text:
How to activate polystyrene with carboxyl group?
Is there a simple (that is one that requires few and not very exotic reagents) chemical reaction to "activate" or "functionalize" the polystyrene of a simple microtitre plate?
In other words, I would like to add carboxyl groups to the surface of a common ELISA plate to then use ethyl-carbodiimide hydrochloride and N-hydroxysuccinimide to immobilize a target on the bottom of the well.

These may be useful if you have access https://www.tandfonline.com/doi/abs/10.1080/03602559.2017.1370109?journalCode=lpte20   https://www.tandfonline.com/doi/abs/10.1080/07366578708078821?src=recsys    https://www.sciencedirect.com/science/article/abs/pii/003238619390483Q

It is certainly possible to nitrate polystyrene powder with nitric & sulfuric acid mixture and reduce the nitro groups to aniline with $\ce{SnCl2/HCl}$ as described in here1. How effective this would be for a microtitre plate I cannot say, but chemistry is an experimental subject.
Reference

Zhao, H. & Guo, J. & Bian, L. & Wei, Z. & Deng, K.. (2009). Synthesis of nitro and amino linear polystyrene. Hecheng Shuzhi Ji Suliao/China Synthetic Resin and Plastics. 26. 22-26. (link)


text:
Adsorption of Cu2+ using banana peels for pH 2,2.5,3,3.5,4
so basically I did my experiment which is the title. This is my procedure and results:

Make 0.05M CuSO4
pipette 50ml CuSO4
change to respective pH values using HCl
Add 2g of oven-dried banana peels
wait for half an hour
Filter the solution into 100ml volumetric flask
add 5ml of 2M HCl to change all replicates to approx. pH 1
top up to 100ml mark using deionized water
mix well
put solutions into cuvettes and place the into the UV-vis spectrometer to measure absorbance
compare it with the calibration curve to determine concentration

my 入max was 811.
For those wondering why I made all the replicate the same pH after filtration, it was so that the pH at the end would not affect the absorption value, because I found out that pH changes the colour of the solution.
results:

edit: so the first reading is a blank, the 2nd reading is the absorbance of the pH 2 solution after it has been made to pH 1, the 3rd reading is the absorbance of the pH 2.5 solution after it has been made to pH 1, the other readings are for pH 3,3.5 and 4 (in order). The K*abs value doesn't matter.

calibration curve:

Does this look right? Cus I can't find any research papers that do this. The ones that I found do pH 2,3,4,5,6 and use extremely small concentrations of CuSO4, something like 0.0002M. Even if u assume the trend will be the same, my results don't follow the same trend, which is that the concentration of CuSO4 left in the solution after the experiment decreases as pH increases. Basically it means that at a higher pH, the banana peels can adsorb more Cu2+ ions.

You do not give us enough information. For example, the first image reports 6 absorbance measurements, 1 to 6. But are they related to the same sample ? Are they related to different samples ? What are their meanings ? What should we do with theses values ? What is the meaning of K*Abs being always zero in this image ? Why is there no information about pH ?

ahh ok. I'll put more info

@Maurice I added some explanation below the first picture. To ans ur qn about what to do with those values, it is in steps 10 and 11

I would say that the absorbance does not depend on the pH value, and that the unknown sample has a concentration of 0.02 (unit unknown)

@Maurice that's not really what I'm asking. I want to know if the results look ok? cus i found this journal that says my trend (in the excel table) should continue going down.
https://www.geomatejournal.com/sites/default/files/articles/227-234-3c-anwar%20hossain.pdf

@ Brien. Are the samples measured in the first table identical or different ? Is it the same sample measured at different moments ? Is it a test for checking if the technique is reliable ?

@Maurice The samples in the first table are different. The first reading is a blank(just deionised water). The subsequent readings are the samples with different pH. They are in the order of pH 2,2.5,3,3.5,4.

@ Brien. The absorbance does not seem to be related to the $p$H. The absorbances values are changing randomly. If you repeat the same series of measurements with another series of samples, do you obtain the same values ? I doubt it. I have a feeling that the only differences between these absorbances are random. And that that all may be presented by : $0.25 ± 0.02$.

@Maurice I've done some research and quite a few studies have said that the concentration of cu2+ left in the solution should decrease as pH increases. They don't use uv-vis though, cus the concentration they use is extremely small. However, if they did use uv-vis, they would have gotten decreasing absorbance as pH increases. But I don't know if that applies to the concentration I'm using

@ Brien. Who is "they" who "don't use UV-Vis" ? You have modified the pH between 2 and 4. Why ? And I repeat : Have you tried to repeat the same series of measurements, just to see if your experimental values are reproducible ?

@Maurice "they" are the people who have done the paper/journal that i linked. I modified my pH between 2 and 4 because i need 5 data points, and the first time i asked a question on chem stack exchange, i asked if it was possible to get pH 5&6 CuSO4 and u replied no. I have not repeated it, cus this is actually graded, so there's like a fixed amount of time that i can go to the lab. But I might try to see if i can go again. But for now i just wanted to know if my results were even remotely correct

@Maurice i also wanted to know if my experimental procedure was ok. cus i can't find any procedures online that i could follow even though there are quite a few experiments like mine, that change pH, but they use a different heavy metal solution and different biosorbent

source: Stack Exchange

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text:
Possible structure if a compound like AB4 is polar?
I was asked the following question in JEE Mains (2nd September, Shift 1) 2020,

If a compound AB4 is polar (having non-zero dipole moment), then it's structure could be
(A) Square planar
(B) Rectangular planar
(C) Square pyramidal
(D) Tetrahedral

My thoughts

As this compound has only 4 ligands, therefore it cannot be a square pyramidal.
It can neither be a tetrahedral nor a square planar bcoz, it would be non-polar then.

So, we are left only with "rectangular planar", which I'd never heard before. How to move on?

I don't know how one would get to square pyramidal with 4 ligands (I can imagine a weird case...) but that's the only answer that's not overly symmetry so as to eliminate any chance of being polar. I was expecting see-saw.

Strictly interpreted none of those geometries would be polar (unless they mean trapezoid not rectangle). Trigonal pyramidal would be but that isn't what they said.

My guess is that by square pyramidal they mean that the ligands form the square base of the pyramid and the central atom is the apex, which would make it the correct answer.

Trigonal bi pyramidal was one option i think....

@sai-kartik: I don't think so, bcoz if that had been present, I wouldn't ask this question then :P (also, TBP needs AB5 formula)

@RahulVerma what about Lone pair on AB4  (see-saw) ?

@sai-kartik: it would be see-saw then, and it would be polar, yes. But, I don't remember it...

https://en.wikipedia.org/wiki/Sulfur_tetrafluoride.

Given all $\ce{B}$ are equivalent in terms of AXE notation, for $\ce{AB4}$ the following coordination environments are possible [1, p. 177]:

square-planar $(D_\mathrm{4h},$ SP-4);
square-pyramidal $(C_\mathrm{4v},$ SPY-4);
tetrahedral $(T_\mathrm d,$ T-4);
see-saw $(C_\mathrm{2v},$ SS-4).

Table IR-9.3 Polyhedral symbols, geometrical structures and/or polyhedra

Option (A) square planar implies a centrosymmetrical environment and such molecule cannot be polar.
The derivative of (A), option (B) rectangular planar, as well as option (D) tetrahedral also refer to the environments that posses zero dipole moment.
Speaking of (C), I think you are misinterpreting the geometry of square pyramidal coordination.
It's a square non-coplanar arrangement with a non-zero dipole moment commonly seen in period 4 transition metal porphyrins $(\ce{MN4}$ fragment) and also lead(II) oxide and tin(II) oxide $(\ce{MO4}$ fragment).
According to Brecher [2, p. 1944], the structure of bis(N,N-dimethyldithiocarbamato)lead(II) would be yet another example:

Sample depiction of a compound known to exist in square pyramidal configuration:

Since a see-saw option is not present, this leaves only one answer, option (C).
Reference

IUPAC. Nomenclature of Inorganic Chemistry, IUPAC Recommendations 2005 (the “Red Book”), 1st ed.; Connelly, N. G., Damhus, T., Hartshorn, R. M., Hutton, A. T., Eds.; RSC Publishing: Cambridge, UK, 2005.
Brecher, J. Graphical Representation of Stereochemical Configuration (IUPAC Recommendations 2006). Pure and Applied Chemistry 2009, 78 (10), 1897–1970. DOI: 10.1351/pac200678101897.

By square planar, I'd assumed the structure of XeOF4 and that ceased me to think of other types of square pyramidal. So, what would be the hybridization of this kind of square pyramidal, or is it unhybridized?

@RahulVerma Since the majority of examples for the four-coordinate square pyramidal environment are for the metal complexes, I wouldn't fixate on hybridization (see Why is it wrong to use the concept of hybridization for transition metal complexes?).

It's square pyramidal.  What you miss is there are only four ligands, but the central atom itself forms the apex of the pyramid.  In practice, though, such a molecule likely distorts to a seesaw shape which could still be polar.

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text:
Electron configuration of iron(I) cation
What is the electron configuration of $\ce{Fe+}$ cation?
\begin{align}
\ce{Fe+} &\!:~ [\ce{Ar}]\mathrm{(3d)^6(4s)^1}\label{chm:1}\tag{1}\\
\ce{Fe+} &\!:~ [\ce{Ar}]\mathrm{(3d)^7(4s)^0}\label{chm:2}\tag{2}
\end{align}
What I've been told is that \eqref{chm:2} is correct since the $\mathrm{4s}$ electron jumps to $\mathrm{3d}.$
Which representation is correct?

https://chemistry.stackexchange.com/questions/27662/what-is-the-electronic-configuration-of-feii-ion

Isolated in a vacuum, in the absence of external fields, the first configuration is correct - $[\ce{Ar}]\mathrm{(3d)^6(4s)^1}$ is the ground electronic state of the iron(I) cation $\ce{Fe^{+}}$. More specifically, the ground state also has the term symbol $\mathrm{^6D_{9/2}}$.
NIST's Atomic Spectra Database has compiled the ground state electronic configurations of almost all possible ions of almost every known element (including such exotic things as $\ce{U^{58+}}$), obtained from a combination of experimental data and theoretical calculations. For example, search for iron, and you will be presented a table with all ground state electronic configurations, from the neutral atom $\ce{Fe^{0}}$ to the hydrogen-like $\ce{Fe^{25+}}$. You can find a description of the output data here.
I'm sure you'd like a reason for why $\ce{Fe^{+}}$ has the first configuration you suggest and not the second, but there is almost certainly no explanation which is both simple and correct. Unfortunately the predictive power of the aufbau principle is somewhat overstated, even when taking into consideration qualitative corrections. Depending on your notions, there are arguably several dozen "unexpected" electronic configurations, across all possible ions of all elements. The worst offenders are the heavier elements (say, the sixth period and onwards), which are rarely (if ever) covered in early chemistry classes.


text:
The quartic electronic state of octahedral Fe(III): How make crystal field splitting diagram?
Using simple arguments from crystal field theory, how do I rationalize the quartic state of octahedral Fe(III)? I am not able to in a logical way distribute the 5 d electrons in such a a way as to end up with three unpaired electrons in the normal splitting diagram for octahedral complexes.

Fe(III) has $5 d$ electrons. If the $\Delta_O$ value is small, the five $d$ levels are all occupied by $1$ electron. This is a stable situation.

Do you mean the quartic excited state? start with low spin $t_{2g}^5$ and excite one of the paired $t_{2g}$ electrons to $e_g$.


text:
Behavior of Natural Abundance and Mass Spec Ratios
The presence of sulfur in a mass spec is 96% and 4%. I believe this to be because the natural occurrence of sulfur isotopes are within these proportions.
However, chlorine behaves in a 66%/33% ratio in mass spec, but the most abundance isotopes of chlorine are 75%/25%. Why is this so?

See also chemguide mentioning 3:1 atomic ions and 9:6:1 molecular ions, based on webbook.nist.gov

What sort of compounds are you using ?

Or in more general terms: ratio (66:33 relative to each other) vs percent (75% and 25% relative to their sum).

Side note: isotopic distributions may vary both along time and place.  This is why IUPAC's atomic weight for Li seems to be less and less precise today and compared to other elements. Isotopic fingerprints are used in geology not only to tell age, but origin of the samples (example) e.g.for the analysis of gemstones as well as in forensics (example).

You are probably mixing natural abundance (NA) and relative abundance (RA).
In mass spectrometry RA is a more valuable parameter as it can be directly obtained as the $y$-coordinate of a plotted mass spectra: the most abundant ion (isotope) corresponds to the base peak, which is always $100\%.$
In other words, RA reflect isotope ratio, not NA.
For the isotopes of the elements RAs can easily be derived from NAs via normalization; however, the problem of finding RAs of the various isotopic molecular species is a bit less trivial [1].
The following table contains compiled data for NAs [2, p. 1-12] and RAs [3, p. 89] for both elements you've mentioned:
$$
\newcommand{\d}[2]{#1.&\hspace{-1em}#2}
\begin{array}{lllrlrlr}
\hline
Z & \text{Isotope} & & &\text{Mass}/\pu{u} & \text{NA}&\hspace{-1em}/\% & \text{RA}&\hspace{-1em}/\%\\
\hline
16 & \ce{^{32}S} & \ce{[E]}      & \d{31}{9720711744(14)} & \d{94}{99(26)} & \d{100}{000} &\hspace{-1em} \\
   & \ce{^{33}S} & \ce{[E + 1]}  & \d{32}{9714589098(15)} & \d{0}{75(2)}   & \d{0}{789} \\
   & \ce{^{34}S} & \ce{[E + 2]}  & \d{33}{96786700(5)}    & \d{4}{25(24)}  & \d{4}{433} \\
\hline
17 & \ce{^{35}Cl} & \ce{[E]}     & \d{34}{96885268(4)} & \d{75}{76(10)}    & \d{100}{000} \\
   & \ce{^{37}Cl} & \ce{[E + 2]} & \d{36}{96590260(6)} & \d{24}{24(10)}    & \d{32}{399} \\
\hline
\end{array}
$$
References

Margrave, J. L.; Polansky, R. B. Relative Abundance Calculations for Isotopic Molecular Species. J. Chem. Educ. 1962, 39 (7), 335. DOI: 10.1021/ed039p335.
Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data, 97th ed.; Taylor & Francis Group (CRC Press): Boca Raton, FL, 2016. ISBN 978-1-4987-5429-3.
Gross, J. H. Mass Spectrometry: A Textbook, 3rd ed.; Springer International Publishing: Cham, Switzerland, 2017. ISBN 978-3-319-54397-0.

So 66% / 33% is wrong, it should be 100% / 33% or 75% / 25% if you round a bit. For sulfur with the low abundance of S-34, ratio or fraction are similar, at about 4%.


text:
How is nomenclature decided in this case (formyl vs oxy)?
I was asked the IUPAC Name of the following compound in JEE Mains (2nd September, Shift 1) 2020.

There were two names occuring to me:

5-formyl-2-methyl-hex-2-enoic acid
(2,5)-dimethyl-6-oxo-hex-2-enoic acid

I finally chose (2), but the temporary answer key says (1). Can someone please tell which is the correct one and why?
Note: As I realised after receiving a comment, I've accidentally made the double bond have (E) configuration. This is an unintended error due to the website I used to construct the compound's structure. Please note that the original compound did not have any stereochemistry indicated

Neither one. There is no (E) designation for the double bond. Possible help: https://chemistry.stackexchange.com/questions/117065/difference-between-oxo-and-formyl

@user55119: Actually, the OP drawing had (E) config. The original question asked in JEE had no specific config. It should be mentioned in the post, I think.

I think the simple answer would be to first define the parent chain. As there are no rings, there is no need to leave the aldehyde group out of the parent chain. So the option with formyl should be incorrect.

According to the new recommendations of IUPAC nomenclature rules, when choosing the senior parent structure, you have to first consider the senior parent structure, which has the maximum number of substituents corresponding to the principal characteristic group or senior parent hydride in accord with the seniority of classes (functional groups). Then, if there is a choice, choose the principal chain, which has the greater number of skeletal atoms. Next in order is the chain, which has the greater number of multiple bonds, and then has the greater number of double bonds (This is according to Blue Book in 2013, which was displayed in 5 steps you have mentioned from Wikipedia).
Accordingly, your most senior function is carboxylic group, which would have the highest priority:

Then, there is a choice to choose the principal chain, which has the greater number of skeletal atoms. Both choice have six carbons, but the numbered chain indicated in the image has additional function, thus it is the principle chain. You also need to consider the stereochemistry if they are indicated. Here, there is a double bond in trans-orientation. That should be included in the name. Although, $\ce{C}$-5 is a chiral carbon, its stereochemistry is not indicated so you can leave it alone. Therefore the correct IUPAC name is: (2E)-2,5-dimethyl-6-oxohex-2-enoic acid.

Just as a sidenote, the original question had no stereochemistry indicated. It's just that, I couldn't find any online tool to draw the compound without indicating strereochem

"but the numbered chain indicated in the image has additional function, thus it is the principle chain" Could you please cite an official source saying so, because to challenge the answer key, a verified source citation id compulsory.

It is in the blue book. For instance given answer, 5-formyl-2-methyl-hex-2-enoic acid, is also carbon 6 chain. When you have choice to choose, you go with the one with most functional groups. Methyl and halo groups are valued equally but oxo group has a priority.

Further, in case of formyl group attached to carbon-4, then you could use the name (2E)-4-formyl-2-methylhex-2-enoic acid (also, between methyl and hex does not have a hyphen). And, given answer is wrong by not giving stereochemistry of the double bond, anyway.

I couldn't find the point saying "when both chains have same number of carbons, choose the one with more functional groups" in blue book. Could you please locate it for me?

Read this accepted answer. Find out why it is not a heptane derivative.

Aldehyde is taken in parent chain because a point in the Blue Book of IUPAC prioritises aldehyde due to it being the principal group. Here, that's not the case. The principal group is -COOH

"Formyl" and "oxo" are used to denote when the aldehyde group is a substituent/carbon skeleton component respectively. Seeing as the aldehyde group is clearly in the main chain, the "oxo" prefix must be used and therefore the second option must be the correct one. Regardless of this, seeing as the carboxylic acid group must have the highest priority, the aldehyde must be numbered 6, so we can rule out the first option.

Actually what the second option does is that it takes the methyl group on C-5 into the parent chain, hence rendering the -CHO as a substitution. This leads to the formyl name on C-5. But, the question is how do we decide which is supposed to be the parent chain?


text:
Why the formation of a fog is observed when triethylamine is added?
In the procedure for the synthesis of N-anisoyl-2-pyrrolidinone that was given to me, it is written that when triethylamine is added, a "fog" is observed inside the flask.

Triethylamine is added to neutralize the hydrochloric acid that is generated during the reaction, forming triethylammonium chloride.
I searched for similar procedures involving an acyl chloride and an amine, but the formation of a "fog" is never mentioned.
A quick search on the internet gave me no clue.
In my experience I have never witnessed the formation of a "fog" with this kind of reactions.
Since I cannot enter the lab due to the COVID-19 restrictions, I cannot perform the reaction now, can someone help me to understand if and possibly why in this case a "fog" is formed?

What happens if you put two small beakers with conc. HCl and conc. NH3 next to each other?

@Karl I edited the post, I apologize for my mistake.

I'm not sure what you wrote is actually wrong, just the alternative is more usual here, and should imo ring a bell. :-)

You are adding Et3N to a solution that contains HCl as a reaction product. That means there will be HCl vapour above the solution. The Et3N is fairly volatile so there will be Et3N vapour around the addition stream/droplets and this will react with the HCl vapour as it is added to give triethylammonium chloride. This gives you the fog.


text:
Arrhenius acids and bases
According to the Arrhenius definition of acids:
An Arrhenius acid is any species that increases the concentration of
hydrogen ions in its aqueous solution.
Hydrogen Chloride gas ionises in water to give hydrogen and chlorine ions. Thus, it increases the concentration of hydrogen ions in the solution. So is hydrogen chloride gas considered to be an Arrhenius acid or is the aqueous solution of hydrogen chloride, i.e., HCl(aq.) gas an Arrhenius acid?
If the aqueous solution is an acid then why is hydrogen chloride gas referred to as an ‘acidic’ gas?
Similarly, is NaOH(s) an Arrhenius base or is NaOH(aq.) an Arrhenius base?

The thing you're missing is that substances rather act as acid or base, depending on situation, then are an acid or base. Both HCl and NaOH do act as Arrhenius acid and base in aq. sol. Not so much in solid or gaseous phase.

As a complement to Mithoron, I would add that pure gaseous $\ce{HCl}$ is $NOT$ an acid. For example, it does not react with solid and anhydrous calcium carbonate according to : $$\ce{CaCO3(s) + HCl(g) ->  no reaction }$$ As soon as some droplet of water exists, HCl reacts according to : $$\ce{CaCO3(s) + 2 HCl(aq) -> CaCl2 + H2O + CO2}$$ Of course $\ce{HCl(aq)}$ should be rewritten as made of ions $\ce{H+ + Cl-}$ or even $\ce{H3O+ + Cl-}$


text:
Molar conductivity in compounds showing coordination isomerism
Question

Which of the following complex has lowest molar conductivity?

$\ce{[Co(NH3)6][Cr(NO2)6]}$
$\ce{[Cr(NH3)5(NO2)][Co(NH3)(NO2)5]}$
$\ce{[Cr(NH3)4(NO2)2][Co(NH3)2(NO2)4]}$
$\ce{[Co(NH3)5(NO2)][Cr(NH3)(NO2)5]}$

Answer

 3. $\ce{[Cr(NH3)4(NO2)2][Co(NH3)2(NO2)4]}$

My approach
My textbook only has examples of single centered complex ions, unlike what seems to be coordination isomers in the question. Normally, I see stuff with hydrates and ammine ligands which we predict by the number of active anions in the primary valency. But I have no clue on how to proceed with this question and the website does not provide any solution/ reference.

Hint: Check the ion charges.

Assuming NO2 to contribute -ve 1 charge, I can see that the third option is +1 -1 coordination in primary valency, while others have higher charges. So since this particular compound having the least charge separation it should have the least molar conductivity, is my logic right?

If there are ions A+ and B++ of about same size in the same electrostatic field, which would move faster ? Is conductivity proportional to mobility and charge ?

B++, since it is conductivity is proportional to charge, but I think hydration effects and charge/mass ratio is also involved.

They are minor to the charge effect, that is higher than proportional. As the ion is faster AND carries bigger charge.


text:
Powdered Dipropylene Glycol Monomethyl Ether
Entrepreneur here, background in electronics, so thank you in advance for your patience with me.

Question:
Is Dipropylene Glycol Monomethyl Ether available as a powder? If not, is there a powdered alternative?
Application:
My invention, for everyday people to use inside their homes on steel, plastic and aluminum, requires two liters of diluted liquid degreaser used at room temperature. On lay assumptions pertaining to design, logistics and safety, I am intent on offering this as powdered "just add water" degreaser. Dishwasher detergent does work, but a liquid degreaser sold by Harbor Freight Tools, diluted 50:1 works significantly better -and- it's MSDS describes overall threats on par with dishwasher detergent. It contains:

Sodium MetaSilicate Pentahydrate - readily available in powder form

Tetrasodium EDTA readily available in powder form

Dipropylene Glycol Monomethyl Ether Unable to find in powder form

I could not determine the constituent percentages. (sub-question: is there a way find this out?)

Its a liquid at room temp https://en.wikipedia.org/wiki/Di(propylene_glycol)_methyl_ether

EDTA does not belong into municipal wastewater. Forbidden in houshold detergents > 1% in e.g. Switzerland.

Thanks Waylander and Karl. Environmental concerns are another key element - this degreaser must be able to be safely disposed of or neutralized then disposed of. International sales are hoped for, so thanks Karl for bringing it up. Thanks Waylander, I figured the powder question might be like asking for powdered water for room temperature use.


text:
Assigning R/S configuration

Question:
Assign the R/S configurations to the carbons which are attached to different rings of this molecule (carbons 1 and 3 according to IUPAC numbering):

My Solution
I first started out by numbering the carbons. The one at the top would be 1 and the bottom chiral carbon would be 3. Next, According to CIP rules, I have included the phantom atoms by replacing the double bonds while deciding the priority order. Moving on, at carbon 1, My priorities would be:

$\ce{-OH}$
Cyclopentene ring (on the left)
Cyclopropene ring (on the right)
Carbon 2

This gives me a counter-clockwise turn and since the least priority group is on the vertical line (away from us) the configuration I assign would be S.
Following a similar procedure for carbon 3, The priorities would be :

Bottom 5-member ring (contaning $\ce{O-O}$ bond)
3 member ring on the right
5 member ring on the left
Carbon 2

Again, this gives me a counter-clockwise turn and since the least priority group is away, I would assign the configuration: S.
The answer key, however, says that the answer is 1R, 3S.

Where am I going wrong?

Of the oxygen containing rings, doesn't the one on the bottom have the lowest priority because there is only one oxygen atom on the carbon attached to the rest of the molecule, whereas for the rings on the left and the right, there are two?

I have included the phantom atoms by replacing the double bonds while deciding the priority order. - you did it wrong, as Karsten just mentioned

The configurations are 1R, 3S. For C1, the order is 1, 3, 2, 4 using your numbers. One switch gets you from S to R. For C3 the order is 2, 3, 1, 4. Two switches here keeps the S assignment. You were correct for the wrong reason. I'm preparing and answer but the drawing takes some time.

@sai-kartik: You may want to look at this link for help with digraphs: http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/isomers/RS14272/pinene.html

@user55119 hey, i followed the link you provided and tried to get as much as possible. But i have a confusion, which one has priority over real atom and duplicate atom, as in a case of (1R, 5S, 6R)-5,6-Dimethylcyclohex-2-en-1-ol, it was written that C,C,H has priority over C,(C),H. That duplicate atom was considered for double bond. I think real atom has priority there, can you please tell whether I am right or wrong?

Arpit Raj Choudhary: A real atom is the same as a duplicate atom as to the  atomic number. The difference lies at the next level. In  (1R, 5S, 6R)-5,6-Dimethylcyclohex-2-en-1-ol The real methyls have [H,H,H] attached while the duplicate methyls of the double bond have atoms of at. no. zero attached, i.e., [0,0,0]. Clearly, every real atom outranks a duplicate atom. (C) < C and [0,0,0] < [H,H,H]. I hope this clears things up and that you enjoyed the tour.

@Arpit Raj Choudhary: Forgot the ping. See my response above.

@user55119 thank you so much, this helped a lot.

I have separated C1 and C3 into unique structures. For C1, it is clear that the oxygen of the hydroxyl group has the top priority while the methylene group has the lowest priority. A digraph is constructed such that we delineate two paths around each of the unsaturated rings. Thus, non-duplicate atoms C3 and C5 terminate with their respective duplicate atoms C(3) and C(5). The cyclopentene ring has locants in sphere 3 of {(7),7,H} which is a tie with {(4),4,H} in sphere 3 of the cyclopropene ring. But the locants {8,H,H} of the cyclopentene lose to {(3),3,H} of the cyclopropene ring. The priorities are O>cyclopropene>cyclopentene>methylene. C1 has the (R)-configuration. continued...

A different digraph for C3 is constructed by dissecting the three rings into a total of six pathways, two for each ring. The rings are designated with red, blue and black dots. The red and blue rings take precedent over the black ring because each one has locants {O,O,H} at the branchs while the black ring has {O,C,H}. To distinguish between blue and red, we move to the next atoms on the chain. Blue has two oxygens while red has two carbons. The priorities are blue>red>black>methylene. C3 has the (S)-configuration.

Related issues:

What are the CIP rules for cyclic substituents?
How to check for geometrical isomerism in cyclic compounds?
More on Rings and Digraphs


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How to determine the shape of hybridized atomic orbitals in VB theory?
From diagrams, it's rather obvious how $sp$ orbitals are hybridized - the hybrids are just a composite of the $s$ and the $\pm p_{(x)}$ orbitals. However, $sp^2$ orbitals are not just composites of $s, \pm p_x, \pm p_y$, and the same goes for $sp^3$.
This incongruity leads to my question.
I am aware that the general shapes are trigonal planar and pyramidal, respectively. However, I recall learning of a more mathematical approach that used the wave function (I think?)
In my notes, I found this formula.
$$\Phi_i=\sum_{j=1}^nc_{ij}\phi_j, i=1,2,3...n$$
My understanding is that $n$ is the number of atomic orbitals (and thus also the number of orbitals). I believe $c$ is a constant, and $\phi$ is the orbital.
Later, for $sp^2,$ I have the following formulas:
\begin{align}
\require{cancel}
\Phi_1 &=c_{1,1}\cdot s+ \cancelto{0}{c_{1,2}\cdot p_x} + c_{1,3} \cdot p_y &&= c_{1,1}\cdot s+c_{1,3} \cdot p_y\\
\Phi_2 &=c_{2,1}\cdot s+ c_{2,2}\cdot p_x + c_{2,3} \cdot p_y &&= c_{2,1}\cdot s+ c_{2,2}\cdot p_x + c_{2,3} \cdot -p_y\\
\Phi_2 &=c_{3,1}\cdot s+ c_{3,2}\cdot p_x + c_{3,3} \cdot p_y &&= c_{3,1}\cdot s+ c_{3,2}\cdot -p_x + c_{3,3} \cdot -p_y \\
\end{align}
I have some recollection that the wave functions zeroes out for particular quantum number values, but I can't remember how it all fits together with this equation.
I couldn't find this formula online, so I'm hoping that it might make sense to someone here.
Thanks for the help!

Referring to MathJax: the align environment will flush to the right at odd ampersands and to the left at even, ergo r&l&r&l&r.

What we call "Hybridization" is really just a mathematical transformation of an approximated wavefunction. Your basic summation formula describes how to build some sort of customized wavefunction by forming a linear combination of some set of "basis" orbitals; you will find this equation over and over if you look up "basis set expansion". Now, of course you can't just use any random coefficients $c_{ij}$ of your liking for this custom wavefunction; you have to adhere to some physical rules. Most importantly, the overall energy that the wavefunction represents must not change. I won't go into detail about the constraints, but if you are interested, google "unitary transformation".
For orbital hybridization, you first have to find a set of basis orbitals from a physically sound starting wavefunction. If you only look at an isolated atom, then atomic orbitals are obviously a useful choice. From there, you can form combinations of those orbitals (by some criteria that really aren't uniquely defined) while adhering to the rules of the unitary transformation. In the end, you will arrive at your preferred hybrid orbitals that fulfill your chosen criterion as best as they can while keeping the wavefunction energetically equivalent to the original one. Some of the coefficients $c_{ij}$ in your equations may come out to be 0, typically for reasons of symmetry. For example, an $sp^2$ hybrid that is oriented along the $y$ axis of your coordinate system would obviously only have contributions from the $p_y$ orbital, not from $p_x$ and $p_z$. But the exact values of the cofficients will generally depend on the specifics of your problem, most importantly the definition of your coordinate system. So you really can't say per se which coefficients will drop out to be 0 and which won't -- it's one of those "it depends" situations.

General Thoughts
Unfortunately, hybridization is a concept that is so misleading that I would consider its general use in chemistry to be outright physically incorrect. People invoke it to suggest and visualize bonding situations in molecules and to explain the resulting molecular shapes, but unfortunately always seem to forget some very basic undergraduate quantum chemistry in this process.
Hybrid orbitals are linear combinations of eigenfunctions to different energy eigenvalues of the underlying one-electron Hamiltonian; think $s$ and $p$ orbitals. As such, they are not solutions to the time-independent Schrödinger Equation anymore, which in turn means that they are time-dependent functions. This implies two things:
1.) Hybrid orbitals do not have a fixed shape; what you see visualized is just a single snapshot in time. The orbital shapes periodically "morph" back and forth and can at other points in time look nothing like those suggestive localized "bonding" orbitals anymore. In fact, the hybrid orbitals can even change places among each other over time if the symmetry of the system permits.
To better drive this point home, here are some (semi-quantitative) visualizations of the time development of the electron densities in the bonding and non-bonding orbitals in water; the C-C σ and π "banana" bonds in ethylene; and an sp³ hybrid in methane. In each case, the orbitals transform over time into something that is actually quite the opposite to their original "purpose": Bonding electron densities shift outwards to resemble something of anti-bonding character, and symmetry-equivalent orbitals migrate from one bonded pair of nuclei to another. At each point throughout these videos, one could stop the animation and take a freeze-frame of the orbital at that time, and it would still describe the exact same "localized electron pair" that was used in the beginning to explain how electron density is confined with a certain purpose to some specific region of the molecule.
2.) Hybrid orbitals do not have defined energies, since they are not solutions to the time-independent Schrödinger Equation. If you were to do any sort of "energy measurement" on a hybrid orbital, you would end up with a Schrödinger's Cat situation where you observe the energies of the constituent atomic orbitals with certain probabilities that are connected to the expansion coefficients $c_{ij}$.
3.) Hybridization is not stringently mathematically defined if one regards it as an orbital localization procedure. Different for concepts exist for this localization process, which can yield qualitatively different results (see, for example, the discussion about "rabbit ears" lone pairs in water). This is also what ultimately gives rise to the weird fractional mixtures of s-p hybridization ratios that one sometimes finds discussed in the literature.

"MO and VB Theory are ultimately equivalent"
Quantum chemistry deals with two major theories to describe the electronic structure of molecules: Molecular orbital (MO) theory, which gives rise to canonical MOs that can generally be delocalized over an entire molecule and are thus very unappealing from a classical chemical perspective; and Valence-Bond (VB) theory, which attempts to reconcile the idea of isolated chemical bonds and electron pairs with quantum mechanics by postulating the existence of the localized hybrid orbitals that are in question here.
Now, in almost any discussion on the topic, it is at some point claimed that the two theories are ultimately equivalent and related to one another only through mathematical transformations. This is technically correct, and one can arrive at the total overall wavefunction and energy of the molecule either way. However, it is again also very misleading, because it suggests that the two theories sort of "meet in the middle" if you apply them in some strict, physically rigorous way. But what this statement actually means is that appropriately constructed hybrid orbitals can be used as a basis set for developing static wavefunctions with valid energy solutions to a one-electron molecular Hamiltonian; and that, when taking the totality of all hybrid orbitals together, you arrive at the same wavefunction as MO theory. In other words, "accurate" VB theory is simply MO theory with extra intermediate steps! The two theories are equivalent in the sense that at the invoked "limit of correctness", VB theory becomes MO theory; the latter never has to budge in the process, as it is a physically and mathematically sound treatment of the molecule to begin with.

Experimental (Counter-)Evidence
Up to this point, all arguments have been made from a purely theoretical perspective. The one concession that can be readily made to VB theory is that a combination of all hybrid orbitals again yields the total MO wavefunction for the molecule. As chemists, however, we are very often not just interested in a "holistic" description of molecules as a whole, but rather its individual electrons (or electron pairs). In that case, this equivalence is of no help. What we would really like to have is a direct look into whether the orbital hybridization concept holds up as a model in physical reality for the constituent electron pairs inside a molecule. In other words, can we interrogate electrons as to "what kind of orbitals they are in"?
Usually, this question immediately prompts responses that hinge on Photoelectron Spectroscopy (PES), where one measures the photon energies required to eject electrons from a molecule. An exemplary case of this is again the methane molecule: MO theory predicts a 3:1 ratio of two energetically distinct orbital types in the valence shell, whereas VB theory postulates its equivalent hybrid orbitals. The actual photoelectron spectrum reveals two distinct peaks at roughly a 3:1 intensity ratio, which is often taken as direct evidence for the veracity of the MO picture. VB proponents typically counter these claims by invoking involvement of the ionic product state of methane in the spectrum, and by requiring that it is not just one hybrid orbital that contributes to the spectrum, but a combination of all four of them (which, again, is simply the reconstruction of MOs).
Since the interpretation of these findings is contentious, they don't serve us well to find an answer as to whether hybrid or molecular orbitals are actually observable in any way. Luckily, there is another method we can use to investigate the nature of bound electrons, called Electron Momentum Spectroscopy (EMS). In essence, this technique allows the measurement of the momentum distribution of valence electrons in the initial state of the molecule without significant involvement of the ionic product, thus addressing one of the criticisms of the PES experiments. When compared to MOs or the extremely similar Kohn-Sham orbitals (KSOs) from Density Functional Theory (which are essentially MOs with inclusion of some electron correlation), it can be seen that the measurement data is almost perfectly explained by MOs or KSOs, but qualitatively incompatible with hybrid orbitals. (1,2)
Obviously, these observations lend strong support to the idea that orbitals can indeed be interpreted as a physically meaningful constructs for describing the behaviour of specific electrons in a molecule. In order to be observable in this way however, they must be eigenfunctions to the molecular Hamiltonian, which -- with the arguments above -- directly explains why MOs (or KSOs), but not hybrid orbitals, are able to describe these orbitals.

From my own perspective, the concept of orbital hybridization thus has a very convincing body of both theoretical and experimental evidence speaking against it, or at least against the way that it is usually invoked to explain molecular shapes and chemical bonding. As mentioned above, you can use VB theory and its hybrid orbitals in a sense that makes them physically rigorous; but it just means that you're actually back to doing MO theory again.
Unfortunately, at least some of the points I have addressed above -- especially the time dependence and lack of energy eigenvalues -- seem to rarely ever be mentioned when it comes to discussions of hybrid orbitals. This is undoubtedly because these topics are simply too advanced for a lot of the teaching contexts that hybrid orbitals appear in (although even high-level disputes in quantum chemistry journals generally fail to address these points too). But especially with regards to chemical education, we must ask ourselves whether teaching a concept "not  because it is correct, but because it is simple" is really an appropriate thing to do. Obviously, the answer should be no; and I think that chemistry as an entire discipline needs to abandon the idea of hybrid orbitals in the same sense, and for the same reasons, that we abandoned Bohr's atom model.

"an sp2 hybrid that is oriented along the y axis of your coordinate system would obviously only have contributions from the p_y orbital, not from p_x and p_z." Did you mean to say "sp", or am I misunderstanding?

Great points in your General Thoughts section!

Well, both an sp or sp2 hybrid along the y axis wouldn't have x or z components. sp3, too as a matter of fact. Simply because if they had px or pz orbitals mixed into them, they wouldn't be oriented purely along the y axis. But again, that also depends on the coordinate system, and if you change that coordinate system, the designations of your orbitals may also change.

Short question: can I use the concept of hybridisation to rationalise the fact that methane (or chloromethane) have a tetrahedral like shape? In spite of your big warning, I would say yes, I can.

Well then I'd prompt you to demonstrate how. Neither do the orbitals that you propose have a fixed shape, so you can't point to them and say "See, these bonding electrons like to be between the C atom and this H atom", nor do they give you useful energy values to characterize the energetics of your bonds. The only way hybrids work to rationalize the "static" geometry of a molecule is if you use the superposition of all hybrids so that they become time-independent functions, in which case by definition you're back at MO theory.


text:
Why is the units of rate constant dependent on the order of the reaction?
In many textbooks, it is written that

For order (m + n), the rate constant, k has units of mol1−(m+n)·L(m+n)−1·s−1

However, by the Arrhenius equation, k can be expressed as

k = PZexp(-Ea/RT)

and its unit doesn't seem to change with the order.
I think if reaction velocity, v is defined as

v = k([A]/cref)m([B]/cref)n

like the definition of equilibrium, the unit of k becomes mol·L−1·s−1 and it looks more reasonable. Am I wrong?

The rate has units $\displaystyle \frac{concentration}{time}$ and is $\displaystyle \frac{d[C]}{dt}$ and the right hand side of the equation has always to have the same units as the left hand side, so for first order reaction consuming C;
$\displaystyle \frac{d[C]}{dt} = -k_1[C]$
and the rate constant $k_1$ therefore has units $\displaystyle \frac{1}{time} $ usually $\mathrm{s^{-1}}$ and for second order
$\displaystyle \frac{d[C]}{dt} = -k_2[C][C]$
the units of rate constant $ k_2$ are $\displaystyle\frac{1}{time \times\,concentration}$ which is $\mathrm{dm^3\,mol^{-1}\,s^{-1}}$.
The units for other examples follows in the same way.
In the Arrhenius equation as the exponential is dimensionless the other terms must have the units of the rate constant, which means units of PZ in your notation. These units must always depend on the order of reaction as described above and if they don't then there is an error somewhere.

By the Arrhenius equation, $k$ (rate constant) can be expressed as,

$$ k = \rho Z \times \exp\bigg(-\frac{E_a}{RT}\bigg) $$

where $\rho$ (aka. steric factor, or probability factor) is defined as the ratio of $A$ (Arrhenius contant or pre-exponential factor) and $Z$ (collision frequency) (Ref. 1).
And, units of $A$ are same as that of units of $k$ (Ref. 2), hence the units of $\rho Z$ is same as the units of $A$, or $k$.
Reference

Steric factor - Wikipedia: para 1
Pre-exponential factor -Wikipedia: para 2


text:
Aldol Condensation Product selection
It is known that the enol form double bond tends to attack the electrophilic carbon centre in an aldol reaction. But what will be the majour product in this reaction.
The latter one may be enol attack but forming a 5 membered ring. The former apparently is more stable forming a six membered ring but the enol form is less stable having lower no. of hyperconjugating structures(since greater number of hyperconjugation structures directs to more stablity.) So, is the keto form actually involved in aldol in this case, contrary to general mechanism? Moreover what will be the majour product?

In example 1 the alternative product is a 4-membered ring which is disfavoured because of ring strain. In example 2 the alternative is a 7-membered ring which is disfavoured relative to [5]. Remember that in each case you are getting the thermodynamic product as these reactions are reversible.

I think that stability of the 6 member ring will pull the reaction forward..


text:
Solubility Product, and it's relation to solubility
In order to remove temporary hardness of water, it is boiled.
On boiling, while Ca(HCO3)2 gives CaCo3(ppt.) , Mg(HCO3)2 gives Mg(OH)2(ppt.).
According to my book precipitation of Mg(OH)2 occurs due to it's high solubility product.
But doesn't a higher solubility product mean a higher solubility? (Since solubility product is the product of solubilities of each component)

Can you provide the exact full quote from the book?

it should be lower solubility product.

Is it an error in the book?

Yes. it is a mistake.


text:
Using the right equipment for successful detection of Novichok
Apparently, Navalny’s biomaterials were flown to Moscow's Nii Sklif lab which didn't find any traces of the poison. But they used an American Agilent Technologies GC-MS [1], while Bundeswehr toxicology lab reportedly used an order of magnitude more sensitive, and more expensive, Bruker's apex-Qe FTMS (no official source for this as "Bundeswehr didn’t disclose the details of their work") which successfully detected novichok.
How did German scientists outplay their Russian counterparts in this case and could it be attributed to simply using a more sensitive equipment with perhaps a better MS method?

And why would a Russian lab want to find evidence for the Russian state poisoning somebody?

@JonCuster It equally may be speculated if / which part(s) of the analysis and therapy disclosed (when) to a / the public, too.  Recalling the dioxine case of Victor Yushchenko, maybe was an exception (2009Lancet1179 or 2011TocicolSci310).

@JonCuster, had they found it, they wouldn't have released him, and they might not even disclose the results. But instead they were so openly certain he was clean of any traces as to allow him to be transported to the scrutiny of German labs. So the question is should they upgrade their hardware.

@rych Equally plausible speculation, given the scale of research (e.g., entry in wikipedia including details like the yields e.g., in the Polish edition): #1 hardware is not the problem (in Russia), #2 neither is it the know-how, likely including ADME data (in Russia), #3 Mr Nawalny was permitted to leave Russia when the chance to identify the substance (and / or) its origin by a non Russian lab was assumed to be insignificant.  Because of #3, #4 someone else is now sweating bullets.

Related: Differences in Novichok agent mass spectra

Lacking additional information, answers likely will be speculation-based only.   To mention a few factors to consider:
Stating GC-MS analysis only states, that there is technique to separate compounds by different rate of retention over a stationary phase (gas chromatography, GC) combined (a.k.a. hyphenated) with an other technique to analyze fractions sorting at different times by mass spectroscopy (MS). Mass spectroscopy typically shatters the molecules into charged fragments which, by their mass/charge ratio, behave differently in electromagnetic fields.  However, depending on the setup, the separation of the fractions by GC may be better, or worse.  On the side of mass spectroscopy, there are multiple methods to ionize and fragment molecules and to record these.  It is a problem that the parameters of running a mass spectroscopy may alter the shape of mass spectra quite a lot; signals may vary in their relative intensity, or sometimes be below the detection limit altogether.  It need not be the engineering advantage among corporations, of Agilent's technology over the by  Bruker (if there is?).
In the line of smaller benchtop mass spectrometers typically mounted behind a GC, larger sector field mass spectrometers, and Fourier mass spectrometers, there is a significant increase of resolution to record tiny differences in the masses of the fragments  the experiment of mass spectroscopy generates.  This may help to attribute the signals to the correct analyte, especially if there already is a suspicion to search for a selection of compounds only.
Routine GC-MS used for forensics and IOC doping screenings, for example compare the mass spectra recorded for a sample against previously recorded trustful references.  (Different to IOC doping screening, it is not evident if the Russian government shares / shared such standards here, though.)   Depending on the compound to identify, though, the analysis does not need to stop at the level of a GC-MS.  Both other types of chromatography may have been used now (e.g., HPLC) to separate the compounds; or multiple mass spectrometers coupled into a tandem (MS/MS) to observe not only the initial fragments obtained by mass spectroscopy, but equally their further fragmented «daughter radicals», too which equally rises the chance to identify your compound in question.
Maybe the Germans did not trace the poison itself, but metabolites judged as sufficient evidence for earlier reported, related compounds.  Especially with small amounts of sample material, the success of an analysis equally depends on every step of sample preparation; here likely (incomplete listing): what was the delay between Nawalny's intake of compound X and obtaining a sample by him (metabolism), the nature of the sample (tissue, blood, hair, etc.), delay and conditions of storage of the specimen to get it to the lab (unwanted sample decomposition), concentration and digestion protocols in the lab to eventually submit the sample to the mass spectrometer; the people's experience and availability reference data / techniques.


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Why are branched alkanes used as fuels instead of straight chain alkanes?
My textbook states that branched alkanes are thermodynamically more stable than straight chain alkanes. Of course, iso-octane has an enthalpy of formation of $\Delta H_\mathrm f$= $\pu{-259.34 kJ mol-1}$ and n-octane has an enthalpy of formation of $\Delta H_\mathrm f$= $\pu{-249.73 kJ mol-1}$ which indicates that iso-octane is relatively more stable than n-octane at room temperature. When they are used as fuels, both produce energy and the enthaply of combustion of iso-octane would be less than enthalpy of combustion of n-octane. So, n-octane should be more efficient fuel than iso-octane. But in petrol, iso-octane has more composition than n-octane. Why?

See Octane_rating: pre-ignition and knocking

Related: https://chemistry.stackexchange.com/questions/99333/why-do-highly-branched-alkanes-have-higher-octane-numbers-than-their-correspondi?r=SearchResults

Because petrol engines work differently to diesel engines and the property differences matter more than the absolute energy of combustion
Your basic assumption is wrong: diesel engines use mainly straight chain hydrocarbons and it is petrol engines that prefer the unsaturated or branched hydrocarbons.
The absolute amount of energy produced in engines is not the most important factor in the fuel (at least when comparing hydrocarbon fuels). What matters is how they burn and how the particular engine type triggers that combustion.
Petrol engines use a spark and want a volatile fuel that can form a flammable vapour easily. Diesel engines trigger combustion by compressing the fuel-air mixture to the point where it combusts from the heat of compression. These modes of operation have different requirements. If spontaneous combustion happens in a petrol engine (before the spark) it damages the engine and lowers the power. So petrol uses a hydrocarbon mix that doesn't spontaneously ignite at typical engine pressures: this drives the choice to hydrocarbons that are more volatile but less likely to combust from pressure alone. Volatile unsaturated hydrocarbons have those properties. Diesel needs hydrocarbons that can combust without a spark at a given temperature. Saturated hydrocarbons fit this requirement (you can, simplistically, think of this as a product of the higher accessibility of their carbon chains to the oxygen in the air). See How or Why does diesel/kerosene have a much higher 'flash point' but lower auto/self-ignition point than gasoline/petrol? for more details.
Diesel engines tend to be slightly more efficient than petrol engines, partially because of the effect you note (though the situation is really far more complex and many features other than energy output are usually far more important).
But your overall observation that branched chain or unsaturated fuels are more common applies only to petrol engines.

The main reason for the high quantity of iso-octane is to prevent Engine Knocking. The Wikipedia page for Octane explains this quite well:

The octane rating was originally determined by mixing fuels from only heptane and 2,2,4-trimethylpentane (a highly branched octane), and assigning anti-knock ratings of zero for pure heptane and 100 for pure 2,2,4-trimethylpentane. The anti-knock rating of this mixture would be the same as the percentage of the latter in the mix. Different isomers of octane can contribute to a lower or higher octane rating. For example, n-octane (the straight chain of 8 carbon atoms with no branching) has a -20 (negative) Research Octane Rating, whereas pure 2,2,4-trimethylpentane has an RON rating of 100. Some fuels have an octane rating higher than 100, notably those containing methanol or ethanol.

A fuel made of only octane would have such a high enthalpy of combustion, that it would combust spontaneously in the cylinders, causing knocking, as seen by the RON number of -20 for it. Thus, iso-octane content is high because it reduces knocking and prevents engine wear.


text:
Does fractionation use enough heat to destroy nutrients in Coconut Oils when seperating the MCT oils?
We have recently been giving MCT (medium-chain triglyceride) oil to our dogs in place of Coconut Oil as from what I understand the majority of the benefits of Coconut Oil comes from the MCT's themselves and by doing this you can avoid Lauric Acid, which has 12 carbon atoms and can behave like both classes of fatty acids (MCT's and LCT's (long chain triglycerides)).
Apparently only 10% of the saturated fat in Coconut Oil has MCT's:

A small percentage of the saturated fat in coconut oil, about 10%, is made up of these less harmful saturated fatty acids, but virtually all the rest of coconut oil’s saturated fat is made up of the long-chain varieties that send LDL soaring.

However I have heard claims as well that Coconut Oil is 2/3 MCT, but I would assume this comes down to Lauric Acid and whether or not it is classified as an MCT or an LCT; the aforementioned articles states it is an LCT but others state it is an MCT.
I am getting off track here, so I digress.
Based on the information I have found it is best to avoid Coconut Oil and instead give straight MCT oil to avoid LCT's (including Lauric Acid), however after speaking with the manufacturer of the MCT oil we are using, I asked them if the oil was "cold-pressed" and was told whilst they did cold-press it they also have to use fractionation to separate the oils.

To get MCT the process is a little different from other oils. First it starts with organic coconut oil which is cold pressed. Then the oil goes through  the process of fractionation, which is done by steam. There are no solvents used in this process, it's just heat. It needs to be heat because whole coconut oil needs to have its fatty acids broken to extract the medium chains (the MCT part). Coconut oil is a mix of long chains and medium, and you can't separate them using a cold press, as it won't be enough to break the fatty acids bonds. Once the fractionation is done we are left with the long chain fats (which we use in our soaps and personal care ranges) and the MCT.

The heat from the fractionation isn't enough to change the structure of the fats, it's just enough to split the long chains away from the medium. I hope that's clear! It's always a little tricky to understand first go - it's almost like separating the milk solids from melted butter, or curds from whey, if you can imagine that?

Now, obviously the reason you want a cold-pressed oil is because you don't want to destroy the nutrients in it - apparently using too much heat can destroy things such as antioxidants, vitamin E and oleic acid.

Cold pressed retain healthy antioxidants that are otherwise damaged by being exposed to heat. Antioxidants help combat free radicals that cause cell damage in the body. Most cold pressed oils are rich in vitamin E, which has anti-inflammatory and healing properties. They are also rich sources of oleic acid that help boost your immune system. All these oils when cooked on low heat are excellent oils; however, they lose all the nutrients once exposed to too much heat.

Based on the response from the manufacturer and the statement above that "low heat" should be fine, generally, does the process of fractionation retain all the nutrients in the oil?


text:
How are organic compounds with radioactive atoms synthesized?
We often see reactions such as these in textbooks to highlight how reaction mechanisms work:

How are the radioactive substrates required for these mechanisms synthesized in the first place? There seems to be no easy way of just 'substituting' a radioactive atom in place of a non-radioactive one (especially for Benzene-14C, it might be easier to make a radioactive sodium salt using $\ce{^{24}Na}$). Also, Once the substrate is synthesized, how is it separated and purified? I'm assuming that the physical properties of these radioactive substrates would be very similar to their non-radioactive counterparts, which would make it very difficult to do so.

The current methods employed for synthesis fall into the following three categories1:

Chemical synthesis
Biochemical methods
Isotope exchange reactions

Chemical Synthesis:
Most chemical syntheses involving $\ce{^14 C}$ are done with Grignard reagent and $\ce{^14 CO2}$ or $\ce{^14 CN-}$. As an example, Benzene-14 is synthesized as such2:

Basic preparatory methods such as the above have been worked out long ago. An example of a more complex synthesis would be preparation of 18F-Fluorodeoxyglucose:

Biochemical methods
A few examples of these, as mentioned in (1), are:

High yields of [U-14C]sucrose, up to 70% based on $\ce{^14 CO2}$, are obtainable in detached leaves after a few hours photosynthesis using tobacco or Canna indica leaves for example.
DNA labelled either with $\ce{^14 C}$ or with $\ce{^3 H}$ is prepared by feeding E.coli with the correspondingly labelled thymine precursor. Similarly, by feeding [$\ce{^35 S}$]sulphate to E.coli or to yeast cells, isolation and hydrolysis of the proteins yields L.[$\ce{^35S}$]methionine and L-[$\ce{^35 S}$]cystine in radiochemical yields of 30% and 10% respectively.

Another example is the synthesis of C-14 labelled Phenylpyruvic acid3

Isotope exchange reactions
These are fairly common for atoms that can easily be exchanged, such as replacing $\ce{^23 Na}$ for $\ce{^24 Na}$, which was mentioned in the question. Another common mechanism for synthesis of Hydrogen isotope compounds is Hydrogen-Deuterium exchange. A similar mechanism also allows hydrogen-tritium exchange4.
Purification of radiolabelled compounds
This is mainly done through Chromatography1, as the amount of radiolabelled substrate required is generally quite small. High-performance liquid chromatography is used to separate the compounds. The chemical identity of the compounds is  confirmed by co-chromatographic behaviour with the unlabelled compound.
For further information, refer to the references (1) and (3), which go into a lot more detail with regards to the synthesis of these compounds.
References:

Anthony Evans, E. “Synthesis of Radiolabelled Compounds.” Journal of Radioanalytical Chemistry, vol. 64, no. 1–2, Mar. 1981, pp. 9–32. doi:10.1007/BF02518337.
Schmid, K., et al. “A Simple Method of Preparing Labeled Benzene.” Advances in Tracer Methodology, edited by Seymour Rothchild, Springer US, 1966, pp. 37–44. doi:10.1007/978-1-4684-8625-4_5.
Atzrodt, Jens, and John Allen. “Synthesis of Radiolabeled Compounds for Clinical Studies.” Drug Discovery and Evaluation: Methods in Clinical Pharmacology, edited by Hans Gerhard Vogel et al., Springer Berlin Heidelberg, 2011, pp. 105–18. doi:10.1007/978-3-540-89891-7_12.
Englander, S. W., and J. J. Englander. “[3] Hydrogen—Tritium Exchange.” Methods in Enzymology, vol. 49, Academic Press, 1978, pp. 24–39. doi:10.1016/S0076-6879(78)49005-6.

You asked and answered the question at the same time? Hmm? The chemical syntheses of monolabeled benzene do not explain how specifically labeled chlorobenzene, in the question, is prepared.

@user55119 I was doing some research for the question, and then realized I had done enough research to write an answer :) I'm still open for a better answer, or additions to this answer, though.

Some more papers for you: here, here and here.

For specifically C-labeled toluene and benzoic acid, see: https://science.sciencemag.org/content/109/2820/35

@user55119 You have e.g. in the second synthesis a specifically labelled cyclohexanone. You take a dive into your OC textbook, and see how you can get to chlorobenzene without making plain benzene on the way.

@Karl: My bad! Totally missed the bis-Grignard reagent example. I had actually gone back and read the paper cited previously by me.  The work was instrumental in J. D. Roberts landmark proof of the benzyne mechanism. My apologies to Aniruddaha Deb, also. Two upvotes.


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Estimate the relative intensities of absorption given the split in energy levels

Estimate the relative intensities at $\pu{25 °C}$ of absorption originated in the ground state and the first excited state when the energy levels involved are separated by:
(a) $\pu{10000 cm^-1};$
(b) $\pu{1000 cm^-1};$
(c) $\pu{10 cm^-1}.$

Here, the given numbers are the wave numbers of the EM radiation.
I have a problem understanding the term “relative intensities”. What does it imply? My professor has told me to calculate it using the population of states using the formula:
$$\frac{N'}{N} = \exp\left(\frac{-\Delta E}{k_\mathrm B T}\right)$$
But my problem is that the term $N'/N$ is the ratio of the number of particles in excited state and the ground state. How does it relate to relative intensities?
I'm fairly new to spectroscopy so I might have used few terms in the wrong fashion.

The relative intensities are proportional to the number of particules. So try to calculate N'/N for the three values of energy levels. You will immediately obtain the result your teacher would like you to get.

@Maurice Thanks for help. Can you give some text where this concept is described in detail? Hope you don't mind.

You don't need a textbook. You have the formula. Only thing to do is transform the unit $cm^{-1}$ in frequency unit (Hertz), then in energy unit (Joule), and divide the result by $kT$, where $ k$ is Boltzmann constant. Hopefully you know the corresponding numerical values.

@Optimus IF truly needed, see textbooks about physical chemistry, because Boltzmann constant and distribution (statistical thermodynamics) / Boltzmann statistics. Nuclear Magnetic Resonance spectrosccopy (NMR) and optical pumping of LASERs seek to go beyond what heat taken alone would offer (population inversion).  Note however, this detail of understanding may be required only if you study e.g., chemistry, but is presented to you typically in or even past year three.

Do you know the equation $E = h\nu = \frac{hc}{\lambda}$?

@MathewMahindaratne yes ofc. Every one studies it in high school.


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Can I use activated carbon electrod as a reference electrod in a pH meter instead of using murcury?
Activated carbon is less expensive than mercury . They are recyclable . Mercury is toxic. Nowadays activated carbon is being used in supercapacitors,Li-ion battery.

Reference electrodes for pH measurements are made of silver/silver chloride electrodes.

Just for curiosity - what do you think is the potential of the activated carbon electrode ?

Activated carbon may be relatively quickly contaminated by foreign substances coming from the analyzed solutions.

Electrodes are made from graphite of glassy carbon, not activated carbon.

Struggling to understand the chain of thought present in the question.

Reference electrodes must involve a well defined redox system with reproducible potential.
Mercury is an essential part of the redox system of the calomel reference electrode $\ce{Hg|Hg2Cl2(s)|Cl-}$:
$$\ce{2 Hg <=> Hg2^2+ + 2 e-}$$
$$\ce{Hg2^2+ + 2 Cl- <=> Hg2Cl2(s)}$$
and cannot be removed nor replaced, unless you want to use a difference electrode.
A similar, more often used reference electrode is the silver chloride electrode, $\ce{Ag|AgCl(s)|Cl-}$, being safer, simpler and more practical:
$$\ce{Ag <=> Ag+ + e-}$$
$$\ce{Ag+ +  Cl- <=> AgCl(s)}$$
This electrode can be very compact and is frequently integrated to measurement electrodes, forming a full electrochemical cell, like the pH "glass" electrode.
If a glassy carbon were used, it would be inert electrode material, similarly as platinum in the primary hydrogen reference electrode. It would need a well defined redox system established within the solution.
Note that Li-Ion cells use graphite, not activated carbon, forming lithium graphite intercalate.

I do give you partial credit for contemplating the employment of a form of a carbon-based electrode!
Here is a related extract from a 2019 review published in Frontiers in Chemistry:

Boron-doped diamond (BDD) electrodes present several notable properties, such as the largest potential window of all electrode materials (especially in anodic potentials), low background and capacitive currents, reduced fouling compared to other electrodes, mechanical robustness, and good stability over time. On the other hand, flow-based systems are known as well-established approaches to minimize reagent consumption and waste generation and with good compromise between sample throughput and analytical performance (mechanization of chemical assays). This review focuses on the use of BDD electrodes for electrochemical detection in flow systems, such as flow injection analysis (FIA), batch injection analysis (BIA), high performance liquid chromatography (HPLC), and capillary electrophoresis (CE). The discussion deals with the historical evolution of BDD, types of electrochemical pre-treatments (cathodically/H-terminated or anodically/O-terminated), cell configurations, and analytical performance. Articles are discussed in chronological order and subdivided according to the type of flow system: FIA, BIA, HPLC, and CE.

This is evidently a technical area of specialization and, yes, doping even your active carbon large surface area electrode may prove interesting.


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Is a tautomer's stability compared with refernce to its initial compund
Considering the following example while comparing the enolic content in the product, am I supposed to compare its stability with reference to its initial state or am I supposed to compare it between the final products obtained?

Given are keto form of keto-enol tautomerism of reference compounds. Answers are given according to the percentages of enol forms. Your duty is to find which enol form is dominant.
In cyclohexa-2,5-dienone, the enol form (phenol) dominate 100% in solid or in solutions, regardless of the solvent because of the aromaticity (100% B).
In the case of pyridones, however, the dominance form of the compound depends on which phase it is in. For instance, Ref.1 states that:

In the gas phase, there is a preference of the hydroxy form in both 2- and 4-hydroxypyridine, while in a nonpolar solvents such as cyclohexane and chloroform, the two tautomers exist in comparable amounts. However, the two equilibria are shifted entirely to the right in favor of the oxo-form, 2- and 4-pyridone, respectively in polar solvents as well as in the crystalline state.

In general, pyridones' (e.g., 4-pyridone) predominance strongly over their tautomeric hydroxy forms (e.g., 4-hydroxypyridine) can be rationalized by considering the mesomerism of the alternative forms. The charge-separated form of 4-pyridone (far left; negative charge on more negative oxygen), being aromatic, is greatly preferred over the nonaromatic, charge-separated form of 4-hydroxypyridine (far right; negative charge on relatively more positive nitrogen).

Thus, it is safe to say %B > %A and hence you have the correct answer.
Note: There are compilations about this type of keto-enol tautomerism. The freely available Tautobase for example is provided as searchable database in DataWarrior. Also read answers in this question.
References:

Jiali Gao, Lei Shao, "Polarization Effects on the Tautomeric Equilibria of 2- and 4-Hydroxypyridine in Aqueous and Organic Solution," J. Phys. Chem. 1994, 98(51), 13772–13779 (https://doi.org/10.1021/j100102a051).
Oya Wahl, Thomas Sander, "Tautobase: An Open Tautomer Database", Chem. Inf. Model. 2020, 60, 1085–1089 (doi 10.1021/acs.jcim.0c00035).


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Vapors of Aqueous solutions
I'm dealing with a common photoresist developer TMAH (tetramethylammonium hydroxide) and attempting to design a vapor sensor for a sealed chamber. Now I havent been able to confirm what exactly is evaporating from the solution causing it to smell.
I assumed that tetramethylammonium hydroxide would equilibrate with trimethylamine and methanol, and the liberated non-ionized species would evaporate before shifting back into their respective ions.
My question is whether those species as individual molecules would evaporate, or do droplets of the ionized solution evaporate together? And if droplets of water/TMAH solution evaporate, would water absorb energy, throwing off the reading made by a photoionization detector?

Are you just looking for total vapor pressure, or the vapor pressure of a certain component?

You may want to read the Wikipedia article on TMAH.  While it does decompose, it may not produce the methanol you are expecting.  Also, heat is also involved in this process, so you may need to heat and, in the process, vaporize much of the water to get this decomposition to go.

TMAH, in common with many other TMA salts containing simple anions, decomposes on heating into trimethylamine.[1] Dimethyl ether is a major decomposition product rather than methanol.[2] The idealized equation is:

$\ce{2 NMe4+OH− → 2 NMe3 + MeOMe + H2O}$

Such a rearrangement suggests an irreversible reaction.  The Wikipedia article also indicates that TMAH is not obtained by combining trimethylamine with methanol anyway.  Instead, a metathesis between a tetramethylammonium salt such as the chloride and an alkali hydroxide is used.
Cited References:
1.
Lawson, A. T.; Collie, N., "The action of heat on the salts of tetramethylammonium". J. Chem. Soc., Trans., 1888,53, 624-636. doi:10.1039/CT8885300624.
2.
Musker, W. Kenneth. (1964). "A Reinvestigation of the Pyrolysis of Tetramethylammonium Hydroxide". Journal of the American Chemical Society. 86 (5): 960–961. doi:10.1021/ja01059a070.

Hi, Thank you for this post. Im embarrassed that I did not see that information while first doing my research. This is just fine, as the vapor sensor is sensitive to NMe3 and MeOMe.

In regards to the vaporization, do you expect them to vaporize as individual molecules, or hydrated species?

They would vaporize as individual molecules.  You may, however, have to heat and vaporize the water to get the hydroxide concentrated enough to react with itself.  I will add this.


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Lucas test in the presence of EWG like -CN

Does the given compound give Lucas test?

My teacher says due to the presence of a strong electron withdrawing group the carbocation formed is unstable.
But I say if it undergoes $\mathrm{S_N2}$ mechanism then an electron withdrawing group should increase its reactivity. Why then does this compound not give the test?

Do you think Lucas reagent (ZnCl2, dil. HCl) will favor SN2 reaction pathway?

@RahulVerma Yes I do. Is it correct. Or do they undergo only SN1 reaction

Thats a cyanohydrin not a standard alcohol

@ishwarbb Side note, for future reference: It is very welcome to illustrate the structure of organic molecules with a sketcher.  You may both improve the representation further, as well demonstrate you understand the structure of organic molecules, if the three atoms $\ce{N#C-C}$ are depicted aligned in a straight line.  Sketchers often offer to «clean the structure» for you with a click on a dedicated button.

@Buttonwood ok I will keep that in mind

Lucas reagent follows SN1 pathway. $\ce{Zn}$ is a lewis acid that tends to bond with $\ce{-OH}$ and form a partial positive charge leading to the formation of a carbocation.
According to Hammond Postulate, the stability of the transition state should allow successive formation of product or intermediate. The stability of the positive charge induced on the carbon if stabilised in the transition state should therefore stabilise the carbocation. The rate determining step is the formation of the carbocation.
Since tertiary carbocations are the most stable the carbocation is formed quite easily and the product formation happens quickly.
Similarly, the primary and secondary alcohols, need to form a primary and secondary carbocation, which is not stable. Hence the reaction is slow. Zinc forms a weaker bond with $\ce{-OH}$ and induces a partial charge on them, thus not forming the whole carbocation.
So I think that in such cases where the carbocation is not stable, the reaction is slow, and may not have a a very well defined SN1 or SN2 classification i.e. truly borderline reaction may not occur, due to accompanying facets like solvation and ion pairs.


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Composition of arcanite (potassium sulfate) agricultural grade. Potassium content of random drawn sample?
I bought potassium sulfate from this source:
Potassium sulfate granulate
I'm using it as a potassium additive in hydroponics. The composition is listed as K2O = 50%, S = 18%. This equals about 41.5% K (50* (mass of K / mass of K2O) Calculated from mass percentages (mass of K2 / K2SO4) potassium sulfate should contain about 44.9% K and and 18% S. So based on this calculations I concluded that the potassium content is within the margin of error tolerable for fertilizer application. (About 3.4% for potassium)
However after further reading I came across an article which suggested that raw materials should be as homogeneous as possible because otherwise a sample drawn randomly could have huge variations in the content of the chemicals involved. How high could the deviation be with respect to potassium?f I think the source material is arcanite
EDIT: The security data sheet states that the purity of the granulate is guaranteed to be between 90-99%

The deviation can be in range from negligible to huge, depending on the level of nonhomogeneity and locality of the sample.
Note that routine analysis of potentially non-homogenous sources involve sophisticated sampling and sample averaging schema, so the result is representing number.

I have found the security data sheet and it states that the purity is between 90-99%. How can the variation be huge, when it is a nearly pure substance? Couldn't the variation then be 10% at max?

There can be time variation between lots, based on variability of used raws.

So basically I should look for the maximal variability of K2SO4 in arcanite by a literature review?

The point is, it may not be the pure mineral of whatever internal purity, but minor external impurities may be present in variable amount. It is fertilizer, after all, not substance of a lab purity.

If I assumed its purity to be 100%. Then dosing the nutrient solution according to this value. The content of potassium & sulfur can then only be lower but not higher. Is this reasoning correct? I don't mind lower values but higher can be detrimental.

K2SO4 cannot contain more than 100% of itself. But if there is present other compound with higher percentage of potassium, like KCl, then effective K2O content CAN be higher than of 100% K2SO4.

It also states that the max. concentration of chlorine is 1%. Based on your objections of variation between lots the chlorine content could also be higher for some lots but is only on average 1%. Or does this only apply to the main ingredients? (Like K or S?)

Chlorine is kind of guaranteed to be below 1%. So you can count with max K2O content as if it was pure K2SO4.

Thanks you helped me a lot with your knowledge!

The picture of the product (2-4 mm granules) on the website looks very homogeneous.

If you consider the product acceptable, except for possible inhomogeneity, you could examine smaller samples or homogenize larger portions. Dissolving in water seems the easiest way, and could provide a further check on purity/homogeneity. $K_2SO_4$ is soluble in water about 10% by weight. For a small scale test, dissolve a 100 gram sample of your Kalisop in 1 liter of warm (or hot) water and stir to complete dissolution. The specification says “vollständig wasserlöslich” - completely water soluble. If a small amount doesn’t dissolve, you could filter it off and weigh it. Check as many samples as you need to to assure yourself of homogeneity or consistency of the material. Also check pH.
Agrarshop specifies 1% as a chloride maximum: “ Praktisch chloridfrei”; this comes out to about 2.1% (max) impurity if $KCl$. Any precipitate could be a calcium sulfate.
If you are satisfied with the product but want to go the extra mile, you could homogenize 1 kg in 10 liters of water in a 20 liter bucket. Or scale up even further, especially if the product is indeed, completely water soluble.
The source of the $K_2SO_4$ could be arcanite mineral but that is relatively rare. On the other hand, 1.5 million tons of potassium sulfate was made in 1985, typically by the reaction: $ KCl  + H_2SO_4$ —> $HCl + KHSO_4$, followed by heating: $ KCl + KHSO_4$ → $HCl + K_2SO_4$. The cost of Kalisop is about 1.36 US dollars per kg, which could support a large-scale manufacturing process, but it also depends on the cost of $KCl$.

Thanks for the input. I dissolved it and found only very little impurities, looked like sand. That's why I assumed it is arcanite, but could obviously just be impurities from manufacturing / storage..Do you know if dissolved K2SO4 expires?

K2SO4 in water has no chemical reactions available. However, dilute solutions might support biological growths and get cloudy or clotty. Still good enough for plants that are already growing in dirt. The clots might fowl up a delicate pump, but the chemicals will be stable.


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Anthracycline nomenclature - doxorubicin

Why $\ce{-OCH3}$ is Number 3? I think that $\ce{-OCH3}$ Appropriates Number one.

Do you have a reference to support your thought process?

I am not good at English.
I just want to know why that replacement is number three.

The red numbers shown in the picture are not related to nomenclature. Clearly, the locant for the methoxy group is '1', and there is no locant '13'.

According to both PubChem and Wikipedia, the IUPAC name of Doxorubicin is (7S,9S)-7-[(2R,4S,5S,6S)-4-Amino-5-hydroxy-6-methyloxan-2-yl]oxy-6,9,11-trihydroxy-9-(2-hydroxyacetyl)-4-methoxy-8,10-dihydro-7H-tetracene-5,12-dione, which probably is the correct naming. To answer your question about why $\ce{-OCH3}$ is number 4 (actually it is number 4, not 3 as you indicated), I put both numbering and and corresponding names in following image:

I think you have enough knowledge to understand why numbering in $\bf{A}$ is preferred. For instance, the principle functional group is 5,12-dione, which does not change in both numbering. Yet, for the numbering of two major substitutions in tetracene nucleus are changed: it is (7,9)- versus (8,10)-. Further, complete locant set is (4,5,6,7,9,9,11,12) versus (1,5,6,8,8,10,11,12). In addition, it is (7H) versus (10H). I hope you got the point.

I didn't downvote, but if you are OK with some critics, then there are some things I would point out: 1. Neither Wikipedia nor PubChem can be considered trustworthy sources for the nomenclature. 2. Ideally, one should be able to present information without relying on colors or substantial amounts of text on the illustrations (such post appears less professional and are a barrier for color blind and overall visually impaired users). 3. The question is borderline HW and is also based on a false premise.

@ andselisk♦: I do believe you and your points are well taken. I have habit using color to emphasize my points (you'd realize if you look at my publications). I'd keep those in my mind in future perspectives. Only I can't stand is pointless downvoting, that's all my complain about. Thanks for the concern.

Downvoted because the explanation why A is preferred over B is not correct. It is not about (7,9)- versus (8,10)-. The parent structure of A has eight substituent prefixes (4,6,7,8,9,9,10,11) that are all considered together, not just (7,9) since there are no “major substitutions” in nomenclature. A is actually preferred simply because the locant 7 for the indicated hydrogen is lower than 10. (…)

(…) Anyway, the name of the parent structure actually is 7,8,9,10-tetrahydrotetracene-5,12-dione and not 8,10-dihydro-7H-tetracene-5,12-dione, which makes the comparison of the indicated hydrogen in A and B irrelevant from the start.


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Does changing the electrolyte concentration in the electrolysis of water affect the speed that hydrogen gas is produced at the cathode?
I am currently completing a Year 12 Extended Practical Investigation. Due to the lock-down restrictions we are unable to complete our own practical experiment and have been given sample answers with made up lab conditions.
The experiment I chose was to investigate the difference of time taken to produce 25mL of H2 gas in an electrolytic cell when the concentration of the potassium hydroxide electrolyte was changed.
The lab conditions given were a pressure of 101kPa, a temperature of 26.0 celcius and in addition the current in the system was kept at a constant 0.600A throughout all trials of different concentrations.
When analysing the results there was a trend that showed the higher the concentration of the potassium hydroxide the faster the 25mL of H2 gas was produced. However, Through using the formulas Q=It, Q=n(e-)F and the electrochemical half equation for H2 production it seemed that due to the current and the amount of Hydrogen being formed both being constant the time taken to form the hydrogen should also remain constant.
I have found a write up for the exact experiment on https://www.qcaa.qld.edu.au/downloads/senior-qce/sciences/snr_chemistry_19_ia2_asr_high.pdf however when it comes to talk about the effect of increasing the KOH concentration it seems that there is no mention that the current was kept constant.
I would like to know if the trend shown should have occured and why.

With decreasing electrolyte concentration, either the needed voltage raises at constant current, either the passing current decreases at the constant voltage.
Both happens because at lower concentration, a higher potential gradient across the electrolyte is needed to maintain the same current, as the smaller ion count requires  ions to move faster for the same current..
This potential gradient and the resulting potential drop acts against the external voltage. It pushes forced electrode potentials closer to their equilibrium potential. The net electrode reactions are then slower.

Interesting question. Remember for future calculations that in electrolysis either voltage can be kept constant or the current but not both. Now if you were actually doing this experiment you would have noticed that in order to maintain the same current in different KOH solutions you would have needed different voltages.
So the hint is: What is changing when you are changing the KOH concentration?
Answer: The resistance of the solution. Think whether low KOH solution will have high resistance or low resistance.
The ideal situation is that you have high current and low resistance at a given voltage. The voltage should be above the decomposition voltage of water.


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Can chemical energy from food be stored in a battery?
Suppose you built a machine that digested food in a similar way as humans do. Would it be possible, in principle, to extract the chemical energy from the digested food, turn it into electricity, and store it in a battery that powers the machine itself? Doesn't matter if it's practical or efficient, I just need to know if it's not something totally impossible. (I asked the same question on Electrical Engineering SE and someone suggested this SE would be more appropriate.)

A human body is such machine, which proves that yes, it is possible, even though it may be beyond our current technical capabilities.

Biowaste processing plants do exactly this.

Humans can ride a bicycle connected to a generator that can charge a battery...

It is probably possible to design an electrode which "respires" but without the oxidative phosphorylation step and could receive electrons from the reduced coenzymes, NADH and FADH2, produced from the reaction, and then we could pair it with another electrode which reduces O2 to H2O. I mean, if microbial fuel cells (which uses bacteria) and fuel cells which uses the oxidation of organic compounds are possible, I don't see how a fuel cell which uses food could not.

See https://en.wikipedia.org/wiki/Enzymatic_biofuel_cell.and http://large.stanford.edu/courses/2014/ph240/ho2/

As a side note, yes, I know that the human body is a machine that does that, but I thought it was obvious that I meant a human-made machine.

Does a person eating a bunch of hamburgers pedaling an exercise bike hooked up to a battery charger count? There's probably a more elegant and direct way to do it though but that would involve molecular engineering (aka an artificial synthetic lifeform). And if you have the tech to do that you can also do the reverse: make a lifeform that uses electricity as a power source as synthesizes sugar and other organic molecules from the power when you supply it with hydrogen, carbon, nitrogen, etc.

Three ways to turn chemically stored energy into electricity:

a fuel cell: While that works fine for hydrogen or methanol, I'm not sure if sugars are possible, and general "food" won't work. Of course you can chemically convert food into hydrogen or alcohol, or into sth. very specially designed to power a biofuel cell.
dry, burn, run a turbine or steam engine with attached generator, very obvious and simple.
direct conversion into mechanical energy via an artificial muscle, which powers a generator. Horribly complicated (bio-)chemistry, theoretically possible, because our body does it, too.

Olympic track cyclist versus 700 W toaster: https://www.youtube.com/watch?v=S4O5voOCqAQ&fbclid=IwAR0bcckfWY25I5T5Lt3R9eB2Yvm4j-xarjP1qQEkw6gq1RLWHNPOro4k1F8. ;)

@EdV I had seen that, yea. The energy spent during running a marathon is barely enough to cook a simple meal. ;)

You would have to find a system that could extract chemical energy from lipids in a similar fashion to biological systems.


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Experiment to demonstrate Dalton's Law of Multiple Proportions
Is there an experiment that I can do in the lab with middle or high school students to demonstrate Dalton's Law of Multiple proportions ?
For example, the reaction of $\ce{C}$ and $\ce{O2}$ to give $\ce{CO}$ and $\ce{CO2}$.
Is there an experiment where I can demonstrate that $\ce{CO}$ and $\ce{CO2}$ have these proportions ?

Because of toxicity, I'd avoid middle or high school student experimentation with $\ce{CO}$.
You might consider $\ce{H2O}$ and $\ce{H2O2}$, if you stick to 6% (20 vol) or weaker. There is the complication of working with a dilute solution, though, in getting precise measurements on proportions.
Another possibility is demonstrating the various oxides of iron. However, there is another issue: you can produce fairly pure pure red iron(III) oxide, $\ce{Fe2O3}$, by the electrolysis of sodium bicarbonate, $\ce{NaHCO3}$ with an iron anode, but rather than pure black iron(II) oxide, $\ce{FeO}$, thermal decomposition of iron(II) oxalate in an inert atmosphere, makes a mix, wustite (a nonstoichiometric compound) that is primarily $\ce{FeO}$.
Chromium compounds, e.g. oxides, of varying proportions might make an interesting demonstration because of obvious color changes, but the toxicity of chromium compounds precludes use by students, and even safe disposal of the demo material may be problematic.
BTW, an interesting effect is observed in podzol soils, as found in pine forests. If there is iron in the soil, the reducing nature of freshly-exposed soil makes it look greenish; but after exposure to air, the iron is oxidized to a reddish color (I've observed the change in less than 15 minutes.).


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Meaning of Partial Half-Life for Two Parallel Reactions
Consider two parallel reactions:
$$\ce{A   ->  B }\tag1$$
and
$$\ce{A ->  C}\tag2$$
What is the meaning of partial half life of the equation $(1)$ and that of equation $(2)$ with respect to above reactions?
It seems to me that it means the half life when one of these reactions are taking place individually, one-at-a-time, and not simultaneously. For example, partial half life of the equation $(1)$ is when half of $\ce{A}$ converts to only $\ce{B}$ and not to $\ce{C}$.

As you have written it A must go to both B and C with rate constant $k_1+k_2$: A cannot choose and both B and C appear with rate constant $k_1+k_2$. If you know the rate constants then you can choose to define the half life but that is just an alias for the rate constant anyway so why bother ? (The yield to B is $k_1/(k_1+k_2)$ if that is what you are actually looking for.)

The partial reaction half-life relates to the reaction speed constant by the same way as for the "normal" reaction half-life.
If there are 2 parallel reactions of the first order, $\ce{A -> B}$ and $\ce{A -> C}$, and if there is the reaction rate for the former:
$$\frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t}=k_{\ce{B}} \cdot [\ce{A}]$$
then the partial half-life w.r.t. $\ce{A -> B}$ is:
$$t_{1/2,\mathrm{B}}=\frac{\ln{2}}{k_{\ce{B}}}$$
Analogically the similar for the other reaction:
$$\frac{\mathrm{d}[\ce{C}]}{\mathrm{d}t}=k_{\ce{C}} \cdot [\ce{A}]$$
$$t_{1/2,\ce{C}}=\frac{\ln{2}}{k_{\ce{C}}}$$
The overall half-life w.r.t. $\ce{A -> X}$ is then
$$t_{1/2}=\frac{\ln{2}}{k_{\ce{B}}+k_{\ce{C}}}=\frac{1}{\frac 1{t_{1/2,\ce{B}}}+\frac 1{ t_{1/2,\ce{C}}}}$$
----  Responses to comments
The partial half-life is the extrapolated time after which all $\ce{A}$ would have decayed, if it had been decaying by the current and constant rate of given reaction and only by this reaction. But the main meaning is as a kind of reciprocal value of the reaction constant.
If we draw the chart of the partial reaction rate, then it's tangenta at $t=0$ will cross $x$-axis at the reaction partial half time. It is the same as if it was the only reaction. It does not say half of $\ce{A}$ decays in this partial half time.
If $\Delta t \ll \min{(t_{1/2})}$, then it does not matter it is just partial halftime, as we can neglect decay of the other parallel reactions.

Does this term "partial half life" carry any actual physical meaning?

Sir, " if it was decaying by constant rate of given reaction and only by this reaction."  -- This statement makes me think that we are talking in hypothetical sense. Are we?

The ratio of concentration and its (total or partial ) time rate is a very real parameter. See A update.

Are you not assuming that the rate constant A to B is the same as that B to A and similarly for C in making your final half life $t_{1/2}$?  This is not true unless by numerical accident. The reaction studied is A to B and C and not the reverse.

Why should I ? Reversibility of reactions has not come into account at all. E.g. We can consider 2 different modes of radioactive nuclide decay.

@porphyrin But I have fixed the sign, which may have confused you.

yep , thats better :)