text:
Why does allosteric binding produce a sigmoidal curve?
Allosteric binding is where the enzyme can be regulated through having ligands bind onto somewhere that is not the active site. This will then induce a conformational change on the active site, hence preventing binding and catalysis.
Often after a long pathway, the final product can act as the modulator and bind at the allosteric site of the first enzyme.
My question is why does the curve now change to look sigmoidal instead of hyperbolic?

I get that allosteric binding should affect the shape - but this is not what I expected. What I do not understand is why does the velocity decrease so sharply at low concentrations of substrate?
Allosteric binding should mean that at small amounts of substrate, there should be very little amounts of reaction. Thus, there are small amounts of product and hence very little allosteric binding. Thus the reaction should go at full speed, looking just like the diagram on the left. It is only at higher concentrations of substrate that the modulating effect would kick in.
Where is my reasoning wrong?

Are you aware that there are different types of allosteric effector, namely activators and inhibitors? See for instance https://en.wikipedia.org/wiki/Allosteric_regulation

Also, that looks like a very crude cartoon meant to emphasize the change in the shape of the response. Maybe you should examine equations for the theoretical response?

The curve on the right describes only a special case of allosterism in which the binding of the substrate to the allosteric site increases the activity of the enzyme. The velocity vs [substrate] plot is sigmoidal because when the enzyme binds substrate at the allosteric site, the active site of the enzyme works better. It is very common for people who do not understand the basic chemistry of allosterism to generalize this idea to all allosteric enzymes, but this is quite wrong. The sigmoidal plot is pretty much extrapolated from the binding of oxygen to hemoglobin, one of the first cases of allosterism described in the literature, but only a FEW allosteric enzymes behave this way.

This plot describes the dependence of the reaction velocity of phosphofructokinase on [ATP], one of its two substrates. This is very old data (Frieden et al, JBC, 1976) and this is clearly not a sigmoidal curve! ATP binding to the allosteric site decreases the enzyme velocity! It is unfortunate that most "instructors" do not bother to check the math nor the literature.

Why does allosteric binding produce a sigmoidal curve?

There is no tight link between allosteric binding and deviation from simple Michaelis-Menten kinetics.
Cooperative behavior of multi-domain proteins

What I do not understand is why does the velocity decrease so sharply at low concentrations of substrate?

The simplest model for the sigmoidal dependence of enzyme rates on substrate concentration (i.e. positive cooperativity) is a multi-subunit enzyme with multiple substrate binding sites. The enzyme has two states, both individually displaying Michaelis-Menten kinetics. However, the one with more efficient catalysis (larger $k_\mathrm{cat}$) is favored at higher substrate concentration. This is similar to hemoglobin switching to a higher affinity conformation as soon as some binding sites are occupied. This conformational change leads to a higher slope of rate vs. substrate concentration (or for hemoglobin, binding vs. ligand concentration) at the mid-point of the transition, allowing larger response of enzyme rates to substrate concentration (i.e. more sophisticated regulation). For an example, see ATCase. The enzyme has 6 active sites in different subunits, with conformational change explaining cooperativity. Complicating matters, it also has allosteric sites binding to other ligands, for additional regulation.

I get that allosteric binding should affect the shape - but this is not what I expected.

The image is misleading. For the cooperative effect in the model above, you expect binding of the substrate elsewhere (in another active site). Allosteric binding of a non-substrate ligand in an allosteric site generally does not switch the kinetics from Michaelis-Menten to non-Michaelis-Menten.
Cooperative behavior with a single active site
Glucokinase is an enzyme that shows cooperative kinetics (sigmoid rate vs substrate plot) without having multiple subunits or binding sites. This behavior is explained by slow conformational changes (in effect "memorizing" the there was a substrate recently). This means that when the number of reactions per time interval reaches a certain threshold, the enzyme will be faster because it is already in the "fast" conformational state. For a paper on possible mechanisms for this type of regulation, see doi:10.1016/j.bioorg.2011.11.001


text:
Is there any proof of law of mass action?
I was studying chemical equilibrium and my teacher taught me the law of mass action given by Guldberg and Waage. But is there any proof of this law? I did not find the proof anywhere. I searched all over the internet. The proof is not in my textbooks, not did my teacher tell me anything about the proof. It would be of great help if someone could provide the proof. I just want to know from where the law came as I do not feel satisfied just rote learning a formula and not knowing from where it came.

What "proof"? It was originally found just by weighing substrates and products, that you can do yourself. Now underlying molecular structures are known, so what, you want "proof" that atoms exist?

I just surfed through Wikipedia and there's an entire 'History' section describing how all of this was found out. Sigh.

Mithoron and Desai please don't be angry. I am just a poor student who is trying to understand chemistry. Please forgive me if I asked something wrong.

You cannot "prove" things about the natural world the way you can prove mathematical theorems. You can only say that something is consistent with every experimental result we have so far and consistent with other "laws" that have been identified.

We are not angry. We understand the eagerness. Moreover we are much more friendly than we seem. We are all just learners at different levels. We discourage effortless questioning and not the questions. Teaching new members the rules of the site is a way we encourage self-confidence, presentability and etiquette. And that's one reason we may seem a little intimidating compared to other SE members. Moreover, structuring your questions allows self-analysis and improved thought process while improving readability/comprehension.

You may also want to look through other posts on the site dealing with the topic: https://chemistry.stackexchange.com/search?q=law+of+mass+action

If it's a law. There is no proof.

@joaquintorrents  Distinction between law and rule in science is often vaque. Most such "laws" are not fundamental like conservation laws or even mathematical theorems. They cannot be proven, just can be confirmed or refuted agreement with experimental values. Only their mathematical consistency can be proven.

source: Stack Exchange

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

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text:
Balancing redox reaction of zinc and hot diluted nitric acid
I need to balance the following reaction:
$$\ce{Zn + HNO3 -> Zn(NO3)2 + NH4NO3 + H2O}$$
I assigned all oxidation numbers:
$$\ce{\overset{0}{Zn} + \overset{+1}{H}\overset{+5}{N}\overset{-2}{O_3}\longrightarrow \overset{2+}{Zn}(\overset{+5}{N}\overset{-2}{O_3})2 +\overset{-3}{N}\overset{+1}{H_4}\overset{+5}{N}\overset{-2}{O_3} +\overset{+1}{H2}\overset{-2}{O}},$$
but I'm having troubles finding the half reactions. I know
$$\ce{Zn -> Zn^{2+} + 2 e-}$$
is one half reaction, but what about the other? Since ammonium nitrate has nitrogen atoms in two different oxidation states, what do I do with them? Do I add them to see if the nitrogen is or it is not balanced?

Think a bit: none of the hydrogens or oxygens change oxidation numbers. But some (not all, obviously) of the nitrogens drop from +5 to -3 in oxidation numbers. Those are in the ammonium ions.

You just need to know that 42=18, so 4 Zn for 1 ammonium. That simple.

The oxidation state of $\ce{Zn}$ as calculated by you is incorrect. The correct oxidation state of $\ce{Zn}$ is $+2$ as it is associated with two mono negatively charged $\ce{NO3-}$ ions.
Thus the unbalanced reaction with correct oxidation states would be:
$$\ce{\overset{0}{Zn} + \overset{+1}{H}\overset{+5}{N}\overset{-2}{O_3}\longrightarrow \overset{+2}{Zn}(\overset{+5}{N}\overset{-2}{O_3})2 +\overset{-3}{N}\overset{+1}{H_4}\overset{+5}{N}\overset{-2}{O_3} +\overset{+1}{H2}\overset{-2}{O}}
$$
Now the half-cell reactions are as follows:
$$
\ce{Zn -> Zn^{2+} + 2e-}\\
\ce{NO3- + 10H+ + 8e- -> NH4+ + 3H2O}\\
$$
Thus the balanced iconic reaction would be:
$$\ce{4Zn + NO3- + 10H+ -> 4Zn^{2+} + NH4+ + 3H2O}\\\\$$
Now adding nine $\ce{NO3-}$ spectator ions to the ionic reaction gives us the standard molecular reaction.
$$\ce{4Zn + 10HNO3 -> 4Zn(NO3)2 + NH4NO3 + 3H2O}\\\\$$
What did you learn from this?
Always try to write the half-cell reaction in ionic form and only involve the ions which undergo oxidation state change. Don't bother yourself with the spectator ions at first. After you balance the equation in an ionic  form adjust it to the molecular form by adding relevantly needed ions. Identify the oxidizing and the reducing species is the key.

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text:
How to generate sulphuric acid vapour to test an analyzer?
We have a UV gas analyzer we want to test to see if it reads gas phase $\ce{H2SO4}$. How can we safely produce less than $\pu{5 ppm}~\ce{H2SO4}$ as a spike test to see if the analyzer sees the spectra?

Perhaps heat anhydrous calcium sulfate mixed with silica. That generates sulfur trioxide plus calcium silicate. The sulfur trioxide plus water vapor gives the sulfuric acid. Maybe just use fresh Drierite (TM) and silica gel (the chromatographers always have it).

This PDF from psu.edu  contains graphical data about H2SO4 partial pressure at various temperatures


text:
onset of oxidation for cyclic voltammetry
I have plotted a cyclic voltagram for my different polymer blends and I am trying to work out the onset of oxidation. picture below is my graph

from this website: https://www.ossila.com/pages/cyclic-voltammetry
There is a diagram which shows how the straight line should be placed so that it intersects the horizontal line, and thus the onset of oxidation can be found

However I am being told to draw my lines like this:

The procedure I am being told to do, doesnt make sense to me, I dont think the data from that would be useful. I would only get values for the Eonset that are very similar if I were to do it the way I'm being told. From my data, theres a clear difference in oxidation slope, so wouldn't it make sense to plot the diagonal line against that slope, rather than a small section near the beginning? From my 5 cyclic voltagrams, it would make much more sense if I were to do it as the website has told me to. There are more pictures on google which do it that way too

The answer by Farooq is detailed and clear. But as a further help, as I suppose you are on something practical, I leave a comment. Basically all CVs are typical for a solid film and quite irreversible. I would say that they show the same behaviour. If your aim is to link ox pot to the % shown, that is a hard task bcz the plot does not show one (or a simple one). If they would be my data, the polymer is insensitive to the parameter changed. Forced to analyse the data in depth, than you are right, take the steeper slopes. But to which intercept? This is a real experiment, not a book....

In other words, the first slope can also be the onset of real polymer oxidation and not (just) baseline current. Run also a blank for that purpose...

Also, don't expect to much change in EC because they are blends! Polymers are molecules, their oxidation potential will be basically unchanged EXCEPT for (in principle) conformational change and or counterions diffusion (the latter affecting the shape rather than onset, see it as a source of overpotential needed and a kinetic brake). In a paper of mine, those data would be treated in a vague way (no effect, as expected, and the like......).

@Alchimista , thank you for your comments, im an undergrad student and havent been taught much on CV, so my knowledge is more general. I am forced to use cv data as I havent been given anything else (cannot enter labs due to pandemic). Could I comment on the slope changes for the blends? where 20% provides increased charge mobility/ion diffusion and 50% there is sort of a degradation of the mobility?

well that right the kind of speculation one can do. But which one? I should know all the details and other results....But you got what kind of discussion we are about....

Im not quite sure what info I can share with you, however I am mixing a polymer with PEO (0%,10%,20%,30%,50%) and its scanned at 50mv/s. Im discussing my results, however I feel like my work is not detailed due to lack of data, so I'm out here just trying to put some words together

In short,:from those CVs only, you can't say anything except describing in words what it is already in the plots

Please don't apply the reversible CV diagram on something which is clearly a non-classical cyclic voltammogram. When you experimentally don't get the shape as predicted by the theory, it means it that particular theoretical model does not apply in your case.
Look at three different types of voltammograms in this picture. Does your CV look like a reversible one? If not, then you were comparing an apple with an orange.

There is a very nice article which is relevant to you case:
Practical Aspects of Cyclic Voltammetry: How to Estimate Reduction Potentials When Irreversibility Prevails,  Journal of The Electrochemical Society, 2019,  166(5):H3175-H3187
This figure is relevant:

Examples of cyclic voltammograms showing chemically reversible and irreversible oxidation with designated potentials: E a = anodic potential, E c = cathodic potential, E (i) = inflection-point potential, E (p/2) = half-peak potential, E (e) = edge potential.

Thank you for your answer. However I'm being told that I should treat my graphs as reversibel as oxidation and reduction is observed, even though it doesnt look like the reversible picture youve put.

Whoever is telling you that must provide the "proof" of reversibility. It does not look reversible at all.


text:
How to I neutralize the smell of (VERY) spoiled milk
Let's say I have a cup full of milk that had been at room temperature for months (apparently my son was running an experiment...). The smell started seeping out of the makeshift lid, and it's horrible. I have no background in chemistry but I imagine there might be a way to neutralize the smell before getting rid of the contents.
I'm worried if I just flush it down the pipes it might linger and make the whole house smell. Also, ideally I would like to salvage the cup because it's a very nice handmade glassware (he was running a fancy experiment).
Again, this has been there for months so I hope there isn't any dangerous bacteria still around, and any advice to get rid of the smell and clean the thing would be enormously appreciated.

Ew just flush it already. Nothing's going to happen. And about the glassware just wash it many times with hot water and a good dish soap. If you can't take the smell while disposing it spray perfume or something while covering your nose. Hopefully there's nothing nasty growing in it which BioSE could be aware of. There is bacteria in it for sure but don't worry you won't turn into a StanLee character. Moreover, avoid direct skin contact. The smell could stay on skin plus the bacteria are unpredictable.

That was a nice spook on the profile @Desai :) And lol "nothing nasty in it which biose could be aware of"

@Buraian You mean the "I know who you are?"

I was the culprit who upvoted this question because I thought it was funny and I wanted more members to read it which becomes a high possibility when upvoted. Little did I know...

The disgusting smell is probably butyric acid. Try adding a base: For example, washing soda (sodium carbonate), or perhaps drain cleaner (sodium hydroxide, be careful with this one, this is very corrosive to people/eyes), or even ammonia. This should turn the butyric acid into butyrate ion, which should be much less disagreeable. Then flush down the drain.

Even baking soda (sodium bicarbonate) will work, though washing soda is better.

I would be careful not to mix cleaners, e.g.  ammonia and bleach. Just in case: https://www.doh.wa.gov/youandyourfamily/healthyhome/contaminants/bleachmixingdangers


text:
Balancing reaction between copper and nitric acid given the ratio of formed nitrogen oxides
Problem

$\ce{Cu}$ reacts with $\ce{HNO3}$ according to the equation
$$\ce{Cu + HNO3 -> Cu(NO3)2 + NO + NO2 + H2O}$$
If $\ce{NO}$ and $\ce{NO2}$ are formed in 2:3 ratio, what is coefficient of $\ce{Cu}$ when equation is balanced with simplest whole numbers?

Answer

 9

Question
I am getting the coefficient of $\ce{Cu}$ as $2,$ which contradicts the answer:
$$\ce{2 Cu + 9 HNO3 + 5 H+ -> 2 Cu(NO3)2 + 2 NO + 3 NO2 + 7 H2O}$$

What keeps you from providing your solution ? If you do not show what you did, others cannot tell what you did right or wrong.

For eventual writing and formatting of chemical or mathematical formulas or equations, see how to use  MathJax with  mhchem

n * 2 = 2m * 3 + 3m * 1 = m * 9 //
n=9 versus m=2 //
9 Cu versus 4 NO + 6 NO2 // 18e versus 12e + 6e

Can you please edit the question to include the balanced equation which lead you to believe that the answer is 6. This insight might help us to find your mistake.

Who would have thought that hannah montana left their singing career and is now doing chemistr

Your equation is not balanced in charge.

What do you know about oxidation states?

The final arbiter of formal correctness of chemical reaction enumeration  are laws of mass and charge conservation. If total counts of charge and atoms of every element are not the same on each side, the equation is wrong. Reaction enumeration by following these laws may be troublesome, as general solution leads to resolving a set of linear equations. That can be useful for educational purposes, or rather to implant terror and despair in eyes of young chemistry students.
Chemists do not go that way. They usually use various heuristics to speed up the process, enumerating equations in several steps. For redox reactions, the heuristic is inventory of exchanged electrons, which is tracked by formal oxidation numbers (ON). These numbers are rather just arbitrary convention, but extremely useful.
For enumeration of your reaction ( with expressed ONs of relevant atoms):
$$\ce{Cu^{0} + HN^{+V}O3 -> Cu^{+II}(N^{+V}O3)2 + N^{+II}O + N^{+IV}O2 + H2O}$$
the first step is matching components undergoing redox change.

Copper changes ON from 0 to +II.i.e. by 2.
Nitrogen changes ON from +V to +IV ( $\ce{NO2}$), or from +V to +II($\ce{NO}$), i.e. by 1 resp. 3.
There is given $\ce{NO/NO2}$ ratio 2:3, so there would be as products $\ce{2 NO + 3 NO2}$, or their multiples.
$\ce{2 NO}$ exchange $2 \cdot 3 = 6$ electrons.
$\ce{3 NO2}$ exchange $3 \cdot 1 = 3$ electrons.

For simplicity, we can take formally $\ce{2 NO + 3 NO2}$ as a single entity, created from $\ce{HNO3}$ by exchanging totally 9 electrons.
Now comes the "magical" formula $9 \cdot 2 = 2 \cdot 9$.
9 Cu exchanging 2 electrons each gives 18 electrons totally.
2 ( $\ce{2 NO + 3 NO2}$  ) exchanging 9 electrons each has accepted 18 electrons.
So the intermediate result is:
$$\ce{9 Cu + HNO3 -> 9 Cu(NO3)2 + 4 NO + 6 NO2 + H2O}$$
This is enough for the question answer, but we have to finish the equation. It may sometimes happen some coefficients would get fractional, so all would get multiplied to have all integer.
We need supply enough nitrogen for the salt and oxides: $2\cdot 9 + 4 + 6 = 28$
$$\ce{9 Cu + 28 HNO3 -> 9 Cu(NO3)2 + 4 NO + 6 NO2 + H2O}$$
We need to create water from hydrogens:
$$\ce{9 Cu + 28 HNO3 -> 9 Cu(NO3)2 + 4 NO + 6 NO2 + 14 H2O}$$
And we are done. We can formally check the item conservation:
$\ce{Cu}$: $9=9$
$\ce{H}$: $28=28$
$\ce{N}$: $28=28$
$\ce{O}$: $3\cdot 28=84=54+4+12+14=84$
charge: $0=0$
For a quick check,we may just check items we were not enumerating before, like just oxygen and charge.

Thanks for detailed answer, I got where I went wrong I have $\ce{5H+}$ on LHS I.e +5e charge, but on RHS charge is 0, right !

Always include in questions the procedure leading to the questioned result. Without it, it is hard to help.

This is an interesting reaction for without specifying the ratio of NO to NO2 there are essentially an infinite number of solutions. So just for interest if a 1:1 ratio is needed then the coefficients are  2 , 6 , 2 , 1,  1,  3, in the order above, a ratio of 9:1 NO:NO2 the coefficients are 14 , 38, 14, 9, 1, 19 and 1:9 are  6, 22 , 6, 1 , 9 , 11 so more nitric acid relative to Cu favours NO2 production.

As @Poutnik has already very well described the intuitive way to solve such problem so I will try to describe a standard method for such problems.
First analyse the reaction and find out the species which are being oxidised and reduced by finding the change in their oxidation state. Initially ignore all other ions which are not included in the redox reaction.
In this reaction copper is being oxidised from its $0$ state to $+2$ state. Whereas the nitrate ion is being reduced to it's $+4$ and $+2$ states from its $+5$ state.
Now try to form half cell reactions for the oxidation and reduction parts of the reaction.
In this example oxidation of copper and reduction of nitrate can be described as follows:
$$
\ce{Cu -> Cu^{2+} + 2e-}\\
\ce{3(NO3- + e^-) + 2(NO3- + 3e^-) + 14H+ -> 3NO2 + 2NO + 7H2O}
$$
Note, I have used coefficients $3$ and $2$ in the reduction half cell as the question clearly states that $\ce{NO2}$ and $\ce{NO}$ are produced in the ratio of $3:2$.
Now add these half reactions by multiplying them with coefficients in such a way that there is no net electron gain or deficit.
In this case the coefficients or as @Poutnik says magic numbers/components of magic equation are $9$ and $2$.
Multiplying the half reactions with "magic numbers" and than adding them gives the following balanced ionic reaction:
$$\ce{9Cu + 10NO3- + 28H+ -> 9Cu^{2+} + 6NO2 + 4NO + 14H2O}$$
Now to get the molecular reaction we can add $\ce{18NO3-}$ ions to the reaction to form $\ce{HNO3}$ and $\ce{Cu(NO3)2}$ respectively.
This gives us the final equation:
$$\ce{9Cu + 28HNO3 -> 9Cu(NO3)2 + 6NO2 + 4NO + 14H2O}$$


text:
Ratio of carbon monoxide and carbon dioxide when carbon and oxygen are made to react

$\pu{12 g}$ of $\ce{C}$ reacts with $\pu{64 g}$ of $\ce{O2}$ to give a mixture of $\ce{CO}$ and $\ce{CO2}$. Find amount of $\ce{CO}$ and $\ce{CO2}$ at the end of reaction.

What I tried was to let x moles of carbon react to give $\ce{CO2}$ and y moles react to give $\ce{CO}$.
$$\ce{ \underset{x}{C} + \underset{x}{O2} -> \underset{x}{CO2}}$$
$$\ce{\underset{y}{2C}} + \underset{y/2}{\ce{O2}}\ce{ -> \underset{y}{2CO}}$$
Solving for $x + y = 1$ and $x + 0.5y = 2$ , we get $x=3$ and $y= -2$ which is meaningless.
I followed the method used in this answer.
Can anyone tell me where I went wrong?

Comments are not for extended discussion; this conversation has been moved to chat.

This is a bad question. The answer is indeterminate since the problem doesn't define the conditions for completion.  Some mixture of CO, CO2 and O2 would seem likely. The only way to have a definite "solution" for the problem would be to make an explicit assumption that all the carbon goes to CO2 since O2 is in excess.

The answer that you got is meaningless because it is wrong.
You missed one reaction here. $\require{cancel}\ce{CO}$ can further react with $\ce{O2}$ to give $\ce{CO2}$
$$\ce{2CO + O2 -> 2CO2}$$
Now we have three reactions to consider,
$$\ce{2C + O2 -> 2CO} \tag{1}$$
$$\ce{2CO + O2 -> 2CO2}\tag{2}$$
$$\ce{C + O2 -> CO2} \tag{3} $$
(1) and (2) can be combined here to give:
$$\ce{2C + O2 + \cancel{2\ce{CO}} + O2 -> \cancel{\ce{2CO}} + 2CO2}$$
Which gives:
$$\ce{C +O2 ->CO2} \tag{3} $$
These reactions progress in the following manner, (1) + (2) = (3).
Now writing the three equations and using the above statement(which means we can find the value for just (1) first), we get that $\pu{1 mol}$ of $\ce{C}$ reacts with $\pu{0.5 mol}$ of $\ce{O2}$ to give $\pu{1 mol}$ of $\ce{CO}$. We still have $\pu{1.5 mol}$ of $\ce{O2}$ left, so the reaction goes further.
Now onto (2), $\pu{1 mol}$ of $\ce{CO}$ reacts with $\pu{0.5 mol}$ of $\ce{O2}$ to give $\pu{1 mol}$ of $\ce{CO2}$ and we still have $\pu{1 mol}$ of $\ce{O2}$ left which is in excess.
You can use this same method for this question as well


text:
Find the structural compound from carbon and proton NMR spectrum

The proton and carbon NMR spectrum datas for compound W (201 g/mol) are
shown below. Based on Diagram 1 and Diagram 2, draw the structural formula of compound W according to the type of hydrogen and carbon.

I have extracted the information from the NMR spectrum into a table.

Hi, welcome to chemistry.SE! Could you please add some more information about what you have done so far to solve the question? Maybe mention what are your guesses about the structure of the compound.

Not enough information to predict the structure. Need to have at least integration of proton NMR. Only one thing is for sure. It is a 1,4-arene.

@ShoubhikRMaiti I'm so sorry, that's all the information I have:(  I really don't know how to predict the structure based on the information given. There is no integration ;(

What table are you referring to? And I agree with the answers that it is a horrendously bad assignment. Not only it's nearly impossible to draw a meaningful conclusion, but it also contained several typos and illiterate formatting.

Interesting Problem and quite a nailbiter.
I will try to sort some of my ideas, even thought this won't be a full answer.

This looks like a question from an exam. Did you have any reactions during study with hydrochloric acid and propanoyl chloride (acid chloride of propionic acid). So maybe look into different reactions of acid chlorides.
As @MethewMahindaratne mentioned, it is hard without integration but for carbon we dont need integration. We have 7 Carbons, some hydrogen, pretty sure some oxygen and for sure some nitrogen. Why? Because only nitrogen is able to produce odd numbered masses. If we have CH compound, there will be always an even number of hydrogens and carbon has a mass of 12  so overall its even. When we create double bonds or add oxygen, the amount of hydrogen is decreased by 2 so it remains uneven. Only with nitrogen, we can add 1 hydrogen per nitrogen, therefore we have 1 or 3 or 5 etc. nitrogens in it.
The proton g is also quite specific. This high shift in proton NMR usally comes from a carboxylic group. So we have $C_7O_2H_xN_y$.The idea of the carboxylic group is also supported by the C atom at 175 which is a really high shift.
With a molecular mass of 201 g/mol this leaves quite some but not endless options. We can take of 7 x 12 (for the seven carbons) and 2 x 16 (for oxygen carboxylic group) and 1 (for hydrogen carboxylic group) leaving us with a mass of 84 split along oxygen, hydrogen and nitrogen.
This is a lot. On the bottom line it is mentioned, that W consists of certain things. halogenides which usally make compounds heavy are not included there.
Appart from the carboxylic group we are left with 6 carbons which could give the idea of a benzene. But I think it is not benzene because we could only attach $NH_2$ or $OH$ groups onto it and even with 5 $NH_2$ groups we get not enought mass. In addition, aromatic C atoms have high shifts usally 120-150. But we only have 3 others (next to our carbon from the carboxylic group) so not enought for a six membered ring.
But also not enought for just one alkene which I would prefer. Alkene shift ist around 110-140 and fits well.
Now that is all I could provide without crazy speculations.

I can throw in some suggestions but they are realy wild and you need to look in deeper.

The big peek a looks a bit like some alkohols I worked with so maybe a hydroxylic group?
Maybe Carbon Atom u contains 2 or even 3 peaks? (the line looks thicker than the others)= that would contribute to the idea of a substituted benzene ring and provide more carbons so we could add other groups.
Could $H_f$ be two dupletts instead of a quartett so that could come from a double bond?
With integrals you could look at the singuletts and determine wheather it is an amine (singulett with an integrall of 2)
Maybe my whole carbon acid idea is wrong and it is an ester or something.
If we are allowed to include hallogens, then it is pretty sure a benzene with a carboxylic group maybe hydroxylic and a hallogen atom.
If you have more information, maybe you can use the sdbs data base https://sdbs.db.aist.go.jp
I was not able to find an entry that fits.

Hope this brings the project forward.I am very happy to hear, what the compound was when you get a result.

Wow. This is a very bad question (the exam/assignment, not the OP's). There is no way that you can categorically determine the structure of this based only on the information given. This would have to be open-book providing access to chemical shift tables in the very least. Peaks are poorly labelled and what's more, I believe the molecular mass is incorrect. Anyway, let's break it down into a thought process and see what we come up with based on the information given.
What we know from first inspection: these are simulated spectra - no solvent is defined (this makes it very difficult to make predictions on labile peaks). No field is defined, so coupling constants cannot be used. Either we have to use chemical shift mapping only, or are expected to make some guestimates based on peak heights and linewidths to determine numbers of C/H. But we assume that the information provided can lead to a definitive structure.
From the 13C spectrum-
we have 7 labelled sites - but is the peak labelled u actually two closely adjacent peaks?? It's a lot thicker than the other lines. I think so, and the gut feeling looking at the aromatic region of the 1H spectrum suggest a 1,4-disubstituted benzene. So I'm going to say 8 sites. The intensities (one of the joys of simulated 13C spectra is that peak intensities are quantitative) show that t and s each contain 2 equivalent carbons. So, we get 10x carbons.
From the 1H spectrum-
we have a minimum of 7 sites - but again we make an assumption that the peaks at f arise from two signals - so probably 8 sites. Let's break the signals into 2 groups: those that are broad and those that are narrow. For the narrow peaks, we can estimate integrals to be 3a:1b:1e:2f:2f (if we say f is actually 2 lots of signals). For the broad peaks, we get 1c:2d:1g. So, all up: 3:1:1:2:1:2:2:1 which means 13x 1H
Now, let's work out a molecular formula based on C10H13XnYm... yuck. I'd like to think that for a problem question like this, any inclusion of a non-standard nucleus would be made abundantly clear. You can work out a range of formulae based on this mass by hand, but it's quicker using a tool like https://www.chemcalc.org/mf-finder. Whack in 201, limit it to C10H13, and you come up with junk. Add some flexibility around numbers of C and H, and you come up with mostly non-integer DBE, so alarm bells are ringing. So, let's leave the molecular mass and look closer at the NMR fragments.
From the 1H spectrum: peaks at c, d and g are labile, and in slow exchange, so probably not connected to 13C. Chemical shift information is largely useless without knowing the solvent for labile peaks. For instance, an -OH peak comes at about 1.6 in chloroform, but around 3.5 in dmso. For -NH2, the differences are even larger.
So, now let's pair up the carbons and protons to see what fragments we might have:
a/p     -CH3. Could be just about anything: Ar-CH3, C(O)-CH3 are key possibilities
b/q     -CH. Doublet.... q is connect to r
e/r     -CH. Doublet.... r is connected to q
f/s     -CH. Looking awfully like a 1,4-Ar
f/t     -CH. Looking awfully like a 1,4-Ar
u and v are probably quaternary 13C-1H
c, d and g are probably labile X-H
g could be just about anything, but in the absence of any other information I'm going with something simple like a carboxylic acid. That would tie in with the chemical shift of v.
the key map pairs are b/q and e/r, as they have quite distinct 13C/1H shifts with (probably) vicinal coupling. It looks small enough to be vicinal. So, this to me looks very like a 1,2-diol. But the broad peaks in the 1H spectrum suggest a ratio of 1:2, so perhaps a 1,2-olamine.
So, I think we've got the following groups:

How do we put them all together. Well, honestly, if you've got no other information to go on, then you could put them together any way you like and it would be close enough. I think it's too close to know whether the olamine is Ar-C(OH) or Ar-C(NH2). But we can probably guess that we have Ar-CH3 rather than Ar-COOH because the quaternary peak would shifted upfield if it were a benzoic acid (due to increased elecrton density on that quaternary).
So, armed with that, I plugged in the following molecule into ChemDraw and got it to spit out a 13C prediction:

and I got:

Well, at least I know what prediction software your examiner uses. Identical to your question. The 1H prediction is not identical, giving the following in dmso:

You'll notice the labile peaks are indeed all over the place. But as 1H chemical shifts are not good guides for accurately determining structure alone, I am happy to ignore these. Structure determination software uses 13C chemical shifts and connectivities to determine structure, and rarely even considers 1H chemical shifts. So we have a molecular formula of C10H13NO3 with a mass of 195. I believe this is an error in the paper, although will be very interested to see another answer that can map the 13C shifts and get a mass of 201.
Hope this is helpful.


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Is it the solution energy or the reduction potentials (or both) that drives a galvanic cell?
I have received conflicting explanations for why the redox reaction in a galvanic cell occurs.
Explanation 1:

Some say that it occurs because the anodic metal dissolves more easily than the cathodic metal, and as such, the anodic metal will dissolve more quickly, becoming a positive cation that releases electrons (oxidizes). At the same time, in the cathodic half-cell, the positive cations in the solution will solidify on the surface of the cathode, due to rather wanting to be solid than an aqueous cation. However, these cations joining the cathode will add more positive charge than negative charge, and as such, the cathode will be positively charged. This will create a voltage between the electrons, and attract the electrons from the negatively charged anode, producing a current.
This means that the energy produced by a galvanic cell happens the system has less enthalpy when the anodic metal finds itself in a solution complex rather than a crystalline complex AND the cathodic metal finds itself in a crystalline complex rather than a solution complex.

Explanation 2:

Some say it occurs because of the reduction potential difference. The enthalpy change of an oxidation is always positive, and the enthalpy change of a reduction is always negative. If the cathodic ion's reduction has negative enough enthalpy change to make up for the positive enthalpy change of the anode's oxidation, then the reaction will happen. If the anode oxidized and the cathodic ion reduces, the resulting system will in total be of less enthalpy. Therefore, the redox reaction happens.

So, both explanations involve the fact that the energy of the system is lower if they react in the respective ways, but it has different explanations for exactly which process it is that involves the ultimately negative enthalpy change. Is it the solvation of the anode + the solidification of the cathode  ion that produces the negative enthalpy change (and therefore the energy to be used), or is it the oxidation of the anode + the reduction of the cathode ion that produces the negative enthalpy change? Or is it a combination of both?
Looking at types of corrosion, there can happen a redox-reaction without an electrolyte (dry corrosion), which means solvation isn't necessary for redox in general, but perhaps is needed for a redox reaction to happen when the reactants aren't in physical contact.

Electrode metals can be completely inert and system can still form galvanic cell. See the vanadium redox battery, used for industrial scale energy storage, based on Conductor|V^3+/V^2+||VO2^+/VO^2+|conductor.

@Poutnik Not sure why the trait of inertness is relevant to your comment, but given how all the substances in your cell reactions are ions, that would mean they're all dissolved from the get-go. That means that it's not the solvation energy causing the energy generation in that cell, correct? Rather just a redox reaction? Is this proof that the second explanation is correct, or an example of how it can be correct?

I reacted because  quoted "...'Some say that it occurs because the anodic metal dissolves..... " // Solvation energy participates on ion chemical potential, therefore on Gibbs energy of redox reaction and therefore on redox potential. by other words, dfiferences in solvation energies of ox and red forms participate on the redox potential.

@Poutnik I'm sorry but I don't quite get the formulation in your second comment (and possibly first). Could you reword it? Potentially make an answer?

You may imagine the galvanic cell as a water dam with a power plant. The redox potentials of electrode systems define the height of the water drop. The solution or solid component energy defines the multiple of that height drop and amount of water in the dam, i.e. equivalent of total energy.

@Poutnik Ah I think I understand. The solvation energy is a component of the reduction potential, along with things like differences in ionization energies, stability of electron orbitals, etc?

Yes. because it is closely linked to the well known equation from electronics $\mathrm{d}E=U \cdot \mathrm{d}q$


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How does electrolyte impurities affect the voltage in a voltaic cell?
I'm currently doing my FYP assignment on electrochemistry (predominantly on voltaic cells). As I was conducting my research I managed to come off with this topic as stated in the title, however, I am not too keen with the term electrolyte impurities. So, my question is what are electrolyte impurities? Can anyone give a brief explanation with an example? And how does the presence of this impurities affect the voltage in a voltaic cell?

How can we answer this question? Every impurities can interfere whit everything. The words probably mean impurities that can be in the electrolyte for EC (so they can be with the salt or with the solvent) or some electrolyte in the measured devices etc.

Yes! that's what I meant, sorry if I word it wrongly. What would be the impurities in an electrolyte for EC? An example would be great.


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Why is there no exothermic crystallization peak visible during cooling of PLA and no endothermic melting peak visible during a second heating of PLA?
I am doing lab research on DSC measurement (Heat-cool-heat) for understanding melting characteristics of polymers (PLA, PA6 and PE). During cooling i saw exothermic peak and during second heating i saw endothermic peak on PE and PA6. I don't understand why there is no crystallization peak during cooling and no melting peak during second heating on PLA. I have posted DSC measurement graph of PLA done by DSC Q-series. The figure is shown in exothermic up formation and crystallisation is exothermic process and melting is endothermic process. Any guidance will be really helpful.

Evidently some structure is not reforming during the cooling step. Perhaps it's not recrystallizing?


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Can PTFE be dissolved by aqua regia?
At the moment I am cleaning crucibels in aqua regia. I was wondering, whether I could stir it with a standard PTFE stirring bar or not?
I am not sure if it is needed or if the normal flactuation in the liquid is enought but usally stirring makes things better.
There is no doublebond in it to be attacked by Cl-radicals and C-F is one of the strongest bonds so I could not think of any reaction that could occur. But maybe there is some crazy side reaction I am not aware of.

There is no problem. Stir away!

See for instance https://www.calpaclab.com/teflon-ptfe-compatibility/

@Buck this is Nitric/sulfuric not nitric/hydrochloric acid

I know but see one of the answers which provides a general reply to your question. PTFE is resistant to a broad range of acids.

You can skip stirring unless doing a kind of cleaning race.

While I generally agree that there is an answer in the linked question that also answers this question, it can easily be missed when only looking at the question an top answer. (I know; who wouldn't read the whole thing?!) So maybe it would be beneficial to quote the relevant portions here and link to the original for context. (Or write a completely new answer.)


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Intermolecular interaction Van Der Waals
In the empirical demonstration of the Van der Waals equation at Khan Academy, they made a demonstration of the real volume, for example, it was said that it has to be larger than the ideal volume, since the size of the molecules is not disregarded, hence it would be:
$$ V_{\text{ideal}} = V_{\text{real}} + nb $$
Where, $ b $ was said to be related to the size of the molecule.
But why don't I also include intermolecular interaction? Depending on it, the volume may even be smaller.

How would interatomic interactions affect the volume? Can you please elaborate on how you think interatomic interactions affect the volume?

Interatomic interactions will affect the pressure and not volume.

@NisargBhavsar Intermolecular interactions can cause attraction or repulsion, changing the volume, right?

Atomic interactions can't change the volume of a gas. Just assume if they did decrease the volume than wouldn't there be vacuum spaces inside a gas.

If you see the full van der Waals equation, the parameter a is associated with the correction for intermolecular attraction.

@NisargBhavsar I disagree. In an attraction for example, there would still be vacuum spaces, they would just be a little bit smaller

You are confusing vacuum with interatomic void.

@NisargBhavsar what would be the interatomic void?

Let's say it is mobile, a piston, the kinetic energy (with the addition of heat, say) overcomes the interatomic attraction, and the volume increases. If you increase the attraction while increasing the kinetic energy, which is bizarre, you would even have that he said

@NisargBhavsar To say that interaction does not change volume is absurd. If it were in this logic, the compressibility factor Z would always be greater than 1, and guess what, even if the real gas does not have a negligible volume, the compressibility factor is often less than 1, due to interactions

Compressibility and Volume are two different quantities. Compressibility does depend on interatomic interactions. But that doesn't mean volume depends on interatomic interactions.

@NisargBhavsar z is real volume/ideal volume

*Volume ratio at the same pressure and to assure that you will have to include $a$. $Z= \frac{(P+ \frac{an}{V^2}) (V-nb) }{nRT}$. And I just noticed your equation you mentioned is also incorrect.

$V_{\text{ideal}} = V_{\text{real}} - nb$ is correct. Refer this question

Let us continue this discussion in chat.

I think you got something backward: the volume V is always "the real volume", no matter what gas you have. But to obtain an expression that looks like the ideal-gas you can make the substitution $V_{ideal}=V_{real}-nb$.

Intermolecular attraction can reduce the volume, in the same way that increasing the pressure can lower the volume (Boyle's law).
One way to see the effect is to take the derivative of the volume, as given by the van der Waals (vdW) equation, with respect to the attractive parameter $a.$
The vdW equation is
$$\left(p+\frac{a}{\bar{V}^2}\right)(\bar{V}-b)=RT\tag{1}$$
Ignore for simplicity the excluded volume term $b,$ so that
$$p\bar{V}=RT-\frac{a}{\bar{V}}\tag{2}$$
Then the derivative of the volume wrt $a$ is
$$\left(\frac{\partial V}{\partial a} \right)_{p,T} = \frac{\bar{V}}{a-p\bar{V}^2}=-(RT)^{-1}\tag{3}$$
In essence what this result says is that the volume decreases as you increase the parameter $a.$


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How do I comment on the Gibbs energy values standard and reaction in different situations?
I have the following graphs:

Taking into account the following graphs, how do I comment on the Gibbs energy values
standard and Gibbs energy of reaction in different situations (a, b and c), and how do I relate the values of the latter to the spontaneity of the reaction? I'm really lost here.

The graph is a little bit confused, because there exist different sorts of Gibbs energy. i
It can be the Gibbs energy of the rectant. It can be the Gibbs energy of the products. And it can also be the total of these two Gibbs energies.
First let's discuss the Gibbs energy of the reactant. In the beginning of the reaction, the Gibbs energy of the reactant is high. But when the reactants are consumed, their Gibbs energy decrease, because Gibbs energy is not like enthalpy which does not change if the concentration changes. No ! Gibbs energy of a substance decreases at low concentrations. So the curve of the Gibbs energy of the reactant decreases all the time, during the reaction. The value of the final point is not important.
Now let's discuss about the Gibbs energy of the product. At the beginning of the reaction, there is practically no product yet. The Gibbs energy of the product is very low. As the reaction goes ahead, the amount of the product increases. So the curve of its Gibbs energy increases. The final point of this increasing curve is rather important. It may be higher or lower the initial value of the Gibbs energy of the reactant. If it is lower, it means that the reaction Reactant - Products is spontaneous. If it is higher, It means that the reaction Reactant - Product is not spontaneous.
But what is most important is the sum of these two curves. Because the sum of the Gibbs energies of reactants + the Gibbs energy of the product, well , this sum has a minimum somewhere. And the position of the minimum gives the values of the equilibrium constant of the reaction. It is $b$ on your picture.

Here is a labeled and annotated version of your graph (from OpenStax Chemistry).


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What (non toxic) light sensitive materials do not burn away at high heat?
Specifically related to photoceramics, I am wondering what photosensitive materials exist that can be bonded to glass at high heat in a kiln (a microwave kiln to be more exact).
By non-toxic I mean a material that does no give off hazardous fumes, so silver, cyanotype and dichromates would be out of the question as far as I am aware.
The process I had in mind is, coat on glass, expose image via contact print using UV, then kiln fire to make the image permanent with the glass.
Edit:
Essentially im after a non-toxic light sensitive material that can survive/is stable when it is bonded to glass. This could be at a vitreous enamel temperature, or a low melting point frit temperature.

There are a number of types of photosensitive glass which change in appearance and chemical reactivity upon exposure, and of photochromic glass, which undergoes reversible change in color with radiation.
If you are using this as a surface applique on ceramic, one key factor would be having thermal expansion coefficient compatible with the ceramic, or the object will crack on cooling. Consult the literature from a manufacturer, such as Schott or AZO Materials.
There are also techniques for making a photosensitive layer on ceramics, such as Pyrofoto, that do not involve applying glass. A small Pyrofoto kit is ~US$20.
BTW, you might find that microwave oven kiln adapters do not reach high enough temperature to be useful in this work -- literature states a limit of ~900°C, cone 010.
You might be better served with alternate technologies.

Stencil: Lazertran, for example, makes decals that can be made in an ink-jet or laser printer. Of course, that requires inks fusible onto ceramic. JETEC offers inks that adhere to ceramics and are rated to 1,000 °C.
Direct Photoetching: A laser can blast off a surface layer from coated ceramic. More powerful lasers can change the surface of the ceramic, so no kiln is needed to fuse the artwork.
Photochemical machining: Use photoresist to define areas of a ceramic to be retained or to be etched away by chemicals or by an air-abrasive system. This is used to produce micromechanical machines (MEM) and integrated circuits.

BTW, to see some astounding art in glass, often using etching to remove various colored layers of glass, se the works of Preston Singletary. In addition to hot glasswork, he uses various subtractive techniques to produce some spectacular pieces.

A photosensitive layer on ceramics is what im after. Pyrofoto is a dichromate (dichromates are now banned in europe), Are there other layerable types? Or am I stuck with the pyrofoto option?

You may be better served by alternate method, which I'm adding to the above answer

Old method (and quite expensive at present) would be to apply silver gelatin ("Liquid Light") to unfired porcelain, enlarge or contact print onto the silver gelatin, then replacement tone the silver image developed out with iridium chloride. Iridium will replace virtually all of the silver. Refer to pre-1990s 20th century cemetery portraits on headstones, typically on a bronze plate with a hinged cover. This commemorative technique was quite popular in the US among the Eastern European immigrants.
Modern technique is ceramic printing. Google / youtube search that phrase.


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Splitting of atomic orbitals due to Stark effect
I am performing graduate level research on ZEKE Photoelectron Spectroscopy and central to this form of spectroscopy is Rydberg states and the Stark effect. The Stark effect is caused by an electric field polarizing the species, thereby inducing a non-zero dipole moment. This causes degenerate states to split. However, say for the hydrogen atom's $n = 2$ state, there is 4-fold degeneracy: $\mathrm{2s}$, $\mathrm{2p}_x$, $\mathrm{2p}_y$, and $\mathrm{2p}_z$. When an external electric field is applied to the system, two spectral lines, each red and blue-shifted, are observed for the $\mathrm{2s}$ and $\mathrm{2p}_x$ orbitals. Each has an azimuthal quantum number equal to $0$.
Why are only these orbitals shifted? I do not understand why each of these orbitals are split; is it because each is centered on the $z$-axis? Is what I described only apply to the linear Stark effect, and the $\mathrm{2p}_x$ and $\mathrm{2p}_y$ orbitals are shifted according to the quadratic Stark effect? If so, what is the difference between the linear and quadratic Stark effects?
I have read several research papers about the matter, but I don't have a strong understanding of math involved in quantum mechanics. Typically, when the author explains the Stark effect using Dirac notation, I quickly become confused.

Presumably, your grad level studies are under the supervision of whomever is supporting the research, so ask them to recommend suitable reference publications. This is something research mentors/advisors/directors are happy to do: they want you to succeed. And taking a course or two might help with such things as the Dirac notation! Best of success!

A proper understanding will require some quantum mechanics (especially angular momentum). A good reference for the Stark effect in Rydberg states is Gallagher's book. Note also that there are several things happening here as well: the levels split and there are transitions between the levels that are governed by selection rules.


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Is it possible to make a concentration cell without ion-permeable membrane?
I struggle to understand the use of ion-permeable membrane (or salt bridge). https://en.wikipedia.org/wiki/Concentration_cell

If i pour two solutions with different concentration, why do you need the junction between solutions ?
Like this :

According to my readings on Galvani's potential the electrode should be polarised with the potential of the solution their in. Hence you can calculate the concentration of unkown solution C2 without junction ?
In this page https://en.wikipedia.org/wiki/Salt_bridge it is said that the salt bridge is needed to avoid having one solution depleted of electrons and another full. Why the connection of electrode isn't creating equilibrium between solutions ?

You have no complete circuit. At best, there is an ultra-tiny and very short lived capacitive charging. And it is a bad capacitor at that: two beakers with a big air gap. Search around this site for salt bridges and answers asking why they are necessary. This one keeps coming up.

Suppose one electron has left one of these solutions and gets to the other one. The solution rorm which it is coming becomes positively charged, and so will be reluctant to accept that a second electron would leave. Further more, the second solution becomes negatively charged, and will repell any second electron from coming. So no current could be established between the two solution. There may be a small tension, but no current. It is not a cell.

Please see What happens inside salt bridges?

Please avoid asking same question on multiple SE. I had a conversation with you on Physics SE . If you believe that this question cannot be answered on Physics SE than delete it from there or try migrating the original question to Chemistry SE.

You must have heard about charge balance and what happens when charge disbalance occurs. 10 km long CG+ lightning needs charge that easily passes during just 1 minute of cell operation.


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Iron-II to iron-III change in complex formation

We have an iron dust sample, which is treated with hydrogen chloride in aqueous solution. The solution enters in a reaction with a sodium hydroxide powder, from which we obtain a precipitate. Could we solubilize the precipitate with excess of ammonia? Note that, at the end of the reaction, qualitative chemical analysis identified the ion $\ce{Fe^{3+}}$, but not $\ce{Fe^{2+}}$.

First reaction is, obviously:
$$\ce{Fe + 2HCl -> FeCl2 + H2 ^}$$
Then, we have:
$$\ce{FeCl2 + 2NaOH -> 2NaCl + Fe(OH)2}$$
We observe that we have, now, only $\ce{Fe^{2+}}$ in the aqueous solution, not $\ce{Fe^{3+}}$. Precipitate solubilization means coordinative compounds formation. If for iron we have a similar reaction to the one with copper ion:
$$\ce{Cu(OH)2 + 4NH3 -> [Cu(NH3)](OH)2}$$
We would have only $\ce{Fe^{2+}}$. So, is there any redox process happening in background? Or a physical process (non-chemical)?

First iron(II) hydroxide $\ce{Fe(OH)2}$ is a precipitate made in aqueous solution, and it does not react with ammonia. Second, the iron(II) hydroxide $\ce{Fe(OH)2}$ is a green substance which is extremely sensitive to the oxygen of the air. In a couple of minutes, it gets brown, due to the formation of iron(III) hydroxide $\ce{Fe(OH)3}$ according to $$\ce{4 Fe(OH)2 + O2 + 2 H2O -> 4 Fe(OH)3}$$ If water is absent, the reaction may produce another compound of iron (III) $$\ce{4 Fe(OH)2 + O2 -> 4 FeOOH + 2 H2O}$$


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Effect of +M and +R effect in basicity of 4-methoxy aniline
We know that methoxy group exerts a +M effect, as a result of which basicity should increase even after resonance as +M effect is considered to be stronger than resonance effect.
So by this reason 4-methoxy aniline must be more  basic than cyclohexane amine, but this is not true.
I am not able to understand what factor should make cyclohexane amine more basic than 4-methoxy aniline.

Cyclohexane amine is aliphatic whereas p-methoxy aniline is aromatic.

To get real insights you should compare 4-methoxy aniline with aniline.

Since you mentioned the $\mathrm +M$ effect, I'm assuming you're familiar with resonance and aromaticity.
This is a structure of cyclohexane amine:

Here, the lone pair of electrons present on the Nitrogen atom make it a Lewis Base (an electron donor).
This is a structure of 4-methoxy aniline:

Here the lone pair of Nitrogen is delocalized due to resonance. Hence it's not readily available for donation. That's why 4-methoxy aniline is a weaker base.

I would like to know the reason as to why this answer was downvoted. Thanks.

Thanks, for the answer, but isn't  resonance effect considered  to be less dominating than +M effect, in the second there's is a +M effect due to methoxy group, which should make it basic?

I think you should revise what $\mathrm+M$ and resonance mean, since it does not 'dominate over resonance' :). Due to resonance, the lone pair on Nitrogen isn't local to it, which means that even though there is a methoxy group, 4-methoxy aniline is still a weaker base than cyclohexane amine.


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Measuring energy required or generated in gasification
The typical equation used to measure heat is as follows:
$q = mcΔT\\
\text{Where, }\\
q = \text{heat} (\pu{J}) \\
m = \text{mass} (\pu{g}) \\
c = \text{specific heat capacity} (\pu{J g^-1 K^-1})\\
ΔT = \text{change in temperature} (\pu{K})$
This is meant to measure the change in thermal energy (i.e. net potential energy of a system).
Gasification is a process that takes hydrocarbons (often fossil fuels) and produces carbon monoxide, hydrogen, carbon monoxide, and several other substances (as a gaseous mixture) by submitting it to a gasification agent. Sometimes oxygen is used, though other agents can be used; nonetheless, combustion does not take place. The mixture is known as syngas.
The process occurs in settings where temperature is greater than $\pu{700 °C} (\pu{973.15 K})$. However, I am not entirely certain how energy is measured — I imagine we are using changes in enthalpy.
If anyone has knowledge of this subject, please enlighten me. Gasification and syngas are espoused as alternatives to recycling and incineration (since the latter are not always environmentally friendly or economically feasible).

The cost will depend substantially on the starting material : coal and food waste would have significant differences. Depending on a specific process , carbon monoxide is commonly produced which is dangerous..

Are you aware that the equation you wrote doesn't apply to chemical reactions?

@ChetMiller Yes, I am aware. It measures the quantity of heat transferred. It’s helpful when dealing with intermolecular forces, but not with chemical reactions. That is why I brought up enthalpy—and possibly entropy. I am no expert on the subject, but I am trying to become more of an expert, so that’s why I came to stack exchange :)


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Stability in 3-4 membered conjugated ring
We have 3 products one is cyclobuta-1,2 diene cyclobuta-1,3 diene   and the other is cyclo prop-2-en-one, we want to find the stable product out of the 3. So,the third  one due to high angle strain, must be unstable, however this is not true as per datas.  Even some of the resonance structures of cycloprop2-enone(the one with positive charge on O atom are highly unstable), so I am not able to understand what makes the third one more stable than the first and second products.
Please help.

I have edited the question, hope it is more understandable can it be reopened again please

One of cyclopropenone's resonance structures has a positive charge in the 3-membered ring (and a negative charge on oxygen) making it aromatic (increased stability) according to Huckel's 4n+2 rule.

Please don't edit and practically change the question because those who have already answered do not get a notification.

Can anyone please tell why was it downvoted? What improvement should I make in the post? Thanks.

In an allene system(two double bonds next to one another) the substituents on either carbon try to be in the plane perpendicular to each other. For reference see this image:

But as we know that a cyclobutane ring is completely planar, it will prevent this phenomenon and will produce a ring strain even more than that in cycloprop-2-en-one.
Now you might wonder why does an allene system tend to behave in such a manner. For that you can refer to @ringo's answer on Why are the two CH2 groups in allene perpendicular to each other? .
Looking up the 3d structures of these compounds on pubchem can also help you.
2-Cyclopropen-1-one
Cyclobuta-1,2-diene

Thanks, for the answer, but then what makes cyclobuta-1,3-diene  less stable than 2 cyclo propen- one, in this all the carbon atoms have less angle strain than the first one

@green Yes, angel strain is the key here.

I didn't understand, so cyclobuta-1,3 diene  should be  more stable than 2 cyclopropenone  as angle strain is less in the former one  but this is not true, as per the datas .

@green In the question you mentioned cyclobuta-1, 2 diene and not cyclobuta-1, 3 diene

Sorry, but I didn't remove cyclobuta-1,2 diene  it is still there, I just added a new product, your answer is still helpful, thanks. I was just wondering what could be reason for cyclobuta-1,3diene


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Why can't the energy of an electron exceed 0 eV?
Today we were learning about atomic structure, our teacher introduced the mathematics of the 'Niels Bohr Model of Atom', and calculated the value of total energy $\frac{-13.6z^2}{n^2} \pu{eV}$. He explained that when the value of $n$ tends to infinity, the value of total energy tends to $0$. Mathematically I understood this.
I couldn't understand, or comprehend, how or why the energy of an electron can't exceed $\pu{0 eV}$? What is the reasoning behind it; why is it not possible and what is the physical significance of it.
For the sake of the argument, what would it look like if the total energy of an electron were to exceed $\pu{0 eV}$?

Energy at infinity is assumed to be zero. We need some refer point to measure the work done which is stored in the form of energy.

So what you mean to say is that its' possible for the total energy to be >0 but then the electron would no longer be a part of the Atom and would behave as an independent sub atomic particle. Did I get it right?

Energy of an electron, in rest and infinitely far(=far enough) from electrostatic source is considered as zero by arbitrary convention for practical purposes. It is the same as for classical gravity. Obviously, the total energy of a free electron is E=sqrt((mc^2)^2 + (pc)^2), but this is omitted in atom model context.

it would be helpful if the downvoters explained why they are downvoting.

The value of the energy in the Bohr model is zero when the quantum number is infinity  because that is the limiting value of the Coulombic potential at large distances, and because the electron is assumed bound to the nucleus (the atom is stable), which constrains the value of the total energy.
The energy of a stationary hydrogen-like atom is described by the equation
$$ U(Z,n) = R_\mathrm{E} \frac{Z^2}{n^2} $$
where $R_\mathrm{E} = \pu{-13.6 eV}$. It can equally well be written as
$$ U(Z,n) = \frac{-Zk_e e^2}{2r_n} \tag{1}$$
where $r_n$ is the allowed orbit radius in the Bohr model, which illustrates how the energy becomes more positive (goes to zero) as the average distance between the nucleus and the electron increases.
The Bohr model assumes the universe consists of just one electron and nucleus. It is implicit in the model that the electron is bound (the electron and nucleus can never get far enough apart to escape their mutual attraction). According to the virial theorem the total energy in the above equations is one half of the Coulombic potential energy and the negative of the kinetic energy. That means the expectation values of the potential and kinetic energy for a given quantum number are fixed, and that the value of the total energy is bounded by the value of the potential energy (see Eq. 1):
$$U(Z,n)=\frac{\left<U_{\mathrm{Coulomb}}\right>}{2}$$
The zero limiting value of the Coulombic potential can be interpreted as indicating that the interaction between the particles vanishes at large distances. On the other hand, the zero limiting value of the total energy can be interpreted as indicating that the reference frame of the atom is at rest. If the atom is assumed to have kinetic energy than this can be represented as a non-zero  offset of the total energy. In that case the energy converges to this threshold as the particles move farther apart.
In addition, in practice external fields might interact with the atom, contributing to the potential. It is possible for instance to superimpose a uniform linear potential on the Coulombic potential. The solutions (wavefunctions) for the new perturbed atom are not identical to those of the original isolated atom. Suffice it to say that if you now pull the particles apart far enough  they will eventually be free to follow trajectories that are no longer bound.
The following figures illustrate (roughly) the energies of an atom (for Z=1) before and after superimposing an external linear potential, plotted against the quantum number or the orbit radius in the original Bohr model. In the case of particles starting at rest, what determines whether particles will remain bound or fall apart is relative values of the potential. It is the slope of the potential that determines forces that drive particles together or apart. In the last figure, the (slight) downward slope at large r as the radius increases indicates that particles at rest starting at this internuclear distance are no longer bound.
Hydrogenic atom in Coulombic potential...

... and perturbed by linear potential

An electron can also be excited by a photon into the continuum of non-quantized states with energy greater than zero, but then the energy is no longer described by the Bohr model. The excess energy (greater than the ionization energy) amounts to the kinetic energy of the dissociated (nonbound) state (which is alluded to in the other answer).
You might also want to read these other posts.

Might be a silly question but until and unless stated we will assume that the system we are studying in a system the consists of only that atom and nothing else, right?!

Yes, the Bohr model assumes that the universe consists of only one electron and a nucleus .

The OP can understand, mathematically, why the energy of the electron in the Bohr atom is zero at infinity, so let's avoid the mathematics in order to assist the imagination.
In the (small?) universe which consists of just a single hydrogen atom (plus all necessary equipment to measure potentials and distance), it is necessary to cause a separation of the electron and proton, but this equipment is never described. It consists of two very tiny tweezers each connected to a remote controlled motor which grabs the electron and proton and moves them slowly apart so their attractive force and energy can be measured. Of course, when the separation gets to infinity, nothing further can be done, and the energy equals zero, or the ionization energy of a hydrogen atom, whatever you have selected as reference.
However, if you are inclined to be somewhat mischievous, and adjust the separating device to speed up, you could supply so much energy to the electron (the proton will be the center of mass of your tiny universe) that its velocity does not trend toward or  decrease to zero, but continues to carry it away from the proton. Where to? Well, it could just bounce around in that universe, totally unbound to the proton, or it could penetrate into the next universe (you know, like photoionization kicks an electron totally out of the metal it started from).
The question resembles Zeno's Paradox (Achilles and the tortoise https://en.wikipedia.org/wiki/Zeno%27s_paradoxes) in that it only considers the range of energies up to the unbinding or ionization of the electron.

Funny enough I was thinking along the same lines and also thought of Zeno's paradox, that perhaps you can describe the process as equivalent to bodies in a gravitational orbit, one receiving enough energy to escape. But I think this is contrary to the principle of the Bohr model. To deal with excitation into a free state (a system with escape velocity) you need a different model altogether. The reason is that I don't see where this extra kinetic energy of the electron fits into the Bohr model.


text:
After formation of the Grignard reagent, would it react with Et2O or methylpyridinium?
I know that a Grignard would form, but im confused as to whether it would react with Et2O or with methylpyridinium. If it reacts with methylpyridinium, what would the mechanism be like?

Grignard reactions are often carried out in diethyl ether as it is unreactive to them.
It is well established that organometallic reagents, including Grignard, react with Pyridine-N-oxide at the 2-position to give dihydropyridines or ring-opened products depending on work-up1.
Based on this reactivity and these references2,3, I think is likely that the benzyl Grignard will attack the 2-position of the pyridinium species. The actual product will depend on subsequent work-up, but the question setter probably expects to see the N-methyl-2-benzyl-dihydropyridine as product.
References

Hans Andersson, Xiaoyang Wang, Mikael Björklund, Roger Olsson, Fredrik Almqvist, "Reaction of pyridine N-oxides with Grignard reagents: a stereodefined synthesis of substituted dienal oximes," Tetrahedron Letters 2007, 48(39), 6941-6944 DOI: 10.1016/j.tetlet.2007.07.161
Mikio Takeda, Arthur E. Jacobson, Ken Kanematsu, Everette L. May, "4-Methoxy-1-methylpyridinium iodide. Grignard products and transformations thereof," Journal of Organic Chemistry 1969, 34(12), 4154-4157
DOI: 10.1021/jo01264a087
Edward V. White, "The Reactions of Organometallic Reagents with Pyridinium Salts," Ph.D. Dissertation;  University of New Hampshire, NH, 1968 (pdf)


text:
Excitation of electron in Niels Bohr's atomic model
Professor taught us that an electron gains or looses only those energies which are equal to difference in two energy levels. That is $E_1 + \Delta E = E_2$ or $E_1 + \Delta E = E_3.$
What if we give, say, $\pu{11.3 eV}$ and the remainder is $\pu{-2.2 eV},$ so the electron jumps to $n = 2$ where the energy is $\pu{-3.4 eV}$ and the remainder $\pu{-1.2 eV}$ is released?
I'm really confused with the working here.

Have you thought about why atomic absorption spectra contain just narrow lines and not wide bands ?

Yes, that's because we have specific 'energy levels' and not continuous i.e. only for n=1,2,3.... and not n=1,1.1,1.2.....
and because the energy levels aren't continuous the spectra isn't continuous but discrete? Am I right?

Yes, and also atoms absorb photons of just specific energy levels, equal to differences between atom discrete allowed energy levels.

This is a very good question. It is very important to specify the form in which atoms are, isolated, in molecules, in lattices. I hope we can write a nice answer. But at a basic level, and for what is named Absorption (leading to Absorbance) in isolated atoms you need a perfect (indetermination apart) match.

..as molecules have band-wise spectra and condensed phases even like continuous spectra.

I'm confused about the transitions you are suggesting. Could you explain more clearly what you mean in the middle paragraph. I understand that providing 11.3 eV would drive the $n=1\rightarrow n=2$ transition and would place the atom in the energy state with E=-2.2eV. I don't understand the second part about -3.4 eV etc.

@BuckThorn What I mean to say is. I've give it more energy than required to excite the electron from n =1 to n=3 but not enough for it to jump from n=1 to n=4. So why does it come back to n=1 instead of jumping to n=3 and releasing the excess energy in the form of radiation. Do you understand what I mean to ask?!

It can't reject the excess energy because it can't absorb it in the first place. You need to understand that photons can provide all energy they have or none. There's no absorb half reject other half.

I understand your question better now, but you need to dive into fairly advanced quantum mechanics to get an answer, for instance: https://royalsocietypublishing.org/doi/pdf/10.1098/rspa.1927.0039  One talks of a "resonance" condition between the radiation and the atom.

Another relevant topic is that of transition rules for specific excitations, which seems to be exactly what you are asking about, see eg: https://en.wikipedia.org/wiki/Selection_rule

@NisargBhavsar So let me get this straight. So it's just like how Bohr explained it. We have a ladder that the electron can climb up to. It can either make its way up by stepping on the next or can go down by stepping down, but only on the steps and not in between. So either give me the required amount of energy or don't. Right?!

Quantum mechanics received its name from the concept of the "quantum" which is Latin for "how much" or "amount" and is used to mean "small discrete piece or fragment". Classical physics prior to the development of QM described the energy of bodies as a continuum of possibilities. The novelty introduced by QM was the idea that within the continuum only a subset of the possible energies are in fact allowed, corresponding to allowed states. This means by extension that the energies of a body in different possible states differ by discrete rather than continuous amounts. This discreteness becomes increasingly evident for small and confined systems, where the wave-like properties of matter become more evident, as demonstrated for instance by the self-interference of the electron wavefunction in the double-slit scattering experiment.
Electrons in atoms fulfill all of the conditions for observation of QM phenomena. The particles involved are subatomic (very small) and are confined (by a Coulombic potential). The result is that electrons in atoms exist in a discrete number of allowed states, each state defined by a fixed energy and wavefunction that can be interpreted as the probability distribution of the electron.
When an atom is excited by a photon (a discrete light particle with a fixed amount of energy) it absorbs the entire photon. All of the photon's energy is acquired by the atom. However, such an excitation implies a transition between possible states of the atom. It is necessary that the energy of the photon match the difference in energy of the atomic states. In the case where the two states are bound (if an electron is not excited into the continuum of ionized free states) the result is that only a subset of photon energies can be absorbed by the atom.

As noted by Shoubhik R Maiti, photons can also be scattered by atoms and in the process transfer only part of their energy to the atoms. However the principle and meaning of quantization of electronic states remains the same. Electronic states are quantized (discrete) and so only particular amounts of energy can be absorbed (quanta corresponding to differences in energies of discrete levels).

+1 This is a good answer. I would also add that there are some cases where it's not necessary to have a metastable excited state, for example Raman scattering. A molecule can absorb photons to go into a virtual electronic state (that does not exist as a measurable state—uncertainty principle) and then releases a photon back, going into an actual excited state which exists for some time.

@ShoubhikRMaiti Thanks, that would invalidate some of what I wrote in the last paragraph, although I was thinking of absorption, not scattering. You might want to write an answer - if not I will add a caveat to my own to incorporate your comment.

No I think what you have added to your answer is fine. Yes scattering is a two-photon process so it's quite unusual and different from the normal one-photon absorption or emission etc.


text:
Conversion of Benzene to 9-Bromo 9,10 Dihydroanthracene

I tried to proceed through following steps:
Step-1:
Step 2:
Step-3:
Step-4:
I agree with the fact that yield might not be good. But is this whole reaction possible? Am I going correct? A better method and the corrections to this method are appreciated!

Wait: how, assuming non-nuclear chemistry, do you convert Cl (step #3, 1,2-dichlorobenzene, coupling mediated by sodium) into C?

This is probably the first time I see four different drawing styles in one post:)

@andselisk In the longer run, I'm more concerned about (still) drawing the Thiele (?) ring in benzene's structure.  Too easy to lose track of counting the $\pi$-electrons present, not only for first time learning about either electrophilic or nucleophilic aromatic substitutions.  Even more so for e.g., naphthalenes when some drawings coalesce these individual Thiele rings across bonds into something in a shape like a toilet seat.

@andselik Haha..Some of them I had to copy paste from google while I had to make some of them on my own..didn't have any one source :P

Where do the two methylene groups come from in the step from 1,2-di-chlorobenzene to di-hydroanthracene? Step 3 looks OK

@Buttonwood I completely agree, good point about Thiele rings. I think you mean a so-called horseshoe bond.

Consider the Friedel-Crafts reactions of phthalic anhydride

@Waylander I am Sorry, I have edited step-3, I took another compound "1,2 Chloromethyl benzene" I hope it is correct now!

I think you would be better off starting from two equivalents of benzyl chloride. Those can react with each other in Friedel-Crafts fashion, yielding 9,10-dihydroanthracene. Use one equivalent of NBS afterwards and you obtain your product.

@TheRelentlessNucleophile Thank you, It was helpful!

OP has shown research effort and it should not get downvoted.

@andselisk While very descriptive, the horseshoe is not the form I thought about, because it lacks crossing a bond.  I made a photo for the chatroom here, as young as by about 2002.

What you are looking for is known as the Wurtz-Fittig reaction - wikipedia article , JACS paper
In this proposed reaction I think you will get a complex mixture (also known as a mess), but you may be able to find some of your desired product in it. Good luck with trying to isolate it cleanly!
If you want a straightforward preparative route then look at the F-C reaction of phthalic anhydride with benzene, the reduction of the resulting 2-benzoyl benzoic acid to 2-benzyl benzoic acid and cyclisation with polyphosphoric acid. This product can be reduced to the dihydroanthracene by Zn/HCl

Thankyou, but since I am a 11th grader, can you please help how do I carry out cyclisation process after reduction of carboxylic group?

You could try treating the resulting ortho-benzoylbenzoic acid with $\ce{SOCl2}$ and then carry out another acylation. But this process will for sure be low-yielding because of the deactivating nature of the acyl radical. Just start from benzoyl chloride. It's the easiest way to synthesize anything "anthracene-like" from a single aromatic nucleus.

@TheRelentlessNucleophile Thank you so much! :)

The reagent of choice for the cyclisation of the benzyl benzoic acid (reduction product from benzoyl benzoic acid) is polyphosphoric acid


text:
Which one will be the major product when multiple sites for lone pair attack is available?
I have been asked to find the major product of the following  reaction:

Clearly, there are two possible sites for the lone pair of $\ce{NH_3}$ to attack, and thus two possible products.

The carbon that is further from the $\ce{-Me}$ should be more electrophilic than the one that is closer to the methyl group (due to the positive inductive effect of the methyl group). With that logic, the product $(i)$ should be the major product. But, according to the book, product $(ii)$ will be the product formed.
Why is product $(ii)$ the major product?
How do you determine which one will be the major product in this kind of situations?

This would be more obvious if you had a strongly electron-withdrawing group (e.g. nitro or CF3)  instead of the mildly electron-donating Me. In that case it would be clear that the carbonyl para to the EWG group would have electron density drawn away from the C, increasing the polarity of the C=O bond and making it more reactive to nucleophiles.
Instead we have a mild electron donation from the Me which makes the C=O bond slightly less polarised and reducing the electrophilicity. Thus the meta C=O, which sees less effect from the Me, is slightly more reactive and NH3 attack there will be the major product (though not by much).

One thing I want to ask, when methyl is substituted at the benzene ring, it also shows hyperconjugative effect (basically resonance) , and that hyperconjugation won't operate at the meta position and only at para position increasing electron density there, though inductive effect is there, so how do we decide which effect is dominating?

You run an experiment, either in the lab to find the actual yields or on the computer to model the electron densities at each site.

So is it experimental that hyperconjugation is negligible here? And we really cannot compare between inductive effect and hyperconjugation in such types of compounds using some arguments?

A mixture at best.


text:
Basics of wave-mechanical model of atom
Why is it that for the $\mathrm{d}$ subshell we have  $\ \ \mathrm{d}_{xy}, \ \ \mathrm{d}_{yz}, \ \ \mathrm{d}_{xz}, \ \ \mathrm{d}_{x^{2}-y^{2}},$ and $\mathrm{d}_{z^{2}}$ orbitals only?
Why aren't there $\mathrm{d}_{y^{2}-z^{2}}$ or $\mathrm{d}_{y^{2}}$ or other orbitals similar to that?
Perhaps it is because we have an empirical evidence that there are only 5 orbitals in $\mathrm{d}$-subshell, in that case, why are $\mathrm{d}_{x^{2}-y^{2}}$ and $\mathrm{d}_{z^{2}}$ orbitals named in this particular fashion only? (as the axes are arbitrary and there could have been a different choice of axes also)

Well, the choice is indeed somewhat arbitrary.

Just armchair speculating, it may be related to the fact that within cartesian<>polar coordinate transform, axis x,y,z are not equivalent. The angle $\phi$ relates to x and y projections, the angle $\theta$ relates to z projection..

Yes, completely arbitrary physically, but biased to simpler math. Because of the funny toroid around the dz^2, the resultant electron density if all are filled is still perfectly spherical. There is no actual bias in the density along one axis. In order to discuss orbitals, it's easiest if we all use the same orthonormal basis set. We could have chosen a different one.

These orbitals represent the angular part of the wavefunction. The solution obtained directly from solving the Schrödinger equation produces equations containing complex numbers so cannot be drawn on normal $xyz$ axes and are hard to visualise.
The angular part of the wavefunction is given by functions called Spherical Harmonics these usually are given the symbol $Y$ and require two quantum numbers $\ell$ and $m_z$ making $Y_\ell^{m_z}$.
What is done to make the orbitals real functions, as opposed to complex ones, is to take linear combinations of the spherical harmonics, thus choosing the $z$-axis then the $\mathrm{d}_{z^2} = Y_2^0 = N(3\cos^2(\theta)-1)$) where $N$ is the normalisation constant. In cartesian coordinates ($xyz$) rather than polar ones (which are in $r,\;\theta,\;\phi$), $\displaystyle Y^0_2= N\frac{3z^2-r^2}{r^2}$ so you can see where the $z^2$ comes from and $r$ is the radial distance.
The other orbitals are also linear combinations e.g. $\mathrm{d}_{xz}=(Y_2^{1}+Y_2^{-1})/\sqrt{2}$ and in cartesian coordinates $\displaystyle Y_2^1 \sim \frac{z(x+iy)}{r^2},\; Y_2^{-1}\sim \frac{z(x-iy)}{r^2}$ so you can see where the notation $zx$ comes from and by making a linear combination the sum is no longer a complex number ($i=\sqrt{-1}$).
For completeness $\mathrm{d}_{yz}=(Y_2^{1}-Y_2^{-1})/\sqrt{2},\; \mathrm{d}_{x^2-y^2}=(Y_2^{2}+Y_2^{-2})/\sqrt{2},\; \mathrm{d}_{xy}=(Y_2^{2}-Y_2^{-2})/\sqrt{2}$  and in cartesians $\displaystyle Y_2^{\pm2}\sim \frac{(x\pm iy)^2}{r^2}$.


text:
Does spoiled milk have something else in it besides lactic acid that gives its characteristic taste and odor?
Even slightly spoiled milk has a taste that I don't like, while plain yogurt tastes either great, or at least nowhere near as bad. A dozen sources (all of them) say that the spoiled odor and taste of spoiled milk and also of yogurt is caused by the presence of lactic acid. Does spoiled milk have something else in it besides lactic acid that gives it the taste that I dislike that yogurt (which has plenty of lactic acid it it) almost never has. Or is there more than one kind of lactic acid?
Is the taste of spoiled milk due to lactic acid and one or more amino acids?
https://foodsafety.foodscience.cornell.edu/sites/foodsafety.foodscience.cornell.edu/files/shared/documents/CU-DFScience-Notes-Milk-Flavor-Defects-Sensory-Eval-04-10.pdf has this to say:
"CHARACTERISTICS OF MILK OFF-FLAVORS...
Typical Milk:
No criticism. Very little distinct odor, pleasantly sweet and clean with no aftertaste...
Acid:
Basic taste sensation. Sour, tart, may cause tingling sensation on tongue. “Cultured milk” or “sour” odor may be present. Cause: Growth  of  lactic  acid  producing  organisms  such  as Lactococcus lactis,  due  to  poor refrigeration, especially when temperatures exceed 70°F (21°C)."
Like everyone, the Cornell author explains the spoiled taste and odor of spoiled milk by referencing only lactic acid.
https://www.nationalgeographic.com/culture/article/when-good-milk-goes-bad is no different when talking about slightly spoiled milk, as opposed to the much later very spoiled moldy milk, which involves yeasts and molds and ...danger.
https://www.reddit.com/r/chemistry/comments/j15maj/what_does_lactic_acid_smell_like/ says pure lactic acid does not smell bad. Aha! On the other hand, it does say whether or not it smells like spoiled milk. What a pity.

I found out recently (with a lot of help from the folks at Seasoned Advice -thanks guys) that what I used to call 'sour milk' a lot of people call 'spoiled milk', and what I call 'fermented milk products' a lot of people call 'soured/sour milk products'. I think this has been a cause of quite a bit of talking at cross purposes involving me, both here and at Seasoned Advice. I have edited the question accordingly.

Comments are not for extended discussion; this conversation has been moved to chat.

Probably. Lactic acid bacteria produce other volatile compounds such as acetaldehyde and diacetyl that could be perceived as off-flavors. Both where fermentation is the objective and in the case of spoilage, the VOC profile can vary with the identity of the responsible bacteria but also of the milk and its treatment.
For instance, in Ref. 1 they discuss the effect of different lactic acid bacteria:

During fermentation, S. thermophilus and L. delbrueckii ssp. bulgaricus generate lactic acid and a variety of volatile organic aroma compounds. These compounds impart particular flavors to fermented milk (Beshkova et al., 1998). The overall flavor of any fermented milk product is formed by a large number of these volatile compounds. Among these compounds, carbonyl compounds and organic acids such as acetaldehyde, diacetyl, and acetic acid play a crucial role and can be used to evaluate the flavor quality of dairy products (Zha et al., 2015). Ott et al. (1997) found that acetaldehyde provided the typical flavor of dairy products such as yogurt and buttermilk. Erkus et al. (2013) also reported that diacetyl contributed the characteristic sour buttery flavor to some fermented dairy products.

The offending chemical could also be vinegar (acetic acid). Again from Ref. 1:

Acetic acid has been found previously as a volatile in yogurt by Panagiotidis et al. (2001). High levels of acetic acid contributes most to the tart flavor of yogurt and is particularly associated with a “vinegary, pungent, acidic” taste (Panagiotidis et al., 2001).

If you search for "chemical profile of spoiled milk gc" you'll find a wealth of references discussing the effect of different milk (due to species, handling or forage), milk treatment, and spoilage bacteria on the VOC profile.
In addition, what makes the milk taste bad may depend on the bacteria and on you. That makes narrowing the answer down to a particular compound impossible.
References

T. Dan, D. Wang, R.L. Jin, H.P. Zhang, T.T. Zhou, T.S. Sun, Characterization of volatile compounds in fermented milk using solid-phase microextraction methods coupled with gas chromatography-mass spectrometry, Journal of Dairy Science, Volume 100, Issue 4, 2017, Pages 2488-2500, ISSN 0022-0302, https://doi.org/10.3168/jds.2016-11528.

Diacetyl is an interesting compound. In grad school, one fellow in our research group needed to use it and he could not smell it at all. Me and the 3 other grad students were stunk out by the overwhelming movie theater popcorn smell. Unforgettable.

I think Table 1 from this paper is also useful

Maybe my experience with spoiled milk is different than most people.  In my experience, it definitely smells different than vinegar, lactic acid, or diacetyl.  Definitely seems more like short-chain fatty acids that are responsible for the sour rot smell.

@CurtF. Table 1 from that paper http://www.marsiliconsulting.com/pdfs/j_ag_food_chem_2000_48-8.pdf and the whole paper in fact is great. I learned so much about milk and how it spoils. Everyone should read it. On the other hand, it doesn't explain why sour milk seems to me to always taste exactly the same, as if the sour milk taste was caused by a single chemical or single type of bacterium.

@CurtF. I also smell a rancid odour, like when fat are involved. Must be some kind of aldehyde. Can the acids be freed? They else smell not that good...

Raw milk compared to pasteurized milk when spoiled I think would tell us a lot.
Raw milk sours slowly from the day it exits the cow, the lipids separating, with a sour yoghurty kind of smell that gets stronger and stronger until it gels into a sour yoghurt.
Pasteurized smells like nothing at all for a week or so and and then suddenly smells like something created by an antichrist from another planet.  I mean it screams "What kind of a maniac would even touch this let alone try to drink it"
Whatever the processes happening, I would imagine that they are very different bacterial/fungal and chemical processes.


text:
Lone pairs of electrons and covalent bonding
Every time I saw a covalent bond structure, I notice that there are always lone pairs or no electrons left in the outer shell of the atoms. Is it compulsory for covalent bonds to have lone pairs or no electron in outer shell left to form ?
Thank you.

Well, check out NO!

There are odd-electron species.

@NisargBhavsar So, you won’t take NO for an answer? ;-)

@Ed V My comment was directed to OP.

@NisargBhavsar I know! I was punning in my first comment and my second was a play on words. This is an excellent site, but sometimes dryer than $\ce{P4O10}$.

@EdV I cordially invite you to [chat], there's them puns. Sometimes.

And $\ce{H-H}$?

For many of the compounds that you encounter in a first-year chemistry course, yes, this is the case, as satisfying the octet rule usually results in stability through full s- and p-orbitals. However, this is definitely not a guarantee
Some common examples include nitrogen compounds due to its odd number of electrons in its ground state, such as $\ce{NO}$ with a bond order of 2.5.

More generally, compounds with unpaired electrons fall into the class of radicals, which make up a deep set of reactions. Radicals are formed through the homolytic cleavage of bonds (one electron transferred to each bonding constituent), and this cleavage occurs most often through the input of light or heat. These cleavages are promoted when the product can stabilize the free-electron through resonance, similar to the stabilization of a carbocation.
Radicalization of chlorine through light (light promotes electron to antibonding orbital, which is unstable and causes a homolytic cleavage):

Radicalization of benzoyl peroxide (resonance stability of radical due to conjugation):

Radicalization of an azo compound (formation of an especially strong $\ce{N2}$ triple bond):


text:
Are there scales that record weight continuously?
I would like to monitor the weight of something continuously (e.g. once a minute) and access the data easily via a phone app.  Is there something like that?
There are numerous options for similar devices if you want to measure ambient temperature and humidity (e.g. https://www.amazon.com/SensorPush-Wireless-Thermometer-Hygrometer-Android/dp/B01AEQ9X9I).
But all my searching for scales with similar functionality have failed.  I've found many lab scales and kitchen scales, and body weight scales.  Only the body weight scales interact with apps on your phone, but they do not record continuously.
If you know of one, or just have better google-fu than me and can suggest a more precise search term, any pointers would be appreciated.
[ In case you are curious about the application (which is not chemistry), I want to measure water transpiration and evaporation in a bonsai tree, and measuring the weight of the tree in its pot is my proxy for water usage.  This has worked well with manual measurements, but I want to go to the next level. ]

Try “data logging balance”. I had a colleague who used to insist that fish have scales, chemists have balances.

The load cells that are used in many cheaper balances certainly put out a weight-dependent voltage continuously, so it should be eminently possible to do this.  I know folks who have made their own data logging balance by hooking up such a load cell to an Arduino or Raspberry Pi...

@edv Thanks for the clarification. I didn't know if a balance implied the old-fashioned mechanical balance.

@CurtF. I have an arduino lying around.  Project time, maybe!

You could use a normal balance and then record the display with the smartphone camera. I did that once for a project, it's not fancy, but quite simple.

@ShoubhikRMaiti  Clever!  And hilarious.  Nothing will stop you, obviously.  I want to do this continuously, and outside, so I can't dedicate my phone to it.  ;)

@ShoubhikRMaiti No, not at all sarcastic.  I like the idea.  It's perfect for a short term solution.  Sorry if my tone suggested otherwise.

Oh, ok alright then, sorry. Good luck with your experiment :-)


text:
Which Half-Reaction to Use?
I was hoping for some help with choosing half-reactions for an electrolytic cell. This was the problem that sparked the question (Question #6 from the 2020 USNCO Part II):

The correct answer is
$$2\ce{Cu^2+ + 2H2O (l) -> 2 Cu (s) +  O2 (g) + 4H+},$$
which fits in with the details about gas evolution. However, my question if we were not given these details, and rather just the solution and table of half-reactions, how would you determine the correct combination of half-reactions? Is it always the reaction that has the greatest value of $E^{\circ}$? And does this change when a current is being passed through the solution?

It would be better to cite the text of the question and not including it as an image. Images cannot be searched and often enough have not the best readability. (Also it is sad to see that USNCO has such a high regard for correct typesetting.)

Well I'm not completely sure but I'll try to answer.
To answer your first question, yes, it is almost always certain that the half reaction with the greater $E^\circ$ will determine the correct combination. But, as always there are instances when the above is not always true due to reaction factors.
For instance, in the electrolysis of a $\ce{NaCl}$ solution,
we have the following possibilities at the anode:
\begin{align}
\ce{Cl^- (aq) &-> 1/2Cl (g) + e-}, &
E^\circ &= \pu{-1.36 V}\\
\ce{2H2O(l) &-> O2 (g) + 4H+ (aq) + 4e-},&
E^\circ &= \pu{-1.23 V}
\end{align}
It would suggest that the correct half reaction is the latter, but practically speaking, the $\ce{O2 (g)}$ formed forms a layer of bubbles over the metal surface of the electrode which slows down the reaction quite a lot, and so to increase the rate, an overpotential[1] or bubble potential is applied, but that makes the former reaction kinetically favorable, leading to the oxidation of $\ce{Cl-}$ and not $\ce{H2O}$ as expected.
Besides some exceptions like the above, it is always the half reaction with the greater $E^\circ$.
To answer your second question, if we look at the expression for $E^\circ$,
$$
\Delta G^∘=-nFE^\circ
$$
and $\Delta G^∘$ only depends on temperature, so unless the current causes the solution to heat thereby changing its temperature, the value of $E^\circ$ will not change.
References

Wikipedia: overpotential