text:
Failure Cases of Adiabatic Approximation
We may exactly write the wavefunction of a molecular system as
$$
\phi(r, R, t) = \sum_{k = 0}^{\infty} \chi_k(R, t) \psi_k(r, R)
$$
where $r$ is the coordinates of electrons, $R$ is the coordinates of nuclei, $\phi_k(r, R)$ is the $k$-th eigenfunction of the electronic Hamiltonian for the electrons and $\chi_k$ is a corresponding nuclear wavefunction.
We then get that the time-dependent Schrodinger equation for the nuclear wavefunctions are
$$
i \hbar \frac{\partial \chi_k}{\partial t} = \left[ - \sum_I \frac{\hbar}{2M_I} \Delta_I + E_k(R) \right]\chi_k + \sum_l C_{kl} \chi_l
$$
where $I$ denotes a specific nucleus, $E_k$ is the $k$-th eigenvalue of the electronic hamiltonian and $C_{kl}$
is a coupling function defined by
$$
C_{kl} = \int dr \psi_k^* \left[- \sum_I \frac{\hbar^2}{2M_I} \Delta_I \right] \psi_l
+ \sum_I \frac{1}{M_I} \left[\int dr \psi_k^* \left[ -i\hbar \nabla_I \right] \psi_l \right] \left[-i\hbar \nabla_I \right]
$$
We may make the adiabatic approximation and assume that the coupling matrix is diagonal such that
$$
C_{kk} = - \sum_I \frac{\hbar^2}{2M_I} \int dr \psi_k^* \Delta_I \psi_k
$$
I am curious in what real life situations this approximation causes an issue. I know with the full Bohn-Oppenheimer approximation where the coupling is completely ignored there are many cases where this causes issues and I know there are many cases where issues arise when the approximation is made to only consider the ground state $k=0$ but I am not sure what possible issues could arise by ignoring the off-diagonal coupling functions. Are there any chemical reactions in non-extreme, non-relativistic situations that are not adequately described by just a diagonal coupling matrix?

You need to give an example. But I would comment that if you are in an diabatic representation you will need off-diagonal terms to describe coupling, but if in the adiabatic case these are already included as these diagonalise the full Hamiltonian and describe the case when nuclei move slowly enough. (The usage is confusing though as sometimes diabatic is also called non-adiabatic although this is also used to describe transitions between adiabats as in Landau-Zener crossings.)

With a diagonal coupling matrix, you will not have any transitions between the electronic states of your molecule. The inclusion of a diagonal Born-Oppenheimer correction does not change anything in this regard, though it improves the description of each adiabatic state by itself. So every reaction that explicitly involves an electronically excited state (all of photochemistry) can not be described in the adiabatic approximation. Reason being that after your initial excitation to that excited state, the molecule will never be able to return to the ground state because of the missing non-diagonal couplings. For example, if you want to calculate reaction rates for a photochemical reaction, knowledge of the off-diagonal couplings $\int\psi_k^*\nabla_I\psi_l dx$ is essential.


text:
How can I store reagent under inert gas?
I need to store some aliquots of 2,3,5-tribromothiophene. I searched for how to store reagent under inert gas, And I found this : https://avantilipids.com/tech-support/faqs/protect-from-oxidation
In summary, it says:

aliquot the reagent as desired into vial.
blow a gentle stream of inert gas into the vial for few seconds.
cap and seal the vial and store in fridge.

Is this enough? Somehow I feel that this method is a little bit clumsy. Does really just blowing inert gas few seconds ensure inert atmosphere?

It depends on how carefully it is done. And depending on the way you close the vial, the cap, etc. I can't write an answer for this. But consider asking a colleague to be there for quick action. Longer you bubble better it is. Ar is better than N2. Etc. Or a vacuum/inert line is on place and a branch of it is for storage. For prolonged storage or relatively small sample vials of thin glass ready to be molten-sealed do exist (nothing expensive, just glass "bubble" with a thin open input. You can purge them through a needle using Ar and quickly seal. Think of the vials for injection drugs

This particular compound will solidify; not on bench, but in the fridge. Beside the larger effect by temperature, this equally will lower a bit the rate of decomposition. Don't forget to put a legible label on the container.

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text:
Is Avogadro's constant equal to 1?
The mole is defined to be exactly $\mathrm{mol}=6.02214076\cdot 10^{23}$ entities. Avogadro's constant is defined as $N_A=6.02214076\cdot 10^{23}\mathrm{mol}^{-1}$.
Then why is it not usual to write
$$N_A=6.02214076\cdot 10^{23}\cdot (6.02214076\cdot 10^{23})^{-1}=1?$$
It doesn't make sense to me.

One mole of anything contains exactly 1.0 moles of that thing. Important as it is, this fact is somewhat too trivial to be stated more often.

@Ivan Neretin This is not the point of this question at all

It’s not a silly question, but what you’ve proven is precisely that: 1 mole of stuff has exactly 1 mole of stuff in it. In a way, we would probably be in trouble if it wasn’t 1.

You have mixed up Avogadro number $N_0$ with Avogadro constant $N_\mathrm{A}$, and used the constant to define itself, which is incorrect mathematically. Analytical form of your suggestion is as follows:
$$N_\mathrm{A} = N_0\cdot\pu{mol^-1} \label{eqn:1}\tag{1}$$
$$\implies \pu{1 mol^-1} = N_\mathrm{A}\cdot N_0^{-1} \label{eqn:2}\tag{2}$$
Plugging \eqref{eqn:2} in \eqref{eqn:1} (even though it makes no sense mathematically) one gets
$$N_\mathrm{A} = N_0\cdot N_\mathrm{A}\cdot N_0^{-1}\tag{3}$$
Or, since both $N_0$ and $N_\mathrm{A}$ are non-zero by definition,
$$1 = 1\tag{4}$$
This is correct, but useless.

Please explain how I used the constant to define itself. The definitions I wrote were taken from Wikipedia (I changed "particles" to "entities", though).

And what you're doing here is very different from the thing in question. You use one equality only and substitute it into itself. Of course you get $1=1$. But I used two equalities. 1) $\mathrm{mol}=6.02214076\cdot 10^{23}$, 2) $N_A=6.02214076\cdot 10^{23},\mathrm{mol}^{-1}$. Taken together, they imply $N_A=1$.

There is a circular definition $\pu{mol}\to N_\mathrm{A}\to\pu{mol}.$ The quantities you are using in numerical form are, as I said, Avogadro number $N_0$ with Avogadro constant $N_\mathrm{A},$ correspondingly. You are "taking them together" against mathematical principles.

From Wikipedia: "The mole is defined as exactly $6.02214076\times 10^{23}$ particles." Also from Wikipedia: "Avogadro constant is defined as $6.02214076\times 10^{23},\mathrm{mol}^{-1}$." As you can see, there is no circularity involved. First, we defined mole, then we defined $N_A$.

You write $\mathrm{mol}\to N_A\to\mathrm{mol}$. That is not correct, as $N_A$ was not used in the definition of a mole. We defined mole just as a number of particles. And then we used $\mathrm{mol}$ to define $N_A$.

Numerical value of $N_\mathrm{A}$ is used in the definition of the mole, $N_0$ and $N_\mathrm{A}$ are inseparable entities, at least in context of current definition.

The number $6.02214076\cdot 10^{23}$ exists without any reference to the symbol $N_0$. We use $6.02214076\cdot 10^{23}$ to define $\mathrm{mol}$. Then, we define $N_0$ as $6.02214076\cdot 10^{23}$.

In other words, the number $6.02214076\cdot 10^{23}$ is defined purely in terms of mathematics.

This is not true. The number you for some reason return to all the time is "is the fixed numerical value of the Avogadro constant" (SI Brochure). Look, let me ask you straight: don't you think it's strange that according to your proposition the same constant has two vastly different values, one which is dimensionless?

The constant has one and only one value. $N_A=6.02214076\cdot 10^{23},\mathrm{mol}^{-1}=1$. Indeed, the equality is correct if $\mathrm{mol}=6.02214076\cdot 10^{23}$. Nevertheless, I managed to find an article resolving this "apparent issue (my question)". https://pubs.acs.org/doi/pdf/10.1021/acs.jchemed.9b00467 You may also find it interesting.

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text:
Corrosive potential of molten ionic salt
I was looking at this video about fourth generation of nuclear reactors and one of developing concepts is molten salt reactor (MSR) where melted ionic salt replaces water as coolant and also acts as a solvent for nuclear fuel. This has many advantages regarding safety and efficiency but one question always arises - corrosion of steel tubing with high temperature molten salt. The response of one developer was that this questions is based of a false extrapolation of everyday experience where we see car metals are attacked when exposed to snow/road salt. He says that the mixture of water and salt has the greatest corrosive potential and in fact, pure molten salt is less corrosive not only from the mixture but also than the pure water. Is that generally true for ionic salts (salt in question is sodium chloride) or is it maybe true only for specific salt/temperature range?

Water is corrosive because of $\ce{H+}$; molten salt doesn't have that.

@IvanNeretin So, you think his claim is false?

No, I think the claim is true. Pure molten NaCl must be less corrosive than water.

A seriously researched subject ;  Try NACE ( National association of Corrosion Engineers)  references if interested . Molten salts are also used to heat-treat steels .

Does he state which kind of salt is picked? Probably not organic ionic liquids. Don't forget that there equally is the sodium-cooled approach (e.g., French Superphénix) of fast breeders ...

Indeed, the larger problem is what happens if the pipes cool too much with molten salts in them...

Molten salt reactors use specific mixtures of salts (often LiF/BeF2 mixtures). They have been widely researched so their corrosive potential is well known (or the lack of it). Besides, the comparison is with reactors using high pressure steam (PWRs) and that is not a system with zero potential to corrode the pipework. Given this, there is little concern that well-chosen salts will have high corrosion potential.

@matt_black Well, that is true for thermal reactors. The reactor in question is actually fast reactor and uses molten sodium chloride.


text:
The ways to determine if an ionic compound is paramagnetic or not

Find which of $\ce{Na2O2}$ and $\ce{CsO2}$ is paramagnetic.
(Adapted from JEE advanced paper-1 2016, q.22)

The given solution involves removing the cations and taking the anionic part, i.e. $\ce{O2^2-} $ and $\ce{O2-}$, and then calculating the MO diagrams of the anions.
However, why is it that we don't have to consider the cationic part for magnetic behaviour? The calculations I had in my textbook were more direct because I'd have binary compounds of form $\ce{X-Y}$, then I'd take MO of the whole compound and see if it has odd or even electron, and see if the total compound has a bond order $\gt 0$.
Why do we only have to look at MO diagram of anion to determine if an ionic compound is paramagnetic or not in this case?

The cationic part has the electron configuration of an inert gas, hence it does not matter. These are alkaline metals, after all.

So, the method described is not a general procedure? @IvanNeretin

With transition metals, you might want to consider the MO of the whole compound, like you said.

To put it another way you always have to look at the whole system, but the ionic model factors it into two (or more) parts which can be considered independently. Na+ and Cs+ are "clearly" diamagnetic being isoelectronic with noble gases, so in such cases you only need consider the anion.

I just can not understand how this question was marked as needs more focus...

As mentioned by @IvanNeretin and @IanBush in the comments, the cation in this case can be neglected as they have noble gas configuration, and therefore are diamagnetic.
In general, for an ionic solid, it is possible to consider the magnetic effect of the ions separately. As there is no covalent bond, it isn't particularly necessary to look at the molecular orbitals (MO) of the system. This is true, even for many transition metal compounds, such as magnetite ($\ce{Fe3O4}$), which has $\ce{Fe(II)}$ and $\ce{Fe(III)}$ centres. The magnetic behaviour of the solid can be reasonably explained by considering each of the iron ions as a separate magnetic dipole. The oxygen stays as oxide ($\ce{O^2-}$) which has no unpaired electrons, and is diamagnetic.
There are some interesting cases, where covalent molecules show magnetic behaviour due to unpaired electrons. In those cases, you have to consider the MO of the whole molecule, there is no way to consider the individual atoms as magnetic.
For example, this compound:

This is a spontaneous charge transfer salt that is produced when you react decamethyl ferrocene with tetracyanoethylene under the right conditions. The ferrocene moiety gives up an electron to the $\pi^*$ orbitals of tetracyanoethylene. The tetracyanoethylene radical anion formed is magnetic due to the unpaired electron. The iron is in $\ce{Fe(III)}$ state and is also magnetic.
In solid state, the ions are close enough for the spins to couple, and the solid becomes a ferromagnet below $\pu{4.8 K}$.
Incidentally, this is the first compound isolated where part of the magnetism is coming from unpaired electrons that are in p-orbitals (i.e. MOs formed from p-orbitals). All previously known magnetic materials had d-block elements.

If one ion is diamagnetic and another is para, then what is the magnetic prop of compound in total?

@Buraian Paramgnetism is almost always stronger than diamagnetism, and overrides that effect. I use the word "almost" because I am not completely sure, but I have never come across any instance where diamagnetism was stronger than para. Think about it—paramagnetic materials also have paired electrons, if their effect was stronger than the one or two unpaired electrons, there wouldn't be any paramagnetic material!!

KMnO4 is an example where the paramagnetic and diamagnetic terms are comparable - but this is not the usual unpaired electron paramagnetism, which as you say (always?) trumps the diamagnetic term - https://www.tandfonline.com/doi/abs/10.1080/00268976000100311 https://en.wikipedia.org/wiki/Van_Vleck_paramagnetism


text:
Some salts lower the vapour pressure of water - do any raise it?
Plain old NaCl salt can slightly lower the vapour pressure of water for a given temperature. Other salts such as CaCl₂ lower the vapour pressure to a larger degree.
Do any salts, or compounds of any type, do the reverse, and raise the vapour pressure of water? Is this provably impossible/possible?

Consider why the salts  lower the vapor pressure. Now consider how the opposite effect could happen.

@Mithoron Isn't that the melting point, not boiling point?

Chemistry is very complicated and I'm probably missing certain edge cases, but I think the answer is almost certainly 'no'.
Salts lower the vapour pressure of water because intermolecular forces attract the salt ions to the water, raising the energy needed for water molecules to escape.
In order for water's vapour pressure to rise, the salt would need to (on average) repel the water ions, the typical term for this is an insoluble salt.
Alternatively, something in the gaseous mixture could attract the water molecules, but this is going to be difficult, since the defining feature of gases is their minimised inter molecular forces.
I'd note this is not a rigorous proof, but I cannot see a way around it. Perhaps the minimal amounts of insoluble salts that dissolve in water due to the statistical nature of chemistry raise the vapour pressure?

You should probably look at understanding colligative properties.
Effectively non volatile salts lower the vapour pressure by taking the place of water. Colligative means only depending on the amount not the nature. Just the fact they are there, rather than interacting is what counts. Calcium chloride has more of an effect as there are more ions, 2 vs 3 for sodium chloride.
If you need to know more try here.
https://chem.libretexts.org/Courses/University_of_Arkansas_Little_Rock/Chem_1403%3A_General_Chemistry_2/Text/13%3A_Solutions/13.04%3A__Colligative_Properties

That is an interesting question. The vapor pressure of water at a given T is defined by its chemical potential which is effectively the amount of water present. Dissolving anything in water, or anything for that matter, lowers its concentration and its chemical potential, thereby decreasing its vapor pressure. At first blush it seems that the vapor pressure must be lowered. However, there exist low boiling azeotropes where the VP of at least one of the components is raised so there is a mechanism by which VP can be higher than normal. This can be accomplished by physical means, the best example being reverse osmosis. Applying pressure to a water solution raises its chemical potential compared to pure water. This phenomenon must exist in membranes especially in living cells. What is needed is a solute that increases the pressure on the water molecules, a hydrophobic hydrogel, an oxymoron, but who knows?


text:
Choosing reaction stoichiometry in organic synthesis
Reading papers about organic synthesis, I see many cases like setting one material as 1.2 eq. I presume that reagents used more than 1 equiv are relatively inexpensive or less reactive than other reagents. (Am I right?) But relative percentage of catalysts is still a mystery. For example, when I searched for Buchwald–Hartwig amination, I saw various mol% of catalysts, ranging from 0.5 mol% to 10 mol%. What criterion or rule of thumb is used? Is this just a result from experimental trials varying mol% of catalysts?

There can be no "rules of thumb" that work across the gargantuan variety of possible chemical reactions. The best you can do is to look for what worked in similar reactions in the literature. If the reaction is completely novel, then the only solution is to experiment and test what works. A lot of chemistry is, after all, an experimental not a theoretical science.

A minor sidenote: Setting one material at 1.2 equivalent is not always done just because it is cheap, but it is usually done to make sure the material it reacts with completely reacts away.

The group that initially publishes a new method will usually run a series of test reactions using different loadings of catalyst, different relative equivalents and different reagents to deterine which combination gives the most optimal results. For example, if n-BuLi is a reagent in a new method, it could be that the optimal amount of n-BuLi is $\pu{1.2eq}$ to ensure complete deprotonation of the starting material; however, it could also be that $\pu{0.95eq}$ of n-BuLi are ideal to prevent overreaction and side reactions. The same testing is typically performed for most reagents and thus you will end up with reactions that work best with, say, $\pu{3eq}$ of amine base although only catalytic amounts of base would be needed according to the proposed mechanism.
The original publication will often contain one or more optimisation tables that detail which combinations and variations were tried (including leaving out the catalyst to prove the reaction is indeed catalysed).
Once a later group uses the previously published results by Previous Group to apply a reaction to their own synthesis, they will typically simply use the best conditions they can find. Sometimes, there may be a publication that uses $\pu{5mol\text{-}\%}$ while another uses $\pu{10mol\text{-}\%}$ for essentially the same transformation. A synthetic group may try whichever they found first, whichever gave the higher yield, whichever they trust more or both.
Then, a reaction may not work as intended in the later synthetic works; the product will be there but the yield will not be sufficient or a side reaction may occur. In those cases, synthetic groups will often just guess which reagent is insufficient in the published method and up that reagent. For example, if the published reaction uses $\pu{3 mol\text{-}\%}$ catalyst but the later group only achieves $\pu{30\%}$ yield, they may just re-run the reaction with $\pu{9 mol\text{-}\%}$ catalyst and hope that pushes the yield to $\pu{80\%}$ or more. If it does: they score. If not: ‘well, maybe if we triple the amount of base …’

The most precious component of the reaction is the starting material you made yourself so the materials used in excess are usually the bought ones, unless you are at the first step of a sequence.
In some cases the % catalyst will be the result of running a series of experiments to establish how much was needed - these studies would normally form part of the published paper or would be referenced if published separately. In other cases it may be "this is what we did and it worked so we carried on using it".
There can be considerable variation between batches of catalyst depending age, supplier etc which are out of your control unless you are making your own (and your own experimental technique may introduce variability). Even then you may fail to achieve the reported yields (and this opens a whole other debate).
I would suggest looking at the published procedure that is closest to the reaction you are trying to run and using their conditions as your starting point but be prepared to alter them as each reaction is different.

Not mentioned yet by the other answers is this: Many (not all) organic reactions lead to a chemical equilibrium, and only substantially increasing the relative amount of one / of multiple reagent(s) compared to an other, or / and removal of one / of multiple reaction products renders these reaction synthetically useful.
Recall e.g., esterification reactions where a distillation (Dean-Stark apparatus) constantly removes water from the reaction mixture. Or, the reversible introduction of $\ce{-OH}$ / $\ce{-NH2}$ groups in the course of the Bucherer reaction.  Depending on the concentration of ammonia, either the formation of the naphthol 1 or the if naphthylamine 6 is favourable:

(source)
A bit related to the answer provided by @Waylander is the observation that many catalysts are expensive.  This may be by their elemental composition, or their cost of fabrication.  This is true for many transition metal-based catalysts (you count by unit of money per mole), e.g. Grubbs' catalysts for olefin metathesis.  Or, see e.g., Friedel-Crafts acylations, where the stoichiometric use of Lewis acids like $\ce{AlCl3}$ in may be ok at lab scale.  For a manufacturer, they however may represent a heavy cost factor after the reaction for their safe hydrolysis and the subsequent neutralization of waste water.  Thus, you seek for (reasonable) short, yet reliable efficient syntheses with conversions requiring low stoichiometric equivalents of reaction partners.  Process chemistry (one representative journal) is a whole branch within chemical research to scale bench chemistry to large scale, often using different solvents / reagents / reactions to perform eventually same transformations safely (a relevant question on chemistry.se).


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Which is a more accurate scale for a chemistry high school lab - a triple beam balance or a digital balance and why?
I would like to ask the experts of this chemistry community their opinions on which balance is more accurate and effective to use in a high school chemistry lab- a triple beam balance or a digital balance. I would appreciate your rationales for your answers as well. Thanks!

I would say digital scales would be more robust, what is needed in high school environment.

on the other hand, making people use a beam balance will surely teach them more.

Consider a different perspective, seen e.g., in mechanical or civil engineering: what are the specifications the device in question must fulfill to be good enough; what is a nice have, or irrelevant? Note these parameters e.g., what is the accuracy yo need in the experiments ahead, then look out for a balance.

Having used a triple beam and chain weights in college, I doubt a modern HS student could reliably handle them.

Orthogonal suggestion: The United States penny weighs 2.5 grams; the nickel, 5 grams. Along with UK coinage, sub-gram accuracy can be achieved using a simple balance and a modest investment of under 20 USD. I haven't explored completely, but I suspect milligram tolerance can be achieved with US and UK coins alone. Add coinage from other countries and you're all set.

Recall that Theodore W. Richards got a Nobel Prize on "in recognition of his accurate determinations of the atomic weight of a large number of chemical elements." Guess which year was that? 1914!! Digital balances did not exist then. All high accuracy balances were mechanical balances then.
The point is that you can have a very crappy digital balance or a triple beam balance (= low price, no certification). All the accuracy of balances depends on their calibration with known weights. The accuracy of those known weights, depends on how they were calibrated by national and international standardizing agencies.
It all depends on the purpose. For organic synthesis work, a triple beam balance would be fine where ultrahigh accuracy of measured weight is not a big deal. What is being weighed and what is the level of accuracy required? How many accurate digits are required?
In short, both digital and triple beam balances are fine for routine work. We cannot say one is more accurate than the other. After all, the digital balance is also requires mechanical movement (i.e., it also has moving parts which you cannot see).

I would say durability is more important than accuracy in high school lab context( and the former affects the latter during the time too) . :-) I remember in my high school days in 1980-84 that our very classical mechanical scales were frequently out of order. Students are the ultimate realization of the Murphy's law saying even things that you cannot imagine be broken can be broken.

@Poutnik, You have not probably seen how modern students misuse digital balances :-) Students don't respect the instruments anymore. I had recently got a digital balance calibrated and one PhD student left a heavy object on it overnight.

I have not said digital ones would survive. I have just said they may survive longer. :-)

The worst part is trying not to drop the calibrated weights on the floor… Ive experienced even expensive digital scales going out of calibration just in a few days. Learned the importance of checking them daily.

In pharma, I was told that they calibrate balances daily. No chances for any mistakes.

@M.Farooq yes we do… every morning

Cheaper digital balances ( the ones that are as precise as triple beam balances) are self- calibrated according to the manual or can be calibrated with one 200g standard weight. How accurate can that be really??

@Suse, Quite accurate, it is even better if they have an internal weight for calibrated.

A digital balance is much easier to use.  I guarantee some of your HS students are going to have huge errors when using a triple beam because they don't know how to read it.   That's not necessarily a downside—you might want to give them training in the skill of using and reading a mechanical balance. OTOH, if you want them to focus on other aspects of the lab, or are concerned about time constraints, then a digital balance (which can be used much more quickly) would be the way to go.
As far as accuracy goes, at the entry level for calibrated quality scales (~\$150), you can get a digital balance with NIST-traceable calibration and the following specs (Cole-Parmer Symmetry Compact Portable Toploading Balance, 300g x 0.01g):

Triple beam balances typically only have 0.1 g readability. To get 0.01 g readability in a beam balance, you'd need to add finer adjustment, with either a dial or an extra beam.  These get pricer (~$250+).  Here's a quadruple-beam balance with 0.1 g readability, though without specs for repeatability or linearity: Ohaus 311-00 Cent-O-Gram Overhead Mechanical Balance, 311g x 0.01g.
But for a HS lab, you'll probably want to spend more like $30/scale. There the only way to determine the relative accuracy of the triple beam and digital balances will be to buy a set of calibrated weights and check for yourself. [Regardless of what claim the vendors make.] [And note that there will probably be a lot of inter-unit variation at that cost level, where some will be much more linear than others.]

Calibrating a balance with standard weights is irrelevant, if the actual weights used by the students are contaminated and/or corroded by who knows what. Of course contamination may occur during a student's experiment, so it is not even a systematic measuring error.

"Calibrating a balance with standard weights is irrelevant", no it is not. It is a norm. First, because students don't calibrate balances. It is done by a certified technician annually or a qualified teacher and there is an elaborate procedure to do it. Standard weights are highly protected so the chances of contamination is very low.


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How much CO2 is produced when burning 1GJ of Natural Gas
Background
I am trying to determine how many kg of $\ce{CO2}$ are released when burning 1 GJ of natural gas.
What I've tried
I have very little chemistry knowledge, but google led me to these two sources (Source 1:Energy Contents and Source 2:Combustion of Fuels. Based on that, I've assumed the following;

Natural gas has energy content of 48.2 MJ/kg
Which means 1 GJ of natural gas corresponds to 20.75 kg (1000/48.2)
Natural Gas has a 75 % carbon content

So I've (incorrectly) assumed that burning 1 GJ of natural gas will produce ~15.5 kg of $\ce{CO2}$. (0.75 * 20.75 kg)
Problem
Obviously, I am wrong, since that second source (Source 2:Combustion of Fuels) has written in the table, that;

1 kg of natural gas releases 2.75 kg of $\ce{CO2}$.
1 GJ of natural gas will produce 50 kg of $\ce{CO2}$.

How does that work? what am I missing?

What is natural gas?

Let me solve the first half of it for you, and see if this gives you enough background to solve the second half on your own:
Natural gas is methane ($\ce{CH4} (g)$). The balanced chemical equation for the combustion of methane is:
$$\ce{CH4 (g) + 2O2 (g) -> CO2 (g) + 2H2O (g)}$$
We know it's balanced because the total number of atoms of each element on the left side is the same as that on the right.  E.g., there are four atoms of oxygen on the left, and four on the right.
Each mole of $\ce{CH4}$ weighs 12 g (for the carbon) + 4 g  (for the 4 hydrogens) = 16 g. Each mole of $\ce{CO2}$ weighs 12 g (for the carbon) + 32 g  (for the 2 oxygens) = 44 g.
Since we produce 1 mole of $\ce{CO2}$ for each mole of $\ce{CH4}$, we produce  44 g of $\ce{CO2}$ for every 16 g of $\ce{CH4}$.  Thus burning 1 kg of natural gas will produce:
$$\require{\cancel} 1 \,\ce{kg} \,\cancel{\ce{CH4}} \times \frac{44\, \cancel{\ce{g}} \,\ce{CO2}}{16 \,\cancel{\ce{g}} \,\cancel{\ce{CH4}}}  = 2.75 \,\ce{kg} \,\ce{CO2}$$

Brilliant, thanks!! The equation you shared was the missing link, I never realized how heavy Oxygen was (16 g/mol).

Happy to help.  The way I look at it is not that oxygen is unusually heavy (it's still one of the lighter elements), but that hydrogen, which is the most numerous component of methane, is unusually light (it's the lightest element), which is what makes CH4 so light relative to CO2.

Natural gas is methane, but only after processing.

@RodrigodeAzevedo  Not quite. A more accurate statement would have been as follows:  "The term natural gas (NG) can refer to either refined or raw (unprocessed) NG. While the former is the more common usage, both are acceptable. One can typically infer which meaning is being used from context. E.g., the OP was asking about CO2 production from combustion of NG, from which one can infer he meant refined NG, since far more CO2 is introduced into the atmosphere from the combustion of refined than raw NG.  That was theorist's interpretation, which is why he equated NG it to methane."

@theorist Yes, but raw natural gas is also flared. Without a caveat, it's easy for people to assume this answer applies also to raw natural gas.


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Confusion in D/L nomenclature

With regard to carbohydrates, it is known that in a Fischer projection, if the $-OH$ group is on the right side of the lowest priority asymmetrical (chiral) carbon, then we name it as D and if it is on the left side we name is as L. But my question if in the left compound of the image, i.e. D-glyceraldehyde, why have we kept the $-CHO$ group on the top of the vertical line and the $-CH_{2}OH$ group on the bottom of the vertical line ( or alternatively just rotate the molecule by 180 degrees), because if I exchange both the groups at the vertical position with each other, and both the groups of the horizontal position with each other, then we get the same molecule, but now -OH is on the left so it will be L !? Where am i wrong, how do we know which groups will be on top and which on the bottom, in the Fischer projection. Need help.

The carbon with the highest oxidation state is kept at the top.

I think you should read https://chemistry.meta.stackexchange.com/questions/3709/are-you-confused-by-confusing-question-titles-including-the-word-confusion?r=SearchResults&s=1|30.9232

Does this answer your question: https://chemistry.stackexchange.com/questions/16914/what-do-the-stereochemical-labels-d-and-%e2%88%92-mean-in-tartaric-acid/59447#59447 ?


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Different products with varying aluminium chloride concentration in a reaction of propiophenone

How does the product change with varying $\ce{AlCl3}$ concentration?

Answer: In excess, meta bromination is observed and if not, then alpha hydrogen is substituted by ketone halogenation mechanism

I am trying to understand why the answer is so using mechanisms. I know that the first step of brominating involves the attack of di bromine bromine onto aluminium chloride, and then, the attack of the benzene's pi electrons onto the one of the bromines, the reaction ends with removal of a hydrogen by the remaining bromine- aluminium chloride molecule.
In the ketone halogenation mechanism, it begins with an enol with negative charge formation into attack of bromine molecule. And finally the product.
In the first case, we have that the Bromine is made more electrophilic and in the second we have that the compound is made more nucleophilic. So, how does concentration control which pathway the reaction takes?

If you have less than 1 eq of AlCl3 then there will still be molecular bromine in the reaction mixture for the enol to react with giving alpha bromination

Oh so is the 'excess' relative to bromine?

Yes. It is reasonable to assume that the bromine reacts preferentially with any AlCl3 present 1:1. When all tthe free AlCl3 is consumed the residual bromine will react with the enol.

Perhaps a noob question, but in the wikipedia for the ketone halogenation reaction, the first step involves acetic acid/ OH attack, I see neither here, so how does the reaction begin with the enolate? @Waylander

This is not any ketone, there is a greater enol content because of the adjacent benzene ring

Hmm right but how is the first enol formed? @Waylander

It is always there in equilibrium with the ketone


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Relation between standard silver-silver chloride electrode potential and solubility product of silver chloride
I learned that $E^\circ_\ce{Cl^-|Ag,AgCl}, E^\circ_\ce{Ag^+|Ag}$ and $K_\mathrm{sp}$ of $\ce{AgCl}$ are related as
$$\boxed{E^\circ_\ce{Cl^-|Ag,AgCl} = E^\circ_\ce{Ag^+|Ag} + \frac{RT}{F}\ln K_\mathrm{sp}}$$
I was given this proof:
For an $\ce{Ag^+|Ag}$ electrode in non-standard conditions, using the Nernst equation
$$E_\ce{Ag^+|Ag} = E^\circ_\ce{Ag^+|Ag} - \frac{RT}{F}\ln\frac{[\ce{Ag}]}{[\ce{Ag^+}]}$$
As $\ce{[Ag^+][Cl^-]}=K_\mathrm{sp}$ and $\ce{[Cl^-]} = \pu{1 M} \ \text{(standard state)}$, so $\ce{[Ag^+]} = K_\mathrm{sp}$. Also $\ce{[Ag] = 1}$ as pure solids have activity of one.
So,
$$E_\ce{Ag^+|Ag} = E^\circ_\ce{Ag^+|Ag} - \frac{RT}{F}\ln\frac{1}{K_\mathrm{sp}} = E^\circ_\ce{Ag^+|Ag} + \frac{RT}{F}\ln K_\mathrm{sp}$$
What I don't understand is the next step. In the proof, $E_\ce{Ag^+|Ag} = E^\circ_\ce{Cl^-|AgCl,Ag}$ and thus the boxed equation above is proved. But why is $E_\ce{Ag^+|Ag} = E^\circ_\ce{Cl^-|AgCl,Ag}$?

Because both the reactions involve reduction of $\ce{Ag+}$

@napstablook Then why is the non-standard potential one reaction equal to standard potential of the other?

First of all you made a slight mistake one of the equations is wrong
First equilibrium reaction,
$$\ce{Ag(aq)+ + e- <=> Ag(s)}$$
Writing the Nernst equation for this reaction
$$\ce{E_{Ag|Ag^+} = E_{Ag|Ag^+}^0 - \frac{RT}{F}ln\frac{\ce{[Ag]}}{\ce[Ag^+]}}$$
Second equilibrium reaction,
$$\ce{AgCl(s) + e- <=> Ag(s) + Cl-(aq)}$$
Writing the Nernst equation for the second reaction
$$\ce{E_{AgCl|Ag,Ag^+} = E_{AgCl|Ag,Ag^+}^0 - \frac{RT}{F}ln\frac{\ce{[Ag][Cl^-]}}{\ce[AgCl]}}$$
Since the activity of solids are considered to be 1
$$\ce{E_{Ag|Ag^+} = E_{Ag|Ag^+}^0 - \frac{RT}{F}ln\frac{\ce{1}}{\ce[Ag^+]}}$$
$$\ce{E_{AgCl|Ag,Ag^+} = E_{AgCl|Ag,Ag^+}^0 - \frac{RT}{F}ln\ce{[Cl^-]}}$$
When you short the wires of the two cells potential of the two cells equals
$$\ce{E_{Ag|Ag^+} = E_{AgCl|Ag,Ag^+}}$$
So
$$\ce{E_{Ag|Ag^+}^0 - \frac{RT}{F}ln\frac{\ce{1}}{\ce[Ag^+]}} = E_{AgCl|Ag,Ag^+}^0 - \frac{RT}{F}ln\ce{[Cl^-]}$$
Simplifying the math here gives your equation,
$$\ce{E_{AgCl|Ag,Ag^+}^0 = E_{Ag|Ag^+}^0 + \frac{RT}{F}ln([Ag^+][Cl^-])}$$
Since we know
$$\ce{AgCl(s) <=> Ag+(aq) + Cl-(aq)}$$
$$\ce{K_{sp} = [Ag^+][Cl^-]}$$
The equation is,
$$\ce{E_{AgCl|Ag,Ag^+}^0 = E_{Ag|Ag^+}^0 + \frac{RT}{F}ln(K_{sp})}$$

Thanks, this cleared up some things for me. However, I'm not sure what you mean by "shorting" here. How exactly do you short the wires attached to the two half-cells?


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Why do we use external pressure to calculate work done by a gas in case of compression instead of using internal pressure?
According to formula work done is (Resisting pressure X dV)? Then in case of compression, isn't the resisting force is internal Pressure? Then why we the formula for compression is same as expansion which is (External Pressure X dV)?

By Newton's 3rd law, the internal force exerted by the gas on the inside face of the piston is equal to the external force exerted by the inside face of the piston on the gas.  If the expansion or compression is reversible, the internal force can also be determined by the ideal gas law, which applies at essentially thermodynamic equilibrium.  But, if the expansion or compression is irreversible, the internal force cannot be determined using the ideal gas law.  So the only alternative is to use the force manually applied externally (if this is known).


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Friedel–Crafts acylation of substituted anisole
I am attempting to form a synthetic route to a tetrasubstituted benzene compound. In the scheme, there is a step that requires the formation of a carbon–carbon bond in the C-6 position of a methoxy group in a trisubstituted benzene ring (there is a silyl protected phenol in the 3 position and an alkyl group in the 4 position). Instead of the actual structure I may not disclose, below, I drew a structural prototype in the conversion intended:

I would like to complete the transformation via a Friedel–Crafts acylation because I have heard that anisoles, though not phenols, are suitable for performing Friedel–Crafts reactions. However I have also read that aluminum chloride can be used to demethylate anisole derivatives. I would like to know if I need to worry about demethylating the methoxy group and, more interestingly, if I do not, why?

A picture would greatly help.

One of the advantages of organic chemistry is that we've formulae to depict what is going on / is planned to happen.  While the reaction inserted is a mere place holder for to have some idea about the substitution pattern already installed, please adjust the drawing to the structures actually in mind.  Learn how to use a sketcher (a survey), or use a service like this, or insert a drawing by hand (in decreasing preference for chemistry.se).

This is a very activated aromatic system so you don't need as strong a F-C catalyst as AlCl3. ZnCl2 or TiCl4 are milder and will not demethylate your anisole.

Thank you for the replies, and I am sorry for not providing a picture in the original post.

@user114654 If there not patent application or publication of said structure intended, replace the drawing with one which depicts the situation better.  Or camouflage the structure a well enough (different alkyl group, other protecting group, etc) to depict the principle.  By the way, $\ce{SnCl4}$ would be an other, less strong Lewis acid for such an activated electron rich system.

Hello buttonwood, I believe that the current drawing is satisfactory for my purposes. The only other functional groups present are silyl ethers on the acid chloride.  My professor told me that the compounds should be kept secret, so I was vague in my description. I hope you understand.

@ Waylander: At refluxing temperature in toluene, anisole polymer has been demethylated quantitatively according to here.

If you do not need demethylation, you cannot use $\ce{AlCl3}$ in this reaction. Anisole with o-electron withdrawing substituent will demethylate at room temperature (see here).

@user114654 If you cannot find suitable F-C conditions I would suggest iodinating. Iodination should occur in the 6 position. You can then do a low temperature Li-Hal exchange and quench with acetyl Weinreb amide to give the required product

@user114654 Fair argument, I slightly edited the question to account for this.

Why worry about demethylation; remethylate. If you have access to the free phenol, consider a Fries rearrangement followed by methylation.

I think it is clear that you want to acylate a trisubstituted benzene compound. It is also clear that the ring consists of two oxygen atoms, one of which is methoxy group. That makes the acylation easy because two oxygen atoms make the benzene nucleus more active. Same time, it makes the acylation difficult with $\ce{AlCl3}$, a strong Lewis acid, which is known to demethylate methoxy groups on the ring under special circumstances even at room temperature (Ref.1 and 2). Thus I urge you that you should not use the common and convenient $\ce{AlCl3}$ catalyst in this Friedel-Crafts reaction.
However, there are milder conditions available for convenient acylation of aryl nucleus. One such condition is use of Rare Earth metal trifluoromethanesulfonates (triflates) such as $\ce{Sc(OTf)3}$ and $\ce{Y(OTf)3}$ (Ref.3 and Ref.4). For instance, when 1,3-dimethoxybenzene is reacted with two equivalents of acetic anhydride in the presence $0.2$ equivalent of $\ce{Sc(OTf)3}$ in nitromethane, 2,4-dimethoxyacetophenone is obtained in 89% yield at $\pu{50 ^\circ C}$ within $\pu{4 h}$ (this is awfully similar to your compound, 2-methoxy-5-methyl-4-trisopropylsilylacetophenone).
References:

Zhen-Ting Du, Jing Lu, Hong-Rui Yu, Yan Zu, and An-Pai Li, "A Facile Demethylation of Ortho Substituted Aryl Methyl Ethers Promoted by $\ce{AlCl3}$," Journal of Chemical Research 2010, 34(4), 222–227 (DOI: https://doi.org/10.3184/030823410X12708998015900).
Doo-Jin Byun, Kil-Yeong Choi, and Sang Youl Kim, "Synthesis of Phenol Group Containing Polyethylenes via Metallocene Catalyzed Ethylene-Allylanisole Copolymerization," Macromol. Chem. Phys. 2001, 202(7), 992-997 (DOI: https://doi.org/10.3184/030823410X12708998015900).
Atsushi Kawada, Shuichi Mitamura, Jun-ichi Matsuo, Takehiro Tsuchiya, and Shu Kobayashi, "Friedel-Crafts Reactions Catalyzed by Rare Earth Metal Trifluoromethanesulfonates," * Bulletin of the Chemical Society of Japan* 2000, 73(10), 2325-2333 (DOI: https://doi.org/10.1246/bcsj.73.2325).
Shu Kobayashi, "Method of friedel-crafts acylation of anilides," European Patent Office 2007, EP 1 359 141 B1 (Application number: 02715709.8; International application number: PCT/JP2002/000020; International publication number: WO 2002/055483).


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How would you test for ppm of lead in a powder?
Looking to test the ppm for lead in a powder that is a food item.
No experience in chemistry.
Possibly may need to test for other heavy metals.

Can you add more context? What instruments you have access to?

I have zero instruments but could purchase some basics from a surplus store. I don't have great access to a workspace...

No the instruments for detect metals in food will cost a fortune. You need to contact a professional lab.

Yes! Here are EPA-recommended kits and vendor contact information.
Google is your friend. You are in Canada, so I'm uncertain if your regular pharmacies carry such kits (many of ours do in the US).
Start with home kits - they have you mail-in a specimen - and graduate if needed based on LDL (lower detection limits).
You should be able to get a good idea (maybe binary) about contaminants, which you can then follow-up with more expensive but very worthwhile tests involving more advanced analytical techniques.

The article you linked is titled "Lead Test Kit Buying Guide," but it doesn't recommend or link any products. It's just very general info about conducting at-home lead testing, mostly on paint.

@SusieDerkins - Good point, edited to include the EPA recommended products which contain links to the vendors.

Thank you, it looks like these kits are only for paint though.

@SusieDerkins - What exactly are you interested in testing? These tests are general in the sense that a powdered (dry) substance can be used, regardless of source. Paint is the most common one. What is your intended use case and what have you found in your searches?

My use case is exactly the same as the OP's. I want to test a food substance that is a powder for lead content. I'm not aware of any documentation that says food samples can or should be run in paint test kits. My searches have only turned up kits for testing paint samples.

I'm assuming that dehydrating, grinding, and otherwise prepping the food sample are not on the table here; so you have a few options, first of which is a commercial lab with the requisite equipment and analytical procedures. The second, if the food has been ingested or you are concerned that someone has been consuming lead-containing food, is/are blood/urine tests which can be done via regular physicians. That said, the latter doesn't confirm from whence the lead came; but if you are willing to spend the money, that's definitive.


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Why does increasing subshell cause ionization energy to decrease?
I know that increasing subshell decreases ionization energy; for example, going from Beryllium to Boron. I do not understand why this is.
The answer my textbook (Chemistry 2e on OpenStax) gives is that the higher subshells (eg $\ce{p}$) has more energy than the lower subshells (eg $\ce{s}$). I do not understand why a higher energy subshell require more energy to remove (ie more ionization energy)? Is it as if the higher energy subshell is less stable, and thus has lower ionization energy?

Yeah, that doesn't make sense.  Ionization energy does increase in going from Be to B, but it's not because you're going to a higher-energy subshell (which you are), it's because you're staying in the same shell but increasing the effective nuclear charge.  Could you provide the exact quote from your textbook?

2p electrons have higher energy ( and lower ionization energy ) than 2s electrons within the same atom context, but these energies shift in the different element atom context. So 2s/2p( if excited ) electrons in Be have higher energies ( lower ionization energies ) than respective  2s/2p electrons in B.

I think in order to understand this phenomenon you have to consider what is happening during an ionization. During this process we are removing a single electron from our molecule or atom. The electron we are going to remove will be the one which is easiest to remove (the one we require the least amount of energy to remove).
Having this in mind, we can take the knowledge that the electron shells keep increasing in energy so e.g. 2s < 2p. "Higher in energy" means that they are less bound to the nucleus (that's just the way we usually define our energy scale).
Thus when combining these two, we can understand why the ionization energy decreases when ionizing atoms that have electrons in higher shells.
The probably more interesting question here is probably why the energy of the shells keeps increasing. To break it down the reason is that electrons in higher shells have (on average) a larger distance to the nucleus. As the electrostatic potential (which is is the main interaction between electron and nucleus) is proportional to the inverse distance (search for "Coulomb potential" if you want more details), this means that these electrons just feel less and less of the attraction from the nucleus which causes them to be more easily removed.
Another effect that plays into this is called "shielding" and it essentially means that the electrons from the lower shells will shield the nucleus' positive charge to some degree and thus the outer electrons the ones in the higher shells) will feel less attractive force to begin with.
Note that when comparing different atoms you also have to take into account that these have different nuclei with different charges so the comparison between atoms (in contrast to different shell within a single atom) can be a bit more complicated than this view alone and might require to crunch the actual numbers but I think conceptually this can help explaining the trends we are seeing throughout the PSE.


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What's the "Charge" in Steric Number Formula
I was reading about the Steric Number Formula here.
There, I came to know that the

Steric Number $N=\frac{V+M \pm I}{2}$ where $V = n(\ce{e-})$, the number of valence electrons of central atom, which is equal to the group number according to the old IUPAC system, $M = n(\text{atom})$, the number of monovalent atoms directly bonded to it, and $I$ is the number of positive or negative charges present (subtract it if the charge is positive, and add it if the charge is negative).

I was trying to apply the formula to $\ce{HNO_3}$ with lewis structure:

Image from here
From this: $N = \frac{5}{2}$ since there is no net charge on the compound and there is no monovalent atom bonded with $\ce{N}$. $N = \frac{5}{2}$ is not possible.
Maybe, I think the charge means, the charge on the central atom, but then again $\ce{N = \frac{4}{2} = 2}$ which is not right. (It's actually $3$)
Also, my friend says that the steric number is the half of

original valence electrons in central atom +
number of monovalent atoms in the entire compound
minus net charge on the entire compound

in which case, the formula would hold right as $V = 5, M = 1$ (as we now consider $\ce{H}$) and $\ce{I = 0}$ which would be $3$ and correct.
I would be glad if some explained what the "charge" represents. If the "charge" represents the net charge on the central atom, then we'd still need to make Lewis Structure and get the charge, then this technique would be not so easy again?
Thanks!

I think the simplest formula to use for steric number is - number of sigma bonds + number of lone pairs. So here you'd have number of sigma bonds = 3, lone pairs = 0, so net = 3

Hi @TRC, I agree with your method - It's correct. But, I want to figure out, what is wrong with the formula that I've stated above? Thanks

There is no real source for the post you have linked. There probably is not a unique definition on how to use that. The whole concept has very, very little value. If you want an easy to understand an follow model that deals with predicting the molecular structure, use VSEPR. But don't fall into the trap of explaining everything with hybridisation.

Hi @Martin-マーチン , I agree hybridisation is older than VSEPR, but I want to know hybridisation well because there are several questions in school exams that ask us the hybridisation of molecules.

Have you considered the $\ce{-OH}$ bond to be a monovalently attached group to the central N atom? considering $\ce{HNO3}$ as $\ce{N(O)2(OH)}$ might help..

Hi @SafdarFaisal can you please elaborate on why $\ce{OH}$ bond is also to be considered? Is it always so? Thanks

You are indirectly using the formula already stated by TRC though I am trying to make your formula accurate and unambiguous (especially because of the efforts you have put into the question).
$$N=\frac{V+M-T-I}{2}$$
N=Steric Number of atom X (say).
V=original (ie when an atom is isolated (not experimentally possible for all atoms)) valence electrons on the central atom.
M=Number of atoms directly bonded to X through a single bond.
T=Number of atoms directly bonded to X through a triple bond (use the example of $\ce{HCN}$ to check this).
I=Number of positive charges (formal charge) present on X atom (take I negative if the charge is negative).
Don't consider/take into account 'D=Number of atoms directly bonded to X through a double bond.'
Relating it to $\sigma +$lone pair formula:
$$\sigma=M+T+D$$ $$LP _{\text{(Remaining on X after bonding)}} =\frac{V-M-2D-3T-I}{2}$$
As per Buraian's suggestion, you may apply this formula for all the examples available here. Scroll down and find AXE method in that see the second table last row.

Hi, so in $\ce{HNO_3}$, the steric number of $\ce{N}$ equals $\frac{5 + 0 - 0 - 1}{2} = 2$? It should be $3$? Thanks

@Jay Thanks for the clarification. I wonder if there's a simpler formula for it. After all, we still have to draw the Lewis Diagram to calculate?

@Jay Do you have any idea if the formula that my friend mentioned is correct or not? I have checked the formula on multiple cases and it seems to work fine. Thanks

@MangoPizza your friend's formula will go incorrect with atoms bonded through triple bond eg:$\ce{HCN, C2H2, N2}$..... Please delete previous comments to keep space.

I should have read what you said before carefully, I have deleted my answer and reposted it here


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What is the mechanism of temporary bleaching of flowers by SO2 ? Can H2S be used?
There are many claims on the web, explaining that the bleach action of Chlorine gas in discoloring of flower proceeds as follow and require moisture for it :
$$\ce{ Cl2 + H2O <=> HOCl + HCl <=> HCl +nascent O}$$
And this nascent Oxygen reacts with the $\pi$ conjugated molecules of flowers, discoloring them.
Similarly, temporary discoloring by SO2 was explained as a result of nascent Hydrogen, something like :
$$\ce{ SO2 + 2H2O <=> H2SO4 + 2 nascent H }$$
And this nascent Hydrogen can also discolor the flowers.
My colleague and I had a debate where I claimed if $\ce{SO2}$ can discolor the flower so will $\ce{H2S}$. And if $\ce{SO2}$ can produce the "nascent hydrogen", so will $\ce{H2S}$ because it is a better reducing agent than $\ce{SO2}$.
I searched the web and didn't found much information either backing the nascent Hydrogen hypothesis or discoloring by $\ce{H2S}$ or by $\ce{SO2}$. Though I found one or two articles explaining how the nascent hydrogen hypothesis came and how it got discredited. Modern theory can explain all the reactions of such "nascent hydrogen" by assuming $\ce{H2}$ molecules at high temperatures, proving that they are nothing but non-equilibrated states of H2 which are just formed.
So, my question is :
Can we claim that if one reducing agent like $\ce{SO2}$ can discolor a flower, other strong agents like $\ce{H2S}$ will surely discolor it?

See these questions: 1. https://chemistry.stackexchange.com/questions/147506/why-is-bleaching-with-chlorine-permanent-but-with-sulphur-dioxide-temporary 2. https://chemistry.stackexchange.com/questions/131162/is-there-any-evidence-any-evidence-at-all-that-nascent-hydrogen-actually-exi

All one needs to know about nascent hydrogen is in the first paragraph of corresponding Wikipedia article: "the concept of nascent hydrogen is discounted, and even ridiculed."

@andselisk I already read some literature on nascent hydrogen and why this got discredited. The main question is can H2S instead of SO2.

The critical point in the understanding of the bleaching process is the property of the dye molecule itself. Textile dye molecules or natural pigments are large conjugated systems that absorb light in the visible region (400 to 700 nm). Therefore, any chemical reagent which destroys this conjugation or destroys the dye molecules into smaller soluble units without significantly destroying the substrate (cloth, plastic, flowers, unfortunately even food is bleached in some countries!! etc.) gets the label of a bleaching agent. This includes chlorine-based bleaching agents, peroxide-based bleaches, enzymes, and reducing agents.
It is not necessary that the bleaching mechanism is well understood. The reducing bleaches probably destroy the carbonyl group if it is present in the dye molecule. See for example Bleaches and Sterilants by Marianna A. Busch, Kenneth W. Busch, in Encyclopedia of Analytical Science (Third Edition), 2019. Will hydrogen sulfide reduces a dye? You can try Googling along these lines. H2S is a relatively weak reducing agent so I don't think it can destroy/reduce organic dyes.
In older books, from the 1920 to-40s, you will see an extensive usage of the word "nascent" terminology, and some parts of the world which have not updated their high school chemistry syllabus in the last 50 years still use this terminology in their school books. Their university syllabs are very good but nobody bothered to change school and college texts. However, old works are not to be ridiculed as Wikipedia claims; they should be respected as they represent the thought process of our learned scholars to explain the phenomenon which is not fully understood as yet. I already addressed the current usage of nascent terminology. It still occasionally appears in respected journals. See this previous post Is there any evidence, any evidence at all, that nascent hydrogen actually exists?

"Some parts of the world which have not updated their high school chemistry syllabus in the last 50 years" - unfortunately, really familiar for me...


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Difference between nucleophilicity of phenoxide ion and acetate anion in protic/aprotic solvent
As far as I could think, the phenoxide ion is comparatively larger as compared to acetate ion, so it should undergo less solvation in presence of a protic solvent and thus its nucleophilicty should be more than that of acetate ion (in a protic medium). This is precisely what my solution sheet says. Also, acetate ion has more dispersed electron density due to it being a much better electron withdrawing group than phenoxide ion, so combining both of these points, phenoxide ion should be a overall better nucleophile than acetate ion. But from here:

acetate ion: In the acetate ion, the methyl group is an electron donating group. The negatively charged oxygen atom increases the electron density around the carbonyl carbon Thus, the acetate ion is more nucleophilic than the phenoxide ion.

From this answer, I do seem to understand that there is always more to nucleophilicity than can be estimated with electronic and size related effects. But still, such questions is frequent in my textbook. I need to understand as how exactly should I compare relative nucleophilicity of given nucleophiles here. Also, if the solvent is changed to a aprotic solvent, would it lead to any difference here?

Negative charge in case of acetate is highly delocalized on the two oxygen atoms. It would be quite a poor nucleophile - much like how $\ce{HSO4^-}$ never acts as a nucleophile whenever used in aqueous medium. Acetate is not as bad as $\ce{HSO4^-}$, but still, definitely weaker than phenoxide.

Also, in general, if the nucleophilic atom is same (here oxygen in both species), then order for nucleophilic and basic strengths are same.


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Regarding the oxidation of Manganese(II) ion by Peroxodisulphate ion
The following information is mentioned in my textbook:

In the laboratory, a manganese(II) ion salt is oxidized by peroxodisulphate to permanganate.
$\ce{2Mn^2+ + 5S_2O_8^2- + 8H2O -> 2MnO4- + 10SO_4^2- + 16H^+}$

The manganese ion does get oxidized,  but as the oxidation number of sulphur in $\ce{S_2O_8^2-}$ and $\ce{SO_4^2-}$ is the same, i.e., equal to +6, $\ce{S_2O_8^2-}$ doesn't seem to undergo reduction. For a redox reaction, both oxidation and reduction are necessary to occur simultaneously. But in this equation, only oxidation seems to be taking place.
I think it might have to do with the peroxy-bond in the peroxodisulphate ion, but I am not entirely certain. Is it because - oxidation state of two oxygens (of $\ce{S_2O_8^2-}$) on the reactant side is -1, while all the oxygen atoms on the product side have their oxidation state equal to -2, so there is a net reduction? Could anyone please confirm if my understanding is correct?

Review how the oxidation number is defined and you have your answer.  // simple hint: Do you need any electrons to convert peroxodisulphate to sulphate ? Are peroxocompounds generally oxidation agents ? How is managed oxidation e.g. by hydrogenperoxide in oxidation number context ?

@Poutnik Thank you so much, I see it now; $\ce{S_2O_8^2^- + 2e^- -> 2SO_4^2-}$ - gain of two electrons, so reduction is indeed taking place

It seems OP understands Poutnik's rationalization on the reduction step, so this question does not need an answer. However, I feel it is not completed because OP's statement of:

[...] but as the oxidation number of sulphur in $\ce{S2O8^2−}$ and $\ce{SO4^2−}$ is the same, i.e., equal to +6, [...]

This statement is not quite true if you just do the calculations and conclude from that as OP initially did. For benefit of other novice chemistry readers, I think it is better to explain what's going on in this situation in detail.
Suppose  the oxidation number of sulfur in peroxodisulphate ion $(\ce{S2O8^2−})$ is $\alpha$. Then if you do usual calculations, it is like:
$$ 2 \cdot \alpha + 8(-2) = -2 \ \Rightarrow \ \therefore \alpha = \frac{16 - 2}{2} = +7$$
This means there is a reduction happening. However, this calculation is also erroneous. Let's look at the Lewis structure of the peroxodisulphate ion, which is a particularly interesting one:

The trick here is that you should be able to notice that $\ce{S2O8^2−}$ ion has two oxygen atoms that are bonded via a single bond, which is known as a peroxide linkage. This linkage will influence the average oxidation state of oxygen in $\ce{S2O8^2−}$ ion. The two oxygen atoms in peroxide linkage will have a $−1$ oxidation state, similar to the same oxidation state that oxygen has in all peroxides. The rest of oxygens in the anion have usual $−2$ oxidation state, which implies that the average oxidation state of oxygen in $\ce{S2O8^2−}$ ion is:
$$\frac{6(-2) + 2(-1)}{8} = -\frac74$$
Thus,  if we redo the calculations:
$$ 2 \cdot \alpha + 8(-\frac74) = -2 \ \Rightarrow \ \therefore \alpha = \frac{14 -2}{2} = +6$$
This did not imply a reduction when we look at the oxidation number of sulfur in $\ce{SO4^2−}$ ion. If it is $\beta$ in $\ce{SO4^2−}$, then,
$$ \beta + 4(-2) = -2 \ \Rightarrow \ \therefore \beta = 8 - 2 = +6$$
Yet, if you know the Lewis structure of the peroxodisulphate ion, you would realize there is something going on for oxygen atoms in $\ce{S2O8^2−}$ ion, which cannot be explained by these calculations. Therefore, it is always better to conclude by the facts found using the redox half reactions. Consequently, I admit that I also prefer to do usual redox half reactions if I find no change in oxidation states doing usual calculations, because that gives no place to make erroneous calculations. Accordingly, for the reduction half reaction:
$$\ce{S2O8^2− + 2 e- <=> 2SO4^2- } \tag1$$
This is the reduction half reaction. The equation $(1)$ clearly indicates the $\ce{S2O8^2−}$ ion has undergone a reduction during the complete reaction. The corresponding oxidation half reaction would be:
$$\ce{Mn^2+ + 4H2O <=> MnO4- + 8H+ + 5 e-} \tag2$$
If you combine the equations $(1)$ and $(2)$ in order to cancel the electrons, you get:
$$\ce{2Mn^2+ + 8H2O + 5 S2O8^2- -> 2MnO4- + 16H+ + 10 SO4^2-} \tag3$$
This is the given equation. When you consider the Standard Electrode (Reduction) Potentials $(E^\circ)$ in aqueous solutions at $\pu{25 ^\circ C}$, you would find for reaction $(1)$ it is $\pu{2.01 V}$ and for reaction $(2)$ it is $\pu{-1.51 V}$ (oxidation potential). Thus, for the completed reaction $(3)$ at standard conditions, $E^\circ_\mathrm{Rxn} = 2.01 + (-1.51) = \pu{+0.50 V}$, implying it is spontaneous at standard conditions.

There were 4-5 minor grammar errors (probably typos or keyboard autocorrect) which I edited. Please rollback if the changes are not appropriate.


text:
Which oxygen in ester group is more nucleophilic?
According to my understanding, the filled orbital (lone pair electron for this) is more nucleophilic (higher energy) if:

it has more $\mathrm p$ character (since p orbital is the furthest from the nucleus, which has the lowest stabilizing effect between $\mathrm s$, $\mathrm{sp}$, $\mathrm{sp^2}$, $\mathrm{sp^3}$, and $\mathrm{p}$)
the atom has a negative charge

Drawing the resonance formulae about an ester group

and according to the internet, the oxygen in C=O bond is much more nucleophilic than another oxygen since:

it has $\mathrm{sp^3}$ character
it is negatively charged

However, what I am thinking is:

another oxygen also bears $\mathrm{sp^3}$ character (left resonance structure)
the resonance structure on the right is minor compared to that on the left (according to what I have been taught, more charges means less major resonance structure), so the magnitude of partial-positive in that oxygen should not be high

Therefore, won't both oxygens in the ester group comparably nucleophilic?

Because of the depicted resonance structure, you rightly stated, that the carbonyl group is more nucleophilic (since no negative charge can be "resonanced" on the non-carbonyl oxygen). The other thing is, non-carbonyl oxygen has $\mathrm{sp}^2$ hybridization - with its $\mathrm{sp}^2$ lone pair it can stabilize antibonding sigma orbital of the carbonyl (on the carbon) and with its lone pair in p orbital it can stabilize antibonding pi orbital of the carbonyl (again on the carbon) - and that is why it can even donate the lone pair to this antibonding orbital in the resonance structure you are showing.
So, because of the resonance structure plus donation of electrons from non-carbonyl oxygen (thus lacking the $\mathrm{sp}^3$ character), the carbonyl oxygen should be more nucleophilic.


text:
Why is the term "nascent nitrogen" still used in metallurgical industries?
Continuation of Why do nitrogen-containing organic compounds give N2 as the combustion product? (comment converted into question)

Nascent state is an obsolete theory in chemistry and is discredited. However, nascent nitrogen is still relevant in metallurgical domain especially in the "nitriding process". Basically, nitriding is a process of diffusing nascent nitrogen into the surface of steel, iron and other metals like chromium, aluminum etc. (since ordinary nitrogen is inert at treatment temperatures in the range of 500-600 °C). Ammonia acts as a source of nascent nitrogen because at treatment temperature, it dissociates into nitrogen and hydrogen which cracks on the metal surface and then diffuses into the metal to form respective metal nitrides which creates high hardness in the surface of the nitrided component. This nitrided layer is called the "white layer".
Relevant links:

https://www.asminternational.org/documents/10192/1849770/06950G_Chapter_1.pdf
https://www.boconline.co.uk/en/images/Principles%20of%20Nitriding%20and%20Nitrocarburising_tcm410-114390.pdf
https://www.bodycote.com/technical-glossary/nitriding/

Question: Why is the term nascent nitrogen still used although the theory has become obsolete? Isn't atomic nitrogen a better term?

Most likely because of their long history of using it with a specific metallurgical meaning which, while based on the old chemistry notion, has acquired a life of its own. They don't flow 'atomic nitrogen' into the ovens, the flow ammonia that ultimately gives the nitrogen. And, details of the exact process occurring on the surface are of passing importance as long as it happens fast enough. So, I'd say that the metallurgical term is distinct from, albeit originally based on, the old chemistry term.

I doubt there's actual atomic nitrogen in any reaction, that isn't conducted under extreme conditions.

I do not remember ever using the term "nascent nitrogen" altough I have dealt with nitrided steels and nitrogen in titanium. The "whitelayer" of nitrided steels is generally undefined composition, poorly understood. And considered the use of N as an alloy element adding pitting resistance to stainless type steels.

Nilay Ghosh, I removed the ridicule wording from that Wikipedia essay because none of the two articles ridiculed the nascent hydrogen theory but rather showed that there is no need to invoke a species may not exist.

The simple answer is probably that metallurgists are not chemists and stick with imprecise terminology that no chemist thinks means anything because, in the context of metallurgy, they it means something specific and the actual chemistry behind it isn't relevant.

As we discussed in the previous query Is there any evidence, any evidence at all, that nascent hydrogen actually exists?, the nascent terminology is still in use even in good ACS / RSC journals, which are pretty selective. However, science does not rest on authority; seeing the usage of nascent (gas) does not make it an absolute truth whoever is the author. This type of wording can creep in ocassionally. The idea of nascent hydrogen, oxygen, chlorine, nitrogen was not discredited recently, and people have been arguing against or in favor of it for a long time.
I think in modern terminology, the word nascent is simply a placeholder term when the reaction mechanism is precisely not known or clear. Sometimes, it was unnecessary to invoke a nascent "gas" (esp. hydrogen) in some clear-cut cases. For example, borohydride reduces arsenic ions to arsine-this method is used in very sensitive analytical procedures, and people thought it was nascent hydrogen making this reduction. Not so.
One intriguing example is that if an electric discharge is passed through pure nitrogen, it remains pretty reactive for a while! It ignites in contact with iodine or sulfur, whereas ordinary nitrogen does not even bother to react with them. See Bakerian Lecture on Chemically Active Form of Nitrogen.
Now one cannot replace nascent nitrogen with atomic nitrogen because nascent nitrogen does not need to be in a single atomic state. It could be an excited nitrogen molecule or a singly ionized atom or could be a nitrogen radical. Who knows? I only see people putting forward hypothesis without any convincing experiments. Unless there is a proof, people will continue to use dummy placeholder as nascent nitrogen. Atomic nitrogen would be too restrictive-hence the placeholder term "nascent nitrogen" in metallurgy too.


text:
Expansion of a gas under isothermal and reversible conditions
What is the reason for the statement - "When the expansion of an ideal gas is carried out under isothermal and reversible conditions, the internal energy does not change, i.e., $\Delta$U=0?"
A relation with the equation ($\Delta$H=$\Delta$U-P$\Delta$V)  or ($\Delta$U = q + w) would be helpful
U = Internal Energy; H = Enthalpy; V = Volume; q = Heat gained by the system; w = Work done by the system.

For ideal gas change in internal energy(Delta U) is given by U2-U1=nCv(T2-T1). In isothermal process T1=T2. Hence U2=U1.

Note that it applies rather to a perfect gas which is an ideal gas with the constant heat capacity.

Irreversible isothermal expansion also gives $\Delta U=0$

Internal energy is an equilibrium physical property of an ideal gas (or any other substance), independent of any process.  So it can't depend directly on q or w.  Historically, it was observed experimentally that, for gases that approached ideal gas behavior, U was a function only of temperature (not pressure or volume) and that $\Delta U$ depended only on the temperatures of the two end states.  Later, after the 2nd law of thermodynamics was developed, it was shown mathematically that, for a substance that obeys the equation of state PV=nRT, U=U(T).


text:
Change in entropy in reversible and irreversible processes
The change in entropy for a reversible process is given as-
$\Delta$S = $\frac{q_{rev}}{T}$
Where qrev is the heat supplied isothermally and reversibly.
The change in entropy for a irreversible process is given as-
$\Delta$S = $\frac{q_{irrev}}{T}$
Where qrev is the heat supplied isothermally and irreversibly.
Could someone please explain these equations while giving the meaning of qrev and qirrev? I didn't really get a proper understanding of them.

For a reversible process that is not isothermal, you need to integrate between the initial and final states of the system:  $$\Delta S=\int{\frac{dq_{rev}}{T}}$$where $dq_{rev}$ is the differential amount of heat transferred across the boundary of the system during the process and T is the temperature.
The change in entropy for a system experiencing an irreversible process is not $q_{irrev}/T$ or even $\int{\frac{dq_{irrev}}{T}}$.  To get the entropy change for a system that experiences an irreversible process, you need to devise (i.e., dream up) an alternate reversible process path between the same initial and final states as the irreversible process, and evaluate $\int{\frac{dq_{rev}}{T}}$ for that reversible process path.

"For any reversible process carried out change in entropy (S2-S1)=Q(rev)/T." -Is there a derivation for this, or is it just the definition?

Somehow, back in the day, Clausius recognized observationally that the integral of dq/T is the same for all possible reversible paths between the same two end points (there are an infinite number of reversible paths).  That means that the quantity is a function of state (like internal energy or enthalpy).  He called this function of state the entropy.

For any reversible process carried out change in entropy
(S2-S1)=Q(rev)/T. If Temperature varies then you have  to  integrate the differential changes in entropy. Q(rev) is the amount of heat transferred to the gas in this "reversible" process. I hope there is no confusion till now. Next consider an irreversible process. Let initial state of the system be A and final state be B. Now the problem  with irreversible process is that during such a process all parts  the system is not in equilibrium with each other. In other words you will not  be  able to find out a particular equation representing the process. So in this case you can't find out work done by integration and hence you would not be able to apply first law of thermodynamics to find heat exchanged. So we exploit the intensive property of entropy. An intensive property is one which depends only on initial and final state of the system and does not depend on the path. So it may occur to you that entropy change in the irreversible process will be same as that in a HYPOTHETICAL REVERSIBLE PROCESS CONNECTING THE TWO STATES A AND B. Now in this reversible process you can easily find out the heat exchanged during an infinitely small step  during which temperature can be assumed to be constant. Thus you can integrate and find total entropy change. This method is valid for any process not only for isothermal. In isothermal since temperature is constant you need not integrate and the problem becomes  easy


text:
Good sample references for TGA, DSC and DTA thermal analysis
I am trying to estimate kinetic parameters for my liquid decomposition using thermal analysis. After doing a literature survey, I have decided to use the following -

thermogravimetry (TGA)

differential scanning calorimetry (DSC)

differential thermal analysis (DTA)

Each method will be tried with 3 different constant heating rates. My expected output from this analysis is

If multiple steps are taking place

prefactor $A$ and activation energy $E_a$ (Arrhenius equation) for each step

Reaction model for my process.

And products (my products will be in gaseous form) from these will be analyzed using FT-IR spectroscopy to support my conclusions in case of multi-step kinetics. The problem is  I gave one sample for DSC to a 3rd party company but I am unsure about the accuracy given by the company. What I decided was to give them a known sample whose DSC, DTA, and TGA results are common and fairly complex but known properly, so I can test their method of doing.
If anyone could suggest any common materials which will be available in any chemistry labs whose thermal analysis output I can use as reference it would be very helpful.
Any help will be appreciated.
P.S My liquid decomposition will be accompanied by

Mass loss

Heat release (Intially endothermicity as the solution will be in water and then exothermicity)

I am puzzled by the proposed course of action. You selected three techniques where you perhaps do not have expertise and you want to test the accuracy of a  third party company by sending them substances that have been well characterized by others. How will you know if they are acceptably accurate?

A well managed/certified lab routinely checks its instruments with traceable references to meet the standards in the field (a listing: ISO 11357-1, ISO 11357-4, DIN 53765, DIN 51007-1 or ASTM E1269). Cheap and easy (compared to indium) for DSC is sublimed benzoic acid (or others listed here), but what after?

It might be more efficient to ask them to display their expertise recording these data for you, e.g., an accreditation, ongoing collaborations with clients other than you, publications, etc.

@EdV So my thinking was that if they provide same results as the known substances then I can be sure that they are doing the process right. What do you think can be a more better procedure?

@Buttonwood Thank you for your valuable inputs, so I am a researcher in a university (IIT Bombay) and all these analytical techniques are run by PhD students, so it will be a little tricky to ask them about their expertise directly as I am heavily dependent on them for characterization.  So that's why I thought of taking this politically correct route.  If you could suggest some alternative so as to how I can proceed, I would be grateful.

I think the suggestions by @Buttonwood are better. The issue with “same results” is how same is same? Frankly, if I was the third party analysts, I would be wondering why I was being sent some simple test samples. Is it because the customer does not trust me or is it a trick of some sort, i.e., a mixture designed to trip me up? By the way, I have not voted at all. Best of success whatever you decide!

I copy and paste Ed V comment

@obiwan Continuation in the chatroom The Periodic Table.


text:
Identifying radial probability distribution curve given $n$ and $l$ value

The correct radial probability distribution curve for the hydrogen atomic orbital with principal quantum number,$ n = 3$ and azimuthal quantum number, $l = 1$ is: ($4πr^2ψ^2 $= radial probability density function and r = radial distance from the nucleus)

Ans (3), source

Clearly, we are dealing with $3p$ orbital, now we know that there is one angular node and one radial node. This leads me to think there would be two minima for the graph, however, in the correct answer 3., there is only one minima. So, I suppose either the angular node or radial node was considered.
But, even then, supposing there is only one minima, how do we further distinguish between 3 and 4?
I would appreciate answers/ comments which would point me to the ideas I need to understand to solve this.

The form of the distribution always has the form of fig 1, 2 or 3  not 4. This follows because the $r^2$ in the probability increases rapidly with distance $r$ but is eventually curtailed by the exponential part in the wavefunction.

Do all orbital have such an exponential factor attached in it's mathematical representation? @porphyrin

@Buraian Kindly Do not use math jax etc. In the title as it is not possible for the search engine to find the question. Thanks!

yes, a decreasing exponential such as $e^{-ar}$ in the radial part that is. You can look up the equations in your phys chem textbook.

Here is a 3D diagram of the 3p orbital. You get the corresponding radial probability by considering probability at a certain distance in any direction.

These images are actually hard to look at. I mean The wave function should tail off towards 0. Either the quality of the image is so terrible, or they only spend five minutes on hastily preparing it.

I hardly ever link to it anymore, but this one seems to have the maths just right: "The Wavefunctions" by David M. Hanson, Erica Harvey, Robert Sweeney, Theresa Julia Zielinski, LibreTexts is licensed under CC BY-NC-SA.

The wave function of the H atom have a number of nodal surfaces equal to the first quantum number. But these nodal surfaces are not always spheres, as you think. They can be planes or still different surfaces.
In general the quantum number $n$ gives the total number of nodal surfaces. The second quantum number $l$ gives the number of nodal non-spherical surfaces of the particular atomic orbital. As a consequence, $n -l$ gives the number of nodal spheres, that you are calling "minima" in your drawing.
Let's talk about the atomic orbitals with $n = 3$, as it is your desire. The $2$nd quantum number can be $0$, $1$, or $2$.
If $n = 3$ and $l = 0$, the orbital is $3s$ : there are one nodal sphere at infinite distance, and two nodal spheres near and around the nucleus. And there is only one such orbital, as the third number $m$ is equal to zero.
If $n = 3$ and $l = 1$, the orbital is called $3p$, and it is your problem. As the third number $m$ can be $-1$, $0$, or $+1$, there are three such orbitals $3p$, called $3p_x$, $3p_y$ $3p_z$. For each orbital, there is $l = 1$ nodal plane per orbital, and $n - l = 2$ spherical nodes in each orbital, one being at infinite distance, and one near the nucleus. There are three such orbitals, with the same number of spherical and planar nodes : one has a nodal plane perpendicular to the $Ox$ axis, the second perpendicular to the $Oy$ axis, and the third perpendicular to the $Oz$ axis.
If $n = 3$ and $l = 2$, the orbitals are called $3d$. As the third quantum number $m$ can have 5 values, being $-2$. $-1$. $0$, $+1$ or $+2$, it makes five $3d$ atomic orbitals. These $5$ orbitals have all $n - l = 1$ spherical node, which is at infinite distance from the nucleus. Remains two nodal surfaces per each one of these 5 orbitals. For the first three of these five orbitals, the $2$ non-spherical nodal surfaces are two mutually perpendicular planes, namely $Oxy - Oyz$, then $Oxy - Oxz$, and $Oyz - Oxz$. The two other $3d$ orbitals must have two nodal surfaces which are neither spheres, nor planes. They may be conical, but they are difficult to explain without drawing.
This long development explains why the distribution curve of $3p$ orbitals have two minima (one at ∞), so that your figures $3$ and $4$ are equally valid. To explain why Fig. 3 is to be preferred, you should use the radial wave function for $3p$ orbital, which is $$\ce{R(r) = constant · (4 - {r/a}) ({r/a}) e^{-{r/2a}}}$$ When calculating $\pu{R(r)}$ for different values of $\pu{r/a}$ going from $1$, to $2$, $3$, $4$, etc. up to $10$, you will find that $\pu{R(r)}$ takes the following values :  $$ 1.82, 1.47, 0.67, 0.00, 0.41, 0.59, 0.63, 0.586, 0.50, 0.404$$ This shows clearly that the maximum value of $\pu{R(r)}$ must be near the nucleus, and before the minima observed at $\pu{r/a = 4}$. Now if you multiply this value by $\pu{r^2}$, to get your wanted distribution curve, you find that $\pu{R(r)·r^2}$ takes the following values : $$\pu{1.82, 5.89, 6.03, 0.00, 10.25, 21.24, 30.87, 37.5, 40.5, 40.4}$$ This shows clearly that your distribution curve has a maximum far away from the nucleus, and far away form the minima (observed at $\pu{r = 4a}$).
As a consequence, your curve $3$ has to be preferred.

Hmm not exactly, I felt tyberius has written an answer closer to what I wanted

This is a nice and well written answer but I suggest writing comments ("is this what you are looking for?") before posting such a lengthy answer that does not address the question. Because of that last sentence this might be removed as a comment.

I think it should be $n-1$ and $n-1-l$ for the node values

@Buraian. The number of nodal surfaces is $n$ and $n-l$ if the infinite is counted. It is $n-1$ and $n-1-l$ if you count the finite surfaces.

Why doesn't angular node not appear?

These plots are with respect to $r$, so it will only show radial nodes. An angular node would appear in a plot with respect to $\theta$. So for 3p, you will only have one node.
To distinguish between options 3 and 4, you just need to realize that larger $n$ will have it's maximum further from the nucleus.

The second point, does that property have a name/ can you suggest where I could read more about it? (The larger n rule)

is R only the radial part of the probabiltiy i.e: is there constribution from azimuthal and polar part as well?

@Buraian I don't know if this property has a name, but you can recognize this pattern by visualizing the orbitals. You can see that as $n$ increases, you get a sequence of $n-1$ shells with gradually increasing size as you move away from the nucleus. The radial distribution function already integrates over the azimuthal/polar parts, so you are just seeing how the distance affects the probability. Its basically averaged over all the azimuthal/polar angles.


text:
What is the difference between structure and morphology in polymer blends?
When talking about polymer blends there is a difference between its structure and morphology, but I am not sure what it is. These two terms are used interchangeably in materials I use to study, but I don't see what is the difference if any.
Quote from my materials: "Properties of polymer blends are determined by component interactions and blend morphology." The next sentence: "Blend structure is determined by miscibillity of components."

You better elaborate more about that...

Morphology and structure can be very ambiguous terms, meaning just about anything (shape, tacticity, molecular conformation, macromolecular or crystal arrangement). It would be nice to see a good ("canonical") answer from somebody knowledgeable.

You might want to check the IUPAC Compendium of Polymer Terminology and Nomenclature.

They appear to be used as synonyms in this context. It is a reference to the structure-function paradigm in the context of polymer blends, where chain structure has a role to play in determining miscibility and therefore phase structure.

In short:  it may be used as a question of scale.

A polymer blend is a physical mixture of two (or more than two) polymers.  As such, you may compare it with granite, mainly consisting of grains of feldspar, quartz, and mica put together; here, different chemical compounds are present in one material and they retain their individual local chemistry.

This contrasts to a copolymer (e.g., polyesters, polyamides) like PET, or nylon; here, the constituents are chemically bound together and form a new compound.

Even for the (carbon based) polymers said to be crystalline, or to contain small local volumes with some regular, crystalline order, it is rare for polymers to have a unit cell-like local pattern which repeats itself in a predictable fashion multiple thousand times in the three directions of space by simple translation as observed, e.g., for $\ce{NaCl}$.

Thus structure (in the narrow sense) may be used to describe the spatial arrangement of polymer's atoms at a small scale (say a up to few $\pu{10 Å}$).  On the other hand, in comparison of the former, morphology is about shape/form/organization at larger scale of length (even if still microscopic) as e.g., for spherulites, domains, grains including properties observable if many molecules are in one place e.g., to yield a rough (vs. a smooth) surface.

Yes, this may be true. It is a question of scale.


text:
Why does polymerisation of n glycine molecules give n water molecules?
Referring to this webpage, Condensation polymerisation of glycine
A simple counting argument would tell us that:
2 glycine molecules react, releasing 1 water molecule
3 glycine molecules react, releasing 2 water molecules
4 glycine molecules react, releasing 3 water molecules
in other words,
n glycine molecules react, releasing n-1 water molecules
So, Why does polymerisation of n glycine molecules give n water molecules?
Is it because the author assumes that the polyglycine molecule forms a loop (end-to-end) instead of a "straight chain"?

n/(n-1) converges to 1 for large n. If you do not know n, but you know it is large, you can safely count with 1:1 glycine:water approximated ratio.

Don't call it polymerisation, but polycondensation.

Because the product as pictured is not the complete polymer; it includes only $n$ units (called "mers") in the middle of the chain and not the complete polymer which has more than $n$ mers. The missing mers include the ends that retain extra hydrogen and oxygen atoms and would make the number of eliminated water molecules one less. Think of the case $n=10$ but the actual polymer has 30 units mers with only the middle 10 mers shown in the quoted product. The 10 mers shown eliminate 10 out of the 29 water molecules that emerge from the complete 30-mer chain including the proper ends.
Textbooks often do this because, supposedly, the ends are negligibly small portions of the long chain. Whether they truly are negligible in a chemical sense is something I will leave to users specializing in polymers or especially in proteins.


text:
Molar mass question
Molar mass of oxygen is 16g/mol , but O2 is 32g/mol , it is shows that 32 gram per 1 mol ,but if we add 1 mol oxygen and one mol of oxygen , shouldn't it be 32 gram per 2 mol for dioxygen ?
it doesn't make sense to me , it is like 2candies/dollar + 2candies/dollar = 4candies/dollar
And we pay a dollar for 4 candies , shouldn't it be 2 dollars for 4 candies ?
Maybe this is a stupid question but i need an explanation

You will need to add some context to explain why that first equation makes any sense at all. Without that it makes no sense to start with.

One oxygen atom plus one oxygen atom yields one dioxygen molecule, so 1 + 1 = 1? That is at the root of your question.

There are two layers to this question. The first is purely mathematical, and from the mathematical point of view, units are to be treated the same as any other number or variable. That is, they follow the same rules as plain old numbers $1$ or some unknown variable $x$. Thus, purely from a mathematical point of view,
$$\pu{23 g/mol} + \pu{23 g/mol} = \pu{46 g/mol},$$
exactly like how one might write $23x + 23x = 46x$ in an algebra course. This is not all that interesting. The more interesting part is the physical interpretation of this equation. Let's start with a scenario in which you can't quite write this.
The molar mass of a substance—let's say the molar mass of $\ce{Na}$, which is $\pu{23 g/mol}$—is an intensive property. What that means is that the molar mass doesn't depend on the size of the system. So let's say you have two lumps of sodium, and you combine them into one big lump. The molar mass of sodium doesn't change when you do this: the molar mass of sodium is $\pu{23 g/mol}$, regardless of how big the lump is.
This is pretty analogous to your price of candies analogy. Suppose each small lump of sodium was $\pu{1 mol}$ exactly: that would mean each lump weighs $\pu{23 g}$. Then if you add them up, the total mass is $\pu{46 g}$, but you have a total of $\pu{2 mol}$, which gives you the same old $\pu{23 g/mol}$ at the end.
But, it is not appropriate to write this as $\pu{23 g/mol} + \pu{23 g/mol} = \pu{23 g/mol}$. The real calculation that we did above should be expressed as this, where $M$ indicates molar mass, $m$ indicates mass, and $n$ indicates amount of substance (which is the proper term for "number of moles"). Since $M = m/n$, we have that
$$\begin{align}
M(\text{big lump)} &= \frac{m(\text{big lump)}}{n(\text{big lump})} \\[8pt]
&= \frac{m(\text{small lump 1}) + m(\text{small lump 2})}{n(\text{small lump 1}) + n(\text{small lump 2})}  \\[8pt]
&\color{red}{\neq} \frac{m(\text{small lump 1)}}{n(\text{small lump 1})} + \frac{m(\text{small lump 2)}}{n(\text{small lump 2})} \\[8pt]
&= M(\text{small lump 1}) + M(\text{small lump 2}).
\end{align}$$
Note the inequality sign, which follows simply because $(a + b)/(c + d) \neq (a/c) + (b/d)$. So it is mathematically quite unsound to write that $\pu{23 g/mol} + \pu{23 g/mol} = \pu{23 g/mol}$.
Note that this doesn't mean that you can't add up molar masses. From a mathematical point of view, it is perfectly valid to do so. The problem lies not with the maths, but with its applicability to the physical problem at hand.

The situation in which we want to add up molar masses is a completely different one, and it occurs when you are trying to calculate the molar mass of a molecule. Let's say, NaCl.* Then, we can sort of write the following; and the "sort of" qualifier will become clearer later.
$$\begin{align}
M(\ce{NaCl}) &= M(\ce{Na}) + M(\ce{Cl}) \\
&= \pu{23 g/mol} + \pu{35.5 g/mol} \\
&= \pu{58.5 g/mol}.
\end{align}$$
Now, the reason why this works goes back to the maths earlier. Let's say, just like before, we have two small lumps, each containing $\pu{1 mol}$: but the first lump is sodium, and the second lump is chlorine. (This is all hypothetical, of course.) And then you combine them into one big lump of $\ce{NaCl}$. What is the molar mass of $\ce{NaCl}$? We have:
$$\begin{align}
M(\ce{NaCl}) &= \frac{m(\ce{NaCl})}{n(\ce{NaCl})}.
\end{align}$$
The numerator can be decomposed into $m(\ce{Na}) + m(\ce{Cl}) = \pu{58.5 g}$, because the mass of two things is just the sum of those two things. But, the denominator cannot be decomposed into $n(\ce{Na}) + n(\ce{Cl})$, because $\pu{1 mol}$ of $\ce{Na}$ plus $\pu{1 mol}$ of $\ce{Cl}$ gives you not $2$, but only $\pu{1 mol}$ of $\ce{NaCl}$!
Ergo, we have $n(\ce{NaCl}) = \pu{1 mol}$, and so:
$$\begin{align}
M(\ce{NaCl}) &= \frac{m(\ce{Na}) + m(\ce{Cl})}{\pu{1 mol}} \\[8pt]
&= \frac{m(\ce{Na})}{\pu{1 mol}} + \frac{m(\ce{Cl})}{\pu{1 mol}} \\[8pt]
&= \frac{m(\ce{Na})}{n(\ce{Na})} + \frac{m(\ce{Cl})}{n(\ce{Cl})} \\[8pt]
&= M(\ce{Na}) + M(\ce{Cl}).
\end{align}$$
Note that this time we can split up the fraction into two, because $(a + b)/c = (a/c) + (b/c)$. Aha, so that's the origin of this equation. But note that we didn't purely do mathematics to get to this point: we also invoked our knowledge about moles, in that $\pu{1 mol}$ of $\ce{Na}$ plus $\pu{1 mol}$ of $\ce{Cl}$ is equal to $\pu{1 mol}$ of $\ce{NaCl}$. We then used that to simplify our original equation, so that at the end of the day we could write
$$M(\ce{NaCl}) = M(\ce{Na}) + M(\ce{Cl}).$$
This can of course be generalised to molecules with more than 2 atoms. So the validity of this final equation we have here, rests upon the fact that 1 mole of every component adds up to give you 1 mole of the composite system.

* Yes, yes, $\ce{NaCl}$ isn't a molecule. I know that.