text:
Pressure vs volume plot for real gas and ideal gas
Background
Recently I have been studying about the states of matter and came to the topic of ideal gases and real gases, and the laws related to them. While studying it from my textbook, I saw the plot of pressure vs volume for real gas and ideal gas. The problem is that the interpretation I am getting from the graph is quite contrary to what I have learnt till now (this might be because of some conceptual misunderstanding of mine. Sorry if the question looks foolish).
Question Explained

In the above image of the plot, there is a point where real gas behaves the same as ideal gas (where the curves intersect). From this point, as we decrease the volume in the graph we find that pressure exerted by the real gas is greater than that by the ideal gas. This seems to be contrary to what I have learnt, which is as the volume of a real gas decreases the pressure exerted by a real gas becomes lesser than ideal gas at the same volume (due to increased intermolecular attraction). In reality even this isn't the case as decreasing the volume much further causes the intermolecular repulsion to overcome the attraction at some point and hence increase the pressure. But as far as I know the graph has been made with the use of the van der Waals equation (I checked it myself via plotting the graph in Desmos and it looked the same as shown above). But the assumption with which the equation (and hence the graph) were derived say something different. The equation I'm talking about is $P_\mathrm{ideal} = P_\mathrm{real} + \dfrac {an^2}{V^2}$, and as far as I can comprehend it says pressure by an ideal gas is greater than pressure by a real gas at a given volume. Hence the question:
Why does the equation $\left(P_\mathrm{ideal}=P_\mathrm{real}+\dfrac {an^2}{V^2}\right)$ and the graph (shown above) say totally different thing about the pressure at a given volume?
Please do point out if I am going wrong anywhere.
Thanks in advance!
Here in this crash Course video the same logic as mine is given for decrease in pressure for real gas.

Could you please provide a reference or link to the source of that figure. It is misleading and has appeared in numerous other posts. It would be great if we could track down the culprit.

@BuckThorn Please see pg. 14 https://ncert.nic.in/ncerts/l/kech105.pdf

Summary
At sufficiently low volumes, the excluded volume effect causes the van der Waals (VDW) pressure of all gases to exceed the ideal gas (IG) pressure.   At higher volumes, the VDW pressure may or may not cross below the IG pressure.  This depends on the temperature, and the relative size of the VDW a and b terms. If the temperature is sufficiently low, and if the a term is sufficiently large relative to the b term, you will see a crossover.  Otherwise, you won't.
Notes:
(1) The OP asks for a discussion of VDW gases, and thus most of this analysis concerns the predictions of the VDW model.  However, the curves presented in the OP's screenshot are for real gases and thus, at the end, I discuss their behavior.
(2) Below the VDW critical temperature (which equals $\frac{8a}{27bR}$, and is thus  gas-specific), the VDW isotherms (the curves on a  $p-V$ graph) begin to show oscillations.  Those oscillations are beyond the scope of this question, but they should be noted for completeness.  This answer assumes we are above the VDW critical temperature for each gas.
The VDW model
The VDW model assumes the gas molecules are hard spheres that interact only attractively.  Unlike the case with the attractive forces, there are no long-range repulsive forces in the VDW model.  Indeed, there are no repulsive forces at all between the gas molecules (except insofar as they are hard spheres that can't interpenetrate).
Instead, the molecules are assumed to have a physical volume, and the effect of this volume is to reduce the free volume left for the gas.  The volume of one mole of molecules (or atoms) is given by the VDW "b" term.
This should help illustrate how the physical volume of the gas particles affects the pressure:
Suppose you are at a volume where the hard spheres themselves take up half the container.  If there were no attractive forces, the pressure of the VDW gas would be double that of the ideal gas, because only half as much free volume is left for the VDW gas, such that p will be a monotonically decreasing function of V.
Physical explanation of difference between pressure of VDW gas and ideal gas
At very low volume (low relative to the volume taken up by the hard spheres), the molecular volume term dominates, and thus p_VDW > p_IG (where IG = ideal gas).
What happens at higher volumes, however, is gas- and temperature-specific.  And thus whether or not there is a crossover is gas- and temperature-specific.
Case I:  Lower temperature
At a sufficiently high volume, the pressure-increasing effect of molecular volume becomes less than the pressure-reducing effect of the attractive force, and thus p_VDW < p_IG.  Hence there is a point at which the VDW and IG curves cross over.
Again, let me emphasize that what constitutes "lower" vs. "higher" temperature is gas-specific. For instance, while $T = 273 \, \ce{K}$ is in the low-temperature regime for $\ce {N2(g)}$, it's in the high-temperature regime for Ne.   Thus, at this temp, you would see a crossover for $\ce {N2(g)}$, but not for Ne.
More specifically, $\ce {Ne(g)}$, which has a small a value relative to b, loses its crossover behavior above $149.8 \, \ce{K}$, while $\ce {N2(g)}$ doesn't lose its crossover behavior until the temperature exceeds $425.8 \, \ce{K}$.
Case II: Higher temperature
As explained above, as the volume increases, the effect of molecular volume decreases.  However, the intermolecular spacing also increases, lessening the average attractive force between the particles.  Further at, high temperatures the kinetic energy of the gas is higher, thus making the effect on pressure of intermolecular attraction relatively less.  If the attractive force is sufficiently low, and the temperature is sufficiently high, then the attractive force is never able to overcome the molecular volume effect.  Hence, for these gases at these temperatures, p_VDW > p_IG at all volumes.
Note: the above discussion is referring to the VDW model, not necessarily to how real gases actually behave.
Mathematical explanation of difference between pressure of VDW gas and ideal gas
We can also see this from the mathematical structure of the VDW equation:
$$p=\frac{RT}{V_m-b}-\frac{a}{V_m^2}$$
In the limit as $V_m \rightarrow b$, $(V_m-b)\rightarrow 0$, causing the first term to dominate over the second.
As $V_m$ gets large, whether the attraction can win out over the excluded volume is determined by the magnitude of T, and the relative values of a and b.
In the limit as $V_m \rightarrow \infty$, for all gases, the VDW equation reduce to the ideal gas equation:
$$p=\frac{RT}{V_m}$$
Plots, Case I, lower T ($\ce{N2(g)}$ at $\pu{273K}$)
Comparative pressure vs. volume plots of VDW and ideal gas
Now let's take a look at what the actual $p-V$ graphs look like, using $\ce {N2(g)}$ as an example.  The VDW a and b terms for $\ce {N2(g)}$ are:
$$a=1.37 \ce{ \frac{bar L^2}{mol^2}}$$
$$b=0.0387 \ce{ \frac{L}{mol}}$$
Assuming $T=273 \,\ce{K}$, we can determine crossover point by setting pIdeal = pN2VDW (the VDW pressure for $\ce{N2(g)}$), and solving for $p$.  The crossover occurs at $V = 0.108 \,\ce{ L}$.  But if we plot  pIdeal and pN2VDW from about 0.1 L to 10.0 L, we find they are nearly on top of each other:

Thus to see the crossover, we need to focus on the region around $V = 0.108 \,\ce{ L}$:

Plot of p_VDW–p_ideal vs volume,  using $\ce{N2 (g)}$ as the VDW gas
We can further improve our understanding by plotting the difference between pIdeal and pN2VDW vs $V$.  This allows us to see, directly, how the deviation between the VDW equation and the ideal gas equation changes with volume.
Let's start at high volume, and move to the left.  At high volume, the attractive term dominates.  As the volume decreases, the molecules get closer together, so the attractive term has even more effect, increasing the difference between pIdeal and pVDWN2. This continues until $V=0.154 \,\ce{L}$ [I determined this minimum by setting the derivative of (pN2VDW - pIdeal) equal to zero, and solving for $V$.]  Beyond this, the increase in pressure due to the diminishing free volume becomes greater than the decrease in pressure due to the increasingly strong attractive forces.  Eventually, at $V=0.108 \,\ce{L}$ the effect of the hard sphere volume just balances that of the attractive forces, and the curve crosses above the zero line.

Plots, Case II, higher T ($\ce{N2(g)}$ at $\pu{500 K}$)
Assuming $T = 500 \,\ce{K}$, if we solve for the crossover point as we did above, we get a non-physical answer of  $V = -0.222 \,\ce{ L}$.  This indicates the curves don't cross over.  We can see this from the plots:

Real Gases
The easiest way to see how the isotherms for real gases compare with those for an ideal gas is to look at the measured compressibility factor, $Z$:
$$Z=\frac{p V_b}{R T}$$
$Z = 1$ for an ideal gas.  Thus, for a given $V_b$ and $T$, $Z>1 \implies p_{real} >p_{ideal}$, and visa-versa.
From https://en.wikipedia.org/wiki/Compressibility_factor#/media/File:Z_Overview.png, we have the following diagram for $\ce{N2(g)}$.  It appears to show behavior similar to what the VDW equation predicts— at lower temperatures, we see a crossover from
$p_{real} > p_{ideal}$ at high $p$ (low $V$), to $p_{real} < p_{ideal}$ at low $p$ (high $V$).
However, as the temperature increases, $p_{real}$ reduces its tendency to dip below $p_{ideal}$ (see red curve).  However, a direct examination of the numerical data would be needed to determine if, at high $T$, there is no crossover at all.

Can even at very high pressure $V_m$ $\ce{->}$$b$ ?

+1 for your effort but I am confused as there are contradictory answers. Few confusions: (1) From the VDW equation you mentioned, when $V_m$$\ce{->}\infty$, it reduces to ideal gas expression but from the graph, $P_{real}<P_{ideal}$. $P_{real}=P_{ideal}$ point is at a much lower $V_m$. (2) Assuming the graph do cross, can we say that at very high $V_m$, a real gas is more $perfect$ than an $ideal$ gas?

It looks like there is something wrong here: at volume 0.25 dm$^3$/mol the vdw pressure for nitrogen is 85.56 bar and ideal gas  90.8 at 273.15 K using your constants.  You seem to have used a V high temperature in your plots, are you sure the constants used apply when this is the case ? Also look at the scales in you crossover graph they appear to show no difference $\delta x=\delta y=0$ ? What value for gas constant did you use?

You sometimes write 'van der Walls' instead of the correct 'van der Waals'. If you again edit it, you might want to change that.

Please have a look at the MathJax Guides here and here. You are using it wrong. The macro \ce{…} needs the curly braces to work properly, your way of writing it \ce N_2(g) is equivalent to \ce{N}_2(g), however \ce{N2 (g)} is what you want. Don't use it for units either. Incorrect: a=1.37 \ce{ \frac{bar L^2}{mol^2}} Correct: a = \pu{1.37 bar L2 // mol2}.

@Apurvium (1) According to the VDW model, the pressure goes to infinity in the limit as  Vm -> b.  Physically, it's much more complicated, because gases aren't hard spheres, and they will typically become solids before extreme pressure pushes them into a repulsive regime. I.e., energetically, gas interactions are always attractive, and it's entropy that keeps them apart.

@Apurvium  (2)For all the plots that go to higher volumes you can see, from the RHS of these the plots that, as Vm increases, the VDW and IG curves get closer.  The curves never exactly overlap on these plots, because they only become the same as Vm->infinity, and you can't plot to infinity.   The only point at which the WDW and IG curves show identical pressure is at the crossover (if there is a crossover).

@porphyrin Thanks, I should have caught that. Turns out it's tricky to use units when plotting in Mathematica, so I just made my quantities dimensionless, which eliminated the scaling error (other functions, like Solve, handle dimensioned units without issue). Also caught another error. Please take another look.

Can we conclude a final mathematical expression between $p_{real}$ and $p_{ideal}$? The one which is written in the question or assuming $V>>nb$, $p_{real}=p_{ideal} +(n^2/V^2)(bRT-a)$ as posted by Lllt?

@Apurvium  Yes, I agree with the expression dervied by Lllt for the high-volume regime ($V_b>>b$).  Assuming you are above the VDW critical temperature ($\frac{8a}{27bR}$), if there is a $T$ below which $a<bRT$ then, at high volumes,  the attractive forces will win, and you will see a crossover.  Otherwise, you will not, because the VDW pressure will always be greater than the IG pressure.

@Apurvium The accurate expression is the one which I wrote in my answer initially, however using certain approximations I tried to explain the graph mathematically.

@Apurvium  See the note I added at the end about the behavior of real gases. They also show a crossover at lower temperatures.

The lower right part of graph indicates that real gases exert less pressure than ideal gas for a fixed volume. This happens because the ideal gas molecules are not attracted by surrounding molecules (therefore if an ideal gas molecule move at velocity $V_1$, It will strike the wall with same velocity, provided it doesn't collide with another molecule) while real gas molecules face attraction from surrounding molecules as they move towards the wall ( thus if a real gas molecule moves with a velocity $V_1$, it will strike the wall with somewhat less speed as nearby molecules' attractive force will slow it down)
In the upper left part of graph, you can see that to exert same pressure , we need more of a real gas then an ideal gas or similarly at a certain volume of both , the pressure exerted by real gas is more than that of ideal gas. This can be explained by fact that molecular size of ideal gas particles is negligible while that of real gas particles is considerable. If you take equal volume $V$ of both, the ideal gas has occupied complete volume $V$ while real gas has occupied effectively somewhat less volume (which is $V'=V-nb$). Therefore even if we have same number molecules of real gas as that of ideal gas, real gas molecules have lower volume to move into thus they exert more pressure.
The Van der Waals equation of state is
\begin{align}
  p &= \frac{nRT}{V-nb} -\frac{an^2}{V^2}\\
    &= \frac{\frac{nRT}{V}}{1-\frac{nb}{V}} - \frac{an^2}{V^2}\\
    &= \frac{nRT}{V}\left({1-\frac{nb}{V}}\right)^{-1}
       -\frac{an^2}{V^2}\\
    &= \frac{nRT}{V}\left({1+\frac{nb}{V}}\right) - \frac{an^2}{V^2}
       & \text{assuming } nb &\ll V\\
    &= \frac{nRT}{V} + \frac{n^2bRT}{V^2} - \frac{an^2}{V^2}\\
    &= p_\mathrm{ideal} + \frac{n^2}{V^2}(bRT-a)
\end{align}
Depending on whether $bRT$ is less than or greater  than $a$, you have $p$ less than or greater than $p_\mathrm{ideal}$.
Also note that  when $V$ is very large the second term can be neglected and in that case real gas tends to ideal conditions.

Please compare your answer with the equation also for both situations, i.e. large and small volumes.

@Apurvium If repulsive forces become dominant then why should pressure be less than $P_{ideal}$, it should be more

Yes, you are right, I will repost.

The $T$ is constant & I think the $V$ is volume of container. The only way we can increase the $P$ is by reducing the volume of container by a piston or so. When we do so, the repulsive forces also become significant for real gas. Now, the pressure exerted at the walls of the container should be more than $P_{ideal}$. This is not according to the equation between $P_{ideal}$ and $P_{real}$ but yes, according to the graph. At high volume the situation is vice versa.

$\frac{1}{1-a} \ne 1+a$  //  $\frac{1}{1-a} \approx 1+a$ for $|a| \ll 1$. So  there is the condition $nb \ll V$

@Poutnik yes, Thanks for pointing out, I edited my answer

(1) Here, the assumption is $V>>nb$ for all cases (which I don't know, but may be true). (2) When $bRT>a$, $P_{real}>P_{ideal}$ and when $bRT<a$, $P_{real}<P_{ideal}$. Is there any intuitive explanation for this? The units of $a$ and $bRT$ are $\pu{J m3 mol-1}$, which does not give any insight. (3) When $V$ is $very~very~large$, $P_{real} <P_{ideal}$ from the graph but $P_{real}$$\ce{->}$$P_{ideal}$ from the equation.

Please note that the symbol for pressure is a lower case $p$.

It's an assumption , certainly not true for all cases

intuitive explanation is already given in above paragraphs, I try to frame mathematical explanation also

Yes $P_{real}<P_{ideal}$ and tends to or approaches $P_{ideal}$

When the pressure tends to infinity, the volume of the ideal gas tends to zero, and the volume of the real gas does not. It tends to the proper volume of the molecules, which is greater than zero. This is exactly what you see on the upper part of the picture.
And when the pressure is extremely low, the volume increases but not as much as calculated with the ideal law, because there is some small attraction between the molecules. This is exactly what you see on your picture if you draw a horizontal line at its lower part

At very large volumes and low pressures, the behavior of a real gas should approach that of an ideal gas.

Sorry but I have a confusion. If that's the case then aren't the both graphs (real as well as ideal) showing the fact of non zero volume (reason being that both look like a rectangular hyperbola to me)?

The ideal gas is a hyperbola. The real gas has a non analytical curve.

@maurice At a constant $V$, if we compare the pressures vis-a-vis the equation between $p_{ideal}$ and $p_{real}$?

Related

The ideal gas pressure is always above the 'real' gas pressure as given by the van-der Waals eqation
$$p=\frac{RT}{V_\mathrm{m}-b}-\frac{a}{V_\mathrm{m}^2},$$
where the volume is the molar volume $V_m$. The molar volume varies from say $0.1$ to $\pu{5 dm^3/mol}$ and by experiment constant $a$ in units $\pu{dm6 bar mol-2}$  is numerically greater than $b$ in units $\pu{dm3/mol}$ and the gas constant $R=\pu{0.08314 dm3 bar mol-1 K-1}$.
At large volume the second term tends to zero and the first to $RT/V$ the ideal gas value. At small volumes the second term becomes larger and negative and reduces the pressure from the ideal behaviour. Thus the ideal and real curves do not cross and the ideal is always greater than the real.
The interaction between molecules ensures that the real pressure is lower than in the ideal case. (You can imagine this in a hand waving sort of way by imagining that a small fraction of molecules form dimers so that the expected pressure is not measured because some molecules are paired up.)
Thus the figure in the question is wrong on a number of points.

I'm adding some notes here about the vdw equation as there is confusion in the comments.
First it is clear that the van-der-Waals equation can produce negative pressures. It is an empirical equation that produces largely qualitative results. All that has to happen to get negative pressures is that the second term exceeds the first in the equation above. Negative pressures are, of course, unphysical and this shows that this equation has to be used with great care otherwise completely misleading results will be obtained.
Second, at large volumes the $V_m-b$ term is smaller than $V$ because $b$ is positive, so the pressure will exceed that given by the ideal gas but the effect is small. A typical molar volume of $0.25\;\mathrm{ dm^3/mol}$ vs a  typical value of $b=0.02\;\mathrm{dm^3/mol}$, so small as to be physically meaningless but mathematically possible, what I mean is that we would normally take $V_m-b \to V_m$ at large $V_m$.
The important point about the two curves in the figure is that the interaction between molecules, (due to polarisability/ dipoles ) makes the experimental pressure less than that of the ideal, e.g. methane at 273.15 K pressure is 78.6 bar by experiment but 90.8 by ideal gas law and 72.9 by van-der-Waals. The van-der-Waals eqn. tries to model the experimental results and does so quite well but should not be pushed to the mathematical limits as here it fails.

Your answer is definitely enlightening ,are you sure that those curves never cross each other (I am asking because the figure taken above is from the national book of India)

They never cross, you can see that in the limit when V is very large the ideal gas law is produced to that they tend to the same value. Try this with some numbers, say at 300 K for methane  a=2.3 , b=0.043 in the units I gave and V=1 vs V=10 etc.

I am afraid the national book of India is wrong !

@Maurice please see figure at pg.14 https://ncert.nic.in/ncerts/l/kech105.pdf

@porphyrin It is true, the van der waals gas curve is always below the ideal gas curve

@Theorist see my comment above, you seem to have used a v high temperature which would be the effect of using 8.314 instead of 0.08314 dm$^3$bar/mol/K for the gas constant

Correcting my earlier comment: Sometimes the curves do cross over, sometimes they don't. As $V_m$ gets large, whether the attraction can win out over the excluded volume is determined by the magnitude of T, and the relative values of a and b.

I don't have the rep to downvote this, but it is plain wrong. While the second term indeed pulls the pressure below that of ideal gas, the first term produces pressure higher than that of ideal gas (RT/V), because the denominator is smaller. See the answer by "theorist" for a practical example.

@TooTea It is not 'plain wrong' as you write, the curves are not sketched but  calculated using the vdW and ideal gas equations using experimental constants for methane. I have explained with some extra text in my answer why taking the vdw equation to its limits leads to an incorrect understanding, i.e. the difference between strict mathematical usage and using a scientists physical understanding of what the equations are describing.

@porphyrin Sure, for methane at ambient temperatures, the curves never cross. What's wrong with your answer is that you then generalize this to all gasses. See the answer by "theorist" for counterexamples where the curves do cross for real gases under non-extreme temperatures.

@Tootea, When you compare the ideal gas law with the empirical vdw you can get negative pressures and crossings etc. What one is doing  is only comparing two equations nothing else - no physics/chemistry is involved. The original question showed real gas vs ideal gas I used vdw to model a real gas because I don't have  any real gas data at hand.  Within the limits I used what I said is correct. It is true that I should have added that the vdW eqn can give negative pressures so is limited in what it can predict. The main point is that real molecule interact with one to reduce pressure vs ideal.

@porphyrin My point is that at high temperatures and/or densities, the repulsion always wins (at least for real gas, even though the vdW parameters for some gasses might be unable to capture this) and thus the real pressure becomes higher than ideal (compressibility factor exceeds 1). This is inevitable because ideal gas means no repulsion, no molecular volume at all. If you compress any (supercritical) gas hard enough, at some point the molecular volume will overcome any other interaction (Pauli repulsion always trumps everything with shrinking interatomic distance).

@TooTea,  quite so, I agree, but this is at an extreme.  The figure in the original question must relate to the gas at sort of normal conditions one might meet as its a from a general textbook, it is very misleading for students otherwise, i.e. to show extreme case (as it does) and ignore more general case.  My comments related to that general position, interactions between molecules lead to a reduced pressure vs ideal. From this position you can then go on to extremes.

Mathematically, the curves of real and ideal gas equations can intersect.
$(P + \dfrac{an^2}{V^2})(V - nb) = nRT$
Plugging $P= \dfrac{nRT}V$ gives  $V = \dfrac{anb}{a-bRT}$.
Thus the curves will intersect (in the first quadrant) if $a>bRT$, otherwise they won't.
Graphs intersect:

Graphs don't intersect:

Desmos


text:
Mechanism involved in bromine water test

$$\ce{CH3CH=CH2 ->[Br2] CH3-CHBr-CH2Br}$$

In the reaction mechanism, why does the nucleophile attack at the 2nd carbon atom? Shouldn't it attack at the 1st one, so that in the transition state, the carbanion is relatively more stable?

Which site is the more stable carbocation in the transition state?

There is no carbanion in this reaction mechanism.

There are no carbanion in this reaction.
In bromine water, $\ce{Br2}$ is partially made of $\ce{Br+}$ and $\ce{Br-}$ ions. Then $\ce{Br+}$ attacks the double bond, producing a carbocation, with $\ce{Br}$ at the outer $\ce{C}$ atom, and the + charge on the central $\ce{C}$ atom. This is the most stable carbocation. Then it will react with the $\ce{Br-}$ ion which will be attached on the central $\ce{C}$ atom.
It also should be mentioned that a secondary reaction may happen where $\ce{H2O}$ reacts with the transition state carbocation. As a result, the molecule $\ce{CH3-CHOH-CH2Br}$ is produced.

Your explanaion is okay but mechanism is wrong. Did you hear about cyclic bromonium ion mechanism?

Of course I know the bromonium ion mechanism. I have not mentioned it, because I think it might have been avoided here.

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text:
Does reaction of alkali and alkaine earth metals' reaction with water an acid-base reaction?
In my chemistry notes the teacher gave me the note that reaction of alkali, alkaline earh metals' reactivity with water, acids(e.g. H2S) increases down the group and bases(e.g.NH3) decreases down the group,
And also I have some doubts that
-alkali metals such as Li first react with water and produce oxides first and those oxides react with water and produce hydroxides as end product, is this true?
-and normally water has H2O molecules, (H3O+,OH- ions at small concentration). And when alkali metals react with water do they react with H2O molecule or H3O+ molecule?, If someone could give me a reaction mechanism of an alkali metal with water?
-do alkli, alkaline earth metals act as bases when reacting with water?
Any help would be most greatfull

Na and K react with the O-H bond of the water molecule. This removes the H atom that is reduced to an H atom and then to a H2 molecule. It is not an acid/base reaction. It is a redox reaction. Na or K atom is oxidized at the end. No metal oxides are produced as intermediating state.
Alkaline and alkaline earth metals act in the same way.

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text:
What happens to air at 1000 degrees Celsius
I was looking at the graphs for the density of air at different temperatures (link below) and I was wondering if air would not react with itself at high temperatures? I read that oxygen and nitrogen react at 1600 degrees Celsius, but even below that average temperature (because of the Maxwell Boltzmann distribution) I’m guessing the reaction takes place? Also would it just catch fire with contact with anything in the environment like carbon / almost anything? I’m wondering about the hypothetical use in hot air balloons with very high temperatures and low volumes.
https://www.engineeringtoolbox.com/air-density-specific-weight-d_600.html

Radiation loses increase proportionally to the fourth power of the temperature (Boltzmann law), so your baloons would be extremely inefficient

Are you aware 1000 Deg C hot surface thermally radiates about 150 kW/m2 ? Even if surface itself was 500 Deg C, it would be 20 kW/m2. If we consider the mass 500 kg, and lift about 1 kg/m3, it would have radius 4.9 m and surface 305 m2. That is 6.1 MW for 500 deg C surface. It would need impossibly huge power source and the useful load would be baked in a huge grill.

For just 200 Deg C surface, it would be cca 2.8 kW/m2 and 866 kW.(orvsone 10-15% less after subtracting background radiation) Still very mighty power. Some light aircraft would be more energy efficient.

At $\pu{1000^oC}$, oxygen and nitrogen will not react substantially or probably even noticeably if at all. Besides that, whatever organic material you are using for a hot air balloon will most definitely ignite immediately. The average temperature that most organic materials char and eventually ignite is $\pu{270^oC}$. At almost double the kinetic energy at $\pu{1000^oC}$, most things will almost instantaneously ignite. Note that in air, temperature diffuses rather quickly so even if the fire or whatever you are using to get to $\pu{1000^oC}$ will probably be significantly colder when it reaches anything flammable. In fact, most fuel used in normal hot air balloons burn at over $\pu{1000^oC}$, but it diffuses very quickly that the average temperature of the air in the balloon is around $\pu{110^oC}$

Perhaps to make clearer use of terminology it is better to say "Note that hot air diffuses rather quickly so even...."?


text:
Is there any difference between medical and cleaning ozone?
What are the main differences between medical ozone (auto-hemotherapy, …) and cleaning ozone (for example, the kitchen cleaning devices for vegetables and food). I noticed that they can produce a lower amount of ozone per hour while the medical ones have bigger amount of production.

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I see as the only difference the ozone concentration and the gas mixture purpose. The ozone itself is the same, there is no medical nor cleaning ozone. Breathing ozone containing gas is harmful. It is a potent gas initiating the SO2 imission damage of spruce trees, and is the key component of the oxidative smog of the Los Angeles type, that can cause defoliation of trees.

“Medical ozone” is a mixture of 0.05 to 5 vol% of ozone produced from medically pure oxygen (usually via silent electrical discharge) and 99.95 to 95 vol% of oxygen.
Using air for medical-grade ozone production is strictly forbidden due to formation of hazardous nitrogen oxides $\ce{NO_x}.$
Note that ozone therapy is still considered an alternative medicine.
Ozone used for cleaning doesn't have such constrains and its concentration and production method vastly depends on application area and regulatory standards therein.
For example, for water treatment Russian federal standards prescribe the ozone concentration in the ozone-air mixture from 15 to 22 mg/l based on the required efficiency and level of pollution of the water source, and average contact time of 5 to 15 min.
Prolonged exposure or increased ozone concentration during water treatment may lead to unwanted effects such as problems with coagulation and extensive phenols formation.


text:
Why dried, closed coconuts go through carbonation?
Some dried, closed coconuts go through carbonation:

The white meat of the coconut becomes disintegrated and smeary
The liquid becomes carbonated with a bit sour-salty tasted coconut water (which reminds the taste of a water and baking soda solution) and a grey to blue color
If the grey-blue liquid given enough time (inside a dried, closed coconut) it could become yellowish and possibly contaminated (I have no idea if the yellow color is due to further carbonation or bacteria)

Given that a dried, closed coconut is a dry and sealed by hard, fibrous shell,
Why dried, closed coconuts go through carbonation?

Hello to mods; if you think this is a better fit for biology SE, please don't hesitate to migrate. I assumed here it will be better due to biochemical process.

I can recall the bad taste of very old coconut water. It does get fermented. Sugars are converted to carbon dioxide and sugars to organic acid.

Hint 1: what causes champagne to become carbonated? Hint 2: What is the etymology of the English word enzyme?

@DrMoishePippik: Are you 100% sure that yeast causes carbonation of coconut meat and liquid? When I search yeast and coconut, all I find is articles saying that coconut oil can be used as a remedy for yeast infections. There seems to be evidence for destroying yeast, but no evidence for containing yeast. Do you have a source?

@user1271772  actually yeast can ferment coconut oil see the link in my comment

@JohnDoea see this link : https://www.sciencedirect.com/science/article/pii/0740002086900092

why does it say modified by organicgem and i dont see any modifications?


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Comparison between solubility of amines and alcohol
Why are alcohols more water soluble than amines for similar molecular weight.
I know this is a duplicate question . The reason i posted it again was that on the original question i posted my answer and requested others to correct me if i am wrong so that i could deeply understand the concept. So I am posting my answer here again please comment on it.

I think it may be due to van der waals force (v.w.f). For both alcohols and amines the v.w.f acting between them and water molecules is keesom force i.e. dipole dipole interaction. In such case v.w.f is directly proportional to molecular mass , but if the molecular mass is also same then v.w.f is directly proportional to surface area of molecule which is inversely proportional to branching. Hence since amines have more branching than alcohol they have less surface area and hence less van der waals force. Due to weaker v.w.f amines are less soluble than alcohol.
Please correct me if I am wrong

If respected people have the time to close my question could they please get some time to solve my query

What is the question ?

Why are alcohols more water doluble than amines for similar molecular weight?


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Why does the colour of the same compound (e.g. magnesium nitride) differ?
When I am searching the colour of $\ce{Mg3N2}$ I came across photos showing different colours: greenish-yellow on Wikimedia,
and some other just yellow on some Google Images results.
I am eager to know what causes some samples to be bit greenish.

Maybe due to impurities, the color differs.

Particle size and polymorphism would also affect the color of a solid.

Privacy note: be very cautious when you post a link to the Google search results. Not only the address string looks messy, it also contains private info such as your device and browser names in plane text. There are several browser extensions that fix that and create a short nice-looking URL without using services like tinyurl; at the moment, I personally use ClearURLs.

Minor differences in color are likely either due to impurities or due to intrinsic crystal defects such as f-centers. For example, $\ce{ZnO}$ is often a shade of yellow, rather than pure white, due to those defects. In alkali halides, h-centers are similar defects giving rise to colors. You can easily create color centers in $\ce{NaCl}$ with a spark coil in just a few minutes.
Another possibility is that there are mixtures of different allotropes, or polymorphic forms. An extreme example is transparent diamond vs. black graphite. BTW, diamonds having varied colors due to impurities (e.g. nitrogen) and to intrinsic defects are useful in studying quantum phenomena, as well as gems.


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When aluminium is ignited in atmosphere, will it form only aluminium oxide or also aluminium nitride?
We know when lithium is burnt in air, it forms lithium nitride and lithium oxide. I am wondering if this applies to aluminium (other metals) also?

Magnesium also forms the nitride as well as the oxide when burning in air.

According to the paper by Gromov and Vereshchagin [1], it is indeed formed when burning aluminium, but you could have predicted this simply by considering the standard enthalpy of formation of aluminium oxide $\Delta_\mathrm{f}H^\circ (\ce{Al2O3}) = \pu{−1675.7 kJ/mol}$ which will be liberated upon ignition and this energy helps in breaking the strong nitrogen-nitrogen triple bond (bond energy is $\pu{−945 kJ/mol})$ and finally this would form aluminium nitride, which also produces some energy (standard enthalpy of formation is $\pu{−318 kJ/mol}),$ but the combustion here is not a simple one it requires high temperature as mentioned in the paper, I think this is to sublimate $\ce{Al2O}$ which will react with nitrogen and give rise to aluminium nitride.
As HYDR0GEN stated, this combustion isn't simple like throwing a match on aluminium sheet, it needs high temperature. According to the work by Trunov et al. [2, p. 38], as the size of aluminium particles decrease the needed temperature for combustion decreases and that's why researchers who wrote the paper used superfine aluminium powder:

Figure 5. Experimentally determined temperatures of aluminum
  ignition as a function of the used sample size.

References

Gromov, A.; Vereshchagin, V. Study of Aluminum Nitride Formation by Superfine Aluminum Powder Combustion in Air. Journal of the European Ceramic Society 2004, 24 (9), 2879–2884. DOI: 10.1016/j.jeurceramsoc.2003.09.020. (PDF)
Trunov, M. A.; Schoenitz, M.; Dreizin, E. L. Ignition of Aluminum Powders Under Different Experimental Conditions. Propellants, Explosives, Pyrotechnics 2005, 30 (1), 36–43. DOI: 10.1002/prep.200400083.

When citing third-party resources, make sure to provide a complete human-readable reference and not just "this link" to make your post robust against link rot. Also, always cite the original publication, not a reproduced rip-off of a questionable quality where there is typo in the axis name ("Diamter") and illiterate numbers formatting. I took a liberty to replace the illustration with with the one from original paper.

@andselisk thanks for help, it's my first answer

When the aluminum [powder] is ignited, it will cause a reaction between the aluminum and oxygen in the atmosphere. The required temperature for the aluminum oxide to be formed is almost 3000+ °C.
$$\ce{4 Al + 3 O2 -> 2 Al2O3}$$
In addition, aluminum nitride can't be formed when the aluminum powder is ignited in atmosphere, because the aluminum nitride starts to decompose into aluminum and $\ce{N2}$ at a high temperature under atmospheric pressure.

Aluminium needs high temperature to be ignited but that's why the researchers who wrote the paper used ultra fine powder of aluminium to be ignited easily . See this link :https://www.researchgate.net/figure/gnition-temperatures-of-different-samples-of-aluminium-samples-reproduced-from-Trunov-et_fig3_240497487


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Does water in big cities get heated (boiled) for treatment?
I live in Bangkok and read how water comes to customers here:
I understand from the chart that water go through Thon Buri treatment plant (west Bangkok) and Mahasawat treatment plant (east Bangkok) before being pumped and transferred to the customer.
In the chart I clicked the links for these two treatment plants and read information in other parts of the site and understood water is treated at least either by chloride, fluoride and sulfate based chemicals but it wasn't clear to me if water is purified from bacteria by heating (boiling).
Does treated water in big cities get heated (boiled) to destroy bacteria as well as proteins, and if so, where does all the amino acid residue go to?

The most important parts in freshwater treatment is clarification (get the turbidity, caused by microscopic impurities, down, by sedimentation with the help of clarifiers). That already takes care of the greatest part of bacteria and other single cell organismns. Then comes filtration, and the tiny rest that still comes through there is killed via chlorine (or, more modern, ozone).

No. Heating usual water would be extraordinarily expensive. In all water treatment plant, all over the world, bacteria are destroyed by adding chlorine gaz Cl2 or similar substances delivering Cl2 when dissolved in water. The bacteria are killed by chlorine. So the customer drinks water containing dead bacteria. It does not hurt.

I don't want to be gross, but today I saw a man spitting on a paved road. I told myself, if someone spits or pisses in water reservations, shouldn't these be boiled at least from an aesthetic perspective? I guess I must go for "mineral water" (whatever that means); maybe these are indeed boiled...

@JohnDoea Isn't it against the law to spit on a paved road, aren't there heavy fines or stringent punishments for such violaters?

@Zenix I can't speak about the law of every state but I can tell you that I would prefer boiled plus well chemically treated water on not...

Chlorination is very oldschool (creates lots of toxic chlorinated compounds, ozone is used instead in modern installations), and before chlorination or other chemical treatment, you need to filter the water so good it would be drinkable already (the pipes would still foul over time with this water, hence the chlorination).

@johnDoea - LOL, you do realize where the waste from  fish, birds and turtles in a river goes, don't you?

@Karl In fact, ozonization is more aggressive than chlorination by creation a bunch of very suspicious compounds.

@Poutnik the problem is less acute toxicity but more environmental stability. have you got a link?

@Karl unfortunately not. Many of my knowledge comes from era the http protocols has not existed yet, or was not easily available. :-)

@JohnDoea Mineral water is definitely not boiled.

@Poutnik that is not less depressing for me --- in small countries like Cyprus, Jordan, Israel, Armenia, Tunisia so forth, I would assume residents are likely to put many body liquids into water reservations. Perhaps desalinated tap water are the best bet for me personally.

As mentioned above, boiling all the wastewater would be incredibly expensive. One technique not mentioned in the other answers is that the wastewater treatment center has portions of infrastructure where the water sits in open air pools for a certain amount of time. The UV rays from the sun kill some bacteria.

Joe, shouldn't such water be covered to prevent more bacteria to enter than to be destroyed as well as to prevent dirt - I mean, shouldn't there be some transparent (glass?) covering?

Any bacteria to enter from the air or particulate probably wouldn’t be pathogenic as would be some species in the wastewater. It probably comes down to cost though, the pools are rather large

They also bubble O2 in to kill anaerobic bacteria

UV irradiation for water sterilisation is done, but surely not in the open with sunlight. What a nonsense. You are aware what a perfect habitat for algae that would be?

Bubbling with oxygen (air, rather) is a common wastewater treatment method. For freshwater, you want 99.999% of the bacteria out, not just dead.

@Karl I am referring to wastewater treatment and if you look into it what I have said is true. The water is not stagnant but is rotated constantly

The original question is about freshwater. And for wastewater treatment, the irradiation by sunlight is completely irrelevant in terms of "killing bacteria". You want the algae to grow in wastewater treatment, because they eat a lot of unwanted compounds in the water and can easily be filtered out afterwards. Algae need light.


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How can an Organic chemist know the chemical formula of a natural product she isolated from an organism?
As a non scientist I several times wondered how can an Organic chemist successfully isolate a single molecule which a natural product of an organism;
Whether a plant-part/mushroom-part/algae-part or a coral-part or even from an animal or human organ in autopsy (neurotransmitter/hormone and so forth).
I also wondered how could she go further to know the natural product's (exact?) chemical formula and give such newly discovered molecule a name.
Given that molecules are such small entities, way smaller than the smallest uicellular organism and are hard to be photographed uniquely with current microscopal technologies (AFM/STM); it is unclear to me How could she know if a plant part contains a certain molecule (say caffeine or nicotine) so precisely that she could tell:

This is what I will now name Caffeine → not Theobromine
  or any other Methylxantine → just Caffeine, pure and exact and unequivocal.

Practical examples for my problem are:

How did organic chemists isolated THC from various Cannabis plant-parts and gave it a formula and a name, although today it is known that Cannabis plant parts has tens of cannabinoids?
How did organic chemists isolated Ganodermadiol in some Ganodrma mushroom part/s and gave it a chemical formula and name and claimed it is similar to steroidal hormones?
How did they know they weren't wrong?

My question
How can an Organic chemist know the chemical formula of a natural product she isolated from an organism?

Because they perform analysis such as LC-MS or NMR using accredited methods. For the high-precision structural models there is single-crystal XRD and neutron diffraction.

How do you know about AFM and STM, but not methods mentioned by Andselisk? Not to say such questions were already asked and also tended to be too broad https://chemistry.stackexchange.com/questions/32240/how-can-an-inexperienced-chemist-determine-the-chemical-structure-of-a-molecule

@Mithoron I saw about AFM/STM in Wikipedia article molecule; if the same question was asked then it should be marked as duplicate.

https://chemistry.stackexchange.com/questions/34048/can-molecular-structure-be-determined-without-modern-characterization-methods https://chemistry.stackexchange.com/questions/771/how-were-elements-and-molecular-formul%c3%a6-determined-in-the-old-days

Similar question https://chemistry.stackexchange.com/questions/86063/how-do-scientists-figure-out-the-chemical-structure-of-a-given-compound was closed as duplicate of https://chemistry.stackexchange.com/questions/57784/has-anyone-even-taken-a-picture-of-a-molecule-to-confirm-the-geometry-predicted still it's somewhat different...

This is so broad it should be trimmed to be good dupe target. https://chemistry.stackexchange.com/questions/7269/atomic-properties

How can an Organic chemist know the chemical formula of a natural product she isolated from an organism?

There are two levels of answers. One is historical and one is modern. Historically, determining the chemical formula for had been a trivial job for most small or medium sized molecules. You do a combustion analysis, determine how much carbon dioxide, and water were produced. By doing other classical element detection tests, you would determine if it had halogens, nitrogen and sulfur. These were the most common approaches. The determination of molecular weights was done by classical physical chemistry experiments.
Today you would still do combustion analysis and elemental analysis as a first step and determine the CHNS ratios. Then you run a mass spectrum of the compound, and assess the molecular weight of the compound.
Determining a structural formula may take a long time and others have already mentioned in the comments. Today advanced NMR techniques, high resolution mass spectrometry, X-ray of crystals (if they exist), all help in structure determination. These things have become routine now.
Nobody routinely determines the molecular structure by AFM or STM as far as I know.

Determining the structure of a medium sized molecule used to be a take years of determined work before xrd and nmr.

Yes, because only chemical methods existed. I can only praise those early German organic chemists of the 18th and 19th century who did extensive organic reactions to determine the structure of a molecule by bits and pieces.

Working out the structure of a compound (organic, organometallic or organomain group) is not a simple matter. It is something which can be very intellectually taxing. I have done it many times in my life and I can tell you that for some thing you can do it in a matter of minutes while for other things it can take weeks, months or even years)
The first problem is that you need to know that your substance is sufficiently pure that the peaks in spectra are due to your compound of interest rather than some impurity.
For an organic compound if it is a pure substance which you have plenty of, my advice is normally.

Get an infrared spectrum of the substance so that you can identify some of the functional groups. Look for multiple bonds, you can look for evidence of ketones, amides, esters, carboxylic acids, nitriles, metal carbonyls, benzene rings and alkenes. In my experience it is sometimes hard to see the C C stretch for acetylenes but you might get lucky. This will give you a list of mental jigsaw bits.
Get a proton NMR of the substance, this will either be a great help or close to useless. If you are lucky you will be able to identify parts of the structure. For example if you have a phenyl group, a methyl close to TMS and two multiplets which appear to be all coupled to each other. Then using the 3J couplings you could tell if you have a propyl group.

Alternatively if you had a isobutyl group then the integration heights would be different and the coupling pattern different, for the isobutyl you would have doublets for the methyls instead of the triplets you see in the propyl.
You should also look for signs of symmetry, for example if you have benzopheone (diphenyl ketone) then you will have seen in the IR a ketone (which is slightly lower than 1700 as a result of conjugation) and only one carbon / hydrogen group. As a result of symmetry both phenyls are magnetically identical.
Equally if you have 1,3,5-trichlorobenzene you will have a single singlet in the aromatic part of the proton spectrum. Compare this with the multiplets you would have for 1,2,4-trichlorobenzene.
If you see a very large H H coupling then it is possible you have a ring or some other thing where the conformation is locked. This would be a substance with a 2J coupling. Note this it could be a great clue. For example in the structure of some penicillin derivatives you can see 2J couplings where two different hydrogen on the same carbon have different magnetic environments.
Look in the proton for anything which is not usual, for example cyclopropanes have a very distinctive shift which is close to zero ppm or even at - ppm.
If the proton was little use or you can not understand it then my advice is to try some more things.
If the proton NMR is very congested, then I would suggest running it again in a different solvent. Commonly people use CDCl3 for NMR. I always say that if the spectrum in chloroform is impossible to understand as a result of being congested then I run a sample dissolved in D6 benzene. The change of solvent will alter the solvation and the proton environments oftein change in their chemical shift, this can make some areas of the spectrum more easy to understand. In this way you may be able to get a clearer look at it and as a result you might be able to understand the structure better.
Another trick which makes proton NMR more simple is to use double irradation, this can allow you to make the spectrum more simple by removing one coupling effect at a time. This will make the spectrum more simple. If it solves you problem then be thankful otherwise note down any new insights you have into the substance and move on.
I would also say that getting a carbon-13 NMR can be a great help, this will help you understand the substance more. You will learn quickly how many unique carbon centres there are in the substance.
You should also consider the question of will a 2D NMR experiment help you, for example you can use a COSY or a CH correlation. For NOSEY you need to exclude all oxygen from the NMR sample. For this you will need to use deoxygenated chloroform and work on a vacuum line and seal up the tube with no air getting inside. This will take some skill and time to do.
If you are stuck you should consider getting a mass spectrum of the substance, this can give you an idea of how large the thing is. Be careful with mass spectra. It is sometimes possible to get a nice spectrum off a tiny impurity which is present in the sample while the main compound does not give a strong nice spectrum.
The mass spectrum will give you an idea of the formula, you can get the formula weight. One thing which is worth doing is to look at the isotope splitting pattern of the molecular ion. For example bromine is present in the form of two isotopes with about equal abundance so for a monobromo compound you should expect a doublet. If you can see such splitting in the molecular ion then it is a good sign you have a heavy atom in the substance. Keep in mind iodine only has a single stable isotope.
Next trick I have is to use a sodium fusion test, now this test has been banned in a lot of universities as undergrads has made loud bangs or started fires with it. But when used with care it is perfectly safe. I like the test as a method of detecting nitrogen, sulfur or halides in an organic substance. Heat a SMALL pellet of sodium until it melts. Then add some of your thing and carefully reheat. There is a great debate on how best to do this, and it takes some practise. Heat the mixture up until it glows and plunge it into cold water. It may go bang, pop, fizz etc when you do this. Then boil the water mixed with broken glass and filter it.
Test the filtrate for halide with silver nitrate, if it goes cloudy then add ammonia. If the solid redissolves then it is likely to be a chloride while if it stays cloudy then it is likely to be either a bromide or iodide.
Test the filtrate by adding ferrous sulfate, boil it and then add acid. If it goes blue on adding the acid then you have made Prussian blue and you had some nitrogen in the organic substance.
If you can detect either thiocyanate or sulfide in the water then you had sulfur in the organic compound.
Add these new facts to the list of facts you have and see if you can work out a structure which makes sense for the organic molecule. Sometimes you can work out several possible structures based on NMR and IR. For example toluene and methyl benzoate will look very similar in the proton NMR.
Next take the possible structures and see if you can distinguish between them by molecular weight and which heteroatoms you have found in them. For example diphenyl sulfide and 4,4'-dihydroxybiphenyl would have the same molecular weight. But one will have sulfur in while the other will not. Also the fine detail in the mass spectrum on the molecular ion will help sort these two out.
Now if you have stared at it for a while and you still can not solve it, then it might be time to get a UV / vis spectrum. This is good when you have a conjugated system present in the molecule. It can be used to give you an idea of the size of a pi system.
If you see any strange couplings in the proton NMR spectrum where you can see one thing which couples to something which can not be seen then you might have a phosphorus atom present. Some years ago a student showed me a strange NMR spectrum with a doublet they could not understand. I spotted that the sample contained some HMPA, the doublet was in the right place for HMPA.
If you have phosphorus or another heteroatoms which couples such as selenium or platinum then you can try getting more NMR. For example consider the phosphorus spectrum you would have if you had a Pt binding to a triphenyl phosphine and to a dppe (1,2-bis diphenylphosphino ethane) and a chloride. You would expect to have three different phosphorus environments. Also three different coupling constants.
If you have a Pt with two PPh3 ligands and a dppe, then you would have two P environments and a very complex splitting pattern. Keep in mind also that the Pt-195 will couple with the P nuclei giving sidebands on the P signals.
The key thing to do is to keep gathering more data and trying to look at the clues from the different bits of the data to see if it will fit together to give a structure. If you have multiple structures then you can attempt to use physical properties to distingish between them. If the melting points are very different then you can use a copy of the rubber book, a melting point rig and your sample to work out which it is.
In the old days the gold standard was to take an authentic sample of the thing and mix it with your suspected sample of the same thing. Then to go and do the melting point at the same time as pure samples of the authentic sample and your sample. This is the idea of a mixed melting point. If all three samples melt sharply at the same temperature then you have identified your thing.
Now if you do not have a winner, then there are several things you can do. You can try to form derivatives or degrade your compound in some predictable way. This might make new compounds which can be identified with greater ease. This will give you clues to what you have. For example treatment with ozone can cut up a molecule by cleaving alkenes, this will make it more easy to spot some fragments.
Now if you can grow a crystal of your thing, then my advice is to try for single crystal diffraction methods. Just be careful of the rouge crystal problem, even if you get a pretty picture of the atoms you will need to check that the spectroscopic properties of your bulk sample are right.
You can get some nasty shocks, I once handed in crystals of a organic phosphorous sulfur compound to discover to my horror that the crystals were of some sulfur (S8) which was present as an impurity. The S8 is impossible to see with NMR or IR, so I thought that I was handing over a pure sample to the crystallographer. That was embarrassing.
There are other tricks you can try but I think for most things that the rather long post I have given will sort most things out.

This question cannot be answered in such a Forum. Because there is no unique answer. The method for giving a formula is a huge process. It takes hours and days and weeks to the university students and professors to be able to answer this question. Sometimes a given process is specific for a given product, and not applicable to the next substance. It is a long and hard job. The chemist must perform a lot of tests and counter tests, each one giving only a part of the answer. Sometimes the substance may be irradiated, oxidized or hydrolyzed (treated by water), and by analyzing the irradiated, oxidized or hydrolyzed product, you can infer the structure of the original product. It is usually a long and hard process.
And of course, the chemical obtained after such a first treatment must often be then treated by an acid or chlorine or soda, then in series with acetone, acid and water, then with zinc powder, then with acid again, and so on. The number of successive operations may be higher than twenty.


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How would I synthesize ethyl 2‐oxocyclohexane‐1‐carboxylate from cyclohexanone?

I thought the 1st step would be deprotonation of a H atom to form the enolate. Then an aldol reaction, but the product I end up with is the alcohol, not the ketone.

Well, you need suitable reagent, to get such product...

Read this: https://pubs.acs.org/doi/abs/10.1021/ja01853a065.

You could use ethyl chloroformate (EtOC(O)Cl) to make the desired product, I would suggest that you react the cyclohexanone with sodium ethoxide in cold ethanol before adding the electrophile to it.

As you will see if you look at the reference mentioned by @Mathew-Mahindaratne,   the standard conditions for this transformation is NaH/KH in THF at reflux with diethyl carbonate (EtOCOOEt) with an acetic acid workup.
The Organic Syntheses procedure here uses dimethyl carbonate, but will be equally applicable for diethyl carbonate.

Diethyl carbonate can be sluggish in this reaction. The classic procedure is to use the more reactive diethyl oxalate instead and then decarbonylate carbon monoxide from the codensation product by heating over crushed soft glass. This reaction was discovered by taking the melting point of a solid condensation product in a soft glass melting point capillary.


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Is organic chemistry mechanisms limited to organic chemistry?
I am curious to know the mechanism of each reaction. Organic chemistry is satisfying for me because there is a mechanism behind each reaction which make a sense but when I Google for the mechanism of inorganic reaction I do not find any good results. I applied an organic reaction mechanism on an inorganic reaction and it worked as shown in the picture. By summarizing "Can we apply organic reaction mechanisms to inorganic reactions?"

Is it a stupid question?...lemme know to delete it.

It would be nice to see your proposed mechanism (try to reduce image size or try VPN if your ISP is known for blocking stuff), but in general the question boils down to what makes sense and what doesn't. Mechanisms work fine for molecular substances, but are quirky when one tries to use them for network solid, which prevail in inorganic chemistry. BTW there is an entire book devoted to the mechanisms in inorganic chemistry: Ghosh and Berg's  Arrow pushing in inorganic chemistry.

If I were you, I would refrain from assuming your question is stupid. If you did the research and gave it a good thinking, but you still cannot come up with the solution or rationalization, it is stupid not to ask.

Related: Inorganic chemistry reaction mechanisms.

Actually I googled a lot and can not found anything related...so I thought maybe my question is soo easy that there is no need to answer it.

@andselisk uploading still not working...By the way it was the reaction of CO2 + H2O -----> H2CO3

Mechanism: H+ attacks on CO2 ...carbocation HCO2^+ formation occur....in the 2nd step OH^- attacks on carbocation and leads to the formation of H2CO3....Can you please input mechanism in my question by editing it?

My gut feeling says that we can apply organic reaction mechanisms to discrete molecules reaction in inorganic.

You should be able to it by yourself. If you still cannot sort out image uploading, just append the comments to your question by [edit]ing.

I approved the edit, but I think you should really check what's going on with your device or internet provider. Probably clear the cache and delete cookies for *.stackexchange.com site, then log back in.

I am using stack exchange app

Official SE app for Android hasn't been updated in 2.5 years and doesn't support many features, see e.g. Meta, so I would recommend not to use the app and stick with the web browser.

Mechanisms of hydrolysis, of certain common reactions and of organometallic compounds is known to me. But i mostly mug up the reactions

@andselisk bundle of thanks for  recommendation.


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Question On Balancing
I recently worked on a lab which involved the following reaction:
$$\ce{HCl(aq) + CuO(s) -> CuCl2(aq) + H2O(l)}$$
In my description of the reaction I wrote the following
"The equation shown above represents a multi–step reaction between copper (II) oxide and hydrochloric acid in aqueous solution. To predict the products, we must recall that a metal oxide turns into a metal hydroxide when placed in water. The initial copper ion had a +2 charge, so the hydroxides must have a net charge of –2. This is accomplished through two hydroxyl groups. Once copper (II) hydroxide has formed, it may react with hydrochloric acid. These reactants undergo a special type of double replacement reaction known as a neutralization reaction. In essence, the metals combine ($\ce{CuCl2}$) and water is formed"
From here, I was wondering how exactly to balance the equation. Specifically, do I have

$\ce{CuO(s) + 2 HCl(aq) -> Cu(OH)2(aq) + 2 HCl(aq) -> CuCl_2(aq) + 2 H2O(l)}$
$\ce{CuO(s) + 2 HCl(aq) -> Cu(OH)2(aq) + 2 HCl(aq) -> CuCl2(aq)  + H2O(l)}$

It seems like I can "cancel out" a water molecule when going from middle reactants to final reactants. Is this true?

CuO is very poorly soluble in water and there is no intermediate copper hydroxide formation in HCl solution. The proposed description looks like a Rube Goldberg machine to me. Do you have any experimental evidences to support suggested "mechanism"?

Your equations are not balanced (count the atoms!). First balance, then cancel out. Then you'll get a single result. Besides that, as @andselisk said, there's no reason to assume intermediate copper hydroxide.

As andselisk points out, CuO is poorly soluble, but the hydroxide is listed in the CRC Handbook (blue, insoluble, no numbers), so it can be assumed to exist under some conditions. The Reference Book of Inorganic Chemistry, Latimer and Hildebrand, 3rd ed, 1951, p110, mentions Cu(OH)2 being formed in solution by addition of hydroxide, cold, to Cu++ as green or bluish-green (in hot solution, black CuO is formed). Cu(OH)2 is a very weak base; K = 5.6 e-20 (this must be Ksp).
Now, your experiment: you start with black CuO and wind up with green CuCl2. Because the blue/bluish-green Cu(OH)2 color will be so faint because of its low solubility, I'm going to suggest that under ordinary concentrations of HCl (~0.1 - 1 M), you will not be able to detect the very fast appearance and disappearance of Cu(OH)2 in solution. However, with a large excess of CuO and very dilute HCl, you might, under some conditions, be able to get some Cu(OH)2 into solution if you did it delicately. But I wouldn't think this would be easy. If you wanted to see Cu(OH)2, the way to go would be to add cold NaOH solution to a cold copper salt solution.
Now, for your equations: #1 is not balanced: you have two total hydrogens on the left and four on the right.  #2 is balanced right and left, but the middle has too many hydrogens.
I think you overthought the reaction. It might be two-step, but that intermediate step is very difficult to investigate and can be swept under the carpet. But its better to overthink and walk back than to underthink and miss the boat.


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As adjectives, not in a math context: 'different' vs 'differential'
I looked up  differential, but still don't savvy why the books below use 'differential' rather than 'different' and vice versa.
Advanced Organic Chemistry: Part A: Structure and Mechanisms by Francis A. Carey, Richard J. Sundberg. p 211.

Electroanalytical Chemistry: A Series of Advances
edited by Allen J. Bard. p 195.

Tietz Textbook of Clinical Chemistry and Molecular Diagnostics - E-Book by Carl A. Burtis, Edward R. Ashwood, David E. Bruns. p 519.

I'm voting to close this question as off-topic because this belongs on an English Language site. The dictionary entry linked has the correct definition.

@JonCuster I disagree.  It is true the dictionary has the correct definition.  But the OP's challenge was, I believe, specifically in how to understand the definition in the context of its application to chemistry.

"Differential" is a compact way of saying "difference between the" (or its equivalent).  For instance, in your first example, where the author writes "Chromatographic separations result from the differential interactions of the enantiomers...", what's meant is  "Chromatographic separations result from the difference between the interactions of the [two] enantiomers...".
I.e., chromatographic separation results from the fact that there is a difference between how the two different enantiomers react with the stationary phase (the solid column packing material).
The first definition given in your link applies here: "Of, showing, or depending on a difference; varying according to circumstances or relevant factors."

Still, I would have written "from the different interactions of....". At least for the first example given the question is more suitable for English SE. The other instances refer to the techniques not the different behavior, so they might have wanted to say that more specifically, like in differential scanning calorimetry etc.  English is not my tongue, though.

I would use differential when there is a quantitative (perhaps subtle) difference, not necessarily a quantitative difference (for the chromatography, if I say different interactions, I would expect different types of interactions, but if I say differential, I would mean different chromatographic behavior because the strength of the interactions are not the same).

@KarstenTheis  I would describe that as a personal choice, since the term can be applied to both quantitative and qualitative differences.  As an example of the latter, consider "differential diagnosis":  https://en.wikipedia.org/wiki/Differential_diagnosis


text:
Is boiling tap water a must to destroy human mucus, proteins, fats, sugars and nucleic acids it may contain?
I understand that humans consume tap water which are chemically treated after pumped from natural or artificial reservations.
Sometimes people contaminate these reservations by spitting, pissing and shitting in them (for example, parents hold babies while swimming in a lake)
Pumped water are likely to be filtered and then chemically treated by:

ozone-based substances
fluoride-based substances
sulfate-based substances
chloride substances
Bubbling o2 oxygen allotrope into the water
Other substances
Some combination of substances
Radiation (probably artificial UV in a closed water container)

My problem
Because tap water aren't boiled by water suppliers, they contain dead saliva bacteria and other human cells, but they might also contain:

Human Mucus
Human proteins, fats, sugars and nucleic acids

My question
Is boiling tap water a must to destroy human mucus, proteins, fats, sugars and nucleic acids (and also human derived dead bacteria) it may contain?

You appear to be obsessed with the idea that boiling is necessary to destroy certain possible water contaminants. This is wrong on at least two counts (as previous answers to similar questions have explained): 1) boiling does not destroy those contaminants; and 2) other treatment methods are already completely adequate to ensure safe water.

@matt_black I am not asking about safety, but on purity from human derived materials such as saliva, urine, feces, dead bacteria derived from them, human proteins, human fats, human sugars and human nuclei --- everything the rest, everything non human, is fine with me. Do you understand my intention now?...

Please help me to ask a better question if you understand me better now.

Yes I get disgusted from the think that I might drink another human's saliva or piss in my tap water.

Boiling water is for me just a mean, not an end...

Your personal disgust reaction is irrelevant to what cities can justifiably do to provide people with clean water. Plus you seem to believe that boiling does something it doesn't achieve. Distillation might, so feel free to drink distilled water. But there are plenty of standards for what clean water supplies can  contain and current treatment methods keep the things you seem to worry about at a unmeasurably low levels.

@matt_black again, I am willing to take personal action to remove any described-human materials from my water; if boiling chemically treated water doesn't help, would would?...

I just reiterate that drinking distilled water is not healthy and should not be done. In case a new reader land here and read assertions dictated by exhaustion :)

@Alchimista your comment is in my opinion uncommunal and should be flagged as offensive.

I don't think so. I just refer to the nice comment by @matt_black in which is written "feel free to drink distilled water" assuming a kind of fatigue as ground for an assertion he would have not made. Anyway you can flag the comment and let others decide. This is not my job, just a way to spend break time learning and sometimes help others. In the specific case you got help already. We can't change answers just because you are obsessed with mucus etc. And mucus or urine in a lake even cease to be mucus and urine, to add another chemical consideration.....

@Alchimista here, by the exreme, offensive uncommunal statement just because you are obsessed with mucus, you just prove my point that was based upon assertions dictated by exhaustion :).

Boiling water will destroy living organisms like bacteria. It will not destroy chemical substances like sugars, proteins, fats and acids, which are not living species. It will probably transform their structure, coagulate the proteins, hydrolyze some sugars or fats. But this is not exactly what could be understood as a destruction.

Can I even destroy these in home kitchen?

Yes. As discussed in previous posts, boiled tap water is good and safe enough for drinking. Remember that you are drinking tap water not water from sewerage. You did not mention your country of residence. Most municipal corporations can provide you water analysis report.

I wrote I live in Thailand --- I meant to total destruction...

Do you know that there are good quality domestic water filtration system which can be attached to tap water source. Water goes through a series of filters, then activated carbon cartridge (which absorbs all organics) and eventually through a UV lamp. This is good enough to destroy/remove, yes "total destruction" of all disgusting things you can $ever$ think of. Whatever disgusting is coming to your mind, the answer is that activated carbon and UV lamp with remove or destroy it.

You can perform a very simple experiment to convince yourself that boiling will not destroy fats or sugars. This experiment is being performed countless times every day (in a slightly modified version) in kitchens all over Europe and probably other parts of the world.
Take a pot of water, pour in a shot of vegetable oil (e.g. sunflower oil) and add a spoonful of sugar. Boil the water. Let it cool down. You should be able to observe both the drop of oil (a second phase as it is immiscible with water). Furthermore, if you taste the water it will taste sweet. This is enough to show that simple oils (a type of lipids or fats) and sugars are not destroyed in solution or emulsion with water at boiling temperatures.
Your question proceeds to answer itself thenceforth.

Thank you, I have upvoted your nice answer.

You are laboring under a misunderstanding that most non-chemists have when it pertains to chemistry. You are unaware of the absolutely massive scale of tinyness that things are in.
Let's say you go to the sea and fetch a glass of water. Scoop it up. Now, you decide to have a contest with your friend. You have a magical bag that can contain anything in whatever amounts.
So you set off trying empty the water glass, one molecule at the time...
...and your friend sets off trying to empty the oceans of the world with one glass of water at the time.
You keep the same pace, the number of glass or molecule per minite.
You will lose, and with eons upon eons. You will not finish before the universe dies, though your friend migh.. In a $20$ cl glass of water there will be approx $4$ moles of water which is, if I typed it correctly, about $24920000000000000000000000$ molecules
Boiling water, purifying it, cleaning, scrubbing, leaching - none of these create pure water without contaminants. There is only a question of how many contaminants you can handle. 1 per 10000 or 1 per 10000000 makes the difference
Boiling water for 10 minutes brings the number of organic still living beings in the water down to a level where it does not harm your body. (normally - there are some parasites more durable than others - and harmful prions will be unaffected). It will still contain dead organic beings, mucus, shit, piss, toxic metals, harmful toxins - but in quantities your body can handle.
As long as tap water is measurably drinkable (potable, from french) the water works doesn't really care about the level of contaminant (of course they do, but if it is drinkable, it is drinkable, and the bar is set very low for that)
So no, boiling water is not strictly necessary as long as the level of contaminant is below whatever level of that contaminant is perceived as safe. And making the water chemically pure, as in totally free, of that contaminant requires a lot more than boiling, and is rarely done - and if you want water that clean you usually make it from it's constituent atoms or distill it repeatedly.
Don't be afraid of minute quantities of contaminants - you have been exposed to them since birth. Fun fact, if your eyes sting when swimming in a public swimming pool, it is likely ammonia (di) chloride, which is likely from urine. Enjoy!

@JohnDoea That is not to be found, no. If you, as per admission, have the understanding of the tinyness of things, this should be apparent to you. There will always be stuff that is not the right stuff in stuff. It is all about dosage. Accept it or die. (Not a threat, just a fact of life, without water you die). Also, bottled water is usually way more contaminated than tap water, unless you are in a 3rd world country. Or Flint, Michigan.

Nice answer. That is what I meant commenting on basically the same question days ago (mentioning omeophaty).


text:
Oxidation number of Oxygen in peroxides: -2 or -1?
My textbook says that the oxidation number of Oxygen in peroxides is -2, but when I searched up the internet, I found it to be -1. It could probably be a misprint in my book, but I am still not sure. Not only that, the oxidation number of Hydrogen in hydrides is said to be +1 in my book, when it's supposed to be, I believe, -1? Please help me out on this.
My book says:
" The oxidation number of oxygen in most of its compounds (except peroxides, superoxides and suboxides) is -2. In peroxides (e.g., H2O2, BaO2, etc.), the oxidation number of oxygen is -2.
The oxidation number of oxygen in superoxides (e.g., KO2) and suboxides (e.g., C3O2 ) depends upon the nature of the compound. Exceptions also occur when oxygen is attatched to a more electronegative atom. For example, in the compounds like OF2 and O2F2, the oxidation numbers of oxygen are +2 and +1 respectively."

Oxygen has an oxidation state of -1 in peroxides; Hydrogen has an oxidation state of -1 in saline hydrides like $\ce{NaH}$

The textbook might be referring to the charge of the peroxo group $\ce{O2^2-}$. I suggest adding the exact quote from the textbook alongside with a complete reference and examples of chemical substances.

Ok, I will quote what my book says, just a minute..

There used to be a definition which assigned these numbers, it has changed somewhat recently, see here: Gold Book. For some more see this question: Electronegativity Considerations in Assigning Oxidation States

I have added the exact quote from my book

So, should I follow what my textbook says? I guess not?

Seems like a typo in the book to me.  One slip of my fat fingers on my cell phone, and suddenly there are 9 planets again instead of 8.

It must be a typo (the book would otherwise contradict itself). The correct oxidation number in pedoxides (and superoxides) is -1


text:
Why doesn't Oxygen form coordinate covalent bonds?
We know that an Oxygen atom has two lone pairs. Why doesn't oxygen then form coordinate covalent bonds using those since it has nothing to lose?

Of course it does, why anyone would think otherwise?

Take a deep breath, and think about how that oxygen reaches your brain bound to the iron atoms of heme in the hemoglobin in the red blood cells.

https://chemistry.stackexchange.com/questions/67220/how-can-oxygen-have-three-bonds?rq=1

Oxygen very much does form bonds in which both electrons come from the oxygen atom.  Examples include:
1.
The $\ce{H3O^+}$ ion at the center of the solvated proton in aqueous acids, also available as salts of some of the strongest acids such as $\ce{(H3O)(ClO4)}$
2.
Carbon monoxide, with its triple rather than double bond.
3.
Ozone, in which the oxygen atoms form a bent chain and one of the three bonds must have both electrons coming from the middle atom.  The resulting charge separation makes ozone one of the few one-element substances with a nonzero dipole moment.

The number of complex compounds that are surrounded by water molecules is quite big , a few examples of water being a ligand include [Cu(H2O)4]2+ or [Co(H2O)6]2+.
Anyway oxygen is indeed less eager to donate its electrons because its electronegativity is considerably high, nitrogen for example works better as a ligand (actually the first complex compounds to be observed and classified were with NH3 as ligands).But the examples where O donates his electrons can be found quite readily.


text:
Reaction between acrolein and hydrazine
When acrolein is heated with hydrazine we get dihydropyrazole. But I can't see how. First I think hydrazine should add on acrolein to form corresponding hydrazone, but what will happen next? Also is my first step correct?

If your first step is correct is the resulting hydrazone a Michael acceptor?

Count the number of atoms on the chain in the resulting hydrazine, starting from the carbonyl carbon as 1. You'll see the last nitrogen is 5. This is a setup for a nice 5-membered ring. So, the next step after hydrazine formation is dehydration to give dihydropyrazole.

Your suggestion of resulting corresponding hydrazone by the reaction between acrolein and hydrazine is correct. It is normal hydrazine-hydrozone reaction you perform in idetifocation of aldehyde and ketone carbonyl groups. Usually, E-isomer is preferred (Ref.1). E- to Z-isomerization can be occurred in presence of trace amount of acid (Ref.1). I expect that may be the case during constant heating. As a result, Z-isomer can undergo intramolecular Michael addition as depicted in following mechanism to give dihydropyrazole:

References:

Shigehisa Uchiyama, Masanori Ando, Shohei Aoyagagi, "Isomerization of aldehyde-2,4-dinitrophenylhydrazone derivativesand validation of high-performance liquid chromatographic analysis," Journal of Chromatography A 2003, 996(1-2), 95–102 (https://doi.org/10.1016/S0021-9673(03)00542-9).

The reference you cite did not deal with acrolein but rather the hydrazones of saturated, aliphatic aldehydes. While the cyclization you propose is a 5-endo-trig cyclization (Baldwin's Rules), it is not precluded since there is no competitive cyclization, i.e., 5-exo-trig. However, conjugate addition of hydrazine to the acrolein hydrazone would permit a preferred 5-exo-trig cyclization with regeneration of hydrazine by elimination.

@user55119: To my knowledge, hydrozone formation is preferred over Michael addition with hydrazene. OP is a novice to organic chemistry so I don't want to confused him. I just gave a reasonable mechanism for formation of dihydropyrazole. Also, Baldwin's rule do not apply to all cyclizations. There should be sme exceptions. Don't you think? :-)

I did account for exceptions. That is why I said " it is not precluded since there is no competitive cyclization". Baldwin's Rules apply to competitive processes. Just another possibility.


text:
What is the correct term of a chemical formula for very large molecules if not just "chemical formula"?
I know that many molecules, wihch I as a non chemist, might define as "small" if their two dimensional diagram can be sketched on an A4 page, are often formulated;
That is → by one or more processes given a chemical formula;
Maybe after being isolated from a natural deposit or organism.
For example, here is such formula for vitaminB12 of the type hydroxocobalamin:
C62H89CoN13O15P

My problem
Whenever I examined a wiki article about a protein molecule, I understood it as too big to be represented in a "chemical formula", let alone in a two dimensional diagram on an A4 page, such as those I have described.
I though for example:

As these are isolated molecules after all, they do have a "chemical formula" and should be standardly represented somehow.
Such large molecules doesn't have to be proteins → maybe proteins and other molecules in similar size have different ways to be represented.

My question
What is the correct term of a chemical formula for very large molecules (such as proteins), if not just "chemical formula"?
From prior research I understand a "protein sequence" is the standard parallel for "chemical formula" in the context of proteins but I don't know about a standard parallel to such large molecules in general.

Chemical formula is chemical formula, not much to say about it — whether it's one of the largest proteins, titin $(\ce{C169719H270466N45688O52238S911})$, polyoxomolybdate $(\ce{(NH4)12[Mo36(NO)4O108(H2O)16] · 33 H2O})$ or ammonia $(\ce{NH3}).$ I suspect your confusion arises from mixing up molecular formula with its abbreviated forms and various graphical representations of structural formulas.

There can't be and isn't such thing in general. There are, however, certain local approaches in the areas which are important to us for some reason or another: protein sequence, DNA sequence, etc.

What does JohnDoea want to change ? Why is he not satisfied about the word "Chemical formula" ? Chemical formula can be applied to very large molecules. Where is the problem ?

@Maurice there is no problem and I am totally satisfied with the word - I used my own words then to find the parallel or shortcuts for very large materials where a chemical formula line is very long. Anyone is welcome to communally suggest an edit to try to make the question more reachable and shorter.

Chemical formulas of the type you describe are just as valid for large molecules as for small ones. The reason they are not commonly used is that the number of possible molecules described by a given formula goes up exponentially with the number of atoms.
Thus, it is much more common to describe larger molecules with more precise terms. In the case of polymers like proteins or nucleic acids, the primary sequence of the constituents (amino acids or bases) is often used if space allows. Alternatively, a protein or nucleic acid  might be referred to by a common name with which a sequence is associated.
Even for small molecules, IUPAC formal names or common names are used in preference to molecular formula in many cases, because the names provide information about the atom connectivity that is absent in the molecular formula. Other representations that include connectivity information are forms such as SMILES, which is a standardized system for representing connectivity in text.
In the scientific literature, names of molecules are often accompanied by pictorial representations that provide atom connectivity information in an easily understood format. For very large molecules like proteins whose connectivity cannot be easily represented in a small image, pictorial representations are more often used to show secondary and tertiary structural features.

Nice answer as the question could attract various and reasonable comments that do not remove the doubt of the asker. This goes to the points.


text:
Reimer-Tiemann reaction on indole
According to me net reaction should be at $\ce{C3}$ when Reimer-Tiemann reaction is done on indole (see image), since in the resonance structure, negative charge is on this carbon.

But, the answer given is on $\ce{C2}$, and the solution given is

In pyrolle, the most electrophilic site is $\ce{C2}$. Think in a similar way.

But, I can't either find fallacy in my reasoning, not do I convince with the solution.
TL;DR Where will dichlorocarbene attack on indole (am not interested in ring expansion).

The paper by Order and Lindwall [1] agrees with you that the R-T reaction gives indole-3-carboxaldehyde.

Since its isolation in 1903 by Hopkins and Cole (1), 3-indole aldehyde has been investigated very little. It was first prepared by Ellinger (2) from indole through the use of the Reimer-Tiemann reaction. This method was improved on later by Boyd and Robson (3).

References

Order, R. B. V.; Lindwall, H. G. 3-Indole Aldehyde and Certain of Its Condensation Products. J. Org. Chem. 1945, 10 (2), 128–133. DOI: 10.1021/jo01178a006.

What about the explanation given in the solution?

Indole is not pyrrole

Also note that the answer refers to C2 as "most electrophilic", which is true. But in the first step of the R-T reaction, the aromatic ring acts as a nucleophile, so the reaction happens at the most nucleophilic position (C3) rather than the electrophilic position (C2).


text:
How do we weigh one C atom?
As far as I get it, we did weigh one carbon (12,6-isotope) atom. Then we said that electrons have nearly zero weight and protons weight nearly the same as neutrons. Hence we took the weight of that one atom and divided it by twelve and this way we got the mass of a proton/neutron:
$w_\mathrm {n,p} = 1.6726 \cdot 10^{-24}\ \mathrm g$
Because that isn't easy to calculate with, we introduce a number which multiplied with the weight of a proton/neutron gives us one gram:
$N_\mathrm A = w_\mathrm {n,p}^{-1} = 1\ \mathrm{mol}$
So far so good. Now hydrogen weights 1 g/mol and carbon weights 12 g/mol. Two questions to that:

How did we actually measure the weight of one carbon atom in the first place?
Why is the molar mass not precisely 1 and 12?

1.) did you check wikipedia? Its explained thoroughly there. 2.) nuclear binding energy, $E=mc^2$

Your equation for the Avogadro constant is not correct.

This is a question worth debating and discussion. First of let us realize the fact that we are used to looking at relative atomic weights or masses (if someone wants to be too accurate). No Table lists the absolute atomic masses of atoms in the periodic table, i.e., through out the history we have to assumed either the weight of hydrogen atom is exactly 1.0000, or oxygen-16 had a weight of 16.0000 or lately the weight of carbon-12 atom is exactly 12.0000 units.  So let us forget the picture that someone actually made a measurement, took a single atom of hydrogen or carbon and weighed by some fancy means. These are all man made references and numbers. We could have taken the weight of C-12 as 100, and all the atomic masses would change proportionally.

So far so good. Now hydrogen weights 1 g/mol and carbon weights 12 g/mol. Two questions to that: How did we actually measure the weight of one carbon atom in the first place?
  Why is the molar mass not precisely 1 and 12?

See above. You have to choose a reference. Fix hydrogen to 1 or carbon as 12 exactly but not both. Now you may ask how these numbers came up? You have to look at the question historically, when Avogadro's number was not used. If you understand German, have a look at Handbuch der Anorganischen Chemie by Abegg. There is long history how the number 12  got associated with carbon.
https://archive.org/details/a2handbuchderano03abeguoft/page/12;

You must realize that one proton free, lonely, has not the same mass as when included in a nucleus. A proton looses a small part of its mass when entering a nucleus. This mass loss is transformed into energy through the Einstein law : E = mc2.


text:
General justification for steady state approximation
I want to understand the assumptions that have to be fulfilled in order for the steady state approximation to work. I did some research myself and was able to deduce assumptions of the form $k_1\gg k_2$ for some systems, by solving the system exactly and taking the limiting case in which the exact system reduces to the approximative system. (Later I found this answer doing the same thing Meaning of steady state (kinetics)) This approach will not work for most systems though, as no closed exact solution exists for most non-linear differential equation systems.
I want to develop a clear chemical intuition in which case it is justified to use the approximation. Most literature seems to just tell the reader that it's fine to use the approximation in the reaction thats discussed. This is unsatisfying. From a mathematical standpoint this approximation is all but trivial.

This steady-state approximation is always wrong. It is only justified by common sense. In chemistry, you will never be and stay exactly at a point where the rate of the reaction producing a compound is equal to the rate of destruction of this compound. But you may imagine being "not too far" from this ideal position. That is the trick used for understanding the kinetics of some chain reactions.

Obviously it's never exact but I want to develop an understanding to what kind of processes it's an acceptable approximation. Surely there must be papers comparing numerical solutions to steady state approximations and surely there must be 1 or 2 handy rules for estimating weather it's OK to use steady state approximation based on the underlying chemistry.

Whether it’s okay is going to depend also on what you are using the result for, so one can’t make general rules. A good way to get the feel for the inaccuracy is to plot product vs time for SS approx and actual. To simulate actual, you don’t need an explicit solution, just iterate the differential equations at short time intervals

Related: https://chemistry.stackexchange.com/q/110794

In addition to k1>>k2, there is also the condition k2t>>1, because it only holds after a brief induction period (initially, the rate of change of the intermediate is greater than 0, but by the time k2t>>1, it has become 0). You can see that from this image:https://chem.libretexts.org/@api/deki/files/50967/Reaction2a.bmp?revision=1&size=bestfit&width=356&height=318
Note that initially, d[B]/dt>0, but it quickly becomes 0. This condition often isn't mentioned because the induction period is typically very short.

I would add that in terms of chemical intuition, it's about how unstable/reactive the intermediate formed is. The approximation is valid if the intermediate is an unstable species which therefore reacts as soon as it's formed (examples might include a free radical), whereas if the intermediate is relatively stable (like a benzylic cation stabilised by resonance) then it will be less accurate.

Let me know when (or if) you want the bounty awarded and to whom. I assume I have to do the actual awarding (first time for me), but you decide to whom the bounty goes (if anyone). We still have two days before the deadline, so just tell me what you want to do by then or earlier.

@EdV right now I would give it to Jeffrey Wang, for the valid answer, that there is no general justification. His answer give a nice and relevant example of a case analysis that leads to the approximation in a special case.

@Andrew I think this approach should also be made into an answer, so it is easier for others to find. Obviously this is also no general justification though.

Thanks! I will award Jeffrey Wang the bounty in 10 minutes!

Thanks for drawing attention on the question! @EdV

No very satisfying justifications are available unless you do a case-by-case analysis. But PSSA does not come from common sense. In general, pseudostationary state approximation (PSSA) is justified by the theory of differential equations.
In fact, the pseudostationary state concentrations represent the zeroth approximation
to a set of particular solutions of the correct chemical kinetic equations, which are called principal solutions. After the initial period, the concentrations of the intermediates approach their principal solution rather than their pseudostationary state values.
Therefore, the accuracy or usefulness of the pseudostationary state assumption depends upon the magnitude of the differences between the principal solution and the pseudostationary state concentrations.
OK, that sounds useless. But after doing some maths (like finding the principle solutions), some (obviously very ambiguous) rules are indeed justified mathematically.

Induction period should be short. (As you can see PSSA solutions cannot satisfy the initial conditions)
Destruction of the intermediates should be fast.

For example, in $\ce{A ->[$k_1$] B ->[$k_2$] C}$ if $k_2\gg k_1$ then the conditions above are met while if $k_1\gg k_2$, they aren't.

By the way, PSSA originates from Michaelis-Menton theory for enzyme kinetics. PSSA in it is well justified while the conditions above are not necessarily met. Consider
$$
\ce{E + S <=>[$k_1,k_{-1}$] ES ->[$k_2$] P}.
$$
Let $[\ce{S}]=s,[\ce{E}]=e,[\ce{ES}]=c,[\ce{P}]=p,e+c:=e_0,s+p+c:=s_0$. Furthermore, let $\tau=k_1e_0t,u(\tau)=s/s_0,v(\tau)=c/e_0$. The kinetic equations can be written as
$$
\begin{cases}
\frac{\mathrm du}{\mathrm d\tau}=-u+(u+K-\lambda)v\\
\color{red}{\varepsilon}\frac{\mathrm dv}{\mathrm d\tau}=u-(u+K)v
\end{cases}
$$
where  $\lambda=k_2/\left(k_1s_0\right),K=\left(k_{-1}+k_2\right)/\left(k_1s_0\right),\color{red}{\varepsilon=e_0/s_0\ll 1}$. Therefore, PSSA can be applied (i.e. setting $u-(u+K)v=0$, justified by perturbation methods) without loss of much accuracy. This example indicates that more often than not, a case-by-case analysis is required.

Finally, I think to use PSSA safely, one should do some experiments to check (numerical simulations, real experiments, etc.).

Thank you for you input. This is close to what I personally would have answered. There is no general justification and care should be taken when applying PSSA. Eventhough many introductionary courses just take it as given, that the approximation is fine, there are cases where it is probably not.

In the scheme $ A \overset {k_1} \longrightarrow B \overset {k_2}\longrightarrow C$ the exact solution for B is $B=\displaystyle \frac{k_1A_0}{k_2-k_1}\left( e^{-k_1t}-e^{-k_2t} \right) $. To find out how quickly the steady state is reached and by how much it deviates from the true value, the steady state is compared to the exact value by taking a ratio of concentrations.
Species B follows $\displaystyle \frac{dB}{dt}+ k_2B=k_1A_0e^{-k_1t}$  and letting $dB/dt=0$ forms our steady state value $B_{ss}$.   The ratio to measure the extent of deviation will be defined as $\displaystyle R=\frac{B}{B_{ss}} = \frac{k_2}{k_2-k_1}\left( 1-e^{-(k_2-k_1)t} \right)$ and as a measure at steady state we let $R=1$.
Initially when $t=0,\; R_0=1$ and at long times $\displaystyle R_{\infty} =  \frac{k_2}{k_2-k_1}$, so the difference between $R_0$ and $R{\infty}$ indicates the extend or deviation of steady state from the true value.
If, for example, $k_2 = 5k_1$ then $R_\infty-R_0=1/4$ or a 25% deviation. When $dB/dt=0$, species  B has reached its maximum value which takes time $\displaystyle t_m =\frac{1}{k_2-k_1}\ln(k_2/k_1)$ and in this example this takes $\ln(5)/(4k_1)$ seconds, i.e. this time is needed before steady state can be invoked.
This analysis indicates that for steady state to apply $k_2 \gg k_1$, i.e. the steady state applies under conditions that B disappears rapidly once it is formed.  For other schemes the same general principles apply but the analysis will be far more complicated.