text:
Equilibrium Constants for Diatomic Gases at High Temperatures
I'm new here and I haven't taken chemistry beyond high school, so please forgive the ignorance. I am reading a book about hypersonic aerodynamics, and I am trying to connect some chemistry knowledge into my reading. I have four reactions listed in my book:

$$
\begin{align}
\ce{O2 &-> 2 O} &\quad (\pu{2000K} < T < \pu{4000K})\\
\ce{N2 &-> 2 N} &\quad (\pu{4000K} < T < \pu{9000K})\\
\ce{N &-> N+ + e-} &\quad (T > \pu{9000K})\\
\ce{O &-> O+ + e-} &\quad (T > \pu{9000K})
\end{align}
$$

Now, the book doesn't go into the chemistry here really at all. I know that some minor dissociation happens at cooler temperatures, and there will still be some diatomic oxygen/nitrogen at higher temperatures too. But I'm trying to answer for myself just how much is around at those temperatures, and how much they change in the ranges that they list. Assuming I remember correctly, I need the dissociation constants for this. However, I'm having trouble finding them.
NIST chemical reaction searching tool doesn't seem to have the information I'm looking for, and Wolfram|Alpha (which generally helps) is not understanding what I'm trying to tell it… so I'm at a loss. My chemical engineer family member mentioned I could find it in the CRC Handbook of Chemistry and Physics, but unfortunately we can't seem to find our copy. My chemistry-major friend in college knew that it was possible to get the numbers, but he couldn't quite remember where I should start.
Either way, I'm looking for the equilibrium coefficients of:
$$
\begin{array}{clr}
\hline
\# &\text{Reaction} & T/\pu{K} \\
\hline
1 & \ce{O2 -> 2O} & 2000 \\
2 & \ce{O2 -> 2O} & 4000 \\
3 & \ce{N2 -> 2N} & 4000 \\
4 & \ce{N2 -> 2N} & 9000 \\
5 & \ce{N -> N+ + e-} & 9000 \\
6 & \ce{N -> N+ + e-} & 11000 \\
7 & \ce{O -> O+ + e-} & 9000 \\
8 & \ce{O -> O+ + e-} & 11000 \\
\hline
\end{array}
$$
Can anyone either provide those values or help me find them somewhere? Even the reaction equilibrium constants that I can find are only listed at a single temperature (and never this high, because these temperatures are… uncommon).

related https://chemistry.stackexchange.com/questions/124505/thermal-decomposition-of-steam-to-hydrogen-and-oxygen-at-what-temperature-is-ha https://chemistry.stackexchange.com/questions/133766/is-silicon-dioxide-decomposition-possible

I would suggest that you post this in physics. Astrophysicists would be interested in this type of data. I do vaguely recall Saha equation in atomic spectroscopy related to ionization equilibrium in at very high temperatures. You will not find this data in ordinary chemistry or web. Not everything is out there on the web.

@M.Farooq I will take a look in the Physics Stack Exchange -- thanks for the tip. I'll leave it here too, though, in case someone has more information (even on just to calculate it at a nonstandard temperature). I will look into the Saha Equation, I appreciate the suggestion. Glad that I wasn't just ignorant at not being able to find it!

And thank you, @andselisk, for reformatting. I couldn't figure out how to use LaTeX commands after ten minutes of trying, so I just gave up.

@Publius No prob. I would, however, strongly encourage you to get to know MathJax and LaTeX common macros if you plan to do a substantial amount of scientific writing. Feel free visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. For LaTeX there is TeX.SE. And remember that MathJax ≠ LaTeX, they only share some command names.

@andselisk oh, I do know LaTeX! That's what I typeset my homework in actually. For some reason, it wasn't liking my LaTeX code when I put it into the window; does it compile when I hit "post?" Or is there a button to compile LaTeX code that I just missed? I was able to typeset LaTeX code just fine when I posted on the TeX SE, but for some reason it wasn't having it here.

@Publius I'm really glad you are using LaTeX. Unfortunately, MathJax is not a full-featured LaTeX distro and works entirely differently. It's just a JavaScript library which happens to share some macros names with LaTeX, so don't expect full compatibility and packages support. Packages must be rewritten from the ground up to be used with MathJax, this, for example, happened with mhchem, but there is no siunitx in MathJax.

@Publius So, no, when you hit "Post" button nothing is going to change as MathJax renders the result nearly in the real time (unless there are hundreds of formulas).

@andselisk Gotcha, I understand now I think. Thank you so much for the tip!


text:
What is the opposite of a catalyst and examples?
I remember a while ago I heard of something that slowed reactions and was effectively the opposite of a catalyst. What is this called? Also would appreciate examples and an explanation of their uses just for curiosity. (I think the one I heard of was used to prevent spontaneous combustion)

Also there is "poison" which reduce the effect of catalyst.

Whatever answer you'll get there is not a real opposite of a catalyst. That would mean to force the reaction to take a more difficult path. Obviously, that is not possible from a mechanicist point of view. Still, the overall effect of reducing the reaction rate is a true thing.

The opposite of a catalyst is essentially an inhibitor. From IUPAC's gold book Gold book

inhibition:
The decrease in rate of reaction brought about by the addition of a substance inhibitor), by virtue of its effect on the concentration of a reactant, catalyst or reaction intermediate. For example, molecular oxygen and p-benzoquinone can react as 'inhibitors' in many reactions involving radicals as intermediates by virtue of their ability to act as scavengers toward these radicals.

Then there is a term called negative catalyst, and there are various opinions as to how they work.  Journal of Chemical Education

source: Stack Exchange

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

sort: votes


text:
How does total depth affect the colloid gradient?
When a colloid reaches diffusion–sedimentation equilibrium, it creates a gradient where the solute concentrates toward the surface or the bottom of the vessel based on its density relative to the solvent even though it never completely separates.
This got me wondering, if you were to compare two vessels of different heights containing the same colloid solution, would the taller vessel have the same gradient slope, resulting in a greater difference between the concentration at top and bottom of the vessel, or would the slope become more shallow resulting in the same difference in concentrations, or would it be something in between?

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


text:
Biodegrability of straight chain and branched chain detergents
From NCERT Chemistry: Textbook for class XII, Part II [1, p. 461]:

Main problem that appears in the use of detergents is that if their hydrocarbon chain is highly branched, then bacteria cannot degrade this easily. Slow degradation of detergents leads to their accumulation. Effluents containing such detergents reach the rivers, ponds, etc. These persist in water even after sewage treatment and cause foaming in rivers, ponds and streams and their water gets polluted.
These days the branching of the hydrocarbon chain is controlled and kept to the minimum. Unbranched chains can be biodegraded more easily and hence pollution is prevented.

Why are straight chain detergents generally biodegradable while branched chain detergents are non biodegradable? Basically, how does branching affect biodegradation?
Is there any more detail/ better explanation than this or explanation from any other view point?
References

NCERT. Chemistry: Textbook for Class XII. Part II, Reprint: August 2019 Bhadrapada 1941; M. Siraj Anwar, Shveta Uppal, Arun Chitkara, Bibash Kumar Das, R.N. Bhardwaj, Mukesh Gaur, Series Eds.; National Council of Educational Research and Training: New Delhi, 2007. ISBN 978-81-7450-716-7. (PDF)

According to Scott and Jones1:

... problems arise with branched alkyl chains, a side chain methyl group or a gem-dimethyl-branched chain cannot undergo β-oxidation by microorganisms and must be degraded by loss of one carbon atom at a time (α-oxidation).

* text by the authors, highlighting and hyperlinking by me.
A study by Whyte et al2 on the degradation of alkanes by Rhodococcus yielded similar findings:

After 28 days of growth at 5°C in MSM containing 0.1% diesel fuel, Rhodococcus sp. strain Q15 degraded almost all the n-alkanes (C10 to C21) and, to a lesser degree, many of the branched alkanes found in diesel fuel.

P.s: both of these articles are open access, and explain the biodegradation process quite well. You can refer to them directly if you need more detail or specific examples.
References:

Scott, Matthew J., and Malcolm N. Jones. “The Biodegradation of Surfactants in the Environment.” Biochimica et Biophysica Acta (BBA) - Biomembranes, vol. 1508, no. 1–2, 2000, pp. 235–51. doi:10.1016/S0304-4157(00)00013-7.
Whyte, L. G., et al. “Biodegradation of Variable-Chain-Length Alkanes at Low Temperatures by a Psychrotrophic Rhodococcus Sp.” Applied and Environmental Microbiology, vol. 64, no. 7, 1998, pp. 2578–84. PMID: 9647833


text:
While calculating the effective nuclear charge, why don't we consider effect of outer electrons on inner ones?
When calculating the effective nuclear charge for an electron, why do we only consider the repulsive effect of the inner electrons on the outer electrons? The outer electrons also repel the inner ones (towards the nucleus) and so, when determining the effective nuclear charge for inner electrons, don't we need to account for this effect as well?

The outer electrons just don't penetrate the orbitals of the inner shells enough to have a "significant" effect. That isn't to say that there isn't any effect. This is chemistry where a few significant figures is generally good enough.

The effect of outer electrons may be also within the error ranges of the model, as there are already multiple simplifications in place to be ever able to compute anything. Be aware that even a classical 3-body gravitationally bound system cannot be solved exactly.

You´ve got a bit of a wrong assumption here. If the electrons were in actual spherical shells, the outer electrons would indeed have zero influence.

As chemists, we are typically more directly concerned with the effective nuclear charge on outer electrons than inner electrons, because it's only the outer electrons that interact with other atoms and thus determine an atom's chemical properties. [Though the effective nuclear charge on inner electrons will affect their radial probability distributions and thus create a second-order effect on the effective nuclear charge of outer electrons.]
Nevertheless, this is an interesting question.
When we are calculating the effective nuclear charge seen by electrons in an atom (say, for outer electrons) (which directly impacts, for instance, their ionization energies), we start by doing so for isolated atoms, i.e., atoms in the absence of an external field.  Adding an external field (from the presence of other atoms, especially bonding atoms) is an added complication we won't consider here.
And in isolated atoms, the probability distributions of all electrons are spherically symmetrical.
So let's consider an inner electron in an isolated atom.  According to Gauss's Law, any charge that is spherically distributed about the nucleus, and is farther from the nucleus than that electron, "cancels itself out" (it integrates to zero), i.e. it has no net effect on the effective charge felt by the electron.
So an inner electron is only affected by outer electrons to the extent that the outer electrons spend some time closer to the nucleus than the inner electron.  [Again, when they are further from the nucleus than the inner electron, they have no effect on it, because of the spherical symmetry of their probability distributions.]
Based on the relative probability distributions of inner and outer electrons, the percent of the time this happens (outer electrons being closer to the nucleus than inner ones) is low.  Having said that, when it does happen, it will cause the inner electrons to feel a reduced nuclear charge.  I.e., to the extent that outer electrons affect inner electrons (in isolated atoms), they don't push them towards the nucleus, they push them away!
Hence there is no "reverse shielding" effect by outer electrons on inner electrons (again, in isolated atoms).  There is only a (small) shielding effect.

Good job. I guess I would say that the whole treatment is approximate and that we focus on things that are most important, since time is finite (at least our time on earth :-))


text:
Data analysis in TCSPC for fluorescene decay (reconvolution with measured IRF)
I am trying to understand the data analysis of fluorescence decay counts measured by TCSPC technique, particularly with reconvolution with measured IRF.
I am able to get the fitted counts (given by the software according to the model only in the case of tail fitting and not in the case of reconvolution.
Tail fit
The basic model is a multi-exponential decay with y-offset(background-offset).
$$
I(t) = A + \sum_{i}^{N}B_{i}\textrm{e}^{\frac{-t}{\tau_{i}}}
$$
Here $B_{i}$ and $\tau_{i}$ are the pre-exponential and lifetime due to the $i$th component respectively and $A$ is the y-offset.
The fitted parameters for a 2 exponential model ($I(t) = A + B_{1}\textrm{e}^{\frac{-t}{\tau_{1}}} + B_{2}\textrm{e}^{\frac{-t}{\tau_{2}}}$) are.
A   =   11.63431
T1 (ns) =   0.221457836
T2 (ns) =   2.69572
B1  =   6618.36
B2  =   644.1506

Time per channel (ns) = 0.026128421
Data range = 402:1289

Peak count  =   10000
Total count =   190155

Here I am able to get the fitted counts in the following example by using these parameters. For example in case of bin no. 405.
|    | Bin | Time (ns) | DecayCount | FittedCount |
|----|-----|-----------|------------|-------------|
| 0  | 401 | 10.47750  | 9464       | -           |
| 1  | 402 | 10.50363  | 8462       | 6531.378    |
| 2  | 403 | 10.52975  | 7298       | 5870.6414   |
| 3  | 404 | 10.55588  | 5758       | 5282.8124   |
| 4  | 405 | 10.58201  | 4800       | 4759.7832   |
| 5  | 406 | 10.60814  | 4069       | 4294.3481   |
| 6  | 407 | 10.63427  | 3396       | 3880.1036   |
| 7  | 408 | 10.66040  | 2908       | 3511.3585   |
| 8  | 409 | 10.68652  | 2649       | 3183.0549   |
| 9  | 410 | 10.71265  | 2315       | 2890.698    |
| 10 | 411 | 10.73878  | 2171       | 2630.2929   |

I am able to plug in the fitted values of the model parameters in the formula to get the fitted count values for any time $t$.
t =  10.58201035 - 10.47749666 = 0.104513682

FittedCount = 11.63431 +
  (6618.36 * exp(-0.104513682/0.221457836)) +
  (644.1506 * exp(-0.104513682/2.69572)) = 4759.783

Reconvolution fit
Here the model is.
$$
I(t) = L(t)\otimes \sum_{i}^{N}B_{i}\textrm{e}^{\frac{-t}{\tau_{i}}}
$$
$$
F(t) = A + B.I(t+\Delta)
$$
Here $B_{i}$ and $\tau_{i}$ are the pre-exponential and lifetime due to the $i$th component respectively, $L(t)$ the instrument response function (IRF), $B$ is the amplitude scaling factor, $\Delta$ is the shift parameter and $A$ is the y-offset.
A   =   11.41109
T1 (ns) =   0.191460122
T2 (ns) =   2.626401242
B1  =   0.2411146
B2  =   0.007343028
Shift (ns)  =   -0.158856098

Time per channel (ns) = 0.026128421
Data range = 398:1289

Peak count (Decay)  =   10000
Total count  (Decay)    =   190155
Peak count (IRF)    =   10000
Total count  (IRF)  =   101269

How to get the fitted counts in the case of reconvolution fit ?
| -  | Bin | Time (ns) | IRFCount | DecayCount | FittedCount |
|----|-----|-----------|----------|------------|-------------|
| 0  | 397 | 10.37298  | 9743     | 5665       | -           |
| 1  | 398 | 10.39911  | 10000    | 7982       | 9757.034    |
| 2  | 399 | 10.42524  | 9776     | 9420       | 9106.409    |
| 3  | 400 | 10.45137  | 8918     | 10000      | 8363.679    |
| 4  | 401 | 10.47750  | 7445     | 9464       | 7626.371    |
| 5  | 402 | 10.50363  | 5922     | 8462       | 6899.12     |
| 6  | 403 | 10.52975  | 4538     | 7298       | 6221.999    |
| 7  | 404 | 10.55588  | 3327     | 5758       | 5582.207    |
| 8  | 405 | 10.58201  | 2450     | 4800       | 5004.482    |
| 9  | 406 | 10.60814  | 1605     | 4069       | 4481.801    |
| 10 | 407 | 10.63427  | 1191     | 3396       | 4012.029    |

Specifically, in the calculation, How to

get the amplitude scaling factor (in addition to the amplitudes) and
shift the measured IRF using the shift parameter.

I have tried multiplying by the total IRF count value to get the $B_{i}$s corresponding to a tail fit, but still not getting the fitted counts as in the output.
Update
Here are the comments regarding scaling of $B_{i}$s by the scaling factor $B$ in few fitting software program documentations.
http://www.nanoer.net/d/img/1-FASTissue6.pdf

If the fit was performed in the Reconvolution mode, $B_{i}$ values are
scaled by the integrated IRF, i.e. if the same decay would be analysed
with a 10 times bigger IRF, the numerical values of the $B_{i}$ values would
be reduced by a factor of 10. If one determines the total number of
counts of the IRF, and then multiplies the $B_{i}$ value with this number,
this will result in a value that will be similar to the $B_{i}$ value for a
tail fit, extrapolated to the channel of the peak of the IRF.

https://www.horiba.com/fileadmin/uploads/Scientific/Downloads/UserArea/Fluorescence/Manuals/DAS67_Manual.pdf

The magnitude of the $B_{i}$ values returned in the fit are dependent
on whether or not reconvolution is used in  fitting  the  data.
Without  reconvolution  the  values  are  in  counts  and  can  easily
be  linked  to  the  peak  number  of  counts  (ie  for  a  single
exponential  its  magnitude  should  be  ~  the  count  value  in  the
start  channel).  The  value  after  reconvolution  is  not  in
counts,  but  still  reflects  the  amount  of  a  particular
emitting species.

The shift parameter is part of the fitting and is used because different filters/wavelength etc may be used when measuring the irf vs fluorescence so this time delay is adjusted by the convolution process to get a best fit. It has no physical meaning otherwise. The amplitudes $B$ are relative to one another;  the absolute values usually do not matter (as fluorescence yields are not often measured via tcspc), but you can normalise if you wish to get x% B1 and y% B2 etc.

@porphyrin I understand why shift is used fitted for a reconvolution fit and how $B_i$s are different in case of tail fit vs reconvolution fit. I want to know how to get the fitted counts at a time $t$ by using the fitted parameter values in the model, as I can do with tail fit.

I don't quite understand, if you use the fitting parameters and convolute the function, formed from two exponentials plus background, with your instrument function your should get the experimental counts.

@porphyrin Yes, I should be able to get that. But the problem is with the $Bi$s. They are scaled with the amplitude scaling factor $B$, which I am not able to figure out. When I try to fit with other software, I am getting the values for the amplitudes B1 = 16781.47 and B2 = 580.40. The relative amplitude from above are B1 =  0.2411146 and B2 =  0.007343028. But also in both cases the normalised pre-exponential values are identical (B1 = 0.97 and B2 = 0.03).

It looks as though the background has not been included in the convolution: it is 11.4 but B is 0.24 & 0.007 so something is wrong here. The background should be far smaller , approx 6600 times comparing the fit without and with convolution. Its best to convolute with background because subtracting from a  poisson distributions is never a good idea since negative values are impossible.

The low values of $B_{i}$s is due to scaling factor $B$ (Check the comments in the update above), not because background was not incorporated into the convolution.

Your background should be included as a parameter, especially as this can affect the lifetime of any long lived species especially when amount is small, i.e. long lifetime and background can 'interfere' with one another. You should look at the covariance matrix to check this.  BTW, if the convolution uses FFT you must ensure that the data is as close to zero as possible at the start and end of the dataset. (FFT assumes a repetitive signal series so any discontinuity is seen as a feature and messes up analysis, but this is not your present problem)


text:
What are the differences and similarities of Sugar Alcohol and the Alcohol in Alcoholic Drinks?
Question part 1: In particular of Sugar Alcohols, I ask about Sorbitol.
I am a 100% strict teetotaler. I do not care if a small dose will not intoxicate me, I am absolutely anti-alcoholic.
Recently I ate a few of some treat, first I checked the ingredients at least four times because I was suspicious.. there was no "regular" Alcohol but there was Soribitol. I really wanted to eat the treat even though from my Web searches I was not totally convinced that Sorbitol is clean.. I caved in and I am regretting it. The treat tasted suspicious.
I really am a serious, strict teetotaler 100%. Please advise me.

Question part 2: If you were an absolutely dead-strict teetotaler.. would you feel bad about eating something that contains Sorbitol?

Sorbitol is not an alcohol that goes against teetotaling since the main culprit would be ethanol and Sorbitol fortunately doesn't contain ethanol in it. If you have consumed ethanol, you are guilty of drinking.

That is interesting @Safdar

What are the other culprits? Please

There is a huge class of compounds designated alcohols because of their chemical structure. Sorbitol is an alcohol by that designation. Ethanol is the only one you should be concerned about

@Waylander Worrying about methanol would also be a good idea, even though the OP isn't going to encounter it.

Methanol is pretty toxic so no-one should be ingesting it.

Pretty much everything organic contains some form of sugar and - if not cared for well enough - may eventually contain some ethanol. Be aware of everything fermented.

Ethanol is the compound that goes against teetotaling. Otherwise, all sugary foods are alcoholic, carbohydrates are alcoholic and what else.
Alcohols are defined as organic hydrocarbons containing the hydroxyl group ( the OH group.). So sugar is an alcohol; it has a hydroxyl group.
Thus, only ethanol is the compound responsible for intoxication and going against teetotalism. No problems with anything else.

Be aware that alcohol has 2 main meanings:

alcohol(1) = ethanol / ethylalcohol / drinking alcohol
alcohol(2) = organic compound containing alcoholic group $\ce{-OH}$ bound to non aromatic carbon atom. Such compounds have usually names ending by -ol, unless empirical names take preference.

Similarly as palm oil is just one of many kinds of fat ( esters of glycerol and fatty acids ), alcohol(1) and sorbitol are just two of many kinds of alcohol(2), which includes also ethylenglykol ( anti-freezing agent ), glycerol and sacharides like glucose, fructose, sucrose, starch, cellulose etc.


text:
Why is 4-hydroxypyridine more acidic than benzoic acid?
4-hydroxypyridine (or 4-pyridone) has a $\mathrm{p}K_\mathrm{a}$ in the ballpark of $3$ to $3.33\:^1$ (in the pdf linked below, the $\mathrm{p}K_\mathrm{a}$ of 1-methyl-4-pyridone has been given to be $3.33$, however I suppose removing a methyl would only cause a slight decrease in the $\mathrm{p}K_\mathrm{a}$ value). This seems an unexpectedly low value to me. 4-pyridone seems more acidic than I expected it to be, since the $\mathrm{p}K_\mathrm{a}$ of other common acids are:$\:^2$
\begin{array}{lr}
\hline
\mathrm{Acid} & \mathrm{p}K_\mathrm{a} \\
\hline
\text{Benzoic acid} & 4.20 \\
\text{Formic acid} & 3.75 \\
\text{Acetic acid} & 4.75\\
\text{Phenol} & 10.0 \\
\text{Carbonic acid} & 6.37\rlap{\:^1} \\ \hline
\end{array}
Now I had expected 4-hydroxypyridine to be more acidic than phenol, because of the negative charge in the former's conjugate base being delocalised onto the electronegative nitrogen, yielding a more stable resonating structure than the latter's corresponding resonating structure. I thought that 4-pyridone's $\mathrm{p}K_\mathrm{a}$ might be somewhere around $\ce{H2CO3}$'s $\mathrm{p}K_\mathrm{a}$, but I was wrong. Also, I wasn't quite sure that it would be acidic enough to have a $\mathrm{p}K_\mathrm{a}$ value any close to other carboxylic acids. But it seems to me that it is more acidic than benzoic acid, formic acid and acetic acid.
Why is this so? I can't see any excetional characteristic in its structure which might impart it such a low $\mathrm{p}K_\mathrm{a}$. This does seem more of an exception to me, as of now, but it would be very helpful if someone clarified and explained such a trend.
References:
$[1]:$ http://gissmo.nmrfam.wisc.edu/static/pKa_compilation_R_Williams.pdf
$[2]:$ https://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/acidity2.htm

https://www.chemicalbook.com/ProductMSDSDetailCB6303828_EN.htm 4-hydroxypyridine has a pKa of 2.8 according to this

@Safdar However, this page from the same website shows a $\rm{pK_a}$ value of $3.2$, so I wan't really sure about the authenticity of the $\rm{pK_a}$ value mentioned there. However, my question still holds, in both the cases, the $\rm{pK_a}$ value is definitely lower than expected.

Why is it lower than expected? The conjugate base is quite stabilized.

@Zhe It is stabilised, but I don't see why it is so stabilised that 4-hydroxypyridine is so much more acidic than phenol.

The following are two resonance forms of 4-pyridone,

So, when 4-pyridone gets de-protonated, the negative charge on nitrogen gets delocalised on more electronegative atom i.e., oxygen, resulting in more stable resonance structure!
Note that, this explanation is identical to that one which is given when we compare $\mathrm pK_{\mathrm a}$ of benzoic acid and phenol.

Ref: Stability of 4-pyridone vs 4-pyridinol

I thought about the answer, and I am still wondering what makes the two equivalent resonating structures of the formate ions less stable then the conjugate base of 4-pyridone. In fact, oxygen is more electronegative than nitrogen, so having a negative charge delocalised on two oxygen seems like a much stable configuration than delocalisation on one oxygen and one nitrogen. Moreover, the resonating structures of formate are equivalent resonating structures, imparting an "extra" stability to the anion.

Can the delocalisation over mere 3 extra carbons justify the increased acidic strength of 4-pyridone?

@FakeMod, extended conjugation and aromaticity in 4-pyridone is what making it more stable than formate anion

But aromaticity isn't new to the 4-pyridone anion. It was there even before deprotonation, so I don't think it can be a driving factor here, however, extended conjugation does seem reasonable. Thanks!

Well, you too have wrong tautomer in there...

Hey @Mithoron, I've drawn resonance structures, they're not tautomers.

Resonance structures of the wrong tautomer... Just like second answer had. One can think this position of proton is no biggie but 4-hydroxypyridine and 4-pyridone are different compounds, even if usually in equilibrium, and OP is thinking primarily about the other one.

@Mithoron, you're right that 4-pyridone != 4-hydroxypyridine. But, the link OP had provided at starting shows 4-pyridone.

First of all let us examine the structural formula of  4-pyridone:

Now, as you can see, when it acts as an acid a negative charge forms on the nitrogen atom. This negative charge is well stabilized as it undergoes extensive resonance which includes oxygen as well. The fact that the negative charge is formed on nitrogen also contributes to the acidity.

"when the hydrogen is abstracted and negative charge comes , it acquires aromaticity": I don't think so. Wasn't it aromatic all along, even before deprotonation? Reference: https://chemistry.stackexchange.com/a/126146/91457

@FakeMod yes , thanku for the correction , it has been updated.

Wrong tautomer you have there...

@Safdar: I changed it back to tautomers. Also, followed  andselisk♦'s instruction to make similar fond in both text and image.

I thought about the answer, and I am still wondering what makes the two equivalent resonating structures of the formate ions less stable then the conjugate base of 4-pyridone. In fact, oxygen is more electronegative than nitrogen, so having a negative charge delocalised on two oxygen seems like a much stable configuration than delocalisation on one oxygen and one nitrogen. Moreover, the resonating structures of formate are equivalent resonating structures, imparting an "extra" stability to the anion.

Can the delocalisation over mere 3 extra carbons justify the increased acidic strength of 4-pyridone?

The pKa value that triggered the discussion must be incorrectly reported in this old document. It corresponds to the pKa of 4-hydroxypyridinium, i.e. expressing poor basicity of 4-hydroxypyridine. 4-hydroxypyridine is also poorly acidic (pKa near 11). The reason is that the 4-pyridone tautomer shown in other answers is highly prevalent in aqueous solution. This tautomer has neither acidic (phenol like) nor basic (pyridine-like) functionality. Correct values can be found in: A. Albert, J. N. Philipps, J. Chem. Soc. 1956, 1294-1304.


text:
Acidity of phenol and carboxylic acid
Carboxylic acid is much more acidic than phenol ($\mathrm{pK}_\mathrm{a}$  difference of ~6). I wonder why? In terms of resonance, there are 5 resonance structures for phenol, and only 2 in carboxyl group.
My thought is 4 out of 5 structures in phenol are minor (lone pair electrons at carbon, less stable), compared to the equivalent resonance in carboxyl group. Is that really the sole answer to the large $\mathrm{pK}_\mathrm{a}$ difference? I am not so convinced.
Although these resonance (4/5) structures are minor in phenol, I think the combined delocalization effect should make it more acidic than a carboxylic acid.

The facts are  what  they are. Carboxylic acids are more acidic than phenol. If I'm a negative charge, perhaps I'm happier on two electronegative oxygens rather than one oxygen and three carbons.

This question is not exactly a duplicate. OP is asking why there is more effective delocalisation and some practical proof for this.

Since you haven't mentioned the carboxylic acid to be measured, I assume the carboxylic acid to be $\ce{HCOOH}$. It has a $\mathrm pK_\mathrm a$ of $3.75$
Now, as you mentioned, phenol (or carbolic acid) has a $\mathrm pK_\mathrm a$ of $10.0$. There is a difference of about $6$ in the two values. This is because of the effective delocalization that takes place in formic acid compared to that of phenol. The best way to check for this is by computing the charges on the atoms in their respective conjugate bases and then check the stabilities of the two.
So first, we check the delocalization on the phenoxide ion.
The Mulliken partial charges on the atoms ($\ce{C-1}$ as carbon connecting to oxygen and we move counter-clockwise. $\ce{H-2}$ signifies the hydrogen bonded to $\ce{C-2}$)
$$
\begin{array} {c}
\begin{array}{lr}
\text{Atom} & \text{Charge} \\ \hline
\ce{C-1} & 0.307 \\
\ce{C-2} & -0.246 \\
\ce{C-3} & -0.204 \\
\ce{C-4} & -0.243 \\
\ce{C-5} & -0.204\\
\ce{C-6} & -0.246 \\
\end{array}
\begin{array}{lr}
\text{Atom} & \text{Charge} \\ \hline
\ce{H-1} & 0.103 \\
\ce{H-2} & 0.100 \\
\ce{H-3} & 0.089 \\
\ce{H-4} & 0.100 \\
\ce{H-5} & 0.103\\
\ce{O} & -0.659\\
\end{array}
\end{array}
$$
Now, we check the delocalization on the formate ion,
$$
\begin{array}{lr}
\text{Atom} & \text{Charge} \\ \hline
\ce{C} & 0.363 \\
\ce{H} & -0.083 \\
\ce{O-1} & -0.639 \\
\ce{O-2} & -0.639 \\
\end{array}
$$
Comparing the two, we can see that the more effective delocalization has taken place in formic acid since the net charge on an individual oxygen atom is less ($-0.659$ in phenoxide ion and $-0.639$ in the formate ion)
Due to this, we can say that the net effective charge is much better delocalized in formic acid which means that the conjugate base is better stabilized. Even though there were five possible resonance structures for the phenoxide ion, there do not contribute to the final resonance hybrid as well as the two equivalent resonance structures for the formate ion.
Since order of conjugate base stability is directly proportional to acidity. Hence formic acid would be more acidic than phenol.
Note: These values were calculated on WebMO with geometry optimization using Gamess DFT B3LYP/3-21G


text:
What is the dimension of Avogadro's constant?
What is the dimension of Avogadro's constant ($N_\mathrm{A}$).
On Wikipedia it says it is dimensionless, but in Nigel Wheatley's article (pdf) On the dimensionality of the Avogadro constant and the definition of the mole it says it is $\mathsf{N}^{-1}$.
\begin{align}
\text{Number of particles} &= N_\mathrm{A}\times \text{Amount of substance}\\
[\text{Amount of substance}] &= \mathsf{N}\\
[\text{Number of particles}] &= \mathsf{1}
\end{align}
If $[N_\mathrm{A}] = \mathsf{N}^{-1}$ then the above equation follows, otherwise it doesn't. Is this the right value?

See https://chemistry.stackexchange.com/a/108506/79678

Mildly related question about the value of $N_\mathrm{A}$ (not the units): What is the correct value of the Avogadro constant? And how was it derived?

The dimension/unit of the Avogadro constant $N_\mathrm{A}$ is actually $1/\mathsf{N}$ or $\pu{mol-1}$. It is shown that way in your first reference as well, i.e.
$$ N_\mathrm{A} = \pu{6.02214076E23 mol-1}. $$
What is "dimensionless" is the Avogadro number or the number of atoms / molecules in a single $\pu{mol}$ of that substance, sometimes written as $N$ or $N_0$.
The numerical value is the same, but the concept is different. Basically the relationship between the two is:
$N_0 = (\pu{1 mol})\times N_\mathrm{A} = \pu{6.02214076E23}$
See also the answer discussed in the comment by Ed V, where the new definition of the constant is explained: Effects of Changing Avogadro's Constant.

The dimensions of Avogadro's Constant is $1/\mathsf{N}$. In SI units, this is $\pu{1/mol}$.
Your unit analysis is correct. I have seen this type of question before, I believe the confusion is caused by people normally saying particles per mole, and "particles" is a "phantom" unit (similar to radian). But either way, the SI units are definitely $\pu{1/mol}$.


text:
Szabo and Ostlund's calculation of repulsion integral on page 172 via A.41
Returning to Szabo & Ostlund after a hiatus I tried to recreate his STO-3G calculation for HeH+ which begins on p. 168. It all went fine except for the two-electron, two-center integrals. I employed Szabo's values for these on p. 172 and was able to get his result for the energy and so on.
What I was not able to do is duplicate his p. 172 values for the 2-e integrals from his expression on page 416 (A.41). For example, on page 172, he gives $(\phi_{1}\phi_{2}|\phi_{1}\phi_{2}) = .1773.$
There are other approaches to this problem and I have looked at a few, but I would like to understand it as Szabo presents it. Because I was able to get the nuclear repulsion terms and so on I think my expressions for the contracted gaussians are fine but am misinterpreting some aspect of (AB|CD) in A.41.
If anyone is familiar with this particular calculation, and could show the proper calculation of (say) (12|12)...This is a very picayune question in the sense that I am interested in the particular calculation in Szabo: how to get 0.1773 on p. 172? Thanks for any illumination.
Edit
The Fortran program alerted me to a confusion. The "contracted" gaussians, using (11/11) as an example, look like:
$$G_a= .444635(2a_1/\pi)^{3/4}e^{-a_1|r-R_A|^2}+.535328(2a_2/\pi)^{3/4}e^{a_2|r-R_A|^2}+.154239(2a_3/\pi)^{3/4}e^{a_3|r-R_A|^2}$$
Regardless of how we do the integral it is:
$$\int G_a^2 \frac{1}{r_{1,2}}G_a^2$$ and we are raising the coefficients associated with $G_a$ to the 4th power and they come out of the integral. If we could do this integral in the usual naive way using linearity we would have 81 terms (some repeated) involving constants like:
$$(d_id_jd_kd_l)(2a_i/\pi)^{3/4}(2a_j/\pi)^{3/4}(2a_k/\pi)^{3/4}(2a_l/\pi)^{3/4} $$ where $i,j,k,l$ range from 1 to 3 respectively, where the $d_i$ are the coefficients $0.444365,$ etc.
That is why on p. 420 in the fortran program the quantities D1 include weights as factors, and a product of four D1's are pre-multiplied on p. 421 by the constants calculated in "TWOE" on p. 423. Using Fourier transforms to do the integral does not change the constants.
In short, the program is doing a lot of work here. It should still be possible to exhibit (11/11) easily but no wonder A.41 is not used in practice--it represents no economy over the fortran.

If you don't wind up getting an answer here, you could try moving it to Matter Modeling SE

@Tyberius: Thank you for the suggestion.

FWIW, I found a 2012 post by someone who checked the fortran77 and posted substantially Szabo's numbers for the STO-2G, so I do trust those. Admirable tenacity. https://techreport.com/forums/viewtopic.php?f=20&t=83633&p=1136478#

what's the constant you are off by for the (11|11) and (22|22) cases? It may clarify the error.

@daniel Your link should be squirreled away in https://chemistry.stackexchange.com/questions/37303/resources-for-learning-chemistry!

@daniel, let's talk in chat: https://chat.stackexchange.com/rooms/111279/reproducing-results-from-szabo-ostlund

Restricting attention to the integrals representing "self-interaction " of helium in HeH+ we have to find coefficients of the integral,
$$\int A\cdot A\frac{1}{r_{12}}\cdot A  \cdot A$$
in which for an STO-3G calculation $A=a\cdot g_1+b\cdot g_2+c\cdot g_3$ with $g_1$ a gaussian and $a$ a coefficient formed as on page 153 sec. 3.203. The coefficients will fall out as a sum of 81 terms, some of which are identical:
$$(a+b+c)^4 = a^4+4a^3b+6a^2b^2+4ab^3+b^4+4a^3c+12a^2bc+12ab^2c+4b^3c+6a^2c^2+12abc^2+6b^2c^2+4ac^3+4bc^3+c^4.$$
That there are 81 terms may be checked by viewing the fortran at page 420 and 421. The expression for V1111 is formed by pre-multiplying by D(i)D(j)D(k)D(l) in which i,j,k,l range from 1 to 3. The coefficients DDDD do not appear explicitly in A.41.
The 12 terms in $ab^2c$ and so on may represent different quantities, and care must be taken to keep the accounting correct because (using notation defined below)
$$12ab^2c\cdot 2\pi^{5/2}/((a_1+a_2)(a_2+a_3)\sqrt{a_1+a_2+a_2+a_3})$$
$$\neq 12ab^2c\cdot 2\pi^{5/2}/((a_2+a_2)(a_1+a_3)\sqrt{a_1+a_2+a_2+a_3}) $$
Letting $\zeta_1 = 2.0925$ (page 170) we have:
$a_1=0.109818\cdot\zeta_1^2,~~ a_2=0.405771\cdot\zeta_1^2,~~ a_3=2.22766\cdot\zeta_1^2; $
$a =0.444635\cdot (2a_1/\pi)^{3/4},~b=0.535328\cdot(2a_2/\pi)^{3/4},~ c= 0.154329\cdot(2a_3/\pi)^{3/4}$
The sum represented in the appendix A.41 for the 2-center 2-electron integral
$$ \int He \cdot He \frac{1}{r_{12}}\cdot He\cdot He$$ is given below, and in Mathematica I get 1.307238. Substituting in the constants for hydrogen I get about 0.77466. These are Szabo and Ostlund's values on page 172. They are a good test because they avoid calculation of the error function used in cross terms and they are close to results for the integration of Slater orbitals using spherical coordinates, which are $(5/8)\cdot \zeta. $
The calculation exhibits the coefficients in A.41 explicitly and is in essence the fortran code for V1111.
$ a^4\cdot 2 \pi^{5/2}/((a_1 + a_1)\cdot(a_1 + a_1)\cdot\sqrt{4\cdot a_1})+$
$4 a^3b\cdot 2 \pi^{5/2}/((a_1 + a_1) \cdot(a_1 + a_2)\cdot\sqrt{3 a_1 + a_2})+ $
$ 2 a^2 b^2\cdot 2 \pi^{5/2}/((a_1 + a_1) (a_2 + a_2)\cdot\sqrt{2 a_1 + 2 a_2}) +$
$ 4 a^2 b^2\cdot2 \pi^{5/2}/((a_1 + a_2)\cdot(a_1 + a_2)\cdot \sqrt{2 a_1 + 2 a_2}) +$
$4 a b^3\cdot 2 \pi^{5/2}/((a_1 + a_2) (a_2 + a_2)\cdot \sqrt{a_1 + 3 a_2}) +$
$ b^4\cdot 2 \pi^{5/2}/((a_2 + a_2) (a_2 + a_2)\cdot \sqrt{2 a_2 + 2 a_2}) +$
$ 4 a^3 c\cdot 2 \pi^{5/2}/((a_1 + a_1)*(a_1 + a_3)\cdot \sqrt{3 a_1 + a_3}) +$
$ 4 a^2 b c\cdot 2 \pi^{5/2}/((a_1 + a_1)\cdot(a_2 + a_3)\cdot \sqrt{2 a_1 + a_2 + a_3}) +$
$ 8 a^2 b c\cdot 2 \pi^{5/2}/((a_1 + a_2)*(a_1 + a_3)\cdot \sqrt{2 a_1 + a_2 + a_3}) +$
$  4 a b^2 c\cdot 2 \pi^{5/2}/((a_1 + a_2) (a_2 + a_3)\cdot \sqrt{a_1 + 2 a_2 + a_3}) +$
$  8 a b^2 c\cdot 2 \pi^{5/2}/((a_2 + a_2) (a_1 + a_3)\cdot \sqrt{a_1 + 2 a_2 + a_3}) +$
$ 4 b^3 c\cdot 2 \pi^{5/2}/((a_2 + a_2)\cdot(a_2 + a_3)\cdot\sqrt{3 a_2 + a_3}) +$
$ 2 a^2 c^2\cdot 2 \pi^{5/2}/((a_1 + a_1)\cdot(a_3 + a_3)\cdot \sqrt{2 a_1 + 2 a_3}) +$
$ 4 a^2 c^2\cdot 2 \pi^{5/2}/((a_1 + a_3)\cdot(a_1 + a_3)\cdot\sqrt{2 a_1 + 2 a_3}) +$
$ 8 a b c^2\cdot 2 \pi^{5/2}/((a_1 + a_2)\cdot(a_3 + a_3)\cdot \sqrt{a_1 + a2 + 2 a3}) +$
$ 4 a b c^2\cdot 2 \pi^{5/2}/((a_1 + a_3)\cdot(a_2 + a_3)\cdot \sqrt{a_1 + a_2 + 2 a_3}) +$
$ 2 b^2 c^2\cdot 2 \pi^{5/2}/((a_2 + a_2)\cdot(a_3 + a_3)\cdot \sqrt{2 a_2 + 2 a_3}) +$
$ 4 b^2 c^2\cdot 2 \pi^{5/2}/((a_2 + a_3)\cdot(a_2 + a_3)\cdot \sqrt{2 a_2 + 2 a_3}) +$
$  4 a c^3\cdot 2 \pi^{5/2}/((a_1 + a_3)\cdot(a_3 + a_3)\cdot \sqrt{a_1 + 3 a_3}) +$
$ 4 b c^3\cdot 2 \pi^{5/2}/((a_2 + a_3)\cdot(a_3 + a_3)\cdot \sqrt{a_2 + 3 a_3}) +$
$ c^4\cdot2 \pi^{5/2}/((a_3 + a_3)\cdot(a_3 + a_3)\cdot \sqrt{4 a_3})$
Some copy-and-paste Mathematica code --don't know if it works in Wolfram Alpha.
z11 = 2.0925; z21 = 1.24;

a1 = .109818*z11^2; a2 = .405771*z11^2; a3 = 2.22766*z11^2;

a = .444365 (2 a1/Pi)^(3/4); b = .535328 (2 a2/Pi)^(3/
     4); c = .154329 (2 a3/Pi)^(3/4);

a^4*2 Pi^(5/2)/((a1 + a1)*(a1 + a1)*Sqrt[4 a1]) +
 4 a^3 b*2 Pi^(5/2)/((a1 + a1) (a1 + a2)*Sqrt[3 a1 + a2]) +
 2 a^2 b^2*2 Pi^(5/2)/((a1 + a1) (a2 + a2)*Sqrt[2 a1 + 2 a2]) +
 4 a^2 b^2*2 Pi^(5/2)/((a1 + a2)*(a1 + a2)*Sqrt[2 a1 + 2 a2]) +
 4 a b^3*2 Pi^(5/2)/((a1 + a2) (a2 + a2)*Sqrt[a1 + 3 a2]) +
 b^4*2 Pi^(5/2)/((a2 + a2) (a2 + a2)*Sqrt[2 a2 + 2 a2]) +
 4 a^3 c*2 Pi^(5/2)/((a1 + a1)*(a1 + a3)*Sqrt[3 a1 + a3]) +
 4 a^2 b c*2 Pi^(5/2)/((a1 + a1)*(a2 + a3)*Sqrt[2 a1 + a2 + a3]) +
 8 a^2 b c*2 Pi^(5/2)/((a1 + a2)*(a1 + a3)*Sqrt[2 a1 + a2 + a3]) +
  4 a b^2 c*2 Pi^(5/2)/((a1 + a2) (a2 + a3)*Sqrt[a1 + 2 a2 + a3]) +
  8 a b^2 c*2 Pi^(5/2)/((a2 + a2) (a1 + a3)*Sqrt[a1 + 2 a2 + a3]) +
 4 b^3 c*2 Pi^(5/2)/((a2 + a2)*(a2 + a3)*Sqrt[3 a2 + a3]) +
 2 a^2 c^2*2 Pi^(5/2)/((a1 + a1)*(a3 + a3)*Sqrt[2 a1 + 2 a3]) +
 4 a^2 c^2*2 Pi^(5/2)/((a1 + a3)*(a1 + a3)*Sqrt[2 a1 + 2 a3]) +
 8 a b c^2*2 Pi^(5/2)/((a1 + a2)*(a3 + a3)*Sqrt[a1 + a2 + 2 a3]) +
 4 a b c^2*2 Pi^(5/2)/((a1 + a3)*(a2 + a3)*Sqrt[a1 + a2 + 2 a3]) +
 2 b^2 c^2*2 Pi^(5/2)/((a2 + a2)*(a3 + a3)*Sqrt[2 a2 + 2 a3]) +
 4 b^2 c^2*2 Pi^(5/2)/((a2 + a3)*(a2 + a3)*Sqrt[2 a2 + 2 a3]) +
  4 a c^3*2 Pi^(5/2)/((a1 + a3)*(a3 + a3)*Sqrt[a1 + 3 a3]) +
 4 b c^3*2 Pi^(5/2)/((a2 + a3)*(a3 + a3)*Sqrt[a2 + 3 a3]) +
 c^4*2 Pi^(5/2)/((a3 + a3)*(a3 + a3)*Sqrt[4 a3])

Added cut-and-paste Mathematica code. Editing to z11 = 1.24 will give (22/22).


text:
Formal charge in a co-ordinate covalent bond
In coordinate bond, Does the donor atom have positive formal charge and the acceptor atom have negative formal charge?
For example, in carbon monoxide the oxygen atom donates its lone pair of electrons and has positive formal charge and the carbon atom accepts this lone pair and has negative formal charge. The same thing happens in $\ce{NH3BF3}$, the nitrogen atom donates its lone pair of electrons and has positive formal charge and the boron atom accepts this lone pair and has negative formal charge.
Do the molecules which have coordinate bond follow the octet rule?

In coordinate bond, Does the donor atom have positive formal charge and the acceptor atom have negative formal charge?

Not always. e.g. In a reaction $\ce{NH3 + H+ -> NH4+}$, the lone pair of nitrogen is donated to the proton. As a result, the nitrogen has a positive formal charge, but the formal charge of the hydrogen is $\pm0$.

Do the molecules which have coordinate bond follow the octet rule?

Not always. e.g. It is difficult to determine whether metal atoms are following octet rules in metal complexes.


text:
Is methylchloride reaction with water SN1 or SN2
I wanted to know if methylchloride reacting with $\ce{H2O}$ would be SN1 or SN2.
Methyl halide would not react using SN1 because methyl carbocation is very unstable but it would also not react using SN2 because $\ce{H2O}$ is polar protic.
Please correct me if I am wrong. I think in this problem there would be no reaction. What are your opinions?

Just because the medium is polar protic, it does not necessitate SN2 not happening.

Just $\ce{H2O}$ would give you a very slow SN2 reaction. Adding a base eg aq $\ce{KOH}$ would be faster.

@Aniruddha Deb , So is it correct to say that primary substrate in the presence of polar aprotic solvent, in general, will give a very slow SN2?

No, the nucleophile is $\ce{OH-}$, whose concentration will increase if you add a base, and since SN2 is bimolecular, the rate will increase. Nothing to do with solvent here.

@Aniruddha Deb, Thankyou, now I have understood


text:
Potassium tetraoxalate preparation

How is potassium tetraoxalate produced industrially? In fact How I can limit the reaction of $\ce{H2C2O4}$ and $\ce{KOH}$?
Or is there any other route for this purpose?

In fact How I can limit the reaction of H2C2O4 and KOH?

If you are to react potassium hydroxide with oxalic acid in 1:1 mole ratio, you will get potassium hydrogenoxalate ($\ce{KHC2O4}$). Its hydrate form exist at a specific temperature range. This salt contains the hydrogenoxalate anion and hence known as acid potassium oxalate. But if you add an excess of concentrated oxalic acid and maintaining a temperature of below $\pu{50 °C}$, the much less soluble potassium tetraoxalate ($\ce{K+[C2HO4]^−.C2H2O4}$) forms and precipitates out of solution since it is sparingly soluble in water. So your target compound depends on the concentration of the reagents and the temperature.

Or is there any other route for this purpose?

There is a patent which claims to have made potassium oxalate from potassium formate.


text:
Calculating the amount of oxygen in a glass using candle burning time
I would like to calculate the amount of oxygen or concentration of oxygen in a glass or conical flask using candle burning time.
I am thinking of a simple way to calculate this based on the fact that the atmosphere contains $21\,\%$ oxygen. But I am not sure if this would give correct values.
Here is what I am thinking about:
I could do a candle and glass experiment and count the time for burning of candle.
First I will take a glass with atmosphere air and then keep it inverted on burning candle and count the time for burning.
Next I will take a glass with oxygen-enriched air (with unknown concentration) and then keep it inverted on burning candle and count the time for burning.
Here is a simple formula using cross multiplication, I could use
$$x = \frac{t_\mathrm{OEA}}{t_\mathrm{air}}\cdot 21\,\%,$$
where $x$ is the concentration of oxygen-enriched air, $t_\mathrm{OEA}$ is the time taken to burn in oxygen-enriched air, $t_\mathrm{air}$ is the time taken to burn in atmospheric air.
Is this possible or is there a way to do this without having to buy oxygen sensors?

Your idea implies the rate of burning is independent on oxygen content, what is not true.

@Poutnik , Thanks for reply, I am implying that the time of burning will depend on oxygen concentration in glass, as other ingredients such as fuel(candle wick) is in sufficient quantity

You might want to think about the horrible Apollo 1 tragedy. As @Poutnik says, the burn rate is not independent of oxygen content and a linear relationship cannot simply be assumed.

@EdV , Thanks for explaining, so my assumption was wrong, a linear relationship cannot be made.
If you could help me with a formula using which concentration of oxygen could be calculated, that would be great help.
I just want to somehow get the concentration of oxygen without buying expensive oxygen sensor

I would help if I could, but I do not know a good alternative that does not involve titration, etc. But there are many excellent chemists here, so maybe one of them will post a solid answer. I hope that happens!

@EdV Thanks for your good words, I would like to tell you what I was working on.
I wanted to create a DIY Oxygen Concentrator, which will help save Covid patients.
I just wanted to see the output oxygen concentration of this concentrator, so that I could modify it a bit, to increase efficiency.
You can have a look at the prototypes created by ETH Zurich at here:
http://oxygenator.geprojects.tech/
Maybe you could make one for yourself, if you are interested :)

Then perhaps a Thermal conductivity detector, with air versus oxygen-enriched air. These detectors are used in gas chromatography, but there the reference gas is hydrogen or helium. Still would need calibration and linearity would need to be investigated. You might want to contact a gas chromatography company and see if they have any insights or willingness to help. Best of success!

Early chemists used something similar to achieve the goal you seek. But the way to do it is to burn the candle to completion and observe and observe the volume of gas that disappears from the inverted vessel (water will rise as oxygen is consumed). (it gets a little more complex as a key product of burning is also a gas but one much more soluble in water so one can refine the experiment to give a good idea of the volume lost, which is the volume of oxygen).

Thanks @EdV for your idea, So what I will be using is a thermistor (these are very cheap) to detect the temperature difference of O2 given out by concentrator . I really never want to know the exact concentration, just wanted to know that whether the concentration of O2 increased or not after doing modifications in the prototype. So I just have to see the temperature of O2 before modification in concentrator and after modification, to know whether there is a increase in concentration or not
Just let me know if this idea wouldn't work
Thanks once again

Thanks @matt_black for your valuable comment, I ended up using the idea suggested by EdV

@matt_black but yours and John's suggestion definitely helps me for calculating the approximate concentration of oxygen, without titration

I think you will be better off measuring the volume change after complete burning.
A common demonstration  was to place a candle in a dish of shallow water and light it, then put a clear cup over it and see how much water was sucked into the bottom as the oxygen is consumed. utilizing the same effect in a more controlled way perhaps would be a straightforward way to get a rough oxygen concentration.
When calculating it remember that 2 oxygen molecules are replaced by a single co2 molecule and you will want your water deep enough so that when the heated air expands it doesn't escape out the bottom and account for temperature differences either by doing the math or waiting for the temp to be the same.
Fully burning steel wool and measuring the gain in weight due to the oxygen getting incorporated into solid iron oxides might also work.

One–not two–oxygen molecules produce one carbon dioxide molecule. But that can be absorbed in suitable water solutions.

For burning hydrocarbons like candles it's 2 to one. It is 2O2 + CH4 (or other hydrocarbon) -> CO2 + 2H2O

Actually for methane it is 3 oxygens in total but only one gives a gaseous product. For wax, that will be closer to 2 because the overall formula of the burning compound is closer to CH2, and one of the products is water, a liquid. You are right that you have to account for this to get the loss of oxygen gas, but the fact one product is also a gas is a complication (often ignored, though).

Yeah, you would have to work it out for your specific fuel, or just do a batch of atmospheric air and reverse engineer the ratio. It sounds like for their purposes of an oxygen concentrator they will only need some fairly course measurements as they would expect fairly large changes in oxygen concentration and this method should work fine.

You might get better results using titration. Pick a substance that reacts fairly completely with atmospheric oxygen, and measure how much needs to be added to remove all the $\ce{O2}$ from a known volume.
For example, you might use washed, fine, steel wool (washing is needed o remove processing oils) to remove oxygen, and methylene blue with alkaline glucose to detect any remaining $\ce{O2}$.
You might also try the Winkler test to measure dissolved oxygen in water.


text:
Plasticisers and other impurities from polyethylene laboratory bottles?
Does polyethylene (PE), widely used for wash/squirt bottles in labs, contain soluble compounds such as plasticizers that can accumulate in organic solvents (e.g. alcohols, acetone) after prolonged storage? I noticed an odd, "plasticky" smell from old propanol/ethanol from such bottles and the bottles, once emptied, also give off a similar smell. This smell is not present in fresh, analytical grade solvents or new, unused bottles.
Can plastic bottles be a source of e.g. bisphenol A and get absorbed by the skin, when applying alcoholic hand rub from such bottles?
It has been shown that plastic labware can contaminate their content with various autofluorescing compounds.
PS.: I have the exact model, but I'm not sure if one is allowed to post links to specific products/manufacturers here...

Polyethylene (polythene, to some) may have plasticizers, such as adipates, terephthalates, or phthalates, but they are not inherently needed in all cases. Hopefully, the supplier of those alcohols does not bottle them in cntainers with plasticizers, but a wash bottle, purchased separately, might very well have them added to the plastic, particularly because a squeeze bottle must be very flexible.
It is also possible that the polymer itself degraded over time, and with exposure to light, yielding ethylene monomer and other scission products. Compare the odor of the alcohol with that of ethylene, a faint sweet and musky odor (or "odour" if it's ethene).
As to which plasticizer is used, if any, you'd need to consult the bottle manufacturer. You could substitute a glass wash bottle, e.g. a gas wash bottle with aspirator tube, or a PTFE pipette pump, if you're concerned about contamination, e.g. because your measuring phthalates.
However, if your concerned about personal health, this is not the correct forum.

Thanks! Yes, "sweet" and "musky" is a very good description of the smell, even somewhat rancid. And yes, the health question is definitely of interest, although it's "downstream" of the above chemical question (no impurities, no harm). For analysis, glassware is definitely the way to go, but the PE wash bottles are frequently used for washing/rinsing (even for sanitizing hands against SARS-CoV2)

BTW, most "squirt" bottles of ethanol-based hand sanitizer I've seen are polyethylene terephthalate, "PETE" or "PET". Whether that sheds phthalates into the cleaner is a good question... you might test for that, and publish your findings, if you have the facility to do so.


text:
What grade of sodium fluoride is needed for preservation of urine for drug and alcohol testing?
We are preparing DUI test kits and the distributor we purchased our sodium fluoride vials from is no longer in service.  We are having difficulty finding a new supplier.  Would food grade work for this application?  Food grade can be purchased online and we can fill the vials ourselves eliminating the issue.  Only .2ml is needed for the application.  For the process, the officer opens the lid on the plastic vials and drops the sodium fluoride into the urine specimen, then seals it back up and ships it off to be tested.  Would the pharmaceutical grade make a difference in the ability to preserve the urine?
Thanks for your help

The grade for conservation is rather irrelevant. But is very relevant for the analysis, so it should be p.a. grade ( per analysis, try Sigma or Merck ) or similar. Farmaceutical grade could be good, but there should be run blank tests for pure water samples ( or  pure urine ). Or making parallel analysis for old and new fluoride batches.


text:
Sign of change in enthalpy and change in entropy
Why is it wrong to assert that the change in entropy and the change in enthalpy must always have the same sign?
What makes me think that they must have the same sign is the fact that every reaction invariably comes to equilibrium under suitable conditions; and so we have the corresponding temperature equal to $ΔH/ΔS$ (setting $∆G = 0$ in the equation $ΔG = ΔH - TΔS).$

Why, it's trivial. Before the reaction comes to equilibrium, it is not in equilibrium, and then $\Delta G\ne0$.

Do you mean the changes in entropy and enthalpy?

@IvanNeretin What I meant is that we can determine (at least theoretically) the temperature at which a reaction comes to equilibrium by using the equation I mentioned in my question. And since temperature is a non-negative quantity, it leads me to think that the change in entropy and the change in enthalpy must have the same sign.

@ChetMiller Yes, my bad.

Temperature is not the only variable affecting $\Delta G$. That being said, yes, when the reaction is in equilibrium, $\Delta H$ and $\Delta S$ must have the same sign.

@IvanNeretin Yes, I get that. Okay, that answers my question, then. Thank you! Excuse me for going off-topic now, but doesn't this have significant implications? Like it is sometimes possible to easily guess if a particular reaction is accompanied by a huge increase in entropy (for example, when the number of moles of gaseous products largely outnumbers that of gaseous reactants), while it is not that easy to guess the change in enthalpy. So, is it safe to use the fact that the change in entropy and that in enthalpy must have the same sign to predict if a reaction is exothermic or endothermic?

TL;DR
In general the entropy of reaction can be written as
$$  T\Delta _r S=  \Delta_r H  + RT\log \left(\frac{Q_e}{Q}\right)  $$
At equilibrium $Q_e=Q$ and
$$  T\Delta _r S_e=  \Delta_r H   $$

Consider a simple reaction that behaves ideally (occurs under ideal solution conditions).
If it is carried out at constant T and p we can write
$$\Delta_r G = \Delta_r G^\circ + RT \log Q \tag{1}$$
where Q is the reaction quotient.
But we can also write that
$$\Delta_r G = \Delta_r H - T\Delta_r S\tag{2a}$$
and
$$\Delta_r G^\circ = \Delta_r H^\circ - T\Delta_r S^\circ\tag{2b}$$
Equation (1) can then be written as
$$\Delta_r G = \Delta_r H^\circ - T(\Delta_r S^\circ-R \log Q) \tag{3}$$
Matching terms in equations (2a) and (3) we have that
$$\Delta_r H = \Delta_r H^\circ \tag{4a}$$
and
$$\Delta_r S = \Delta_r S^\circ - R\log Q \tag{4b}$$
When the reaction is at equilibrium $Q=Q_e$ (the reaction quotient is then equal to the equilibrium constant, here written $Q_e$) and $\Delta_r G = 0$ which means, combining equations (2a) and (4a) that
$$ T\Delta _r S_e =  \Delta _r H^ \circ \tag{5} $$
and
$$ T\Delta _r S^ \circ = T\Delta_r S_e + RT\log Q_e  \tag{6}$$
so that
$$  T\Delta _r S=  T\Delta_r S_e + RT\log Q_e  - RT\log Q \tag{7a} $$
or
$$  T\Delta _r S=  \Delta_r H ^\circ + RT\log\left(\frac{Q_e}{Q}\right)  \tag{7b} $$
Now compare equations (5) and (7b). Equation (5) holds at equilibrium and says, sure enough, that the reaction entropy and enthalpy are equal in sign at this point in the reaction coordinate. However, equation (7b) - which is the more general expression - says that $\Delta_r S$ can in fact differ in sign from $\Delta_r H^\circ$, depending on the magnitude of the reaction quotient Q. It turns out that while the enthalpy of a reaction in an ideal solution is a constant, the entropy of reaction can be tuned by modifying Q.

Note my notation is not entirely orthodox, but was chosen in part to make the derivations and meanings clear.

Thanks a lot for helping me out! @BuckThorn


text:
Why is the electrode potential for a cathode positive?
Electrode potential is the potential difference between the electrode and the surrounding electrolyte.
If the potential of the cathode is positive this means that the potential of the electrode is greater than the potential of the electrolyte, which means that the electrons will flow from the electrolyte to the cathode.
Shouldn't it be the other way around? shouldn't electrons flow from the cathode to electrolyte meaning the electrode potential should be negative?
Similarly, we can say that the electrode potential for an anode is positive.
Can someone point out the faults in this reasoning?

Electrode potential is NOT the potential difference between the electrode and the surrounding electrolyte.

"An electrode potential is strictly the potential difference developed at the interface between the electrode and the solution" ---wikipedia

Then there is electrode potential(1) and electrode potential(2).

what does that mean?

It means 2 different potential refererence levels. 1 = SHE, 2=electrolyte. but the latter is quite hard to define, determine or reproduce.

Possible duplicate: Why do positive ions go to the cathode?

The question: Why is the electrode potential for a cathode positive?

Since question didn't directed to what kind of cell, the answer is it is not always positive:

As M. Farooq clarified, cathode is where reduction is happening in a cell. Thus, in Galvanic cell it is positive as the reduction reaction, $\ce{Y+ + e- -> Y}$, happens. The electrons needed for this purpose came from the anode. However, in an electrolytic cell, electrons from an outer source (e.g., a battery), flow to the negative end where the reduction reaction, $\ce{M+ + e- -> M}$, happens. Thus, the cathode is negative here.

Edit: I agree with  Poutnik's comment below. Accordingly, it may be better to say cathodes are more negative than anodes in electrolytic cells and more positive than anodes in galvanic cells. As more positive does not mean positive and more negative does not mean negative. The convention of marking positive and negative cell contacts is good enough for every day life, but it says nothing about the sign of electrode potentials either with respect to free electron or SHE (e.g., relevant half-cell reduction reaction's potential may carry positive or negative sign, based on the corresponding reaction).

It may be better to say cathodes are more negative than anodes in electrolytic cells and more positive than anodes in galvanic cells. As more positive does not mean positive and more negative does not mean negative. The convention of marking positive and negative cell contacts is good enough for every day life, but it says nothing about the sign of electrode potentials ( either wrt free electron either wrt SHE ).

@ Poutnik: I have edited the answer accordingly. It's a very good point. Thank you for checking.

See also my last comment to the Maurice below my answer. I had in mind rather physical measurable potentials that do not flip signs. There can be cells with both electrode potentials negative or both positive wrt SHE. And all are positive wrt the potential of a free electron.

@Poutnik, very important point that ALL are positive wrt to free electron!!

Electrode potential is NOT the potential difference between the electrode and the surrounding electrolyte.
The electrode potential is conventionally the relative potential to the reference standard hydrogen electrode(SHE) with conventionally assigned potential 0 V.
The "absolute" potential of the  SHE wrt a free electron potential is then estimated as $\pu{+4.44 \pm 0.02 V}$.
Cathodes are more negative than anodes in electrolytic cells and more positive than anodes in galvanic cells. But more positive does not mean positive and more negative does not mean negative. The convention of marking positive and negative cell contacts is good enough for every day life, but it says nothing about the sign of electrode potentials ( either wrt a free electron in vacuum either wrt SHE ).
The electrode potentials can be positive + negative, or both positive or both negative wrt SHE. They are always positive wrt a free electron.

so it is the potential difference between the electrode and the hydrogen electrode?

Yes, unless it is explicitly stated you consider the potential difference between the electrode and the electrolyte.

my textbook says:

"a potential difference develops between the electrode and the electrolyte which is called electrode potential"

is my textbook lying to me?

Rather you may have missed the context. But potential of the SHE  0.0 V does not mean the potential difference electrode-electrolyte is zero.

is the potential difference between the electrode and electrolyte negative if it is a cathode then?

Rather, the cathode potential is lower than its equilibrium potential, so it is the source of electrons to reduce electrolyte components.

Let us continue this discussion in chat.

All the sign conventions about electrode has been changed "recently". In my Handbook of Chemistry dated 1961, the standard electrode potential is + 0.7628 V for Zn, and - 0.346 V for Copper. So this is exactly the opposite of the present values. Minigame may have an old book.

@Maurice There is need to distinguish formal potential of the electrode half reaction, bound to the change of Gibbs  energy, that switch the sign with the reaction direction,  and the physical potential of electrode wrt SHE, that does not switch the sign.

I think your source of confusion is the name cathode and anode. Anode and cathodes have nothing to do with the electrostatic sign. They have to do with the processes.
Cathode, whether electrostatically positive OR negative, is the electrode where reduction is occurring.
Anode, whether electrostatically positive OR negative, is the electrode where oxidation is occurring.
Hope that clarifies your confusion about the signs.

Electrode potential is the potential difference between the electrode and the surrounding electrolyte.

This is not the electrode potential but is rather called interfacial potential difference.
Mathew wanted to elaborate this further on "Electrode potential is NOT the potential difference between the electrode and the surrounding electrolyte". I think this an electrochemist's unfulfilled dream to find the absolute potential difference, i.e., we dip Cu electrode in a copper solution and we measure the potential difference between the two. I think one can improvise what Poutnik is saying. The tabulated electrode potentials, found commonly, are not the potential differences of metals and their solutions. The reason is that it is impossible to measure such a hypothetical construct.

Make sense. However, I'd include the definition of electrode potential as well.

Also, @Poutnik says "Electrode potential is NOT the potential difference between the electrode and the surrounding electrolyte." What is your take of this?

@MathewMahindaratne, I edited the post.

I think AChem does a great job explaining the meaning of cathode and anode. (It has to do with the process occuring, not the electrostatic potential).
The anode always gives electrons (oxidation).
The cathode always accepts electrons (reduction).
The postive electrode is the one with the higher potential than the negative electrode. Since the potentials are all defined relative to a theoretical standard electrode (see Poutnik's answer), this designation may or may not match the actual sign of the potential.
When the reaction proceeds spontaneously (galvanic cell), the negative electrode is the anode. Electrons flow from the negative electrode to the positive electrode.
When the reaction must be driven by an external energy source (electrolytic cell), the negative electrode is the cathode. Electrons are forced into the negative electrode from the positive electrode.


text:
Is benzocyclobutadiene aromatic or anti-aromatic?
My teacher told that benzocyclobutadiene is both aromatic and anti-aromatic. I do not understand why. Can anyone explain?

Benzocyclobutadiene has an aromatic benzene ring as well as an anti-aromatic cyclobutadiene ring. This gives it the characteristics of both aromatic and anti-aromatic compounds.
Now, according to Cyclobutadiene and its related compounds, [1,p 180]:

Benzocyclobutadiene, the monobenzo derivative of cyclobutadiene, is intermediate in structure between the unstable parent compound, cyclobutadiene and it stable dibenzo analog biphenylene. Benzocyclobutadiene and its derivatives resemble cyclobutadiene rather than biphenylene in being known as only transient reaction intermediates.

Reading this, we can say that benzocyclobutadiene resembles cyclobutadiene more than biphenylene and so has a dominant anti-aromatic character. This means that like cyclobutadiene, the cyclobutadiene part of the ring undergoes dimerisation to form a more stable compound.
Note: The preview of the book referenced has an entire chapter based on just benzocyclobutadiene. If interested, you can go through it for more information
References:

Cyclobutadiene and Related Compounds; M.P. Cava, M.J Mitchell; Chemistry Department, Wayne State University, Detroit 1967. ISBN 978-01-2164-450-5. [Preview]

Reading this you cannot say it's dominating the aromaticity. The NMR calculations referred in the book shows the alkene like character at cyclobutadiene promoting Diels Alder. You've manipulated the statements well. FYI, Cyclobutadiene can be NON aromatic as well depending on the adopted geometry. As a member of 21st century you may perform the calculations yourself at any chem-computer and find the aromaticity of phenyl ring and alkene-city of cyclobutadiene

@AnindyaPrithvi If that is the case, you could submit your own answer and prove this is wrong. Unfortunately, I am currently short of such resources as you state and if a substantial proof can be given, I would be more than happy to upvote it. I stated It would be more prevalent as similar to cyclobutadiene it seems to behave more as an intermediate than be present as a compound. And so assumed it would be more antiaromatic than aromatic. There may be flaws centered around my lack of complete knowledge, so as I said above please formulate your answer. Thanks for your concern regarding my ans.

Unfortunately I do not have the time to compile resources as of now and present a full sourced answer. I only pointed out the incorrectness of your assumption using available NMR analysis, which maybe edited appropriately. As a valuable resource, you can just read about how NMR points at aromaticity and alkene nature, that will also be beneficial to point out the inaccuracy without computational/Lab analysis.

https://pubs.acs.org/doi/10.1021/ja00168a059


text:
Stubborn CuAAC reaction as derivatization for ESI-MS
Problem description
I have a substance that is extremely non-polar and therefore does not ionize in the ESI source, which makes it impossible to quantify small amounts. The substance contains only two functional groups, which are a point of attack for possible derivatization. One is an alkyne, and the other is a hydroxyl group. Both groups are attached to the same carbon. Since the substance is to be detected from blood plasma, I decided against derivatization at the hydroxyl group, since numerous side reactions with proteins, which also carry a lot of hydroxyl groups, are to be expected. The alkyne, however, is a good point of attack and is suitable for bioorthogonal reactions. A professor once said about the copper(I) catalyzed alkyne-azide cycloaddition (CuAAC) that this is a reaction that "any fool could manage". I have been trying to get the reaction going for a few days now, but so far, I have not been able to detect any product in LC-MS... Decide for yourself what this says about me as a chemist.
Since I have followed several approaches from papers, I slowly ask myself if my substance is not suitable for a CuAAC. Since the research project and the substance are new and unpublished, I cannot provide a complete structural formula here. But this much can be revealed: It is essentially a steroid, which carries a hydroxy group and an alkyne on carbon C17.

What was done
I performed the reaction with the pure substance (i.e. without blood plasma) according to Presolski et al., but I did not use THPTA nor aminoguanidine. All concentrations were used as described in the paper and the reaction was performed at 40°C according to the suggestion under "Troubleshooting". The azide was dissolved in DMF. Thus, there were 10 µL DMF (5%) in the buffer solution.
I also performed the reaction with the pure substance according to Hong et al.. Again, the proposed concentrations were used but neither THPA nor aminoguanidine were used.
I also tested different variations of those protocols with much longer reaction times (8 to 24 hours), more or less temperature, with and without pH 7 buffer.
As far as I know, the 4-substituted 1,2,3-triazoles formed by CuAAC should be extremely stable, even to metabolic degradation (Agalave et al.), so I do not believe that they decomposed so quickly that they were all destroyed by the measurement. To avoid ion suppression, I purified the reaction after the one-hour reaction time by RP-SPE, using the SigmaAldrich/Merck protocol.
Questions

Does anyone have experience with CuAAC reactions that do not want to run?
Which parameters should I at best adjust to make the reaction successful?
Is it possible that a hydroxide in alpha-position disturbs the reaction? I haven't found anything on this yet.
What advice is there to make the reaction successful even at very low (< 10 pg/mL) substance concentrations?

Update as of 24/07/2020
The reaction was, in fact, successful when applying the already cited protocol by Presolski et al. and using a DMF-water-ratio of 3:1 and incubating at 60 °C for 2 h. Unfortunately, though, the intensities are pretty weak which means that the reaction turnover is not great. The peak of my original substance is at 9.6 × 107 (50 µM concentration) and the intensity of the derivatized product, which should be way better ionized, is 4.2 × 107.
New question

How to get a better turnover while maintaining a reasonably short reaction time?

Interesting question.  I once got RuAAC to work when derivatizing internal alkynes for LC-MS, and CuAAC should be easier than RuAAC, since the Ru catalysts are very unstable to air/water.  I'd recommend trying a Cu-chelating azide like picolyl azide, which should be more reactive.

I'd also get a test alkyne that you can run your CuAAC protocol on.  If you can make test alkyne-derived triazoles, then the problem is with your molecule.  If you can't, then the problem is with your protocol (at least).

I have no idea what "R" is in your structure or how the 5/6 rings are fused together or what the relative stereochemistry is at the carbon with the alkyne/alcohol. There may be an issue where the alkyne is in the "pocket" of the 5/6 fusion, and that steric bulk is really slowing the reaction.

How to get a better turnover while maintaining a reasonably short reaction time?

To speed up the reaction, you should look at changing (1) the reagents, (2) the catalysts, or (3) the reaction conditions.  In particular you might be dealing with poor solubility for your alkyne substrate if you are doing the reaction in aqueous conditions.
1. Reagents
You could try the picolyl azides I mentioned in the comments in lieu of the 4-acetophenyl azide you originally tried.  These azide reagents chelate copper significantly, thus increasing reaction rates since proximity of two of the three components of the reaction (\ce{Cu+} and aizde) is enhanced by the chelation effect.
2. Catalyst
There are a variety of exotic $\ce{Cu+}$ organometallic complexes that work better than the "standard" copper salts.  An example publication is Sebest et al. 2018, which reports that the N-heterocyclic carbene complex $\ce{[CuI(Mes‐6)]}$ (Mes-6 being an NHC ligand pictured below) worked well at loadings down to 0.05 mol% in their trial reaction.

This might be overkill if you are not using the ligand THPTA used in your original protocol.  Maybe you could just use that?
3. Conditions
You an always heat more, but I think a more important variable is your solvent.  Is your reaction solvent one that dissolves all the species, including alkyne, azide, and copper(I) complex?  Instead of aqueous conditions, can you try acetonitrile or 100% DMF?  Or n-butanol?
Additionally, the Presolski reference you give mentions that high chloride concentrations (~0.2 mol/L) can compete with ligand for copper ion.  Plasma has about a 0.1 mol/L chloride concentration, so maybe it's an issue?  Probably speaks to the need to use a good copper ligand like THPTA.
Bottom line
If it were me, I'd try getting some THPTA or other ligand to improve $\ce{Cu+}$ solubility in organic solvents, and trying the reaction in non-aqueous solvents where your steroid is likely to be soluble.  I'd also get the picolyl azide if my lab could afford it.  If that doesn't work, and I still wanted to improve things, I guess I'd look at increasing temperature or using more exotic NHC ligands for the copper.


text:
Why am I getting higher conversion values of isopropanol dehydration over a zeolite catalyst than n-propanol dehydration?
The two process are run separetly but under the same condition, same activation of the catalyst, same reaction temperature, same flow.
The initial conversion of n-propanol is ~50% and isopropanol ~70%.
My expections was lower conversion values of isopropanol, because of it's branched structure it has a more contrained diffusion to the acid sites of the zeolite.

Stability of carbocation intermediate?

What about concentrations?

Since isopropanol has higher vapor pressure I run it under slower flow. So the concenrations should be equalized.

In a typical acid-catalyzed dehydration, the reaction goes by an SN1 mechanism, and the rate-limiting step is
$$\ce{R-OH2+ -> R+ + H2O}$$
The rate of this step is typically a function of the stability of the cation product. In your case, the secondary cation formed from isopropanol is more stable than the primary cation formed from n-propanol.
I'm using "typical" and "typically" here, because these are results from homogeneous reactions, and it sounds like you are doing heterogeneous catalysis with a solid zeolite. That can certainly change the reaction behavior, but in your case it appears that the zeolite is not selective for one molecule over the other, and the trend is the same as what is observed for homogeneous systems.

I was expecting the zeolite to be more selective towards n-propanol since iso-propanol is a branched molecule. The zeolite I'm using in the reaction has a very small window 0.38 nm to the cage that contains the acid sites.

You expect using zeolite might selectively allow n-propanol over isopropanol into the cage of zeolite. While it may be true, 0.38 nm window kind of zeolites might have activity on the external surface as well. For example, there are several research articles available where sugars like glucose and fructose, which are quite big compared to n-propanol or isopropanol, can be dehydrated over SAPO34, SAPO44 zeolites with chabazite structure (also 0.38 nm window).[1-2] And if you consider the reaction at external surface the previous answer by Andrew is spot on.
References for sugar dehydration over SAPO34
https://www.sciencedirect.com/science/article/pii/S1385894716312396
https://www.sciencedirect.com/science/article/pii/S0926669019302973
https://pubs.rsc.org/en/content/articlehtml/2013/ra/c3ra43197e


text:
Equilibrium constant for heterogeneous equilibria having aqueous as well as gaseous reactants
Suppose we have a heterogenous equilibrium :
$$\ce{A(aq) +B(aq) <=>  C(g) +D(aq)}$$
Which equilibrium constant is used here?
Both pressure and concentration terms are there. So, according to me, both $K_\mathrm{p}$ and $K_\mathrm{c}$ can be written separately. If that is the case, can we relate both of them by $K_\mathrm{p} = K_\mathrm{c} (RT)^{\Delta n}$?

This explains the different activity constants

Let us continue this discussion in chat.

Does this answer your question? Which equilibrium constant is appropriate to use?

Your conception of the equilibrium constant is flawed. $K_\mathrm{p}$ is preferably used when the reaction only contains gaseous and solid components whereas $K_\mathrm{c}$ is used primarily for  a reaction taking place in water.
In reality, however, both of these equilibrium constants are special cases of equilibrium. This is because in the actual scenario, what we use to calculate the equilibrium constant is a concept known as activity.
$K_\mathrm{p}$ and $K_\mathrm{c}$ are said to be special cases of $K$ since, in $K_\mathrm{p}$, the activity is expressed in terms of the partial pressure of the components and in $K_\mathrm{c}$, the activity is expressed in terms of concentrations of the products and reactants.
In case of reaction as stated above:
$$\ce{A(aq) + B(aq) <=> C(g) + D(aq)}$$
The equilibrium constant would be expressed as follows:
$$K = \frac{a_\mathrm{C}\cdot a_\mathrm{D}}{a_\mathrm{A}\cdot a_\mathrm{B}}$$
Now, for the aqueous components, the activity is expressed in terms of their concentration in the solvent. For the gaseous component, we express it in terms of its partial pressure. Substituting these conditions, we get:
$$K = \frac{P_\mathrm{C}[\mathrm{D}]}{[\mathrm{A}][\mathrm{B}]}$$
An example of where this is used is in electrochemistry.
A galvanic cell may be made which has a heterogenous reaction in which solid, aqueous and gaseous components may be given.
The Nernst equation used to calculate the EMF of such a  cell is given as:
$$E_\mathrm{cell} = E^\circ_\mathrm{cell}-\frac{2.303RT}{nF}\log_{10}Q$$
Here, $Q$ is the reaction quotient. Using this, we calculate the  EMF taking all the activities into consideration.
Note: The activity of a solid or the solvent is taken to be $1$ in such cases.


text:
Computational chemistry software to generate files of cartesian coordinates and zmatrix of molecules
Is there any software that I can provide bond lengths, bond angles and torsion dihedrals of a molecule and generate an output file in the format of cartesian coordinates or zmatrix?
Example:
I would have input similar to the one below (which i provide all bond lengths, bond angles and torsion dihedrals of a molecule)
BOND LENGTHS
     2 1 Si O2 1.63
     3 1 Si O2 1.63
     ...
BOND ANGLES
     2 1 3 Si O2 Si 148,482
     1 2 4 O2 Si C3 107,426
     ...
TORSION ANGLES
     3 1 2 4 -161,552
     3 1 2 8 -41,721
     ...

And I would like to have an output like this:

My idea is to vary some of this information and generate several Gaussian software inputs to calculate energy points at specific bond lengths, for example, without having to manually edit the molecule structure in this software.
I know that Open Babel software can do something like that, but I don't know what type of input to use.

OBabel can write in Fenske-Hall Z-Matrix format, but it cannot read it —going from internal coordinates to cartesian coordinates is very uncommon. It sounds like your starting data would need to be processed anyway. The algorithm (called Natural Extension Reference Frame (NeRF) algorithm) to do the conversion is a few trigonometric functions that would be very straightforward to implement with numpy and rdkit in Python.

Thanks for the comment @MatteoFerla! I know OpenBabel, I tried to post a topic in the software group but without success (it seems that it is stopped). My idea is to take a structure of a molecule and vary, for example, the bond length, and thereby calculate some energy points in the Gaussian software. The only way I know how to do that today is to manually change the structure of the molecule in softwares like Avogadro or GaussView. I would like to know if there is any way to generate the inputs of the distorted molecule in the bond without doing this manually by editing its structure.

What software group? The problem with internal coordinates is that they are redundant meaning that there are more internal coordinates than internal degrees of freedom. So there is no 1 to 1 transformation. A z matrix is non redundant however so you can use that to vary your structures. Scans along internal coordinates are frequently performed using z matrices

@EmersonPL If you are going to calculate the rotation or another change in geometry by Gaussian you should use Opt=ModRedundant. And you don't even need to provide Z-matrix. Also it is possible very easy to calculate barriers in Schrodinger's Jaguar.

Which atom is O2?

This is not the answer to your question, but it is a comment about an approach to the aims, based on the comments. If coding is an option, RDKit is a good option, specifically adding constraints (restraints) to the ForceField and optimising. Given a molecule:
from rdkit import Chem
from rdkit.Chem import AllChem

acetate = Chem.MolFromSmiles('CC(=O)C')

Normally the 3D conformer is generated with:
AllChem.EmbedMolecule(acetate)

And it stops there. However, then you can manually control the force field —either UFF (more elements) or MMFF (classic choice).
ff = AllChem.UFFGetMoleculeForceField(mol)
ff.Initialize()
print(ff.Minimize()) # this tells you if it failed.

Within this, you can add constraints, such as:
ff.AddDistanceConstraint(0, 1, 2., 2.1, 1_000_000_000)

Note that the weight needs to be set absurdly high for absurd lengths. To check it work you could do:
dm = Chem.Get3DDistanceMatrix(acetate)
print(f'{dm[0,1]} Å. {ff.CalcEnergy()} kcal/mol')


text:
What is the difference between Cyclic Process and Reversible Process?
I want to ask the difference between Cyclic Process and Reversible Process.
Consider this graph:

If we have a process that goes from A to B then go back to A through the same path, then will it be called a Cyclic Process or Reversible Process or Both?

The key concept in a reversible process is that the system is in equilibrium with it's surroundings at every instant. A cyclic process is different: in a cyclic process, the initial and final state are the same.
The graph that you have drawn consists of a reversible isothermal expansion(AB) and a reversible isothermal contraction(BA). Overall, the process A->B->A is a cyclic process made up of two reversible processes.

I would add that a cyclic process does not have to be reversible.  The only requirement is that the initial and final states of the system are the same.


text:
Grignard with nitrile: why is the imine protonated twice?

Steps 2 and 3 in the picture show the imine being protonated twice to form the iminium ion in step 4. I'm a little confused as to why the second protonation occurs; does the imine after step 2 still have basic tendencies?

The imine is a stable compound, if you work up carefully you can isolate it. This mechanism assumes that you want to make the ketone so it shows the hydrolysis to the ketone as well as the Grignard addition.

@Waylander that makes sense, thank you!


text:
Solubility of ortho- and para-nitrophenol in benzene
Which one between o-nitrophenol and p-nitrophenol is more soluble in benzene? And what is the reason? If the solvent had been polar, hydrogen bonding would have provided a lot of explanation, but what about non polar solvents?

Same thing$ $...

The nitrophenols have completely different physical behavior based on the position of nitro group:
$$
\begin{array}{c|ccc}
\hline
\text{Compound}  & \text{Melting point} & \text{Boiling point} & \text{Water solubility at } \pu{25 ^\circ C}\\
\hline
\text{2-Nitrophenol}  & \pu{43-45 ^\circ C} & \pu{215 ^\circ C}  & \pu{2 g/L} \\
\text{3-Nitrophenol}  & \pu{89-95 ^\circ C} & \pu{278 ^\circ C}  & \pu{13.5 g/L} \\
\text{4-Nitrophenol}  & \pu{113-114 ^\circ C} & \pu{279 ^\circ C}  & \pu{16 g/L} \\
\hline
\end{array}
$$
This different behavior is due to having intramolecular hydrogen bonding:

Due to these intramolecular H-bonding on 2-nitrophenol, its $\ce{OH}$ group is not readily available to form a hydrogen bond with solvent water. Hence 2-nitrophenol is sparingly soluble in water while 3- and 4-nitrophenol are soluble due to intermolecular H-bonding with water.
Similarly, because 3- and 4-nitrophenols contains much more intermolecular H-bonding than that of 2-nitrophenol, they have large difference in boiling and melting points than that of 2-nitrophenol (need extra energy to break intermolecular H-bonding; The intramolecular H-bonding increases the volatile nature of the molecule comparatively with their isomeric siblings that cannot have intramolecular H-bonding, thus effecting the relevant boiling points).
The intramolecular H-bonding effects oppositely in the solubility in nonpolar solvents such as benzene. Compounds involved in intramolecular H-bonding (kind of chelation) become non-polar (not many interaction forces intermolecularly). As a consequent, these compounds are soluble in nonpolar solvents (oposite to only sparingly soluble in water) whereas their meta and para isomers are less soluble in non-polar solvents but more soluble in water due to intermolecular H-bonding (such as water).

Good answer but your answer tells that Ortho nitrophenol will be more soluble than para and meta and it contradicts with the first answer!?

Not really. For example, check percent composition at $\pu{40 ^\circ C}$ at all three isomers in @Aniruddha Deb answer. That was about 90% for 2-nitrophenol. To get the same composition, you need to increase the temperature to about  $\pu{90 ^\circ C}$ for 3-nitrophenol and  $\pu{105 ^\circ C}$ for 4-nitro-isomer. Aniruddha Deb's interpretation is incorrect.

It turns out that the opposite of what happens in polar solvents takes place when a non-polar solvent is used. At the same temperature, o-nitrophenol is more soluble in benzene than it's m and p isomers.
Sidgwick et al.1 did a study of this and obtained the following results. (Note that the solvent they used was toluene and not benzene, but they are similar enough that the same principles would apply.)

EDIT: The original version of this answer wrongly interpreted the graph and stated that o-nitrophenol would be less soluble in benzene than the other isomers. That was an error on my part. Thank you to Mathew Mahindaratne for pointing it out!
References:

Nevil Vincent Sidgwick, William James Spurrell, Thomas Ellis Davies, “CXXXII.—The Solubility of the Nitrophenols and Other Isomeric Disubstitution Products of Benzene,” J. Chem. Soc., Trans. 1915, 107, 1202–1213. doi:10.1039/CT9150701202.

Good reply. But yours and 2nd one contradict each other?!

@Aniruddha Deb: Please see my comment to OP's question under y answer. According to chart, solubility should be higher for 2-nitrophenol than that of other two at all temperatures. Accordingly, would you correct your answer.

@MathewMahindaratne I edited the answer. Thank you for pointing it out :)


text:
Confusion with definitions of mass
What is the difference between:
atomic mass
relative atomic weight
relative isotopic weight
standard atomic weight
atomic weight
relative atomic mass
I am told by Wikipedia that:
relative atomic weight = atomic weight
and a the start of the article it says:
For a single given sample, the relative atomic mass of a given element is the weighted arithmetic mean of the masses of the individual atoms (including their isotopes)
so standard atomic weight = relative atomic weight
But at https://en.wikipedia.org/wiki/Relative_atomic_mass#Standard_atomic_weight#Definition it says:
It is a synonym for atomic weight, though it is not to be confused with relative isotopic mass. Relative atomic mass is also frequently used as a synonym for standard atomic weight and these quantities may have overlapping values if the relative atomic mass used is that for an element from Earth under defined conditions. However, relative atomic mass (atomic weight) is still technically distinct from standard atomic weight because of its application only to the atoms obtained from a single sample; \textbf{it is also not restricted to terrestrial samples, whereas standard atomic weight averages multiple samples but only from terrestrial sources}
I am told here that:
The standard atomic weight (Ar, standard) of a chemical element is the weighted arithmetic mean of the relative atomic masses (Ar) of all isotopes of that element weighted by each isotope's abundance on Earth.
But in this video I am told this is the definition of relative atomic mass.
Here I am told:
Atomic mass (ma) is the mass of an atom. A single atom has a set number of protons and neutrons, so the mass is unequivocal (won't change) and is the sum of the number of protons and neutrons in the atom.
Atomic weight is a weighted average of the mass of all the atoms of an element, based on the abundance of isotopes. The atomic weight can change because it depends on our understanding of how much of each isotope of an element exists.
In other words:
Atomic weight = standard atomic weight
Also, which of these definitions is used to convert grams of a substance into moles

Related: Quick and simple explanation of molar mass, molecular mass and atomic mass

I have a bad feeling that the terminology is not properly standardised and that chemists are terminologically confused. However, this may not matter much in practice as the specific meaning is almost always obvious from context. I'm happy to be proved wrong, though.

https://chemistry.stackexchange.com/questions/38082/what-is-the-difference-between-molecular-mass-average-atomic-mass-and-mola

@matt_black, I agree that the way atomic masses or atomic weights are taught in general  chemistry is very confusing and superficial.

Thanks these websites clear some of the ambiguity

Relative Atomic Mass vs Atomic Weight
Relative Atomic mass ($m_a$) is the measurement of the mass of an atom relative to carbon-12. It's dimension is M.[1] It can be measured in Da/u, kg or g.* [3]
$$m_a \approx \mathrm{Mass\ Number} \times {1\ \mathrm{u}}$$
except for carbon-12 where they are the same.
Atomic weight is:
Relative atomic mass vs Standard atomic mass
Relative atomic mass or Relative atomic weight is the weighted average of the isotopes of an element according to their abundance in a given sample compared to the atomic mass of carbon-12. It's dimension is M/M or 1. [2] [4]
Standard atomic mass is the same except that the samples are restricted to terrestrial samples normally considering/averaging results from many different samples or results given by the CIAAW.
Relative Isotopic Mass vs Relative Atomic Mass
Relative Isotopic mass considers the mass of an isotope compared to the atomic mass of carbon-12. This is also dimensionless.
*Universal mass constant vs kg , g ,etc
The universal mass constant ($m_u$) is defined as $\frac{1}{12}$ of the mass of carbon-12 which defines 1 Da or 1 u. All measurements in Da are comparisons to the mass of a carbon-12 atom. Since the discovery of Avogadro's constant, we know: [3]
$$ 1\ {\rm{Da}} = 1.660\ 539\ 066\ 60(50)\times 10^{-27}\ \mathrm{kg}$$
Molar Mass vs Molecular Mass (or Formula Mass)
Molar Mass is the grams of a substance in 1 mole of it measured in gram per mole. It's dimension is $\mathbf{M N^{-1}}$.
Molecular mass or molecular weight is the sum of the atomic masses of all the component atoms of each element measured in u. It's dimension is M. Formula mass or formula weight is the sum of the atomic masses of component ions (not sure of the difference between this and molecular mass - in terms of quantity).[2]
[1] see the answer to this Q by M Farooq
[2] What is the difference between "molecular mass", "average atomic mass" and "molar mass"?
[3] Mass number, (relative) atomic mass, average mass
[4] Quick and simple explanation of molar mass, molecular mass and atomic mass

If you are confused, it is not a surprise because chemists remained confused for 200 hundred years. Welcome to the world of real science.
Atomic weight is the older term for atomic mass. Modern day hair splitters would say that mass and weight are different things but historically in the old chemical literature, they meant one and the same thing.
Nobody can repeat two centuries of work here, but briefly, atomic mass of an element meant that it is the mass of any atom that combines with the mass of one oxygen atom.
Now the key question is which leads to opening the Pandora box. What is the mass of one oxygen atom? People had to choose a number, because nobody knew the mass of oxygen atom. Some said, let us fix it to 200, some fixed it to some other number. This arbitrary "mass" of oxygen atom kept on changing, finally, someone said, let us fix it exactly to 16.000000000... (ad infinitum) units.
Notice that I did not need any crutches of moles or grams per mole. This is all modern day stories fed to fresh minds. We are still in early 1920s.
Now once the mass of oxygen is fixed, you can relatively easily assign an atomic weight of other elements with respect to oxygen
For example, atomic mass of hydrogen was determined to be 1.008 which is the mass which would combine with exactly 16 units of oxygen.
Now you can understand that chemists never  talk about absolute atomic masses, they are all relative to some element, so it is relative atomic mass.
Today, the reference is carbon-12 (after 1960s) and there are many more nuances which may not be of interest at this moment.
Think of relative atomic weights like this. Suppose I want to assign an "atomic mass" to you, but I do not know your mass. However, I know your friend's mass which is arbitrarily assigned as 100 lb. Using your friend as a reference we can assign an atomic mass to you indicating whether you are heavier or lighter than your friend.


text:
What is the enthalpy change of solution of oxygen and water?
What is the enthalpy change of solution of oxygen dissolved into water?

I am willing to edit the question but I have no idea about what is meant by "details or clarity" in this case. Is the problem that the solution can not be diluted infinitely?

At $\pu{25 ^\circ C}$ ($\pu{298.15 K}$), it is reported to be $\pu{(-12.06 ± 0.04) kJ mol-1}$ (Ref.1).
Reference:

S.J. Gill, I. Wadsö, "Flow-microcalorimetric techniques for solution of slightly soluble gases. Enthalpy of solution of oxygen in water at $\pu{298.15 K}$," J. Chem. Thermodyn. 1982, 14(10), 905-919 (https://doi.org/10.1016/0021-9614(82)90001-5).

What could be wrong with the question?