text:
Explanation of rusting of iron through metallic bonding

In the diagram , colour representations :
[O - oxygen atoms , red dots - electron ]
I will explain rusting of iron due to metallic bonding as I have understood from online. Please correct me if I’m wrong.

In the picture , metallic bonding is happening between Fe and Fe atoms. Electron present then circulates around the different Fe atoms.
After some time , oxygen atoms jump in and the electron of Fe is given to O which forms $Fe_2O_3.$ .Then , what happens is that Fe atoms have formed a new bond.
New molecule I.e $Fe_2O_3.$ formed drifts apart from other atoms. Therefore , there is a loss of Fe atoms. This loss is knows as Rusting.
My questions from this explanation are :

Is my explanation correct ?

Why does rusting take time to happen ? Why doesn’t it happen immediately ?

My answer : It depends on the time taken for Fe to form bond with O atoms.
Please do tell me how can I improve the quality of my question. This way , I can post better questions and grow along this site.
Thank you.

No clear question, no clear problem

Rust is not Fe2 O3. It is various hydrated oxides and hydroxides with several formula and crystallographic patterns  Fe2 O3 develops at elevated temperature ( dry ), and fretting corrosion at room temperature

You better remember to use proper punctuation and formatting in the future. Don't waste time on this one.

Greg is quite correct: no clear question. But sometimes you don't know enough to define a question clearly. The implicit question is: "Is this mechanism helpful or even correct?" The second, clear, question, is why does rusting take time to happen.
I think the answer to the first, implicit, question is that it is a start, but far from helpful and it has some erroneous elements (not the Fe, of course, but the "O" should probably be written as O$_2$). Then, as blacksmith37 points out, rusting as it occurs in nature involves H$_2$O. Other things get involved too, like chloride ion, which accelerates rusting. And pH. It begins to get so complicated that even a knowledgeable chemist would find it difficult to pose a clear question that could be answered in a kiloword or less. The whole anti-corrosion industry continually seeks answers to questions like this.
Answering the second question is a bit easier: everything takes time. Why? Well many reactions of things we deal with in real life have a certain stability, otherwise we wouldn't have them to deal with. For instance, liquid helium has to be protected (insulated) to keep it around, otherwise it vaporizes so rapidly that it goes away in no time at all. Iron, on the other hand, resists air and water for quite a while because those iron atoms in their sea of electrons (the OP got that image fairly well) don't see any better (more energetic) use for those electrons. Now along comes a water molecule, and maybe the iron finds a way to bind to the oxygen and loosen the bond to (one of the hydrogens in the H$_2$O. Maybe. All reorganizations take some energy to move atoms around - break this bond, double that one, then release to another atom, etc. We would call that an activation energy: the greater it is, the less likely the reaction is to occur - or, it just takes more time. Thank goodness - otherwise it would be like making bridges out of sodium. You couldn't get one up before it disappeared from oxidation!

Hi sir. I have edited my Q. Is there anything I should change ?

@SrijanM.T: Start with my answer, but realize that many steps, many different reactions, take place. Look at some models and formulas that are already on the internet, e.g., at https://byjus.com/chemistry/rusting-iron-prevention/. Look up some articles on rusting (Wikipedia, and others). Start your investigation by using some description or model that already exists, and build up on that. The model in your question is very elementary; it doesn't allow for a detailed mechanistic explanation. You have begun by asking a simple question; now, keep digging! Keep searching! Mg, Cu, Al, Ti, Ag, Pt...

Ok @JamesGaidis Thank you. I’ll surely do that. I have a small request that I am not able to ask any Q on this site now. If you see my Q & Ans now. I have improved and got good upvotes as well. If you can give this Q or any other Q an upvote so I can further ask Q. I’d be really grateful. Thanks

Be more specific: what question(s) are you proud of? Give me a link.

https://chemistry.stackexchange.com/a/153055/104564 @JamesGaidis

https://chemistry.stackexchange.com/a/152743/104564

First 2 are answers which I do am proud of.

https://chemistry.stackexchange.com/q/146720/104564 . This one is a Q which I think for someone who has started mole concept. It is really helpful to understand this concept.


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Why can't sodium borohydride reduce carboxylic acid and ester groups?
Sodium borohydride ($\ce{NaBH_4}$) is a reducing agent used in various organic reaction to reduce aldehydes, ketones and acyl chlorides.
Why can't sodium borohydride reduce carboxylic acid groups and esters? Why can it reduce acyl chlorides but not esters, both which are derivatives of the carboxylic acid group?
I read somewhere that it had to do with the oxidizing property of sodium borohydride, but I'm not sure if it is correct or how it explains the same.

It forms the acyloxyborohydride which is a stable species, and indeed a reducing agent cf sodium triacetoxyborohydride.

Think about the way the borohydride ion can react with carbonyls, mechanistically. This should make you think about the electrophilicity of carbonyls, and how bad carboxylic acids (and esters, to some extent) behave as electrophyles. Now, for the carboxylic acid, try to think about the acidity of this species and how this could affect the borohydride ion.

NaBH4 is very slow to reduce esters but LiBH4 (NaBH4 + LiCl in solution) will get the job done.

Thank you. I understand what all of you mean.

@NilayGhosh Yes, it does. That  answer makes more sense, thanks.

The reason for this is because OH and OR groups are much more electron-donating. Typically, when you have an aldehyde or ketone, the electronegative electron is slightly more negative and thus the carbonyl carbon is more positive. However, with an OH or OR group attached (i.e in carboxylic acid and esters), the group can push electrons towards the carbon since they are electron-donating groups. This makes the carbonyl carbon less positive and less open to nucleophilic attack.

But won't that effect be the same regardless of the reducing agent? I think https://chemistry.stackexchange.com/questions/38747/why-nabh4-does-not-reduce-carboxylic-acid-while-lialh4-does might be right...

Yes LiAlH4 can reduce carboxylic and esters because it is stronger than NaBH4. I just answered why NaBH4 can reduce aldehydes and ketones but not carboxylic acids or esters

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What is the mechanism of formation of alkali metal superoxide?
I was studying about alkali metals. I can get how oxides and peroxides are formed.
In alkali metal oxides, both of the bonds of the dioxygen molecule first undergoes homolytic cleavage, then the nascent oxygen gets two electrons from two metal atoms.
In alkali metal peroxides, one bond of dioxygen molecule undergoes homolytic cleavage and this results in formation of O-O with one unpaired electron on each oxygen then 2 metal atoms donate their electrons to each of the oxygen.
But how superoxides are formed?

Assigning electrons to specific atoms in the O2 species is not really accurate.  I would begin by looking at how the electrons are shared among both atoms in the molecular orbitals of this system.

" I can get how oxides and peroxides are formed." Do you? You got it wrong, there's always reduction first and knowing that, the superoxide mechanism would be simplest for you.

I thought the whole nascent oxygen theory has been debunked.

Getting to know the superoxide anion radical
Hayyan et al. (2016) have studied and characterized the formation of the superoxide anion radical, $\ce{O2^.-}$. This chemical entity forms as a result of the reduction of diatomic oxygen
$\ce{O2 + e^- -> O2^.-}$
There is some conflicting data on the reactivity of this radical, however it can be isolated:

A few studies have reported that $\ce{O2^.-}$ is relatively reactive. In contrast, a previous study showed that $\ce{O2^.-}$ or hyperoxide, as some scientists name it, is not highly reactive despite being a free radical. The 17-electron $\ce{O2^.-}$ is one of the isolable paramagnetic main-group ions and one of the many intermediate ORR species.

This chemical species can be synthesized in a myriad of ways. It is actually an intermediary in the mitochondrial respiratory chain and it plays a really important role in the activity of respiratory and immune defense systems.

Superoxide can be generated not only enzimatically (as it has been previosuly mentioned), but also electrochemically, photochemically and photocatalitically. Usually, it is either hydrogen peroxide or the peroxy radical used as target-molecules for the synthesis of this species.
Carrying out the synthesis of superoxide photochemically - a mechanism based on solvated electrons
As mentioned earlier, crossing the boundary between diatomic oxygen and superoxide can be done in mutliple ways. Photochemically, this leads to the formation of solvated electrons in the reaction mixture. One electron can attack one oxygen molecule, forming superoxide. Holroyd et al. have concluded in 1978 that photochemical composition of water is the main process leading to the formation of $\ce{e^-(H2O)n}$. The hydroxyl anion $\ce{OH-}$ that forms after the primary dissociation can absorb energy and disproportionate into hydroxyl radical $\ce{OH^.}$ and a solvated electron.
Chemically generating the species
Superoxides of alkali metals have been heavily documented, as well as their synthetic pathways. These alkali cations (such as potassium, sodium and even rubidium and cesium are referred to as superoxide carriers in this context). Zhdanov et al. (2004) have studied the kinetics of synthesis of potassium superoxide. The chemical environment in which this reaction was carried out is alkaline solution of hydrogen peroxide.
$\ce{H2O2 + HO- -> HO2- + H2O}$
Apart from the hydroperoxide species (which involves proton exchange between hydrogen peroxide and the hydroxide species), there has been observed an exchange leading to the formation of an intermediate adduct:

Being a weak acid, hydrogen peroxide enters, under certain conditions, into an exchange reaction with potassium hydroxide to give an intermediate adduct $\ce{K2O2*H2O}$ (potassium peroxide diperoxohydrate).

$\ce{KOH + 3H2O2 -> K2O2*2H2O2 + 2H2O}$
It is this intermediary that disproprotionates upon mild heating into the superoxide species:
$\ce{K2O2*2H2O2 ->[40^o C] 2KO2 + 2H2O}$
This is the most widely used method for obtaining these salts. It is different from the photochemical/electrochemical methods, since the superoxide radical is only generated in situ (if, at all).

Recently, an in situ method was proposed for preparing superoxides by integrating $\ce{H2O2}$ with aqueous $\ce{KOH}$ or sodium hydroxide ($\ce{NaOH}$). $\ce{NaO2}$ was generated by mixing $\ce{NaOH}$ and $\ce{H2O2}$ in aqueous media under ambient conditions (eq 3). $\ce{NaO2}$ is generated by the rapid formation of alkali peroxide, followed by its decomposition in excess $\ce{H2O2}$. However, this method did not provide alkali superoxides in a powder form. By contrast, it was stated that, in this in situ method, $\ce{O2^.-}$ can be generated in solutions for diverse applications.

Conclusions
Two different mechanistic pathways are known for the synthesis of superoxide-based alkali metal salts. One of them involves directly the $\ce{O2^.-}$ species (obtained electrochemically, enzimatically or photochemically) which gets to react with the alkali metal's exposed surface.

Sodium metal, coated in a mixture of superoxide, peroxide, oxide, hydroxide and possibly carbonates
The other mechanism involves the disproportionation of the alkali peroxide-hydrogen peroxide adduct, freeing the superoxide species in the reaction mixture (apologies, but I have found no concrete mechanistic details for this rather bizarre disproportionation).
References

https://en.wikipedia.org/wiki/Sodium_superoxide
Hayyan m., Hashim M.A., Al Nashef I.M., Superoxide Ion: Generation and Chemical Implications, Chem. Rev., 2016, 116 (5), pp 3029–3085.
Sawyer, D. T. Superoxide Chemistry, McGraw-Hill,
D. V. Zhdanov, M. A. Ul’yanova & Yu. A. Ferapontov, A Study of the Kinetics of Synthesis of Potassium Superoxide from an Alkaline Solution of Hydrogen Peroxide, Russian Journal of Applied Chemistry
Richard A. Holroyd and Benon H. J. Bielski, Photochemical Generation of Superoxide Radicals in Aqueous Solutions, Contribution from the Chemistry Department, Brookhaven National Laboratory, Upton, New York 11 973, Received March 9, I978

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text:
Reaction of but-2-ene with Cl2/CCl4
I have often studied the reaction of Br2/CCl4 with But-2-ene which results in formation of

But I wondered what would be formed if Cl2/CCl4 reacts with But-2-ene. Does it lead to the formation of
This:
Or This:
Or something else which I never expected

What have you attempted to do with regards to the answer to this question? What makes you think it is something different?

Because although allylic halogenation reactions don't take place with Br2 direclty (NBS is usually used for the same for low concentration) I've seen allylic halogenation reactions taking place directly with Cl2 probably at a high temperature

And thats what made me wonder which course will this reaction take. Standard addition or allylic halogenation

High temperatures lead to radical reactions. $\ce{Cl2/CCl4}$ is a standard addition reaction leading to vicinal dihalides, no high temp here.

@SafdarFaisal Ok thanks got it. One more thing I noticed was that in allylic halogenation reactions of Cl2, at high temperatures ofcourse, it was in gaseous state and not dissolved.

$\ce{Cl2}$ becomes a gas at $\pu{-34.6 ^\circ C}$, what is expected at high temperatures?


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Are the orders of reactants with respect to a reaction different for different starting concentrations?

If the reaction is repeated with $\pu{2 M}$ ethyl iodide the pyridine concentration decreases as shown below. Give the rate law of the reaction in terms of pyridine and ethyl iodide. Explain your reasoning.
$$
\begin{array}{l|rrrrr}
\hline
t/\pu{min} & 0 & 20 & 40 & 80 & 120\\
\hline
[\ce{C5H5N}]/\pu{M} & 0.1 & 0.0746 & 0.0559 & 0.0312 & 0.0175\\
\hline
\end{array}
$$

I have found the rate constant and found that the orders of ethyl iodide and pyridine are both $1$ with respect to the reaction when their initial concentrations were both $\pu{0.1 M}.$ I would have thought this would be the case for any set of initial concentrations.
I would like to confirm this. In that case, I am thinking the only thing different would be the rate constant. But how would I go about finding that?
Could I plot the data, interpret the half life and go from there? Or is my approach totally wrong?

the order of a reaction is independent of starting concentration, if it were so then order would have kept changing at every moment sinc each moment can be considered as a starting point.

When the concentration of one reactant is much higher than the other, then it can appear to be zero order in the high concentration reactant because the percent change in that concentration is so small that it has a negligible effect on the overall rate. But it is still first-order in that reactant even if it's not observable under those conditions

This makes sense, thank you both. The rate constant would be a different value though wouldn’t it?

Disregard that. I read it over and realise that wouldn’t make sense

A reaction's rate does depend upon the consumption of its reactants, and the manner in which the reactants interact. Let's consider the following reaction: $$\ce{A + B -> final products}$$ The rate of consumption of  reactants $$\dfrac{-d[A]}{dt}= k[A][B]$$ and $$\dfrac{-d[B]}{dt}= k[A][B]$$
It seems that this reaction is a second order reaction since the rate order appears to be:$$\text{Rate} = k[A][B] $$
As @Andrew astutely pointed out in the comments, a reaction can be of a lower order if one of the reactant's concentration is much higher than the other's.
Let's suppose, that one reactant is in great excess (say $10 \text{M of B vs } 0.01\text{M A}$)
So what we're observing now is that the reactant B is barely consumed, yet the reaction terminates since reactant A, the limiting reagent is exhausted.
In other words, the rate equation looks something like this: $$\text{Rate}=k'[A]$$, where $k^{'}=k[B]_o$ .Now, the reaction appears to be a first order reaction. This is known as a pseudo first-order reaction, since the rate of the reaction ultimately depends only on one reactant.
An example that I found here:
$$ \rm{CH_3Br + OH^- \rightarrow CH_3OH + Br^-}$$

[...]"Imagine we had an initial concentration of CH3Br of 100 μM and and an initial concentration of OH- of 10 mM.   Now even if all of the CH3Br has reacted the concentration of OH- will be essentially unchanged.  Therefore during the course of the reaction, the concentration of OH- will be essentially constant.  This makes the reaction "like a first order reaction", thus the name pseudo-first order."

Effectively, the rate now is: $$\ce{rate= k'[CH_3Br]}$$
This can also be mathematically represented in the integrated rate law:
$$\dfrac{1}{\ce{[OH-]_0 - [CH3Br]_0}}\ln \dfrac{\ce{[OH-][CH3Br]_0}}{\ce{[CH3Br][OH-]_0}} = kt \quad \ce{where [OH-]\neq[CH3Br]}$$
Now, since $\ce{[OH-]_0 \gg [CH3Br]_0}$, we can write the equation as:
$$\dfrac{1}{\ce{[OH-]_0 - [CH3Br]_0}}\ln \dfrac{\ce{[OH-][CH3Br]_0}}{\ce{[CH3Br][OH-]_0}} \approx \dfrac{1}{\ce{[OH-]}}\ln \dfrac{\ce{[CH3Br]_0}}{\ce{[CH3Br]}}=kt $$
Remember, however that the rate and order of a reaction usually depends upon its mechanism.
For further reading, check here and here as well.


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Reasoning for redox reactions to be classified as such
Could anyone help me out solving this question regarding redox reactions? I am fairly new to this topic.

Which of the following is a redox reaction?
(1) $\ce{NaCl + KNO3 -> NaNO3 + KCl}$
(2) $\ce{CaCO3 -> CaO + CO2}$
(3) $\ce{Mg(OH)2 + 2NH4Cl -> MgCl2 + 2NH4OH}$
(4) $\ce{Zn + 2AgCN -> 2Ag + Zn(CN)2}$

I know that the answer here is option (4) but could someone help me understand why this is so? I see a change in oxidation in the other three equations but it is not redox.

why do you say it is not redox when you "see" a change in oxidation?

I suspect you do not really know what is an oxidation number

never mind @Maurice and Safdar. I finally (after much research and frustration) found out that redox reactions don't depend on the change in the presence of oxygen. you know, indian textbooks are very incomplete in terms of content and you could only get full understanding after watching some khanacademy

$\ce{Zn + 2AgCN -> 2Ag + Zn(CN)2}$
Zinc changes its oxidation state from $0$ to $+2$, and silver changes its oxidation state from $+1$ to $0$. So, zinc is reducing agent and silver is oxidising agent.
$\ce{Zn^0 ->[-2e^-] Zn^{+2}}$ (oxidation);
$\ce{2Ag^{+1} ->[+2e^-] 2Ag^0}$ (reduction).
Other reactions are just equations of ion exchange without changing the oxidation states of chemical elements.


text:
Which of these can undergo decarboxylation reaction readily

Which of these carboxylic acids can undergo decarboxylation readily?

According to what I know beta-keto acids undergo decarboxylation easily and readily due to formation of a stable intermediate. So my general answer was option (A) however the answer key states that (A),(B) and (D) are correct. I am not able to understand, based on my current theory what is the reason for (B) and (D) to undergo decarboxylation readily.

With respect of picolinic acid (D) have a look at this   https://en.wikipedia.org/wiki/Hammick_reaction

Why do $\beta$-Ketoacids give decarboxylation reactions readily?

@SafdarFaisal https://chemistry.stackexchange.com/questions/53879/why-are-beta-ketoacids-better-at-decarboxylation-than-gamma-or-delta

On the decarboxylation of the picolinic acids: https://chemistry.stackexchange.com/questions/86718/rate-of-decarboxylation/86747#86747

You have said that  β−ketoacids undergo decarboxylation easily and readily due to formation of a stable intermediate. That intermediate is the carbanion on the alpha carbon. So, any species that can stabilize that carbanion can undergoe decarboxylation easily. In (B), the carbanion formed is stabilized by the double bond beside it so it undergoes decarboxylation easily.
(D) is a special reaction as referenced to in this answer.

The only thing I felt was whether stabilization by double bond is "enough" for formation of carbanion to occur but it probably is. Thanks

Won't the Aromaticity of the cyclo pentene ring be lossed?


text:
Which process would separate the mixture of toluene, cyclohexane and water?
I think that answer would be:

Toluene and cyclohexane do not dissolve in water (since they are nonpolar), so I would first separate the water from the mixture with a separator funnel.

A mixture of cyclohexane and toluene would be separated by fractional distillation. Cyclohexane, which has a lower boiling point than toluene, would be eliminated from the mixture.

Is there anything wrong in my answer? Would there also be anything to add?

Sounds good to me.

Cyclohexane and toluene may probably be difficult to separate by distillation. Toluene is not very different from benzene. And i know that benzene and cyclohexane are forming an azeotropic mixture (59% benzene) which boils at 77°C, that is lower than the boiling points of benzene and of cyclohexane. So I would not be surprised to learn that toluene and cyclohexane are making an azeotropic mixture. This should be checked.

Cyclohexane and toluene would not make azeotrope. Fractional distillation would do the trick.

Need to check ternary phase diagram of cyclohexane:toluene:water.
There are azeotropes water:toluene at 89 °C (0.49:0.51) and water:cyclohexane at 69 °C (30:70).
Hence, first separate water if you can on the basis of density (layer separation or extraction). And then separate cyclohexane and toluene by fractional distillation. In fractional distillation, you need to repeat the vaporization–condensation cycle to achieve your desired quality.

Could you please append the phase diagram you are referring to?

The first step will be to separate the water from the cyclohexane; a separatory funnel could be used, or the organic layer could be suctioned off with a pipette.
Toluene freezes at -95ºC; cyclohexane freezes at +6.46ºC. Dry ice (solid CO2) sublimes at -78ºC (Ref 1).
Get some dry ice, put the cyclohexane-toluene mix (in a test tube) into the dry ice, wait a little bit, pour off the toluene and let the cyclohexane melt. The separation will be ~90-95% complete. (The separation could be better on an industrial scale.)
The reference also describes some cooling bath temperatures above -78ºC. The very lowest temperature may not be needed, and higher temperatures might give better crystallization (i.e., large crystals of cyclohexane rather than a fine powder or a clunky solid mass).
Ref 1. https://en.wikipedia.org/wiki/Cooling_bath

Distillation
The most effective and easiest way to separate the mixture of toluene and cyclohexane is simple distillation. Cyclohexane has boiling point is about 81°C, and toluene’s boiling point is 111°C.
References:

http://origin-www.thermoscientific.jp/content/dam/tfs/ATG/CAD/CAD%20Documents/Product%20Support%20Bulletins/NMR%20Support%20Documents/pS45-pS80-Simple-Distillation-of-Cyclohexane-and-Toluene.pdf
https://www.sas.upenn.edu/~dbalmer/chem502/Labs/lab2_distillations/balmerexp2.pdf

Simple distillation is not the most effective but easiest. If you look at the NMR of the first distillate, it is 1:2 mixture of toluene to cyclohexane. The residue is 8:5 mixture of toluene to cyclohexane. I made my point.


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Chemical properties of dicopper (I) tetraiodomercurate (II)
This is a continuation from my  previous post.
Where can I find information on the chemical properties of $\ce{Cu2[HgI4]}$? I found a mention that in reaction with $\ce{Na2S2O3}$ in presence of acid it gives $\ce{HgS}$.

I came from your previously asked question. While it was a good idea to create a separate question, you should link your original question here lest some users feel you haven't done enough research.

Check II of this answer


text:
How does the blue color of starch solution change depending on temperature and what is the cause for this?
I was doing a project for my school and im not sure the following answer is correct:
"After adding iodine liquid to starch solution the colour turns blue. By heatig the mixture loses its colour. "
I am not sure if that is fully correct, i also don't know what even is the cause for this change.
If anyone can check the answer and explain what would the cause be, i would be very grateful. Thanks!

Yes it is correct. Iodine and starching solution are making a deep blue complex at room temperature. The reaction is part of an equilibrium, and it is slightly exothermic.  When heating an equilibrium at a higher temperature, the endothermic back-rection is favored. So this complex is destroyed and the solution contains only starch and iodine. The reaction is reversible. The blue complex is produced again when cooling the solution.

Thanks a lot for answering, so if i understood correctly the blue color disappears because the iodine comes out of the starch helix while heating ?

You have perfectly understood.

Not entirely reversible, as starch slowly hydrolyzes, and faster at higher temperature, breaking into simple, clear, sugars.


text:
Toxicity of Mercury (II) Iodide and dicopper (I) tetraiodomercurate (II)
This is a continuation from my previous post.
Could you please give me some info on the toxicity of compounds $\ce{HgI2}$ and $\ce{Cu2[HgI4]}$? I found on sigmaaldrich.com that these are very dangerous but could not find any information on the dosages for skin contact and swallowing.

You should check the msds of the compounds. For e.g. HgI2:. https://fscimage.fishersci.com/msds/13820.htm

Following the @Nilay Ghosh recommendation, I found an SDS sheet for $\ce{Cu2[HgI4]}$ and there is SDS for $\ce{HgI2}$.
For $\ce{Cu2[HgI4]}$:

Acute toxicity estimate Oral (Expert judgment): 5,1 mg/kg
Acute toxicity estimate Inhalation (Expert judgment) 4 h: 0,051 mg/l
Acute toxicity estimate Dermal (Expert judgment): 5,1 mg/kg

For $\ce{HgI2}$:

LD50 oral rat: 18 mg/kg
Acute toxicity estimate Inhalation (Expert judgment) 4 h: 0,051 mg/l
Acute toxicity estimate Dermal (Expert judgment): 5,1 mg/kg


text:
Why are enthalpy of combustion and molar mass of organic compounds proportional?
I noticed that when you graph heat of combustion against molar mass, you get a completely straight line with a correlation of 1. I know that the longer the chain is, the stronger the LDF get, so it is expected to see an increase in heat of combustion if molar mass increases. Though, I don't know why the curve is completely straight.
You can see heat of combustion against molar mass of alkanes in the graph below.

Data source: Organic Compounds: Physical and Thermochemical Data

What is LDF? Where did the data come from?

@BuckThorn London Dispersion Forces.

What are the units of the quantities in your data table?

@KarstenTheis Since I am showing a relationship between two variables which can be found even in the question's title, I do not think units or error calcuations are relevant as long as there is consistency within the data.

You don't need the units to show linearity, but you do need the units when comparing the order of magnitude of the combustion enthalpy to the enthalpy of breaking London dispersion forces. And you need the sign to by consistent (these reactions are exothermic, and breaking London forces is endothermic).

The dominant effect is simply the number of carbons in the molecule. In simple hydrocarbons, more mass = more CH2 units and each of those adds a fairly constant unit of energy when fully combusted to CO2 and H2O.

The biggest contributor to the heat of combustion is the difference in covalent bond energy of reactants and products. The intermolecular forces that have to be overcome for combustion of liquids (with products all gases) are an order of magnitude lower (adapted from https://chemistry.stackexchange.com/a/125013):
$$
\begin{array}{lrr}
\hline
n\text{-Alkane} & T_\mathrm{b}/\pu{K} & \Delta_\mathrm{vap}H(T_\mathrm{b})/\pu{J mol-1}\\\hline
\text{Methane} & 112 & 8176\\
\text{Ethane} & 184 & 14640\\
\text{Propane} & 231 & 18832\\
n\text{-Butane} & 273 & 22390\\
n\text{-Pentane} & 309 & 26352\\
n\text{-Hexane} & 342 & 28850\\
n\text{-Heptane} & 372 & 31800\\
n\text{-Octane} & 399 & 33972\\
n\text{-Nonane} & 424 & 36910\\
n\text{-Decane} & 447 & 38750\\
\hline
\end{array}
$$

[OP] I know that the longer the chain is, the stronger the LDF get, so it is expected to see an increase in heat of combustion if molar mass increases.

Some of these alkanes are gases, so there are no dispersion forces to overcome. Also, you got the sign of the contribution wrong. If a reactant has higher LDF than the products (which have none), this would make the reaction less exothermic. So you would expect lower magnitudes for the larger alkanes, opposite of what is observed.
Instead, the heat of combustion scales with the number of carbons (see e.g. this table and graph) because the number of $\ce{C=O}$ and $\ce{O-H}$ bonds made (and the number of $\ce{C-C}$ and $\ce{C-H}$ bonds broken) also scales with the number of carbons.

See Table I, "Alkane Heats of Combustion".  All the heats of combustion can be expressed as about 148 kcal/"mol -CH2-", barring some small details such as ring strain.  But when you express them per mole of the alkane, well, the longer the alkane, the more -CH2-s, the more enthalpy, in nearly direct proportion.  The enthalpy per mass burned remains about the same.


text:
Glycerin yield upon saponification of beeswax
I am making a custom grease (castor oil & sodium stearate) that will be suspended in beeswax. I have found that saponifying some beeswax for use in the grease results in higher operating temperatures after the grease is melted with the wax. I cannot have any glycerin in the final product, so I do not yet know how to properly make my own soap [at home].
The first reason I cannot have any glycerin, is because this will be used as a bullet lubricant for cast lead bullet, and I am concerned that the glycerin may form acrolein in the final application. The second reason is because the hygroscopic properties of glycerin are also bad. For more information, look up Felix World Famous Lube.
As far as I can tell, there are no glycerides in beeswax but I cannot find data on this topic. If there are no glycerides in bees wax, then I do not believe it is possible to yield glycerin in the saponification process of only beeswax.
I am considering a trans-esterifcation process, just to remove any possible glycerin, but I may need butanol (due to the high saponification temperature of beeswax) which is expensive to the point that I would rather do something entirely different. I may be able to use ethanol if I slowly titrate the ethoxide into the molten beeswax. I also have many open questions here, since bees wax is more than just fatty acids.
So, does beeswax contain any glycerides?

It appears that there are no triglycerides in beeswax  http://www.pharmacy180.com/article/beeswax-302/     https://www.researchgate.net/publication/304012435_Beeswax_Production_Properties_Composition_Control


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Using the steady state approximation in the decomposition of nitramide to determine rate law

$\ce{O2NNH2 (aq) ->[\mathit{k}_1] O2NNH- (aq) + H+ (aq)}$
$\ce{O2NNH- (aq) + H+ (aq) ->[\mathit{k}_{-1}]  O2NNH2 (aq)}$
$\ce{O2NNH- (aq) ->[\mathit{k}_2] N2O(g) + OH- (aq)}$
$\ce{H+ (aq) + OH- (aq) ->[\mathit{k}_3] H2O(l)}$
We assumed the first two steps were fast and gave a rapid pre-equilibrium, while the third step that generated $\ce{N2O}$ was slow. The last step is very fast.
(i) For this question, use the steady state approximation for [$\ce{O2NNH-}$], assuming it is a reactive intermediate, to derive an expression for the rate law.

 The solution given is
 $$\frac{\mathrm d [\ce{N2O}]}{\mathrm d t} = k_\mathrm{obs}\frac{\ce{[O2NNH2]}}{\ce{[H+]}}$$

My attempt at the solution is as follows:
$$\frac{\mathrm d[\ce{N2O}]}{\mathrm dt} = k_2\ce{[O2NNH-]}$$
Using the steady state approximation,
\begin{align}
\frac{\mathrm d[\ce{O2NNH-}]}{\mathrm dt} = k_1[\ce{O2NNH2}] - &k_{-1}[\ce{O2NNH-}][\ce{H+}] - k_2[\ce{O2NNH-}] = 0 \\
\ce{[O2NNH-]} &= \frac{k_1[\ce{O2NNH2}]}{(k_{-1}\ce{[H+]} + k_2)} \\
\frac{\mathrm d[\ce{N2O}]}{\mathrm dt} &= \frac{k_2k_1[\ce{O2NNH2}]}{(k_{-1}\ce{[H+]} + k_2)}
\end{align}
I have done this so far, but this doesn’t seem right to me. I’ve been looking over at it for seemingly ages but I can’t see where I am going wrong or if I have misunderstood something about the steady state approximation. Any insight would be greatly appreciated.

https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference
This should help for future reference.. I'll fix the text up this time. Take care that you format the text so it is easier for people with text readers.

You must next assume $k_2$ is small wrt $k_{-1}[H^+]$  to get the answer given.

How have you come up with your equation for the production of N2O, the one just under "Using the steady state approximation". The last term, −k2[O2NNH−], is telling me that the rate of production of N2O decreases if there is more O2NNH−, do you think this can be right? Also apply the steady state approximation to the fast equation - the whole point is that the quick ones reach steady state compared to the slow one.

That should be for the rate of [O2NNH-], apologies someone edited this and I don’t know how to change it

@MarinaCalder - ah, OK, that makes sense now. I've submitted an edit correcting, hopefully it will get accepted soon. In that case the earlier comment by porphyrin has the answer for you


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Bad repeatability of GC-FID meassurements
I am new to GC-FID and I am struggling with calibration curve. And I am strugling with the repeatability of measurements. One vial measured 5× in a row provides results which differs even 2 orders of magnitude. Do you see any mistakes in my procedure below?
I need to be able to quantify about 12 substances, mostly glycoles. I use methanol as solvent. I determined their retention times and prepared a calibration solution of these substances. All of them were clearly recognisable from each other in the chromatogram.
Since I need to quantify the substances, I prepared 10 solutions of the calibration substances. I need my results in % (m/m), therefore my calibration series was 10 g of solution + 10 g of methanol, 9 g of solution + 11 g of methanol, 8 g of solution + 12 g of methanol, … , 1 g of solution + 19 g of methanol. I mixed every one of them and fill the vials to the brim. I used only unused septa, put the vials in the refridgerator sealed with parafilm until they are measured.
Every sample was measured 5× in a row and the consequent measurements differed even 2 orders of magnitude. There was no trend in the signal intensity. I am not used to such a behaviour in spectroscopy measurements.
Do you see any mistakes in my procedure? Is there an error in method parametres settings?
Thank you for any help.

It looks to me like some random technical deviations, possible in sampling ( volume? bubbles in needle ?) or random leaks of gas flow, perhaps due mechanical shocks or vibrations. Try to obtain extensive stats of these errors.

Thank you @Poutink. I suspect device error, but since I am complete beginner to (practical) chromatography I was not entirely sure.

Try to make some stat analysis, if values seems to have some statistical distribution, or if there are many outliers. // BTW, I am not going to do any pouting, :-)  Poutnik is a Slavic word ( a pilgrim or a wanderer ).

OK, I will try to measure one of the solutions 50× that should be enough. Sorry for typing error. I know the word Poutnik. I am also Czech. :-)

«Bad repeatability» may refer to many different outcomes.  Regardless of the manufacturer of your GC, however, nothing prevents you to look up the troubleshoot guides, e.g., by gcms.cz, papers and videos e.g. by Agilent

A deep dive may be books specialized on all sort of pitfalls, e.g. this.

Thank you both. It seems it was really something with sampling. With the help of the troubleshoot guides, I tried to remove and clean the syringe. Then I obtained the same result five times in a row. It seems I can continue with the calibration process.


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Spontaneity and nature of attack of fluorine gas on aluminum
What is the nature of the reaction of attack of fluorine gas on aluminium metal? Is it spontaneous in nature?
I have studied reactions of halogens on aluminium, but it had no information about fluorine's attack, and I would like to know about this reaction.

Aluminium has a tenacious oxide layer so the reaction of gaseous fluorine with aluminium metal is firstly the reaction of fluorine with aluminium oxide.

So finally $\ce{F2}$ will react with $\ce{Al2O3}$? Also will $\ce{F2}$ react with fresh $\ce{Al}$ metal surface.

I have edited the question and made it more readable, getting a good idea on English sentences and writing would be great since it was very hard to read before. Your title should reflect the question that you want to ask. Hope you get an answer for this question and enjoy learning :D.

Aluminium has a tenacious oxide layer so the reaction of gaseous fluorine with aluminium metal is firstly the reaction of fluorine with aluminium oxide.

Use $\ce{} for chemical compounds

What do you mean by "spontaneous"? If you mean "exergonic", I recommend to use that term.

First and foremost! Do not do this reaction unless you are properly trained and have the appropriate safety equipment. Fluorine is one of the most dangerous substances out there, and one of the presumed products, $\ce{OF2}$, is also terrifying.
If you can calculate the free energy change for a reaction, you will know whether it is spontaneous. To do this, you need to have more information than you currently provide. We need to know the products of the reaction in order to write a balanced chemical equation. Then we can look up appropriate thermodynamic values for each substance in the reaction in order to calculate the free energy change. Below is what I presume to be the balanced equation for the reaction of fluorine with aluminum oxide:
$$\ce{Al2O3 + 6F2 -> 2AlF3 + 3OF2}$$
Using the NIST Chemistry Webbook, we can look up the enthalpies of formation and standard entropies of each substance so that we may calculate the enthalpy and entropy changes.
$$\begin{array}{l|c|c|c|c|c|c}
& \ce{Al2O3} & \ce{F2} & \ce{AlF3} & \ce{OF2}  \\ \hline
\Delta_f H^\circ\ \mathrm{(kJ/mol)} & -1675.7 & 0 & -1510.4 & 24.52 \\
S^\circ\ \mathrm{(J/K)} & 50.92 & 202.8 & 66.5 & 247.46 \\
\end{array}$$
We can use these data to calculate the free energy change of the reaction:
$$\Delta_r G^\circ = \Delta_r H^\circ - T\Delta_rS^\circ$$
If the free energy change is negative at that temperature, the reaction is spontaneous.

But does it go kinetically?  I can see it with aluminum hydroxide, but the anhydrous oxide may be hard to attack unless heated.

Quoting Housecroft and Sharpe, second edition, section 12.4 "Reactions of Al with halogens at room temperature ... give the Al (III) halides ..."

A person without proper training does not have access to fluorine gas. ;)

@Karl - I would like to believe that to be true everywhere in the world, but I cannot guarantee it.

@Karl Couldn't you extract it from fluoridated water with electrolysis?

@nick012000 Per https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4547570/ the target concentration for drinking water fluoridation is 0.7 mg/L ($1.84 \times 10^{-5}$ mol/L), so I don't think that's practical.  (I'm not sure whether the article means 0.7 mg of $\ce{F-}$ or 0.7 mg of $\ce{NaF}$; if it's $\ce{NaF}$, the molarity is even lower.)

@zwol Looking up the LD50 of fluorine gas, it looks like 100ppm is enough for a lethal dose (https://www.ncbi.nlm.nih.gov/books/NBK220011/), and according to Wolfram Alpha, 1 ppm of air is equivalent to 1 mg, so you'd probably be able to produce enough for a lethal dose by electrolysing 150L (approximately one bath tub) of fluoridated water at those concentrations. Obviously, this isn't something anyone should actually do.

@nick012000 Electrolysis of a "bathtub" of fluorinated water seems to me unlikely to produce fluorine given fluorine will spontaneously react violently with water to produce oxygen (or ozone if there is excess fluorine which there surely won't be here) and HF.

What's your reasoning for the reaction you wrote down rather than $\ce{2Al2O3 + 6F2 -> 4 AlF3 + 3O2}$?

@Jan Oxygen reacts with fluorine? Because everything reacts - if it hasn't already reacted - with fluorine?

@Karl Ah, good point. Fluorine is probably in excess after all. That's what I get for morning comments ^^'

Oxygen does not react with fluorine unless you introduce something like BF3 into the mix, or with an electrical discharge https://en.wikipedia.org/wiki/Fluorine#Reactivity I have no reference to support me, but I would expect Jan's reaction to be right


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Which d orbitals are involved in "d3s hybridisation"?
I was reading about hybridisation and came across a point saying that $\mathrm{d^3s}$ hybridisation involves only $\mathrm d_{xy}$, $\mathrm d_{yz}$, and $\mathrm d_{zx}$ orbitals and not $\mathrm d_{x^2-y^2}$ nor $\mathrm d_{z^2}$. Why is that so?

Please read https://chemistry.stackexchange.com/questions/76726/why-is-it-wrong-to-use-the-concept-of-hybridization-for-transition-metal-complex

What is d3s hybridisation and in which compound have you seen it?

@NisargBhavsar [MnCl4]- has Mn(+3) which is $\mathrm{d^4}$. I don't think d3s hybridisation is possible, it has to be sp3. Besides, hybrdisation is not useful for transition metals complexes.

@NisargBhavsar But CrCl4- has Cr(+3) which is d3 so only two d orbitals are free, and you can't get d3s. Hybrdisation theory is good for main group elements; for transition metal complexes, it's not accurate and often gives wrong predictions or explanations. It has been mostly superseded by MO theory.

Please note that the labels of the d orbital are not $x - y$, etc. they are $xy$. Products, not differences.

@orthocresol, the other ones are fine, but wikipedia (see the orbitals table) says it's $d_{x^2-y^2}$. Sorry if it's stupid; I don't really know how the orbitals actually come about anyway.

@HarryHolmes Yes, that’s the one exception, and that’s why I kept it as such in the original post.

First off, I'll reiterate what's mentioned in the comments - that hybrid orbitals are not a particularly good approach for transition metal complexes. But with that said, if you are going to apply this approach to a tetrahedral transition metal complex, here's how it works. For simplicity, I'm going to assume that only sigma bonding is relevent. You can do the same analysis including pi bonding, but it's more complicated.
Short answer: only the off-axis d orbitals have the right symmetry to match the ligands in a tetrahedral point group.
Long answer: The mathematical basis for constructing both delocalized and localized hybrid bond orbitals comes from group theory. If you haven't gotten to group theory in your studies, this explanation might not make a lot of sense, but at least you'll have to starting point to look deeper.
A tetrahedral molecule with four identical ligands is said to be in point group Td, which is just an abbreviation of "tetrahedral". Every point group has symmetries associated with it based on what operations can be performed that return the original structure. In point group Td, these are named A1, A2, E, T1 and T2.
When we build molecular orbitals, we combine the orbitals on the four ligands and look for possible symmetry groups. Then we look at the symmetry of the orbitals on the metal center and match the metal orbitals and "ligand group orbitals" based on symmetry. (An alternative approach is to look at the symmetry transformation of the bond vectors, which is described here: https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Modules_and_Websites_(Inorganic_Chemistry)/Advanced_Inorganic_Chemistry_(Wikibook)/01%3A_Chapters/1.09%3A_Td_Molecular_Orbitals. The result is the same.)
Making a long story short, the tetrahedral ligand group orbitals for a sigma-bonded complex fall into two symmetry groups. One orbital, in which all four ligand orbitals have the same phase, is A1. A second group of three orbitals in which the ligand orbitals are pairwise in opposite phases matches the T2 symmetry of the Td group.
On the metal center, our valence atomic orbitals are 4s, 3d and 4p. The 4s orbital matches the A1 symmetry, while the 4p and 3d (xy, yz and xz) match the T2 symmetry. The 3d(x2-y2) and 3d(z2) orbitals, on the contrary, match the E symmetry from group Td. Since the E symmetry doesn't match either of our ligand group orbitals, we can't use these two orbitals for bonding.
Since the 4s orbital is the only match for the A1 symmetry, we must involve that orbital in bonding. For the T2 symmetry, however, we have two choices. We can either use the three 4p orbitals and ultimately make four sp3 hybrids, or we can use the three T2 symmetry 3d orbitals and end up with d3s hybrids. The sp3 hybridization is much more familiar to most people because they have learned it for organic compounds such as methane, but that's not the right choice here. If we consider the relative energy levels of the orbitals on a transition metal, we see that 3d is lower than 4p and so is closer to the ligand orbitals (which are lower in energy than both). Furthermore, the off-axis d orbitals are pointed more directly at the ligands in a tetrahedral complex (which are also off-axis), so this intuitively makes more sense as well.
Finally, to your question about $\ce{CrCl2O2}$, that molecule belongs to point group C2v, so you can apply the analysis above to that point group to get the allowable hybridizations in the same manner as for the Td case.

+1 But I would also add that the choice of which 3 d-orbitals are used for hybridisation, is entirely dependent on the choice of axes. In a free atom, all 5 d-orbitals are degenerate. It is only when you bring in the ligands and choose the x y z axes that you get a differentiation.

@ShoubhikRMaiti That's only half true. The orbitals are indeed degenerate in the absence of ligands, and the axes in that case are completely arbitrary, but as soon as we assign a point group, that designation is done with respect to specific axes related to the symmetry operations, so the axes are no longer arbitrary. And we have specific conventions for coordinate systems of molecules that are used in group theory, so proper application of convention will always yield the same three d orbitals in the T2 group.

Yes, but hybridisation is not usually taught with symmetry arguments. My point is that, if someone started reading hybridisation and asked why those 3 d-orbitals are used, the answer should be that it's because of the choice of axes. The choice of axes is a convention but convention in itself is arbitrary. Any three perpendicular axes could be chosen, but one of them has been decided to be the right one and that's an arbitrary choice.


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Unit cell of hcp structure

It is given in my book that the structure in figure-1 is the unit cell of hcp structure.
So why then is the structure in figure-2 not the unit cell as we can repeat the structure in figure-2 to get the original hcp structure?
Also the structure in figure-1 is not a Bravais lattice so how then can it be called a unit cell?

A unit cell is just the smallest representation of a mineral that can be said to be repeated without any change in orientation or configuration, your second 'unit cell' does not repeat in the same configuration but needs to be oriented at different angles.

@Safdar Faisal  figure-1 is not a bravais lattice so how then can it be called a unit cell?

It can't. Your book uses the old convention, which needs to be cancelled, spat upon, and forgotten.


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Finding the charge passed in electrolysis of water using volume of hydrogen released

During the electrolysis of water we have gotten $\pu{10 dm3}$ of hydrogen. Temperature was $\pu{0 °C}$ and pressure was $\pu{101325 Pa}$.
What charge has passed trough the electrolysis article/joint?

I don't know how to find the charge while knowing pressure, temperature and volume. How do I proceed with this question?

I have found a formula for charge that goes "Q=CxV" and I do have volume, but how do I find C (capacity), I don't remember doing that in school, but I might be mistaken.

$PV = nRT$ and $\ce{2H+ + 2e- -> H2}$ are your best friends.
Step 1: Find the number of moles of hydrogen gas released
Use the formula: $PV = nRT$, where $P =$ Pressure exerted by the
gas, $V =$ Volume of gas, $n = $ number of moles of gas, $R = $ universal gas constant, $T = $ temperature of gas.
Using this formula with $P$ in atm and $V$ in liters we get:
$$n =\frac{PV}{RT} = \pu{0.0044 mol}$$
Step 2: Find the number of farads of charge used
We know that $$\ce{2H+ + 2e- -> H2}$$
This means that for every mole of hydrogen released, $\pu{2 moles}$ of electrons are used. A farad is the charge on $\pu{1 mol}$ of electrons and so number of farads of would be two times the number of moles of electrons. So charge would be $\pu{0.0088 F}$.
Step 3: One farad = 96500 coulombs
Therefore, $\pu{0.0088 F = 850.32 C}$ and that should be the answer.


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Enantiomers or identical structures?

What is the relationship between the following pair of molecules? Is the underlined answer correct? Explain.

I'm getting mixed answers. Some say these are identical molecules. Other say these are enantiomers. Which is correct and why?

The correct answer is "This picture won't let you know", and the answer they want from you is "Enantiomers".

Nevertheless if one draws either the left hand structure, or the right hand structure in ChemDraw's test page and later fetches the individual SMILES strings (first lasso selection, than structure -> get smiles), the output about the two apparently differs: C[C@H]1CCC[C@@H](Br)C1  on the left, bromine on R, methyl on S; vs Br[C@H]1CCC[C@@H](C)C1  right, bromine on S, methyl on R configurated carbon centre.

See if you can access a model kit (example), then build the two molecules and try to superimpose them.  It takes time to train the eye.

Draw the hydrogen atoms on the chiral carbon atoms too, then that should help you see.

Draw the wedge/dash thing, it should help even more readily.

Remove the "and" between the two structures and replace it with a mirror. Does that answer your question?

@Dani51: Please do not attach wordy documents as images.

What is the relationship between given pair of molecules?

They are enantiomers as given underlined answer.

How did you know that?

You know that by assigning $R/S$ configuration (following Cahn-Ingold-Prelog rules) on each chiral carbon. That is the easiest way to identify the enantiomers. If you have one chiral center in the molecule and its CIP-rotation is $(S)$, then its enantiomer (mirror image) has  CIP-rotation of $(R)$ and vise versa. If you have two chiral centers in your molecule of interest and its CIP-rotation is $(S,S)$, then its enantiomer (mirror image) has  CIP-rotation of $(R,R)$. If your comparing molecule has CIP-rotation of $(R,S)$ or $(S,R)$ instead of $(R,R)$, then they are not enantiomers. They are called diastereomers. In your given two molecules (1-bromo-3-methylcyclohexane) where $\ce{C}$1 and $\ce{C}$3 are chiral carbons (two chiral carbon system). First assign $R/S$ configuration on each molecule:

For convenience to visualize, I avoid the chair conformation but put the relevant chair conformation below each molecule (e.g., the planar molecule to the left is the right structure in the question). The priorities of groups are given in red for the $\ce{C}$1 center and in powder blue for that of $\ce{C}$3. The molecule to the left has $(1S,3R)$-configuration while the molecule to the right has that of $(1R,3S)$. Since there is no plane of symmetry in each molecule, they are enantiomers (mirror images of each other).

Is it possible to draw the chair conformation from the planar drawing directly without seeing the R,S configuration (or vice versa)? Because for me, it takes time to first find out their configurations individually, then draw the chair form, and then again verifying their configurations... basically what should I take care of, to not draw the planar drawing of the left molecule as the chair form of the right molecule? Or in other words, how do I show that the structures you have drawn on left/right are indeed equivalent?

I don't have that capability at home. Visualize from top view of the molecule to the left (chair conformation) and rotate your molecule $180^\circ$ clockwise. Do the same for the molecule to the right and rotate it $180^\circ$ anti-clockwise. You get two chair confirmation of mirror images.

Enatiomers are "optical isomers which are mirror images of each other"
For any compound to show optical isomerism disymmetry is the thumb rule.
For checking disymmetry compound should have dissymetric center(s).
If we consider the given structures they have 2 chiral centres each and now we need to draw the fisher diagram and find R/S configuration of each chiral  carbon and then compare like for example if Compound 1 has  the configuration (1R,3S) and compound 2 has (1S,3R) then they are enantiomers.

What is a "fisher" diagram? You don't even need a Fischer structure! Just do this: https://chemistry.stackexchange.com/questions/99721/twisting-stereoisomers-with-rings-to-determine-r-s/99750#99750

I'd appreciate whoever downvote this answer give the reason for downvoting.

@Mathew I didn't downvote, but I can give you a couple of reasons why some would. 1) The definition of enantiomer (in quotes without source) is not quite correct. 2) Some are quite allergic to rules of the thumb. 3) Applying a Fischer projection is completely unnecessary to find the configuration of the stereocenters. 4) Spelling and grammar. | meta: It's a 'fastest gun in the west' situation. | I still think it could have been helpful with some cleanup. With your answer that hardly seems worth it.

@Martin: Actually this is not my post. I thought this shouldn't get downvoted but should have given good reason to correct as you did. I wasn't agree with this answer and post my own yesterday. I appreciate your reasoning.


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What is the amount of UV light required to make fluorophores glow and what is the minimum amount of electricity needed to achieve this?
So I have heard that fluorophores glow under UV light, so I was wondering what is the minimum amount of UV light needed to make the fluorophores glow and what would be the minimum amount of electricity required to make that much amount of UV light.
Also, is there any other type of light that makes fluorophores glow but requires a lesser amount of electricity?

There is no minimum, per se, though you will encounter a limit of detection issue as you decrease the UV excitation light intensity. As for electricity, there are various electrically powered UV light sources, e.g., UV LEDs, UV fluorescent lamps, UV lasers, etc. Their efficiencies vary greatly.

@EdV Could you suggest to me a device that releases UV light and is small, light, inexpensive, and uses less electricity than an LED or at least a similar voltage to it?

What specific chromophore are you interested in? You may be able to tailor the light source specifically to the absorbance wavelength of your chromophore to reduce power consumption by the lamp.

@Andrew I don't have an exact fluorophore or chromophore, but rather a mixture of 3 fluorophores, quinine, rhodamine B and fluoroscecin.

The recording of the signal seems underappreciated, but it is part of the overall equation, too. How sensitive is your device to record the released radiation in this spectral range? Do you observe by naked eye? A human behind a microscope?  A normal (video) camera mounted on a microscope? A Peltier-cooled ICCD camera? Is the sample excitation (and thus fluorescence) continuous (say, a deuterium lamp), or a pulsed laser? For how long of exposure time may you integrate signal intensity per frame to improve the S/N ratio which would be good enough (living cells vs. dead sample) for you?

@Buttonwood I am no genius but I can only say that it will be viewed by the naked eye and will probably be used about 7-8 hours a day. I am trying to create a type of light out of this.

@Aster17 So your interest is to harvest the fluorescent light like in a lamp, and not as a fluorescence marker in microscopy.  Well, then you (equally) search for the (photochemical) quantum yield which is a value on occasion reported (example, here) to see the ratio of photons emitted by the dye over photons injected to excite the dye.

@Buttonwood I am sorry but I am really unable to understand any of it, so if you could ease on the words a bit, I would really appreciate it.

@Buttonwood I am unable to the quantum yield of quinine, but the quantum yield of Rhodamine B is 0.7 and fluorescein is 0.79, but if quinine sulphate works like quinine (gives a blue glow in UV light), then its quantum yield is 0.546.

The efficiency depends on the solvent, and on the wavelength of excitation, too like here; quinine.  There are instances water (like in tonic water) and a commercial black light source suffice. Yet for me, I'm afraid at present your question too much open ended.

@Buttonwood My solvent is water if you need to know. I just need a small and inexpensive source of UV light that uses less electricity than LED.

A single photon is all that is needed to cause a molecule to fluoresce or phosphoresce, and photomultipliers can detect single photons. As to the amount of electricity, there are endless possibilities, but sunlight is free.

@porphyrin But there is no sunlight in the night!

@Aster, its always sunny somewhere :)

@porphyrin But not inside a house.

It appears you are wanting to invent a white light source, with inexpensive UV excitation of blue (quinine), green (fluorescein) and red (rhodamine) fluorophores in liquid solution. It needs to work at night, and UV LEDs require operating voltages higher than you want. So this is something aimed at benefitting people who would otherwise have no light at night. Is my understanding correct? If so, I think white LEDs, with rechargeable battery power source, and something like a hand crank generator or solar cell to charge the battery, might be better.

Yes, but I am trying to achieve a source of light that uses a very low amount of energy, so if there are any other types of light forms I can use that use less electricity. So anything like that would be seriously helpful.

So now you need to get real and do the numbers: low energy in means low light energy, i.e., intensity in watts, lumens or whatever, as output. If your power input is 1 watt, for example, then you will get less than 1 watt of light output. So what is the minimum light output you have as a goal? Once you have that, you can start working backwards. But you also need to be more specific: what power source(s) are potentially worth consideration?

anything that is in the middle of 100-300 milli-amps as input would be good. As for how much I want to get, I can fix that by adding more fluorophores if they are falling low. As for output, the amount an LED light gives is good enough for me.

So it appears that a white light LED flashlight (aka “torch”) is all that is needed. I have half a dozen or so of them. They are relatively inexpensive, use 3 or 4 1.5 V batteries, use individual white discrete LEDs or COB LEDs (those yellow things). You can also buy inexpensive UV LED flashlights, if you really want to use a liquid solution of fluorophores and excite with UV light. All I can really suggest is experimenting and seeing what works best. As for the current, 200 mA times 4.5 V is 0.9 W, which is low power and easily achievable: even a USB port can do that.

How long do those UV flashlights last and can I use AC current for them?

Let us continue this discussion in chat.

I'd like everyone to remember that comments are not intended for extensive discussions. Please use chat rooms instead and please be aware that comments will be deleted.


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Process parameters of typical chemical processes
I am researching to opportunity of heat recuperation in the chemical industry be means of high temperature heat pumps. However, I am having some trouble with finding suitable applications. I would like to find overviews of processes (e.g. drying, distillation, thermoforming...) that clearly indicate the heat demand of the process with the corresponding temperature at each state. I also need the temperatures and amount of heat released by cooling processes or waste heat sources (the heat pump will transfer this heat to the heat demanding process).
I searched a lot in literature but I found few suitable cases. Does anyone happen to know about certain papers or websites where I can find typical temperature and heat demand/released values of chemical processes ?
A nice example is illustrated in this figure


text:
Substituting K2CO3 for KOH in Soap
Wood water obtained by running water through hardwood ashes has a high quantity of $\ce{K2CO3}$, historicaly this was called lye.  Yet soap recipes today seem to use $\ce{KOH}$ or $\ce{NaOH}$ as lye.
$\ce{KOH}$ is stronger than $\ce{K2CO3}$, so how would someone determine how much more $\ce{K2CO3}$ to substitute when $\ce{KOH}$ is called for in a recipe
I am using the method in the link to make lye, and the lye produced is mainly a combination of $\ce{K2CO3}$ and a little $\ce{Na2CO3}$, this should be sufficient to create soap
https://www.wikihow.com/Make-Lye

Hi, welcome to Chemistry.SE! Please note that you can render the chemical formulae with MathJax, using $\ce{...}$

$\ce{K2CO3}$ is not strong enough to make soap with oils, whatever its amount or concentration. And don't forget that ashes contains only $10 - 20$% $\ce{K2CO3}$. The main constituant of wood ashes is $\ce{CaCO3}$ which is present at $40 - 70$% . There is also $10 - 15$% $\ce{SiO2}$ and $5 - 10$% $\ce{MgCO3}$ in wood ashes

Historically, K2CO3 was not called lye, but potash . As lye was referred KOH, today mostly NaOH, both historically produced by process called caustification. ( $\ce{K2CO3 + Ca(OH)2 -> CaCO3 + 2 KOH}$ )

well then how come in both https://www.motherearthnews.com/homesteading-and-livestock/how-to-make-soap-from-ashes-zmaz72jfzfre and https://www.wikihow.com/Make-Lye it is referred to as lye

Chemists are a little picky about chemical names, including common language imprecise names such as “lye”. But your question has not been answered yet, so someone with soap making experience may yet answer.

There are sometimes differences what substances are called lye versus what solutions are called lye. That can be source of big confusion. The relative "strength" is not fixed number, it is concentration dependent. Generally, KOH is used in such amounts/concentration K2CO3 cannot replace it for the same effect. Count with saponification taking long time, unless you prepare "caustic potash" from "potash".


text:
Diffusion of multiple gases at once
Graham’s law of diffusion states the rate of diffusion between two gases is inversely proportional to their molar masses. Now, this is helpful when we are focused on two gases. But that is rarely the case in a beaker or even the atmosphere.
So how do you measure the rates of diffusion of many gases moving at once? Do you measure each rate of effusion (inversely proportional to molar mass) and just order them, then use stoichiometry to determine which one will be dispersed quickest based on the amounts present?

If the gases can be regarded as ideal then the diffusion of each one is treated as independent of any other.

@BuckThorn So a few comments: (1) an ideal gas is that which meets the criteria for kinetic/particle molecular theory; (2) if treated as independent of any other, then its diffusion rate can be compared in an ordered list; (3) once you have that list together, you can compare each to one another. Is that all correct? As a result, that’s why we have Graham’s law — because if effusion is inversely proportional to the molar mass, then you would just divide the inverse of the molar mass of one to the inverse of the molar mass of another.

If I understand you correctly, you seem to be wondering if the relationship is transitive, and the answer would be yes.

@BuckThorn Yes, it’s a transitive relation. Assuming the gases are ideal, then we order them in a list. If we can order them in a list, then we can compare them. And that’s how we got Graham’s law, which clarifies things, thanks.

If the gases in a mixture are ideal then their diffusion is independent*.
The relationship described by Graham's law is transitive, therefore the relative effusion rate of multiple substances with respect to a reference substance also provides a ranking of their relative diffusion coefficients (an ordered list).
*Outside of condensed or supercritical phases it seems unlikely that two fluid substances would display sufficiently strong interactions to alter the relative order of their diffusion coefficients.


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

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

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

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

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


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How thick is the colored oxide layer on heat treated stainless steel?
When stainless steel is welded, or just heat treated, it will develop a thicker oxide layer, which in some cases is is rainbow colored, but can become so thick as to appear almost black. From the colors (diffraction effects, I assume), the nicer-looking layers must be on the order of a wavelength of light. Is there any more numerical information on their thickness?
Follow-up question: The colored oxide layers can be removed with a pickling paste (based on HCl) or a passivating cleaner (based on HNO$_3$ plus HF), but the resulting surface is much rougher than the original. Well, it is observably rougher if the original surface before heat treating was mirror-smooth. We call the new surface satinized. Is there any simple way to estimate surface roughness/surface area of the satinized surface?
Further follow-up: would the surface chemistry be expected to change due to the surface area increase (perhaps like shiny Pt vs platinized Pt (~340 mV): https://en.wikipedia.org/wiki/Platinum_black

You may have already found this reference: https://core.ac.uk/download/pdf/288376834.pdf

@Andrew 's source, Figure 2, seems like it might be compared to an ultramicrotome section reference chart from Radboud University's EM class.  But I don't know which nodules or layer(s) you're comparing, or whether other factors like refractive index affect the spacing distance for structural color.

@Andrew Thank you very much!! Great reference!

@Mike Serfas:  Great color chart! Thank you!

I understand that they are related issues, but you might want to break up your question into one about structure and another about reactivity.

@Buck Thorn: I will. Yep, once I started asking, I couldn't stop.

Not very thick. Carbon steel develops the same colors at much lower temperature/ times . Gold to pink to blue to purple , stainless shows the colors better because the base metal is brighter . It is caused by the thickness of oxides , a matter of a few layers of oxide molecules ; unfortunately I cannot give you a number in microns and can't remember which book it is in.


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Why is 295 nm a good wavelength to check absorbance when testing for aspirin?
Take aspirin (acetylsalicylic acid) and hydrolyze it with NaOH. This results in sodium acetylsalicylate and water. I think the sodium is held on with an ionic bond, so it is really salicylate dianions that are the important thing in the solution.
This is then diluted to several different concentrations. Using UV-VIS, the absorbance of these solutions are measured at 295 nm. Then a calibration line is plotted, that is, absorbance verse concentration. It will be a (pretty much) straight line according to the Beer–Lambert law. You can use this calibration line to calculate the concentration of a similar, but unknown, solution from its absorbance.
Why was 295 nm a good choice of wavelength to measure at?
The only idea I currently have is that at 295 nm there is a distinct difference between the absorbance of acetylsalicylic acid and salicylic acid. Apparently, the salicylic acid has a strong response at 295 nm, whereas the acetylsalicylic acid does not. However, I am not sure this is exactly relevant, since I'm not sure I have salicylic acid in my solution.
If I am on the right track here, then I have a follow-up question:
Apparently the lambda-max for salicylic acid is 302 nm. So, why would 295 nm be used in this experiment instead of 302 nm? Or, would anything between, say, 295 nm and 310 nm be just as good?

Searching for “aspirin absorbance spectrum” results in spectra showing 296 nm as the peak absorbance wavelength for salicylic acid and 229 nm for aspirin. Try it! And the peaks are broad, so being off a couple of nm just results in slightly lower sensitivity.

What is "UV-VIS"? Ultraviolet and visual [wavelengths]? Or the name of an instrument or a particular measurement method?

@PeterMortensen It is an analytical chemistry technique based on the absorption spectra of ultraviolet and visible light: https://en.wikipedia.org/wiki/Ultraviolet%E2%80%93visible_spectroscopy

Please consider giving the green checkmark (or upvote or both) to the most helpful of the posted answers. It encourages people to put some thought and time into crafting answers that are factually correct, relevant, understandable and likely to be of benefit to those, in future, who encounter the question and accepted answer. It is a small reward for those who volunteer their considerable time, effort and experience to aid others and they might well look favorably at future questions from the same person. Thanks for considering this!

I think this was basically an educational experiment. Your analytical method assumes that the hydrolysis was complete (this is the key requirement of a successful calibration curve). Your experiment did not include a test to assess the completion of hydrolysis. Acetylsalicylic acid begins to hydrolyze in alkaline solution to salicylic acid and basically it is an indirect method of detection.
You want to analyze acetylsalicylic acid but you are detecting salicylic acid. Guess where does this method fail? It will fail if you had a mixture of acetylsalicylic acid and salicylic acid in the beginning and you were asked to quantify both.
As EdV explained to you, being 1 nm off from the lambda max is not a big deal in UV-Vis spectra of solutions. Even the monochromator may be slightly off!

This is a good answer, but I had wanted to extend it with Woodward's rules.  They seem to explain the aspirin absorption well enough, but unfortunately I didn't think of the reason why the salicylic acid absorption is so much higher.

Good point. I did not try to apply Woodward's rule. Your point could be a separate question.

thank you for the great answers everyone! that helped a lot.


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Assignment of the 13C NMR spectrum of 4'-bromoacetophenone
I was doing a 13C NMR prediction of 4'-bromoacetophenone on chemdraw but I can't figure out why positions 5 and 5' appear further downfield than positions 4 and 4'.

I know both Br and the carbonyl are deactivating groups, with Br being ortho/para directing and carbonyl is meta directing. However, the mesomeric effect of both groups should "shield" position 5 but this is not the case. I'd appreciate an explanation that'd help me understand that.

Maybe this isn't the answer you are looking for, but $\ce{^13C}$ NMR shifts are often not amenable towards analysis via electron-withdrawing / electron-donating effects—in my experience this is especially so for aromatic rings. Any such analysis is quite unreliable and can lead to wrong conclusions if you try to use it for assigning peaks.
The numbers that ChemDraw predicts are based on empirical tables of chemical shift corrections. These are experimentally observed chemical shift changes that occur at each position of the benzene ring upon substitution. For a ring with multiple substituents you can approximately add these up. See e.g. Hans Reich's website for such tables.
See also long's answer to "Assigning the 13C NMR spectrum of bromobenzene and 4-bromobenzophenone": although most of the discussion is on the ipso carbon (C6 in your labelling), the general point is the same: it's often not trivial to rationalise or predict chemical shift changes in aromatic rings.
[There might be more sophisticated theoretical analyses out there based on diamagnetic and paramagnetic shielding arguments, but I'm unfortunately not aware of these.]


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How to make silicic acid, potassium silicate or calcium silicate?
I am a student of tropical agriculture. I am taking the class of agronomic chemistry and plant nutrition. I want to make use of silicate source I have at home for e.g. zeolite, wood ashes, diatomaceous earth, silica sand. I have hydroxide of calcium, potassium and sodium. Can I use any of these to make silicate for use on plant nutrition ?

You can make the silicates by fusing the hydroxides with silica. Potassium silicate fertilizer grade are successfully produced by direct fusion of silica ($\ce{SiO2}$) and potasium compounds($\ce{KOH}$ and $\ce{K2CO3}$) in furnaces at temperatures up to melting point of mixture. The fusion temperature reaches around 1350 °C.
$$\ce{SiO2 + 2 KOH ->[\Delta] K2SiO3 + H2O}$$
$$\ce{SiO2 + K2CO3 ->[\Delta] K2SiO3 + CO2}$$
Calcium silicate is prepared by reacting calcium hydroxide and silica. This paper1 discussed the reaction at 30-85 °C.
References

https://www.matec-conferences.org/articles/matecconf/pdf/2016/21/matecconf_bisstech2016_01021.pdf
REACTION BETWEEN SILICA AND CALCIUM HYDROXIDE SOLUTIONS. I. KINETICS IN THE TEMPERATURE RANGE 30 TO 85°, S. A. Greenberg, The Journal of Physical Chemistry 1961 65 (1), 12-16, DOI: 10.1021/j100819a005

You should add that the potassium silicate obtained by fusion of $\ce{KOH + SiO2}$ is extremely difficult to dissolve into water. This can only be done with hot water and under pressure.


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IUPAC name of cyclopentanol derivative
Is the name of the following compound 3-(ethenyl)-5-(prop-1-enyl)-2-(prop-1-ynyl)-4-(prop-2-enyl)cyclopent-2-en-ol?

The problem I am facing is that how do I arrange 5-(prop-1-enyl), 2-(prop-1-enyl) and 4-(prop-2-enyl). In which order do I write these in the IUPAC name?

The relevant rule for the order of citation of detachable substituent prefixes in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows.

P-14.5 ALPHANUMERICAL ORDER
(…)
Alphanumerical order is used to establish the order of citation of detachable substituent prefixes (not the detachable saturation prefixes, hydro and dehydro), and the numbering of a chain, ring, or ring system when a choice is possible.
Alphanumerical order is applied as follows in organic nomenclature. Nonitalic Roman letters are considered first, unless used as locants or part of a compound or composite locant, for example, ‘N’ or ‘4a’ (see P-14.3), or in an isotopic descriptor. (…)

P-14.5.4 When two or more prefixes consist of identical Roman letters, priority for order of citation is given to the group that contains the lowest locant(s) at the first point of difference.

Therefore, the correct name for the compound that is shown in the question is
3-ethenyl-5-[(1E)-prop-1-en-1-yl]-4-(prop-2-en-1-yl)-2-(prop-1-yn-1-yl)cyclopentan-1-ol
since the order is ‘ethenyl’ > ‘propenyl’ > ‘propynyl’ and ‘prop-1-en-1-yl’ > ‘prop-2-en-1-yl’.
The fact that ‘4-propenyl’ is lower than ‘5-propenyl’ is irrelevant.

ChemDraw names the above compound as (E)-3-allyl-2-(prop-1-en-1-yl)-5-(prop-1-yn-1-yl)-4-vinylcyclopentan-1-ol . What's the reason for this?

@LightYagami Ask ChemDraw developers.

@andselisk Does that mean it's not reliable?

@LightYagami No, that literally means ask ChemDraw developers. It's a proprietary program and nobody else knows how they implement name generation. Once you receive their answer and compare it with the Loong's, then you can draw a conclusion for yourself.


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Does iron(VI) fluoride exist?
Why does $\ce{FeF6}$ not exist? There are hexavalent iron compounds, so that is not the problem. There is the $\ce{[Fe(CN)6]^3-}$ ion, so there doesn't seem to be a steric problem.

Maybe reading about bonding characteristics of Potassium Ferrate may help.

No, it doesn't, as far as it is possible to prove a negative. Nor do CrF6, MnF7 or CoF5. And note your steric argument is not correct, Fe3+ is appreciably bigger than Fe6+, as far as those terms have any meaning in chemical environments.

Not an answer. But $\ce{FeF3}$ as a crystal already is something like $\ce{FeF6}$, in the sense that each iron is bound to six fluorine atoms; the lattice is linked via $\ce{-F-}$ bridges. The existence of $\ce{[Fe(CN)6]^3-}$ is more of an argument for $\ce{[FeF6]^3-}$ than $\ce{FeF6}$.

Food for thought: The dimer of iron(III) fluoride doesn't exist(unlike iron(III) chloride), but the anion, $\ce{[Fe2F6]^-}$ does exist consisting of $\ce{Fe^2+}$ and $\ce{Fe^3+}$ ions and it form salts like $\ce{[NH4]Fe2F6}$.

Besides steric factors related to the small size of the transition metal core, we could be seeing an electronic effect described in this answer.  Iron is fairly early in the transition metal series, so when pushed beyond the $+3$ oxidation state it has few $d$ electrons in the central core.  As explained in the referenced answer, this makes the iron strongly pi-electron accepting, and therefore more favorable for combining with a stronger pi-donor ligand such as oxide instead of fluoride.

Thank you!  So: what about FeO2F2  or FeOF4..?

You may want to look at https://en.wikipedia.org/wiki/High-valent_iron.  Organic compounds do allow a wider variety of ligands, but all the examples given there (with iron in oxidation states of +4 to +6) still involve pi-donor ligands from Groups 15 and 16, primarily oxygen and nitrogen.

@ThomasForster Not exactly FeO2F2 but there are similar types of compounds: $\ce{Fe3(OF)2}$, $\ce{Li2Fe4OF8}$, $\ce{FeSbO2F2}$.

@nilay good finds, but no Fe(VI).  They are just Fe(II) and Fe(III).

This shows that iron can't achieve +6 o.s. that easily.


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Reactivity order among Interhalogens
I was reading about Interhalogens in JD Lee and in the book it is stated that (without any reasoning) the reactivity order for few of the interhalogens is :-
$\ce{ClF_3 > BrF_5 > IF_7 > ClF > BrF_3 > IF_5 > BrF > IF_3 > IF}$

$\ce{ClF_3 > BrF_3 > IF_3}$
The above is the order for the interhalogens of type $\ce{AX_3}$ which I can understand by the fact that stability would be directly proportional to the polarisability of the compound. But the interhalogens are more reactive than halogens due to poorer overlaps in the first place and going by that logic the order should be reverse. But here that is not the case, Why is that?

$\ce{IF_7 > IF_5 > IF_3}$
Going by the polarisability logic, this order does not makes any sense.
According to me this may be explained by the steric factors, as the number of fluorine atoms around the central atom increases, repulsions between the substituent atoms increases so much that it’s stability gets affected. But the above statement is just a speculation. So what would be the correct reasoning here?

$\ce{ClF > BrF > IF}$
The polarisability argument should be overcome by the poorer overlap at least here according to me but again that is not the case. Why?

[Here by poorer overlaps I mean, We know that for a strong bond, orbital overlaps should take place between the orbitals having little difference in their energies]
And my last question is it possible to theoretically determine the given below order with basic arguments(that is without any data), I suppose it comes out of net considerations of polarisability, overlap, steric factors.
$\ce{ClF_3 > BrF_5 > IF_7 > ClF > BrF_3 > IF_5 > BrF > IF_3 > IF}$
Tl;dr (A) What is the reasoning for the given below reactivity orders?
1.$\ce{ClF_3 > BrF_3 > IF_3}$
2.$\ce{IF_7 > IF_5 > IF_3}$  3. $\ce{ClF > BrF > IF}$
(B) Is there a way to determine given below order theoretically with basic arguements?  $\ce{ClF_3 > BrF_5 > IF_7 > ClF > BrF_3 > IF_5 > BrF > IF_3 > IF}$
[Note that in the book no specific mentioning has been done that is, it has not been mentioned reactivity with respect to what]
References:-
Concise Inorganic
Chemistry by JD Lee (Fifth edition)


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Does optical isomerism fall under the category of stereoisomerism?
I have been asked to find the total number of dichlorination products (including stereoisomers) for the following reaction:

I have figured out nine possible structures out of which two are optically active.
Should I count the additional six isomers (three for each of the optically active structures - RR, RS, SS)?
I am confused about whether to count the optical isomers or not.
Please provide clarification on this.

There are no additional six isomers. You've already counted them all.

@IvanNeretin Won't there be three different R-S structures for each of the optically active structures (because of the two chiral carbons)?

Indeed there will, and you've already counted them all. Your "optically active" are RR and SS, and your "meso" is RS.

@IvanNeretin oh, I just realized. This is now clear to me! This was a rather silly confusion.

The answer to the literal title question is, by the way, YES.


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Why is inversion temperature equal to twice of Boyle's temperature?
Can you please give me the proof of the following equation?
$$T_{i}= \frac{2a}{Rb}$$
where $T_{i}$ is inversion temperature for a Joule-Thomson experiment on a van der Waals gas. And why is it twice of that of Boyle's temperature?

Asking this way is not how this site works. Before asking – think, search, read, think, think more. If you still fail to find the answer then ask, but together with explicit summary of ideas a/o results you got until then. Questions that are like duplicates, or are without explicit solving attempts or thoughts,  are usually closed. You get what you paid for. Users  pay here by their own effort.

Is this supposed to be the inversion temperature for a JT experiment on a van Der Waals gas?

Yes, your are correct.

Please go through the tour of the site (if you haven't yet done so)  to understand some of its rules...

https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange/142#142

On this site "homework" means basic textbook problems. Please edit and add for instance that this refers to a van der Waals gas and JT experiment etc... help us help you. Also, show effort. Why can't you solve this yourself?

Because I am just a Grade 12 student and I don't have books which show the derivation.

At the Joule-Thomson inversion temperature
$$\mu = \left(\frac{\partial T}{\partial p} \right)_H = 0$$
It can be shown that
$$ \left(\frac{\partial H}{\partial p} \right)_T=-\mu C_p \tag{1}$$
(this can be derived from the differential form $dH = \left(\frac{\partial H}{\partial p} \right)_Tdp + \left(\frac{\partial H}{\partial T} \right)_pdT$)
Also, the thermodynamic equation of state (in terms of enthalpy) is
$$\left(\frac{\partial H}{\partial p} \right)_T=V-T\left(\frac{\partial V}{\partial T} \right)_p$$
(this can be derived starting from the definition of the enthalpy and the second law,
$dH = -Vdp + TdS$)
which means that at the inversion temperature (see Eq. 1)
$$\left(\frac{\partial V}{\partial T} \right)_p=\frac{V}{T} \tag{2}$$
(consistent with the fact that the gas behaves ideally at this point).
For a van der Waals gas
$$(p+\frac{a}{\bar{V}^2})(\bar{V}-b)=RT$$
so that (for one mole)
$$\left(\frac{\partial V}{\partial T} \right)_p=\frac{V-b}{T}\left(1-\frac{2a(V-b)^2}{RTV^3} \right)^{-1}$$
so that at the inversion temperature (see Eq. 2)
$$1-\frac{2a(V-b)^2}{RTV^3} =\frac{V-b}{V}=1-\frac{b}{V}$$
or
$$\frac{2a}{RT}\left(\frac{V-b}{V} \right)^2  =b$$
If $b\ll V$ it follows that
$$T_{\mathrm{inv}}  =\frac{2a}{Rb}$$
which is twice the Boyle temperature of a van der Waals gas.


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Are CaOCl2 and Ca(OCl)2 same or different?
Are the two compounds same? Also, which one is considered bleaching powder? Please explain their differences on the basis of their structures also.

Apart that the formulae are certainly different.

People do not hesitate to ask questions like "Are 1 and 2 the same or different ?", if it is in context of chemistry.

CaOCl2 is incorrect notation as it implies Calcium in +4 oxidation state which does not occur.

@Waylander As a summary formula, it would be correct, unless IUPAC prefers other element order, like CaCl2O.

$\ce{CaOCl2}$ is a calcium chloride hypochlorite. There are two different types of anions: chloride $\ce{Cl-}$ and hypochlorite $\ce{ClO-}$. So its structure is:

$\ce{Ca(ClO)2}$ is a calcium hypochlorite (also called bleaching powder or bleach). There is only one type of anions: hypochlorites $\ce{ClO-}$. So its structure is:

They are not covalent compounds. Note that bleaching powder is complex mixture of calcium hydroxide and basic calcium chlorides and hypochlorites. https://en.wikipedia.org/wiki/Calcium_hypochlorite

These "structures" are nonsense and have little to do with reality.

Of course, these structures are drawn by me just for clarity.

Could you please send here the right structures of these compounds?


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properties of solution
I had homework with the following question, Eugenol is the active ingredient in the oil of cloves used to relieve toothache. Calculate the boiling point of a solution in which 0.17 grams of eugenol, (C10H12, O2) is dissolved in 10.0 grams of benzene. (Kb benzene= 2.53 0C/m; Tb(solvent) = 80.100C). Show your solution.
I solved it like this but it's wrong, I would like to know your thoughts.

@Martin Hmm, I think closing this Q as duplicate without clarification, when a newcomer tries to improve the closed question erroneously in a new post, may left the OP very confused.

Tried rather to edit the original question, the best if you write your solution attempt as text,  instead of picture, and then ask for its reopening.

@Poutnik It is the same question, from a moderational point of view there is only one more option: deletion. And that is harsher, I guess.

@Martin I do agree it is the same question to be closed as duplicate. :-) I have meant  there is always an option to post a short comment.

@Poutnik I have not much time to enjoy this site, so I chose to handle this quickly and without further comment. I did this because I saw that you have posted a well meaning comment on the original question. I believe all the information about the site is easily accessible before you hit the post button, so I do not think special care is necessary after the fact that the poster has ignored most of the warnings. There is unfortunately more that I need to get to, so excuse my brevity.


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Does syn-dihydroxylation form racemic mixture?
I have been asked to determine the nature of the product for the following reaction:

And as I recognized this reaction to be a case of syn-dihydroxylation, my obvious response was that the product formed will be meso.
But, in the solution to the problem, it says it makes a racemic mixture.

Shouldn't both the $\ce{-OH}$ be on the same side and form a meso-compound? And what does the $(sym)$ stand for?

The very fact that 'syn' has been written as 'sym' should've heightened your suspicions regarding this answer, which is blatantly wrong. Cold KMnO4 will produce a meso compound in this case. This is a very standard reaction, and it is surprising that a book messed this one up. I strongly suggest you change the book/material you're studying from.

May I ask, however which book this came from?

@C_Lycoris It's not from any standard organic chemistry book. The book is called "Problems in Organic Chemistry", and is basically a chapter-wise question bank for the JEE exam.

Referring to the product through its Fischer structure alone was an obvious red flag. Ditch this textbook.

Oxidation  of alkene to 1,2-syn-diol by cold basic $\ce{KMnO4}$ is a common classification test for alkene in undergraduate organic chemistry laboratory courses. OP had correctly recognized that this reaction is a case of syn-dihydroxylation, and OP's conclusion of the product to be meso is also correct since the alkene in hand is a cis-but-2-ene. The schematic representation of the proposed mechanism is given below:

Source of the scheme
Since but-2-ene structure is planer and a prochiral molecule, it has two faces (Si- or Re-face; above or below the plane) to get attacked by a reactant, here it is $\ce{KMnO4}$. The above mechanism displays the attack by only one face. The oxidation gives butan-2,3-diol by syn-attack on that face. The product has two stereocenters and the stereochemistry of that product is $(2R,3S)$. As indicated in the diagram, the product also has a plane of symmetry, and hence, the $\ce{C}$2 stereocenter is the mirror image of the $\ce{C}$3 stereocenter. Therefore, the product is a meso-compound.
If the $\ce{KMnO4}$ attack is done on the opposite plane, the product would be $(2S,3R)$-butan-2,3-diol, if numbering has maintained the same. This is essentially the same meso-product, as we discussed above.
If the starting substrate is trans-but-2-ene, the $\ce{KMnO4}$ attack from one face gives you $(2S,3S)$-butan-2,3-diol, while the $\ce{KMnO4}$ attack from opposite face gives you $(2R,3R)$-butan-2,3-diol. Since possibility of attacking from the Si-fase and/or Re-face is $0.5$ (or $50\%$), you get essentially a racemic mixture.
Based on above facts, OP's textbook description of the reaction is erroneous, not only on stereochemistry, but also on the notation of the reaction (it is syn- not sym-addition).


text:
Epoxidation for dienes
I have been asked to determine the major product for the reaction:

From the reagent given, one can tell that this is an epoxidation reaction. But, as only one mole of the peroxyacid is given, epoxidation can only take place in one of the two bonds.
How do you determine on which double bond the epoxidation will take place?

It should be one equivalent, not one mole.

MCPBA preferentially reacts at the more electron-rich alkene, in this case the more substituted alkene according to this entry on chemistryportal.net here and this review RSC Advances


text:
Prepare solution of salt in ethanol
The test for anions can be done with a platinum wire and a Bunsen flame.
My textbook says that it can also be done by preparing a solution of the given salt in water and ethanol and spraying it onto the Bunsen flame.
How do I prepare this solution in terms of how much water, ethanol and salt ?

Are you sure about anions ?

Ethanol-based hand sanitizer is readily available, so add a dash of salt, spritz it into a flame and behold the beautiful sodium D lines around 590 nm (yellow). Or just use cheap vodka and add the salt. Or add a little water to ethanol and add the salt. This is one experiment that is guaranteed to work.

You don't even need the alcohol... almost anything with sodium, whether salt, salt water, or MSG from the kitchen, displays the bright sodium spectral lines when put in a flame.


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Staggered Conformation Classification
I was working on an assignment I had a question about conformations of a molecule about a single bond.
Given a random molecule like this with multiple large substituents on one side of the $\ce{C-C}$ bond (in this case, $\ce{CH3}$ & $\ce{C2H5}$), is the molecule considered to be anti when the largest substituents on either side of the $\ce{C-C}$ bond are opposite of each other? (i.e. would the $\ce{CH3}$ need to be an $\ce{H}$ atom to be considered in the anti conformation)?


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Comparison: Esterification with acyl chloride and alcohol vs Fischer esterification
I would like to know if there are more than mentioned pros and cons with the different methods in making an ester, and also ways to maximize product yield (comparing the two).
Fischer- Pros: cheap, easy, non-toxic (or less at least) to environment/laborants. Cons: worse yield and also slow
Acyl chloride- Pros: Faster, more yield. Cons: More dangerous, harder, and more expensive.
A way to maximize yield for Fisher esterification is to increase concentration to any reactants, alternative to remove the forming water.
I have not really found how I can maximize it for the acyl chloride-method however.

Depending on the reactants, Fisher esterification can also be inefficient giving rise to a range of side products from acid dehydration of the alcohol

Give Emil Fischer credit for his eponymous esterification, not Fisher.

The acyl chloride-primary alcohol reaction's yield should be near quantitative. The by-product, HCl, is a gas.


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Why do terminal alkynes have lower pKa values than non-terminal alkynes?
We were asked in class which have the lower $\mathrm{p}K_\mathrm{a}$ values: terminal alkynes or non-terminal alkynes. The teacher told that terminal alkynes have the lowest $\mathrm{p}K_\mathrm{a}$ values without really explaining why.
Please provide some explanation on why do terminal alkynes have lower $\mathrm{p}K_\mathrm{a}$ values than non-terminal alkynes?

It has to do with hybridisation of the resulting anion, see https://chemistry.stackexchange.com/q/32341/16683

This post may be helpful. https://chemistry.stackexchange.com/questions/124896/c%e2%88%92h-bond-length-in-organic-compounds/124909#124909

Maybe I am not understanding your question, but non-terminal alkynes don't have a proton that can be removed so which pKa are you considering?

@ShoubhikRMaiti There's no such thing as proton that can't be removed.

@Mithoron You do know that's not how I meant it, right ?

@ShoubhikRMaiti I know what you meant and protons in non-terminal alkynes have pKa too.

@Mithoron Yes of course, I was just confused because those protons would not usually be referred to as alkyne protons obviously. So the question basically boils down to why alkyne protons are more acidic than alkene and alkane protons.

But some internal alkynes have no protons at all.  I don't think they would have a pKa.

$\mathrm{p}K_\mathrm{a}$ is an indicator of acidity constant and it is defined as the negative decimal logarithm of the constant itself: $\mathrm{p}K_\mathrm{a} = -\lg K_\mathrm{a}.$ So, (non-formal) the higher is the dissociation, the lower is $\mathrm{p}K_\mathrm{a}.$
Terminal alkynes have general formula $\ce{R-C#CH},$ where $\ce{R}$ is a radical. The hydrogen atom which is attached to sp-hydridised carbon atom can easily dissociate from it: $\ce{R-C#C- + H+}.$ The resulting carbanion is stable enough to favour the dissociation, therefore the lower is the $\mathrm{p}K_\mathrm{a}.$ On the contrary, in non-terminal alkynes dissociation is not favourable due to non-stable resulting carbanion.
As a result, the $\mathrm{p}K_\mathrm{a}$ of terminal alkynes is lower in comparison with that of non-terminal alkynes.

Non-terminal alkynes (aka internal alkynes) don't even have a hydrogen atom to potentially lose!   What could possibly be acidic about them?