text:
How much 'stronger' is water vapor than carbon dioxide, molecule-per-molecule, over the course of a day, in terms of global warming potential?
I have read, many times, that water molecules are far more effective than CO2 molecules at trapping heat...
But, the 'relative G.W.P.' value of water vapor (or its 'radiative forcing') is not listed anywhere I can find, and the stated reason (if one is stated) is that water vapor has always been in the atmosphere in substantial amounts, of course, and if it has increased in prevalence recently it is because of increased CO2 levels, not the other way around....
So what is the amount of solar energy trapped/re-radiated by a water vapor molecule, over a very short time period, in direct comparison to a CO2 molecule?
I am very curious....

Generally compounds are assigned a global warming potential over a 100 year period, probably because the lengthy averaging facilitates understanding and simulation (and likely increases accuracy/precision). It may not make quantitative sense to lower that period to a single day, in which case nobody would bother assigning such a value. Apparently there is already criticism of even the 20-year GWP being too short a timespan to consider.

@NicolauSakerNeto These 20 or 100 year averages do afaik simply take into account the lifetime of the compound in the atmosphere. That is imo not what OP was asking for. (If I'm right, OP should make that more clear, i.e. strike "over a day".)

I think what you’re asking is the relative value of the integrals of the IR absorbance spectra at the bottom of this article? https://www.acs.org/content/acs/en/climatescience/greenhousegases/properties.html

@Karl I understand, and it could be that OP is asking for something like what Andrew suggests. All I mean to say is that, with regards specifically to GWP, the lifetime of water vapour in the atmosphere is a major factor to consider, as its mean residence time is only ~10 days (compared to ~5-200 years for CO2), and therefore water behaves rather uniquely (not even counting the radiative effects of cloud formation...). Making a simple comparison of the relative infrared absorption intensities of CO2 and H2O may be somewhat confounding without context.

Generally is said, water makes 75% of the GH baseline and CO2 25%. Considering lifetime of H2O in atmosphere does not make much sense in GH effect context, as the water content is dynamically kept about the same in global long term view, slowly increasing with the global warming.

P.S.: As the GH effect baseline is roughly +30 deg C or litttle more, water part of GH effect makes about +22-23 deg C and CO2 part some +7-8 deg C, what about corresponds to the global mean temperature raise with raising CO2 level 0.03% to 0.04%.

It does not make much sense to calculate it for a single molecule, as there are multiple very variable factors. It would be always an average calculated from the global level and amount of substance in atmosphere.

@BuckThorn This may have been a mistake; now that I've read everything carefully this may have been a good question for Earth Science SE but it's not really about Chemistyr, is it. The OP was "Last seen 11 hours ago" so they may not have given up entirely...

@KurtHikes do you want to edit and clarify your question based on comments? If not, I'll just award the bounty to the closest approximation of an answer if one is posted.

I found the following useful: http://www.realclimate.org/index.php/archives/2005/04/water-vapour-feedback-or-forcing/


text:
Relative strength of intermolecular forces

My attempt at a solution:
First I look for hydrogen bonds:
The only compounds with hydrogen bonds are the second and fourth ones, so they have the strongest intermolecular forces. The second must have stronger intermolecular forces since it harbors three hydrogen bonds, while the fourth only has room for two.
Then I look for dipole-dipole interactions:
The first compound has the potential for a dipole-dipole moment. It would be the third strongest.
Then I look for potential for London-dispersion forces:
The third compound and The fourth experience London-dispersion forces. But I'm stuck on discerning which is stronger.
So, my working solution is: 3, 5, x, 4, x

HINT: Induced dipole forces depend upon surface area of the compound.

I would say that your analysis of first, second and fourth one is correct, as H-bonds are stronger than dipole-dipole interactions. However, you made a mistake while filtering compounds having dipole-dipole interaction. The $1$ and $3$, both have dipoles!
The $\ce{C=O}$ bonds in $\ce{CO2}$ are polar and hence they possess dipoles, but unlike a water molecule, $\ce{CO2}$ shape is linear, and therefore the equal dipoles cancel each other, resulting in a non-polar molecule as seen in the following figure,

That's the reason $\ce{CO2}$ possess dipole-dipole interactions whereas $\ce{F2}$ doesn't as it has only one bond which is non-polar!
Image taken from courses.lumenlearning.com: Polar Molecules

See https://chemistry.stackexchange.com/questions/54115/quadrupole-moment-of-a-molecule

@KarstenTheis: So, acc. to you, CO2 isn't non-polar, but possess a quadrupole (two dipole moments sitting back to back in opposite dir, para 5)? But, what's the difference b/w being a quadrupole and non-polar? Both have zero dipole moment, isn't it...

There is a interaction that is weaker than dipole:dipole: https://chemistry.stackexchange.com/questions/42946/can-nonpolar-molecules-exhibit-dipole-dipole-forces

@RahulVerma even the molecule is non-polar there are vdW interactions present. You can take example of Ar-dimer systems.

source: Stack Exchange

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

sort: votes


text:
Oxidation of Unsymmetrical Ketones using Selenium Dioxide
I have searched the whole internet but I could only see examples where the authors of the article took $\ce{R1}$ and $\ce{R2}$ as the alkyl groups around the carbon. This is regarding the oxidation of carbonyl compounds (I am specifically searching about ketones) using $\ce{SeO2}$. What I understood was $\ce{C=O}$ got introduced on the alpha-carbon and two hydrogens from that carbon were lost. I would like to know the product obtained on reacting pentan-2-one with $\ce{SeO2}$. Will the carbon having 3 alpha-hydrogens become $\ce{C=O}$ or will the other one having 2 alpha hydrogens become $\ce{C=O}$?
References:

Selenium dioxide oxidation of ketones and aldehydes. Evidence for
the intermediacy of .beta.-ketoseleninic, K. Barry Sharpless and
Kenneth M. Gordon Journal of the American Chemical Society,
1976, 98 (1), 300-301 DOI: 10.1021/ja00417a083
https://www.adichemistry.com/organic/organicreagents/seo2/selenium-dioxide-seo2.html

See Riley Oxidation

You find some insight here.

The mechanism of Selenium Dioxide (Riley) oxidation proceeds via the enol tautomer of the ketone (see scheme - image from Wikipedia). If the ketone is assymetric and can enolise to either side then the reaction should give the two di-ketones in proportion to the amount of either enol formed though I cannot find any reference where this has been experimentally demonstrated. In the OP's case of 2-pentanone, a mixture of the two possible products will be formed.

why would it not make a double bond as  it does for 1,4 diketones?

@https://chemistry.stackexchange.com/questions/33034/why-doesnt-the-seo2-oxidation-of-ketones-stop-at-the-hydroxyketone-stage: It's depend on the several factors including structure of starting material, solvent, and pH. You can find an answer to your question here.

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


text:
How to identify whether a molecule is planar
While solving for plane of symmetry of molecules I came across 1,3-dichlorocyclobutane (1) and I realized it does not have any plane of symmetry because we don't know whether hydrogen and chlorine are in same plane. How to judge which plane do the substituents belong to?
Similarly for compound 2, but in compound 3 it is known that both the planes of the rings are perpendicular. Is there some kind of rule for it?

First you must unknow a number of things that you already know. 1,3-dichlorocyclobutane (any isomer) does have a plane of symmetry. This has nothing to do with being planar, though.

@IamAlita You can do it yourself using free 3D molecular editors like MolView or  Jmol (which will also find and draw symmetry elements for you). I think there is not much to elaborate in a "show, don't tell" case like this.

There are levels upon levels of different approximations to reality, and jumping two levels up does not do the learner any good. For the purposes of this question, consider the 4-membered cycle planar.

@IvanNeretin do you mean that we dont need to know whether the molecule is planar or not in order to determine plane is symmetry?

It might be easier if instead of 'plane of symmetry' you used 'mirror plane' instead as the atoms do not need to be in the plane but appear identical when viewed in a mirror. Molecule 1 has 3 such mirror planes.

@Saniya To find out mirror plane, we need to know the geometry of molecule - all of it. Consequently, we'll know everything that follows. The molecule does not have to be planar, though.

@I am Alita  For simplicity  and to help the IP I assumed the molecule is as drawn with a flat ring.

The concept of a plane of symmetry is easier to understand with molecules which are planar by itself.  Benzene, for example, has such a plane which may accommodate simultaneously all carbon and hydrogen atoms:

However, planarity of molecules is not a requirement to identify such planes.
Even if a construction of the cyclobutane molecule with model kit illustrates you the non-planarity of the molecule

(source)
you drawing depicts it like it were planar; because you defined drawing plane where carbon atoms alternate to be either closely above, or below this level in common.  For compound 1 illustrating 1,3-dichloro cyclobutane, however, the symmetry analysis is much affected by the configuration of the two special carbon atoms.  Either you find the two chlorine atoms on the same side of this plane (syn)

or you find them directing to different sides (anti):

The interactive training set by Symmetry@Otterbein may be a helpful resource for you.  Note that the representation of compound 2 is a simplification, and the symmetry attributed strongly depends on the underlying geometry.


text:
Generating hypochlorous acid for disinfecting water using HClO2
I use a product that contains $\pu{700 ppm}$ of $\ce{HClO}$ to disinfect water at $\pu{1 ml}$ per $\pu{1 l}$ and let it rest for about $\pu{2 hrs}$ before I use it so there is no $\ce{HClO}$ left because I will mix it with enzymes and rhizobacteria which could be destroyed by the $\ce{HClO}$.
I found a device (Eco One Electrolyzed Water System) that can generate $\ce{HClO}$ at $\pu{200 ppm}$ using just water and $\ce{NaCl}$ at a $\mathrm{pH}$ of $5$, which would be much cheaper in the long run.
Will this have the same disinfectant effect when I add $\pu{3.5ml}$ per $\pu{1 l}$ water of the $\pu{200 ppm}$ solution and should I also let it rest for $\pu{2 hrs}$?

In a closed container such as a spray bottle, the concentration of hypochlorous acid decreases about $1\%$ per day.

Are there any buffers or stabilizers that I can add to the generated $\ce{HClO}$ to slow down this process which are safe and environmentally friendly when the water is used for watering fruit plants?

Why not to use the commercial stock solution of NaClO ? As HClO solution would contain NaCl as well.

Looks like that NaClO is not that easily available in my region, but I found that I could buy NaClO2 25% (which is also used for water treatment in my region), would this be safe to use? I also thought about using H2O2, but I could not find any information how long it takes for the H2O2 to disintegrate. Most important to me is that it will not leave any toxic substances in the water and that I can add enzymes and rhizobacteria after a fixed amount of time (e.g. 2 hours like for HClO) so that they get not destroyed and to remove most of bacteria / spores.

I have never thought that the "liquid bleech" can be not easily available, unless shops are empty because of coronavirus. Chlorites will be more stable than hypochlorites.

I only found products that also contain other chemical like Na2CO3 for private customers and found reports that it's not safe to use it with plants, but I'm no expert in (bio)chemistry. I found some chemical suppliers which don't list NaClO right now, maybe because of coronavirus.

Like poisons, all is matter of dosage. Na2CO3 dosage comparable with the active chlorine dosage is harmless. Eatable pretzels are being soaked in NaOH to get nice dark brown and shiny crust.

Ok thanks, I will test if it works in different concentrations (I just found that some people use it at 0.5-3ppm) and also check how the plants react.

Or, you can alkalize the hypochlorous acid solution.

when the water is used for watering fruit plants - what's wrong with using normal tap water for that?


text:
Sucrose skeletal formula - missing Carbon and Hydrogen
I was reading the Wikipedia article about Sucrose, when I noticed that the skeletal formula was missing some $\ce{C}$ and $\ce{H}$ atoms. The chemical formula being $\ce{C12H22O11}$, I counted the $11$ Oxygen parts. But there seem to be only $3$ $\ce{C}$ atoms and $14\, \ce{H}$ atoms, if I counted correctly. My chemistry lessons date back about $15$ years, so I'd like to understand where the missing carbon and oxygen is hidden within the skeletal formula.

Every point in the bond is a carbon atom.. H is just added to satisfy valency after.. You could read up on the bond-line formula of a compound..

Ah, that makes sense. And the "bold" parts are CH ions then? That would account for the missing Hydrogen.

This Libretexts page should help.

Any formula is a condensed knowledge that must be unpacked by applying certain rules. For example, the mathematical formula for "eleven" is 11. One who does not know the rules might wonder why there are only two sticks and not eleven. Same thing here.

Since your chemistry lessons date back about 15 years, I'd loke to refreash your organic chemistry knowledge. You may need to be aware that carbon can make maximum if 4 single bonds. Therefore, when you have a line structure of organic compound like sucrose (see below), keep in mind that an each corner represent a carbon atom and appropriate hydrogen atoms, which are not shown. For example, if a corner contains only three bonds (as shown in most of sucrose molecule), then the forth bond is $\ce{C-H}$ bond:

I have put those $\ce{C-H}$ at each corner, which are representing them. However, the anomeric corner of fructose molecule already has 4 bonds. Therefore, that corner represents only $\ce{C}$ as indicated in turquoise color. Therefore, the molecular formula of sucrose can be counted as $\ce{C12H22O11}$.


text:
How do I find miller indices for a plane whose intercepts are fractions of the lattice constant?
[I'm talking with respect to cubic lattices alone.]
For instance, if a plane has $x,y,z$ intercepts $a/2,a/2,a/2$ (where $a$ is the lattice constant) the miller index would be $[2\space2\space2]$. The book I'm referring to says that for fractional intercepts, the indices do not have to be reduced to the smallest whole numbers(hence $[2\space2\space2]$).
But, miller indices are supposed to represent a set of parallel planes and I can't think of any other plane with the index $[2\space2\space2]$
Then I saw this:

The image(3rd row, 3rd image) shows the set of $[2\space2\space2]$ planes.
I'm guessing that one of them has intercepts $a/2,a/2,a/2$. What are the intercepts of the other planes, and how are they all $[2\space2\space2]$?
Or, if my method of calculation is wrong, how else do I calculate miller indices for fractional intercepts?
PS: I've only just started to learn this concept, so it's possible that my understanding of miller indices is fundamentally flawed.

note: the same question was asked on https://physics.stackexchange.com/questions/575940/how-do-i-find-miller-indices-for-a-plane-whose-intercepts-are-fractions-of-the-l

The plane closest to the origin has the intercepts you mentioned. The rest have intercepts that are integer multiples of this. In the figure, the origin is on the bottom left of each cell. As the figure shows, the [1 1 1] planes correspond to every second of the [2 2 2] planes because the latter is a multiple (by 2) of the former.

Higher indeces are associated with a lower distance of the planes and with a higher resolution (i.e. larger Bragg angle) of the corresponding reflection.

"the [1 1 1] planes correspond to every second of the [2 2 2] planes" - Shouldn't every second of the [2 2 2] planes be indexed [1 1 1], since miller indices have to be scaled to the smallest whole numbers? Or can a plane have more than one miller index?

"The rest have intercepts that are integer multiples of this." - So a plane with intercepts $(5a/2,5a/2,5a/2)$  would be indexed $[2 \space2 \space2]$?

@ElonTusk See my edited answer


text:
Can non-deuterated solvents be used for 13C-NMR (and in fact other nuclei)?
Basically, my question is can you used a non-deuterated solvent for 13C NMR? I'm thinking that there might be some problems with locking?

On top of long's answer, I would just mention that you can use nondeuterated solvents even for ordinary 1H NMR. (This is already implied in the answer.) Ideally, you'd pick a solvent which doesn't overlap with your peaks of interest; and the solvent suppression techniques (routinely used with H2O samples) will be useful too.

If using a deuterium lock, then normally you would require, as minimum, about 10% deuterated solvent. Without any deuterated solvent, it is not possible to obtain a spectrometer lock (unless using a 19F lock channel, but that is unlikely, and is a hardware requirement).
It is, however, perfectly feasible to acquire spectral data for all nuclei without any deuterated solvent, and hence, without a spectrometer lock. When running without a lock, you need to turn the lock sweep off, to hold the spectrometer field constant. This does not provide the same level of field stability as a lock, but is more than adequate for short-medium 1D experiments that do not require very high resolution. Using gradient shimming, it is also possible to shim on non-deuterated solvents and achieve the same lineshape specifications.
Running long experiments (such as an overnight 13C{1H} experiment), without a lock will give you linewidths that are slightly broader than usual; the consequences of this are that (i) your resolution is reduced (rarely a problem with 13C), and (ii) your signal-to-noise will be reduced (and therefore longer experiment times may be required...leading to further broadening and longer experiment times...)
Not claiming to be the first to report about this technique, the Hoye group published two papers in Organic Letters (2004OL953 and 2004OL2567) with examples of application of no-D $\ce{^1H}$-NMR spectroscopy, covering sample preparation and data acquisition, too.  Especially the second one, outlining how to determine the concentration of LDA or Grignard reagents by addition of 1,5-cyclooctadiene as internal standard may be a reliable and faster alternative to the titrations once you are used shim manually for samples of this type.  But speak and obtain clearance in advance with your NMR manager before running your no-D sample; some interfaces to the spectrometer tend to be less puzzled to record NMR with a lock after your sample (e.g., elder Varian), than the more modern (e.g., Topspin Bruker) to prevent unnecessary delays in the queue of samples following yours.

In groups where recording NMR spectra by your self is permitted / assumed, recording no-D 1H-NMR may be something as normal as recording your «normal» 1H-NMR spectrum.  Thus I thought the addition of the two Hoye papers were beneficial to your answer given.

It is indeed possible and you will indeed have problem with locking. My solution is to use sealed capillaries with a deuterated solvent inside. This procedure allows you to lock without having the solvent in your mixture.

This is an interesting approach, can you provide additional details or citations for those of us not familiar with the technique? Does it require specialized NMR tubes? Can this help in situations where peaks are lost due to rapid H-D exchange?

@dissemin8or You may use already a normal NMR tube with a capillary containing your e.g., deuterated trifluoroacetic acid.  Depending on the sensitivity of the NMR spectrometer, a about 5 cm partially [80%] filled, but flame sealed (e.g., by a lighter) melting point capillaries already may do the job.  This isn't too much weight added and may be used multiple times.  Indeed, this will prevent H-D exchange completely. (H-D exchanges may be slowed in dry DMSO-d6, too.) Schematic example

@dissemin8or I use normal NMR tubes (5mm) and capillary for melting point determination closed on one side (dimensions 1.55 mm diameter, 75 mm lenght) i fill them in a way that no air is contained in the measuring zone but you have to leave a bit of space for flame seal them. As already mention you can add up to two of them (in this way you even reduce the volume of solution needed in your measurements), my experience is that with C6D6 one is more than enough but when you go for a solvent with less deuterium content like CDCl3 you most likelly need two.


text:
API for molecular classification
We would like to identify an application programming interface (API) which can be fed a SMILES string, PubChem CID, IUPAC name or equivalent, which returns a set of strings corresponding to molecule classification e.g. ketone.
Direction to any such interface, API, tool or instructions on how to access this info through PubChem or ChemSpider would be greatly appreciated.

What is API, SMILES and CID ?

Edited question.


text:
Metal coordination in enzymes: What type and number of bonds form between amino acid residues and Fe(II)?
My understanding: Transition (d-block) metals, as found in intracellular proteins, are limited to about 10 elements including amongst others Fe, Ni, and Cu. Due to the reductive environments inside cells, they are usually/expected to be found with an oxidation state of +2 (?). However, the number of bonds that coordinate the metal, assuming that you know that the oxidation state is +2, is not completely certain (?).
Confusion about bond and coordination: As the d-orbital of e.g. Fe(II) is not completely filled it can be considered somewhat reactive/unpredictable, and we cannot for sure say what coordination it will have in an enzyme (?). Or, is there some sort of general assumption we can make about its coordination based on its oxidation number (i.e. +2)? As I was looking for literature on the matter I stumbled on the chemistry stack exchange answers regarding ‘bonds in coordination complexes’ which used Fe(II) as an example of how Pauling’s Principle could be applied to simplify how metals are coordinated.
A brief summary of this answers would include that coordinating residues can be thought to contribute ½ electron to the coordination of Fe(II); and thus, Fe(II) would typically have four coordinating interactions while Fe(III) would have six coordinating interactions, in order to cancel the charge.
Summary of question(s): If my statements above are correct, then it could be generally assumed that all +2 metals in enzymes, such as Fe(II) , are coordinated by four groups? Furthermore, the coordinating bonds are considered dative (i.e. not entirely ionic)?  Very speculatively, could it also be assumed that Fe(II) would typically be in a tetrahedral complex (why/why not)?
*update based on comments: Octahedral coordination of Fe(II) was mentioned as classical example in heme enzymes, while it is also known that non-heme enzymes can e.g. coordinate Fe(II) with three residues and a water molecules in a tetrahedral conformation. It could therefore be reasoned that coordination of the metal (e.g. tetrahedra vs octahedral) is related to enzyme function and evolution over time, and not so much the oxidation state of the metal?

Pauling's electroneutrality thingy is not really widely used anymore; it's a very rough rule of thumb and one should not draw generalisations from it. Fe(II) is very commonly octahedral and the factors that determine tetrahedral vs octahedral (& other shapes) coordination can be quite complicated. Hemoglobin is the most common example of a Fe(II) metalloprotein and the coordination geometry is octahedral (although strictly speaking, only one of the bonds is to an amino acid residue – the 'proximal histidine').

Thanks, so this add to the fact that transition-state metals cannot be so early understood, and that one should not assume simple 'rules of thumb' for e.g. Fe(II).  Im still a bit confused about coordinatin: I know (non-heme) enzymes with Fe(II) that have three amino acids and a water molecule coordinating it (tetrahedral), while heme-enzymes with Fe(II), like you mention, are octahedral. There seems to be no real pattern – it is more a case-by-case scenario that has to be experimentally validated?

I think sometimes it is more unambiguous. For example, Zn(II) is pretty much always tetrahedral, as far as I know. But Fe(II) can be a bit of a chameleon, even outside of proteins. I'm rusty on the (very fascinating) bioinorganic chemistry details, but I do suspect the function of the protein can sometimes lend a clue as to why certain geometries, etc. are adopted. After all, the metal environment (and indeed the choice of metal) have been "selected" through many years of evolution.

Functional evolution in terms of metal coordination in enzymes is a good point, I will update the question a bit with this reflection. Thanks!

One reason for the variety is that macromolecules like proteins can enforce a coordination geometry that is not what you would expect for small ligands coordinating the metal in solution. But most of the protein metal-binding sites fall into one of a small number of classes (eg 4Fe-4S or 2Fe-2S) types, so if you read some reviews of metalloenzyme structures, you can learn the common ones quickly.

It's further complicated by the fact that many metalloproteins are redox active, so the oxidation state of the metal changes between two states while still coordinated to the protein

It seems like the answers is ‘no’, there is no general rule that can realistically be applied. General metalloprotein review are also too general (enzyme superfamily level). It will always depend on the function of the specific enzyme of interest. As you state, redox active enzymes will even change the state of the metal during catalysis and probably have complexed ways of coordinating the metal to facilitate the reaction. The most systemic approach is then to narrow the question to a specific enzyme classes and fold-type, otherwise it is simply too complicated to rationally deduce.


text:
quantum self-consistent ab initio lattice dynamics QSCAILD
I approached the quantum self-consistent ab initio lattice dynamics (QSCAILD) method by referring to the article: https://arxiv.org/abs/2006.12867. I have created a latex text   in which I outline the method as I understand it and ask questions in the less clear points. I would like to be able to understand, even in an intuitive way, the fundamental principles on which the method is based. I hope that among you there is someone who knows this method and can help me to clarify.
I have added my method below:

The QSCAILD package is a python library that computes the effective interatomic force constants at second and third order in a crystal including anharmonic effects. The approach is based on quantum statistics and on a self-consistent procedure based on a harmonic model. \newline
We take a system consisting of $ N $ ions (atoms with their core electrons) and describe them through a harmonic Hamiltonian $ H_h $:
\begin{equation}
  H_h=\sum_{R,\sigma}\frac{P_{R,\sigma}^2}{2M_{\sigma}}+\frac{1}{2}\sum_{R,\sigma}\sum_{R',\sigma'}U_R\sigma\Phi^{\sigma \sigma'}(R+b_{\sigma}-R'-b_{\sigma'})U_{R',\sigma'}
\end{equation}
the subscript $ \sigma $ indicates the atoms of the base, R indicates the cells of the Bravais lattice, $ b_{\sigma} $ the atomic positions relative to the Bravais lattice, $ U_{R, \sigma} $ the displacements of the atoms , $ P_{R, \sigma} $ the momentum of the atoms, $ M_{\sigma} $ the atomic masses and $ \Phi ^ {\sigma \sigma '} $ the second order interatomic force constants.
We define the probability $ \rho_ {H_h} (u) $ to find the system considered in the configuration in which each ion $ i $ is shifted by $ u_ {i \alpha} $ from its equilibrium position along the Cartesian direction $ \alpha $.
This probability (Gaussian distribution) is proportional to:
\begin{equation}
  \rho_{H_h}(u) \propto e^{-\frac{1}{2}u^T\Sigma^{-1}u}
\end{equation}
where $ \Sigma (i \alpha, j \beta) $ is the quantum covariance (what is that?) for atoms $ i, j $ in the Cartesian directions $ \alpha, \beta $:
\begin{equation}
  \Sigma(i\alpha,j\beta)=\frac{\hbar}{\sqrt{2M_iM_j}}\sum_m\omega_m^{-1}(1+n_B(\omega_m;T))\varepsilon_{mi\alpha}\varepsilon_{mj\beta}^*
\end{equation}
$ M_i $ is the atomic mass of the i-th atom, $ \omega_m $ is the phonon frequency of the $ m $ mode, $ \varepsilon_m $ is the corresponding eigenvector (called polarization vector and describes the direction in which the ions move) and $ n_B $ is the Planck distribution at temperature T:
\begin{equation}
  n_B=\frac{1}{e^{\frac{\hbar \omega_m}{kT}}-1}
\end{equation}
is a function that tells me the average number of phonons as a function of the temperature $ T $, the frequency $ \omega_m $, the constants $ \hbar $ and $ k $ which are the reduced Planck constant and the Boltzmann constant respectively.
The Planck distribution is a special case of the Bose-Einstein distribution, i.e. the one in which $ \mu $ = 0 (zero chemical potential).
In fact, the Bose-Einstein distribution has the form:
\begin{equation}
  n_B=\frac{1}{e^{\frac{\varepsilon-\mu}{kT}}-1}
\end{equation}
Let's see the two main steps in which we can divide the method: \newline
1)The $\Sigma$ matrix is calculated from the frequencies of the phonons and the eigenvectors (which frequencies and which initial eigenvectors do I choose at the first cycle to start the procedure?) and is used to generate a random set of atomic shifts $\{u_{i\alpha}^{(n)}\}$ with the subscript $i$ indicating the atoms and the subscript $\alpha$ indicating the Cartesian direction in the supercell (is the supercell a non-primitive cell?) and the superscript $n$ indicating each random configuration calculated.
(How exactly are $ \{u_{i \alpha}\} $ atomic shift sets generated?)
2)The forces $ \{f_{i \alpha}^{(n)} \} $ (calculated by DFT) acting on each atom of the supercell, generated by the displacement sets $ \{u_{i \alpha}^{(n)}\} $, are used to fit the interatomic force constants of a potential model, using least squares minimization.
You take the quantity:
\begin{equation}
   S=\sum_{k,n}(s_k^{(n)})^2
 \end{equation}
where $ s_k ^{(n)} = f_k^{(n)} + \sum_l \Phi_{kl} u_l^{(n)} $.
To implement least squares minimization the condition must be satisfied:
\begin{equation}
  \frac{\partial S}{\partial \Phi_{ij}}=0
\end{equation}
for each $i,j$
\begin{equation}
  \sum_n(f_i^{(n)}-{f_{H_{hi}}^{new}}^{(n)})u_j^{(n)}=0
\end{equation}
$f_{H_{hk}^{new}}^{(n)}=-\sum_l\Phi_{kl}u_l^{(n)}$
is the harmonic force.
Within the limit of an infinite number of configurations we have for all i, j:
\begin{equation}
  \int_{{R}^{3N}}[(f_i^{(n)}-{f_{H_{hi}^{new}}^{(n)}})]\rho_{H_h}(u)du=0
\end{equation}
When self-consistency is reached, equality holds: $ H ^ {new} = H $. (Is $ H $ the real Hamiltonian of the system that I am approximating with a harmonic Hamiltonian through a self-consistent process? Did I understood the philosophy of the method? How do I know $ H $? Since I put $ H ^ {new} = H $ it is obvious that I need to know $ H $ in order to determine when this equality is satisfied!)
If we include the third order force constants in the least squares equation, the self-consistent solution does not change:
\begin{equation}
  s_k^{(n)}=f_k^{(n)}+\sum_l\Phi_{kl}u_l^{(n)}+\sum_{lm}\Psi_{klm}u_l^{(n)}u_m^{(n)}
\end{equation}
By placing the condition:
\begin{equation}
  \frac{\partial S}{\partial \Phi_{ij}}=0
\end{equation}
you get:
\begin{equation}
  \sum_n(f_i^{(n)}-{f_{H_{hi}^{new}}^{(n)}})u_j^{(n)}+\sum_{lmn}\Psi_{ilm}u_l^{(n)}u_m^{(n)}u_j^{(n)}=0
\end{equation}
In the limit of an infinite number of configurations we have:
\begin{equation}
 \int_{{R}^{3N}}(u_l^{(n)}u_m^{(n)}u_j^{(n)})\rho_{H_h}(u)du=0
\end{equation}
By placing instead:
\begin{equation}
  \frac{\partial S}{\partial \Psi_{hij}}=0
\end{equation}
you get:
\begin{equation}
  \sum_n(f_i^{(n)}+\sum_{lmh}\Psi_{hml}u_l^{(n)}u_m^{(n)})u_i^{(n)}u_j^{(n)}+\sum_{ln}\Phi_{hl}u_l^{(n)}u_i^{(n)}u_j^{(n)}=0
\end{equation}
In the limit of an infinite number of configurations we have:
\begin{equation}
  \int_{R^{3N}}[(f_i^{(n)}-{f_{H_{hh}^{new}}^{(n)3rd}})u_i^{(n)}u_j^{(n)}]\rho_{H_h}(u)du=0
\end{equation}
(If I am approximating my system with a harmonic Hamiltonian ($ H_h $), where do I get the $ \Psi $ matrix of the third order constants with which I can start the minimization procedure above?)
(Sorry if the questions may seem silly, but I have never approached lattice dynamics before (I have only the main bases). I understood that in the model proposed here the potential in the harmonic hypothesis is developed in the Taylor series in which the oscillations are small around the equilibrium positions of the atoms with respect to their interatomic distances. From this supposition the self-consistent method starts with which we try to describe the anharmonic behavior of the system of interest, finding an always better estimate at each cycle of the force constants per second and third order. From what I understand this approach allows to calculate the anharmonic part of the potential even when this is large, which is not possible with the perturbation theory which requires that the anharmonic part be small, that is a perturbation of the unperturbed system .)
If anyone could, even in a small part, answer my questions I will be very grateful. I would also like to know where are the errors in this scheme that I propose hoping that there are not too many and that there is something meaningful. Thank you!

Full LaTeX syntax isn't supported here, only a subset of it (essentially the maths parts). You can probably see that your preamble isn't doing much.

Can you narrow down what is your question?

I'd like to know if it is correct to say that to model the system I use a harmonic Hamiltonian i.e. an expandable potential in Taylor series.
Solving the least squares problem I find the force constants.
This is similar to the SSCHA method in which the variational principle is applied: the free energy gradient = 0 is set and minimized with respect to the harmonic Hamiltonian $ H_h $, i.e. with respect to the force constants.
So I find the second and third order force constants that make H_h = H.

I also wanted to know by which equation I find the displacement vectors $ u_ {i \ alpha} ^ {(n)} $ using the statistical probability $ \rho_{H_h} $.

I've taken the liberty of removing the incompatible parts of the latex document attempting to preserve the original as much as possible. Is this addressed to CSE community or is it to someone else (The end with parantheses)?

Actually at first I wanted to put my personal questions in brackets about the parts of the method that were not clear to me, then in the end I used them in a confusing way ... they have no other meaning


text:
What compounds are produced when you heat CuO and sugar?
I'm wondering on what compounds are produced when you heat CuO and sugar?

Under what conditions? In solution or just heat the solids together?

Brown goo, then black goo. And no, it is not going to be much more specific than that.

Any own ideas? And a context?

If there is enough copper oxide, all you obtain is metallic copper, $\ce{CO2}$ and water. This mixture ($\ce{CuO}$ + organic substance) was the standard way of analyzing any new organic substance in the $19$th century. The formula of the compound was derived from the weight of $\ce{CO2}$ and $\ce{H2O}$ collected. This is is still a standard question to be solved in high school chemistry courses all over the world.

That is, if there is enough copper oxide and if you heat well enough.

@ Ivan Neretin. OK. I have edited your comment.

Justus Liebig's (1803-1873) development of microanalysis: https://www.chemistryworld.com/opinion/classic-kit-liebigs-kaliapparat/3004924.article


text:
Quantum calculation of free energy change in neat conditions
I am slightly baffled by a seemingly simple situation: I want to calculate the free energy of association between a molecule $\ce{A}$ and the solvent $\ce{B}$ using quantum methods (e.g. DFT).
From what I've understood, the basic way to do this in most situation is to optimize the geometry of each compound, then perform frequency calculations to obtain free energy corrections to the SCF. Let the SCF plus the corrections for specie $\ce{X}$ be $\ce{G_X}$. This yields the free energy of each compound at 1 atm, which can be converted into the free energy at 1 M via a factor of 1.89 kcal/mol. Thus, for the complexation $\ce{A + B \rightleftharpoons AB}$, the free energy of complexation would be
\begin{align}
\Delta G_{comp}^0 &= (G_{AB}^0 + \pu{1.89 kcal/mol}) - (G_{A}^0 + \pu{1.89 kcal/mol}) - (G_{B}^0 + \pu{1.89 kcal/mol}) \\
&= G_{AB}^0 - G_{A}^0 - G_{B}^0 - \pu{1.89 kcal/mol}
\end{align}
However, in the case of association between molecule $\ce{A}$ with concentration $\ce{C_A}$ and solvent $\ce{B}$ with concentration $\ce{C_B}$, how could one obtain the free energy of complexation?
There is the equation $\Delta G = \Delta G^0 - RT \ln\left(\frac{C_{AB}}{C_A*C_B}\right)$, but the "initial" concentration of the adduct is null. There seems to be a $RT \ln(C_B)$ term floating around in the literature, but I'm not too sure where it's coming from and how to handle this situation rigorously.

One needs to distinguish between the standard free energy (of reaction) for a process, $\Delta _r G^\circ$, and the free energy in general, $\Delta _r G$. You show how to use quantum methods to compute $\Delta _r G^\circ$. This tells you how much the free energy changes if you take 1 mole of each reagent at 1 M concentration and generate one mole of complex at 1 M concentration. At other concentrations one uses
$$\Delta_r G = \Delta_r G^0 + RT \ln\left(\frac{C_{AB}/C^\circ}{(C_A/C^\circ )(C_B/C^\circ)}\right) \tag{Eq. 1}$$
for the reaction $\ce{A +B<=>AB}$ (assuming all components behave ideally; $\pu{C^\circ= 1 M}$ is the standard concentration). In fact, you use an analogous expression to compute the free energy difference between 1 atm (at $\pu{0^\circ C}$) and 1 M concentrations for each reagent or product:
$$\begin{align}\Delta_r G(\pu{1 atm}\rightarrow\pu{1 M}) &= RT \ln\left(\frac{C_X(\pu{1 M})}{C_X(\pu{1 atm})}\right)\\ &= RT \ln\left(\frac{\pu{1 M}}{(\pu{1e-3 \pu{L/m^3}})\cdot(1/(RT/\pu{1 atm})}\right)\end{align} $$
You use analogous relations (usually in the form of Eq. (1)) if you intend to compute the free energy change for one mole of complex formation when reagents and products are at particular concentrations (information which you then have to provide).
If you are interested in computing the equilibrium concentration of complex given specified concentrations of A and B (assumed constant) then you solve for $C_{AB}$ in Eq. (1) after plugging in values for $C_A$ and $C_B$, remembering again that you are assuming your solution behaves ideally.


text:
How can an element have two outermost shells incomplete?
As I read my book, I saw its written that the Transition metals have two outermost shells incomplete.
But , as I had read earlier, that shells are filled in stepwise manner.
So how is this possible?
And why don't the outermost electrons then don't jump in the inner shells?

This occurs because the shells follow Hund's rule. Notice that  shells are filled in order of the letter's capacity per number, not in the order of numbers. This is why the heavier elements may have one or two shells incomplete as their valency is shared between multiple numbered shells.

Because "shells" are a simplification, and become a supersimplification with transition metals.

I'm afraid this is really too broad. Hundreds of questions here would have to be compiled to provide a satisfactory answer. I recommend progressing a bit further in your course, which would not explain everything but would make them less confusing.


text:
Book Recommendation for Molecular Orbital Theory
I posting my question here since I have seen similar questions but I haven't found the specific answer I am searching for. I am seeking textbook recommendations on Molecular Orbital Theory with comprehensive sections and explanations on diatomic molecules. I want to learn how to construct the molecular orbital energy-level diagrams and to write the electronic configuration for diatomic molecules.
I have seen several books and papers such as Mulliken's papers, Spectra of Diatomic Molecules book by Herzberg, Molecular Quantum Mechanics by Atkins, and other modern books, etc but unfortunately, I haven't found sections that treat other than simple diatomic molecules. To be more specific, homonuclear diatomic molecules and Hydrides (i.e. LiH, BH, CH, ...) in the case of heteronuclear diatomic molecules. So I am looking for books that treat molecular orbital theory more than simple diatomic molecules (moderate and heavy heteronuclear diatomic molecules) and I haven't found any!
Any help in this regard is much appreciated.

Did you try this: Resources for learning Chemistry?

Orbital interactions in chemistry by Albright Burnett and whangbo

The whole business becomes messy quite quickly. MO schemes/diagrams become very hard to construct for anything beyond two participants. Your best bet, if you want to learn more than the qualitative stoff organic chemists love so much, is to learn quantum chemistry. CO is already a real head scratcher, have a go at these: How can the dipole moment of carbon monoxide be rationalised by molecular orbital theory? How to rationalise with MO theory that CO is a two-electron donor through carbon?

If you are interested in an applications, look at MO theory in the context or organic reactions, which is a part of chemistry where the theory works well.  Check out Fleming's Frontier Orbitals and Organic Chemical Reactions:
https://onlinelibrary.wiley.com/doi/book/10.1002/9780470689493 .


text:
Terminologies used in isomerism
What is the difference between the terms constitutional isomers, stereoisomers, geometric isomers?
I know cis and trans-1,2-dichloroethane shows cis trans isomerism and 1,1 dichloroethene, 1,1,2 trichloroethene do not show cis trans isomerism rather show stereoisomerism.
In both the cases the compounds are not superimposable. So what is the difference between cis trans isomerism and stereoisomerism?

You'd better not do that again. Instead, just delete the old question and ask a new one.

@Anusha I guess because OP edited the old question to make a completely new one instead of asking it as a new question.. Also, formatting was poor,

Does this answer your question? Difference between conformational, constitutional, and structural isomers and the same and different molecules

It is preferred if you can post separate questions instead of combining your questions into one. That way, it helps the people answering your question and also others hunting for at least one of your questions. Thanks!

Among different isomer classes, cis- and trans-isomerism falls under the category of diastereomers where two compounds are neither superimposable nor mirror images of each other. I believe the other type of isomerism you're referring to here is structural isomers (constitutional isomers), which are in fact not stereoisomers, but are defined by compounds that have the same molecular formula, but different bonding patterns. Stereoisomers differ from this in that they have the same bonds between the same atoms, but they are arranged differently in space.

In the example given for structural isomers, 2-fluoropropane has the fluorine atom bonded to the 2nd carbon on the parent chain, while 1-fluoropropane has it bound to the 1st. Neither of these compounds have stereoisomers, because they do not have special properties such as chiral centers or double bonds that can create 2 or more forms of the same compound that have differing chemical properties. The examples that you gave in your question all have cis- and trans-isomerism, because the double bond present prevents it from rotating freely, and thus the full accurate model of the compound cannot be fully explained without including a "cis" or "trans" that indicates whether not the groups on either side of the double bond are on the same side or opposing sides.
Hope this clears things up a bit.
Edit: Looks like I made a mistake in reading your examples. 1,2-dichloroethane does not have cis- or trans- (geometric) isomerism, as it doesn't have a double bond that restricts rotation.  1,1-dichloroethene and 1,1,2-trichloroethene do not have isomers either. Although they both have a double bond, there are no two forms of the compounds with the same bonding patterns that are non-superimposable.

1,1 dichloroethene, 1,1,2 trichloroethene dont show cis trans isomerism rather stereoisomerism but what is the difference

@Saniya None of them have isomerism. If you're looking for an explanation of enantiomers, one of the other subclasses of stereoisomers, they are created by swapping any two substituents of a chiral center such that the end result is a mirror image of the original. Conformational isomers, a subclass of diastereomers, do not apply in the case of double bonds as they deal with single bonds.


text:
How or Why does diesel/kerosene have a much higher 'flash point' but lower auto/self-ignition point than gasoline/petrol?
This is a very common question, but I have yet to find a good theoretical answer to both questions (diesel has higher flash point BUT lower spontaneous ignition temp.)

Flash point and autoignition temperature measure different things: gasoline and diesel are optimised for different ignition conditions
First, the standard definitions.
Flash point
Flash point is the lowest temperature at which a liquid can gives off vapor to form an ignitable mixture in air near the surface of the liquid.
Autoignition point
The autoignition temperature or kindling point of a substance is the lowest temperature in which it spontaneously ignites in a normal atmosphere without an external source of ignition...
The first point to make on the differences between diesel and gasoline is that the liquids are designed for different properties in engines.
Petrol/gasoline engines need fuels that vaporise easily to give a mixture that is ignited by a spark. So they want a mixture that is easy to ignite but that does not explode spontaneously in the engine (which is bad for the engine and upsets the control of when the exploding fuel happens). So a low flash point is good but a low autoignition point is bad.
Diesel engines work differently. The explosion is not triggered by a spark but by compression of the fuel mixture which heats it and leads to a spontaneous explosion in the mixture driven by the heat of compression. So a diesel engine wants a specific autoignition temperature and one that is low enough to create reliable fuel ignition.
So the properties of gasoline and diesel are there by design as that is how the engines they feed are meant to work.
The chemical properties that create these properties are, broadly, different mixtures of hydrocarbons. Gasoline tends to be more volatile and contains far more unsaturation (aromatics and branched hydrocarbons). Volatility is necessary so the mixture can be vaporized easily to make a flammable mixture. Diesel tends to have many more straight chain hydrocarbons which are more susceptible to attack by oxygens in the air. Diesel also tends to be much less volatile than gasoline as volatility is much less important to creating an effective mixture in the engine. This has the side effect of increasing the flash point for diesel.
So design intent and chemical properties make the metrics different.


text:
Are food calorie values really integers?
According to what I've seen, fat has a listed caloric value of 9 Cal/g, while carbs and protein have listed caloric values of 4 Cal/g.* Are these numbers exact, or have they been rounded? And if they have been rounded what are the exact numbers? One hit said that the numbers are rounded, but not by how much.
*These are "food calories" (aka "large calories" or "Calories" with a capital "C").
1 food calorie = 1000 "small calories" or "gram calories" = 1 kcal = 4.184 kJ
See: https://en.wikipedia.org/wiki/Calorie

I have burnt in my memory from deep past rather 4.3 kcal/g for "carbs" and 9.1 kcal/g for fats. So yourvalues are definitely rouded up, covering the value span across particular food components.

They're not exact numbers. These numbers aren't exact for three reasons:

Each type of carb, protein, and fat has a different caloric value.  These are overall averages for each class.

Even if you were dealing with a single pure compound, the value couldn't be exact because there is individual variation in how much of that compound is metabolized based on digestion, absorption, etc.

Even if #1 and #2 were not issues, the number still couldn't be exact, because it is a measured value, and measured values are never exact.  Only counted numbers, and values that are exact by defintion (e.g., the speed of light in SI units) are exact.

To get a better idea of how these values are determined, and the issues and uncertainties associated with measuring them, I'd recommend reading through:
Food and Agriculture Organization of the United Nations. FAO food and nutrition paper. Chapter 3:  Calculation of the energy content of foods - energy conversion factors. Food and Agriculture Organization of the United Nations, 1979.
Here is a direct link to chapter 3:
http://www.fao.org/3/Y5022E/y5022e04.htm

Are you saying that, for example, saturated fat may have a different caloric value to unsaturated fat, and that refined carbs may have a different caloric value to complex carbs, etc?

Not only that, but a saturated triglyceride containing three 16:0 fatty acid chains will have a (slightly) different caloric value per gram than one containing, say, two 16:0 chains and one 14:0 chain.  And, likewise, different carbs will have different caloric values.  Whether such differences, particularly if they are small, are nutritionally significant, is an entirely different question.  But you weren't asking about nutritionally significant differences, you were asking if the numbers are exact.

Thx for you input. I edited the question heading, after realizing that it missed the mark slightly.

What I am interested in is if the average between the different sub-classes of fat was rounded up, or down, to 9, and with how much?

I added a link that should help give you a picture of the situation.

@ToddMinehardt  Nope, I'm not off by a factor of 1000. First, it's understood from context that these are nutritional Calories (=1000 calories).  The OP's mistake was just in not capitalizing the "C". Second, I never specified either "calorie" or "Calorie" in my answer. Third, and most importantly, anyone who carefully thinks this through will see that your "gotcha" is what's immaterial since, logically, his questions—about whether these values are exact, and whether there are within-class differences between these values—are independent of how the unit itself is scaled.

My question is a follow-on to another question I recently posted on the Biology site. I would be very grateful if you could have a look at it and tell me what your thoughts are. Unfortunately the question is closed so one can not answer it, but a little thoughtful comment here or there would be much appreciated.

@Constantthin  The effects of low-fat vs low-carb diets is a very complicated and unresolved issue in the field of nutrition.  I took a quick look through the primary source (https://www.cell.com/cell-metabolism/fulltext/S1550-4131(15)00350-2) cited in the popular press article you linked and, IME, to the extent they are seeing differences it is not because they used an incorrect figure for the caloric value of fat.  If you read the primary source you can see there are many metabolic differences created by the different diets, and that's much more likely to be the explanation.

Whatever the explanation is for this phenomenon, would you agree that one ought to inflate the number of consumed fat-calories when counting calories to lose weight? Like with 10%-20%? Thus, should one make 1000 fat-calories to read 1100 or 1200?

@Constantthin: What's the point of counting calories if you're just going to make arbitrary adjustments? That's not counting, that's guessing badly on purpose.

@Constantthin  Simply put, no.  I don't know how many studies have been done on the effect of diet compositon on weight loss, but it's enormous—when I entered "diet composition and weight loss" in Google Scholar I got 2,520,000 results.  And their findings are all over the map.  So here you want to take the result from just one study and, based on this, adjust the current consenus figures for the caloric value of fats vs carbs. The problem is that you could probably find another study with the opposite result; so do you adjust these again, in the opposite direction, after reading that one?

I have just read through the original study-notes and concluded that although a low-fat diet leads to fat-loss, it doesn’t necessarily lead to weight-loss. Because on the opposite low-carb diet muscle-mass is lost, which leads to weight-loss. So there is a difference between fat-loss and weight-loss., apparently.

Yes. Now you're starting to see the some of the complexity.  Weight change is due to changes in fat, muscle, and water.  Anytime you use up carbs you temporarily lose a lot of weight, because while the body can store nearly pure fat, it can only store carbs as a combination of about 25% carbs and 75% water.  So a lot of weight loss at the start of a diet is loss of water due to loss of carbs. Conversely, with aerobic training, weight will increase (in a good way) because of increased blood volume. And with weight training, weight will  increase (in a good way) because of increased muscle mass.

This is completely out of the realm of chemistry, but the best advice I've heard is the best diet is the one that doesn't feel like a diet, i.e., the one on which you feel the best and the one that you can stick to.  If you've got that sorted, the particular dietary compositon is of secondary importance.

@theorist to be fair, this convention (Cal = 1000 cal) is absolutely horrible, and the interpretation of the order of magnitude shouldn't depend on the context. If the question is wrong, it should be edited, especially now that it's on HNQ. Finally, it does seem relevant to know if a value is ~4 or ~4000 when asking if it's an exact integer. The exact value could be 3998, as an example. Good answer, BTW, I'm only referring to the above comment.

@EricDuminil  (1) Agreed it should be edited; just did so.  (2) Agreed, one convention would be better.  But both are in such widespread use, there's nothing to be done.  E.g., to express everything as cal or J, you'd need to re-do every nutrition label in the US. And since they are in such widespread use, most are aware of both and can thus adjust based on context. (3) My comment was in reply to a comment by another poster, which he's since deleted (so you can't see the context), in which he insisted the fact that the scaling wasn't properly specified made both the question,and my answer,....

... "immaterial".  That's not just unfair, it's false. That's why I replied as I did. (4) The choice of scaling really isn't relevant to the question of whether a number is exact.  That turns purely on the inherent nature of the value itself, and is entirely independent of the scaling factor used for the associated unit. All the integers you mentioned: (4, 4000, and 3998)—and, indeed, any integer value—could be either exact or inexact.....

... Ask yourself this: Is there any way—any way at all—in which my answer would need to be different if the questioner expressed those values as cal instead Cal? [I.e., if he instead referenced 9000 cal/g for fats, and 4000 cal/g for carbs and proteins.] The answer is clearly no, demonstrating the compete irrelevance of the scaling factor to the question at hand. (5) Thanks for the positive comment on my answer :).

It doesn't "take one calorie to burn", you gain one calorie by burning.
Now noting obviously what everybody else has said about selection of the type of fat affecting the precise amount of energy available from one gram (which in practice will be measured in a calorimeter), but what this is really saying is that by metabolising some specified fraction of a gram of fuel you can do (approximately) one calorie of work, e.g. raising the temperature of water by some specified amount or moving your muscles in such a way that they lift a weight by some specified number of centimeters.
So no, it's not an integer: it's an approximation to the amount of work you'd have to do to prevent the food you've just eaten being stored as flab, and to get an accurate figure you'd have to do an assay of an identical portion of that food taking into account both its chemical composition and its digestibility (e.g. an increased amount of fibre in the specimen might not change the result of the physical calorimetric test, but might change the availability of the nutrients to the digestive system).


text:
how to remove acetone free nail polish remover from a thinkpad laptop keyboard?
I had dried spraypaint on my laptop keyboard and I tried my oxx studio brand nail polish wipes
(ingredients; propylene glycol, dimethyl adipate, dimethyl glutarate, dimethyl succinate, caprylic/capric triglyceride, parfum, propylene, glycol propyl ether, phenoxyethanol, benzyl alcohol, ethylhexylglycerine)
the wipes did get off the spray paint but it left the keys being sticky...
How do I get rid of the stickiness?


text:
Why can't Grignard reagents react like Organolithium does (with acids)?
I'd read yesterday that Grignard reagents can't give ketones on reaction with carboxylic acids. It is because of the fact that Grignard reagent is a "super-base", and it will deprotonate the acid to form a carboxylate salt, rather than doing a $\ce{Nu-}$ attack.
However, after watching a video on Youtube, on the reaction of organolithium with acids (example reaction below),

I wondered,

After the acid-base reaction, why couldn't Grignard reagent (when taken in excess) do a $\ce{Nu-}$ attack on the carboxylate salt, just like organolithium does and then form ketone? Just like it does with an ester, acid chloride, etc.

Does this answer your question? Why are organolithium and organomagnesium compounds nucleophilic in nature?

@Safdar, No. I couldn't think how a difference in Nu nature can prevent the grignard reagent from attacking the carboxylate salt?

After formation of salt, the $\ce{O-}$ makes carbon way less electrophilic due to the distribution of electron density, so we need a much better nucleophile.

Two ideas come to mind for how a Grignard reagent could differ from an organolithium compound. Both are related to the bulkier nature of the Grignard molecule.

Magnesium forms a more ionic bond than lithium.

In this hypothesis, the reason the organolithium compound attacks the carboxylate is that we have significant covalent bonding between one oxygen and the lithium, so we don't really have an ion: the carbonyl function is preserved with the non-lithiated oxygen. If the $\ce{MgX^+}$ moiety from the Grignard reagent were bonded more ionically, then the Grignard substrate would behave more like a true carboxylate ion with poor electrophilicity. The larger size of the Grignard fragment (including the halogen atom) combined with only one net positive charge (rather than two as on a bare magnesium ion) could promote such greater ionic character.
The diagram below shows how the extra atom could promote more ionic character in the Grignard product (using bromine as the halogen). Assuming magnesium bonds covalently primarily with its $3s$ orbital, that orbital forms a three-center four-electron bond with the more electronegative neighbors, which gives an ionic contribution to each individual linkage. With the corresponding lithium compound a similar delocalization does not occur because the lithium is bonded only to a single atom.

Note that a similar delocalized bonding model can be rendered for the Grignard reagent itself, thus enhancing the ionic character of the Grignard reagent's carbon-magnesium bond.

The ketone precursor with a Grignard reagent is sterically hindered.

When an organolithium compound reacts, two $\ce{OLi}$ groups become attached to the original carboxylate carbon and these are later hydrolyzed to a carbonyl oxygen. With a Grignard reagent, two bulkier $\ce{OMgX}$ groups would have to be attached to form the ketone precursor, creating more possibility for unfavorable steric conditions.


text:
What has happened with the pH-meter?
The glass sensor has became red inside like something melted. I stored it in a liquid with pH that mentioned in the manual. What is that and why could that happen, how often that happens with pH-meters?
(10 bucks pH-meter from China)

Be more precise ! What is the meaning of "red like something melted" ? Which "liquid" did you use to store your electrode ? How long did you store it ? What is the "pH mentioned in the manual" ? What goes wrong, the pH-meter or the glas sensor ?

@Maurice some reddish substance appeared on the inner of the glass sensor (pretty sure it wasn't like that). I used liquid with pH=4, I stored it for about couple of weeks. The meter can't be calibrated anymore, always the same value.

@Buttonwood I uploaded images.

This is still not precise enough, what is the liquid you used? You should disassemble the electrode and clean it from the inside if possible. These glass electrodes sometimes have built-in salt bridges maybe it got contaminated there, but based on the picture the damage is more likely due to the long soaking. Probably something absorbed on its surface (glass membrane on the bottom), hindering its functionality. It is most likely you have to buy a new one, this one seems hopeless. You can try soaking it in dilute mineral acids. Also these liquids are for calibration not for storage.

It may be colloidal silver, coming from the Ag/AgCl electrode. Who knows ? Does it modify the performance of the pH-meter ? If it does not, why don't you ignore this trouble ?

@Maurice forgot to say, it does affect the performance, as you can see on the display it's always shows 1 since even if I turn calibration screw.

@Andrew Kovács I left it in one of the liquids used for calibration as it has pH=4 (distilled water with calibrating powder), because in the manual it is said to store the device in a liquid with pH=4.

There must be something that you do not mention. The display simply cannot always show $1$.

@Maurice really, how so?

I've never heard of storing a pH meter in a pH 4 solution, but I haven't used that many. I've always stored the pH probe in a KCl solution.

@user137 what is pH of your KCl solution, why KCl?

I assume the pH is neutral. The concentration is 4 M KCl. That's always been what the manufacturer suggested.

@user137 I've just found a mentioning that instead of KCl a solution with pH=4 can be used, the description under the 3rd link: https://duckduckgo.com/?t=canonical&q=store+ph-meter+in+distelled+water&ia=web

@Andrew Kovács I added a new picture of disassembled device, is it salt bridges you mentioned ? Look like those red stuff spilled from the upper part, because there is yellow trail going from the top.


text:
Oxidation of ethanol upon heating in a test tube

The diagram shows an experimental set-up which can be used in several different experiments.

Which processes could be demonstrate by using the above apparatus?
1. Oxidation of ethanol (liquid X)
2. Dehydration of ethanol (liquid X)
3. Cracking of paraffin (liquid X)

The answer is 2 and 3. I understand that for 2, ethene is produced which is a gas and for 3, cracking of paraffin will produce short-chained hydrocarbons which have low boiling points.
However, I expected 1 to be an answer as well because oxidation of ethanol will produce ethanal which has a boiling point of 21 °C. So shouldn't 1 be an answer as well?

An oxidant is needed if you want to oxidize ethanol into ethanal. And there is no oxidant in this picture.

@Maurice wouldn't you need a catalyst for the dehydration of ethanol as well (e.g. concentrated sulfuric acid)?

Silica is an acidic surface

@Waylander Oh, I see... and alumina also acts as a catalyst for the dehydration of alcohols. My bad for assuming that oxidants were added. Thank you


text:
Comparing nucleophilicity of oxygen and nitrogen in acetylation of 4-aminophenol

Here, the answer has $\ce{N}$ acting as a nucleophile making the answer (C).
I thought that $\ce{O}$ would be more nucleophilic, considering the fact that $\ce{O}$ has two lone pairs, one in conjugation with the ring, the other more readily available, compared to $\ce{N}$, where its only lone pair is in conjugation with the ring.
Why is nitrogen more nucleophilic?

How good an acylating agent is acetoxyphenol?

I guess it's because of the fact that –NH2 is a better EDG than –OH, hence it would experience a lower e-density. However, I'm not sure.


text:
How does the surface tension of water decrease with an increase in the concentration of NaCl?
Can anyone help me with the theory of surface tension? My chemistry experiment is related to the surface tension and concentration of NaCl. I am confused about my hypothesis. On the internet, I found that increasing salt concentration also increases the surface tension of the water. But in my experiment, surface tension is decreasing with the increase of NaCl. So can anyone help me with the theory behind this?
With the help of needle and water drops, I was able to calculate the value of surface tension of water for each concentration of NaCl.

Suspect contamination of your equipment or the salt. Even a tiny amount of detergent, for example can greatly decrease surface tension. https://www.youtube.com/watch?v=jHWuJXYsYag

The surface excess of a salt (NaCl) is negative because salts prefer to be hydrated.  So there is less salt at the inferface than in the bulk phase.  The Gibbs adsorption equation says the change is surface tension is the opposite of the surface excess - so for a salt solution, the change is positive and the surface tension increases.

Since adding NaCl should increase the boiling point it should increase the surface tension.
This has been explained previously by others here:
How does NaCl (or any inorganic salt) increase surface tension?


text:
Comparing reactivity of 1-chloroethane and 1-chloropropane in an SN1 reaction
Beginning with a reference to a well known question.

Order by reactivity towards SN1 reaction.

According to some facts about SN1 reactions, rate depends on:

Stability of carbocations
Better leaving groups
Electron donation

If we consider point $1$, this rounds down the above question to a previously asked question.

Unfortunately the second answer to the above question raises doubt about the validity of the first in all cases. (The kinetic and thermodynamic stability)
Since my question is a bit different and asks about SN1 reaction I would propose considering the following:

The number of $\alpha$ hydrogen in the first compound is one more than the second.
The inductive effect of the ethyl group on the carbocation formed at the second compound is more than at the first.
Using hyperconjugation mechanics, we say that when the $\ce{H}$ atom is aligned with the Carbocation's $\mathrm p$-orbital, proper donation of electron takes place.

(So even though both compounds have unequal number of $\alpha$ hydrogen, the stability effect would be same?)
Since the hyperconjugative effect failed me, I would resort to orbital analysis and induction. An explanation of this point is given below.

Explanation for point $4$:
Since the C+ structure resemble the following :

We can say that the $\%s$ character of $\ce{^+C-C}$ bond in both compounds will increase. In turn the effective p character along $\ce{C-C+}$ would increase. This would imply that the $\%\mathrm s$ character of $\ce{C-H}$ bonds and $\ce{C-C}$ bond (in case of the second compound) will increase. We know that the size of carbon would be $\equiv$ to itself, so increasing the $\%\mathrm s$ more along $\ce{C-C}$ would be a great idea. This would imply that the $\mathrm p$ character of $\ce{C-H}$ bonds would be greater in the case of the second compound relative to the first, implying more donation effect.

So, to sum up my question, considering the above points can I say that the rate of reactivity towards SN1 will be greater for the second compound, i.e. $\ce{C3H7Cl}$?
Opposition or Alliance if also supported by experimental evidence would be more helpful.

Additional recurrence in the form of questions in tests or books:

A variation of the above question has also been asked in a national level entrance exam, JEE MAINS 2020, 4 September, Second shift. (Only instead of specifying Sn1, there was a reagent namely $\ce{HI}$).

The question is as follows, the objective being to find product B:

GRB Organic chemistry by Himanshu Pandey, Chapter: Halides, question number 66 to 76

I'm afraid you just wasted 100 rep. Why? First, that's another question hitting the wall of usability of various "rules of thumb" like inductive effect. Second, that's hardly experimentally verifiable because these compounds react via SN2 not SN1.

@Mithoron Firstly, I never said to use the thumb rule! (perhaps not clear, but now it is) Secondly, I said experimental evidence is appreciated ...that doesn't mean it is necessary for a post to qualify as the answer, third, I would've made it even greater if I had the rep. Also, why do you say these compounds react solely via SN2

Well, one important thing is SN isn't, in general, purely SN1 or SN2 and there's lots of things influencing that. If you conducted the reaction in such ways that it was more of of SN1, results would still vary depending on case. BTW, what is the meaning of last sentence in this post?

@Mithoron yes I am aware that it won't be a pure SN1. But, how would the result vary depending on the case, both are  apparently small. Last sentence in the post about "alliance and opposition" is in reference to my way of the justification using point 4. If it were a question about C+ stability, what would you have said(which is more stable)?

Did you looked at the solution, it says it'll proceed through SN2 pathway. It is due to the fact that degree of initial carbocations formed by cleavage of HI are comparable.

Okay, I got my fault. The second step is indeed dehydration. But, still I would say that it is unrelated to your question, as the first step happens through SN2 pathway, and not SN1.

@RahulVerma Please do not tamper questions without a rigid proof. see here

Hey, I didn't intent to tamper your question. Instead, I was trying to improve it by removing the unrelated ref. That referred question is indeed cleavage by HI, but the pathway is SN2. Also, the link you mentioned vague too (for the question).

To make things clear, 1) I'm following MOC for a long time. 2) I'd said that "pathway is SN2" specifically for the referred question, not in general. Finally, arguing is also a part of science, don't feel frustrated, take a deep breath and enjoy CSX ;) Yo!

I am not sure I understand this correctly, but is the reason your question for below 3. the idea that at a time only one H can be donating its electrons using hyperconjugation? If it's the case, the problem should be resolved considering molecules have so much energy,rapid rotation is always occurring about these c-c bonds, the ch3 is like a ceiling fan whose α H spend small amounts of time one at a time helping, so having 2 α H means less time receiving electrons so lesser stability. Having an extra α H is so beneficial, that the orbital character ideas you mention pale before it.

Consider the following assumptions to somehow force these substrates to proceed for $\ce{S_N1}$ reaction mechanism:

Low temperature

Polar protic solvent

Low concentration of nucleophile w.r.t. reactant

Now, in an $\ce{S_N1}$ reaction, the formation of carbocation is must.
The primary factor that determines the reactivity of organic substrates in an $\ce{S_N1}$
reaction is the relative stability of the carbocation that is formed.$\ce{^1}$
According to your linked question: 1-propyl cation would likely be more stable than the ethyl cation.
Hence, your deduction that "the rate of reactivity towards $\ce{S_N1}$ will be greater for the second compound, i.e. $\ce{C3H7Cl}$" is correct.
P.S.: The 1-propyl cation can rearrange itself to a more stable 2$\pu{^o}$ carbocation, but that is not part of RDS to decide the reactivity order.
Reference

Solomons, Fryhle, Snyder, 12E, Page 264

You avoided all of my question and bounty description. I specifically asked about checking the 4. th step, and also, the second answer on that link contradicts your claim thermodynamically. How did you check the relative stability, that is the question.

@IamAlita (1) I concentrated on your question, not what asked in JEE paper recently. (2) The answers on the link compares purely the stability of carbocations with no consideration of any reaction. And propylium is more stable than ethylium, thermodynamically. But here, we cannot consider possible rearrangements of carbocation (as mentioned in 'P.S.'). (3) In your 4th point, you are unnecessary making a simple thing complicated. (4) This is not a practical question and theoretically also won't lead to  anything.

Did you check if my reasoning applies to other questions as well? Of course you didnt considering the last line of the above comments.If you'd please, check GRB Advanced problems in Organic Chemistry(main and advanced), --> Halides, question 66 to 76. quoting propylium is more stable than ethylium, thermodynamically , how would you conclude such a thing in a competitive exam or a series of 10 questions with least access to such "experimental material" which is not practical?  The P.S. part would be unnecessary because I am not considering rearrangements, just given a race between two ---

Which of the participants will be finished off first. Also, if I had to eat the linked answer raw, I would not have created a question which supplies my logic, either you disprove it using a counter statement or an experiment.


text:
Cell voltage with different n factors
I’m a bit confused regarding cell potentials. If I have two half  reactions each involving a different number of electrons, there are two ways I can see to calculate the net emf of their cell:

Simply add the potentials across the two theoretical half cells, that is, treat each half reaction as a voltage source in series
Calculate the value for the net cell reaction as a whole, that is, make consideration for the energetics of the reaction

As a more specific example, say I have a metal $\text M$, and $$\ce{E^0_{M^{3+}/M}=2V, E^0_{M^{2+}/M}=1V}$$ and would like to figure the standard cell potential, say $x$ of $\ce{E^0_{M^{3+}/M^{2+}}}$.
With Method 1 we have $$x= \ce{{E^0_{M^{3+}/M}}-{E^0_{M^{2+}/M}}}=2-1=\mathbf{1V}$$
With Method 2 we have $$\ce{\Delta G^0_{M^{3+}/M}=-3*F*2=-6F, \Delta G^0_{M^{2+}/M}=-2*F*1=-2F}$$
$$\implies \ce{\Delta G^0_{net} = -6F - (-2F) = -4F}$$
$$\implies x=\frac{-\ce{4F}}{\ce{-1*F}}=\mathbf{4V}$$
How do I rectify these inconsistent values? Does it make any difference here whether I say I am looking for $\ce{E^0_{M^{3+}/M^{2+}}}$ or $\ce{M|M^{2+}(1 molar)||M^{3+}(1 molar)|M}$? What if I use an inert electrode like $\ce{Pt}$ in the latter representation?
Further, if I have another metal $\ce Y$ for which $\ce{E^0_{Y^+/Y}}=\ce{3V}$, which is a stronger oxidizing agent? $\ce{M^{3+}}$ or $\ce Y^+$?
In various sources it seems some use method 1 while others use 2. In this answer on CSE it seems 1 is the right one. But I maybe missing some conceptual context.
Thanks!

Cell potential is not additive but free energy is.. and so the second method is accurate/correct.. Try using method 2 in that reaction and it would be correct as well.

Conceptually related: https://chemistry.stackexchange.com/q/136774/95133

The first method is actually derived from the second one. In case of "different-elements" Galvanic cell, the first method works, but in case of disproportionation, it doesn't.

@RahulVerma Whats the difference in such a cell? It is the same physical process, so why should there be different rules? And does this have any thing to do with the electrodes—I.e. if I use an inert electrode will it behave like a normal galvanic cell ?
And I did figure where the first method comes from, it’s rather the order of operations, if you will, that I find confusing.
@ Safdar But won’t that also give a mathematically different result since in one case it is a simple difference and the other is a weighted average of sorts ?

Take any example of "different elements" cell (say Cu-Ag Galvanic cell), and then use the second method to calculate E(cell). You'll always end up with same values from both methods.

Simply speaking, the first method is wrong. You can't use it.. It just so happens that it works for redox reactions, but it is not general..

@Safdar, I'd say that the first method isn't wrong, it's just specific to "different elements" cell. No?

@RahulVerma It is a specific case, it just happens to work.. You can't use it as a general method and so I would say it is wrong (from a purely conceptual frame).

@RahulVerma Is that really true? By method 2 you get a completely different mathematical expression, in this example ${E^0_{\text{cell}}={(-F\cdot0.8)-(-2F\cdot0.34)\over -2F}=0.06V}$ unless I’ve got Method 2 wrong, but if I have then it works for disproportionation also $\dots$

If I've understood your question properly, you are asking - in the condition $E^\circ_{\ce{M^3+/M^2+}} =\pu{4 V}$, $E^\circ_{\ce{M^3+/M}} =\pu{1 V}$, $E^\circ_{\ce{Y+/Y}} =\pu{3 V}$ - whether $\ce{M^3+}$ is a stronger oxidising agent than $\ce{Y+}$ irrespective of it going to $\ce{M \text{or} M^2+}$.
This would not be true since $\ce{Y+}$ has a higher reducing potential than $\ce{M^3+/M}$ but not $\ce{M^3+/M^2+}$, and so $\ce{M^3+}$ would be a stronger oxidising agent than $\ce{Y+}$ where it converts into $\ce{M^2+}$ but not when $\ce{M^3+}$ converts into $\ce{M}$.

The method $1$ that you suggest is a very specific case that can only be used for a general redox reaction. By this, I mean a reaction as follows for a cell having it cell notation denoted as $[\ce{N(s)|N^{y+}(aq)|| M^{x+}(aq)|M(s)}]$
\begin{array} {rlllc} \require{cancel}
\ce{N^y+ + ye- &-> N } &E^\circ = E_n^\circ &\quad|\times(-x)\\
\ce{ M^x+ +xe- &-> M}  &E^\circ = E_m^\circ &\quad|\times(y)  \\  \hline
\ce{yM^x+ + xN  +\cancel{(xye-)}  &-> yM + xN^y+ + \cancel{(xye-)}} &E^\circ = E_{\text{cell}}^\circ  \\
\end{array}
So here, if you apply method $1$, you get $E^\circ_\text{cell} = E^\circ_m -E^\circ_n$
Let's see whether this works using method $2$ (the free energy varient). Now, we know that $\Delta G$ is additive, $\Delta G^\circ =nFE^\circ$ and that when we multiply the total reaction by a value, we also multiply the $\Delta G$ by the same number. Using these properties we find the $\Delta G$ values for the two reactions.
For just one mole of M being reduced , we get
$$\Delta G_{\text{m}}=xFE^\circ_m \tag{1}$$
For just one mole of N, we get
$$\Delta G_{\text{n}}=yFE^\circ_n \tag{2}$$
Now, according to the above cell reaction, the $\Delta G_{\mathrm {cell}} = y(\Delta G_m) + (-x)\Delta G_n$ which would be equal to
\begin{align}
\Delta G_{\text{cell}} &= (-x)yFE^\circ_n + (y)xFE^\circ_m \\
&= xyF(E^\circ_m -E^\circ_n) \tag{3}
\end{align}
Now, the value of $\Delta G_{\text{cell}}$ in terms of $E^\circ_\text{cell}$, we get:
$$\Delta G_{\text{cell}}=xyFE^\circ_\text{cell} \tag{4}$$
Now substituting this value of $\Delta G_\text{cell}$ into equation (3), we get
\begin{align}
\cancel{(xyF)}E^\circ_\text{cell} &= \cancel{(xyF)}(E^\circ_m -E^\circ_n) \\
\implies E^\circ_\text{cell} &= E^\circ_m -E^\circ_n
\end{align}
For a case where you have the same compound disproportionate, this question and its answers should be enough.


text:
How to change pH of CuSO4?
I'm currently very desperate to know this. I'm doing an experiment about adsorption/chelation of Cu ions in CuSO4 using banana peels using the UV-Visible spectroscopy. How do I change the pH of 0.1M or 0.05M of CuSO4? My main problem is with making the CuSO4 more alkaline/basic or whatever u call it. My base pH of the CuSO4 is about 4.0 or 4.1. I've tried making a pH 5 and pH 6 buffer solution then adding it to the CuSO4 solution, but so far it's only made it more acidic.
These are the calculations:
pH 5
Weigh 5.10575g of potassium hydrogen phthalate(C8H5KO4)
Dissolve it in a 100cm3 beaker
Transfer the solution into a 250cm3 volumetric flask and fill it with distilled water
Measure 5cm3 of 2M NaOH and add it to a 100cm3 volumetric flask and fill it with distilled water
Add 100cm3 of 0.1M potassium hydrogen phthalate and 45.2cm3 of 0.1M NaOH into a 250cm3 beaker
pH 6
Weigh 3.40225g of KH2PO4
Dissolve it in a 100cm3 beaker
Transfer the solution into a 250cm3 volumetric flask and fill it with distilled water
Measure 5cm3 of 2M NaOH and add it to a 100cm3 volumetric flask and fill it with distilled water
Add 100cm3 of 0.1M KH2PO4 and 11.2cm3 of 0.1M NaOH
But I realised the phosphate buffers will cause a precipitate. So I tried using a citric acid + sodium citrate buffer. I think the calculation was something like 88.5cm^3 of citric acid + 11.5 cm^3 of sodium citrate (or it could be the other way around). But I don't think that was a buffer solution, cus when I added the slightest amount of deionized water, the pH increased. I got it from https://www.sigmaaldrich.com/life-science/core-bioreagents/biological-buffers/learning-center/buffer-reference-center.html#citric2.
Someone please help this is experiment is basically graded, so I'm really desperate.
edit: I made the copper sulfate using CuSO4.5H2O if that helps

Copper(II) phosphate is insoluble in water. You could try acetic acid/acetate (pKa = 4.76). pH 6 is very close to neutral i would rather avoid it.

Also this could be useful for you. http://microscopy.berkeley.edu/Resources/instruction/buffers.html

Your major problem is you are using way too high concentration of your $\ce{Cu^2+}$ solution. The maximum concentration of solutions used in the given article is $\pu{0.00787 M}$.

Alas, your problem has no solution, because of the risk of producing $\ce{Cu(OH)2}$ at pH $5$ and $6$. The solubility product of $\ce{Cu(OH)2}$ is $2.2·10^{-20}$. At pH $5$ and $6$, the concentrations of the ion $\ce{OH-}$ are respectively $10^{-9}$ and $10^{-8}$ M. This means that, at pH $5$ and $6$, the maximum molar concentrations of the ion $\ce{Cu^{2+}}$ are $2.2·10^{-2}$ M and $2.2·10^{-4}$ M respectively. So you will never be able to obtain a $0.1$ M solution of $\ce{CuSO4}$ at pH $5$ or $6$.

how about 0.05M?

For 0.05 M, it is the same. 0.05 M is greater than the limit of 0.022 M.

I saw one article or paper that said they did it up till pH 6 then after that they got precipitate. https://www.geomatejournal.com/sites/default/files/articles/227-234-3c-anwar hossain.pdf

also, do u know what's the highest pH I can get for the solution then?

one more thing, can u think of any reason y my pH 6 buffer made the CuSO4 more acidic?

No. I don't know the maximum value of the pH for copper sulfate solutions. The problem is that, if you want to make precise calculations, you have to use activities and not concentrations. All my previous calculations are made with concentrations. They are probably a little wrong. As you know, insoluble products are always more soluble than the values obtained from the solubility product, specially with substances producing doubly charged ions like $\ce{Cu^{2+}}$.

Sorry. I cannot tell you the reason why a buffer does not give its pH value, without knowing the composition of this buffer.

It was citric acid and sodium citrate. Both at 0.1M

Citric acid is a weak acid. Lets call it $\ce{H3X}$. In aqueous solution, it doe not produce many ions $\ce{H+}$. But in presence of copper ions $\ce{Cu^{2+}}$ it reacts producing a complex $\ce{CuX^-}$ and $3$ ions $\ce{H+}$. Three ! It becomes a strong acid. The equation is $$\ce{Cu^{2+} + H3X -> CuX^- + 3 H+}$$

ohhh nice. no wonder my solution became acidic


text:
Why is heroin a more potent drug than morphine, despite having a similar structure?
The structures of heroin and morphine are quite similar, with heroin being formed by acetylation of morphine:

Why is heroin so much more potent than morphine, when their structures are so similar? What biochemical effects lead to heroin's prevalence as a recreational drug, and consequently, the much higher number of fatalities from overdosing on heroin as compared to morphine.
This Source has an attached reference1, which claims the following:

It has been reported that Diamorphine is twice as potent as morphine .
...
We found no significant differences between the groups over a 24-hour period in the degree of euphoria, nausea and vomiting or sedation, as assessed by subjective visual analogue scores, or in pain relief, nausea and euphoria as assessed by verbal rating. There was, in addition, no significant difference in the requirements for antiemetics between the two groups. Both groups demonstrated relatively low nausea scores. The mean amount of morphine demanded was approximately twice that of diamorphine.
...
The scientific basis for the alleged superiority of diamorphine over morphine is dubious and since diamorphine is metabolised to monacetyl morphine, and morphine itself, it would be surprising if there were a marked difference. Very few studies authenticate this claim of superiority and the present study is no exception.

This does not explain the prevalence of heroin compared to morphine. Over 15,000 people died in the US from heroin overdose2, whereas the number for morphine is quite less. Are the reasons behind this purely sociological (easier access to heroin compared to morphine), or is there some chemistry that I am missing out on here?
References:

Robinson, S. L., et al. “Morphine Compared with Diamorphine.: A Comparison of Dose Requirements and Side-Effects after Hip Surgery.” Anaesthesia, vol. 46, no. 7, July 1991, pp. 538–40. doi:10.1111/j.1365-2044.1991.tb09650.x.
CDC Heroin Overdose data. Link.

The answer to this question lies in the pharmacokinetics of these two drugs: the acetyl groups cause Heroin to be 200 times more lipid soluble than Morphine1. Another example of a common drug would be Fentanyl, which is 580 times more lipid soluble than morphine, with a structure very similar to that of Morphine.
Increased lipid solubility allows these drugs to pass more rapidly across the blood-brain barrier into the Central Nervous System, giving them more analgesic potency and a more rapid onset of action than morphine (the peak blood level of heroin when used intravenously is around 5 minutes2, whereas that of fentanyl is 1.7 minutes3).
As for the sociological aspects, the high potency of these drugs acts as a positive feedback loop. The heroin/fentanyl dose is smaller than morphine, which is preferable for drug smugglers, as they have to smuggle less of the drug, reducing their risk. This explains the high potency and usage of heroin as a recreational drug as compared to morphine.
References:

Mahesh Trivedi, Shafee Shaikh, Carl Gwinnut: "Pharmacology of Opioids", Updates in Anaesthesia, 2007, pp. 118-124. Link.
Oelhaf RC, Azadfard M. Heroin Toxicity. In: StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; Jan 2020. Available from: https://www.ncbi.nlm.nih.gov/books/NBK430736/
Fentanyl Citrate Injection, USP. FDA Label.

A nice compilation of information, but there is a simple fact missing. Heroin is a prodrug. It has psychoactive properties itself, and is metabolized to morphine. Thus in equimolar amounts it gives the same as morphine plus its potency before breaking down.

@AndrewKovács While true, this is irrelevant because the active ingredient can't have an effect if it can't get to where it acts. Potency is about speed of action and the amount reaching the site of action.

The only other thing worth mentioning is that the WHO Pain Pyramid of analgesia explicitly has diamorphine as an end point. It is very rarely prescribed; almost always for palliative (end-of-life) care and when literally everything else hasn't worked. Morphine, on the other hand, is incredibly commonly prescribed, not just IV but orally (gel oromorph; liquid solutions) and similar. The vast majority of heroin users are addicts, not after analgesia: their doses, routes, and lack of medical training often leaves them found dead in the community – the difference is more than acetyl groups alone.

Another thing to mention is "codeine", the monoacetyl form of morphine. It is probably the least potent of all opoids though it contains an acetyl group. It is   less potent than morphine(morphine is 3x more potent than codeine) and as such, it is used as a prescribed-only painkiller.

@NilayGhosh Codeine is actually the monomethyl form of morphine, not the monoacetyl.

Yes, extra monomethyl group. Need to brush up my biochem :( . Nevertheless, the three structures are more or less same; having no carbon, 1 extra carbon and 4 extra carbons in the three structures respectively.

"As for the sociological aspects, the high potency of these drugs acts as a positive feedback loop." I get the sense you don't understand what the phrase "feedback loop" means. Potency being involved in a feedback loop means that potency causes something that in turn causes potency.

@AndrewKovács According to Wikipedia's summary of: Inturrisi CE, et al.(1983). "Evidence from opiate binding studies that heroin acts through its metabolites". Life Sciences. 33: 773–6, which is available at https://en.wikipedia.org/wiki/Heroin#Pharmacology, heroin itself has relatively low affinity for the u-opiod receptor. Thus it appears heroin's direct effects are low relative to those of its morphine metabolite.

A minor issue. Fentanyl is structurally very different and completely unrelated to morphine and its derivatives. It has a similar effect on the body, hence it is classed as an opiate despite not being derived from the same chemical skeleton.


text:
Which salt is a surfactant and lowers the surface tension of the water?
I am doing my experiment on the surface tension of the water. How to show the decrease of surface tension of water with the increase in the concentration of salt? Is there any salt that lowers the surface tension of the water?

Related (though it doesn't quite answer your question): https://chemistry.stackexchange.com/questions/49570/does-nacl-reduce-the-surface-tension-of-water/49584#49584

This one answers your question: https://chemistry.stackexchange.com/questions/134948/effect-of-solutes-on-surface-tension/134970#134970. Technically, I guess Sodium Lauryl Sulphate would count as a salt (it's a cation-anion pair).

Soap is a salt as well.

Does this answer your question? Effect of solutes on surface tension

Cetyltrimethylammonium bromide (CTAB) and its cheaper chloro homologue are just two examples of cationic detergents commercially available and frequently used e.g., for hair shampoos.  They are called cationic because beside lipophilic alkyl groups their head group is positively charged:

(source)
Note, there equally are anionic and nonionc detergents, too.
There are multiple approaches to demonstrate their effect to lower the surface tension of water.  Just a qualitative illustration is placing carefully a paper clip on water which will sink after the addition of a detergent:

(source)
If you are up to quantify the effect, the measurement of the contact angle between the liquid and the solid is a frequently used approach.  Again, the Journal of Chemical Education (2017JChemEduc230) published a lab class showcasing the relevance of surface tension e.g., when spraying leaves of lettuce.  A second nice demonstration I recall from this journal was to record the rising meniscus in pasteur pipettes (2016JChemEduc1647) in variation of the concentration of sodium dodecyl sulfate (a popular anionic surfactant).


text:
Tautomerism in 3,5‐dimethyl‐4H‐pyrazole
How would tautomerism take place in 3,5‐dimethyl‐4H‐pyrazole?

I am aware of nitro–aci nitro and keto–enol, but I haven't seen such a compound before. I recognize that a hydrazone formation was done but nothing more. From what I was able to reason, an alkadiene is formed but that is unstable, so tautomerism shouldn't be possible. How does this compound show tautomerism?

Replace the N’s with O’s and remove the single bond between them. Do you know how to draw the tautomer of that? Well yes you do, so draw it. Then replace the O’s with N’s and add the single bond back.

You might want to try tautomerism with a simpler, similar compound first eg pyrrole, then move on to making the tautomers for this compound.

You are lucky, because not only the pyrazole motif, but the complete structure drawn by you is part of the Tautobase database

with a preference for the second structure drawn.
By the way, the values of this database are accessible to the public on GitHub, and since January 2020 implemented as (equally [sub]structure) searchable data file within DataWarrior.  Thus, you may train your chemical eye on 1680 tautomers.
Tautobase: An Open Tautomer Database, Wahl, O.; Sanders, T. J. Chem. Inf. Model. 2020, 60,1085-1089; doi 10.1021/acs.jcim.0c00035