirable byproducts?

See this publication and references therein.

The short answer is Yes, Hydrogen Peroxide will remove Chloramine from water.
There may, however, per my sourced review of the underlying chemistry, be also some immediately unfavorable by-products relating to chemical formations, and longer term storage/use of the treated Chloramine/$\ce{H2O2}$ may also present biological hazards as well.
To understand why, one must note that Chloramine ($\ce{NH2Cl)}$ in water undergoes hydrolysis forming Ammonia and Hypochlorous acid, the $\ce{HOCl }$ being the main disinfecting agent (paralleling the action of Chlorine in water) and the reputedly rationale of employing $\ce{NH2Cl}$ to disinfect water. Albeit, although Chloramine is decidedly weaker than aqueous Chlorine, it serves as a longer lasting stable provider of the active/unstable Hypochlorous acid in water:

$\ce{NH2Cl + H2O = NH3 + HOCl}$  (See Eq (1) here)

$\ce{Cl2 + H2O = H+ + Cl- + HOCl}$

The addition of a Hydrogen Peroxide to Hypochlorous acid is actually a fast reaction resulting in a powerful disinfectant action as a consequence of radicals and even transient Singlet Oxygen formation, but unfortunately, it completely destroys/removes the Hypochlorous acid.

$\ce{ H2O2 + HOCl = O2 + H2O + H+ + Cl- }$ (See Eq(4) here)

So if one adds Hydrogen Peroxide to both sides of the opening $\ce{NH2Cl}$ reaction cited above, one would expect the main residual product to include Ammonia with the removal of $\ce{HOCl}$. However, more recent chemistry, a 2008 article in the field of Atmospheric Science “Hydroxyl radical from the reaction between hypochlorite and hydrogen peroxide”,  suggests that the above reaction proceeds first  with the introduction of the powerful hydroxyl radical. Now, in the further presence of ammonia, this could introduce the amine radical and as a consequence similar unfriendly products as reported in the photolysis of such systems.  For example, this work: “Photolysis of Mono- and Dichloramines in UV/Hydrogen Peroxide: Effects on 1,4-Dioxane Removal and Relevance in Water Reuse”, as to products, to quote:

photolysis of chloramines produces inorganic nitrogen breakdown products, including nitrate, ammonia, gaseous nitrogen, and organic nitrogen products (Figure 6)...Overall, ammonia accounted for 60−75% of the nitrogen products. The production of ammonia was likely via the breakdown of NH2• that was transformed into the intermediate N2H4 and then NH3 [38]. The initial formation of nitrite was possible via decomposition of NH2•; however, the subsequent oxidation of nitrite by chloramine generated nitrate [39].

where sunlight induced radicals (particularly, the hydroxyl and amine radical) can apparently foster unfriendly nitrite, nitrate and even traces of hydrazine ($\ce{N2H4}$).
To review what has occurred with the addition of Hydrogen Peroxide is the removal  of the Hypochlorous acid and the introduction of Ammonia and transient powerful radicals. Now, if we consider the case of what occurs when we simply boil water containing dissolved Oxygen, it loses its disinfecting power and soon becomes contaminated with mold and bacteria.  In the current case, similarly, one should likely expect the further action of Oxygen and bacteria acting on Ammonia, in the absence of disinfecting $\ce{HOCl}$, correspondingly allowing for such growth, which in the presence of Ammonia, may  lead to nitrate. This problem has been well documented as occurring in well water where problematic high nitrate levels can kill cattle and, at low concentrations, even human infants appear to be very susceptible resulting in nitrate poisoning (see CDC’s “Nitrate and Drinking Water from Private Wells”).
My related recommended reading includes: "Occurrence, genotoxicity, and carcinogenicity of regulated and emerging disinfection by-products in drinking water: a review and roadmap for research", which notes: "Disinfection by-products (DBPs) are formed when disinfectants (chlorine, ozone, chlorine dioxide, or chloramines) react with naturally occurring organic matter, anthropogenic contaminants, bromide, and iodide during the production of drinking water."
So if even one wishes to dismiss/minimize potential problematic initial chemical compounds, there appears to be a good parallel between how to safely use freshly boiled water and water containing Chloramine that has been treated with Hydrogen Peroxide thereby removing the protective Hypochlorous acid.

So the quick answer is: "For drinking water, boil it after; for plant-food water, use as is."?

Yes, boiling after treating with H2O2 (but that is safe H2O2 for consumption as adding stabilizers to H2O2 can make it unfit for internal cosumption). On plants, a source: "Chloramines make the water acidic which over time can change soil pH. This may result in nutrient tie-up and create yellowing (chlorosis) problems in many plants. Chloramines prevent the absorption of other nutrients which also may lead to yellowing." So, H2O2 may make the NH2Cl water 'better' (not sure if I would eat those plants). Ref: https://www.natureswayresources.com/DocsPdfs/chloramine.pdf

Chloramine does react with hydrogen peroxide and the reaction kinetics has been studied in a paper linked by Todd Minehardt. Hydrogen peroxide is indeed used to remove chloramine disinfectants because it affects further water treatment (See the PDF linked below for more details). The proposed mechanism for the initial step of the reaction between monochloramine is as below:

The abstract says:

Rate constants for the mono- and dichloramine-peroxide reaction were on the order of $\pu{10^{-2}M^{-1}s^{-1}}$ and $\pu{10^{-5}M^{-1}s^{-1}}$, respectively. The reaction of trichloramine with peroxide was negligibly slow compared to its thermal and photolytically-induced decomposition. Arrhenius expressions of $\mathrm{ln(k_{H_{2}O_{2}-NH_{2}Cl})= \frac{(17.3±1.5)-(51500±3700)}{RT}}$ and $\mathrm{ln(k_{H_{2}O_{2}-NHCl_{2}})= \frac{(18.2±1.9)-(75800±5100)}{RT}}$ were obtained for the mono- and dichloramine peroxide reaction over the temperature ranges $\pu{11.4-37.9}$ and $\pu{35.0-55.0 ^\circ C}$, respectively.

Reference

McKay G, Sjelin B, Chagnon M, Ishida KP, Mezyk SP. Kinetic study of the reactions between chloramine disinfectants and hydrogen peroxide: temperature dependence and reaction mechanism. Chemosphere. 2013 Sep;92(11):1417-22. doi: 10.1016/j.chemosphere.2013.03.045. (pdf)
Photolysis of Mono- and Dichloramines in UV/Hydrogen Peroxide: Effects on 1,4-Dioxane Removal and Relevance in Water Reuse, Samuel Patton, Mariano Romano, Vincenzo Naddeo, Kenneth P. Ishida, and Haizhou Liu
Environmental Science & Technology 2018 52 (20), 11720-11727
DOI: 10.1021/acs.est.8b01023

See also chloramine T that is usually used for water desinfection and PubMed: Aqueous chloramine T solutions as disinfectants: chemical composition, reactivity and toxicity In aqueous solutions of Chloramine T (CAT), caused by dissociation, hydrolysis and disproportionation processes, seven different kinds of molecules emerge (HOCl, OCl-, R-NCl-, R-NHCl, R-NCl2, R-NH2, and R-NH- [R = CH3-C6H4-SO2]).  ......

Nilay: Would you kindly answer this part of the question; "or does it produce undesirable byproducts?"

@AJKOER Yes: https://chemistry.stackexchange.com/a/63531/17368


text:
Calculating Enthalpy of formation versus Calculating Enthalpy of a reaction not occurring at standard conditions
My understanding of Hess's law and its use in determining the change of enthalpy of a reaction has been challenged recently.
Up until recently, I thought that the only way to calculate a chemical reaction's change of enthalpy was to use its intermediate chemical reactions and their associated enthalpies (usually provided in a question). So, I would just re-arrange the intermediate reactions so that when combined the compounds and elements would be located on the appropriate side of the yield sign (matching the main reaction). Additionally, I would alter their enthalpies at the same time. Finally, I would add all of the intermediate reaction enthalpies and their sum would be the correct change in enthalpy of the main reaction.
However, I learned that when calculating the change in enthalpy of formation you do not simply add all of the other intermediate reaction's change in enthalpy. Instead, after arranging the other reactions appropriately and altering their mole coefficients and making the same changes to their enthalpies, you are supposed to add the reactants together and the products together and then subtract the reactants from the products. Why? This seems to ignore the fact that energy and enthalpy are state functions. Why must I combine the enthalpies of formation in this particular way if enthalpy is path independent?
I am only studying chemistry at a college level. Though at the college level, I am only studying the basics of chemistry. Please consider this when replying to my question. This community is very smart and seems to be, generally, at a level of understanding far above my own. So, again, please do your best to explain the answer at my level.
Thank you.

When I attempt to solve this question using the first method I get $\pu{-1.09E4 kJ}$. The correct answer is $\pu{-1.2E4 kJ}$.

Both methods will give you the same answer whether you alter the reactions and add or just subtract the formation enthalpy of reactants from the products. You can try for an example.

I recently tried to apply the first method I described to a Enthalpy of Formation question, it did not work. I want to share a picture of the question but this comment box won't let me paste it.  The best I can do is. . . But, an example would be Find the change in enthalpy for the following reaction: S + O2(g) --> SO2(g). Then the following two equations are provided: S + 3/2O2 --> SO3 (DeltaH = -395 kJ); 2SO2 _ O2 --> 2SO3 (Delta = -198.2 kJ). An example of Enthalpy of formation would be: Using a table determine the standard enthalpy of formation of NO2 (4NH3 + 7O2 --> 4NO2 + 6H2O)

You can actually edit your question to include all this info and also an image.

As already mentioned by me in the comments, both the approaches are exactly the same.
To explain this point, I would like to find the numerical value of the question posted by you in the comments using both methods.

By Manipulating the Equations:
Given:
$$
\begin{align}
\begin{array}{ c c}
\ce{S + 3/2O2 -> SO3 } & ΔH = \pu{-395 kJ}  \\
\ce{2SO2 + O2 -> 2SO3} & ΔH = \pu{+198.2 kJ} \\
\end{array}
\end{align}
$$
To find:
$$
\begin{align}
\begin{array}{ c c}
\ce{S + O2 -> SO2 } & ΔH = \pu{?? kJ}  \\
\end{array}
\end{align}
$$
Method:
We can manipulate the given reactions in the following way to obtain the result we want:
$$
\begin{align}
\begin{array}{ c c}
\ce{S + 3/2O2 -> SO3 } & ΔH = \pu{-395 kJ}  \\
\ce{SO3 -> SO2 + 1/2O2} & ΔH = \pu{+\frac{198.2}{2} kJ} \\
\hline
\ce{S + O2 -> SO2} & ΔH =\pu{ \frac{198.2}{2} -395 = -295.9 kJ} \\
\end{array}
\end{align}
$$
By this method we get, $ΔH =\pu{-295.9 kJ}$.

By using Enthalpy of Formation:
Given:
$$
\begin{align}
\begin{array}{ c c}
\ce{S + 3/2O2 -> SO3 } & ΔH = \pu{-395 kJ}  \\
\ce{2SO2 + O2 -> 2SO3} & ΔH = \pu{+198.2 kJ} \\
\end{array}
\end{align}
$$
To find:
$$
\begin{align}
\begin{array}{ c c}
\ce{S + O2 -> SO2 } & ΔH = \pu{?? kJ}  \\
\end{array}
\end{align}
$$
Method:
We can see that we need to find the enthalpy of formation of $\ce{SO2}$ in this question, given the enthalpy of formation of $\ce{SO3}$ and enthalpy of reaction of a reaction including $\ce{SO2}$ and $\ce{SO3}$.
For, $\ce{2SO2 + O2 -> 2SO3}$ we can say:
$$
\begin{align}
\Delta H_r &= 2\Delta H_{f}^\ce{SO3} - 2\Delta H_{f}^\ce{SO2} - \Delta H_{f}^\ce{O2}\\
2\Delta H_{f}^\ce{SO2} &= 2\Delta H_{f}^\ce{SO3}- \Delta H_{f}^\ce{O2} - \Delta H_r\\
\Delta H_{f}^\ce{SO2} &= \Delta H_{f}^\ce{SO3}- \frac{\Delta H_{f}^\ce{O2}}{2} - \frac{\Delta H_r}{2}\\
\Delta H_{f}^\ce{SO2} &= \pu{(-395 +  \frac{198.2}{2} - 0) kJ}\\
\Delta H_{f}^\ce{SO2} &= \pu{-295.9 kJ}\\
\end{align}
$$
By this method also we get, $ΔH =\pu{-295.9 kJ}$.

This shows how both the methods are same. The former is a more visual way of thinking about the problem while the latter is a more mathematical approach.

Additional Calculations after the question was updated
Now you want to find $\Delta H_r$ of the following reaction:
$$\ce{2C8H18 + 25O2 -> 16CO2 + 18H2O}$$
Using the $\Delta H_f$ of octane, carbon dioxide and water.
You can rearrange the enthalpy of formation reactions in the following way to get the desired outcome:
$$
\begin{align}
\begin{array}{ c c}
\ce{2C8H18 -> 16C + 18H2 } & ΔH = \pu{(269 \times 2) kJ}  \\
\ce{16C + 16O2 -> 16CO2} & ΔH = \pu{-(394 \times 16) kJ} \\
\ce{18H2 + 9O2 -> 18H2O} & ΔH = \pu{-(286 \times 18) kJ} \\
\hline
\ce{2C8H18 + 25O2 -> 16CO2 + 18H2O} & ΔH_r ={(269 \times 2) -(394 \times 16) -(286 \times 18)}\pu{  kJ} \\
\end{array}
\end{align}
$$
Upon solving we get $\Delta H_r=\pu{-1.0914E4 kJ}$. Even if we would have used the second method we would have gotten the same result.
This is the correct answer to the question and not $\pu{-1.2E4 kJ}$.

The solution you posted later in the question is incorrect because it assumes that the enthalpy of formation of octane is positive whereas it is provided to be negative.

I edited my initial question above to now include a picture of the exact question I am struggling with. Can you give the question a try?  Thanks for your help so far.

If you find the answer helpful you can upvote it and accept the answer.

Thank you for your sustained effort in explaining the procedure of solving changes in enthalpy.  I just upvoted you. Before I accept your answer, can you take a look at the explanation of the answer I just added to my original question? What do you think about it?

The photo you uploaded is only half the process can you please upload the complete image.

My bad. The entire explanation is there now.

@Haley The solution you posted later in the question is incorrect because it assumes that the enthalpy of formation of octane is positive whereas it is provided to be negative in the question. Does this answer your query?

No. To me it looks like the explanation used Hess's law to rearrange octane's equation so that octane is a reagent and then adjusted the reaction's mole coefficients, which is why it is written as a positive value in the explanation. It did the same to the other two reactions. Why would it use some of Hess's law but not all?

@Haley Because the solution is wrong. If it is from someone of your school than you should raise an issue or if it is from a random website than don't trust that site from now on.

The question is from a website, but I asked a professor and was told that the explanation was correct. I'm going to give you the answer. Thank you.


text:
Why can element gallium be obtained through the electrolysis of sodium metagallate
In the textbook element gallium is produced by the electrolysis of sodium metagallate , which is obtained as the leftover solution of the alkali treatment of bauxite .
But why is the metagallate anion , negatively charged , going to be reduced on the cathode , which is also negatively charged ? Won't the Coulomb repulsion prevent the reaction ?
Or generally can an anion be reduced on the cathode ?
We shall use the iron anode in the electrolysis treatment of Cr(VI) waste water to have chromium reduced in the solution by ferrous ion rather than the electrons on the cathode ,which is unnecessary if bichromate can be reduced directly in the electrolysis .

Anions travel away from a cathode just partially, until there is built the counter potential gradient due charge displacement. When both gradients cancel each other, the migration stops. That happens when there is no ongoing electrolysis, e.g. if too small external voltage is applied.
When there is ongoing electrolysis, the balance is continuously disturbed, forming an additional charge displacement electrostatic gradient leading to continuous ion migrations.
So there are anions at the cathode and cations at the anode and can be reduced or oxidized.
Note that chromium electroplating is done with $\ce{Cr^{VI}}$ , not $\ce{Cr^{III}}$, reportedly because the latter is often kinetically inert.


text:
Why is a branched alkane more stable than the straight-chain isomer?
My text book says that branched alkanes are more stable than their straight-chain isomers, but does not give any explanation.
Is this merely a fact, or something that can be logically rationalized ?
A thermodynamic cycle is welcomed, and an intuitive clarification is even better.

Maybe thinking in terms of entropy might help?

See these posts: https://chemistry.stackexchange.com/questions/153936/why-is-the-enthalpy-of-formation-of-2-2-dimethylpentane-larger-than-that-of-2-2 ... https://chemistry.stackexchange.com/questions/139763/why-are-branched-alkanes-used-as-fuels-instead-of-straight-chain-alkanes ... https://chemistry.stackexchange.com/questions/80324/why-the-boiling-and-melting-points-of-n-alkanes-are-higher-than-those-of-iso-or ... https://chemistry.stackexchange.com/questions/42426/standard-enthalpy-of-formation-for-branched-alkanes

See this also: https://chemistry.stackexchange.com/questions/123093/deciding-the-order-of-heat-of-combustion-of-isomeric-alkanes/130811#130811

More stable in what respect?


text:
Is it correct to class these two compounds as enantiomers?
This question is from GRB Problems In Organic Chemistry For Jee Advanced.

Identify the relation between these two compounds:

(a) Homomers
(b) Enantiomers
(c) Diastereomers
(d) Different compounds

The answer given is (b) Enantiomers.
However, I feel that it should be (c) Diastereomers.
Enantiomers are non-superimposable mirror images of each other. But as far as I can see, these two compounds are not mirror images of each other, and should be diastereomers.
What is the correct answer to this question?

First of all, they are poor drawings! If you assume that the uppermost group, Ph or H, is syn to the methano bridge, then they  are diastereomers.

Bicyclo[2.2.1]heptane (red box) has two planes of symmetry. It is achiral.  Placing a phenyl ring at $\ce{C2}$ in A and B removes the symmetry and renders both structures chiral. A has the phenyl syn to the methano bridge while B has the phenyl syn to an ethano bridge. They have the same atom connectivity but different arrangements in space. They are diastereomers of each other.
Now "slide the phenyl group forward to $\ce{C3}$". This operation gives the enantiomers of A and B, namely, C and D, respectively. If you are not convinced, rotate C and D about a vertical axis in the plane of the page by $180^\circ$ in a counterclockwise direction such that $\ce{C1}$ and $\ce{C4}$ change positions, C and D will mirror A and B, respectively. Note that the R/S descriptors of enantiomers mirror one another but those of diastereomers do not.


text:
How to find the empirical formula from mass percentages of oxides
Say you've given an element X and it makes two oxides. The first oxide is $X_\mathrm{4}O_\mathrm{6}$ and the mass percentage of oxygen is 43.7%. In the second oxide, the mass percentage of X is 43.7%.
Can someone say if my calculations are correct to find the empirical formula of the second oxide.

$$m_{X_\mathrm{4}O_\mathrm{6}} =  \frac{16*6*100}{43.7} = 219.7g$$
$$m_{X_\mathrm{4}} = 219.7 - 96 = 123.7g$$
$$m_X = \frac{123.7g}{4} = 30.99g$$
$$n_X = \frac{43.7\%}{30.99} = 1.443\ \ \ \ \ \ \ \ \ \ \ \ n_O = \frac{56.3\%}{16} = 3.519$$
$$\frac{1.443}{1.443}\ \ \ \ \ \ \ \ \ \ \ \ \ \frac{3.519}{1.443}$$
$$1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2$$
Therefore the oxide $XO_2$

Good practice is starting with symbolic algebraic formulas and keeping it this way until all formulas are ready to finally put there literal numbers.
It helps focusing on principles and improves orientation, spotting mistakes, the text reusability and its permanent value.

What keeps you from checking it yourself, e.g. verifying what is X percentage context in XO2 ?

Yup I was so dump I wasn't rounding the number properly thanks found the answer X2O5. Thanks

The first oxide is $\ce{P4O6}$ and not $\ce{XO2}$ as Dulsara stated.
The second oxide is $\ce{P4O10}$ as Dulsara said.
My final question is : Why publish a question where its author knows the answer ?

Sorry about that. First when I got XO2 I did in fact check the mass percentages and didn't get the correct answers and thought my approach was wrong for the entire question and later found out that it was just a rounding problem. Because I round 2.5 as 2 at first.

Here's the answer if anyone is wondering.
$$m_{X_\mathrm{4}O_\mathrm{6}} =  \frac{16*6*100}{43.7} = 219.7g$$
$$m_{X_\mathrm{4}} = 219.7 - 96 = 123.7g$$
$$m_X = \frac{123.7g}{4} = 30.99g$$
$$n_X = \frac{43.7\%}{30.99} = 1.443\ \ \ \ \ \ \ \ \ \ \ \ n_O = \frac{56.3\%}{16} = 3.519$$
$$\frac{1.443}{1.443}\ \ \ \ \ \ \ \ \ \ \ \ \ \frac{3.519}{1.443}$$
$$1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2.439$$
Then you have to round it by multiplying the number by 2
$$2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4.878$$
$$2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 5$$
Therefore,
$X_2O_5$


text:
Dissolution of sugar in water and its temperature dependence
As far as I can remember, I was taught in 6th grade that warm water would dissolve sugar with greater ease than cool water. In 11th grade, I was taught that Hydrogen Bonding is the reason behind compounds dissolving in water, and was later taught that at higher temperatures hydrogen bonds tend to break. Now, I realize that the two kind-of contradict each other as if sugar were to dissolve easily at a higher temperature, then it should form hydrogen bonds more easily at higher temperatures, however, higher temperature weakens the hydrogen bonding between sugar (say sucrose) and water and that must inhibit dissolution. So what actually is going on?

We can start by watching two videos showing how sugar and salt are dissolved in water: https://www.youtube.com/watch?v=fwjvwoFHTbg and https://www.youtube.com/watch?v=14iv71ZTRTo. Particularly, the second video is fun to watch, where molecular dynamics simulation is used to show how water molecule solvates an ion at the corner of a NaCl particle.
To reconcile the contradiction, you need to consider two points here, one being thermodynamics and the other being kinetics. Thermodynamics is about the change of state functions of a system, whereas kinetics is about the rate of the change with time involved.
You can first consider thermodynamics. Sugar crystals or granules put into water are the initial state and the sugar molecules dissolved in water forming a homogeneous solution are the final state. Since sugar can be dissolved in water, it must be true that the final state is more stable than the initial state, even at elevated temperature. You can do a simple experiment at home to verify this is the case. Using thermodynamics terminologies, $\Delta G = \Delta H - T\Delta S$, where G is Gibbs free energy, H is enthalpy, S is entropy, T is temperature, and the symbol $\Delta$ means changes, the greater stability of the final state than the initial state means $\Delta G$ is negative. This means that the dissolution of sugar in water is thermodynamically favorable. The hydrogen bonding between sugar molecules and water molecules may be weakened with increasing temperature, and therefore $\Delta H$ may be less negative at a higher temperature. However, due to the loss of hydrogen bonding the final state could be more disordered, and therefore $\Delta S$ may be more positive with increasing temperature, which can contribute to make the $\Delta G$ more negative favoring the dissolution of sugar at a higher temperature.
Then you need to also consider kinetics. Your statement that "warm water would dissolve sugar with greater ease than cool water" has to do with kinetics, where water molecules have higher energies to move around to dissolve sugar molecules more quickly from the initial crystal/granule state into the final homogeneous solution state. More specifically, water molecules at the solid/liquid interface carry sugar molecules from the solid surface into the liquid phase, especially from the crystal corners and edges where the chemical potentials of sugar molecules are higher than at basal planes. Additionally, at a higher temperature, water molecules have higher diffusion coefficients to move from the liquid bulk to the solid/liquid interface to dissolve sugar molecules and then the dissolved sugar molecules also have higher diffusion coefficients to move from the solid/liquid interface to the liquid bulk. Generally speaking, at higher temperature molecules have higher energies and move faster to speed up the dissolution process.
Therefore, you need to consider both thermodynamics and kinetics to get a more comprehensive understanding of this problem. This is my simple answer with only some handwaving arguments, which may not be complete and others may provide corrections and/or additional explanations.

Thank you! Pretty comprehensive, I was previously looking at it from a very narrow perspective.

Your statement " ΔS is positive, which can contribute to make the ΔG more negative favoring the dissolution of sugar at a higher temperature." is incorrect. The entropy change does not contribute in any way to the temperature dependence of the equilibrium. Only the enthalpy change matters. See the derivation of the van't Hoff equation for validation.

I believe it does. Please do not just look at the differential form of the van't Hoff equation. You need to look at the linear form of the van't Hoff equation. If you plot lnK as a function of 1/T, the slope is affected by $\Delta H$ but the intercept is affected by $\Delta S$, so both $\Delta H$ and $\Delta S$ contribute to affect the equilibrium as temperature is changed.

@Josiah_H  Andrew is correct.  The temperature dependence of the equilibrium constant is determined only by $\Delta H$.  It should also be noted that your picture is oversimplified, since both $\Delta H$ and $\Delta S$ are temperature-dependent (though their temperature-dependencies cancel out, so we can ignore them when it comes to determining the temperature dependence of $\ln K_{eq}$).  Thus, for the purposes of addressing your picture, let's assume $\Delta H$ and $\Delta S$ are constant.   What determnes the temperature dependence is the slope ($\Delta H$), not the intercept ($\Delta S$).

@Josiah_H I.e., think about two different simple linear equations:  $y = 3 x + 1$ and $y = 3x +2$.  The amount by which $y$ changes when we increase $x$ from, say,  $5$ to $6$, is the same for both equations.   The difference in intercept between the two only changes the vertical offset.  The $x$-dependence (analogous to the $T$-dependence) is the same in both

You are both correct and wrong. You are correct in that only $\Delta H$ affects the "temperature dependence" of lnK. I fully agree with that. However, in my answer to the question, I was discussing the effect of temperature to $\Delta G$ through $\Delta S$. Therefore, both $\Delta H$ and $\Delta S$ will affect $\Delta G$. This can be seen from the linear van't Hoff plot, i.e., lnK as a function of 1/T. Suppose two processes have the same slope, meaning the same temperature dependence and the same $\Delta H$. If they have different intercept $\Delta S$, then they will have different $\Delta G$.

You see the issue is that the topic was accidentally changed from my discussion on "temperature effect on $\Delta G$ through $\Delta S$" to "temperature dependence of the equilibrium". In your example given by the two linear equations, are the two y values the same? No, they are different.

I think I see your confusion.  This should clear it up:  At constant pressure, the temperature-dependence of $\Delta G$ is determined solely by $\Delta S$, and the temperature-dependence of $K_{eq}$ is determined solely by $\Delta H$.  When you are asking the extent to which a reaction is favored or disfavored by a change in temperature, what you are really asking, physically, isn't what happens to $\Delta G$; rather, you are asking what happens to $K_{eq}$.  Thus  $\Delta H$ is the physically meaningful attribute to consider when asking about the temperature dependence.

And this is definitely wrong, since here you are talking purely about the equilibrium constant, not $\Delta G$:  "You need to look at the linear form of the van't Hoff equation. If you plot lnK as a function of 1/T, the slope is affected by Δ but the intercept is affected by Δ, so both Δ and Δ contribute to affect the equilibrium as temperature is changed."  And this isn't thermodynamics, it's just math -- only the slope (delta H) affects how the value of the function (ln K) changes as you change the independent variable (1/T).  Changing the intercept (delta S) just changes the offset.

And to address your specific math statement:  You write: "In your example given by the two linear equations, are the two y values the same? No, they are different."  That's right, the two y values are different. But, if you are asking about the x-dependence, it doesn't matter, because it's the same for both equations.  E.g., suppose x changes from 10 to 20.  In the first equation, y changes from 31 to 61. In the second equation, y changes from 32 to 62.  I.e., in both equations, even though the intercepts are different, y changes by the same amount (30) when x is increased from 10 to 20.

Sorry, I don't have time for debate. I just offered my help to a student by saying something off the top of my head, instead of writing it rigorously as we normally do with research articles. As I said at the end of my answer, this is my simple answer with only some handwaving arguments. If you feel necessary, you are welcome to revise or write another answer. When you write, it would be helpful to talk about the comparison between reaction quotient Q and equilibrium constant K, which serves to indicate which way the reaction shifts and which side of the reaction is favored (see libretexts).

As far as I can remember, I was taught in 6th grade that warm water would dissolve sugar with greater ease than cool water.

That is true. Sometimes, the sugar in honey crystallizes. If you carefully heat up the honey, you can get it back to a liquid. So there is experimental evidence that sugar is more soluble as you heat up the solution.

In 11th grade, I was taught that Hydrogen Bonding is the reason behind compounds dissolving in water, and was later taught that at higher temperatures hydrogen bonds tend to break.

Sometimes, water interacts with solutes through hydrogen bonds (in the case of sugar). Sometimes, it interacts through ion-dipole interactions (in the case of ionic solutes). If you stick with sugar, you should know that sugar molecules in a sugar crystal also interact through hydrogen bonds.
It is true that hydrogen bonds tend to break at higher temperatures because the hydrogen bonds restrict the movement of particles, and particles move "more" at higher temperature. However, because all the molecules interact via hydrogen bonds in the water/sugar system (i.e. water with water, sugar with sugar, water with sugar) this does not let you predict how solubility will change with temperature.

Now, I realize that the two kind-of contradict each other as if sugar were to dissolve easily at a higher temperature, then it should form hydrogen bonds more easily at higher temperatures, however, higher temperature weakens the hydrogen bonding between sugar (say sucrose) and water and that must inhibit dissolution.

There is a misconception there. The hydrogen bonds do not tend to break at higher temperature because they get weaker. They break because the particles move faster (have more kinetic energy).

So what actually is going on?

The dissolution of sugar is an endothermic process (the temperature drops as you mix in the sugar). If you increase the temperature in an endothermic process, you favor the product. Hydrogen-bonded or not, some substances dissolve in an endothermic process and others in an exothermic process. The former will be more soluble in warmer water, while the latter will be less soluble in warmer water. There is no good explanation for this at the level of this answer.

... dissolve sugar with greater ease ...

The answer focuses on solubility. "Ease" might also refer to how fast the sugar dissolves. The higher the temperature the faster the process, and the finer the solid sugar (i.e. confectioners sugar instead of rock candy), the faster the process.


text:
Converting pyrite to iron sulfate
I was reading a Jules Verne book recently, in which the following chemical reaction was done:

He placed a layer of branches and chopped wood, on which were piled some pieces of shistose pyrites ... This done, they set fire to the wood, the heat was communicated to the shist, which soon kindled ... Then new layers of buised pyrites were arranged so as to form an immense heap, the exterior of which was covered with earth and grass, several air-holes left.

(The shistose pyrates consisted of a mixture of coal, iron sulfide, and aluminum oxide)
After 10 days, the $\ce{FeS2}$ in the pyrite mixture was converted into iron sulfate. However, my experience with roasting iron pyrite involves sulfur dioxide vapors coming off and leaving behind iron. How would the iron sulfate be created?

Note, chemical information may be advantageously formatted using on ChemSE with mhchem. Take moment to familiarize with this. You are encouraged to use it in the body of questions, answers, and comments. Because it is something special not all web browsers understand well, do not use it in the title of questions or answers.
An addition of the source you quote would be a benefit for the future readers, too.

This may be of interest to you: https://chemistry.stackexchange.com/questions/142837/vital-force-theory/142838#142838

Typically, roasting sulfides leads to the generation of sulfur dioxide and metal oxide.  In the case of pyrite, this may be described by
$$\ce{4 FeS2  + 10 O2 -> 2 Fe2O3 + 8 SO2 }$$
The fire is needed to initiate the reaction.  If well maintained, the heat generated by roasting may sustain the reaction.
However, and especially for marcasite as an other form of $\ce{FeS2}$ (polymorphism), it is known that exposure to humidity and air suffices to oxidize the sample in some parts, leading to the generation of small amounts of sulfuric acid.  Because sulfuric acid is a stronger acid than $\ce{H2S}$, the former may displace the later.  Eventually, the sample literally gets eaten away to yield $\ce{FeSO4}$ which is described as «pyrite decay».
The same source refers to a Conserv O Gram «Storage Concerns for Geological Collections» (April 1998, Number 11/2) by the National Parks' Museum Management Plan.  This publication mentions $\pu{60 \%RH}$ as critical threshold (though, likely at ambient temperature, without statement how fast this process is).  Possibly, the process observed in the museum doesn't happen in the kiln Verne described.

Yep, when moisture and oxygen are plentiful, "acid mine drainage" forms easily from the aerobic oxidation of iron pyrite to iron(II) sulfate.  Iron(II) sulfate is highly water soluble, at least in acid.  It is further oxidized by air to insoluble iron(III) oxides and hydroxides, but the abiotic reaction is slow at low pH.  However, even at low pH, the reaction can be effected by specialized microbes.  Under the right conditions, this can lead to thick scums of biofilm that grow at the acid/air interface in molar concentrations of iron and pHs of as low as zero or below.

Wet-oxidation of pyrite lead to iron(II) sulfate.
$$\ce{2FeS2 + 2H2O + 7O2 ->[100 °C] 2FeSO4 + 2H2SO4}$$
iron(II) sulfate formed is eventually further oxidized to iron(III) sulfate, iron(III) oxide and iron(III) hydroxide. Note that the reaction is not that simple as it proceeds through many Fe(II) and Fe(III) intermediates:

There are different proposed reaction mechanism for this reaction. You can look up here, here and here.
References

Pyrite Oxidation Mechanism by Oxygen in Aqueous Medium
Egon Campos Dos Santos, Juliana Cecília de Mendonça Silva, and Hélio Anderson Duarte
The Journal of Physical Chemistry C 2016 120 (5), 2760-2768
DOI: 10.1021/acs.jpcc.5b10949
https://pubs.usgs.gov/of/1995/0389/report.pdf


text:
Why can't the Fajans' rule and lattice energy theory be used to explain the difference in thermal stability of magnesium and calcium carbonates?
$\ce{CaCO3}$ is more thermally stable than $\ce{MgCO3}.$ I do understand that a larger cation would be able to stabilize a larger anion to a greater extent, but why can't we use Fajans' rule or lattice energy to explain?

(1) Thermal Stability:
$\ce{CaCO3(s) ->[\Delta] CaO(s) + CO2(g)}$
$\ce{MgCO3(s) ->[\Delta] MgO(s) + CO2(g)}$
Because of smaller $\ce{Mg^2+}$ ion, it will polarise the $\ce{CO3^2-}$ ion more than $\ce{Ca^2+}$ ion. So the removal of $\ce{CO2}$ will be easier. See this explanation.
Fajans' rule compares the covalent character in ionic compounds. For which, it compares the polarising power of cations if the anion is same. Hence, this rule in a way, explains the thermal stability of these two compounds.
(2) Fajans' Rule:
The covalent character is less in $\ce{CaCO3}$ than $\ce{MgCO3}$.
(3) Lattice Energy:
The lattice enthalpy of an ionic solid is defined as the energy required to completely separate one mole of a solid ionic compound into gaseous constituent ions.
$\ce{CaCO3(s) ->[\Delta] Ca^2+(g) + CO3^2-(g)}$
$\ce{MgCO3(s) ->[\Delta] Mg^2+(g) + CO3^2-(g)}$
$\ce{MgCO3(s)}$ has more lattice energy but it is less thermally stable than $\ce{CaCO3(s)}$. Hence, it cannot explain the trend in thermal stability because of different products obtained in (1) and (3). More on this is here.


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How do we write spin multiplicity for Mn(2+), Mn(7+)?
Mn has atomic number = 25
Since it is an exception to electronic configuration , unlike having = $\mathrm{3d^7}$ , it has electronic configuration = $\mathrm{3d^5,4s^2}$.
Formula for spin multiplicity = 2|S| + 1.
Spin multiplicity of Mn = 2(0) + 1 = 1 since there are no unpaired electrons.
I’m having difficulty with the further ions multiplicity.
For $\ce{Mn2+}$ , there are 5 unpaired electrons in d sub-shell. The 2e- of $\mathrm{4s^2}$ are lost.
M = 2(5/2) + 1 = 6. Now , according to my textbook. Mn = Mn2+ in terms multiplicity which I do not see as correct.
M for $\ce{Mn7+}$ ,
Electron configuration =$\mathrm{2s^2}$, $\mathrm{3d^0}$
Therefore , here I think it should be 2(2/2) + 1 = 3.

You may find useful these links for text formatting ( not to be applied to titles ):
notation ,  chem/math formula/equation formatting  and
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@Poutnik I will edit my Q in some time accordingly as it can be useful.

Sure, no hurry because of me, just for you to be aware for future cases.

Spin multiplicity = $2S+1$, where $S$ is the total spin angular momentum. Now $S = \frac{n}{2}$ where $n$ represents total number of unpaired electrons. So now we can write spin multiplicity = $n+1$.
Now coming to your question,

$\ce{Mn}$ has 5 unpaired electrons in it. Therefore, its spin multiplicity $(S) = 5+1 = 6$.
Similarly, $\ce{Mn^{2+}}$ has 5 unpaired elections in it. Therefore, its spin multiplicity $(S) = 5+1=6$.
$\ce{Mn^{7+}}$ has zero number of unpaired electrons. Therefore, its spin multiplicity $(S) = 0+1=1$.

You might want to have a look at the formatting guide posted as a comment to the original question.

Consider also the usage of it's versus its.