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Is it safe to combine sodium hypochlorite (bleach) sodium triphosphate (TSP) for cleaning?
I've heard this combination can be particularly effective at cleaning biofilms, perhaps because the TSP is a surfacant, allowing bleach to "attack" more layers of the biofilm.
However, I'm also aware one should be cautious when mixing anything other than water with bleach, as there are a wide variety of reactions that can occur. But I've seen nothing about bleach and TSP. Should it be safe?

Dissolved in water, trisodium phosphate $\ce{Na3PO4}$ produces ions $\ce{PO^{3-}4}$ that react by hydrolysis with water, according to $$\ce{PO^{3-}4 + H2O -> HPO^{2-}4 + OH^-}$$ So such a solution contains the ions $\ce{OH-}$, and these ions exist also in the  bleach. Indeed, bleach main ingredient is sodium hypochlorite $\ce{NaClO}$. But its solution contains significant amounts of $\ce{NaOH}$ which produce the ions $\ce{OH-}$. As both solutions (bleach and trisodium phosphate) contains the same ions, they can be mixed without any trouble. Furthermore the hypochlorite ion $\ce{ClO-}$ does not react with the phosphate ions $\ce{PO^{3-}4}$ and $\ce{HPO^{2-}4}$ .

Actually, the more complete answer is that Na3PO4 is a recommended stabilizer for bleach.
Also, per a ScienceDirect source, to quote:

Trisodium phosphate hydrolyzes to produce hydroxide ions. Phosphate acts as buffer...'coordinated phosphate pH control.'

where pH control prolongs the shelf life of chlorine bleach, as supported by the statement (same source):

One down-side of chlorinated products is a limited storage stability. This instability reduces the chlorine benefit which the detergent can deliver, but also reduces the benefit of the other active ingredients which are destroyed by bleach action.

Further, more advanced radical-based bleaching is suggested by the comment:

For dyeing with vinyl sulphone dyes, dyebath is prepared with trisodium phosphate (2%) and dye.

So, the scavenging of light (or transition metal contamination) induced radicals (which can lead to unwanted chlorate formation), is likely also an advantage of employing Na3PO4, in my opinion.


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IUPAC nomenclature of cycloalkadienes with substituents and alkenes on the same carbons
Consider the following organic compounds:

Compound $(1)$ has two double bonds, which I'll label $C1$ and $C4$ such that the methyl group is attached to $C1$ and thereby assign lowest numbers to all functional groups and substituents. Officially its name is 1-methylcyclohexa-1,4-diene.
Generally we try to give just enough information to be able to reconstruct the original compound, but not too much. With that in mind, I feel that compound $(1)$ should be named simply 1-methylcyclohexa-4-diene:

Since it ends in -diene, it's understood that there are two double bonds.
Since it ends in -4-diene, it's understood that one of them begins at $C4$.

If the second double bond were located after $C4$, the numbering is backwards; instead of, say, -cyclohexa-4,6-diene, it should've been named -cyclohexa-1,3-diene.
If the second double bond were located at $C2$ or $C3$, then priority was given to the methyl substituent over the alkene functional groups. Rather than calling it 1-methylcyclohexa-2,4-diene, it should've been named 5-methylcyclohexa-1,3-diene.
Therefore, the only option left is for the second double bond to be at $C1$, and therefore specifying 1-methylcyclohexa-1,4-diene is extraneous.

Following a similar line of thought, compound $(2)$ ought to be named just 1-methylcyclohexa-3-diene, rather than its proper name of 1-methylcyclohexa-1,3-diene.
Taking this logic to an extreme, the third compound should be named 1,4-dimethylcyclohexadiene, instead of its proper name of 1,4-dimethylcyclohexa-1,4-diene, because the same logic used above to determine that the second double bond is located at $C1$ should be sufficient to determine the locations of both double bonds in this case, where they both have substituents!
Why in all of these cases, and others like them, do we specify both locants of the alkenes, when one (for compounds $(1)$ and $(2)$) or none (for compound $(3)$) is sufficient?

The question is based on a false premise:

Generally we try to give just enough information to be able to reconstruct the original compound, but not too much.

This isn’t true. The prime aim of IUPAC nomenclature is clarity and the construction of unambiguous names.

Three logicians walk into a bar
The bartender asks: “Do you all want a drink?”
The first logician says: “I don’t know.”
The second logician says: “I don’t know.”
The third logician says: “Yes.”

IUPAC nomenclature doesn’t work like that. Generally, the complete information about a structure is explicitly given by the name and does not rely on any implied information or hidden logic involving other structures.
The fact that “1-methylcyclohexa-2,4-diene” would actually be called 5-methylcyclohexa-1,3-diene does not change the preferred name for 1-methylcyclohexa-1,4-diene.
Nevertheless, there are a few exceptional cases where locants are omitted, but only when there is no ambiguity. For preferred IUPAC names, the nomenclature rules explicitly stipulate when locants are omitted. In most cases, if any locants are essential for defining the structure, then all locants must be cited for the structure.
Also note that your third example is clearly wrong even when following your unorthodox logic. Your name “1,4-dimethylcyclohexadiene” could mean 1,4-dimethylcyclohexa-1,4-diene as well as 1,4-dimethylcyclohexa-1,3-diene or even 1,4-dimethylcyclohexa-1,2-diene, which are all correct IUPAC names.

I second @Loong's statement about IUPAC, the system based the nomenclature on rule that avoid any two person from coming to a different structure derived from the same name.
Generally the functional group is your starting point based on priority and goes sequentially with priorities until you complete the name. Exceptions are still there for some basic compounds. In your example, all nomenclature based on alkene group plus methyl group, this is your number 1, the direction of numbering depends on the direction (location) of the first alkene.

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Are there any "catalysts" that decrease the rate of reaction
Catalysts are things you can add to a reaction that provide an alternate reaction pathway for the reaction that decreases the overall activation energy of the reaction meaning the reaction is faster but the catalyst isn't consumed in the reactions. Are there any materials, in any reactions where a similar material can be added to a reaction that will decrease the rate of reaction by providing a more favorable reaction pathway but one that takes longer.
Maybe it isn't possible for a reaction to be more favorable but require a higher activation energy and this is why this kind of "pseudo catalyst" doesn't exist.

We'd probably call such a catalyst an inhibitor, and it probably work by increasing the activation energy instead of decreasing it. Perhaps a mutated enzyme could work, one that still binds its substrate with high affinity, but was not capable of catalysis.

In an electrochemical cell setting, one can accelerate the electrochemistry by increasing the surface area of the cathode relative to the anode.  It follows that modifying the cell with inert material (less anodic than the current anode and less noble than the existing cathode) that decreases the surface area of the cathode could be an inhibitor. For example, create a partial CuO layer on an existing Cu cathode (aka, poison the cathode).

Well, strictly speaking, a catalyst can only be a catalyst if it lowers the barrier as that is part of its definition.

@AJKOER Certainly, but how do you do that?

The poisoning of 'catalyst' occurs frequently with a vehicle's catalytic converter. In my reference to a Cu cathode, just remove part of it, wash and heated it in air till its glows red (Cu->Cu2O->CuO) inducing a visible alteration in color. Replaces the part and the battery cell performance changes.

Continuation of my comment: Relatedly, if one is trying to perform an anodic dissolution of say Tungsten metal, in say HOCl (this is a battery cell-based electrosynthesis) with a noble cathode like Pt, using a large surface area Pt mesh is much more efficient.

@AJKOER If you take the system apart anyway, you might as well cut off half the cathode.

Well, such a "negative catalyst" cannot increase the activation energy of already existing path.
An alternative with the higher activation energy would not be significantly taken. You can detour a river making it flow easier by a side channel, but not if you lead it across a hill.
It could act differently, as an  inhibitor, blocking the low activation energy path by reaction with the key intermediate product or existing catalyst, or by competitive inhibition.
Many redox reactions, including food deterioration, are catalyzed by traces of transient metals acting as redox catalysts, typically copper. An additive, forming metal complexes like EDTA, in large extend inhibits these catalyzed reactions.

With a pipe, if the hill is less than ten meters? Sorry.  ;-)

@Karl You would need to suck water up to this 10 m first. BTW, is it still a hill, according to English cartographic terminology ?

Possibly. Let's say your reaction is
$$\ce{D -> E}$$
Your inhibitor could catalyse an additional reaction
$$\ce{C <<=> D}$$
If the equillibrium of that reaction was well on the left side, that would slow down the original reaction quite a bit.
Alternatively, it could catalyse a reaction $\ce{D->B}$, where B had an additional pathway
$$\ce{E <- B <- D -> E}$$
, which might be slow.
This is all however very hypothetical. I cannot name an example for either.
Generally, if you find that something acts as an inhibitor, that means the reaction was already catalysed, and your inhibitor either permanently poisons the catalyst, or reversibly converts it into an inactive form.

Equilibrium arrows are <=>. <-> are for tautomerization/resonance.

@Poutnik And you can even give them a bias <<=> <=>>. +1

sure, I use it often.  See mhchem

Karl: +1, your answer seems to incorporate my example of cathodic poisoning in a battery cell (cited in my comment above).

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Azimuthal Quantum Number
In the radial equation of hydrogen atom the differential equation is described by

But why is l taken to be integer. I know the principal quantum number n correspond to energy levels so that's why it's taken as integer. Why should azimuthal quantum number be taken as integer though

Do you know what $l$ is and whether number $1$ is an integer?

l is the angular momentum

But why is it an integer always when solving for hydrogen we use laguerre polynomials and solve the radial equation but then we substitute l as an integer in l(l+1)

When $\ell$ is not an integer, you'll get bad solutions.

Please don't cross-post your question. Instead, I'd suggest to grasp a more complete understanding on what $l$ is (e.g. Wikipedia). The short answer would be to look at the title: the word "quantum" kind of implies an integer.

@andselisk yeah I get that intuitively but I need a rigorous math explanation

@IvanNeretin what do you mean by bad solutions like can you elaborate

How rigorous an answer do you want? Your explanation for $n$ being integral, although correct, doesn't exactly sound very rigorous either. There is a lot of maths behind that. It falls out of the Schrodinger equation, much like how the condition for $\ell$ does, too.

If you want the maths, though, I honestly suggest checking out a QM book. There are lots of great ones out there. I'm sure many of them have the same derivation that show how $n, \ell, m$ have the conditions they do.

can you suggest which QM book has it

In pretty much any QM curriculum, harmonic oscillator predates the hydrogen atom. It is then that you've first encountered some number ($n$) which miraculously ends up being an integer. It is then that you should have asked yourself why is it so. The reason is the same: for non-integer $n$ you'll get bad (unphysical) solutions.

@zeeman $\ell$ Is NOT an angular momentum, it is just the integer quantum number determining the discrete values of orbital angular momentum.

@IvanNeretin when azimuthal quantym number is not an integer what bad solution do I get( like are the eigenvalues unreal) or like what do you mean by bad solution

You'll get solutions which are infinite series instead of polynomials, and non-integrable at infinity.

@IvanNeretin oh okay. Is there a textbook which talks about this

There are two parts to the answer to your question. (a)  When solving a differential equation the boundary conditions must be specified (this is not always emphasised enough in textbooks), and (b) these conditions are determined by the physics of the problem via the postulates of quantum mechanics, including the nature of the wavefunction. Quantum mechanics (QM) has no derivation; we accept its postulates and QM survives only because experiment has so far always confirmed it.
As an example of quantisation consider the particle in a box. Here there are all sorts of solutions to the differential equation, i.e Schroedinger's equation. However, as the walls are infinitely high the wavefunction must have zero amplitude here and this means that the only solutions that remain and are physically realistic are when quantum number $n$ is quantised. The fact that the wavefunction must be zero at each wall are the boundary conditions.
A similar situation occurs on a particle on a ring where the angular momentum quantum number $l$ is quantised only because we insist, by applying the postulates of QM that the wavefunction repeats itself exactly each $2\pi$ round the circle. This is in effect the boundary condition.
In the H atom the equation is far more complicated but the general conditions imposed on the wavefunction remains the same; the imposition of boundary conditions, as determined by QM, results in the generation of integer quantum numbers so that theory matches experimental data.

you said by applying the postulates of QM that the wavefunction repeats itself exactly each 2π round the circle but isnt this the boundary condition for n like 2×pi×r=n lambda. Can you explain a bit more on how it extends to azimuthal quantum number and imposition of boundary conditions that force integer quantum numbers

The boundary condition on $\psi$ is that $\psi (0)=\psi(2\pi)$ and that the gradient $d\psi/d(\theta) $is the same at $0$ and $2\pi$


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Reaction of mixed carbonyl compound with ethylene glycol in acidic medium

I Know that there is cyclic ketal or cyclic acetal is formed when carbonyl compound reacts with ethylene glycol. But I am confused here because I am unable to decide what will be the major product  formed here because there are two possibilities, i.e. cyclic ketal and cyclic acetal.
Can anyone help me in reasoning what will be the major product and why? Also, I want to know the role of dry $\ce{H+}$ in this reaction.
I think the cyclic ketal should be formed as major product because the oxygen of ketone will be more willing to donate lone pair to $\ce{H+}$ and react. Is it correct?

I presume you are implying a 1/1 molar ratio of ketoaldehyde and ethylene glycol. Are you asking which carbonyl reacts faster (kinetic) or which aetal/ketal is more stable (thermodynamic)?

Yes,ketoaldehyde and Ethylene glycol are in 1:1 Molar ratio.I want to know which among acetal or ketal will be formed as major product in this reaction,so Kinetic based i.e which carbonyl will react faster and form stable product.

@Ashutosh aldehydes are more susceptible to nucleophilic substitution so if I’m not wrong, the aldehyde will form the acetal here.

You are Right ,but isn't this reaction Electrophilic addition (due to H+) I don't see why it would follow Nucleophilic Substitution or Nucleophilic Addition path .

Complicated by the possibility of intramolecular cyclisation

The bis-ketal is selectively cleaved with dry FeCl3-silica liberating the ketone and leaving the acetal intact. https://www.thieme-connect.com/products/ejournals/abstract/10.1055/s-1987-27833.

Frankly,I am unable to comfortably understand what do you mean . Can you (or anyone) explain in a simpler language? (I am a high school student)

Why is no one answering :(


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Raman signal of black matte object
This is the raman spectrum of the black cap:

I was expecting it to be Polypropylene but the peaks didn't match in wave numbers. Anyone got idea what kind of plastic has that peaks on the side of 1500?

When I tried the raman analyzer on the clear housing.

I got this raman spectrum.

Which matches that of Acrylonitrile butadiene styrene see:https://en.wikipedia.org/wiki/Acrylonitrile_butadiene_styrene

Original message:
I was trying to identify the material of this sediment filter cap whether it is really Polypropylene but i couldnt get a good signal but only noises no matter the exposure and time (my raman spectrometer is 532nm.

Is it really hard to get raman signal out of black matte object, maybe it cant reflect the laser?
I was able to get good raman signal of the blue casing though but not the black cap.

".. maybe it cant reflect the laser?" Because it is black? Wow.

If you haven't realized from the above comment, I'll clarify. Black objects absorb all wavelengths of visible light.

Well, most probably there's a paint or coating there, so, no wonder...

But even if there is paint or coating..it should get raman signal of the paint right? Also in raman, the laser just excited the molecules so even if its black, it should work the same right? The spectrometer is working bec i tried cyclohexinex alcohol, water, PET, it all works except the black cap. Any ideas?

I think that @dval98 is right here; as the scattering and absorption are in competition, if all the photons are absorbed, less are available to produce the Raman signal, thus increasing sensitivity by lowering spectral resolution could help, otherwise you may have no choice but to seek out a more sensitive instrument.

@Karl Nothing is perfectly black.

@gerrit For a Raman spectrometer, black is black. ;)

@Karl, What a misconception that black objects cannot have a Raman spectrum! If this were nobody would be able to collect Raman spectrum of carbon or its allotropes.

@M.Farooq That's what OP posted above. ;)

The comments pretty much sum up whats going on, but here is a more in-depth explanation to clear up any confusion for the OP.
There is a lot more behind this but here is the (very) basic principle:
Raman spectroscopy uses monochromatic light to excite molecules. These photons interact inelastically with the molecule (i.e. they gain or lose kinetic energy) and are then scattered. Some detector analyzes the scattered light and determines vibrational energy levels by calculating how much kinetic energy the photon has gained/lost which corresponds to the difference in energy of the vibrational modes.
When a black object is hit with a laser, the photons are absorbed. The absorbed energy is released as heat (This is why black cars are significantly hotter inside than white cars when left in the sun). Maybe some scattering does occur, but who knows if it is detectable. There are some articles on trying to determine the Raman spectra of black objects [Here is one].
Do some research; look at the Journal of Raman Spectroscopy if it peaks your curiosity. I'm not saying it is impossible, but it might require a lot of effort and understanding of the fundamental principles of this technique.

What you see is the spectrum not of the polymer, but the dye.
Something that is black in VIS gives a weak spectrum (check the extremely high noise compared to your other spectrum of a clear plastic) if you irradiate it with visible light, and it also has a pretty boring NIR spectrum.
If you do a quick google search for "carbon black raman spectrum", you will however recognise the double hump around 1500/cm.
And if you're really keen, you can accumulate spectra overnight or over a weekend, and perhaps the polymer spectrum comes out of the noise.


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Why aren't the pi electrons delocalized in cyclopropene?
I was searching on the internet whether cyclopropene is aromatic or not. I found out that the electrons are not delocalized, causing the molecule to be non-planar and so forth, a lot of reasons.
But I don't know why I just don't seem to fall for it. My question is, if there are 4n+2 pi electrons, why aren't they delocalized? And what exactly stops them from being delocalized? I have tried but failed to come to a conclusion about the reasoning. Any help would be greatly appreciated.

Your question is hinged on a false premise. There are two pi electrons, and of course they are delocalized between two atoms. This is a phenomenon known as double bond. You must have heard of it before.

I might be making a mistake, but I'll just repeat what my instructor told me. He told me that if the delocalized pi electrons of a cyclic compound can be denoted as 4n+2 where "n" is an integer, then the compound can be called aromatic. And in the case of cyclopropene, if the pi electrons were delocalized then there would be 2=4n+2, giving n=0, an integer. Hence the question,

Well, there is an important condition missing here. Cyclopropene is weird, let's put it aside.  Imagine cyclohexene: a thing made of four -CH2- and one -CH=CH-. It is surely cyclic. It surely contains 4n+2 electrons (namely, 2). Is it aromatic? What do you think?

You seem to be asking why there is no delocalisation of the pi electrons in the 3-membered ring. The answer to this is very obvious: There is no free continuous path for the delocalisation of electrons over the ring to occur. Yes, there are 2 carbon atoms with overlapping p orbitals but the third one is bonded to two hydrogen atoms and does not have an available p orbital to overlap with the p orbitals of the other two atoms.

Yup, they aren't.  With the saturated carbon atom in the ring, you can't get a delocalized ring of pi electrons that you would need to apply Huckel's Rule.
A more rigorous analysis in Ref. [1] give us a curious conflict between theoretical and experimental results for this molecule.  The theoretical calculations suggest that delocalization of the pi bond occurs through overlap with the $\ce{C-H}$ antibonding orbitals, making the molecule somewhat aromatic after all.  But the experimental results referenced in the paper do not support thys prediction.  With delocalized pi bonding the theory predicts that the $\ce{C-C}$ to the methylene carbon should be shorter than the $\ce{C-C}$ bonds in cyclopropane, but the experimental data show essentially the same bond lengths.  Moreover, the quoted experimental dipole measurements do not agree with the modeled electron shift towards the methylene carbon.  The experimental evidence, therefore, is that even with the possibilities offered by molecular orbitals pi delocalization fails in cyclopropene.
Reference
1.
L. Radom, W. A. Lathan, W. J. Hehre, and J. A. Pople, "Molecular Orbital Theory of the Electronic Structure
of Organic Compounds. VIII. Geometries, Energies, and
Polarities of C3 Hydrocarbons", Journal of the American Chemical Society 93:21, 5339-5341 (1971).


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Do formic acid and phosphorus pentoxide react at room temperature?
If I add phosphorus pentoxide to formic acid at atmospheric pressure and 20°C will they react? At what rate will CO be produced?
I have not been able to find a reference discussing this.

I have no particular knowledge about this, but P4O10 dehydrates many stable substances. Formic acid is not stable, very slowly decays to H2O and CO, so it is adviced not to keep it very tight closed. So I predict it may be vivid to violent reaction.

Using sulphuric acid instead is the laboratory method to produce CO.

Also, P2O5 is a stronger dehydrating agent than H2SO4.

As noted, the dehydrating action of H2SO4 on HCOOH liberates CO gas. To quote a source:

The sulfuric acid removes the elements of water (i.e., H2O) from the formic or oxalic acid and absorbs the water produced.

As P2O5 is at least as dehydrating as Sulfuric acid, expect the same per the reaction:
$$\ce{P2O5 + 3 HCOOH -> 3 CO (g) + 2 H3PO4}$$

Note that P4O10 does not get hydrated  to H3PO4, but rather to polymerized metaphosphoric acid (HPO3)n.


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H NMR signals in 1,4-dichlorobutane spectrum
I have come across H NMR spectrum of 1,4-dichlorobutane and I have noticed that signals there are different to what I would intuitively expect. Instead two triplets there are two multiplets of more complicated structure. I wonder why that is. Looks to me like some higher order spectrum, but I'm not sure. I don't think that is the matter of spectrometer frequency. I wonder if that is some instance of magnetic inequivalence (at the first glance in 1,4-dichlorobutane there are two sets of methylene groups with enantiotopic hydrogen atoms). Can someone clear that out for me?

The pairs (C1/C4 and C2/C3) of CH2's are chemically equivalent but magnetically non-equivalent. Here is a perfect answer: http://u-of-o-nmr-facility.blogspot.com/2009/08/500-mhz-1-h-nmr-spectrum-of-butane.html

If it were only to the spectrometer's field strength (and eventually, the resonance frequencies of the nuclei), then AIST's entry about the same compound (here, peak data available, too) possibly could be "better" (i.e. more like the anticipated pattern) because it was recorded at about 400 MHz.  Yet equally, the resolution (data points per ppm), design of the probe head (e.g., what receiver coil is closer to the sample) may interfere on top to user55119's comment.

Yes, your intuition is correct. The enantiotopic protons are the key.

Is this spectrum further complicated by the $^2J$ coupling to the chlorine atoms, or is their relaxation time too short to make any difference? Their $\gamma_0$ is about a tenth of that of hydrogen, so I'd guess the coupling constant would be ~1 Hz?

@Karl Suggestion: Why not posting this as a new question, e.g. H-Cl coupling in NMR spectroscopy?  Both chlorine isotopes seem to be as similar receptive as e.g., phosphorous (ref Cl, ref P), thus it may yield an interesting broader insight ChemSE strives for, too.

@Buttonwood Gammazero is similar, but phosphorus is spin 1/2, both chlorine isotopes are 3/2. Dont want to get downvoted because of a lack of research effort, and also don't feel like studying tonight. ;-)


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Forcing atoms closer together
Is there a method in chemistry to alter the energy of atomic bonds by forcing atoms closer together in a smaller volume?
Similar to how pressure squeezes atoms closer together.

Do you have any own idea about that? What kind of substance do you think could be made to shrink?

In short, chemistry does not operate in the same terms as you use.

There are many ways in which bond lengths can be changed, and these will be accompanied by changes in bond energy.  I will give examples for four different classes of substances, though they may not all count as "forcing" the bonds to be shorter in the way in which you mean.  You can look these over, and perhaps clarify your question:
Glasses: Toughened glass is made by rapidly cooling glass, which creates internal stresses in which the interior is under tension (increasing bond lengths), while the surface is under compression (reducing bond lengths).  So while the changes in bond length are still a result of pressure, the pressure is internally, rather than externally, applied.
Molecules: You can get the molecule to undergo a chemical change.  For instance, changing ethane to ethene will shorten the carbon-carbon bond by converting it from a single bond to a double bond.
Metals: Different metallic crystal structures will have different densities, and thus likely also different average bond lengths (the change in density is not due purely to a change in bond lengths, but also to changes in atomic arrangements, including the number of nearest neighbors, next-nearest-neighbors, etc.). Thus you can alter bond length by inducing the substance to change from one crystal structure to another.  As an example, I'll quote from https://www.conservation-wiki.com/wiki/Tin_pest :

Solid tin exists in two allotropic forms, white and grey, each of which contains different chemical properties. White tin has a body centered tetragonal crystalline structure and a density of 7.28, while grey tin has a cubic structure and a density of 5.84 (Faulkner 1965). This transformation is generally accepted to occur around 13C (Gilberg 1991:3). Below this temperature, grey tin is more stable, while white tin is more stable at temperatures above 13C.

Covalent network solids:  Essentially the same effect as what was described for metals, above, except that the transformation involves making and breaking covalent bonds.  A good example is the transformation between diamond and graphite; though note that, in this case, the structure changes from 3D (diamond) to 2D (graphite)
Here's a graphic by Wikipedia contributor Nath321 illustrating this: https://en.wikipedia.org/wiki/Atomic_spacing -- though here the comparison is between a 2D structure:


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Detecting impurities in polyethylene glycol 3350
I would like to test samples of polyethylene glycol 3350 ($\ce{H−(O−CH2−CH2)_n−OH}$ with an average molecular weight of 3350, so n is around 75) to determine if there are any impurities and if so the rough concentration of them. Some likely impurities that may be present are:

ethylene oxide
diethylene glycol
ethylene glycol

My goal is to test samples from different sources to find out which one has the lowest level of impurities.
If there happens to be an easy way for me to test this at home using supplies that I would be able to buy online for < $1000 that would be great. More likely, I am interested in finding out what equipment would be used to test this so that I can try to find a lab that would do this testing for me.
Level of detection:
In a sample of 10 grams, I would want to detect levels at least as low as the following (though detecting even lower levels would be preferable):

ethylene oxide: 0.07 mg
diethylene glycol: 1 mg
ethylene glycol: 4 mg

If there is a way of detecting arbitrary impurities rather than just these specific ones I'd be interested in knowing that too (mass spectrometer? $\mathrm{^{13}C \; NMR}$? these are things I've heard of but am not sure what would be applicable here).

Extraction and GC-MS or maybe LC-MS, if you really need to measure these specific impurities
I don't know a lot about this area, but you are looking for very low amounts of impurities.  For ethylene oxide, for example, your desired detection limit of 0.07 mg in a 10 g sample corresponds to 7 parts per million, or 7 ppm.  Your limits for the glycols correspond to 100 and 400 ppm.
I'm not an NMR expert, but I think NMR will not be remotely  sensitive enough for this purpose.   Even if pure, NMR on a 0.07 mg sample is challenging and far from routine.
The best route forward might be a selective extraction of these impurities.  You should chose a solvent that does not dissolve polymeric PEG at all, or as little as possible.  This can be a challenge, as PEG is soluble in water, alcohols, and even dichloromethane.  Ethylene oxide is a relatively nonpolar gas, so you may be able to dissolve it in hexanes or a similar nonpolar solvent.  Unfortunately the glycols are unlikely to be soluble in hexanes.
So for ethylene oxide, say you extract 10 g of sample with X mL of hexanes, and say X is ~100 mL.  The resulting solution would be 700 ug/L of ethylene oxide at your desired detection limit.  That should be detectable by GC-MS.
For the glycols, supposing it is somehow possible to find a solvent that selectively dissolves the impurities but not the polymer, you could then derivatize the solution of glycols with a reagent like TBDMS and analyze them by GC-MS.  Alternatively, you may be able to analyze them without derivatization by LC-MS, but sensitivity of electrospray-based LC-MS for aliphatic alcohols is limited.  GC-MS is probably the way to go...if you can find a solvent for the selective extraction.
If no selective solvent is available, you might be able to analyze the glycols by directly derivatizing the solid PEG sample with a reagent like TBDMS in a liquid solution of some (non-selective) solvent.  The derivatization will convert the glycols into trialkylsilyl ether derivatives.  These may be lower polarity than PEG or more volatile than PEG such that they could be selectively removed and analyzed after derivatization.
Do you really need these specific impurities?
Maybe thermogravimetric analysis (TGA) would work for you.  There are variants of TGA where a mass spectrometer samples the gas phase above a material while it is slowly heated and decomposed.  If you have a sample of PEG you know is "pure enough", you could use that as a reference material, and then spike it with known amounts of contaminants to verify that these contaminants result in a perturbation to the TGA signal, or to the TGA-MS signal.
You may or may not get a precise numerical estimate of the mass fraction of your target impurities, but you will probably get a sense of whether a given 10 g sample of PEG is pure "enough" as measured by similarity to known reference material.
Analytical chemistry is hard
No matter what, you'll need to do a lot of methods development work to verify the appropriateness and robustness of your method.  E.g. what's the best value of X to use for selective extraction?  Can you for sure detect the impurities at your target levels with your chosen procedure?  You'll probably need to spike reference material with known amounts of impurities to make sure.  And is the response of your analysis method or instrument linear with respect to impurity mass fraction?  If so, over what range?
These are things that analytical chemists spend a great deal of time fretting about.  It's not easy.
Good luck!


text:
Entropy change as temperature changes under isobaric conditions
everybody:
I have this question that is definitely a challenge for me and I was looking for some help. I did manage to do part a but I'm not sure if it's correct.
Part b is even harder for me. I'm not sure how to answer that question. From what I know, higher temperature should mean higher entropy but I actually did put the numbers into the equation that I used for a and the answer for it is lower. I'm not sure what's going on.
I hope my work is clear to follow. If not, let me know politely and I will clear things up.

You have to use absolute temperatures, not centigrade.

Ok I can fix that but what about part b? You have any comments about that?

Hint: What is the relation of dS and dQ ?

@Poutnik dS = dQ/T

Do not stay at that, apply it in your mind. You know from school that for 2  same numbers, the one divided by the bigger number gives a smaller rusult.

I'm thinking lower if I simply apply 75 in the denominator. It makes the ratio smaller as opposed to 50.

Shouldn't it be the same since entropy is a state function?

What is the average absolute temperature over each of the changes?

@ChetMiller 285.65

Temperature difference is not a state.

I was talking about entropy when I brought up state functions.

I asked for the average temperature over each of the changes.

@ChetMiller ((25+50)/2)= 37.5 and ((50+75)/2)=62.5

@ChetMiller So just plug those values in for T and compare?

So if you approximate the entropy change as the amount of heat divided by the average absolute temperature, and the amounts of heat are the same, which entropy change is larger?

The 2 to 50 range would be larger. Ok, that makes sense. I'm not sure how he expected us to not calculate and answer that.

Note that T in thermodynamics means absolute temperature, not temperature of Celsius nor Fahrenheit scale.

Again,  school math, the same number divided by bigger number .... You need not to calculate anything.  Is dQ/T1 > dQ/T2 if T1 < T2 ?

You just did do it without calculating it.  Now confirm it with the accurate calculation.

For part (b), you can approach the problem generally, noting that since for a reversible isobaric process $$dS = \frac{dq_p}{T}= \frac{C_pdT}{T}$$ an equal change in temperature dT will result in a smaller change in entropy dS at higher T assuming constant $C_p$.
You might also proceed as follows. The difference in entropy changes for the first and second steps is
$$\Delta S(T_1 \rightarrow T_2) -\Delta S(T_2 \rightarrow T_3)=C_p\log\left(\frac{T_2}{T_1}\right)-C_p\log\left(\frac{T_3}{T_2}\right)\\=C_p\log\left(\frac{T_2^2}{T_1T_3}\right)\\=C_p\log\left(\frac{T_2^2}{(T_2-\Delta T)(T_2+\Delta T)}\right)\\=C_p\log\left(\frac{T_2^2}{T_2^2-\Delta T^2}\right)\\=-C_p\log\left(1-(\Delta T/T_2)^2\right)$$ Then since $(\Delta T/T_2)^2>0$ it follows that $\Delta S(T_1 \rightarrow T_2) -\Delta S(T_2 \rightarrow T_3)>0$,
which means the entropy change for the first step is greater than for the second.


text:
Rusting in every other bar
I have noticed that, when a structure with bars (such as a gate or a grate) develops rust, this sometimes happens differently across the bars. In particular, it seems that the bars are alternatively clean and covered in rust.
The photos below (taken by me) illustrate this effect (I have highlighted the rusted bars with red arrows). In both cases, from inspection, all bars appeared to be made from the same material and covered in the same kind of protective paint (the rusted parts were exposed to the elements).
My theory is that perhaps the rusting process activates some sort of "circuit" between bars, such that one bar protects the next one from rusting.
Is this possible?

Usually it happens like this: a customer wants to save money and orders metal fence as cheap and as light as possible. After a while, they realize the fence is not dense enough to serve the purpose (say, their cat is pregnant again), and extra bars get welded in the gaps (the welds on the first photo are different for every second bar). Apparently, older bars are likely already corroded, but there is also a chance that the new ones hasn't been painted or are made of an alloy of a worse resistance — then the new bars would rust faster.

Long story short, I doubt there is some complex electrochemistry behind this phenomenon other than poor planning and design decisions.

@andselisk your theory is an effective application of Occam's razor :-) However, I am not entirely convinced. In both cases, the part did not seem to have been reworked after installation. In particular, notice that in the second photo, the "cleaner" bars seem to be part of the original design and difficult to install at a later time (although of course it is possible that they have been replaced later...)

In the second case they might as well use bars made of different alloys with different corrosive resistance. They definitely used different forging technics and metal profiles to make the straight bars and trapezoidal sections.

It is unlikely that each bar protects the next from oxidation. If each bar is the same material and was put outside (I.e. exposed to moisture and other environmental factors) then they should generally rust at the same time. Additionally, if all the material is the same, the resistance should be about equal in each bar and electron flow would not be favored in one bar over the other.

Did anyone else notice the noose in the background? Cant tell if this post is a joke now

It seems pretty clear that second sets of bars were added and had different surface conditions from the originals.

@andselisk you are right that "different forging technics" were used... it would be interesting to see how this affects the corrosion resistance

@dval98 Not a joke, but good catch :-) I think that the grate was used to protect access to a well of some sort, hence the rope (I suppose there was a bucket below)

@dval98 my theory was that, once corrosion starts in a bar, this could change its resistance and make it more prone to further corrosion, in a self-reinforcing loop. I now believe this was a naive idea and your observations are more realistic

The hypothesis that corrosion self-accelerates is almost certainly correct, regardless of the formation of a circuit between different bars. However, this effect alone seems insufficient to explain the observed pattern. Disregarding some potential selection bias (would you have noticed and photographed a fence with randomly-corroded bars? I don't think so!), having a perfectly alternating sequence of over a dozen corroded/non-corroded rods by pure chance would appear to be something of a statistical anomaly.

Rebar for concrete is frequently stored in heaps, and rusts. It actually develops a stronger bond if it is not too rusty, because the rebar surface is rougher and better able to adhere to the hardened cement. It's interesting to see a pile of old rebar with a fresh delivery on top - silvery, not a trace of rust - on top of a pile of rebar that looks almost disgustingly rusted. And then you think, yeah, but the stuff on the bottom will have a better bond.
The key word in the observation about the fences is "sometimes".
I'm going to guess that the maker of the fence in the top picture also had a pile (or two) of iron bars, and some were rusty from being stored long or outdoors, or were a teensy bit different from some other pile of iron bars. This fence maker, knowing how his fence weathers, has to make it as uniform as possible, so he takes one bar from one pile, then a bar from the other pile... He prepares them identically, primes them identically, paints them identically... But then nature takes over and the most corrodable bar develops a spot, which spreads until it is all corroded-looking, before the more corrosion-resistant bar has even gotten started!
And then you come along and take a picture just before the other bars begin to corrode! This experiment is in the initial stages, not yet at equilibrium. Check back on the fence every year or two - I believe things will even out in time.

James Gaidis: Is your argument that the starting bars where in a different state of oxidation, and in time electrochemically corroded as a continuation of the original differential in corrosion chemistry?  As such, this could be considered, in part, an electrochemistry based argument with initial bar selection arising from an arbitrary design decision.

James Gaidis (+1): On further thought, I like your explanation.

@James Gaidis: your explanation seems very reasonable. My only objection is on "Check back on the fence every year or two - I believe things will even out in time": while the first fence was something I noticed passingly during a vacation, the second one is close to where I live and work and has been like that for a good 10 years. But maybe the period required to even things out is longer...

So the easiest non-chemical answer is to search for fence corrosion in google
As you will see a quick glance is showing that you probably have taken a non significant sample in your investigation since there is plenty of fences where you have a uniform corrosion.
Indeed there can be also more chemical explanation related to corrosion. But before doing that we need to define corrosion in this case. We are referring to galvanic process where your metal deteriorate through oxidation and usually form it's oxide. So we need a metal and oxygen and your case also humidity is a factor. This is explained visually here here
towhere you have the formation of an oxide . The anodic site on the surface of the metal is often related to an impurity or a defect in the metal (can be a lattice defect).
In the second photo you posted, if you look at the exposed surface of the bars there is indeed a difference in the shape of the fences that probably also causes different deposits of humidity and so in principle it is possible that the anodic site has a preferential "macro" site with the consequent texturing you see.
The first case you presented looks more tricky. Indeed it can be that the bars were welded in different time or painted with different paint (and also some of them pre-treated in comparison with the others) or just restored before the others. Indeed if corrosion "starts" in one site due to its weakness what you have is that the corrosion will insist in that point before spreading to other parts and that's the idea. This principle is used also in cathodic protection were you intentionally couple a sacrificial anode with a nobler metal knowing that the corrosion will proceed till it has consumed the anode.
(So indeed there is a circuit since all corrosion process involves one, but that would have happened also in the case of a more uniform corrosion)

I would like to propose an alternative to those dismissing the advent of possible complex electrochemistry behind the phenomenon and asserting instead perhaps just poor planning and/or design decisions.
In this regard, an interesting 2011 article, 'An Fe3O4–FeO–Fe@C composite and its application as anode for lithium-ion battery', discussed here, to quote:

X-ray diffraction, Raman spectrometry, scanning electron microscopy with energy dispersive spectroscopy, transmission electron microscopy and galvanostatic cell cycling have been used to characterize the structure and electrochemical performance of the as-prepared Fe3O4-FeO-Fe@C composite. The electrode shows a stable and reversible capacity of over 600mAhg−1 at a current of 53mAg−1 for up to 60 cycles, which could be ascribed to the unique double-layer core-shell and embedded structures.

Also, per another reference discussing the same electrode, to quote:

Here, we report the synthesis, characterization and evaluation of the electrochemical properties of pristine Fe3O4 nanoparticles synthesized from a base catalyzed method mechanically mixed with electrospun Polyacrylonitrile (PAN) derived carbon fibers as simple composite material. The Fe3O4/C electrodes are characterized by galvanostatic cycling experiments using currents as high as 2000 mA g−1, revealing high capacity values and capacity retention, together with a very good capacity recovery during the rate capability experiment.

The picture appears to display a red pigment, possibly Fe3O4 created at elevated temperatures. Other bars, resemble Fe/FeO.
So speculatively in the current case, an alternate proposition that if the iron metal fence was exposed to say a brush fire, heating the iron and also depositing fine carbon, this may have resulted in a series of Fe-FeO@C and Fe3O4@C electrodes naturally developing over time (aka, yes, some possible complex electrochemistry).

I don't see how a composite electrode for Li-ion batteries synthesized from iron(II) sulfate and glucose has anything to do with a rusting fence. Also, why would anyone apply brush fire to every second bar? Besides, neither of the sources support this theory in any way.

Andselisk: The point is Fe3O4 makes a good anode. However, as a provided source notes, its creation requires some heating, hence fire exposure to the whole fence. There is also an apparent interesting electrode composite arising from added carbon. Note, activated carbon, for example, naturally develops zones of + and - charge buildup. The fence is subject to wind, for electrostatic charge creation,...., bottom line, a possible outline for 'some complex electrochemistry' at work. There is, per the counter-argument, equivalently no support for the renovation of the fence explanation.

How is any of this related to the question which is about why every second bar is corroded differently? With a stretch, the carbon build-up could explain some zone effects, but the rest makes in my opinion no sense in a context of the question.

Andselisk: Agree, a weakness of the argument. My allusion to charge zones, implying self-selected anode/cathodes among the bars is the only idea.  However, the explanation of some design selection per the answer by James Gaidis, could possibly be a stronger reply, which stills does not refute completely an electrochemistry footprint.


text:
Devarda's alloy procedure
Hello dear I have a small question
I want to do kjeldahl distillation for a 100 mL sample of water after reduction with devarda's alloy to measure the concentration of nitrate nitrogen... I want to know what is the quantity of alloy must be added to 100 mL water?

Take 10 mL of your solution. Add an equal volume of NaOH 1 M. Heat to 100°C and test if the gas has the odor of ammonia. Or check if the gas has a basic reaction by introducing a small and wet indicator paper in the gas getting out of the hot solution (without teaching the glass container). If yes, it means that there is some ammonium salt in the solution. Heat until this gas production has stopped. When this evolution has finished, add 0.1 g Devarda's alloy, and repeat the same test : odor, or reaction of the gas with a small and wet indicator paper. If the gas is basic, it means that the hot solution contained nitrate or nitrite ions.

The OP is interested in quantitative approach.

Thanks dear but I am not doing a qualitative test for nitrate I am doing Kjeldahl distillation with devarda's alloy that's what I am asking for the quantity of alloy needed for 100 mL of water and the exact procedure.


text:
Bonding in d10 metals such as Zn
I recently learned about the MO bonding model in metals, where many metal atoms form lots of orbitals close in energy that form bands. However, I can't understand the bonding in d10 metals. It seems to me that if they form MOs, then the bonding and antibonding orbitals will both be filled, resulting in net zero bonding. The d10 metals obviously bond anyway, so what is wrong with my reasoning?

https://chemistry.stackexchange.com/questions/55305/what-type-of-bonding-is-there-among-d-block-metals https://chemistry.stackexchange.com/questions/47553/how-many-d-eletrons-are-in-the-outer-shell-of-following-compounds

https://chemistry.stackexchange.com/questions/91487/can-we-classify-all-the-d-block-elements-as-transition-metals https://chemistry.stackexchange.com/questions/124026/how-is-zn-not-a-transition-metal

You miss the fact that the $p$ orbitals lying immediately above the occupied orbitals are also valence orbitals.  These unoccupied orbitals then allow the formation of unoccupied antibonding molecular orbitals whereas the corresponding bonding MOs are occupied.  Such a $p$ orbital contribution is then responsible for both the partially covalent bonding of these elements in compounds and the metallic bonding of the pure elements in their condensed phases.
In the case of metallic bonding in the elements, we see this effect in the alkaline earth metals as well as the Group 12 metals.  In the case of alkaline earth metals, if these were bonded only through the $s$ orbitals they would not be bonded at all; but the real metallic bond structure mixes in the otherwise unoccupied $p$ orbitals (or $d$ orbitals in the heavier elements). This gives more net bonding (as measured by hardness and by melting and boiling points) than that in the alkali metals that precede them.

Is it really still that simple near boiling temperature? I was always assuming these models  work near absolute zero...


text:
Aromaticity of different compounds
I'm new here and I'm still learning, I understand that there four rules for aromaticity:

cyclic
conjugated double bonding (every atom in the ring should be sp2)
huckle number should be 2, 6, 10, 14, 18, ..18
planar ring

The 2nd question should be 4 electrons and is antiaromatic, but how about the first question?

https://chemistry.stackexchange.com/questions/76142/which-rings-of-this-heterocyclic-compound-are-aromatic

Cytosine is aromatic, there are 6 electrons in the ring. Two pairs of pi bonds(thus 4 electrons) and a lone pair on Nitrogen(the one which has H written below it). The other Nitrogen (which is forming a pi bond) in the ring, does have a lone pair, however it is "localised". That lone pair cannot participate in resonance, and hence it is not counted when considering aromaticity.

I put 6, but I got wrong answer, that is why I'm asking here, is 6 or 10? @Kaiser

My answer would be to refuse to answer this case.
The 4n+2 rule applies only if the electrons are conjugated in a cycle -- nothing more, nothing less.  If you have pendant pi bonds in the conjugated system, like the carbonyl group in cytosine and many similarly structured compounds, you don't have the right conditions to apply the 4n+2 rule at all.
Interior conjugated atoms, as in pyrene, also mean you don't have specifically conjugated cycle.

Cytosine is proven to be aromatic, how do they calculate the electrons inside the ring, it seem they add 2 if there is HN: present as hetero, but not the case in NH2, I want to know exactly how many pi bonds and electrons inside the cytosine ring. @OscarLanzi

The issue, of course is not whether it's aromatic.  No doubt it is.  Rather the issue is whether we can really apply the 4n+2 rule that they rendering in many a textbook.  No doubt we can't;  IUPAC recommends the rule only for single, fully symmetric rings with no additional conjugated atoms.

hmm..that's interesting...thank you for pointing this out!, would the answer be 10 electrons in resonance? @OscarLanzi

Yes there would be ten electrons in resonance.  But without them being in a ring, the figure being 4n+2 is not really a predictor of aromaticity.  Cyclopropenone has a pendant carbonyl group and shows aromatic character too -- with four conjugated electrons.

Thank you so much for these tips, I know now that Huckel rule has its limitation like Lewis acid base concept, or octet rule in carbon, so when we have pendant carbonyl group or methylene such in fulvene, do we consider the conjucated electron to be zero because there is a missing conjugation fulvene https://en.wikipedia.org/wiki/Fulvene @OscarLanzi


text:
If x and y are 2 extensive properties of a system then is x/y always intensive?
I have a case to contradict my own question but in few books, it is given that x/y is always intensive. So
I got confused. My case is as follows :
Let a wire be our system whose resistance is R,
area of cross section is A,
length is L
and constant of resistance be ρ.
V,R are extensive
V(volume) = AL
R  = ρ(L/A)
So R/V = [ρ(L/A)]/AL = ρ/A^2
ρ is a constant but, A depends on the shape and size of system. So ρ/A^2 depends on shape and size of system. So it is an extensive property.
Therefore x/y can be extensive as well.
But in the below picture which is of G.R.B Op Tandon physical chemistry, it is given x/y is always intensive. Where did I make a mistake ?
Please note that I did not understand this is related more to physics or chemistry, so I asked it here as
chemistry is where I have first come across this problem. If I should ask this in the physics website, please tell me and I will do so.

Perhaps the book is right. It depends on what is written before the assertions that you have posted...

Extensive properties are those that are directly proportional to the mass of the system, under the restriction that all intensive properties are held constant.
The restriction that "all intensive properties are held constant" means we are not changing the inherent nature of the system; rather, we are only changing its size.
For instance, if we keep all the intensive properties constant, and double the mass of the system, all the extensive properties will double.
Thus extensive properties all have the form $X=nZ$, where n is the number of particles in the system, and Z is some intensive property.  [We can use n in place of mass because the restriction that all intensive properties are held constant means that the composition doesn't change.]
Consequently, if we have two extensive properties, X1 and X2, their ratio is:
$$\frac{X1}{X2}=\frac{nZ1}{nZ2}=\frac{Z1}{Z2}$$
Since the n's cancel out, the ratio of two extensive properties becomes the ratio of two intensive properties.  And since intensive properties are independent of n, their ratio is also independent of n.
Based on the above definition of extensivity, volume is only extensive in 3-dimensional systems, because it's only in 3-D systems that $V\propto n$.  Area is only extensive in 2-D systems, because it's only in 2-D systems that $A\propto n$.
I have never before considered resistance in the context of extensivity, but I believe the reason you are running into trouble is that resistance depends both on the size of the system and its geometry.  Given that, I would conclude that resistance is neither intensive nor extensive.
You could make resistance extensive by constraining the geometry, specifically by constraining A to be constant, in which case resistance would be a 1-D extensive property that depends upon L: $R \propto L \propto n$.
However, I don't believe you could turn resistance into an extensive property by constraining L to be constant, since then you would have: $R \propto A^{-1} \propto n^{-2}$.

Thanks for the reply, you helped me a lot.

@RyugaGod  Happy to help.  And thanks for the question -- it was fun to think about!

The conductance (inverse resistance) describes a non-equilibrium property, the current transported through a medium divided by the applied voltage:
$$\frac{1}{V}\frac{dQ}{dt} = R^{-1} = \Lambda$$
where $\Lambda$ is the conductance.
You might say that an applied voltage is an intensive property independent of geometry, and could write instead
$$V = R\frac{dQ}{dt} = \frac{dQ/dt}{\Lambda}$$
Both conductance (or resistance) and current depend on the geometry (dimensions) of the sample, but clearly something cancels when dividing these extensive properties because we obtain an intensive property (voltage).
On the other hand, conductivity is often assumed to be an intrinsic property of a material independent of geometry, defined as
$$\rho = \frac{E}{J}$$
where $E$ and $J$ are the magnitude of the local electric field and the current density in the material (the vector properties $\vec E$ and $\vec J$ are here assumed collinear, more generally $\rho$ is a tensor).


text:
Pink gatorade turned bluish purple by thick clear liquid
If I had two teaspoons of pink strawberry gatorade and one teaspoon of a clear thick liquid and when mixed together it turned bluish purple in a chemical reaction, what might the thick clear liquid be?

"Quick questions" without elaboration effort are not very welcome, and may be closed. Always think about and search for possible answers thoroughly before posting a question.

Someone had me drink a toxic chemical under false pretenses. I didn't consider proper website etiquette. I do apologize.

Generally, many dyes, natural or artificial, change color with changing pH ( acidity ).  Natural anthocyan dyes are usually about red in acidic, about blue/violet in alkaline solutions. You can e.g. check red beetroot juice when washing soda is added. So that clear solution was probably an alkaline solution. These reactions are reversible, if pH is returned to a out initial value, the initial color will return.

That is most likely the case and plus 1. I want to add that bleach can also cause colour changes or various colours to appear based not only on the pH but on oxidative reactions. In most cases the latter processes results in discolouration but sometimes not.

@Alchimista But as bleach is alkaline, if discoloration did not occur, the colour change would happen even for the pH change itself.

Well, sure. Just pointing to a different mechanism not (entirely at least) based on pH.

It does not have to be a full-fledged alkali to generate this color change effect.  Alkaline earths work quite well (and less hazardously) even with only limited or sparing solubility.  For a neat-looking demonstration with magnesia (or more accurately, the hydroxide it produces in water), see the picture and do the experiment here.

Alkaline vs acidic is the only info I can gather from the chemical reaction I saw?

Like that, based on the only info we were provided with. For more, we would need the fortune-teller crystal ball to tell.


text:
Why is Gibbs free energy different for different processes?
Under constant pressure and temperature ∆G=∆H-T∆S.
Let us take a system with initial state (1) and a final state (2).Now let us change the state of the system from 1 to 2:

Case 1) Reversibly by adding heat Qrev.So now the entropy change
∆S=Qrev/T.Also the enthalpy change ∆H=Qrev.So the Gibbs free
energy change is 0(plug in the variables).

Case 2)  Irreversibly by adding heat Qirr.But since the state change in
both the processes is same, entropy change ∆S is still Qrev/T
(Since entropy change is a state function).
Similarly, enthalpy change is also a state function so ∆H=Qrev
again so the Gibbs free energy is again 0.

.Why is this happening? I feel that the Gibbs free energy in the
second case should be Qirr-Qrev.But I am not able to get that mathematically.
Where am I wrong?

The difference would apply to the surrounding, not to the system. Surrounding entropy would be higher for the irr. case.

Like entropy, Gibbs free energy of the system is a function of state, independent of the process.

Hey Chet miller I saw your other answers on Gibbs energy and they were really good .Could you explain to me a bit more about Gibbs energy and give me some resources?

It isn't clear to me what you are looking for.  In terms of processes in closed and open systems, see Denbigh.  The real value of Gibbs energy, in my judgment, is not in terms of processes, but in using it to quantify interphase multicomponent phase equilibrium and chemical equilibrium involving chemical reactions.  To understand this, you need to understand the thermodynamics is mixtures.  This is covered very nicely in Fundamentals of Chemical Engineering Thermodynamics by Smith and Van Ness, starting in Chapter 10.

Gibbs free energy is zero at EQUILIBRIUM

@chemeng No, it is not zero at equilibrium, it has just minimum there.

@poutnik ok if we’re going to be this technical. I’ll explain it in clear English $$\Delta G=0$$ at equilibrium, G itself is at a minimum . Good thing I was here to say it, you’re all welcome

@Chemeng It was rather essential than  this technical. $\mathrm{d}G=0$ and $G=0$ is a big difference.

I believe they’re both called Gibbs though(seems like a flaw in thermo and there are allot of them) I mean look at inexact differentials and this is what they accept as the basis of their field

@Chemeng  The flaw is probably somewhere else. V and dV are both volumes as well, but nobody says it is the same. 1 kg of graphite is in equilibrium, so dG=0, yet it's G is enormous.  It is quite difference between $G=H-T \cdot S$ and $\mathrm{d}G=-S \cdot \mathrm{d}T + V \cdot \mathrm{d}p + \sum \mu_i \cdot \mathrm{d}n_i$. They manage inexact differentials right, there is just a practical issue with inexact differential symbol, you need to copy or insert the extra nonhandy unicode symbol.


text:
KMno4 is the strongest oxidizing agent. So why does it tend do have +2 oxidation state with reaction with acids?
Manganese has an oxidation state of +7 in potassium permanganate. When reacted with sulphuric acid, how does it tend to have +2 oxidation state by forming $\ce{MnSO4}$ along other products ?
In acid, there is already tendency to donate proton and not electrons. Yet it forms +2. While in basic medium Manganese forms +7 to +6.
As I think, it shouldve formed +2 in basic by gaining 5 electrons. Where have I gone in my opinion?

KMnO4 is not the strongest oxidazing agent. Reaction with H2SO4 does not produce MnSO4, but oily and highly unstable Mn2O7.

i doubt your answer. im sure it does produce mnso4

I admire your  confidence, but I have doubts if it is backup up by knowledge. Yes, if Mn2O7    is reduced, or decomposes, it may finally end as MnSO4.

OTOH, Mn(II) in alkaline environment forms Mn(OH)2. It is reducing agent, quickly reacting with dissolved oxygen, what is used in analytic chemistry.

KMnO4 is reduced in strongly acidic environment by reducing agents ( like oxalic acid) as this:
$$\ce{2 MnO4- + 5 (COO)2^2- + 16 H3O+ ->[catalyzed Mn^2+] 2 Mn^2+ + 10 CO2 + 24 H2O}$$
It means, acids themselves do not convert it to Mn(II), unless they have reducing properties as well, like hydrochloric acid.
In alkaline solutions, manganese tends to reach Mn(III,IV) oxidation state, if oxidized from Mn(II), or Mn(VI,IV) states, if reduced from Mn(VII).


text:
What are the properties of water in glaciers?
Does water from glaciers have many minerals (hard water) or would it be considered pure or distilled water?

"Quick questions" without elaboration effort are not very welcome, and may be closed. Always think about and search for possible answers thoroughly before posting a question.
Chemistry SE site expects to explicitly elaborate the question by at least basic textbook/online search, writing down what was found/understood/tried/failed.
Written effort attracts elaborated answers.

Pftt, you could just as well ask the same about water in a river.

Water of glacier origin is at the beginning very close to distilled water.
As it comes into contact to sedimented dust and rocks/minerals in a glacier, or flowing from it, it gets contaminated by solid particles ( glacier rivers are  very turbid ) and eventually by some dissolved minerals.
But even then, mineral content is still very low, compared to other water sources.
Consider Earth water cycle as a giant distillation apparatus, contaminating water along all its path from the source to the glacier deposite. These contaminants are solid, liquid or gaseous of both natural and human origin.
For many technical cases, it case be used after filtration insteadof distilled water, similarly as rain water. But for serious lab usage, it is too contaminated with undefined trace contaminants.

Thank you for your explanation! Could you recommend some literature for me to learn more about the similarities between water of glacier origin in comparison to distilled water, thank you!

My understanding is that the water in glaciers constitutes over 2/3 of the designated 'freshwater' on earth and is an integral part of the so-called water-cycle.
Per U.S. Geological Survey, to quote:

Just because water in an ice cap or glacier is not moving does not mean that it does not have a direct effect on other aspects of the water cycle and the weather...Even though the amount of water locked up in glaciers and ice caps is a small percentage of all water on (and in) the Earth, it represents a large percentage of the world's total freshwater. As these charts and the data table show, the amount of water locked up in ice and snow is only about 1.7 percent of all water on Earth, but the majority of total freshwater on Earth, about 68.7 percent, is held in ice caps and glaciers.

And further on the age of the freshwater itself:

Short-term storage might be days or weeks for water in a lake, but it could be thousands of years for deep groundwater storage or even longer for water at the bottom of an ice cap, such as in Greenland.

As such, the captured freshwater from the atmosphere found in glaciers is apparently representative of especially CO2 and existing pollutant concentrations.
So on the question: "would it be considered pure or distilled water?", that depends on the age of the water and in what region the glacier resides as select places have experienced significant air quality problems.
For example, see this 2019 work: Air Pollution in the Hindu Kush Himalaya, indicating troubling visible signs.


text:
Use an Ozone generator to get rid of fungus in the flat?
I was told that I can use an Ozone generator to get rid of fungus (resulting from not enough fresh air I suppose). But I didn't find any reputable source for that claim. Would it work, Chemists of Stackexchange? What should we pay extra attention to?

Wow, now that's a particularly bad idea...

Here is a good commentary, in my assessment from the National Ozone Association, on which I will review the chemistry latter, to quote:

A debate has swirled about for years about the use of ozone generators for mold infestations.  Some mold experts call ozone "useless" for mold treatment, while others strongly insist that ozone will kill mold...The truth will not please either side, but it is clear that high PPM ozone does kill mold and harms mold spores so they cannot reproduce.  However, ozone treatments alone are not enough to treat a mold problem.  Not only is mold a health concern, but even dead mold and mold spores can be a health threat...The reason that ozone generators are not a full mold solution is that mold is a kind of SYNDROME.  Mold does not happen unless there are several factors working in sympathy to provide the conditions mold growth requires.  So, an ozone treatment may be like a running a water pump in a basement with a broken water pipe...

Further, an important point, in my opinion, to continue quoting:

So, killing mold with an ozone generator has merit, and it is a proven mold-killing process.  I want to add, that we are not talking about many of the ozone generators on the market today.  90% of of these are simply too small to drive the PPM levels needed for an effective ozone treatment...Obviously, ozone levels strong enough to kill mold means you must do this work in an uninhabited/vacated building...We have seen some documentation by groups claiming that ozone treatments that are below EPA or OSHA permitted levels (for occupied areas) are not effective.  We agree.  You can't have both.

So, high levels of also toxic to humans O3 is required to kill mold, but the gas is not effective on mold hidden behind walls and such, and does not address underlying causative conditions leading to mold formation. Some level of physical removal and exorcism is still likely required as dead mold and mold spores remain potential health issues.
A review of the chemistry of ozone reveals it to be complex, as is noted, for example, in this work: 'OZONE {O3} IN DRINKING WATER TREATMENT a brief overview...', which details possible reaction pathways with ozone, to quote:

Ozone can attack as a dipole, molecule, electrophilicly, or as a  nucleophilic agent. Ozone reacts in water, or any aqueous solution, in two ways: DIRECTLY as molecular ozone via three mechanisms; (slow and very selective), forming aldehydes, ketones and carboxylic acids, cyclo  addition (+  & -); on unsaturated bonds, as a dipole. ozonide  >  carbonyl  >  hydroxy-hydro  peroxide  > carbonyl  &  hydrogen  peroxide electrophilic (+);  on molecular sites with strong electronic density. aromatics  (phenol  &  aniline) nucleophilic (-); on molecular sites with an electronic deficit, usually on carbons carrying electron-withdrawing groups. INDIRECTLY via radicals formed as it decomposes in water; A few such radicals are as follows; hydroxyl radical, .OH,  a  main reactive ingredient hydroperoxide radical,  .HO2 superoxide radical ion, .O2-, ozonide radical ion, .O3- The indirect reactions produced by molecular ozone are limited only by the various radicals it produces, which varies with the initial water quality. The worse the water quality problems are, the more ozone can potentially rise to the occasion.

Also:

Even with the ozone used up quicky, the  radicals it forms will continue reacting. One indirect example of ozone molecules contacting  water $\ce{ (O3 + H2O  = O2 + .OH + .OH)}$...

Technically, the last reaction is not accurate as are some comments in the article (like the hydroxyl ion is a water cleaning agent, it is primarily the hydroxyl radical, but OH- does serve an initiating role, see equations below). My technical assertion on the corrected reaction(s) are as follows:
$\ce{O3 + hv -> O2 + O }$
$\ce{O + H2O  ->  .OH + .OH }$
Source: GoogleBook
So, the net reaction of ozone with H20 in the necessary presence of UV light is:
$\ce{O3 + H2O + hv -> O2 + .OH + .OH}$
Note, hydroxyl radicals can be created by the action of water on ozone, but the reaction is pH-dependent and is not given by the above equation. In fact, in water with a pH above 7, the reaction can proceed as follows:
$\ce{O3  +  OH-   ->  .O2-  + .HO2}$
$\ce{.HO2  =  H+  +  .O2-}$
$\ce{.O2-  +  O3  -> O2  +  .O3-}$
$\ce{.O3-  +  H+   =   .HO3}$
$\ce{.HO3  ->  .OH  +  O2}$
Source: See, for example, Ozonation of Tris-2-Chloroethyl Phosphate (TCEP) in Water by  Michael J. Votruba.
The above approximately (as there are more possible reactions) implies a net reaction in alkaline water of:
$\ce{2 O3  +  OH- ->  .O2-  +  .OH  +  2 O2}$
where pH cited as 8.5 or higher in applications.
but, in actuality, intermediate reactions and subsequent products are also possible with more ozone.
Note, in atmospheric chemistry, where we are working with aerosols, due to a change in the dielectric constant for the medium (which impacts pKa, see Wikipedia the hydroperoxyl radical), the superoxide is found mostly in the form of $\ce{.HO2}$. The implied approximate net reaction would then be:
$\ce{2 O3  +  H2O  ->  .HO2 +  .OH  +  2 O2}$
So evidently, the chemistry of ozone can address mold when in contact (especially in the presence of UV light), but many transient (albeit, very powerful) radicals are very short-lived, and this is, by no means, a permanent solution. To the latter, new mold resistance wallboard installed following treatment is likely advisable as is natural sunlight lighting fixtures and a de-humidifier as water presence is also a causative agent for mold growth.


text:
Still struggling with Huckel rule and aromaticity
I'm still learning huckel rule and aromaticity in different compound.
For the 1-methoxy-1,3,5,7-octatetraene, is it correct that it has 8 electrons inside the cycle, but since they will be 8, then it is antiaromatic?
And for the Imidazole, is it 6 electron inside because the N: doesn't contribute to the pi, or is it 8?

Why would someone downvote a question? I'm new here in chemistry.stackexchange and I'm still learning...can someone explain?

https://chemistry.stackexchange.com/questions/27099/when-do-lone-pairs-on-nitrogen-contribute-to-aromaticity https://chemistry.stackexchange.com/questions/33332/why-is-cyclooctatetraene-non-planar-but-the-cyclooctatetraenide-anion-planar

I’m voting to close this question because there are two parts to the question, and both are duplicates (see Mithoron's comment). If this were just one question, we could close it as a duplicate, but because there are two, we can't.

I'm voting otherwise.

Advice
Take a closer look at the lone pairs and double bonds that participate only and only in the delocalization of electrons.
1-methoxy-1,3,5,7-octatetraene
As you can see, 1-methoxy-1,3,5,7-octatetraene has four double bonds in an octagonal cycle. The lone pairs of the adjacent oxygen atoms, even though they directly influence the delocalization by means of the resonance effect, aren't actually part of the cycle and cannot be included amongst the pi electrons of the double bonds. Therefore, your molecule is antiaromatic. Its structure is not planar, in fact it looks something like this (the picture below includes only the cyclic backbone):

For aromatic molecules, planarity is absolutely necessary. But Huckel's rule which I'm sure you're familiar with, provides the reasoning behind the planarity of the molecule.
The imidazole
As for the imidazole, your assumption is absolutely correct. Here we get to know a scenario similar to pyridine:

The lone pair does not participate in the aromatic cycle. This can be easily checked by taking a look at the hybridization of the nitrogen atom. If the lone pair were to be part of the aromatic system, it should have been perpendicular to the other orbitals of the cycle. However, this is not the case. The 2p orbitals of the nitrogen are actually involved in the aromatic cycle, while the aforementioned paired electrons get to be embedded in an sp2 orbital, placed perpendicular to the plane of the cycle. This actually has significant consequences on the basicity of pyridine.
On the other hand, in order not to interrupt the aromatic cycle, the lone pairs of the other nitrogen atos must come into play when counting pi electrons. Thus, the molecule is aromatic.

Thank you @TheRelentlessNucleophile I just want to make sure for 1-methoxy-1,3,5,7-octatetraene it is nonaromatic, not antiaromatic (because antiaromatic should retain planar feature), and it is nonaromatic because it is not flat like the figure shows.

In case of 1-methoxy-1,3,5,7-octatetraene, the answer would be zero for the number of electron, right? because since it not planar, then it conjugation will be accounted from because resonance is not happening anymore, please correct me if I'm wrong? @TheRelentlessNucleophile

@user432797 Yes you're right.

I turned out that @TheRelentlessNucleophile is right in everything even with antiaromatic categorization

Apologies, I didn't see the comments you had posted after my answer. If a molecule has an uninterrupted network of pi electrons and 4n pi electrons, it is usually antiaromatic. Because they lack the stabilization that would otherwise be present in the aromatic molecules thanks to the pi orbital that delocalizes over the whole molecule (see Huckel's rule), the molecule stabilizes itself by adopting non-planar configurations.

Rule of thumb: uninterrupted network of pi electrons, planar molecule, 4n+2 electrons, extended pi orbital that delocalizes over the whole molecule: aromatic. Violate one rule and the molecule is rendered either antiaromatic (4n pi electrons) or non-aromatic. Now, I lack the appropriate knowledge to tell you exactly why do we bother with antiaromatic molecules in the first place, but let's just say that, as far as I know, if you were to draw the Frost circles in a Huckel diagram, their HOMOs would be at the ground energy level, rendering the orbitals non-bonding.

Also, keep in mind that several of these molecules prefer to exist as diradicals, instead of completely stable molecules

thank you @TheRelentlessNucleophile ! Can you refer me to the reference that you used? I still don't get why my question was downvoted!

Your question was downvoted becausw it looked like homework and also because of lack of details involving YOUR work in answering the question ans sharing YOUR knowledge. Please refer to the rules of ChemSE.


text:
Example of an achiral molecule without a plane of symmetry or inversion center?
Some websites, such as this textbook, mention how generally a plane of symmetry or inversion center in a molecule is enough to consider it achiral. Nonetheless, this source clarifies

but if these symmetry elements are absent the molecule should be checked carefully for an  S$_n$  axis before it is assumed to be chiral.

Furthermore, Wikipedia notes the definition of chirality as "a molecule which has no S$_n$ axis for any value of n is a chiral molecule."
Thus, are there any molecules lacking plane of symmetry/center of inversion but that are achiral due to presence of an axis of improper rotation? Are there are also examples of vice-versa, where a molecule has a plane of symmetry/center of inversion, but lacks an axis of improper rotation and is thus chiral?
edit: This post does cover some definitions which is helpful, but the point of this question was to see an actual example.

You have it all backwards. A plane of symmetry means no chirality, dead sure. Ditto for an inversion center.

Part 1

Are there any molecules lacking plane of symmetry/center of inversion but that are achiral due to presence of an axis of improper rotation?

Yes, although such examples are very rare. In general, the point groups $S_{2n}$ $(n \geq 2)$ are where you should look. These molecules have a $S_{2n}$ rotation axis, but no plane of symmetry, and no inversion centre.
Molecules in the $S_6$ and $S_8$ point group are are exceedingly rare, and generally need to be constructed specifically to satisfy these criteria. The best examples, therefore, come from the $S_4$ point group. You can find several examples on these websites:

http://gernot-katzers-spice-pages.com/character_tables/S4.html
http://csi.chemie.tu-darmstadt.de/ak/immel/tutorials/symmetry/index7.html#Sn

In particular, one of the nicest examples is 2,3,7,8-tetramethyl-spiro[4.4]nonane from the second link. In the graphic below, I've taken the liberty of using four different coloured atoms in place of the methyl groups. The reader should therefore bear in mind that the third image is the same thing as the first, even though the colours are different.

The XYZ coordinates that I used for the pictures above are as follows. You can download these and view them in software such as Avogadro or ChemCraft, although I don't think you can perform a "reflection" per se (I manually swapped the colours between the second and third pictures to achieve this effect).
C         -2.32545        1.12231        0.91714
C         -2.19292        2.47894        0.24035
C         -0.83350        2.34003       -0.44571
C         -1.75438        0.10430       -0.11277
H         -3.31782        0.88874        1.35769
H         -1.62382        1.17900        1.77968
C         -0.98281        0.99078       -1.13371
H         -1.65567        1.19844       -1.99603
H         -0.05915        0.56089       -1.57585
C         -2.87426       -0.73718       -0.79187
C         -0.83536       -0.95866        0.55714
C         -2.33811       -2.16138       -0.80738
H         -3.27791       -0.35456       -1.75318
H         -3.74616       -0.78244       -0.10108
C         -1.65652       -2.23999        0.55910
H         -0.36215       -0.67792        1.52187
H          0.00882       -1.17785       -0.13479
B         -1.40058       -2.29939       -1.82101
H         -3.13035       -2.92594       -0.95624
O         -2.60331       -2.18962        1.57227
H         -1.04117       -3.15429        0.69961
H         -2.24954        3.33056        0.95160
F         -3.19594        2.63169       -0.70627
N          0.17857        2.27493        0.50135
H         -0.60168        3.16745       -1.14998

Part 2

Are there are also examples of vice-versa, where a molecule has a plane of symmetry/center of inversion, but lacks an axis of improper rotation and is thus chiral?

An improper rotation $S_n$ is defined by a rotation about $360/n$ degrees, followed by reflection in a plane that is perpendicular to that rotation axis.
A plane of symmetry ($S_1$) and an inversion centre ($S_2$) are special cases of an improper rotation ($S_n$). It is easier to convince yourself of the $S_1$ case: according to the definition above, $S_1$ means rotation through $360^\circ$ followed by reflection in a plane. Since rotation through $360^\circ$ obviously does nothing, this is the same as a reflection in a plane.
So, the answer to this question is no. if a compound has either a plane of symmetry or an inversion centre, that automatically means it has an improper rotation axis.

Neat, thanks @Karsten! That is a lot better than the figure. I was hoping it would be on chemtube or something, but alas.

https://i.imgur.com/PZ7M8UB.png For this molecules, does this count as a plane of symmetry? I feel it can be symmetrical when viewed in the animation.

Hmm, in that frame, you are right that it certainly looks like it could be a mirror plane. I am not sure now whether it is just the angle which conceals some details, or whether that is the wrong conformation to begin with. The first figure I attached was only one of many conformations, albeit the most stable. If you look at that one, you can see that the CH2's next to the oxygens are not accurately preserved by the mirror plane you're proposing.

@ebehr: I changed the example to something that you may find easier to visualise (well, at least, I found it easier...!).

More on this example: https://chemistry.stackexchange.com/questions/95979/number-of-stereoisomers-of-tetramethylspirononane/96027#96027


text:
What is the mechanism of benzyl ether hydrogenolysis?
The mechanism for the hydrogenation of alkenes or alkynes using Pd/C is fairly well explained to involve the absorption of hydrogen onto a metal's surface followed by the addition of that hydrogen into the π* orbitals; however, I am having a hard time understanding or finding any literature on the mechanism for the simple hydrogenolysis of benzyl ethers.
I would not expect this to involve the same surface chemistry as the hydrogenation of alkenes/alkynes due to the geometry of the C–N and C–O σ* orbitals; however, the catalyst used, Pd/C, is heterogeneous, so I don't know what other mechanisms could be possible. I found a paper1 that reported the mechanism for hydrogenolysis involving a homogenous catalyst was an oxidative addition followed by reductive elimination; however, I am not sure if this is the same mechanism that applies to benzyl ether cleavage with Pd/C.

Martínez, A. G., Alvarez, R. M., Aguirre, J. A., & Subramanian, L. R. (1986). Mechanism of hydrogenolysis. Part 1. Catalytic hydrogenation of vinyl and aryltrifluoromethanesulphonates. J. Chem. Soc., Perkin Trans. 1, (0), 1595–1598. doi:10.1039/p19860001595

While the mechanism you show seems like obvious thing, it's very sketchy and perhaps even incorrect. Heterogeneous mechanisms are problematic...


text:
Molecular Orbital Theory and s-p mixing
I understand the bulk of MO theory , the formation of the bonding and anti bonding MOs etc. However, the topic about s-p mixing is quite puzzling to me. As a general rule of the thumb, I was taught that diatomic molecules up to N2 would exhibit s-p mixing, while the molecules past that threshold, O2 and above, would have insufficient amount of s-p mixing for the bonding MOs to be swapped. So my question is for heteronuclear molecules , eq CO , NO etc, how would I know whether s-p mixing occurs for that molecule? Ive found some sources online stating that molecules with <= 14 electrons would show s-p mixing, but this doesnt work for the NO molecule as it has 15 electrons and apparently exhibit s-p mixing


text:
Are these molecular structures of cyclodecapentaene identical?

I think that the above structures are identical compounds as I doesn't exist in the shown form but it would exist naturally in the II form, so these should be identical.
But these are mentioned to be geometrical isomers in a book that I was reading. Even if I assume that I existed in it's given form, then how can they be geometrical isomers? In I, there are two trans alkenes, three cis alkenes and in II, all five are cis alkenes.
So, are these compounds isomers or identical?

Both I and II (and there are more) are mere projections of a 3D molecular structure onto a 2D plane; shadows of the Plato's Cave, or an impression of a Flatland's citizen of an object from our world. Cyclodecapentaene does possess geometrical isomerism, and I and II are indeed isomers, but to get an idea how it works you need to think in 3D and think stability in terms of angle strain.

Seriously and chemistry let apart, you can't compare existing X to non existing X' and claim there to be identical.

Well, you can tell that two groups are trans vs all-cis and that cis-trans is isomerism, so what's the point of the question?

If you make spring-and-ball models of both compounds, you can see that the two compounds are entirely different, and both are contorted.

The all-cis model on the left has all its hydrogens pointing outward, so the contortions of the carbon chain involve just carbon-carbon twisting to come close to 120 degree angles. The model of the cis-trans molecule on the right has two inward-pointing hydrogens (all the rest point out) which are sterically interacting so much that they introduce strain that is not evident from the carbon-only diagram in the original post. (If I put in all the hydrogens, the image is too cluttered.)
The fact that one is all cis and the other cis-trans certifies that the molecules are different, but the 2-D drawing can seem vague. The NMR spectra of these two compounds would be different. I can imagine the molecule on the left flipping all around (the model sure did!), but the molecule on the right seems pretty stuck.
Thinking in 3-D is easy after you've done it many times. But when you are starting out, it is helpful to make molecular models. And even if you've done it a million times, there are situations where your mental image is so fluid that you need to solidify it with a model, no matter how inadequate the model is. And then you mentally add onto the solid model.


text:
Comparing acidic strength of ortho and para hydroxybenzoic acid
In orthohydroxybenzoic acid intramolecular forces operate which I believe tend to lock the position thereby making acidic H not available for release whereas in parahydroxybenzoic acid there in no intramolecular force so h is relatively more available so I believe that para should be more acidic than ortho but the answer is reverse

Why is ortho-hydroxybenzoic acid more acidic than its para-isomer? this qo. Though talks about it but does not resolve my query

Please structure your question, it gives the reader a headache. It would say it is one sentence, except it isn't, it is a typed down stream of thoughts.

I think the below scheme depicts well how the intramolecular hydrogen bonding more favours the collapse of the carboxylic O—H bond in the orthohydroxybenzoic acid than in the parahydroxybenzoic acid.

This is correct but the dictionary is quite personal. One has to say that the conjugated base in the ortho isomer is stabilised by a H bridge bond, involving a relatively stable 6 members ring.


text:
Finding final pH of the buffer solution without applying Henderson equation

A is beaker with $\pu{0.1 M}$ $\pu{25 cm^3}$ solution of $\ce{NH3}.$ B is beaker with $\pu{0.1 M}$ $\pu{5 cm^3}$ $\ce{HCl}.$ Now the solution of A and B is mixed. What will be the final $\mathrm{pH}$ of the mixture? $K_\mathrm{b} = \pu{3.3E-5}.$

I believe the mixture will become a buffer solution and we should use Henderson equation to solve this.
$$n(\ce{NH3}) = \pu{0.1 mol L^-1} × (\pu{25E-3 L}) = \pu{2.5E-3 mol}\tag{1}$$
$$n(\ce{HCl}) = \pu{0.1 mol L^-1} × (\pu{5E-3 L}) = \pu{5E-4 mol}\tag{2}$$
$$\mathrm{pOH} = \mathrm{p}K_\mathrm{b} + \log\frac{n(\ce{NH4Cl})}{n(\ce{NH4OH})}\tag{3}$$
$$\ce{NH3 + HCl -> NH4Cl}$$
$$n(\ce{HCl}) = n(\ce{NH4Cl}) = \pu{5E-4 mol}\tag{4}$$
$$n(\ce{NH3}) = n(\ce{NH4OH}) = \pu{2.5E-3 mol} - \pu{5E-4 mol} = \pu{2E-3 mol}\tag{5}$$
$$
\begin{align}
\mathrm{pOH} &= \mathrm{p}K_\mathrm{b} + \log\frac{\pu{5E-4 mol}}{\pu{2E-3 mol}}\\
  &= 4.4819 - 0.602\\
  &\approx 3.88 \tag{6}
\end{align}$$
$$\mathrm{pH} = 14 - \mathrm{pOH} = 14 - 3.88 = 10.12\tag{7}$$
However, the book from which I am solving this problem suggests that this problem should not be solved by Henderson equation without providing any reason. Can anyone tell me why this solution should not be considered a basic buffer? I would really appreciate some opinion on this.
P.S. The problem was given in Bangla. I tried my best to translate it. This is an admission question of KUET-2019 Bangladesh. My textbook has this question, but I'm not sure about the solution. Please note that this is an admission question of previous year (2019).

The HH equation should yield a reasonable result in this case. The precision of the measurements is sloppy. // One reason to solve without HH equation would be to see if you understood how to solve the general problem without the HH assumption. Can you solve problem without using HH?

No, I can't solve the problem without HH assumption. The solution in my textbook says that the pH should be 1.819 which is way of what I got. Can you please explain how this could be solved without HH assumption?

No way the pH could be acidic given the problem statement so a pH of 1.819 is ridiculous.

You can't have an appreciable amount of $\ce{H+}$ floating around in an ammonia solution. So essentially all (to two significant figures) the $\ce{HCl}$ reacts to make $\ce{NH4+}$. However some of the $\ce{NH3}$ will react with water to make $\ce{NH4+}$ too. So the solution should be slightly more basic that what you'd get from HH equation.

In my understanding, the task author wanted to force you to solve the task by alternative way, not by the H. equation.

As the amounts of substance in the final solution are known to be $n(\ce{NH3}) = \pu{2 mmol},$ and $n(\ce{NH4^+}) = \pu{0.5 mmol},$ you may simply use the definition of the constant $K_\mathrm{b}:$
$$K_\mathrm{b} = \frac{n(\ce{NH4^+})[\ce{OH^-}]}{n(\ce{NH3})} = \frac{\pu{0.5 mmol}\times [\ce{OH-}]}{\pu{2 mmol}} = \pu{3.3E-5}$$
from where $[\ce{OH-}],$ $[\ce{H+}]$ and $\mathrm{pH}$ can be quickly obtained:
$$[\ce{OH-}] = \pu{1.32E-4 mol L^-1}$$
$$[\ce{H+}] = \frac{10^{-14}}{[\ce{OH-}]} = \pu{7.57E-11 mol L^-1}$$
$$\mathrm{pH} = -\log[\ce{H+}] = 10.12$$

There is an accepted answer for this question. I think there is an another way to solve the problem using basic principles.
Beaker A contains $\pu{25 cm^3}$ of $\pu{0.1 M}$ $\ce{NH3}$ solution while beaker B contains $\pu{5 cm^3}$ of $\pu{0.1 M}$ $\ce{HCl}$ solution. Since $\ce{HCl}$ is a strong acid, all of $\ce{HCl}$ ($\pu{0.1 \times 0.002 mol} = \pu{0.0002 mol}$) in the beaker B has been totally dissociated:
$$\ce{HCl  + H2O -> H3O+ + Cl-}$$
Thus, when the solutions in A and B are mixed, $\pu{0.0002 mol}$ of $\ce{NH3}$ will react with $\ce{HCl}$ to give $\pu{0.0002 mol}$ of $\ce{NH4Cl}$, which is also completely ionized in final solution:
$$\ce{NH4Cl (aq)  -> NH4+ (aq) + Cl- (aq)} \tag1$$
Since final volume of of combined solutions is $\pu{0.027 L}$, initial concentrations in final solution are:
$$[\ce{NH4+}] = [\ce{Cl-}] = \frac{\pu{0.0002 mol}}{\pu{0.027 L}} = \pu{0.00741 M} $$
$$[\ce{NH3}] = \frac{\left(\pu{0.0025 mol} - \pu{0.0002 mol}\right)}{\pu{0.027 L}} = \pu{0.0852 M} $$
The amount of $\ce{NH3}$ remain in the final solution $(\pu{0.0023 mol})$ would be partially dissociated according to its $K_\mathrm{b} $:
$$\ce{NH3 + H2O  <=> NH4+  + OH- } \quad \quad K_\mathrm{b} = \pu{3.3E-5}\tag2$$
If the amount of $[\ce{NH3}]$ dissociated to receive the equlibriun is $\alpha \ \pu {M}$, then the final concentrations at equilibrium are: $[\ce{NH3}] = \pu{(0.0852 - \alpha) M},$ $[\ce{NH4+}] = \pu{(0.00741 + \alpha) M},$ and $[\ce{OH-}] = \pu{(\alpha) M}.$ Apply these values to $K_\mathrm{b} = \pu{3.3E-5}$ for the equation $(2)$:
$$K_\mathrm{b} = \pu{3.3E-5} = \frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]} = \frac{(0.00741 + \alpha)(\alpha)}{(0.0852 - \alpha)} \tag3$$
It is safe to assume $\alpha \lt \lt 0.0852,$ the equation $(3)$ can be rewritten as:
$$\alpha^2 + 0.00741 \alpha  - 0.0852 \times \pu{3.3E-5} =  0\tag4$$
When solve for $\alpha$:
$$\alpha = \pu{3.89E-4}$$
Since $\alpha = [\ce{OH-}]$, $\mathrm{pOH} = -\log (\pu{3.89E-4}) = 3.41$
Thus, the $\mathrm{pH}$ at equilibrium would be:
$$\mathrm{pH} + \mathrm{pH} = 14.0 \ \ \Rightarrow \ \ \mathrm{pH} = 14.0 - 3.41 = 10.59 $$
Note that the equation $(4)$ could be simplify by further assuming $\alpha \lt \lt 0.00741 = \pu{7.41E-3}.$ However, you can see that assumption is not entirely correct by looking at calculated $\alpha = \pu{3.89E-4}$.
For comparison, let's see what is the answer if you have used Henderson-Hasselbalch equation for the weak base $\ce{NH3}$:
$$\mathrm{pOH} = \mathrm{p}K_\mathrm{b} + \log \left(\frac{[\ce{NH4+}]}{[\ce{NH3}]}\right) = -\log (\pu{3.3E-5}) + \log \left(\frac{\pu{0.00741 M}}{\pu{0.0852 M}}\right) = 3.42 $$
Thus, $\mathrm{pH} = 14.0 - 3.42 = 10.58 $.


text:
Nitration vs Nitrolysis
I am interested in nitrating an aromatic ester in a solution of dichloromethane. I found the following paper that describes creating a nitric acid solution in dichloromethane using KNO3 and sulfuric acid:

Strazzolini P, et al. "Nitric Acid in Dichloromethane Solution. Facile Preparation from Potassium Nitrate and Sulfuric Acid." Tetrahedron Letters 42 (2001) 1387 - 1389.

This paper describes both the nitration of Benzyl Methyl Ester and the nitrolysis of Benzyl Tertbutyl Ester and Phenyl Tertbutyl Ester using their nitric acid in dichloromethane method.
But I am having trouble understanding how they controlled the reactions to get nitration of one ester and nitrolysis of another ester. The reaction conditions seem similar, but the concentration of nitric acid in the nitrolysis reactions is about twice as high as in the nitration reaction. Could the difference be in the tertbutyl group vs the methyl group on the ester?

Tert-butyl esters are routinely cleaved by anhydrous acid conditions by loss of the tBu+ cation. TFA in DCM is a common system for this. Neat formic acid, HBr/AcOH and PTSA in refluxing benzene or toluene are also used (see Greene - Protective Groups in Organic Synthesis (2nd ed) p246).
Methyl esters usually need aq. acid for cleavage.

Good to know, the compound I want to nitrate is usually dissolved in a TFA/DCM mixture, so if it can survive that, it can probably survive with the addition of nitric acid.

If you are happy with the answer please check the accept box. A quick google search for nitration of methyl benzoate produces a good number of hits with lab procedures, including one done in neat sulfuric acid. The only concern would be your workup  if you quench into water.


text:
What happens when FeF3 is reacted with benzene?
My text book says that halobenzenes can be prepared by aromatic substitution on benzene ring by using Fe or FeX3 as Lewis acid but this method isn't useful for preparation of fluoro benzene because of it's high reactivity.
My question is what will be the product if FeF3 is reacted with benzene, I tried searching a bit on internet but couldn't find anything helpful.

Halides are only catalysts in Friedel-Crafts halogenation. Without halogen there's no reaction.

Ferric halides plus elemental halogen (Cl2, Br2) give the functional equivalent of Hal+. The functional equivalent of F+ cannot be made this way.

See: https://chemistry.stackexchange.com/questions/21604/electrophilic-aromatic-substitution-with-fluorine

Iron catalyst will not be needed as Fluorine will react directly with benzene explosively,  the probable product formed would be hexaflurocyclohexane , which can predicted by looking at similar reaction of chlorine with benzene in presence of UV rays which is mentioned in NCERT class 11 chemistry  Textbook.

(Replace $Cl$ with $F$  and since fluorine is more reactive UV light is not needed.)

With fluorine, depending on conditions, I wouldn't be surprised to find the product is carbon tetrafluoride (amongst other things)

I strongly suspect you would generate a complex mixture (and a great deal of heat)

It may be possible that a lot of products are formed but I tried to compare reaction with a similar known reaction so the predicted product can also be formed,  maybe this type of question involves experiments too.

From what I remember of organic a general rule of thumb is the following reaction: organic + F2 -> mess + bang