text:
What causes HPLC peaks to look like gauss curves?
I was wondering how the form and curvature of HPLC peaks come to be. As I understand it, the mobile phase which contains the analyte, runs through the column. Many times, the distribution equilibrium is established inside the column, so the analyte is held back in the column. Finally, in the detector, peaks like a gaussian curve like this are recorded:

What causes the form of a peak to look like a gauss curve? I suppose more analyte is detected at the halfway point of the peak, but why?

This must be described in many textbooks and internet sites focusing on chromatography, as it is foundation of all chromatographic techniques. // Simulate multistage extraction with shifting mobile phase steps. It turns out it follows binomial distribution, converging to Gauss curve.

You are asking 2 different questions. "What causes the form of a peak to look like a gauss curve?" : as Poutnik said, this is well explained by many sources, but in essence it's down to diffusion. Imagine the initial sample as a very short cylinder on top of the column, which would result in a narrow ~rectangle when detected. As the solvent carries this cylinder through the column, the molecules in it also diffuse in all directions, resulting in a random distribution, which translates to a ~gaussian peak.

"I suppose more analyte is detected at the halfway point of the peak, but why?". That's an approximation, but is justified for the same reasons as above. If the shape resulting from the above mentioned processes is ~gaussian, the mode and mean ~coincide. The mean occurs at 'half time', and so ~does the highest concentration.

Even in this picture it's quite visible it's not Gauss curve.

@Mithoron "Why to spoil a good story by the truth?" :-} We know the world is not ideal and there are always some effects not  involved in education models. // There are 3 factors, Diffusion, prevalent at slow linear mobile phase speed, analyte transfer delay, prevalent at high speeds, and speed independent factor due dispersion of mobile phase passed distance. // Peak distorsion is often caused by nonlinear distribution between phases.

P.S.: E.g. for typical reversed phase HPLC column with silicagel substituted by C18 chains, there may be residual polar OH groups, capturing analyte more strongly then nonpolar C18 chains. This leads to trailing peaks. Another trailing reason may be wrong or badly maintained sample injection, with trailing sample insertion. This is more common in GC.

I'm with Poutnik on this. One can analyse/model physical processes at any required degree of accuracy, and as pointed out, there are textbooks and sources that explain (and account for) the chromatographic process in very intricate detail. Perhaps the tone of the question seemed to indicate that the OP was after some semi-qualitative explanation for the approximate shape of the HPLC peak, and diffusion processes are the answer to that question. If the OP had asked for the exact mathematical treatment of the shape of HPLC peaks, one would be right to point out that it's not a gaussian.


text:
Very Little Astatine
Astatine is the penultimate halogen in the periodic table with atomic number 85. It is very much radioactive and would vaporize itself by its own radioactivity before being collected. But there is something that Wikipedia says which is difficult for me to comprehend. It says

Astatine is the rarest naturally occurring element. The total amount of astatine in the Earth's crust (quoted mass $2.36 × 10^{25}$ grams) is estimated by some to be less than one gram at any given time. Other sources estimate the amount of ephemeral astatine, present on earth at any given moment, to be up to one ounce (about 28 grams).

My question is how can we even predict that the amount of astatine on the planet at any time must be less than 1 gram or about 28 grams if we haven't ever had any samples to investigate the element? Or are there any indirect means to do the same?

From 4 major decay series. When the series decaying is in steady state, the amounts of formed isotopes are proportional to their half times.

It's not a matter of samples. Estimation is from abundances of other elements and decay times. @Poutnik ninja! ;)

Does that mean that over very longer times, the amount of it remains more or less constant?

"at any given time" means just that...

@DrMoishePippik  $\ce{^{209}Bi -> ^{205}Tl + ^{4}He}$ //   Isotopes_of_bismuth

@DrMoishePippik Or the other way round: $$\ce{^{215}At -> ^{211}Bi + ^{4}He}$$ Half of astatine is already converted to more stable bismuth (having longer half-life) before we try to concentrate it.

From the Episodic Table website, read also this astatine story or listen the embedded dramatized audio version.

It's true that astatine is radioactive and it will vaporize by its own radiation. Not to mention its very low half-life (8.1 hours), meaning it will lose half of its mass before you even try to detect but there are novel ways to detect and measure astatine:

by using its ionization potential (the amount of energy needed to remove one electron from an atom, turning it into an ion or a charged particle).

Physicists at CERN's ISOLDE (Isotope Separator On Line-Detector) Radioactive Ion Beam facility created artificial isotopes of astatine (atoms with different numbers of neutrons than those occurring in nature) by shooting beams of energetic protons at a target of uranium (which has 92 protons and electrons). The collisions created a shower of new particles, some of which were astatine. The physicists then shined laser beams of varying wavelengths at the atoms to ionize them. By isolating the astatine ions, and checking which wavelength of laser had created them, the researchers determined astatine's ionization potential to be 9.31751 electronvolts (the ionization potential of hydrogen, for example, is 13.6 electronvolts) [Ref. 1]

By bombarding a heavy element with some light mass (e.g. bombarding bismuth isotope with alpha particles using a cyclotron confirmed the existence of  Astatine-211 --> Quantity obtained: 0.001 mg).
$$\ce{^{209}_{83}Bi + ^{4}_{2}He -> ^{211}_{85}At + 2^{1}_{0}n}$$
through radioactive decay chain. Astatine is one of the isotopes formed from uranium decay, plutonium decay etc.

Through this detection, scientist were able to estimate how much astatine would be there at a specific time.

Little bit of historical info.:
Astatine was so hard to detect and produce that scientists who first created it in 1939 had to resort to a trick. They created a tiny bit of astatine within a sample of bismuth by bombarding the bismuth with particles from a cyclotron. They then fed the whole thing to a guinea pig. After a few hours of digestion the guinea pig’s iodine-hungry thyroid gland had filtered and concentrated the astatine. It remains the only element discovered by a nonhuman.

Fun Fact: Its name comes from the Greek "astastos" meaning "unstable."

Reference

https://www.scientificamerican.com/article/physicists-measure-earths-rarest-element/
https://www.azom.com/article.aspx?ArticleID=7907
https://www.sciencehistory.org/distillations/tiny-productions
https://www.npr.org/sections/krulwich/2013/07/12/201481293/the-hardest-thing-to-find-in-the-universe

It seems for Astatine, the ratio of amount of artificial creation to natural occurrence is the greatest for any element which occur both ways.

@Pscionexam2003 According to Ref.2 "It is estimated that less than 30 g of astatine is present in the Earth's crust, only very few µg of astatine have been produced artificially so far and elemental astatine has not been viewed by the naked eye due to its instability.". Although a crude estimation, the naturally occurring astatine quantity is never fixed as it transforms to different isotope all the time or evaporates by its radiation. You won't find it in its elemental form as any radioactive ore would contain some atoms of astatine in it.

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text:
At what temperature is ammonia liquid at 1.5 bars?
I would like to know at what temperature ammonia condenses from gaseous form into a liquid, at a pressure of 1.5 bars. I know that at normal 1 bar pressure, ammonia is a liquid between -78 and -33 degrees Celsius, but don’t know the math to work out what the temperature would be at 1.5 bars, and can’t find the answer on the internet. It sounds silly, but try typing the above question into a search engine and you’ll see what I mean.

See https://www.engineeringtoolbox.com/ammonia-pressure-temperature-d_361.html

Stack Exchange sites are not forums, and the CoC has little to do with what a on-topic question is.

I hope they know that 1.5 bars is 23 psia. Not everyone is "bilingual".

Using NIST Reference Fluid Thermodynamic and Transport Properties Database (REFPROP) – NIST Standard Reference Database 23, Version 9, I found the saturation point of ammonia at equilibrium for a pressure of $p=1.5000\ \mathrm{bar}$ at a temperature of $T=-25.210\ \mathrm{^\circ C}$.
If you still want to "try typing the above question into a search engine", try this one: https://www.wolframalpha.com/input?i=saturation+ammonia+at+1.5+bar which also gives $-25.21\ \mathrm{^\circ C}$.

I found these two sources which seem useful in determining the behavior of ammonia at 1.5 bar (21.76 psia). Seems like it melts around 196 K and boils around 250 K (perhaps a little lower) at that pressure. Here's a phase diagram that shows this behavior (1.5 bar is 150 kPa, and you can see that the boiling point will be just under 250 K while the melting point will be around the same temperature as the triple point, just under 200 K). This can then be verified in the table of data provided below (the source where I found the table also included the triple point temperature, 196 K, written above).

Table from: https://www.engineeringtoolbox.com/ammonia-gas-liquid-equilibrium-condition-properties-temperature-pressure-boiling-curve-d_2013.html
Graph from: https://www.researchgate.net/figure/Pressure-temperature-phase-diagram-of-ammonia-The-dashed-line-is-extrapolated-58_fig1_281290526
edit: Thank you Martin for pointing out that 1.5 bar is 150 kPa. That was a silly mistake.

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text:
Is the enthalpy change of a reversible adiabatic expansion any different from enthalpy change of a irreversible adiabatic expansion?
Some sites say that adiabatic expansions of ideal gases in general have $\triangle H =0$ , whereas some say it is not.
I tried to find $\triangle H$ by considering a reversible adiabatic process where Initial conditions are $P_1, V_1$ and $T_1$ and final variables are $P_2, T_2, V_2$ also it implies $PV^\gamma = k$, hence $\triangle U = \frac{P_2V_2 - P_1V_1}{\gamma-1}$, $\triangle (PV) =P_2V_2 - P_1V_1$ and since $\triangle H$ = $\triangle U$ + $\triangle (PV)$ , $\triangle H =\frac{\gamma (P_2V_2 - P_1V_1)}{\gamma -1}$
What is wrong with this proof?

Please do not post the same question twice. Make edits to the first instead. Please visit the help center for more information.

"Some sites say ..." Really? Which ones? Please provide references in support of such statements.

$\Delta H = 0$ statement for adiabatic expansion  ($Q=0$) of an ideal gas would be true for expansion against vacuum in Joule-Thomson effect scenario ($W=0$), as  $\Delta T = 0$,  $\Delta U = 0$ and $\Delta (PV) = 0$

Are we talking about a closed system or can we also include an adiabatic flow through an open system?

For a closed system, do you think it is possible to have an adiabatic reversible process between the same initial and final states as an adiabatic irreversible process?

Is the enthalpy change of a reversible adiabatic expansion any different from enthalpy change of a irreversible adiabatic expansion?

Enthalpy is a state function, and its changes $\Delta H$ are independent of the path from state $1$ to state $2$. Thus, an adiabatic reversible process or adiabatic irreversible process, will yield an identical $\Delta H$.

So some sites say that adiabatic expansions of ideal gases in general have $△H=0$ , whereas some say it is not.

The enthalpy change for an ideal gas only depends on temperature. The only way that this change is zero is proved below
\begin{align}
  \Delta H = \int_{T_1}^{T_2} c_p^\pu{ig}(T) \; dT = 0 \therefore
   \boxed{\Delta H = 0 \Leftrightarrow T_1 =T_2} \tag{1}
\end{align}
In consequence, ideal gases suffer an isenthalpic process if and only if the process is isothermal. Adiabatic reversible or irreversible processes have a change in temperature, and thus, cannot be isenthalpic.

What is wrong with this proof?

The proof is fine. Remember it is only valid for:

An ideal gas.
It has constant-pressure and constant-volume heat capacities, so that $\gamma$ is also constant.

However, if you pay attention, there was no need to invoke any reversible/irreversible adiabatic process. It is a general expression, I will show you another way
\begin{align}
  \Delta U &= \int_{T_1}^{T_2} c_V^\pu{ig} \; dT \\
  \Delta U &= c_V^\pu{ig} \int_{T_1}^{T_2} \; dT \\
  \Delta U &= c_V^\pu{ig} (T_2 - T_1) \\
  \Delta U &= \frac{c_V^\pu{ig}}{R} (RT_2 - RT_1) \hspace{1 cm} (\text{use that $R = c_P^\pu{ig} - c_V^\pu{ig}$})  \\
  \Delta U &= \frac{c_V^\pu{ig}}{c_P^\pu{ig} - c_V^\pu{ig}} (RT_2 - RT_1) \\
  \Delta U &= \frac{1}{(c_P^\pu{ig}/ c_V^\pu{ig}) - 1} (RT_2 - RT_1)
  \hspace{1 cm} (\text{Use ideal gas law}) \\
  \Delta U &= \frac{1}{\gamma - 1} (P_2V_2 - P_1V_1) \tag{2} \\
\end{align}
And by the definition of enthalpy, using Eq. (2)
\begin{align}
  \Delta H &= \Delta U + \Delta (PV) \\
  \Delta H &= \frac{1}{\gamma - 1} (P_2V_2 - P_1V_1) + \Delta (PV) \\
  \Delta H &= \Delta (PV)\left(\frac{1}{\gamma - 1} + 1\right) \rightarrow
  \Delta H = \left(\frac{\gamma}{\gamma - 1}\right)\Delta (PV) \tag{3}
\end{align}
So, it was not necessary to specify how state $1$ goes from state $2$. Nevertheless, it respects Eq. (1)
\begin{align}
  \Delta H &= \left(\frac{\gamma}{\gamma - 1}\right) (P_2V_2 - P_1V_1) \\
  \Delta H &= \left(\frac{\gamma}{\gamma - 1}\right) (RT_2 - RT_1) \\
  \Delta H &= \left(\frac{R\gamma}{\gamma - 1}\right) (T_2 - T_1) = 0 \therefore T_2 = T_1 \tag{4}
\end{align}

Nice derivation. +1 // BTW the OP is not a proof, as it skips many deriving steps, taking the first equation for Delta U from thin air, with trivial following steps.

You could limit it to perfect gas = ideal gas with constant heat capacities.

@Poutnik Thanks Poutnik! Yes, those are the needed conditions to arrive to that equation.


text:
How to convert SMILES to 3D (with higher accuracy than RDKit)?
I am interested in assigning 3D coordinates to (atoms in) some 10K molecules that I have, currently represented by SMILES. This is because as have been shown by many chemoinformatics papers, 3D and 2D structures together can better represent a molecule.  This is also emphasized in large-scale molecular representation learning challenges, where 3D structures are estimated from DFT.
However, it seems time-costly to compute DFT with publicly available libraries (e.g. pyscf), or maybe I am missing something.  And the faster approach used by RDKit or OpenBabel (force field-based, e.g. MMFF, UFF) generates less accurate 3D coordinates (yes, I did notice this post).  I have been thinking of some ways: (1) find some possible conversions of my compounds to nominal identifiers (e.g. DrugBank ID) and extract 3D structures from those databases; (2) just run pyscf; (3) find some papers that predicts 3D structure from SMILES (does this kind of papers even exist?)
I'm not from a chemistry background but just started to look into interesting problems in computational chemistry.  Would appreciate you experts' suggestions!

Welcome. For part 1, while I don't know DrugBank in particular, most of these databases use RDKit, Open Babel, or OMEGA (a commercial product) to generate coordinates using force fields. One challenge, if you're not from a chemistry background, is to know that for most molecules, there are many energetically-accessible conformers (e.g., you need multiple 3D geometries).

You don't have to go full dft, you can try semiempiric models. But you have to start somewhere for that, too. Also the larger the molecules get, the more important Geoff's comment becomes. Here's a practical recommendation though: xtb.

Even though I'm on board with this question (is on topic here), you'll might find a better audience at [mattermodeling.se]. If you want, we can push it there; please do not cross post the question.

What physical/chemical/biological property of the molecules / of the material is the property of interest the model should learn? Or are there multiple to account for? Perhaps there already are data compilations available (see an earlier post to narrow the question. Is the interest in the conformations of global energetic minima (per molecule), or equally for more shallow minima (which, depending on the property in mind, can be more interesting / more relevant)?

@GeoffHutchison Thank you for the suggestions!  I have sent an inquiry to DrugBank since they didn't provide any information re. the source of 3D structures.  I am aware of conformers, and I'm happy to inject those equivariances into my encoder, so that as long as there is one valid (not sure how to define this chemically) 3D structure it should be fine.

@Martin-マーチン Thanks for the suggestions!  Would look into that after doing a search in databases as Buttonwood suggests.
I'm not super familiar with these two substacks but 'd be happy to move this to Matter Modeling if that's more suitable!  It looks like you folks who answered my question are crossing these two substacks :)

@Buttonwood Totally agree.  I'm interested in essentially modeling protein-ligand docking, which is inherently 3D.  In this context, many drugs/compounds that are included in datasets such as PDBBind are only represented by SMILES.  It might be possible to do a search in other 3D databases based on SMILES, though I am not sure whether such golden-standard 3D databases exist in the chemistry space.

@jasperhyp For docking, the reduced representation by SMILES lacks some crucial information; the molecules' conformation (how groups of atoms are tilt around bonds) by coordinates. But then, the most stable one of an isolated molecule (e.g. DFT computed) still can be a different one vs. if a fitting molecule (could already be water, an ion; the ligand/the protein) is close enough. The consistent assignment of SMILES (see some formats e.g. here) however can help to identify the molecules in question.

I'd say that the way to approach this problem greatly depends on what you want to do and why you need the 3D structure.
What grade of precision do you need and what do you need to measure?
Molecules move all the time unless crystallized. Do you need the crystal structure or do you need a representative conformation that is often assumed in a solvent at room temperature?
If you just need to represent the molecules in 3D in a conformation that "makes sense" for small molecules, a force field based method will generally work and produce conformations that work well enough; otherwise you can try some semiempirical methods, those are quite fast and cheap as well.
DFT will allow you to predict with better precision the conformation such a molecule would assume in vacuum at 0K (unless you go through several passages to try and model something else).
A third option could be using experimentally measured crystal structures, though I don't know how comfortable it is to go finding and cleaning up 10,000 of them.
I would not use databases containing computed structures such as DrugBank as you'd have no control over the process nor the possibility to check the quality of the results.
You may obtain quite different conformations using the different methodologies. Your model may work on some molecules because they have a particular conformation and fail on others that would normally have a similar one, however also have another different one that comes out slightly favorable in the calculations. 3D QSAR is particularly complex because of this and due to the difficulties in encoding molecular information in a comparable way. In a three dimensional object it's much more difficult to look for similarities and dissimilarities.
Regarding DFT computation time, that greatly depends on the sampling method you use to search the conformational space. Performing a steepest descent/BFGS may give you results in a couple minutes when we're talking about small molecules. However that won't work when searching for a global minimum, I'd say that computing 10,000 structures using DFT and with no access to a large cluster where to parallelize the computations would take too much time.
Since you're new to chemistry; I'd advise you to stay away from 3D QSAR for the time being. It does provide its advantages, but carries along a large quantity of complexities which can just be skimmed over when working with 2D structures.
EDIT:
I just noticed in the website you link they use models to predict the structure of a molecule as modeled by DFT. In that case you necessarily have to use DFT to obtain the structure. It appears as if they were suggesting to use the 3D structure as input to the model; but I doubt that's the case, they evidently meant that you should train a model on 3D data so that it will later be able to predict other 3D structures from the 2D information.

Thank you for the very insightful answer!  I'm not sure how precise I need actually -- I am working on a protein-ligand docking problem and it seems natural to model 3D molecular structure in addition to 3D protein structure.  I am pretty sure I need just a conformation since I can inject equivariances into the molecular encoder.

"If you just need to represent the molecules in 3D in a conformation that "makes sense" for small molecules, a force field based method will generally work and produce conformations that work well enough" -- this sounds very encouraging!  Perhaps I'll start with those structures.  Thanks!  Re. the drugbank comment, I am not sure if the 3d structures for drugs are sort of more experimentally-based since many drugs are pretty well-studied.  Have sent an inquiry to them.  Totally agree that 3D QSAR might be too hard, but I guess my use case is 3D inherently so that might make sense to include.

For the website's use of model predictions -- I noticed this "3D molecular structures provided. We additionally provide 3D structures for training molecules. These structures are calculated by DFT and are obtained together with the HOMO-LUMO gap."  Does thi mean they still used DFT?

I'm not sure what you're trying to do with protein ligand docking; if you're predicting the conformation of the ligand it would make more sense to use a docked conformation during the training. You definitely can use 2D otherwise. I'm not sure what you mean by injecting equivariance; molecular conformations can change a lot depending on the conditions.
In that website they build a model which predicts the minimum DFT conformation. They do indeed compute it with DFT for the training data, there's no way around it. Unfortunately, they do not explain HOW they obtain such structures...

@jasperhyp - it looks like they're using geometries from PubChemQC which were calculated with B3LYP DFT. One warning is that these are not necessarily the low-energy or best binding conformations. PubChemQC does not include conformers.

I guess I am not really concerned about conformation but more about the binding site (on protein) & affinity, which is sort of more "abstract".  Perhaps it could be fine to still use 2D, or just use DFT although the actual binding conformation could change.  By "equivariance" I mean modeling the molecules in a way that all conformers of one molecule would have the same representations.  Might not be the most ideal way but given the more abstract task maybe this is fine.

@GeoffHutchison Thanks for the information!  I guess I can start with these DFT-based conformer-unaware 3D structures and see if they actually add some values to using only 2D molecular graphs!

@jasperhyp if all conformations have the same representation; then you're not really providing any conformational information to your model, thus I don't see how that would improve the model.
Having to go through the effort of optimizing the structure through DFT can take time, you should also confirm that the effort is worth it due to speed or efficiency when compared to other methods such as molecular docking or molecular dynamics.
If your model requires significant amounts of preprocessing time, why not just use those methods?


text:
How do I make a 6-mercapto-1-hexanol (MCH) aqueous working dilution?
I am using 6-mercapto-1-hexanol (MCH) as part of the immobilization procedure for forming a self-assembled DNA monolayer on a gold surface (see the first figure of this paper for a good visual representation). Most protocols I see online and in the literature say to use 1 mM aqueous MCH. For example, this paper says "...mercaptohexanol (MCH) was co-adsorbed onto the DNA-modified gold electrodes by exposing the electrodes to a Tris buffer solution (10 mM Tris, 50 mM NaCl, pH ) 7.3) containing 1 mM MCH for 1 h", and this paper says "The slides were also exposed to a 1.0 mM aqueous solution of 6-mercapto-1-hexanol (MCH) for 1 hr".
However, I am struggling to make this working dilution. As far as I know, MCH always comes as a liquid, so I tried first making a stock 10 mM solution by mixing 1 mg of MCH liquid with 745 µL of water. The MCH appeared to be immiscible in water, however, forming distinct droplets in the water that rise to the top without mixing.
How do I make a working or stock dilution of MCH in water that isn't biphasic?

6-Mercapto-1-hexanol is only slightly soluble in water - 2.74g/L according to this source http://www.thegoodscentscompany.com/data/rw1631661.html   which equates to ~20mM so it should be possible. Have you tried heating it or sonicating it to get it into solution?

A very recent publication (Ref.1), stated that 6-mercapto-1-hexanol (MCH) was selected as the spacer thiol for its ability to:

Block nonspecific adsorption to the surface;
Dissolve in aqueous solution; and,
Avoid interference in the hybridization reaction to surface bound DNA as the six-carbon chain of MCH is the same length as the methylene group spacer in thiol-linked ssDNA.

Thus, it's pretty sure the solubility of MCH is satisfactory enough to do such extended research in literature. According to TMIC, one of manufacturers of 6-mercapto-hexan-1-ol (molar mass: $\pu{134.24 g mol-1}$), MCH is slightly soluble in water. In that website, MCH's water solubility is listed as $\pu{2.35 g L-1}$. And also, it is a very weakly acidic compound based on its $\mathrm{p}K_\mathrm{a}$, which is listed as $10.2$ (hence, it'd stay in non ionic form in both phosphate buffer and Tris buffer)). Accordingly, the concentration of its saturated solution at $\pu{20 ^\circ C}$ (assuming the solubility is given at this temperature) would be $\pu{0.0175 mol L-1}$ or $\pu{17.5 mM}$. That is above the required concentration for your stock solution.

Reference:

Mingfu Chen, Thuy T. Nguyen, Nitinun Varongchayakul, Chloé Grazon, Margaret Chern, R C. Baer, Sébastien Lecommandoux, Catherine M. Klapperich, James E. Galagan, Allison M. Dennis, and Mark W. Grinstaff, "Surface Immobilized Nucleic Acid–Transcription Factor Quantum Dots for Biosensing," Advanced Healthcare Materials 2020, 9(17), 2000403 (26 pages) (DOI: https://doi.org/10.1002/adhm.202000403).


text:
how I can separate nickel and chrome in single
How I can separate chrome and nickel from a nickel sulphate,nickel chloride and chrome salt separately??when these mixed up in a single solution.Please tell me how to do it using simple house hold items that does not cost too much. Please any one can help me as I am new to this field of chemistry.
Ok let me explain what happened. I want to do some electroplating. So for that I need chemicals to do nickel and then end chrome plating. I asked a guy in other city to send me chemicals that need for plating. He sent the chemicals but he used low grade packing that didn't survive the journey by general mail as in general mail they use very rough handling. So all mixed up but some did remain save and don't how eact quantity mixed up as parcel with torn from different places and it was some what losing it contents.
The parcel was containing nickel sulphate, nickel chloride and chromic salt. So i don't want to through away expensive chemicals with trying to save much as possible damage happened due to negligence of my supplier.

Provide more context to your question. // People with need to separate nickel and chrome in solution are seldom limited to household stuff.

Before answering this question, it would be useful to know the nature of the "simple house hold items" that may be used to get to the goal : separate nickel $\ce{Ni^{2+}}$ and chromium $\ce{Cr^{3+}}$ ions. Because I am afraid this separation is not possible with "simple" products.
The usual method for analyzing these ions require the following steps. First a solution is made with about $0.4$ g of the initial solid mixture  dissolved in $20 - 30$ mL water. Add $\ce{NH3}$ solution up to pH $8 - 10$. Add then a solution of colorless ammonium sulfide $\ce{(NH4)2S}$ (or sodium sulfide $\ce{Na2S}$) in feeble excess. Heat to boiling in a hood : odor ! Filtrate on paper. Nickel and Chromium make a precipitate of $\ce{NiS}$  and $\ce{Cr(OH)3}$  which remains in the filtrate.  Forget about the filtrate.
Add $5 - 15$ mL of $2$ M $\ce{HCl}$ to the precipitate. Stir the mixture for one minute. Filtrate the whole : Nickel ions remain unsolved as a precipitate. Chromium hydroxyde $\ce{Cr(OH)3}$ is dissolved as $\ce{Cr^{3+}}$ ions. This allows the separation of nickel and chromium. This is probably the goal of the present message.
If the experimenter wants to prove that the last precipitate contains nickel, and that the solution contains chromium, more selective tests must be done. But more complicated chemicals must be used.
To prove that nickel is in the precipitate, one should add $4 - 6$ drops concentrated nitric acid $\ce{HNO3}$, then heat at boiling temperature. This transforms $\ce{NiS}$ into $\ce{Ni^{2+}}$ ions. Take a small part of the obtained solution And add enough $2$ M $\ce{NH3}$ solution to get a basic pH. Add a couple of drops of dimethylglyoxime $1$% in ethanol. A red precipitate is produced if the solution contained nickel.
To prove the presence of $\ce{Cr}$ in solution, some $2$ M $\ce{NaOH}$ has to be added to the solution. Filtrate the greenish precipitate containing $\ce{Cr(OH)3}$. Take a small part of the precipitate. Add $2$mL $2$ M $\ce{NaOH}$ and $0.5$ g $\ce{PbO2}$. Heat to boiling temperature. Dilute with $10$ mL water. Acidify with $2$ M $\ce{H2SO4}$. Add then a few drops of diphenylcarbazide $0.2$% in ethanol. A purple reaction proves the presence of chromium.

Common household ammonia would possibly quantitatively precipitate Cr[OH]3 while maintaining Ni++ in solution as the ammonia complex. Another messier possibility is to oxidize Cr+3 to chromate with bleach and peroxide in base and filtering off nickel hydroxide [must check the potentials first] Reduce the chromate back with honey [invert sugar glucose + fructose both reducing sugars] and vinegar.

Or you could use vodka to reduce the chromate. Those household chemicals have much potential.

can i uses sodium carbonate to react with nickel sulphate to from ppt and then can filter that out????

Adding a carbonate ion to a solution made of both $\ce{Ni^{2+}}$ and $\ce{Cr^{3+}}$ ions will produce a mixed precipitate made of a mixture of nickel carbonate and chromium hydroxyde. Nickel and Chromium are not separated with this method.

@jmchemist. You are right. Thank you for the idea.  It is possible to separate nickel and chromium with ammonia $\ce{NH3}$. Adding ammonia to a solution contiaining both ions $\ce{Ni^{2+}}$ and $\ce{Cr^{3+}}$ will produce a precipitate of $\ce{Cr(OH)3}$ and a solution containing the blue and soluble $\ce{[Ni(NH3)6]^{2+}}$ ion.

can i use ammonia chloride in the reaction or i have to get pure ammonia as it's expensive here in my country

Chromium ions will not react with ammonium chloride or any ammonium ions origin to produce chromium hydroxide. But chromium ions do react with hydroxide ions $\ce{OH-}$ that are present in ammonium salt solutions if some $\ce{NaOH}$ has been added to the $\ce{NH4Cl}$ containing solution.

can i use NaOH in solution to ppt both but u guys might know that chromic salt will dissolve with excess NaOH while Nickel sulphate will remain undissolved. Am I right here?????

@Naveed AA. Yes. $\ce{NaOH}$ in solution makes a precipitate of $\ce{Cr(OH)3}$ with Cr salts. This precipitate is soluble in an excess of $\ce{NaOH}$ with formation of sodium chromite $\ce{NaCrO2}$. So an excess of $\ce{NaOH}$ in a solution containing both $\ce{Ni^{2+}}$ and $\ce{Cr^{3+}}$ produces a precipitate of $\ce{Ni(OH)2}$ without chromium. Filtration allows a separation of the ions. The precipitate may be filtered and should first be washed by a $\ce{NaOH}$ solution before using water. When diluted,  the chromite gets partly hydrolyzed and transformed back into $\ce{Cr(OH)3}$.

I think it much very cheaper then Ammonia right????

I don't know the prices of soda and ammonia

How much  NaOH do i need for 2.3kg of Nickel sulphate and nickel chloride (2.2+1.1KG).Chromic acid around 1.1KG

First remark : What is the meaning of "nickel chloride (2.2+1.1KG)". Is it equal to 3.3 kg nickel chloride ? Second remark : Chromic acid is not a salt of chromium. Chromic acid can be $\ce{H2CrO4}$ or $\ce{CrO3}$ or $\ce{H2Cr2O7}$. These substances are not chromium salts. Whatever the formula used, they do not produce $\ce{Cr^{3+}}$ ions in water. So they do not react with $\ce{NaOH}$ or ammonia $\ce{NH3}$ as described previously. They do not produce a precipitate with both reagents ($\ce{NaOH}$ or $\ce{NH3}$). If you add $\ce{NaOH}$ to your mixture, nickel will precipitate alone.

sorry i mean 2.3 kg nickel sulphate and 1.1kg of nickel chloride

I rewrite your question : How much NaOH do I need for destroying 2.3 kg of Nickel sulphate and 1.1 kg nickel chloride. Chromic acid around 1.1 Kg  Your mixture contains 2300/154 = 14.9 mol $\ce{NiSO4}$ and  1100/129 = 8.53 mol $\ce{NiCl2}$. In total 23.43 mol nickel. To destroy it, the double (in moles) of NaOH is necessary, so 46.86 mol, which weighs : 46.86 mol x 40 g/mol = 1874 g $\ce{NaOH}$. To obtain this final value, the nickel salts have been calculated as anhydrous compounds, which is a special supposition, as the commercial substances are hydrates with 7 or 6 $\ce{H2O}$.

pls not to bother you. Can u write in KG how much NaOH in total?

$1874$ g NaOH is $1.874$ kg NaOH

pls tell how much water do i need to dissolve 100g NaOH???

At $0$°C, the maximum solubility of $\ce{NaOH}$ in water is $42$ g in $100 g$ water. So to dissolve $100$ g $\ce{NaOH}$ you need a minimum of $238$ g water. Personally I would need a little more, at least $250$ g water.


text:
Why is the distinction from "coordinative bond" useful at all?
There are 3 types of coordinative bonding, complex formation at the electron donor, complex formation at the electron donor acceptor and complex formation at the electron acceptor, so what is the difference between coordinative bonding and covalent and ionic bonding?
Holleman (the German "Bible" of Inorganic Chemistry in the 102nd edition) even writes that there is basically no difference, so why is there this concept at all?

In other sources I find the explanation that coordinative bonds only refer to those where the electron pair comes from a single atom (donor), but Hollemann (whom I normally trust a lot) even writes:

Coordination bonds can formally originate half from Z and half from L.

The other two cases (whole of L or Z) agree with the well-known definition of coordinative bonds...

...in which the two electrons derive from the same atom.

https://en.wikipedia.org/wiki/Coordinate%20covalent%20bond?oldformat=true

This can pretty easily become a bottomless pit. It's good you trust hw, it's a great book. Not for beginners though. The description of chemical bonding is a hot button topic. Essentially, there are two types: ideally covalent and ideally ionic. And there is a spectrum between. A coordinate bond often is more covalent than ionic. There are no clear lines and the ideal forms don't exist. The rest is a bit ideology and what's easiest to understand. There, I hope I didn't make your question worse...

@Martin-マーチン Something between covalent and ionic should rather be polar covalent, I have the diagram in mind which shows the "ionic character" depending on the electronegativity difference, but then I just don't know where the concept of "coordinative" bonding should find place.

Just a different label, a coordinate bond usually is polar covalent. That's true for almost all bonds though. It easily finds its place with bond formation. And with complexes and with ligands and back bonding and lots of other stuff. It's really descriptive for the chemical reaction happening. Think for example dissolution of a salt. Also: ionic character does not depend electronegativity. That's just a nice rule of thumb.

Heavyhearted I'll be closing this question. I think it steers in the right direction, but ultimately, I cannot be answered by fact. It'll always be a discussion and there will always be opposing sides. If I had an idea to remove this from the question, please trust me, I would have used it to keep it open. In its current form, it's too attractive to create opinionated answers.


text:
Can someone help me with the experimental tabulation or spread sheet data on enthalpy, temperature, and pressure for aluminum and oxygen combustion
I need an experimental tabulation or spread sheet data of enthalpy, temperature, and pressure tabulated in columns for aluminum and oxygen combustion under non standard conditions.
Any references to books might help. Please I have tried checking it up I can't find any performed experiment whose data have been tabulated for further research purpose. I found something similar on NEST but it was for one bar pressure and enthalpy was a function of temperature. But I need for all temperature and pressure so I can interpolate the data for my research purpose.

Somehow doubt you need "all temperature and pressure".

Please don't double post, even if the previous post was closed: https://chemistry.stackexchange.com/questions/173203/can-someone-help-me-with-a-tabulation-on-enthalpy-of-combustion-of-aluminum-and The reason it was closed is to ask that you edit the post to clarify what you have done. Also, "all temperature and pressure" is not very meaningful. It would help if you narrowed down what you need.

The chemical reaction is as follows
$$ 4\ce{Al (s)} + 3\ce{O2 (g)} \rightarrow 2\ce{Al2O3 (s)} $$
We will abuse the fact that enthalpy is a state function. The process is illustrated in the following figure. The change we want is in red, but we will do it in blue:

First Step-Reactants The enthalpy differential for a pure fluid of constant composition is
$$ dH = c_P dT + \left[V - T\left(\frac{\partial V}{\partial T}\right)_P\right]dP
   \tag{1} $$
Given that we have two separated non-interacting phases, we can attack both species separately. Respectively:

Solid $\ce{Al}$: we will assume it is an incompressible fluid. If you find data of the volume expansivity $\beta$ in order to evaluate the partial derivative, then use it. If not, just get the density at room temperature and consider it constant. Integrating Eq. (1) yields
\begin{align}
  \Delta H_{1} = \int_{T_1}^{T_0} c_\ce{P,Al(s)}(T) \; dT + V_\ce{Al(s)}(P_0 - P_1) \tag{2}
\end{align}
Gaseous $\ce{O2}$: this is a non-polar, light gas, so the virial equation of state up to the second coefficient will suffice
\begin{align}
  Z &= 1 + \frac{BP}{RT} \\
  \frac{PV}{RT} &= 1 + \frac{BP}{RT} \rightarrow V = \frac{RT}{P} + B \tag{3} \\
\end{align}
Remember the virial coefficients are only function of temperature.

We will divide the enthalpy change of oxygen in two substeps. The first substep starts on $P_1$ and ends on $P_0$ at constant $T_1$. Eq. (1) yields
\begin{align}
  \Delta H_2 &= \int_{P_1}^{P_0} \left[\frac{RT_1}{P} + B(T_1) -
   T_1\left(\frac{R}{P} - \frac{dB}{dT}(T_1)\right)\right] \; dP \\
   \Delta H_2 &= \int_{P_1}^{P_0} \left(\frac{RT_1}{P} + B(T_1) -
   \frac{RT_1}{P} + T_1\frac{dB}{dT}(T_1)\right) \; dP \\
   \Delta H_2 &= \left[B(T_1) + T_1\frac{dB}{dT}(T_1)\right]
   \int_{P_1}^{P_0}  \; dP \\
   \Delta H_2 &= \left[B(T_1) + T_1\frac{dB}{dT}(T_1)\right]
   (P_0 - P_1) \tag{4} \\
\end{align}
The second substep starts on $T_1$ and ends on $T_0$ for constant $P_0$. Eq. (1) yields
$$ \Delta H_3 = \int_{T_1}^{T_0} c_\ce{P,O_2(g)}(T) \; dT \tag{5} $$
Finally, the enthalpy change for the "reactant" step which I'll call $ \Delta H_R $, is the sum of Eqs. (2), (4) and (5)
\begin{align}
  \Delta H_R =& \int_{T_1}^{T_0} c_\ce{P,O_2(g)}(T) \; dT +
  \left[B(T_1) + T_1\frac{dB}{dT}(T_1)\right] (P_0 - P_1) + \\
  &\int_{T_1}^{T_0} c_\ce{P,Al(s)}(T) \; dT + V_\ce{Al(s)}(P_0 - P_1) \tag{6}
\end{align}
Second Step-Reaction We on purpose made the reaction to happen at the conditions where the standard enthalpy of formation of the species are available. This one is easy, let me call this step $\Delta H_\pu{RXN} $
\begin{align}
  \Delta H_\pu{RXN} &= \sum_j \nu_j \Delta_\pu{f} H^\Theta(j,\pu{state}) \\
  \Delta H_\pu{RXN} &= 2\Delta_\pu{f} H^\Theta(\ce{Al2O3},s) \tag{7} \\
\end{align}
Since oxygen in the gas state and aluminum in the solid state are in their most stable form.
Third Step-Products This is exactly the same as we did for aluminum, i.e., Eq. (2) but changing names. I will call this step $\Delta H_P$
\begin{align}
  \Delta H_P = \int_{T_0}^{T_2} c_\ce{P,Al2O3(s)}(T) \; dT + V_\ce{Al2O3(s)}(P_2 - P_0) \tag{8}
\end{align}
Conclusion The enthalpy change of all the process is just the sum of Eqs. (6), (7), and (8), which I will resume because the equation is too large
$$ \boxed{\Delta H = \Delta H_R + \Delta H_\pu{RXN} + \Delta H_P} \tag{9} $$
This gives you the enthalpy change as a function of pressure and temperature for this particular chemical reaction.
References

Density of aluminum.
Density and standard enthalpy of formation of aluminum oxide.
Specific heat capacities for the three components, in their respective state, can be found on NIST.
The second virial coefficient for oxygen can be obtained by statistical mechanics, given that its intermolecular potential is well established. I will give you the correlation needed. The Lennard-Jones potential is
\begin{equation}
  \Gamma(r) = 4\varepsilon\bigg[\left(\frac{\sigma}{r}\right)^{12} -
              \left(\frac{\sigma}{r}\right)^6\bigg] \tag{10}
\end{equation}
and for $O_2$ the parameters are $\sigma = \pu{3.433E-10 m}$ and $\varepsilon = \pu{1.560E-21 J}$. The second virial coefficient is given exactly by the following equation
\begin{equation}
  B(T) = 2\pi N_A \int_{0}^\infty \bigg[1 - \exp\bigg(-\frac{\Gamma(r)}{k_B T}\bigg)\bigg]r^2 \; dr \tag{11}
\end{equation}
Coupling Eqs (10) and (11), and solved for many temperatures $T$, gives the value of $B$ at these temperatures. However, the function has no primitive, so the calculation has to be done numerically. The results and the correlation that I've just did are shown below, along with the derivative that is also needed:

As you see, there is virtually no difference. The function is the following:
\begin{equation}
 B(T) = 26.168 + 7.0889\cdot 10^{-3}T - \frac{8728}{T} - \frac{1.009 \cdot 10^6}{T^2} \hspace{1 cm} (\pu{50 K} \leq T \leq \pu{1000 K})
\end{equation}
with $T$ in $K$ and $B$ in $\pu{cm3/mol}$.


text:
Hydrolysis of polyamides using solid acid catalyst
The neutral hydrolysis of polyamides (eg. nylon 6,6) does not go to completion as the reaction is reversible. You can get around this by carrying out the reaction in an acidic/basic medium. With acid catalysed hydrolysis, the final amine product is protonated. This prevents it from reacting with the carboxylic acid formed.
My question is, will using a solid acid catalyst (eg. Zeolites) allow the reaction to go to completion as with a homogeneous catalyst (eg. HCl)?

With an acid catalyst only a proportion of the amine ends up protonated. Soundsl ike you need more than 1 eq of acid to protonate all the amine that is formed.


text:
post-treatment of silicon surface after silanisation
I recently attempted to make a silicon oxide surface hydrophobic using trimethoxyoctylsilane and methanol as a solvent in ambient air conditions. I left the silicon in the solution for 24 hours, but after rinsing the surface with ethanol and pure water, the surface is still not clean. (I even rinsed it with methanol, water, and ethanol.)I have two questions:

Will I receive a hydrophobic surface if I reduce the silanisation time from 24 hours to just 5 minutes? I read somewhere that a maximum of 5 minutes is enough to achieve a hydrophobic surface.

How can I effectively clean the silicon after silanisation to receive a clean surface?

Surface should be treated before silanization.


text:
Ethylene Glycol in Wall Primer
No scientific background here!
Due to a chemical sensitivity, and general concern for healthier indoor air quality in a home I am renovating, I opted to choose "greener" alternatives in terms of building materials as I do have a toddler at home. Unfortunately, the primer that was used on our walls, albeit low VOC (11.4g/L), it does contain a solvent.... ethylene glycol. EG is a pretty nasty chemical if what I have researched is correct.
Based on my limited knowledge, EG is a slow evaporating solvent. My concern is how long it will linger even after the primer dries. Generally, I know the expected cure time is 2-4 weeks. With this particular ingredient, from a scientific/chemist point of view, how long will EG take to evaporate, and will exposure to it be ongoing after the "curing" period? If so, approximately for how long, or is it infinite? Thanks!

While ethylene glycol is really bad to drink, it is hard to inhale enough to cause health effects unless you have lots of open liquid around. The Derived No Effects Level (DNEL) for consumers is 7mg per cubic meter. Given the slow evaporation rate it will prove hard to reach anywhere near such levels with any reasonable air flow.

Thank you Jon. I'm assuming that it will also, after said amount of time, evaporate and dissipate as well, regardless of slow evaporative rate or not....thanks for your response!


text:
Energy of d(x²-y²) and d(xy) orbitals the same with respect to trigonal symmetry?
In the orbital splitting diagrams of trigonal planar and trigonal bipyramidal complexes the d$_{x² - y²}$ and d$_{xy}$ orbitals have the same energy. I have always assumed that this is because they have different, co-prime symmetry — $C_4$ for the d orbitals, and $C_3$ for the ligands — and that therefore the overlap of the orbitals with the triangularly spaced ligands would be the same, and furthermore, that the choice of orientation of the coordinate system with respect to rotation of the x and y axes within the xy plane, around the z axis was arbitrary.
I "calculated" this, and it turns out to not be true. I used GeoGebra to calculate the area of overlap of a square and a triangle, both centred at the origin, expecting it to be independent of the relative rotation of the objects around the origin.
Here's a picture:

Here's the calculation: GeoGebra document/project.
Details of how I made the construction are given below.
I want to know why these orbitals have the same energy if the overlap integral with the ligands is different.  Or am I setting up this calculation incorrectly?  I know that the orbitals have non-uniform density within the lobes, so the interaction goes up rapidly as they align with the ligand orbitals, but I would think that you could still make the symmetry argument, or else the choice of orientation of the x and y axes with respect to z is not arbitrary, which I think would cause a cascade of problems and choices to question.
____ The construction ____
To make the square:

Made point A at (1,0).
From A, made point A' rotated by 90º around the origin.
Constructed a regular polygon with 4 sides (poly1) using AA' as one side.

To make the triangle:

Made point B rotated from A by angle "a", which is adjustable.
Made B' rotated from B by 120º.
Constructed a regular polygon with 3 sides "poly2" using BB'.

Found the intersection points of the sides of the square and the triangle: intersect(poly1,poly2).
Calculated the area of the polygon enclosed by those points:  area(F,G,K,J,I,H).
"a" is an adjustable slider about halfway up, the area is down the bottom. I couldn't figure out how to put them near each other so that you can look at the area while changing the angle. I also couldn't figure out how to graph the area as a function of angle, but I assume that it would have to have $C_{12}$ symmetry since it must have both $C_4$ and $C_3$ symmetry.  GeoGebra looks pretty and seems powerful, but I find the interface incredibly obtuse


text:
Calculating the enthalpy of reaction at non-standard concentration for reactions in aqueous phase
Standard conditions for a chemical reaction in the aqueous phase, for instance, a chemical reaction in solution, consider that reagents should be at a 1 M concentration. I was wondering how to calculate the $\Delta$H for other concentrations. Let's say we have the next reaction:
$$\ce{2HCl(aq) + Na2CO3(aq) -> 2NaCl(aq) + CO2(g) + H2O(l)}$$
It resumes to:
$$\ce{2H+(aq) + CO3^{2-}(aq) -> CO2(g) + H2O(l)}$$
To determine the $\Delta H^0$ we can perform:
$\Delta H^0$ = $\Delta H^0_{f}$ $\ce{CO2(g)}$ + $\Delta H^0_{f}$ $\ce{H2O(l)}$ - 2$\Delta H^0_{f}$ $\ce{H+(aq)}$ - $\Delta H^0_{f}$ $\ce{CO3^{2-}(aq)}$
Which results in: - 3 kJ/mol
But this is assuming standard conditions and thus 1 M concentrations for the reagents in the aqueous phase. How can I calculate the $\Delta H$ for other concentrations, for instance, when mixing 100 mL of 0.2 M $\ce{HCl(aq)}$ with 50 mL of $\ce{Na2CO3(aq)}$ 0.3 M?

Using Hess law and tabulated or measured enthalpies of dilution.


text:
Why can't the change in a crystal structure be due to the rotation of octahedra?
I have the following information about barium titanate ($\ce{BaTiO3}$).
It adopts the perovskite structure and
has a phase transition from a cubic to a tetragonal phase at 120 °C.
The space group of the cubic phase is Pm3m.
The space group of the tetragonal phase is P4mm.
Just above 120 °C the cubic cell parameter, a = 4.01 Å;
at room temperature, the cell parameters of the tetragonal phase are a = 3.992 Å and c = 4.034 Å.
Apparently the phase transition occurs because of a displacement of the $\ce{Ti}$ atom; why can it not be due to rotations of the $\ce{TiO6}$ octahedra?

It could, like wagging the dog instead of wagging the tail.

@Karsten Apparently it cannot, I just don't understand why. Perhaps the pressure and temperatures are insufficient to cause a large enough tilt for it to undergo a phase change.

The crystal structure of BaTiO3 is governed by barium and oxygen ions, titanium is slightly smaller that the vacancy it occupy. Thus it is best to describe changes as shifts in Titanium position.

Below is a depiction of the two unit cells (source)

Apparently, the entire structure stretches in one direction (and scrunches in the other two), with the c-axis 2.5% longer than the other two in the tetragonal structure (source). As a consequence, the octahedral oxygen configuration is now a distorted octahedron. In addition, the titanium is no longer centered in the octahedron, leading to a dipole moment along the c-axis (source).
However, all these movements are translations along the c-axis. They preserve the symmetry of the tetragonal space group. If you were to rotate the (distorted) oxygen octahedra, the tetragonal symmetry (which includes mirror planes perpendicular to a and to the a+b diagonal) would break down.
In comparing different views of the unit cell, be aware that some authors place the titanium on the origin (with barium in the center), while others choose to place the barium on the origin (with titanium in the center).

Why can't the change in a crystal structure be due to the rotation of octahedra?

It could, but not for the transition to the tetrahedral crystal form. There are less symmetric crystal forms under other conditions, and these would not have the symmetry constraints.
UPDATE: The cubic structure seems to have some added complexity, with 8 possible sites for a disordered titanium atom in the cubic structure.

Source: doi:10.1103/PhysRevB.70.134107


text:
Reaction of methyl triphenylbenzoate with concentrated sulphuric acid
In the following reaction, compound Q is obtained from compound P via an ionic intermediate.

What is the degree of unsaturation of Q?
This question was asked in JEE ADV 2020

I thought that the ester would undergo hydrolysis to form a carboxylic acid but the product Q is

which has degree of unsaturation = 18
Does this happen because of the phenyl groups?
What is the mechanism for this reaction and does this have a name?

The conditions are concentrated sulfuric acid so there is no water present to hydrolyse the ester. If it was dilute sulfuric then you would get the hydrolysis. This reaction proceeds by protonation of the ester group, loss of MeOH creating R-C=O+ cation which then does an electrophilic attack on an adjacent Ph group. This is generally called acid-catalysed aromatic acylation.


text:
How to find the initial rate of a rate plot?

In the graph above, how do I find the initial rate of $$\frac{d[A]}{dt}$$
My balanced chemical equation is $$\ce{3A + B -> 2C}\\[1em]$$

Just draw a straight line that overlays the first few points of the purple curve. (It doesn't have to be super accurate. Or at least, if you wanted something that was super accurate, you probably shouldn't be doing it graphically.)

Can you expand on that? I watched a video and they said to draw a tangent line through zero, and to pick two points on the tangent. Once I find the slope of those points, do I multiply it by -1/3?

Sorry, you mentioned drawing a straight line for the purple curve? I thought I was supposed to draw a line for the blue curve since we're finding the initial rate of [A]?

The slope of the blue line will give you the initial rate of change of the concentration of A. If you want the initial rate of the reaction, you do have to multiply by -1/3.

To get the most accurate results, simply draw a line from the first two point of the graph and find its slope. This can also be done numerically. Say the first two observations are $[A]_{t_0}$ and $[A]_{t_1}$ at times $t_0$ and $t_1$, respectively:
$$
\left(\dfrac{d[A]}{dt}\right)_{t_{0}} \approx \dfrac{[A]_{t_1}-[A]_{t_0}}{t_1 - t_0}
$$
Note: we are approximating the initial rate at $t_0$ with the average rate between $t_0$ and $t_1$. The initial rate will be slightly more negative than the average rate we have calculated.

Reference: ChemLibreTexts


text:
What household objects are made of metals with a high degree of purity?
I've looked into this briefly and from what I've understood, the kind of copper that's used to make electric copper wire is generally of a high purity (well above 99%). What are some other household objects that may be primarily composed of highly pure/unalloyed metals?

Depends on your threshold for purity. Some computers have shells made of aluminium (some suggest the alloy used by Apple is designated 6061 which is >98% Al). Al foil is purer than this. Some batteries may contain relatively pure lithium, some small cast objects might be largely zinc, jewellery can be silver, platinum or gold.

As you mention, electrolytic copper, used in electronics, is >99.9% Cu.
Household aluminum foil is ~99.8% Al, though it migh be alloyed for greater strength.
You might find fairly pure lead used for plumbing or ammunition, but it is often alloyed to make it harder. A very rough purity test can be performed with different grade hardness pencils.
Fairly pure tin is used for specialty solder, e.g., for items that come into food contact. Ordinary solders were Sn/Pb alloys, where lead has been replaced with Bi or other lower melting point metal now.
Mercury of ~99.9% purity might be found in old thermometers and "silent" electric switches.
Tungsten wire in lamp and valve (radio tube) filaments was fairly pure, but is now "doped" with oxides, including thorium, to make it stiffer and increase electron emission.
One might have a titanium tooth implant, or tantalum joint implant, but I don't know the purity of those materials.
Jewelry made of pure Au, Ag or other such metal is rare, since they would lose shape quickly, even take fingerprint impressions. Perhaps precolumbian New World artifacts? Stronger Pt might be used, but is difficult to find >90% pure. One might find a tiny amount of pure Au in gold lettering, or in the radiation emitter in a smoke alarm. Yes, a radio-emitter is added to the gold foil, but <<0.1%.

Gold jewelry is often alloyed with base metals for wearability, so we see things like 14 karat instead of tge pure 24 karat.

@OscarLanzi, exactly, which is why precolumbian ornaments were mentioned, because they might not have been alloyed. BTW, I've used gold so soft that pressing (my somewhat calloused) finger into it left an impression!


text:
Hydrogen bonding of water molecule
When we say that energy of hydrogen bonds in liquid water is $\approx \pu{6.3 kJ/mol}$,
does it mean that if we divide this number among $\pu{1 mol} \approx \pu{6.022E23}$ molecules, the result is we need $\approx \pu{e-23 kJ}$ for breaking each single hydrogen bond between 2 molecules of water?

Note that 1 mol of water contains n/2 mol of hydrogen bonds, where n is the average number of hydrogen bonds for 1 water molecule.

It means you need this much energy to break every hydrogen bond in one mole of hydrogen bonds in water, where the number of hydrogen bonds does not have to correspond to the number of water molecules. Every water molecule can form multiple hydrogen bonds; the Wikipedia page оn H-bonding claims about 3.5 H-bonds per water. I am not sure about your value for hydrogen bond energy, the same Wiki page gives about 20 kJ/mol.


text:
Acetic acid reacted with Fehling's solution
During our experiment, we reacted acetic acid solution with Fehling's solution, expecting nothing to happen. Much to our surprise, the blue faded and the solution turned reddish-brown. Has this happened to anyone  before? What might be a good explanation for this?

Are you talking about vinegar or an acetic acid solution (pure)?

@AChem acetic acid solution. I'll change the question.

The reddish-brown colour indicates copper(I)oxide which suggests your acetic acid has a reducing agent in it. Check the purity of your acetic acid.

@jphys I agree with Waylander. If the acetic acid was made by oxidizing (ethyl) alcohol, it may well have some alcohol or acetaldehyde (an intermeduate) in it. These are party time for Fehling's solution.


text:
What is the meaning of "proto" in nomenclature?
"Proto" is generally used to describe "first", "foremost" or "earliest form of (something)" indicating something primitive that transforms into something known or common. In chemical terminology,  it can be used to denote the first of a series of compounds, or the one containing the minimum amount of an element. But I still cannot understand its usage. Consider below cases:

Protocatechuic acid - it is the common name for 3,4-dihydroxybenzoic acid. So, why not "catechuic acid"? Is there any "catechuic acid" in the first place?
(+)-proto-quercitol - It is one of the stereoisomer of quercitol.  Other stereoisomers: https://pubs.acs.org/doi/10.1021/acs.jafc.8b01584. Also including (−)-gala-quercitol: https://pubs.acs.org/doi/10.1021/acsomega.9b02986. Is it a common practice to give such prefixes to stereoisomers?

For 2, see Fletcher Jr, Hewitt G., Laurens Anderson, and Henry A. Lardy. "The nomenclature of the cyclohexitols and their derivatives." The Journal of Organic Chemistry 16.8 (1951): 1238-1246.

For (1), I think the proto- sense is different. OED has a detailed paragraph on the chemical use of proto in at least two senses. I searched "proto- in chemical nomenclature organic compounds" and the first result explains it all (1909)

Just a guess, but I think it means "primitive form of" in this sense. The compounds catechin and epicatechin might be regarded as non-carboxylic elaborations of protocatechuic acid.

Catechin is a flavanoid isolated from acacia. I presume protocatechuic acid is the product of oxidative degradation of catechin. Hence, the use of the term proto.

According to the Merck Index, catechin is identically catechuic acid.

Most of the names of organic compounds from the 19th century have little or no structural meanings. Organic chemistry and structural understanding of molecules was very limited at that time. The name protocatechuic acid has a German origin from Protocatechusäure (ca. 1860s), and this also means Catechusäure (catechuic acid) also existed. However, elemental analysis and qualitative tests were well understood then with pretty good accuracy. Imagine, we are in the 1860s, Catechusäure (now catechin) has an elemental analysis of $\ce{C15H14O6}$. Protocatechusäure must be a simpler acid because the elemental composition is $\ce{C7H6O4}$. Testing for acid behaviors was easy.
I assume that the OP is from South Asia, so he might be familiar with paan (betel leaf) which is eaten with lime, kat-tha (an astringent bark extract of a certain Acacia tree), and betel nuts in weddings or other occasions. The word catechu comes from India. Today we know that Catechusäure does not have carboxylic groups, but given its astringent nature, it must be an acid for the early organic chemists.
For the second question, I already gave the reference in the comments. This time is a different era of 1950s, almost 90 years later. Structural understanding has considerably increased. This time, proto- has a specific implication*. See Fletcher Jr, Hewitt G., Laurens Anderson, and Henry A. Lardy. "The nomenclature of the cyclohexitols and their derivatives." The Journal of Organic Chemistry 16.8 (1951): 1238-1246. These authors mainly introduced some of these terms (not all). These include, allo-, cis-, epi-, gala-, muco-, neo-, proto-, scyllo-, talo-, and vibo-.
*See the table here in Wikipedia talk (?) archives under Diastereomerism: https://en.wikipedia.org/wiki/Wikipedia:Reference_desk/Archives/Science/2011_May_31


text:
Does thorium(IV) oxide has a "fluorite" like structure?

The fluorite structure shows 8:4 coordination. Which of the following solutions has fluorite-like lattices?

$\ce{CaF2}$
$\ce{SrCl2}$
$\ce{BaF2}$
$\ce{ThO2}$?

My book tells me only $\ce{CaF2}$, $\ce{SrCl2}$ and $\ce{BaF2}$ will have fluorite like structure. But should not $\ce{ThO2}$ also show fluorite like structure?

Wikipedia:  Thoria [ThO2] exists as two polymorphs. One has a fluorite crystal structure. This is uncommon among binary dioxides.

What do you mean?

When in doubts about statements or explanations of lectures or textbooks, always try first to verify the facts in independent available offline or online sources, before you decide to ask.
// This way you prevent creation of redundant info  if existing answer or info could be easily found.

Well I already searched and also used bard AI.But still had some confusions

Not being mentioned cannot be distinguished from not being done. You should explicitly list your findings within the question while being still confused about them.

Does this mean ThO2 is not a fluorite like lattice?

Let us continue this discussion in chat.

Please include more information. "My book tells me..." What book? Have you searched anywhere that ThO2 has fluorite like structure and other compounds do not? What is your thought process to challenge the answer?

Thorium dioxide very much does have a fluorite structure. True, such a structure is more difficult to realize with oxides than with fluorides, for it requires a cation that is suffuciently large while having four positive charges. Thorium is one of the few metals that can make these constraints; cerium is another.
The fluorite structure also appears in compounds of magnesium with Group 14 elements, most commonly $\ce{Mg2Si}$. In this case the four-coordinate element is the electropositive one (magnesium) rather than the electronegative one (fluorine in fluorite or oxygen in ceria/thoria), so the term "antifluorite" is often used to indicate the reversed bond polarity. Used currently in alloying of aluminum, magnesium silicide along with its (anti)fluorite counterparts having heavier Group 14 elements has been considered for thermoelectric applications.

here you can find the crystallographic date for $\ce{ThO2}$, which claimed to have fluoride type structure. The following reference provides supporting thermal expansion data for binary oxides with the fluorite, $\ce{MO2}$, and the antifluorite, $\ce{M2O}$, type structures. It states that:

This paper gives regression data for a modified second order polynomial fitted to the expansion data of, and percentage expansions for dioxides with (a) the fluorite and antifluorite structure: $\ce{AmO2, BkO2, CeO2, CmO2, HfO2, Li2O, NpO2, PrO2, PuO2, ThO2, UO2, ZrO2,}$ and (b) the rutile structure: $\ce{CrO2, GeO2, IrO2, MnO2, NbO2, PbO2, SiO2, SnO2, TeO2, TiO2}$ and $\ce{VO2}$. Reduced expansion curves for the dioxides showed only partial grouping into iso-electronic series for the fluorite structures and showed that the 'law of corresponding states' did not apply to the rutile structures.

Thus, the author of your book have been mistaken to tell you that only $\ce{CaF2, SrCl2,}$ and $\ce{BaF2}$ have fluorite like structure. However, there are much more binary metal oxides may have fluorite like structure including your compound of interest, $\ce{ThO2}$.
Reference:

D. Taylor, "Thermal expansion data. II: Binary oxides with the fluorite and rutile structures, $\ce{MO2}$, and the antifluorite structure, $\ce{M2O}$," Transactions and Journal of the British Ceramic Society 1984, 83(2), 32-37 (ISSN: 0307-7357).


text:
How does the synthesis of [18]annulene proceed stepwise?
Finding good literature on the synthesis of [18]annulene is challenging, as most sources are well old and quite archaic, so I was hoping some explanation in regard to the mechanism of synthesis.
My biggest question is in regard to the formation trimeric intermediate formed by Glaser coupling of  1,5-hexadiyne. It is not clear to me why this reaction seems to be "templated" towards forming a ring, instead of forming a linear structure formed by the terminal carbon's. In any case, I could not find any reason why this reaction could not be used for forming larger or smaller rings than being 18 carbons long, but I must be missing something.
Here is a good basic synthesis template: OrganoSynth

For readers: See annulene group of unsaturated hydrocarbons.

Have you references for this?  A reaction scheme would be useful

https://www.orgsyn.org/demo.aspx?prep=cv6p0068

Should have added one, like Mithoron did. In any case, you can see how they are from the 60’s (The references)

Did you see the yield of [18]-annulene in the Org. Syn. preparation? 0.54%! Perhaps there is considerable polymerization.

Perhaps there is, perhaps the synthesis is complicated. As I said, the literature is sparse and relatively old. or at least from what I could find.


text:
How to write the balanced chemical equation for the system in a rate plot?
I have the rate plot below:

I know that $\ce{A}$ and $\ce{B}$ are the reactants, and $\ce{C}$ is the product. The textbook says that the balanced equation is $$\ce{3A + B -> 2C}\\[1em]$$ However, I am not sure how to determine those coefficients based on the graph. Any help is greatly appreciated.

Which textbook?

(6.5 - 2 ) : ( 4 - 2.5 ) : ( 3 - 0 ) ->  3 : 1 : 2

Cite your references and workings on the problem while posting it on CSE for a better chance at receiving an answer and fewer downvotes.

By stoichiometry, the number of moles of species $j$ is related to the extent of the reaction via
\begin{align}
  n_j &= n_{j0} + \nu_j \varepsilon \hspace{3 cm} (\text{divide by $V$}) \\
  \frac{n_j}{V} &= \frac{n_{j0}}{V_0} + \nu_j \frac{\varepsilon}{V} \\
  C_j &= C_{j0} + \nu_j \varepsilon_V
  \rightarrow \nu_j = \frac{C_j - C_{j0}}{\varepsilon_V} \tag{1} \\
\end{align}
where we assumed that the volume of the reacting mixture does not change as the reaction take place, i.e., $V\approx V_0$. This is a good approximation for liquid-phase reactions, where the temperature does not change much, so that the density is almost constant. We also defined for convenience an extent of reaction per unit volume $\varepsilon := \varepsilon_V / V$
Apply Eq. (1) to each component
\begin{align}
  \nu_A = \frac{C_A - C_{A0}}{\varepsilon_V} \tag{2} \\
  \nu_B = \frac{C_B - C_{B0}}{\varepsilon_V} \tag{3} \\
  \nu_C = \frac{C_C - C_{C0}}{\varepsilon_V} \tag{4} \\
\end{align}
Dividing Eq. (2) with Eq. (3), and Eq. (4) with Eq. (3), and using the values shown in your plot, we obtain the ratio of the stoichiometric numbers
\begin{align}
  \require{cancel}
  \frac{\nu_A}{\nu_B} & = \frac{C_A - C_{A0}}{C_B - C_{B0}} =
  \frac{(1.2 - 6) \; \cancel{\pu{M}}}{(2.4 - 4)  \; \cancel{\pu{M}}} = 3
  \rightarrow \nu_A = 3 \nu_B \tag{5} \\
  \frac{\nu_C}{\nu_B} &= \frac{C_C - C_{C0}}{C_B - C_{B0}} =
  \frac{(3.2 - 0) \; \cancel{\pu{M}}}{(2.4 - 4)  \; \cancel{\pu{M}}}  = -2
  \rightarrow \nu_C = -2\nu_B \tag{6}
\end{align}
And finally we write the chemical equation using Eqs. (5) and (6)
\begin{align}
  \require{cancel}
  \nu_A A + \nu_B C + \nu_C C = 0 \\
  3\cancel{\nu_B} A + \cancel{\nu_B} B - 2\cancel{\nu_B} C = 0 \\
  3 A + B - 2 C = 0 \rightarrow \boxed{3A + B \rightarrow 2C}
\end{align}

Good post. But, if I imagine I am a beginner, I am not sure if I would thank you or would be rather running to hills screaming. :-)

@Poutnik Hello Poutnik. You say that this is too advanced when you are in the first course of general chemistry at the university?

I think what you ask is more important than where you are. // Initially, I wanted to provide example of Calvin and Susie exchanging fruits. (Calvin & Hobbes)

@Poutnik Sorry Poutnik, I didn't quite understand... what do you mean?

1st part: I answer simply to simple questions and vice versa. Equal question can be asked by skilled high school freshman and ehm, simple college freshmen. // The question is so simple that it can be colourfully illustrated by basic arithmetic task from elementary/basic school, not too far from the Calvin level.

Calvin has got  4 apples, 3 pears and 2 oranges. After exchanging fruits with Susie, he has got 1 apple, 2 pears and 4 oranges. In what ratio has he exchanged his fruits? :-) :-)


text:
Reading Graduated Cylinders for a non-transparent liquid
A graduated cylinder is meant to be read with the surface of the liquid at eye level, where the center of the meniscus shows the measurement line. Is it also true for any non-transparent or a colored liquid? There're some diverse information in different sources.

Illustration source

Please edit to expand on “There're some diverse information in different sources”.

https://reefcentral.ru/forum/topic/51927-kak-pravilno-nabirat-vodu-dlia-testov/
in Russian language, second reply

Graduated cylinders are not designed to deliver accurate volumes, so one does not worry about upper or lower meniscus issues with transparent or dark liquids. The best way is to actually measure the volume of the dark liquid by mass, if this is a concern. This can be tested by weighing the nominal volume delivered by graduated cylinder. One should know the density of the liquid. Make sure the cylinder has a marking which says TD (=to deliver) or Ex (which also means to deliver).
For example, for a 500 mL good quality cylinder (class B) may have an error of 4 mL. This error is quite larger than what one might expect from meniscus errors. This is pretty large for accurate and analytical chemistry measurements.

The volume of the meniscus can be considered as negligible. So the difference in volume between upper and lower meniscus is negligible. Please refer to one of these conventions and stay there. This is specially true when measuring differences of volumes, for example when a certain amount of liquid had been removed from a burette for a titration.


text:
Why is nucleus coupling neglected to explain electronic states but LS coupling only rises because of?
Here, with LS coupling I mean coupling between total orbital momenta and S total spin momenta and treating eletrons in light atoms with Russell-Saunders approximation. I found a bit paradoxical that LS coupling happens because of a positive nucleus charge Ze - and its dependence follows Z^4 - but the coupling between the spin momentum of the nucleus and the eletrons' is alltogether neglected in at least this approximation. My reference is Hollas, Modern Spectroscopy, 4th Edition, pg. 208, cap. 7 (Electronic Spectroscopy). That is, why is nucleus spin momentum weak?
Also, what are the evidences that support the efficiency in Russell-Saunders approximation for lighter atoms (ll coupling strong, ss appreciable, j weak) and jj- coupling for "a few states of heavy atoms" (j strong, jj appreciable, ll and ss weak)?
It is curious that the extent of the coupling j apparently inverts from light to heavy atoms, given these 2 approximations.

Nucleus spin momentum is not weak. It's magnetic momentum is, being reciprocal to mass. Angular momenta interact with each other via their magnetic momenta.


text:
Why is Fe3O4 a non-stoichiometric compound?
Fe combines with O in a whole number ratio 3:2. Even the ions Fe2+ and Fe3+ are in the ratio 1:2. Then, why Fe3O4 is a non-stoichiometric compound?

Review the guides Asking and How to ask. Not following the guidance may lead to lack of satisfying answers, misunderstanding, objections, question down-voting or even question closure.

Stoichiometric compounds have composition with fixed and exact atomic ratio in small integer numbers.
Nonstoichiometric compounds have their real composition expressed by atomic ratios in rational numbers.
There is also a gray zone, where the compound is not exactly stoichiometric, but the deviation is small in considered context and we use for it a stoichiometric formula that fits it "good enough". Many metal oxides fall in this category.
The fact a compound has mixed oxidation numbers like $\ce{Fe3O4}=\ce{Fe^{II}Fe^{III}2O4}$ or $\ce{Pb3O4}=\ce{Pb2^{II}[Pb^{IV}O4]}$,  so it cannot be expressed by a single integer oxidation number, does not alone make the compound non-stoichiometric. But deviations in composition,  leading to rational atom ratios, do make such a compound non-stoichiometric.
Iron oxides are famous for that, even if $\ce{Fe3O4}$ is generally closer to stoichiometry than $\ce{FeO}$ or $\ce{Fe2O3}$. That is general phenomena for transition metals with multiple close oxidation states and oxide properties. There are vacancies or extra atoms and changes in oxidation states in frequent lattice defects.
E.g. the formal FeO(s) is usually a non-stoichiometric oxide with real composition $\ce{Fe_{0.84−0.95}O(s)}$. There to produce its stoichiometric variant too:
Wikipedia:

Stoichiometric $\ce{FeO}$ can be prepared by heating $\ce{Fe_{0.95}O}$ with metallic iron at $\pu{770^{\circ}C}$ and $\pu{36 kbar}$.

Other typical cases of non-stoichiometric compounds are

manganese oxides
mixed oxide ceramics, e.g. high temperature superconductors
Li-ion cell cathode materials
"hydroxides"/"oxidohydroxides"/"hydrated oxides" of transition metals
hydrides of transition metals

Then is Fe2O3 non stoichiometric?

I have thought it was clear. Its ideal composition is not. Its real composition is. Same as for FeO, Fe2O3, MnO, Mn3O4, MnO2. They at special cases and careful preparation may approach stoichiometry, but will not achieve it exactly.

Although I agree with the notion that mixed valence compound can be non stoichiometric at times (due to its deviation from ideal integer ratio of atoms and it is very much a case when you think those compounds are adduct of two small compounds, also non stoichiometric at times and so in practical cases, the atoms doesn't always reach integer values, e.g ($\ce{Fe3O4}$ = $\ce{FeO.Fe2O3}$ but for simplicity we give them a stoichiometric formula), it's better to classify them separately. Not all non stoichiometric compounds are ionic (so don't have mixed valence). E.g. Palladium hydride.

@NilayGhosh I have not said nor assumed all nonstoichiometric compounds are ionic. See also related closed question. The answer was intended to address the specific iron oxides case and not as general classification, which you correctly commented in the follow up question the user could find it himself. I would added there metal hydrides too, if remembered in the time of writing.

The exact formulation of $\ce{Fe3O4}$ is indeed stoichiometric and requires a fixed amount of $\ce{Fe}$ and $\ce{O}$ to react as stated in the question statement. However, usually, this won't be the case; if you try to synthesize stoichiometric $\ce{Fe3O4}$, chances are that you will find some deviation from this formula due to defects and varying amounts of $\ce{Fe(II)}$ and $\ce{Fe(III)}$. Therefore, iron oxides are usually characterized as nonstoichiometric. As pointed out in the comments, stoichiometric iron oxides have also been synthesized.
Nonstoichiometric compounds are known to exist in various, sometimes simple and sometimes not, such ratios and the composition may be affected by various factors such as synthesis conditions, reactivity etc. Common examples of such compounds are:

Iron oxides
Iron sulfides
Palladium hydrides
Tungsten oxides

Do note that the nonstoichiometric behavior does have applications such as catalysis, hydrogen/sulfur storage, ion conductance, and more.

why was this answer downvoted?

@ananta I suspect your answer is downvoted because you first sentence assumes that the iron oxide is always exactly stoichiometric which avoids the question rather than answering it. it might be better to say, if the formulation is exact, then it isn't non-stoichiometric but the substance often isn't depending on the exact conditions of its creation.

@matt_black It could be also said the formula Fe3O4 itself is stoichiometric, but real composition may not be, if it statistically significantly deviates from the given atomic ratio.

@ananta The point by matt_black is also good. You are assuming that iron(II,III) oxide is always stoichiometric. If you were to take the formula literally then yes, iron(II,III) oxide is stoichiometric but it oxidizes in air leaving behind $\ce{\gamma-Fe2O3}$ and you are left with a $\ce{Fe2O3-Fe3O4}$ mixture which is not at all stoichiometric. Although you may call it partially oxidized magnetite, it is nonetheless magnetite and better to formulate it $\ce{Fe_{3−δ}O4}$ with $\ce{\delta}$ values reaching upto 0.02.

@matt_black thank you for the suggestion. I made some changes to the answer. Hope this clarifies the discrepancy.


text:
How does MO theory explain the effect of directing groups on aromatic compounds?
From what I could find, this is usually explained using resonance. For example,

from Wade's Organic chemistry, 9th edition, p. 823.
but I can't figure out how to see this through the MOs of the molecule. If the lowest energy MO is two toruses above and below the aromatic ring, then in this case, the addition of the electrophile E sort of "breaks" that torus, but isn't the remaining orbital just two "C shaped" densities, in each case equally capable of delocalizing the positive charge?

Please include a citation for the image you are using. It is using horrible notions. There are no "especially unstable" resonance structure. That's just bogus terminology and inverting the whole concept of resonance.

Disclaimer: all calculations were performed using MolCalc, an excellent web-based tool to quickly calculate quantum mechanical properties of a molecule.
Debunking the Misconceptions

Lowest energy MO is not two toruses above and below the aromatic ring, it is the two $\mathrm{1s}$ orbitals of $\ce{O}$ in the nitro group of nitrobenzene at -551.12 eV (essentially not interacting).

The torus didn't disappear upon substitution of the nitro group. The toruses are still present at -13.27 eV.

The torus didn't break. Even upon substitution of the -chloro group at the para position, the toruses can still be found at -14.35 eV.

Explanation using Molecular Orbitals Theory
According to ChemLibreTexts, -$\ce{NO2}$ is a deactivation group, reducing the electron density at ortho and para positions, as explained by the resonance structures as well, but this is not so easily seen in the molecular orbital diagrams.
These are the second and third occupied molecular orbitals of nitrobenzene (the first is localized on the nitro group):

The second highest occupied molecular orbital shows no electron density at the para position and the third highest occupied molecular orbital shows low electron density at the ortho position.
Caution
Electrons are not actually filled exactly according to the Aufbau principle, since there is always some thermal energy. Furthermore, the interaction is far more complex than singly pointed out molecular orbitals. Moreover, the calculations by MolCalc (being a fast tool) are not very accurate and you need to use more sophisticated software tools and analyze transition states to get a clear picture.


text:
what happens to electron after beta minus decay
In the beta minus decay, electron is emitted from the nucleus. It depends on the previous nuclei(that was before beta decay) how much energy emitted electron will have.
After this event, Internal bremsstrahlung could happen, where electron could be slowed down which would cause the photon emission.
on chatgpt, I read the following:

The probability of a high-energy beta particle interacting with the
electromagnetic field of the nucleus is higher because it can
penetrate deeper into the region surrounding the nucleus where the
field is strongest

I don't understand this why low-energy beta particle wouldn't be able to do the same as high-energy. I'm also curious when Internal bremsstrahlung would NOT happen in beta decay and where does the emitted electron go ? could it be captured by the atom or it will leave the atom ?
I'd appreciate the thorough explanation of what things happen actually. (Only interested in beta minus)

-1 for quoting artificial stupidity.

I agree, so it's stupidity right ? I never trust it but I thought now what if it told me the truth. so can I remove that part from the question ?

Yeah. Maybe paraphrase if you ever needed to mention it. I heard some issue about license for reusing this "content" on top of its general unreliability.

chatGPT is not an authetic source of information.

Despite the poor reference cited, the question is legitimate and deserves an answer.

Quoting NuclearPower.com:
"[B]elow 1 MeV... the energy loss due to bremsstrahlung is very small. Radiation loss starts to become important only at particle energies well above the minimum ionization energy. At relativistic energies, the ratio of loss rate by bremsstrahlung to loss rate by ionization is approximately proportional to the product of the particle’s kinetic energy and the atomic number of the absorber."
As for the final resting place of the electron, it depends on the structure of the material in which it finally slows to thermal speed. For example, in crystalline solids, such as alkali halides, F-center defects can be produced, coloring the crystal. An F center is a type of crystallographic defect in which an anionic vacancy in a crystal lattice is occupied by one or more unpaired electrons. See halide salts being colorized with a spark coil.
In electrically conductive media, the electrons can flow, effectively creating an electrostatic atomic battery  (which should be  cell, rather than battery, if you have just one) or betavoltaic device.

Thanks for the answer. So, emitted electron can either attach to the orbital or even leave the atom, right ?

Likely, they have too much energy to stay in the emitting atom's vicinity, leaving it with a net positive charge... which would eventually be equalized in a conductor. There are isomeric transition/internal conversion radionuclides that rearrange the nucleus and emit only γ rays, staying neutral (unless they self-photoionize). BTW, you might remove the ChatGPT reference to improve the question.


text:
Why is black powder used in modern military munitions?
I was surprised to read in this recent WSJ article about a black powder shortage:

Today, [black powder] is a specialty commodity with few commercial applications—mostly for rocket hobbyists—but it’s still used in more than 300 munitions, from cruise missiles, to bullets for M16 rifles, to the vital 155mm shells.  In each case, a small amount of black powder is used to detonate a more powerful explosive packed in the same bullet or missile. A 155mm shell for a howitzer, for example, will use half an ounce of black powder, lodged next to 26 pounds of a more powerful explosive.

That's describing black powder being used as a detonator/primary explosive, which I had always understood it could not do because it is strictly a "low explosive" meaning it can't develop a supersonic shockwave (VoD).  I have never seen black powder listed as a primary explosive.  So is the article mistaken about its use in modern military explosives?  Or does black powder serve some purpose other than detonation in those munitions?

Explosive booster?

@Mithoron I've never heard of it used in that capacity either.  That wikipedia article lists all the boosters I've heard of and does not mention black powder..

It is pretty easy to assume the WSJ used the wrong terminology. It seems to confuse the nature of the propellant with the nature of the detonator, for example. Propellants are usually guncotton-based and this has nothing to do with the initiator of the explosion for shells containing explosives designed to blow up on arrival. Bullets, for example, don't have explosives designed to detonate on arrival, shells sometimes do. The WSj is confused on this.

I think black powder is used as a combustion initiator, to quickly raise temperature and pressure to the point where the main smokeless powder charge burns efficiently. BP is not pressure sensitive for its burn rate where nitrocellulose based propellants are very sensitive, burning faster at higher pressures. BP is a propellant, 'low explosive' with a relatively constant burn rate dependent on surface area. It is not a primary or secondary explosive; it does not detonate by shock propagation. I remember seeing that naval guns have a small pouch of black powder attached to the back of the smokeless powder bags. It would seem that 155mm artillery shells are similar.


text:
What is the structure of quinine?

I am very confused on what the inner lines in the hexagons mean, are these representing double bonds? If yes, does that mean quinine has 6 double bonds? Also what would be considered a central atom that has at least 2 atoms bonded to it and how would I find the angle.

You have heard about quinine, but not about benzene and generally aromatic compounds?

I have but not very familiar with them. This is more of an independent study than a full chemistry class

Those 2 condensed rings with N atom alone  is chinoline, derivate of naphthalene, which is aromatic hydrocarbon. The rings are often drawn with 3 double bonds, what is incorrect, as the bonds are delocalized. That means neither of the ring bonds is single nor double, they are something between. We call such a ring the aromatic ring.

I see, that makes sense, how would I find double bonds in a structure like that? I know how when its in lewis structure but not like this. Also what would be considered a central atom that has at least 2 atoms bonded to it and how would I find the angle.

For bond angles, see hybridization at each point. If there is strain it would be slightly different from original.

If you originally determined 6 double bonds, then you know how to do so.

The easiest way to understand it is that there are certain molecules in which the electrons move around very freely. This means that they're not localised between the nuclei which have been bonded together, but, due to a resonance, they can travel almost freely all around the ring (given an approximately equal distribution of electron density). A very simple way you can tell something is resonant is if you can draw the electron distributions in different ways without changing the molecule itself, e.g.:

So, the rings in the quinine molecule above are aromatic rings (as Poutnik mentioned). When someone says that something is 'aromatic' in chemistry, it means that these molecules are stabilised by these resonance effects. In high school, they told us to think of them as 1.5 bonds.
And, to be honest, molecular geometry is a lot more complicated than 'finding the angle' to something. That being said, if you look up VSPER, that might help? I'm not quite sure what you mean by a 'central atom', though...
Note: This is a highly simplified answer, but I'm assuming (no offence) that you need a high school-level introduction, rather than a university-level lecture.
Also, that picture is a lot bigger than I thought it would be... sorry :s

Yes, this helps a lot thank you. By central I mean its connected to an atom to the left and right or like both sides

There are so many atoms in quinine which fall into that category, so the question is a bit ambiguous...


text:
To what extent should a question such as: "how many enantiomers are there?" take geometric isomerism into account?
Pretty simple question here.

How many enantiomers are there for the structure $\ce{CH3CH=CHCH(CH3)CH2CH3}$?

On the one hand, there is just one chiral centre so one would suggest $2$. On the other hand, there is some E/Z cis/trans isomerism at play leading to four distinct stereoisomers.
I made a sketch:

If I'm not mistaken, $(A)-(D)$ are distinct molecules. Let's say we favour $(A)$. Would we say that the only enantiomer of $(A)$ is $(B)$, or would we say that all of $(B),(C),(D)$ are enantiomers of $(A)$?
I ask because I have been hearing some confused messages from my chemistry teacher and for a question as specific as this, I could not find a satisfactory online discussion. I would personally count them all as different enantiomers since it is truly a different arrangement of atoms in space and it is reasonable to suppose that all of them have slightly different optical properties.
What would you say?

I think there is a H missing in the formula. / On the matter at hand, I'd say there are two sets of enantiomers (a and b, c and d). Everything else would be ambiguous at best. Enantiomers always come in pairs, so the answer is pretty much always two and only differs in the number of sets.

@Martin-マーチン There is a missing $H$ yes. Thank you for sharing your thoughts

Please [edit] the question to fix the error.

A chiral atom has four different groups attached. Enantiomers refer to a comparison between two stereoisomers. So enantiomers are steroisomers that are non-superimposable mirror images (=chirality). While enantiomers contain chiral centers in the molecules, but not all stereoisomers of a molecule are enantiomers of each other. It's such an uncomfortably boggling jargon - always drawing them helps !

@bonCodigo So you seem to be suggesting only $(A),(B)$ and $(C),(D)$ are enantiomers.

FShrike That's correct but @Martin-マーチン would be a better explainer for you.


text:
Why are Sonogashira reactions moisture sensitive?
Exemple, aryl halide + TMS-acetylene in NEt3 + Pd[PPh3]2Cl2 or Pd[PPh3]4:
The reaction does not form water at any time so its presence wouldn't shift the equilibrium. Neither TMS-acetylene or the aryl halide are subject to react with water. This would signify that the Palladium catalyst is moisture sensitive. Yet, Palladium can not be oxidized by water as oxidative addition of water isn't possible (OA of OH extremely rare) and beta hydrogen elimination to form Pd=OH seems impossible?
Hope someone will be able to enlighten me.

Quenching of the copper acetylide possibly

The addition of CuI as co-catalyst allows to lower the reaction temperature (often ambient instead of say $\pu{100 ^\circ{}C}$) and to use a mix of THF (for solubility) and aqueous ammonia as solvent, possibly passing a Cu acetylide.  See e.g. 2002ChemLett756 and 2005BullChemSocJpn160. The conditions equally suit a carbonylation (2003OrgLett3057, a visual abstract).


text:
How to perform the Bray-Liebhafsky Reaction
I am trying to complete the Bray-Liebhafsky reaction in the lab so I can use it in the future for demonstrations.
The above Wikipedia link quote:

A fundamental property of this system is that hydrogen peroxide has a redox potential which enables the simultaneous oxidation of iodine to iodate:
$$\ce{5 H2O2 + I2 -> 2 IO3− + 2 H+ + 4 H2O}$$
and the reduction of iodate back to iodine:
$$\ce{5 H2O2 + 2 IO3- + 2 H+ -> I2 + 5 O2 + 6 H2O}$$
Between these two reactions the system oscillates causing a concentration jump of the iodide and the oxygen production. The net reaction is:
$$\ce{2 H2O2 -> 2 H2O + O2}$$
necessitating a catalyst and $\ce{IO3-}$.

I am trying to set it up in a very simple manner, in a beaker with a stirrer and hotplate. From what I have seen from reading numerous papers it should just need a heated solution of sulfuric acid and potassium iodate and then be mixed with another solution of hydrogen peroxide and the reaction should proceed.
However, in my attempts so far I have been unable to get the desired colour changes. How do you run this reaction?
There seems to be lots of information on how to run both the BZ and BR oscillating reactions but I have been unsuccessful in finding meaningful information for this one.
I am wondering if I need an indicator to be able to visibly see the oscillations?
UPDATE: Current experimental procedure.
a 50 mL solution of 0.125M potassium iodate and 0.24M sulfuric acid was made and heated to 65 degrees in an oil bath. After half an hour 1.3 mL of 30% w/w hydrogen peroxide was added.
I have also tried other varying concentrations of iodine/sulfuric acid and hydrogen peroxide with no success as yet also I have done this at temperatures of 25, 50 and 60 degrees.

I suspect there is no colour change without some indicator that shows strong colour change when iodine is present.

Try different starting points, different ratios, different mixing order.

The reaction equally is known as iodine-clock. Did you venture out references like wikipedia, or -- because it is appealing -- in the Journal of Chemical Education  (some of the hits here may address it in detail)?

@Buttonwood I will look into it thankyou. I was under the impression that the iodine clokc was slightly different as it uses sodium thiosulfate. However, I notice they use starch as an indicator. Perhaps I can do that with this one.

@matt_black any ideas on what a suitable indicator may be? I have seen starch used in similar reactions so this could be suitable perhaps?

It seems to me the used molar ration peroxide : iodate about 2 : 1 is far too low, if iodate is said to be catalyst. it does not have IMHO enough space to  cycle between iodate and iodine. You may also use starch solution for small iodate concentrations.

@HeathDimsey Your are right, the reagents are different, the reaction is a different one. With "An increase in temperature reduces the cycle in the range of hours." (in the article about the Bray-Liebhafsky reaction), my recollection clearly was about the iodine clock where the states "pitch black/dark blue" and "almost colorless" alternate much faster (initially about a minute; though over time, it slows down / changes are less and less sharp).

@HeathDimsey Starch is notoriously known indicator for free iodine, forming intense dark blue even for small iodine concentrations. Search for practical preparation, as it may be tricky

@Poutnik By recollection from school I assumed the dark blue complex requires both $\ce{I2}$ and $\ce{I^-}$ to form $\ce{I^-_3}$ which then enters the (water soluble) starch - as e.g. on first page preview of a 1978JACS3215 paper, or a ChemistryViews snippet by 2016. The equations above however do not show yet a $\ce{I^-}$.  Is there a publication that iodine alone (now) is capable to form said complex, i.e. to revise/extend the scope of the test?

Hm, You are right it needs I3^-. I am not sure if there will be enough equilibrium I- to form enough I3^- and how equilibrium is shifted by forming adduct with starch. Would have to be tried.

In a paper by 2017, Cervellati and Greco provide some context about said discovery.  Bray's first publication by 1921 for research up to 1917/18 already describes how the initial concentration of sulfuric acid affects the rate of reaction:

Normality (as e.g. $\pu{0.0916 N}$ about trace III in figure 1) is a description of concentration no longer used (for quite some time, actually).  In contemporary terms – because sulfuric acid can provide two protons – a solution of $\ce{H2SO4} (\pu{2 N})$ is a solution of $\ce{H2SO4} (\pu{1 M})$, or $\pu{1 mol/L}$.  Within this context, the initial concentrations about trace III in figure 1 are: $\ce{H2O2}$ ($\pu{190 mmol/L}$), $\ce{KIO3}$ ($\pu{94 mmol/L}$), and aqueous $\ce{H2SO4}$ ($\pu{45.8 mmol/L}$).
(1) Cervellati, R.; Greco, E. Periodic Reactions: The Early Works of William C. Bray and Alfred J. Lotka. J. Chem. Educ. 2017, 94, 195–201. doi 10.1021/acs.jchemed.6b00342.
(2) Bray, W. C. A periodic reaction in homogeneous solution and its relation to catalysis. J. Am. Chem. Soc. 1921, 43, 1262–1267. doi 10.1021/ja01439a007.  There equally is an open access copy of this seminal paper on zenodoo (link).