text:
Can I increase the melting point of paraffin wax over 100 degrees celsius
Can I increase the melting point of paraffin wax over 100 degrees celsius at room temperature and pressure.

"…over 100 degrees celsius at room temperature…" — not sure how this makes sense. Also, it's  degrees Celsius, and a question is usually supposed to end with a question mark.

I think you need to try to read about polymer processing and polymer chemistry, look at processing methods for the cross linking of HDPE and LDPE. I think that the question might look like a super simple one but I think it is a hidden gem. It touches on radiation chemistry, radicals and polymer processing.

I'm rolling the question back to the original state because the edit conflicts with OP's intent (which already was unclear enough) and appears to be produced to retrofit an existing answer.

Well if you were to cross link the parafin wax then it would be possible to increase the melting point.
Sadly I doubt if you will have the equipment needed, if you were to put a parafin wax in a glass ampoule and then stick it is a gamma irradation machine such as a gamma cell 220 and give it a mighty dose of gamma rays. I think you would need megagray doses then you would change the parafin into a different material. You would have a combination of chain breaking and chain cross linking. This will alter the properties of the parafin.
You can regard parafin as a shorter version of high density polyethene, it is well known that radiation will cross link polyethene. The cross linked polymer is much harder to set fire to than the untreated product. It also is harder to melt.

The simplest answer is NO ! By definition, a paraffin is a mixture of rather heavy alcanes (more than $\ce{20 C}$ atoms), with some alkenes. If the gamma irradiation is able to polymerize alkenes, it is much more efficient in breaking $\ce{C-C}$ bonds, producing a mixture of low molecular substances, some of which are alkenes. Mixtures have melting points lower than their constituants. So the obtained irradiated mixture will have a lower melting point than the original paraffin wax.

I think you are wrong, if you read A.J. Swallow's book on organic radiation chemistry or Radiochemistry and Nuclear Chemistry (G Choppin, Jan-Olov Liljenzin, J. RYDBERG, C. Ekberg) you will find that the irradation of alkanes / polyethene does cause molecules to grow. I have irradated alkanes myself and then done GCMS, I found that the size of the molecules goes up. Decane will tend to form C20 alkanes

It is right that dimerization occurs in the radiolysis of alcanes. But degradation occurs simultaneously producing plenty of simpler hydrocarbons, and also some gasous Hydrogen $\ce{H2}$. The resulting mixture contains a lot of dimers, of course, but it also contains lighter hydrocarbons in $\ce{C2, C3}$ , etc. The resulting mixture is melting at a lower temperature than the original paraffin.

Surely crosslinking with peroxides would be more realistic than gamma rays


text:
Chemistry: The Central Science 14th edition, chapter 3, Design an experiment problem

I am new to chemistry as well as StackExchange network, so any advice is appreciated.
I tried to answer it myself, but I am sure it is not completely correct and there are certainly alternative solutions. There are also points I would like to know which I listed at the end of my solution.
Here's what I came up with:
a) We assume $S+O_2 \rightarrow SO_2$. The equation is balanced and $1 \space mol \space S≏ 1 \space mol \space O_2 ≏ 1 \space mol \space SO_2$ thus, $0.1 \space mol$ $O_2$ is needed for $0.1 \space mol$ $S$ to react.
b) We assume $S+O_2 \rightarrow SO_3$. Balancing the equation gives $2S+3O_2 \rightarrow 2SO_3$ and $2 \space mol \space S≏ 3 \space mol \space O_2 ≏ 2 \space mol \space SO_2$ thus, $0.15 \space mols$ $O_2$ is needed for $0.1 \space mol$ $S$ to react.
c) Equipment given is an analytical balance, which have an accuracy of $0.0001$ to $0.00001$ grams. Since we know the molar mass of Oxygen (gas) is $31.999$ g and that of Sulfur is $32.065$ grams, we can weigh them on the balance to use correct number of moles. (For practical purposes, we can approximate both of these quantities as $32$ grams)
d) At room temperature, Sulfur Dioxide is a colorless gas meanwhile Sulfur Trioxide is a crystalline solid. Their physical properties can be used to identify them. I am unsure about the instruments.
e) I would use $1 \space mol \space S$ and $1 \space mol \space O_2$. If the assumption a) is true, then the reaction will result in one mole of Sulfur Dioxide. If no solid is left on the reaction vessel then assumption a) is supported. If the assumption b) is true, then $O_2$ will be the limiting reagent and products of the reaction will be $\frac{2}{3} \space mol$ Sulfur Trioxide and $\frac{1}{3} \space mol$ leftover Sulfur. Sulfur is also a solid in room temperature and if these two solids weigh 64 grams (that is, the weight of reactants) then assumption b) is supported.
The last case is when the product is a mixture of $SO_2$ and $SO_3$, let's call it assumption c). Balanced equation then becomes: $4S+5O_2 \rightarrow 2SO_2+2SO_3$. Since Oxygen is still the limiting reagent, we expect products of the reaction to be $\frac{2}{5} \space mol  \space SO_2$, $\frac{2}{5} \space mol  \space SO_3$ and $\frac{1}{5} \space mol \space S$.
Points I would like to know:

What are the ways to distinguish between Sulfur Dioxide gas and Oxygen gas besides the impractical (downright foolish) idea of inhaling the resulting gas and checking for poisoning symptoms?
What could I have used the furnace for?
If I carry out the experiment, will I get Sulfur Dioxide or can initial conditions like temperature affect the results?

You cannot use $1$ mole $\ce{O2}$, because $1$ mol of any gas has a volume of about $25$ liters at room temperature and pressure. No laboratory vessel has a volume greater than $1$ liter. The furnace is here to heat the sulfur. Sulfur does not burn at room temperature. It has to be heated to about $200$°C to start burning. The amount of oxygen will be easier determined by knowing the volume of the vessel, and deriving the number of moles by $\pu{n = pV/RT}$

Sulfur is burned by the Megatonne to make sulfuric acid. All the conditions have been worked out and can be found in the literature or if you live in Florida, USA by a trip to the Bone Valley phosphate mines where they [used to] make sulfuric acid to convert phosphate rock into phosphoric acid and gypsum [with yellow cake as a side product].
This is a hellish chemical system for a newcomer to chemistry to even consider. It involves consecutive reactions, the handling of gases, and has always been a chemical engineering nightmare. Sulfur is easily burned to give SO2 but further oxidation to SO3 is difficult and requires a complicated reaction or catalysis. What I am trying to tell you is that any reaction sequences or analysis techniques that might be dreamed up would probably have no use in reality and would be useless in learning any chemistry. It took a lot of chemists a lot of work to work it out. One other comment: Equations such as you propose for SO2 +SO3 are not allowed. They are two separate equations and if there is a reaction between the two the stoichiometry is dependent on the various amounts; you don't get one SO2 for every SO3.
If you must tackle this! the approach should be to convert the sulfur to SO2 by burning in excess O2[not air, N2 might cause complications, more research to do]. [If SO3 were to be formed there would have to be excess of O2 so no need to use less or even stoichiometric amounts, just burn all the sulfur].  Then research the various methods of converting SO2 to SO3. Now you might need an analysis technique. I would recommend Raman spectroscopy on the gas phase. This will easily analyse for SO2, SO3, and O2and probably even for residual S8. BTW the analogy between the oxidation of carbon to CO and CO2 and Sulfur to SO2 and SO3 is at best a very weak one. The energetics are very different.

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What is local and global aromaticity?
I was browsing through some research papers, I came across the terms locally and globally aromatic. What does it mean? I was thinking locally aromatic meant it was conjugated in a specific space not the whole molecule, i.e benzene joined by a big alkyl group and ending in another phenyl group. But that didn't seem like a satisfying definition to me, I googled around the internet and only found this image.

Peeks, M. D.; Jirasek, M.; Claridge, T. D. W.; Anderson, H. L. Global Aromaticity and Antiaromaticity in Porphyrin Nanoring Anions. Angew. Chem., Int. Ed. 2019, 58 (44), 15717–15720. DOI: 10.1002/anie.201909032.
And now I am even more confused about what is going on, how did adding electrons all of a sudden make a locally aromatic structure into a globally aromatic / antiaromatic structure?

Tbh, the meaning of 'global' vs 'local' seems to be clear / intuitive enough from the graphical abstract. I think the more interesting part is the 'aromaticity' - there are lots of definitions of aromaticity out there, the question is which they are applying. From the manuscript I gather that they are using the presence of a ring current to define aromaticity. But I only skimmed it; I'm sure you can get more information by reading it, and the references therein, carefully.

@orthocresol but how did adding electrons make the ring current global?

From the diagram, it appears that local and global aromaticity are related to how the aromatic coupling is arranged in the macrocycle. Local aromaticity is characterized by aromatic coupling within individual rings or regions with little conjugation between. Such structures are planar within their aromatic rings/regions, but generally have nonplanar geometry in-between. Global aromaticity involves aromatic coupling throughout the macrocycle, with strong conjugation between rings and an extended planar structure.
Adding electrons can alter the mode of aromaticity if the electrons enter orbitals that are predominantly bonding between rings (converting local to global) or in the reverse fashion predominantly bonding within rings (converting global to local).
This effect can be seen in a simpler system, [a,e]dibenzocyclooctetraene:

Zhu et al. 1 considered the reduction of this molecule from the neutral state to the dianion using alkali metals. The neutral molecule is clearly aromatic only locally within the benzene rings, with a nonplanar structure in the intervening cyclooctatetraene ring. When the molecule is reduced, it becomes fully planar suggesting that aromatic coupling has become globalized. (Figure from Ref. 1)

We can use the Hückel model to show semiquantitatively how this change comes about. Using this model we can calculate the pi-bond order between each pair of neighboring atoms, obtaining the following for the neutral molecule:

We have nearly a full pi bond across the mirror plane between the benzene rings, while the neighboring benzylic linkages have much reduced pi bonding. This suggests that in the neutral molecule the aromatic coupling is predominantly localized to the benzene rings, agreeing with the locally planar but globally nonplanar molecular geometry shown above.
When the alkali metal is added and two additional electrons are inserted, they go into an orbital that the Hückel model identifies as bonding in the benzylic linkages but antibonding in most other linkages, particularly in the bridges that close the benzenoid rings and the previous nearly-isolated pi bonds in the middle of the structure. The bond orders are thereby shifted:

The newly occupied orbital, selectively bonding in the benzylic linkages, strengthens conjugation there while weakening the bonds that close the smaller rings. The aromatic coupling has shifted away from the individual benzenoid rings to cover the full macrocycle, becoming global.
Reference

Yikun Zhu, Zheng Zhou, Zheng Wei, and Marina A. Petrukhina (2020). "Two-Fold Reduction of Dibenzo[a,e]cyclooctatetraene with Group 1 Metals: From Lithium to Cesium", Organometallics, 39, 24, 4688–4695.
https://doi.org/10.1021/acs.organomet.0c00688.

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Can shop-bought caustic soda be used to measure titratable acidity of wines?
When making wines, it's occasionally necessary to measure the wine's 'titratable acidity' (TA). To do this, a 0.1N solution of NaOH is titrated with a wine sample. 0.1N NaOH solution can be purchased, but I'd like to know if I can make it myself using household caustic soda.
I want to make 100ml of 0.1N NaOH solution, which I understand involves adding 0.4g of NaOH to 100ml of water.
I have household caustic soda (as used for cleaning drains), but I have no idea whether this is pure NaOH. It only needs to be 'pure enough' for the purposes of getting an accurate TA reading from my wines.
How would I determine whether the caustic soda I have in my cupboard is pure enough for the purposes of measuring TA? I have a pH meter, and I know that 0.1N NaOH should be pH 13; does this mean that so long as the resulting solution is pH 13 then the solution is suitable for measuring TA?
(Note: in order to measure TA, I would take a 15ml sample of wine, and keep adding 0.1N NaOH until the pH = 8.2 The TA would be the amount of NaOH added (in ml) divided by two. A typical TA is 6-9 g/l.)

In our quantitative analysis course, students carefully prepared sodium hydroxide solutions from reagent grade solid sodium hydroxide. This requires care and following a valid preparation protocol: it is easy to mess up. Then they standardize their sodium hydroxide solutions against a primary standard acid, typically potassium hydrogen phthalate (KHP). Only then are their sodium hydroxide solutions acceptable for use in the titration of acids in samples.

Commercial caustic soda is usually pure enough for making up  $0.1 M \ce{NaOH}$ solution. But don't try to measure the $p$H of the obtained solution. A $0.1 M \ce{NaOH }$ solution does not have a $p$H exactly equal to  $13.00$, because of the activity effect. The relation $p\ce{H} = - log[\ce{H^+}]$ is only valid at concentration $\ce{[H^+] < 0.01} M$. At higher concentrations, the concentration should be replaced by the activity.

I do not see how this could be a personal medical question.// As Homework is considered literal homework, exams, self-study questions, puzzles, worked examples etc.

Homework is often selected as a reason to close by people who feel a question could be answered with minimal research by the OP. In essence your question is covered in introductory analytical chemistry courses, within the topic of acid/base standardization, and a typical textbook will cover the topic satisfactorily.

@BuckThorn I think the question is rather specific, bypassing general topic of acidobasic volumetry, covered by textbooks. It is rather for out of lab context of cheap, available chemicals, questioning being pure enough to make orientation/screening quality titration with acceptable accuracy.


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Would a 0.3 micron filter be effective in blocking formaldehyde fumes produced by curing cyanoacrylate?
I’m looking into purchasing a mask that blocks particles larger than or equal to 0.3 μm in size. I’m hoping it blocks the formaldehyde fumes produced by curing cyanoacrylate.
How effective such mask would be, i.e. would the size of formaldehyde fume particles be greater than or equal to 0.3 microns?

Don't know.  Formaldehyde could be present as a gas rather than an aerosol, in which case the particles would be molecule-size.   I'd thus get a mask that fits either of these cartridges:   https://www.3m.com/3M/en_US/p/d/v000057398/  or https://www.3m.com/3M/en_US/p/d/v000152051/.  Depending on the conentration, you might want to seal your eyes off (using goggles) as well.

Thank you. Do you know if the half-masks these cartridges are used in are a tight seal compared to, say, the classic paper and cloth masks we’ve seen through the pandemic? I only ask bec it is crucial that my breath is not able to leak out the top of the mask near my eyes. Is it a tight/snug fit round the nose bridge?

These masks are made of a rubbery material that should seal to your face better than cloth or plastic, assuming (a)if you're  a guy, you're clean-shaven; and (b) the mask fits well. Fit is critical. If you were in industry or academia, you'd undergo a fit test to check this (https://www.3m.com/3M/en_US/respiratory-protection-us/support/center-for-respiratory-protection/fit-testing/). On your own, you'll need to use your best judgement (which in your case at least means you can't smell any formaldehyde). You'll just need to order some on AZ and try them. The 3M 5203 Medium fits many faces well.

Generally, put in questions enough relevant details and background to avoid needed clarification or possible misunderstanding.

@theorist so sorry to bother you again, but would either of these filters be suitable in blocking out formaldehyde fumes? I’m in Aus so I’m having to use the Aus site for faster delivery. Thank you so much for your help

@theorist https://www.3m.com.au/3M/en_AU/p/d/v100561013/, https://www.3m.com.au/3M/en_AU/p/d/v000057497/?preselect=8745506+8784387

@theorist does the fact the first link says “A1 form” mean anything?

You can see from their respective descriptions that 3M says they're effective against formaldehyde. The question is what percent of the fumes they block out and/or up to what level of formaldehyde are they safe to use with. For this more detailed info., an answer to what "A1 form" means, and also to get a link to a table showing what masks these cartridges work with, I would recommend contacting 3M directly for guidance. You can find their contact info. on their website.

And of course you should also look into having good ventilation. That, plus the mask, will be better than the mask by itself.  And for eye protection from fumes, if you're not going to wear a full face mask, swim goggles will work if you don't need to wear them for an extended period of time (too long and they get uncomfortable and/or fog up).  Just get some "no-tears" gentle shampoo (baby shampoo works), put a tiny drop onto the inside surface of the goggles, and gently buff dry; you can do this with your thumbs (no water needed). That will work as an effective anti-fog agent.


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Stereochemistry of nitrosyl chloride addition to an alkene
My book states addition of $\ce{NOCl}$ to propylene results in 2‐chloro‐1‐nitrosopropane:

Would it follow syn or anti addition?
According to Mallya and Nagendrappa [1], the reaction would follow syn addition:

Cycloalkenyltrimethylsilanes add nitrosyl chloride stereo- and regio- selectively. The addition is syn, with chlorine and nitrosyl group adding to C-1 and C-2 respectively.

I am unsure whether it's applicable to my problem. Moreover, I was under the impression that the addition would be anti due to steric repulsions.
Reference

Mallya, M. N.; Nagendrappa, G. Nitrosyl Chloride Addition to Cycloalkenyltrimethylsilanes. An Unexpected Reversal of Regiochemistry of Addition, and Synthesis of β-Aminosilanes. Arkivoc 2007, 2006 (14), 155–174. DOI: 10.3998/ark.5550190.0007.e17.

You might be better with a cyclic alkene such as cyclopentene as it is possible to have cis and trans products which will give clues to how the reaction occurs

Syn or anti addition is difficult to resolve with propene because the final product admits rotation about the single bond. However, if the rotation is constrained or the relative positions of the functional groups are marked by chirality, then we see anti addition. See this summary of reaction mechanisms, step 41.

More is involved than steric favoritism here. When the nitrosyl cation adds to the alkene, it picks up an electron pair on the nitrogen which can then back-bond to the carbocation center (cf. the bromonium ion formed with electrophilic bromination). The electron-accepting orbital is then antibonding with respect to this linkage and thus displaced to the back side of the carbon. So the chloride ion will attack the carbon there to give the anti product.

The most important fact about this question is the stereochemistry of nitrosyl chloride $(\ce{NOCl})$ addition to an alkene. The OP has wondered whether it follows syn- or anti-addition.
The answer is not simple one word answer. According to the Ref.1, the nitrosochlorination of alkenes is shown to possess features common to other reactions of electrophilic addition, yet the stereochemistry of the addition depends on the structure of the alkene. For example, the addition of $\ce{NOCl}$ to norbornene follows cis-stereochemical course while that to $\Delta^9$-octalin follows trans-stereochemical course. However, the Ref.2 suggests that stereochemical outcome of the reaction also depends upon solvents used in the reaction. For instance, the stereochemistry of the addition of nitrosyl chloride to cyclohexene in liquid sulfur dioxide is not the same when the identical reaction is performed in $\ce{CH2Cl2}$ or in  $\ce{CHCl3}$ (Ref.2).
In later studies of same reaction in liquid sulfur dioxide, the trans-stereochemical out come of the dimeric product has established by NMR spectral data (Ref.3).
Structural influence on this reaction was also established on the $\ce{NOCl}$ addition on 3,4,6-tri-O-acetyl-D-glucal, which has predominantly showed cis-addition (Ref.4 & 5):

The structural effect is also evedence in oxymercuration: The oxymercuration of D-glucal has given the corresponding cis-product while that of its 3,4,6-triacetate has given the corresponding trans-product (Ref.6).

References:

P. P. Kadzyauskas and Nikolai S. Zefirov, "Nitrosochlorination of Alkenes," Russian Chemical Reviews 1968, 37(7), 543 (DOI:
https://doi.org/10.1070/RC1968v037n07ABEH001665).
Masaji Ohno, Masaru Okamoto, and Kenkichi Nukada, "Dependence of steric course of nitrosyl chloride addition to cyclohexene on solvents," Tetrahedron Lett. 1965, 6(45), 4047-4052 (DOI: https://doi.org/10.1016/S0040-4039(01)99612-1).
B. W. Ponder, Thomas E. Walton, and William J. Pollock, "Addition of nitrosyl chloride to cyclohexene," J. Org. Chem. 1968, 33(10), 3957-3959 (DOI: https://doi.org/10.1021/jo01274a065).
R. U. Lemieux, T. L. Nagabhushan, and S. W. Gunner, "Synthesis of 3,4,6-Tri-O-acetyl-2-oximino-α-D-hexopyranosides," Can. J. Chem. 1968, 46(3), 405-411 (DOI: https://doi.org/10.1139/v68-06).
R. U. Lemieux, T. L. Nagabhushan, and I. K. O'Neill, "A new approach to the synthesis of 2-amino-2-deoxyglycosides," Tetrahedron Lett. 1965, 5(29), 1909-1916 (DOI: https://doi.org/10.1016/S0040-4039(01)89472-7).
Philip T. Manolopoulos, Morton. Mednick, and Norman N. Lichtin, "Oxymercuration of D-Glucal and Its 3,4,6-Triacetate," J. Am. Chem. Soc. 1962, 84(11), 2203–2210 (DOI: https://doi.org/10.1021/ja00870a036).


text:
What is Dornic acidity and how to measure it?
Context: I work with dairy product (cheese making) and in order to improve the quality of my process, I have documentation (in French) which refers to Dornic acidity over time.

This graph represents the optimal acidity range over time for the production of cheese and reduce the risk of certain bacterial infections. The Y axis is the Dornic acidity and the X axis is the time in hours from coagulation of milk with rennet.
Problem:

What exactly is Dornic acidity?
How do I measure it? (At low cost preferably)
I am familiar with pH and I am already equipped with pH meters. Is there a pH-to-Dornic correspondence? (And if not then why?).

I hope my compound questions are acceptable as they really are closely related.

From https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3568963/: "Dornic acidity is a way of expressing the acidity degree of milk by measurement of the titrable acidity."

1 Dornic degree (1°D) is equal to 0.1g of lactic acid per liter. chal. Interesting is Dornic acidity seems to have FR and DE but not EN Wikipedia pages.

Different regions have different measures of acidity: https://www.dairyscience.info/newcalculators/acidity/

I would think a regular pH meter would do poorly with milk and even worse with cheese production because of the liquid fats would interfere.

The acidity of milk can be expressed in various ways. There are three main scales:

Soxhlet Henkel degree: Number of ml of 0.25 N $\ce{NaOH}$ required to neutralize 100 ml of milk
Thörner degree: Number of ml of 0.1 N $\ce{NaOH}$ required to neutralize 100 ml of milk
Dornic degree: Number of ml of 0.11 N $\ce{NaOH}$ required to neutralize 100 ml
of milk

One ml of 0.11 N $\ce{NaOH}$ = 0.01 g lactic acid i.e 1 Dornic degree corresponds to 0.01g of lactic acid per 100 ml of milk or 0.01%. Therefore, number of Dornic degrees divide by 100 gives the content of lactic acid in percent. The measurement is done through titration with phenolphthalein as indicator. During analysis, the milk is diluted to several degrees which can be used to make comparison of various degrees of acidity. Table of comparison is as follows:

These scales are very important during cheese making. Cheese makers wants to reduce the acidity of milk from 0.3% (30° Dornic) down to 0.16% (16° Dornic). The calculation below determines the total amount of acid contained in 100 litres of milk:
$\mathrm{0.3-0.16 = 0.14~kg~of~acid}$
$\mathrm{(30 ^\circ D - 16 ^\circ D) \times 100~kg = 1400^\circ D = 140~g~of~acid}$
Neutralization will be done with the fact that 40 g of $\ce{NaOH}$ will nautralize 90 g of lactic acid so the quantity of $\ce{NaOH}$ needed to neutralize the above amount of acid would be:
$\mathrm{\frac{40 \times 140}{90} = 62.2~g~of~NaOH}$
Reference

Robinson: Modern Dairy Technology: Volume 1 Advances in Milk Processing by R. Robinson, Springer Science & Business Media, 2012
The Oxford Companion to Cheese by Catherine W. Donnelly, Oxford University Press, 2016
The Technology of Making Cheese from Camel Milk (Camelus Dromedarius) by J.P. Ramet, Food and Agriculture Organization of the United Nations
Food & Agriculture Org., 2001

How would titration be effected once the milk solidified in to curds? The document from which I got the graph suggests a measurement of acidity over 24 hours after adding rennet to the milk, by which time the cheese would already be in solid form and have drained. It implies a measurement should be possible during that stage too. But then is it possible to do titration with a solid?

I am sure, they would do filtering. Impossible to titrate a curd directly.

I was given an answer to the measurements of acidity post-coagulation of curds, and the answer to that is the measurement is made from the remaining whey. I was under the impression one could not assume the whey and the cheese would share the same acidity but apparently that is the practice. Source: private discussion with a French microbiologist who studies fungal and bacterial growths on cheese.

Interesting term...never heard of it nor it is common in English chemical analysis literature. It turns out that it is an ancient method from >100 years. The unit is named after a person. The original definition from a 1900 document is ""D" indicates acidity in degrees Dornic, the number of cubic centimeters of a NaOH solution containing 4.445 g NaOH per liter required to neutralize the acidity of 100 cc of milk." Journal of Dairy Science,  1925, Volume 8, pg 172.

It is a simple acid-base titration, phenolphthalein is used as an indicator.

"To be considered fresh, a milk must have an acidity less than or equal to 18 °D. Dornic acidity is the result of the natural acidity of milk (linked to its richness in proteins and minerals). It is an indicator of the degree of milk preservation. Naturally the lactose contained in the milk gradually degrades to lactic acid by the bacteria. The less milk is fresh, the more it contains lactic acid." from Adv. Biores., Vol 9 (2) March 2018: 166-177.

In a complement to AChem's informations, the titration should be carried out until the measured $p$H value is equal to $8.30$, even if there is no visible $p$H leap at this $p$H value. .

That is not answer to part 3. It is however a reasonable response to the main question.

@user2617804 There is no general relation between pH and acid content. There would have been such a relation if milk had been nothing but a lactic acid solution. For milk, one can construct an empirical chart pH versus Dornic acidity.

Thanks. So no PH-°D conversion. May I ask for guidance on practical references to processing titration? I'm not quite familiar with the process and pointers would be much appreciated thanks.

From https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3568963/ I read: "To measure Dornic acidity we added a drop of alcoholic solution of 1% phenolphthalein as an indicator of the color change point in 1 mL of milk. We titrated 0.01 mL of sodium hydroxide N/9 (NaOH), until the sample changed color from white to light pink and the color change was maintained. Each 0.01 mL of NaOH necessary for the sample to change color accounts for 1° of Dornic acidity (°D). "  Is that in sum the whole methodology?

" It turns out that it is an ancient method from >100 years. The unit is named after a person" I tried searching but couldn't find exactly when the scale was first introduced nor who invented it. Is it really a person who invented it? Who is Dornic?

@NilayGhosh, Yes, it is indeed a French person. See German Wikipedia https://de.wikipedia.org/wiki/Grad_Dornic and machine translate it.

Thanks. I was looking in English WP, hence couldn't find anything :)


text:
How do I determine if a nucleophile attacks the electrophilic atom directly or indirectly?
How do I determine if a nucleophile attacks the electrophilic atom directly or indirectly? For instance a nucleophile could attack the carbon in a molecule with a carbonyl directly, but in another example the nucleophile might only attack the nucleophilic atom after attacking a hydrogen first (deprotonation). Thank you in advance!

Hard-Soft Acid-Base Theory provides rules of thumb to predict which electrophilic atom a given nucleophile will attack. But pattern recognition/memorization is also useful.

Ultimately only experimental data is definative. One can make predictions but until they are backed by experiment they remain "best gueses".


text:
Electrolysis with salt water on steel plates creates horrible black liquid
We are doing an electrolysis experiment using distilled water with sea salt on clean stainless steel plates.  A horrible black gunk has formed in the water, very quickly. What is this?

I don't believe this is simply the brown colour that others are seeing on their plates.  The black liquid appears after just 90-120 minutes in the bath.  We are using 4V at 2A in this.  And we definitely do not see this when we use KOH as the chemical.

Not an exact duplicate, but see the following post for an answer that should work with your question: https://chemistry.stackexchange.com/questions/50683/why-are-electrolysis-cell-plates-turning-a-brown-color?rq=1

I don't believe this is simply the brown colour that others are seeing on their plates. The black liquid appears after just 90-120 minutes in the bath. We are using 4V at 2A in this.

Just count the number of electrons that you provided to the electrodes during 2 hours: 120 x 60 x 2 A=14,400 Coulomb. 1 Coulomb = 6,240,000,000,000,000,000 electrons! Do this multiplication with 14,400 Coulomb. Also rest assured that each electron caused a chemical reaction, because it has to if you forced this high current. What is there to react during electrolysis? Water, salt, and steel plates. When all of them are decomposing, the products react together in a mixed compartment- you get this reddish gelatinous precipitate of mainly iron (III) hydroxides because the cathode side is alkaline.
This precipitate cannot appear all of a sudden after 120 minutes, the electrode corrode slowly with the passage of time and the electrode (anode) continuously dissolves as current is passed through it.

Thanks.  No, it doesn't appear "all of a sudden", but builds up over the time.  It just seems like a heck of a lot for just two hours.

And stainless steel is not pure iron. It is an alloy of iron containing in principle 18% nickel and 8% chrome, These elements are also oxidized by the chlorine produced by the electrolysis, and transformed into hydroxides which get mixed with iron(III) hydroxide

The anode is oxidized before chloride ion is oxidized

@foolishmuse, It appears bulky because of the number of electrons you passed through the electrode! It is simply huge. In electrochemistry, the amount of products is proportional to the number of electrons that passed through the circuit. 2A is a large large number.

The applied current drives sodium ion to the cathode (negative pole) where hydrogen is evolved and the surrounding medium becomes more alkaline.
The more interesting - in the sense of catastrophic - area is at the anode, where the very high applied positive voltage oxidizes water to oxygen and chloride to chlorine, while simultaneously demolishing the passive film on the stainless steel. At this point, the elemental chlorine which was just produced can attack the steel, instantly reverting to chloride ion, which is oxidized to elemental chlorine, but I repeat myself.
Another thing that happens at the anode is reduction of pH, because if you take oxygen away from water, all you have is hydrogen. Now steel should react with water, and it will, faster with a catalyst like chloride ion, probably much faster with an applied current of 2 amps, and even faster with elemental chlorine present, and even faster with the equivalent of hydrochloric acid in the solution.
And then, the temperature: 8 watts (4V x 2A) in a tiny little dish (what - about 50 mL? - couldn't be more than 100 mL). In 90 minutes, you are putting in over 10 kilocalories of heat (granted, some energy goes out as the hydrogen and oxygen). So the temperature got to what? About 90º-100ºC? This would guarantee a rapid reaction, lots of chlorine odor, and alkaline and acid mists spraying out into the lab. (Each electrode will produce its own mist, and much will probably escape without being completely neutralized.)
The black gunk is a mix of iron, chromium (and possibly nickel) chlorides, oxides and hydroxides. with some of the sodium from the sea salt. It's essentially mushy mixed rust. It's an efficient method of destroying stainless steel.
The more I ponder this experiment, I wonder if there were any injuries. Please don't scale up!

It is probably a mix of Fe++, Ni++ and Cr+3 oxides and hydroxides. This is a common trick by RO system salespeople to convince people how terrible their water is. They electrolyze a nail, the more ions the more gunk that forms [the water usually is terrible]. My explanation for no precipitate with stainless steel with KOH electrolyte is that the amphoteric Chromium is most easily oxidized to the soluble chromite ion in the basic solution. A more careful chemical analysis is in order. It is important to have a purpose for a chemical experiment to answer specific questions. This appears to be playing with chemicals with no defined purpose or reasons. So back to the planning stage, what is the purpose of the experiment, what variables must be controlled etc. [Water purity, solute, anode composition voltage to name a few].

Warning! In the process of this experiment, you most possibly have created some amount of toxic, carcinogenic, and environmentally harmful waste. Read further for answer to your question and more details.
All the nasty buildup is probably composed of carbon that was originally in the steel, mixed with some amounts of insoluble oxides and/or hydroxides of steel's constituting metals.
Stainless steel is a particularly bad choice for an electrolytic anode, because it contains chromium (at least 11%, often much more) and nickel. Both nickel metal and soluble nickel salts are toxic and carcinogenic. Chromium gets mobilized into the solution by being oxidized to trivalent $\text{Cr}^{3+}$ ions which, under favourable conditions, could get oxidized further to especially toxic and carcinogenic hexavalent forms $\text{CrO}_4^{2-}$ and $\text{Cr}_{2}\text{O}_{7}^{2-}$ (chromate and dichromate anions). It is absolutely unacceptable and illegal to dispose of such waste by pouring it down the drain. If EPA knew about any such incident, they would be on that person like flies on fresh manure.
From https://www.sciencedirect.com/science/article/abs/pii/S0360319920301403 (the article uses $\text{Cr}^{6+}$ for hexavalent chromium -- no such cation exists in standalone form, though; it is just a convenient way of denoting $\text{CrO}_4^{2-}$ and $\text{Cr}_{2}\text{O}_{7}^{2-}$):

Alkaline water electrolysis is one of the easiest methods used to produce hydrogen, offering the advantages of simplicity and low cost. [...] Stainless steel (SS) is considered one of the least expensive electrode materials for alkaline electrolysers, since it is relatively chemically stable and has a low overpotential. Nevertheless, SS anodes do not withstand high concentration alkaline solutions because they undergo a corrosion process. If the electrolyser operates at a voltage of up to 1.6 V, it can generate Fe3O4 and hazardous hexavalent chromium (Cr6+) at the anode. Hexavalent chromium is generated when the chromium-containing stainless steel electrodes undergo an electro-oxidation process.

In order to neutralize any potential hexavalent chromium you need to add a large amount of reducing agent, such as sodium thiosulfate -- it will reduce the toxic and carcinogenic hexavalent chromium to the less harmful trivalent. Nonetheless, while much less dangerous, trivalent chromium is still somewhat toxic and cannot be poured down the drain; not only that, but nickel still remains a problem.
Pour your waste into robust plastic containers, like those white, one gallon "jugs" in which household bleach is sold, and clearly label them as hazardous. From then on, you can pay a chemical waste company to have this waste disposed of for you, or you can store them in some safe and isolated place where absolutely no children or animals have access to, until you can have them properly disposed of.
PS If someone spells "chromium" as "chrome", or claims that any amounts of sodium could be obtained in aqueous electrolysis, I would highly suggest against taking any advice from such a person.

You are somewhat exaggerating ! It is impossible to get chromium(VI) compounds by oxidizing metallic chromium in an acidic solution.


text:
How to calculate <ab|ab> integrals to carry out Schwarz pre-screening of molecular integrals
During writing my own two-electron integration library I faced an issue of dealing with integrals that have very small values beyond the C/C++ double data type. Basically, most integrals turn out to be NaN. So, the prescreening is apparently necessary.
Referring to this thread I see how the prescreening is done, but I have just a little concern. Is there a simple method/algorithm of calculating (ab|ab) and (cd|cd) integrals or am I supposed to calculate them as if they were regular 4-index integrals? Maybe there's a transformation that can be used to calculate them as 1-electron integrals? I lack the mathematical education and am trying to learn things on the fly.

NaN is not the same as a very small value, it is typically  the result of an "illegal" mathematical operation. If this were one of my own codes I would be very suspicious of a bug, not underflow. Why do you think these NaNs are indicating very small values?

I checked by manual calculations. A few values that appeared NaNs are something like 6E-78. I did consider it as a bug, but the math I coded doesn't seem wrong.

I would strongly recommend comparing the results you are getting with a recognized library that should give the same results. https://github.com/sunqm/libcint is one possibility. https://github.com/evaleev/libint is another. General code to calculate 4 centre Gaussian integrals is not trivial, and having a robust testing environment is essential.

I know about that and will test, but not until it's a working library. I'm quite confident that NaNs are caused by the out-of-range arithmetics.
The question about implementing the Schwartz inequality is still open and important.

Please start testing before it is a "working" library. When I did an integral implementation (in Python) I did never encounter underflow problems. It was always just a bug in my programming.


text:
Why don't secondary alcohols react with HBr/HI via SN2?
In polar protic solvents, $\ce{Br-}$ and $\ce{I-}$ are considered to be very good nucleophiles. Now, here the solvent is alcohol which is polar protic. So, why don't secondary alcohols react with $\ce{HBr/HI}$ via $\mathrm{S_N}$2 and go by $\mathrm{S_N}$1 instead?
Notably, primary alcohols do undergo $\mathrm{S_N}$2, and secondary carbons can definitely undergo $\mathrm{S_N}$2 as well, so something must be special about water as a leaving group that makes us say that secondary alcohols primarily undergo $\mathrm{S_N}$1.

The hydroxyl group is not a good leaving group. Thus, the reaction is not simple substitution reaction. It is acid catalyzed and go through a cabocation intermediate based on the alcohol syructure (e.g., a primary alcohol does not go through carbocation intermediate).

@MathewMahindaratne Yes, but both SN1 and SN2 can occur after protonation of alcohol( as we know because SN2 seen in primary and SN1 in tertiary)...now, in secondary alcohol SN2 should occur because Br- and I- are strong nucleophiles?

You said it is in alcohol medium so alcohol is also a nucleophile. Weak but has plenty. So product is a mixture. Nonrtheless, the secondery alcohol reaction is whether SN1 or SN2 depends also on solvent. If it is polar protic SN1 is predominate due to stabilization of carbocation intermediate.  If it is polar aprotic, SN2 predominate.

@MathewMahindaratne But aren't Br- and I- strong nucleophiles in polar protic? So does that mean all nucleophiles(no matter how strong) will show SN1 in polar protic?


text:
Electrolysis when E_cell = 0
I'm a bit confused by the electroplating of Zinc onto Copper, with Zinc (anode) and Copper (cathode) electrodes in a Zn(NO3)2 solution.
The reduction reaction is: Zn2+ + 2e- ---> Zn (s) (E_rxn = -0.76 V)
The oxidation reaction is: Zn(s) ---> Zn2+ + 2e-  (E_rxn = 0.76 V)
Thus, in the overall reaction, all elements "cancel out" and E_cell = 0 V. Does this mean the reaction occurs spontaneously (although electrolysis is generally non-spontaneous)? Also, what is the overall cell reaction? I'm confused about this and can't seem to find any clarifying information online.
Thank you!

You don't mention which electrode is the positive pole and which is the negative. Both choices are possible. But the chemistry is not the same. Anyway it is not a good idea to electrolyze a zinc nitrate solution. Zinc sulfate would be a better choice. Electrolyzing zinc nitrate produces side reactions due to the nitrate ion, with the reduction of nitrate ion to $\ce{NO2^-}$ ions and even to $\ce{N2O}$.

Sorry about that, although I believe it could be inferred from my equations! I just updated my question: zinc is the anode (positively charged) and copper the cathode (negative charge). Ok, I will think about using zinc sulfate instead.

You are confused because you are mixing two different concepts. Half-cells belong to the field of potentiometry when no electric current is passing through the circuit. So that calculations you do above, do not apply in electroplating.
Electroplating is just the opposite and it belongs to the field called voltammetry. As the name indicates, you play with "volts" and "ammetry" i.e., the current. One variable is fixed and the other is varied. You can have constant current electrolysis or a constant potential electrolysis. Both cannot be arbitrarily controlled. Herein are forcing a non-spontaneous or even a spontaneous process to proceed by providing electrons from an external source like a battery. The moment electric current passes through a solution, electrolysis must occur. Nature does not provide a choice to matter in that case. Either the solvent like water will decompose, or the cation or the anions must undergo a redox process and if there is no choice, the electrode material itself has to participate in a redox process. The key point is that electroplating needs a threshold voltage.

Thank you! How would the threshold voltage you mention be calculated?

@cim, This information can come from a voltammetric experiment. You scan the voltage in a given solution and observe current changes. The moment redox process starts, current rises rapidly.


text:
Understanding isosteric enthalpy (heat) of adsorption
I am trying to understand what is isosteric heat of adsorption. Based on van't Hoff equation:
$$ \left(\frac{\partial \ln K}{\partial T}\right)_θ = \frac{ΔH^\circ}{RT^2} $$
and $ΔH^\circ$ is defined as the isosteric enthalpy of adsorption. The equilibrium constant for adsorption is:
$$K = \frac{a_{ads}}{a_g \cdot a_s}$$
and it is given by:
$$K = \exp\left(-\frac{ΔG^\circ}{RT}\right)$$
What is the point of differentiating with respect to temperature while keeping surface coverage $θ$ constant. I mean equilibrium constant depends only on $T$, that is $K=f(T)$.
I can't understand even how we are able to force surface coverage to have a constant value.

Not saying that's an easy part we can force the surface coverage to have a constant value. This answer may be only partial since I am only showing the result in gas phase. I haven't done the calculation for liquid, but probably we can find relation involving molar concentration, though this does not make it easy to keep coverage invariant.
The isosteric heat of adsorption of compound $c$ is the difference between the partial molar enthalpies of the gas phase and in the adsorbed phase. Thus, we have:
$$\Delta H_c^i = H_c^{gas}-H_c^{ad} \tag{1}$$
At equilibrium you have equality of the chemical potential of $c$ in both phases. In the gas phase we have:
$$\mu_c^g = \mu_c^{g,*}+RT\ln\left(f_c^g\right)\tag{2}$$
For the adsorption phase, the expression of the enthalpy is quite direct (thank you Gibbs-Helmholtz). We just derive the equality of chemical potential with respect to temperature and at constant amount of $c$ in the adsorbed phase. We note this fixed amount $x_c^{ad}$.
$$H_c^{ad}=\left(\frac{\partial\left(\frac{\mu_c^{ad}}{T}\right)}{\partial\left(\frac{1}{T}\right)}\right)_{x_c^{ad}} \tag{3}$$
Now comes the "tricky" part (at least at first). For the gas phase phase we can't directly do that because we chose a constant amount of $c$ in the adsorbed phase, thus gas enthalpy can't be in relation with gas chemical potential and the fixed amount $x_c^{ad}$, i.e.: $$H_c^{gas}\neq\left(\frac{\partial\left(\frac{\mu_c^{gas}}{T}\right)}{\partial\left(\frac{1}{T}\right)}\right)_{x_c^{ad}} \tag{4}$$.
However, we know from thermodynamics how to express the chemical potential of a gas, see $(2)$. f denotes fugacity which is what you would consider for real gas. You would then use an advance EOS to determine it. We can however keep it simple and assume ideal gas for the purpose of this answer. So, we have:
$$\left(\frac{\partial\left(\frac{\mu_c^{gas}}{T}\right)}{\partial\left(\frac{1}{T}\right)}\right)_{x_c^{ad}}=H_c^{gas, *}+R\left(\frac{\partial\ln\left(f_c^{gas}\right)}{\partial \left(\frac{1}{T}\right)}\right)_{x_c^{ad}} \tag{5}$$
$(1)$ can then be rewritten as:
$$\Delta H_c^i = H_c^{gas}-\left(H_c^{gas, *}+R\left(\frac{\partial\ln\left(f_c^{gas}\right)}{\partial \left(\frac{1}{T}\right)}\right)_{x_c^{ad}}\right) \tag{6}$$
Considering ideal gas law, fugacity reduces to molar gas fraction $y_c$ times overall pressure (fugacity coefficient is one for ideal gas). I consider here a pure gas so $y_c$ does not even matter. We are left with:
$$\Delta H_c^i = H_c^{gas}-\left(H_c^{gas, *}+R\left(\frac{\partial\ln P}{\partial \left(\frac{1}{T}\right)}\right)_{x_c^{ad}}\right) \tag{7}$$
At low pressure, i.e. ideal gas, $$H_c^{gas} \approx H_c^{gas, *} \tag{8}$$
Living us with:
$$\Delta H_c^i=RT^2 \left(\frac{\partial \ln P}{\partial T}\right)_{x_c^{ad}} \tag{9}$$
You asked, how do we keep the amount absorbed constant? Well, it's not easy-peasy but here $P$ and $T$ are independent variables so as you change $P$ you can adjust $T$ to keep $x_c^{ad}$ constant and vice-versa.
I hope this makes more sense.


text:
Data for Madelung Constants
Where can I find data for the Madelung constants for various ionic compounds? Specifically, I am looking at silicon tetrachloride, yttrium (III) chloride, zirconium (IV) oxynitrate hydrate, niobium (V) chloride, molybdenum (V) chloride, hafnium tetrachloride, tantalum (V) chloride, and tungsten hexachloride.
Is there a database that has already-determined Madelung constants? Thanks.

Why do you want Madelung constants for this set? Apart from the first two (and even then it's debatable for Zr4+) the ionic model is going to be an extremely poor representation of such systems.

I wanted the Madelung constants because I wanted to calculate the lattice energy using the Born-Lande equation. Am fairly new to this kind of chemistry, sorry if I am misinterpreting some terms. Would there be an alternate method to determine the lattice energy?

The Madelung constant is the signed sum of the reciprocal lattice distances so you must have the x-ray structure and then from the unit cell and using symmetry calculate numerically all the distances, possible many hundreds until the summation converges.


text:
Why chlorate(V) is one of the primary products of disproportionation reaction between chlorine gas and hot concentrated NaOH?
Cold diluted $\ce{NaOH}$ reacts with $\ce{Cl2}$ producing hypochlorite:
$$\ce{2 NaOH(aq, dil, cold) + Cl2(g) -> NaClO(aq) + NaCl(aq) + H2O(l)},\tag{R1}$$
whereas hot concentrated $\ce{NaOH}$ yields chlorate(V):
$$\ce{6 NaOH(aq, conc, hot) + 3 Cl2(g) -> NaClO3(aq) + 5 NaCl(aq) + 3 H2O(l)}\tag{R2}$$
The explanation I gathered is that the reduction potential of $\ce{NaOH}$ increases with temperature, making it a stronger oxidizing agent causing further oxidation of $\ce{Cl}$.
If that is the case, why $\ce{Na\overset{+5}{Cl}O3}$ and not chlorite $\ce{Na\overset{+3}{Cl}O2}$ or perchlorate $\ce{Na\overset{+7}{Cl}O4}$? What is so special about chlorate(V)?

The explanation you have gathered is not correct. $\ce{NaOH}$ is not an oxidizing agent. The conversion of hypochlorite to chlorate is a very complicated reaction so there is no direct formation of chlorate as your direct equation shows. Instead, this is a third order reaction (at least in an electrochemical cell) and if you are seriously interested in the mechanism, start with Wanngård and Wildlock [1] and go to the earlier references therein. I am sure the heat accelerates the collisions of hypochlorite ions to form chlorate ions.
The other part of the question as to why there is no intermediate chlorite is relatively easy, at least from an electrochemical point of view. If you are familiar with Frost diagrams, then it clearly shows chlorite is very unstable with respect to disproportionation. Look at the red curve (for low pH) and blue curve for alkaline pH. On the $x$-axis we have the oxidation number and on the $y$-axis we have the free energy. Lower the value, more stable the species is—thermodynamically. Chlorite ion is very slightly raised in that red curve as compared to its adjacent neighbors, it means that it is thermodynamically unstable and chlorine would like to stay either as hypochlorite and chlorate but not as chlorite. Similarly, the highest oxidation state species is on the extreme right, which means that is also less stable than chlorate.
It is then obvious that there are relatively two choices for chlorine atoms—stay as chloride ions which it does all the time (sea water for millions of years is still chloride) but under oxidizing conditions, the next stable choice is hypochlorite, and then under even more harsh oxidizing conditions, chlorate but not chlorite.

Figure 5.18 A Frost diagram for chlorine. The red line refers to acid
conditions $(\mathrm{pH} = 0)$ and the blue line to $\mathrm{pH} = 14.$ Note that because $\ce{HClO3}$ and $\ce{HClO4}$ are strong acids, even at $\mathrm{pH} = 0$ they are present as their conjugate bases.

Image source: [2, p. 185].
Bonus trivia: Mars' soil has huge amounts of perchlorate salts and this was discovered by an electrochemical sensor's mistake sent to Mars. It was sent to detect nitrates but nobody checked that the same sensor also responds to perchlorate. So natural formation of even perchlorate is not out of the blue. It is just that Earth's atmosphere is not too oxidizing... otherwise if there were no atmosphere, all sea salts would have met the same fate as Mars. Solar radiation would have fried seasalt to perchlorate.
References

Wanngård, J.; Wildlock, M. The Catalyzing Effect of Chromate in the Chlorate Formation Reaction. Chem Eng Res Des 2017, 121, 438–447. DOI: 10.1016/j.cherd.2017.03.021.
Shriver, D.; Weller, M.; Overton, T.; Armstrong, F.; Rourke, J. Inorganic Chemistry, 6th ed.; W. H. Freeman: Oxford, 2014. ISBN 978-0-19-875717-7.

wow thanks for the explanation, i did not expect some old A levels MCQ question to open such a can of worms.

@monke, Just curious what was the explanation provided by the A-level book or how does your teacher explain it?

Solar radiation cannot convert chloride to perchlorate. It can supply the energy for a less favorable redox reaction.  If that is the case there should be evidence of the reduction half of the reaction. It would be interesting were the perchlorate salts something like ferrous perchlorate; they seem to be calcium and magnesium. We really know very little about  Mars chemistry

@AChem,"with hot NaOH,ClO3- and not ClO- is formed " that is all the answer key stated. My teacher just told us to memorise

The following is an addon to the existing answer.
Frost diagram is indeed a simple and intuitive visual aid helping to identify which oxidation states of an element are most stable in solution.
However, the textbook Frost diagrams, like the majority of thermodynamic data, are usually reported for 25 °C.
Since OP is asking about boiling hot solution, we need to compile the Frost diagram for that ourselves.
The standard reduction potentials alongside with their temperature coefficients $\mathrm dE^\circ/\mathrm dT$ have been collected and reported by Bratsch [1].
The temperature dependence of reduction potentials is approximately linear between
$\pu{0 °C}$ and $\pu{100 °C}:$

$$E^\circ_T = E^\circ_{298} + (T - 298.15)\left(\frac{\mathrm dE^\circ}{\mathrm dT}\right)_{298}$$

For the further reference I made a tableau with calculated $(\pu{60 °C}$ to $\pu{100 °C})$ potentials for the range of elevated temperatures for the half-reactions of interest:
\begin{array}{llrrrrr}
\hline
& \left(\frac{\mathrm dE^\circ}{\mathrm dT}\right)_{298} & & & & & \\
\text{Half-reaction} & /\pu{mV K^-1} & E^\circ_{298}/\pu{V} & E^\circ_{333}/\pu{V} & E^\circ_{343}/\pu{V} & E^\circ_{353}/\pu{V} & E^\circ_{363}/\pu{V} & E^\circ_{373}/\pu{V} \\
\hline
\ce{Cl2(g)}/\ce{Cl-}       & -1.248 & 1.360 & 1.404 & 1.417 & 1.429 & 1.442 & 1.454 \\
\ce{ClO^-}/\ce{Cl2(g),OH-} & -0.90 & 0.420 & 0.389 & 0.380 & 0.371 & 0.362 & 0.353 \\
\ce{ClO2^-}/\ce{Cl^-,OH-}   & -1.267 & 0.786 & 0.742 & 0.729 & 0.716 & 0.704 & 0.691 \\
\ce{ClO3-}/\ce{Cl2(g),OH-} & -1.35 & 0.465 & 0.418 & 0.404 & 0.391 & 0.377 & 0.364 \\
\ce{ClO4-}/\ce{Cl2(g),OH-} & -1.322 & 0.446 & 0.399 & 0.385 & 0.372 & 0.358 & 0.345 \\
\hline
\end{array}
Using this, one can make a Latimer diagram for the oxidation numbers (O.N.s) of interest, which I'm going to make in a tabular form for $\pu{25 °C}$ and $\pu{100 °C}:$
\begin{array}{rrr}
\hline
\text{O.N.} & E^\circ_{298}/\pu{V} & E^\circ_{373}/\pu{V} \\
\hline
-1 & -1.360 & -1.267 \\
0 & 0.000 & 0.000 \\
1 & 0.420 & 0.353 \\
3 & 2.146 & 1.958 \\
5 & 2.325 & 1.819 \\
7 & 3.122 & 2.413 \\
\hline
\end{array}
Plotting this data gives a Frost diagram for chlorine at room temperature and boiling hot solution.
Note a more pronounced local minimum for O.N. 5 for the hot solution, which implies tendency of both chlorite and perchlorate to comproportionate forming chlorate $\ce{ClO3-}$ as the more thermodynamically stable oxyanion.
Chlorite always lies above the line connecting two adjacent species and would tend to undergo disproportionation at room temperature and much faster (negative slope) in hot solution.

Reference

Bratsch, S. G. Standard Electrode Potentials and Temperature Coefficients in Water at 298.15 K. J Phys Chem Ref Data 1989, 18 (1), 1–21. DOI: 10.1063/1.555839.

Thanks for adding high temperature information. The good news is that the trend is the same.

Latimer Diagrams may also helpful to find unstable species with respect to disproportionation. I'd list Latimer Diagrams for chlorine in acidic and basic solutions given in here:
$$\text{In acid: } \ce{ClO4- ->[\pu{1.19 V}] ClO3- ->[\pu{1.21 V}] HClO2 ->[\pu{1.65 V}] HClO ->[\pu{1.63 V}] Cl2 ->[\pu{1.36 V}] Cl-}$$
$$\text{In base: } \ce{ClO4- ->[\pu{0.36 V}] ClO3- ->[\pu{0.35 V}] ClO2- ->[\pu{0.65 V}] ClO- ->[\pu{0.40 V}] Cl2 ->[\pu{1.36 V}] Cl-}$$

Accordingly, Latimer diagrams provide a very easy way to determine if disproportionation is spontaneous: If  the voltage to the right of the species in question is greater than the voltage to the left of the species, the species is unstable with respect to disproportionation.

Thus, $\ce{ClO2-}$ is not stable with respect to disproportionation in basic medium as $\pu{0.65 V} \ \gt \ \pu{0.35 V}$.


text:
Why covalently bonded molecules have strong bonds within the molecule?
While studying the chapter carbon and it's compounds in NCERT page 60
I came across this line:

Covalently bonded molecules are seen to have strong bonds within the
molecule but intermolecular forces are weak. This gives rise to low
melting and boiling points of these compounds

While I do understand that the atoms in the covalent bond are held together by the attractive force between the shared electron pairs and the nucleus of the atoms in the molecule , I have failed to understand why covalently bonded molecules have strong bonds within  the molecule?
So I searched on the internet and went through some books but couldn't find any understandable answer. Could someone please explain why do covalently bonded molecules have strong bonds within the molecule but weak intermolecular forces?

Explain what about it? The division between inter- and intramolecular interactions is somewhat arbitrary. Some covalent bonds are way weaker then other, some interactions usually considered weak can become quite strong. To get a good answer you need to know what you really what to know.

Intermolecular forces, also called van der Wals forces, are rather weak, because they are due to electric interactions between charges which are far away from one another. The nucleus of an atom attracts strongly its own electrons ; it also attracts electrons belonging to other atoms, farther away. But the electric attraction force decreases with the square of the distance. So the attraction between the nucleus of any atom $\ce{A}$ included in a covalence $\ce{A-B}$ is still strong with the electrons of "its" $\ce{B}$ atom, because $\ce{A}$ is next to $\ce{B}$. But the attraction is weak with any other $\ce{A}$, $\ce{B}$ or $\ce{X}$ atom if this atom belongs to another molecule, because of the distance.

Could someone please explain why do covalently bonded molecules have strong bonds within the molecule but weak intermolecular forces?

Intermolecular bonds are weaker (or more fleeting) than intramolecular bonds by definition. The intramolecular bonds define a molecule. They persist when a molecule diffuses in solution, or when a molecule changes from one physical state to another (or from one phase to another). On the other hand, the intermolecular bonds can break and reform during these processes. When an intramolecular bond is formed or broken, we say there is a chemical reaction.

Covalently bonded molecules are seen to have strong bonds within the molecule but intermolecular forces are weak. This gives rise to low melting and boiling points of these compounds

This statement makes sense if we compare and contrast ionic compounds and molecular compounds. For the simplest examples, ionic compounds will have higher melting and boiling points than molecular compounds. For example, the only type of bond present in the ionic compound sodium chloride is the ionic bond between sodium and chloride. In order to melt sodium chloride, you have to break these bonds. On the other hand, to boil a molecular compound like acetone only the fairly weak intermolecular bonds will break (in this case dipole-dipole interactions) while the covalent bonds remain, ensuring that acetone as a liquid and as a gas has the same molecular structure.


text:
Are fluorine atoms less polarizable than the table predicted?
According to this table, per-fluorocarbons like CF4 are slightly more polarizable than analogous hydrocarbons like CH4 (2.824 vs 2.448). However, polarizability is measured under uniform electric field which is never encountered in reality. Instead, molecules carry only tiny amounts of dipole moment. Because the dipole field decays quickly with distance, the induced dipole interaction should be highly sensitive to distance. Considering that per-fluorocarbons are substantially bigger but have similar polarizability, their interaction should be much weaker than hydrocarbons. This paper also claims that per-fluorocarbons interact much more weakly with hydrocarbons, which is why they don’t mix despite both being nonpolar.


text:
Comparing reactivity of 3-bromocyclohexa-1,4-diene and 5-bromocyclohexa-1,3-diene towards SN1 reaction

Compare reactivity of 3-bromocyclohexa-1,4-diene (P) and 5-bromocyclohexa-1,3-diene (Q) towards $\mathrm{S_N1}$ reaction.

My teacher claims that the reactivity of P is more than that of Q. He mentions that the carbocations formed are the same (they are the same hybrid), and further the bond dissociation in P will be more than that in Q due to the two adjacent alkenes facilitating the dissociation in P as compared to Q. He mentions the name “field effect”, but I may have heard this wrong. He also adds that Q is more stable due to a conjugated diene and will show less reactivity.
My friend (studies from a different teacher) says that this is wrong, as the carbocation from P will be cross conjugated, while the one from Q will have an extended conjugation, so will be more stable, and so Q will be more reactive than P. He says that the immediate carbocation matters and that you don't look at the hybrid.
Which one is correct and why? A full explanation on this problem would be appreciated.

Disclaimer : I don't know whether my answer is correct or not.
This answer is purely based on my theoretical knowledge but not on the experimental evidences.

I feel that P reacts faster than Q. Let $E_\mathrm{P}$ and $E_\mathrm{Q}$ be the absolute enthalpies  of P and Q respectively. As resonance is possible in Q, we have $E_\mathrm{Q}<E_\mathrm{P}$. Now, let P' and Q' be the transition states formed during the dissociation of $\ce{C-Br}$ bond. And $E_\mathrm{P'}$ and $E_\mathrm{Q'}$ be the absolute enthalpies of P' and Q' respectively. Here are the transition states:

Now let us compare $E_\mathrm{P'}$ and $E_\mathrm{Q'}$. During the transition state the tendency of stabilizing the partial positive charge is more for P' than Q'. This is because, the leftmost pi bond in Q' can either involve transition state or can be in resonance with bottom-most pi bond. But in P' any of the two pi bonds can involve in the transition state. So the positive charge stabilization is more in P'. So we have $E_\mathrm{P'}<E_\mathrm{Q'}$. So the activation energy required for P $(E_\mathrm{P'}-E_\mathrm{P})$ is less than activation energy required for Q $(E_\mathrm{Q'}-E_\mathrm{Q})$.
And the formation of intermediates is given below:(Here I showed the resonance hybrid of the carbocations formed)

So more energy is required for Q than P for the formation of the above intermediate $(\because E_\mathrm{Q}<E_\mathrm{P})$. This analysis shows that formation of intermediate form P is move favorable than form Q. And so the reactivity of P is more than Q.

Thank you very much! The point on transition state and comparing enthalpies makes sense. By the way, is field effect operational here (as I mentioned in the question, I might not have heard this correctly but still)?

@ultralegend5385 Field effect has nothing to do with this question. It is a type of effect similar to inductive effect. The more dominating effect here is resonance effect. You can verify the article on "field effect" in Wikipedia. Field effect is rarely used and important in few cases where inductive, resonance effects cannot explain the experimental evidences.

According to Wikipedia,

Cross-conjugation is a special type of conjugation in a molecule, when in a set of three pi bonds only two pi bonds interact with each other by conjugation, while the third one is excluded from interaction. Whereas a normal conjugated system such as a polyene typically has alternating single and double bonds along consecutive atoms, a cross-conjugated system has an alkene unit bonded to one of the middle atoms of another conjugated chain through a single bond. In classical terms, one of the double-bonds branches off rather than continuing consecutively: the main chain is conjugated, and part of that same main chain is conjugated with the side group, but all parts are not conjugated together as strongly.

So there is no cross conjugation in the cation formed by removal of Bromine from P .
One further statement made wrong by the friend is that the immediate carbocation matters. This answer, though about benzene, clears the misconception that hybridisation is in any way dependent on time.
In terms of resonance structures, both the compounds are eqivalent
According to Wikipedia,

A field effect is the polarization of a molecule through space. The effect is a result of an electric field produced by charge localization in a molecule. This field, which is substituent and conformation dependent, can influence structure and reactivity by manipulating the location of electron density in bonds and/or the overall molecule. Field effects are relatively weak, and diminish rapidly with distance, but have still been found to alter molecular properties such as acidity

As there is no localisation of charge in this case, field effect is non-operational (localisation of charge is absence of resonance structures with charge present on different atoms in each).
Coming to the question, as both molecules would be equally stable in carbocation state, the only thing that can be done is to compare the relative stability of the original compounds: the one lesser stable in neutral state is more likely to undergo the reaction.
It is important to know the Inductive effect and $sp_3$ and $sp_2$ hybridisation to compare the molecules.
In P, the carbon atom bonded to benzene has two $sp_2$ hybridised carbons (with 33% s character) and the a bromine molecule adjacent, all three of which are electron-withdrawing, making the carbon more electron-deficient and thus, making the molecule more unstable than Q, which has only one $sp_2$ carbon adjacent to it. Thus, P will be more reactive
*By Jmancus0 - Own work, CC BY-SA 4.0, https://commons.wikimedia.org/w/index.php?curid=67323330

What your friend stated is absolutely true. During an $SN^1$ reaction, the slowest and hence rate determining step (RDS) is the formation of carbocation $C^+$. The stability of the $C^+$ determines whether the reaction is feasible.
You may want to refer to https://goiit.com/t/stability-of-resonating-structures/22797
Basically extended Resonance stabilized compounds are more stable than cross Resonance stabilized structures.
It’s like the $C^+$ in P has two different paths to go which interfere with each other, whereas in Q, the resonance is continuous and stabilising, in an ‘extended’ manner.
Honestly, it’s essentially the fact that the easier and more extensive the resonance, the more stable is the $C^+$ and hence the reaction proceeds in a better way.

From my knowledge, resonance structures are not real, so in both P and Q, the cation should exist as it's hybrid right, which is same in both cases? Also, can you provide a link to some more "trustable" source?

@ultralegend5385 I have learnt this, from books and through teachers. Maybe you can see a video or two on this topic, I know some videos but they’re in Hindi.

And also, in the answer upvoted the person is not sure about the correctness of it, whereas I am 100 % sure, and my answer is downvoted, just wanted to say.

I did not downvote your answer first of all. Secondly, that answer seems plausible and the writer also helped my confusion in the comments with logic and source. The other answer also has sources as Wikipedia. But you did not reply to my query and your source isn't reputed enough.

@ultralegend5385 as you wish, I tried my best to help, as I had learnt this from my teachers, so couldn’t really cite a source. Anyways the main thing is that your doubt has been cleared :)


text:
Calculating Band Gap Energy When Absorption Spectra Shifts with Concentration?

I am not a chemist. However, from some experimental work I have been doing, I have found that the intensity of the spectrum changes with dilution, this makes sense. However, when calculating the band gap energy through Tauc's equation for band gap (ie: taking the line of the energy vs $(ahv)^p$ x intercept) will be affected by the dilution. The line of best fit on the energy graph gets less and less sloped resulting in a changing bandgap. A
If the band gap is an intrinsic property of the material, how can this be a trustable method for measuring bandgap if it is affected by concentration?

Hint: What is alpha? Is it absorbance only? Also see the nice paper "How To Correctly Determine the Band Gap Energy of Modified Semiconductor Photocatalysts Based on UV−Vis Spectra" Phys. Chem. Lett. 2018, 9, 6814−6817

The absorbance is related to concentration by a very famous approximation called Beer's law.

You will notice that the spectrum does not change with dilution other than in magnitude as expected. You should find or measure the fluorescence spectrum, the  'band gap' is fairly accurately the mid point between longest absorption and shortest emission peaks, when converted to wavenumbers

For anyone with a future question like this: to calculate bandgap from absorbance (not emission spectrum as some have suggested) you must normalize the spectrum before applying tauc.


text:
Why is ammonium ion net positively charged?
I know that this might look a very dumb question but I have been spent my days to think about it and couldn't understand and finally decided to ask.
Consider Sharing of electrons to form covalent or coordinate bonds , elements can complete their octet by the method of sharing and unlike ionic bonds, they don't have to lose their electrons i.e. they only share and do do not lose electrons . The shared electron ( or a pair) is counted on both atoms then. For example, in Cl2 molecule, even though each chlorine atom had 7 electrons before, but after sharing, they both have 8 electrons  and thus complete their octet. This is only possible if we count the shared electron in both the atoms while checking their octets.
NH3  +  H(+)   =   NH4(+)
Consider an ammonium ( denoted as NH4+ ) , after bond formation, within the molecule, let's count for doublet of hydrogens. We find that each hydrogen had a complete doublet. Let's count for nitrogen, again, we see that nitrogen had 8 atoms and hence an octet ( the shared electron pair is also counted while counting for octet, as we do in chlorine molecule) . Now, my question is that if each hydrogen and nitrogen are having their valence shells filled, then from where an extra positive charge comes and people call it a cation .Positive charge comes when there is a deficiency of electrons , but in ammonium ; no atom is positively charged and hence the molecule.It would have been a cation if the nitrogen was positively charged, but clearly nitrogen has not lost it electrons, it had only shared . So why is there a positive charge on the ammonium?

All your confusions will vanish if you recall that atoms also contain protons-positively charged entities. You are focusing on the valence shells only. The charge on a molecular species reflects this net balance.
Start from an electrically neutral ammonia molecule, which simply means that the number of electrons and protons are balanced in that molecule. Count the total number of electrons and protons in ammonia molecule. You add a hydrogen ion to it via chemical reaction. The hydrogen ion nothing but an hydrogen atom stripped on an electron. What is left? A single proton.
Once you added a proton to the neutral ammonia molecule, you disturbed that original charge balance. There is a shortage of a single electron in the new molecule so formed and this denoted by the plus sign on the ammonium ion.

@Shiva Enough is to remember some bonds are "2=1+1" and some "2=2+0".


text:
Why doesn't an acid-base reaction occur in basic amide hydrolysis instead of nucleophilic acyl substitution?
In the reaction of Grignard reagent with a primary and secondary amide, nucleophilic acyl substiution doesn't happen. The reason given is that R- is a strong base as well as strong nucleophile, an acidic H is present in amide and finally, acid-base reactions are much faster than nucleophilic reactions. However, in basic amide hydrolysis, nucleophilic acyl substitution is able to occur despite OH- also being a strong base and nucleophile. Why doesn't the acid-base reaction occur here?

OH- is nowhere near as strong a base as a Grignard. Look up the pKa of a typical amide proton.

@Waylander AFAIK, amides are weak acids as well as weak bases. OH- is a(relatively) strong base and so the amide should react like an acid. Even, if the reaction is slower than that with Grignard, shouldn't it still be much faster than SN2th?

The pKa of a typical amide is around 20, OH- is not stronger enough to deprotonate it


text:
Why are amides weaker acids than water?
Based on a a recent question
I asked on ChemSE, the answer to my question was that amides are weaker acids than water with a typical pKa value of ~20. Although, this answers my question, it got me wondering why amides are weaker acids than water. After deprotonation, the negative charge formed on the amide is delocalized by resonance and on two electronegative atoms (N & O). After deprotonation in water, the negative charge is localized on a single electronegative atom (O). So why are amides weaker acids than water? Is it linked to the fact that, on average , the negative charge is on a more electronegative atom in OH- than it is in the conjugate base of amide? Isn't resonance delocalization a bigger factor?

Consider that even if amides are very weak acids, they are much stronger acids than ammonia.

How is ammonia relevant to amides? An amide is in one formal way just ammonia with 1 H substituted by acyl.  $\ce{H-NH2}$ versus $\ce{H-NH-CO-R}$


text:
Can anisole be nitrated only with nitric acid?
Is it possible to nitrate anisole using only concentrated nitric acid? Or is it necessary to use sulfuric acid or acetic anhydride? Could you please help me with some references?

Welcome to ChemSE! Your question will likely be considered as homework and be voted down. You need to make some effort to answer the question yourself. Take a look here about homework policy. https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange.

related https://chemistry.stackexchange.com/questions/152250/why-is-the-ortho-product-major-in-the-nitration-of-anisole-with-nitric-acid-and

Yes, it is possible to nitrate anisole (methoxybenzene) using not only concentrated nitric acid $(\ce{HNO3})$, but also dilute $\ce{HNO3}$ (Ref.1). The nitration of anisole in aqueous nitric acid in the concentration range 25.18-31.38 mol% at temperatures in the range $293$-$\pu{328 K}$ has followed the pseudo-first-order kinetics and the values of rate constant and activation energy have been reported.
The mechanism for the nitration is given as follows:
$$\ce{HNO3 + H3O+ <=> NO2+ + 2H2O} \tag1$$
$$\ce{NO2+ + ArH <=> encounter pair} \tag2$$
$$\ce{encounter pair <=> benzenonium ion ->[fast] products} \tag3$$
The author has stated that the activation energy of the rate determine step decreases with increasing acid concentration. This is a good indication that the rate of nitration is faster in concentrated nitric acid. The regioselectivity of the nitration (o-, p-, and m-selectivity) using different nitrating agents has been discussed elsewhere (Ref.2).
When only the fuming nitric acid in the solvent is used, the nitration
starts due to the autoprotonation of nitric acid and generation of a
nitronium cation $\ce{NO2+}$ (Ref.3):
$$\ce{HNO3 + HNO3 <=> H2NO3+ + NO3- <=> NO2+ + + H2O + NO3-} \tag4$$
The extended details of nitration of aromatics including anisole are given in Ref.4.

References:

David J. Belson, "Kinetics of nitration of anisole in aqueous nitric acid," Ind. Eng. Chem. Res. 1991, 30(7), 1485–1488 (DOI: https://doi.org/10.1021/ie00055a013).
George A. Olah, Henry C. Lin, Judith A. Olah, and Subhash C. Narang, "Variation of isomer distribution in electrophilic nitration of toluene, anisole, and o-xylene: Independence of high regioselectivity from reactivity of reagent," Proc. Natl. Acad. Sci. U S A 1978, 75(2), 545–548 (DOI: https://doi.org/10.1073/pnas.75.2.545).
Joanna Adamiak, "Controlled nitration of anisole over $\ce{HNO3/PO4/MoO3/SiO2/solvent}$ systems," Journal of Molecular Catalysis A: Chemical 2015, 407, 81–86 (DOI: https://doi.org/10.1016/j.molcata.2015.06.025).
J. G. Hoggett, R. B. Moodie, J. R. Penton, and K. Schofield, In Nitration and Aromatic Reactivity; Cambridge University Press: London, United Kingdom, 1971 (ISBN: 0-521-08029-0; ISBN 978-0521080293).

I am a little fuzzy about the given mechanism. In water solution the electrophile could actually be molecular HNO3 or even NO+ from some acid getting reduced. See this answer and comments to it.

Yes, it is possible to nitrate anisole with just nitric acid. Following is the excerpt from a 1948 paper:

The nitration of anisole leads to formation of a mixture of nitroanisoles and nitrophenols in proportions depending on conditions. It is here shown that nitration of anisole in dilute nitric acid solution is nitrite-catalyzed and that the nitroanisoles are probable the intermediates in the reaction.
[...]
More precisely, anisole have been found to react rapidly with 40-50% nitric acid containing a little nitrous acid, but fails during many hours to react with nitric acid containing urea. The principal products when 40% nitric acid is used is nitrophenols; when more concentrated (60% nitric acid) used, then the principal product are nitroanisoles. The quantitative results were reported in table I:

They went ahead and tried to nitrate substituted anisoles e.g. p-nitroanisole
Ref.: The Mechanism of the Nitration of Anisole, R. M. Schramm and F. H. Westheimer, Journal of the American Chemical Society 1948 70 (5), 1782-1784
DOI: 10.1021/ja01185a036


text:
Why is a strongly-ortho directing substituent preferred over a strongly-meta directing substituent?
When we were taught electrophilic aromatic substitution at our school, our teacher made us write the "rules" of di-substituted benzene concerning the decision of the substitution position based on the substituent type. One rule states that in the case of a strongly ortho-directing group and strongly-meta directing group, the ortho-directing group controls the location where the substitution will take place. Why is that so if that is the case? According to me, a strongly ortho-directing group would be a great activator and a strongly meta-directing group would be a great deactivator, hence the effect should cancel each other and the substitution could be oriented considering other factors like steric hinderance.
Apologies if there is any orthodox or "awkward" language usage here, I'm primarily a high-school student.

At a given temperature  the rate of reaction of o,p-directors>benzene>>>m-directors. m-Direction is, in effect, a default position. In the nitration of benzene the sequential introduction of nitro groups becomes more and more difficult at a given temperature. Look up the Friedel-Crafts reaction. Notice that nitrobenzene is often used as a solvent for acylating other aromatics. It doen't get acylated. Investigate the mechanism of electrophilic aromatic substitution to get a better understanding of the reasons for this behavior.


text:
Reaction of 1,4-Dichlorobenzene & Hydroxylamine
I'm looking for reactions that would attack the chlorine in dichlorobenzenes. Wikipedia mentions a reaction with hydroxylamine, yet Googling, I haven't found exact said reaction. Can anyone provide it and why it would attack the chlorine?

Don't capitalize element names unless it is at the beginning of the sentence.

Sukhov et.al tried to aminate chlorobenzene and they found the products to be isomeric chloroanilines and chlorophenylenediamines but I am not if they went ahead and tried to aminate di/trichlorobenzene but I am pretty sure it would lead to mixture of isomeric aniline/chloroaniline/mixed di/tri-chloro-di/tri-amino compounds or even di/triaminobenzene. The mixture will contain so many isomers and mixed substituted product that it will be hard to pinpoint where is the actual attacking point. The following is the abstract of their paper:

The process of electrochemical amination of chlorobenzene using hydroxylamine and the Ti(IV)/Ti(III) mediator system is studied in aqueous solutions of 6–16 M sulfuric acid containing acetic acid or acetonitrile. The substitution products in these media are isomeric chloroanilines and chlorophenylenediamines, with the exception of 5-chloro-1,3-isomer, phenylenediamines, and aniline. The overall current efficiency of amino compounds corresponding under the experimental conditions to the efficiency by hydroxylamine can reach 71%. Para- and ortho-chloroanilines are obtained in 6 M sulfuric acid and 11.1 M acetic acid with the weight fraction and current efficiency of 97 and 51%, respectively.

Ref.: Lisitsyn, Y.A., Sukhov, A.V. Electrochemical Amination of Chlorobenzene in Aqueous-Organic Solutions of Sulfuric Acid. Russ J Electrochem 54, 886–892 (2018). DOI: 10.1134/S1023193518130268


text:
Phenol's resonance structure doesn't match NMR
I've asked a question before and quickly realised that the resonance/mesometric effects played a large role in determining the C-13 NMR spectrum of phenol, shown below.

This is the resonance structure of phenol:

But given this, I would have expected the carbons at $b$ to be more shifted than the ones at $d$. This is because although they both have partial negative charges, the inductive effect of the hydroxy group on $a$ should make $b$ more deshielded than $d$.
Why is this?

now it seems that you're accounting only for the mesomeric effect and not the induction. Consider the combined effects, especially with regard to how the positive charge distributes.

@Andrew If I account for the inductive effect, shouldn't that would mean that d should be more shielded than b? This is because b is closer to the positive oxygen which will be withdrawing electron density. So why isn't this the case?

I think the answer is gonna be more complicated than just inductive/resonance effects --- 13C shifts are not very straightforward, it's tempting to think of them like we do 1H shifts, but the underlying reasons for the chemical shifts are rather different


text:
Quantity vs quality in determining acidic strength
Consider the following order of acidity based on pKa values:-
ethanoic acid > phthalimide >phenol >water >acetamide
Looking towards similar questions on stack exchange such as this
, the recurring theme is that both quality(the electronegativities of the atoms on which charge is delocalized)as well as quantity of resonance(the number of resonating structures) should be seen when determining the stability of conjugate base and quality is given precedence over quantity. However, this doesn't seem to work here. Let me categorize the conjugate bases of the aforementioned compounds:-

Ethanoic Acid- Delocalisation over 2 oxygen atoms
Phtalimide- Delocalisation over 2 oxygen atoms and 1 nitrogen atom
Phenol-Delocalisation over 1 oxygen atom and 3 carbon atoms
Water-Localised over 1 oxygen atom
Acetamide- Delocalisation over 1 oxygen atom and 1 nitrogen atom

Now, based on the linked question, we get ethanoic acid > phenol > water where we have seen quality of resonance first then quantity. However, this doesn't work for phtalimide which should be placed above carboxylic acid and also acetamide which should be placed above water and possibly even phenol.
So how would you explain this order?

I think you have this whole thing backwards. From the experimental acidity, we give these rule-of-thumb explanations, which fit to varying degrees. You shouldn't expect the other way around to work in general. New chemistry students get too caught up in these rules and overestimate their explanatory power and precision.

@NicolauSakerNeto Agreed, but there must be some explanation given for experimental results, right? Is the explanation in this case too advanced(like based on the schridinger equation) or can it be explained by basic principles?

Indeed. The accurate computation of substance acidities purely from first principles, especially outside of the gas phase, is still a matter of active research.

https://chemistry.stackexchange.com/questions/94870/why-is-ethanoic-acid-more-acidic-than-phthalimide?rq=1

(...) quality is given precedence over quantity. However, this doesn't seem to work - doesn't it? See you're comparing apples and oranges here. N acid is not O acid. If imidic acids were major tautomers and not amides then your logic could been OK.

BTW this is pretty much your earlier q. all over again https://chemistry.stackexchange.com/questions/165706/why-are-amides-weaker-acids-than-water

@Mithoron so is the atom to which H is directly connected to of greater importance than all the atoms upon which the charge resonates?

More precisely it depends which mesomeric structure structure is the major contributor. You were assuming there were equal, while apparently the one with charged nitrogen is more important.

@Mithoron Why is the one with charged nitrogen more important when the charge is more stable on oxygen?

I assume it the matter of stronger C=O then C=N


text:
Why do we sometimes change the sign for standard electrode potentials?
Question:

In this cell, the copper is oxidized and $E_{cell} = \pu{+1.15 V}$.
$$\begin{align}
\ce{Cu^2+(aq) + 2e− -> Cu(s)} & &(E &= \pu{+0.34 V})\\
\ce{Mn^3+(aq) + e− -> Mn^2+(aq)} & &(E &= ??)\\
\end{align}$$
Calculate the value of the standard electrode potential for the
$\ce{Mn^3+(aq)|Mn^2+(aq)}$ half-cell.

So the answer is given as $\pu{+1.49V}$.
The answer stated that the equation would be $(+1.15)-(-0.34) = 1.49$, but why are we changing the sign for the electrode potential for copper? Shouldn't it remain as $\mathrm{+0.34}$ so the manganese half cell would be $\pu{+0.81 V}$?

Try to search the site and internet for the keywords, I bet you will find multiple interesting hits. In a way, it is analogical to question "Which river bank is the left bank?". On your left, or on your left if looking downstream?

I am very curious to see Mn3+(aq) /Mn2+(aq) half cell.

It is given that copper is oxidized, so the required electrode potential will become -0.34 V. Electrode potential is an intensive property and Gibbs energy an extensive one. So we can equate the change in Gibbs energy for the net cell reaction to that for the sum of oxidation and reduction cell reactions.

The answer is right, but I find the double negative pretty unintuitive. The question tells you that
$$\begin{align}
\ce{2Mn^3+ + Cu &-> 2 Mn^2+ + Cu^2+} & E^\circ_\mathrm{cell} &= \pu{+1.15 V}
\end{align}$$
(This is the balanced reaction between $\ce{Mn^3+}$ and $\ce{Cu}$; it specifies that $\ce{Cu}$ is the species being oxidised. The balancing isn't hugely important, it's just there to make the equation correct.)
The $E^\circ_\mathrm{cell}$ for this reaction is the difference between the two half-cell reduction potentials. This much you should be familiar with:
$$E^\circ_\mathrm{cell} = E^\circ(\ce{Mn^3+}/\ce{Mn^2+}) - E^\circ(\ce{Cu^2+}/\ce{Cu})$$
so rearranging,
$$\begin{align}
E^\circ(\ce{Mn^3+}/\ce{Mn^2+}) &= E^\circ_\mathrm{cell} + E^\circ(\ce{Cu^2+}/\ce{Cu}) \\
&= \pu{1.15 V} + \pu{0.34 V} \\
&= \pu{1.49 V}.
\end{align}$$

But this method will not work in case the electrons are not cancelling... Because electrode potential is an intensive property. E.g., in this link https://chemistry.stackexchange.com/a/155900/55479 related to Cu.

@Apurvium sorry but I really don't get what you are saying. The fact that they are intensive is precisely what lets us just subtract one reduction potential from the other, without accounting for stoichiometry. If you use Gibbs energies to work it out, you will reach the same answer just with a lot more hassle. I am curious to see how you would come up with something different.

My point was, when the combination of two half-cell reactions gives a full cell reaction, then the stoichiometry will not matter because $E$ is an intensive property. But in case, two half-cell reactions gives another half-cell reaction, then we have to go by Gibbs energy method. E.g., link in my previous comment.

@Apurvium Oh, I see: yup, that’s correct. But that’s not the case here.

The standard redox/electrode potential for a half-reaction
$$\ce{ox^{m} + n e- -> red^{m-n}}$$
is defined within thermodynamics as
$$\Delta G^°_\mathrm{r}=-nFE^°_\mathrm{r}$$
for a formal reaction
$$\ce{ox^{m} + }\frac n2 \ce{ H2 -> red^{m-n} + n H+ }$$
"American school"  flips the sign of the standard potential value when the opposite direction of reaction is considered, to formally honour the thermodynamic relations. By other words, it is an alternative equivalent way to express the Gibbs energy of redox reactions, that flips the sign with direction too.
"European school" considers the potential value conventionally always for the forward reaction and such values are tabelized as standard redox potentials.
The former approach is cleaner from the thermodynamic point of view.
OTOH, the latter approach is in agreement with measured electrode potentials against SHE, with tendency to be oxidized/reduced and is less vulnerable to confusions.
It can be reconciliated if there are formally strictly distinguished redox and electrode potentials. But even then.....
The problem has 2 analogies:

Left river banks are those on your left

always (naive sense)
if looking downstream (geographic convention).

Anodes/cathodes are electrodes where oxidation/reduction occurs

always(physical chemistry)
if in galvanic mode in context of rechargable cells (engineering cell design convention). For Li-ion cells, cathodes are always the oxide ones and anodes the graphite ones.

This is how I applied what I learnt in my undergraduate freshman chemistry class to answer the given question:
In the given cell, the copper has been oxidized and the $\ce{Mn^3+}$ has been oxidized. The given $E^\circ_\mathrm{cell} =\pu{ +1.15 V}$. The reduction half-reactions can be given as:
$$\begin{align}
\ce{Cu^2+ (aq) + 2e− -> Cu(s)} & & E_\ce{Cu^2+/Cu} = \pu{+0.34 V}\\
\ce{Mn^3+(aq) + e− -> Mn^2+(aq)} & & E_\ce{Mn^3+/Mn^2+} = x \ \pu{V }\\
\end{align}$$
Although both half-reactions are given as reductions to show the correct reduction potentials, in the total redox reaction, $\ce{Cu}$ is actually oxidized in anode. However, the $E^\circ_\mathrm{cell}$ is defined as:
$$E^\circ_\mathrm{cell} = E^\circ_\mathrm{cathode} - E^\circ_\mathrm{anode}$$
where both $E^\circ_\mathrm{cathode}$ and $E^\circ_\mathrm{anode}$ are given as reduction potentials with correct signs. Thus,
$$E^\circ_\mathrm{cathode} = E_\ce{Mn^3+/Mn^2+} = x \ \pu{V }  \ \text{ and } E^\circ_\mathrm{anode} = E_\ce{Cu^2+/Cu} = \pu{+0.34 V}$$
$$\therefore \ E^\circ_\mathrm{cell} = E^\circ_\mathrm{cathode} - E^\circ_\mathrm{anode} \ \Rightarrow \ \pu{1.15 V} = x - \pu{0.34 V} \\ x = \pu{1.15 V}  + \pu{0.34 V} = \pu{1.49 V} $$
In this way, we avoid both double negative and sign reverse.


text:
Conductometric titration curve plotting
In the conductometric titation of HCl vs NaOH, the conductance initially decreases, reaches a minimum at equivalence point, and then the slope of the curve becomes positive:

Source: https://chart-studio.plotly.com/~ramesh.ammanamanchi/25/conductometric-titration-curve/#/plot
How can this be modelled like I could do for pH?

Edit: This is my attempt. Is it correct?

The site expects that you write explicit compact summary of your prior effort to answer the question, based on your knowledge and on searching for existing related  info or answers. It would prevent others to tell you what you already know or what you could easily find yourself.

I don't know where to start. Will Kohlrausch law be applicable here?

The start is to understand underlying chemistry.. Modelling comes after. You must know, what you are modelling. Ongoing reactions and conductivity related laws.

You should use your effort to solve this problem before we can intervene. You should also think about following facts: Must use dilute aqueous solutions; The ion mobility is dependent on temperature so make sure temperature didn't increase during the titrarion (strong acid/base reaction is exothermic); Also, if your model is for weak acid or base, their ka or kb is also temperature dependent.

Also, what is the $y$-axis'  reading? I mean what is $\frac{1000 - a}{a}$ representing?

Just from looking at the image without distracting information like correct labels on the graph it should be almost painfully obvious: linear fit on the left and the right, where they intersect is the equivalence point.

From a technical point: please include everything necessary in the question and don't rely on external source. Links become bringen and then this question becomes useless. Hiding a link under the word source is also not appropriate. Please include a proper reference to the site it comes from. Also make sure you are allowed to use the image.

Expressed in $\pu{\Omega ^{-1} cm^2 mol^{-1}}$,  the ionic conductance of usual ions like $\ce{Na+, K+, Ca^{2+}, Cl-, SO4^{2-}}$ are between $50$ and $80$. But the two ions $\ce{H+}$ and $\ce{OH-}$ have a rather high conductivity : $350$ for $\ce{H+}$ and $198$ for $\ce{OH-}$. This property explain the change of the conductance observed in the titration of $\ce{HCl}$ solution by $\ce{NaOH}$.
In the beginning of the titration, [$\ce{H+}$] is large. So the conductance of the solution is large. When adding $\ce{NaOH}$, the ion $\ce{H+}$ is replaced by the ion $\ce{Na+}$ which has a poor conductivity : the conductance of the solution decreases.
At the equivalence point, the conductance is low, as it is only due to the ions $\ce{Na+}$ and $\ce{Cl-}$. If more $\ce{NaOH}$ is added after the equivalence point, the aded ions $\ce{OH-}$ aren't destroyed and stay in the solution. So its conductance increases. This is what you have observed experimentally.

The dotted plot is for addition of 10M NaOH to 1L of 1M HCl and each mole of NaOH increases volume by 0.1 L. The other one is linear where I didn't consider any increase in volume of titrate on addition of titrant.

Desmos


text:
intramolecular h-bonding in maleic acid
While studying organic chemistry, I came across two practice questions given by my professor:
Compare the acidic strength of:-

Maleic acid and Fumaric acid
o-nitrophenol and p-nitro phenol

And the answer and explanation were as follows-

while for the maleic acid qustion (my query is only pertaining to the first dissociation so kindly ignore the rest):

to summarize the second attachment, what is meant is that maleic acid's conjugate base is stabilized by the hydrogen bonds formed between the carboxylate ion and carboxylic acid group.
If we were to approach the questions along the lines of the nitro phenol question, fumaric acid should've been more acidic as the intramolecular h-bonds ( with substantial bond energy of about 21 KJ/mol) in maleic acid would cause some hindrance in the disocciation the proton but that isn't the case, why is that so?

In the nitrophenol case, H bond stabilizes the o-nitrophenol acid that becomes less acidic.
In the dicarboxyethene case, the H bond stabilizes the carboxylate of maleinic acid, what makes it less basic conjugate base, therefore the acid is more acidic.