text:
What is the structural formula of a monomer of hydroxypropyl methylcellulose?
I'm studying the properties of the polymer hydroxypropyl methylcellulose (HPMC) with the static scattering technique. I'm trying to interpret the molar mass obtained with a Zimm plot, that is $\pu{(3.08\pm 0.47)\times 10^{6} g mol^-1}.$
In particular, I would like to calculate the number of unites in a polymer chain, but I don't understand from the PubChem reference  what is the formula of single monomer. At the website above, I've found the formula $\ce{C56H108O30},$ but at the same site there is a molecular structure with a lesser number of carbons (about 30).
What is the formula for a HPMC monomer?

It's not a natural polymer and I wonder how could you not guess the unit from the name of the substance.

Yes it' s true! Should I start from a single glucose molecule adding the hydroxypropylic and  methilic groups?

Indeed from the name as it is there is formally no way to draw the monomer. This is because the reference is given to the polymer. We only know that a polymer called cellulose has some hydroxypropyl and methyl groups here and there. One working on it will certainly know, but strictly formally one can't.

Just to say that I am not surprised by the question. Nothing more, as this is quite normal in cellulose derivatives at least.


text:
Is it the solution energy or the reduction potentials (or both) that drives a galvanic cell?
I have received conflicting explanations for why the redox reaction in a galvanic cell occurs.
Explanation 1:

Some say that it occurs because the anodic metal dissolves more easily than the cathodic metal, and as such, the anodic metal will dissolve more quickly, becoming a positive cation that releases electrons (oxidizes). At the same time, in the cathodic half-cell, the positive cations in the solution will solidify on the surface of the cathode, due to rather wanting to be solid than an aqueous cation. However, these cations joining the cathode will add more positive charge than negative charge, and as such, the cathode will be positively charged. This will create a voltage between the electrons, and attract the electrons from the negatively charged anode, producing a current.
This means that the energy produced by a galvanic cell happens the system has less enthalpy when the anodic metal finds itself in a solution complex rather than a crystalline complex AND the cathodic metal finds itself in a crystalline complex rather than a solution complex.

Explanation 2:

Some say it occurs because of the reduction potential difference. The enthalpy change of an oxidation is always positive, and the enthalpy change of a reduction is always negative. If the cathodic ion's reduction has negative enough enthalpy change to make up for the positive enthalpy change of the anode's oxidation, then the reaction will happen. If the anode oxidized and the cathodic ion reduces, the resulting system will in total be of less enthalpy. Therefore, the redox reaction happens.

So, both explanations involve the fact that the energy of the system is lower if they react in the respective ways, but it has different explanations for exactly which process it is that involves the ultimately negative enthalpy change. Is it the solvation of the anode + the solidification of the cathode  ion that produces the negative enthalpy change (and therefore the energy to be used), or is it the oxidation of the anode + the reduction of the cathode ion that produces the negative enthalpy change? Or is it a combination of both?
Looking at types of corrosion, there can happen a redox-reaction without an electrolyte (dry corrosion), which means solvation isn't necessary for redox in general, but perhaps is needed for a redox reaction to happen when the reactants aren't in physical contact.

Electrode metals can be completely inert and system can still form galvanic cell. See the vanadium redox battery, used for industrial scale energy storage, based on Conductor|V^3+/V^2+||VO2^+/VO^2+|conductor.

@Poutnik Not sure why the trait of inertness is relevant to your comment, but given how all the substances in your cell reactions are ions, that would mean they're all dissolved from the get-go. That means that it's not the solvation energy causing the energy generation in that cell, correct? Rather just a redox reaction? Is this proof that the second explanation is correct, or an example of how it can be correct?

I reacted because  quoted "...'Some say that it occurs because the anodic metal dissolves..... " // Solvation energy participates on ion chemical potential, therefore on Gibbs energy of redox reaction and therefore on redox potential. by other words, dfiferences in solvation energies of ox and red forms participate on the redox potential.

@Poutnik I'm sorry but I don't quite get the formulation in your second comment (and possibly first). Could you reword it? Potentially make an answer?

You may imagine the galvanic cell as a water dam with a power plant. The redox potentials of electrode systems define the height of the water drop. The solution or solid component energy defines the multiple of that height drop and amount of water in the dam, i.e. equivalent of total energy.

@Poutnik Ah I think I understand. The solvation energy is a component of the reduction potential, along with things like differences in ionization energies, stability of electron orbitals, etc?

Yes. because it is closely linked to the well known equation from electronics $\mathrm{d}E=U \cdot \mathrm{d}q$

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

sort: votes


text:
Conversion of Benzene to 9-Bromo 9,10 Dihydroanthracene

I tried to proceed through following steps:
Step-1:
Step 2:
Step-3:
Step-4:
I agree with the fact that yield might not be good. But is this whole reaction possible? Am I going correct? A better method and the corrections to this method are appreciated!

Wait: how, assuming non-nuclear chemistry, do you convert Cl (step #3, 1,2-dichlorobenzene, coupling mediated by sodium) into C?

This is probably the first time I see four different drawing styles in one post:)

@andselisk In the longer run, I'm more concerned about (still) drawing the Thiele (?) ring in benzene's structure.  Too easy to lose track of counting the $\pi$-electrons present, not only for first time learning about either electrophilic or nucleophilic aromatic substitutions.  Even more so for e.g., naphthalenes when some drawings coalesce these individual Thiele rings across bonds into something in a shape like a toilet seat.

@andselik Haha..Some of them I had to copy paste from google while I had to make some of them on my own..didn't have any one source :P

Where do the two methylene groups come from in the step from 1,2-di-chlorobenzene to di-hydroanthracene? Step 3 looks OK

@Buttonwood I completely agree, good point about Thiele rings. I think you mean a so-called horseshoe bond.

Consider the Friedel-Crafts reactions of phthalic anhydride

@Waylander I am Sorry, I have edited step-3, I took another compound "1,2 Chloromethyl benzene" I hope it is correct now!

I think you would be better off starting from two equivalents of benzyl chloride. Those can react with each other in Friedel-Crafts fashion, yielding 9,10-dihydroanthracene. Use one equivalent of NBS afterwards and you obtain your product.

@TheRelentlessNucleophile Thank you, It was helpful!

OP has shown research effort and it should not get downvoted.

@andselisk While very descriptive, the horseshoe is not the form I thought about, because it lacks crossing a bond.  I made a photo for the chatroom here, as young as by about 2002.

What you are looking for is known as the Wurtz-Fittig reaction - wikipedia article , JACS paper
In this proposed reaction I think you will get a complex mixture (also known as a mess), but you may be able to find some of your desired product in it. Good luck with trying to isolate it cleanly!
If you want a straightforward preparative route then look at the F-C reaction of phthalic anhydride with benzene, the reduction of the resulting 2-benzoyl benzoic acid to 2-benzyl benzoic acid and cyclisation with polyphosphoric acid. This product can be reduced to the dihydroanthracene by Zn/HCl

Thankyou, but since I am a 11th grader, can you please help how do I carry out cyclisation process after reduction of carboxylic group?

You could try treating the resulting ortho-benzoylbenzoic acid with $\ce{SOCl2}$ and then carry out another acylation. But this process will for sure be low-yielding because of the deactivating nature of the acyl radical. Just start from benzoyl chloride. It's the easiest way to synthesize anything "anthracene-like" from a single aromatic nucleus.

@TheRelentlessNucleophile Thank you so much! :)

The reagent of choice for the cyclisation of the benzyl benzoic acid (reduction product from benzoyl benzoic acid) is polyphosphoric acid

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Oxidation number of oxygen atoms in oxygen difluoride
My textbook says:

Oxygen is assigned an oxidation number of $+1$ in oxygen difluoride. The number assigned to each Oxygen atom will depend upon the bonding state of oxygen but this number would now be a positive figure only.

Can there be multiple bonding states for oxygen in oxygen difluoride?
What can be the oxidation state of each oxygen atom in oxygen difluoride?

Oxygen difluoride $\ce{OF2}$ has a single oxygen atom. You must mean dioxygen difluoride $\ce{O2F2}$.
The rules are the same as for hydrogen peroxide.
Both oxygen atoms in $\ce{O2F2}$  are equivalent, so their oxidation number is +1.


text:
Ramachandran plot Phi(ϕ) Psi(ψ) dihedral angle Convention for Zero, Positive and Negative value- old and new
Update: possible duplicate: What is the precise definition of Ramachandran angles?. Question modified.
G.N. Ramachandran et al, in their own work (PMC) (DOI), did not used phi(ϕ) and psi(ψ) as we use today. They used phi(ϕ) and phi-dash(ϕ') defined as follows:

The conventions we have adopted for the standard configuration ϕ = ϕ' = 0°) and for the positive sense of rotation for the two angular parameters ϕ and ϕ' are indicated in Fig. 1.

Where Fig. 1 is:

Also they had adapted a very different values, 0 to 360 degrees for ϕ and ϕ'. They provided the Ramachandran plot like this:

This looks very different from the Ramachandran plot we see nowadays because:

This (Ramachandran's original) graph looks flipped from todays.

The zero degrees value of present day is perhaps different from Ramachandran's Zero degrees. Because the current day zero degree correspond to Ramachandran's 180 degree (of original graph).

So my question is basically:
On todays convention; how we define the zero value, positive value and negative values of phi and psi dihedral angle? and how does the new and old convention relate?
Prior research:
I did not find anything on intensive internet search. I got one graphic without any reference

I am doubtful about this diagram because its Zero degrees seem to be the zero degree of Ramachandran's original plot.
Many thanks in advance
Update: As the user @Buck Thom said

The meaning of angles with value of zero is the same.

But sill I looked both old and new graph and found what is labelled as 0 degrees in the new graph, has been labelled as 180 degrees in the old graph as follows

@BuckThorn Thank you. I did not mean 360 degrees I mean 180 degrees. Actually I tried to manually reconstruct the Left handed alpha helix, by considering the Web clip (3rd picture). Turned out I made a right handed helix. The confusion gone when I followed Ramachandran's original paper. Seemingly the 2 conventions are very different.

Related: https://chemistry.stackexchange.com/questions/114794/what-is-the-precise-definition-of-ramachandran-angles

Evidently the old $\phi$ is measured from an angle oriented $180^\circ$ relative to the new convention. This implies that the amide H ($H_N$) rather than carbonyl carbon was used as the reference fourth atom used to define the dihedral, since these atoms are related by a $180^\circ$ rotation. The old choice of atom is opposite on the chain to the modern convention of using the heavy backbone atom (carbonyl carbon). You can obtain angles in the new convention by subtracting $180^\circ$ ie $\phi \rightarrow \phi - 180^\circ$.

The meaning of angles with value of zero for $\phi$' aka $\psi$ is the same. However the values between $180^\circ$ and $360^\circ$ in the original Ramachandran plot can be shifted below zero, as $180^\circ\rightarrow -180^\circ$ and $360^\circ\rightarrow 0^\circ$, or more generally $\psi \rightarrow \psi - 360^\circ$.
By the way inspection shows that this is the correct interpretation. See for instance the following diagram. The extended (beta/turn/pII) region is usually larger than the alpha region. The beta region sits at large values (below  $180^\circ$) of $\psi$, whereas the narrower alpha region is close  to  $0^\circ$ (at negative angles):

In Ref. 1 Ramakrishnan and Ramachandran explain the differences in the  conventions:

At a recent conference of some of the representative workers in this field held in Bethesda, it was decided to denote the two dihedral angles about the bonds N-Ca and C.-C' by $\phi$ and $\psi$ respectively, the sense of rotation being the same as that adopted here. The fully extended chain, with N-H and C'=O trans with respect to one another, is to be taken as the standard conformation with $\phi$=$\psi$=0.
It is readily seen that the new $\phi$ is the same as the old $\phi$, but that $\psi$=$180^\circ$+$\phi'$ (and $\phi'$=$180^\circ$+$\psi$). All the data reported here are thus readily converted into the ($\phi$,$\psi$) coordinates. The diagrams in Figs. 2, 3, and 6 have to be shifted up by $180^\circ$, or half the total length, along the vertical direction. In particular, the right-and left-handed alpha helices will have ($\phi$,$\psi$) equal to ($133^\circ$,$123^\circ$) and ($227^\circ$,$123^\circ$). As before, a helix with ($-\phi$,$-\psi$) will be inverse to one with ($\phi$,$\psi$); i.e., it will  be of opposite sense, but having the same number of turns per unit.
This paper was finalized well before this meeting, and so the older conventions are adopted here. However, it is proposed to use the new notation in the following papers in this series.

References
1.Ramakrishnan C, Ramachandran GN. Stereochemical criteria for polypeptide and protein chain conformations II. Allowed conformations for a pair of peptide units.  Biophysical Journal, 01 Nov 1965, 5(6):909-933. DOI: 10.1016/s0006-3495(65)86759-5.

"The meaning of angles with value of zero is the same" Then why 0 of new graph is corresponding with 180 degrees of old graph? I am not getting this.

It's because in a circle zero degrees is the same as 360 degrees. Or, alternatively, you can move clockwise say and call the angles positive, or counterclockwise, and call the angles negative.

Not 360 its 180 i think I failed to express what I wanted to say. let me update the question

This is the clearest answer yet. However looking for some atomic diagram showing the frame of reference, and if possible, reference, since it is a convention question.

The Ramakrishnan et al convention also seems different because it says " It is readily seen that the new ϕ is the same as the old ϕ" where as we can see the newest phi is 180 degrees different from old phi

I agree, that is confusing!  My current assessment is that they confused the symbols somehow (seems unlikely though). In any case the explanation in my answer shows the difference between the different coordinates, even if that quote from the article is not easy to decipher.

The paper compares two historic definitions. With the current definition main chain torsion angles are 180 degrees for an extended conformation, not zero like in the historic reference.

@KarstenTheis Thanks, I understand that there are differences, but I manage to still not understand the difference between $\phi'$ and $\psi$. The bottom figure in my answer shows a configuration with $\psi=0$. Would you agree? The caption calls this $\phi'=0$ The text says that at $\phi'=0$ the conformation is extended, but it clearly is not by modern standards, or?

I should have written in my answer "Ramakrishnan and Ramachandran attempt to explain the differences in the conventions". I am a little surprised that I am still confused by the text. In the text they say the meaning for $\phi$ is unchanged, while $\phi'=\psi+180^\circ$. OTOH I think I understand the standard definition of the dihedrals and that this matches what I wrote in my answer.

I tried to find out what the Bethesda conference decided on the the 1960s, and whether it is different from the current definition, but did not succeed. So maybe the footnote in the Ramakrishnan and Ramachandran paper is inaccurate, or the community decided to switch the atoms defining phi and psi later on. With the sources at hand, it is impossible to tell.

@KarstenTheis I would say the question remains to some degree unanswered, at least the part "how do the old and new conventions relate?"

I've attached a figure to show the dihedral angles. Hope this helps.

Part of a peptide chain. The four atoms defining the $\psi$ (N to N) and $\phi$ (CO to CO) torsion angles are shown on the right. As these chains are drawn as if completely extended, the angles are $180^\text{o}$. When the main chain atoms are eclipsed, the angles are zero. The 6 atoms in the shaded areas are in the same plane.
Ramachandran plot together with the structure for bacteriorhodopsin.

Ramachandran plot of the $\phi$ and $\psi$ torsion angles in the protein bacteriorhodopsin (pdb 1FBB), which contains extensive $\alpha$ - helices. Most angles cluster around the values $\phi = -56^\text{o}, \psi = -53^\text{o}$ typical of an $\alpha$ - helix. A regular or ideal helix would have angles $-60^\text{o}$ and $-50^\text{o}$. The area of $\beta$-sheet structure, of which there is very little in this protein, is in the top left corner bounded by approximately $-55$ and $+60$ degrees.
The structure of the protein shows extensive helical structure. The retinal chromophore, which is positioned in the centre of the column of helices, is shown also.

Thanks for your answer but can you please add how do I define phi=zero? on which conformation the phi would be called zero? similarly for psi, and other positive and negative values?

My questiom is even more basic. Its about how we are counting phi and psi; from which frame of reference.

I think this is what you asked, in the top figure 'As these chains are drawn as if completely extended, the angles are 180 degrees. When the main chain atoms are eclipsed, the angles are zero. The 6 atoms in the shaded areas are in the same plane.'

as to positive negative you will just have to try with a structure such as the one above. If you use python I have code that will draw diagrams and can addd maths  about dihedral angles if needed.

I dont know python but I am trying different online simulators. although recognising C, N , O is difficult in most of these simulators.

The protein data bank has a good simulator.

This is not really an answer. This is some conclusions or assumptions.
1. The ϕ (old) and the ϕ (new) dihedral angle:
It is quite clear that the graph look is unchanged along horizontal axis.

Where ϕ (old) corresponds to ϕ (new) as follows:
ϕ (old)   .................   ϕ  (new)
0 deg    ......... ..........   -180 deg
180 deg  ....... ..........    0 deg
360 deg  .... ..........   +180 deg
Conclusion:
ϕ (Old) - 180 degrees = ϕ  (new)
or, ϕ (old) = ϕ (new) + 180 degrees.
2. The ϕ' (old) and the ψ (new) dihedral angle
Note that diagram is scroll shifted.

Where ϕ' (old) and ψ (new) are related as followes:
ϕ' (old) ..... .....  ψ (new)
0 deg ............   0 deg
180 deg ........... 180 deg, -180 deg.
360 deg ........... 0 deg
It means we do not have a frame shift between ϕ' (old) and ψ (new) , although there is the difference in direction of counting the angle, from where the negative sign comes.

[OP] On today's convention; how we define the zero value, positive value and negative values of phi and psi dihedral angle?

In general, you have four atoms (1, 2, 3, 4) in a chain to define a torsion angle. You rotate around the bond connecting atom 2 with atom 3 to change the torsion angle (bond angles and lengths stay constant). The conformation where 1 and 4 are closest is called zero degrees (cisoid). The conformation where 1 and 4 are furthest apart (transoid, extended conformation) is called 180 degrees. In both cases, all four atoms are in a single plane.
To measure the torsion angles of other conformations, you consider the plane through atoms 1,2,3 compared to the plane through atoms 2,3,4. The angle between these planes is the torsion angle. For a given angle, there are two conformations related by mirror symmetry. The right-handed one is given the positive torsion angle while the left-handed one is given the negative one (see figure in OP's question).
For a mathematical treatment with figures, see https://www.math.fsu.edu/~quine/MB_10/6_torsion.pdf
For the protein main chain, there are multiple ways to define a torsion, e.g. N-CA-C-N or N-CA-C-O. Switching the definition will lead to a shift of torsion angles by 180 degrees. The current convention, however, is to use N-CA-C-N and C-N-CA-C, and this is reflected in the current version of the Ramachandran plot.
The authoratative reference for the definition of torsion angles in general is the IUPAC gold book, citing DOI:10.1351/pac199668122193.

The authoratative reference for which atoms are used to define phi and psi seems to be Richardson, J.S. (1981). "Anatomy and Taxonomy of Protein Structures". Advances in Protein Chemistry. 34: 167–339.

The Cisoid and Transoid is a good way to understand.


text:
Is “double dot” (“··”) notation used for a double radical?
I wondered if this was a typo in the notes given to me. I haven’t come across notation like this before:

Branching — steps in which number of radicals increases (can lead
to explosion)
$$\ce{H^. + O2 -> OH^. + O^{..}}$$

Alone, the oxygen atom has always two separated uncoupled electrons. It never lasts a long time like this, but It has two doublets and two independent electrons in the outer shell. This makes a total of $6$ electrons. On the other hand, the molecule $\ce{O2}$ is a double radical, with one free electron on either Oxygen atoms..

Ahh, that makes sense. Thank you so much for this

It's call diradical or biradical, just saying ;)

TL;DR: The double dot notation is reserved for lone pairs.
Diradicals are depicted either with the appropriately placed superscripted dots within a structural formula (e.g. $\ce{^.O-O^.}),$ or superscripted multiplier–dot pair (e.g. $\ce{O2^{2.}}).$
The recommended notation for monooxygen 1,1-diradical is $\ce{O^{2.}}.$

First, let's sort out terminology.
The recommended name for “double radical” is diradical [1, p. 1333]:

diradicals:
Molecular species having two unpaired electrons, in which at least two different electronic states with different multiplicities [electron-paired (singlet state) or electron-unpaired (triplet state)] can be identified.
E.g. $\ce{H2C^.-CH2C^.H2}$ propane-1,3-diyl (trimethylene). See biradicals.

The term biradical, although being synonymous, is less preferable [1, p. 1322]:

biradicals:
Although this term has been recommended in the past for diradicals, specialists working in the field prefer the latter term.

So, we are dealing with a diradical.
You are probably familiar with a low-energy oxygen 1,2-diradical commonly depicted as $\ce{^.O-O^.}.$
What if we are dealing with a 1,1-diradical?
Note that two dots next to each other signify a lone pair (remember water $(\ce{H-\overset{\Large .\!\!.}{\underset{\Large .\!\!.}{O}}-H})$ or ammonia $(\ce{:\!\!NH3})?)$ and when being used for the depiction of diradicals will inevitably cause confusion.
This seeming contradiction is resolved if you follow guidelines provided in section IR-4.6.2 Formulae of radical from Nomenclature of Inorganic Chemistry – IUPAC Recommendations 2005 (Red Book) (emphasis mine) [2, p. 66]:

A radical is an atom or molecule with one or more unpaired electrons.
It may have positive, negative or zero charge.
An unpaired electron may be indicated in a formula by a superscript dot.
The dot is placed as a right upper index to the chemical symbol, so as not to interfere with indications of mass number, atomic number or composition.
In the case of diradicals, etc., the superscript dot is preceded by the appropriate superscript multiplier.
The radical dot with its multiplier, if any, precedes any charge.
To avoid confusion, the multiplier and the radical dot can be placed within parentheses.
[…]
Examples:
[…]

$\ce{O2^{2.}}$

[…]

$\ce{NO^{(2.)-}}$
$\ce{N2^{(2.)2+}}$

Here you can also find nomenclature for this species [2, p. 319]:

$\ce{O^{2.}},$ oxidanylidene, monooxygen(2•)

Following these recommendations, I would suggest to rewrite your scheme as follows:
$$\ce{H^. + O2 -> HO^. + O^{2.}}$$
References

Moss, G. P.; Smith, P. A. S.; Tavernier, D. Glossary of Class Names of Organic Compounds and Reactivity Intermediates Based on Structure (IUPAC Recommendations 1995). Pure and Applied Chemistry 1995, 67 (8–9), 1307–1375. DOI: 10.1351/pac199567081307.
IUPAC. Nomenclature of Inorganic Chemistry, IUPAC Recommendations 2005 (the “Red Book”), 1st ed.; Connelly, N. G., Damhus, T., Hartshorn, R. M., Hutton, A. T., Eds.; RSC Publishing: Cambridge, UK, 2005. ISBN 978-0-85404-438-2.


text:
Calculating pH of methanoic/formic acid (HCOOH)
When calculating the pH of methanoic/formic acid (HCOOH) why do we not multiply the given molarity of the acid by 2 as there are 2 moles of H+ ions, as seen in chemical formula HCOOH?
However when it comes to calculating the pH of sulfuric acid (H₂SO₄), we do multiply the given molarity by 2 as there are 2 moles of H+ ions.

The reason for this is that the hydrogen atom attached to the carbon will not be lost. Only the hydrogen attached to the oxygen will be ionized. In fact, in a solution of formic acid, not all molecules will be ionized anyways. Only very strong acids will fully ionize. Even in the case of sulfuric acid, not all molecules will fully ionize and lose the second proton. It is actually a simplification.

Hmm, philosophically speaking, if it's incorrect, is it actually an assumption? Maybe "simplification" is more appropriate?

I say assumption because it was stated that 'there are 2 moles of H+ ions' given off per mole of sulfuric acid which technically isn't fully true because ka2 of sulfuric acid is only 1.2*10^-2. I will change it though.


text:
Finding mass percent through molality of potassium nitrate solution
Question

An aqueous $\ce{KNO3}$ solution has a molality of $\pu{4.16 m}$ and a density of $\pu{1.08 g/mL}$.  Calculate the percentage by mass $\ce{KNO3}$ of the solution.

My Approach
I know that molality ($b$) is defined as:
$$b = \frac{n_\mathrm{solute}}{m_\mathrm{solvent}}$$
In order to the mass of solution I multiplied $\pu{(1.08 g/mL)}\times\pu{1 mL}$, since, $m_\mathrm{solution} = m_\mathrm{solvent} + m_\mathrm{solute}$
However, I don't know where to use this information. I am also not sure what I to do with the $\pu{4.16 m}$ since I don't have moles of solute or mass of solvent. I am thinking that I have to assume some number here, but I might be wrong about this. Any help will be appreciated.

Density is excessive input, not needed to calculate conversion. All you need is molar mass of KNO3. Density would be needed for calculation from molarity.

I am not sure you even need the density in this case.
Say you have $\ce{1kg}$ or $\ce{1000g}$ water (the solvent). That would mean that there is $\ce{4.16 mol}$ of $\ce{KNO_3}$.
Since we can figure out the molar mass of $\ce{KNO_3}$ = $\pu{101.11g mol-1}$, the mass of $\ce{KNO_3}$ in the sample must be $(4.16)(101.11)= \pu{420.62 g}$.
We already assumed that we had $\ce{1000g}$ of water, the mass percent of $\ce{KNO_3}$ is then equal to: $$\frac{420.62}{420.62+1000
} \approx 29.6\%$$

You can now upgrade your formatting skills and go beyond using just $\ce{}$, try using $\pu{}$ for physical units.


text:
What are the gaseous outputs of electrolysis of Water along with NaOH or KOH?
Theoretically electrolysis of water yields Hydrogen and Oxygen in gaseous form.
But to enhance conductivity we have to add an electrolyte like $\ce{NaOH}$ or $\ce{KOH}$.
Does this produce any other gases along with Oxygen? Does $\ce{Na+}$ or $\ce{OH^-}$ ions also come out as some form of gas? Or practically speaking can people breathe the oxygen which comes out of electrolysis?
What are the gaseous outputs of electrolysis of Water along with $\ce{NaOH}$ or $\ce{KOH}$?

Per your third question: it's done that way in submarines and on the ISS. Also, you might want to break-up this question into more concise ones. There is a chance this might be close for lacking focus.

@ToddMinehardt all I want to know if can we breathe electrolysis generated oxygen safely. I'm concerned about electrolyte generating toxic gases. You are welcome to edit. Thanks a lot

@ToddMinehardt what electrolyte do they use in Submarine

@NilayGhosh Thank you so much. Regarding the oxygen shortage in India, can people use electrolysis on distilled water to create breathable Oxygen??


text:
Why is a C-N bond weaker than a C-C?
As far as I could remember, in the same period, a bond to a more electronegative atom implies in a more polar and stronger bond, due to the greater ionic character. However the bond dissociation energy for the C-X series follows the order: C-F > C-O > C-C > C-N. The same for H-X: H-F > H-O > H-C > H-N.
Why are the bonds to nitrogen weaker than expected? (Or, more speficically, weaker than bonds to carbon, where no dipole is formed)
Values of average bond enthalpies I based my question off:

Bond
BDE (kJ/mol)

H-F
565

C-F
484

H-O
463

C-O
360

H-N
388

C-N
305

H-C
412

C-C
348

Sources: Atkins' Chemical Principles, Housecroft's Inorganic Chemistry, http://butane.chem.uiuc.edu/cyerkes/Chem104ACSpring2009/Genchemref/bondenergies.html

By the way, I found https://chemistry.stackexchange.com/questions/85767/why-nitrogen-bond-is-stronger-than-carbon-bond but I don't think the answer is satisfactory

I think it has to do with Nitrogen being half-filled which is a rather stable state.

Where did you get your BDE data from? The series you've presented have little in common with the values reported in 97th edition of CRC handbook of chemistry and physics (p. 9-73):
$$
\begin{array}{lr}
\hline
\ce{A-B} & D^\circ_{298}/\pu{kJ mol^-1} \
\hline
\ce{C-F} & 513.8 ± 10.0 \
\ce{C-O} & 1076.63 ± 0.06 \
\ce{C-C} & 605.03 ± 0.28 \
\ce{C-N} & 749.31 ± 0.14 \
\hline
\ce{H-F} & 569.680 ± 0.011 \
\ce{H-O} & 429.74 ± 0.03 \
\ce{H-C} & 338.72 ± 0.11 \
\ce{H-N} & 358.8 ± 0.2 \
\hline
\end{array}
$$

I'm sorry, but 1000 kJ/mol for a C-O bond is absurd, that may be the value for carbon monoxide. I took a look at Housecroft and Atkins Chemical Principles, and I will update the question with values

Of course the table is for diatomic molecules. I'm not sure it makes sense to compare average BDEs for random polyatomic species. Not only the values are less accurate, but also the difference in bond energies for the same bond for for various compounds may vary in hundreds $\pu{kJ mol^-1}.$ I suggest you either compare series of equivalent polyatomic molecules, or stick with diatomic ones from the table I provided. Otherwise it's no different than comparing average patients' temperatures at the hospitals (persons in the morgue and the ones with fever contributing to theaverage 36ish °C).


text:
What is the correct reaction mechanism?
I came across this question:

I propose two things to happen:

It can undergo Aldol condensation

or

it can undergo SN by replacing $\ce{-Cl}$ with $\ce{-OH}$.

But apparently, both these mechanism give the wrong product. I’m unable to find out what is going on.

I see what went wrong. Such questions are often confused with aldol or cannizarro as they have a carbonyl group and a base is given.
Actually this is a completely different name reaction called Favorskii rearrangement. The key to identify it is that its applicable to α-halo ketones.
In Favorskii rearrangement, the base takes α-hydrogen from that side of ketone which is opposite to the side having the halogen. It then forms a 3 member cyclic ring as an intermediate product. The product formed is a carboxylic acid derivative, mostly ester.
You can learn more at organic-chemistry.org or wikipedia


text:
How to model a molecule of benzene surrounded by 100 molecules of water with molecular dynamics?
I have experience calculating molecular properties (mostly DFT). However, I would like to spend some time on molecular dynamics methods. I would like to ask your advice to direct me towards a program with which to calculate systems such as the annealing of a benzene molecule surrounded by a hundred water molecules.

This question from matter modeling SE might be helpful: https://mattermodeling.stackexchange.com/q/6143

Do you mean simulated annealing? You can use Gromacs (https://www.gromacs.org/). Gromacs is quite easy to use. Also have a look at this post: https://mattermodeling.stackexchange.com/questions/4341/finding-parameters-for-a-molecule-in-various-forcefields/4374#4374 I have done MD simulations of small molecules in gromacs, so I can write an answer on that if you want. I haven't done annealing so I can't really write about that.

Nisarg Bhavsar, thanks, I will dive a little bit into NAMD and see if I can get something running, I see there are some tutorials on youtube, for example, https://www.youtube.com/watch?v=IArpsQsZ95U seems good! Shoubhik R Maiti thank you for pointing out Gromacs, also here I could dive into maybe looking at some tutorials will be useful: https://www.youtube.com/watch?v=rYZ1p5lXNyc&t=1s.

@PaulPerkele In my personal experience, learning gromacs was easier than learning NAMD (in fact right now I am trying to learn that). However, you may find the other one easier. Yes youtube tutorials are helpful. Do keep in mind though that for small molecules, you usually get parameters/topology from servers like https://www.swissparam.ch/ or from softwares (because you need general forcefield). As an aside, if you are responding to a comment on this site, make sure to tag them with an @


text:
Does 3% hydrogen peroxide neutralize household bleach?
Wiki and lots of sources all say the same thing. "1 part of 3% hydrogen peroxide to 10 parts water". But I am not trying to neutralize water I want to neutralize laundry bleach. It could mean add 1 part hydrogen peroxide to 10 parts water and then use that to neutralize your bleach product but without knowing the  concentration of sodium hypochlorite that the mixture was intended for that would not make sense either. The directions are a complete mystery no matter how you look at them. So I will ask in the following manner to make it clear.
We have a gallon of 3% sodium hypochorite household bleach.  How much 3% hydrogen peroxide should be added to neutralize it and what will the neutralized product be that remains? The chemical reaction formula would be nice but I might be asking to too much here.

Does this answer your question? How to calculate the volume of oxygen in a reaction between hydrogen peroxide and bleach using this apparatus

Out of curiosity, is there any particular reason to not just use up laundry bleach by using it in the laundry? After all, white clothing dinginess, like rust, never sleeps.

Never mix bleach under any circumstances or for any reason. If you want to get rid of it call your county's  hazardous waste disposal department  and they can give you directions of where you can take it for safe disposal.  You can also call the manufacturer of the product and they can give you directions for disposing of it in the toilet if it is under a certain amount AND copious amounts of water are used with multiple flushes. They can give the specifics.

Never mix bleach with anything. As far as hydrogen peroxide goes the combination can explode from what I read. To dispose of it call your county hazardous waste department.

......or use it in your laundry.

Wiki and lots of sources all say the same thing. "1 part of 3% hydrogen peroxide to 10 parts water."

When talking about aqueous solutions, the concentrations can be expressed as percent by weights, $\% (w/w)$, or by volume, $\% (v/v)$, or by weight in parts of solute by weight in 100 part of solution by volume, $\% (w/v)$. The above statement in the question means if you take 100 part of the solution by weight (e.g., $\pu{100 g}$ of solution), it contains 3 part of hydrogen peroxide and 97 parts of water by weights $\left(3\% (w/w)\right)$. It also can mean that it may contain 3 part of hydrogen peroxide to 97 parts of water by volumes $\left(3\% (v/v)\right)$ if the lebel doesn't say it is whether $\% (w/w)$ or $\% (v/v)$. The third kind is 3 part of hydrogen peroxide by weight to 100 parts of solution by volume $\left(3\% (w/v)\right)$, which is common for prepared chemical reagents to use in the laboratories.
Commercial $\ce{H2O2}$ solution is usually in $\% (v/v)$. However, when prepared as a chemical reagent, it is $\% (w/v)$. That means $\pu{100 mL}$ of $3\% (w/v)$ $\ce{H2O2}$ solution has $\pu{3 g}$ or the molarity of the solution:
$$3\% (w/v) = \frac{\pu{3 g} \ \ce{H2O2}}{\pu{100 mL} \ \text{soln}} = \frac{\pu{30 g} \ \ce{H2O2}}{\pu{1000 mL} \ \text{soln}} = \frac{\frac{\pu{30 g} \ \ce{H2O2}}{\pu{34.01 g mol-1}}}{\pu{1.0 L} \ \text{soln}} = \pu{0.882 mol L-1}$$
On the other hand, since commercial $\ce{H2O2}$ solution is usually in $\% (v/v)$, and density of pure $\ce{H2O2}$ is $\pu{1.450 g mL-1}$ at $\pu{20 ^\circ C}$ (Wikipedia):
$$3\% (v/v) = \frac{\pu{3 mL} \ \ce{H2O2}}{\pu{100 mL} \ \text{soln}} = \frac{\pu{30 mL}\cdot \pu{1.45 g mL-1} \ \ce{H2O2}}{\pu{1000 mL} \ \text{soln}} = \frac{\frac{\pu{43.5 g} \ \ce{H2O2}}{\pu{34.01 g mol-1}}}{\pu{1.0 L} \ \text{soln}} = \pu{1.28 mol L-1}$$
Household bleach is aqueous $\ce{NaClO}$ $(\pu{74.44 g mol-1})$ solution, sometimes included certain percent of $\ce{NaOH}$. If you have $3\%$ solution, you can do the same calculations as above to find out the correct molarity if the solution is given whether $3\% (w/v)$ or $3\% (v/v)$ (and it is without $\ce{NaOH}$). Usually, it is always given in $\% (w/v)$. Sometimes, the default densities may be provided instead of percentages by the manufacturer, which may differ for the different manufacturer since the manufacturer may change the density by adding salt or other chemicals to the solution. Therefore, it is hard to calculate molarity using provided default density if you don't know what else other than $\ce{NaClO}$.
Let's consider the solution you have is $3\% (w/v)$ $\ce{NaClO}$:
$$3\% (w/v) = \frac{\pu{3 g} \ \ce{NaOCl}}{\pu{100 mL} \ \text{soln}} = \frac{\pu{30 g} \ \ce{NaOCl}}{\pu{1000 mL} \ \text{soln}} = \frac{\frac{\pu{30 g} \ \ce{NaOCl}}{\pu{74.44 g mol-1}}}{\pu{1.0 L} \ \text{soln}} = \pu{0.403 mol L-1}$$
The reaction is:
$$\ce{NaClO + H2O2 -> NaCl + H2O + O2}$$
Accordingly, $\pu{1.0 mol}$ of $\ce{NaOCl}$ will react with $\pu{1.0 mol}$ of $\ce{H2O2}$. Therefore, the volume $(V_\ce{H2O2})$ of $3\% (w/v)$ $\ce{H2O2}$ needed is react completely with $\pu{1.0 L}$ of $3\% (w/v)$ $\ce{NaOCl}$ can be calculated using $M_1V_1 = M_2V_2$ as follows:
$$M_\ce{H2O2}V_\ce{H2O2} = M_\ce{NaOCl}V_\ce{NaOCl} \ \Rightarrow \ V_\ce{H2O2} = \frac{M_\ce{NaOCl}V_\ce{NaOCl}}{M_\ce{H2O2}} \\ =  \frac{\pu{0.403 mol L-1}\cdot \pu{1.0 L}}{\pu{0.882 mol L-1}} = \pu{0.457 L}$$
Therefore, since you have one US gallon of bleach, which equals to $\pu{3.785 L}$, you need $0.457 \times 3.785 = \pu{1.73 L}$ of $3\% (w/v)$ $\ce{H2O2}$ solution.
Your final solution is $\frac{0.403}{1.457} = \pu{0.277 mol L-1}$ $\ce{NaCl}$ solution (of course, only if manufacturers did not add additives).

The equation is: $\ce{NaClO + H2O2 -> NaCl + H2O + O2}$
A $3\%$ sodium hypochlorite contains $\pu{30 g}$ $\ce{NaClO}$ $(\pu{0.40 mol})$ per liter. So $\pu{0.1 mol}$ $\ce{NaClO}$ is contained in $\pu{250 mL}$ bleach.
A $3\%$ hydrogen peroxide contains $\pu{30 g}$ $\ce{H2O2}$ $(\pu{0.88 mol})$ per liter. So $\pu{0.1 mol}$ $\ce{H2O2}$ is contained in $\pu{114 mL}$ of your hydrogen peroxide solution.
So mix $\pu{250 mL}$ bleach $+$ $\pu{114 mL}$ of your hydrogen peroxide solution, or any multiple or submultiple, like $1$ liter bleach and $\pu{456 mL}$ peroxide solution.
You'll obtain a dilute salt solution.


text:
What makes an electrolyte strong - solubility or dissociation?
I'm having some confusion related to the terms mentioned in the title. Suppose we have a strong electrolyte but with a very low solubility.
Therefore a given amount of it will not be able to produce lots of ions because less of it will go into solution and dissociate, so what do we call this type of electrolyte overall: a weak electrolyte as it didn't produce many ions or a strong electrolyte as whatever that did dissolve produced as many ions as it can dissociate into?

Not dissolved part does not count. If you mix 100 mL of water and 100 kg of NaCl, does it make it weak electrolyte,because majority is not dissolved ?

@Poutnik if you dissolve 100kg of NaCl in 100L of water then also it will be a strong electrolyte and solution will be fairly conducting but if you dissolve equimolar amount of electrolyte of type OP considers in question in same amount of water, it will not be a fairly conducting solution , In that case  how can you say that the considered electrolyte is strong?

Does this answer your question? How can an insoluble compound be a strong electrolyte

@Lllt  By evaluating molar conductivity, related to total solution concentration. Strong electrolytes have molar conductivity corresponding to ion conductivities at given concentration. Weal electrolytes have molar conductivity much lower. // The mistake is evaluation of conductivity instead of molar conductivity. By this way 0.000001 M NaCl would be weak electrolyte as well.

@Poutnik therefore we can conclude that the electrolyte of the type that OP considered in solution shall be a strong electrolyte as whatever amount of it is dissolved it will completely break into ions

@Lllt Yes, similarly as listed in the referred duplicate question.

See also: https://chemistry.stackexchange.com/questions/131070/is-there-any-difference-between-getting-dissolved-and-getting-dissociated?rq=1 and https://chemistry.stackexchange.com/questions/27863/can-a-strong-electrolyte-be-sparingly-soluble?rq=1&noredirect=1

@Poutnik do you have an example of substance which is strong electrolyte in one solvent and weak electrolyte in other solvent though it dissolves in both

@Lllt You should post it as the question.


text:
Why don't we consider the attached hydrogens in inductive effect?
Consider the molecule 1-chloroethane

Let carbon attached to $\ce{Cl}$ be $\ce{C1}$ and other one be $\ce{C2}$. While learning inductive effect I was told that $\ce{Cl}$ due to high electronegativity pulls the $\ce{C1–Cl}$ bond electrons towards itself leading to development of partial positive charge on $\ce{C1}$ and partial negative charge on $\ce{Cl}$ .Now this $\ce{C1}$ which has became a little electron-deficient tries to pull electron of the adjacent adjacent carbon $\ce{C2}$ leading to development of a further smaller partial  charge on $\ce{C2}$. I have following doubts:

Will $\ce{C1}$ only pull electrons from $\ce{C2}$ or from the hydrogen atom attached to it also. I think it does pull, but then why don't we take that pull into consideration?

When $\ce{C1}$ pulls electrons from $\ce{C2}$, the partial positive charge developed on it due to the pull of electrons by the chlorine atom will be reduced a little but why do we then show $\ce{C1}$ to have same partial positive charge?

Kindly see this and use mathjax again, Thanks! https://chemistry.meta.stackexchange.com/questions/86/faq-how-can-i-format-math-chemistry-expressions-on-chemistry-stack-exchange

The effect exists but is very small and insignificant.

Is not the inductive effect taken as the relative effect of atom/group, compared to hydrogen ? By other words, a bound hydrogen is taken as the reference zero by definition.

@Poutnik do you mean the same as  first line in the given link https://en.wikipedia.org/wiki/Inductive_effect#Relative_inductive_effects

@Poutnik therefore you want to say partial charge developed on $C_2$ by $C_1$ will be relative to hydrogen?

@Lllt No, I want to say that inductive effect of the methyl group or the chlorine atom is compared to inductive effect of hydrogen, if it was there instead of them.

@Poutnik therefore a more electronegative atom than hydrogen will pull electron more towards itself will said to have -I effect and a less electronegative atom than hydrogen will not pull electrons more towards itself than Hydrogen and will said to be electron donating or having +I effect

@Lllt It is more general, as it applies to atomic groups as well. E.g. Even if carbon is more electronegative than hydrogen, a methyl in acetic acid pushes electron density toward the carboxyl carbon, compared to the hydrogen atom of formic acid.

@Poutnik Ok I understand that but is my above statement correct and complete if I incorporate atomic groups in them as well

I will show you how I understand this effect. I like to think that partial positive charge on $\ce{C1}$ attracts the electrons from $\ce{C2}$ because of the electrostatic force. With this model your questions can be answered this way:

Yes, $\ce{C1}$ will pull electrons from every adjacent atom, including hydrogen. It may pull them, however, to less extent from hydrogen because it has less electrons than carbon (or rather alkyl group), and smaller charge will result in weaker attraction according to our model.
What you are saying is right. The reason why we still write just $\delta +$ instead of some kind of $0.7\delta +$ is mostly convenience. This writing can be used in organic chemistry to show mechanisms, for example, and in this case we don't need super-accuracy in determining the magnitude of the charge itself (like it could be in some physical model), we just need to show that it is positive. It will be different in every case (molecule) and calculating the exact charge seems like waste of time, but we still are aware of the inductive effect, although don't explicitly show it.

thankyou for replying. I completely understood the answer of my second question. I'm having few doubts related to first one but ,why does small size of hydrogen cause Carbon to pull electrons to a less extent from it

Oh, sorry, I am formulating my sentences not in the best way. Actually, what I misinterpreted as two things is just one — carbon has more electrons. It is the same as the first one because larger atoms have more electrons in them (this is also how dispersion forces are explained). And what you have is that higher charge means greater attraction and electrons are pulled with greater force from carbon than from hydrogen.

@Azamat Hydrogen has less electrons than carbon, why will this fact cause carbon to pull electrons from it to a less extent

@Lllt, it's the electrostatic force. The law is $F = k \ \frac{q_1 q_2}{r^2}$. Let's say that $q_1$ is the charge of $\ce{C1}$ carbon, the force now depends on the charge $q_2$. If there are more electrons around, this charge is greater, therefore the force is also greater. There are more electrons at the carbon (because it has another hydrogens from which to take them), so the force is greater. However, I don't think that this gives a huge difference between the carbon and hydrogen, and teachers should actually note the effect on hydrogen too.


text:
Doubt while using Le Chatelier's principle
Consider the following reaction
$\ce{H2}+\ce{I2}\rightleftharpoons\ce{2HI}$, for this reaction $\Delta H_r<0$
Now suppose I introduce more amount of $\ce{H2}$ in it, therefore temperature of system increases as total pressure increases. Now by Le Chatelier's Principle, since concentration of $\ce{H2}$ increases , therefore reaction moves in direction where it is consumed i.e. in forward direction as said my book but one thing that troubles me is that when we introduce more $\ce{H2}$ the temperature of system increases, therefore reaction should move backward contradicting the fact given in my book.
Where I'm getting confused?
I think that when my book says that equilibrium shifts forward as we introduce more $\ce{H2}$, it must also say that temperature of system remains constant.

For complex changes with more, mutually opposing factors, quantitative evaluation may be needed. But my guess is, the change of reaction quotient has much bigger influence than change of temperature and therefore of equilibrium constant ( see van't Hoff equation

I introduce more amount of $\ce{H2}$ in it, therefore temperature of system increases as total pressure increases. Why do you think that the temperature is increasing? I believe you are missing that $n$ is also changing.

@nisargbhavsar The OP may mean adiabatic compression by injection of hydrogen.

@nisargbhavsar do you mean to say that pressure increases due as number of  moles of $\ce{H2}$ increases keeping temperature constant

I think in problems with Le Chatelier's Principle we change only one thing and other factors are held constant. In this case, if you want to include more factors at a time, I think more information should be included in the problem. For example, it could be that the reaction is preceding inside a container with a piston, so it's possible to leave temperature constant by increasing the volume (moving the piston).

The principle says temperature increase would push reaction toward left side, while adding a reactant would push the reaction toward right side. It does not say which effect is stronger. You need quantitative computation for that.

You could introduce more gas isothermally.


text:
Hydrogen as a reference in inductive effect
What does it mean when we say that I-effect of hydrogen is taken as zero ?
I think that it means that partial displacement of sigma bond electrons between two atoms of different electronegativity is measured with respect to hydrogen i.e.
Relative partial displacement of $\sigma$ bond electrons in polar bond =
actual partial displacement of those electrons- displacement of electrons that would happen if hydrogen was a part of bond
Is my reasoning wrong? If so, why?

See: https://en.wikipedia.org/wiki/Inductive_effect It is relative inductive effect where we consider hydrogen as reference.

@0.00 in that case only what does it means when we say that hydrogen is reference?


text:
Effect of d-orbital electron shielding on atomic radius
In a standard book it is given:

"Atomic radius of Gallium is less than that of Aluminium. This can be understood from the variation in the inner core of the electronic configuration. The presence of additional 10 d-electrons offer only poor screening effect for the outer electrons from the increased nuclear charge in gallium. Consequently, the atomic radius of gallium (135 pm) is less than that of aluminium (143 pm)."

In another chapter, it is given:

"As we move along the period in 3d series, we see that nuclear charge increases from scandium to zinc but electrons are added to the orbital of inner sub shell, i.e., 3d orbitals. These 3d electrons shield the 4s electrons from the increasing nuclear charge somewhat more effectively than the outer shell electrons can shield one another. Therefore, the atomic radii decrease less rapidly."

In the first para, it is given that the d-electrons shield poorly, but the opposite is given in the second but it is unlikely that the book is wrong. So what am I missing?

See Slater's rules for an empirical evaluation of shielding.

Both sections do not contradict themselves, as they address different shielding aspects.

3d electrons give worse shielding of 4s/4p electrons than 1-3s and 2-3p electrons.
3d electrons give better shielding of 4s/4p electrons than 4s/4p electrons themselves (mutually).


text:
Exchange energy of d6 configuration
In NCERT Chemistry book, it is given as:

Exchange energy is responsible for the stabilization of energy state. Exchange energy is approximately proportional to the total number of possible pairs of parallel spins in the degenerate orbitals. When several electrons occupy a set of degenerate orbitals, the lowest energy state corresponds to the maximum possible extent of single occupation of orbital and parallel spins (Hund's rule). The loss of exchange energy increases the stability. As the stability increases, the ionization becomes more difficult. There is no loss of exchange energy at $\ce{(d^6)}$ configuration. $\ce{Mn+}$ has $\ce{(3d^5) (4s^1)}$ configuration and configuration of $\ce{Cr+}$ is $\ce{(d^5)}$, therefore, ionization enthalpy of $\ce{Mn+}$ is lower than $\ce{Cr+}$. In the same way, $\ce{Fe^2+}$ has $\ce{(d^6)}$ configuration and $\ce{Mn^2+}$ has $\ce{(d^5)}$ configuration. Hence, ionization enthalpy of $\ce{Fe^2+}$ is lower than the $\ce{Mn^2+}$.

But, there are 5 electrons of same spin in $\ce{Fe^2+}$. So, the number of exchanges would be $\mathrm{^5C_2}$ = 10. Then why is there no loss of exchange energy in $\ce{Fe^2+}$ ?

Possibly the loss of degeneracy in presence of ligands is being discussed.
The 5 d-orbitals of Fe which were degenerate will turn into 3 lower energied orbitals and 2 higher energied orbitals in presence of a ligand.

"4 unpaired electrons of same spin" - They don't have to be unpaired. There are 5 electrons of the same spin.

But still why is it zero?

Because you're looking at an ionisation process, and thus the key question is the difference in exchange energy between $\ce{Fe^2+}$ and $\ce{Fe^3+}$.

I used the same logic for ionization energy of Fe(II) and Mn(II), but I am unable to understand why is the exchange energy of Fe(II) is zero.

Where does it say that it should be zero? It says that there is no loss of exchange energy (upon ionisation).

I think I misunderstood this statement, "There is no loss of exchange energy at (d6) configuration". Can you explain what does it mean?

As @orthocresol points out, the key is that you need to compare the exchange energy before vs after the ionization process. Anything that is unchanged by ionization cannot affect ionization energy. This is a general idea for a lot of aspects of chemistry: close the thermodynamical circles carefully, since so much information can be gained from them.
So, let us compare the Fe$^{2+}$ to Fe$^{3+}$ ionization. Both ions have 5 electrons with the same spin,[1] leading to the same exchange energy. Additionally, Fe$^{2+}$ has an electron with the opposite spin, which does not contribute to exchange energy. All in all, no difference in exchange energy, so no effect of the exchange energy in the ionization energy. To reiterate: it is not that the exchange energy is zero, it is the before vs after difference (so, the effect) which is zero.
Now let us see what happens with the Mn$^{2+}$ to Mn$^{3+}$ ionization. Here we are going from 5 electrons with the same spin to 4 electrons with the same spin,[1] so here we are losing exchange energy between the states before and after ionization, and therefore there this is something extra to pay in the ionization energy.
[1] For simplicity of the argument, we will be assuming "high-spin" configuration, as is implicit in the question.

probably the exchanges happening in the same orbital contribute to the exchange energy. if we say that an electron is moved from 3d to 4s orbital then it is said to be excited(Exchanged is not suitable to be used in this case).
so when you ask me to talk about the Fe2+ ion here is its electronic configuration
**1s2 2s2 2p6 3s2 3p6 3d6**
so in this case as for stability, the electron tends to excite to the 4s orbital and hence we call this the excitation of electron and not exchange.
I hope this would clear your doubt.
cheers!!
edit1: you may feel doubted why the electron excites from 3d to 4s so as it gets 3d5 configuration which makes it more stable and more exchanges contribute to more stability.


text:
When does bimolecular nucleophilic substitution reaction take place in phenyl group containing compounds?
We were studying SN Mechanism, and our professor gave us a question to identify the SN, mechanisms which would happen in the given compound.
They were $\ce{CH3Cl}$, $\ce{PhCH2Cl}$, $\ce{Ph2CHCl}$, $\ce{Ph3CCl}$.
So while attempting the given problem, I chose SN1 for both $\ce{Ph2CHCl}$ & $\ce{Ph3CCl}$,as they would face a lot of hindrance. But our professor told us that they both can also perform SN2 mechanism. Why would they both perform SN2 mechanism despite there being a lot of hindrance present?
My Reasoning: Although $\ce{PhCH2+}$ would be resonance stabilized but the hindrance present is too much for SN2 to proceed.

Your own reasoning, based on search and analysis, is supposed to be present to avoid the question closure for lack of own effort.

Honestly, it’s completely ridiculous to suggest that trityl chloride could undergo SN2, and also very disappointing to see such things being taught as if they were true. I’d wager that diphenylmethyl chloride also almost certainly favours SN1. With secondary halides it tends to be more nuanced, but here the groups are pretty bulky and the SN1 carbocation intermediate is pretty stable.

That line of logic is pretty useless. These molecules may also fall apart randomly into three phenyl radicals, a chlorine atom, and a carbon atom. The chances of that happening are basically zero as well. One could insist that that’s possible, which may well be true, but you wouldn’t learn anything useful from that assertion. Insisting that trityl chloride can undergo SN2 is technically correct, but pointless.

Any reaction capable of undergoing in two or more mechanisms has major and minor pathways. In your question the major pathway is obviously SN1 due to high steric hindrance and the resonance-stabilized carbocation, and the minor pathway is SN2. Since, your professor said they can ALSO perform SN2 mechanism, I think he was trying to say that the reaction doesn't only react via the SN1 pathway, but also via the minor SN2 pathway. However, the major pathway is definitely SN1 due to high steric hindrance and resonance stabilized carbocation.


text:
What is the difference between structure assignment and structure prediction?
I am absolutely new to structural bioinformatics (only started last week). I am working on the secondary structure assignment/prediction (actually I am not sure) of proteins using machine learning.
When it comes to protein analysis, most of the books or research papers talk about protein structure prediction. Only a handful of books or articles talk about protein structure assignments.
What is the difference between structure assignment and structure prediction?

I suppose these resources do not speak about these terms out of context. Is this context the same or different for both terms ?

My comment was a hint for you to finish your homework of elaboration of the question. // Your own reasoning – based on searching, reading and thinking – is supposed to be present to avoid the question closure for lack of own effort.

https://rostlab.org/papers/2008_rev_assignment2/paper.html#DSSP

https://pubmed.ncbi.nlm.nih.gov/8749853/

Assignment refers to a statement that a particular amino acid residue
(1) has a given backbone conformation described by dihedral angles falling within certain limits
(2) based on the local conformation of a series that includes adjacent amino acids, the residue can be said to be part of a secondary structure such as one of various types of helix, sheet or turn (or simply lacking specific structure, ie intrinsically disordered).
The basis for the determination of dihedral angles in NMR is to pool together various spectroscopic variables and use these as constraints during molecular dynamics simulations of the protein, in order to generate an ensemble of possible structures, then to evaluate the dihedral angles in those structures. The data usually includes NOEs, which are signal enhancements observed when two nuclei are close and provides a sensitive constraint on the magnitude of their separation. NOEs are labelled based on the atoms involved, including the distance between the residues (in the primary sequence) to which each atom belongs. Certain patterns of NOEs are consistent with particular secondary structures and can provide qualitative evidence of such structure before applying the more elaborate analysis previously described. Other valuable data include heteronuclear chemical shifts and various J-couplings. A high-quality set of NOEs and complementary data can in principle constrain the conformation of the protein down to a non-redundant choice.
It is possible for (1) and (2) to be inconsistent, that is, to assign an amino acid to particular secondary structure, but for that particular amino acid to reveal, on the basis of spectroscopic data and/or simulations, conformations inconsistent with that secondary structure. In that case the assignment is said to violate the data (or vice-versa), making the assignment suspect if not falsifying it, and calling details of the analysis into question. The quality of an ensemble is assessed on the basis of a statistical analysis of the agreement between the data and the assigned secondary structures.
Secondary structure prediction is a term used to refer to the use of known correlations between secondary structures and observables, and can be said to include using spectroscopic or crystallographic data to estimate what the structure of a protein is like, including how flexible. Assigning residues to particular secondary structure is a form of prediction, but when the input is experimental data and the degree of confidence is sufficiently high the process is not usually called prediction. Prediction involves an input of data, possibly sparse and usually including parts or all of the primary sequence, obtained for a protein of interest, into a prediction algorithm which takes such inputs and returns a list of possible secondary structures ranked by likelihood of being accurate. The algorithm is developed (a step in some cases called training) by optimizing the agreement of the output obtained for proteins of known structure.

Is "Assignment" a part of the "Prediction"-process?

Structure assignment is relating e.g., a measured absorbance frequency (like a triplet in $\pu{^1H NMR}$), or a structural feature (like $\Phi$ and $\Psi$  in a Ramachandran plot) to a structure already known.  It equally may be based on a putative / postulated structure; then, you compare e.g., how a simulated NMR spectrum is similar to the experimentally recorded one.  If the two spectra match 1:1, the putative structure likely is a correct one for this NMR experiment.
Structure prediction departs knowing e.g., only a sequence of $\alpha$-amino acids, e.g. Ala-Leu-Cys, and offering a guess about their spatial arrangement (e.g., folding).  Its foundation may be more or less educated / reasonable, which is why one typically infers from already known relationships of sequences of amino acids, and their 3D structure already established, e.g., by NMR, or X-ray crystal diffraction.  Machine learning (ML) calls this stage of establishing a model as training.  Aside from comparing with other algorithms, you eventually benchmark your  ML algorithm in multiple runs with different training sets (e.g., sequences) to watch the outcome of the predicted structures; for one, comparison with experimentally determined data intentionally excluded from the training set now offer a check about the correctness of the results.  For two, this equally offers to estimate the likelihood a folding  predicted in silico to equally occur in vitro e.g., reporting a range for a particular $\Psi$ between two adjacent amino acids.

I am working on the secondary structure assignment/prediction (actually I am not sure) of proteins using machine learning.

Secondary structure assignment is an automated method that defines the secondary structure (helix, sheet, loop) based on a known three-dimensional structure. Secondary structure prediction refers to taking the primary structure (the sequence of amino acid residues in a protein) and predicting the secondary structure.
There are also some experimental methods that help to characterize secondary structure in the absence of a 3D structure. Circular dichroism spectroscopy gives an estimate of the secondary structure content. Assigned chemical shifts of C-alpha and C-beta atoms from NMR experiments give a residue-by-residue prediction (or characterization) of secondary structure.

What is the difference between structure assignment and structure prediction?

Without the qualifier "secondary", I am unfamiliar with structure assignment. If you google "protein structure assignment", most hits will talk about secondary structure assignment. NMR resonances are assigned to protein primary structure (i.e. which residue gives rise to which signal).
Structure prediction most often refers to tertiary structure prediction. However, you can also predict secondary structure, protein-ligand structures (docking studies) and quaternary structure.

I need a layman's explanation of "protein secondary structure assignment". Adding a diagram or picture would be much appreciated.

Secondary structure refers to the conformation of the main chain in a protein, and the hydrogen bonds between the carbonyl oxygen and the amide hydrogen that are observed. The example below shows that the two secondary structure elements called beta sheet and alpha helix differ in the conformation of the main chain and in their hydrogen bonding pattern.

Source: https://commons.wikimedia.org/wiki/File:Alpha_beta_structure_(full).png
To assign a secondary structure based on an atomic model, you would measure the main chain torsion angle and deduce hydrogen bonds from distances of hydrogen bond donors and acceptors. Often, the secondary structure assignment is shown juxtaposed with the primary sequence and sometimes with other annotations. Here is an example screen shot. The highlighted Cys30 is part of a helix (red shade) and makes a disulfide bridge with Cys115 (see additional annotation).

Source: https://www.rcsb.org/sequence/1LYS
Different software packages usually agree on the presence of helices and sheets, but sometime have small differences in the exact start and end of the secondary structure elements (see e.g. here). It is also possible to assign secondary structure from electron density or cryo-EM maps, see e.g. here.

Is "Assignment" a part of the "Prediction"-process?


text:
Does amide ion prefers to act as a base than a nucleophile?

In this question, I think that amide ion, being a good nucleophile should attack on the terminal carbon through  $\mathrm{S_N2}$ mechanism to form compound (c), but the answer given is (b), and the solution:

But I am not able to understand the kinetic factors that favour the amide ion to act as a base rather than a nucleophile. I think it is much easier for a strongly nucleophilic amide ion to act as a nucleophile through a one step $\mathrm{S_N2}$ reaction (polar aprotic solvent will also favour it), rather than first acting as a base and then the carbanion acting as a nuceophile. The $\ce{C-Cl}$ bond is polar, which also increases the chances of an $\mathrm{S_N2}$, but the $\ce{C-H}$ bond is not that polar.
So, my question is, will (b) be the major product, and if yes, than what are the kinetic factor that favour it in comparison to (c)?
(Note: The reaction is assumed to be under kinetic control.)

Your mistake here is to think of amide anion as strongly nucleophilic. It is a very strong base and the protons next to the sulfonyl group have a pKa ~30. It is also poorly soluble in DME

Please note that it's written $\mathrm{S_N2}$ $\mathrm{S_N2}, not $\ce{SN^2}$ $\ce{SN^2} (the latter means sulfur and nitrogen squared).

In general, acid base reactions are very fast. See https://www.masterorganicchemistry.com/2012/05/21/acid-base-reactions-are-fast/

I second what @mcole said. In general, acid-base reactions are faster than nucleophilic substitution. Which is also the reason why Grignard reagent cannot be used for nucleophilic addition on aldehydes/ketones which have acidic hydrogen like $\ce{-OH}$, $\ce{-COOH}$ and so on.

After acid base reaction has occured, the substitution by carbanion will be faster than amide ion since intramolecular substitution is faster because of greater chances of attack. Am I right?


text:
Does 1,4-dimethylcyclohexane have a chiral center?
My textbook says that 1,4-dimethylcyclohexane (cis or trans) does not have a chiral centre, since two groups on the tertiary carbon (that are the part of the ring) are identical, and thus the compound cannot be called a meso compound. But it also says that it has two stereogenic centres (C-1 and C-4), since interchange of groups around them creates a geometrical isomer.
The IUPAC definition of chiral center is:

An atom holding a set of ligands in a spatial arrangement which is not superposable on its mirror image. A chirality centre is thus a generalized extension of the concept of the asymmetric carbon atom to central atoms of any element, for example N+abcd, Pabc as well as Cabcd.

This definition of chiral center is also equivalent to (for coordination number 4) saying that if exchange of any 2 ligands on the carbon atom creates a stereoisomer, then that carbon is chiral center. So according to this, 1,4-dimethylcyclohexane has two chiral center and can be called a meso compound.
But my textbook says the opposite. So, my question is, which one is correct?
(My basics regarding chirality and stereochemistry is clear but the confusion in this question arose as I think that my textbook considered that since the two groups on the tertiary carbon seem to be identical, so it is not a chiral center, but I think it is wrong.)

There are only 2 stereoisomers here, RR and SS are identical, and RS and SR are identical.

@NisargBhavsar if you are able to define R and S, then the carbon must be chiral, and that's what my question is about.

No time for an answer, but see pseudo-asymmetric centre. Google, or searching on SE, will throw up many descriptions.

pseudo-asymmetric centres are created when 2 of the 4 components of a carbon are same structurally but have different chirality correct me if I am wrong, @orthocresol but I don't think "pseudo-chirality" applies in this particular case.

@napstablook That definition is correct and applies precisely here. To go into why would be way beyond a comment, but in short, you need to construct the full digraph which involves "breaking up" the ring: there is a similar example in https://chemistry.stackexchange.com/a/35511/16683.

I would still argue that that definition is improbable here keeping in mind that both the carbons would be pseudo-asymmetric, I think the condition clearly wants to apply chirality on other carbons in 'expanded structures'  rather than depend on "pseudo-chirality" of another carbon to call it pseudo-asymmetric. But it is entirely possible that I am misinterpreting the definition.

@NisargBhavsar: The descriptors should be r/s, and not R/S, because both centers are stereogenic and achirotopic.

As it is said in the definition of IUPAC, for a carbon atom to be chiral it should be attached to 4 different groups:

In case of 1,4-dimethylcyclohexane, you will see that the two groups on the right and on the left are identical. So in this case it's not Cabcd, it's Cabbc. If you alter one group (let's say the one on the right), these groups will no longer be identical, so you will have 4 non-identical groups attached to a carbon atom, and this atom will become a chiral centre.

Meso-compounds are compounds that have chiral centre but they also have a plane of symmetry so that they are still superimposable with their mirror image. Following is the example of a meso-compound from McMurry's book (Ref.1):

So a meso-compound must have a chiral centre by definition, but if a compound has a chiral centre, that doesn't necessarily mean that it's a meso-compound. And the answer to your question is no, 1,4-dimethylcyclohexane doesn't have any chiral centre.

References:

John E. McMurray, In Organic Chemistry, Ninth Edition; Cengage Learning: Boston, MA, 2016, pp. 133 (ISBN-13: 978-1-305-08048-5).

Keep in mind that you should provide all details of the references. Look at my editing as an example, which is in ACS format.


text:
Claisen condensation of methyl benzoate and ethyl propionate
I am planning to preform the Claisen condensation between a benzoate ester (methyl benzoate is the only one I can get) and a propionate ester (ethyl propionate is my only option). The idea is to prepare this ketoester:

Will this work with ethyl instead of methyl propionate?

If ethyl propionate is your only option, then your product is ethyl ketoester. In the scheme, you have used methyl propionate and got methyl ketoseter. Yes, it work with ethyl esters as well, but there is a possibility to get transesterification.

What solvent is this in? Run it in MeOH and you will get the methyl ketoester

To be honest, it doesn't matter what the ester is. Well, my stoichiometry will be a bit off, but I am just hydrolyzing the ester anyway.

And I will likely be using MeOH.

The ester is likely to decarboxylate after hydrolysis. Is that your intention?

It is my intention; I plan to make propiophenone


text:
How do the terms in a rate law relate to effectiveness and number of collisions of reactants?
In chemical kinetics, we use rate law to find the instantaneous rate of reaction.
Let us consider a simple reaction
$$\ce{aA + bB -> products}$$
where the rate law is
$$\mathrm{rate} = k[\ce{A}]^a[\ce{B}]^b$$
I think that the rate constant tells us about the effectiveness of collisions for the reaction at a particular temperature; and the concentration terms raised to the powers of stoichiometric coefficients represents the total number of collisions.
Is my reasoning correct? If not, what is the correct explanation?

Sounds about right. Then again, reactions are rarely simple, and the actual rate law might differ from $k[A]^a[b]^b$, but that's another story.

@IvanNeretin I am not sure how concentration terms raised to the powers of stoichiometric coefficients represents the total number of collisions, could you please prove some mathematical approach to figure this out.
Also, I know about the complex reaction and how it differs but I want to know the basic idea behind the rate law thus assuming it is a simple reaction.
Thanks for your clarification.

Rate laws such as this only show how many reactant species are needed to produce the product according to mass balance. They tell us nothing about how the reaction actually happens. The actual mechanism for any reaction has to be determined by experiment. You can imagine that if, say, $a=b=2$  the chance of four solute species coming together at the same small instance and in the same place is so absolutely tiny, that it just does not happen. What tends to happen is that two species form an intermediate and this reacts again and so on. (There are whole books on this topic)

@RodrigodeAzevedo Because it works but I am unable to understand how?

I have mentioned it is a simple reaction meaning elementary.

As others have pointed out, the power terms are not necessarily equal to the stoichiometric coefficients. In a single-step (elementary) reaction - yes, the concentration terms represent the number of collisions that can occur. Think of it as probability - the probability of 2 species of A coming together at a spot is proportional to $A^2$.

@TRC but the probability is always less than 1 but concentration can be anything.

I'm saying it's proportional to the concentration. The proportionality constant gets included in the rate constant.

Yes I know that but I need a mathematical proof for the same.

Try examining the Arrhenius Equation and see if that gives you more information about the rate constant.
In general, the rate of reaction (in terms of the reactants) is a function of how many successful collisions occur per unit time. There are also unsuccessful collisions. You can also write the rate of reaction in terms of the products. Since the products are not always 1:1 with the reactants, this wouldn't necessarily equate to the number of successful collisions.
In gas phase reactions, inert species can influence the rate of reaction. Since they aren't colliding with the reactants to form the products, the concentrations can't always be equal to the number of collisions.

Thanks but need a more clear explanation

Your reasoning is not correct. The concentration terms raised to the powers of stoichiometric coefficients do not represent the total number of collisions. An inert species does not participate in the reaction but can be included in the rate law. Therefore, the concentrations raised to the stochiometric coefficient can't possibly represent the number of collisions. I think you are on the right track with the rate constant. You should read about and examine the Arrhenius Equation, I think this may give you more insight about the rate constant.

Refer to this article from chem.libretexts.org for more information about collision theory.

I got the concept with the rate constant but can't understand what concentration term really means.


text:
Linear algebra and Hess's law?
I am a high school student taking chemistry who was tasked to solve a large set of Hess's Law problems. Naturally I tried to find a way to get my computer to carry out this repetitive piece of work for me.
I tried to model the problem mathematically and figured out that Hess's Law problems basically had me finding the correct linear combinations.
Suppose we have a problem like this
\begin{align}
\ce{4 NH3 + 3 O2 &->6 H2O + 2N2} &\qquad &\Delta_\mathrm{r}H_1 = \pu{-1516 kJ/mol} \tag{1} \\
\ce{2 H2 + O2 &-> 2 H2O} &\qquad &\Delta_\mathrm{r}H_2 = \pu{-572 kJ/mol} \tag{2} \\
\end{align}
And our goal is to try to find the enthalpy of formation of ammonia. How would I express this as a matrix and solve it to find the linear combination. I assume we have to use the row reduce echelon method here. Just not sure how.

Searching for  linear algebra tutorial gives a lot of interesting links.

A generally useful and efficient method is the Gauss elimination method, getting the triangular  matrix. Many software platforms, including Excel worksheet  functions, provide matrix and vector operations to directly solve the set of n linear equations of n variables. But for n=2, the manual substitution approach is easy.

It gets set up as the part of the Gauss' algorithm itself. You nulify the numbers below the diagonale by substrating (multiplied ) other rows.

Formulate the equation of the Hess law for each reaction and then transform it to the algebraic equation set.

Both. Coefficients for the same reaction on the same row of A. Coefficients for the same variable ( unknown Delta H ) to the same column of A. Absolute terms of f(x)=0 to the separate column b.

Maybe this video will help you. https://www.youtube.com/watch?v=BvcX1n35U5A The resulting system can be easily solved by inspection or with the help of matlab, maple. You could also make a small program to do everything.

To convert this into a generic linear algebra problem you'd rewrite it in the form $Ax=b$ where $A$ is a matrix of stoichiometric coefficients of size $m \times n$; $x$ is a vector of length n of unknown fitting parameters (the heats you want to determine) and $b$ a known vector of length n.
For the example problem the set of equations can be written in matrix form as
$$\left[\begin{array}{rrrrr} 0 & -4 & -3 & 6 & 2\\ -2 & 0 & -1 & 2 & 0 \end{array}\right]\left[ \begin{array}{l} \Delta_\mathrm{f} H(\ce{H_2(g)})\\\Delta_\mathrm{f} H(\ce{NH_3(g)})\\\Delta_\mathrm{f} H(\ce{O_2(g)})\\\Delta_\mathrm{f} H(\ce{H_2O(l)})\\\Delta_\mathrm{f} H(\ce{N_2(g)}) \\  \end{array}\right] = \left[ \begin{array}{c } \Delta _\mathrm{r}H_1\\\Delta_\mathrm{r}H_2  \end{array}\right]$$
Since many of the terms in the vector of heats on the left-hand-side are known and equal to zero this can be rewritten as
$$\left[\begin{array}{rrlll} -4 & 6 \\ 0 & 2 \end{array}\right]\left[ \begin{array}{c } \Delta_\mathrm{f} H(\ce{NH_3(g)})\\\Delta_\mathrm{f} H(\ce{H_2O(l)})   \end{array}\right] = \left[ \begin{array}{c } \Delta_\mathrm{r} H_1\\\Delta_\mathrm{r} H_2  \end{array}\right]$$
(alternatively the original equation can be written as a $5 \times 5$ array by including additional trivial equations defining the heats of formation of the diatomic gases).
Solution of the above system should be easy, for instance in MATLAB:
A = [-4 6; 0 2];
b = [-1516 -572]';
x = A\b

The problem can of course be solved by inspection. The second equation can be solved for $\Delta_\mathrm{f} H(\ce{H_2O(l)})$, which can be inserted into the first to solve for $\Delta_\mathrm{f} H(\ce{NH3(g)})$.

 >>> import numpy as np; A = np.array([[-4, 6], [0,2]]); b = np.array([-1516, -572]); np.linalg.solve(A,b)```


text:
Why is the concentration of gases denoted in form of partial pressure?
In most of the chemical laws of chemistry involving concentration of products or reactants like equilibrium constant, reaction quotient, Henry's law, Nernst equation etc, I often see a similar note stating concentration of gas to be taken as partial pressures of that gas or something similar to that. So, Why is the concentration of gases denoted in form of partial pressure ?
Does it have any relation with Dalton's law?
I don't know why this equation has been closed and it is not the duplicate of this question Please review again.

My guess would be that pressure is much easier to measure for a gas. Of course once you have the pressure, temperature and ideal gas equation would give you the concentration too. Also, concentration term, if replaced by partial pressure, will give you $\ce{K_p}$, not $\ce{K_c}$. Both terms are different.

Considering gas pressure is much more common than gas concentration. There is the relation for ideal gases $p=cRT$. The point is, many gas behaviour quantitatively depends rather on partial pressure than on concentration.

In addition to what TRC states, it can be mentioned that, in thermodynamics, concentration is not a fundamental parameter. Pressure is fundamental : it exists in the definition of the work, which is the very beginning of thermodynamics. $K_\mathrm{p}$ can be directly obtained , and has the correct dimension. On the contrary,$K_ \mathrm{c}$ is indirectly obtained, and the concentration have to be divided by a reference value of concentration

Your question is based on a false premise that concentrations and partial pressures are interchangeable or identical entities, which is not true. Fundamentally different physical quantities cannot "take form" of one another. The question I marked as a dupe is not a literal duplicate, but it explicitly reveals this flaw and the accepted answer by Martin can be used to answer this question of yours as well.

Are you familiar with the derivation of the equation for the equilibrium constant of an ideal gas phase reaction in terms of the standard change in Gibbs free energy?


text:
How the concentration of anodic half cell increases in galvanic cell?
According to my textbook

Concentration of anodic half cell increases and that of cathodic half decreases when two electrodes are joined by a wire.

I have also read that it is connected with a Salt bridge which prevents the flow of the electrolyte ions and supplies the required cation or anion (from the inert electrolyte in the salt bridge) and overall its electrically neutral . Then how they can state that concentration increases in one half cell only ,that is both cells are provide with sufficient ions ? Please correct me if I am wrong somewhere in understanding the concept.

Look at the Daniell cell here. The zinc metal anode is oxidized to zinc ions, so the zinc ion concentration in the anode reservoir increases as the cell provides electrical power. Copper ions get reduced at the cathode, so the copper ion concentration is reduced in the cathode reservior.

@Ed V okay sir , I understood . I was considering anions too , I thought that in cathodic cell there will be accumulation of anions which will increase the concentration of anions there .

Cations and anions must be always balanced. Any tiny disbalance would cause strong potential gradient leading to rebalancing.


text:
How to sustain iodine vapor in a big container (V=200 L)?
I am lost in a quest on how to have a big container in which I can have iodine vapor at a known temperature.
We intend send iodine into an air flow, and observe fluorescence of iodine in this flow. This method is known as planar laser induced fluorescence. The challenge lies in the fact that, we need to be able to send iodine above a certain concentration for about several minutes. The idea is to have a large volume of iodine vapor ready, so that we do not run out of iodine vapor during the experiment and we do not need to wait too long between experiment sessions for iodine to sublimate.
In order to do that, I came up with the idea that we will have a big volume of container, and we will heat this container. The bigger the container, the more iodine vapor mass will be ready for us to send to the experimental chamber. Otherwise, if we had a smaller volume, iodine would deplete quicker and I would have to wait for sublimation.
So, with this purpose, I want to do be able to get a container, and I am trying to come up with a way to heat it up. What kind of a barrel, or container of some other type would be suitable for me, and how could I heat it so that it is around 80-100 C?

Your message is not clear. You speak of an air flow. Well ! Do you need a heated container with motionless air + iodine or will you produce an air jet ? On the other hand, iodine reacts with most metals, producing iodides. Have you taken this into account ?

Well, the idea is to first heat container long enough so that there's saturated iodine in the container. Then I will generate flow into and out from the container, there will be an air jet.
And about the reaction with metals, yes I thought about it, that's why I was thinking of a plastic barrel.
Have I made it clear or is something still missing?

You should describe the expt. in more detail. What sort of air pressure, what sort of I2 concentration, flow rate, expanding into vacuum or not etc. It would be better to do this dynamically, pass gas (air or O2 or N2?) over reservoir of solid iodine at  temp to get vpr pressure you need.

Why don't you pre-heat the air rather?

Usual plastics are good solvent to iodine $\ce{I2}$. So I would not recommend using  a plastic container for iodine vapor. The best container would be in glass, as metals are oxidized by iodine vapor, and plastics absorb iodine.


text:
Variation of vapour pressure with application of external pressure
Many times study of fact that with pressure of a substance only depends on the temperature and nature of the substance recently I study about dew point and bubble point.
Then I came up with this question to which I didn't find any on Google :-
Suppose we have a system as shown in figure:

In the above system the piston is touching the liquid surface that is there is no vapour currently.
Now we slowly move the piston up, keeping the temperature same and the vapour starts forming in the container.

Now the question is how will this pressure applied by the vapours of the liquid so form (i.e. the vapour pressure at equilibrium) will vary with the change in external pressure at the same temperature ?
Will the vapour pressure be equal to the saturated vapour pressure of that substance at that temperature  ? or if it changes will it not contradict the statement used in almost all articles over the net teaching vapour pressure , stating that vapour pressure of a substance only two points on temperature and nature of that substance?

As the approximation, the saturated vapour pressure does not depend on the total pressure, only on the substance and temperature.  (*)
If there is nothing but the vapor in the gaseous phase under the piston then the amount of vapor and its pressure accomodate to external pressure. (**)

If the external pressure is kept greater than saturated vapor pressure for given temperature, all vapour will eventually condense and released heat will dissipate, assuming constant temperature is maintained.
If the external pressure is kept lower than saturated vapor pressure, all liquid will eventually evaporate and needed evaporation heat will be extracted from surrounding.
If the external pressure is by chance equal to the saturated vapor pressure, liquid and vapor will maintain the equilibrium with the same vapour pressure as the external one.

The piston stays in rest if and only if the net force on it is zero, therefore if actual vapour pressure ( saturated in equilibrium ) is equal to the total external pressure caused by external, non-vapor based forces acting on the piston. Otherwise it starts moving toward the lower pressure.
There is assumed the lifting of the piston is slow enough for liquid to have time to establish saturated vapour pressure at any time. Fast piston lifting can end with higher volume and less than saturated vapor pressure, which may not be possible to be topped again due lack of remaining liquid.

(*) - Finer analysis tells us that for gas+vapor mixtures, liquid saturated vapor pressure increases little with the total pressure. This is proportional to liquid molar volume and such an increase is usually negligible. Unless temperature and pressure are near the liquid critical point where it can make the difference.
(**) - Atmospheric pressure is just one of possible sources or components of external pressure on the piston. Atmospheric pressure can be

the only part of external pressure, like in idealized scenario with a free moving, massless, frictionless  piston
missing, e.g. if the system is placed in vacuum
a part of total external pressure, like if the weight caused by the piston mass cannot be neglected and/or if explicit mechanical external force is applied.

What is the relation between vapour pressure and the external pressure ? Are they equal ? Can we apply Boyle's law here ?

Does the piston move if they are not equal?  Boyle's law is for ideal gases, and as approximation for real "permanent gases. It is not applicable for vapor, if increased pressure above saturated vapor pressure causes condensation. Even before condensation starts, there is already big deviation from this law.

What if the question states to consider vapor as ideal gas will Boyle's law be valid then ? I don't think that it is necessary for piston to move when the pressure is unequal and the reason I'm saying that is because of a particular case in which we move the piston all the way up until all the liquid become vapour at that point the sum of external pressure and vapour pressure should be equal to atmospheric pressure ( maybe ) ?

Such assumption does not make sense, ideal gas never condenses, what is the direct opposite to being a vapor.

No, external pressure being sum of atmospheric pressure and eventually additional pressure would be equal to saturated vapour pressure, assuming lifting the piston is slow enough. (Fast moving would end with higher volume and less than saturated vapor pressure.)

So you say vapour pressure > atmospheric pressure

No, I say you should read properly and think independently. In the piston scenario, vapor pressure relates to external pressure. Atmospheric pressure is just one of possible sources of external pressure on the piston. It can be the only part, can be missing or part of total external pressure.


text:
Can phosphoric acid be used to substitute sulfuric acid in the synthesis of thymolphthalein?
I am planning to synthesize thymolphthalein, a pH indicator. I am wondering if phosphoric acid can be used to substitute sulfuric acid in the preparation. I know it can catalyze Friedel-Crafts reactions, which are a part of the synthesis. Any advice would be appreciated.

It would be appreciated if you could attach an image of the normal synthesis involving sulphuric acid. Would help those who are not aware of the compound. Thank you :)

If you use polyphosphoric acid (PPA) I think it will work. PTSA will also probably work

What is your reasoning for, what is your reasoning against the substitution of the strong Bronsted acid $\ce{H2SO4}$ by phosphoric acid (i.e., review of the literature)?  By the way, the waxy polyphosphoric acid may be considered, too (an elder review, a current application).

I do chemistry in a workspace with wooden benches, and sulfuric acid would eat through them quickly. Thanks for suggesting those, I can make or get both of those.

But the question is: would straight phsophoric acid work?

@Jacob Are you asking about tthe commercially avialable 85% phosphoric acid solution or the 99% crystalline material? The 99% material might do it, I doubt the 85% solution would.

The phenol will be more reactive than a normal benzene, by being a phenol + the alkyl groups in addition; this with using an acid anhydride may suggest less need (quantity / strength) of catalyst / less harsh reaction conditions here.  However, consider the  concurrent esterification (phenol-OH + carboxylic acid), too.

The reaction is a friedel crafts reaction, and it forms water. The reaction to form the thymolphthalein is irreversible, but esterification is. Also, phenols are sterically bulky, so this is not favored.

Thanks, all! Polyphosphoric acid does indeed work to catalyze this reaction. I was successful in making it.

@Jacob If you find your answer, kindly self-answer it. It completes your post.

I used polyphosphoric acid and it worked.


text:
Find the ion with mass number 37 if the ion contains 11.1% more neutrons than the electrons

An ion with mass number 37 possesses one unit of negative charge. The ion contains 11.1% more neutrons than the electrons in its nucleus. An ion with a mass of 37 has % negative charge compared to % positive charge.

With the aid of Periodic table, I doubt if the given ion is Chloride $Cl^{-}$, but I am not sure. My attempt is as follows:
Let the ion be $X^{-}$, and let the number of electrons be $x$, then the number of protons will be $x-1$ as the ion carries one unit of negative charge.
Now I know that $A=(2 \times Z)$ if $Z$ is even and $A=(2 \times Z) + 1$ if $Z$ is odd, where $Z$ and $A$ are atomic number and mass number respectively. Therefore, $x-1=(2 \times Z) + 1$.

Let the number of electrons in the ion carrying a negative charge bex
Then Number of neutrons present
$$
\begin{array}{l}
=\mathrm{x}+11.1 \% \mathrm{of} \mathrm{x} \\
=\mathrm{x}+0.111 \mathrm{x} \\
=1.111 \mathrm{x}
\end{array}
$$
Number of electrons in the neutral atom $=(x-1)$
Number of protons in the neutral atom $=(x-1)$
Therefore,
$$
\begin{array}{l}
37=1.111 x+(x-1) \\
2.11 x=38 \\
x=18
\end{array}
$$
Therefore no of protons $=$ atomic no $=x-1=18-1=17$
Therefore symbol of the ion is ${ }_{17}^{37} \ce{Cl}^{-1}$

You are overly complicating the solution with even or odd atomic numbers.
It is by definition that $Z+N=A$ where $A=$ atomic mass number, $Z=$ atomic number, and $N=$ number of neutrons. Thus,
$$Z+N=37 \tag1$$
If the number of electrons in the ion is $E$, then
$$E=Z+1 \tag2$$
because the ion has a negative charge (otherwise $E=Z$ for a neutral element).
And,
$$N=(1+0.111)E \tag3$$
This is from the condition.
You have three equations and three variables. If you solve the three linear equations, you will know the value $Z$. This defines the element. Look, it up in the periodic table. $A$ is already given and you have a negative charge. Remember mass number does not define the element, only $Z$ does.
Your intuition about chlorine is right. All you have to do now is to prove it.


text:
Can a complex reaction have more than one rate determining steps?
I was solving some questions related chemical kinetics and encountered one question with multiple correct answer , the question is :

For a complex (multistep) reaction,
(a) the overall rate of reaction is the rate of slowest step.
(b) the overall molecularity has no significance.
(c) molecularity and order may or may not be same.
(d) the number of rate determining steps may be more than one.

The question have all the four options as correct answer. There's no doubt about first three options, but about the fourth one I want to ask ,
How a complex reaction have more than one rate determining steps? And if it has then how the overall order and molecularity of the reaction may be given as for complex reation we look at the slowest step and with the help of slowest step we decide the order and molecularity .
Then with multiple RDS how it is possible ?

I am not really sure, but maybe its possible that two steps are equally or similarly slow? You can identify a step as rate determining only when its rate is appreciably slower than the other steps.

@S R Maiti So we can take any of the slowest step to determine the order and molecularity ?

No, I think if there are multiple steps which are slow, then you would have a very complicated rate equation. Sometimes you can derive these by steady state approximation. Other times it might not even be possible to write a rate law. For example the decomposition of HBr has a very weird looking rate law with 5/2 exponents and a denominator etc. The order would be undefined in that case. And molecularity is not even defined except for elementary reactions, so there is no question of molecularity for multi-step processes.

There is always one step that will be the slowest even if the difference is miniscule. By definition there can only be one slowest step. I think someone above mentioned that there could be a case where two steps have identical velocities but it's unlikely.

For a sufficiently complicated overall reaction which hides a large interconnected reaction network, the concept of a "rate determining step" may effectively not exist (at least not in any way that we might be familiar with from more simple cases), such that sentence (d) kind of loses meaning. I briefly talk about it in this past answer and the second half of this one.