text:
Activities and molar concentrations
I would be really happy if someone would clarify the term "chemical activity" and its implications for me, and also how it is related to the concept of molar concentration. My textbook fails at explaining this clearly and, sadly, the Internet is filled with vague definitions and scientific verbiage that's too difficult for me to grasp. Also, why are activities included in various formulae, and what are their uses?
I never quite actually understood what are activities and how do they differ from molar concentrations, nor how are they used. In addition, when working with the Nernst equation, why are we allowed to compute the activities of the reactants and the equilibrium concentrations of the products? This is the stuff that really baffled me and I'd be highly appreciative if you could clarify it.

Thermodynamic activity

Activity (a) is a pure mathematical construct used in thermodynamics and molarity (M) is a practical concept which can conveniently carry out in the lab. Someone can tell you to prepare 0.02 M Na$^{+1}$ solution but not 0.02 a Na$^{+1}$ solution. You should rather ask, how activity and concentration related? If you open a physical chemistry text, the definition provided there is abstract and good for regurgitating in an exam only.
Before activity can be explained in a less abstract way, one has to introduce another concept called fugacity. I will quoted original wording to the inventor- G. N. Lewis the person who gave the concept of activity to the world. It is always good to see the original work to see what was going on in the mind of the inventor.
Quote from his paper:

The idea of fugacity is thus evolved from the use of vapor pressure as a measure of escaping tendency. When a substance is in equilibrium with its vapor, the fugacity, in order to fulfil the laws of escaping tendency, must be the same i both. The fugacity of a substance is therefore equal to its vapor pressure if the vapor behaves like a perfect gas. Speaking in terms not very precise, we may say that the fugacity of a substance is equal to the vapor pressure that the substance $x$ could have if its vapor were a perfect gas. It has been shown in the preceding paper that for a given substance in a given state the fugacity is a definite property of which the numerical value can in most cases be readily determined, and which is well suited to serve as an exact
  measure of the escaping tendency.

The unit of fugacity is pressure as you should have guessed by now. Now come to activity. Lewis wanted to express an equivalent concept for solutions so that its unit was same concentration. He defined activity as:
$$ a =\frac{fugacity}{RT} $$
with the following properties:
Besides these we shall use the following definitions of the activity, as per Lewis:

When the activity of a substance is the same in two phases, that substance will not of itself pass from one phase to the other.
When the activity of a substance is greater in one phase than in another, the substance will pass from the one phase to the other, when they are
brought together.
The activity of a perfect gas is equal to its concentration. The activity of the solute in a perfect solution, at constant temperature and pressure, is proportional to its concentration.
We shall see that these statements suffice to define the activity of a substance in any state, and except in unusual cases enable us to calculate its numerical value.

Most thermodynamic equations are derived assuming a perfect gas or ideal solution. However, for real a solution the chemical potential $\mu_i=\mu^0_i +RT\ln(x_i)$ may not be accurate for the solvent where $\mu^0_i$ is the chemical potential of the liquid solvent at 1 atm and $x_i$ is the mole fraction. The solute may, similarly, not follow ideal behaviour.  The activity is introduced to allow us to use the familiar equations but now as $\mu=\mu^0_{liq} +RT\ln(\gamma x_i)$ where $\gamma$ is the activity coefficient thus $\mu_i=\mu^0_{liq} +RT\ln(a_i)$ where $a=\gamma x$. Thus we can think of activity as effective concentration or effective pressure (now called fugasity) relative to its standard state.  The activity coefficient is unity in the ideal case. The activity has to be measured experimentally because $\gamma$ depends on exactly what intermolecular interactions exits as any temperature and pressure and cannot be easily calculated.  As an example, naphthalene dissolved in benzene behaves almost ideally and the activity coefficient $\gamma = 1$ but when dissolved in hexane this is not the case and $\gamma=2.5$. A positive deviation from Raoult's law indicates that the vapour pressure is larger than that expected from perfect behaviour and so the activity is larger than the mole fraction and $\gamma >1$.

Activity is a concept created to make theory match the measurements. For example, the pH is a concept that can be measured with a glass electrode or a Hydrogen electrode. At high dilution, the measured tension of the cell, or the measured pH is exactly described by the formula pH = - log ]H+]. At higher concentration, the measured tension, or the measured pH is not exactly -log[H+]. Some chemists say that the measured pH is the log of a sort of "apparent" H+ concentration, that is called "activity". The activity may be considered as a sort of concentration calculated by dividing the number of mole of H+ by the volume of the free water, and not by the total volume of the solution. Free water is the water which is not attached around the ions of the solute by electrostatic forces. The free water occupies a smaller volume than the whole solution. This volume is nearly impossible to determine, so that we prefer speaking of activity, being $\ce{10^{-pH}}$, and that is all.


text:
How do "basic" amino acids (lysine, arginine) act as base under physiological conditions?
Why do we consider amino acids with high side chain pKa's basic, yet with a higher pKa, less of the amino acid will exist in basic form at equilibrium compared to acid form at physiological pH?
What will the concentration of the base be given the high pKa of these amino acid side chians. The answer is that the higher the pKa, the lower the concentration of the base, since with 1) with a higher pKa, then 2) you will have a lower Ka (pKa = - logKa) and therefore 3) a lower concentration of base, due to Ka = the conjugate base (times H3O+) / acid, then the numerator, the conjugate base will be  lowered in concentration with a lower Ka. This can also be confirmed with the Henderson-Hasselbach equation, where pH = pKa + log [A-]/[HA] and since pkA is increasing will make pKa > pH, meaning pH - pKa = a negative number = log [A-]/[HA], where [A-]/[HA] would then equal 10^negative number, meaning the concentration of the acid will be greater than the concentration of the base.
My question is, when you look at titration curves, we know that parts of an amino acid will act more as a base, than an acid, at higher pKa's. Meaning, if you have the amine group of an amino acid, with a pkA of 9-10, it doesn't want to give up it's hydrogen, until the pH is pretty high (10-12). Why are amino acids with high overall pKas considered basic, yet, when you look at a molecule's pKa at a high pKa, less of the molecule will exist as the base at equilibrium (meaning it is wrong to say that a molecule with a high pKA will exist primarily it is basic form)?

I am unable to follow your question, please edit it and make it more clear

As basic aminoacids are considered those with 1 carboxyl and 2 alkalic groups ( amino group like lysine or guanidine like arginine ). The respective conjugate acids of alkalic groups have obviously higher pKa than  carboxyls.

I edited the question for clarification. Also, I understand that basic amino acids will have higher pKa. I am asking why is that the case, when less of it will exist it in it's basic form at a higher pKa at equilibrium and more of it will exist in its acid form.

I will paraphrase the question:

Why are the amino acids arginine and lysine called basic? Why are the amino acids glutamic acid and aspartic acid called acidic?

This nomenclature makes sense for situations where the amino acids are dissolved in a non-aqueous solvent. In that case, the side chains will be neutral and have nitrogenous base groups and carboxylic acid groups, respectively.
When the amino acids are dissolved in water at neutral or physiological pH, however, the name does not make sense because they have already acted as base (Arg and Lys) or as acid (Glu and Asp), and are now the conjugate acid and conjugate base, respectively. For glutamic and aspartic acid, it is common to refer to them as glutamate and aspartate (the name of the conjugate base). For arginine and lysine, even when protonated, those names are usually used nonetheless.

[OP in comments:] So, in physiological situations (aka in the human body), how do arginine, lysine, and histidine act as bases, if they exist in their protonated form, considering how they are often used in enzymatic catalysis (same question asked of acidic amino acids in their deprotonated forms)?

Histidine
Histidine is a common acid/base catalyst. Its pKa is close to 7, so an appreciable fraction is protonated and an appreciable fraction is deprotonated. Often, this amino acid will act as acid and as base in different steps of the same mechanism (e.g. RNase).
Lysine
Lysine is not a typical acid base catalyst. It has roles in binding negatively charged ligands (such as ATP), acting as a "positive charge on a stick". It also acts as a nucleophile, forming e.g. Schiff's base with aldehydes and ketones. Lysine's pKa in the context of a folded protein varies considerably (see e.g. https://www.pnas.org/content/108/13/5260), so in some contexts, it will be an efficient acid/base catalyst.
Arginine
Arginine is not a typical acid base catalyst. Like lysine it has roles in binding negatively charged ligands but different from lysine, it can act as a bidendate hydrogen bond donor (e.g. making a nice salt bridge with aspartate or glutamate with two hydrogen bonds). It also has a role in cation-pi interactions, stacking its planar guanidinium group on aromatic rings. This web page has more on the functional role of arginine: http://www.russelllab.org/aas/Arg.html (and there are similar pages for the other amino acids on that site)

So, in physiological situations (aka in the human body), how do arginine, lysine, and histidine act as bases, if they exist in their protonated form, considering how they are often used in enzymatic catalysis (same question asked of acidic amino acids in their deprotonated forms)?

@Needhelpwithchemistry I updated my answer to address your follow-up

source: Stack Exchange

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

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text:
Why is solubility often listed as a chemical property? Isn't the action of dissolving just a physical change?
I am making a list of basic physical vs chemical properties.
I have been taught that when salt dissolves in water it is simply a special kind of mixture called a solution.
I have been taught that mixtures do not create a chemical change or reaction.
But I found a lot of high quality chemistry websites that list solubility as a chemical property of compounds.
Why isn't solubility a physical property if dissolving is simply forming a mixture and not causing a chemical reaction?

Because it is not simply a mixture,as dissolving is a kind of a complex chemical reaction.

because it is very useful to know.

It may be considered as a physical process, or a chemical process. It is a chemical process when the individual ions or molecules get surrounded by water molecules adsorbed at the periphery of the ion.

I am taking a guess here, but it is possible that solubility is put in the category of a chemical property simply because chemists deal with it a lot more than the physicists do. Same with pH.

Please consider giving the green checkmark 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!

The classification of physical and chemical changes has done more harm to students than good. This topic was taught in the 19th century but it still lingered in modern general chemistry textbooks which serve to repel good students from taking up chemistry as their field. For example, if we call the charge of electron red, and the charge of proton blue, will it change any fundamental thing?
You can try to understand physical vs. chemical change as follows: If you are changing the physical form (liquid, solid, gas) of one or more component, then you can safely label it as a physical change provided you can bring the original form back.
You can melt NaCl crystals into a liquid, this is a physical change because we can cool it and bring back the original material.
In the same way, solubility is just a property of matter. It is a very complex phenomenon. Whether you call it physical or chemical it does not matter in real life (most scientists will not worry about it and nobody will ever ask you these questions once you pass this course-so much so for the utility of this concept). For example, HCl is very soluble in water, but also reacts with water in such a way that it protonates water. Dry HCl gas is "neutral" but its solution in water is a powerful acid. There is no simple way to extract the gas back out of water. This solubility involved a chemical change.
On the other hand, salt NaCl when dissolved in water, it will not chemical react with water per se and you can recover the salt by evaporation. Should I call this a chemical change or a physical change? It does not matter.

Chemical actions can be reversible though, and phase changes can also be understood as chemical changes, as they involve the breaking and formation of bonds

It seems like gardenhead's point further supports this answer. Many things that were traditionally considered physical changes such as melting of covalent-network solids like SiO2 certainly involve breaking of covalent bonds. The traditional definition seems somewhat unworkable.

In other words, to be more rigorous, a physical change is a $subset$ of chemical change. A physical change is a type of chemical change which involves change of the form of a substance and it is reversible.

The distinction between physical and chemical change is actually not as cut and dried as one might think.  Sure, what happens to copper in nitric acid is very different from what happens to sugar in water.  But what about, say, the process that occurs when a steel strip is cleaned (basically removing carbon from the surface) in an atmosphere containing carbon dioxide?
In this process the carbon "reacts" with the carbon dioxide according to:
$\ce{C(s) +CO2 <=> 2CO}, \Delta H>0$
where we note the equilibrium; the gas can become "saturated usually with a lot of the carbon dioxide "solvent" left over.  As indicated by the sign of the enthalpy change, this is an endothermic reaction, and more carbon reacts upon heating.  If we were to allow the strip to cool the carbon could redeposit (it may also do so on cooler parts of the furnace).  From that point of view the process begins to look more like sugar + water (which is also reversed by cooling) than copper + nitric acid.  Moreover, we could interpret the carbon monoxide as carrying the carbon in a gas-phase solvated form, from which the carbon can be un-solvated again if precipitation is made favorable (e.g. by cooling), again something we would associate more with a physical change than the chemical one that the reaction above describes.
So is dissolution a physical or a chemical change?  Depending on the system it could be either ... or both.

Consider a solution of $\textrm{NaCl}$ in water and another solution of $\textrm{AgNO}_3$ in water. If you mix these two solutions, $\textrm{AgCl}$ precipitates out. If you do not consider solubility as a chemical process/change, then it is very hard to explain this.
A mere physical process should be no more than juxtaposition of four different kind of ions in the above mixed solution. Why should two ions choose to combine and opt out of this free democratic existence?
A chemical process OTOH, is conventionally understood as one in which bond formation and/or bond cleavage takes place. Exactly what has transpired in the above process. Isn't it?

One simple way to look at solubility as a chemical property may be to ask...What will it dissolve in?
You're right. When salt dissolves in water you get a solution.
But, salt does not dissolve well in oil.
There are, in fact, mnemonics to help people remember all the different rules for solubility.
Here's a link to a few solubility rules: https://m.youtube.com/watch?v=AsCLuLS-yZY
Physical properties are much more cut and dry.  When you reach a certain temperature the substance melts. When you add even more heat the substance will eventually boil.
I hope that helps to give you a way to think about solubility as a chemical property.

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text:
What software was used to create these organic reaction scheme figures?
I am aware that software such as ChemDraw, ChemDoodle or MarvinSketch can be used to draw skeletal formulas and the like. But what can I use to draw (in a similarly professional way) reaction schemes, including the skeletal formulas, arrows, reagents/conditions above the arrows etc. such as

I think there's a question on it, see this page.

All of these packages and many others allow for reaction schemes to be drawn as well as just the molecular structures.

$\mathrm\LaTeX$ to the rescue
I reproduced these reaction schemes with $\mathrm\LaTeX$ using the following particular packages:

chemfig for drawing 2D chemical structures (note that I enabled sans-serif font for the labels like the original did);
chemmacros for the typesetting of chemical formulas and chemistry-related notations;
siunitx for handling physical quantities (here, temperatures);
stix for loading STIX fonts, which in my opinion are the best choice for technical writing due to canonical and coherent appearance of the glyphs and great coverage for special symbols.

All of this is free for any type of use, works on all major operating systems used in production and the source code is available.
Result

Source code
% !TEX program = pdflatex

\documentclass{report}
\usepackage[T1]{fontenc}
\usepackage{stix}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{mathtools}
\usepackage{siunitx}
\usepackage{chemmacros}
\chemsetup{
    modules = all,
}

\newcommand{\bndlen}{1.7em}
\newcommand{\bndlenshort}{0.5}
\newcommand{\bndlenshorter}{0.75}
\newcommand{\bndlenlonger}{1.5}
\newcommand{\bndlenlong}{1.75}
\newcommand{\lnwidth}{0.8pt}
\newcommand{\cramwid}{0.3em}

\usepackage{chemfig}
\setchemfig{
    angle increment = 15,
    atom sep = \bndlen,
    bond offset = 1pt,
    double bond sep = 3pt,
    cram width = \cramwid,
    compound sep = 5.0em,
    arrow coeff = 1,
    arrow offset = 12pt,
    scheme debug = false,
    bond join = true,
    chemfig style={line width=\lnwidth},
    arrow style={line width=\lnwidth},
}
\renewcommand*\printatom[1]{\ensuremath{\mathsf{#1}}}

\begin{document}

\schemestart
    \chemfig{MeO-[::30]*6(-=-(-~--[::-60]CHO)=-=)}
    \arrow{->[\parbox{5cm}{\centering
        i) \ch{Hg^2+}, dil.~\ch{H2SO4}\\
        ii) \ch{AgNO3}, \ch{NH4OH}
    }][
        iii) \ch{Zn}-\ch{Hg}, conc.~\ch{HCl}]}[,2.5]
    \textbf{Q}
    \arrow{->[\parbox{5cm}{\centering
        i)~\ch{SOCl2}\\
        \hphantom{i)~}pyridine
    }][
        ii) \ch{AlCl3}]}[,1.5]
    \textbf{R}
    \arrow{->[\parbox{5cm}{\centering
        \ch{Zn}-\ch{Hg}\\
        conc.~\ch{HCl}
    }]}[,1.5]
    \textbf{S}
\schemestop

\bigskip

\schemestart
\chemfig{(-[::120]*5(---=-))-~-(-[::-60]*6(---=--))}
\arrow{->[
        i) \ch{H2}, \ch{Pd}-\ch{BaSO4}, quinoline
    ][
    ii) dil.~\ch{KMnO4}~(excess), \SI{273}{\kelvin}]}[,3.5]
\textbf{P}
\schemestop

\bigskip

\schemestart
    \chemfig{H_3C-[::-30](-[::-60]Br)-[::60]-[::-60]Br}
    \arrow{->[\parbox{5cm}{\centering
        i) alc.~\ch{KOH}\\
        ii) \ch{NaNH2}
    }][
        iii) red hot iron tube, \SI{873}{\kelvin}]}[,3]
\schemestop

\end{document}

While the chemistry-related syntax requires some training to understand in reading (and eventually, in writing), I was able to copy-paste the code above in one of the LaTeX-online windows of learn-latex.org* to successfully replicate the representation.  So possibly overleaf.com does so like cocalc.com did, too. *) This primer still lacks a translation into Russian.

Yes you can draw schemes like the one in your question with them.  In no particular order of importance:

For one, it is a matter of knowing what the programs may offer and choosing your tools accordingly.  A side-by-side comparison (e.g., this one) may help for this.  Equally, it is helpful to test programs listed (e.g., here) for their improvment since last time of consultation.  Don't hesitate to test programs with a different philosophy than the mainstream ones (e.g., winplt) you may see in the working group across the aisle.
Familiarize yourself with the documentation of the software.  Chemdoodle's manual is more than 300 p.,  replicate Chemdraw tutorials like the ones by Pierre Morrieux / ChemdrawWizzard (e.g., here).  Then, test them with own examples, draw the same molecules with either of them to spot similarities and differences.
You equally may use more than one program, too.  As e.g., the .mol2 format is a widely understood format, you could build the molecules in question with the SMILES string from a database in one program, and save the intermediate in this format.  Subsequently, you may finalize drawing and exporting the image as .pdf in an other.
Often, it is a matter of style, litteraly.  Journals have their style sheets (e.g., JACS & Org. Lett. differ from Synthesis, or Helvetica Chimica Acta, or RSC).  Of course you may use these standards even if you don't publish now in  these.
Equally, working groups may have, sometimes inherited / influenced from the PI's time as a post-doc, their style adjusted bond widths, fonts, colors.  So you recognize quickly e.g. Nicolaou's highlighting cyclic systems.  Or the systematic color scheme used in Kurti / Czako «Strategic Applications of Named Reactions in Organic Synthesis» (a review) to guide the reader about reagents, bonds formed, etc. pp.  You may define a personalized style file as well.

You can do all of what you stated on ChemDraw.
Skeletal drawings using the appropriate tools on the sidebar, and for text, there's a box where you can type in reagents and superscript/subscript letters. The same box which you would use to add a different element too (I think, can't access ChemDraw just now to double-check sorry).
For neatness, there is the align tool to make the reagents and reaction arrow look uniform (as well as align the arrow with the reactants/products).


text:
Reaction of alkenes with AlCl3/HCl
I stumbled upon an interesting organic chemistry problem:

Previously I thought $\ce{AlCl3}$ would only be used with benzene in acylation and alkylation, but apparently it can also be used in reactions with alkenes. Because of this, I am wondering what the product of the reaction above will be. My first thought was that there is an addition of $\ce{HCl}$ to the alkene and then the $\ce{AlCl3}$ would react in some way, though this is very likely incorrect. This left me wondering what the role of $\ce{AlCl3}$ is in this particular reaction and whether the $\ce{HCl}$ is just a catalyst.

This has an answer for using $\ce{AlCl3/HCl}$ in alkylation of benzene using alkenes. I believe this method is a shortcut without using alkyl chloride as an alkylating reagent. Thus, I think alkyl chloride might be the product (or alkyl may be polymerized).

I was doing this problem too and I thought that since AlCl3 is a Lewis acid, it should coordinate with one of the fluorines. This makes it a good leaving group so the chloride attacks beta to the carbon with fluorines. Then addition of HCloccurs on the resulting alkene, so that Cl is added to the carbon with the fluorine in regular markovnikov fashion.

are you doing IChO from the US or another country?


text:
Where does the beta notation come from in the alpha 1,2 beta glycosidic bond (in sucrose)?

I am confused as to where the beta is coming from (in the alpha-1,2-beta glycosidic bond in sucrose). I understand that the alpha is from the anomeric carbon of the glucose, but I don't see how the beta is part of this
To me it seems as though the hydroxyl on the anomeric carbon of fructose (i.e. th OH below the plane of C-2 in fructose) causes it to be an alpha form, correct? If so, then why is it called a beta linkage?

The figure in your post is not that clear. See the section on Sucrose here. https://chemistry.stackexchange.com/questions/115942/how-are-glycosidic-linkages-formed

@M. Farooq: The video you recommended gives the wrong loss of water for the formation of maltose from glucose. Your second response answers the OP's question.

@user55119, I deleted the link. The video was of low quality anyway.


text:
Putting sanitiser in PVC bottles
can I put hand sanitiser in PVC bottles or must the bottles be PET.  With the shortage of bottles I am keen to know if a 70% ethanol or isapropyl based sanitiser will distort or soften a PVC bottle in the short term or long term?

I can also obtain small  LDPE bottles (4)   Are these safe?

A don't see major problems in storing it in a PVC bottle,   PVC is not soluble in ethanol, neither in isopropanol. However, in this site they rate isopropyl alcohol as A while ethanol as C (with possible swelling in the long term).


text:
Why does percent ionization of a weak acid decrease when the initial weak acid concentration increases?
By the rules of the equibrilium, as the concentration of the acid increases, the reaction will want to dissolve more. So why does the percent ionization of a weak acid or base decrease when the initial weak acid concentration increases?

If you wrote the equilibrium equation and enumerated it for 2 different nonionized acid concentrations, you would know the answer faster than by posting the question and waiting for answer.

Ostwald Dilution Law might help..


text:
Why do stable compounds have a lower heat of combustion?
My brain somehow doesn't accept that more stable compounds require have less enthalpy. Isn't it like, if a compound is stable, it requires more energy to break the bonds because the compound doesn't want to react. However does have a less stable compound more energy so it doesn't require so much energy to break the bonds?

Your reasoning looks right, but your title doesn't match your question, so I can't quite understand what the question really is about.

I mean if the heat of combustion is the energy breakout when a bond is released, speaking of alkanes, why do cyclopropanes have greater heat of combustion than cyclobutanes although cyclobutane is more stable?

No, heat of combustion is not just that. There is a huge elephant in the room.

See addition content in my answer.

To answer the additional comment:
Heat of combustion is related to energy released by forming bonds of products and energy spent by breaking bond of reagents.
Cyclopropane has more of stress on its bonds, that are therefore weaker and less energy was released by their forming.
By burning, less energy must be spent to break them, while energy released by forming products is the same ( per  CH2 ). Therefore the specific heat of combustion in MJ/kg is bigger.

Chemical stability and stored chemical energy are 2 independent parameters.
Stability relates to energy needed to release the stored energy.
Typical fossil fuel like coal has energy typically about $\pu{40 MJ/kg}$. Explosives like TNT have typical energy $\pu{4 MJ/kg}$, so about 1 tenth.

It has a direct mechanical analogy:
Stored potential energy is related to how deep can the centre of object mass fall.
Stability is related to energy needed to make the object to fall.
If you stood on 1 foot on an edge of a small stone, you would have relatively low potential energy, but low stability. All, what is needed, is a light touch.
If you stood on both feet on a floor of tourist watch tower,  you would have relatively high potential energy, but also high stability. To make you fall, the tower would have to be destroyed or you would have to be thrown over the fence or through a window.

Oh. I get the point now. I think it is the minus sign in the representation that is confusing me. If the enthalpy is big, less energy has released. Isn't it?

Heat of combustion had the opposite sign than reaction enthalpy. For fuel burning, the former is positive, the latter is negative.


text:
How many hydrogen bonds can methanol make?
I believe that it can form 2 hydrogen bonds, one with oxygen and the other with hydrogen but a classmate said that oxygen can form 2 hydrogen bonds? Is this true? I can't find anything on google to support or refute that claim.

I don't know whether this can be proven or disproven. Hydrogen bonds are not as stable as covalences. They can interchange constantly. But well ! In principle, the Oxygen atom of CH3OH has two unused doublets. Each doublet can attract one H atom from foreign molecules.

It depends of the state of the sample, the pressure applied, and the criteria of hydrogen bonds applied (e.g., Phys. Chem. Chem. Phys., 2016, 18, 2736, doi 10.1039/c5cp06583f; https://www.researchgate.net/publication/287358732_Structure_Prediction_of_the_Solid_Forms_of_Methanol_An_Ab_Initio_Random_Structure_Searching_Approach).  Using both a distance criterion (CH-O < 2.6 A) and an angular one (> 130 degree), their model predicts an an increase of H bonds with growing higher pressure, though.

Water makes four hydrogen bonds in ice, so methanol should be able to form three when bonding partners are available.

Although this was an MD study (and might be said to be somewhat dated), quoting from Matsumoto and Gubbins (Ref. 1):

Most of the molecules are in the two-bonding state, but a
  small fraction of molecules are in the one- or three-bonding
  state, as already pointed out in previous studies. 7
  ,8

According to this study most of the methanol molecules form 2 H-bonds but ~4-7% form 3.
References

Matsumoto, M., Gubbins, KE. The Journal of Chemical Physics 93(3):1981-1994 (doi 10.1063/1.459076).

You would do good by pointing out first that yes, technically a molecule can make three H-bonds, but no, in the bulk compound you can't expect much more than 2 per molecule, for purely arithmetic reasons.

@IvanNeretin There is an publication by Etter on my table (Acc. Chem. Res. 1990, 23, 120-126, doi 10.1021/ar00172a005).  With its illustrations about hydrogen bonding patterns, e.g. of p-terephthalic acid, trimesic acid, and in figure 1 (p. 125) about 2-aminopyrimidine and succinic acid co-crystal [https://www.ccdc.cam.ac.uk/structures/search?pid=ccdc:1257588&sid=IUCr], I speculate your comment could be improved. For clarity, I suggest replacing "you can't expect more than 2 per molecule" by the more explicit / specific "you can't expect more than 2 per molecule of methanol".

@Buttonwood Yeah, sure, but a comment can't be edited. It is the answer that is intended to be improved by the joint power of our comments.

@IvanNeretin By arithmetic I suppose you mean there aren't enough donors and acceptors for all molecules to have 3 H-bonds.Luckily the MD simulation was self-consistent. The average number of H-bonds per molecule in that study was below 2 (see Table II).

@BuckThorn Exactly.

If we consider n methanol molecules such that intermolecular H bonding exists between all( value of distance has value that allows intermolecular H bonding).
One molecule has 1 O atom that bonds with n-1 H atoms. This case is for n atoms. So, number of bonds should be n(n-1). The H atom on 1molecule is counted as this bond is formed by other molecules having O.


text:
Latent blood detection with fluorescein
I need help with the reaction between fluorescein and blood in the method of detecting latent blood stains.
I find it hard to find any literature on this subject.
As far as I am concerned, the fluorescein is reduced into fluorescin in a stock solution with powdered zinc, distilled water and sodium hydroxide pellets.
$\ce{Fe^2+}$ (found in the latent blood) is oxidized into $\ce{Fe^3+}$ with hydrogen peroxide.
When fluorescin is reacting with iron ions in the haemoglobin, it will be oxidized back to fluorescein.
Can anyone help me with this? Maybe even on a molecular level.

Fluorescin(left), a non-fluorescent compound, reacts with iron in residual blood samples to produce highly fluorescent fluorescein(right).

The reference$\ce{^{[2]}}$ from which the following excerpt is taken describes the chemistry, preparation of reagents, and application to forensics.

Fluorescin is the oxidized colorless form of fluorescein. Fluorescin reduced with hydrogen peroxide is immediately oxidized to fluorescein in the presence of blood-associated proteins and iron ion found in hemoglobin. Fluorescein fluoresces when viewed using blue light and orange or yellow filters, whereas fluorescin will not fluoresce under the same lighting conditions.

Literature Resource
An excellent starting point to consider for searching for relevant literature is PubMed.
References

Image sources - PubChem
Google Books - Complete Crime Scene Investigation Handbook, Everett Baxter, Jr.
Latent-Prints.com - Fluorescein Detection of Latent Bloodstains
The Presumptive Reagent Fluorescein for Detection of Dilute Bloodstains and Subsequent STR Typing of Recovered DNA
A Comparison of the Bluestar and Luminol Effectiveness in Bloodstain Detection


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What could be a high school experiment that involves changing a condition to get the highest yield of a certain product?
I am looking for ideas for a chemistry experiment and I am interested in doing one that has to do with the optimization of percentage yields. What could be a high school experiment that involves changing a certain condition to get the highest yield of a certain product?

You could try the so-called Job's experiment. You prepare in advance 100 mL of two solutions of reactants which are supposed to have the same molar concentration. For example : CaCl2 0.1 M and Na2CO3 0.1 M. You take 9 test tubes in the same rack. Write 1 on the first, 2 on the second tube, etc. up to 9 on the last tube.
Now, with a plastic pipette, you put 1 mL of the CaCl2 solution in test tube 1, 2 mL CaCl2 in test tube 2, 3 mL of CaCl2 in test tube 3, etc. up to 10 mL in the tube 10. Then, you take another plastic pipette, and transfer 9 mL of NaCO3 solution in the tube 1, 8 mL of Na2CO3 in test tube 2, etc. up to 1 mL in tube 9. You mix thoroughly all tubes. You must see a white precipitate that settles in the bottom of  all tubes after a while. If you measure the height of the deposit, you will find that the highest amount is in the tube 5, if the stoichiometry is 1 + 1. If the mot important deposit is in another tube, it means that the stoichiometry is 1:2, oe 2:3, or any other ratio.
You may change and use any other combination of compounds. You will find that in the mixture CuSO4 plus NaOH, the stoichiometry is not 1:2, as you might suspect, as the precipitate is not Cu(OH)2, but 3Cu(OH)2·CuSO4.

9 or 10 tubes ?

[EDIT] A rapid photolysis experiment centered around B-12 ( a photocatalyst, see, for example, this article and also here) plus some dissolved B-Complex, where the removal of the solution coloration with ordinary sunlight can be witnessed in hours, may be of interest.
One can have several clear plastic contains that have varying amounts of tape limiting light exposure to demonstrate the impact of sunlight. I would take pictures to document the comparative relative transformation over time in solution coloration.
An alternate longer term experiment, based on the article 'Photosynthesis: Response to high temperature stress' in Journal of Photochemistry and Photobiology B: Biology, to quote from the paper:

Temperature is continually increasing due to global warming and has become a serious threat to crop yield. Temperatures above the normal optimum are termed as heat stress (HS) which disturbs the normal cellular homeostasis leading to retardation in growth and development and even death in extreme conditions...Although the plant growth is controlled by a multitude of physiological, biochemical, and molecular processes, photosynthesis is a key phenomenon, which contributes substantially to the crop yield...Photosynthesis is highly sensitive to high temperature stress and is often inhibited before other cell functions are impaired [4].

So, the high school experiment would be to display over time, the effect of light/heat on a living plant with implications to global food production.


text:
How many significant figures should I report calculating volume concentrations?
Let's say a $\pu{300mL}$ solution of $\ce{CuCl_2}$ was diluted to $\pu{900mL}$. I am then asked to calculate the concentration of the new solution. My solution to the problem is as follows:
$$\begin{align}M_1 \times V_1 &= M_2 \times V_2\\[0.5em]
\pu{300mL} \times \pu{4M} &= M_2 \times \pu{900mL}\\[0.5em]
M_2 &= \pu{1.33M}\end{align}$$
Now I'm wondering how many significant figures my final answer should have. From what I know, I have to refer to the least significant measurement, which happens to have one significant figure. Should my answer be 1 instead?

I've tried to clarify your question, but  I don't see what has to do with Stoichiometry...

The data are not clearly defined. For example, M1 is suddenly taken as 4 mol/L. Where does it come from ? Then M2 is 1.33 without any unit . What does it mean ? Further more :  What are the errors on the given volumes (300 mL and 900 mL) Are they ± 1 mL ? We cannot guess them ...

While you cannot ever assume the precision or accuracy of measurements when it is not given directly, most of the time measurements between 250 mL and 1000 mL have a precision of 5 mL and an accuracy of ±5 mL; however, you should always check your graduated cylinder (or volumetric flask) to determine the exact accuracy and precision of your measurements.

RANT - I'll go off on a bit of a tangent here. It seems that it is common to give answers for problems in most books to 3 significant figures regardless of how the problem is setup. To me "4 molar" has one significant figure. If the book's author wanted three significant figures then the problem should specify "4.00 molar."

I don't think that this question should be closed, the user attempted to answer and explained its considerations

We cannot say how many significant figures you should be reporting because:

We cannot be sure about the precision of your volume measurements; and
We cannot be sure about the precision of the original solution’s concentration.

Concerning volumes: most of the time when working with such volumes in the lab, I would use measuring cylinders. These come with a rather poor precision but are adequate for most purposes. However, considering we need $\pu{300ml}$ and $\pu{900ml}$, we would probably want to work with very large cylinders so the error would be in the range of $5{-}\pu{10ml}$. It is possible to use much more precise equipment: in analytical lab courses, we would use pipettes that measure exactly $\pu{25.0 \pm 0.1 ml}$ and dilute using measuring flasks at a similar level of precision.
If, as it seems from the values you plugged, the initial solution was $\pu{4M}$ (1 significant figure), I would argue the result should be given to 1 significant figure too – introducing a large error. Arguably, a proper error propagation calculation would probably give you a better precision in the final result.

The point about volumes is that "$\pu{3.00\times 10^2 ml}$" and  "$300. \ \pu{ml}$" have three significant figures, but "$\pu{300 ml}$" is ambiguous since there could be 1, 2, or 3 significant figures.

The concept of significant figures is not always clear, and the rule that you find in many textbooks can be misleading.
If you read the international standard e.g. Evaluation of measurement data — Guide to the expression of uncertainty in measurement you will hardly find a comparable definition of a significant figure.
In fact, the answer here, in my opinion, depends on the context. If you are adding solution with a precision of one millilitre. Would be correct to report $\pu{1.333M}$ because if you added, for instance, $\pu{0.301 mL}$ you would register $\pu{1.337M}$. So I would say that the correct answer is $\pu{1.333 M}$.
Eventually, it depends on what are you going to do with that solution and you might not need such precision.

-1 ignoring the molarity of the solution, the volume contains at most 3 significant figures. "1.333 molar" would have four. Inventing extra precision is exactly what significant figures is trying to avoid.

@MaxW thanks a lot for your feedback, I have to ponder more about this subject but given the precision of one millilitre, I am still not sure that in practice is not better to preserve also the third digit after the comma because excessive truncation I think it eventually leads to biases.

In my days in high school and college a table of 4 digit log tables or a slide rule was the norm. 3 digits was all you got. Now with a calculator I carry two extra digits in intermediate calculations and round off the final result to the desired number of digits. This will reduce round off errors in the overall calculation but round-off errors can't be prevented entirely. Using the 20 digits that a calculator might have is just to tedious.

@MaxW you are right, thanks for your comments!

One more thing. The guide you pointed to uses more sophisticated methods for error propagation.  However the Wiki article on Significant Figures gives a good explanation.

Short answer: Without knowledge of the partial errors, expressing the final concentration as 1.33 M or 1.333 M is fine.
Not so short answer
The initial point is to know the significant digits of your input values.
E.g. what are significant digits of the 4 M solution ? 300 mL ? 900 ml ?
The are 2 major rules for 4 arithmetic operations and significant digits:
1/ For addition/subtraction, the result has the last significant digit as the number with less absolute accuracy:
$4.005 + 3.2 \simeq 7.2$ ( takes the last significant digit place from the latter )
1/ For multiplication/division, the result has the count of significant digit as the number with less relative accuracy:
$2.0 * 3.211 \simeq 6.2$ ( takes the significant digit count from the former )
Longer answer:
As a general rule, if there is given a function:
$$y=f(x_1,x_x,..) \tag{1}$$
calculate its differential form:
$$\mathrm{d}y=\dfrac {\partial f}{\partial x_1} \cdot \mathrm{d}x_1 + \dfrac {\partial f}{\partial x_2} \cdot \mathrm{d}x_2 + ... \tag{2}$$
Take the given or estimated values of $\mathrm{d}x_\mathrm{i}$ as estimation of the current or maximal error of independent variables $x_\mathrm{i}$ , calculate $\mathrm{d}y$ and then decide the valid digits.
If you evaluate rather standard deviation errors or confidency intervals, you have to evaluates the square, implying there is no correlation between the input variable errors.
$$\mathrm{d}y=\sqrt{ {\left(\dfrac {\partial f}{\partial x_1} \cdot \mathrm{d}x_1 \right)}^2 + {\left(\dfrac {\partial f}{\partial x_2} \cdot \mathrm{d}x_2\right)}^2 + ...} \tag{3}$$

For the example in the question, we have:
$$c_2 = c_1 * \frac{V_1 }{ V_2} \tag{4}$$
and by differentiation:
$$\mathrm{d}c_2 = \mathrm{d}c_1 * \frac{V_1 }{ V_2} + c_1 * \frac{\mathrm{d}V_1 }{ V_2} - c_1 * \frac{V_1 }{ V_2^2} \cdot \mathrm{d}V_2 \tag{5}$$
If we plug in the numbers for $c_1, V_1, V_2$, replace $\mathrm{d}$ by $\Delta$ and apply rather the absolute value for error cumulation:
$$\Delta c_2 = \left|\Delta c_1 \cdot \frac{ \pu{0.3 L} }{ \pu{0.9 L}}\right| + \left|\pu{ 4 mol/L} \cdot \frac{\Delta V_1 }{  \pu{0.9 L } }\right| + \left|\pu{ 4 mol/L} \cdot \frac{\pu{0.3 L} }{({ \pu{0.9 L})}^2} \cdot \Delta V_2 \right| \tag{6}$$
Estimate the error of initial concentration $c_1$, of the initial and final volumes $V_1$ resp. $V_2$.
The $\mathrm{d}c_2$ is the error of the final concentration. Plug in the numbers. The order of the final error should be the last valid/significant  digit.
Be aware  the concentration info 4 M says nothing about its accuracy, in contrary to e.g. 4.00 M.
In case of the SD or confidency intervals, do the same but with the squares as above.
$$\Delta c_2 = \sqrt {{\left(\Delta c_1 \cdot \frac{ \pu{0.3 L} }{ \pu{0.9 L}}\right)}^2 + {\left(\pu{ 4 mol/L} \cdot \frac{\Delta V_1 }{  \pu{0.9 L } }\right)}^2 + {\left(\pu{ 4 mol/L} \cdot \frac{\pu{0.3 L} }{({ \pu{0.9 L})}^2} \cdot \Delta V_2 \right)}^2} \tag{6}$$

Think you went down the rabbit hole. The OP specifically asked about significant figures, not error prorogation in general.

"The order of the final error should be the last significant digit"


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What is diamagnetic in diamagnetic dilution?
I'm reading a paper where the authors study a peptide solution where each peptide may have two spin labels attached to them. They put the peptides on the surface and study them using a diamond sensor.
They perform the experiment in two ways. First, they put only peptides with attached spin labels. Then, they dilute the solution with peptides that do not have spin labels, in a ratio 1:10.
They refer to this dilution as "diamagnetic dilution". I do not know why it's called diamagnetic. I've tried to look for a definition but the term only appears in highly technical papers that I can't understand very well.
Does anyone know why this is called "diamagnetic dilution"?
Ben
Edit: The article I'm referring to is this one:
https://advances.sciencemag.org/content/3/8/e1701116
But my question is general.

The term diamagnetic dilution (in general) implies that the material used for dilution has all its electrons paired, so we will not see its response in an electron spin resonance experiment or any other experiment which can sense unpaired spins.

In other words, the molecules used for dilution do not have paramagnetic labels and are thus purely diamagnetic. Seems a bit peripheral as a description, but makes sense. Thanks.


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Law of equivalence
100g mixture of nitrates two metals a and b is heated to  constant weight containing corresponding 50g oxides of the metals
{i.e METAL NITRATES(100g) --->OXIDES(50g) +(some nitrogen oxide gases) } .
The equivalent weight of a and b are 103 and 31 respectively .What is the percent composition of a and b in the mixture?
In this question book has given that
equivalent weight of nitrate of a= equivalent weight of a + equivalent weight of nitrate ion
And similarly,
equivalent weight of oxide of a= equivalent weight of a + equivalent weight of oxide
How to find EQUIVALENT WEIGHTS of nitrate and oxide.Can you tell me how should i calculate it or where am i going wrong?Or can you just solve this question?

How can the weight of pure metals decrease to 50 gram upon heating? Recheck your problem. Also show your attempt at the problem.

This part is problematic "100 gram mixture of two metals a and b is heated to constant weight of 50 gram containing corresponding oxides of the metals." Weight of two pure metals a and b cannot $decrease$. Recheck the wordings of the problem.

Questions are supposed to be improved and asked to be reopened,  instead of posting a new question. That is against the site policy.

100g of metals A and B or 100g of their nitrates ? Be equation/equivalent weight you may mean molar (equivalent )mass.

Yes, it is most likely nitrates but the student must realize that there is a mistake in the wordings.

The problem has been created by a rather old teacher who has not yet accepted the mole concept. He prefers using the old theory of equivalent, which was in use in the 19th century, and cause a lot of confusions when dealing with redox equations. I would guess A is probably lead Pb, with an equivalent 103.6, and B is probably copper Cu with an equivalent weight is 31.7. But what is worse : Guest 2020 is not able to calculate the weight of the nitrate ion, even if he knows that it is 14 + 3x16. This shows that he does not know what is an atomic and a molecular mass.

Sadly, this is equivalents & normality is still taught in Indian schools I think he is more confused by the fact that formula weight of nitrate is the same as its molecular weight and why oxide's formula weight is divided by 2.

I have checked it .This is the problem and made corrections accordingly. If someone knows the correct answer or could correct me where i have written wrong or where i am getting wrong then they could rather than !!COMMENTING!!. And if you know the answer then give it and tell me why the molecular weight is same as equivalent weight??

If you spent little  of time by a quick search, you would save yourself the time spent by writing your both questions ( first closed ) : https://en.wikipedia.org/wiki/Equivalent_weight
A part of the site policy is not answering some questions even if the answer is well known, but commenting or closing them. Your question needed additionally a clarification. If what you wanted was quick answer for a trivial or homework-like question, post rather on Quora, they answer such questions.

Note that you have already solved on the photo what you have asked for. The rest is just solving 2 linear equations for 2 variables ( Search for it if needed ).

Finally your problem reads correctly! Do you recall how equivalent weights are calculated for redox species and anions? What is the charge on NO$_3^-$ ? It is -1, so equivalent weight is "molecular weight"/ absolute value charge. = Anything divided by 1 stays the same.
Coming to the oxide, O$^{2-}$, the charge is -2. So equivalent weight is 16/2=8.
Keep in mind that equivalent weights are highly dependent on the reaction to reaction. Don't apply these rules without understanding the nature of the reaction. Hope it is clear now.
As an exercise, calculate the equivalent weight of KMnO$_4$, when it reduces to Mn$^{2+}$ in acidic medium. This time, you have to check how many electrons are involved.

Ok now i understand where i was wrong thanku very much. And i think the answer of the exercise should be (158/5).

Yes, this is correct.


text:
How to model the Belousov-Zhabotinsky (BZ) (in Maple)
I need to model the Belousov-Zhabotinsky reaction (or a different oscillating one). I found out about the Oregonator and the Brusselator reaction and now I want to model them in Maple (or similar software). In this post the rate equations are given. On Scholarpedia the rate equations are scaled (I do not know what that means), but I still don't have clue how to model this. I have very basic knowledge of rate equations and modelling reactions (it is a project at the end of secondary school). I do know how to differentiate and find primitives of a lot of functions if that's needed.
I would be very grateful if someone could explain to me how scientists use rate equations and if someone could show me how to enter the Oregonator (and/or Brusselator) in Maple of similar software. Preferably also that I can change the starting concentrations (and the temperature if that's possible).
Many thanks!

Meh, you just need a script written in language of specific software. To write one you need to learn a bit about it's syntax, and, if you write it properly, soft is gonna numerically solve the equations for you.

https://chemistry.stackexchange.com/questions/118675/modelling-the-oregonator

Use function dsolve. Take a look at the examples. All you need to provide is the vector field — that is, the right-hand side of each differential equation — and the initial conditions.

@RodrigodeAzevedo Thanks, my laptop automatically changed it bc it didn't recognise the word.

Thanks! I'm now looking for the needed rate constants and concentrations. I'm also exploring the dsolve. If a new question arises, I'll let you know.

Answering one's own question is encouraged. When you are done with your project, please consider posting an answer here, with graphs and Maple code, so that others can build on your work and do not have to start from zero.


text:
Science of soap bubbles
On the web there are tons of different recipes for a solution to make soap bubbles.
Most of them use only three ingredients:

water
dish soap
glycerin

But their ratios widely change among the sources.
Right now I'm using: 20:5:2 in volume.
The bubbles come out, but not as I desire.
Compared to a commercial solution, mine are bigger but last shorter. End if I constantly blow with some strength, the commercial solution produce literally hundreds of small bubbles (diameter between 1 and 4 cm), while mine will just make a couple of middle-sized bubbles and then the membrane will break.
The goal is to make a solution that allows to:

make big bubbles if the blow is weak
make a lot of small bubbles if the blow is strong
maximize the duration of the bubbles

I will appreciate any hint in this direction, but I'm also interested to understand how to tune the ratios of the 3 ingredients. I mean, given the requirements above (I guess they are the most common!) how would you calculate the ratios?
Of course I can experiment. But I want to learn the theory behind.


text:
Why is substitution of halogens to alkanes a chain reaction?
During the propagation step 2, the alkyl radical always react with the halogen molecule according to my textbook. However, isn't the alkyl radical react with the halogen radical formed in the initiation step much better since it lowers the potential energy?
If there is a chance that the 2 radicals combine, won't the chain reaction terminate rapidly as less and less halogen radical is formed?

That's indeed possible, and that's indeed one of the ways how the chain terminates. But it happens very rarely, because the radicals are very few and far between.

You must realize that the number of atoms created in the first step is smaller than one percent of the whole number of Halogen molecules. So the first reaction produces a tiny amount of HCl and a tiny amount of an alkyl radical. The probability that such a radical touches the second halogen radical is extremely small. It will collide much more frequently another halogen molecule, and produce a new halogenated compound plus a new halogen radical, etc.


text:
Why thermodynamic state variables of a sytem are only defined at thermodynamic equilibrium and not during any other instant?
As you have read the title so my question is why are thermodynamic variables only meaningful when the thermodynamic system is in thermodynamic equilibrium and is always meaningless in any other case.

I disagree with the premise. The Gibbs free energy is still the Gibbs free energy whether the system is in equilibrium or not. And the difference between non-equilibrium and equilibrium is the driving force to get to equilibrium.

Variables such as T and p are not well-defined when a system is fluctuating. Thermodynamic state variables are defined for systems that are static or quasi-static (in the latter case, transitioning very slowly - and therefore reversibly - through a series of equilibrium points). The main reason is that macroscopic fluctuations are absent in true equilibrium states. Each phase in such states is homogeneous. The argument is thus somewhat circular: thermodynamic states are homogeneous states that have fixed properties. The only way to guarantee fixed properties is by being at equilibrium.
Note that it is possible to obtain thermodynamic parameters by imposing restraints that trap a system in a local free energy minimum, and thereby prevent the system from transitioning to a global free energy minimum.


text:
Why are crystalline solids anisotropic?

Why crystalline solids are anisotropic whereas amorphous solids are isotropic?

Is it because in crystalline solids, X, Y and Z dimensions vary, but so does in amorphous.
Please write as simple as possible.

Because a crystal is, well, a crystal. There are defined lattice positions.

It is anisotropic vs. isotropic. Isotonic is entirely different.

In amorphous, there are no X, Y and Z.

@MaxW sry its isotropic...

To qualify as «crystalline», there must be a minimal unique pattern (unit cell) to construct the solid only by applying regular, periodic translations. You may predict the outcome of the translation both in proximity, as well as in far distance reliably.  In an amorphous state, there is no such predictability.
For two, anisotropic means that a vectorial property depends on the orientation within the sample.  In the case of an amorphous sample (as contrasting to crystalline) like glass there is no orientational difference.  As an example, the speed of light in glass is independent on the direction you look at the sample; thus, glass is an isotropic material.
In a crystal, however, the speed of light depends on the relative orientation of the sample, thus crystals generally are anisotropic.  Depending on the symmetry of the underlying Bravais lattice, there may be two, or three special directions of observation along those vectorial properties are equal to each other (e.g., uni- and biaxial crystals in birefringence).  Then, these directions of observation are isotropic to each other; allong all other directions of observation, however, the crystal's vectorial properties are anisotropic.


text:
Melting point of crystalline solids

Why crystalline solids have sharp melting points but melting point of amorphous solids vary within a range?

Is it because the interaction energy is equal between atoms of crystalline solids so they require equal energy to break the bond or is it because the atoms of crystalline solid are similar and are equally distant so energy gets transferred to every molecule equally so all the bonds break simultaneously.

Remember that temperature is a macroscopic quantity, telling the heat content of system. So if the crystalline solid is at its fusion temperature, and every atom in the crystals have equal interaction energy with each other (ignoring all defects of lattice), the whole solid should fuse simultaneously, thus having sharp melting point.
On the other hand, amorphous solids don't have uniform lattice and interaction energies of atoms/molecules may differ. So even at the same temperature, the whole solid might not melt, so we say amorphous compounds melt at range of temperatures.


text:
Cooking IN CO2 gas?
Quick theoretical question:
What if I replaced water with dry ice in my pressure cooker?  Assuming I used a calculated amount to prevent an explosion...
Would It be able to cook?
Would it dehydrate the food to any degree?
Would it penetrate into the center?

If you put the dry ice on the bottom and let the gas flow over the food it could freeze dry the food to an extent, but not cook it.

SAFETY - Trying this would be a very bad idea. The metal vent in the pressure cooker would get cold enough to freeze water. A plug of ice could block the vent and cause a dangerous pressure build up -- think explosion! There is a safety release typically on a pressure cooker, but it is uncertain if an ice build up could occur simultaneously on that too. Why fool around with something that could explode?!? There are safer and quicker ways to dehydrate food.

In short, your pressure cooker will (almost) become an equivalent of the Boston Marathon bomb.

Of course I would not try this.  The idea is not to freeze dry the food nor to dehydrate it any more than deep frying would.
My question remains, maybe I needed to be more specific as to say that I would turn the machine on.  Could the heat from the machine heat the sublimated CO2 fast enough to melt the potential icing problem in the vent and subsequently cook the food?

The idea is that I could cook food under pressure with a gas, since a pressure cooker will not build pressure with just air.
Essentially an air fryer but under pressure.

Why CO2? Just don't put any water in the pressure cooker and you would get the same effect. Which won't, most likely, be good. The reason why a pressure cooker works is that it allows water to boil at a significantly higher temperature which is why it cooks food faster. With no water the hot gas will generate the same pressure but there will be no liquid to warm up and heat the food: you'd basically be pressure frying which has no advantage over normal frying.

No, this would not work because of the way pressure cookers cook
The reason why pressure cookers work to cook food faster is that water boils at a higher temperature when under pressure. This extra temperature cooks food faster that water boiling at 100 C. The pressure is merely a means to ensure that higher temperature.
Typical pressure cookers regulate the pressure to about twice normal atmospheric pressure. This increases the boiling point of water to about 120 C. The cooking is also enhanced by the fact that, when pressurized, most of the gas inside will be water vapour which has a much higher specific heat capacity than most other gases so cooking uncovered food more effectively than most alternative vapours at the same temperature.
Not adding water defeats this mechanism. Heating a pressure cooker without water will generate the pressure (even if you don't add CO2, the air will heat quickly and generate pressure). But that pressure can't go above the regulated level (unless you block the release valve, which just creates a bomb, not a cooking vessel). And there will be no fluid in the cooker to spread the heat around and it is likely that you would just see uncontrolled temperatures in the parts of the food in contact with the surfaces with no way to transfer that heat to the rest of the food. So you will, basically, be frying your food and gaining no benefit from the pressure while being unable to see what is happening (which is not usually a good way to fry things).
Using solid CO2 won't make much difference except to make the process take longer because part of the vessel starts really cold.


text:
Reducing pressure in a binary liquid solution
Consider the following diagram for a binary liquid system comprised of A and B, where A is the more volatile component.

My professor says

Consider an isopleth from $a$ to $a_4$. At the start, the solution is in pure liquid phase at state $a$.
As we reduce the pressure, the solution starts to boil at $a_1$. The composition of the liquid solution is denoted by the point $a_1$ which is the same as the overall composition of the solution since the solution has only just started to boil. The composition of the first trace of vapor formed is $a_1'$.
As the we reduce the pressure, the solution continues to boil to reach point $a_2''$. We may draw a tie line to find the composition of the liquid $a_2$ and the composition of the vapor $a_2'$. At every pressure where the solution exists as a mixture of liquid and vapor, we may use a tie line to find out its composition in the liquid and vapor phase.
Upon reaching $a_3'$, the solution has just been completely changed into vapor phase and the last trace of liquid has the composition $a_3$. After all the solution is vaporized at $a_4$, the composition of the vapor is again the overall composition of the solution.

I am slightly confused about we are able to make the pressure drop from $a_1$ to $a_3'$. My reasoning is: since the liquid and vapor are in equilibrium, shouldn't enough liquid keep on evaporating to maintain a constant vapor pressure?

Consider a system with a variable volume (e.g., a syringe, but on a larger scale). As the volume increases, liquid becomes vapor without changing the amount of material present.
In a system where there is only one component, changing the volume of the container will change the fraction of material in the vapor vs the amount of material in the liquid. But as you concluded, the vapor pressure will remain constant (assume constant temperature, since it was not mentioned), because there is an equilibrium between liquid and vapor for the one material in each phase.
However, when there are two components of different volatility, the vapor composition varies, depending on how much vapor there is, compared to the fraction left as liquid. So the equilibrium is no longer between a vapor and a liquid of the same composition.
It might be easier to visualize the change going on in the liquid/vapor if you change the dimension of the vertical axis to read 1/Volume. Then the vapor over the solution is (intuitively) richer in A, always (again, intuitively), until the liquid is all gone, when the vapor attains the original composition. Then, BTW, you realize that the pressure was changing while you changed the volume, and pressure might be a more relevant parameter.


text:
How does the heat capacity of a calorimeter interact with the specific heat capacity of the water inside it?
I saw this basic calorimetry question and it included the heat capacity of the calorimeter (1550 J/°C) but it also included that there were 150 g of water inside the calorimeter as well (4.184 J/(G°C). I saw that in order to calculate the total q, you add q of calorimeter to q of water. I'm having a hard time envisioning why you have to add the two instead of taking only the q of the calorimeter (isn't q of water included in the q of calorimeter?). The question stated that ∆T was equal to +1.211°C.

No, the heat capacity of the specific calorimeter being used must be determined experimentally. You can look up the heat capacity of water in a table.

Think of the calorimeter as the container.  Its heat capacity is fixed.   Then you add to this the heat capacity of the water, a figure that can vary depending on how much water you've poured into the calorimeter.

Oh ok, so let’s say that there’s a temperature change, would that temperature change be equal to both the water and the calorimeter?


text:
Fixing incongruities in PES Scan
I tried to parameterize a particular x-y-z-t dihedral angle of novel molecule using a Quantum Mechanical scan, but I got the strange result below.

The energy changes sharply around 130° and $E_\text{dihedral}(0°) \neq E_{dihedral}(360°)$.
I performed PES scan in gaussian09 and psi4 and I got the same results in different softwares.
I'm uncertain what approach to take to fix the scan. I saw some parameterization articles which didn't perform PES scan 0-360 degree. Other authors performed scan in the range of 0-60 degree or 0-120 degree and I didn't predict I will face a result like this. What is going wrong with my scan and how can I fix the issue? Would performing single point energy for each angle instead of PES scan help?

What are the units on the figure?

I'm voting to close this question as off-topic because I'd bounce it back too if I was the referee. The plots should be the same at 0/360 and the discontinuity adds doubt to the methodology.

of course you can vote to close this question... but this is not off-topic. your strange comment is off-topic. Nobody asked you "hey if you were referee what would you do".  why are you saying this? Are you trying to get attention? this is a computational chemistry problem (an important one) and we found a solution. TO say "should be equal" is not a solution. You should be more respectful to your olders, teenager.

Karsten Theis, Mathew Mahindaratne, Todd Minehardt, Mithoron, Jan,Let me make a criticism, please: It is really funny to mark an article about to be published in an academic journal with 3.5 impact as a "zero-effort homework". lol :) But I respect that. I hope someday you will get nobel prize.

Even without seeing your molecule, I can say the referee is very likely correct. If you think of doing a scan of the dihedral of hydrogen peroxide, by symmetry the path from $\theta=180\to360$ should just be the reverse of $\theta=0\to180$. In your case, I don't know the symmetry around the dihedral, but at a bare minimum the energy should be the same after a $360$ degree rotation. The fact that it isn't suggests that some scan step (probably that point around $130$) irreparably distorted the molecule geometry so that additional steps didn't bring it back towards the initial geometry.
Generating the scan point by point may help, but you still have to ensure the rotations you are making are physical and aren't distorting the molecule too much. Sharp changes in the energy between steps are a sign that something is going wrong.
You don't mention this detail in your question, but its also important to consider whether what you want is a rigid scan (where you just change this angle) or a relaxed scan (where you change this angle and then allow the rest of the structure to relax).

I found the solution !.  moderation can close this topic.  but the solution has nothing to do with rigid or relax.

You could add an answer, it may help other users with a similar problem.

In regards to your previous comment, you did also ask whether the referees statement was reasonable. As I said, generating the scan steps point by point could help, but you still have to make sure you aren't unduly distorting the rest of the molecule to the point where most of the change in energy is due to the distortion rather than the changing dihedral. Making sure that ~130 geometry doesn't, for example, cause atoms to nearly overlap would help. You can also do two scans from 0 to -180 and 0 to 180 so that distortions along the way are less likely to propagate to later steps.

Thanks Tyberius . I will also try two-step method. (0-180 and 0 180)


text:
Enantiomers and Flipping chiral centres

Hello,
Normally, it seems like enantiomers have inverted R/S configuration at every chiral centre. However, in the pair below which are clearly enantiomers (note the methyl and benzyl groups are going in and out of page) as mirror image of one is superimposable on the other, the configuration of the carbon in the main cycle attached to the nitrogen seem to have the same stereoconfiguration in both structures.
I suspect this has to do with the "prochiral" / meso nature of the molecule without the extra groups on the nitrogen, but have never heard of such exceptions.

These are diastereomers, not enantiomers. Incidentally, neither molecule is chiral. Both are meso. Even with the groups on the nitrogen.


text:
Chemical equation to represent the hydrogen peroxide scavenging experiment, used to determine antioxidant ability of tea
I am currently investigating the antioxidant ability of tea using hydrogen peroxide scavenging. Basically, I will look at absorbance of the tea samples with and without the addition of hydrogen peroxide at 230nm using a UV-Vis spectrophotometer and the difference between both is the amount of antioxidants which were able to reduce the hydrogen peroxide. However, although I've been looking theough many papers, none show me the chemical equation representing this reaction. Could someone help me with this?

Is this a standard analysis?!? // Your proposed experiment seems to have problems. (1) Tea has a mixture of chromophores, thus there is no specific reaction. (2) Why do you think that the chromophores in tea are the only compounds capable of reducing hydrogen peroxide? (3) Why would all  chromophores have the same molar absorptivity at 230nm?

I start by quoting a source (which also goes into great detail on explaining the underlying mechanism of chromophores excitation):

Photosynthesis is initiated by the absorption of light by the chromophores, which excites the molecules from the ground state to an electronic excited state.

So, simply:
$\ce{Chromophores + hv -> *[Chromophores] -> [Chromophores](e-,h+)}$
with the creation of an electron and an electron hole. Now, absence the H2O2, the action of the electron-hole on OH- on water creates an hydroxyl radical:
$\ce{OH- + h+ -> .OH }$
Also, in the further presence of H2O2, a path to the so-called one-electron reduction of hydrogen peroxide from the action of a formed electron (e-) on H2O2, resulting in an additional hydroxyl radical:
$\ce{H2O2 + e- (aq) -> .OH (aq) + OH- (aq) }$
Now, as the tea acts acts as a radical scavenger, it reacts with the hydroxyl radical usually via an H atom extraction reaction forming water. For example:
$\ce{RH + .OH -> .R + H2O}$
The hydroxyl radical continues to attack the chromophores until there is just simple colorless products like CO2 and H2O. In time, the tea solution becomes colorless.
Now, in the combined chromophores and H2O2 experiment, the UV absorbance by tea solution should measurably display more decline with time due to the increased hydroxyl radical concentration.


text:
Turning of a paddle wheel when exposed to light
So, we observe that in the discharge tube experiment when a very light paddle wheel is placed in the path of cathode rays, it starts turning due to the particle nature of the rays.
My question is if the same paddle wheel is placed in light, why does it not start turning ?
Light is also of particulate nature (in some cases- made up of photons).
Is it due to the less mass of photons?
Thanks.
Edit:-
Thank you, particularly @Poutnik for giving me such great answers.
I just read an article while researching that the paddle wheel didn't turn because of the transfer of momentum by the cathode rays but due to changes in pressure of the gas around the paddles due to heating, which is much same as that to Crookes radiometer.
This fact was discovered later, much later after the experiment.
This solves my problem that even in the cathode ray experiment, it wasn't the trasfer of momentum that moved the paddles, because the actual pressure of the rays is quite weak (again as suggested by @Poutnik). And since the ray can't turn the paddles, the light no way can.
And yes, I calculated the momentum of sunlight and that transferred by cathode rays to see for myself that movement of the paddle by each of these is not possible.
Thanks everyone once again.

Looking for Crookes radiometer?

Yes. Pretty much that only. Thanks, I didn't know of it.

Read the fine print.  Crookes radiometer doesn't work due to momentum transfer but because of heat absorption.

Yes, I read that now. Is there any device which works like the paddle wheel in light, due to momentum transfer. Is it possible ? If no, then why ?

Yes.  See the section "Projects operating or completed" under the Wikipedia article on solar sails: https://en.wikipedia.org/wiki/Solar_sail#Planetary_Society_LightSail_Projects

Be aware for electrons from cathode rays it is different. The may exert sufficient force by their momentum, see answer update.

Yes. But the fact is that they don't . The paddle wheel never turned by the force of the electrons.

The discharge tube experiment is always performed with some suitable gas(preferably hydrogen)in the tube. And, if there is no gas, then the paddle wheel doesn't turn.

But the electrons and ions come from gas ionization. In this case, gas plays very different role. BTW, the question suits more to the Physics site.

Ok whatsoever, I have got my solution now.

Not the complete answer, but a short analysis of expectable forces:
The energy and momentum of light are related by equation $E=p \cdot c $.
Then force caused by light $F=\frac{\mathrm{d}p}{\mathrm{d}t}=\frac{\mathrm{d}E}{c \cdot \mathrm{d}t} = \frac Pc$
Therefore the light power 30 W (some cca 50-60W fluorescent bulb or cca 35-40 W LED bulb ) would cause direct  force cca  100 nN, equivalent of cca 0.01 mg.
Or twice as much if reflected.
Crookes radiometer shows the wheel is turning in the opposite direction than it should be, if caused by the light pressure.
The motion is caused by uneven heating and expansion of gas around the wheel, causing pressure differences in the opposite sides of paddles.
When pressure is too low, the gas there is not enough and there remains just light pressure, that is too weak to turn the wheel.

Quoting Effect observations from article.
The effect begins to be observed at partial vacuum pressures of several hundred pascals (or a few torr), reaches a peak at around $\pu{1 Pa}$ ($\pu{7.5e−3 torr}$) and has disappeared by the time the vacuum reaches $\pu{1e−4 Pa}$ ( $\pu{7.5e−7 torr}$) (see explanations note 1). At these very high vacuums the effect of photon radiation pressure on the vanes can be observed in very sensitive apparatus (see Nichols radiometer) but this is insufficient to cause rotation.

Nichols radiometer can register the true light pressure by mirrors on torsion balance.

For cathode rays = electrons
If there is
potential difference $U$
current $I$
elementary charge $e$
electron mass $m_e$
charge $Q$
Area $A$
pressure $P$ (avoiding confusion)
momentum of 1 Coulomb  $p$
electron momentum $p_e$
then for kinematics:
$$eU=\frac 12 m_e.v^2$$
$$v=\sqrt{\frac{2eU}{m_e}}$$
for momentum:
$$p_e=m_e.v=\sqrt{2m_eeU}$$
$$p=\frac 1e \cdot p_e=Q \sqrt{\frac{2m_eU}{e}}$$
for force and pressure
$$F=dp/dt= I \sqrt{\frac{2m_eU}{e}}$$
$$P=F/A= \frac IA \sqrt{\frac{2m_eU}{e}}$$
for $\pu{380 V}$ and $\pu{1 A}$ it gives force about $\pu{66e-6 N}$, what is equivalent to weight of the mass about $\pu{6.6 mg}$.
This is a model case. For a particular scenario, the geometry and electrostatic forces would play significant role.

But crookes radiometer doesn't work due to the momentum transferred by the protons. Right ?

Right, it does not, even if it is often incorrectly thought it does. And you mean photons not protons.

Yes.i get the point. So, @Poutnik, I wanted to ask you that if a light paddle wheel contained in a vaccum is kept in light, then it doesn't rotate. My question is that why does it not rotate ?

light pressure is too weak and there is no gas to push it.

'light pressure is too weak', can you please explain that in your answer ? Thanks.

it is already there.

So, you give the example of the bulb. If we take sunlight, then the pressure must be huge, then why doesn't the paddle turn in sunlight ?

No, it is not huge. Full sunlight is about 10W/dm3. If a paddle has area e.g.  2 cm2, it gets just 200 mW.

Ok, so you mean that even the sunlight is not enough to turn the paddle, even if there's a gas inside the chamber ?

NO. We were talking now about light pressure turning the wheel. That will not turn it. It will be turning, just not because of light pressure

Just if you can explain me one thing:-

The cathode rays successfully turn the paddle wheel due to high pressure. Can you give me some mathematics to take out the pressure exerted by cathode rays? Thanks.

Let us continue this discussion in chat.

Thanks for the formulae. I will shortly make an edit in my answer which solved my entire problem. Thanks for inputs from you. Your answer is what gave me a good info on the topic.

Please see my edit for the solution.


text:
Non - aromatic nature of quinones
I have read that quinone systems are conjucated but not anti-aromatic. I had read the related questions:

Is the aromaticity broken in some resonance structures of para-nitro-aminobenzene?
Is a quinone ring aromatic?.

In one of those answers, I read a point which said that it was non aromatic:

[...] and you must be able to draw double bonds - internal to the ring
  - from each ring carbon atom. [...]

I don't get what this means as I had never heard of such a thing in Hückel's rule. I did read elsewhere that the system must be completely conjugated which is true in this case but due to cross conjugation the electrons are not delocalised throughout the system? I don't get why cross conjugation affects the delocalisation of the electrons this much? What is the proper explanation for this?

The reason you must be able to draw double bonds within the ring is the $4n+2$ rule applies rigorously only when the conjugated atoms are all in such a ring. If you are drawing double bonds to adjacent atoms off the ring then you don't have all the conjugated pi electrons in the ring anymore, and then there is no longer a $4n+2$ rule for aromaticity or a $4n$ rule for antiaromaticity.
In some systems it is possible to render contributing structures that have all the conjugated electrons in the ring and meet the $4n+2$ rule, in which case you can assert that this contribution is enhanced by aromaticity and the properties of the compound affected accordingly.  For example, the aromatic contribution to cyclopropenones leads to enhanced polarity of the carbonyl group and increased basic character of the oxygen atom, compared with ordinary ketones.

Illustration source: https://commons.m.wikimedia.org/wiki/File:C3H2OResonance.png.  Drawn by WP user Smokefoot.  Link to License
Quinones, however, cannot be given such a structure without leaving the highly electronegative oxygen short of an octet, so they lack any significant aromatic contribution.


text:
Why are polynuclear carbonyl complexes coloured?
We know that polynuclear carbonyls are colored and their intensity increases with increase in metal ions. But why does the color arise?
I am aware that this is due to some transition between electronic energy levels; but, what sort of transition takes place?
Is it due to transition between orbitals that are largely localized on the ligands i.e. CO, or is due to transition between orbitals localized on metal frame work?
I have tried gathering some information about the same and I have found something called metal to ligand charge transfer (MLCT) that is responsible for colors in complexes, but I am not sure whether this is applicable in this case or not.

Related: Colorless nature of (mononuclear) carbonyl complexes


text:
What does the intramolecular aldol condensation of 6-oxoheptanal form?
I know that when 6-oxoheptanal is treated with NaOH, an intramolecular aldol condensation will take place. But will a five-membered ring or seven-membered ring be formed?

I know that the five-membered ring is more stable in comparison to the seven- membered ring, but as per the mechanism of aldol condensation the first step is removal of acidic hydrogen (which is also the slowest step i.e. the rate determining step {if I am not wrong}).
So if I remove the acidic hydrogen on C-7 (the terminal methyl group), then the (primary) carbanion formed will be more stable than if I were to deprotonate on C-5.
So I think that the major product should be the seven-membered ring.
Am I right? Or is my reasoning incorrect?

You need to remember that every step in this process is reversible.

Related: https://chemistry.stackexchange.com/questions/48622/why-are-7-membered-rings-less-likely-to-form-than-5-and-6-membered-rings

B.Anshuman has given a short answer, but I felt more explanation is needed to understand this situation. This reaction can be done in kineticlly controlled conditions (e.g., using $\ce{LDA/THF}$ as a base at $\pu{-78 ^\circ C}$) or in thermodynamically controlled conditions as in this case. Each case may give totally different major products. For example, alkylation of 2-methylcyclohexanone gives 6-alkyl-2-methylcyclohexanone as the predominant product under kinetic conditions (1. $\ce{LDA/THF}/\pu{-78 ^\circ C}$; 2. $\ce{R-Cl/THF}/\pu{-78 ^\circ C}$, alkyl chloride is $1^\circ$ or $2^\circ$). However the same starting material would gives 2-alkyl-2-methylcyclohexanone as the amjor product under thermodynamic condition such as $\ce{NaOEt/EtOH}/\gt \pu{25 ^\circ C}$.
In above example shows that the thermodynamic conditions allow equilibration of the two possible enolates (which are derived from $2^\circ$ and $3^\circ$ carbanions); the enolate with greater alkyl substitution on $\ce{C=C}$ would form in greater concentration (Zaitsev-like). The question in hand can also be treated like the same because the situation is similar ($1^\circ$ vs $2^\circ$ carbanions).
The literature has shown that the evidence for carbanion formation might not be the rate determining step (Ref.1). The abstract if the reference is given below for your benefit for reasoning:

Rate and equilibrium constants have been determined for both the aldol addition and the elimination steps in the intramolecular condensation reactions of 2,5-hexanedione, 2,6-heptanedione, 1-phenyl-1,5-hexanedione, and 5-oxohexanal. The overall thermodynamics are similar for cyclization of 2,5-hexanedione and 2,6-heptanedione; conversion of 2,5-hexanedione to the corresponding enone is actually more favorable, but the cyclization of 2,5-hexanedione is 2400 times slower than that of 2,6-heptanedione. As expected on the basis of intermolecular analogs, the addition step is less favorable and slower for 1-phenyl-1,5-hexanedione, and the addition step for 5-oxohexanal is more favorable though similar in rate to that for heptanedione. Detailed analysis of the kinetics and equilibrium for all of these compounds, as well as 2-(2-oxopropyl)benzaldehyde, in terms of Marcus theory, leads to the same intrinsic barriers for the intramolecular reactions as were seen previously for the intermolecular reactions. This means that rate constants for intramolecular aldol reactions should be predictable from the energetics of the reactions and that the effective molarity can be calculated. Methods for estimating thermodynamic quantities for reactants and products of these reactions have been examined.

Thus, my conclusion is formation of 5-membered ring is predominated here to give 1-acetylcyclopentene as the major product among three theoretically possible products.
Reference:

J. Peter Guthrie, Junan Guo, "Intramolecular Aldol Condensations:  Rate and Equilibrium Constants," J. Am. Chem. Soc. 1996, 118(46), 11472-11487 (https://doi.org/10.1021/ja954247l).

Intramolecular-Aldol reactions involves heat, as you have mentioned, and the base used (NaOH) isn't much affected by steric factors. Therefore, the preferred product should be thermodynamically controlled product. Moreover, as you commented about the formed carbanion being less stable in case of 2° carbon, actually the carbanion is converted to an enolate then the reaction proceeds further. So, the 5-membered cyclic ring should be the major product.

The carbanion isn't converted to an enolate; it is an enolate.

The base extracts the hydrogen from the carbon, carbon accepting the electron. So though temporarily, the carbon must have an excess negative charge, i.e a carbanion, which becomes an enolate. Please explain, if otherwise.

Nope; you deprotonate a carbonyl, you get an enolate. The carbanion that you're thinking of is a resonance form. It doesn't exist as an independent entity. Not even transiently.

Enolate can also be considered as a one of the resonating structure, so actually, none of them should be an individual identity. Sorry for taking your time.

Well, okay; let's call it a "resonance hybrid which looks almost like an enolate", or RHWLALAE for short. When you deprotonate a carbonyl compound, you get a RHWLALAE; there's no carbanion intermediate or transition state. So we can rephrase my original comment: the carbanion isn't converted to an enolate; it is a RHWLALAE. If you accept that the RHWLALAE is close enough to an enolate, then we can call it an enolate, but if not we can agree to call it a RHWLALAE. But that doesn't affect my point about using the word "converted" - it's not right to say that.

[...] To be a bit more helpful: my concrete suggestion would just be to say that the rules for carbanion stability don't apply to enolates, even though enolates have a (minor) resonance form in which the negative charge is on carbon. I mean, your answer is otherwise correct in its analysis; this is a minor phrasing issue which I would prefer to be fixed, or at least pointed out.

Yeah, I understand. My teacher had told us to check the stability of carbanion for more stable product, from then I had wondered whether to check the stability of enolate or carbanion? As in this question, 2° carbon leads to more stable enolate, while less stable carbanion. Usually, the nucleophilic addition reaction of formed enolate decides the major product, but what if it is same for 2 different products?

Mmm, more substituted enolates are generally more stable. In this regard they behave more like alkenes rather than carbanions, which is consistent with the relative importance of the O–C=C vs O=C–C resonance forms. See e.g. Clayden 2ed p 599 (most (under)graduate organic texts will discuss it). But this is only important if your enolate is being generated under thermodynamic conditions. If you use LDA (for example) the enolate formation is kinetically controlled, and you need to look at deprotonation transition state stability, not enolate stability. It can get very complicated very quickly.

The rate determining step is nucleophilic addition reaction for aldol condensations, but you said that we check the stability of deprotonated transition state. Why?

I said "if you use LDA". There are many kinds of aldol condensations. Some, like the one in the question with NaOH and heat, are determined by product stability. Others, using other conditions, are determined by other factors, and can lead to different products. It's not a situation where you can have one rule that covers every aldol reaction. Anyway, this isn't the best place to have a discussion about things that are unrelated (or tangentially related) to the question itself.


text:
Energy decomposition analysis. Are there schemes that actually splits up the steric term?
In quantum chemistry, there are many types of energy decomposition analysis (EDA) which partitions the interaction energy between two or more fragments into steric, polarisation, exchange and orbital energy terms
For example, in a Morukuma type decomposition method employed by ADF, the interaction energy is partitioned into electrostatic, pauli and orbital energy, which the last term can be further decomposed by Natural Orbitals for Chemical Valence (ETS-NOCV) to show the $\sigma,\pi,\delta$ contributions
In ALMO-EDA in Qchem, the energy is instead decomposed into electrostatic, pauli, dispersion and polarisation by using fragment electrical response functions (FERF) as an ortogonalisation scheme
Both methods were generalised to be applicable to multiple fragments in some more recent parts of the literature since 2014
Despite extensive work and examples on decomposing the electronic term, be it polarisation, orbital interactions or similar, it seems there isn't an example where the steric term is decomposed into contributions for each fragment. For example, considering a trimer of molecules, all such EDA schemes always report the steric or pauli repulsion term as the sum of contribution of all fragments, thus there is no information on how much sterics each fragment contributed in the complex.
Attempt to recover individual steric contributions of each fragment using the intuition that pauli repulsion in most EDA schemes are computed based on how much molecular volume is excluded in the overlapping region of the fragments and assuming that those depletion regions are additive, found that compared to the full 3 fragment EDA, the sum of steric terms for the pairwise fragments is off from the total steric by around 100-400 kJmol-1, suggesting there are higher order interactions not accounted for such as two fragments depends on the third one at the same time.
Is there a reason why it is not meaningful to decompose the steric or pauli term into fragment contributions?


text:
Can bond rotation take place in Pinacol-Pinacolon rearrangement
I was given a question of pinacol pinacolon rearrangement of butan-2,3-diol where the methyl groups were anti to the respective hydroxy groups. Can the C2-C3 bond be rotated to allow hydrogen to be anti to the leaving hydroxy group, but I have heard that mechanism of this rearrangement is concerted. Therefore I can't decide the major product.

Yes,it can be, as in this configuration, the antibonding sigma orbital of the leaving group will be parallel to the lone pairs of the other hydroxyl group

Sorry, I didn't quite understand how this will lead to bond rotation. Can you explain a little more?


text:
No resonance in Anti Aromatic compounds
My teacher commented on Anti aromatic compounds as being less stable compounds as it doesn't undergo resonance inspite of having π-bonds/electron pairs in conjugation. I didn't understand why so. Can you help?

As according to Molecular Orbital theory , when anti- aromatic compound is formed, the electron goes to anti bonding, and is unpaired even, which reduces it's stability.,
Or we can say that in anti- aromatic,the plane of electron cloud twists there bonds between carbons, making it unstable, thus resonance is not possible.
Antiaromaticity is a characteristic of a cyclic molecule with a π electron system that has higher energy due to the presence of 4n electrons in it. Unlike aromatic compounds, which follow Hückel's rule ([4n+2] π electrons) and are highly stable, antiaromatic compounds are highly unstable and highly reactive. To avoid the instability of antiaromaticity, molecules may change shape, becoming non-planar and therefore breaking some of the π interactions.
Greater the delocalisation energy(resonance energy) of a compound, more stable it is. The resonance energy can be calculated using Huckel Molecular Orbital Theory(HMOT). Antiaromatic compounds have zero resonance energy hence are unstable.
According to HMOT,
Delocalisation energy=( Pi electron binding energy) - (total energy of isolated double bonds in a classical structure)


text:
Why is the OO-H dissociation is unlikely for the Caro's Acid?
Let us consider the Caro's Acid: $\mathrm{H_2SO_5 \equiv SO_3H-OOH}$.
It is widely known that the hydrogen tied with the $\mathrm{-OO-}$ group is pretty much unlikely to dissociate:
$$ \text {Terminal dissociation: }\mathrm{SO_3H - OOH \rightleftharpoons H^+ + [SO_3-OOH]^-} \\
   \text{An unlikely step: }\mathrm{[SO_3-OOH]^- \rightleftharpoons H^+ +[SO_3-OO]^{2-}} $$
The first time I encountered that fact I was presented a pretty inconclusive piece of argumentation: it was stated that $ \mathrm{-OO-} $ group is way more electronegative than $ \mathrm{-O-} $, which somehow brings about an obstacle for the dissociation.
Unfortunately, it was a long time ago and I cannot remember the exact details. I stumbled upon this problem not so long ago. Having a better background in chemistry, I attempted to apply the Sanders' electronegativity balance theory, but the result was pretty inane in terms of our chemical reality: $\chi_{eq}(\mathrm{H_2SO_4}) \approx 2.89894; \chi_{eq}(\mathrm{H_2SO_5}) \approx 2.96803 $. In other words: $ \chi_{eq}(\mathrm{H_2SO_5}) > \chi_{eq}(\mathrm{H_2SO_4}) $.
What am I doing wrong? Is it possible to tweak my attempt to use Sanders' theory? Is there any other approach to this problem?

It's just not delocalised there, that's it about it.

@Mithoron well, that’s a good point, but the localization analysis is, in fact, just another way to look at electronegativity computational models. I mean, the resulting electronegativity does depend on the electron-nucleus relations (i.e. ionization and electron affinity energy). So, to be honest, this does not really answer my question. I may have worded it a bit clumsily: I’m interested in the computational aspect of the problem (and the application of the Sanders’ theory, to be more precise).

So the question: why is it a peroxide instead of e.g. an oxide?

Hmmm... What is the "Sanders' electronegativity balance theory" you are referring to? I believe most of us in the community are not well-versed with this theory. Perhaps, you could provide a reference for us to look at so that we can learn about this theory?

Based on this reference (https://dash.harvard.edu/bitstream/handle/1/11913975/jacob-chemistry-of-hydrogen.pdf?sequence=1), the first $\ce {pK_a}$ of the acid is 0.4, which is higher than that of sulfuric acid's corresponding first $\ce {pK_a}$ value of around -3. It is not surprising due to the fact that there is a greater extent of delocalisation of the negative charge in the sulfate conjugate base compared to in the peroxymonosulfate conjugate base. Note that the 2 $\ce {O}$'s in the peroxy group are not able to participate in resonance.

@TanYongBoon, sorry for withholding the reference for quite some time. You may be interested in the 3rd segment of this paper: https://deepblue.lib.umich.edu/bitstream/handle/2027.42/37977/560250117_ftp.pdf; also, I believe Joel Seblin described it in one of his books - I'll add it in case I manage to get hold of it.

One possibility is the formation of an intra-"molecular" hydrogen bond between the peroxy-bonded hydrogen and a third oxygen atom.  The structure shown in Wikipedia for the acid suggests this interaction.  The involved hydrogen is curled up into the five-membered ring formed by such a hydrogen bond, while the other hydrogen atom is freely exposed and ready to launch.  Potassium peroxymonosulfate corroborates this; it shows only the non-peroxo hydrogen dissociated.
Hydrogen peroxide does not form this sort of intramolecular hydrogen bond, and as we expect from induction considerations the proton dissociates from this molecule more readily than from water.