text:
Gas chromatography flame ionization detector (FID) - why hydrogen gas?
Why is $\ce{H2}$ used in an FID, apart from the fact its combustion does not contaminate the flame?
In other words, is the temperature of the flame important or critical? Is the stability of the temperature?

What alternative would you suggest ? E.g.a methane-air flame in FID would be like e.g using methylchloride as the makeup gas in ECD.

@Poutnik One alternative would be electric arc ionization. Another extreme UV. What is special about H2?

It would be even much worse than methane. The key is to keep as low baseline ionisation as possible. That does not happen with methane flame and definitely not with arc or UV ionisation. With the arc, there would be additionally very high noise, several orders higher than useful signal.

The key to the answer is understanding how FID works.
The hydrogen flame has a minimal flame ionisation, what is needed for the low signal baseline.
Incoming organic molecules from the HPGC column create in the flame a lot of ions and increase the flame electric conductivity.
Using alternatives causing higher ionisation would decrease FID sensitivity that is not great even at ideal conditions, compared to e.g. ECD.

Another fundamental aspect of using hydrogen in gas chromatography is the so-called van Deemter curve. The curve shows the linear velocity of the gas on the x-axis and plate heights on the y-axis. It can be shown that by using hydrogen one can obtain the best efficiency as compared to any other gas.
And of course, hydrogen is required for burning the organic molecules exiting the column. It is a clean flame (no carbon background).


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Why does the -OH group in β-naphthol direct the incoming diazonium salt towards the ortho position?

In the following reactions, the major product W is

I was doing this question which asks me the product of a diazotisation coupling reaction between diazonium chloride and β-naphthol. I know that the diazonium salt will act as an electrophile and compounds like phenol will direct it towards the para position, but I am not able to understand why the diazonium salt attaches itself to the ortho position of the -OH group in this compound.
The salt can also attach itself to the other benzene ring opposite to the -OH group. Also there are two ortho positions with respect to the -OH group. Can somebody please explain this to me?
This is the answer to the question:

It's sort of a process of elimination.  The electrophilic ion could potentially attack anywhere, but:

The hydroxyl group is activating, so the electrophile will prefer that ring.  You are left with three open positions.
One of those three is meta to the hydroxyl group and so less favored, as activating substituents are generally ortho/para directing.  Hydroxyl follows that rule.
Finally -- electrophilic substitution in a naphthalene system is kinetically favored next to the other ring.  See here for example.

We have one position that's on the same ring as hydroxyl, ortho or para to that hydroxyl group, and next to the other ring.  That is the position shown in the book.

Coupling reaction of β-naphthol with benzene diazonium is an example of electrophilic aromatic substitution.
If the electrophile attacks at alpha position ,then two resonance structures 1 and 2 .Both 1 and 2 are aromatic.

If the electrophile attacks at gamma position ,only one resonance structures 3 , with aromatic ring is possible and 4 is not aromatic.

Therefore attack at alpha position is the major product.
UPDATE :
The resonance structures will have a structure in which all atoms possess complete valence electrons.

Reference: Pielhop, T.; Larrazábal, G. O.; Studer, M. H.; Brethauer, S.; Seidel, C.; Rudolf von Rohr, P. Lignin repolymerisation in spruce autohydrolysis pretreatment increases cellulase deactivation. Green Chem. 2015, 17 (6), 3521–3532 DOI: 10.1039/c4gc02381a.

Technically, each possibility has one more contributing structure containing $\ce{=\overset{+}{O}H}$.  These structures require placing a positive charge on an electronegative atom but enable all atoms to have full valence shells.

I know it's some old jee stuff.Usually they ask for the most stable product. Think of the hydrogen bonding at ortho.This should stabilise the transition state.Draw the meisenheimer complex.It's low temperature therefore extra bonds provide extra stable transition state and the activation energy of reaction is lowered.It's kinetically favoured product as expected. Thermodynamics would check for steric factors and give para product. For kinetics/low temp look for any kind of stabilization which can lower energy of transition state .

You have two positions ortho to the hydroxyl group bit only one is listed as the major product.  Why would you pick that one?  See https://en.wikipedia.org/wiki/Naphthalene#Reactions_with_electrophiles and incorporate what that says into this answer.

Ok what about this, draw resonance structures on both sides(just to say: only one side that is drawn above it is permissible).

I would have posted an immage but it is not possible.

source: Stack Exchange

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

sort: votes


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Can somebody explain the low temperature Schottky anomaly for two level systems?
Statistical thermodynamics question. Can somebody explain the low temperature Schottky anomaly for two level systems? Why is this anomaly not seen in all materials?

The Schottky anomaly is a general feature of two levels systems. The heat capacity is defined as $\displaystyle C= \left( \frac{\partial U}{\partial T} \right)_\Delta$ where $U$ is the internal energy and the energy gap between the two levels is $k_B\Delta$, $k_B$ being the Boltzmann constant.
The gradient of the internal energy is thus important. At constant energy gap and at very low temperatures  only the lowest level is populated and the internal energy is small, and heat capacity small as the internal energy increase is small, i.e gradient $U$ vs $T$ is small. However, as the temperature is increased the upper level begins to be populated and so the internal energy rapidly increases, then the gradient $U$ vs $T$ is large and so is heat capacity.
When both levels are almost populated equally, which occurs at very high temperature, no more energy can be absorbed and the gradient $dU/dT$ effectively becomes constant and so the heat capacity becomes small and tends to zero.
Similar behaviour is observed when there are three or four etc i.e. a small  number of levels. In the 'normal' case of a molecule there is effectively no upper limit on the number of levels that can be populated as temperature increases, and the slope of $U$ vs $T$ has a constant finite value which leads to $3R$ as the heat capacity per mole at high temperatures.

The figure shows the internal energy (in units of $k_B\Delta $)  and heat capacity (in units of $k_B$) vs $T/\Delta$. To make the plots more general $T/\Delta$ is plotted as this ratio occurs in the final equation.
The equations are:
partition function $ Z=1+e^{-E/k_BT} = 1+e^{-\Delta/T} $ when energy gap $E=k_B\Delta $.
The Internal energy is $\displaystyle U=k_BT^2\left( \frac{\partial \ln(Z)}{\partial T} \right)_\Delta$.
The heat capacity is $C\displaystyle=\left( \frac{\partial U}{\partial T} \right)_\Delta= k_B\frac{ \Delta^2}{T^2}\frac{e^{\Delta/T}  }{(1+e^{\Delta/T})^2}$

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


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Why is a weak base more able to deprotonate a strong acid than a weak acid?
Very basic question, but I'm rather confused.
Say I want to separate a carboxylic acid $(\ce{RCOOH},$ $\mathrm{p}K_\mathrm{a}~5)$ from a phenol $(\ce{ArOH},$ $\mathrm{p}K_\mathrm{a}~10)$ via acid-base extraction. Apparently I am able to use sodium bicarbonate $(\ce{NaHCO3}),$ a weak base, as a way to deprotonate the stronger acid, carboxylic acid. This allows the carboxylic acid to dissolve into the aqueous phase in an ether extraction, as shown in the far left of the attached diagram, while the phenol would remain in the organic phase (A at the bottom corresponds to the carboxylic acid, B is the phenol, C is the amine).
The question is:
Why does a weak base 'prefer' to deprotonate a stronger acid? Is it because, being a stronger acid, the carboxylic acid will have more anions in solution compared to the weaker acid, so the (relatively few) cations of the weak base will be more likely to find carboxylates than phenoxides?
If this is the case, could we still use a strong base, but just in a very very low concentration and have the same effect as using a weak base at a higher concentration?

The picture is a bit confusing: Phenol should $\ce{PhOH}$ and nothing else. Sometimes, $\ce{Ar}$ is used to denote an aromatic compound, but that does not seem to be the case here.

@TAR86, agreed, I've updated the diagram with a full screenshot including the question and solution, in case it helps any. In any case I believe they just want to get the point across that a weak base can be used to extract a strong acid, and conceptually I'm trying to understand why.

The key is that water is the great equalizer and mediator. Let's deal with weak and strong base first: You are right that a sufficient quantity of a strong base would have the same effect. In this particular case, the hydrogencarbonate will separate into $\ce{CO2}$ and water on protonation and we can assume that most of the carbon dioxide will bubble out. So it will functionally not make a difference whether you started with $\ce{NaOH}$ (a strong base) or $\ce{NaHCO3}$. This is a part of the equalizer part: in an aqueous solution, you cannot have a stronger base than $\ce{OH-}$, because any stronger base will just deprotonate a readily available water molecule.
To some extent, even weak bases will do so in an equilibrium. To understand how a proton gets from an acid to a base, we have to understand the structure of water: A water-based solution like the one from the question is a huge network of water molecules engaged in hydrogen bonding with each other and mainly electrostatic bonding to ions. A when a proton "moves" through this network, it will basically jump from one water to the next, and that water will give up a proton to the next etc. (This is the reason for the huge mobility of protons in water as measured by conductivity experiments.) Thus, an acid and a base such as $\ce{RCOOH}$ and $\ce{HCO3-}$ do not have to come close to each other to effectively transfer a proton between them.
Thus the question "How does the base know to deprotonate the strong acid instead of the weak acid?" loses its basis. In an equilibrium, the stronger acids and bases will deprotonate and protonate more, respectively, than their weaker counterparts because that's what acid/base strength means.

I see... so it's similar to electrons jumping from atom to atom through a solid wire, but in this case it's protons moving in all directions in a fluid, rather than a solid. I've read strong acids and bases are better conductors of electricity. As the protons jump from the 'better conductor,' because it deprotonates more easily, the stronger acid becomes negatively charged and separates into the aqueous solution. Correct...?

@chompion That's basically correct.


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Does the current of an electrolytic cell increase if the surface area of the Cathode is increased
I am currently in high school and will be conducting an experiment looking at the effect of surface area on electroplating. I will be changing the surface area of the Cathode during my experiment but am unsure whether the amp and voltage would be controlled or dependant variables.

For a given experimental electroplating setup, there is likely to be an optimal current density at the cathode. This is the current that flows divided by the exposed (to solution) area of the cathode. So current and cathode area are two obvious control variables, and, of course, there are others. Best of success with your project!

Just remember two major rules in electrolytic electrochemistry:

If you fix voltage, current cannot be controlled. This is called constant potential electrolysis
If you fix the current, voltage cannot be controlled. This is called constant current electrolysis

Now your electrolysis cell has resistance:
(i) It will depend on electrolytes
(ii) Surface area of the electrodes
(iii) Distance between the electrodes
So let us know your results and findings. Plot a graph systematically by choosing condition 1 or 2.

The voltage won't vary as it depends on the materials used as active electrodes and their respective standard reduction potentials.
The current will vary with plate area due to varying resistance of the cathode plate.


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converting SMILES (or .mol2) to SMARTS
The conversion between different chemical structure formats, e.g. from .mol2 to .xyz, by openbabel may tied to conditions to be met.  One example of such a filter is to allow such a conversion only for data sharing a certain scaffold, e.g.
obabel -imol2 AHECAS.mol2 -oxyz AHECAS.xyz -s "c1ccccc1-[#16](=[#8])-[#6]-c2ccccc2"

So the program allows SMARTS as constraining input parameter.  In difference to SMILES, (apparently, based on babel -L formats) openbabel however does not offer a conversion of any format into SMARTS and my source to generate them is the on-line PubChem Sketcher.
Question:  Aiming to generate SMARTS locally (offline), I know Marvin by ChemAxon may export a structure into SMARTS.  However, is there a portable alternative (with lesser footprint than a complete new GUI sketcher), preferentially allowing a conversion in a pattern like SMILES2SMARTS (or mol2toSMARTS) from the terminal / by a script?

Similar thread on Openbabel forum: Conversion of SMILES to SMARTS String

Please note that babel is the old interface. The recommend obabelinstead now. See https://openbabel.org/docs/dev/Command-line_tools/babel.html#babel-vs-obabel

The thread indicated in the comment by @andselisk dates back to July 2014. It may be condensed to

"A substructure querying with openbabel requires a good SMARTS string. In case you don't have one, often a good SMILES string may be used instead of a SMARTS string, too."

A check for a rather simple molecular structure of (benzylsulfinyl)benzene

there is more than one way to describe it in either form of a SMARTS, or a SMILES string.  Consequently, the strings per se sometimes differ from each other.  Consequently, recognizing the relevant entries in the collection of .mol2 with openbabel sometimes failed.
Below I summarize (status: 2019-May-15) a few experiences: the string offered by the software, its type, and source; eventually, if it worked (i.e., entries retrieved), or not (i.e., entries not retrieved) in a small test set of 419 model data and 7 relevant hits.
|-------------------------------------+--------+--------------------+--------|
| entry                               | type   | source             | status |
|-------------------------------------+--------+--------------------+--------|
| c1ccccc1-[#6]-[#16](-c2ccccc2)=[#8] | SMARTS | pubChemSketcher    | works  |
| C1=CC=CC=C1C[S](C2=CC=CC=C2)=O      | SMILES | pubChemSketcher    | fails  |
|-------------------------------------+--------+--------------------+--------|
| O=S(Cc1ccccc1)c2ccccc2              | SMILES | ACD ChemSketch     | works  |
|-------------------------------------+--------+--------------------+--------|
| C1=CC=C(C=C1)CS(C(C=CC1)=CC=1)=O    | SMILES | ChemDoodle         | fails  |
|-------------------------------------+--------+--------------------+--------|
| O=S(C1=CC=CC=C1)CC2=CC=CC=C2        | SMILES | ChemDraw Pro12     | fails  |
|-------------------------------------+--------+--------------------+--------|
| O=S(CC1=CC=CC=C1)C1=CC=CC=C1        | SMILES | Marvin JS          | fails  |
|-------------------------------------+--------+--------------------+--------|
| O=S(Cc1ccccc1)c2ccccc2              | SMILES | Peter Ertl JSME    | works  |
|-------------------------------------+--------+--------------------+--------|
| c1cccc(c1)CS(=O)c1ccccc1            | SMILES | openbabel, default | works  |
| c1ccc(cc1)CS(=O)c1ccccc1            | SMILES | dito, universal    | works  |
| c1ccc(cc1)CS(=O)c1ccccc1            | SMILES | dito, Inchified    | works  |
| O=S(c1ccccc1)Cc1ccccc1              | SMILES | dito, canonical    | works  |
|-------------------------------------+--------+--------------------+--------|

PubChem Sketcher: version 2.4, https://pubchem.ncbi.nlm.nih.gov/edit2/index.html
ACD ChemSketch:   Freeeware version 2015
ChemDoodle:       version 8.1.0, 2019 and webcomponents,
                  https://web.chemdoodle.com/demos/sketcher/
ChemDraw:         version 12 Pro, (then) CambridgeSoft, 2009
Marvin JS:        ChemAxon, v19.11.0, on-line test at
                  https://marvinjs-demo.chemaxon.com/latest/
                  Its SMARTS (very differently written) equally failed:
                  O=S([#6]-[#6]-1=[#6]-[#6]=[#6]-[#6]=[#6]-1)[#6]-1=[#6]-[#6]=[#6]-[#6]=[#6]-1
Peter Ertl JSME:  https://peter-ertl.com/jsme/JSME_2017-02-26/

Addition:
The approach mentioned in the question basically calls obabel (or òpenbabel) as an iteration per file.  Especially if the data set is accessed more than once to identify different structures, and / or consists of multiple thousand models, this may be tediously slow.  Instead, it is worth to "bundle" the individual information
obabel *.mol2 mymols.sdf --unique

creating a container like a multi-model .mol2 (this even works with .xyz files as input).  The .sdf may be indexed
obabel mymols.sdf -ofs

by fingerprints (cf. chapter 5 of the manual).  Now, very quickly the names of the files about a structure matching the query substructure are identified, e.g.
obabel mymols.fs -ifs -s"O=S(c1ccccc1)Cc1ccccc1" results.smi -xt

and stored in results.smi.  Consequently, any further work may focus on them.

"If you deploy either a SMARTS, or a SMILES string on Linux' bash" If you enclose the string in '' to make it literal, you do not have to escape anything.


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Degradation of carbon anode in electrolysis of copper sulfate
I'm electrolysing a copper sulfate solution with a 12 volt DC supply using two carbon electrodes. The anode is reducing in mass and becoming thinner. Is there any way I can overcome this problem?

Is the electrode rather mechanically desintegrated , e.g. by the internal oxygen creation ? Or, is the material rather disappearing ? I remember home experiments from my childhood, and the carbon electrode from a battery cell created a carbon dust.

The carbon electrode has mechanically disintegrated into a dust at the bottom. I would like to know whether I can use a steel anode instead because I do not have access to a platinum electrode. My question is “will a steel rod not disintegrate or corrode?” Thanks.

I suppose stainless steel will survive. Try rather the steel plate, to keep low current density.


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Does 4-methylbenzaldehyde give haloform and aldol reactions?
The following question was asked in JEE Advanced 2017:

The answers given are option A and B. I was able to get B as the correct answer but A look like the wrong option to me.
My approach for option (A) and checking the reactions for Q:

Since (Q) of option (A) is able to give both haloform and aldol reaction, it should be incorrect. The reason I say that aldol reaction makes the option incorrect is that cannizaro reaction is given by aldehydes or ketones which have non enolisable hydrogen atoms. But aldol reaction requires enolisable hydrogen.
Can anyone confirm what should be the correct answer and why is my approach wrong if the answer given is right.

Q gives Cannizzaro NOT haloform. Check formulas of Q and S. A and B are correct.

First, Cannizzaro reaction is not given by ketones (your statement says ketones give cannizzaro reaction).
Counter to your analysis, the following reaction will not occur since $\ce{H}$ bonded to methyl group is not sufficiently acidic for hydroxide ion to attack. Refer to $\mathrm{p}K_\mathrm{a}$ values given by Mathew Mahindaratne.

Under the given conditions  P on ozonnolysis gives formaldehyde and 4-methylbenzaldehyde. Both are nonenolizable aldehydes capable of undergoing Cannizzaro reaction. Both compounds do not undergo haloform reaction.
Scheme 1

Under the given conditions R gives acetophenone and formaldehyde. Acetophenone gives haloform test and it also undergoes aldol condensation, but not cannizaro reaction.
Scheme 2

Choice B also meets the conditions given.
Thus choices A & B are correct.

Please note followings: (i) R in Scheme 2 is not denoted; (ii) R does not gives acetone upon ozonolysis (I corrected it to formaldehyde in the text); and (iii) On last diagram, iodoform products  are idoform and sodium salt of benzoic acid. I'd like you to correct those errors.

@MathewMahindaratne agreed R does not give acetone but formaldehyde..It was a TYPING  error..I have corrected haloform product as sodium salt of benzoic acid.

@MathewMahindaratne i was trying to focus on why 4-methyl benzaldehyde does not undergo haloform reaction which Groverkss wasnt convinced.

The Cannizzaro reaction is a chemical reaction that involves the base-induced disproportionation of two molecules of a non-enolizable aldehyde to give a primary alcohol and a carboxylic acid (Wikipedia).
The haloform reaction requires a methyl ketone as the substrate.
Two compounds (P and R, respectively) in choice [A] give you 4-methylbenzaldehyde, $\ce{C8H8O}$ (Q; a non-enolizable aldehyde) and acetophenone, $\ce{C8H8O}$  (S; a methyl ketone) upon ozonolysis (byproduct from each compound is formaldehyde): 4-Methylbenzaldehyde can undergo Cannizzaro reaction but not haloform reaction (your given mechanism for bromform is unacceptable). Acetophenone, an enolizable ketone, easily undergoes haloform reaction but not Cannizzaro reaction (it is not an aldehyde).
Also, two compounds (P and R, respectively) in choice [B] give you 3-methylbenzaldehyde, $\ce{C8H8O}$ (Q; a non-enolizable aldehyde) and acetophenone, $\ce{C8H8O}$  (S; a methyl ketone) upon ozonolysis as well (byproducts from P and R here are acetaldehyde and acetone, respectively): 3-Methylbenzaldehyde can undergo Cannizzaro reaction (it is also a non-enolizable aldehyde) but not haloform reaction. Acetophenone, as pointed out earlier, undergoes haloform reaction but not Cannizzaro reaction (it is not an aldehyde).
Thus, both choices are correct.
Please also note that the compounds representing P and R in choices [C] and [D] will also give you a non-enolizable aldehyde and a methyl ketone upon ozonolysis. Each of these non-enolizable aldehyde and methyl ketone set shows the ideal reactivity towards Cannizzaro and haloform reactions as described earlier. However, they simply don't have the identical molecular formula as required.

I'm unable to understand why the hydrogen on the methyl part of 4-methylbenzaldehyde are non enolisable. As I see, they act similar to alpha-H in a ketone due to conjugation with carbonyl group. Am I wrong?

Note that $\mathrm{p}K_\mathrm{a}$ of methane and toluene is about 48 and 41 (about 56 and 43 in DMSO). That of diphenylmethane is about 34. What I'm trying to say is 4-aldehyde group should be lowering the $\mathrm{p}K_\mathrm{a}$ of tolyl group, but not that much when compared to that of acetophenone ($\mathrm{p}K_\mathrm{a}$ is about 25 in DMSO). Thus, heloform reaction won't work with 4-methylbenzaldehyde.


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Unknown glassware please help, vshaped with a small bulb on one side, hollow but narrow to both sides!
Looking to try an identify some glassware we have found in a cupboard. This is just the start of it...
1 item here, v-shaped, small bulb but hollow, narrow entrance and exit...

In advance, I'd suggest to post one piece of equipment per question and make the title of each question more descriptive, e.g. "Name for an Y-shaped glass tube with a bulge" or something along these lines. Dimensions and comments on the construction details are also welcomed.

As for the question, this somewhat looks like a viscosimeter.

No idea what it is, but I don't think that it is a viscometer.

@MaxW Yeah, this Y-shape would make little sense for a visco; that's why I used "looks like" :) (Cf. "A policeman attempted to rob me" and "A person who looked like a policeman attempted to rob me".)

This was a volumetric pipette for sure, which was converted into a home-made crude viscometer. One can see the remaining brown ring on the left arm. One end of the pipette was elongated to reduce the diameter. Fill the bulb with water and see how quickly it drains. If you share the background that would be better to id i.e. whose cupboard was that?

Unfortunately we have no idea who's cupboard it was... We did wonder if it was a homemade item... and we have a fear we have a lot more upon emptying boxes yesterday. We found more of these items that we can definitely see the fill line and volume from a pipette. At least if we have an idea what it is a crude item of it may be possible to use/offer to academics. Many thanks!

Rather than a viscometer it could just be a nice homemade trap for an air-sensitive reaction that is producing gas (e.g. fermentation). The reaction would be in a vessel attached to the left side, and liquid would be in the lower part of the U shape. Produced gas would escape, but no gas would come in as long as the pressure is higher in the vessel than outside. The bulb is there so that if a burst of gas pushes the liquid up out of the trap (or if liquid escapes from the reaction vessel), it falls and collects in the bulb rather than going out the tube and contaminating whatever is out there.


text:
Reason Behind Decrease In Potential Energy
Atoms combine to gain stability. They gain stability when they lose energy. In my book, it's given that the potential energy of the system is reduced during chemical combination. But I am eager to know: how is the potential energy of the system reduced?
I want the theoretical reason.

What do you mean by "How?" – quantitatively (e.g. how change in PE is calculated) or qualitatively (e.g. how PE changes according to mechanism)? As of now, the question seems to be too broad and unclear.

'How PE changes according to mechanism' is what I meant.

The potential energy that changes during reactions is electrostatic energy (+/- attraction, +/+ or -/- repulsion) from the charged particles (electrons and protons) within the atoms. The theory is described by Coulomb's Law (involving the potential energy of charged particles):

http://guweb2.gonzaga.edu/faculty/cronk/CHEM101pub/energy.html
Generally, this comes from the attraction of electrons to nuclei and the repulsion between electrons and between nuclei. When atoms/molecules/the "system" becomes more stable, we will see either an increase in the attraction present (electrons are closer or less shielded from nuclei) or a decrease in repulsion (electrons or nuclei are further apart).

http://ch301.cm.utexas.edu/atomic/#bonding/bonding-all.php
More advanced theories will involve the changing of the allowed potential energy levels of the electrons either within atoms (atomic orbitals) or within molecules (molecular orbitals). Whichever theories are used, the net result is that there is more attraction and/or less repulsion in a more stable/lower potential energy state.
Interestingly, by the first law of thermodynamics, if the potential energy decreases, another type of energy must increase (so you don't end up with less energy than you started). It is normally thermal energy that increases and then is transferred as heat. This process ends up being one the most important ones in understanding why some processes happen and some do not.

how is the potential energy of the system reduced?

To be honest, there is no specific mechanism to decrease in potential energy.
Potential energy is merely a mathematical function that scientists came up with to describe energy changes in physical systems after rigorous measurements and calculated through equations backed by experimental results. These equations, assume certain facts making up the postulates of the theories we apply to predict behaviour of atoms. Based on that we predict the behaviour of molecule, interaction with other molecules.
The postulate involved is that potential arises from the interaction between particles and at infinite distances, such interactions are negligible, so potential energy is zero. As the particles or atoms, whichever one we are dealing with come closer, the interactions get stronger producing positive or negative potential energy based on the nature of those interactions which are experimentally measured by energy absorbed by the system or energy released by the system respectively.
Take the simple example of hydrogen atoms combining to form a chemical bond. The change in energy is -218 kJ/mol and that is all. We have atomic theories, most accurate one being those of quantum mechanics but, in the end, there is no answer to that why. It is just the way the universe works.
PS I am not sure whether the term  postulate is better or axiom.

The mechanism is complicated by quantum chemistry aspects, but the principle can be tracked down to the low level classical electrostatic energy analysis, or even to the classical gravitational orbit energy analysis.
An object $O$ is on a circular orbit around a massive central object $C$. This $C$ puts the attractive force $F=A/r^2$ on the object $O$ (this is common for gravity and electrostatic force) .
The $O$ has:

potential energy $E_\mathrm{p}=-\frac Ar$
kinetic energy $E_\mathrm{k}=\frac A{2r}$
total energy $E=E_\mathrm{p} +E_\mathrm{p}=-\frac {A}{2r}$

If the object is orbiting along an elliptical orbit, the values of kinetic and potential energies are just the mean values, with the total energy as their sum being constant.
The potential energy is equal to the mechanical work to be done to free the object from the force field of the central force.
The quantum chemistry complicates it a lot, but with  great simplification the following is valid:
Exothermic reactions releasing energy get the energy from energy of valence electrons. These electrons form during the reaction "lower orbits" ( molecular orbitals) with less energy. This less energy means more energy would be needed to release them as free electrons.
The above mentioned molecular orbital has 3 main meanings:

It is a wave function as a particular solution of electron wave equation
Determines probability of electron occurrence in particular location
Determines quantum state of the electroncluding, but not limited to, its energy.

The quantum aspects put specific limitations, mainly these:

Only particular values of electron energy and of some other quantities are allowed
Electrons do not have classical orbits, nor particular position nor velocity.


text:
Relevance of strain energy, total energy and stability of molecule?
Would anybody please highlight the difference between total strain energy of a molecule, total energy of a molecule and what we can infer about the stability of the molecule using the total strain energy of the molecule as well as the scope of using such inferences made.

Would anybody please highlight the difference between total strain energy of a molecule, total energy of a molecule

Strain energy is usually defined as the sum of all angle and steric strains that a certain molecule possesses, compared to an unstrained analogue molecule.
For example, here is a table that compares various cyclic alkanes to cyclohexane:

wikipedia
In this way, stain energy can be thought of as a positive contribution to potential energy that destabilizes a molecule above the stability that its ideal bonding would suggest.
Total energy of a molecule could be defined in lots of ways, often it is done as a $\Delta H_f$, which could be positive or negative, and compares the potential energy of a molecule (i.e. how strong the bonds are, stronger means lower potential energy) to the elements from which it is formed. However, this will also take into account any intermolecular forces present and these are not inherent to a single molecule, but to a collection of molecules in standard states.
We could also define the energy of a molecule by thinking of the sum of the bond enthalpies, which define how strong the bonds are compared to unbonded atoms (specifically by breaking bonds homolytically, forming radical species, in the gas phase). The larger the value, the stronger the bonds are and the lower the potential energy of the molecule.
In computational chemistry, the total energy of a molecule can theoretically be calculated using quantum mechanical means. This is done using energy minimization software. There is a good post here about the theory in this process. This value will take into account the bond enthalpy/strength and strain within a molecule.

and what we can infer about the stability of the molecule using the total strain energy of the molecule as well as the scope of using such inferences made.

In general, the greater the strain energy and the higher the total energy, the less stable/more reactive the molecule.

Are there any rules to determining the analogue molecule used for comparison?


text:
Why do most explosives have symmetrical chemical structures?
I was just turning the pages of my chemistry dictionary, when I found the structure of TNT (Trinitrotoluene):

...there after I got interested in its symmetry as an explosive, and started looking for structures of other explosive compounds and found that almost all of them have very symmetrical structures.... Here are some that I found:

RDX

HMX: High Molecular Weight RDX (cyclotetramethylene-tetranitramine)

PETN [Pentaerythritol tetranitrate (PETN)]

Heptanitrocubane

HHTDD (hexanitrohexaazatricyclododecanedione)

TATB, triaminotrinitrobenzene or 2,4,6-triamino-1,3,5- trinitrobenzene

There are many others which can be found here: wikipedia/chemical-explosives
I wonder why they generally tend to be so symmetrical (as compared to any other random non-explosive molecule)?

Most of the small molecules are symmetrical.

What those compounds have in common is a lot of $\ce{NO2}$ groups. They are trying to repeat the same structural element as often (and densely) as possible. So there is a lot of repetition which kind of implies high symmetry.

I did up voted the answer by @Poutnik but it can also be that there is no a straight answer to your Q, at least as formulated. Do not identify symmetry in itself as a source of instability. I up voted the answer because it goes point by point indeed.

Along with symmetry, steric hindrance is also a major similarity to all of the above mentioned molecules. It seems the compounds tend to be highly reactive in order to relieve their steric strains.

Before discussing symmetrical nature of explosives, we might ask, "Why do they contain so many nitro groups?"
The answer is in estimation of energetic properties that relies on the empirically derived Kamlet and Jacobs equations 1 .
In these equations the heat released by the decomposition, the number of moles of gas produced, and the molecular weight of these gases are all critical factors. Density too is crucial. The more molecules of a high-energy material that can be packed into the limited volume of a shell or rocket the better.
E.g.: Nitrocubanes carrying five or more nitro groups contain enough oxygen to oxidize all constituent carbon and hydrogen atoms to gaseous CO, CO2, or H2O. Each of these, along with N2, "explodes" from the solid to 12 gaseous molecules. The expansion from the dense solid to a lot of gas (much expanded by the released heat) produces the desired effect in propellants and explosives.
If one examines the synthetic route, then

RDX is synthesized with treating Hexamethylenetetramine or methenamine, also known as hexamine or urotropin with fuming nitric acid. Urotropin is a SYMMETRIC molecule synthesized from formaldehyde and ammonia.2

PETN (Pentaerythritol tetranitrate) production is by the reaction of pentaerythritol with concentrated nitric acid. Pentaerythritol is prepared via a base-catalyzed poly-addition reaction (aldol) between acetaldehyde and 3 equivalents of formaldehyde, followed by a Cannizzaro reaction with a fourth equivalent of formaldehyde to give the final product.3
A SYMMETRICAL precursor pentaerythritol is used.

HMX is obtained by a two-stage nitrolysis of hexamethylenetetramine (HMTA) in acetic acid, using the solution of ammonium nitrate in nitric acid. A SYMMETRICAL precursor is used.4

Heptanitrocubane is obtained from nitration of TNC. TNC is synthesized from cubane, which is SYMMETRIC.5

HHTDD (hexanitrohexaazatricyclododecanedione) is essentially an open analogue of the cyclic nitroamine cage compounds such as CL-20.[6]

Summarizing
Synthetic routes used in production lead to symmetrical compounds in which more molecules of a high-energy material can be packed into the limited volume of a shell or rocket leading to greater density. For a given explosive, the detonation pressure is proportional to the square of its density.
References
1 http://www.ch.ic.ac.uk/local/projects/wang/discussion1.html
2 https://www.organic-chemistry.org/namedreactions/delepine-reaction.shtm
3 https://en.wikipedia.org/wiki/Pentaerythritol
4 http://www.chemikinternational.com/pdf/2012/01_2012/CHEMIK_2012_p58-63.pdf
5 http://www.ch.ic.ac.uk/local/projects/wang/nitro5.html
[6] https://en.wikipedia.org/wiki/HHTDD

The main reason to synthesize symmetric molecules are related to their synthesis:

Some symmetry is due to the nature of the molecular structure itself, like RDX and HMX. This is related to tricyclic structure of hexamethylenetetramine. Similarly, for Tri/tetramers of peroxyacetone. For these structures, they are created rather spontaneously (HMX with some help) than by directed multistep synthesis.
Some symmetry is due to reaction thermodynamics and kinetics, that does not prefer asymmetrical structures. E.g. TNT is produced easily by direct toluene nitration, as methyl and nitro groups direct coming groups to 2,4,6 positions, forming the symmetric molecule. An asymmetric 2,3,4-TNT is not created this way, its synthesis would be much more complicated, expensive and product may not be safe to produce.

Many asymmetric molecules are not stable enough to be synthesized, or are not chemically stable enough to be kept.

For the keeping stability, Picric acid and its salts are famous for their disintegration products making the explosive oversensitive.

Some symmetry is due low control over the reaction quotient, combined with the advantage of reaction undergoing all steps. See nitroglycerine being trinitrate ester, PETN being tetranitrate ester.

If they are symmetrical, their instability is evenly distributed over the molecule, so they are relatively stable, compared to eventual asymmetric variant, considering their explosive potential.
That allows their synthesis, safe and cheap production, safe manipulation, safe aging.
If they were asymmetrical, their instability would be localized and concentrated to some molecule locations and the molecule would be much less stable.
Additionally, at big asymmetry of bigger molecules, there is question, why to keep the "non-explosive molecule part"?
Consider 2,3,4,5,6-pentanitrodiphenyl versus 2,3,4,5,6-pentanitrotoluene.
Another example of an asymmetrical unstable explosive is the renowned $\ce{NI3 . NH3}$
Summary
I would summarize the answer as kinetic and thermodynamic instability, together with synthesis complexity, discriminates existence and production of asymmetric explosives, comparing to much broader group of non-explosive compounds.
Additionally, the symmetry origin is often coined in symmetry of molecules chosen to be converted to explosive, or to be the structural basis for explosive.
As their symmetry makes promise to form an efficient explosive. E.g. pentaerythritol is perfectly symmetrical alkohol with 4 OH groups to be nitrated.

Hello Poutnik, I have a doubt. Had these symmetrical structures been adding to the stability of the molecule, then that would rather mean that the molecule being more stable would prefer not to react! ...and if that happens, it would no longer remain an explosive (explosives are supposed to react violently, and should be expected to be slightly unstable). That doesn't seem to be a valid reason. Thanks anyways for your answer.

I have meant relatively more stable, with high activation energy, but at the same time high Gibbs energy. That means not stable in thermodynamical sense. There are 2 independent properties of explosives. Their stability against accidental explosion and their brisance when exploding. Professional explosives must be stable enough, otherwise no professional would work with them. There is an explosive I forgot its name, used for nuclear charge detonators, that is so stable it is on the edge to be called explosive at all. Additional reason for symmetrical molecules is, they are easier to synthesise

Yes, I do partially agree with you, but I was expecting a more detailed answer that could clearly illustrate why it is so, and how exactly does symmetry help, and how well does the instability balance against the stability you proposed....maybe a bit of a rigorous scientific explanation to an interesting question! Didn't mean to hurt. Thanks.

P.S.: as reaction tvermodynzmics and kinetics usually prefer the energy rich functional groups to be evenly distributed.

You know, the question is rather broad, deserving a broad answer. Particular structures and their symmetric and asymmetric variants have to be evaluated by the case basis. Also, experimental behaviour is not easily determined theoretically. So, let wait on an explosive expert. Unfortunately, I do not know anybody at Synthesis Semtín factory near Pardubice, where Semtex was invented.

What do you mean by "eventual"?

Eventual as possibly not existing, impossible or hard to synthetize safely in large volumes or at all. OP should provide particular examples to discuss.
E.g. TNT is produced easily by direct toluene nitration, as methyl and nitro groups  direct coming groups to 2,4,6 positions. 2,3,4-TNT synthesis would be very complicated, expensive and product may not be safe to create.

Hello Poutnik, the question is NOT "why specific molecules are symmetrical", it is rather "Why explosive molecules tend to be more symmetrical AS COMPARED TO other molecules IN GENERAL. Meaning if I randomly pick up 10 different chemical explosives and 10 different non-explosive molecules, why is the incidence of symmetry more in explosive molecules?" So, one potential reason that I could get from your answer is that: asymmetric molecules being more unstable would simply not form. Still let's see if there are some other experts of the pyro!

It would be good to update the question.


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Why when I add jam to my tea it stops producing thin "membrane" on top?
Thin "membrane" is created if I leave cup of tea for hour or more. And it usually sticks to the inner sides of the cup. But I noticed that when I drink tea with jam or syrup then the cup always clean inside as if the syrup contains some kind of acid that remove that tea-thingy.

I suspect your tap water is pretty hard and this layer is floating limescale particles.

@andselisk It happens with my cup of tea too, only that I don't add water, boil the milk, and add the tea leaves.

Hold on, so you observe a similarly looking "membrane" on top of just milk (no water)? Or you add milk to the tea? Also, a photo might help.

@andselisk but this layer emerge only in tea. It smells like tea, it has similar color. Also I was sure that everyone saw it :/

So, you make a tea from milk and tea leaves (no extra water) and the layer is there? I saw this limescale layer where people with hard water supply don't use reverse osmosis, that's it.

@andselisk looks like I've found the answer https://cooking.stackexchange.com/a/78500. You was right, it is all about hard water

Well, even though you haven't answered half of my questions, since you figured it out, I suggest you can answer your own question: feel free to start with my comment then:).

Are you sure that it is not milk fat?

The parallel process to deposition of limescale ( that is inhibited by acidic jam) is deposition of products of oxidation of epigallocatechine gallate(EGCG) and similar compounds, that are getting adsorbed on limescale particles and cup walls.

Epigallocatechin gallate (EGCG), also known as epigallocatechin-3-gallate, is the ester of epigallocatechin and gallic acid, and is a type of catechin.
It is found in high content in the dried leaves of green tea (7380 mg per 100 g), white tea(4245 mg per 100 g), and in smaller quantities, black tea (936 mg per 100 g).During black tea production, the catechins are mostly converted to theaflavins and thearubigins via polyphenol oxidases.

EGCG is a polyphenol, vulnerable to oxidation on neutral/alkalic environment, forming insoluble product of intense brown color.
Acidic environment, caused by addition of jam or lemon juice, stabilizes the compound.
Minor natural content of ascorbic acid in fruit juice is protected by abundance of organic acids by similar way.

"It does similar thing to ascorbic acid" Do you mean "It does a similar thing as ascorbic acid [does]"/"It does a thing similar to what ascorbic acid does"?

No. I mean high abundance of organic acids ( citric, tartaric, acetic, malleic ) protects oxidation of minor ascorbic acid, that gets oxidized much slowly in acidic environment. OTOH, if ascorbic acid gets alkalized, it quickly oxidizes by oxygen, turning quite clear solution to intense yellow. But of course, abundance of ascorbic acid would protect EGCG as well.

As @andselisk suggested the problem lied in hard water. And calling that thingy "layer" instead of "membrane" helped to google the problem properly: cooking.stackexchange.com >, www.teamuse.com >
"The scum on the top of the tea is due to hard water (ie calcium carbonate) deposits combining with the tea and reacting with oxygen."
To dissolve that layer we need to rise acid level: add lemon, jam, syrup or even make stronger tea.


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Identifying weaker base
Can you tell me the reason behind the answer to this:

Which is the weaker base?
(a) o-cresol; (b) Trifluoro-o-cresol.

Answer: (b) trifluoro-o-cresol.
I'm not a student in any institution. So it's not homework and it was in the chapter reaction mechanism. I'm studying for a medical exam. So basically I'm on my own. Just found this in an old book, but it offered no hints.
Is it the hydrogen bonding between H and F? That's my best guess, because I think it's a case of ortho effect exception. In ortho effect it doesn't matter which group comes at o-position. The basicity decreases regardless of the group. But if memory serves me right, I think my teacher said there were some exceptions. So that's my best guess.

The weaker base is the stronger acid. The electron withdrawing effect of the 3 F atoms makes the phenoxide of tri-F- cresol more stable so it is more acidic than cresol.

The stronger acid has the weaker conjugate base. If the (deprotonated) conjugate base is more stable, it is easier for the (protonated) acid to donate a proton.

Simply check the order of acidic strength and reverse it to find basic strength as a weaker base will be stronger acid and vice versa. Among them trifluoro-o-cresol is more acidic because after the removal of acidic hydrogen the negative charge thus generated is stabilized by the −I effect of three fluorine atoms.
Whereas in the former the $\ce{CH3}$ molecule provides +I effect which destabilizes the negative charge. So, trifluoro-o-cresol which is a more stronger acid is thus the weaker base among them.


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Calculating molecular weight of an acid based on given values of mass, volume, pH and pKa
This is what I thought was an easy problem but according to a given answer I am doing it wrong:

$\pu{11 mg}$ of a monoprotic acid was dissolved in $\pu{0.5 l}$ of water. The resulting $\mathrm{pH}$ of said solution was measured to be $5$. What is the molecular weight of said monoprotic acid? $\mathrm{p}K_\mathrm{a} = 6.$ $(t = \pu{25°C})$

My solution
Given values are:
$$m = \pu{0.011 g} \qquad V = \pu{0.5 l} \qquad \mathrm{pH} = 5 \qquad \mathrm{p}K_\mathrm{a} = 6$$
Formula used to calculate molecular weight:
$$M = \frac{m}{n}$$
Formula used to calculate concentration of acid in the solution, that is, the (hopefully correct) formula for calculating the acid equilibrium constant:
$$K_\mathrm{a} = \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}$$
Formula used to calculate molar concentration:
$$C = \frac{n}{V}$$
Putting them together:
$$M = \frac{m}{\frac{[\ce{H+}][\ce{A-}]}{K_\mathrm{a}}\cdot V} \label{eqn:alpha}\tag{α}$$
Concentration of $\ce{H+}$ and $\ce{A-}$ based on $\mathrm{pH}$ value:
$$[\ce{H+}] = [\ce{A-}] = 10^{-5}$$
Calculating the acid equilibrium constant from $\mathrm{p}K_\mathrm{a}$:
$$K_\mathrm{a} = 10^{-6}$$
Result:
Entering the above vaules for $m,$ $V,$ $K_\mathrm{a},$ $[\ce{H+}]$ and $[\ce{A-}]$ into equation \eqref{eqn:alpha} yields the molecular weight of $\pu{220 g mol-1}$. This is answer is wrong, supposedly. Did I make any mistakes?

I find your lack of faith disturbing.. 200 it is.

Sorry, it was meant to say 220. I re-edited the post, it now says 220 g/mole which is the result I get when I plug in the values. 200, however is the correct answer that I was given. Could my math simply be wrong? I am literally plugging in those number into ($\alpha$).

You don't have enough significant figures to say 220. Half a liter is a very coarse measurement for volume.

That's it! That completely slipped my mind. Thank you!

@Zhe What is the role of sigfigs here?

220 is too many significant figures given the number of significant figures that are in the statement of the problem. That doesn't mean everything else was right, but this was certainly one issue with answering 220 here.

@Zhe I think you have to be careful with significant figures in these toy problems. If I write $\frac{1}{2}$ in a formula, is that one significant figure? If the pKa is given as 6 and the pH as 5, does that mean we have less than a significant figure (maybe just answer in terms of order of magnitude, i.e. molar mass is on the order of hundreds to thousands of g/mol)?

@KarstenTheis Yes, pH 6 and 5 have less than 1 significant figure. I don't take significant figures as gospel (since you should better handle error with error bars), but it is frequently a cause of confusion in such problems. This is simply something I noticed while looking superficially at the statement of the problem and the proposed answer. Hence, I added a comment instead of writing an answer.

Ok here's what you're doing wrong.
When you write $$C = \frac{n}{V}$$
$C$ is meant to be the concentration of the monoprotic acid dissolved initially, not at equilibrium. Plugging in the equilibrium concentration is disasterous, as it is bound to be different from the actual ratio of amount of substance in moles of $\ce{HA}$ to the volume of solution in liters.
This approach will better suit the problem:
$$\ce{HA \ <=> \ H+ \ + \ A-}$$
$[\ce{HA}]_{\mathrm{initial}}=\frac{m}{M}.\frac{1}{V}$
Assuming $x$ moles/liter to be consumed at equilibrium,
$[\ce{HA}]_{\mathrm{eq}}=\frac{m}{M}.\frac{1}{V}-x$
Now $x$ moles/liter each of $\ce{H+}$ and $\ce{A-}$ would have appeared.
$[\ce{A-}]_{\mathrm{eq}}=[\ce{H+}]_{eq}=x$
$x$ happens to be $10^{-5}$, as $\mathrm{pH}$ is given to be 5.
$K_a$ would be $10^{-6}$, as you have already mentioned.
$$K_a = \ce{ \frac{[H^+]_{eq}[A^-]_{eq}}{[HA]_{eq}}=10^{-6}}$$
Plugging in, I get
$$10^{-6}=\frac{10^{-5} \times 10^{-5}}{\frac{0.011}{M \times 0.5}-10^{-5}}$$
Simplifying a little, I get
$$M=\pu{200 g/mol}$$


text:
Is hydrolysis of polypeptides and polysaccharides "anabolic" or "catabolic"
When a polysaccharide or polypeptide is hydrolyzed into mono-saccharides or amino acids, the building blocks can be oxidized to release energy.  The oxidation is considered to be catabolic since it reduces the building blocks to simple compounds: carbon dioxide, water, ammonia, and releases energy.

Is the process of hydrolysis that breaks up polypeptides and polysaccharides a net endothermic or exothermic process?
Do the free amino acids and monosaccharides have more or less stored energy than the polypeptide or polysaccharide that they were broken down from?
Is it proper to call the isolated process of "hydrolysis" of proteins and polysaccharides "catabolic"?
Are protein synthesis, glycogen synthesis, (and triglyceride formation), by dehydration synthesis processes that require energy or release energy.  I think that they release energy which is semantically interesting since protein and glycogen synthesis are the main examples of anabolism in the body but may actually release energy which is a key component of the definition of catabolism.  Even if the energy released from protein synthesis is not generating ATP directly, wouldn't the heat produced conserve ATP in the long run.

They all consume ATP, so they are typical examples of anabolic reactions. For protein synthesis, here is a more detailed answer: https://biology.stackexchange.com/questions/40964/how-much-nucleoside-triphosphate-is-required-to-form-one-peptide-bond-during-pro

1) Is the process of hydrolysis that breaks up polypeptides and polysaccharides a net endothermic or exothermic process?

Under physiological conditions, it is a process that goes forward, i.e. the Gibbs energy is negative. As a consequence, it can happen outside of cells in the absence of ATP. When we eat, the hydrolysis of polysaccharides starts in our mouths, while the hydrolysis of proteins occurs under harsher (acidic) conditions in the stomach and continues in the intestine (slightly basic conditions).
For warmblooded animals like us, exothermic or endothermic is less important, but you could look it up.

2) Do the free amino acids and monosaccharides have more or less stored energy than the polypeptide or polysaccharide that they were broken down from?

Less Gibbs energy, as the hydrolysis is spontaneous. However, this Gibbs energy is not captured (different from downstream oxidation of building blocks to $\ce{CO2}$, which is coupled eventually to synthesis of ATP).

3) Is it proper to call the isolated process of "hydrolysis" of proteins and polysaccharides "catabolic"?

If your definition is just breaking down complicated molecules into simple ones, then you would call it catabolic. If you also require that Gibbs energy is captured in a coupled reaction, then you would classify these reactions as digestion, and the downstream oxidation to carbon dioxide as catabolism.


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Butanone or butan-2-one?
According to IUPAC rules, am I right in assuming that butanone is sufficient, because there's only one possible molecule? I keep noticing butan-2-one, so I want to make sure.

For strict IUPAC, butanone is not sufficient; the more specific name is required.  For common usage, it is clear enough for most organic chemists.

Butanone can only be one ketone. As per IUPAC it has to be named aButa-2-one.Pentanone can be Pent-2-one or Pent-3-one.

The preferred IUPAC name is butan-2-one. This is explicitly mentioned in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book).

Generally, the complete information about a structure is explicitly given by the name and does not rely on any implied information. Nevertheless, the practice of omitting locants when there is no ambiguity is widespread in general nomenclature. For preferred IUPAC names, however, locants are omitted only in a few exceptional cases described in Subsection P-14.3.4 of the Blue Book:

P-14.3.4.1 terminal locants
P-14.3.4.2 the locant ‘1’
P-14.3.4.3 where there is only one kind of substitutable hydrogen
P-14.3.4.4 when no isomer can be generated by moving suffixes or prefixes from their position to another or by interchanging them between two different positions
P-14.3.4.5 when all substitutable positions are completely substituted
P-14.3.4.6 when all substitutable hydrogen atoms have the same locant.

None of these cases applies to butan-2-one. Therefore, the locant cannot be omitted in the preferred IUPAC name.

What if one argues ad P-14.3.4.4, that -one suffix cannot be moved to 1 (4) position, as it would become butanal?


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I live in an area with a company that claims they've made soft carbonated drink out of lactic acid and milk, how true is it?
I live in Iran, recently a massive cooperate company has produced a soda/carbonated soft drink,  under the brand name "luckydo" which they claim it's beneficial for your health and each 2 glasses equal the 1 glass of milk in terms of calcium, I've searched the whole internet there are no products similar to this at all.
I quote, this is how they claim they've made the product
"Luckydo is a soda (carbonated soft drink) based on milk!, Fresh milk enters our factory, proteins and fat are separated and all that remains is a liquid that contains calcium, nutrients and vitamins available in the milk's liquid are then combined with different flavors, lactic acid product of fermented milk and water are then added, and alas it's carbonated."
Here is the picture of the product.

My question is that, is this true?
if so why no other company on earth has tried or claimed to have such
product ?
is it really good for health ?

because I plan on buying lots of these, if what they say is not fake for commercial purposes and true.

There's something called whey: https://en.wikipedia.org/wiki/Whey .In some countries yoghurt-like drinks are sold in a bottle. But this doesn't look or sound like whey or yoghurt. For one thing it's translucent. This sounds like soda pop + calcium + other additives. Impossible to tell from the information provided if it's good for you. Probably not. Best to consult a dietitian. Also, you may want to try the cooking SE site.

Whey Is very different to yoghurt drinks. If purified(getting rid of whey proteins) their is mainly just lactose(or residuum of it if processed by fermentation as sour whey), lactic acid, salts and other  soluble compounds. It is usually considered healthy, but not frequently directly used.

How does it taste / feel?  Might answer question w

Nothing impossible anyway. They recycle waterly part of milk. The fact they claim health benefits is quite common in advertising.  Likely a mere claim but this does not necessarily makes the product unhealthy itself.

Yes, thank you, it's pretty much like soda, tastes like soda and looks crystal clear. ofc not impossible but then if it is as they claim why no other company on earth has tried to make this "healthy" soda before ? it's just hard to believe there is clacium in that thing.


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Reverse Cannizzaro reaction to make aldehydes?
Is it possible (and practicable) to reverse Cannizzaro reaction (disproportionation of aldehydes to carboxylic acids and alcohols) to produce aldehydes from carboxylic acid and alcohol without need to use neither chromate or $\ce{LiAlH4}$?
For example: 1,4 butanediol + 1,4 butanodioic acid - > 1,4 butanedial

What reagents you were suggesting to be able to undergo reverse Cannizzaro reaction?

In case it makes a difference I'll edit the question to include an example

That'd be ideal.:-)

I think it's impossible. Somehow alkoxide would need to be strong enough donor of H- that carboxylate would accept it - definitely not the case you mention.

@Mithoron: I agree with you. It need a $\ce{H-}$ source to start with and that would react with alcohol first as acid-base reaction. Thus, no reverse Cannizzaro reaction is possible.

@MathewMahindaratne That's not my point. Alcohol and acid need to be fully neutralised in the first place; that's why I mentioned specifically their anions.

No, this is not practicable because aldehydes are higher in free energy than either alcohols or acids.  Thus, the equilibrium constant of the Cannizaro reaction lies strongly against the aldehydes.
The case of formaldehyde may be even more extreme than the example reaction you gave, but it is what I could find data for.
The eQuilibrator database lists the equilibrium constant for the reaction of
$$\ce{2 HCHO + H2O <=> CH3OH + HCOOH}$$
as $$K'_{eq} = 3.4 × 10^{13}$$
This is for each reactant at 1 mM in concentration, at pH 7 and an ionic strength of 0.1 mol / L.  Thus, you'd need a teramole or two of formaldehyde to get one mole of acid/alcohol.  The thermodynamics for other aldehydes may be slightly more "favorable" but you aren't going to get 13 orders of magnitude of favorability for any "regular" pair of aldehyde and acid/alcohol.

Maybe under scrupulously anhydrous conditions with continuous removal of trace water you could get the backwards reaction to go a tiny tiny bit, since lack of water would shift the equilibrium constant.  But probably still not to a practicable degree.


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Sample: How to calculate mass of nanopaticles with given density and Vol %
I'm trying to calculate the wt% and the mass of the nanoparticles in a ferrofluid solution.
Attached is data from the supplier:
**EMG 605 Specifications and Physical Properties**
- Appearance:   Black-brown fluid
- Carrier Liquid:   Water
- Nominal Particle Diameter:    10 nm

- Saturation Magnetization (Ms): 220 Gauss or 22 mT (SI)
- Viscosity @27°C:  <5 cP or <5 mPa·s
- Density @25°C:    1.18 g/cc or 1.18x10^3 kg/m3 (SI)
- Initial Magnetic Susceptibility:  0.24    or 3.02 (SI)
- :Magnetic Particle Concentration  3.9 % vol.
- pH:   9-10
- Nature of Surfactant: Cationic

Note the site requests posters show their attempt to solve homework type problems. I answered anyway (hopefully correctly).

Is this ferrofluid solution $\ce{Fe}$ nanoparticle in water?

Defining the following

Density of solution: $\rho = \pu{1.18 g/cc}$
Density of water: $\rho_w$
Density of nanoparticles: $\rho_{np}$
% (v/v) n.p.s: $p_v=3.9$%
% (w/w) n.p.s: $p_{wt}$

Then solve for $\rho_{np}$ in the following equation (you'll need to look up missing value of water density):
$$\rho = \frac{p_v}{100}\rho_{np} + \frac{100-p_v}{100}\rho_{h2o}$$
With $\rho_{np}$ you can compute $p_{wt}$
$$p_{wt} = 100\frac{p_{v}\rho_{np}}{p_{v}\rho_{np}+(100-p_{v})\rho_{w}}$$
As noted in the comments, the resulting density of the particles ignores the effect of the added cationic surfactant. Also, the solvent density is assumed that of water, free (not particle bound) detergent or other additives assumed to be negligible.

Up voted but it is assumed that the surfactants don't change the liquid density much. And I think is a reasonable assumption. If OP has the sample s/he can validate the result, at least roughly, by weighing a V of it.

@Alchimista Thanks, and yes, that's a good point, I thought of the surfactant and ignored it. I could estimate concentration from monolayer coverage (but this would of course be particle bound) or estimated cmc, just for the fun of it, and if I find some time I'll add that. I'm guessing added surfactant won't affect the density too much, but it should be straightforward enough to estimate the uncertainty from that.

I am not sure if I get whats going on, is there a way I can share my data sheet? I want to determine the number of nano-particles in 1 grams similar to what I determined for my milled fibers

@user642407 Depending on the document in question you can insert an image (photo or screenshot) to your question, or include a link. If you can provide more information about how you are stuck on this question that might help get it reopened (no guarantees though).


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Purification of Solid Phase Synthesized Peptide
I began working in a newly setup lab, and I'm having a dilysine (K2) peptide synthesized by solid phase method. I plan to cleave the peptides from the resin with a cleavage cocktail of TFA. Unfortunately, our freeze-drier is missing a part and not working then. We also don't have a rotary evaporator yet. We have our HPLC setup, but do you think it is a good idea to take the resulting solution of cleavage (after filtration to remove beads), which contains Boc, Fmoc, TFA, and possibly other impurities, directly to the HPLC?
I've read that HPLC sample should be dissolved in initial composition of my HPLC solvents, but it is in the case of having a crude dry sample.
It is my first time doing a peptide synthesis so I appreciate your ideas.

If you are a using reversed-phase (RP) column, it is okay to directly inject your crude sample in the running HPLC solvent system.  But, make sure to filter the injecting sample through a membrane filter to remove minute particles (from glass beats, etc.) from the original sample after dissolving it in suitable chosen solvent. Please also note that your chosen solvent for peptide separation usually contain small percentage of TFA. Therefore, you may need at least a rotary evaporator  for final concentration.

The "dissolving" part is a bit confusing to me because I already have a cleaved peptide + TFA solution, is dissolving just the same as mixing with water? (since my initial solvent composition is going to be water, linearly increasing acetonitrile over time).

I believe you don't have dry sample. Usually you dissolve your dry sample in HPLC mobile phase you supposed to use. In this case I'd add enough water to dilute TFA if needed. Sample should be clear solution (do not let it precipitate when adding water). Since you are using RP, using water to inject did not effect the separation much (volume should be small as possible).


text:
Methyl Orange Indicator-Reason for small amount of NaOH
I used methyl orange as my indicator to titrate a weak acid:

At first, I added 3 drops of methyl orange to a solution of the weak acid, $\ce{CH3COOH}$, and the solution turned red. Then, after adding a few drops of the strong base $\ce{NaOH}$ solution, the solution became yellow. Why are only a few drops of $\ce{NaOH}$ needed to make the solution yellow?
In addition, I then added a few drops of the strong acid $\ce{HCl}$, and the solution became red again. Why are only a few drops needed here as well?

If you add an equimolar amount of NaOH and HCl, you have effectively dissolved a bit of table salt in your solution, and the pH is back where it was. Check the concentrations of your weak acid and your HCl and NaOH solution, and do the math.

Using methyl orange for titration of weak acids is wrong idea.
The indicator has range $\mathrm{pH}=3.1~\mathrm{(red)}  - 4.4~\mathrm{(yellow)}$.
If acetic acid with $\mathrm{p}K_\mathrm{a}=4.75$ is to be determined, the indicator starts to turn toward yellow, when majority of acid is not titrated yet.
The concentration ratio $\frac {\ce{[CH3COONa]}}{\ce{[CH3COOH]}}$ is $0.022$, resp. $0.45$ for full red resp. full yellow. $99.9\%$ of the acid is titrated at $\mathrm{pH}=7.75$.
Therefore, indicator shows a change after a small addition of $\ce{NaOH}$. A small addition of $\ce{HCl}$ returns things back.
Methyl orange is suitable for titration of strong or weak bases by strong acids, turning yellow to red.
Much more suitable and frequently used indicator for weak acid titration is phenolphthalein with range $\mathrm{pH}=8.2~\mathrm{(clear)}  - 10.0~\mathrm{(violet)}$
Aside of suitable indicator $\mathrm{pH}$ range, the transition from weak color(clear, yellow) to intensive colour (red, violet) is easier to catch than vice versa.


text:
Looking for a non-volatile liquid less viscous than water
I have tried searching everywhere for a stable, non-volatile, bio-safe, low viscosity (< .8cST) fluid. Everything I have come across that is low viscosity is inflammable - acetone, ether, silicone oil, fuel, etc.
I have found some confusing things in my research. Acetone has a viscosity 3x less than water but weighs 2.5 times more. I would like to know how water weighs less but flows slower. Google says viscosity is from low intermolecular forces, but I do not know how to quantify those forces or how to find chemicals that match.
I am trying to make a science project that shows the principle of siphoning, but I want a lower viscosity liquid to show how viscosity effects flow rate. I am working with glass right now and PVC tubes, but I can change my materials based on whatever low viscosity liquid you can find.
Also if anyone knows anything about siphoning, viscosity or material chemistry please let me know if there is a database I can search or if you know chemical that is less viscous than water and meets my specifications.
Is there something I can add to water to lower the viscosity aside from heating? I do not want to have boiling water in my hands. I hear there are surfactants that lower the viscosity of water, but google says surfactants increase viscosity.

Why do you say acetone "weighs 2.5 times more" than water? Its density is less than 80% that of water. Could you not do your demonstration comparing water and something more viscous than water?

Low viscosity and non-volatility are rather mutually exclusive requirements, as the former requires low intermolecular interactions, while the latter the direct opposite.

Welcome to the site! This sounds like an interesting and fun problem! I look forward to seeing what kind of answers others come up with. I did notice you seem to have several questions nested in this question. Would you be willing to post each question separately? It would help break down your problem by its parts and help you get the answers to all your questions.

The Mw of water is smaller than that of acetone, yes. Water is protic, it hydrogen bonds extensively (self-associates) which explains the viscosity and high b.p. despite low MW. Acetone is polar aprotic, no h-bonds.

Have you checked out “viscosity reducing additives”? Long ago, I used to add polyethylene oxide (“PEO”) to water to make “liquid banana peel” for fun in the freshman dorm.

There is one class of available liquids (some readily available from chemical suppliers eg perfluorohexane) that might fit the requirement: perfluorocarbons.
They have similar boiling points to hydrocarbons with the same number of carbons (so room temperature liquids with > 5 or 6 carbons). And they typically have low viscosity when compared to their hydrocarbon equivalents which, in turn, then to be less viscous than water.
This class of compounds were once investigated as blood-substitutes as the solubility of oxygen in them is very high. And some are used in other medical procedures today. Which suggest they are not acutely toxic.

I guess you could go with Peppermint essential oil. Its density is 0.898 g/mL at 25 °C. It obviously is volatile to some extent if left uncovered however its viscosity is considerably lower than water. Hope this helps you out!


text:
Mass of H2 in a balloon
Question

A balloon of diameter $\pu{2 m}$ has a volume of $\pu{4190 dm^3}.$
  It floats since it is given an up thrust equal to the mass of air it displaces.
(Given: average $M_\mathrm{r}$ of air = 29.
  Under conditions of inflation, $\pu{1 mol}$ of $\ce{H2}$ occupies a volume of $\pu{23 dm^3}).$
Find mass of $\ce{H2}$ in balloon.

My method
$$\text{Amount of}~\ce{H2} = \frac{\pu{4190 dm^3}}{\pu{23 dm3 mol-1}} = \pu{182.2 mol}$$
$$\text{Mass of}~\ce{H2}~\text{in balloon} = \pu{182.2 mol} × \pu{2 g mol-1} = \pu{364.4 g}$$
Correct answer: $\pu{182.2 g}$.
I don't understand why the answer is 182.2g

Why are you doubling the mass of H2 in the final step of your calculation? Where does the factor (x2) come from?

Isn't mass = Mr × number of moles?

Sure, I just didn't want to guess your thoughts from the numbers (because the guess could be wrong). It might help to edit the post and add a short explanation and the units.

At first glance your solution looks reasonable. I allowed myself to add missing units – please don't drop them and always carry values and units together.

Thank you for adding the units.That was my bad.

Help me please.why is the answer 182.2?or is the answer sheet wrong?

The answer sheet is wrong.


text:
Equivant mass of double salt
What is the equivalent mass of double salt like dolomite? And what if the double salt is hydrated?
Is a double salt taken as a single unit while equating the gram equivalent in any reaction?
For example, in reaction
$$\ce{CaCO3.MgCO3 + 4 HCl -> CaCl2 + MgCl2 + 2 CO2}$$
is
$$\frac{\text{wt of dolomite}}{\text{eq wt of dolomite}} = \frac{\text{wt of}~\ce{CO2}}{\text{eq wt of}~\ce{CO2}}?$$

$\ce{CaCO3.MgCO3}$ reacts with $\ce{4 HCl}$, therefore $m_\mathrm{eq}=m/4$.
$\ce{CaCO3}$ reacts with $\ce{2 HCl}$, therefore $m_\mathrm{eq}=m/2$.
$\ce{MgCO3}$ reacts with $\ce{2 HCl}$, therefore $m_\mathrm{eq}=m/2$.
$\ce{Na2CO3.10 H2O}$ reacts with $\ce{2 HCl}$, therefore $m_\mathrm{eq}=m/4$.
$\ce{1 CO2}$ of carbonate origin is released by $\ce{2 HCl}$, therefore $m_\mathrm{eq}=m/2$.

As a consequence of Poutnik's answer, the molar mass of the double salt $\ce{CaCO3·MgCO3}$ is $40 + 60 + 24.3 + 60 = 184.3 g/mol$, and the equivalent is $184.3/4 = 46.1 g/eq$.
And if the double salt is hydrated with $n$ molecules water, this would increase the molar mass by $n·18 g$. And this new mass should be divided by $4$ to get the corresponding equivalent mass.


text:
Solubility of PbSO4
The solubility of $\ce{PbSO4(s)}$ increase with the addition of $\ce{H2SO4}.$ Why?
I don't quite understand this. When dissolving $\ce{PbSO4(s)}$ we get the equilibrium equation:
$$\ce{PbSO4(s) <=> Pb^2+ + SO4^2-}$$
I know $\ce{H2SO4}$ is a strong acid, that will dissociate completely (or almost completely), like this:
$$
\begin{align}
\ce{H2SO4 &-> H+ + HSO4-}\\
\ce{HSO4- &-> H+ + SO4^2-}
\end{align}
$$
So as far as I can see, the concentration of $\ce{SO4^2-}$ increases, which should give a lower solubility of $\ce{PbSO4(s)}$ (common ion effect). What am I missing?

$\ce{H2SO4}$ is a strong acid all right, but $\ce{HSO4^-}$ is not that strong.

Thank you. So what will happen? H+ from the acid will react with SO42- from the compound to form HSO4-, which will only partly dissociate? (I.e. the concentration of SO42- will be reduced)

That's right.$!$

AFAIK, lead sulphate forms in concentrated sulphuric acid soluble lead hydrogen sulphate, Pb(HSO4)2, what is supported by protonization of sulphate anion. Kind of calcium carbonate analogy.

How could I know on beforehand that HSO4- was not a that strong acid? I have a table work of pKa values, and I notice that H2SO4 has a pKa value of <0 where as HSO4- has a pKa value of 1.99.

You've just answered your own question.

Thanks! In general, in what range is the pKa value of. strong acids?

According to https://en.wikipedia.org/wiki/Acid_strength?wprov=sfla1, strong acids are conventionally acids with pKa< -2.

@poutnik -2?  Oy!  Nitric acid misses!

It is tight miss.


text:
Unknown possibly handmade glassware- description in information below
Okay, Another random item from the cupboard (unknown users or discipline of life sciences unknown) could be anything up to 40 years old....
This appears to be a 2 neck quickfit adapter (B19 and B14) leading to a long narrow tube which loops back on itself then loops again into a closed vessel. None of the other technicians can remember using this item but some of the items do look certainly altered by a glassblower if not entirely made by a glassblower. Whole item is 30cm in length...
Any thoughts on what it could be used for or what it might be?

Just a guess, but it looks like it might be for preparing and transporting an air-sensitive sample/reaction. The sample is in the bulb and liquid is in the loop to create a trap. With the liquid in place, you could open the top and connect to something else without exposing your sample to very much air.

I agree with @Andrew.   Thanks for sharing OP, I love seeing bits of odd and unique chemistry glassware.

Interesting idea thank you Andrew! I will relay to the team and Academics and see if they can be useful or not!

Michael we have a lot more that I haven't taken photographs of and put up yet so there will be more odd and unique chemistry glassware to come!


text:
What are the primary, secondary and tertiary physicochemical properties of ionic liquids?
In the following sentence

..the ability to modulate the individual components presents an advantageous framework for tuning secondary and tertiary physicochemical characteristics without sacrificing the primary function of the ionic liquids.

What do they mean by secondary and tertiary physicochemical characteristics?
Can somebody give example of this categorization of the physicochemical property?

The definition of generations of physicochemical properties can be attributed to Hugh et al. [1] as recapitulated by e.g. Ferraz et al. [2]:

The search for new and different ILs has led to the progressive development and application of three generations of ILs: 1) The focus of the first generation was mainly on their unique intrinsic physical and chemical properties, such as density, viscosity, conductivity, solubility,and high thermal and chemical stability. 2) The second generation of ILs offered the potential to tune some of these physical and chemical properties, allowing the formation of “task-specific ionic liquids” which can have application as lubricants, energetic materials (in the case of selective separation and extraction processes), and as more environmentally friendly(greener) reaction solvents, among others. 3) The third andmost recent generation of ILs involve active pharmaceutical ingredients (API), which are being used to produce ILs with biological activity.

The passage quoted in the OP stems from an article by Zakrewsky et al. [3] and refers to a study on a third generation IL. Third generation characteristics would in this case include those granting the IL antibacterial, nontoxic and transdermal delivery properties, that is allowing solubilization of drug compounds, minimizing damage to human cells, and disupting bacterial biofilms. In this case, "secondary" properties were not targeted, but these might include (as per the above quote) reduced environmental toxicity.
References

Hough, W. L.; Smiglak, M.; Rodríguez, H.; Swatloski, R. P.; Spear, S. K.; Daly, D. T.; Pernak, J.; Grisel, J. E.; Carliss, R. D.; Soutullo, M. D.; et al. The Third Evolution of Ionic Liquids: Active Pharmaceutical Ingredients. New J. Chem. 2007, 31 (8), 1429–1436. https://doi.org/10.1039/B706677P.
Ferraz, R.; Branco, L. C.; Prudêncio, C.; Noronha, J. P.; Petrovski, Ž. Ionic Liquids as Active Pharmaceutical Ingredients. Chem. Med. Chem. 2011, 6 (6), 975–985. https://doi.org/10.1002/cmdc.201100082.
Zakrewsky, M.; Lovejoy, K. S.; Kern, T. L.; Miller, T. E.; Le, V.; Nagy, A.; Goumas, A. M.; Iyer, R. S.; Sesto, R. E. D.; Koppisch, A. T.; et al. Ionic Liquids as a Class of Materials for Transdermal Delivery and Pathogen Neutralization. PNAS 2014, 111 (37), 13313–13318. https://doi.org/10.1073/pnas.1403995111.


text:
Amount of benzoic acid required to alter pH of a NaOH-solution
I am trying to do a practice problem but I can't figure it out:
How many grams of benzoic acid needs to be added to 0.10 l of a NaOH-solution with a concentration of 1.0 mM to reduce the pH of said solution to 4.00? Assume that the added mass of acid does not alter the total volume of solution. $K_a = 6.5 \cdot 10^{-5}$
My attempt:
Benzoic acid is a monoprotic acid with a molecular weight of 122.1 $\,$ g mol$^{-1}$ and I will assume that NaOH has a 100% dissociation.
I choose to divide the problem into two parts; calculating the substance amout of acid needed to neutralize the OH$^-$ ions separate from the substance amount needed to lower the pH to 4.
Substance amount needed to neutralize OH$^-$, $n_n$:
$\ce{HA_{eq} + H_2O <=> A^- + H^+ + OH^-}$
$K_a = \frac{[H^+][A^-]}{[HA]_{eq}}$
$[HA]_{eq} = [HA] - [H^+]$
$[HA] = \frac{n_n}{V}$
Deriving an expression for $n_n$:
$n_n = \biggl(\frac{[H^+][A^-]}{K_a} + [H^+]\biggr) \cdot V$
$K_a$ is listed as $6.5 \cdot 10^{-5}$, the volume, $V$, is $0.10$ l and $[H^+] = [A^-] = [OH^-] = 1.0$ mM as listed.
Result: Plugging in these values yields the following substance amount: $n_n =1.6 \cdot 10^{-3}$ mol.
Substance amount needed to lower the pH to 4, $n_s$:
Deriving expression for $n_s$ using same method as above:
$n_s = \biggl(\frac{[H^+][A^-]}{K_a} + [H^+]\biggr) \cdot V$
In this case $[H^+] = [A^-] = 1.00 \cdot 10^{-4}$ based on pH = 4.00. ($V$ and $K_a$ same as above)
Result: Plugging in these values yields: $n_s = 2.5 \cdot 10^{-5}$ mol.
The mass of benzoic acid needed can now be calculated using the substance amount $n = n_n + n_s$:
$m = nM$
Final result: Plugging in the values for $n$ and $M = 122.1 $ g mol$^{-1}$ yields $m = 0.20$ g.
This answer is not right and I know I am doing something wrong. Teach me the ways.

If you keep adding benzoic acid until all of NaOH present is neutralized, your solution will consist of sodium benzoate ions, which will undergo hydrolysis to form a solution with pH > 7.
So you have to continue adding benzoic acid. This means that your beaker will now consist of a buffer solution. So, you can use the Henderson-Hasselbalch equation to get the answer directly.

Have you tried pH=pKa + log ( [A-]/[HA]) ?


text:
An unknown glassware piece from our cupboard
We have multiple of these. It is a small glass funnel with splits into 3 spouts from it. Now, we are thinking it might be similar to a cow receiver, however, I have never seen one this small or delicate. It's 11cm in length.
We have put 3 test tubes underneath and ran water through and it does appear to distribute it relatively evenly.
What could it be?

Trident harpoon for glass fish? ;-)

Why does this post have closure votes as homework? If I search google for this image I get results for glass bottles, which is certainly not what is expected.

We have currently been calling it a funnel fork.... but I really have a feeling that isn't what its called!

It might be a cold finger to collect up to three fraction, fitting distillation of a micro amount (mg scale) of a single compound (head, x, tail).

My best guess is a "pis de vache" (i.e. "cow's udder" in French) which is used in distillation to collect several parts of the distillate. Though it is generally more useful during vacuum distillations, which is not the case here. See picture https://www.chemscience.com/categorie-produits.html?cat_id=249

Mad idea: if the "funnel like" ending fits in a chromatographic column, and if the tiny ends keep it anchored to the opening, maybe a normal funnel can be placed between the (so strangely bent) tiny arms, and the whole thing could serve as a kinky funnel adapter for chromatography .-.


text:
Heating 2% solution to get 4% solution
I have a 2% solution of Zinc sulfate in water (in heptahydrate form) that I want to turn into 4% solution instead of having to buy 4% solution separately.
Would heating it to 100 degrees Celsius and waiting for extra water to vaporize be the right way to do it?
Thanks.

The best approach would be to crystallize $\ce{ZnSO4.7H2O}$ and then weigh the required amount and prepare a fresh solution.

How:

Slowly evaporate the solution until it is about one-fifth of its original volume. Caution: Do not boil the solution as it may spit.

Allow the concentrated solution to cool until crystals form.

Filter off the crystals and put the filter paper and crystals on a watch glass and dab dry with another piece of filter paper.

Allow it to dry thoroughly (if you're keen on very accurate concentrations)

I assume by $x\%$ solution you mean $x\%$ weight($\pu{g}$)/volume($\pu{mL}$)
You may proceed to prepare the solution by dissolving $4$ grams in 100 $\pu{mL}$ water. In case you need a different volume, you may use unitary method.
Edit:
Crystallization is certinly more tedious than simple evaporation and volume restoration, but is bound to be very accurate. There might be loss of salt in trace amounts during evaporation, adheration to vessel walls et cetera. But if much precision is not your requirement, you may follow the latter, by all means.

Even though boiling the original solution down to half the volume might create technical issues, once you boil it down to 1/5 the volume couldn't you let it cool and add distilled water until the volume is back to half of the original amount without having to isolate crystals?

@JosephHirsch I suppose that'd work. A lot easier too!

Or might we lose some small trace Zinc Sulfate during evaporation?  I think not if it's slow enough.

@JosephHirsch zinc sulfate isn't that volatile, I believe

I believe that evaporation would work, but for technical reasons, I would evaporate it to below half the original volume, cool it, and then add distilled water until the volume is half of the original level.

Thanks, this is what I will do. I wasn't sure which answer to pick as correct here. I picked William's because it also contains the warning not to boil the solution, which might be useful to anyone who finds this question later.


text:
Calculating melting temperature of water with thermodynamical data
So at zero degrees celsius($\pu{273 K}$), for clean water, we have the equilibrium eq.:
$\ce{H2O(s) = H2O(l)}$
As it is at equilibrium, we know that $\Delta G = 0$. As we are dealing with clean water and clean ice we also know that $\Delta G_\text{standard} = 0$. (From the $\Delta G_\text{standard} =-RT\ln K$ relationship, where $K = 1$ results in $\Delta G_\text{standard} = 0$).
We also have the relationship:
$\Delta G_\text{standard} = \Delta H_\text{standard}-T\,\Delta S_\text{standard}$.
which we can re-arrange to:
$\Delta H_\text{standard}= T\,\Delta S_\text{standard}$.
And we can calculate $T$ which will be our melting temperature.
I found thermodynamical data for the reaction for $\ce{H2O(s)}$ and $\ce{H2O(l)}$, and got the equilibrium temperature to be $\pu{241 K}$! Which seems weird. What is wrong?
EDIT: I know that $\Delta H$ and $\Delta S$ aren't necessarily temperature independent, although I guess that is a prerequisite for the way I have done it?

Could you show us the values for entropy and enthalpy of melting (fusion) and tell us where you found them? Also, how is the title of your question related to the body of your question?

deltaH_fusion = 6 kJ/mole. deltaS = -29 J/ K mole. All data from SI Chemical Data, 6th edition. You are right about the title, I will change it!

I googled +6.02 kJ/mol and +22.0 J/(K mol), both for 273.15 K, see https://en.wikipedia.org/wiki/Water_(data_page)

I offer an alternative solution. We will apply the Clapeyron equation at the point of interest $(P_m,T_m)$ lying in the P-T diagram on the solid-liquid equilibrium
\begin{align}
  \dfrac{dP^\pu{sat}}{dT} &= \dfrac{\Delta_\pu{fus} H}{T_m \Delta V} \tag{1} \\
\end{align}
Here we make our bold assumption: we consider that the solid-liquid equilibrium curve can be approximated by a straight line, in the range of the triple point $(P_t,T_t)$ to the melting point $(P_m,T_m)$; obviously $P_m = \pu{101325 Pa}$. Thus, we can approximate the derivative in the left-hand side of Eq. (1) by
\begin{align}
  \dfrac{P_m - P_t}{T_m - T_t} &= \dfrac{\Delta_\pu{fus} H}{T_m (V_\pu{l} - V_\pu{s})} \\
  T_m &= \dfrac{T_t}{1 - \dfrac{(V_\pu{l} - V_\pu{s})(P_m - P_t)}{\Delta_\pu{fus} H}} \tag{2}
\end{align}
Useful data from Wikipedia and knowing that the molar mass is $\pu{18.015 g/mol}$:

$\rho_\pu{s} = 916.7 \; \pu{kg/m3}$ or $V_\pu{s} = \pu{1.9652E-5 mol/m3}$.
$\rho_\pu{l} = 999.8 \; \pu{kg/m3}$ or $V_\pu{l} = \pu{1.8019E-5 mol/m3}$.
$\Delta_\pu{fus} H = 6008 \; \pu{J/mol}$.
The triple point has the coordinates $P_t = \pu{611.657 Pa}$ and $T_t = \pu{273.16 K}$.

Replacing in Eq. (2) yields
$$ \boxed{T_m = 273.1525 \; \pu{K} = 0.002521 \; ^\circ\pu{C}} \tag{3} $$
Pretty near to the exact value.


text:
Common ion effect
Let $\ce{HA}$ ($x$ molar) be a strong electrolyte and $\ce{HB}$ ($c$ molar) — weak electrolyte. As $\ce{HA}$ is strong electrolyte, it dissociates completely:
$$
\begin{array}{lcc}
\ce{&HA &-> &H+ &+ &A-} \\
\text{initially} & x && 0 && 0 \\
\text{after} & 0 && x && x
\end{array}
$$
As HB is weak electrolyte
$$
\begin{array}{lcc}
\ce{&HB &-> &H+ &+ &B-} \\
\text{initially} & c && 0 && 0 \\
\text{equilibrium} & c - cα && cα && cα
\end{array}
$$
Then
$$K_\mathrm{a} =\frac{cα\cdot (cα + x)}{c - cα}$$
Now my doubt is how can we add the concentration of $\ce{H+}$ together as their volume can be different. Suppose $x = \pu{1 M}$, and $cα = \pu{0.01 M},$ then how can be the resultant concentration be $\pu{1.01 M}?$ Can someone explain in detail?
Edit:
$\alpha$ is the degree of dissociation
Now let us suppose we had a 'x' molar strong electrolyte and a 'c'  having a common ion let us suppose H. Now while writing Ka for the dissocition of weak electrolyte as given in equation
$$
\begin{array}{lcc}
\ce{&HB &-> &H+ &+ &B-} \\
\text{initially} & c && 0 && 0 \\
\text{equilibrium} & c - cα && cα && cα
\end{array}
$$
Then
$$K_\mathrm{a} =\frac{[B] \cdot [H] }{[HB] }$$
Now due to common ion effect we have to consider the total concentration of H we get which gives us
$$K_\mathrm{a} =\frac{cα\cdot (cα + x)}{c - cα}$$
Now how did the concentration of H become $$(cα + x)$$
Let us now suppose instead of x we had a 2 molar strong electrolyte (HA) so the concentration of H that comes from HA is 2 molar. Now for the weak electrolyte let us suppose we get 0.01M as the concentration of H$(cα) $ that comes from HB now according to the equation
$$K_\mathrm{a} =\frac{cα\cdot (cα + x)}{c - cα}$$
We get the total concentration of H is  (cα + x). Using our supposed value we get total concentration of H as 2+0.01=2.01M but how can we just add the two concentration. Just like in if we had a 1M solution1 and a 2M solution2 we would not get the total concentration of the two solutions after mixing as 3M

Are x and c the concentrations before mixing or after? What are the different volumes you are talking about?

Initial concentration

It is a common ion, so you should have a single variable for the $\ce{H+}$ concentration, and it should appear in the equilibrium expression. The protonation state of the weak electrolyte (just like a pH indicator) will be influenced by the pH of the solution. Another consequence is that the concentration of $\ce{H+}$  will be different from the concentration of  $\ce{B-}$.

@Satwik It's not the concentration that we add, we add the moles. If the volume is 1 L, then naturally  concentrations get added.


text:
Is a hydrogen molecule smaller than hydrogen atom?
Recently I came across a textbook of physical chemistry which stated the following:

In the formation of a stable hydrogen molecule, two hydrogen atoms share a pair of electrons between them and two $s$-orbitals overlap along the same axis resulting in the formation of sigma $(\sigma)$ bond. As the shared electrons are attracted by both the nuclei, the size of hydrogen molecule is smaller than that of hydrogen atom.

However, the actual radius of hydrogen molecule is $120~\text{pm}$ which is greater than that of hydrogen atom i.e. $53~\text{pm}$. So, what does the author actually mean when he says that the size of hydrogen molecule is smaller than that of hydrogen atom? Moreover, does it imply that the electron density is maximum at the COM of the hydrogen molecule? Any help would be appreciated.

The hydrogen molecule is not a sphere, so it does not have a radius.

The hydrogen atom is not a hard sphere, so you have to be careful to define what you mean by size and radius.

The figures in this article might be helpful to debunk the textbook statement: https://resonance.is/direct-imaging-wave-function-hydrogen-molecule/

Maybe they meant: As the shared electrons are attracted by both the nuclei, the size of hydrogen molecule is smaller than the sum of the sizes of two hydrogen atoms.

Since an isolated hydrogen atom is spherically symmetric, perhaps the original claim refers to the difference in the effective radius looking along the bond in dihydrogen rather than the radius looking perpendicular to the bond. Is the claim true in that context?

You can define the size of atoms and molecules in various ways. You can, for instance, derive a measure of the size of a hydrogen molecule from the density of solid hydrogen:

Solid hydrogen has a density of 0.086 g/cm3 making it one of the lowest-density solids.

From the above you can derive an effective radius for $\ce{H2}$ of $\pu{2.10 Å}$, which is eerily similar to $4 \times \pu{0.53 Å}= \pu{2.12 Å}$.
However this is a distraction from the main point of the statement in the book, which is simply that concentration of electron density between the two $\ce{H}$ nuclei results in an "end-to-end" distance (the distance, using some measure to determine the edges of the molecule, along the internuclear axis) that is less$^\dagger$ than $\pu{2.12 Å}$, and that the internuclear distance is less than $\pu{2\times\pu{0.53 Å}= 1.06 Å}$ (which it is: the H-H bond length is $\pu{0.74 Å}$). An internuclear distance shorter than twice the lone H radius is consistent with bonding, a process by which electrons and nuclei are allowed to occupy lower-energy regions of the potential than they would when the atoms are far appart.
But the answer to the question is no, a hydrogen molecule is still bigger than a hydrogen atom.

$\dagger$ But just barely. Define the edge of an atom or molecule as the point where the electron density equals that at a distance of $\pu{1 Bohr}$ ($\pu{0.529 Å}$) from the nucleus in a hydrogen atom. Next find the edges in a simple  MO model of $\ce{H2}$ ([3]). The edges are $\pu{1.056 Å}$ from the center of the molecule, that is, the molecule along the longest axis has a diameter of $\pu{2.11 Å}$, barely less than $4\times$ the Bohr radius.

[3]: Pearson, Ralph G., and Palke, William E.. Proc.Natl.Acad.Sci.USA, Vol. 77, No.4, pp.1725-1727, April 1980.


text:
Is there pure iron in nature?
Is there pure iron in nature, would that be not become Iron Oxide (rust) immediately because of the humidity? How do we produce iron ore/dust on the lab/industry?

I suppose it depends on what physical state (phase) you mean. There is probably plenty of molten iron in the interior of the earth. Much of the iron in our lithosphere started and never stopped oxidizing when oxygen-producing photosynthetic organisms evolved. To produce unoxidized iron from the oxide you need to reduce it, for instance with carbon, see eg https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/3_d-Block_Elements/1b_Properties_of_Transition_Metals/Metallurgy/The_Extraction_of_Iron

Previously I used to think that the knowledge of iron metallurgy is safely stored in the books and Wikipedia. Turns out I was wrong. It is kinda scary to think that it may die with me. Then again, why bother. Apres moi le deluge!

@IvanNeretin https://en.wikipedia.org/wiki/Telluric_iron ? ;-)

For @Karl Telluric Iron is either Iron with lots of Carbon, or Iron with Nickel. So I guess Iron really does not exists on its own in Nature. Perhaps only noble metals like Gold or Silver has this property.

@RonPringadi Nope. Silver and gold are always contaminated with each other, + copper, palladium, etc. ;-)

How fast iron rusts is very dependent on its purity. Some relatively pure iron does not rust quickly.

Fairly pure iron (or iron/nickel alloys) are found in meteoric iron. Some of these are surprisingly corrosion resistant... close to stainless steels. One well-known example is a dagger from Tutankhamun's tomb. See JSTOR for some more uses of meteoric iron.
As for iron smelting, there is much information on the web, and I suggest you try some research.

Is it appropriate to describe an alloy as "pure"?

@WilliamR.Ebenezer Depends on your own conception of what makes a material pure.

Based on the comment from @Karl the wikipedia article stated "Type 2 telluric iron also contains around 0.05 to 4% nickel, but typically less than 0.7% carbon" Assuming the rest of 99.25% is Iron, this may be the closest one we have to pure Iron in nature.


text:
Is there a method to separate iron from mercury?
The mercury is contaminated with iron. Is there any way to remove iron from mercury?

Does the question "is mercury magnetic?" strike you?

What physical form does the iron have? Fine powder, shavings, small lumps, an entire I-beam, ...?  And does it matter to you whether the iron retains that form after the iron and the mercury have been separated?

As iron is one of few metals not forming amalgams,

Almost all metals can form amalgams with mercury, the notable exceptions being iron, platinum, tungsten, and tantalum.

..it should preferably contaminate the mercury surface.
I would try, perhaps repeated,  dropping mercury, e.g via punched filtration paper to sulphuric acid bath.
It can be combined with  agitating mercury - acid phase border.
Important can be the origin of impurity and the intended purpose of purified mercury. There is possible the only reasonable way would be distillation, mentioned in the other answer, respecting the warning.

I say this might work, but what has this got to do with the fact that iron does not form an amalgam?

Cleaning mercury surface should be easier than extracting iron from the whole mercury volume.

@karl It is relevant that iron doesn't form an amalgam as, if it did, there would be a homogeneous solution and the iron would be hard to separate without something like distillation. Because iron doesn't dissolve in mercury, physical separation is likely to work.

@matt_black it could be combination of physical and chemical separation.

Well, for physical separation, you can just put a bit of soap water on your mecury and bubble air through it. For chemical separation, mixing with acid and subsequent phase separation might work. The latter however would also work if the metal (say, zinc instead of iron) was dissolvend.

.... but differently.

Non-amalgamation of mercury with iron means that the iron is not dissolved, and therefore floats on the surface as a dross. However, non-amalgamation does not mean total insolubility. If a few ppm of iron needs to be removed, washing with a dilute solution of HNO3 might be best, with much agitation.
There will be some dissolution of mercury to Hg++. That's just a loss due to purification, and can be minimized. Hg++ ion in solution will be reduced by Fe metal in the mercury liquid until there is no more iron in the mercury and there is some slight excess of Hg++ ion in solution. This approach will also remove other contaminants. You will be able to see a bright mercury surface when it is purified.
Other methods have relied on agitation (oxidation) in air and removal of dross, and might suffice, but is probably not as complete as washing with nitric acid.

Interesting! I had not anticipated the redox reaction.

Several decades ago I used a lot of mercury (in polarography with a dropping mercury electrode) and purified it by distillation in a still looking something like this https://www.researchgate.net/figure/Distillation-apparatus-for-mercury-cells-a-and-design-b-del-Campo-et-al-2008_fig1_325075540
However, mercury is toxic, especially in vapour, I strongly recommend against doing this unless you really know what you're doing.