text:
What is the physical meaning of fractional oxidation numbers of elements in compound
Many times I encounter oxidation numbers like -1 in the peroxide for oxygen, -1/2 in superoxide, and in some other compounds some elements have 8/3 or such like that...what is the physical meaning of fractional oxidation numbers and how can i imagine that to be at the molecular level

Oxidation numbers are human invention, they don't necessarily have physical meaning.

https://chemistry.stackexchange.com/questions/52010/is-it-possible-for-one-specific-atom-in-a-molecule-to-have-a-non-integer-oxidati https://chemistry.stackexchange.com/questions/5357/are-fractional-oxidation-states-possible

The main issue here is that you are asking for all atoms to have the oxidation states. One thing that pushes us in this direction is that our specification for chemical formulae does not differentiate in general different kinds of the same atom.
Suppose you have a compound where half the oxygen atoms have oxidation state -1 and the other half have oxidation state -2. This is not captured in our notion of a chemical formula. For example, sodium superoxide is $\ce{NaO2}$. If you want all of the oxygen atoms to have the same oxidation number, your best bet is to average over all oxygen atoms. In this case, the average is $-\frac{1}{2}$.
We can take this a step further. Why should we require that each atom have an integral oxidation state? The logic behind this is that electrons are atomic components and can't be split. But molecular ions and molecules are complicated, and some times you need to share one electron over the entire system. A fractional oxidation state may be the most accurate way to describe atoms of this system.

Okay thank you but please tell how oxidation numbers in different integer are different from others and what is their physical meaning?

One of cases is the element have multiple oxidation states within the compound, that has, or even has not a stoichiometric composition given by small integers.

Oxidation number is a formal chemist invention  of expected charge of an atom, if a molecule or a substance is hypothetically torn apart to particular atoms, according to the conventional electronegativity.
Some compounds would lead, by this mental operation, to sharing an electron among multiple atoms, like superoxides, what formally lead to oxidation state -1/2, as both atoms are
Some compounds have a particular element in a mixed bonding state, leading to different formal oxidation states in distinguished molecules like thiosulphates, or to a mixed, average state, like some oxides as $\ce{Pb3O4, Fe3O4}$.
Some compounds, have non stoichiometric composition, typically solid metal oxides, with  elements of multiple oxidation states, leading to even a rational number of the formal oxidation state. It is usually caused by some vacant oxygen or metal atoms in the oxide matrix.


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Cahn–Ingold–Prelog Rules, cyclic structures, and breaking ties

(3S,5R)-3-ethyl-5-methylcyclohex-1-ene
Does anyone know how one reaches the conclusion that the 5C is R? I tried using the Cahn–Ingold–Prelog priority rules but kept getting confused.

From chem.libretexts.org:

When looking for the first point of difference on similar substituent chains, one may encounter branching. If there is branching, choose the branch that is higher in priority. If the two substituents have similar branches, rank the elements within the branches until a point of difference.

From Wikipedia:

To handle a molecule containing one or more cycles, one must first expand it into a tree (called a hierarchical digraph) by traversing bonds in all possible paths starting at the stereocenter. When the traversal encounters an atom through which the current path has already passed, a ghost atom is generated in order to keep the tree finite. A single atom of the original molecule may appear in many places (some as ghosts, some not) in the tree.

You don't need a digraph for this one. Your structure is A. Remove the double bond and add duplicate carbon atoms to each of the double bond carbons (B). These duplicate carbon atoms each bear three phantom atoms of atomic number zero. Clearly at C5 hydrogen has the lowest priority and methyl is the next lowest. The methylene groups at C4 and C6 are a tie. [There is no need to consider C2 since it comes into play for both routes. It is a wash.] The red carbon of the ethyl group bears C,H,H while the red duplicate carbon at C1 bears 0,0,0. Place your right hand on C5 and point your thumb in the direction of the hydrogen. Now your remaining fingers will point as follows: C4>C6>CH3>H.  C5 is of the R-configuration. See, Twisting stereoisomers with rings to determine R/S.  .

The most difficult thing in the presented problem is to determine the priority of the substituents at C5. User55119 explained the priority well but ended everything again with the notorious "hand rule", which is no less catastrophic than molecular turns. Because it is difficult to understand how to orient the hand.
Knowing the priority of the substituents and the direction of their decrease AS IT SEEN on a picture (CW or CCW), you just need to assign a configuration.
In the above example, the priority decreases counterclockwise, and the configuration should be S, but the lowest priority group (H) is CLOSER to us, so we have to change the configuration to the OPPOSITE and get 5R.

Your suggestion is fine as long as the lowest priority group is above or below the plane of the other three atoms. But how convenient is it if the lowest priority group lies in the plane? As to the notoriety of the hand rule, to each his own. After all, chiral is derived from the Greek for hand.

Good question! The solution is very simple - swap any two substitutes twice so that the younger group is behind the plane of the screen (sheet, board, etc.).
https://orgchem.tsu.ru/ENGLISH/R_S_swap.PNG

In VERY RARE cases of cyclic compounds the lowest priority group is placed in the ring, and swapping would be confusing.

source: Stack Exchange

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

authors: Макс Сергеев

sort: votes


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Methyl iodide in water
Iodide being a good leaving group and water being quite a weak $\ce{Nu-}$, does  $\ce{CH3I}$ undergo $\mathrm{S_N1}$ reaction? I was taught that whenever a protic solvent which can ionize the halide is present, $\ce{C+}$ will form and any type of $\ce{Nu-}$ will attack (both weak/strong).
Also, if not, is there any other reagent which brings about $\mathrm{S_N1}$ pathway?

Methyl generally does not participate in Sn1 reactions because the transition state for making a methyl cation is quite high in energy. You might be able to force it with something like antimony hexafluoride to rip off the iodide.

If you try to propose a mechanism involving CH3+, you will get called out: https://science.sciencemag.org/content/364/6440/eaax7083/

Methyl iodide does not undergo $\mathrm{S_N1}$ reactions. The methyl carbocation is so unstable that there is virtually zero possibility of unimolecular dissociation:
$$\ce{CH3I \not\to \color{red}{CH3+} + I- }$$
For all intents and purposes the SN1 pathway is only operable for secondary and tertiary halides.

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


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Strontium-89 vs strontium-90
Why is radioactive strontium-89 is considered medically useful for bones (along with other benefits) whereas strontium-90 (also being radioactive) is harmful for human body?
How does the addition of one atomic mass has adverse effects on human body?

"Medically useful" is an awfully vague term. A scalpel is medically useful, come to think of it. A scalpel can kill, too.

Strontium 89 (or 90) have zero benefit for human bones or other tissue. Sr-89 is used to treat (=destroy) bone metastasis. Normal, non-radioactive strontium is used for treating osteoporosis.

The half-life of $\ce{^{89}Sr}$ is 50 days compared to a half-life of 29 years for $\ce{^{90}Sr}$. So a large enough dose of $\ce{^{90}Sr}$ to be useful would leave the patient radioactive for the rest of his/her life.

Edited for use of $\ce{^{89}Sr}$. In a comment Karl pointed out that the isotope is used to treat bone cancer.

Not so much detectable, mind you. To treat cancer you want to hit it with a nice big dose of radiation in a short period of time to beat it down.

The physical half-life of Sr-90 is indeed 28.79 years; however, the biological half-time of strontium is shorter. In particular, the removal half-time of strontium from exchangeable bone volume is about 80 d. After ingestion of Sr-90, about half of the resulting total effective dose occurs within the first 4 years after intake.

The specific radioactive properties of an isotope are very different even from other isotopes of the same element. The chemical properties will be very similar.
The issue for using isotopes for medical purposes is that you need a combination of specific chemical properties and the right radioactive behaviour. So one strontium isotope will not be a remote substitute for another.

Both isotopes decay by beta emission, but strontium 90 has a decay energy of 0.546 MeV, while strontium 89 has a decay energy of 1.463 MeV. (CRC Handbook)
The range of the electron from Sr 90 is better suited to putting a strontium atom in the middle of a tumor without irradiating too much of the surrounding flesh.
https://images.search.yahoo.com/yhs/search?p=range+of+beta+particles&fr=yhs-Lkry-SF01&hspart=Lkry&hsimp=yhs-SF01&imgurl=http%3A%2F%2Fwww.alpharubicon.com%2Fbasicnbc%2Fimages%2Farticle16radiological7104.gif#id=1&iurl=http%3A%2F%2Fwww.alpharubicon.com%2Fbasicnbc%2Fimages%2Farticle16radiological7104.gif&action=click


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Posible Pair of Enantiomers

What is the number of possible enantiomers in monochlorination of 2-methybutane?

My approach
When $\ce{Cl2}$ is added to any alkane it prefers $3^\circ > 2^\circ > 1^\circ$.
When it is being added to 2-methybutane, final product (major) formed will be 2-chloro-2-methybutane, which will show 0 chiral centre, hence the maximum number of enantiomers will be
$$2^n = 2^0 = 1$$
My book gave me answer $2$, but no explanation is provided.

You gave a good explanation, the book did not give any - you win! Or they meant the question differently: of all the products that could form, how many are chiral?

@Who, I was wondering what happened to your doctoral title (:D). Btw preference of chlorination is highest at tertiary carbons, that's true, but don't you think it will definitely happen even at secondary and primary centers (albeit to a lesser extent). That's why we say "major product" is the tertiary one, but minor products also form.

Which book are you taking this question from?

Chlorine is less selective then Bromine in free radical reactions(at least here).Even though tertiary radical is most stable as in the post , its product product may not be major.

Partly related: What is the major product on chlorination of 2-methylbutane?

Assuming monochlorination of 2-Methyl Butane ,the following products are possible as shown in the figure below.


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What does “at%” mean and how it differs from “mol%”?
Sorry if my question is a bit obvious, but for some reason I'm getting confused with the percentages in this question:

Concentration of $\ce{A}$ is $\pu{80 at\%}$ and concentration of $\ce{B}$ is $\pu{19.8 at\%}$ for $\ce{AB2}.$ What would be the mass for $\ce{A}$ and $\ce{B}$ individually, if the atomic weight of $\ce{A}$ is $\pu{5 g/mol}$ and $\ce{B}$ is $\pu{10 g/mol}?$

I understand that the molecular weight $M$ for my compound would be:
$$M(\ce{AB2}) = (5 + 2\cdot 10)~\pu{g/mol} = \pu{25 g/mol}$$
How do I include the atomic percentage to get the mass of $\ce{A}$ and $\ce{B}?$
I am just not sure how to use $\pu{at\%}.$ My understanding is that if I assume $\pu{1 mol}$ of $\ce{AB2},$ then I can calculate the mass of $\ce{A}$ by assuming that $\pu{80 at\%}$ is $\pu{0.80 mol}?$
I have another problem where they specify $\pu{mol\%}$ (mole percentage) instead.
So I am getting confused as to how to use $\pu{at\%}$ and $\pu{mol\%}.$

Atom percent is used in X-ray photoelectron spectroscopy or energy dispersive spectrometry. Could you provide some context? Is this an experimental data or just an textbook exercise?

Any kernel with nucleon number 5 decays extremely fast, so we can definitely exclude option of experimental data.

at% = mol% in case matter is atomised.


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Is this balanced redox equation correct?
I'm trying to balance an equation using oxidation numbers
$\ce{MnO4^- + H+ + Cl- -> Mn^2+ + Cl2 + H2O}$
my suggestion:
$\ce{MnO4- + 8H+ + 5Cl- → Mn^2+ + 2.5Cl2 + 4H2O}$
the 2.5 $\ce{Cl2}$ is throwing me off, would like to make sure my methodology is correct.

Could you edit the question and tell us what your methodology is (or how you check whether the equation makes sense)? How many electrons are being transferred, and from where to where? If the 2.5 is bothering you, just multiply all coefficients (including the explicit ones in front of the manganese species) by two.

The balanced equation seems correct. Multiplying by $2$ we get the following:
$$\ce{2 MnO4- + 16 H+ + 10 Cl- → 2 Mn^2+ + 5 Cl2 + 8 H2O}$$
which is also correct.

Actually, if you balance given redox reaction strictly following the half-reactions method where you have to equate the number of transferred electrons, you won't end up with fractional coefficients:
$$
\begin{align}
\ce{\overset{+7}{Mn}O4- + 8 H+ + 5 e- &→ \overset{+2}{Mn}^2+ + 4 H2O} &|\cdot 2 \tag{red}\\
\ce{2 \overset{-1}{Cl}^- &→ \overset{0}{Cl}_2 + 2 e-} &|\cdot 5 \tag{ox}\\
\hline
\ce{2 MnO4- + 16 H+ + 10 Cl- &→ 2 Mn^2+ + 8 H2O + 5 Cl2} \tag{redox}
\end{align}
$$
As with any equation, you sure can divide both parts by any real number which is not zero. Dividing the balanced redox equation above by two yields in your solution.


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Grignard alkene on a carbonyl
I was given the following mechanism to prove today post-lecture:

and my attempt is given below.
Now, when the negative charge on the alkene attacks the carbonyl, the electrons are pushed onto the oxygen - which is understandable. My question is: why do these electrons come back down, rather than just be protonated to give an alcohol?

This is the Weinreb "amide". In your first rxn., Mg is bound to both oxygens. The complex is stable until aqueous workup causes decomposition to the ketone. Your proposal would cause the ketone to react with the Grignard reagent. Ketones are more reactive than amides.

Why don't ketones form geminal diols in water? http://www.colby.edu/chemistry/CH242/15-2.pdf

@KarstenTheis perfect if you can add that as the answer then I can accept it as the answer!

@Karsten Theis: Not at all. I appreciate your concern and thank you for doing the right thing. Cheers! :-)

Nucleophilic acetyl substitution is common among carboxylic acid, acid chlorides, anhydrides, esters, and amides.
Two very common examples are anhydride decomposition and esterification of carboxylic acids. For example, if you look at esterification, carboxylic acid and alcohol first make an addition, giving tetrahedral intermediates 1 and 2. Intermediate 2 then collapses and displaces the water molecule as a leaving group to give the corresponding ester (see the scheme).

Your substrate is an amide. Like all other carboxylic acid derivatives, this molecule also undergoes a nucleophilic acetyl substitution with Grignard reagent, which is a strong nucleophile. The polarized carbonyl group has partially positive carbon, which is a good electrophile.
The nucleophilic carbon on Grignard reagent attaches to carbonyl carbon and gives tetrahedral intermediate as a metal alkoxide complex. But, this alkoxide can collapse to reform strong C=O bond, displacing the amino portion of the amide as a leaving group, in the form of the amide ion, $\ce{R1R2N-}$. This produces the ketone as a final product, assuming you have used only one equivalent of Grignard reagent (Channel reactions of Grignard reagent with esters).

Why does the reaction continue after the tetrahedral carbon forms?

After nucleophilic attack on a carbonyl, the resulting tetrahedral species either gets protonated to form the final addition product, or it is an intermediate on the way to the substitution product. This depends on the type of reaction. For example, a geminal diol will usually dehydrate to form a ketone. There is a nice overview of the different scenarios, including some nitrogen chemistry, in this document.


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Does ethanol "decompose" or "remove" phosgene from chloroform?
A question I was asked was: "Why is ethanol added to chloroform bottles?". I know that chloroform can be slowly converted to poisonous phosgene gas, and that ethanol is added to prevent phosgene levels from rising. However, the options I was given were:

It decomposes phosgene by converting it into ethyl carbonate.
It removes phosgene  by converting it into ethyl carbonate

Is there a difference between the two wordings, and is one more correct than the other? The answer is supposedly "decomposes".

I won't say it decomposes. It either converts to ethyl carbonate or all the phosgene gets consumed to form ethyl carbonate. So, for me, it should be "it removes phosgene".

Wanted to comment but the comment by Nilay Ghosh already gives the answer. https://en.m.wikipedia.org/wiki/Chemical_decomposition

It's a weird way of phrasing it, but I think they are trying to lead you to this
Basically, chloroform is stabilized with ethanol or amylene, but if a sample is dried, it will no longer contain stabilizer. To actually remove phosgene, you can wash it with NaHCO3. This would essentially leave you with entirely harmless products.
This is my best guess as to what they are trying to get you to understand. Please let me know if this helps.

What's your stance? Which option are you suggesting?


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Why are isotopes an issue in reading mass spectra?
It is my understanding that the isotope distribution for carbon and hydrogen, for example, are skewed heavily toward one species, i.e., one isotope accounts for ~99% of the abundance of the element in nature. If this is true (I question my ignorance on this point and others), then, the other isotope peaks should be negligible in the mass spectra or at least fall below the threshold for processing and remain unprocessed peaks.
I am asking this question within the realm of biological compound mass spectra. I read a post here that states that $\ce{Br}$ has near 1:1 abundance for two isotopes. It would explain the need to consider isotopes for those biological compounds with elements such as $\ce{Cl}$ which has ~3:1 ratio for its two most abundant isotopes. So, in light of the fact that $\ce{C, H, N,}$ and $\ce{O}$ make up 96.5% of a biological organism's weight, are isotopes an issue only when elements such as $\ce{Cl}$ or $\ce{Br}$ are constituents of the compound?

You may find these answers useful to look at a bit of the mathematics involved. Isotopes become quite important even for relatively small biomolecules composed of just H, C, N and O (e.g. oligopeptides, oligosaccharides, etc.)

@NicolauSakerNeto I included a link to your answer in my answer. I approximated with the Poisson distribution, you used the exact multinomial destribution, so your answer definitely adds value.

This is what happens to large molecules: https://en.wikipedia.org/wiki/File:Glucagon_mass.gif You need to be able to correctly interpret readings like these.

I rolled back the follow-up question and ask it as a new question: https://chemistry.stackexchange.com/questions/116997/significance-of-isotope-mixtures-for-electrospray-ionisation-mass-spectromety. Otherwise, the answers would have to be edited as well, which gets confusing.

Here is the "periodic table for biomolecules" (leaving out hydrogen, please ignore silicon):

The pie chart shows the isotope distribution. Of the elements C, N O, P, S and H, sulfur has the highest abundance minor isotope (around 4%). On the other end of the spectrum,phosphorous is a mono-isotopic element, with just one stable isotope. There is an interactive version at https://www.isotopesmatter.com/applets/IPTEI/IPTEI.html.
So for small metabolites, there will be little signal due to minor isotopes. However, in a macromolecule with hundred carbons, the 13-carbon isotope (around 1% abundance) does become relevant. The signal for the mass ion will approximately follow a Poisson distribution, with zero or one 13-carbon isotope in the molecule as most probable:

Source: https://homepage.divms.uiowa.edu/~mbognar/applets/pois.html
For smaller numbers of atoms in a molecule, the multinomial distribution is appropriate, as explained here: https://chemistry.stackexchange.com/a/87669/72973

You have already got an answer by Dr. Karsten, but let me add that modern mass spectra have become sensitive beyond imagination.
Secondly, the resolution of mass is no longer a problem. Fourier transform ion-cyclotron resonance mass spectrometer (a mouthful) is extremely sensitive to mass differences. Yes, if you have a low concentration, your main analyte peak may be buried and the isotopic pattern might be hard to distinguish.
However, if the concentration is decent, you should be able to see C-13 (and other isotopes) as stated above. One can find a chart, where the isotopic pattern is given for each element to help in interpreting mass spectrum. Isotopic pattern is actually not an issue but a blessing because it confirms the presence of a certain element in a complete unknown.

Don't refer to answers "above" -- the order that answers appear in depends on user settings and the number of votes answers have received. What's "above" for you today might not be above for somebody else, or for you, tomorrow. Instead, link to the answer you're talking about. (Edit submitted.)

No, isotopes are always an issue, even just for hydrocarbons.
$\ce{^13C}$ may only be 1.1% of most natural pools of carbon atoms, but as molecules get larger and larger, i.e. as they consist of more and more carbon atoms, the probability that at least one of the carbon atoms in a large molecule is carbon 13 becomes much larger.  Molecules with more than ~96 C atoms have an "M1" peak, i.e. a peak in the spectrum 1.0034 Da heavier than the "usual" monoisotopic peak that is more intense than the "M0" monoisotopic peak.  Molecules with more than 180 C atoms have an M2 peak that is more intense than the M1 peak, etc. etc.
The human protein titin has 169,179 carbon atoms.  At natural abundance, the most common mass isotopologue would have 1816 of these atoms labeled.  That is, the most intense peak of just the carbon atoms would be 1.8 kDa heavier than the monoisotopic peak.
The other thing that happens as molecules get larger is that the isotopic distribution gets broader and broader.  Any particular isotopic peak may not be well-resolved from its neighboring mass isotopologues.  And the "monoisotopic" peak may be very weak or completely absent from the spectrum.


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Why is ozonolysis of benzene possible?
I was looking at a question which asked to predict the product of reductive ozonolysis of benzene using $\ce{Zn/O3/H2O2}$  and I predicted them correctly using the cleaving of the double bond shortcut but then I realized that benzene is losing its stable aromatic state to form not as stable products.
So I would like to know why benzene proceeds through this reaction even if it doesn't seem as favourable? Also in that case, why don't reactions like hydrohalogenation or hydroxylation occur on benzene? As I see it, they would involve loss of stability too so what's so special about ozonolysis?

See this . The conditions required are quite vigorous

Then under such conditions, can the other reactions I mentioned also occur?

As per my knowledge, these other two reactions can't be performed under the conditions mentioned in the answer above. Different conditions may exist for hydrohalogenation and hydroxylation of benzene,but I am not aware of them (if they exist)


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How does a pH glass electrode work?
Every source I read mentions the potential difference building up at the glass membrane (according to the Nernst equation). But how is this difference actually measured? Sticking an electrode into a solution doesn't mean that I can measure the potential at a membrane which is some distance away. Hence, there must be some reaction at the electrode itself. Mainly Ag/AgCl electrodes are used, as far as I read. So what causes this reaction?
On the other hand, as I already noted, every source is talking about the ion selective glass membrane. It is selective regarding $\ce{H+}$ ions and doesn't let them pass through? But what does this have to do with the measured potential? The Nernst equation only mentions the different activities of the solutions.

you may find some helpful in formation here: pH probe bulb - what is happening within the glass?

I know this thread and read it already. Didn't answer my questions tho.


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Calculating phase diagram of mixed component natural gases (Bubble and Dew points)
I'm developing a piece of software which calculates and plots the Phase diagram (PT) for a mixed component natural gas.
For reference I have a common used natural gas plotted using a commercial software. See image above. The Bubble curve is purple and ends at the critical point. Dew curves with different quality (liquid phases of 25%. 10%, 5%, 1%) continues after the critical point.
My results, after plotting with Excel seems to be a bit different. I have not calculated the critical point, but the Bubble and Dew curves are similar. See screenshot below.
The blue curve is the first found Bubble pressure at any temperature (within the Bubble range of course). The green curve is the second found Bubble pressure at any temperature. Likewise with the Dew, orange is the first found Dew temperature and red the second found Dew temperature.
What I can't understand is why the second roots of Dew and Bubble both cross each other, and extend inside the phase envelope.
Is it a calculation error? Or do you have to calculate both Bubble and Dew and cut off the curves where the cross each others path, or turn into the phase envelope?

PS. I use Van der Waals mixing rules to mix the gas solution. I use Peng-Robinson EOS.
Otherwise I've been following this approach
https://cheguide.com/bubble_p_flash.html
https://cheguide.com/dew_p_flash.html

Pictures would have helped more than a little.

@IvanNeretin Yes you are right. Now added pictures.


text:
Is molecular mass different than the molar mass?
I thought that molar mass is the mass of one mole of a substance whereas molecular mass is the mass of one molecule of a substance.
But the molecular mass of water is 18.2 u,
which is similar to its molar mass i.e. 18.02 g/mol.
Are these two different and how?

33 kilograms of something and 33 grams of it look pretty similar in writing. They feel different, though.

The molecular mass is the mass of one molecule in amu (atomic mass units).

Related: Quick and simple explanation of molar mass, molecular mass and atomic mass and How are the molar mass and molecular mass of any compound numerically the same?.

That is like saying my car can go 92 mph, and the distance from home to the big city is 92 miles, so those two are similar. However, one is a speed and the other a distance. (And the reason they are similar is because in my world, an hour is defined as the time it takes my car to go to the big city at maximal speed).

Molecular mass is the mass of a single molecule
Molar mass refers to the mass of 1 mole of molecules, ie $ 6.023 × 10^{23}$ of them.
Eg:
 Molecular mass of $\ce{H2}$ is about
$3.32 \times 10 ^{-24}\ \mathrm g$ or about $2\ \mathrm u$.
Here "$\mathrm u$" is just a unit,
$1\ \mathrm u = 1.66 \times 10^{-24}\ \mathrm g$
Whereas molar mass is about $2\ \mathrm{g/mol}$.


text:
Calculate the pH
Calculate the pH of a dissolution of 100 ml of 0.1 M solution of ammonium chloride in the style of 100 ml of 0.1 M hydrochloric acid solution
$\ce{NH4Cl -> NH4+ + Cl-}$
$\ce{HCl -> H+ + Cl-}$

$\dfrac{(\pu{0.1M})(\pu{0.1L})}{\pu{0.2L}} = \pu{0.05M}$

     NH4+   +   H20 <-> NH3 + H3O+

[]   0.05        -       -    0.05

eq[] 0.05+x      x       x    0.05+x

$K_\mathrm{b} = \pu{1.8*10^{-2}}$
$K_\mathrm{b} = \dfrac{(x(0.05+x))}{(0.05-x)}$
From this equation I get $x = \pu{1.799*10^{-5}}$
$\ce{pOH = [H3O+] = 1.3}$
$\ce{pH} = 12.7$

Is this solved right?
Sorry but I don't know how to put it pretty

You need to calculate the contribution of the HCl to the pH.  That will be far more important than the ammonium chloride

the contribution of hcl is already in balance

The NH4 undergoes hydrolysis so an acidic solution results. The $K_b=1.6\cdot 10^{-5}$ for NH3 which gives a pH of $\approx 5.1$. ( Is there a typo in your value?) There is a general method see, https://chemistry.stackexchange.com/questions/60068/how-to-set-up-equation-for-buffer-reaction?noredirect=1&lq=1

The NH4Cl is just distraction. pH of HCl itself is so low that hydrolysis of NH4+ is fully negligible, considering implicit simplification due activity versus concentration. As [NH3] Is by about 8 orders lower than [NH4+], so does [H+] due this hydrolysis. Note that correction for activity by the simplest, Debye-Hueckel equation, is about +0.16 pH. Real one will be somewhat lower, but comparable.


text:
Calculating Mixing Enthalpy in GROMACS
What is the most reliable way to calculate mixing enthalpy of binary solvent mixture using GROMACS?
[EDIT]
I have simulated two-component systems of water and organic solvent with GROMACS. In a research paper from my field of study, authors calculate 'enthalpy of mixing' with the software in order to compare it with experimental mixing enthalplies. Under the Experimental Section of the paper, authors show that they do this according to this term:

A comment added in the text says the following:
"The internal energy, U, of the liquid mixture was obtained directly from the potential energy (see Tables S1
and S2) during the simulation, and yi is the molar fraction of component i."
I am interested in calculating same thermodynamical properties for my systems. I am aware that I can do that with g_energy function in GROMACS, or at least some parts of it (Pressure, Volume, Potential Energy).
My question is:
Firstly, is this an appropriate way of calculating enthalpy of mixing in GROMACS via this equation?
Secondly, how can one specify with the g_energy function which of the liquid's U(or potential energy), P and V is calculated if I initiate the function with .edr files, i.e. how can I choose to calculate i'th component's properties?
Reference for the image and quote:
Aguilera-Segura, S. M.; Di Renzo, F.; Mineva, T. Structures, Intermolecular Interactions, and Chemical Hardness of Binary Water-Organic Solvents: A Molecular Dynamics Study. J. Mol. Model. 2018, 24 (10), 292. https://doi.org/10.1007/s00894-018-3817-2.

BTW, the authors of that paper, according to your image, need an introductory course about scientific typography.

mhchem, could you elaborate? [EDIT] Do you mean there should be a space before gamma (or molar fraction in their case)?

@Koryphae There are several typographical flaws in that formula: 1. textual subscripts (liq, mix) should be upright; 2. minus sign doesn't look like one (vertically misaligned and has tight spacing on both sides); 3. use of asterisk * as multiplication sign; 4. inconsistent notations ("liq,1" but "liq2" without comma). Maybe @mhchem can find even more inaccuracies:) In defense of the authors, they use proper upright subscripts $(ΔH_\mathrm{mix})$ in the text, so it might as well be a journal editor to blame, but overall quality is indeed rather poor.

GROMACS won't be able to calculate the enthalpy of mixing directly but using the energy function in GROMACS you will be able to get energy, pressure, and volume of the simulations.
Going through the paper that you posted it looks like the ran multiple simulation boxes for the different components and mixtures. They did this because you will not be able to get U,P, and V for individual components in GROMACS. So in order to calculate the enthalpy of mixing for a 2 component mixture they ran 3 simulations: one of the mixture, and one of each component by themselves.
Once you have the simulation boxes for the different components and mixtures it should be pretty straight forward to calculate the enthalpy of mixing.
Just a side note: I am always a little nervous calculating thermodynamic expressions using pressure from GROMACS because the pressure will fluctuate so wildly. So if you're going to make sure you're careful about your equilibration run and your NPT equilibration run as well.
Below are the GROMACS manual for energy which you would use to get U,P, and V from your simulation. The next is a tutorial for an NPT run from Virginia Tech and their tutorials are really well written and should help as a starting point and finally the last is a chat between a user and GROMACS developer discussing pressure fluctuations in NPT.
http://manual.gromacs.org/documentation/current/onlinehelp/gmx-energy.html
http://www.bevanlab.biochem.vt.edu/Pages/Personal/justin/gmx-tutorials/lysozyme/07_equil2.html
https://mailman-1.sys.kth.se/pipermail/gromacs.org_gmx-users/2011-May/061370.html

Thank you for your Answer, @dtg67. What is actually puzzling for me, is as far as I can understand, there is an option for g_energy to choose number of molecules (-nmol) for the calculations. If following the formula they suggest and you calculate mixing enthalpies for pure components and the mixture, shouldn't number of molecules in a system for pure components be exactly same i.e. if you have mixture of 2500 molecules then you calculate thermodynamic parameters for 2500 of both pure components as well and then make the calculations according to the formula of mixing enthalpy?

Authors have provided their system compositions in Supporting Information and they have all different sizes for all solvents they use.

Yes you should use -nmol but thermodynamically we don't need to have the same N in all the simulations. Looking at the SI I think that they ran all the simulations at the same volume which is why the Ns are different

Okay, yes I see it now, that the using the equation, it has molar fraction component to it. However, when I calculate potential energy in my pure ethanol system, without specifying with -nmol, I get value of -100 000 kJ/mol. The paper discussed has value of -40 kJ/mol. I guess this is when you specify for "-nmol 1"? But now question arises, what is appropriate -nmol for the mixture? Shouldn't they be comparable?

I can't comment on the differences between the result that your simulation has with literature value but all the -nmol flag does is divide energies by the number that you specify.


text:
Pool Water Electrolysis for Rust Removal
I had an old tool I was trying to remove rust from using electrolysis.  I used a solar panel and chlorinated pool water.  I have had a difficult time online finding what the products of the reaction should be.  After hours in the sun the water turned to an orange reddish color and then almost green.  I had read about doing this with salt water and it producing chlorinated gas, does pool water go through the same reaction?
Thanks & Regards

Hi, and welcome to Chemistry SE.  What voltage are you using, and what does your setup look like?  You haven't described what happened to the tool?  It would also be interesting to do a control experiment where the tool (or something similarly rust) is simply left in the pool water.

I am using a usb solar panel. 5V at 2.4A.  I clipped a usb cable and am using anode and cathode that looks like Allen wrenches and I have them in a plastic cup. It appears like the rust is removed on the tip but on the handles it seems to stay. The tool is for cutting sheet metal, I think it is a mild steel and tip has a bit of a goldish tinge once rust is removed

Rust generally consists of hydrated ferric oxides and hydroxides and is a poor conductor of electricity. During rust electrolysis, the water that seeps underneath the bulk rust and comes into contact with the cathode is split, generating bubbles of hydrogen gas as $$\ce{2H2O(\ell) + 2e- -> H2(g) + 2OH-(aq)}$$ that then simply mechanically loosen up and dislodge the bulk rust from the underlying metal (which gave your solution the "orange reddish" color that you noticed after the electrolysis).

I used a solar panel and chlorinated pool water.

Chlorinated pool water is a rather oxidizing environment containing hypochlorites and chlorinated cyanurates, which may inadvertently lead to further corrosion of your tool via $$\ce{2Fe(s) + H2O(\ell) + 3OCl-(aq) -> 2FeO(OH)(s) + 3Cl-(aq)}$$ as well as the formation of chloride-green rust, which likely gave your solution the green color that you noticed a few hours later as well.

I had read about doing this with salt water and it producing chlorinated gas, does pool water go through the same reaction?

Hypochlorites and chlorinated cyanurates may release chlorine gas at the cathode as $$\ce{2H2O(\ell) + 2OCl-(aq) + 2e- -> Cl2(g) + 4OH-(aq)}$$
Saltwater pools can additionally release chlorine gas at the anode as $$\ce{2Cl-(aq) -> Cl2(g) + 2e-}$$

You remove rust mechanically . You can treat the clean surface with a "conversion" coating ( phosphoric acid  + ?) to slow new rust.

Would chemical / "non-mechanical" means of removing rust (e.g., EvapoRust/HCl) be a bad idea? For example, such as https://www.youtube.com/watch?v=6-MC_ZEXQbw and https://www.youtube.com/watch?v=CQMzUkEKaBo, prior to applying that conversion coating you're referring to afterwards...


text:
Tiffeneau–Demjanov rearrangement products
With reference to the paper McCasland, G. E. Pinacolic Rearrangements of Epimeric Aminocyclanols. J. Am. Chem. Soc. 1951, 73 (5), 2293–2295 DOI: 10.1021/ja01149a110. (ResearchGate link). It states that

cis-2-Aminocyclohexanol on treatment with nitrous acid undergoes a pinacol rearrangement, yielding both cyclohexanone and cyclopentylmethanal as products.

However, it goes on to state that the trans epimer would specifically yield the latter as the major product. Can someone explain this (preferably via mechanism)?

I would suggest that you draw out three-dimensional structures for both starting materials.

Quite a fun one to work out. The first thing to know is that there is a stereoelectronic requirement for the migrating group to be anti to the leaving group ($\ce{N2+}$), for orbital overlap reasons.

(apologies for hand-drawing - doing this on my phone)
You need to consider the chair conformations of both stereoisomers, in both possible ring-flip forms. For the cis stereoisomer, you would expect roughly a 1:1 mixture of ring-flip forms. When the $\ce{OH}$ is axial, the group anti to the LG is H, and the cyclohexanone is formed. Similarly, when $\ce{OH}$ is equatorial, the $\ce{C-C}$ bond is anti and the cyclopentane product is formed.
Now, for the trans-isomer there are two factors favouring the cyclopentane. First, the chair form in which both groups are equatorial will predominate, and this is the stereoisomer that leads to the cyclopentane. Secondly, in the (minor) conformer where the $\ce{OH}$ and $\ce{N2}$ are axial, there is no anti group to migrate according to the normal mechanism. The OH itself is anti to the $\ce{N2}$, and I'm not aware (off the top of my head) of any examples of 1,2-hydroxy shifts. However, I expect some epoxide might be formed via an SN2 pathway.

I took the opportunity to read the McCasland paper1 to which the OP's link referred. The products were isolated as their 2,4-dinitrophenylhydrazones. In the case of the cis-isomer 4, the reaction mixture was "apparently richer" in the aldehyde 3.
As @PCK has argued for the trans-isomer 1 the highlighted bonds are aligned properly for a ring contraction. Clearly, the diaxial conformation of 1 is not a factor.
Assuming that the diazonium group has an A-value similar to that of cyanide (0.2 kcal/mol; OH, 1.0 kcal/mol), the equilibrium constant for the two cis conformations at ambient temperature is 3.8, which favors conformation 4b. If indeed the formation of aldehyde 3 predominates over cyclohexanone 5, then hydride migration must be a slower process than ring contraction. In the event, orbital overlap of the red bonds in 4 is essential.

1) G. E. McCasland, J. Am. Chem. Soc., 1951, 73, 2293


text:
How one can calculate Si-O-Ca/Mg valence angle energy?
I want to perform the relaxed scan (or relaxed surface scan) for the valence angle energy of Si–O–Ca, Si–O–Mg, Al–O–Ca, and Al–O–Mg. I will use DFT (ORCA), probably B3LYP and triple zeta basis set (Def2-TZVP or 6-311++G(2df,2pd)) I'm making reference data to parameterize an emperical force field.
But the problem is metal ions, Mg²⁺ and Ca²⁺, because their coordination chemistry is very different from Si or Al. More precisely, I'm not sure how can I 'terminate' the Mg or Ca atoms.
I was thinking of making simple molecules, like (OH)₃–Si–O–Mg–OH or (OH)₂–Al–O–Mg–OH, just giving one OH to Mg or Ca. In this case, Mg or Ca have 2 oxygen neighbors. I can perform geometry optimization, but not sure if that geometry is proper one to perform valence angle scan or not.
I'm not considering aqueous environment (because this is for condensed phase material), so I'm not planning to use water molecules at current stage. But maybe I need to if there are no other way...

It's not true - coordination for Mg and Ca is 4 or 6, virtually never 2.

@Mithoron Yeah, I also read some papers introducing even 6 ~ 8 oxygen coordination around Mg inside MAS glass, though most stable oxygen number was not 8. Anyway, thanks for the answer.


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Robinson annulation
In the mechanism of Robinson annulation, what is the rate determining step?
I have searched in my book but it is not given. If possible please specify the reason for that step being the RDS.
Thanks in advance

It is a type of Michael addition. Are you aware of the mechanistic details and RDS of this particular type of addition?

For an annulation, the ring formation is usually fast given that it's intramolecular. Even if it's not, the reaction is run at high dilution to make it as fast as possible relative to the intermolecular step(s).

I suspect this would be highly dependent on the reaction conditions and particular substrates being used. Note that all of the steps are reversible.


text:
Vapour pressure of a solute obtained during eudiometric analysis of methane combustion
Problem

In a eudiometer $\pu{14 ml}$ of $\ce{CH4}$ and $\pu{38.5 ml}$ of $\ce{O2}$ was introduced. After explosion and cooling to initial temperature ($\pu{23 °C}$) the volume of eudiometer was found to be $\pu{25 ml}$. Calculate the vapour pressure(in $\pu{mmHg}$) of aqueous solution containing a non-volatile solute (A) at $\pu{23 °C}$. Mole fraction of A is $0.1$. The pressure of eudiometer is $\pu{1 atm}$ and is constant.

Answer

 $\pu{13.68 mmHg}$

Question
I wrote the balanced equation. Then applied Avogadro's law and found that the volume of gases in the eudiometer after reaction is $\pu{24.5 ml}$. So the extra $\pu{0.5 ml}$ must be due to some water which must have converted into gas. Using $pV = nRT$ I can calculate the amount of water evaporated, but I am unable to understand the last part of the question regarding solute A.
Is there some information missing in the question regarding solute A? I assumed it as the final solution after the reaction, but still don't know how to proceed.

The vapour pressure of the solution is equal to the partial pressure of water in the gas phase. I'm not sure how this question was intended. You could get the vapour pressure of water at the given temperature from published data (pure water: 21.1 torr), or you could figure out the partial pressure of water in the solution. A complicating factor is that CO2 is soluble in water - maybe you are supposed to ignore that. If the non-volatile solute makes the solution acidic, not much CO2 will dissolve, but if it is basic, a lot will dissolve.

Could you edit your question to show your work that led to 24.5 mL of product? Are you including water in that, or is that just CO2?

"The pressure of the eudiometer is 1 atm and is constant." "After explosion..." :/

@Karsten Theis. The 24.5ml I have accounted is the volume for the excess O2+ volume of  CO2 formed.

0.5 / 25•760*0.9=13.68, but this is not chemistry.

Could you please elaborate a bit more?

You seem to have put your own question on hold. It appears you might not understand how the site works, but this is going to prevent people from posting a complete answer to your otherwise not-so-bad question. In any event, you already solved the hard part, and if you can decrypt Karsten Theis's comment you should have a complete answer. Hint: Raoult's law says Pv = Ppure*X

thank you. I haven't learnt about raolt's law yet.  I checked Wikipedia about Raolt's law now and I think I now understand Karsten Theis's comment

I'll continue after volume of $\ce H_2O$ vapours = $0.5 ml$
Vapour pressure is equal to the partial pressure of water in gas phase.
$P_{\ce H_2O} = P_{total}$ × mole fraction of $\ce H_2O$
Where $P_{total}$ is $1 atm$ and mole fraction = $0.5/25$
So $$P_{\ce H_2O} = \frac{1}{50} atm = \frac{760}{50} mm Hg$$
Now for the solution of non volatile solute $A$
Using Raoult's law,
$P_{solution}$ = $P_{\ce H2O}$ × { 1 - (mole fraction of $A$) }
So $P_{soln} = (760 × 0.9)/50 = 13.68 mm Hg$


text:
Non-equilibrium gas cap above saturated liquid-gas solution
If I could somehow instantaneously replace the $\ce{CO2}$ gas cap at the top of a soda bottle with a different gas, say $\ce{N2}$, would the bottle pressure increase as a result of the $\ce{CO2}$ de-gassing into the gas cap space?
I've come to this question because I am trying to design a method to measure the dissolved $\ce{CO2}$ content of a water liquid stream. The method involves exsolving the $\ce{CO2}$ into either an evacuated or $\ce{N2}$-charged vessel and measuring the pressure change to correlate dissolved $\ce{CO2}$. By Henry's Law the pressure in an $\ce{N2}$-charged vessel should increase as a result of $\ce{CO2}$ exsolving since the $\ce{CO2}$ partial pressure above the liquid is zero, should it not?
More clarification (hopefully) for those who have marked the question as unclear:
Let's say I have a closed system (soda water in a bottle as an example) with a $\ce{CO2}$ gas cap at a pressure of 10 psig. The gas cap, like in a soda bottle, is small relative to the total volume, say 5%. The liquid (water) in the bottle is fully saturated with dissolved $\ce{CO2}$. Now let's say I somehow instantaneously replace the $\ce{CO2}$ in the gas cap with another gas, say $\ce{N2}$, at the same pressure (10 psig). The system is still closed (we've magically replaced that gas). Will the bottle pressure increase as a result of the $\ce{CO2}$ in solution exsolving into the gas cap space, since the partial pressure of $\ce{CO2}$ there is zero? This of course is hypothethical, but it would have some similarity to how the dissolved gas content of liquids drawn (pumped) from deep within a well could be measured.
Thanks

Guinness beer will do.

Please see my updated description, thanks

@IvanNeretin I had to look it up: http://allaboutbeer.com/man-invented-nitro-guinness/

To the point of the question: yes, it will increase.

Ok, thanks. Love the Guinness angle too...


text:
Are any commercial plastics made from biomass/plant matter?
Are any commercial plastics made from plant matter currently. I expect it can be done in a lab but I am wondering about cost competitive tonnage quantities.

You can buy "bio"ethanol-based PE. If you want to get your own share of responsibility for the destruction of the Brazilian rainwood forest.

Biomass polymers can and must never become more than a niche product for guillible customers, or for specialities.
Any attempt at upscaling these to large tonnages (that can actually make a dent in our oil consumption!) inevitably leads to food shortages and/or depletion of soils and biodiversity.
Same is true for biofuels. Photosynthesis is far too inefficient (max. 3%, compared to 15% for PV, and much more for thermal solar power).
Food production is bad enough for the environment as is. Don't make matters worse.
Actual examples for bio-based plastics are of course cellulose derivates, packaging made of starch, then polyols for polyester or PU, and the notorious sugarcane-ethanol based polyethylene.

While there is merit to your points, this doesn't really constitute an answer to the original question.

Also, bioplastics and biodegradable plastics are independent sets. Some bioplastics are not better biodegradable than conventional plastics, while some plastics have better degradability than some bioplastics.

@Poutnik There was no mention of biodegradibility in the question. As you say, that's very independent of the feedstock source.

@Karl I have not say there was. And that was the reason of my complementing note, as there is widespread mistake "bio-originated plast = bio-(Fast)degradable plast".

The original answer only gave a link to a commercial bioplastic producer.  The answer has been edited to provide more information.

Yes there are.  Here is an example: Biome Bioplastics
European Bioplastics (an industry group) estimates that the global production of bioplastics was about 2 million metric tonnes in 2017.  They forecast that this will increase to almost 2.5 million metric tonnes by 2022.Source
One estimate puts the market value of bioplastics at around 6 billion USD in 2018 and growing to almost 15 billion USD by 2023.  Source
Common sources of bioplastics are starches (like corn), cellulose and aliphatic esters.
Commonly, bioplastics are used to make compostable/biodegradable packaging and films, as well as disposable cups and utensils.

I do not see any info on the website of actual commercial production.   Lots of great customer PR but no production info. 30 years ago Amoco made some stuff out of biomass in the lab , but I am wondering if there is any serious production.

They have bioplastic products for sale, but do not talk about production - that seems standard for a commercial endeavour.


text:
What molecules are produced in carbon arc lamp and how do they produce light?
I am curious about the chemistry of the arc lamp (an indirectly for the incandescent bulbs) thus I would like to ask the following.
It is a common understanding that the electrons pushed inside an arc lamp -- let's assume a carbon one, that is with electrodes in graphite -- will collide on the atoms of the carbon crystal and increase their kinetic energy rising those in the cathode to about 3600 degrees and those in the anode to 4200 degrees. My questions are:

Why is the cathode, where the electrons arrive first and at full energy, cooler than the anode?
What kind of molecules are produced at the electrodes? It is said that carbon evaporates at the electrodes when the current is allowed to flow between them. What is the vapour of carbon? From wikipedia, dicarbon is listed, but what would be the reaction from graphite (pure carbon) to dicarbon?
If dicarbon is produced by evaporation on the arc lamp, would that be produced also in the old incandescent lamps that used carbon filaments, such as that made by Edison?
Does dicarbon react with oxygen? Does it oxidise to carbon dioxide or monoxide?
And finally, how does dicarbon produce light? I imagine that it must be at a high energy state and photons are released with the decay of the electrons to their ground levels, is that correct? Is there some kind of quantitative measure of why arc lamps are more luminous than incandescent or halogen lamps?

Thank you

Hot matter emits light: sun, filament in incandescent bulb, embers. The spectrum depends on temperature (black body radiation).

Could you edit your question to focus on one conceptual question? If you have more than one question, you can ask them in separate questions.

but they are all related to the chemistry of the arc lamp and in particular how it produces light. If I split them, then there would be redundancy...

Answers:

Consider the mass and kinetic energy of an electron or of an ion of nitrogen or oxygen, and the reactivity with carbon.
Good question! It would be interesting to see the relative abundance of allotropes of carbon deposited from a filament in vacuo. There is a metallic glisten to the deposit, so there might be graphite, carbon nanotubes or buckminsterfullerene. $\ce{C2}$ would not be there for long, though.
There are a great many chemical species produced by an arc lamp operating in air. Depending on the gases in which it operates, an arc can produce complex organic molecules. Even something as ornate as a buckyball can be made!
I've not seen the relative abundance of products of an arc lamp, but I assume carbon oxidizes rapidly to the monoxide and dioxide, depending on abundance of oxygen. Charcoal briquets oxidize at a far lower temperature.
Most of the light is black-body radiation, though there would also be some spectral lines produced by recombination, such as from nitrogen. With a spectroscope, you'd see a continuum with a few bright lines added (or dark lines from absorption, as the gases cool).


text:
Antimatter molecules properties
Do the antimatter molecules have the same chemical properties with matter molecules?Should we define electropositivity for antimatter molecules since the chemical bonding would be between positrons?

See https://en.wikipedia.org/wiki/Antihydrogen as a start. I have not seen papers on anti-matter molecules (rather than single anti-hydrogen atoms), but may have missed something.

Yes I know anti hydrogen and I know anti helium as well.Now they are trying to make anti lithium which will be very cool

Well, cool or not, no molecules yet. Given the initial results on anti-hydrogen, one would generally assume that the bonding energies for molecules would be the same.

Molecules of antihelium exist and molecules of antihydrogen also exist.

Cool - do you have a link?

Dont make fun of me ok?Well if they managed to create low energy antihydrogen atoms then they would have made antihydrogen molecules as well .

I'm genuinely curious, since I haven't seen any papers claiming that. And, no, given how few antihydrogens have been made there is no particular reason to believe that they have made molecules. And, while antihelium has been produced, it doesn't make molecules normally.

Many many antihydrogen atoms havebeen  made . And lately there were created antihelium atoms/molecules by colliding gold atoms.

Oops, I retract my comment on antihelium - looks like a handful were observed at RHIC in 2011...

Look, there is a difference between a molecule and an atom - lets keep the terminology straight, particularly on Chemistry SE!

There is no other  "anti"-stuff. I mean, sure, there are certain particles which have come to be known by the moniker of "antiparticles", but this is solely a feature of our area of universe (or maybe of all universe, because of the breaking of symmetry, but that's another question). The interaction of an antiproton and a positron is exactly the same as that of a proton and an electron. By "exactly", I mean "quite exactly, totally, to the very last digit, with all doubts bulldozed over". The antimatter is the same as the matter, only the other way around. No new theory is required.

Thank you Ivan but the properties of anti matter are not 100 per cent known.We dont know only one thing if antimatter has negative mass or regular mass but antimatter having negative mass is considered very unlikely.

A helium atom is not a molecule: https://goldbook.iupac.org/html/M/M04002.html A "monoatomic molecule" is an oxymoron.

@IvanNeretin - to be fair, there has been a lot of effort put into experimental validation of the symmetry between hydrogen and antihydrogen. Now, that was partly because there are some dark matter implications of an asymmetry. But, none of this really is particularly chemistry related.

Orthocresol so noble gases do not form molecules:)

Look:https://www.google.com/url?q=https://en.m.wikipedia.org/wiki/Gas&sa=U&ved=2ahUKEwj9oLW5j-PiAhVOyqQKHWM-D7oQFjAAegQIBhAB&usg=AOvVaw2JdjgZ4gdyWRXZGvZXK0sx

Molecules are smallest non dividable amounts of given substance :-) But if a substance consists of smallest non dividable amounts of chemical element, formally atoms of noble gases as elements are at the same time molecules of noble gases as substances.But as they are identical, we are used and somewhat turned it to convention, not to speak about them as molecules.

With modesty, if antimatter molecules would exist, I think that we can reasonably assume the situation as described in your question. Either you reverse the order, or you call it positropositivity ;)

Haha yes positronegativity

The following is an approximation based on the most accepted theories about cosmology (that I know of).
We currently consider that our observable universe is composed only of normal matter, by opposition with antimatter. Also, matter and antimatter should have the exact same physical and chemical properties.
However, there is a catch! According to the current theories, the Big Bang should have produced exactly the same amount of matter and antimatter, and both should have annihilated together by now. The fact that our universe is composed only(?) of matter implies that there is something fundamentally different between normal matter and antimatter which left an excess of normal matter.
Because we think that there must be a fundamental difference between the 2, we try to isolate antimatter and mesure some of its physicochemical properties to be compared with normal matter so we can determine where the current theory is wrong.

In principle they could be "identical" and a  very small excess of matter would have resulted in just matter left existing. But perhaps there are not enough photons around to allows for this possibility? Forgot the point, if the above is an idea of mine or if it something that my books considered and discarded.


text:
What is dielectric constant and how does it explain the reason why water is one of the best solvents
When i was reading solublity of ionic compounds,they mentioned that polar compunds are fairly soluble in polar solvents and water is better solvent due its high dielectric constant.Can anybody explain what is it and how does it affect solublity.

There is expected some a priori effort, like Wikipedia and/or  Google search.

2 comments about this question: (1) Dielectric constant is an average value but solvents can have different values along their structure; (2) water is a fantastic solvent for polar compounds but maybe water is not such a wonderful solvent and life (as we know it) evolved around water and created mostly water-soluble compounds.

That statement is not exactly wrong, but it also does not explain a lot. I recommend reading about the chemical properties of water and what a dielectric constant ist, separately, the connection will then become clear.

The dielectric constant $\epsilon$ (known also as permittivity) is a measure of the extent to which a substance is polarized under an applied (external) electric field. Polarization amounts to net separation of charge across the substance. A "dielectric" is the common name used to refer to a substance placed between the plates of a capacitor and used to store electrical energy in the form of polarization, that is, as charges separated across the material. A capacitor is the non-chemical analogue of a voltaic cell, that is, a form of storing electrical energy.
A substance with a high dielectric constant is easily polarized. Polarization allows countercharges to be placed around an ion, resulting in Coulombic interactions between solvent and ion, promoting solubilization of the ion by competing with interionic interactions. Similarly, a polar solvent - one with a high dielectric constant (relative permittivity $\epsilon_r>\approx15$) - will solubilize polar molecules by forming stabilizing interactions with the solute that compete with solute-solute interactions.
Water is a good solvent for polar molecules for the above-mentioned reasons. Water is sometimes referred to as the "universal" solvent, but it is generally a good solvent for polar solutes, not apolar ones.


text:
Priority of carbons attached to chiral group, one with double bonded oxygen, one with two oxygens
I have a chiral carbon. Two of the atoms attached to the carbon are other carbon atoms. One of those carbons has a double bonded oxygen and a methyl group attached. The other has two oxygens and a methyl group. Which carbon substituent has higher priority?

Could you edit the question to give an example of such a molecule, and try to apply the priority rules to it? The answer might depend on the context.

The double bonded oxygen counts as two atoms attached. The first is the real oxygen atom with no substituents. The second is a phantom oxygen that is like any other oxygen except it has immediately lower priority than a real oxygen atom.

Given the same atom, those bond via triple bonds have higher precedence over double and double bonds have higher precedence over single. This is exemplified here and, a bit more formal covering all CIP-rules, here.

Welcome to ChemSE.  I edited the answer a bit, because even if the original question leaves space for a little variation (second edit), the original OP eventually seeks an binary answer of 'either stop, or go'.


text:
What is responsible for the smell of incomplete hydrocarbon burning and what sensor can detect it?
After I pass the smoke from partially burned fuel oil through a filter, I can still very strongly detect the smell. It is the same thing you smell from the exhaust when an old car or truck goes by.
Even though I can easily smell it with my nose, none of my sensors can detect it. I've tried $\ce{VOC}$, $\ce{O3}$, $\ce{PM_{2.5}}$, $\ce{CO}$, $\ce{NO2}$, & combustible gas detectors and all come up clear on the filtered exhaust.
So what is left in this gas that is responsible for this strong smell (and the resulting headache) and what sensors can I use to detect it?

In analytical chemistry, sensors refer to analyte specific devices, e.g., glucose sensors, pH sensor etc. Smell on the other, generated by a burning complex mixtures such as fuels can rarely be associated with one compound. People have spent their life (e.g. Alan Marshall) studying the constituents of fuels. There are literally thousands and thousands of compounds.
However, if you were in a research lab investigating odors, you would be using something called as a olfactory gas chromatography technique. There is a chromatography column which separates all the gaseous/volatile components and a scientist puts a nose at the end of the column and records the "response". Later you can identify that particular odor by a mass spectrometer.
In short, there is no sensor per se. You need a instrumentation which would fit in a kitchen sized area to determine what causes that particular odor after burning an extremely complicated mixture such as fuel oil.
Here is how it looks like:

"a scientist puts a nose ..." – Would any nose do? :-)

Nosy noses are not needed in olfactory GC. Actually, you need a highly trained person's nose.

I was hoping someone had already identified the likely culprit(s) in this very common situation. Any thoughts as to even what kind of molecules could be contributing to this smell? Products of incomplete combustion? Compounds from impurities in the original fuel? Other? Interestingly you do not get this smell from a newer and well tuned engine. Thanks!

@bigjosh, have searched Google Scholar for this topic? The moment I search, the second result is a relevant paper. https://link.springer.com/article/10.1007/s00216-009-3071-7. Read the abstract. You can move on from there by using better keywords.


text:
Why does mixing two solutions decrease the concentration of the solutes?

The irreversible liquid reaction
$$\ce{A + B -> C + D}$$
follows an elementary rate law and is carried out
  isothermally in a plug flow reactor (PFR). The volume of the PFR is 500 litres. The concentration of $\ce{A}$ is $\pu{2 mol L-1}$ and concentration of $\ce{B}$ is $\pu{2 mol L-1}$ before mixing. The volumetric flow rate of each stream is $\pu{5 L min-1}.$
The streams are mixed immediately before entering. The reaction rate constant at $\pu{20 °C}$ is $\pu{0.05 L mol-1 min-1}$
  and $E = \pu{20 kcal mol-1}.$ (Note: gas constant $R = \pu{1.987 cal mol-1 K-1})$
Find conversion at $\pu{40 °C}.$

This is what I had:

From the solution given the $c_{\ce{A},0}$ used is $\pu{1 mol L-1}$ which is after mixing. Need some clarification on this. Why is it not $\pu{2 mol L-1}$ before mixing?

From the solution given the Cao used is 1mol/L which is after mixing. Need some clarification on this. Why is it not 2mol/L before mixing

Because it has been mixed. Each substance is now in doubled volume, compared to volume before mixing. If you had whiskey and if you mixed it with the same volume of soda, it wouldn't have the same strength as before, would it ?

alright.. thanks for the explanation.

Comment turned to answer:
Because it has been mixed.
Each substance is now in doubled volume, compared to volume before mixing.
If you had whiskey and if you mixed it with the same volume of soda, it wouldn't have the same strength as before, would it ?


text:
what is the difference between how taste and smell are detected?
Taste and smell are both chemical processes

... parts of the food dissolved in saliva come into contact with the taste receptors.[1] These are located on top of the taste receptor cells that constitute the taste buds. The taste receptor cells send information detected by clusters of various receptors and ion channels to the gustatory areas of the brain via the seventh, ninth and tenth cranial nerves. (https://en.wikipedia.org/wiki/Taste_bud)

, is there any difference in the way taste and smell receptors work?

Seeing, hearing, touching and thinking are chemical processes as well. Should then throw biology sciences out ?

@Poutnik, https://en.wikipedia.org/wiki/Flavor, says there are only three chemical senses

You wrote processes, not senses. Still, you should ask neurologists, not chemists.

Beside biology, neurology and psychology aspect are often neglected when we deal with human as detectors. This is another example along with many about optical phenomena. Ask in Biology SE.

@Poutnik, what is the third chemical sense?

I have not said there are 3 chemical senses.

Proteins are almost certainly the 'detectors'  for both smell and taste, so chemistry is at the heart of the process. For airborne chemicals these will have to diffuse through some form of 'mucus' before reaching the cells in your nose on the surface of which are the receptor proteins. Typically these would span the cell membrane. Interaction of the smell/taste molecule with the protein (in an active site usually inside the protein) may cause some ions to be transported though the protein and so into, or out of the cell, and which eventually gives us the sensation. Clearly the whole process is very complex to say the least. Bacteriorhodopsin works in this way (although process is initiated by light) and this process has been called isomerisation-switch-transfer (see Oesterhelt Biochem. 1977,36,p2) but the idea is generic.
The trigger would be your smell or taste molecule that causes a conformational change in the protein, the proton gradient  produced is then used to drive other chemistry that results after many steps, in a sensation. The very crude figure below gives the essence of the detection idea.

Taste receptors work with water (or saliva, or whatever other liquid that happens to occur in your mouth) and the things dissolved therein.
Smell receptors work with air and airborne chemicals.
Short of that, they are similar in that both are wide-range chemical sensors, and different in that they send different signals to our brain.

Thanks, what is the difference in the signals they send? are signal electrical , chemical or what?

@user157860 Ask rather neurology resources.

Not to answer: not necessarily different by their nature, but by their meaning

That's right: both are nerve signals (which are both electrical and chemical in nature), but they convey different information to our brain.

Both taste and smell are not that well understood yet. Here is a summary of some of the chemistry and physiology.
Types of substances detected
The sense of smell detects volatile substances. As stated in porphyrin's answer, scent molecules dissolve in aqueous solution and then reach the transmembrane receptors. The sense of taste detects substances dissolved in saliva, typically after they are ingested. That way, non-volatiles like sodium chloride or sucrose can also reach the taste buds.
Channels and receptors
Substances are detected by transmembrane proteins that either act as channel or receptor. Channels selectively take up e.g. $\ce{H+}$ or ions, triggering a downstream response in the cell. Receptors bind to a ligand and change conformation, triggering a downstream response in the cell (e.g. G-protein coupled receptors). There is a hypothesis that the taste of "sour" and the taste of "salty" might involve selective ion channels.
What triggers a given taste or smell?
Tasting is a combination of what taste receptors detect with other information (notably from smell of volatiles traveling from the back of the mouth into the nose). There are only a handful of tastes. Each receptor seems to be linked to only one of them. For smell (olefaction), one substance will trigger multiple types of receptors, and one receptor will be triggered by multiple substances. The combination of receptor types firing gives rise to the perception of a certain smell. This (along with the higher number of different receptors compared to the taste buds) explains why we can distinguish more smells than tastes.


text:
Can I safely use a separatory funnel to extract the precipitate from a solution?
Can a separatory funnel be used to collect a precipitate, or would that risk clogging the stopcock on it?
I'd like to recoup the precipitate for filtering without passing the entire solution through a filter (to avoid contaminating it, and also for convenience / saving time).

Most times you'll just jam the stopcock. Filter it first.

@Waylander Thanks for your answer! :)

If your wrote what kind of precipitate you are trying to recover, one could give better answers than "that's likely to go wrong".

@Karl: Thanks for this tip / specific feedback. It's useful to me in the process of learning how to best use this platform. As for the question, I was not thinking of any precipitate in particular but wondering if it would be smart to invest in a reparatory funnel in the prospect of using it for different purposes, and thus wondering if one of these purposes, in addition to separating liquid phase matter, could also be to conveniently extract the small fraction of a solution in which most precipitates are located (generally, with any precipitate). ANSWER CONTINUES HEREAFTER ⤑

Now pragmatically, the most likely precipitates I would be dealing with in an immediate future would be ordinary metal carbonates or metal salts.

And for what purpose? Preparatory? Inorganic analysis?

@Karl Yes, preparatory.

Good thought, but Waylander is correct.
However, you might consider centrifugation to consolidate the ppt, then pour off the supernatant. If you don't have a centrifuge, and the quantity of liquid is not too large (100 mL or less), you could make your own, since it seems that your ppt settles fairly fast.
Put the solution in a tough plastic bottle, cap it, and get a pretty strong string or light rope, tie the bottle SECURELY and twirl it over your head. Maybe outdoors is better for this, unless you are very careful. A few gentle swings (~20) should be enough, if it is going to work at all. I have had some success with this method a few times.
Secure tying is key, since the extra Gs will put a strain on your knots. I did something like (vertical swing, about 4 foot circle) this inside a room with 12 foot ceilings, and the plastic bottle hit the ceiling and splashed (I did not have the cap on). Fortunately, the stain was not very noticeable, but I did lose some of the material.

Don't do that outdoors, unless your material is water-based and environmentally harmless. You don't want to get an invoice for five cubic meters of soil that had to be dug up and disposed as hazardous waste.


text:
How did the tablet become like this?
How did the tablet become like this?
There was a gap in the packet, so I'm guessing it reacted with moisture but is it so?
Edit: Composition of 'Tendocare'- Chondroitin Sulfate and Vitamin C tablets.
Ingredients- Lactose, Chondroitin Sulfate sodium, Binder(468), Collagen peptides type 1, Sodium Hyaluronate, Binder(462) Stabilizer(1202),(1205)and(464), Silicon Dioxide, Magnesium Stearate, Sorbic Acid, Emulsifier.
Contains permitted natural colors and flavor. Appropriate overages added.
NUTRACEUTICAL, NOT FOR MEDICINAL USE.

The post plausible scenario

A bit hard to say with no indication of the tablet's composition. But, yes, many tablets swell and disintegrate when exposed to moisture.

@jeffB I've edited my question in response to your comment.

The most likely possibilities are mechanical crushing or expansion due to absorption of water vapor.


text:
Shouldn't it take more energy to break CO2 compared to CO?
I read in a book that bond dissociation energy of carbon monoxide $(\ce{CO})$ is $\pu{749 kJ/mol}$ and carbon dioxide $(\ce{CO2})$ is $\pu{532 kJ/mol}.$ Shouldn't it take twice as much energy to break $\ce{CO2}?$

532 kJ/mol times two...

I'm voting to close this question as off-topic because your question lacks context. This question needs revision before it is ready for a great answer. Please edit to include how this question came up and how you tried to answer it. This will help writing an answer that is useful for you and for others.

The bonds are not equivalent, therefore they have different bond energy.

Carbon monoxide consists of one carbon atom and one oxygen atom, connected by a triple bond that consists of two covalent bonds as well as one dative covalent bond.It is the simplest oxocarbon and is isoelectronic with other triply-bonded diatomic molecules having ten valence electrons, including the cyanide anion, the nitrosonium cation and molecular nitrogen. In coordination complexes the carbon monoxide ligand is called carbonyl.

Source: Wikipedia page on Carbon Monoxide

The bond dissociation refers to each bond.  Therefore, the number you cited for $\ce{CO2}$ is for only one of the C=O bonds.

Look at it simply. Bond strength is proportional to the number of bonds present. CO has a triple bond which is consisted by 1 sigma and 2 pi bonds. This kind of covalent bond is relatively hard to break compared to a single and double bond.


text:
Dosing leaves with Zinc Nitrate for pollution experiment - Need specific concentration
I am trying to assess the pollution tolerance of an insect using Zn as a proxy. I have a supply of Zinc Nitrate Hexahydrate that I want to use to contaminate dried leaves (the insect's food). The LD50 for my insect is around 2000 micrograms g^-1.
I was thinking I could achieve this by using 0.1 g squares of dried leaf that I would contaminate by topically applying 1 ml of Zn solution. To get my desired concentration, I am thinking that I would dissolve 200 micrograms of Zn into 1 ml of distilled water, so that when I apply 1 ml of this solution to a 0.1 g square of leaf I would have a final concentration of 2000 micrograms/gram of food. But if I wanted to apply only 0.5 ml of Zn solution to 0.1 g of leaf but still have the same final concentration of 2000 microg/g food, I would create a solution of 400 micrograms/ml. And if I wanted to use 1.0 gram of leaf and have a concentration of 2000 micrograms/g, I would make a solution of 2000 micrograms/ml. Is all of this correct?
I am also open to suggestions of how to better contaminate the food in a more consistent way. I thought about soaking leaves in a defined concentration, but then I am worried that some leaves will absorb more or less liquid than others, making the final Zn/g food concentrations unequal across different leaves.
Thank you for any solutions (pun intended) anyone might provide.

Could you please incorporate the comments within and following the posted answer to update your question. It seems the quesiton of what the wt/wt measure of LD50 refers to is not clear.

It seems to me that it will be difficult to get a uniform concentration of Zn in a non-uniform material like dried leaves, unless you measure only in bulk (this pile of dried leaves weighs X grams and contains 2000X micrograms of Zn). Things get more confusing if the insects don't eat all the dried-leaf structure (leaving stems, for example).
I'm also not sure how you'll get all the zinc from 1 ml of liquid onto a 0.1g leaf sample. (And, of course, the leaf won't be dry for a while after that!) I think a more concentrated solution would make your task easier. Your math for adjusting the concentration to match the amount of leaf matter appears correct.
As an aside, is your question confusing LD50 (micrograms per gram of insect) with leaf concentration (micrograms per gram of leaf)? I'm not sure why the concentration in the leaves would need to correspond to the LD50, unless there's something significant about the insects eating their own body weight in dried leaves...?

Thanks for your answer jeffB. The literature on Zn tolerance in insects does a really poor job of explaining the methods employed. I am more or less following the method that the majority of papers use (as far as I can tell), which will make it more likely that reviewers will complain less when I go to publish. For my specific insect, several papers indicate that at 2000 micrograms Zn/g food the insects start to die rapidly. So maybe not an LD50 per se, but there are not any other very good measures out there in the literature.


text:
Things to consider to determine stability of a compound to another?
Some Compounds with higher bond dissociation energy seem to be more reactive towards certain compounds then ones with lower bond dissociation energy , so what are the things you need to factor in to know if a compound will show reactivity with another?

BDE's refer to homolytic bond strength and not heterolytic bond strength. While the BDE of the C-H bond of acetylene is much higher than that of ethane, the pKa of the acetylene C-H bond (~25 kcal/mol) is more acidic than that of ethane (~50 kcal/mol).

First, you need to determine the relative stability of the compounds. Usually, this can be known by comparing the preferred state of the element or compound to its current state. For example, metals commonly want to donate electron so seeing a negatively charged metal should give a hint that the molecule/atom you're dealing with is unstable.
Second, the spontaneity of the reaction is a big help in determining if the two molecules would atleast react (the spontaneity is given by how negative the change in gibbs energy is).
Third, compounds with variety of resonance structures prefer to be themselves because the electron density is delocalized in the molecule.
Lastly, reactivity of the reactants is accompanied by energy. Presence of heat, current, work, etc. usually pushes the equilibrium of the reaction to favor the formation of the product. However, not all reaction have this kind of characteristics especially exothermic ones.


text:
Methods and observations that help you achieve complete decomposition of a compound in decomposition analysis
Different chemicals decompose in different ways to give different observations, but they usually involve heating the compound to high temperatures.
My question is that what are your lab techniques that help you ensure the compound you're analyzing is fully decomposed? what do you usually look for that signifies that your compound has decomposed fully?

Typically you insure complete reaction via the reaction conditions.

Bomb calorimetry falls into this category: To determine the enthalpy of formation of a compound, you combust a known mass of a sample into water and carbon dioxide and then apply Hess's law of heat summation.  To do this successfully, the sample is enclosed in an atmosphere of pressurized pure oxygen gas (hence calling the device bomb calorimeter).

Your opening sentence is not quite clear, or the question for that matter. Do you have a specific technique in mind?


text:
Does cooling a potato change the nature of its carbohydrates?
A talk-show guest of Joe Rogan claimed that cooling a potato after cooking creates "resistant starch" that's better for human consumption
https://www.youtube.com/watch?v=niwqfwA2Lb8
The claim is also made here: https://www.healthline.com/nutrition/cooling-resistant-starch
This seems to contradict everything I know about chemistry, since cooling only slows down all chemical processes.
Is there any validity to the claim? What is this "resistant starch"? Why does cooling a potato after cooking create it?

It doesn't happen so often, but wikipedia seems to explain it quite well https://en.m.wikipedia.org/wiki/Resistant_starch

Does cooling a potato change the nature of its carbohydrates?

Yes, retrogradation is a reaction that takes place when the amylose and amylopectin chains in cooked, gelatinized starch realign themselves as the cooked starch cools.
Resistant Starch:

There is weak evidence that resistant starch can improve fasting glucose, fasting insulin, insulin resistance and sensitivity, especially in individuals who are diabetic, overweight or obese. In 2016, the U.S. FDA approved a qualified health claim stating that resistant starch might reduce the risk of type 2 diabetes, but with qualifying language for product labels that limited scientific evidence exists to support this claim. Because qualified health claims are issued when the science evidence is weak or not consistent, the FDA requires specific labeling language, such as the guideline concerning resistant starch: "High-amylose maize resistant starch may reduce the risk of Type 2 diabetes. FDA has concluded that there is limited scientific evidence for this claim."

U.S. National Library of Medicine - "Studies on effect of multiple heating/cooling cycles on the resistant starch formation in cereals, legumes and tubers" (non-paywall source), (July 2009), by Yadav BS, Sharma A, and Yadav RB:

Abstract
  'Resistant starch' (RS) is defined as starch and starch degradation products that resist the action of amylolytic enzymes. The effect of multiple heating/cooling treatments on the RS content of legumes, cereals and tubers was studied. The mean RS contents of the freshly cooked legumes, cereals and tubers (4.18%, 1.86% and 1.51% dry matter basis, respectively) increased to 8.16%, 3.25% and 2.51%, respectively, after three heating/cooling cycles (P< or =0.05) with a maximum increase of 114.8% in pea and a minimum of 62.1% in sweet potato (P< or =0.05). Significant positive correlations were observed between the RS content and amylose (y=0.443x-5.993, r=0.829, P< or =0.05, n=9) as well as between the percentage increase in RS and insoluble dietary fiber content (y=2.149x-24.787, r=0.962, P< or =0.05, n=9). A differential scanning calorimeter study showed an increase in the T(0), T(p), T(c) and DeltaH values of the repeatedly autoclaved/cooled starches. The intact granular structure was also observed disappear, as studied using scanning electron microscopy.


text:
What does it mean for chemical reaction network to be in equilibrium?
What does being in "equilibrium" mean for a set of chemical reactions? If we have some number $n$ of $\ce{A}$ molecules and $m$ of $\ce{B},$ and the following reaction where $\ce{A}$ and $\ce{B}$ can reversibly bind:
$$
\begin{align}
\ce{A + B &->[$k_\mathrm{f}$] AB} \\
\ce{AB &->[$k_\mathrm{r}$] A + B}
\end{align}
$$
Suppose $k_\mathrm{f} = 2,$ $k_\mathrm{r} = 1.$ Can this system be "in equilibrium" and what would it mean? Does it mean all reactions have to be reversible and that as $t \to \infty$ go on at the same rate? In that case it seems that this system cannot be in equilibrium, since $k_\mathrm{f} \neq k_\mathrm{r}.$

k$_f$ and k$_r$ are rate constants. You have to multiply them by concentrations to get rates. So the rates can be equal at equilibrium even if the rate constants are not.

and so at equilibrium $\ce{ k_r [AB]=k_f[A][B] }$ making $\ce{K_e=k_f/k_r=[AB]/[A][B]}$ which means that species A, B and AB all exist.

@porphyrin is it correct to say that any set of such chemical reactions will always have at least one equilibrium? if not what is an intuitive example of one without it, if we assume conservation of mass?

No it is not a rule, reactions do not need to have an equilibrium step, $A \to B$ or $A\to B \to C$ type reactions are common.

Ignoring the math, "in equilibrium" means the forward and reverse reactions are proceeding at the same rate, so there is no gross change in the amount of reactants. In a closed system, if initially one reaction was faster, then it's product will start to dominate, and, by the law of mass action, the reverse action will "catch up" because there is now a larger supply of those reactants.
A simple physical analog:

Balance a ruler on two fingers, one finger near the end of the ruler, the other finger inset from the end towards the middle.
Gradually move your fingers together.

No matter where you start, the fingers will come to rest near the middle of the ruler.


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Reaction between copper and silver(I) nitrate in aqueous medium

What is the reaction between $\ce{Cu}$ and $\ce{AgNO3}$ taking place in aqueous medium?

The answer given was:

"Only" oxidation of $\ce{Cu}.$

My concern is: how is only oxidation of $\ce{Cu}$ happening, as redox will occur in pair?
So either $\ce{Ag}$ will reduce, or $\ce{NO3-}$ will. But, since $\ce{Ag}$ has higher reduction potential, it won't. What will happen?
What does $\ce{NO3-}$ reduce into, if it does at all?

The context of other answers would help to understand, why the answer was formulated this way.

Why, according to the corresponding SEPs the reaction $$\ce{Cu(s) + 2 AgNO3(aq) → Cu(NO3)2(aq) + 2 Ag(s)}$$ is plausible and in fact takes place (sometimes called a "snowy tree" reaction). However, you are correct, the given answer makes no sense.

@andselisk Sorry , I got confused with SEP of Ag and Cu, You are right , Thanks a lot :-)

@andselisk Also does$\ce{ NO3-}$ , gets reduced or oxidised ?? What does it reduce or oxidise to??

Nothing happens to nitrate (N is already fully oxidized) and you should try quite hard to get it reduced under those conditions.

@user232243 No prob, keep in mind you can answer your own question (feel free to use my comment as a starting point, no need to cite me though) if you understand it now.

@EdV Sure, zinc is a much better reducing agent that copper. Nice addition to the question!


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Column length for baseline separation
I have a GC-chromatogram with 4 retention times and peak widths and a column length.
How can I calculate the column length that I need in order to get baseline separation for all components in the sample?
I know I can calculate the resolution between the peaks and that it should be equal or greater than 1.5 for baseline separation, but in what way should I use the equation of Rs in order to calculate the best column length, using all retention times for the components?
Thanks in advance!

Resolution depends on three factors (i) efficiency N (ii) alpha or selectivity $\alpha$ and (iii) retention factor k, which the same as the capacity factor in older texts.
Guess which factor is dependent on the column length? It is only the efficiency. If you recall the resolution equation:
$$R_s=\frac{\sqrt{N_2}}{4}\left[\frac{\alpha-1}{\alpha}\right]\left[\frac{k_2}{k_2+1}\right]$$
you need the desired resolution, say 1.5, you already know $\alpha$ and ks from the experimental retention times. All you need is to solve N. You would also know that N scales linearly with the column length. With these hints you can estimate the necessary column length.
Other ways to increase resolution, if it were a research problem: You would use a smaller thermal gradient, change temperatures, and lastly change the column length.
In the real world, people would use brute force, i.e. they would change the column, because it's the selectivity which affects the resolution the most. For example, pharmaceutical companies test about 8-16 columns at once (HPLC). The column which separates a given sample with their fixed mobile phases is chosen for further optimization.

Now use simple ratio and proportion. Suppose your shorter 30 m column has an N of 1x, where x is some number. However the the equation above shows you that N must  be 2x. Calculate the corresponding length. You need a column which is 2x30 = 60 m long. I just chose random numbers. Share your chromatograms and calculations if you wish to double check.

Miana, a very important point. All these equations are valid for ISOCRATIC separations only. Make sure that the exercise/question is clear about it. Even experienced people make this mistake.


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Reaction of coloured metal complexes with water and sulfuric acid
Reaction of coloured metal complexes with water and sulfuric acid
I recently started researching the coloured compounds made from transition metals focusing on one with oxidation state II, so Mn,Fe,Co,Ni,Cu and Zn with all of them in hydrated metal sulfate complexes.
I've done some research and found why they have their colours. However I still do not know what metal ligand complex will my compounds will make when reacted with water and concentrated sulfuric acid?
Knowing the end product would help me in formulating the reaction equation and would be easy to understand for the presentation.

If your question is about the composition, synthesis and appearance for all compounds with $\ce{M(H2O)_$x$SO4}$ formula (M = Mn, Fe, Co, Ni, Cu, Zn), then it seems to be very broad.

@andselisk I just chose this from a few experiments that were available and we aren't expected to go too deep. I need to explain why they have different colours, google helped with that and also show the reaction equations forming the coloured compound which i am  not a 100% sure about. For Fe(II) it would be green and the complex would be $\ce{[Fe(H2O)6]^2+}$ but I don't know what complex the others would make.

Your question is slightly unclear to me: your question states that you want to know what the reaction of metal complexes with conc sulfuric acid is, but comment says otherwise.

If a metal is made to react with sulfuric acid, it will form a sulfate salt. When you dissolve those sulfates in water, you would be studying the color of the aquated metal ion. Sulfate is just a counter ion in each case. Search crystal field theory.
You are right, iron (II) sulfate in water would be light green, Zn would be colorless, Mn would be light pinkish, nickel would be green, copper would be blue. This is the color of aquated metal ion i.e. when the central metal ion is surrounded by six water molecules.

and would the general formula for each of these complexes be $\ce{M(H2O)^2+_$6$}$ where M = Mn, Fe, Co, Ni, Cu, Zn? Crystal Field Theory would then explain why they have different colours?

Yes, you are right.


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How can I figure out if my store bought chlorine is the percentage it says it is?
Chlorine is an important part of maintaining any pool. I've read horror stories of people buying chlorine from pool stores or big box stores and the chlorine being ineffective.
If stored long enough Chlorine can degrade, lose it's strength. If put in the sun i've read that it can speed up that process.
Like most pool owners, I have a pool test kit that tests how much chlorine is in a pool water sample. I specifically own a Taylor K-2006c Kit which test for Free Chlorine using the FAS-DPD Method.
At most pool stores they supply 12.5% Sodium Hypochlorite and 87.5% Other Ingredients.
With my given test kit would it be possible to accurately test whether or not it's actually 12.5% Sodium Hypochlorite or not? If so, how would I go about doing so?
If not; What do you recommend using in order to test the chlorine.

According to this source (https://www.taylortechnologies.com/en/page/142/putting-fas-dpd-to-the-test) the high end for this analysis is 20 ppm free chlorine. This would require such a high dilution of the bleach that the results would be meaningless. Also, welcome to chemistry.SE!

@airhuff I assumed as much, but I am willing to buy another experimental test kit if one is suggested.

Bleach solutions can be analyzed by addition of iodide and back titration with thiosulfite - somewhat difficult for non-laboratory conditions. Kits are available, but expensive (~$250).
A less precise (but probably adequate) method involves measuring the heat evolved as bleach oxidizes acetone. https://www.flinnsci.com/api/library/Download/650c3e8ed0574ed18c7b37b8832cccae It seems that you might be able to modify this process to do a rough quantification (+/- 10%) quickly. It was also discussed briefly on Quora: https://www.quora.com/When-happens-when-acetone-and-bleach-are-mixed
Just realize that if you drop solid hypochlorite into pure acetone, it could explode, catch fire, and make you wish you had never tried it. Creep up on this reaction: add only small amounts of acetone (1-2 grams) to a dilute hypochlorite solution (~5% ?) until you get the feel of how fast and how hot this reaction will go.
Eye protection is mandatory. Also have some water available (like a hose). Just be safe by expecting danger.


text:
About Smoke Grenades and Sugar Rockets
The most common formula for smoke grenades and sugar rocket fuel is $60\%\ \ce{KNO3}$ (potassium nitrate) and $40\%\ \ce{C12H22O11}$ (sugar). This is a 3:2 ratio, but I've also seen 2:1 and 5:3 with similar results.
The formulas I've seen (https://www.scienceforums.net/topic/19364-kno3-and-sugar-smoke-bomb/) are: $\ce{5C12H22O11 + 48KNO3 -> 24K2CO3 + 36CO2 + 55H2O + 24N2}$ and $\ce{C12H22O11 + 24KNO3 -> 24KNO2 + 12CO2 + 11H2O}$.
Based on the first formula, using mole ratios and molar masses, it would be ideal to have $\pu{68.5 g} \;(\pu{0.2 mol})$ of sugar mixed with $\pu{194.1 g} \;(\pu{1.92 mol})$ of oxidizer. This is not even close to the advertised 3:2 ratio, being a 17:6 ratio.
Question 1: Why do you need substantially less potassium nitrate than as indicated by the chemical equation?
Now, in terms of the smoke grenade, an alternate formula I've seen used involves potassium chlorate (oxidizer) and lactose (same as sucrose). This, from what I've read, burns at a lower temperature and can produce colored smoke.
Question 2: Which one produces more smoke than the other (and why)?
Question 3: How can I modify the $\ce{KNO3}$ formula to produce more smoke. Which reactant produces the smoke? Fuel (sucrose/lactose) or oxidizer (K-nitrate/K-chlorate)? From a 60:40 ratio, should I add more sugar to make it like a 40:60 ratio, or more potassium nitrate, therefore 70:30 is better?

I'm not sure the reaction is correct, I've found another source with the following reaction: $\ce{2KNO3(s) + CH2O(s) → 2KNO2(s) + CO2(g) + H2O(g)}$, $\ce{CH2O}$ being an empirical formula for glucose, which is actually $\ce{6 x CH2O}$. The ratio would still be off (about 6:1), not sure why. Maybe the recipes simply work with an excess of sugar?

Well, if there was no air there and you you didn't want any, well, smoke, then putting that much oxidiser would be proper.

@Mithoron Are you saying that more sugar would release more smoke? I kinda figured that because smoke comes from incomplete combustion, by piling on more organic stuff, more smoke would appear.

Either you want smoke or a complete reaction.


text:
intuitive explanation of Wegscheider condition and its violation
The Wegscheider condition restricts the values that rate constants in a set of chemical reactions can take on. can someone please explain intuitively why the condition has to be true, and give a simple example of a set of reactions and rate constants that violate it? why could such a system not exist physically?

The Detailed Balance In Wikipedia page has details on this condition. In a system of reversible reactions it seems to boil down to the Law of Mass Action.

@porphyrin ... and $K_{1+2} = K_1 * K_2$, i.e. Gibbs energy is a state function.

@porphyrin that page is dense and lacks simple example. does wegscheider imply that in $A \rightarrow^{k_1} B, B \rightarrow^{k_2} A$ we must have $k_1 \times k_2 = 1$? what defines a "cycle"? Is this a cycle?

No, the product of rate constants can have any positive value $\gt 0$. If $K^e$ is the overall equilibrium constant for a series of equilibria,$\ce{A<=>B<=>C<=>} \cdots $ then $K^e=K_1^eK_2^e\cdots $ where $K_i^e$ are the individual equilibrium constants, $K_1^e=k_1/k_{-1}=\mathrm{[B]/[A]}$ etc. for rate constants $k_1, k_{-1}$. (Unless you absolutely have to, don't bother with this Wegscheider stuff unless you really want to dig around in the fundamental math of differential equations as applied to chemistry. )


text:
Why electrons move to Cu in galvanic cells?
I'm student who learning about chemistry, and I got problem about understanding why electorns move to Cu in galvanic cells.
Let's think the reason for electrons going to Cu is that the standard reduction potential of Cu is high. It seems strange that Cu does not have a Cu2 + ion to be reduced, but Cu takes electrons.
How about lead? it is a bit strange too that the electrons follow the lead always.

Electrons follow neither copper neither lead. They follow voltage gradient. Galvanic connection of electrodes disbalance their equilibrium potential, causing reduction of $\ce{Cu^2+}$ on copper electrode.

Yeah, Poutnik is right. If you're wondering where from did the $\ce{Cu^2+}$ come, it is present in the electrolyte solution.

In the galvanic cell you are talking about, the copper is the inactive electrode. The activity, the electron production, comes from the more active metal. When a neutral metal atom falls into solution, it will leave is electrons behind:
$$\ce{M -> M^2+ + 2 e-}$$
But those electrons are just a few of the many that are left behind by all the $\ce{M}$ atoms falling into solution. They repel one another, pushing all the way to the copper electrode, where they may find an exit. If the solution contains an easily reducible metal ion, or a $\ce{H+}$ ion, the ion may pick up an electron (or two):
$$\ce{Z^2+ + 2 e- -> Z}$$
or
$$\ce{H+ + e- -> H}$$
followed by
$$\ce{H + H -> H2}$$