text:
Mechanism of decarboxylation of alpha-keto carboxylic acid
What is the probable mechanism for the following reaction?

$\alpha$-Keto acids on heating with conc.$\ce{H2SO4}$ undergo decarboxylation to give monocarboxylic acids:

Also, which of the two carbons (the carbonyl carbon and the carboxylic carbon) is getting eliminated here?
Note: I tried to apply the mechanism for the decarboxylation of $\beta$-keto carboxylic acids in this case too, but couldn't work out one.

Btw which book is it? The number of oxygen atoms on either side is not balanced. So it's definitely not a simple decarboxylation - it involves oxidation too.

It's Organic Chemistry (II) by O. P. Tandon, GRB Publication.

Hi @Nilabja. I voted to reopen your question. Make sure to add your efforts while making the initial draft of your question so that your question won't get closed in the first place and you don't have to go through the effort of making changes and making comments explaining the changes and expanding the comment thread chain. That being said, your previous question was also insta-closed. I don't think the question is bad, so would you like to make some changes?

@NilayGhosh Thank You for your help, the question has been reopened. I would keep in mind the suggestions for further questions. Also, as you asked, I have got satisfactory solution to my problem in the previous question.  So, I wouldn't like to make any more changes to it, at least for now.

All your questions are reall y hard lol

There are two characteristic reactions of $\alpha$-keto acids with sulphuric acid, but none of them produce acetic acid and carbondioxide as the products as indicated in the reaction given in the OP's textbook. Instead, pyruvic acid (the simple $\alpha$-keto acid shown in the textbook reaction) is easily decarboxylated with warm dilute sulphuric acid to give acetaldehyde:
$$\ce{CH3-C(=O)-CO2H ->[dil.H2SO4][\Delta] CH3-CHO + CO2}$$
If the reaction is in conc.$\ce{H2SO4}$ as given in the sought reaction, the products are acetic acid and carbonmonoxide:
$$\ce{CH3-C(=O)-CO2H ->[conc.H2SO4][\Delta] CH3-CO2H + CO}$$
Mechanism of the first reaction is given in here as follows:

Although the mechanism of this reaction is uncertain yet it occurs only in $\alpha$-keto acids, it is  suggested that the $–I$ effect of the $\alpha$-carbonyl group plays an important part in this elimination. You may also find the mechanism of the second reaction in the same site as well:

However, it is also possible for $\alpha$-ketocarboxylic acid to undergo decarboxilation to give corresponding carboxylic acid and $\ce{CO2}$, but need redox conditions similar to pyruate decarboxylation in biological conditions:
$$\ce{CH3-C(=O)-CO2H ->[Oxidation] CH3-CO2H + CO2}$$
For example, this reaction will undergo in the presence of $\ce{H2O2}$ and following mechanism is given (Ref.1):

References:

Antonio Lopalco, Gautam Dalwadi, Sida Niu, Richard L. Schowen, Justin Douglas, and Valentino J. Stella, "Mechanism of Decarboxylation of Pyruvic Acid in the Presence of Hydrogen Peroxide," J Pharm Sci. 2016, 105(2), 705–713 (DOI:  https://doi.org/10.1002/jps.24653).


text:
Dehydration of aldol
In my book it is written that to dehydrate an aldol the product of aldol condensation is heated in an iodine environment. But what is the role of Iodine here and how does it catalyse the dehydration of the aldol?

The aldol is not the product of an aldol condensation but rather the unsaturated carbonyl compound is the result of a condensation. A condensation requires the loss of some small molecule such as water or alcohol. The iodine likely generates a catalytic amount of HI by oxidation of some alcohol in the aldol substrate. Acid catalyzed loss of water occurs to afford the aldol condensation product.

source: Stack Exchange

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

sort: votes


text:
What can be used instead of magnesium to make silicon from sand (in lab)?
I found this post talking about making silicon from sand. They mixed magnesium with sand and then burn them then it create silicon and magnesium oxide. And my question is what can we use instead of magnesium in this experiment (something easier to find at home) to make silicon from sand? Thanks in advance!
P/S: I want to figure out other ways to do this before buying magnesium online :) Just want to try this at home with easy-to-find (at home) chemicals.
This post shows how to make magnesium from magnesium oxide (which I can found in my batteries) but it need silicon to do that :)

The easiest way would be to get already made silicon, from old chips or wafer waste from chip industry. As there is no easy way to produce silicon using home resources.

I just checked ebay and you can buy 5 kg of silicon for 95 USD. A lot safer than getting badly injured trying to make it at home.

If you really are interested in making your own silicon, Wikipedia offers several methods used industrially. None is easy; all require high temperature operation, with furnaces more likely to be seen in an industrial environment than in a typical chemical laboratory, let alone at home.

Carbothermal reduction. Silicon dioxide (silica) is reduced by elemental carbon (coke) in an electric arc furnace. Excess silica is used to prevent contamination with silicon carbide which can also form. Usually this process is not used to make silicon as such, but done with iron also included to make ferrosilicon, used in steelmaking. The ferrosilicon may be used to reduce magnesium oxide (also not trivial in a laboratory setting), as the iron does not interfere with that reaction.

Aluminothermic reduction. Silica is reduced by aluminum metal, as described by Maurice in his answer. Silicon produced in this way is suitable for use in chemical processes. The reaction given by both WP and Maurice is likely a simplified description, as ternary aluminum-silicon oxides might also be formed. This makes the production of silicon more favorable thermodynamically, but takes up some of the silicon in the ternary oxides. As Maurice points out, the aluminum has to be molten. So as with the carbothermic reduction, you need a furnace capable of achieving high enough temperature.

Special methods are needed to generate semiconductor-grade silicon. These actually do not use silica as the silicon source; instead, volatile compounds that can be pre-purified by fractional distillation, such as $\ce{SiCl4}$, are employed. Additional purification by zone melting is needed to get the purity needed for semiconductor use.

Silica $\ce{SiO2}$ can be reduced at high temperature into silicium with aluminum powder, in a reaction like : $\ce{4 Al + 3 SiO2 -> 2 Al2O3 + 3 Si}$. This reaction is exothermic by $\pu{\Delta H = -185 kJ mol^{-1}}$ and exoergic by $\pu{\Delta G° = -178 kJ mol^{-1}}$ which becomes $\pu{\Delta G = -165 kJ mol^{-1}}$ at $\pu{1000 K}$.  It was sometimes used to prepare metallic silicium. But this reaction is rather difficult to start, as it occurs only with liquefied aluminum, so that the mixture must be pre-heated to the melting point of aluminum ($\pu{659°C}$ )

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


text:
Conjugation and resonance
I'm new to Organic chemistry, and I'm just starting to learn about resonance. I learnt that resonance and conjugation are related - if a system is conjugated, with alternating sigma and pi bonds/pi-sigma-lone pair/pi-sigma-free radical/pi-sigma-p orbital/lone pair-sigma-positive charge arrangements in the same plane, we can derive resonance structures by alternating the bonds/radical/charge between atoms. My question is - can resonance occur without conjugation, and if so, what are the rules governing it?

A good example would be a polar covalent bond; such a bond exists as a resonance hybrid of totally covalent and totally ionic canonicals; switching on and off the (line in a Lewis stucture) respectively.
For example, the carbon tetrafluoride molecule* is covalent but has significant contributions from an ionic structure involving "a carbon(IV) cation electrostatistically bound to four fluoride anions" in tetrahedral symmetry- a "no lines, just ionic bonds" canonical.
*which is non-polar but still has polar covalent bonds in it

So basically resonance can even occur on the types of bond (ionic or covalent) as well as  pi-bond and lone pair structures? And are there any other arrangements in which something similar can happen?

Yes it can- both theoretically and empirically.

Are there any other arrangements in which something similar can happen?

OP - Beware that it is difficult to pose a limit on the extent of the delocalisation and or the number of limiting forms, in principle. Then we use criteria. Also, and in reverse, the fact that a system is conjugated does not strictly implies delocalisation. At least you can conceive comformers right tilted so that delocalisation does not occur.

I think this is stretching the concept of resonance a bit thin, but definitely one point of view. It is rather important to understand the points that arise through this view, especially Valence bond methods, and those seemingly opposite to that view, e.g. molecular orbital theory.// Anyway, there is a typo, cation twice, you might want to fix.

@Martin-マーチン fixed the typo you mentioned


text:
What does the change in Gibbs Free Energy refer to?
Usually change is indicated as the difference between some initial and final state. However, for ΔG there seem to be multiple definitions and I am a bit confused. For example, at equilibrium, ΔG = 0 but I am confused as to what the initial and final state between which this 'change' is measured since equilibrium is just a single point. Could someone please explain exactly what this change is measured for? I am only in high school so I haven't learnt more complex thermodynamics so I would appreciate it if it's possible to include a relatively basic explanation.

At equilibrium, G function has minimum with zero gradient. Therefore infinitesimal or at least small deviations from equilibrium lead to zero or negligible change of G.

Have you checked the many posts asking similar questions? For intance: https://chemistry.stackexchange.com/questions/49145/intuition-on-gibbs-free-energy?rq=1 , https://chemistry.stackexchange.com/questions/58138/gibbs-free-energy-change-at-equilibrium?rq=1 or https://chemistry.stackexchange.com/questions/4/gibbs-free-energy-minimum-or-zero

More generally: https://chemistry.stackexchange.com/search?q=gibbs+free+energy+of+reaction

Planckton seems to forget that the Gibbs energy is a value that changes with concentration. Let's consider a reaction $\ce{A + B <=> C + D}$. In their pure state, the Gibbs energies of formation of $\ce{A, B, C}$ and $\ce{D}$ have the values $\pu{G°(A), G°(B), G°(C), G°(D)}$, resspectively.
Let's start from $1$ mole of the pure substances $\ce{A}$ and $\ce{B}$, without any $\ce{C}$ and $\ce{D}$. The reactants are not yet mixed. Their total free energies are $\ce{G°(A) + G°(B)}$. In a first step, these substances must be mixed, even without any reaction. This simple mixture changes their Gibbs energies, because this mixture is spontaneous and not reversible. In the mixture, before the reaction starts, the Gibbs energies of $\pu{A}$ and $\pu{B}$ are $$\pu{G(A)_{t=0}} < \pu{G°(A)}$$ $$\pu{G(B)_{t=0}} < \pu{G°(B)}$$. At this point, there is no $\ce{C}$ and no $\ce{D}$ in the mixture, and their Gibbs energies are zero. $$\ce{G(C)_{t=0} = G(D)_{t=o} = 0}$$ At the beginning, before the reaction starts, the variation of Gibbs energy due to the reaction is $$\ce{\Delta G_r,_{t=0} = G(C)_{t=0} + G(D)_{t=o} - G(A)_{t=0} - G(B)_{t=0} = - G(A)_{t=0} - G(B)_{t=0} < 0}$$  As $\ce{\Delta G_r}$ is negative, the reaction can proceed. So the reaction starts, and both [$\ce{A}$] and [$\ce{B}$] decreases. Simultaneously, [$\ce{C}$] and [$\ce{D}$], being zero in the beginning, are becoming finite.  But as time in going on, the reaction proceeds and $\ce{\Delta G_r}$ becomes : $$\ce{\Delta G_r = G(C)_t + G(D)_t - G(A)_t - G(B)_t }$$ As a consequence, $\ce{\Delta G_r }$ is still negative, but slightly increasing with time, as the concentrations and the free energies of $\pu{C}$ and $\pu{D}$ are increasing, and $\pu{A}$ and $\pu{B}$ are decreasing. After a rather long time, $$\ce{G(C)_{t=∞} + G(D)_{t=∞} = G(A)_{t=∞} - G(B)_{t=∞}}$$ At this point, $\ce{\Delta G_r = 0}$ The reaction is at an equilibrium.
The most important result is the fact that the equilibrium constant of the reaction $\pu{K}$ depends on the values of the Gibbs energies in their pure state, and not in the mixture. This constant depends on the $\pu{G°}$ values, according to : $$\ce{K_r = \frac{[C][D]}{[A][B]} = G°(C) + G°(D) - G°(a) - G°(B) = \Delta G°_r}$$

Thank you for you response. My confusion seems to arise from whether to interpret ΔG as the change in Gibb's energy of the complete reaction vs. the 'instantaneous' diffierence of the gibb's energy between the reactants and products at any given point in the reaction. For example, when calculating ΔG=ΔH−TΔS is this for the whole reaction at a given temeprature or is this at a single point in the reaction?

@planckton. ΔG is valid for both the whole reaction and at a single point in the reaction. The only difference is the index ° which yields an important change. Without this index (°), ΔG defines any point during the reaction. With this index (°), ΔG° only characterizes the difference between the pure reactants $\pu{A + B}$ and the pure final products $\pu{C + D}$.

Enthalpy, entropy and Gibbs energy are extensive properties and also state functions.

Any difference can also be defined as:
Reactants$\ce{->}$Products; $\Delta_r H = \Sigma H_{\text{Products}} - \Sigma H_{\text{Reactants}}$
In this definition, there is no initial and final states but rather all terms are in same state.
Similarly, the change in Gibbs energy and entropy can be defined as:
$\Delta_r G = \Sigma G_{\text{Products}} - \Sigma G_{\text{Reactants}}$
$\Delta_r S = \Sigma S_{\text{Products}} - \Sigma S_{\text{Reactants}}$
Gibbs energy is defined as, $G=H-TS$. Absolute value of Gibbs energy, i.e., $G_{\text {initial}}$ and $G_{\text {final}}$ cannot be calculated but change in Gibbs energy can be calculated as:
$\Delta_r G=\Delta_r H-T\cdot \Delta_r S$, at constant temperature and pressure
Another way of thinking is by writing the differential form:
$\mathrm dG=\mathrm dH-T\mathrm dS-S\mathrm dT$

More general case:
$G$ is defined in terms of $H$ and $S$, absolute value of both cannot be calculated.
In the equation, $\Delta G_{\text{system}}=\Delta H_{\text{system}}-T\cdot \Delta S_{\text{system}}=-T\cdot \Delta S_{\text {total}}$, the change in Gibbs energy is equal to the product of change in total entropy (system + surroundings) from initial to final state and the constant temperature.
Also note that, $\Delta H_{\text{system}}$ is the total energy change and $T\cdot \Delta S_{\text{system}}$ is the unavailable energy. Therefore, $\Delta G_{\text{system}}$ is the available energy.

@BuckThorn Doesn't that change in composition is reconciled when we are calculating the difference between products and reactants?

You are right, you are defining the Gibbs free energy of reaction which is a property of a given fixed state. But I think this might be improved by explaining more clearly what makes this special compared to some other (discrete) Gibbs free energy change.

No, G=H-TS is a general definition. $\Delta G$ can be defined for any thermodynamic process. The difference between the reaction free energy change and the general free energy is similar to the difference between a force and a potential. The free energy change is a partial molar quantity (or rather a difference in partial molar quantities or chemical potentials).


text:
Why Potassium Bicarbonate has a higher solubility than Sodium Bicarbonate?
SODIUM BICARBONATE SOLUBILITY
Wiki

POTASSIUM BICARBONATE SOLUBILITY
wiki

As According to this answer

Now, note that the hydride ion is "hard", having high negative charge
density. While the carbonate ion is "soft", having a lower negative
charge density.
Applying the "hard-soft acid base" (HSAB) principle, we would expect
the "hard" ions to form more stable compounds with each other and the
"soft" ions to form more stable compounds with each other.
Since the heavier alkali metal ions are softer, we would expect them
to form more stable compounds with carbonate. As for the harder
lighter metal ions, we would expect them to form more stable compounds
with hydrides.

As Bonding of Potassium and Biarbonate ions is stronger than bonding between Sodium and Bicarbonate( Aka Lattice energy) but the hydration of Sodium is higher than Potassium so according to my intuition when we dissolve Sodium Bicarbonate in water as its hydration energy is more it can overcome it small Lattice energy and it will get solvated by water further and get dissolved and the opposite should be the case for Potassium Bicarbonate, but Experimental answer is completely opposite? Can it be explained or it is a anamlous behaviour cannot be explained by current science

:/ I guess you didn't consider such details like molar masses...

Predicting correct solubility from the first principles is far from reality. Nobody can look at structure or a compound and tell you its solubility in water. You need experimental information. All those internet stories are just hand-waving stories. Take them with a grain of salt. No pun intended :-)

Note that both salts do not differ only in the cation size, but also in the ionic lattice structure. Even though both sodium bicarbonate and potassium bicarbonate shows hydrogen bonding in the  crystal structure, sodium bicarbonate forms a polymer and potassium bicarbonate forms a dimer.

Related: https://chemistry.stackexchange.com/questions/85312/fajans-rules-versus-lattice-enthalpy/105032#105032

Basically, the solubility is an interplay of both lattice energy and hydration enthalpy (if you are talking about the enthalpy of dissolution). In both NaHCO3 and KHCO3, the lattice energy (represented by the product of charges upon internuclear distance) the charges are same (+1 and -1) so the differentiating factor is internuclear distance.  But due to the large size of the anion and a relatively small difference in the radii of K and Na, this factor doesn't influence the dissolution energy a lot. So the major difference falls in the hydration enthalpy, which is where NaHCO3 wins

@Poutnik  Can you link a paper where i can study about it please

Try simple Googling, as I found it quickly, having remembered I had read it somewhere before.


text:
Role of 'p' Orbitals in Graphite Carbons
In the graphite arrangement of carbons, if we model them according to hybridization theory, the carbons in graphite are sp2 hybridized. This would mean that one s and two p orbitals hybridize, making 3 hybridized orbitals. This clearly would make the molecular structure trigonal planar. However, 1 electron is left. Wouldn't this electron be found in the remaining p orbital that wasn't hybridize? I couldn't find any information about this so I ask it here. Moreover, if this is the case, would this p orbital of a carbon in a layer overlap with the other p orbitals in the top and bottom layers?
If I said something that is wrong, feel free to correct me. Thanks in advance.

I am nitpicking: Atoms cannot hybridize, only the mathematical representations of orbitals can. Hybridisation itself is not a theory but a description. Hybridisation always follows structure, not the other way around. Hence, the sheets in graphite are (in good approximation) planar, that's why the carbons can be well described with sp² orbitals.

By the obsolete concept of hybridization, all carbon atoms in graphite are sp2 hybridized. A total of three (1 - 2s and 2 p-orbitals) lead to this hybridization. Due to the formation of sigma bonds with axial overlap of one of these sp2 orbitals with another sp2 orbital of another centre, the atoms exist as a planar array of hexagonal carbon rings. The odd electron resides in the remaining unhybridized p-orbital of sp2 carbon centre, perpendicular to the plane containing the three equivalent sp2 hybrid orbitals. The last odd electron is not localized in the unhybridized p-orbital, but is rather delocalized throughout the layer, due to lateral pπ - pπ, overlap, causing conjugation of odd electron, throughout the plane. It is a bit like an ethylene molecule with H-C bonds replaced by C-C bonds over an enormous plane.
(This is a reason for the high electric conductivity of graphite compared to other carbon allotrophs and other non metals).

Unhybridized p-orbitals (of C atoms in adjacent planes) do not overlap end to end.
In this diagram, the distance between the two adjacent layers of graphite ~0.34 nm (order of 10-9) >>>>>>> greater than the longitudinal length of a lobe of the perpendicular p-orbital projecting from a plane - in the order of picometers (order of 10-12), so clearly they are not large enough to overcome the immense distance of seperation to overlap.
Moreover, the odd electrons in them are already in conjugation, by pπ - pπ lateral overlap. This will further retard any cross-layer bonding interaction.

This could end with "indeed graphite is a lubricant as for the planes can slide on each others". Plus 1, it seems useful to OP.

I am downvoting this after I read the first sentence. Hybridisation is not an obsolete concept. It is still a widely used and successful (approximate) mathematical description. In a way it is part of the foundations for (quantitative and qualitative) Valence Bond theory. Also - still the first sentence - atoms cannot be hybridised, so an sp² hybridised carbon simply does not exist. If you are going to reject hybridisation, which I can understand, please get the terminology right.

I am also going to object to that order of magnitude comparison. If a bond is 140 pm according to your graphic, and the distance between the sheets is 340 pm, then that is not three orders of magnitude, it is only triple (or a third). There certainly are orbital interactions between the sheets, that is just the nature of the mathematics behind it.


text:
Parameters of NaHCO3 + citric acid + H2O + NaCl mixture potentially impacting functional efficacy in pest control application?
I have been searching for infomation on this application, but have some specific questions that I have not seen directly addressed in the Google/Stack Exchange posts I've come across thus far; below are some posts I've read through):

Baking Soda + Citric Acid + Water = CO2, how much though? [closed]
Chemical reaction to produce 5 liter of gas [closed]
Why doesn't baking soda react with the solid form of an acid?

I was a physics major in my school days and had to give up on chemistry due to scheduling conflicts, so I'm out of my depth here and unsure if I'm making accurate inferences from what I'm reading.  I'd appreciate if someone can clarify on the two uncertainties I have below:

Does the amount of water only affect the rate of reaction between sodium bicabonate + citric acid, irrespective to whatever ration the pair consist of?  In other words, is it correct to assume that given static values/rations of NaHCO3 and C6H8O7, the gross volume of CO2 produced will remain the same regardless of dilution, given sufficient time and assuming the inputs thoroughly undergo the reaction?
Does sodium chloride interact with the reaction in any meaningful manner?  As this will be one of the two mechanisms for lethality, I want to make certain that misuse (over applying, etc.) will not impact the functional efficacy of the CO2 production.

Thank you!

Too much water can dissolve significant portion of CO2. Both too much or little can cause slow release and CO2 dissipation what may or may not matter. // It is better to use the name than the summary formula for citric acid. As it would be unclear, what substance is represented by the formula, even if there was only one substance with such a formula.  To be used in equation, one can use e.g. a formal formula  $\ce{H3A}$ with 3 acidic hydrogens, explaining It means citric acid.

@Poutnik Since the mixture is meant to be used as a chemical pesticide — more specifically, a mosquito attractant station incorporating a mechanical trap + a chemical lethality agent — the "slow" release might be beneficial.  Are there any established percentages or per-UOM data regarding the rate of dissolution by dilution?  Thanks for the tip regarding the naming versus formulaic conventions.  As I'm wholly unfamiliar with the subject, I'd greatly appreciate it if anyone else who knows what they're doing would edit my post with the correct formatting or vernacular.

Note that it is always good to include in the question body enough question background, avoiding misunderstanding and follow up clarification.

Consider an arrangement of citric acid solution, progressively being dropped on solid sodium bicarbonate. That would provide controlled CO2 release.// NaCl does not play significant role, just affecting solubility of other salts.

@Poutnik The citric acid I have on hand is in the form of a white crystalline(?) powder.  For a liquid solution, what would be the concentration?  And thanks for the idea regarding controlled release!

Leave some space for experimentation. I would start with solution 200 g/L. Solubility is almost 60 mass percents of acid in the solution.// If mosquitos are attracted by body heat as well, it may be good to have there elevated temperature.


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Why does the reaction between copper(II) ion and iodide ion proceed even though the E_cell of the set up is negative?

(iii) Another example of a reaction that does not agree with the theoretical prediction is the reaction between $\mathrm{Cu}^{2+}$ and $\mathrm{I}^{-}$ according to the equation below.
$$
2 \mathrm{Cu}^{2+}(\mathrm{aq})+4 \mathrm{I}^{-}(\mathrm{aq}) \rightleftharpoons 2 \mathrm{CuI}(\mathrm{s})+\mathrm{I}_{2}(\mathrm{aq}) \quad E_{\mathrm{cell}}^{\theta}=-0.39 \mathrm{~V}
$$
White precipitate of $\ce{CuI}$ is formed during the reaction. Suggest a reason why the reaction proceeds despite the $E_{\text {cell }}^{\theta}$ being less than $0 \mathrm{~V}$.
The continuous precipitation of $\ce{CuI}$ from the solution will shift the equilibrium position to the right and drive the reaction forward.

Can I please ask what the words in italics mean? Because to my understanding solids are normally not considered in equilibrium calculations as they are assumed to have a constant concentration

It is answered here: http://www.rod.beavon.org.uk/feasibility.htm quite accurately. Read until the end.

The problem is misstated, possibly by the book. The reference given in M. Farooq's comment (http://www.rod.beavon.org.uk/feasibility.htm, repeated here in case the other comment is lost) shows that in fact the negative cell potential applies to the reaction $\ce{2 Cu^{2+} + 2I^- -> 2 Cu^+ + I2}$ with all species in solution, not including the precipitation of the copper(I) iodide.

@M.Farooq so if I understand it right, it is due to the fact that the CuI is extremely insoluble and readily precipitated that leads to basically the Cu+ concentration always being extremely low, which is what shifts the eqm of the reaction between Cu2+ and I- to the right and this enables the reaction to proceed despite having a negative E knot? On a similar note, can I ask where can I read up on the relationship between Ksp and E cell of reactions? Or is it just that I have to infer it based on the question and the context. Thank you.

@CaptainTeemoOnDuty, You are right. Another key point is that all the standard values of electrode potential that you see in textbook tables are tabulated under standard conditions, i.e., all concentrations are 1 M. This is not the case, when copper (I) is precipitating as iodide.


text:
Reference request: book that covers an overview of useful materials
I was wondering if there is a book that has a broad coverage of inorganic materials, like concrete, glass, and steel.
Ideally the book would go into how each material is manufactured and some of the basic chemistry behind it.
I have a background in physics and some undergraduate level chemistry.
Thank you.

https://chemistry.stackexchange.com/q/37303/108386 may be useful

There are two classical sources which are very thorough and authoritative and describe exactly what you want. They are readable by an undergraduate. A trip to the library is needed. No.1 is Kirk Other Encyclopedia of Chemical Technology and the no. 2 is Ulmann's Encyclopedia of Industrial Chemistry. You can find entries like glass, cement, steel and literally tons of more stuff than you will ever need.
You can search the contents online. Google these titles.

Büchel, K. H.; Moretto, H.-H.; Werner, D. Industrial Inorganic Chemistry, 2nd edition; Wiley-VCH.
With 642 pages, this book is not as big as an entire encyclopedia. It covers the chemistry of the manufacturing processes of important inorganic materials, for example, metals (but surprisingly not steel), fertilizers, construction materials (including concrete), pigments, and glass.


text:
Why is the enthalpy change of an ideal gas a function only of temperature?
While teaching us thermodynamics our chemistry teacher told us that enthalpy change of a real gas is a function of temperature, pressure etc. but the enthalpy change of an ideal gas is only a function of temperature. He didn't give us any explanation for this fact.
But I am curious to know why it is so ?
I have tried to answer it myself but cannot verify it.

Let assume hypothesis it depends on other factors as well. How would be this energy stored within an ideal gas?

@Poutnik Sorry I didn't understand what you meant, please elaborate your comment. Assume what ?

The alternative hypothesis to "ideal gas enthalpy depends only on T" is "ideal gas enthalpy depends on T and other factors(e.g. p)". If so, then how?(rhetorical question)

Based on the first and 2nd laws of thermodynamics, it is possible to prove mathematically that, for a gas that obeys the equation of state PV=nRT, the the internal energy U and enthalpy H are functions only of T.  Have you learned about the 2nd law of thermodynamics yet, and are you familiar with the equation dH=TdS+VdP?

@ChetMiller I have studied the 2nd low of thermodynamics today ( and obviously I am very cofused for now regarding order, disorder and entropy ) but I haven't encountered the equation mentioned by you, maybe I get to learn about in upcoming days.

Let's start from the definition of enthalpy change (not going to formal definition because it's very non-intuitive) here's how I like to define it:

Enthalpy change is the amount of heat absorbed ($\mathrm dH\gt0$) or given ($\mathrm dH\lt0$) by the system at constant pressure of system.

$$
\mathrm dH=Q_p
$$
$$
\begin{array}{l}
\therefore \mathrm d U=Q_p+W \\
\Rightarrow Q_p=\mathrm dU-W=\mathrm dU+p_{\text{ext}}\, \mathrm dV
\end{array}
$$
Now as the pressure of the ideal gas system is constant so it must be equal to the external pressure.
$$
\begin{array}{l}
p_{\text{ext}}=p_\mathrm g \\
\Rightarrow p_\mathrm g=\frac{nRT}{V_\mathrm g}
\end{array}
$$
$$
\begin{array}{l}
\Rightarrow p_\mathrm g V_\mathrm g=nRT\\
\Rightarrow p_\mathrm g\mathrm dV=nR\,\mathrm dT=p_{\text{ext}}\,\mathrm dV
\end{array}
$$
The above equation is valid only if the composition of the gas does not change. Now using this equation
$$
\begin{aligned}
\mathrm dH &=\mathrm dU+p_{\text{ext}}\,\mathrm dV \\
&=\mathrm dU+nR\,\mathrm dT
\end{aligned}
$$
Now as we know that $\mathrm dU$ is a function of temperature only for ideal gas (as according to the kinetic theory of gases there is no intermolecular interaction between molecules of ideal gases so they possess kinetic energy only which itself is a function of temperature only) so we can say that
$$\mathrm dH=f(T)$$
That is, enthalpy change of an ideal gas is only a function of temperature given that the composition of ideal gas system do not change.
Please do correct me if I am getting something wrong.

See my comment above.

There's a simpler way to see this.
By definition, $H = U + pV$. We would like to show that, for an ideal gas, $H(n, p, T) = H(n, T)$, or equivalently that $$\left(\frac{\partial H}{\partial p}\right)_{n, T} = 0.$$
Clearly, $$\left(\frac{\partial H}{\partial p}\right)_{n, T} = \left(\frac{\partial (U + pV)}{\partial p}\right)_{n, T} = \left(\frac{\partial (U + nRT)}{\partial p}\right)_{n, T} = \left(\frac{\partial U}{\partial p}\right)_{n, T} = 0,$$ because an ideal gas satisfies $pV = nRT$ and $U = U(n, T)$.


text:
How to identify the wavelength to compare absorption at in spectrophotometry?
I boiled spinach solution (ground spinach leaves homogenized in 90% methanol) in solutions of varying pH, and am using spectrophotometry to determine chlorophyll content. My results show changes in lambda max, and all my trials seem to end up with varying wavelengths. I was planning on picking one wavelength to compare absorption between my samples, but due to these variances, I am unsure of how to choose a wavelength. How should I do this? I will attach examples in the changes of lambda max. (Lambda max is 403.0 nm in sample 1 and 410.5 nm in sample 2.)
Also, a community member suggested that I use a wavelength at which both chlorophyll a and b absorb light (where they overlap on the absorption spectrum), which is at around 450 nm and 650 nm. The peaks in my results do not match these wavelengths, but would it be okay to compare the absorption levels even if they are not peaks?

Are you sure that your two spectra are different form one another ? If you repeat the measurement of the first spectrum, are you sure to obtain  a maximum at $403$ nm ? If you repeat the second spectrum, are you sure to obtain the maximum at $410$ nm ?  Don't you think that both maximums may be identical ? As a scientist, you must know than repeating a measurement never gets exactly the same result. This is part of the so-called statistical analysis.

Those peak shapes suggest you might be saturating your detector.  compare abs at a couple different dilutions to ensure that you are in the linear range for both samples. You might also see a more consistent max wavelength

What type of spectrophotometer are you using? Don't forget that spinach extract is a complex mixture. You cannot do simple spectrophotometry to follow its degradation. Other pigments besides chlorophyll are interfering.

Are you following a tested protocol or lab procedure or is this ad hoc? I hope it is not the latter because there are many ways things could go awry. It might help to measure the absorbance spectrum of something that has a known absorbance spectrum, e.g. a potassium permanganate solution or holmium chloride solution.

Looks ok to me. Chl does have a Soret band at 400nm ish and characteristic transitions at 680-700nm ish so that is ok. Additionally there are carotenes/xanthophylls that absorb in this region that will confuse things.  The size of the absorption is ok (an OD of 2 is ok but not 3) but you do expect variations due to amount extracted.  As you have a mixture take absorption as $\pm 3$ nm ish as being the same. Find the carotene absorption spectrum to check against your spectrum and/or remove carotenes somehow from your sample.


text:
Acid dissociation constant calculated from the ionization percentage
The question goes like this:

An acid (HX) is 25% dissociated in water. If the equilibrium concentration of HX is $\pu{0.30 M}$, calculate the value of $K_\mathrm{a}$.

So I tried as follows:
if I know that the ionization percentage $\alpha = 25\%$
$$\frac{25}{100} = \frac{[\ce{HX}]_\mathrm{equilibrium}}{[\ce{HX}]_\mathrm{initial}} =  \frac{0.30}{[\ce{HX}]_\mathrm{initial}}$$ Thus: $$[\ce{HX}]_\mathrm{initial} = \pu{1.2 M}$$
With this initial concentration I could calculate the concentration variation:
$$[\ce{HX}]_\mathrm{equilibrium} - [\ce{HX}]_\mathrm{initial} = -0.9$$
Thus, knowing that the initial concentration of $[\ce{H+}]$ and $[\ce{X-}]$ is $0$, in equilibrium their concentration would be:
$$0 - (-0.9)= 0.9$$
And finally replacing this in the relation to the acid dissociation constant:
$$K_\mathrm{a} = \frac{[\ce{H+}][\ce{X-}]}{[\ce{HX}]} = \frac{(0.9)(0.9)}{(0.30)} = 2.7$$
But unfortunately, this procedure is wrong and doesn't match the answer of the exercise. Could someone tell me what I'm missing? Or where did I go wrong?

The error is in the  first equation line. Read the task in hand more carefully.

Although this is a homework question, OP's effert to solve it deserve some insight. As Poutnik correctly pointed out, the error is in the OP's first equation (due to not reading the task in hand more carefully, see Poutnik's comment elsewhere).
Suppose the ionization of the given weak acid is as follows:
$$\ce{HX (aq) + H2O <=> H3O+ (aq) + X- (aq)} \tag1$$
Suppose intial concentration is $c$ and $\alpha$ amount of $c$ is ionozed at the equlibrium $(\text{both in } \pu{mol L-1})$. Thus, the equlibrium concentrations of $\ce{HX, X-},$ and $\ce{H3O+}$ are $(c-\alpha)$, $\alpha$, and $\alpha$, respectively. Since 25% of acid is ionized at the equlibrium:
$$\alpha = 0.25c \tag2$$
Since the equlibrium concentration of $\ce{HX},$ is $\pu{0.30 mol L-1}$:
$$c - \alpha = 0.30 \tag3$$
Now you have two unknowns and two independent equations so that using your mathematical knowledge you can solve the problem. The acid dissociation constant is:
$$K_\mathrm{a} = \frac{[\ce{H3O+}][\ce{X-}]}{[\ce{HX}]} = \frac{\alpha\cdot\alpha}{c-\alpha} = \frac{\alpha^2}{0.30}$$


text:
Why does physisorption pass into chemisorption on increasing temperature
I have been told, that on increasing temperature, physisorption passes into chemisorption (with greater Activation Energy, generally). But why does that happen?
Shouldn’t the increment in Temperature decrease the bond strength of attractive forces like van der Waal’s, instead of resulting into the formation of chemical bonds?

Do you have a reference?

Increasing temperature does not affect the potential energy or forces, but makes it more likely that higher energy levels will be populated. So if there is a small physisorption well with a big energy barrier to a v. deep chemisorption, well heating will allow the chemisorption well to become populated.


text:
Why turmeric turns green?
When I added some turmeric solution to a strong base it should have turned red but to my surprise it did change colour but became green and when I added some more base and stirred it turned red! Why did this happen and it is happens quite often with people according to the data I have!

Post some images regarding that because I personally never experienced that. In my experience when I put some soap in turmeric powder solution it turns red watch this https://youtu.be/Olezbt9cxfo. By the way you will have ear some reputation first to use images in your questions.

Tumeric has a beautiful green fluorescence, btw, but that is usually visible in bright sunlight and in UV it is a very bright green.  Were you doing this experiment outdoors?Interesting observation. Is your tumeric adulterated, which is so common thing these days? What base did you add. Can you try a different brand?

No , I did it in kitchen and it only happened once ,never again.But why?

Try a different brand. Maybe your container was contaminated with some pigment. When you cannot repeat an experiment chances are there was a mistake. Was you strong base pure? Which base did you use?

See https://www.compoundchem.com/2016/11/30/turmeric/amp/   The fluorescence visibility may depend on light source, concentration  and angle of observation.

@Poutnik Nice link! I may try this just for fun, since I have everything necessary! Thanks!

I just upvoted so you can post pics if you wish. Using my 405 nm Space gun, I zapped my bottle of turmeric and it fluoresced yellow. I dissolved a little in pure ethanol and it fluoresced green when I zapped it at 405 nm. Next I will add a dash of water and a sodium hydroxide flake and see what happens to the fluorescence.

@EdV It might be possible it works for white LED or (cold white) fluorescent light too.

Ed: If you have cooking oil, let tumeric soak in it. You will see beautiful greenish yellow fluorescence even under white fluorescent light. Long time ago, I planned to use magnifying glass as a poor man's Xenon arc lamp and it works. Even quinine water begins to glow blue with magnifying glass.

@EdV, Heat the baking soda to convert it into sodium carbonate. I think the OP saw the fluorescence at an angle.  The natural fluorescence appears only on the surface.

This is what the fluorescence of turmeric solutions looks like: https://chemistry.meta.stackexchange.com/a/4757/79678. I will post temporary better (unbloomed and in focus) photos if I can get them. The basic solution was obtained by dropping small pieces of partially oxidized calcium metal into the cuvette.

Okay,I used surf for base and  unbranded turmeric.

I saw the images it wasn't that bright it was like a normal grass green.

The photos are for laser-excited fluorescence, which is not what you saw. The odds are you had contamination or an adulterated turmeric. In any event, none of us can guess what caused your irreproducible observation, so that pretty much closes the case. If it ever happens again, you can get photos, document the turmeric specifics, and edit your question.

@MohdSaad, Surf (detergent brand) has plenty of chemicals. This is not a strong base. EdV's results are authentic. It was nice of him that he did experiments for you.


text:
Which double bond will PhMgBr attack and why?

I am confused about which double bond will undergo nucleophilic addition.C=O should undergo but the book says the double bond of the ring.Please tell why

Which book says this?

@Waylander Himanshu Pandey Organic Chemistry(Indian Book)

@OscarLanzi  Interesting argument, except that the reaction gives the 1,2 product (b) as is demonstrated by the references in the answer to the duplicate.

@PrakritishGhosh  get a better book, this one is wrong.

So much for that argument.


text:
Why doesn't NGP happen in the bromination of fumaric acid?

In this question from JEE $2015$, a surprising neighbouring group participation occurs, and the product turns out to be $(3)$. A link for the mechanism can be found here.
However, why doesn't the same COOH group participate in the bromination of fumaric acid (Or maleic acid for that matter)?
We would then not get a meso product.

I thought the answer would be 2,3 why is there a stereoselectivity?

I don't get how the meso thing comes here only one group is being placed right?  For Meso, it is chiral but prevented from being so due to plane of symmetery

@Buraian Consult the link for the mechanism. N$2+$ doesn't create a carbocation here, as that would be destabilised by COOH. It is instead pushed out by COOH, resulting in stereoselectivity.

Two bromine groups are added to adjacent carbon atoms, resulting in two chiral centres.

Neighboring group participation (NGP) in organic chemistry has been defined by IUPAC as the interaction of a reaction center with a lone pair of electrons in an atom or the electrons present in a $\sigma$-bond or $\pi$-bond contained within the parent molecule but not conjugated with the reaction center. The NGP shown in the given reaction create a highly energetic $\alpha$-lactone, but is feasible (Ref.1). As a matter of fact, Ref. 1 has shown participation of not just Neighboring group, but by active carboxylic acid groups on other carbons such as $\beta$ and $\gamma$ to give larger and stable cycles such as $\beta$- and $\gamma$-lactones:

As shown in the above scheme, $\alpha$-lactone is the most unstable as it take few minutes to alkylate. As a comparison, $\beta$- and $\gamma$-lactones take hours to alkylate (all at the carbon shown by the vertical arrow). The experimental rate constants for the nitrosation are found to be $4.8\pm 0.1 \times 10^{-2}$, $1.34\pm 0.02 \times 10^{-2}$, and $\pu{0.89\pm 0.03 \times 10^{-2} M-1 s-1}$ for $\alpha$-, $\beta$-, and $\gamma$-aminobutyic acids, respectively. That means NGP through $\alpha$-lactone is the preferred one if there is a choice. For instance, From the reaction between glutamic acid and dinitrogen trioxide, $\alpha$-hydroxyglutaric acid as well as a $\gamma$-lactone can be formed. It has been described that $\alpha$-hydroxyglutaric acid is formed from glutamic acid when the $\mathrm{pH}$ is increased to 7 after the reaction, while the $\gamma$-lactone remains the end product if the $\mathrm{pH}$ is not subsequently increased (Ref.2):

Thus, when you look at bromination of maleic acid, you'd see there two possible NPG to happen: 1) through $\alpha$-lactone and 1) through $\beta$-lactone. But as seen in above cases, most probable participation should be through the $\alpha$-lactone. If that happens, the product would be the meso-isomer:

However, careful pioneering study this field has shown that it is just through the bromination intermediate (Ref.3), the work of which leads to the bromonium ion intermediate mechanism. The bromonium ion is so strong that the steric hindrance such as the presence of t-butyl group does not alter the outcome (Ref.4)), anti-trans-addition of bromide or other nucleophile (as noted in Ref.3: the new atom will approach one of the carbon atoms from the side opposite to the $\ce{X}$ atom already present). As indicated in the scheme, you'd see the carboxylic group cannot be in $180^\circ$ (look at the Newman projection of intermediate) to bromonium ion to make the necessary participation. Regardless, bromonium ion is much more stable than alkyl azide to be readily replaced by weaker nucleophile such as carboxylic ion.
Nonetheless, as shown in the scheme, even if it is possible to have NGP in the bromination of maleic acid (or fumaric acid), you may not be able to avoid meso-product formation. As shown, maleic acid gives meso-product and hence, fumaric acid would give racimic mixture, if NGP is considered. That would be the same outcome if $\beta$-lactone formation is considered instead of $\alpha$-lactone formation as shown in the scheme.

Note: The fact that brominations with the maleate ion resulted in cis-addition driven by repulsion between the negatively charged carboxylic acid anions being stronger than halonium ion formation is established by the authors of Ref.3.

References:

M. del Pilar García-Santos, Emilio Calle, and Julio Casado, "Amino Acid Nitrosation Products as Alkylating Agents," J. Am. Chem. Soc. 2001, 123(31), 7506–7510 (DOI: https://doi.org/10.1021/ja010348+).
Daniel Pleissner, Reinhard Wimmer, and Niels T. Eriksen, "Quantification of Amino Acids in Fermentation Media by Isocratic HPLC Analysis of Their α-Hydroxy Acid Derivatives," Anal. Chem. 2011, 83(1), 175–181 (DOI: https://doi.org/10.1021/ac1021908).
Irving Roberts and George E. Kimball, "The Halogenation of Ethylenes," J. Am. Chem. Soc. 1937, 59(5), 947–948 (DOI: https://doi.org/10.1021/ja01284a507).
Robert C. Fahey, "Polar Additions to Olefins. II. The Chlorination of Di-t-butylethylene," J. Am. Chem. Soc. 1966, 88(20), 4681–4684 (DOI: https://doi.org/10.1021/ja00972a030).

So to summarise, the reason NGP doesn't occur, is because of steric reasons preventing anti addition, and the bromonium ion being far too stable to be substituted by carboxylate?

Basically yes. But keep in mind that all these are speculations according to known facts.


text:
Why do we have to use anhydrous AlCl3 in Gatterman-Koch Reaction?
In reference to a Gattermann-Koch reaction, such as the one shown below:

Why does $\ce{AlCl_3}$ have to be anhydrous?
Why can't we just use normal $\ce{AlCl_3(H2O)_6}$?
What would the reaction look like if we use $\ce{AlCl_3(H2O)_6}$?

Anhydrous $\ce{AlCl3}$ is used because it reacts violently with water details here and you then no longer have an aluminium compound that will react.
$\ce{AlCl3}$-catalysed reactions occur because it is a Lewis acid definition here and readily accepts electron pairs. Water readily donates an electron pair which stops any other molecule from combining with it. This stops the reaction.

+1 Any Friedel-Crafts-like reaction requires anhydrous AlCl₃, not just Gatterman-Koch. Also, dissolving the hydrate in benzene isn't all that easy.

@TooTea  The purpose of $\ce{AlCl3}$ is to act as a Lewis acid catalyst.  Depending on the electron density of the aromatic system (Lewis basicity), the group to introduce to this ring by subsitution, reaction parameters (solvent, temperature, etc) it may be way too powerful (pyrroles, thiophenes) for a selective reaction and is replaced by less strong ones (e.g., $\ce{FeCl3} > \ce{SnCl4} > \ce{BF3} > \ce{TiCl4} > \ce{ZnCl2}$) to mention a few examples.

@Buttonwood Oh, I see I should have worded that better. What I meant to say is that for any Friedel-Crafts-like reaction (not just Gatterman-Koch),the Lewis acid needs to be anhydrous (no matter if it's AlCl₃ or something else, but the question was about AlCl₃).

@TooTea True.  Equally, neither solvent nor reagents may posses hydrogens easily abstracted by the Lewis catalyst.  Thus careful with amines, alcohols, phenols etc. which may (if not protected) additional Lewis acid ...


text:
Do contiguous monolayer aromatic structures retain the same magnetic susceptibility as the core constituents?
As the synthesis of Porphyrin monolayer strips is achievable, retaining the core aromaticity of the tetrapyrrole; does it still retain the same value of magnetic susceptibility of porphyrin? Does this apply to other systems like these?


text:
Possible dehydration after nucleophilic addition
Here is the question:

 Answer: B

My reasoning was the following:
The ethoxide ion first abstracts a proton from the $\mathrm{sp^3}$ carbon of the cycloalkene. This carbon then attacks the electrophilic carbonyl carbon of acetone, ultimately forming the nucleophilic addition product. On heating, it loses an $\ce{H2O}$ molecule to give the corresponding alkene.
However, the answer given here suggests that the final dehydration step doesn't occur. What is the explanation behind such an activity?

How you imagine that happen in strongly basic environment?

Similar to to last step of aldol condensation to give $\alpha, \beta$-unsaturated ketone, the last elimination step at sought reaction is also supposed to undergo elimination of water molecule through $\mathrm{E1_{cb}}$ mechanism.
To refresh your memory, the definition of the $\mathrm{E1_{cb}}$ mechanism is as follows:

An elimination reaction mechanism featuring carbanion formation followed in the next step by expulsion of a leaving group on a $\beta$-carbon, with subsequent formation of a $\pi$-bond. Leaving group expulsion is the rate-determining step (source).

Keep in mind that the leaving group of the most of $\mathrm{E1_{cb}}$ elimination reactions is a poor leaving groups such as $\ce{OH-}$ ion. Yet, the driving force for the elimination is to give the thermodynamically more stable product such as the $\alpha, \beta$-unsaturated ketone in aldol condensation. Following is the dehydration of a $\beta$-hydroxyketone via the $\mathrm{E1_{cb}}$ mechanism:

The methylene carbon of cyclopentadiene is more acidic because abstraction of one of the two protons gives a carbanion which is aromatic and satisfies $(4n+2)$ rule. Once the reaction of this carbanion with acetone gives the alkoxide of the product $(c)$, the excess base would pick up the second proton from this product to give corresponding carbanion, which is also aromatic. Here, if the dehydration occurs as usual, the resulting product is not thermodynamically more stable than the carbanion (even if the product is conjugated), because the carbanion would lose the aromaticity. Thus, the dehydration of the final step is not possible here. The carbanion would become neutral during acid workup.


text:
Compare density of halogen substituted benzene rings
The correct decreasing order of densities of the following compounds is :

(1) (D) > (C) > (B) > (A)
(2) (C) > (D) > (A) > (B)
(3) (C) > (B) > (A) > (D)
(4) (A) > (B) > (C) > (D)
Question Source: JEE Mains 2021
Official Answer: (1)
How to solve this question practically in exam? Can we compare density assuming packing efficiency approximately similar for given compounds in option and $\rho \propto$ Molar Mass?
This may be helpful but is irrelevant here because we are comparing densities in between cyclic compounds only.

You have completely change the question. Actually this is a new question. I did what I have to do.

I have a feeling this has something to do with dipole moment

@Buraian if $(\rho)^{-1} \propto$ dipole moment then density of C should be greater than D. Assuming your guess to be correct.

I remember reading somewhere that density of phenol was related to packing efficiency and that was related to dipole moment. I spent like few hours searching where I read it yesterday but I couldn't find it :( . Maybe I guess I learned some wrong information.

@Buraian The thing of packing efficiency and dipole moment is about alkenes in cis or trans form and packing efficiency can be determined there in alkene but here there is no easy way to determine packing efficiency; you might have read that in isomerism chapter.

Examining the compounds from a distance suggests these conclusions, with little or no extensive knowledge of chemistry:

B is more dense than A and C is more dense than B, because you know that hydrogen is the lightest element (the least dense, because it is used in blimps) and chlorine is denser (it sinks and was used as a poison gas because it hugs the ground).

D is more dense than C, because bromine is farther down the periodic table than chlorine, and is likely to be denser.

Therefore D is the densest compound. Only one choice puts D as the densest: #1.

It seems to me that the intricacy of the question, and/or the difficulty, lies in the wording: what is decreasing order of densities?
At the very least, it should be obvious that D is the extreme compound, but a confusion between the meaning of increasing and decreasing could suggest that #4 is the correct choice. A little reflection rules that out: the density of D cannot be less than that of A!
The density of 1-bromo 3-chlorobenzene was not available in my search, but for the others, benzene has 0.876, chlorobenzene 1.11, m-dichlorobenzene 1.228, and bromobenzene 1.5. Bromochlorobenzene would probably come in at about 1.75. The actual densities are not at all similar, so the question must have been going after something else.
The effect of packing efficiency, due to dipole moment - in liquids - is probably much, much smaller. One challenging possibility would be that one of the compounds, perhaps C or D, could be a solid - but, no, they are all liquids.
Maybe it was a freebie: you have to work at it to get it wrong.

The bonded radius is moer complicated... but nice that this method gives the answer @Jay

The wanted densities are the ratio of the mass of the molecule by the volume of the molecule.
You may borrow an approximation used in crystallography for an estimate of the trend of the densities and substitute the former

the mass of the molecule is the sum of the atomic masses (tabulated or/and memorized as an approximate)

the volume of the molecule is approximated as the sum of the atomic volumes (tabulated, e.g., Hofmann table mentioned in an earlier answer here)

Thus, using the relevant entries from the Hofmann table
|------------------------------+-------+------+-------+------+------+------|
| element                      |     C |    H |     F |   Br |   Cl |    I |
| average atomic volume [A**3] | 13.87 | 5.08 | 11.17 | 32.7 | 25.8 | 46.2 |
|------------------------------+-------+------+-------+------+------+------|

you obtain
|--------------------+--------+--------+---------+----------|
| compound           |      A |      B |       C |        D |
|--------------------+--------+--------+---------+----------|
| formula            |   C6H6 | C6H5Cl | C6H4Cl4 | C6H4BrCl |
| mass [amu]         |  78.11 | 112.56 |  147.00 |   191.45 |
| volume [A**3]      | 113.70 | 134.42 |  155.14 |   162.04 |
| density [amu/A**3] |   0.69 |   0.84 |    0.95 |     1.18 |
|--------------------+--------+--------+---------+----------|

As stated earlier, this provides a coarse estimate of the trend for the density, which however is not the density of the liquid.  This approximation (if you compare to, e.g. packing of apples in crate) does not consider the gaps between molecules packed which, given the shape of molecules and conformational flexibility of e.g., alkyl chains may be more complex than just close packing of spheres of one diameter/hcp.
Reference Hofmann table
Hofmann, D. W. M., Fast estimation of crystal densities in Acta. Cryst. B58, 2002, 489-493; doi: 10.1107/S0108768101021814.


text:
Why can covalent compounds dissolve in organic solvents?
Why can covalent compounds dissolve in organic solvents? I researched about it a little and I found an article which said:
"Mostly all organic compounds are non-polar in nature. Also, covalent compounds are by nature non-polar.". I'm not sure what it means for a compound to be "non-polar" or "polar" but I'm guessing it has something to do with net charge? Also, what does that have to do with covalent compounds being able to dissolve in organic solvents.

Really just a little. You should research more before asking, otherwise the question may be closed due duplicity and/or not enough of own effort. Polarity and solubility topics have been discussed many times on many places. The latter even in medieval alchemy  ( similia similibus solvuntur ).

There is one factor that is entropy. Itself, it provides a driving force. It can be not enough, so usually it is important the energetic of the molecules in their covalent solid vs that of the molecules surrounded by solvents one. Nothing is totally apolar. Organic solvents have various degrees of polarity too. So it is possible to find some balance. But to pose such a question you should read more on polarity of molecules or part of them. Once cleared, likely you won't even pose this question anymore.

What an interesting way to ask about solubility!
Non-polar vs polar. We usually fret about relatively minor differences in solubility between polar compounds. Here, I'm going out on a limb and calling a factor of two minor.
But non-polar compounds are sometimes quite soluble in organic solvents: 10% or more, even in solvents where the idea of "like dissolves like" has no significance. That's quite a lot when you consider that some chemists are OK with calling a 1% solubility for an inorganic compound "somewhat soluble".
One way (not the only way, for sure) to explain the possible high solubility of non-polar compounds in organic solvents is to look at melting points: the organic compounds of greatest solubility are the ones with lowest melting points, i.e., lowest "lattice energy", if we may use that term, somewhere below 200C. On the other hand, polar inorganic compounds generally have melting points much higher - 500-600C or more (unless already hydrated), and only dissolve at all when their polarity is addressed by the solvent, e.g., water, with its hydrogen and/or oxygen bonding.
And although I dismissed "like dissolves like", "organic solvents" suggests non-polar, and the solutes were non-polar, so the solvent and solute did have some similarity, even if not by virtue of similar structures.


text:
Who makes a better nucleophile? CH3O- or OH-?
My argument is that the +I effect of CH3 group increases the electron density at oxygen atom, making it a better nucleophile

Does this answer your question? Is methoxide a better nucleophile than hydroxide?

When the donor atom is same, a general rule applies: nucleophilicity parallels basicity. Thus, comparing the basic strength of methoxide and hydroxide, it is observed that  methoxide is a stronger base than hydroxide (as it the conjugate base of a weaker acid i.e. methanol when compared to water). Thus, methoxide is a stronger nucleophile than hydroxide.
However, at the same time NMR data predicts that methanol is a stronger acid than water (despite the I+ effect of $\ce{-CH3}$ group). So, in this case, there exists a controversy.


text:
How To Know Which Oxidation Number To Use For Nitrogen
When given a compound like (NH4)2MoO4 and asked to find the oxidation number of Mo, how do I know which oxidation number to sub in for Nitrogen? On the periodic table, Nitrogen has oxidation numbers of $-3, -2, -1, +1, +2, +3, +4, +5$.
I know the process will look like:
(2xNitrogen's oxidation number) + (8x1) + x + (4x-2) = 0
Is there a process to figuring out which number to sub in for Nitrogen's oxidation number?

This problem can be solved by dealing with separate ions. In water, $\ce{(NH4)2MoO4}$ is dissociated in $\ce{2 NH4^+ + MoO4^{2-}}$. Here each ion can be analyzed separately, to get the unknown oxidation numbers.
In the ammonium ($\ce{NH4^+}$} cation, Nitrogen is at oxidation number $x$, and H at $+1$. So that $x + 4·(+1) = +1$, so that $x = -3$.
In the anion $\ce{MoO4^{2-}}$, the same calculation yields $+6$ for oxidation number of molybdenum.


text:
Statistical entropy of perfect gas and indistinguishability
The weight, $W$ (the same $W$ as in Boltzmann formula $S=k\text{ln}W$) of a configuration {$N_0, N_1, N_2$, ...} is given by
$$W=\frac{N!}{N_A!N_B!N_C!...} \tag{1}$$
where $N$ is the total number of molecules in the system and $N_j$ is number of molecules with state $j$.
The denominator is to account for different permutations of indistinguishable molecules. For example, molecule labelled 1 and 2 below have exactly the same energy and are of a same element. If the sample is a perfect gas, molecule 1 and 2 will be indistinguishable. The simple N factorial overcounts by counting such many different permutations which are indistinguishable multiple times, and the denominator in (1) is to correct this. Molecule 1 and 3 are distinguishable; they have different energies.

If we proceed to operate on (1) using Stirling's approximation and through some simple algebraic manipulations, the equation for entropy below can be deduced:
$$S(T) = \frac{U(T)-U(0)}{T}+Nk\text{ln}q \tag{2}$$
Because (2) is derived from (1), (2) has already taken into account of different permutations of indistinguishable molecules, just like (1).
Now here is the question, the entropy of perfect gas is actually given by
$$S(T) = \frac{U(T)-U(0)}{T}+Nk\text{ln}q -Nk(\text{lnN}-1)\tag{3}$$
(3) is derived using canonical partition function and
$$Q = \frac{q^N}{N!} \tag{4}$$
which is true for independent, indistinguishable molecules.
(2) can be reached similarly, but using
$$Q = q^N \tag{5}$$
which is true for independent and distinguishable molecules.
What is wrong with the argument above? I thought (2) has already accounted for the fact that the molecules are indistinguishable, but proven otherwise by canonical partition function, or did I understand the indistinguishability wrongly.

The $W$  you give is the number of ways of placing $N$ distinguishable objects into $r$ distinguishable boxes so that they contain $N_A,N_B\cdots $ each.  Is that what you wanted?

You should also have a look at the Gibbs Paradox. The partition function is divided by $N!$ because the $N!$ permutations of the molecules among themselves do not lead to physically different situations so the number of states summed in the partition function is too large by this amount.

Inspired by user porphyrin.
A common, potentially confusing way of approaching configuration is to think of each energy state as a distinguishable box, and each molecule as a marble.
The number of ways of arranging N marbles in a line is N!. But the exact arrangement of marbles in one box is irrelevant. If molecule 1 and 2 are in state A, they both have energy $\epsilon_A$, and that's all. There is no such thing as 'arrangement of molecules in a state'. State is not a physical entity; if two molecules are in a same state they have the same energy; it is not like the two molecules are really enclosed by a confinement.
For illustration, the two cases shown below are counted separately by N! but they are just exactly the same physically because molecule 1 and 2 have energy $\epsilon_A$, molecule 3, 4, 5 have energy $\epsilon_B$, and so on.

When making the diagram above, the molecules have already been assumed to be distinguishable; they can be labelled with different numbers, and you can tell which one is 1, which one is 2...
$\therefore$ The equation
$$W=\frac{N!}{N_A!N_B!N_C!...}$$
is for distinguishable molecules.
For indistinguishable molecule, the labelling should not be done:

But that gives rise to another question...This will mean that for indistinguishable molecules the W of any configuration is 1, so $S=k\text{ln}W = 0$?


text:
Why are glyosidic linkage of glucose and fructose rings often occurring at the 1' and 4' prime positions?
Here are some example combinations from my textbook:

In all of these combined products, the reactive centres seem to be the carbon at first position/ fourth position from the oxygen in priority. Is there something deep going on here?

Source: NCERT Class 12

The 1' position on glucose is the anomeric carbon, which pretty much has to be involved because of the mechanism of the coupling reaction. The 4' position is common because it is opposite the 1' and leads to a linear polymer, but other positions (esp 6') are also very common. 6' is off the ring, so it allows for more flexibility in the linkage. 2' is quite sterically hindered.

Why is 2' hindered? @Andrew

2' is adjacent to 1', so if you have linkages at both positions (as in a polymer longer than two units), the two linked sugars interfere with each other. But it's not unknown. See Kojibiose and kojitriose, kojitetraose, etc.

There is no need for a prime in these structures. Nucleotides have it to distinguish sugar atoms from nucleotide atoms.


text:
Rietveld Refinement: Calculated Intensities far lower than observed for a small selection of miller indices

I'm performing a Rietveld Refinement in excel of an I4/mcm material CeAlO3. The fit is very good for most peaks but on a few, I'm coming up very short on the intensities. Could this be due to a local instead of global minima and how would one remedy this situation?

In general, rather that referring to your particular case, the starting values can often determine the outcome, so finding a local minimum looks like what has happened. (a) Try fitting each using a random selection of starting points to see if that improves matters. (b) Use another fitting programme rather than Excel which is not really designed for this type of problem.

Without knowing the structure and the expected variance of peaks with refinement, it could well be either a numerical solution problem or a physics problem. While vaguely impressed that you got something to work in Excel, I'm also quite horrified since I don't trust Excel to fit a straight line much less something complex like diffraction intensities.

If future answers to your question don't resolve the problem to your satisfaction, you may attempt a structure solution and refinement with an other program dedicated to PXRD.  If the ones compiled by IUCr here don't fit your needs, give Fox a chance (a more recent review/tutorial, open access).

For peak fitting (esp. PXRD data) and a simultaneous display of experimentally recorded intensity (or the profile fitted) in function of $2\theta$, the open source program Fityk seems to be more suitable than Excel, too.  Consider using this program once up and including the point to simultaneous display of fitted intensity profile and variation (against the experimental intensity) and add this as clarifying detail to your question (a total view and some regions where the modeled profile fits the experiment less well).


text:
Electronic Configuration and Octave Rule
According to Octave rule, a shell cannot hold more than 8 electrons
But according to atomic orbitals,

The maximum number of electrons in the shell with principal quantum number n is 2n2.

How do these rules reconcile?
Appreciate your help.

They reconcile just fine when $n=2$.

@Ivan Neretin Do they only reconcile when n=2? What about when n≠2?

When $n\ne2$, things get... complicated.

The point is, for n>2, the rule 2n^2 is still true, but with delayed filling up of d resp. f orbitals. Therefore n+1 shell starts filling up, when n shell contains just 8 ( s + p ) electrons.

Octet, not octave.

@EdV The student or their teacher is mixing octet rule with the law of octaves (part of the early history of the periodic table).

If you look up the time frame of the so called octave law, it is from the mid-19th century (1860s). At that time, electron did not exist, protons did not exist and the concept of atom was not clear.

According to Octave rule, a shell cannot hold more than 8 electrons

The above statement is blatantly wrong and the textbook teaching this should be trashed. Octave rule was not meant to state this.
Anyway, the formula you quote is correct 2n$^2$.
K is shell no. 1, it can hold 2 electrons
L is shell no. 2, it can hold 8 electrons
M is shell no. 3, it can hold 18 electrons

If exceptions are "forgotten", the teacher can teach to beginners that the octet rule works well "at least" for the $20$ first elements. And these $20$ first elements are the most frequent and most important elements found in elementary textbooks of chemistry. They contain $\ce{C, N, O, F, Na, Mg, Al, Si, P, S, Cl, K}$ and $\ce{Ca}$. These elements and the octet rule are sufficient to develop the basic theory of covalent and ionic bonds for newcomers in the world of chemistry.


text:
How much Na2O (from NaOH) do I add to sodium silicate with a weight ratio of SiO2/Na2O = 2.34 to get the solution to a weight ratio SiO2/Na2O = 1.9?
There is some confusion in industry as sodium silicate is made with NaOH but given a ratio with respect to Na2O (i.e. SiO2/Na2O not SiO2/NaOH). I presume it follows from this equation:
2NaOH -> Na2O + H2O
Looking at a product information sheet of sodium silicate with a weight ratio displayed as "SiO2/Na2O" the actual ingredients in it are:

SiO2 wt% = 30.3%
NaOH wt% = 16.7%
H2O wt% = 100% - 30.3% - 16.7% = 53%
SiO2/NaOH = 1.81

This corresponds to SiO2/Na2O = 2.34
So if I'm given a 100g solution of sodium silicate that has these specifications and wanted to change the solution by adding an amount of NaOH so that the final wt% of NaOH would correspond to a final solution weight ratio of SiO2/Na2O = 1.9, what would be the best approach assuming I can add a mass [g] of a liquid specification 50% NaOH and 50% H2O?
EDIT 1: Here is my working so far.
First off, to convert the equivalent NaOH to Na2O I do the following
Assuming an initial 100 g of sodium silicate and using the above reaction for molar equivalents then we have:
16.7g NaOH * (1 mol / 39.99 g NaOH) * (1 mol Na2O / 2 mols NaOH) * (61.98 g Na2O / mol) = 12.94 g Na2O
The total water in the system is thus 100 g - SiO2 g - Na2O g = 100 - 30.3 - 12.94 = 56.76%.
Therefore, the sodium silicate with respect to Na2O and not NaOH is:

30.3 g SiO2
12.94 g Na2O
56.76 g H2O

##
If I add 7.7 g of 50% NaOH I now have this mass added to my system.

2.98 g Na2O (using same logic above)
4.72 g H2O

Allowing time for stirring, I could then expect my new desired SiO2/Na2O weight ratio to be (30.3 g) SiO2 /(12.94 + 2.98)g Na2O = 1.90
Does this seem correct? Should I take into account any equilibria considerations from the main reaction because H2O is the solvent?
EDIT FROM MAURICE REPLY 1:
Let me be more specific of the small calculation step I missed in between going from the ## symbol above.
To get my desired SiO2/Na2O ratio of 1.9 I rearrange to get the total Na2O required in the system which is Na2O = 30.3 g SiO2/1.9 = 15.947 g Na2O.
Considering the 12.94 g Na2O in the original solution, I need to add (15.947 - 12.94)g Na2O = 3 g Na2O into the final system to achieve my desired ratio of 1.9.
In 100 g of a 50% NaOH solution (with respect to NaOH) the quantities are:

50 g NaOH
50 g H2O

With respect to Na2O, these quantities are:

38.7 g Na2O
(100 - 38.7) = 61.25 g H2O

Thus if I want to add 3 g Na2O from the 50% solution then that corresponds to:
3 g Na2O x (1 g NaOH Solution/0.387 g Na2O) = 7.75 g NaOH solution (slight rounding errors in answers before this edit)

What exactly are troubles in getting the result? The old good mass conservation law and Na2O/NaOH molar mass ratio would do.

@Poutnik Sure, that's probably what i want help with. I'm trying to make a graph where i can plot added NaOH against different weight ratios. I have already done this for 100% NaOH (was hoping to confirm my answer) but not for the 50% liquid caustic which I'm having trouble with.

It is rather a simple algebra at high school level. Readers may hesitate to help you unless you put your resolution attempts to the question and describe your troubles particularly. Solving everything is not preferred way to help here.

@Poutnik, yep that's what I'm having trouble with. Do you think I should edit this post with my attempt or would you recommend me posting another question and deleting this one? I have not ever had any issues on StackOverflow or Math Overflow so pretty surprised that both of my questions were downvoted. Just hoping for some help/guidance.

Improving the question by editing is definitely preferred to reposting. // In contrary to some other Q/A or forum sites, answers on CH SE site are figuratively paid by the user's own effort. When you ask, it is expected you have already searched for and thoroughly thought about the topic, providing explicit compact summary of partial answers or at least ideas or thoughts you have got until then. Effort not shown may be considered as effort not done and such a question may be closed.

Focus on the fact it is rather a mathematical then chemical task, with simple math skills more needed than particular chemical knowledge. The only CH knowledge needed is NaOH/Na2O molar mass ratio, what you already have to recalculate ratios toward SiO2.

@Poutnik thank you for helping me out and for your patience! I've just added my worked example in the edit :)

It seems correct. No equilibrium to be considered. Even the 2 NaOH vs Na2O equivalency is purely formal, without NaOH becoming Na2O. It is "what if all NaOH would be converted to Na2O", without any such real conversion. // A good practice is: Always start with symbolic algebraic formulas and keep it this way until all formulas are ready to finally put there literal numbers.
It helps focusing on principles, orientation, spotting mistakes, improves the text reusability and its permanent value.

Thanks for the encouragement you two :)

Let focus on a generally applicable solution and then on your particular case.

Quantity
Equation
Value
Legend

$M_\ce{NaOH}$

$\pu{39.9971 g mol−1}$
the molar mass of $\ce{NaOH}$

$M_\ce{Na2O}$

$\pu{61.979 g mol−1}$
the molar mass of $\ce{Na2O}$

$w_\ce{SiO2}$
The given value
30.3 %
the reported (equivalent) weight percentage of SiO2 in the stock sodium silicate solution.

$w_\ce{NaOH}$
The given value
16.7 %
the reported (equivalent) weight percentage of NaOH in the above solution.

$m_\mathrm{sol}$
The given value
$\pu{100 g}$
the total mass of the solution to be adjusted.

$r_{\ce{SiO2}/\ce{Na2O}}$
The given value
1.9
the intended final $\ce{SiO2}$/$\ce{Na2O}$ ratio in the adjusted solution.

$r_{\ce{SiO2}/\ce{NaOH}}$
$r_{\ce{SiO2}/\ce{NaOH}}=r_{\ce{SiO2}/\ce{Na2O}}\frac{M_\ce{Na2O}}{2M_\ce{NaOH}}$
$\approx 1.472$
the equivalent final $\ce{SiO2}$/$\ce{NaOH}$ ratio in the adjusted solution.

$m_\ce{SiO2}$
$m_\ce{SiO2}=m_\mathrm{sol} \frac{w_\ce{SiO2}}{100}$
$\pu{30.3 g}$
the (equivalent) mass of $\ce{SiO2}$ in the used stock silicate solution

$m_\ce{NaOH}$
$m_\ce{NaOH}=m_\mathrm{sol} \frac{w_\ce{NaOH}}{100}$
$\pu{16.7 g}$
the  (equivalent) mass of $\ce{NaOH}$ in the used stock silicate solution

$m_{\ce{NaOH}\mathrm{,add}}$
$m_{\ce{NaOH}\mathrm{,add}}=\frac{m_\ce{SiO2}}{r_{\ce{SiO2}/\ce{NaOH}}} - m_\ce{NaOH}$
$\approx \pu{3.88 g}$ (resp. $\pu{7.76 g}$ of 50 % $\ce{NaOH}$)
the  mass of $\ce{NaOH}$ to be added to the stock solution of the mass $m_\mathrm{sol}$ to obtain the desired $r_{\ce{SiO2}/\ce{Na2O}}$ ratio. (For obvious reasons, the value must not be negative, otherwise we are out of luck, as there is already too much NaOH).

Brilliant comment that's very easy to follow! Thank you for making the whole thing clearer for myself and any other readers :)

I have not understood your calculations. Here are mine.
Let me first rewrite your data in the following way. All your solutions have a mass ratio defined by : $\ce{$r$ = \frac{$m$(SiO2)}{$m$(Na2O)}}$ The initial solution is defined by is $\ce{$r_o$ = 2.34}$. And you wanted to obtain a final ratio $\ce{$r_f$ = 1.9}$, by adding $\ce{NaOH}$ and maybe some water, but without changing the amount of $\ce{SiO2}$, i.e. $\ce{30.3 g SiO2}$.
In other words, you wanted a final solution containing $\ce{30.3 g/1.9 = 15.95 g Na2O}$.
In moles, this amount is equal to :  $\pu{\frac{15.95g}{62 ~g/mol} = 0.2572}$ mol $\ce{Na2O}$, which is also equivalent to $0.5144$ mol $\ce{NaOH}$.
But $100$ g of the original solution contains already $\pu{16.7 g NaOH}$, or : $\ce{ \frac{16.7 g}{40 g/mol} = 0.4175 mol NaOH}$ .
As a consequence, you have to add  $\pu{~ 0.5144 mol - 0.4175 mol = 0.0969 mol NaOH}$, which weighs : $\pu{0.0969 \times ~ 40 g = 3.88 g}$ $\ce{NaOH}$.
Sorry. In my original answer, I made a wrong calculation. I have edited it recently for correcting this mistake.
This corresponds to adding $7.76$ g of a $50$% $\ce{NaOH}$ solution, to $\pu{100 g}$ of the initial solution. This is the same result as Hendricks13 in his reply 1.

Hi @Maurice, thanks for the comment! I don't think your answer is right but I couldn't fit my working/explanation in this comment box. Instead, I've edited my original post above briefly; refer to the "EDIT FROM MAURICE REPLY 1" above. I'm very curious about what the answer is now because I could be overlooking something...? Not sure if I have done anything wrong. Keen for more of your input :)


text:
Differences in activated manganese dioxide synthesis
I've looked into preparations of activated manganese dioxide for some time now, and what confused me is the contrast between the original Attenborough method and some of the alternatives.

A solution of MnSO4⋅4H2O (110 g) in H2O (1.5 L) and a solution of NaOH
(40%; 1.17 L) were added simultaneously during 1 h to a hot stirred
solution of KMnO4 (960 g) in H2O (6 L). MnO2 precipitated soon after
as a fine brown solid. Stirring was continued for an additional hour
and the solid was then collected with a centrifuge and washed with
water until the washings were colorless. The solid was dried in an
oven at 100–120 ◦C and ground to a fine powder (960 g) before use.

That is molar ratio MnSO4.4H2O:KMnO4 - 1:8,34

A solution of MnCl2⋅4H2O (200 g) in H2O (2 L) at 70 ◦C was gradually
added during 10 min, with stirring, to a solution of KMnO4 (160 g) in
H2O (2 L) at 60 ◦C in a hood. A vigorous reaction ensued with
evolution of chlorine; the suspension was stirred for 2 h and kept
overnight at rt. The precipitate was filtered off, washed thoroughly
with H2O (4 L) until pH 6.5–7 and the washing gave a negligible
chloride test. The filter cake was then dried at 120–130 ◦C for 18 h;
this gave a chocolate-brown, amorphous powder; yield 195–200 g.

Manganese (II) sulphate monohydrate (140 g) was dissolved in 2.66
liters of water and heated to 60°C. Potassium permanganate (97.3 g) in
1.85 liter of water was added over a period of 15 minutes and stirred at 60°C for 1 hour, until manganese dioxide precipitated out. The
reaction mixture was filtered and the residue was washed with
deionised water until no sulphate ion was present. The solid was dried
under suction for 2 hours followed by drying at 70°C under vacuum to a
constant weight (about 8 days) to give 115 g of a dark brown powder

In the span of 2h, to 250ml of 5% MnSO4 solution add 200ml of 5% KMnO4
solution while stiring at 50-60C. Filter out the precipitate and wash
until the absence of SO42- ions. Dry for 10h at 120C.

Whereas in these examples, it's more closer to 1:1. Why is that, will there be a difference in oxidative abilities? Alternatively, what would be the best way to synthesize this from Mn2+ salts?
Refs.:
Attenburrow, J.; Cameron, A. F. B.; Chapman, J. H.; Evans, R. M.; Hems, B. A.; Jansen, A. B. A.; Walker, T.. 194. A synthesis of vitamin A from cyclohexanone. J. Chem. Soc. 1952, 1094-1111; doi 10.1039/JR9520001094.
Galecki J. Preparatyka nieorganiczna. WNT; Warsaw: 1964. p. 439.

Duplicate?: https://chemistry.stackexchange.com/questions/154231/synthesis-of-active-chemical-manganese-dioxide-cmd-for-primary-alcohol-oxidati#comment317311_154231

@user55119 read more carefully "Whereas in these examples, it's more closer to 1:1. Why is that, will there be a difference in oxidative abilities? "


text:
Are hydrogen bonds between two hydroxyl groups stronger if one is deprotonated?
Hydrogen bonds can form between two hydroxyl groups. Is the hydrogen bond stronger if one of the groups is deprotonated?

Yes, if the one on the left is deprotonated, the hydrogen bond shown will be stronger. The opposite is true for the one on the right.

Thanks. Ah I managed to read up on that, or, understand it since it was pretty easy. The one on the right, if it is deprotonated, that would prevent any bond forming at all with it as the donor, right? Since it can't donate much anymore.

@KarstenTheis An interesting observation. The deprotonated group, it has, relatively speaking, less negative charge. Otherwise, it would be able to hold onto the proton just as well. And since the hydrogen bond acceptor is the group with most negative charge, the deprotonated one would only be favoured because it lacks the proton, not because is has less negative charge in itself.

@NilayGhosh I'm actually looking at the issue from a very unusual angle. So my complete self-answer would include it. The thing is, I think that, if introduced in a "lock and key" setting with a specific type of fit, the deprotonated group might actually be the hydrogen bond donor. This is very counter-intuitive, but, it relies on the groups being forced close to each other to start with. Then, the combined negative charge of the groups, will actually protonate the deprotonated group, and it will be the donor the the more negatively charged other group.

@NilayGhosh What I did was I wrote the answer that was the one I was looking for (plus the simple answer Karsten gave. ) I think it is factually correct. What it lacks is context, but, I know the context already, it just takes a bit of time to communicate it so I left it out. So in this context, I just require it is factually correct even if the scenario seems far fetched, and it seems right to me. It also points people to the fact that the deprotonated group is actually deprotonated because it is "relatively speaking" "less negative", an interesting observation.

The image in the question suggests that the hydroxyl groups would align parallel, which is of course not the case. If you remove a proton from the system (whether or not and how this might be done is another question) you'll have a system, in which dynamic effect will be very important. See the Grotthuss mechanism. So whether or not the bond can be considered "stronger" is highly debatable and depends on the point of vie you would like to take.

To show appreciation to @ KarstenTheis and @NilayGhosh who were both kind and polite. My exact discovery, is that ATP binds the p-loop region in a way where ATP is actually fully protonated. The exact relevance of this, can be understood by catching up with the last decade of advances in biology and the physics of what happens to water at surfaces. Here is a publication of an overview of what I've discovered. If that is "forbidden" to share then delete away I guess, I am not in control of what "Martin" or anyone else does in their "kingdom" here. https://doi.org/10.5281/zenodo.5225148

Others - including me, who actually published on the mechanism of ATP hydrolysis in the p-loop region (you're leaving out switches 1 and 2) of myosin - have demonstrated that ATP is far from fully protonated (experiment and theory). In addition to your aggressive tone, you're convinced you're right. The combination is not a good one.


text:
Can polarizing film reflect light instead of blocking light?
I was thinking about how to electorally switch glass between transparent and reflective just out of curiosity. I thought that maybe you could polarize the reflective glass in the vertical direction and polarize the light hitting the glass in the vertical direction as well, then have a layer of liquid crystals selectively reorient the light into the horizontal direction which will cause the light to be reflected back. Is this possible?

Should be the arrangement of a LCD? Except for the back light replaced by a mirror? It fits better to Physics SE, tough a better answer than my comment is certainly possible here.

Basically except I would need it to also let light through like a window when off so I couldn’t just a mirror behind it.

it should be possible as well. Otherwise active windows do exist.


text:
How to compute the work Done in an adiabatic irreversible expansion?
The question is as follows

What is the magnitude of work performed by one mole of an ideal gas when its volume increases eight times in an irreversible adiabatic expansion if initial temperature of gas is $\pu{300K}$.
$C_V$ for the gas is $\pu{1.5 }R$. ($R=\pu{2 Cal/mol/K})$
(A)$~\pu{900 Cal~}$  (B) $~\pu{450 Cal~}$ (C) $~\pu{675Cal~}$ (D)$~\pu{331.58Cal}~$

I feel like I do not have enough information. Am I missing something? I tried to use the relation $PV^\gamma$=$k$ ,But  did not get much ahead.
I used $\frac {P_1 V_1}{T_1}$=$\frac {P_2 V_2}{T_2}$
i.e keeping the moles constant to find a relation between pressures and temperatures. But this still leaves me hanging as I cannot seem to get rid of variables.
Any more suggestions?
Here is the question directly from source

PV^γ=k is only valid for reversible processes. Try using ΔU=W instead.

Yes, the problem statement should have specified that the work is done against a constant external pressure.

The problem lacks precision. There are many different ways of carrying out an irreversible expansion.

Assume the work is against constant pressure and that the final pressure is that which is worked against.
Then
$$\begin{align} w&=\Delta U = C_V \Delta T= C_V (T_f-T_i) \\ &= -p_\textrm{ext} \Delta V =-p_f(V_f-V_i) \\ &= -\frac{nRT_f}{V_f}(8V_i-V_i)\\ &= -\frac{7RT_fV_i}{8V_i}\\ &= -\frac{7RT_f}{8} \end{align}$$
Solving for $T_f$ and inserting $C_V=\frac32 R$ gives
$$T_f=\frac{C_VT_i}{\frac78 R + C_V}=\frac{1.5R T_i}{\frac78 R + 1.5 R}=\frac{1.5 T_i}{\frac78  + 1.5 }=\frac{12}{19} T_i = \pu{189.47 K}$$
Using the approximation $\pu{R= 2 cal mol-1 K-1}$ results finally in
$$w = C_V(T_f-T_i)= \pu{ 3 cal mol-1 K-1}\cdot(\pu{189.47 K}-\pu{300.00 K})=\pu{-331.58 cal}$$


text:
Migratory aptitude of chlorobenzene vs nitrobenzene in pinacol-pinacolone
I do know the basic order of migratory aptitude in pinacol-pinacolone rearrangement, which is usually said as subject to reaction conditions.
However, I wanted to know how can I compare the migratory aptitude of chlorobenzene vs nitrobenzene in pinacol-pinacolone rearrangement? What are the factors that we are supposed to check when only the benzene substituent is changed?

Higher the electron density in the benzene ring higher the migratory aptitude. So, the chloro derivative will have a higher migratory aptitude than the nitro derivative.

@NisargBhavsar thanks, I heard about this some time ago aswell however is there any rationale behind this logic?

@Ashish I guess you can think about this in terms of the transition state, where there is partial positive charge on both the carbon that the group is migrating from, and the carbon it's migrating to. If there is more electron density on the migrating group, it can donate some of that density to the positive carbons and make the TS more stable. (this is my guess)


text:
Hoffman broamide degradation but with alkoxide as base instead of NaOH

Ans.3

I thought it would be similar to Hoffman bromide reaction, and hence the answer (1) but turns out the answer is (3), suggesting the alkoxide attacks the formed amine. But, I couldn't find any references for the reaction.

This answer is wrong, the answer should be the carbamate 4.
The reaction intermediate is the isocyanate which is captured by MeOH to give the carbamate (std Hofmann mechanism below from masterorganicchemistry.com). For a Hofmann in MeOH see this Org Syn procedure here


text:
Deriving kinetic equations for reversible reactions
Problem

Given the reaction $\ce{A <=>[$k_\mathrm f$][$k_\mathrm b$] B}$ with rate constants $k_\mathrm f = \pu{4E-2 s-1}$ and $k_\mathrm b = \pu{10^{-2} s-1}$. Initially, $\pu{2 mol}$ of $\ce{A}$ and $\pu{0 mol}$ of $\ce{B}$ are taken in $\pu{1 L}$ of solution. Find:

The equilibrium concentrations of $\ce{A}$ and $\ce{B}$.
Time taken for $\ce{B}$ to reach $50\%$ of its equilibrium concentration.
Time taken for completion of $10\%$ of the reaction.

Answer

 1. $[\ce{B}] = \pu{1.6 M},$ $[\ce{A}] = \pu{0.4 M}$
 2. $\pu{14 s}$
 3. $\pu{2.8 s}$

Question
Finding the equilibrium concentrations was a trivial task. However, I cannot find the concentrations of $\ce{A}$ and $\ce{B}$ as a function of time. I wrote the relevant differential equations:
$$
\begin{align}
\frac{\mathrm d[\ce{A}]}{\mathrm dt} &= -4×10^{-2}[\ce{A}] + 10^{-2}[\ce{B}]\\
\frac{\mathrm d[\ce{B}]}{\mathrm dt} &= 4×10^{-2}[\ce{A}] - 10^{-2}[\ce{B}]\\
\end{align}
$$
I am unable to proceed from here, as I have so far only learnt application of chemical kinetics for one-way reactions.

Look up coupled differential equations , http://www.maths.surrey.ac.uk/explore/vithyaspages/coupled.html

@andselisk - I realize my fault in using sec for seconds, but I couldn't spot what had been changed each time in kf and kb (they look similar before and after). Could you please clarify, so that I'll format them properly in my future posts? Thanks :)

@TRC In these notations "f" and "b" are descriptive indices, referring to "forward" and "backward" labels (not variables), which should be upright. See Which symbols are written in roman (upright) font and which are italicized? I cannot find a reference at the moment, but preferred symbols in kinetics are actually $k_\mathrm{fwd}$ and $k_\mathrm{rev}$ ("forward" and "reverse", correspondingly) to avoid ambiguity.

Actually you need concept of coupled differential equations but I suppose you can do a 'trick' to avoid it:
$$ \frac{dA}{dt} =- a A +  b B$$
$$ - \frac{dB}{dt} = - aA + bB$$
Are the form of the two equations, you may subtract the two equation to find $ \frac{dA}{dt} = - \frac{dB}{dt}$, and by integration: $ A(t)+B(t)=C$, setting $t=0$ , we find $A(0)+B(0)=2=C$. Hence, we have the following law governing the equation:
$$A(t) + B(t) = 2$$
The above equation is interpreted 'conservation of moles'.
Using this new equation , plug it into any of the first order diffy eqtn and make it single variable diffy equation which you have standard methods to solve.

Pls note: I took the given equation for dB/dt and multiplied by negative one , a= 4* 10^{-2} and b=10^{-2}

Thank you so much - Really dumb of me not to think of A+B = 2!

I wouldn't say it is dumb , you can't be sure of anything unless you can prove it /experimentally verify ;)

I'll just post what I did after @Buraian's answer for a general case -
Let the reaction be $\ce{A <=>[$k_\mathrm f$][$k_\mathrm b$] B}$
Let the concentration of $\ce{A}$, $\ce{[A]} = a$, and $\ce{[B]} = b$. If initially $c$ moles of $\ce{A}$ and $0$ moles of $\ce{B}$ are taken, then the relation holds: $a + b = c$.
Also, by using equilibrium conditions
$$\frac{a_\mathrm{eq}}{b_\mathrm{eq}} = \frac{k_\mathrm{b}}{k_\mathrm{f}}$$
which gives
$$a_\mathrm{eq} = \frac{k_\mathrm{b}c}{k_\mathrm{f} + k_\mathrm{b}} ,  b_\mathrm{eq} = \frac{k_\mathrm{f}c}{k_\mathrm{f} + k_\mathrm{b}} $$
Now the differential equation gives
$$
\begin{align}
\frac{\mathrm d[\ce{A}]}{\mathrm dt} &= -k_\mathrm{f}[\ce{A}] + k_\mathrm{b}[\ce{B}]\\
\frac{\mathrm da}{\mathrm dt} &= -k_\mathrm{f}a + k_\mathrm{b}b\\
\frac{\mathrm da}{\mathrm dt} &= -k_\mathrm{f}a + k_\mathrm{b}(c-a)\\
\frac{\mathrm da}{\mathrm dt} &+ (k_\mathrm{f} + k_\mathrm{b})a = k_\mathrm{b}c\\
\end{align}
$$
The RHS is a constant, making it a first order linear differential equation which can be solved easily using standard methods.
The initial condition, $a = c$ at $t = 0$ can be used for the particular solution.
The solution comes out to be
$$ a = \left(\frac{k_\mathrm{b}c}{k_\mathrm{f} + k_\mathrm{b}}\right) + \left(\frac{k_\mathrm{f}c}{k_\mathrm{f} + k_\mathrm{b}}\right)\mathrm{e}^{-(k_\mathrm{f}+k_\mathrm{b})t}$$
which can also be written as
$$ a = a_\mathrm{eq} + b_\mathrm{eq}\mathrm{e}^{-(k_\mathrm{f}+k_\mathrm{b})t}$$

This is a classic example of the kinetics of reversible reactions.
From the mechanism  $\ce{A <=>[$k_\mathrm f$][$k_\mathrm b$] B}$ you can write the kinetic equations
\begin{align}
\frac{\mathrm d[\ce{A}]}{\mathrm dt} &= - k_a [\ce{A}] + k_b [\ce{B}]\\
\frac{\mathrm d[\ce{B}]}{\mathrm dt} & =  k_a [\ce{A}] - k_b [\ce{B}]\\
\end{align}
To obtain the dependence of the concentration with time, you must solve this set of differential equations, but its solution is
trivial since both concentrations are not independent
\begin{equation}
    [\ce{A}]_0 + [\ce{B}]_0 = constant = [\ce{A}]_\mathrm{t} + [\ce{B}]_\mathrm{t}
\end{equation}
thus
\begin{equation}
    [\ce{B}]_\mathrm{t} =  constant - [\ce{A}]_\mathrm{t}
\end{equation}
In the case you describe $[\ce{B}]_0 = 0$ so
\begin{equation}
    [\ce{B}]_\mathrm{t} =  [\ce{A}]_0 - [\ce{A}]_\mathrm{t}
\end{equation}
If the initial concentration of the species A is $[\ce{A}]_0 = a$, and $[\ce{B}]_\mathrm{t} = x$, $\mathrm{[A]_t = a -x} $
\begin{equation}
   \mathrm{ \frac{d[\ce{A}]}{dt} = -\frac{d x}{dt} = - k_a (a-x) + k_b x}
\end{equation}
The solution of this differential equation is simple and the result is
\begin{equation}
   \mathrm{ x_{eq} - x = x_{eq} \, e^{-(k_a + k_b) t}}
\end{equation}
where $\mathrm{x_{eq} = |B|_{eq}}$.
You write this expression as a function of $\mathrm{|A|_t}$ and $\mathrm{|B|_t}$ as
\begin{align}
  \mathrm{|A|_t \,=\,} & \mathrm{|A|_{eq} + |B|_{eq}e^{-(k_a + k_b) t}}\\
  \mathrm{|B|_t \,=\,} & \mathrm{|B|_{eq} (1 -e^{-(k_a + k_b) t})}\\
\end{align}
Thus this reaction behaves as a first order reaction with $\mathrm{k_1 = k_a + k_b}$ and with $\mathrm{|A|_\infty = |A|_{eq}}$ and $\mathrm{|B|_\infty = |B|_{eq}}$

Furthermore, once the reaction has reached the equilibrium
\begin{equation}
  \mathrm{v_a}  \,=\, \mathrm{v_b}
\end{equation}
\begin{equation}
  \mathrm{k_a\,|A|_{eq}} \,=\, \mathrm{k_b\,|B|_{eq} }
\end{equation}
so
\begin{equation}
   \mathrm{K_{eq}} \,=\, \frac{\mathrm{|B|_{eq}}}{\mathrm{|A|_{eq}}} \,=\, \frac{\mathrm{k_a}}{\mathrm{k_b}}
\end{equation}
Using the equilibrium equations and the integrated rate equations you can answer the questions.
Bibliography.
This particular mechanism is a classic in chemical kinetics.
You can find more information on general physical chemistry or introductory chemical kinetics textbooks. For example:
T. Engel et al., Physical Chemistry, 3rd Edition, Pearson (2013)
M. J. Pilling and P. W. Seakins, Reaction Kinetics, 2nd Edition, Oxford University Press (1996)
K. J. Laidler,  Chemical Kinetic, 3rd edition, Pearson (1997)

Using SymPy:
>>> from sympy import *
>>> t = Symbol('t', real=True)
>>> M = Matrix([[ -0.04,  0.01],
                [  0.04, -0.01]])
>>> exp(t * M) * Matrix([2,0])
Matrix([[0.4 + 1.6*exp(-0.05*t)],
        [1.6 - 1.6*exp(-0.05*t)]])

In order to find when the concentration of species $\ce{B}$ reaches $50\%$ of its equilibrium concentration, we solve the following transcendental equation
$$1.6 - 1.6 \exp(-0.05 t) = \frac{1.6}{2}$$
and obtain
$$t_{50\%} := 20 \ln(2) \approx 13.86 \approx 14 \,\pu{seconds}$$


text:
What is the format of mol2vec featurize?
When I use the featurize command in mol2vec, I get a .csv file as an output with about 300 columns. But these columns are simply labelled by their number. What feature does each column represent?
Reference:  Jaeger, S.; Fulle, S.; Turk S. Mol2vec: Unsupervised Machine Learning Approach with Chemical Intuition in J. Chem. Inf. Model. 2018, 58, 1, 27–35; doi 10.1021/acs.jcim.7b00616.

For future reference: there was to little information to understand the question right from the get-go, so I added what looks like the program your question is about (I don't know it).  Perhaps a more suitable audience is mattermodelling than chemistry.se.

To quote from a documentation of the program here (still assuming only it were the one of the question):

Subcommand 'featurize'
Featurizes new samples using pre-trained Mol2vec model. It saves the result in CSV file with columns for molecule identifiers, canonical SMILES (generated during featurization) and all potential SD fields from input SDF file and finally followed by mol2vec-{0 to n-1} where n is dimensionality of embeddings in the model.
Example:
To featurize new samples using pre-trained embeddings and using vector trained on uncommon samples to represent new substructures: mol2vec featurize -i new.smi -o new.csv -m model.pkl -r 1 --uncommon UNK
For more detail on individual subcommand run: mol2vec $sub-command --help