text:
[Co(H2O)6]3+ vs [Co(NH3)6]3+ oxidising water

Cobalt(III) complexes cannot be prepared by addition of the ligand to a solution of the metal cation because the aqua cation $\ce{[Co(H2O)6]^3+}$ has no stable existence in solution. It is strongly oxidising with an E° = +1.84 volts for $\ce{[Co(H2O)6]^{3+} + e- -> [Co(H2O)6]^2+}$. It oxidises water according to the equation:
$$\ce{2 [Co(H2O)6]^3+ -> H2O + 2 [Co(H2O)6]^2+ + 2H+ + 1/2O2} (2)$$

Complex formation, however, can modify the standard electrode potentials considerably (for example, E° = +0.10
volts for $\ce{[Co(NH3)6]^{3+} + e^- -> [Co(NH3)6]^2+}$ and therefore, oxidation from Co(II) to Co(III) is easily
accomplished in the presence of coordinating ligands".

Why is it that $\ce{[Co(H2O)6]^3+}$ oxidise water, but $\ce{[Co(NH3)6]^3+}$ does not? I understand how values of reduction energy potential effect the proceeding of a reaction, but what is it about $\ce{[Co(H2O)6]^3+}$ that makes it able to oxidise water more?

There are 2 competing principles.
$\ce{[Co(H2O)6]^3+}$ has  strong oxidation effect, but...
$\ce{[Co(NH3)6]^3+}$ is much more stable than $\ce{[Co(NH3)6]^2+}$.
As the consequence, the former from the next 2 equations is significantly shifted to the right.
$$\begin{align}
\ce{  [Co(H2O)6]^3+ + 6 NH3 &<=>> [Co(NH3)]^3+ + 6 H2O} \\ \ce{[Co(H2O)6]^2+ + 6 NH3 &<=> [Co(NH3)]^2+ + 6 H2O}
\end{align}$$
Therefore
$$\frac{c_{\ce{[Co(H2O)]^3+}}}{ c_{\ce{[Co(H2O)]^2+}}} \ll   \frac{c_{\ce{[Co(NH3)]^3+}}}{ c_{\ce{[Co(NH3)]^2+}}}$$
As both complex pairs keep the same redox potential, their standard redox potentials are therefore very different.

Although the exact proof needs complicated, empirically-derived equations, the essential idea is simple.
Cobalt(III) is a highly charged cation of a moderately electronegative(1.8) atom, so it wants electrons badly.
When bare-naked cobalt(III) is near water, it gets so mad at the highly electronegative, "greedy" oxygen atoms of water's not sharing its lone pairs that it smashes the oxygen-hydrogen bonds of water and "takes away" the bonding pair(s) of electrons of water. Cobalt(III) is reduced to cobalt(II), and the electron-taken-away, zero-oxidation-state oxygens end up pairing with each other to form dioxygen gas- just like how the exceedingly greedy King Louis XVI ended up on the Guillotine and "cut apart".
However, things change when cobalt(III) is near ammonia. Nitrogen is  less electronegative than oxygen (meaning that it likes sharing its lone pair to cobalt more than oxygen), and although nitrogen is still much more electronegative than cobalt, the ammonia molecule doesn't mind forming a dative bond to cobalt, even if that would mean that it would gain a formal positive charge. This is obvious, since the atmosphere around the positively charged nitrogen in the ammonium cation is already less "greedy" than that of the oxygen atom in the oxonium cation- when one replaces one hydrogen in the ammonium cation with the even-less-electronegative-than-hydrogen cobalt, the positively charged nitrogen becomes even less "greedy". (Sure, it does want more electrons under extraordinary situations, but it doesn't want more electrons at ordinary circumstances and hence doesn't need to (and doesn't) "take away" electrons from the oxygen-hydrogen bonds of water.) The nitrogens have 8 electrons each near themselves, the cobalt has 18 electrons near itself, and the hydrogens have 2 electrons each near themselves. Everybody is happy- just like socialist democratic Scandinavia.
Hence, hexaamminecobalt(III) is stable in water while hexaaquocobalt(III) isn't.


text:
Dissolving copper in nitric acid and heating it

A sample of copper metal is dissolved in 6 M nitric acid contained in a round bottom flask. This reaction yields a blue solution and emits a colorless gas. The water is evaporated form the blue solution to leave a blue solid. When the blue solid is heated further, nitrogen dioxide gas, oxygen gas, and a black copper oxide is left. If all of the blue solid were not decomposed into the black oxide during the final heating, what would be the effect on the calculated percentage of copper in the oxide?

I know that there is a redox reaction of $\ce{Cu + H+ + NO3- -> Cu^2+ + NO2 + O2}$, and the black oxide is $\ce{CuO}$.
I have spent a long time thinking about this question, and my reasoning is that since less copper oxide is formed, the calculated ratio between the original mass of copper and the final mass of copper oxide would be too high.
However, the answer says that the calculated ratio would be too low. Why is this?

@M.Farooq The question talks about the incompletely decomposed blue solid, not the incompletely decomposed CuO

The "incompletely decomposed copper nitrate" will be weighed as CuO by the experimenter. The incompletely decomposed salt will have a larger apparent weight w' than the true weight w if the salt were fully decomposed into CuO. % Cu in CuO = mass Cu/ wt of CuO. If we use the larger weight w' in the denominator, what will happen to the magnitude of the fraction?

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What's the most complete basis set for calculating chemical shifts with DFT?
I'm trying to benchmark some NMR chemical shifts. The literature suggests that the pcS-4 basis set will be the most accurate. I was wondering if there were any basis sets that are more complete than the Jensen basis sets? Even for just general basis sets. I'm looking for something similar to the Partridge basis sets, except for DFT.

For the pcS-4 the basis set error is less than the error produced by your density functional. So if you don't choose a really really good and costly functional, the pcS-4 is basically useless. What stops you from benchmarking against a wavefunction theory method?

Do you need benchmark existing methods?

IMHO : "good and costly functional" -> agreed. || "more complete than the Jensen basis sets? "  -> if you are working with an open source(quantum expresso-like) DFT tool, to add a basis set will be well within your reach. || my catch : "which basis set is more complete ?" --> FWIW, unless someone had numerically/qualitatively report a comparison table.. I'm not really convinced that any basis set is better than the other.

Using pcS-4 only tells you how much the error from your basis set is. This doesn't necessarily have any link to the total error, which comes from basis set, DFT functional, molecular geometries, treatment of solvation, and various other aspects of your chosen method. So, if all you want to do is to investigate the effect of basis set while keeping everything else the same, then sure, go ahead and use the pcS-n basis sets. They were designed with convergence to the complete basis set limit in mind, so pcS-4 is probably as close to the CBS limit as you can reasonably get. For a recent (and very thorough) example of benchmarking using pcS-4 (technically the segmented version pcSseg-4), see J. Chem. Theory Comput. 2018, 14 (2), 619–637.
If your aim is just to get the computed shifts as accurate as possible, then CCSD(T) with large pcS-n or cc-pVnZ (or even extrapolation to the CBS limit) is probably what you want. For benchmarking you're not likely to be using any excessively big molecules anyway, so for a one-off calculation I imagine the computer time needed is quite tolerable.
As far as DFT is concerned, as has already been pointed out, if you don't use a good functional then no matter what basis set you use, the shifts will not be great anyway. Hybrid functionals like PBE0 or mPW1PW91 are popular in the literature (and B3LYP of course), but there are some other interesting ones like M06-L (see J. Phys. Chem. A 2008, 112 (30), 6794–6799), and recently double-hybrid functionals have been suggested (J. Chem. Theory Comput. 2018, 14 (9), 4756–4771).
Aside from that, there are also empirical ways of increasing the accuracy of DFT-predicted shifts, which you may or may not want to take into account when benchmarking. A popular one is linear scaling of shifts (the slope and intercept being determined from a given test set). Tantillo has a nice collection of these scaling factors at http://cheshirenmr.info/index.htm. Although some of the work cited can be a bit outdated in such a rapidly-moving field, the methodology itself is still highly relevant.

First and foremost: there is no best or most complete finite basisset (except for trivial examples). The only complete basis is your infinite dimensional Hilbert space.
You may phrase it better like this:
With the current state of computer hardware and in combination with your ab initio method of choice applied to a specific model system this basis set gives results with an acceptable error for your variable of interest and is still computationally feasible.
One good test is always to increase the basis set. If your variable of interest is not affected, you are probably converged with respect to your basis set.
Note that it really depends on your variable of interest. If you want to find equilibrium geometries then hartree fock + smallish basis set is often enough.
For reliable chemical shifts you need to push the basis set and method really hard.
The comment suggested wavefunction based methods, because they can be consistenly improved by increasing the number of excitations. If you can use only DFT, you should check at least two different functionals to see if they agree.

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Thermal stability order of NaF, MgF2 and AlF3
I came up with a question to arrange thermal stability order of $\ce{NaF}$, $\ce{MgF2}$ and $\ce{AlF3}$ and I think the answer is $\ce{NaF>MgF2>AlF3}$ because $\ce{Na+}$ has largest ionic radius among the cations(anion is same) and also NaF has the greatest ionic character. And as we know greater the ionic character, the greater is the thermal stability.
But the answer is exactly opposite that is $\ce{NaF<MgF2<AlF3}$. Where am I wrong?
Do lattice enthalpy also have a role to play in determining the thermal stability of metal chlorides and fluoride? If yes, then how?

I what you understand by thermal stability, I have a feeling it's not what you think it is.

@Mithoron According to the most widely prescribed Chemistry textbooks (published by the NCERT) in India, thermal stability is measured by the melting/boiling point of an element/compound relative to other members in a group/period/family of compounds (and whether something decomposes on heating or doesn't exist stably at room temperature). The terminology isn't very accurate and some books are known to contain errors.

@Kartik you can't trust NCERT just because it is the most widely prescribed chemistry textbook. They are known to contain notorious errors.

@Nilay Ghosh. Who's trusting it? Just clarifying how the context in which the term thermal stability is used in the question came about. Your comment helps that explain even more!

Metal fluoride thermal stability trends are largely dictated by lattice enthalpies; what matters is the relative size of the cation in consideration (the charge/radius ratio - a measure of polarizability - is also a key factor). Smaller cations can better hold small anions like fluoride.
The larger the lattice enthalpy factor, the tougher thermal decomposition becomes.
As atomic size decreases across a period, the order of stability is AlF3 > MgF2> NaF.

But as Al3+ has the smallest size wouldn't three F- electrons result in very large magnitude of inter electronic repulsion, thus decreasing its stability?

@suhridisen Several ionic compounds such as AlF3 have high thermal stability, as exemplified by AlF3's melting point (1290° C). This is due to the electrostatic attraction between the cation and the anion. Even though there are several fluorine electrons around the small Al, the strong attraction between F electrons and the nucleus overcome the repulsive forces. Also, fluorine being highly electronegative instead of having its electrons wandering around the Al ion would keep much of them to itself while Al achieves a stable electronic configuration.


text:
Why enthalpy of adsorption is negative?
I have already read Is adsorption exothermic, and if so, why?  but it didn't answered my query.
To illustrate, let's take a hypothetical vacuum space. In that space, there is a solid and a gas, the gas is getting adsorbed over the solid. In that case, if we'll use momentum conservation, we'll get that the net kinetic energy of system has increased due to positive work done by the internal forces, but heat exchanged from system to surrounding or vice-versa is zero( as there is no outside interaction present. Therefore, dU>0 as U is sum of kinetic and potential energy of system( note that I haven't made the internal force as due to a field with a assigned potential function). We know dH=dU+PdV for processes in which the pressure outside the system is constant( which we have assumed). Therefore, as dV=0( because I had taken a fixed volume of space), therefore dH>0. But in books it is written dH due to adsorption is negative. Also, if I had assigned a potential function for internal forces, what change would have taken place? Please help me out.

In effect, you discuss a system where gas molecules don't interact with walls except for elastic collisions. In such a system, indeed, adsorption won't be exothermic (and consequently, won't happen at all).

In what case, will it be exothermic?( I mean what kind of interaction with wall, like you specify is required)? and what is the change if I introduced potential energy for the internal forces?

An attractive interaction, of course.

I will point out another equation:
$$\mathrm{d}G = \mathrm{d}H - T\,\mathrm{d}S$$
For adsorption $\mathrm{d}S<0$, as molecules go more organized.
Therefore also $\mathrm{d}H < 0$ must be true, otherwise $\mathrm{d}G < 0$, needed for a spontaneous process, would not be achieved.

Its ok but my question why enthalpy is negative according to my model and your answer is justifying enthalpy must negative for adsorption being spontaneous but not how exactly it is.

For the same reason like the condensation from gaseous phase, or a cation  binding an electron. The energy is released by the bonding.

You have deliberately denied for the system to be in equilibrium with the surrounding, so the dH=0 by principle.

Your mistake is in saying that the heat exchanged is zero.  When we talk about a reaction being exothermic, it is based on the condition that the initial and final temperatures of the system are equal, and that, to achieve this, we exchange whatever amount of heat that is necessary.


text:
What are we seeing in this AC salt-water copper-wire experiment?
There's a highly watched video on YouTube by "ElectroBoom" on the effects of passing AC (not DC) current through salt water. The water gets cloudy and at the end of the video we can see one of the copper wires has been eaten away by the process.
Is there really "a reaction of the copper with the salt in water" as the author of the video claims? Or is it a pure physical process where the copper "boils off" the wire?

SAFETY - Using the name ElectroBoom gives me a bad vibe about this. By using alternating current you're getting both hydrogen and oxygen at each electrode. Together the two gases do react explosively. Don't do it...

@MaxW: I take your point about safety, but I haven't heard of AC current causing outright explosions in water. Although if that's the case, it would be an interesting part of the answer.

@MaxW: presumably if the current was high enough it could have the same effect as dropping something very hot in water, which can result in a vapor-related explosion, e.g. https://youtu.be/PDRWQUUUCF0?t=123

Is there really "a reaction of the copper with the salt in water" as the author of the video claims? Or is it a pure physical process where the copper "boils off" the wire?

Yes it reacts, no it doesn't boil. Essentially the wire corrodes at an accelerated rate where during the part of the cycle where the wire is the positive terminal the copper reacts as follows:
$$\ce{Cu^0 + 2OH- -> Cu^2+ + 2e- + 2OH- -> Cu(OH)2 + 2e- -> CuO + 2e- + H2O }$$
(expanded for clarity, not a reaction mechanism)
There is obviously oxygen generated too but I am only focusing on the copper part of the reaction. During the part of the cycle where the wire is the positive terminal the reverse of this reaction does not occur on the other half of the cycle and over time more and more copper will be consumed eventually consuming the entire wire.

The dark brownish color in water seems indeed to match that of CuO much more than that of metallic copper.

@Fizz There may be some $\ce{Cu2O}$ produced as well which is reddish, the video isn't very clear.

Congrats on 1×10⁴ rep:)

@andselisk Thanks - It was quite the grind.


text:
Role of pH in azo C-coupling reaction, continued
In the question, "What is the role of pH in azo coupling reaction of diazonium with phenol and aniline?" asked by @DhrubaBanerjee and answered by @Klaus-DieterWarzecha, I find several ends untied. I considered starting a bounty on the previous question, but since my question is looking for slightly more specificity than just looking at the whole picture, I decided to ask a separate question.
That post discusses why azo coupling of aniline is conducted in acidic medium. However, I am not convinced of a few things.
My query is twofold:

The $\ce{-^+NH3}$ group is inductively deactivating (as mentioned here). That is a turn-off for attacks on electrophiles (like a diazonium cation). But why is the protonation of aniline preferred?

In the case of aniline, the lone pair on the nitrogen atom is the preferred centre for the reaction with the diazonium cation:

This isn't what we want! Protonation of aniline avoids this side reaction.

I do agree with Klaus-Dieter Warzecha that an alkaline $\mathrm{pH}$ would lead to azo N-coupling, but if we were concerned about the electrophilic attack occurring from the nitrogen atom, in my opinion, it should have been acylated at $\ce{N}$, leaving no room for side reactions but also keeping the ring activated (although less than before). I believe the yield would be higher.

If an equimolar mixture of aniline and phenol was added to a reaction vessel containing benzenediazonium chloride in acidic medium, would the major product be phenol's couple product or aniline's couple product?

This Wikipedia article states that phenol's coupling is base-catalyzed. I assume it accelerates the reaction by assisting the formation of phenolate ion. Going by that, the aforementioned conditions will be less favorable for phenol's coupling.
But I find it hard to take in that a protonated aniline could couple faster than phenol. Phenol would still be capable of coupling (only unaccelerated), while aniline should have been weakened by the deactivation.
Hence I would choose phenol to form the major product. I wish to know if my reasoning is correct or not, and if not, an explanation.
I would greatly appreciate any constructive discussion/specific answers.

AFAIK in coupling of aniline first triazene is obtained, which is then rearranged to aminoazobenzene.

@Mithoron Could you give a reference?

https://patentimages.storage.googleapis.com/pdfs/9acc6efb7db031d11d18/EP0052511B1.pdf

re @Mithoron's comment on coupling of aniline, it's discussed in some detail at Does aniline react with diazonium ions at C or N?


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Predict the number and nature of bonds using Molecular Orbital Theory
I found the following question in a high school chemistry problem book:

According to Molecular Orbital Theory which of the following is true:
(A) In $C_2$ molecule both the bonds are $\pi$ bonds.
(B) In $C_2^{2-}$ ion there is one $\sigma$ and one $\pi$ bond.

Both of these are true. But how can one know this using Molecular Orbital Theory?
P.S.: I don't know anything about Quantum Mechanics as I am a high school student. My textbooks only provide a result-oriented explanation. I would appreciate if the answers keep this in mind.

Is answer (B) one or two $\pi$ bonds?

@Blaise (A) is two. $C_2$ violates the one $\sigma$, one $\pi$ rule. (B) is one $\sigma$, one $\pi$.

I presume you can draw the homonuclear diatomic MO diagram for these two species.
As you probably already know, the s orbitals of the carbons form bonding and antibonding orbitals, both of which are filled. (The net effect of this is the “lone pairs” on each carbon.)
The p orbitals for 2 pi bonds and a sigma bond. Due to s-p mixing, the pi bonds are lower in energy and are filled first, and these 2 bonds (with their 4 electrons) are the 2 bonds we see in C2.
Add another 2 electrons to the system and you realise they fill the sigma bonding orbital, which means that the C2 2- anion has 2 pi bonds and a sigma bond (for a total bond order of 3).
Hope this helped.

According to the same book (Resonance Chemistry for JEE Main and Advanced), bond order of $1$, $2$ and $3$ correspond to single, double and triple bond.
This is a rule of thumb and most common molecules, such as $O_2$, $N_2$, etc., seem to follow it.


text:
How can we conduct reactions for insoluble materials?
I am working on metal organic polydehra topic to synthesize new metal-organic cages or other based frameworks. I want to conduct postsynthetic modification for one cage, but the solubility is very poor in most solvents, only methanol can dissolve but it needs large volume.
My question for who in the field of MOFs, COFs, MOPs, and other related chemistries, how can I deal with the problem of insolubility of synthesized materials to conduct postsynthetic reactions? what are the proposed strategies and methods to attain the target modified compound?
Please you can answer based on your experience with reactions for any insoluble materials.

Judging from the work you are doing you should be able to consult the related literature. Solid state reactions are a real thing, if this never happened before in your field bear to reviews in other subfields. I just comment because this empty page attracted my attention.  I only know that one can do quite some by just grinding compounds and eventually microwave assists the process. But I have no idea of which class of compounds are suitable.


text:
How is a ring formed when a carbon atom in an aliphatic chain attached to benzene ring has a positive charge?
This is my approach towards this problem. First off, a positive charge would appear in the C atom at the end of the chain, and then that carbocation would rearrange into a tertiary carbocation.
But I am not being able to figure out what will be the next step. Can the carbocation extract a hydrogen from the benzene ring to form another five-membered chain as the formation of a five-membered rings is very feasible?

Yes, Definitely. The carbocation won't extract proton from Benzene ring, the $\pi$-electron cloud of benzene ring will attack to the carbocation to form the fused five-membered ring and a proton will be lost to rearomatise the system.

You need to review cation rearrangements.

@SoumikDas so should the benzene ring attack from the carbon opposite to the carbon bearing the -och3 group , as then the carbanion formed would be more stable?

Yes there will be electrophilic addition at the para position to form a 5-membered ring.

@PrathamYadav Yes, I think so, as attack from the ortho position of $\ce{-OCH3}$will make lot of steric crowding, attack from para position of $\ce{-OCH3}$ should be more favoured..

You should be familiar with the Friedel-Crafts (FC) alkylation reaction - this is just an example where the nucleophile (aromatic ring) and electrophile (alkyl halide) are connected by a piece of string (the alkyl chain). For me there are a couple of possible products: the FC reaction could occur directly on the primary carbocation, forming a 6-membered ring, or there could be an initial 1,2-hydride shift to generate the more stable tertiary carbocation, where the subsequent Friedel-Crafts would give a 5-ring. Which occurs depends on the relative rates of intramolecular FC vs hydride shift.


text:
Brass - is this corrosion or oxidation and why did it happen?
I have bought a new brass Y-split water valve which is plain brass but inside is a more polished one, it was packed nicely and I opened it 2 days ago, washed it off with water for some reason, then did a bit of drying with a kitchen towel, I then placed it in a plastic container, however it was more of placing it ON a plastic container as I did not seal it or anything.

Today (2 days after, not even 48 hours are passed) I checked it, and on one of the side the brass had these violet/bluish spots (which I thought are reddish/brown spots at first - but at better lightning they are more of a violet/bluish - my photos really do not show their actual colors). So what are they and why did it happen? This valve splitter is used for water so I don't know why this happened if water was the problem.
I tried to scratch it off with paper towels, and then a toothpick - but it did not work. I did not use anything besides water - and it looked perfectly fine 48 hours ago. Similar thing happened to a water aerator - had a perfect new one I actually checked it, put it on the water tap and ran like 3 liters of water and then I wanted to fix something so I got it out and boom it had these brown - reddish spots, but not this dense and this is not that color.
IF this in case is corrosion, my question would be is it possible for the corrosion to happen this fast only from water? Will the corrosion spread and how bad would it be to be used for a washing machine in terms of a health hazard or machine damage (besides leakage)? Again note that the color is not as dramatic as this and the pictures make it looks much more deeper than it is in real life, whereas it looks like there are stains just atop - although they can't be removed.

Related: https://chemistry.stackexchange.com/a/103270/23561

Beware of brass: https://www.proboat.com/2012/07/beware-the-brass/

More colors explained: https://www.canada.ca/en/conservation-institute/services/care-objects/metals/basic-care-recognizing-metals-corrosion-products.html   Metal quality aside - the water and/or the air may be tainted.

https://www.canada.ca/en/conservation-institute/services/care-objects/metals/basic-care-recognizing-metals-corrosion-products.html    mentions sulfur in air. Some water contains sulfur.

Thank you for all the findings, I am not at all knowledgeable at this field, but I could not find a similar photo to mine. You are suggesting it is some kind of corrosion though, and I should not use this valve?

If that happened in only 48 hours, I would return to the store if possible and replace with bronze. If you cannot return, I guess you could use it to water lawn, but would not use it for potable water or in house where a leak could happen.

Yeah, I got it like 2 months ago, opened it 2 days ago, everything was fine and now it is not. For potable water, you mean that I should not drink it if there is corrosion? I read that high-level of chlorine can cause brass to corrode and I know that my place has a higher level of chlorine, although this reaction was pretty fast wasn't it and for a miracle the other side in this Y valve, and most of the valve is untouched.

Do not know about safety in drinking it. So I personally would not until I learned more about what happened.

Related/saving for later:  Identification of the corrosion products formed on copper in sulfur containing environments


text:
For Hess' law type questions, when do you take the inverse of the constant and when do you switch the sign?
For example, when you're given the enthalpy $\Delta H$ of a  reaction and you want to switch that reaction around in order to add it to something else, you switch the sign. Furthermore, when adding two reactions, you add their $\Delta H$ values.
But when you're working with equilibium problems and you have the $K$ of a reaction, if you want to switch the reaction you take the inverse of $K$ and when you add reactions you multiply $K$ values.
When working with cell potentials $E_{cell}$ of a redox reaction, you switch the sign if you switch the equation around, but when you're adding two equations you do not multiply $E_{cell}$ by whatever you multiply the coeffecients on the species by.
Are there any types of patterns other than the three I've described here? Any help is greatly appreciated.

Equilibrium constant
When a reaction is at equilibrium, the species will have concentrations called the equilibrium concentrations. They do not depend on whether the reaction is written in one direction or in the reverse direction, and whether all the stoichiometric coefficients are multiplied by the same value. However, your equilibrium expression will be different (switching concentrations from numerator to denominator if reversing the chemical equation, or changing exponents when changing stoichiometric coefficients). As a result, the value of the equilibrium constant will change while the equilibrium concentrations themselves will remain the same.
Enthalpy of reaction
For a given reaction, the enthalpy change associated with making - say - 1 mol of species A is a given. If you use 1 mol of species A in the reaction instead, the enthalpy change will be of the same magnitude but opposite sign. If I make a different amount of species A, the enthalpy change will change in a manner proportional to that amount. In a similar manner, if I change the chemical equation for $\ce{A -> B}$ to $\ce{2A -> 2B}$, the molar enthalpy of reaction will double as well. The same goes for Gibbs energy or for entropy of reaction.
Cell potential
For a given reaction with certain concentrations of species in the two half cells of a voltaic cell, I will measure a certain cell potential. If I switch the role of reactants and products (in the equation), I will still measure the same potential (the direction of the actual reaction does not change). If I multiply all coefficients in the chemical equation by the same number, it still correctly describes the reaction, but this will not change the potential.
Mathematical relationships
$$\Delta G^\circ = - R T \ln(K)$$
The way logarithms work, if you multiply $\Delta G^\circ$ by a number, you have to raise K to the power of that number. If you add Gibbs energies, you have to multiply equilibrium constants.
$$\Delta G^\circ = - z F E^\circ $$
If you multiply $\Delta G^\circ$ by a number, z (the number of electrons transferred) will also increase by that number. Therefore, $E^\circ $ does not change.
Multiplying all coefficient in a chemical equation by (-1) changes the role of reactants and products. Thus, once you know the rules for changing coefficients by a common factor, you don't need to memorize another rule for reversing the equation.

Energies are additive.
Concentrations, partial pressures, and activities which are components of equilibrium constants are strongly related to probabilities. The probability of two things happening at the same time is computed by multiplying the probabilities, and likewise, the equilibrium constant is computed from products of concentrations/activities.
This is explicitly expressed in:
$$\Delta G^{\varnothing} = -RT \ln K$$
The equilibrium constant on the right is related to the free energy on the left via a log, since
$$\ln (ab) = \ln a + \ln b$$
So summing on the left hand side is related to multiplying inside the equilibrium constant on the right.
The potential $\mathscr{E}$ is related to the free energy, but it is not an energy. Potential is energy scaled per electron, so moving around extra electrons does not change the per-electron value.


text:
Two different electronic configurations for cerium
Depending on the textbook there are two different electronic configurations stated for cerium. On the one hand
$$[\ce{Xe}]\mathrm{4f^15d^16s^2}\quad$$
and on the other hand
$$[\ce{Xe}]\mathrm{4f^26s^2}$$
Are both configurations possible? If not, why so?

There is no need to waste your time in memorizing the differences. Yes, you will hear and read "stories" of this electron configuration and that electron configurations without saying why so. I have yet to find a spectroscopist (asked so many leading ones) who can tell me how these electron configurations were obtained from either experiment or by theory. I bet your teacher will have no clue either. So the question you should ask is how would we experimentally distinguish your first configuration from the other.

The filling of electrons follows Aufbau principle. However, introduction of $\mathrm{d}$ and $\mathrm{f}$ orbitals and resulting poor shielding affects predicted energy order, and it consists last 2 or three shells incompletely filled. The energy associated with this subshells is nearly same, and one of the subshell of outer shell is bigger in size than subshell of inner shell. So, to perform minimum repulsion between electrons, one or two electrons jump to a bigger subshell (with similar energy) so this exceptions occur.
In case of cerium, energy of $\mathrm{4f}$ and $\mathrm{5d}$ subshells are quite similar and $\mathrm{5d}$ subshell is bigger than $\mathrm{4f}$ subshell. So 1 electron in $\mathrm{4f}$ jump to $\mathrm{5d}$ subshell.

The configuration of cerium is
Ce -> 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f1 5d1
https://en.wikipedia.org/wiki/Cerium
The filling of electrons follows Aufbau principle nicely till Z<20, however, introduction of d and f orbitals and resulting poor shielding affects predicted energy order. Lanthanoids have their atomic configuration with a variable occupancy of 4f energy level and a common 6s2 level.The energy level of the 5d sub-shell increases than that of the 4f sub-shell moving across period(57–71). As a result of the low energy of 4f sub-shell electrons occupy these sub-shells before occupying the 5d sub-shell. As we progress from atomic number 57 to 71, an electron is added in 4f energy level with increase in atomic number with the exception of some metals where the electron occupies a 5d energy level.
In case of Cerium, energy of 4f and 5d subshells are quite similar and thus explains the electronic configuration stated above.


text:
Kinetics for E1cb reactions
We were taught that the kinetics of a reaction is determined by the reactions ‘rate determining step’ which is also the slowest step of the reaction. For E1 as well as E2 reactions the slowest steps are easily determined by the mechanism and I derived the rate laws for those.
In the case of E1cb reactions, which step can be taken as the slowest step?
Is it the one in which the acidic hydrogen is extracted by the base or is it the one where the elimination takes place (carbanion intermediate)?

I would think that this depends on the substrate.

See my answer here: https://chemistry.stackexchange.com/q/105485/44877

@TanYongBoon oh ok so the rate is proportional to the [R-X] only right?

@SomV.Tambe Since the rate-determining step involves both C-H bond breaking, wouldn't both base concentration and substrate concentration be factored in?

@TanYongBoon but then why is the reaction known as Unimolecular?

@SomV.Tambe May I know where you heard that from?

@TanYongBoon https://en.m.wikipedia.org/wiki/E1cB-elimination_reaction

I just checked. According to Carey & Sundberg (2007) Advanced Organic Chemistry, the E1cB mechanism involves second order kinetics. And I think it makes perfect sense as E1cB involves both base and the substrate reacting in the formation of the carbanion intermediate. Does that make sense to you?

Did you even read the section of the wikipedia page on chemical kinetics of E1cB? I think there is clearly a line stating that the reaction follows second order kinetics.


text:
Which is easier, the nitration of toluene or the nitration of mesitylene (1,3,5-trimethylbenzene)?
Does the toluene can undergo nitration easier than mesitylene (1,3,5-trimethylbenzene), due to the high repulsion that will occur between $\ce{NO2}$ and the two adjacent methyl groups (in mesitylene nitration case)?

What do you mean by easier? Sufficiency of mild conditions? Faster reaction?

In case of mesitylene one needs to be careful not to oxidize methyl groups - they aren't enough to prevent nitration, but compound is vulnerable to oxidation of side groups.

The kinetics of aromatic nitration depend on the nitrating method, which also bear directly or indirectly on the subject of nitration. For example, aromatic nitration displays second-order kinetics in sulphuric acid (when used in nitrating mixture), and first-order kinetics in nitric acid. Aromatic nitration also displays first-order kinetics in either nitromethane or acetic acid (when used as a solvent) with nitric acid in constant excess. It exhibits zeroth-order kinetics for sufficiently reactive aromatic compounds, and first-order kinetics for sufficiently unreactive compounds used for nitration (Ref.1). The reference predicts that:

The numerical results vary with the conditions, but are not very sensitive to them; so that the following statements may be taken as approximately true over a considerable range of temperatures and nitrating media. Toluene is attacked 24 times faster than benzene, fluoro- and iodo-benzene 6 times more slowly than benzene, chloro- and bromo-benzene 30 times more slowly than benzene, and ethyl benzoate 300 times more slowly than benzene. Qualitatively we know that benzenesulphonic acid is attacked still more slowly, and nitrobenzene even more slowly. Thus the relative effects of substituents in activating or deactivating the benzene nucleus towards this type of attack is expressed by the following sequence:
  $$\ce{Me > (H) > F, I > Cl, Br > CO2Et > SO3H > NO2}$$

Thus, it is fair to conclude that the nitration of mesitylene is faster than that of toluene regardless of the steric effects enforced by neighboring 1,3-dimethyl functions. That conclusion is confirmed by the relative rates listed in Ref.1 & 2 where it is the case in all conditions considered (obs. rates are at $\pu{25 ^{\circ}C}$):
$$
\begin{array}{|c|c|c|c|c|c|}
\hline
\textrm{Compound} & 7.5\% \: \textrm{aq.} & 15\% \: \textrm{aq.}  & 61.01\% \: \textrm{aq.}  & 68.3\% \: \textrm{aq.}  & \textrm{Estimated}  \\
\textrm{nitrated} & \textrm{Sulfolane} & \ce{CH3NO2} & \ce{HClO4} & \ce{H2SO4} & \textrm{Rel. Rate}  \\ \hline
 \textrm{Benzene} & 1 & 1 & 1  & 1 & 1 \\ \hline
\textrm{Toluene} & 20 & 25 & 19  & 17 & 23 \\ \hline
\textrm{p-Xylene} & 114 & 130 & 84 & 38 & 50 \\ \hline
\textrm{m-Xylene} & 100 & 146 & - & 38 & 400 \\ \hline
\textrm{Mesitylene} & 350 & 400 & 78  & 36 & 16000 \\ \hline
\end{array}
$$
Estimated relative rates are from reference 3 (the authors have calculated from the observed value for toluene and the known isomer ratios by assuming the additivity principle).

References:

E. D. Hughes, C. K. Ingold, R. I. Reed, “493. Kinetics and mechanism of aromatic nitration. Part II. Nitration by the nitronium ion, $\ce{NO2^+}$, derived from nitric acid,” J. Chem. Soc. 1950, 2400–2440 (DOI: 10.1039/JR9500002400).
J. G. Hoggett, R. B. Moodie, J. R. Penton, K. chofield, In Nitration and aromatic reactivity; Cambridge University Press: London, United Kingdom, 1971, “Chapter 3: Nitrating Systems: B. Inert organic solvents,” pp. 32–49.
R. G. Coombes, R. B. Moodie, K. chofield, “Electrophilic aromatic substitution. Part I. The nitration of some reactive aromatic compounds in concentrated sulphuric and perchloric acids,” Journal of the Chemical Society B: Physical Organic 1968, 800–804 (DOI: 10.1039/J29680000800).


text:
What is the minimum operating temperature of chemical hand warmers?
I have been looking at chemical hand warmers such as Hot Hands, Grabber and HotSnapZ. Unfortunately, I cannot find a minimum operating temperature for these types of devices. Logic tells us that they won't work at -273°C and will probably work at 0°C, but I'd like to have an idea of where the cutoff lies between them.
From Wikipedia here and here, it looks like they all contain some amount of saltwater. The lowest freezing temperature I can find for saltwater at saturation (26%) is -21°C. Is it reasonable to assume that a frozen hand warmer will not work? Would the other components in the hand warmer lower the freezing temperature even more?
TL;DR
What is the minimum temperature at which a chemical hand warmer can still provide heat?

Here is the MSDS for Hot Hands. The reaction is fine iron powder that reacts with oxygen from the atmosphere to form some sort of iron oxide/hydroxide. See here for example The Chemical Magic of Hand Warmers

The amount of salt is small, so the f.p. depression would not be great. If you want to make your own, CaCl2 would be a better salt for low temps.

See this C&E News Article.

They are often made of the crystallic $\ce{CH3COONa.3H2O}$, that dissolves in its own crystal water, when heated up.
As the cooled down liquid state is supercooled solution, it is hard to predict, at what temperature it would spontaneously solidify.
It will provide heat in any case. But as it warms up itself typically by  $\pu{40-50^{\circ}C}$, it would barely warm itself to room temperature, if it was previously cooled down to e.g.  $\pu{-60^{\circ}C}$.
The air activated heaters, based on oxidation of fine metal powders (thanks @MaxW ), may have problem with too slow initial reaction when too cold. So they may need to be kept at pocket temperature for some time before activation.

The  "HotSnapZ" product, MSDS, is sodium acetate. The other two are iron powder.

I see. I do not have experience with those particular products, having experience with molten hydrates only.

I remember also a catalytic heater, burning vapour of lead-free petrol on fine platinum(?) mesh.


text:
What is the flow of electrons, cations, and anions in a voltaic cell (please see image attached)?

This is an IB question. The mark scheme says that the correct answer is A and only A. But I believe B is also correct. The following is my thought process. Since A is more reactive, it undergoes oxidation to form A2+ ions, while B2+ ions undergo reduction and form B. From this we can deduce that A is the anode and B is the cathode. According to my notes and many sources on the internet, electrons and cations both travel from the anode (A in the image) to the cathode (B in the image). Therefore, A and B would both be correct. Please can someone confirm my thinking or point out any mistakes. Thank you

According to my notes and many sources on the internet, electrons and cations both travel from the anode (A in the image) to the cathode (B in the image).

The idea of the salt bridge is to prevent electrolytes mixing while providing ion flow. When you have a high concentration of inert ions in the salt bridge, cations in the salt bridge will flow into B, and anions in the salt bridge will flow into A. The solution in the salt bridge is not mixed, so it would take a very long time for the electrolyte in A to diffuse all the way to B.
To conclude, ions flow out of the salt bridge into the solutions A and B, but virtually no ion flows all the way through the salt bridge. This tells you that the salt bridge will be depleted after a certain amount of current flows through the wire. Accordingly, electrochemical cells used in batteries usually don't have a salt bridge but a semipermeable membrane instead, or a clever choice of half reactions where mixing of the electrolytes is not a problem (lead acid battery for example, where the only soluble species, sulfuric acid, is common to both half reactions).

It is a bad question because "Metal A is more reactive than Metal B" has no real meaning. It is a vague choice of word, A is more reactive, but with respect to what?
Your logic is correct

Since A is more reactive, it undergoes oxidation to form A2+ ions,
  while B2+ ions undergo reduction and form B. From this we can deduce
  that A is the anode and B is the cathode.

All you have to do now is to deduce the sign of A and B. Since A is losing electrons, where do these electrons go? They impart a negative charge to the electrode A. Electricity flows from the negative terminal to the positive terminal, if we take electrons as the charge carriers.*

In older books or electrical engineering you might see something called as the conventional current. The direction of the conventional current is from the positive pole to the negative pole and the charge carrier is considered to be a positive entity.

The conventional current is still from positive to negative potential, regardless of awareness of the opposite electron current in metallic conductors. And, in case of LHC, the current is based on positive entities.

The word "reactive" to me is opposite of "noble". In my experience with analyzing corrosion issues, zinc is more reactive than carbon steel which is more reactive than copper which is more reactive than stainless steel which is one of the most "noble" (i.e. "unreactive") metals. Zinc, carbon steel, copper, and stainless steel are all members of the "Galvanic Series"; this is an ordered list of metals sorted by the equilibrium cell potentials (a.k.a. "electromotive force" (deprecated); units are volts) of their half reactions.

It is a bit of a tricky question. The trick is in the wording "flow through". The positive ions on the A side do not actually "flow through" the salt bridge from A to B. Rather positive ions exit the salt bridge on the B side to compensate for the cations depositing on the cathode (B). Likewise negative ions exit the salt bridge on the A side to balance the charge of the cations being liberated from the anode (A).
Some explanations also show ions moving into the salt bridge:
Anion moving in.
Cation and anion moving in.


text:
Cis and Trans Isomerism given three different groups around the reference plane
In the following problem I am meant to deduce the molecules that are cis-trans isomers of each other.
This question does not make much sense in my head because I thought that cis/trans isomerism was only applied to configurational isomers if there were only two types of groups bonded to the carbon atoms in the carbon-carbon double bond i.e. the reference plane. Such that, if there are more than two groups bonded to the carbons in the reference plane then E/Z isomerism should be used. However, then I thought that cis/trans isomerism is a subcategory of E/Z isomerism and therefore I tried to work out the solution using the E/Z method of finding configurational isomers by finding the relative priorities based on the atomic numbers for the atoms attached to the carbon atoms on each side. Using this method I discovered, that the solution could only be 'A. X and Z' since X seems to be an 'E' isomer since its high and low priority groups are across from eachother and Z seems to be a 'Z' isomer since its high and low priority groups are on the same side.
Am I on the right track?

Am I on the right track?

Generally, yes. For this question, yes but you have over thought this question.
For starters, any answer with Y can be totally eliminated as Y is neither cis nor trans. Now you are down to answers 'A.' and 'D.'
Here you correctly determined the E and Z configurations of molecules with identical functional groups, but could have saved yourself some effort if you had recognized W and Z are identical molecules that are inversions/180° rotations of each other. Thus it can't be D. and the answer is in fact:

A.

Is the question right in calling them cis-trans or should it actually be E/Z


text:
Interaction of the oscillating magnetic field of electromagnetic radiation with a molecule or atom
Whenever we read rotational, vibrational and electronic spectroscopy, we find details on the interaction of the electric dipole moment of the molecule with the oscillating electric field of electromagnetic radiation. The selection rules are also derived on the basis of transition dipole moment operator on wavefunctions. I was talking to spectroscopists today and asked that why do spectroscopy textbooks (and authors) do not mention anything about oscillating magnetic field of light waves. The magnetic field also carries the same energy as the oscillating electric field. Doesn't that interact with electrons as well?* They said that interaction of the magnetic field with a molecule's electric dipole is very weak and hence nobody mentions about it. Does anyone have a good reference or an equation which shows that the oscillating magnetic field interacts very weakly with the molecule's electric dipole moment? They were mentioning Fermi's golden rule and at this stage I lost them.
__
* Electron spin resonance is an example.

One simple way to intuit the neglect of magnetic field interactions is to note that few molecules have significant magnetic fields to interact with. First order magnetic fields (which would generate a strong interaction with the magnetic field component) are a product are a product of unpaired electrons and these are not that common in stable species. The exceptions are some transition metal compounds and the dioxygen molecule which is a diradical. ESR is an example of a spectroscopic tool that exploits the effect but is often used for exploring short-lived radical intermediates.

The magnetic dipole transition intensity is typically about $10^5$ times smaller than the allowed electric dipole but as this can vary by ten orders of magnitude, second order transitions (with terms $x^2,y^2$ etc ) as well as magnetic dipole transitions can occasionally be important. The electric dipole matrix element  is $\langle a|\hat r|b\rangle$, where $\hat \mu =e\hat r$, the  second order dipole $\langle a|\hat r \hat r|b\rangle$ and magnetic dipole $\langle a|\hat J|b\rangle$ where $\hat J$ is the angular momentum operator
$\displaystyle \sim \langle a|z\frac{\partial}{\partial x} - x\frac{\partial}{\partial z} |b\rangle $.
The transition intensity is the square of these terms and the selection rules differ for each type of transition. A derivation is given over several pages by Steinfeld 'Molecules and Radiation' chapter 1, 2nd ed (publ MIT Press).

Thanks for the reference. Can the electric dipole moment interact with an oscillating magnetic field of the light wave?

Magnetic fields interact with particle spin, electron or nuclear, as in nmr spectroscopy (nuclear Zeeman efffect). I'm not aware of the type of interaction you mention, but that may well just be my ignorance.

I got the book by Steinfeld now. It is pretty intensive. Will check if I can find a qualitative version. Thanks.


text:
Excited state intramolecular proton transfer mechanism
I am trying to explain Excited state intramolecular proton transfer(ESIPT) mechanism through DFT method using Gaussian 09.Generally, in ESIPT mechanism the molecule exist in enol form in its electronic ground state. therefore the calculated energy of the ground state  enol form would be lesser than its ground state  keto tautomer.But, In my case the calculated energy of ground state keto tautomer is less than ground state enol form. moreover,the Potential Energy Surface scanning(PES)[of ground state optimized enol tautomer] was also performed with O-H bond length co-ordinate.the scanning plot has shown least energy of keto tautomer. For ESIPT,can the molecule exist in keto form in its electronic ground state?


text:
Finding simplest formula of a compound with following specifications
I ran into the following problem, wherein I'm asked to find the simplest formula of a compound containing $50 \%$ of element $X$ (atomic wt. 10) and $50\%$ of element $Y$ (atomic wt. 20).
I know about few laws of chemical combinations. I think Law of Definite Proportions may come into play.
$$\text{atomic wt. }X:\text{atomic wt. } Y=1:2\implies XY_2$$
Am I correct in my reasoning? If not kindly point out the flaws, if any? Thanks

The first thing to ask is % of what? If this is mass percent you will get a different answer than you would get if the percentage is expressed in moles.

Could you tell me what the answer would be in each case. The solution is given to be $X_2Y$

$$M=\frac{m}{n}\ \curvearrowright\ m=M\cdot n$$
$$M_X=10\ g\ mol^{-1},\ M_Y=20\ g\ mol^{-1}$$
$$\frac{m_X}{m_X+m_Y}=50\ \%,\ \frac{m_Y}{m_X+m_Y}=50\ \%:$$
$$m_X=m_Y$$
$$M_X\cdot n_X=M_Y\cdot n_Y$$
$$\frac{n_X}{n_Y}=\frac{M_Y}{M_X}=\frac{20\ g\ mol^{-1}}{10\ g\ mol^{-1}}=2$$
$$n_X=2\ n_Y$$
$$\curvearrowright\ X_2Y$$


text:
Stability of molecules by resonance
I am pretty confused by some principles of resonance. Well by my study of electrostatics I know that potential energy of a system of  charges of same nature decreases as we increase the separation between them but same doesn't goes with a system of charges with different nature because their potential energy decreases as we decrease their separation as internal conservative forces do positive work.
I am a newcomer in chemistry so I have a doubt that my book says that energy of a molecule is minimum when they are separated by large distance (under the heading of application of resonance) without caring for nature of charges.This fact goes well for charges of same nature but contradicts the fact for system having different nature of charges.
Please tell me  that why this contradiction exists.

Why not to post what exactly the biok is saying? Nature should be sign,  they is presumably charges, and so on. As far the electrostatic potential is at play, stick to what you have already learned.

In chemical resonance, the same principle holds as in electrostatics:
"In general, the boundary structures of a conjugated system with the lowest energy are those
...
- in which the distance between charges of opposite sign is as small as possible and the distance between charges of the same sign is as large as possible
... ."
Wikipedia: Resonance (chemistry) - Major and minor contributors:
"Energetically unfavourable and therefore less favorable structures are minor contributors. With rules listed in rough order of diminishing importance, major contributors are generally structures that
...
3. carry a minimum of formally charged atoms, with the separation for unlike and like charges minimized and maximized, respectively
... ."


text:
Calculate chemical potentials to find electrode Fermi level
I'm running some simple simulations of the semiconductor-electrolyte interface and I'm having a hard time figuring out the equilibrium condition. Can someone please help me with my electrochemical potentials equation? Here's my thought flow so far:
I'm interested in finding out what is the Fermi level in the SC at equilibrium for OER: $4OH^- = O_2 + 2H_2O + 4e^-$. So I start by writing the condition for equilibrium:
$4\tilde \mu_{OH^-} = \tilde \mu_{O_2} + 4 \tilde \mu_{e^-} + 2 \tilde \mu_{H_2O}$
Since $O_2$ and $H_2O$ have no charge, their electrochemical potential is just their chemical potential ($\tilde \mu_{O_2} = \mu_{O_2}$ and $\tilde \mu_{H_2O} = \mu_{H_2O}$).
Moreover, the $\tilde \mu_{e^-}$ electrochemical potential of $e^-$ in my semiconductor electrode is just the Fermi level $E_F$ I'm looking for. Keeping in mind that $\tilde \mu_i = \mu_i^0 + RTln(a_i)  + z_iF\phi$ and that the activity of water $a_{H_2O}=1$, I arranged the balance equation to:
$E_F = (\mu^0_{OH^-} - \frac{1}{4}\mu^0_{O_2} - \frac{1}{2}\mu^0_{H_2O}) + RTln\left(\frac{a_{OH^-}}{\left( \frac{p_{O_2}}{p^0}\right)}\right) -F\phi_{sol}$.
I can calculate the electrical potential $\phi_{sol}$ at the interface in the solution with the Poisson equation. I can calculate the activity coefficient $a_{OH^-}$ based on the $[OH^-]$ concentration at the interface and of course I know the partial pressure of $O_2$. My problem is how do I determine the chemical potentials at standard conditions from the first parantheses?
$(\mu^0_{OH^-} - \frac{1}{4}\mu^0_{O_2} - \frac{1}{2}\mu^0_{H_2O}) = ???$
Could I related them somehow to the SHE electrode which I know to be approx $4.5\ eV$ vs. vacuum?


text:
Is Gibbs energy change during osmosis positive? If so then why is it spontaneous?

For ideal solution, μA=μA* +RT
  lnxA

ln function is an increasing function, means that the higher the xA the higher the μA......(1)
Now consider osmosis. From Atkins' Physical Chemistry Ninth Edition, page 173,

osmotic pressure, π, is the pressure that must be applied to the solution
  to stop the influx of solvent (into solution across a semi-permeable
  membrane).

Here is my interpretation and thought process:
Had you not exerted π on solution, the μ(solution)< μ(solvent) based on (1) because lnxA is negative for xA<1, so by Second Law, the osmosis is spontaneous and solvent should move from pure solvent (high μ) to solution (low μ). Sounds right.
However, during this process, μ(pure solvent) unchanged (μ is intensive, not affected by amount) but μ(solution) increases due to dilution. Based on (1), for the solution, since xA increases μ(solution) must also increase. Although the process stops when μ(solution)=μ(pure solvent), the effect is that μ increases. μ increases would in turn imply ΔG>0 which means non-spontaneous, but as far as I know osmosis is spontaneous, so where is my misconception?

When I do the math, I get $\Delta G<0$

@ChetMiller How? But I see the μ increases eh

I’m away from my computer now.  I’ll write out my full analysis later.  The final result is dG/dn = RTln x, where n is the number of moles of solvent in the mixture container and x is its mole fraction in that container.

Let the subscript A refer to the solvent and B refer to the solute, and let the subscript 1 refer to the pure solvent container and  2 refer to the mixture container.  Then the free energy of the combined system is given by:
$$G=n_{A1}\mu_A^0+n_{A2}\left(\mu_A^0+RT\ln{\left[\frac{n_{A2}}{n_{A2}+n_{B_2}}\right]}\right)+n_{B2}\left(\mu_B^0+RT\ln{\left[\frac{n_{B2}}{n_{A2}+n_{B_2}}\right]}\right)$$If the number of moles of solvent in container 2 increases by $dn_{A2}$ and the number of moles of solvent in container 1 decreases by $dn_{A2}$, the change in the free energy of the combined system changes by dG.  If we do the math correctly, we obtain:
$$dG=RT\ln{x_{A2}dn_{A2}}\tag{1}$$where $x_{A2}$ is the mole fraction of solvent in the mixture container at any point:  $$x_{A2}=\frac{n_{A2}}{n_{A2}+n_{B_2}}$$Eqn. 1 indicates that the change in free energy is negative (as expected).

Just to clarify something, when you derive eqn.1,

do you assume n of A1 is approximately constant?
why is not μ of pure A (that μ
,A,0) included in dG expression?

Of course I don't assume $n_{A1}$ constant.  If you differentiate my expression and take $dn_{A1}=-dn_{A2}$, the $\mu_A^0$ terms cancel out.

It would have been much easier to do this problem if I initially had just written $$dG=\mu_{A1}dn_{A1}+\mu_{A2}dn_{A2}+\mu_{B2}dn_{B2}$$since $dn_{B2}=0$. This would have immediately led to the same final result.

To evaluate whether the process is spontaneous you have to take the difference of the chemical potentials $\mu$. Solvent molecules go from the pure solvent (reactant) to the solution (product):
$$\Delta_r G = \mu_\text{product} - \mu_\text{reactant} =$$
$$ = (\mu_\text{solvent}^\circ + R T \ln x_A) - \mu_\text{solvent}^\circ < 0$$
Throughout the process, $\mu$(solvent with solute)< $\mu$(pure solvent). As the process goes on, the solution becomes more and more diluted, so $\mu$(solvent with solute) will continuously increase, but never reach $\mu$(pure solvent). So the transport will go on, but the osmotic pressure will decrease as you get closer to equilibrium
In a different example, if one solution is at 1 mM and the other at 2 mM, solvent will go from the lower solute concentration to the higher solute concentration (or you could say from higher solvent concentration to lower solvent concentration) until the solute concentrations are equal and the osmotic pressure difference is zero.

'μ(solvent with solute) will continuously increase', and if μ(pure solvent) does not change, it means your μ difference is positive, so not spontaneous?

$x_{A}$ is strictly less than 1, so $\mu_{\mathrm{product}} < \mu_{\mathrm{reactant}}$. @The99sLearner

@Zhe Does the inequality mean ΔG>0 because potential increases?

I think there might just be some looseness in terminology and notation here. The point is that you're continually diluting, so the mole faction is getting repeatedly smaller. That means that over time the potential becomes more negative thanks to the log, so the overall free energy change is also negative.

To evaluate whether the reaction goes forward you don't compare $\Delta G$ over time but compare the difference in Gibbs energy between products and reactants at each given time point.

@Zhe  Afaik the subscript 'A' in the formula refers to the solvent but not solute. As solvent moves across membrane to solution, the mole fraction of A in the solution actually increases so its( i.e that solvent species') chemical potential also increases. Just like we say pure solvent highest potential, so more diluted higher potential.

@KarstenTheis Good point, but without drawing to molecular terms how do you show that the difference in Gibbs energy between products and reactants at each given time point<0? If your overall ΔG already>0, there must be a step contributes to it>0 right?

"As solvent moves across membrane to solution, the mole fraction of A in the solution actually increases": It increases by an minute amount. If this bothers you, take a very large volume of solution. The solvent molecule goes from pure solvent to a solution where the solvent is at significantly lower concentration, so that's the direction it goes.


text:
What kind of functional groups can give positive silver mirror test?
kindly help me with a mechanism of Tollen's reagent that could explain how it acts as a reagent with organic compounds I wish to understand how we can analyze through structure whether a given compound will give a positive silver mirror test or not.
i found a mechanism that explains it's reaction with aldehydes but again it also gives a positive test for terminal triple bond. (link to that question)
I am unsure whether it could give the test with any other functional group also
link to an article that could give more insight on this would be of great help
p.s. I m really new to asking questions so please do help me in asking better questions and framing them better


text:
Histidine vs Lysine
Among the amino acids, histidine and lysine, which is more basic?
And why?
I tried protonating and checking electronic effects but couldn't reach a satisfactory conclusion.

As their name imply, each amino acid has at least one amine and one acid functional group, but in most amino acids, the basicity of the amine is offset by the carboxylic acid group at physiological $\mathrm{pH}$, and hence considered neutral.
Amines are basic because the nitrogen has an unshared electron pair that can accept an $\ce{H+}$ more readily than that water does. They will even, to some extent, take an $\ce{H+}$ from water, leaving a hydroxide ($\ce{OH-}$) ion. There are three amino acids that have basic side chains at neutral $\mathrm{pH}$, which are named arginine (Arg), lysine (Lys), and histidine (His).
Arginine is the most basic among them because it contains guanidine side group, $\ce{-(CH2)4NHC(=NH)NH2}$, which is basic.
Lysine has two amine groups, which makes it overall basic because of the second isolated amine group $(\ce{-(CH2)4NH2})$.
Histidine, on the other hand, contains imidazole group, which is also basic.
The $\mathrm{p}K_\mathrm{a}$'s of all three side chains are high enough that they tend to bind protons, gaining a positive charge in the process (See below chart for $\mathrm{p}K_\mathrm{a}$ values). The $\mathrm{p}K_\mathrm{a}$ values are in order of
$$\ce{-(CH2)4NHC(=NH)NH2}  > \ce{-(CH2)4NH2} > \ce{-(CH2)-(\text{4-imidazole})}$$
(see the chart), thus, lysine is considered to be more basic than histidine in physiological conditions. Please also note that, in the following chart: $\mathrm{p}K_\mathrm{a1} = \alpha$-carboxyl group, $\mathrm{p}K_\mathrm{a2} = \alpha$-amine group, $\mathrm{p}K_\mathrm{a3} =$ side chain group, and $\mathrm{pI} =$ the isoelectronic point.
$$
\begin{array}{lccccc}
\hline
\textrm{Amino Acid} & \mathrm{p}K_\mathrm{a1} & \mathrm{p}K_\mathrm{a2}  & \mathrm{p}K_\mathrm{a3}  & \mathrm{pI}  & \textrm{acid/base} \\
\hline
\textrm{Aspartic acid} & 1.88 & 9.60 & 3.65 & 2.77 & \textrm{acidic} \\
\textrm{Glutamic acid} & 2.19 & 9.67 & 4.25 & 3.22 & \textrm{acidic} \\
\textrm{Arginine} & 2.17 & 9. 04 & 12.48 & 10.76 & \textrm{basic} \\
\textrm{Lysine} & 2.18 & 8.95 & 10.53 & 9.74 & \textrm{basic} \\
\textrm{Histidine} & 1.88 & 9.17 & 6.00 & 7.59 & \textrm{basic} \\
\hline
\end{array}
$$
I thank University of Calgary, Canada and University of Wisconsin, Madison, WI for relevant data.

I took the liberty to improve this great answer's formatting and code semantics. Please note that \ce{...} is used for chemical formulas only, to put math operators in upright mode \mathrm{...} should be used instead. I also cleared-up the table a little according to ACS style recommendations, see also Small Guide to Making Nice Tables (PDF), which is pretty much a TL;DR version of those.

@andselisk: Great edition! Thank you. You learn something everyday! :-)

Great explanation! I was wondering if there is any way to theoretically(by electronic effects) predict the basic strength without looking at the K values.

@Mathew Mahindaratne , can you reply to my above question please?


text:
Does the prefix of an inverted unit also get inverted?

What is the energy of radiation that has a frequency of $\pu{2.51 \times 10^11 ms-1}$?
(a) $\pu{1.66 \times 10^-19 J}$ (supposedly correct)
(b) $\pu{1.66 \times 10^-22 J}$
(c) $\pu{7.92 \times 10^-37 J}$
(d) $\pu{1.66 \times 10^-25 J}$

My argument was that: since $\pu{Hz} = \pu{s-1}$, then $\pu{ms-1} = \pu{mHz}$. So, I divided $2.51 \times 10^{11}$ by $1000$ and solved the question normally (using $E = h\nu$). But my professor said that since $\pu{m}$ is a prefix, it should follow whatever is in front of it, so $\pu{ms-1}$ will become $1/\pu{ms}$, then you multiply it by 1000 to change it to Hz. Is that the correct way to do it?

I applaud the literal interpretation, but yes the prefix goes with the unit therefore ms^-1 is equivalent to (ms)^-1.

The prefix is part of the unit, in that milliseconds is its own unit. However, in this situation, it seems silly to write $\mathrm{ms}^{-1}$ when you could instead write $10^{3}\ \mathrm{s}^{-1}$.

Yes it does, but to remember it in this form means to sweep a misunderstanding under the rug, which invites troubles in the future. Think of it this way. There are no prefixes. There are just thousands, millions, and other numbers, which you supposedly know how to handle when they turn $(\dots)^{-1}$.

You should read $\pu{ms-1}$ as $(\pu{ms})^{-1}$ and not as $\mathrm{m}(\pu{s-1})$. Your teacher is correct.

[...] since $\pu{Hz} = \pu{s-1}$ then $\pu{ms-1} = \pu{mHz}$

This is incorrect.
If something happens once per hour, it happens less often than when it happens once per minute. This is because an hour is longer than a minute. In a similar manner, if something happens once per second, it happens less often than when it happens once per millisecond. This is because a second is longer than a millisecond.
So $\pu{ms-1} = \pu{kHz}$, i.e. the frequency of something happening every millisecond is larger than the frequency of something happening every second.

This issue is explicitly addressed in BIPM: The International System of Units (SI) as follows.

The grouping formed by a prefix symbol attached to a unit symbol constitutes a new inseparable unit symbol (forming a multiple or submultiple of the unit concerned) that can be raised to a positive or negative power and that can be combined with other unit symbols to form compound unit symbols.
Examples: $$2.3\ \mathrm{cm}^3=2.3\ (\mathrm{cm})^3=2.3\ (10^{-2}\ \mathrm m)^3=2.3\times10^{-6}\ \mathrm m^3$$
$$1\ \mathrm{cm}^{-1}=1\ (\mathrm{cm})^{-1}=1\ (10^{-2}\ \mathrm m)^{-1}=10^2\ \mathrm m^{-1}=100\ \mathrm m^{-1}$$
$$1\ \mathrm{V/cm}=(1\ \mathrm V)/(10^{-2}\ \mathrm m)=10^2\ \mathrm{V/m}=100\ \mathrm{V/m}$$
$$5000\ \mathrm{µs}^{-1}=5000\ (\mathrm{µs})^{-1}=5000\ (10^{-6}\ \mathrm s)^{-1}=5\times10^9\ \mathrm s^{-1}$$

Therefore, $2.51\times10^{11}\ \mathrm{ms}^{-1}=2.51\times10^{11}\ (\mathrm{ms})^{-1}$.


text:
Treatment of 2-Nitrophenol
What are the ways to treat / disassemble 2-Nitrophenol ?
Is there an academic article explaining and detailing the process?

I hope when you ask what are the ways to treat/disassemble 2-nitrophenol, you are asking how to purify industrial waste water from nitrophenol pollutants. If so, there are few research have done in the field. It is well known that nitrophenols are some of the most refractory pollutants present in industrial waste waters. Because of their high stability and solubility in water, the degradation of these compounds to non-dangerous levels has been a very difficult process. One promising method of purification is the photocatalytic degradation of organic and inorganic pollutants in aqueous dispersions. Accordingly, the photolysis of nitrophenols in a homogeneous medium irradiated by UV light has been reported in several occasions. One such research used the photocatalytic degradation of 2-, 3-, and 4-nitrophenol in aqueous titania (anatase; a mineral form of titanium dioxide, $\ce{TiO2}$) dispersions irradiated by near-UV light (Ref.1), the abstract of which states that:

The results are reported of an investigation into the photodegradation of nitrophenol isomers carried out in aqueous titania (anatase) dispersions under various experimental conditions. We investigated the influence of various parameters on the photodegradation rate, such as the concentrations of the catalyst, dissolved oxygen and organic compound, and the presence of various anions in the dispersion. A plausible explanation is proposed for the influence of these parameters, both mechanistic and kinetic factors being accounted for. Gas chromatography and mass spectrometry were used as analytical methods in order to determine intermediate products. Spectroscopic methods (UV-vis and IR) were used to obtain information on the fate of adsorbed and intermediate species formed during the reaction.

The research found out that 2- and 4-nitrophenols were photooxidized with similar photodegradation rates (2-nitrophenol been slightly faster) and with faster rates than that of 3-nitrophenol. A mixture of nitrites and nitrates has been found at the end of these photodegradation runs.
Palmisano and others have also studied the effects of these rate with $\ce{pH}$ (Ref.2) and an increase of the substrate concentration.
References:

V. Augugliaro, L. Palmisano, M. Schiavello, A. Sclafani, L. Marchese, G. Martra, F. Miano, “Photocatalytic degradation of nitrophenols in aqueous titanium dioxide dispersion,” Applied Catalysis 1991, 69(1), 323–340 (https://doi.org/10.1016/S0166-9834(00)83310-2).
V. Augugliaro, M. J. López-Muñoz, L. Palmisano, J. Soria, “Influence of $\ce{pH}$ on the degradation kinetics of nitrophenol isomers in a heterogeneous photocatalytic system,” Applied Catalysis A: General 1993, 101(1), 1991, 7–13 (https://doi.org/10.1016/0926-860X(93)80133-B).


text:
Negative oxidation state
Why doesn’t aluminum have negative oxidation states? In almost all compounds of aluminum  which I know of has positive  oxidation state.
It has valence EC [Ne] 3s2 3p1 , my teacher told it usually forms +1 and +3 oxidation state by completely sharing p orbital electron or both s and p orbital completely(and +2 o.s. is always average of aluminum)
So why can’t aluminum or any 13th group element for that matter accept electron and make its p orbital half filled i.e form compounds with negative oxidation state?

Aluminum does assume negative oxidation states, see the list of elemental oxidation states in Wikipedia. Aluminum assumes a negative oxidation state when forming compounds with more electropositive elements like magnesium.
The magnesium compounds referred to above include $\ce{Mg_{17}Al_{12}}$.  In this compound there is enough magnesium to half-fill the valence $p$ orbital on aluminum and thus attain a $-2$ oxidation state for the aluminum, which is the minimum in Wikipedia's list. (Part of the magnesium would then be below $+2$).  However, this is only a formal designation. Magnesium aluminides are far from ionic in character.

I think you should see https://chemistry.meta.stackexchange.com/questions/4358/whats-the-point-of-providing-a-link-to-the-mobile-versions-of-the-websites

Can you edit url you paste? See my edit. Otherwise you could modify it later, or other users could do it.


text:
Confused between structural isomers versus stereoisomerism
So this question wanted me to find two cyclical compounds that are 'structural' isomers of the molecular formula given in the question.
The two isomers I came up with are: 1,2-dichlorocyclopropane and 1,1-dichloropropane since I thought that structural isomers are molecules with the same molecular formula but different arrangements of atoms.
However, the mark scheme to this question states that the answer is: '(cis or trans) 1,2-dichloropopane', which is confusing to me since I thought that cis/trans isomers are stereoisomers and not structural isomers where stereoisomers are molecules with the same structural formula but different arrangements of atoms in space.
So is the mark scheme correct in it's solution?

Structural isomers have the same molecular formula (1-butene and 2-butene). Stereoisomers are a subset of isomers in that they have the same atom connectivity but not the same spatial arrangement, e.g., (E)-2-butene and (Z)-2-butene.

Alright! But the question wanted me to write down two structural isomers which i assumed i did, but the mark scheme mentions stereoisomers i.e. cis/trans isomers which is confusing to me

For a ring it must be cyclopropane. Now figure out where to place the H's and Cl's.

I agree with you but in a slightly different way. The question is correct, i.e. there are two structural isomers, but as given in your answer, the two isomers are not cis and trans dichlorocyclopropane, because as you mentioned correctly, cis and trans isomers are definitely stereoisomers and not structural isomers.
To answer your question about two different structural isomers, the only two possibilities are two $\ce{Cl}$ either on same $\ce{C}$ atom or on two different $\ce{C}$ atoms, and that's how you get the two structural isomers. So, one should be $\text{1,1-dichlorocyclopropane}$ and the other one should be $\ce{1,2-dichlorocyclopropane}$.
Now in the compound $\ce{1,2-dichlorocyclopropane}$ you can have stereoisomerism (basically diastereomerism) and therefore have cis and trans 1,2-dichlorocyclopropane, but the correct answers are not them, but the two compounds mentioned in the paragraph above, i.e. the two correct structural isomers should be $\ce{1,1-dichlorocyclopropane}$ and $\ce{1,2-dichlorocyclopropane}$.
Here is a picture to demonstrate that..

The last two compounds i.e. $\text{(1R,2S)-1,2-dichlorocyclopropane}$ and $\text{(1S,2S)-1,2-dichlorocyclopropane}$ are basically cis$\text{1,2-dichlorocyclopropane}$ and trans $\text{1,2-dichlorocyclopropane}$ respectively.


text:
Only d orbital electrons for spin only magnetic moment
During my chemistry lessons and self studies, I have come upon many instances where I observed that during calculations of spin only magnetic moment of transition elements only the d orbital electrons are to be counted.  To gain a deeper insight into these instances of "calculations", I would like to expand my question as follow around the same point (i.e. spin only magnetic moment):

Is spin only magnetic moment calculated only for ions? If not, what would be it for $\ce{Cr^0}$: I mean would the $\mathrm{4s^1}$ electron be taken into consideration? If yes, then why?
Are there some other instances apart from $\ce{Cr^0}$ which supports your views?

Any unpaired electron is taken into consideration while trying to quantify spin only magnetic moment.
valence electrons can be unpaired (not always), and inner shell electrons are paired so the latter don’t contribute (since they are fully filled). That’s why usually only the outer shell electrons are considered.
and no it’s not calculated for ions only. You can use it for individual atoms too!
In your example $\ce{Cr^0}$, you can certainly consider that one unpaired electron in the s orbital,

so giving a total of 6 unpaired electron it’s magnetic moment would be $\sqrt{48}$

By using the spin-only formula $M =\sqrt{n(n+2)}$  where  $n$ is the number of unpaired electron and $M$ has the units of magnetic moment.
Caution this formula is applicable for d block elements only.

Another example would be niobium, molybdenum it also has one unpaired electron in the 5s orbital and certainly we consider that electron while quantifying  spin only magnetic moment.

Furthermore, the spin only magnetic moment of molybdenum is very close to that of chromium.
I am telling very close but not same because theoretically calculated magnetic moment don’t usually coincide with experimentally calculated they differ ever so slightly
Sources  -

NCERT Chemistry for class 12 (pg no 221)
Descriptive Inorganic Chemistry by James E. House and Kathleen E. House by Elsevier. 2nd edition (pg no 329)
Shriver and Atkins Inorganic Chemistry, 5th edition (pg no 478)

:The answer is consistent with many questions which I saw over the net, but I want some more 'trusted' sources and references.

NCERT is oversimplified in several parts, don't trust it blindly. You'll end up regretting it later.

It seems that you feel some part of answer is incorrect , could please point it out and also provide counter results/facts obtained from reliable source to support your statement .

@glucose trusted sources is primarily opinion based , none of the resource can guarantee 100% accuracy , as there lots of things humanity  still don’t quite undersrand

I have said previously many times, i am still saying now. don't ever trust NCERT textbook. They are full of notorious errors. I am saying this from my experience as it was my high school textbook and I regret studying from it.

@Chemist Trusted sources include published articles, papers, books (from a good publisher!). Nothing opinion based here.

@Chemist : No. I do not feel the answer to be incorrect but I find it somewhat incomplete without reliable references and sources.

That wasn’t for you @glucose I’m trying to say that not all information given in ncert is completely outdated/disproved theories , these constitute very small portion of ncert. Anyways I checked with other resources as mentioned in the answer

@glucose as requested i went through two other international reference books having high reputation   they all seem to agree with the answer if I have missed something let me know  . If this resolves your query kind consider accepting my answer

@glucose I wanted to include an excerpt from one of references , I couldn’t due to legal reasons ,in the book shriver and Atkins it answers all your queries at one go . Once if you feel that my answer resolves your query kindly consider accepting it

Are there even any complexes with unpaired s-electron? The sources you cite do not have anything to say about $\ce{Cr(0)}$either.


text:
Gravimetric analysis
In a gravimetric determination of phosphorus an aqueous solution of dihydrogen phosphate ion is treated with the mixture of ammonium and magnesium ions to precipitate magnesium Ammonium Phosphate. This is heated and decomposed to magnesium pyrophosphate which is weighed. A solution of $\ce{NaH2PO4}$ yielded 1.054 gram of $\ce{Mg2P2O7}$. What weight of $\ce{NaH2PO4}$ was originally present?
I attempted the question and my answer was coming out to be 1.32 gm where as the real answer is 1.14 gm
P.S: I wanted to post the pic of my solution so that i could be corrected  , but just cant figure out how to upload the photo.
And also can somebody give the answer without using POAC

Maybe this is a case of hydrate vs anhydrous salt...

"And also can somebody give the answer without using POAC". The stoichiometric ratios have to come from a balanced equation, so there is no way to avoid atomic conservation.

The amount of $\ce{Mg2P2O7}$ in 1.054 gram of a pure sample is:
$$ n = \frac{\pu{1.054g}}{\pu{222.57g mol-1}} = \pu{0.004736 mol}$$
There are two atoms of $\ce{P}$ in $\ce{Mg2P2O7}$, but only one in $\ce{NaH2PO4}$. Without considering the two reactions done in detail, we can infer a 2:1 ratio:
$$\ce{2 NaH2PO4 + ... -> Mg2P2O7 + ...}$$
Thus, the mass of $\ce{NaH2PO4}$ is
$$  m_\ce{NaH2PO4} = 2 \cdot \pu{0.004736 mol} \cdot \pu{120.0 g mol-1}= \pu{1.137g} $$

If you use $\ce{NaHPO4\cdot H2O}$ instead, you get the other answer, like @Mithoron mentioned in the comment to the question.

As @KarstenTheis mentioned, the OP must have calculated based on the salt being $\ce{NaH2PO4\cdot H2O}$ for which the mass would be:
$$  m_\ce{NaH2PO4\cdot H2O} = 2 \times \pu{0.004736 mol} \times \pu{137.99 g mol-1}= \pu{1.307 g} $$


text:
What is the theoretical justification for state averaging in SA-CASSCF?
According to the molpro manual, section 7.3:

In order to compute excited states it is usually best to optimize the
  energy average for all states under consideration. This avoids
  root-flipping problems during the optimization process and yields a
  single set of compromise orbitals for all states.

Is see that SA-CASSCF is quite standard in the literature, even in cases when root-flipping wouldn't necessarily cause an issue, and I don't understand why making such compromise to obtain a single set of orbitals makes theoretical sense. Isn't it better to find the set of orbitals and the CI vector that minimizes the energy for each of the roots of interest?

Sorry, I never used that method so do not take my comments seriously. What you wrote seems reasonable to me, I bet it depends on what will you do with those results. I hear that the states provided by that method are orthogonal to each others, which could be advantageous for someone.

The answer is given in my paper of over 40 years ago. Josip Hendekovic, On the energy variation method, Chem. Phys. Letters 90,(1982) 198. Two new theorems are proved, specifically with excited states in mind. On request I will send a copy!


text:
What is the proper laboratory technique to place a solid in a beaker that will be weighed?
I was doing a lab at school yesterday which involved transferring a precise amount of solid reagent into a beaker.
The way my professor told me to do it was to tap the reagent container so that the solid would fall out into the preweighed beaker and then weigh the beaker. However, I always thought that the correct procedure was to weigh a piece of paper, then place the solid onto the paper, reweigh the paper to obtain the correct mass reading, and then use the paper to pour the solid into the beaker.
What are the advantages of using the teacher's way over the way I proposed?

There is no correct or universal weighing protocol per se. The key question is what is the desired level of accuracy? Your teacher's approach is quite cautious but this is not the universal way, as what you did is also common. He/she is apparently avoiding a spatula as well. I personally dislike paper and plastic weighing boats. If you ever had a chance to weigh dry salts or organic crystals on a dry day, these materials charge like crazy and literally all particles fly apart the moment you touch the plastic boat with gloves.
Pharma companies, where a high level of accuracy is required and legal formalities are important, chemists often use weighing funnels. They are not popular in academia (pricey!). In the picture, you see a latest type of balance which avoids static issues. On the top you see a weighing funnel. This one is plastic coated with anti-static material but glass is more common. After weighing, the funnel containing the weighed substance is placed on the mouth of a volumetric flask and washed very carefully. You can't do this with paper.

https://www.laboratorynetwork.com/doc/new-smartprep-weighing-funnel-weigh-in-samples-more-accurately-0001

I am listing some advantages for either method. Depending on the specific task at hand, one or the other (or using a weighing boat or an aluminum weighing dish or yet another methods) might be optimal.
Advantages of weighing paper
It has a lower mass than a beaker, so you can use it in an analytical balance. You can fold the weighing paper and use it as a chute to get the solid into a small vessel. You can put the weighing paper back on the balance to see if any solid is still on it.
Advantages of a beaker
If your next step is to add water to the solid in a beaker, you skip a step. More solid fits into a beaker. It is easier to transport the solid in the beaker. If you are weighing in something that is very dusty (like growth medium) you lose less if you can transfer the powder directly into a beaker.

Many times the decision has to do with what type of balance is available.
Advantages of using a beaker:

Paper cost money.
Paper may adsorb or react with the reagent which has the potential to throw off the measurement.
If the reagent is hazardous, then using paper increases the amount of hazardous waste generated.

Advantages of paper:

Paper is much lighter than a beaker.  This is useful if an analytical balance without sufficient range and/or resolution is available.

For example.  Say a beaker weighs 250g and the reagent weighs 3g.
If the range of the available balance is 300g and the resolution is 0.01g then the beaker method will be the better method.
If the range of the available balance is 5g with a 0.01g resolution the paper method will be better because the beaker will overload the balance.
In these two cases, both methods have the same accuracy.
If a 5g balance with 0.001g resolution is available, then the paper method might be more accurate assuming the paper is not causing issues.
The answer to your question comes down to what type of balance is available.

Use a 10ml beaker instead of one liter for a 3g sample?

@karl You are making assumptions about the sample density.  It is entirely possible 3g sample might not fit in a 10ml beaker.

And what kind of sample did you have in mind for the 1L / 3g combination? (1 L standard lab beaker weighs ~150g, so more like 2L, actually)

The combustion of very light organic matter in the heavier than standard beakers used for combustion experiments exaggerates the need to consider the type of balance.   If the balance can handle the beaker then the professor's method is best.  If the balance cannot handle the beaker then the teachers method is best. A professor is more likely to have the costlier balance with the range and accuracy to handle a beaker than a high school teacher so the reality is both methods were likely proper, but appropriate for the available balance.

Using an unnecessarily oversized beaker to weigh in a substance is Bad Laboratory Practice, period.

@Karl  Good luck fitting your 3g straw man into a 10ml beaker.


text:
Is heating water endothermic?
So I got a problem recently which asked to calculate the amount of energy needed to increase the temperature of 200 g of water from 26 to 51 degrees Celsius. The value I got was 20900 J.
So far I feel like it is endothermic because heat had to be added to the water in order for its temperature to go up, so the water essentially absorbed heat, but I'm really not sure.

Just heating water is not a chemical reaction.

Oh well I got this on a test...weird. The  wording was just to calculate the amount of thermal energy needed to increase the temperature, and then to say if that was endothermic or exothermic. You're saying it's neither then ?

Heating is an endothermic process, but not a chemical reaction.

So far I feel like it is endothermic because heat had to be added to
  the water in order for its temperature to go up, so the water
  essentially absorbed heat, but I'm really not sure.

This is mostly semantics. Such exam questions are annoying when they make you argue about words. The term endothermic and exothermic are reserved for reactions by IUPAC. An endothermic reaction is one where the overall enthalpy change is positive. However one can extend it to endothermic process, which is any process that acquires energy (from external surroundings). Atkin's physical chemistry clarifies it nicely "it must never be forgotten that heat is a process (transfer of energy as a result of temperature difference), not an entity"

Really playing with names. It is what most of the high school books do in half their size.  Think about how much time is devoted to nomenclature of salts and balancing reaction. ...too. Instead of giving solid bases, they reduce to sudoku activity :(

I cannot agree more. Excellent point. There is a beautiful article by Gillespie (VSEPR guy) from1970s. He said chemistry gets the lowest ranks among other sciences and engineering. He meant that chemistry is not choice of the brightest. He was lamenting Canadian system, which is almost universally true.

Exothermic means the enthalpy of a system decreases ($\Delta H < 0$) and endothermic means it increases ($\Delta H > 0$). If you choose your system as just the water, the process is endothermic. This makes sense because you are increasing the thermal energy of the water, and if nothing else changes in the system, the energy has to come from the surrounding.
The question does not specify how you increase the temperature. You could spin a disk on the bottom of your containiner, using friction to turn mechanical work into thermal energy. You could increase the temperature of the water using a microwave oven. Probably, however, you would just place the container with water on a hot surface, and increase the temperature by heat transfer.
If you choose your system to be elemental sodium and your water mixed together, the enthalpy of the system would be constant, and the temperature of the water would increase from the energy released by the chemical reaction. If you choose your system to be just the sodium and the water it reacts with, it is an exothermic process that heats up the surrounding (the water that does not react increases in temperature as a consequence of the exothermic reaction).

Is heating water endothermic?

It depends what you choose as the system and as the surrounding. As an aside, we say "heating up water" to signify we are somehow increasing its temperature, but as the examples above show, this increase in temperature does not necessarily require the transfer of heat.


text:
Boiling point of ethanamide vs propanamide
I just have a question regarding the boiling points of some primary amides. Ethanamide has a boiling point of 222 °C, while propanamide has a lower boiling point of 213 °C. Both amides are capable of hydrogen bonding, however, I would have expected propanamide (the molecule with more electrons) to have a higher boiling due to stronger dispersion forces. The same applies to butanamide which has a boiling point of 216 °C.
Is there anyway to rationalise why propanamide and butanamide have a lower boiling point than ethanamide or why the carbon chain length doesn't seem to affect the boiling point of amides?

The relatively high boiling points for such low molecular weight compounds suggests a primary effect from hydrogen bonding.
Using ethanamide as a reference point, propanamide has a higher molecular weight, which should raise the boiling point a little, but three relatively nonpolar carbons (plus hydrogens) vs two in ethanamide. A lower boiling point would then not be unexpected. Butanamide seems to support this, being a little higher in bp than propanamide by virtue of its higher mw, but still lower than ethanamide because of its lower overall polarity.
But to throw in some difficulty, formamide (methanamide) has a bp of only 210 °C, whereas its low content of nonpolar atoms suggest it should be much higher.
And then there are the melting points, which are difficult to relate to polarity: formamide 2 °C; ethanamide (acetamide) 80 °C; propanamide (propionamide) 80 °C; and butanamide (butyramide) 115 °C.
Formamide could be an outlier because of nitrogen flipping back and forth thru almost-planar nitrogen. Butanamide's high mp is a bit surprising. There could be packing factors in the solid that are favorable.
Maybe too many factors and too small overall effects to be able to make a lot of predictions, even tho we may be satisfied with some explanations after seeing the numbers.


text:
Naming convention for buffer solutions/systems
Is there an official naming convention to follow for buffer solutions/systems? I've noticed that the name of the buffer usually just follows the name of the salt (ex. sodium acetate buffer), but what if the buffer is diprotic?
If I mixed tartaric acid with its double salt, potassium sodium tartrate, would the resulting buffer simply be "potassium sodium tartrate buffer"? Or would it be something like "bitartrate-tartrate buffer"?

I cannot say it is official, but the historical convention is to use the buffer name in sense of the general term for all forms of the buffering substance. E.g. phosphate in the wider sense is the general term for all salts of $\ce{H3PO4}$ ( in even wider sense also for polyphosphates, like in enviromental chemistry).       The particular  buffer active forms are implicitly defined by pH range or the particular value. E.g. the phosphate buffer for $\mathrm{p}H=6.8$  does not contain $\ce{PO4^3-}$ as a major component, but $\ce{H2PO4^-}$ and $\ce{HPO4^2-}$


text:
Compressibility Factor Of Gases
In the general equation of the compressibility factor $Z$, we define $Z$ as
$$Z=\frac{pV}{nRT}$$
Here, what is $p$? Is it $p_\text{real}$ or $p_\text{ideal}$?
Also, what is $V$? $V_\text{real}$ or $V_\text{container}$?

What difference do you see $V_{real}$ versus $V_{container}$ ?

Both quantities are of the real gas.
Note that for the ideal gas behaviour, the compressibility factor $$Z=\frac{V_\text{real}}{V_\text{ideal}}=\frac{pV_\text{real}}{nRT}=1$$
 for any combination of $p, V, T$.
But real gases, in contrary to the ideal gas, have nonzero volume of molecules and there are intermolecular interactions.
The volume of molecules increases $Z$, as it makes the pressure higher, because the collision frequency is higher.
The cohesive forces between molecules decreases  $Z$. As molecule attraction, like hydrogen bonds or dipole interactions, decreases the effective number of molecules and therefore pressure.
This is reflected in the van der Waals equation - probably the simplest state equation of real gases, for not too high pressure:
$$\begin{align}
\left(p + \frac{an^2}{V^2}\right)\left(V - nb\right)&=nRT \\
\left(p + \frac{a}{V_\mathrm m^2}\right)\left(V_\mathrm m - b\right)&=RT
\end{align}$$
$$\begin{align}
Z &= \frac{p\cdot V_\text{real}}{nRT} \\
Z&=  \frac{p\cdot V_\text{real}}{\left(p + \frac{an^2}{V_\text{real}^2}\right)(V_\text{real} - nb)} \\
Z&=  \frac{1}{\left(1 + \frac{an^2}{p \cdot V_\text{real}^2}\right)\left(1 - \frac{nb}{V_\text{real}}\right)} \\
Z&=  \frac{1}{\left(1 + \frac{a}{pV_{\mathrm m,\text{real}}^2}\right)\left(1 - \frac{b}{V_{\mathrm m,\text{real}}}\right)}
\end{align}$$
For $|x|\ll1, 1/(1+x)=1-x$
For not too far from ideal behaviour, we can apply the above approximation.
$$\begin{align}
Z&=\left(1 - \frac{an^2}{pV_{\mathrm{real}}^2}\right)\left(1 + \frac{bn}{V_{\mathrm{real}}}\right) \\
Z&=\left(1 - \frac{a}{pV_{\mathrm m,\text{real}}^2}\right)\left(1 + \frac{b}{V_{\mathrm m,\text{real}}}\right) \\
\end{align}$$
We can therefore also afford to neglect minor terms.
$$\begin{align}
Z&=1 - \frac{an^2}{pV_{\mathrm{real}}^2}+ \frac{bn}{V_{\mathrm{real}}}\\
Z&=1 - \frac{a}{pV_{\mathrm m,\text{real}}^2}+ \frac{b}{V_{\mathrm m,\text{real}}}\\
\end{align}$$
To address clarified scenario, as a the same amounts of the ideal and real gas
are under the same temperature and pressure:
$$V_\text{ideal gas}=\frac{nRT}{p}$$
$$V_\text{real gas}=Z \frac{nRT}{p}$$
There is need to perform iteration for probably the easiest way to get the result.
With the $V_{\rm{ideal gas}}\ $ as the first approximation. The other option is to solve the root of the cubic equation what we probably do not want to.
$$\begin{align}
V_\text{real}&=Z(p, V_\text{real})\frac{nRT}{p}\\
V_\text{real}&=\frac{1}{\left(1 + \frac{an^2}{p \cdot V_\text{real}^2}\right)\left(1 - \frac{nb}{V_\text{real}}\right)} \frac{nRT}{p} \\
V_\text{real}&=\left( 1 - \frac{an^2}{pV_{\rm{real}}^2}+ \frac{bn}{V_{\rm{real}}}\right) \frac{nRT}{p} \\
\end{align}$$
The van der Waals equation can also be expressed in terms of reduced
properties
$$\left( P_r + \frac{3}{V_\rm{r}^2}\right) \left( V_{\rm r}-\frac{1}{3}\right) = \frac{8}{3}T_\mathrm r$$
This yields a critical compressibility factor $3/8$.
The values of pressure, temperature and volume are divided by the
respective critical values of the given gas.

We have p, V, n, T. Now we compare 1 mol of real gas and 1 mol of perfect gas. We set p and T same for both gases, then we inspect their V. eg we set for both gases p=1atm T=298K. p for perfect gas is that p(ideal), p for real gas is p(real), both set to have same value and that 'same value' plugged into that equation. I don't see how p is real or ideal in the equation. It's just a common value that we set experimentally. So, that p is both p(ideal) and p(real), both have the same value. Correct me if I am wrong. I am still a learner.

This all should be in your question elaboration. Your scenario was not clear, neither what quantities are given and what are dependent.

@Poutnik Well, shouldn't that be obvious when we talk about compression factor, which is defined as ratio of molar volume of real gas to that of perfect gas under the same conditions of pressure and temperature? The V is definitely V(real), but I doubt the statement 'Both quantities are taken as real ones.' is correct.

It should be obvious from the question wording, which was rather underelaborated. :-)

Isn't P ideal = P real + an²/V² ?

No, it is not, as the is also $V - n\cdot b$ factor.

Compressibility factor for a gas is defined as the ratio of the volume of real gas to the volume of ideal gas .
Thus
$$ Z = \frac{V_\text{real}}{V_\text{ideal}} $$
Using the ideal gas equation
$$ PV_\text{ideal} = nRT $$
$$ V_\text{ideal} = \frac{nRT}{P} $$
So
$$ Z = \frac{V_\text{real}}{\frac{nRT}{P} } $$
$$ Z = \frac{PV_\text{real}}{nRT} $$
Hence except $V_\text{real}$ all other quantities are ideal.

The answer apart from this answer says that even P is real, I don't know which to trust. can you provide a source to your answer?

P is pressure on container wall. And V is container volume.