text:
Van't Hoff graph, the Ka is increasing at a slower rate as temperature increases

I understand the exponential relationship between Ka and temperature, but I had expected it to be curving upwards(because of the $e^x$), and not like a log graph (where y-values increases slower for every increse in x-values). Can someone verify if this graph is legit? This graph is Ka(y-axis) against temperature in Kelvin (x-axis). I have isolated for $K(T_2)$ in the Van't Hoff Equation.
I did an experiment on the Ka values of acetic acid, and when I graphed it out, it was curving upwards like a parabola. Can someone help explain the inconsistencies between the theoretical and experimental data?

Your expectations do not consider the differential form of the equation d(ln k)/dT=(Delta H)/(RT^2).dT where the slope explicitly decreases with T due 1/T^2. // Note that due semilogarithmic nature of the chart, the vertical axis is geometrically for ln K, in spite of being for K numerically .

But my experiment's graph was also Ka against temperature

Note the 2nd part of my comment.

Because I wish to find the theoretical Ka values at different tempeatures, I have rearranged the Van't Hoff equation.

By other words, the chart is exactly as expected.

Yes, I acknowledge the y-axis of the Van't Hoff equation being ln(K). However, is there a possibility of only wanting Ka against Temperature?

Let us continue this discussion in chat.


text:
Using Thionyl Chloride with tertiary alcohol
I have a question regarding using $\ce{SOCl2}$ to convert an alcohol into a alkyl halide.
The Klein book mentions that $\ce{SOCl2}$ is only used for primary and secondary alcohol; however, in the solutions for one of their problems (Klein), it uses $\ce{SOCl2}$ to convert tertiary alcohol into an alkyl halide.

Is it even possible to use $\ce{SOCl2}$ with tertiary alcohol?

This is not $\mathrm{S_N1}$ or $\mathrm{S_N2}$. This follows $\mathrm{S_N i}$ mechanism. Implying, carbocation formation is absent.

In future please use $\LaTeX$ environments for formatting. Chemical formulae are written in the $\ce{}$ environment. Refer to the edited version (and of course Community Wiki) for more details.

https://en.wikipedia.org/wiki/SNi - Refer.

There is no real barrier against thionyl chloride reacting with a tertiary alcohol. The mechanism described by Ouellette and Rawn[1] does not depend on cleavage of a carbon-hydrogen bond and is even illustrated with a tertiary alcohol substrate.
Hydrogen chloride as the electrophilic reagent may provide a good alternative  with tertiary alcohols because this alternative can proceed through a relatively stable tertiary carbocation (the thionyl chloride reaction does not form a carbocation). Carbocation formation would be more difficult with a primary or secondary alcohol.
Reference

Robert J. Ouellette, J. David Rawn,
"16 - Alcohols: Reactions and Synthesis", Editor(s): Robert J. Ouellette, J. David Rawn,
Organic Chemistry (Second Edition),
Academic Press, 2018,
Pages 463-505, ISBN 9780128128381,
https://doi.org/10.1016/B978-0-12-812838-1.50016-5.

source: Stack Exchange

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

sort: votes


text:
How to determine the dynamic of progression of two simultaneous precipitation (double replacement) reactions?
We have a solution of mineral water with potassium chromate added to it, and want to titrate the chloride ions in the mineral water. We were told to use silver nitrate for this, which would produce the two following reactions:
\begin{align}
        \ce{AgNO3 _{(aq)} + Cl- _{(aq)} & <--> AgCl_{(s)}} + \text{some nitrates} \\
        \ce{2AgNO3 _{(aq)} + K2CrO4 _{(aq)} & <--> Ag2CrO4_{(s)} + 2KNO3_{(s)}}
\end{align}
And namely we interest in:
\begin{align}
        \ce{Ag+ _{(aq)} + Cl- _{(aq)} & <--> AgCl_{(s)} v} \label{AgClX} \\
        \ce{2Ag+ _{(aq)} + CrO4^{2-} _{(aq)} & <--> Ag2CrO4_{(s)} v} \label{AgCrX}
\end{align}
This titration hinges on two facts:

The first reaction will progress to near completion in the forwards direction.
The first reaction will nearly finish before the second one progresses appreciably, so we can observe equivalence by noticing the color of the silver chromate precipitate (brick-red).

I would like to understand why are these two facts true? How do I show that the solubility of silver chromate and silver nitrate causes this behavior?

https://chemistry.stackexchange.com/questions/99507/why-potassium-chromate-can-be-used-as-indicator-in-mohr-method

Consider that solubility of $\ce{AgCl}$ is 1.9 mg/L and of $\ce{Ag2CrO4}$ is 22 mg/L, both at 25 °C.
When just a little $\ce{AgNO3}$ has been added, the $\ce{Ag2CrO4}$ is not concentrated enough to precipitate -- it's still in solution. The $\ce{AgCl}$ falls out of solution at a far lower concentration.
BTW, it's possible to dissolve some of the precipitated $\ce{AgCl}$ by adding more $\ce{Cl-}$, e.g., more $\ce{NaCl}$, forming the soluble $\ce{AgCl2-}$ ion. That principle was used to "fix" photographic images by washing with excess salt water, e.g., when at sea.

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


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Oxidation of substituted cyclic ketone using permanganate

From what I had learnt, reaction of this with $\ce{KMnO4}$ should give a dicarboxylic acid after cleavage of the carbon-carbon bond. But apparently this is the reaction that happens:

Why is there a ketone on one side?
Also, if a dicarboxylic acid WAS formed, which side would the carbonyl group remain with? I've studied for linear ketones on strong oxidation (for eg $\ce{KMnO4}$ or nitric acid) the carbonyl group stays with the shorter alkyl group. Here, what would happen?

The more stable enol determines the direction of the oxidation. If the dicarbxylic acid WERE formed, it be 2-R-adipic acid without a keto carbonyl.

Yes, 2-R-adipic acid is exactly what I expected. Sorry though, I didn't understand what you mean by more stable enol. Are you giving the justification of this: "the carbonyl group stays with the shorter alkyl group"?


text:
Why does the measured CO2, TVOC and CH2O increase rapidly when I cook/bake?
I have a device at home that measures $\ce{CO2}$, TVOC (total volatile organic compounds),
and $\ce{CH2O}$ (formaldehyde) levels, and I conducted similar observations in my friends' house. I noticed a rapid increase in $\ce{CO2}$, TVOC and $\ce{CH2O}$ whenever I bake or cook. I can somewhat comprehend why the $\ce{CO2}$ levels rise, as I suspect it's stored in the water or vegetables, and when I cook them, the $\ce{CO2}$ dissipates.
However, I'm puzzled about the sources of TVOC and $\ce{CH2O}$. Where do these substances originate from?
Edit 1: I use an induction cooking plate (no gas).
Edit 2: I use this monitor https://www.amazon.ca/INKBIRDPLUS-Formaldehyde-Temperature-Multifunctional-350-2000ppm/dp/B097222DTC

You haven't provided the essential info how you cook/bake. But regardless of the energy is based on fossil fuel or electricity, the is no surprise that heating organic matter releases volatile organics. How do you think the smelling of cooking works?

CO2 should increase significantly if you are using natural gas, propane, charcoal or wood.  CO2 could increase just from you being in the room.  Yeast and decomposition of baking soda release CO2 also.

Sorry for that mistake. I will add it to my question (induction cooking plate)

We are excellent CO2 emitters; put the CO2 meter in the car and close the windows etc. drive and watch. Smells are VOCs. the formaldehyde detector may lack specificity or as Dave reports it may just be in food. My observations: sitting in room CO2 rises slowly and continuously, gas stove - CO2 rapidly, air quality decreases. Latest device -Radon detector shows distinct difference between a well ventilated and closed house; am trying to workout source and significance.

Could you list the type of detector you use? Many cheaper detectors report TVOC and eCO2. eCO2 is "effective CO2": calculated CO2 assuming ~400 ppm CO2 baseline + TVOC * a coefficient. This works okay, provided that the sources of CO2 and TVOC in the room are humans (we produce both in a fairly reliable proportion). If that's not the case though, they are completely useless. They are constructed this way because NDIR CO2 sensors are fairly expensive, and TVOC sensors are dirt cheap.

@AI0867 Thank you for the interesting answer. I added the link to the detector.

It's expensive enough that it could contain a real sensor, but given the reviews, name, and lack of datasheet, I suspect most of the money went into the display and design.

I've expanded the comment into a real answer.

Many raw foods naturally contain formaldehyde according to the Hong Kong Center for Food Safety and other sources. Dried Shiitake mushrooms have over 100 mg/kg of formaldehyde, for example.
So formaldehyde in cooking can come directly from the food, from high temperature decomposition of the food, from combustion of cooking fuel, and even from unwanted heating of cabinets and other materials near the stove or oven.
The recent article An Investigation of Concentration and Health Impacts of Aldehydes Associated with Cooking in 29 Residential Buildings compared frying, grilling and boiling.  This study attempts to quantify what you are reporting.  They find increase in formaldehyde and other aldehyde levels from all these cooking techniques.  Unfortunately they don't limit the variables well.  Different foods are used in each technique.  They report different average levels of aldehydes for each cooking technique even "before" cooking.  They don't separate electric data from natural gas data.  You could definitely improve on this study by controlling these factors and have "blanks" like just heating water or just turning on the stove.  Still, the long reference list gives you lots of other sources concerning what's been done before to study these issues.

This is not an answer to the question you asked, but it is a possible explanation for the measurements you observed.
Your measurements may not be reliable
While there are a a lot of things that can be sensed using chips that cost from a few cents up to a dollar, CO2 is not one of them.
"Effective CO2"
There are a lot of cheap sensors out there that pretend to measure CO2. The honest ones among them say they measure "effective CO2", the less honest ones do not.
What they measure is TVOC (total volatile organic compounds), which they then multiply by a coefficient, and add it to the baseline atmospheric CO2 concentration to give you a guess as to the local CO2 concentration.
Given that humans breathe out VOCs and CO2 in a roughly constant proportion, this can work as a reasonable proxy, provided there are no other sources of CO2 or VOCs present. Once you spray some window cleaner though, or light a gas burner, the relation becomes completely invalid, and the number is meaningless.
To actually measure CO2 somewhat reliably, you need a Non-Dispersive InfraRed (NDIR) sensor. This is the cheapest sensor that actually measures CO2, rather than a proxy that may or may not be related in your situation.
Cheap NDIR
NDIR sensors aren't magic though. They measure absorption of one or more wavelengths of infrared light in a tube. CO2 strongly absorbs some of these, so if you know the density of the gas in the tube (temperature and pressure will give you that cheaply), and you measure the absorption along the length of the tube, you can calculate the required concentration to give you that measurement.
However, there are various other gases that have overlapping absorption bands, and components such as your infrared source, the infrared sensor, the filter and other things, might change their characteristics over time. That means you need to recalibrate periodically, and want to measure multiple wavelengths so you can estimate and subtract out the other gases.
The cheapest NDIR sensors (the component itself, not the product you might build around it) are 10-30 dollars. The ones that measure multiple wavelengths will be more expensive. You ideally want the the automatic recalibration to be done on the sensor itself, as it has the most information available, but that will likely make the sensor more expensive again.
If you only have 1 wavelength, all you can do is "Automatic Baseline Calibration". If the sensor doesn't do this, you can also do it on whatever device reads out the sensor.
Automatic Baseline Calibration
Automatic Baseline Calibration (ABC) is basically just picking the lowest value of the last week (or other period) and picking an offset and/or scaling factor to turn that value into 400 ppm.
This assumes that 1) atmospheric CO2 is 400 ppm and 2) the measured area will be empty and well-ventilated enough to approach atmpospheric at least once a week.
These assumptions aren't terrible, but if you're trying to use your sensor to measure a greenhouse, or the variance and/or increase in CO2 in the atmosphere, they're wrong and the sensor is useless.
Also, to do all of this, some part of the device must keep track of these values. With a proper algorithm, this doesn't need to be expensive, but there are probably sensors out there that instead do 'calibration' using a running average of some sort.
Conclusion
Don't buy cheap CO2 sensors. Don't buy CO2 sensors from no-name brands. If the seller won't tell you how the sensor does something, assume that it doesn't, or does it badly/incorrectly. If the product description proudly proclaims something, but it's just one thing in an incoherent pile of others, it's probably just the marketing department copy-pasting 'keywords' that sell well.
Determine what you want to use your sensor for before you buy it, and make sure it's actually appropriate for that purpose. A single-wavelength NDIR sensor that does ABC is probably fine for measuring whether you're ventilating your house well enough, but it can't tell you whether the CO2 levels outside are higher or lower than a month ago.


text:
Question about conservation of amount of substance
if I consider the reaction : $$A \longrightarrow B+C$$
by what we call conservation of the amount of substance, we have $$nA_{0}=nA+nB=nA+nC$$
But for me the conservation says that $$nA_{0}=nA+nB+nC$$, because we have to consider everything in our closed system and we form A and B.
I konw it is false, but I can't figure why we only consider one product (or one reactiv if we have 2 reactivs)
Where is my missunderstanding...?

What is the meaning of the parameter $n$ ? Is it the mass or the amount of substance (in mole) ? For you, it seems to be the same, when you write : $n$A$\ce{_o}$ = $n$A + $n$B + $n$C . This last equation is only valid for the masses, and not for the moles.

I meant the amount of substance.
Why is it not valid for the moles but the first is?
It is not valid because A,B and C do not have the  same molar mass or is it something different?

If something is split into multiple things, the total number of things increases. Amount of substance is a way to count things and with a chemical reaction, the number of particles does change.

Suppose you consider a number $n$ of independent A-families. Each family is made of one parent B and one child C. If now the families are suddenly decomposed into parents separated form children, this "reaction" produces $n$ adults B, and $n$ children C. The number of independent groups (families, parents and children) is increasing, although the total mass does not change. In the beginning, there was $n$ independent groups. At the end of the splitting process, there are no families any more, but $n$ parents, and $n$ children. The number of independent groups has doubled.

Incredible, yes mole is just counting. THANKS, it enlignented me

The general Law is the conservation of MASS-ENERGY. In chemistry this breaks down into two separate Laws because the mass change in most chemical reactions is much too small to measure with the present instrumentation: Conservation of MASS, and Conservation of Energy.  A balanced chemical equation means that the MASS on each side is equal and the ENERGY on each side is equal. Since the mass is contained in the nuclei and electrons of the atoms each side of the equation must contain the same number of ATOMS and the same CHARGE regardless of their arrangement; the number of MOLECULES is NOT conserved. Energy, the oft-ignored reactant or product, is the chemical potential energy on one side of the equation or the release or absorbed energy on the other.
A correct chemical equation is:  A = B + Energy. A and B are EQUAL NUMBERS of ATOMS that are in different molecular arrangements. Energy is the energy released to or obtained from the environment. These values are determined by careful observations and measurements sometimes on paper but eventually in the laboratory. It has taken many chemists, physicists, and other scientists many centuries to get it to what we think is almost right.
You are suffering from a severe case of equationitis; The feat of writing equations before understanding concepts and then writing equations without defining the terms and relating them to what concepts you have defined. Your last premise is correct! Everything must be considered.  I suggest that you start with a very simple equation:  2H = H2 + energy. Start with 2 atoms then escalate to one mole of atoms then 2 moles of atoms. After that is understood you can worry about the mechanism and equilibrium.

Amount conservation must respect the stoichiometry of the reaction.
If the sum of stoichiometric coefficients on both reaction sides is not equal then the total amount of substances is not constant and changes with the reaction progress.
If you cut a heap of apples to halves then the total count of fruit pieces is not constant either.

For the reaction $\ce{A -> B + C}$
and initial respective amounts   $n_\mathrm{A0}$, $n_\mathrm{B0}$, $n_\mathrm{C0}$
and for the reaction completion degree $0 \lt \alpha \lt 1$,
the respective amounts of components follow this schema:
$$n_\mathrm{A}=n_\mathrm{A0}(1-\alpha)$$
$$n_\mathrm{B}=n_\mathrm{B0} + n_\mathrm{A0}\alpha$$
$$n_\mathrm{C}=n_\mathrm{C0} + n_\mathrm{A0}\alpha$$
In summary:
$$n_\text{tot} = n_\text{A} + n_\text{B} + n_\text{C} = \\
n_\mathrm{A0}(1-\alpha) + n_\mathrm{B0} + n_\mathrm{A0}\alpha + n_\mathrm{C0} + n_\mathrm{A0}\alpha= \\
= n_\mathrm{A0}(1 + \alpha) + n_\mathrm{B0} + n_\mathrm{C0} $$

If there were different reaction stoichiometric coefficients, these would be reflected as multiplication coefficients.
For the reaction $\ce{aA + bB -> cC + dD}$
initial respective amounts   $n_\mathrm{A0}$, $n_\mathrm{B0}$, $n_\mathrm{C0}$, , $n_\mathrm{D0}$,
assuming the reactant $\ce{A}$ is in excess,
and for the reaction completion degree $0 \lt \alpha \lt 1$,
the respective amounts of components follow this schema:
$$n_\mathrm{A}=n_\mathrm{A0}-\alpha (\frac ab) n_\mathrm{B0}$$
$$n_\mathrm{B}=n_\mathrm{B0}(1-\alpha)$$
$$n_\mathrm{C}=n_\mathrm{C0} +  \alpha(\frac cb)n_\mathrm{B0}$$
$$n_\mathrm{D}=n_\mathrm{D0} +  \alpha(\frac db)n_\mathrm{B0}$$


text:
Would it be plausible for dark matter to be an undiscovered highly unreactive element with a high atomic number?
I had thought about posting something in physics stack, but I already know it's possible for dark matter to be here on earth, like they say it's all around us...

No, it wouldn't.

@Mithoron oh yea, of course... it isn't plausible for that... guess it's gotta be particles LOL

It's essentially impossible. Our best understanding of dark matter demands its particles to interact weakly through electromagnetism, so they cannot have electrical charges and therefore cannot be chemical elements as we know them.

@NicolauSakerNeto Are you sure of that comment? I think that gets at the crux of it. I think you are in part wrong though. I think we know that it can't interact with normal matter but we have no idea what it even is let alone if it does interactions within it's own realm such as forming chemical elements.

I was going to extend my comment by mentioning "dark sectors" and "dark electromagnetism", which could in principle maybe contain additional "elements". However, a) these could not be chemical elements as we know them and would not go on the periodic table, and b) dark electromagnetism is present only in highly experimental models of dark matter.

If it were true, then creating the element in a particle accelerator would cause it to float away (cause it would stop interacting with e.g. your lab)...?

No, It is more likely all the energy lost from conversion of matter into energy since the formation of the Universe. We only see an infinitesimal amount of energy from nonexistant galaxies Where has all that energy gone

No.
A chemically unreactive element would nonetheless have charged particles, interact with photons, and register electromagnetic emission/absorption spectra. In fact elements at the surface of stars like our Sun are generally in the form of atoms that are not chemically bound, and they are definitely not dark with respect to electromagnetic radiation. True dark matter does not have these properties.


text:
Mechanism of adhesion of particles on solid surfaces
What is the mechanism of adhesion of particles on solid surfaces?
How does anything stick to anything, for that matter?
Why does cigarette smoke stay so stubbornly on our bodies?
Why does perfume also stay so stubbornly on our bodies?
And why do some perfumes stay longer than others on a fundamental level?

See: https://chemistry.stackexchange.com/questions/2384/why-are-smoke-odors-persistent-in-fabrics/9883

A title should not be mere a first sentence of the question. // A good question title should be simple, unique, resembling a magazine article headline or a book title, in plain text, optionally formulated as a question, avoiding being a click bait, saying too much or too little.

All these attractions are probably due to hydrogen bonding between particules and water fixed on solid surfaces.

The general adhesion of particles on surfaces has three major mechanisms:
Mechanical adhesion
This applies mainly to fine, flying dust particles, often categorized as fine particulate matter (PM2.5, size $\le \pu{2.5 \mu m}$). The particles mechanically connect to surface, the best if the surface has rich surface texture. The connection is not dissimilar to mutual adhesion of solid particles in grainy rocks, with or without softer gluing material. The visible part of cigarette smoke falls in this category.
Adsorption
This applies on gases or vapors of volatile liquids or solids, like the invisible part of smoke, or volatile parts of parfumes. Molecules of these compounds attach to molecules of surface typically by van der Waals intermolecular forces involving interactions of electric charges and/or permanenent/induced or randome electric dipoles. Additionally, there are possible relatively strong intermolecular force of hydrogen bonds.
Absorption
This applies to the same scope of particles as for adsorption, but the solid material has additionally the ability to dissolve such particles within its volume, similarly as water is able to dissolve gases. This happens mostly for nonvolatile liquids of soft, organic solids, leading  to higher sorption capacity than adsoption, long retention and slow releasing.
This is also mechanism of long lasting parfumes, where the pleasant fragrances are – after solvent evaporation – still dissolved(absorbed) in the non volatile part of the parfume. The fragrance is then slowly and steadily released, making long lasting effect. Without it, there would be a spike of intense smell, that would relatively quickly disappear.
Traditionally, for the most expensive parfumes, there was often used highly priced ambergris/ambra, a waxy substance found in  the digestive system of sperm whales, or randomly stranded on sea shore. Chemically it is a complex mixture, containing high amount of various terpenoids like  ambrein, ambroxide and ambrinol. Currently it is rarely used, being replaced by synthetic alternatives.


text:
Water of Crystallisation in Borax and Copper Sulphate
The formula for borax is written as $\ce{Na2B4O7.10H2O}$ and also as $\ce{Na2[B4O5(OH)4].8H2O}$.
From the structure, it obvious from where the 2nd formula comes. But the question I have is, what is the water of crystallisation in Borax, 8 or 10?
More specifically, I was to know what's the proper defintion of water of crystallisation? If it is just the number of water molecules present in the crystal structure, not directly bonded to the molecule (Which is what I thought) then the answer should be 8.
But one my teachers said it should be 10, as written in the formula, while the other who used to say 8 couldn't give an explanation to following
"Borax, when heated loses all 10 water molecules. When dissolved in water, all 10 water molecules are lost*. So even though they're bonded to the boron, the 2 water molecules behave the same way as the other 8, chemically"
*Just to mention, the full compound breaks up anyways, so I'm not so convinced with this argument
Another compound which came up in my mind during this was $\ce{CuSO4.5H2O}$, The structure of which is
What would be the water of crystalisation in Copper Sulphate? I'm sure there will be lot more compounds having $\ce{H2O}$ molecules bonded like this. What would be the "Water of crystallisation" in them and what's the proper definition of it?
Unlike Borax, Copper Sulphate doesn't lose all water molecules at once on heating, just opposite of the argument given above for Borax.

Edit:
I found this on wikipedia.
So, according to this the water of crystalisation in $\ce{CuSO4.5H2O}$ should be 1, but according to any website which speaks on this, the water of crystalisation is 5. So what is the actual definition? Or it is just a broad term like mentioned in a comment?

Perhaps "water of crystallization" is too broad of a term. The question clarifies the actual crystal water nicely. Perhaps you could write Na2B4O7.2H2O.8H2O and CuSO4.H2O.4H2O, suggesting two kinds of crystalline water: tight and looser.

Does this answer your question? Structure of Borax

And https://chemistry.stackexchange.com/questions/171016/how-heating-of-copper-sulphate-pentahydrate-blue-vitriol-can-be-explained?noredirect=1&lq=1 with the links therein

@NilayGhosh I was mostly concerned with "Water of crystallization" and what it actually implied. I saw the other question related to borax, and some others, they didn't talk about water of crystallisation directly.
I also found some detail on wikipedia, adding it

A quoted text is better to be presented as a text ( with the quote character > ) than as a screenshot.

https://chemistry.stackexchange.com/questions/171435/do-hydration-as-in-solvation-and-hydration-of-salts-refer-to-the-same-phen


text:
Nitronate anion reduction with TiCl3. Nef reaction
We can do the Nef reaction, by treatment of nitronate anion with $\ce{TiCl3}$. Here is the mechanism from article:

And I can’t understand what is exactly happening in $35\to38$ or $36\to37$. Can you please show precisely with arrows what is going on? Precise mechanism

When we go from 35 to 38 or from 36 to 37, the Ti(IV) species $\ce{TiOCl2}$ is leaving. Since the titanium entered as $\ce{Ti^{III}Cl3}$, this represents a net one-electron reduction.
But wait, there's more. The 36$\to$37 step requires adding another electron to neutralize the positive charge on 36, and if we treat the hydroxyl hydrogen on 38 as protic then that molecule also needs an additional electron. In either case a second $\ce{TiCl3}$ must be introduced at the 36$\to$37 or 35$\to$38 step as the source of this electron, so we have a two-electron reduction of the organic molecule using two $\ce{TiCl3}$ molecules in all. The charge on titanium from the second $\ce{TiCl3}$ molecule is balanced by accepting the chloride ion that the first $\ce{TiCl3}$ molecule surrendered to add to the oxygen earlier in the mechanism.
Below, I add the additional $\ce{TiCl3}$ molecule indicating where the electron from that molecule goes.

Thus we render the overall reaction:
$\ce{(CH3)2C^0H–N^{III}O2 +2Ti^{III}Cl3 -> (CH3)2C^{II}=O + Ti^{IV}OCl2 + Ti^{IV}Cl4 + N^{-I}H2OH}$
where oxidation states (based on expected bond polarities) on the atoms that are oxidized or reduced are indicated. Note that even though the substrate was reduced by two electrons, the nitrogen atom in the nitro group took four leaving the carbon in a higher oxidation state.


text:
Rules for (hydro)chlorination of double and triple carbon bonds
If $\ce{HCl}$ is added in a compound with both double and triple bond the double bond breaks and gets chlorinated.
If $\ce{Cu2Cl2}$ and $\ce{NH4Cl}$ are also present then which bond will get chlorinated, the double or triple bond?


text:
What is the difference between Aluminum Potassium Sulfate and Aluminum Sulfate?
What is the difference between Aluminum Potassium Sulfate and Aluminum Sulfate?  I mix my own photography chemistry and a formula calls for Aluminum Potassium Sulfate. My chemical supplier only has Aluminum Sulfate.  Something tells me I can't substitute the later for the former.  Help please.

Many thanks for the replies!  I failed to mention I'm mixing a fixing bath for films and/or papers.

What a user asks about may confuse and hide what the user wants to do and what the user needs to know to achieve his goal.  A broader context, background, purpose and relevant details could help in the delivering the optimal answer for the goal, instead of for the question.

Photographic gelatin emulsion is relatively soft, and during the development process, soaking in water softens it further. To make it more durable, a hardening agent may be added, though it is not required.
Two chemicals that harden the gelatin (crosslinking or "denaturing" the protein) are:

Formalin (formaldehyde in water) -- toxic, and the vapor is toxic.

Aluminum compounds, including aluminum sulfate, $\ce{Al2(SO4)3}$, and potassium aluminum sulfate, $\ce{KAl(SO4)2}$, AKA "alum" or "potassium alum". Not only are these both relatively non-toxic, but are used in cooking, and are available in the spice section of most supermarkets. Note that there has been some research indicating alum might be linked to Alzheimer's, yet that conclusion is debated, and alum is still used in many locales for baking and pickling. Check the price on the spice shelves -- if less than from you current source, feel free to switch.
"It is the $\ce{Al^3+}$ ion which does the hardening. It really doesn't matter what salt supplies it. It can come from aluminum sulfate or from sodium aluminum sulfate [or from potassium aluminum sulfate, ed.]." By that logic, make the formulation with the same amount of aluminum, not the same total mass of $\ce{Al2(SO4)3}$ as $\ce{KAl(SO4)2}$.

... and make sure you know what hydrate you have because that changes the molar mass.


text:
Rule for which heterocycles are stable?
In medicinal chemistry, is there a rule for which aromatic heterocycles are stable?
I've heard that heterocycles with 3+ adjacent nitrogens for example are unstable. However looking in ChEMBL at the molecules currently in human clinical trials, there are a fair number of 1,2,3-triazoles and tetrazoles, so clearly that rule is not absolute.  For six-membered rings, 1,2,3,4-tetrazine and 1,2,3-triazine don't seem to be used, but neither is 1,2,4,5-tetrazine which doesn't have 3 adjacent nitrogens.  Mixing nitrogen and oxygen, 1,2,5-oxadiazole and 1,3,4-oxadiazole are very rare but do show up a couple of times, but 1,2,3-oxadiazole isn't used.
I'm wondering if there is a general rule about which heterocycles would be stable, including ones with fused rings and a fairly large number of adjacent heteroatoms?  For example (in smiles notation) how about c1nnnc2c1cccc2 or c1cccc2n1nnc2 or n1cccc2n1ncc2?

Or how about something even more unusual, like a pyridinone/pyrone with some extra nitrogens?  c1(=O)[nH]nncc1

https://chemistry.stackexchange.com/questions/171869/stability-of-hexazine https://chemistry.stackexchange.com/questions/42612/stability-of-isomers-of-n4o

"heterocycles with 3+ adjacent nitrogens for example are unstable" - I doubt if there's anything so clear cut in chemistry. You can have even ten or more nitrogens in a row, but the issue what is this "stability" comes up... along with explosions.

@Mithoron Yup, I know stability is relative - I mean medchem stable, as in “wouldn’t spontaneously react with random stuff inside the cell”.  It appears 1,2,4,5-tetrazine isn’t stable in that sense, but tetrazole is.  I’m mainly curious about what happens with similar 5+5, 5+6 and 6+6 fused rings, incl when the bridging atom is a nitrogen


text:
UF4-ZrF4 phase diagram
I am doing physical chemistry tasks in Professor Peter Atkins book. The phase diagram of UF4-ZrF4 system that I found is in the following picture

Question:

Where is the point that the composition x(ZrF4) in liquid is equal to 0.40  (at 900oC)?

Thought: If the point was at the intersect of the x
(ZrF4) = 0,4 line with 900oC line. Then I think that the solution is merely liquid (no solid presents). However, it contradicts with the statement "At 900oC, the liquid solution ... is in equilibrium with a solid solution ..." which indicates that there is solid state of ZrF4

When we are observing the solution below the (L+S) area (x(ZrF4) = 0.40) but above the tie line passing through eutectic point, is it just UF4 crystallizes or both ZrF4 and UF4 crystallizes at the same time?

At 850°C, when the fraction of ZrF4 (x-axis) is equal 0,3 and 0,4. The composition of liquid phase and solid phase remains the same when drawing the tie line. How could this be explained?

It is not a "series" of solid solutions, it is a solid solution (sigh). And the 900C vs 850C points given are in very different parts of the phase diagram, leading to serious confusion , making the 850C discussion irrelevant to the question (sigh). What a terrible question (and I say that lovingly as a phase diagram nerd).

Ad thought:
It is not in contradiction. Notice very different compositions of both phases (0.28, 0.14), compared to $x_\ce{ZrF4} =  0.4$ at $\pu{900 ^{\circ}C}$. The latter will be all liquid until cooled down to $\approx \pu{865 ^{\circ}C}$,  when the solid starts forming with $x_\ce{ZrF4} \approx 0.22$. With progressing cooling, the liquid phase gets progressively enriched by $\ce{ZrF4}$, as the solid is enriched by $\ce{UF4}$.
Ad Q2:
Notice the note continuous series of solid solutions.
In a way, the diagram is equivalent to a phase diagram x-T of two miscible liquids with azeotrope with the minimum boiling point. So the solid is the solid solution containing both.
If the liquid mixture $x_\ce{ZrF4} =  0.4$ at $\pu{900 ^{\circ}C}$ is being cooled down, then:

At $\approx \pu{865 ^{\circ}C}$, the solid solution containing both fluorides starts forming with $x_\ce{ZrF4} \approx 0.22$.
This is enriching the solid phase by $\ce{UF4}$ while the liquid phase  gets enriched by $\ce{ZrF4}$.
When the system is further being cooled, the composition of the liquid phase and the momentarily solidified phase follows the isotherm of the current system temperature.
For the given temperature, these both compositions gradually increase, as lower isotherms cross both composition lines at higher and higher fraction for $\ce{ZrF4}$.
The system converges to its eutectic, where composition of both phases is equal, with $x_\ce{ZrF4} \approx 0.77$ and the melting point $\approx \pu{765 ^{\circ}C}$.
At this composition and temperature, the last remnants of the liquid phase finally solidifies.

Q3: Composition of both solid and liquid phases remain the same, comparing both states, but the ratio of both phases differ. The state with the overall higher fraction of $\ce{ZrF4}$ will have higher ratio of liquid:solid phase.

Just to clarify the phrase "solid solution". Should it mean 2 or more solids in a mixture? Or just miscible solids like alloy?

You do know what solid means and what solution means. then you know what solid solution means = a homogenous solid phase containing 2 or more components. Note that not all alloys are homogenous, having domains of microscopic sizes of different components.

Furthermore, when we cool down more from 865°C, the tie line indicates more enriched ZrF4 solid and ZrF4 liquid at the same time as the %ZrF4 increases to the right of the diagram. Am I making mistakes here?

Let us continue this discussion in chat.


text:
What precipitate forms when adding aqueous sodium sulfate solution to an unknown white solid?
I don't understand why there aren't two solutions to the following problem:

You are presented with a white solid and told that due to careless labeling it is not clear if the substance is barium chloride, lead chloride, or zinc chloride. When you transfer the solid to a beaker and add water, the solid dissolves to give a clear solution. Next an $\ce{Na2SO4(aq)}$ solution is added and a white precipitate forms. What is the identity of the unknown white solid?

Now according to the solubility guidelines BOTH $\ce{BaSO4}$ AND $\ce{PbSO4}$ should be able to form as precipitate, and therefore the answer could  be both barium chloride and lead chloride.
Where am I wrong here?

Without quantitative details it could be both. The expected answer depends on to what details the task author expects you to go.
$\ce{PbCl2}$ solubility is about $\pu{1.1 g/100 ml}$.
$\ce{BaCl2}$ solubility is about $\pu{36 g/100 mL}$.
It is reasonable to expect the used water/salt ratio was low enough to distinguish Ba and Pb. If you have performed the test, would you use too much water to dissolve both $\ce{PbCl2}$ and $\ce{BaCl2}$ and than was not able to tell, if the precipitate is $\ce{PbSO4}$ or $\ce{BaSO4}$ ?

Thank you for answering. There are no further specifications about the problem in the book.
If I understand your answer correctly, I should be able to tell whether it's PbCl2 or BaSO4 by gradually adding water and then matching the "dissolution quantity" of water at which the solid starts dissolving with the solubility information that you provided?

It should be mentioned in the question that "a small amount" of water is added to the white stuff, before testing the solution with $\ce{Na2SO4}$; or about the same volume of water as the volume of the white salt.

As mentioned above, the catch in the question might be "the solid dissolves to give a clear solution." Without specifying the amount of solid and water, it's unanswerable, but if the ratio is above a few g/L, then it can't be the lead salt. If that is the whole question transcribed, it is very poorly worded!


text:
The log K value for Cr-EDTA complex is 23, still Cr(III) cannot be titrated directly
It's related to stability constants.
I don't get the question and where is it taking and wanting to answer.

Be aware, for your future posts, that length of the post converging to zero means probability of obtaining answer is converging to zero as well.

Same question posted 6 years ago: https://www.chemicalforums.com/index.php?topic=91107.0

Where the equilibrium lays and how fast the system gets there are two independent things.
Many chromium(III) complexes are kinetically stable, slow to be formed or changed. This is not addressed by reaction thermodynamics (like the complex stability constant), but by reaction kinetics.
That is the reason why it cannot be titrated directly. Cr-EDTA complex is not being formed quickly enough to perform titration. Metalochrome indicators would signal a false titration end as much of $\ce{Cr^3+(aq)}$ would stay not reacted with EDTA yet.
Sometimes, the chromium(III) salt solutions have to be boiled for the complex formation reaction to occur.


text:
Why is the reaction between the two free radicals below not observed?
All the below reactions are observed:

Why is this one not happening?

Cyclopropylmethyl radical rearranges to homoallyl radical. There may be some kinetic considerations that prevent the self-self coupling. It's possible that the first reaction proceeds through a more complicated mechanism with the same product.


text:
Lewis dot structure vs orbitals
I am currently teaching myself AP Chemistry, so am at a very basic level still.
I am studying chemical bonds now, but am wondering why the orbitals don't make a difference to the way atoms bond.
To explain further:
Oxygen has 6 valence electrons, and so when drawing the Lewis dot structure, you would draw it with 2 pairs of electrons and two single electrons. This seems consistent with the orbitals of an Oxygon atom, because it has 1 pair of electrons in the 3s orbital, another pair in one of the 3p orbitals, and 2 single electrons in each of the remaining 3p orbitals.
This also makes sense to me in terms of and Oxygen atom forming a covalent bond with another Oxygen atom, since the two single electrons in each of the remaining 3p orbitals will "hang out" together, completing the octet rule for the valence shell of both atoms.
But when I looked up the Lewis dot structure for Carbon, I felt confused.
It seems to be drawn with 4 single electrons .
This feels inconsistent with what we know about the orbitals of a Carbon atom.
Given that there are 4 valence electrons, with 1 pair in the 3s orbital, and 2 more single ones in each of two of the 3p orbitals, surely we should draw it as:
1 pair of electrons, 2 single electrons.
But we draw it as 4 single electrons.
I know I am probably being really daft and missing something obvious, but please can someone help explain a bit more about why we draw it this way?
And why the orbitals' pairs don't seem to matter when it comes to bonds?
Thanks

The only thing that matters for the Lewis structure is the number of valence electrons and possibly extra positive or negative charges that subtract/add electrons. Carbon has 4 valence electrons. That is why you draw it with 4 dots. You should also draw oxygen with 6 dots, adding the electrons one side at a time. Then you should consider the electrons on possible partners, for example a second oxygen, and try to complete the octets by sharing electrons and then merge them to bonds and lone pairs.
The concept of Lewis structures is older than orbitals and not strictly related. You will only confuse yourself if you try to derive Lewis structures from orbitals.
Many concepts in basic chemistry are empirically derived rules of thumb; these rules are not strict laws as laws from physics. You will find that many rules that are taught in basic chemistry have exceptions and break down at some point. So be careful when you read about them and try to check when they uphold and when they don't. This point is sadly often understated or completely missing in simple texts and can lead to tremendous confusion if the rules are taken at face value and when one tries to apply them without considering their limitations. You can also have concepts that appear to be contradicting each other. This is also often due to the fact that one concept was originally invented and applied to answer a specific and limited question and never intended to be transferred to other topics/properties.
This can make general chemistry frustrating and is the reason why I personally very much prefer physical and theoretical chemistry.

If you have happened to learn how to write electronic configurations according the shells (not the orbital)
ex: Oxygen(6): 2K 8L 6M

For Lewis Structures: we draw the valence shell electrons (here 6)
For Orbital Structures: we draw the orbitals of valence shell (here of 2s, 4p)
For Compounds, when we draw Lewis structure, we use these valence electrons only, but when drawing orbital structures we find the hybridization and then draw the orbitals.
comment for any doubt. Thanks.


text:
Is it possible to create a solid from PVDF solution?
I am working on a project where I am attempting to line an aluminium tube with PVDF. The finished wall thickness is required to be 4mm thick.
I have investigated such processes as injection moulding and lining, however due to the complex bends and turns of the finished tube as well as the termination of this into a four port bulkhead, it rules these processes out. I have been working on producing a PVDF solution using DMF solvent so I can cast or mould the lining, however when using heat to draw off the solvent I am only left with a membrane as opposed to a solid. I have also experimented heating the solution and holding under vacuum this also produces an inconsistent structure.
I have read numerous studies on PVDF solution casting and the various solvents to use, however they are all focused on producing membranes for piezo electronics.
Any help would be greatly appreciated.

But for the actual problem, Googling "PVDF liner" yields a number of PVDF-lined pipe suppliers.

Many Thanks Jon however liners will not suit my application, hence my reason for wanting to "cast". The pipes terminate in bulk head distribution system and all exposed surfaces will require protection.

Yes, which is precisely the sorts of things that the providers of PVDF-lined pipes are good at - consider a large chemical plant and all the piping.


text:
Boiling point, pressure, and Gibbs free energy

The water boiling process can be represented by the equation:
$$ \ce{H2O(l) → H2O(g)} \tag{1} $$
Since free energy depends on pressure
$$ \left(\frac{\partial G}{\partial P}\right)_T = V \tag{2} $$
and the molar volumes of water are related by the inequality $V_\ce{H2O}(\ce{l}) \ll V_\ce{H2O}(\ce{g})$, evaluate how the boiling temperature of water will change with increasing pressure.

I understand that the boiling temperature will increase, because with an increase in pressure an equilibrium will shift in the direction of smaller amount of molar volumes, according to Le Chatelier's principle. Unfortunately, I cannot find out how to predict an increase in the boiling point of water referring to the statement that $\mathrm{d}G = V_\mathrm{m}\,\mathrm{d}P$ (as far as I understand, Gibbs free energy will experience an increase as well).

Have you looked at the P-T diagram for water?

It is hard to explain! Boiling point is an artifact! it depends on having a flexible container of an inert component, an atmosphere. [Altho a constant force stopper might also work.] For a one component system in equilibrium the Gibbs Phase rule, F = C -P +2, states F = 3-P. This means when 3 phases are present there are no degrees of freedom, a triple point of fixed T, P and chemical activities notice not amounts of the phases just their activities. When 2 phases are present there is one degree of freedom, 1 phase-2 degrees of freedom. Assuming physical possibility, for a liquid -vapor equilibrium this means if one controls either Temperature or Pressure the equilibrium will adjust to establish equilibrium at the new T or P. If both are controlled either one of the phases will vanish or an accidental equilibrium point will be established [the point will fall on the liquid-vapor line in the phase diagram].
Addition of an inert gas to a pure material means additional components. In the case of air it can be considered one component for physical changes: F = 2 -P +2 or F = 4 - P. A This means for 3 phases, a triple point, there is a triple point for each inert gas pressure an envelope of phase diagrams in the 4th dimension. A liquid-vapor equilibrium with 2 degrees of freedom responds to the temperature or the inert gas pressure. Increasing either of these puts more stress on the liquid phase, raising its chemical potential. This means that increasing the temperature OR increasing the inert gas pressure will raise the vapor pressure of the liquid because the chemical potential of the liquid becomes higher than that of the vapor. It seems that this means increasing air pressure increases the vapor pressure and should reduce the boiling point. True the vapor pressure is increased but the external pressure is also increased. This is where the artifact part enters! For a liquid to boil the vapor pressure must be increased enough to push back the higher atmospheric pressure. This means that the boiling temperature must be increased. The net effect is that increased inert gas pressure increases the boiling point. This sort of explains why the effect of changing pressure is less for boiling pt elevation than for freezing point depression; there are two opposing factors: increased inert gas pressure increases the vapor pressure, and the temperature must be increased to raise the vapor pressure enough to push back the atmosphere.
A disclaimer: This analysis is my own. It is based on passages in Daniels and Alberty, Physical Chemistry That I cannot recombine upon rereading the book, and my pondering the phase equation. This analysis describes what I call the 4th dimension of phase equilibria. The pure substance phase behavior is transformed into an envelope of almost parallel diagrams separated in the 4th dimension.

Expressing Eq 2 in differential format as
$$\left( \mathrm{d} \Delta G\right)_T = \Delta V \mathrm{d}P $$ where $\Delta V = V_\ce{H2O(g)}-V_\ce{H2O(l)} > 0$ leads to the obvious conclusion that increasing the pressure increases the Gibbs' free energy of the gas relative to liquid, at constant T. This would suggest that the liquid is preferred at higher pressure (with T constant).
The question however is specifically how the boiling T will change. We know it will increase, but need to show this. In essence we need to derive the Clapeyron equation.
Start from the differential form of G:
$$dG = \left(\frac{\partial G}{\partial P}\right)_T dP + \left(\frac{\partial G}{\partial T}\right)_P dT$$
For the equilibrium system
$$d \Delta G = 0 = \left(\frac{\partial \Delta G}{\partial P}\right)_T dP + \left(\frac{\partial \Delta G}{\partial T}\right)_p dT$$
or
$$ \frac{\mathrm{d}T}{\mathrm{d}P} = -\left(\frac{\partial \Delta G}{\partial P}\right)_T  / \left(\frac{\partial \Delta G}{\partial T}\right)_p\\ \frac{\mathrm{d}T}{\mathrm{d}P} = -\Delta V / \left(\frac{\partial \Delta G}{\partial T}\right)_p$$
From the definition of dG ($= VdP -SdT$),
$$\left(\frac{\partial G}{\partial T}\right)_P = -S$$
so that
$$ \frac{\mathrm{d}T}{\mathrm{d}P} = \frac{\Delta V}{ \Delta S}$$
Since $\Delta S>0$ for evaporation it follows that
$$ \mathrm{d}T = \frac{\Delta V}{ \Delta S}\mathrm{d}P >0$$

you confuse pressures! for a 1component system at T there is a vapor pressure if this vapor pressure is increased the chemical potential of the gas is increased forcing condensation and a new equilibrium at a higher T[heat is released]. Adding an inert gas at constant volume increases the potential of the liquid not the gas this raises the vapor pressure and lowers the T [heat is required] An inert gas or applied force n a non constant volume needs an additional pressure to push back the atmosphere or plunger so inceased external pressure a higher T to supply the vapor pressure.

A phase diagram involves liquid and vapor, liquid and solid, solid and vapor, and at one point liquid, solid and vapor. There is no inert gas present. Addition of inert gas overlays a new phase diagram for each inert gas pressure.

@jimchmst What inert gas? There is no mention of an inert gas either in the problem or in my solution afais. Don't see what point you are trying to make. Have you heard of the Clapeyron equation? It describes the relationship between T,P points on a two-phase coexistence curve.

The atmosphere is the inert gas. the Clapeyron equation describes the relationship between T and the vapor pressure with one component not the boiling point, boiling is evaporation into an unconstrained volume  against an opposing pressure, usually the atmosphere tho it can be a vacuum pump.The vaporpressure curve is monotonic to the critical point, there is no boiling point without an external pressure[inert gas].

@jimchmst I still don't see your point. The question is about the boiling temperature. The boiling Ts are the points on the coexistence line described by the CC equation. I think you attempt to propose a definition for "boiling" as a particular type of event involving inert gases etc, but I don't think that is pertinent here.

@jimchmst I suspect you are trying to make a point about the distinction between the internal pressure of a system, the applied external pressure (or pressure of the surroundings) and the vapor pressure of a particular substance. In this problem that distinction is not relevant. It deals with a single component in a closed system with tunable T and P, where P is the pressure of the system. If there is vapor that is also the pressure of the vapor.


text:
Quantum Mechanics expectation value help
I am working on the following question:

I've so far done the first part of a, finding $H_{so}$ in terms of J L and S. I'm struggling to see how to relate that to $E_{LSJ}$ as it involves the expectation value. I don't see what to integrate.
Edit: is the expectation value just $J^2-L^2-s^2$? That seems to make sense

Please give us more information. Your first sentence is "I have done the first part of $a$" OK. But what is $a$ ? What is the problem ? Are there many parts in $a$ ? Where do you want to go ?  Later on, you say that you don't know "what to integrate". How can we know ? Why would you integrate something ? What for ? To go where ?

The states $|LSJ\rangle$ are eigenstates of the $L^2$, $S^2$, and $J^2$ operators (note: these letters are operators) with the corresponding eigenvalues $L^2$, $S^2$, $J^2$ (these letters are quantum numbers used to label the states).
The long way around is to break the operators down into their Cartesian components. We have that $L^2 = L_x^2 + L_y^2 + L_z^2$ (this is a 3D Pythagoras theorem), and likewise $S^2 = S_x^2 + S_y^2 + S_z^2$, and $J^2 = J_x^2 + J_y^2 + J_z^2$. But also, because $\vec{J} = \vec{L} + \vec{S}$, we have that:
$$\begin{align}
J_x = L_x + S_x; \quad J_y = L_y + S_y; \quad J_z = L_z + S_z.
\end{align}$$
So:
$$\begin{align}
H_\text{so} &= \xi (\vec{L} \cdot \vec{S}) \\
&= \xi (L_xS_x + L_yS_y + L_zS_z) \\
&= \frac{\xi}{2} (2L_xS_x + 2L_yS_y + 2L_zS_z) \\
&= \frac{\xi}{2} [(L_x + S_x)^2 - L_x^2 - S_x^2 \\
&\quad\quad+ (L_y + S_y)^2 - L_y^2 - S_y^2 \\
&\quad\quad+ (L_z + S_z)^2 - L_z^2 - S_z^2] \\
&= \frac{\xi}{2} [J_x^2 + J_y^2 + J_z^2 - (L_x^2 + L_y^2 + L_z^2) - (S_x^2 + S_y^2 + S_z^2)] \\
&= \frac{\xi}{2} (J^2 - L^2 - S^2)
\end{align}$$
and so the corresponding expectation value is
$$\langle LSJ | H_\text{so} | LSJ \rangle = \left\langle LSJ \middle| \frac{\xi}{2} (J^2 - L^2 - S^2) \middle| LSJ \right\rangle = \frac{\xi}{2} (J^2 - L^2 - S^2).$$
The 'short' way, which is basically saying the same thing as above but in a much more concise manner, is:
$$\begin{align}
J^2 &= |\vec{L} + \vec{S}|^2 \\
&= L^2 + S^2 + 2(\vec{L}\cdot\vec{S}) \\
\vec{L}\cdot\vec{S} &= \frac{1}{2}(J^2 - L^2 - S^2) \\
\xi(\vec{L}\cdot\vec{S}) &= \frac{\xi}{2}(J^2 - L^2 - S^2) \\
\end{align}$$
but to convince yourself of the equality going from the the first and second line you need to expand the vectors out into their Cartesian components.
Addendum: Ian Bush has pointed out in the comments that this also assumes that the components of $\vec{L}$ and $\vec{S}$ commute with one another: otherwise, in principle we have that
$$(L_x + S_x)^2 = L_x^2 + S_x^2 + L_xS_x + S_xL_x$$
or
$$|\vec{L} + \vec{S}|^2 = L^2 + S^2 + (\vec{L} \cdot \vec{S}) + (\vec{S} \cdot \vec{L})$$
for the 'short' derivation. Thankfully, because $\vec{L}$ and $\vec{S}$ are different sources of angular momentum, they do fully commute, and so the last two terms in both lines above are equal. Thanks to Ian for the reminder!

Note in the above you have assumed L commutes with S. I have no idea if this subtly is above the pay grade if the OP

@IanBush Thanks for pointing that out! I shall edit. To be honest — I am starting to realise I am very rusty. One day not too far from now these questions may be above my pay grade. :-)

to indicate operators, you can use the command \hat{}, eg \hat{H} gives $\hat{H}$

@Andrew, thank you! I'm aware. I just wanted to follow the style of the question, and I think the hats are usually dropped in the wild anyway (great if you're used to it... potentially very confusing for beginners).


text:
Optical rotation and chirality
I had the same question as in Molecular chirality and optical rotation
In the answer, it says that there can't be such a mirror position where the effect of one molecule be cancelled by other. But I'm so confused.
But don't the first and the third image in the original answer show molecules such that the both in combination could cancel each other's effects?

These two are difficult to compare because in one, we are looking at the discs from the right, and in three, we are looking at the discs from the left. Go back to the answer of the other question and compare the first with the fourth image.

It's a bit like turning a screw so that the head is on the other side.

Thank you so much @Karsten. I understand now, the screw analogy really helped.

Let's just look at the first situation. What you're asking is, in effect, what happens when I run the situation in reverse because for every orientation that looks like the first scenario, there is one that matches the reverse.
But look at the effect down the axis from both sides. In both situations, the plane of polarization has shifted about 45° counterclockwise. So no matter the direction of light in this situation, the sense of rotation is the same, so there is no canceling of the net rotation due to symmetry; the symmetry reinforces the sense of rotation.
This makes perfect sense if you were to look at screw. Regardless of which end you look at it from, for a regular right-handed screw, you always turn to the right (clockwise) to drive it in (away from you).


text:
Derivation of general binding equation (macromolecule with multiple binding sites)
So the last days I'm stuck at trying to solve the an equation - but I just can't (see the screenshot below.

So the idea is that you have a receptor or enzyme with multiple binding sites for a Ligand (let's assume 3 binding sites). They are not influencing each other and have the same affinity.
The equations above show how the microscopic dissociation constants are related to the microscopic dissociation constants and that the same "affinity" of a binding site does not relate to a similar macroscopic dissociation constant for each subsequent binding step.
However, I just cannot see how they get from here

to here

I already tried to solve the calculations in every way I could think of. ChatGPT fails miserably. Also I didn't find any other sources...
Is somebody able to solve it? Or knows a paper/source that solved this step by step? I would be eternally grateful.

$\ce{_AE^A}$ is in equilibrium with $\ce{E}$ and $\ce{A}$ along two paths:
$$\ce{E + 2A <=> E^A + A <=> _AE^A}$$
and
$$\ce{E + 2A <=> _AE + A <=> _AE^A}.$$
So you get the equivalence:
$$[\ce{_AE^A}] = \frac{[\ce{E^A}] [\ce{A}]}{K^{12}} = \frac{[\ce{_AE}] [\ce{A}]}{K^{21}}$$
With these kinds of relationships, you can simplify the expression, cancel all the concentration terms and get the result in the book (after getting rid of fractions by multiplying with the equilibrium constants in the denominators). The math is elementary yet tedious.
$$K_2' = \frac{\ce{([E^A] + [_AE] + [E_A])[A]}}{\ce{[_AE^A] + [E_A^A] +  [_AE_A]}}  \tag{1}$$
$$= \frac{\ce{([E^A] + [_AE] + [E_A])[A]}}{\frac{[\ce{E^A}] [\ce{A}]}{K^{12}} + \frac{[\ce{E^A}] [\ce{A}]}{K^{13}} + \frac{[\ce{_AE}] [\ce{A}]}{K^{23}} }\tag{2}$$
$$= \frac{\ce{[E^A] + [_AE] + [E_A]}}{\frac{[\ce{E^A}]}{K^{12}} + \frac{[\ce{E^A}]}{K^{13}} + \frac{[\ce{_AE}] }{K^{23}} }\tag{3}$$
$$= \frac{K^{12} K^{13} K^{23}(\ce{[E^A] + [_AE] + [E_A]})}{K^{13} K^{23} [\ce{E^A}] + K^{12} K^{23}[\ce{E^A}] + K^{12} K^{13} [\ce{_AE}]}\tag{4}$$
$$= \frac{K^{12} K^{13} K^{23}(\ce{[E^A] + \frac{K^{21}}{K^{12}}[E^A] + \frac{K^{31}}{K^{13}}[E^A]})}{K^{13} K^{23} [\ce{E^A}] + K^{12} K^{23}[\ce{E^A}] + K^{21} K^{13} [\ce{E^A}]}\tag{5}$$
$$= \frac{K^{12} K^{13} K^{23}(\ce{1 + \frac{K^{21}}{K^{12}} + \frac{K^{31}}{K^{13}}})}{K^{13} K^{23} + K^{12} K^{23} + K^{21} K^{13} }\tag{6}$$
$$= \frac{K^{12} K^{13} K^{23} + K^{21} K^{13} K^{23} + K^{12} K^{31} K^{23}}{K^{13} K^{23} + K^{12} K^{23} + K^{21} K^{13} }\tag{7}$$
This is slightly different from the textbook, maybe a typo or a different interpretation of the subscripts. Exercise for the reader...
If you consider the following sequence:
$$\ce{ _AE^A <=>  E^A + A <=> E_A^A <=> E_A + A <=> _AE_A <=> _AE + A <=> _AE^A}$$
you find that the equilibrium constants are not all independent, as
$$ K^{13} K^{32} K^{21} = K^{12} K^{31} K^{23},$$
so there are multiple ways to write this result.
On a different note, if you say the binding sites are equivalent and independent, you can quickly relate apparent and individual binding constants. For the first binding event, you go from one state (all sites empty) to three (one site occupied), hence the factor 3. For the second binding event, you go from three states (one site occupied) to three states (one site empty), so the factor is 1. For the third binding event, you go from three states (one site empty) to one state (all sites occupied), so the factor is 1/3.

Instead of eternal gratitude, I'd be satisfied with a simple check mark if it pleases the OP.

I think I spotted the mistake in my calculations. I counted the concentrations twice. A ligand bound to enzyme at position 1 and 2 compared to 2 and 1 results in the same product, therefore they should be set equal, not add them up. Thanks a lot for helping out.
I know it makes intuitively sense, yet, in order to derive the Higgs equation later, I thought it might be important to understand the maths properly.
You shall have my gratitude and a check mark :)

Just a follow up question, since I still cannot get it right.
Apparantly its wrong to assume that EA2 = EA2(1,2) + E2(1,3) + E2(2,1) + E2(2,3) + E2(3,1) + E2(3,2). Because If I formulate the relation you described (e.g., E2(1,2) = E1*A/K1,2) and plug it into the equation and assume random numbers for it, then K2' = k2/2 and not K2' = k2. So I suppose because E2(1,2) = E2(2,1) I might count it twice. However, how should I plug in that EA2(1,2) = EA2(2,1)? If I set it equal as you described, there is no way to put it into the general equation K2' = ...

Yes, you are counting complexes twice. Putting a glove on your right hand first, then one on your left hand is the same as donning them in the opposite order. In fact, there should be a single symbol for that, not two symbols. On the other hand, it might be more comfortable to put on the left glove with the right hand already gloved than the opposite way, so there are two equilibrium constants for putting on the second glove.

I made my brain hurt adding the trivial math and ... arrived at an answer different from that of the textbook. For the case where all binding constants are equal, it does not make any difference, though.

Now you really have my eternal gratitude. I managed to solve it now and came to the same result as you have, which is almost the same as in the book. Maybe there are some alternative solutions, depending on which relations one uses to solve the equation... anyways, thanks for taking the time!

The 6 equilibrium constants describing going from one to two bound ligands are not independent. This means you could write the equation in different correct ways.


text:
What is the correct name of H2SO2? Sulphoxylic acid or hyposulphuric acid
Given the oxidation state on S, it seems the correct name should be hyposulphurous acid, but I have seen multiple sources call it sulphoxylic acid, or both
What is the reasoning behind this nomeculture

Fairly likely that sulphoxylic acid is the old non-systematic name that chemists got used to using - like acetic/ethanoic acid - and is used in older literature.

Wikipedia names the compound as sulfoxylic acid and the other name provided is hyposulfurous acid

Pretty much any compound can be named in different ways, and it doesn't mean one of them is "correct".

Also depends on locale: US has sulfur, but British likely import all their sulphur.

Wikipedia lists "sulfoxylic acid" first, so that appears to be the preferred name. There are isomers with the same formula $\ce{H2SO2}$ and sulfur in the oxidation state +2, so "hyposulfurous acid" based on the oxygen content may be considered less specific than "sulfoxylic acid".
Unlike what is apparently the case for sulfurous/sulfonic acid and phophorous/phosphonic acid, the most stable isomer of  $\ce{H2SO2}$ seems to be the one with all hydrogens attached to oxygen, thus the sulfoxylic acid isomer.
Many uncommon anions are known with relatively benign alkali-metal cations, but not sulfoxylate. The referenced Wikipedia article reports instead that sulfoxylates have been synthesized with late transition elements (cobalt, copper(I)) or post-transition elements (zinc, thallium(I)). No explanation is given for this apparent preference. Possibly sulfoxylates of these metals draw more interest because they may come up in the roasting of sulfide ores.


text:
Can Ba(OH)2 be used for carrying aldol?
I know that aldol reactions require a weak or a moderately strong base for the mechanism to occur and cannizaro happens when a stronger base is used. But my textbook has used Ba(OH)2 as a base in many aldol reactions as a base, which is even stronger than NaOH. Is my textbook in the wrong or am I missing something?

Aldol reaction assumes presence of enolizable hydrogen on one of reactants. Cannizzaro reaction assumes absence of it.

Barium hydroxide may be a stronger base than NaOH in aqueous, but these aldols are not being done in water. Ba(OH)2 has low solubility in most organic solvents so the aldol reactions occur on the surface of the solid Ba(OH)2.


text:
Can certain spectral lines be common between two spectrums
I have studied regarding atomic spectroscopy in my class and the Rydberg's formula. Suppose if I take $\ce{H}$ and $\ce{He+}$ and an electron jumps from from 2nd to 1st shell in $\ce{H}$ and from 4th to 2nd shell in $\ce{He+}$.
So according to Rydberg's formula
1/lamda(1)= R(1)^2(1/1^2-1/2^2)
1/lamda(2)= R(2)^2(1/2^2-1/4^2)

So both must be equal but I also know that each element has a unique atomic spectrum so how the above equation can be explained.
Please correct my understanding.

Chem+Math formatting reference:
MathJax Basics / Expressions/formulas/equations / Upright vs Italics / Math SE Mathjax tutorial
// MathJax is preferred not to be used in CH SE Q titles.

The original Rydberg formula only works for hydrogen atoms. Even for other hydrogen-like atoms, it won't work because the nuclear charges are different and they affect the actual energy levels. There is an extended Rydberg formula that includes this but it clearly won't give the same wavelengths for similar transitions.

Please define $R(1)$ and $R(2)$ - do you mean the Rydberg "constant" as a function of the nuclear charge, i.e. $R(Z)$?

Also in the second equation I assume you mean 1/2^2, correct?

@matt_black Unfortunately the OP hasn't defined the symbols they use but iff I understand them they have included correctly the effect of nuclear charge in their formula - the quadratic dependence of the Rydberg constant is cancelled by the inverse square of the double quantum number. So actually I like the answer as given, a spectrum is not just one line but the full collection. If you want to quibble the exact frequency will not be the same for any number of reasons, reduced mass is possibly the simplest difference, but at the level indicated the question is a nice observation.

Well spotted, single transitions of different elements can indeed have (nearly) identical energies, i.e. the same emission lines.
However, a spectrum is a collection of emission lines. That collection will be different for $\ce{He+}$ and $\ce{H}$. They may share one (or several) common lines that you indicate, but they will also have many distinct lines and thus a distinct spectrum.
It's a bit like words and letters: "apple" and "argon" share a common letter "a", but they are distinct words because they also contain other letters.
As a matter of fact, Niels Bohr made the same observation as you:

It may be remarked that according to the theory, helium must be
expected to emit a series of lines closely, but not exactly,
coinciding with the lines of the ordinary hydrogen spectrum. (Niels Bohr, 1913, The spectra of helium and hydrogen)

Well, not really. The nuclear charge on different elements means that the same transitions in isoelectronic atoms do not have the same wavelength. This is clear from the use of the version of the extended Rydberg formula that includes atom-specific factors rather than the original formula that only works for hydrogen atoms.

Sure, in the OP's example it's not the same transition that has a (nearly) identical wavelength. It's a different transition (n=4 --> n=2 for He+ vs. n=2 --> n=1 for H).
https://en.wikipedia.org/wiki/Rydberg_formula#For_any_hydrogen-like_element


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Spontaneity of endothermic reactions
I see the Gibbs Free Energy explanation everywhere, but is there an intuitive explanation of how endothermic reactions can happen spontaneously?  It seems intuitively reasonable that heat flows from hot to cold, by considering how more energetic molecules impart energy to less energetic molecules.  But how can this intuition be adapted for spontaneous endothermic reactions?

Are you aware that for closed, isothermic, isobaric systems, $- \dfrac{\Delta G}{T} \le \Delta S_\text{universe}$ ?  // and that for spontaneous processes in isolated systems, $\Delta S \gt 0$ ?

The spontaneous character of a reaction is not only related to the thermic effet. For example, the reaction of $\ce{NaHCO3}$ on a dilute acidic solution is spontaneous and endothermic. This spontaneity is due to the large entropy increase of the reaction, due to the production of the gas $\ce{CO2}$.
Here the variation of enthalpy is positive. But, as the change of entropy is positive and rather important, the change of free enthalpy in the reaction is negative, and the reaction becomes spontaneous.

Would it not be more accurate to say that the spontaneous character of a reaction is not only related to the thermic effect but depends also on entropic factors?

Hi Maurice and Buck, thank you for your replies.  I understand how, formulaically, the entropy is an additional factor to the enthalpy.  However, I'm specifically seeking an explanation from an intuitive angle.  That more energetic molecules collide with, and thereby impart energy to, less energetic molecules, seems to give an intuitive explanation as to why heat flows from hot to cold.  What I'm trying to find is how that intuitive explanation needs to be modified to explain the entropic effect.  Thanks and regards.


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Calculating the solubility of nitrogen in a liquid in high pressures and low temperatures
I'm trying to get nitrogen dissolved into my homebrewed beer. I have a nitrogen tank that I'm going to hook to a keg that's rated for 130 PSI (with a PRV rated at 100 PSI).
Is there any way to reliably calculate the solubility of nitrogen in water, or the amount of dissolved nitrogen in water, in high pressures and low temperatures (up to 80 PSI an -1c)? Does Henry's Law apply here?

130 PSI is not a high pressure, at least not in context of physical chemistry. Solubility of nitrogen can be safely assumed to be proportional to pressure in that range.  //  For carbon dioxide, it would be more complicated due high solubility and several chemical equilibrii.

If tempted to dramatically lower the temperature of the keg, please check the MDMT (minimum design metal temperature) if listed. If not listed, do not go below freezing, if that.

I will propose you one scheme for estimating its value, with the assumptions I think are reasonable. The data employed are found at the end.

The gas phase is assumed to be pure $\ce{N2}$ and we will consider its deviations with ideality, even though the pressure is not that high ($\approx \pu{5.5 atm}$) and nitrogen is a non-polar molecule. The liquid phase is assumed to be a mixture of $\ce{N2 + H2O}$. Since $\ce{N2}$ is barely soluble in water, we will model the non-idealities with Henry's law.
The chemical equilibrium, and thus the molar fraction of nitrogen in the liquid phase turns out to be
$$ \phi_\ce{N2}P = H_\ce{N2,H2O}x_\ce{N2} \rightarrow
   x_\ce{N2} = \frac{\phi_\ce{N2}P}{H_\ce{N2,H2O}}\tag{1} $$
Evaluating Eq. (1) anwers the question, but we need to estimate two factors: (1) the fugacity coefficient of nitrogen in the gas phase $\phi_\ce{N2}$, and (2) the Henry's constant for nitrogen in water as solvent $H_\ce{N2,H2O}$.

$\phi_\ce{N2}$: I will model the gas phase with the virial equation of state up to the second virial coefficient. In this case
$$ \phi_\ce{N2} = \exp\left(\frac{BP}{RT}\right) \tag{2} $$
where the second virial coefficient $B$ is calculated with statistical thermodynamics, using the Lennard-Jones potential for the intermolecular potential function
\begin{align}
    \Gamma(r) &= 4 \varepsilon\left[
    \left(\frac{\sigma}{r}\right)^{12} -
    \left(\frac{\sigma}{r}\right)^{6}\right] \tag{3} \\
    B &= 2 \pi N_\mathrm{A} \int_0^\infty
    \left\{1 - \exp\left[-\frac{\Gamma(r)}{kT}\right]\right\} r^2 \mathrm{d}r
    \tag{4}
\end{align}
Combining Eqs. (3) and (4), we can then numerically integrate to obtain $B$. Then, we go back and calculate Eq. (2).
$H_\ce{N2,H2O}$. This constant is a function of temperature, but only of pressure at sufficiently high values. One important assumption is that I will neglect the dependence of $H_\ce{N2,H2O}$ on pressure, and only take into account the temperature. An empirical equation for this is given in Benson and Krause (1976)
\begin{equation}
  \ln\left(\frac{H_\ce{N2,H2O}}{\pu{bar}}\right) =
  \alpha_2 \left(1 - \frac{T_2}{T}\right) -
  \beta    \left(1 - \frac{T_2}{T}\right)^2 + \ln(1.01325) \tag{5}
\end{equation}
with $\alpha_2$, $T_2$, $\beta$ being fitting parameters.

Fixing the pressure to $\pu{130 psi}$, we can repeat the calculations for many temperatures. The results are shown below:

The minimum in the solubility is due to the dependency of Henry's constant with temperature. When its value is the highest, the tendency of $\ce{N2}$ to leave the liquid phase is the highest, and thus the molar fraction acquires a minimum value. The estimation you want is when the red curve hits the left $y$ axis, which is
\begin{equation}
  \boxed{x_\ce{N2} \approx \pu{1.3 \cdot 10^{-4}}}
\end{equation}

The parameters for the intermolecular potential function for nitrogen are:
\begin{equation}
     \sigma        = \pu{3.667*10^{-10} m} \hspace{1 cm}
     \varepsilon   = \pu{1.378 \cdot 10^{-21} J};
\end{equation}
The constants for the correlation of the Henry's constant are:
\begin{equation}
     T_2 = \pu{162.02 K} \hspace{1 cm} \alpha_2 = 41.712 \hspace{1 cm} \beta= 36.855;
\end{equation}


text:
How do pencil erasers erase on a fundamental level?
One could say we're taking advantage of friction, but I want to dig deeper down to molecular/atomic level: what's the process or chain that happens when one erases from a macroscopic to a more fundamental level?


text:
Dependence of Butler-Volmer equation on n for multiple transferred electrons
I was working on an exercise sheet regarding the RDE results of an irreversible electrochemical reduction, with n transferred electrons. In it, we had to plot the Koutecky-Levich plot at different potentials, and as a result, it was deduced that n=4 electrons were transferred (probably the oxygen evolution half-reaction).
In another task, we had to estimate the rate constant of the reaction. The following relation is known for the reduction regarding the rate constant in terms of overpotential:
$k_{f}=k_{0}e^{\frac{-\alpha F(E-E^{0})}{RT}}$
Strictly speaking, this is valid for one electron processes, but I don't know for certain. Bard and Faulkner gives the general expression as:

My professor as (note it should be reduction not oxidation and a minus in the exponential, z=n):

Anyways, is n incorporated in this expession or not? I got $k_{f}$ from the Koutecky-Levich plot and need to calculate $k_{0}$. And if I used the version that incorporated n, my values would be in the range of $10^{-20} s/m$, which imo is too small, without the n around $10^{-8} s/m$, which seems more reasonable.
Similarly, to calculate the Tafel slope, Bard and Faulkner gives us for multistep electron exchanges with n>1:

Where n is the whole of the electrons transferred (in my case, n=4), f=F/RT. Note that n does not appear in the exponential term, only in the pre-exponential term. We then would need to neglect the oxidation current and calculate the Tafel slope. However, in my lecture, n appeared in the pre-exponential term and in the exponential term.

So, which one is correct for determining the Tafel slope? I would go with the textbook, although I am a bit lost here.

Is the question to determine the Tafel slope or the charge transfer coefficient? The Butler-Volmer equation does not include 'n' in the exponential term, the inclusion by some authors is incorrect and is a consequence of some multi-electron transfers showing this behavior (actually if you work out the individual reactions and write the total kinetic expression it might appear). The Tafel equation is purely empirical, so you just need to obtain how much the current changes per the potential and don't need to correct for the number of electrons.

It is to determine the Tafel slope, the charge transfer coefficient is 0.4

You would have to write the reaction as a set of 1 electron transfers and get the total kinetic expression like this (for which you need the standard reduction potential of each reaction and the charge transfer coefficient for each reaction or some other information on the reaction and, then see what the resulting Tafel slope is.

Note that for cases of math formulas taken from Wikipedia pages, one can click to edit link for a page section, copy  <math>MathJax expression</math> from there and paste it with   replaced by single/double dollar. // Like <math>j = j_0\frac {zF} {RT} (E-E_{\rm eq})</math> converted to $$j = j_0\frac {zF} {RT} (E-E_{\rm eq})$$  displayed as $$j = j_0\frac {zF} {RT} (E-E_{\rm eq})$$

I just checked the Tafel slope for the experiment. It is 149 mV/decade, which would be consistent with the calculated Tafel slope, with alpha=0.4 and no "n" appearing in the formula (147 mV/decade).

Strictly speaking, this is valid for one electron processes, but I don't know for certain. Bard and Faulkner gives the general expression as:

From a theoretical point of view, Butler-Volmer kinetics apply only to a one-step, one-electron transfer. The B-V equation may be derived by extending Eyring's theory to electrochemistry. Your equation seems to be extracted from the second edition of Bark & Faulkner's bible, but I think the first one has this derivation.
From a practical point of view, if you have the academic electrochemical reaction:
$$ \ce{O + ne^- <=> R} \tag{1} $$
nothing stops you from writing the current density of this reaction as follows
$$ j = j_0 \left\{\exp\left(-\frac{\alpha nF}{RT}\eta\right) - \exp\left[\frac{(1 - \alpha)nF}{RT}\eta\right]\right\} \tag{2} $$
where the exchange current density is
$$ j_0 = \color{red}{n} F k_\mathrm{c}^{1 - \alpha}  k_\mathrm{a}^{\alpha} c_R^\alpha c_O^{1 - \alpha} \tag{3} $$
by just adding $n$ on the exponential terms. The reason to add $n$ is to have thermodynamic consistency. Once you set $j=0$ to zero, you recover Nernst's equation. If you do not put that $n$, you will obtain the following expression
$$ E^\text{eq} = E^\circ - \frac{RT}{F} \ln \left(\frac{c_R}{c_O}\right) \tag{4} $$
which is a typical mistake that makes you lose points in a chemistry exam. You may want to visit Chapter 8 of Newman's book (third edition) for more information about these equations
To conclude, it would be a conceptual mistake to not include the $n$.

Similarly, to calculate the Tafel slope, Bard and Faulkner gives us for multistep electron exchanges with n>1:

That part of the chapter is a big dense, so I summarize it to you. Eq. (3.5.11) applies to a generic mechanism of $n$ electrochemical steps, in which:

$n - 1$ of those steps are in equilibrium.
$1$ is not, because it is supposed to be so slow, the so-called RDS.

The resulting current density is Eq. (3.5.11). This equation doesn't have the B-V fashion, because the concentrations still multiply the exponential terms. You need a bit more algebra to arrive at the final expression.
The last equation you posted, is equivalent to my Eq. (2), but still doesn't have the B-V fashion, because the concentrations multiply the exponential terms. The $\color{red}{n}$ you see there has to do with Eq. (3), belonging to the expression of $j_0$, plus lumping the cathodic and anodic rate constants into one, named $k^0$ by your professor.
To conclude, the equation from Bard & Faulkner refers to the general $n$-electron mechanism, whereas the equation from your professor refers to a general B-V kinetics for a multiple-electron transfer taking place in one step. Also, your professor likes the old convention, because the anodic reaction contributes to $j > 0$ and the cathodic reaction contributes to $j < 0$. Eq. (2) in this answer has the opposite convention, which was taught to me to be the one that prevails these days (with the exception in corrosion studies).

That's true, I believe n should be included. However, from the raw data we obtained, the inverse Tafel slope is 6.6 dec/V, which means the Tafel slope is 149 mV/dec. This would be in theory consistent with the expression (-2.303RT/aF), where a=0.4, like given in the exercise sheet. If I apply n=4 to the equation, the calculated result does not match the experimental evidence

@Mäßige You may consult that to your professor. I would like to point out that your Tafel's slope then has a more sublte meaning. It is that which belongs to the RDS in a 4-step electrochemical mechanism. Thus, you cannot associate it to the OER.

Ahh okay so you are saying that even though four electrons are transferred in total, the kinetics are limited by that RDS (whichever of the four steps it might be) and thus appears as around 140 mV/dec like a one step reduction?

And that should also be why the rate constant also appears in the order of magnitude of one step electron transfers

@Mäßige Yes. Since all the other steps are in equilibrium, they contribute zero to the current density. Only the RDS step has a Tafel's slope. In any case, there is more algebraic work to do in Eq. (3.5.11). IMO, I doubt that your professor wanted you to apply this general $n$-electron transfer mechanism. If you make sure, then you can tell me, and I can give further remarks.

Okay thanks a lot for your insight.

@Mäßige I realized another thing. Those $n$ electrochemical steps work for reactions of the type $\ce{O + ne^- <=> R}$. If you are sure that the reaction was $\ce{O2(g) + 4H^+(aq) + 4e^- <=> 2H2O(l)}$ (in acid medium), then I have my doubts, because it doesn't have that form. It is not evident how you would write the OER as a series of $4$ electrochemical steps.

@Mäßige E.g., the approach would work for reactions such as $\ce{Cr^{3+}(aq) + 3e^- <=> Cr(s)}$, but not for $\ce{Hg2SO4(l) + 2e^- <=> 2Hg(l) + SO4^{2−}(aq)}$.

Then it probably is not the OER. The sheet says that the reaction is of the form A(n+) + n e = B


text:
Which group is more ortho/para-directing?
Currently working on assigning the C NMR and H NMR spectra for this molecule (I believe it's called E-E bis(4-methylbenzylidene)acetone ) but I'm a bit confused as to whether the methyl group is more or less activating than the R-group (labelled in image).
Since alkyl groups are directing via hyperconjugation I thought that methyl group would be more activating of the ortho and para positions as it has more C-H σ bonds to donate to a potential empty p-orbital.
If this is the case, are the carbons in position 3 less deshielded than in position 4 so would show a lower chemical shift in NMR spectra?
If I'm being honest I don't think understand how alkyl groups are activating, I was just told that they were and was given a vague description on how they are activating but I can't seem to apply it to different scenarios. I've searched online for different explanations but none quite click with me.

Central group is deactivating, just like analogous formyl.


text:
Is salt formation catabolic or anabolic?
In class we learned that catabolic reactions are exergonic. I later checked that the free energy change of salt formation, $$\ce{2Na + Cl2 -> NaCl}$$ has a negative $\Delta G$. What explains this? Am I even asking the right question?

The terms catabolic/anabolic reactions apply rather in context of  breaking/building larger molecules in context of metabolism. They are not used in context of small inorganic molecular entities.

You would be asking the right question if an organism could metabolize sodium using chlorine as an oxidizer.

The heat of formation of $\ce{NaCl}$ is $\Delta$$\ce{H}$ = - $411$ kJ/mol. This is much. In order for the synthesis of $\ce{NaCl}$ to be spontaneous, the change of free enthalpy must be known and must be < $0$. As this synthesis implies the disparition of a gaz ($\ce{Cl2}$), the reaction is accompanied by a decrease of entropy : $\Delta\ce{S}$ < $0$. The Tables give : $\Delta$$\ce{S}$ = - $180$ J/mol. So $\ce{T$\Delta$S}$ is - $53$ kJ/mol at $25$°C. As a consequence $\Delta$$\ce{G}$ = $\Delta$$\ce{H}$ - $\ce{T}\Delta\ce{S}$ = - $411$ kJ/mol + $53$ kJ/mol = - $358$ kJ/mol which is also negative. So the reaction is spontaneous, as expected.  This is also what you have found. As you are looking for an explanation for the origin of this negative sign, it is due to the high negative value of $\Delta$$\ce{H}$. The negative value of $\Delta$$\ce{S}$ (and of $\ce{T}$$\Delta\ce{S}$) may be important ; it is not sufficient to reverse the sign of $\Delta$H when deriving the value of $\Delta\ce{G}$, through $\Delta\ce{G}$ = $\Delta\ce{H}$ - $\ce{T}\Delta\ce{S}$

Thank you for the answer. Indeed this is above the scope of what we discussed in class. So would you say the formation of NaCl is anabolic since it's combining into a larger molecule? Or do we only use the words anabolism and catabolism in reference to metabolic molecules like glucose? Thank you.

The terms "anabolic" and "catabolic" are used in biology. In chemistry we prefer to speak of broken or made up bonds. In the synthesis of NaCl, one bond is broken ($\ce{Cl2}$) which does not cost much energy. But  the bond in $\ce{NaCl}$ is much stronger, so strong that the synthesis of $\ce{NaCl}$ is rather strongly exothermic. It is so exothermic, that the energy for breaking $\ce{Cl2}$ is so small that it can nearly be neglected.

@Maurice Please do not use \ce{…} for the symbols of physical quantities. This is wrong not only semantically, but also from the formatting standpoint since they are variables and should not be upright. Also, do not invoke several math modes in a row, use one mode per expression: you pollute the code with dollar signs and make spacing all wrong. Since you are loading mhchem by using \ce{…}, consider formatting units properly with \pu{…}, too.


text:
Can plastic items of same type but different rigidity be recycled together?
Quite often we find plastic items of different rigidity which belong to same type. For instance, LDPE is used mostly for films and bags, but sometimes thick and rigid containers are also made.
Can we (after shredding) extrude together two items of same type but different rigidity? Will the rigidity of output material be uniform?

Your "rigidity" isn't really a property of substance itself.

Mechanical properties are (not only) a property of the substance. A plastic bag is so flexible because of the blowing process  that creates the film, of course. But then you don't use the same PE type for film blowing than for other uses.

Yes, certainly.
If it's all LDPE, you get a uniform LDPE afterwards.
During extrusion, large extensional deformations are applied to the melt, which will completely homogenise the material.
Of course you don't get a defined quality ("grade"), unless you have very good control over the relative proportions of different LDPE grades that go into your recycling.
I assume you know that no two LDPEs are quite the same, on a molecular level and also in the resulting mechanical properties. Although they're not nearly as diverse as HDPE or LLDPE.
And please note, film blowing (plastic bags etc.) creates a frozen-in two dimensional orientation of polymer chains in the material. The mechanical properties are totally different then.


text:
Nomenclature of C6H5CO-O-NH2
How to name the following compound using IUPAC nomenclature? I found benzyl carbamate to be very close to my structure, but couldn't find an exact compound online which matches mine.

@Maurice Why exactly do you think there would be no O-benzoil derivative of hydroxylamine?

Let's say it does exist, but is so unlikely that we are compelled to suspect a mistake.

im new to organic chemistry. @IvanNeretin why is it so unlikely?

Because of the bond between O and N.

Sort of cheating, but the CAS number for your compound is 54495‐98‐6 (added as image metadata) so you can learn more about the molecule. ChemSpider might be a good start.

@IvanNeretin  Not only is it a known compound, it is commercially available - as O-benzoylhydroxylamine   https://www.biosynth.com/p/ECA49598/54495-98-6-o-benzoyl-hydroxylamine-hydrochloride

According to the last printed version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the retained name hydroxylamine was used for the construction of the preferred name.

P-68.3.1.1.1 Hydroxylamines
The retained name ‘hydroxylamine’ is a preselected name and designates the structure $\ce{H2N-OH}$. It is a functional parent compound allowing full substitution even, as an exception, on the oxygen atom. (…)

P-68.3.1.1.1.2 Substitution on the oxygen atom of hydroxylamine
Substitution of hydroxylamine on the oxygen atom by hydrocarbyl or acyl groups is
expressed as O-substitution. (…)

$\ce{H2N-O-CO-C6H5}$
O-benzoylhydroxylamine (PIN)

However, this was changed on 3 March 2021. The PIN is now given as aminooxy(phenyl)methanone.
(I am not sure if this still contains a mistake and should actually read (aminooxy)(phenyl)methanone.)

but isnt this some sort of ester? i dont understand

Here it should be some sort of benzoate right? something like aminomethyl benzoate?

@kzmom Hmm, but hydroxylamine is not an alcohol, so it cannot be an ester, AFAIK.

@kzmom One could in principle call it azanyl benzoate, but it's rather an amide analogue made with hydroxylamine instead of ammonia. Something like one tried to make peroxyacid amide.

hmmm my teacher says it is peroxo benzamide

@kzmom Peroxo implies two oxygen atoms - I already mentioned this analogy, but it's useless in nomenclature. Only group of compounds this belongs is  O-acylhydroxylamines.


text:
Any method for measuring hydroxyl and hydroperoxyl radicals in water?
I am looking to irradiate water containing some amounts of iron and would like to measure the amount of hydroperoxyl and hydroxyl radicals produced in real time as a result of the irradiation. Is there any way to do so without very expensive equipment such as an EPR machine?

There are various colorimetric methods to measure peroxide production, and since you plan to have iron in solution, "the chromogenic $\ce{Fe+++}$-xylenol orange reaction, in which a purple complex is formed when $\ce{Fe++}$ provided in the reagent is oxidized to $\ce{Fe+++}$ by peroxides present in the sample," would seem to be ideal for your purpose, as suggested by BioAssay System. Another test uses Amplex™ Red reagent (10-acetyl-3,7-dihydroxyphenoxazine) and a peroxidase to create easily detected light -- but during irradiation, it might be confused with Cherenkov radiation.
As for hydroxyl measurement, if offset by hydronium production, then pH would not change -- but conductivity might. However, measuring such a small change in an otherwise conductive solution with dissolved salts might be tricky. You might test  pH indicators to see if there is any noticeable change, but I'd not be sanguine about that.


text:
Liquid-liquid extraction of sorbicillinol in water with ethyl acetate
My solution consisted of sorbicillinol and water, sorbicillinol being a polar substance. I wanted to isolate sorbicillinol from water using liquid-liquid extraction.
For my solvent I used ethyl acetate. Pouring the ethyl acetate in my separatory funnel resulted in a clear separation of the organic layer (being sorbicillinol) and the aqueous layer (being water and ethyl acetate).
However I don't fully understand the theory behind this. All of these substances are polar, the only difference between the substances being the density. Why did the water go to ethyl acetate. Did it have a stronger polar interaction to it?

According to this article, sorbicillinol  cannot be concentrated without causing decomposition. Thus, I assume your sorbicillinol in organic extract may undergoes reactions like Diels=Alder to give dimer or trimer. It certainly looks more stable in water.

The key point is that the solubility of organic compounds in water, even partially polar ones, often has an upper limit (exceptions include ethanol). For instance, ethyl acetate is soluble in water, but only to a limit (8.3 g/100 mL). If you surpass that limit, adding more ethyl acetate does not significantly change the amount or concentration in the aqueous phase (the water layer, assumed constant in volume).
In addition, solubility typically depends on temperature. For ethyl acetate in water it is lower as you decrease T. You can therefore chill a mixture to reduce the amount of ethyl acetate in water.
Similar principles apply to the solubility of organic compounds such as sorbicillinol in water and in ethyl acetate. In this case the amount of the organic compound may be so low or solubility so high that saturation (as in the case of ethyl acetate in water just discussed) does not occur, but all the same the organic compound partitions between the separate layers (water and ethyl acetate), and the ratio of the concentrations in each layer is approximately constant (the ratio being called the partition constant), regardless of their volumes, the important thing being that two phases form. It means that as you add more ethyl acetate, saturating the water and forming a separate phase, an increasing amount of the organic compound (sorbicillinol) ends up in the organic phase (ethyl acetate). You can add more of this solvent to increase partition away from water and thereby your yield (to a point). That's because the total amount in each phase is the product of concentration and volume in that phase.
From PubChem, the relative partition of the organic compounds in water are:
ethyl acetate XLogP3: 0.7
sorbicillinol XLogP3-AA: 1.8
These values imply that sorbicillinol is less polar (less water soluble, more octanol soluble) than ethyl acetate. Since ethyl acetate is less polar than water, this also implies that sorbicillinol will prefer to be in ethyl acetate. Note also that while it helps if the organic compound highly prefers the ethyl acetate, it is not absolutely necessary. You often wash your reaction mixture more than once to extract the product.

It would say it is rather   (water + sorbicillinol) + (ethyl acetate) ->   (water) + (ethyl acetate + sorbicillinol).
Some substances are more polar than other. Like there is not just white and black, but also the whole continuous scale of gray.
Sorbicillinol polarity is more similar to polarity of ethyl acetate than to water, so the extraction equilibrium prefers the ethyl acetate.
Sorbicillinol  (pubchem.ncbi.nlm.nih.gov):