text:
Volume calculation perchloric acid 70%
HClO4= 100.5, d HClO4=1.8, (ACO)2O=102, d (ACO)2O=1.1, H2O=18, d H2O=1
A) Preparation of 100 ml perchloric acid 0.1 Molar aqueous solution. below is my answer, but what I'm not sure is that the density the problem has given is for 100% pure acid or 70%. below I calculated the density for 70% assuming 1.8 is for 100%.
d_70%=  100/(70 x 1/1.8+30 x 1/1)=1.452
M=10αd/MW=10x70x1.452/100.5=10.11 M
M_1 V_1= M_2 V_2→10.11 x V_1=0.1 x 100→ V_1≅1ml
B) Preparation of 100 ml perchloric acid 0.1 Molar non-aqueous solution in acetic acid. Okay about this one. So all I know is that I need to remove all water in the acid by adding acetic anhydride, and the ratio is 1/1:

I thought about how to calculate the volumes of both acid and acetic anhydride, and maybe I can use the density, or tried to create some equation, but none actually reached a final answer. So I tried this, I'm not sure this actually makes sense:
70/100 x 1/1.8=0.39 ml
M_pure=  ( 0.7/100.5 x1000)/0.39=17.86 M
M_1 V_1= M_2 V_2→17.86 V_1=0.1 x 100→V_(pure acid)=0.56ml
1-0.56=0.44 ml water
0.44ml x 1g/1ml x 1mol/18g x  (1 mol (ACO)_2 O)/(1 mol water)  x 102g/1mol x 1ml/1.1g=2.27ml acetic anhydrade
So 1 ml perchloric acid and 2.27 ml acetic anhydride, up to 100 with acetic acid

Your question is not very clear. Would you please add some details? For example, are you given 70% perchloric acid?

Yes, I have perchloric acid 70%

Your conversion for the density of perchloric acid is wrong. You generally can't convert densities by doing a linear extrapolation. Assume density of 1.8 is for the 70% perchloric acid by weight.

Actual density of 70% perchloric acid is 1.67 g/mL.

@MaxW I see your point. But if I assume 1.8 as density for 70%, how should I know how much water it contains? I thought there was such a point to it. It is like to show the difference between the aqueous and non aqueous environment and in part A I titrate a 70% acid already containing water with water.

@MathewMahindaratne I don't think my professor actually cares for that when designing a problem

Crimson-Ruby - @MathewMahindaratne's point is absolutely correct. I absolutely believe that problems should be realistic to teach some chemistry too. Why just make up a number when you can use the real values?

Crimson-Ruby - The 70% means 70% by weight. Giving a percentage by weight is typical for concentrated acids.

Given:
HClO4= 100.5, d HClO4=1.8
(ACO)2O=102, d (ACO)2O=1.1
H2O=18, d H2O=1
A) Preparation of 100 ml perchloric acid 0.1 Molar aqueous solution.

I had to do some poking around to understand the density number. Without a temperature the density of liquids is nebulous. Wikipedia lists a density of 1.768 g/ml which seems to be for the 72.5% azeotrope.

So just blundering with the information given assume density is for 70% by weight perchloric acid.
Molarity $\ce{HClO4}$
$$\dfrac{0.70\times\pu{1.8 g/ml} \times \pu{1000 ml/L}}{\pu{100.5 g/mol}} = 12.5 M$$
Volume of conc $\ce{HClO4}$ for 100 mls of 0.1 molar
$$\dfrac{\pu{0.1 L}\times \pu{0.1 M}}{\pu{12.5 M}}= \pu{0.0008 L} = \pu{0.8 ml}$$
B) Preparation of 100 ml perchloric acid 0.1 Molar non-aqueous solution in acetic acid.
This is a bit gnarly.
Volumes are not typically additive. So typically an excess of the acetic acid would be made and this would be used to dilute the 0.8 ml of perchloric acid to 100 ml. However in this case the 0.8 ml of perchloric acid contains water too. Also no density is provided for glacial acid acid, nor for the 0.1 Molar perchloric acid in glacial acetic acid. So the problem isn't really solvable as is.
The other wrinkle here is that mixing exact stoichiometric proportions is impossible. If an anhydrous solution was wanted, then an excess of acetic anhydrous would be used or some other means to dry the solution would be necessary.
Never the less, let's blunder through some sort of solution, as poor as the problem is...
Assume that the volumes are additive.

So we'll get the right proportions of the three components but only somewhere near 100 ml of solution, not 100 ml exactly.

$$V_{solution} = V_{anhydride} + V_{water} + V_{\pu{conc acid}}$$
$$V_{\pu{conc acid}} = \pu{0.8 ml}$$
$$\therefore V_{anhydride} + V_{water} = \pu{99.2 ml}$$
However as mentioned before, the concentrated perchloric acid contains 30% water by weight.
$$\pu{g water in acid} = 0.3 \times \pu{0.8 ml}\times \pu{1.8 g/ml} =\pu{0.4 g}$$
Since water has a density of 1 g/ml that is also 0.4 ml.
We were given that (ACO)2O=102, d (ACO)2O=1.1 so for the anhydride the moles per ml is:
$$\dfrac{\pu{1.1 g/ml}}{\pu{102 g/mol}} = \pu{0.0108 mol/ml}$$
We were given that H2O=18, d H2O=1 so for the water the moles per ml is:
$$\dfrac{\pu{1 g/ml}}{\pu{18 g/mol}} = \pu{0.0556 mol/ml}$$
Now since the perchloric acid contained 0.4 ml of water we need to figure out how many ml of acetic anhydride we need to neutralize that water.
$$\dfrac{\pu{0.0556 mol/ml}\times\pu{0.4 ml}}{\pu{0.0108 mol/ml}} = \pu{2.1 ml}$$
So we now need 99.2 - 2.1 = 97.1 ml of water/acetic anhydride mixture in a 1:1 molar ratio. For 1 ml of acetic anhydride the ml of water needed is
$$\dfrac{\pu{0.0108 mol/ml}}{\pu{0.0556 mol/ml}}= \pu{0.194 ml water}$$
Let $x$ be the multiplicative factor
$$x = \dfrac{97.1}{1.194}= 81.3$$
So:
0.8 ml conc perchloric acid
83.4 ml acetic anhydride
15.8 ml water

I loathe that the problem is inferring that volumes are additive.
Looking up some data...
\begin{array}{|c|c|c|c|} \hline
 & Mol. Wt. & Density\ (g/ml) & mol/ml \\ \hline
\pu{Water} & 18.015 & \pu{0.9950 @ 25 °C} & 0.05523\\ \hline
\pu{Acetic anhydride} & 102.09 & \pu{1.082 @ 20 °C} & 0.01052 \\ \hline
\pu{Acetic acid} & 60.05 & \pu{1.0446 @ 25 °C} & 0.01740 \\ \hline
\end{array}

The density data is out of Pubchem. I looked for a while and couldn't find density values at same temperature, but the values are close enough to make the point...

assume 1.0000 ml of acetic anhydride and appropriate water to give 1:1 molar ratio
\begin{array}{|c|c|c|} \hline
 &  ml & g \\ \hline
\pu{Water} & 0.1904  & 0.1894\\ \hline
\pu{Acetic anhydride} & 1.0000 & \pu{1.082 g} \\ \hline
\pu{Totals} & 1.1904 & 1.271 \\ \hline
\end{array}
So density of acetic acid by combining volumes is
$$\dfrac{1.271}{1.1904} = 1.067$$
which gives an error of
$$\dfrac{1.067-1.0446}{1.0446}\times 100\% = +2.1\% $$
Thus the total solution as calculated above would be closer to 98 ml than 100.

This is not an attempt to answer the question(s). However, it is seemingly clear that OP is doing a lab practical, and I'd like to clear his assumption about density of $\ce{HClO4}$ acid.
The series of density ($d$) of $\ce{HClO4}$ acid by percentage at $\pu{25 ^\circ C}$ have been published in 1941 (Ref.1):
$$
\begin{array}{cccc}
\% \ \ce{HClO4} \ (w/w)  & d\text{ at } \pu{25 ^\circ C} \ (\pu{gcm-3}) & d\text{ at } \pu{30 ^\circ C} \ (\pu{gcm-3}) & \text{Conc. at } \pu{25 ^\circ C} \ (\pu{molL-1}) \\
\hline
0 & (0.99707) & (0.99568)  & - \\
2 & 1.00828 & - & 0.2007 \\
4 & 1.01975 & - & 0.4059\\
6 & 1.03154 & - & 0.6158\\
8 & 1.04355 & - & 0.8307\\
10 & 1.05591 & 1.05388 & 1.0507\\
20 & 1.1228 & 1.11999 & 2.2344\\
30 & 1.20002 & 1.19651 & 3.5821\\
40 & 1.29073 & 1.28658 & 5.1372\\
45 & 1.34252 & - & 6.0113\\
50 & 1.39937 & 1.39435 & 6.9620\\
55 & 1.46134 & - & 7.9974\\
60 & 1.52766 & 1.52177 & 9.1203\\
65 & 1.59628 & - & 10.3242\\
70^a & 1.6644 & 1.658 & 11.5928\\
\hline
\end{array}
$$
$^a$ The values are from Ref.2.
Some values in the publication was avoided for convenience (total of 27 density values for 27 percentages not including 0% have been reported). The authors have given best fitting empirical equation as:
$$d = 4.9593 \times 10^{-7}x^3 + 2.30045 \times 10^{-5}x^2 + 5.62796 \times 10^{-3}x + 0.99707 \tag{1}$$
where $d$ is density at $\pu{25 ^\circ C}$ and $x$ is percentage of $\ce{HClO4}$. When plotted using an excel spread sheet, the same data set gives me a parabolic relationship with $R^2 = 0.9996$ agreement. The relevant equation is:
$$d = 7.0 \times 10^{-5}x^2 + 4.8 \times 10^{-3}x + 0.9971  \tag{2}$$

This scientific data shows two facts:

The relationship between percent $\ce{HClO4}$ and its density is not linear. It is most probably parabolic.
The density of $70\% \ \ce{HClO4}$ solution is:
$$d_{70\%} = 7.0 \times 10^{-5} \times 70^2 + 4.8 \times 10^{-5} \times 70 + 0.9971 \approx 1.676$$

According to the empirical formula of Ref.1, the density of $70\% \ \ce{HClO4}$ solution can also be calculated:
$$d = 4.9593 \times 10^{-7} \times 70^3 + 2.30045 \times 10^{-5} \times 70^2 + 5.62796 \times 10^{-3} \times 70 + 0.99707 \\ = 1.674$$
The both calculated values have good agreement with the experimental value of $\pu{1.67 gcm-3}$ (Ref.2 has given a little deviated value of $\pu{1.6644 gcm-3}$ for the $70\%$ solution at $\pu{25 ^\circ C}$).
If you have done for the similar calculations for 100% solution, you'd have ended up with the value, $\pu{2.286 gcm-3}$ from equation $(1)$ and $\pu{2.177 gcm-3}$ from equation $(2)$ for $d_{100\%}$. Either value is much bigger than the given value of $\pu{1.8 gcm-3}$. However, as MaxW pointed out in his comment, extrapolation is not a great idea for this kind of data sets. Based on MaxW's graph in one of his answers, either values of $\pu{2.286 gcm-3}$ or $\pu{2.177 gcm-3}$ are not acceptable. Hence, it is safe to say that it would be acceptable to consider $d=\pu{1.8 gcm-3}$ for $d_{100\%}$ as OP originally did.
Note: I have also calculated the concentration of each solution in terms of $\pu{mol/L}$ using density values. Based on that values, it seems $1\%$ solution is exactly equal to $\pu{0.10 M}$ (at least for 2 significant figures).
References:

Aaron E. Markham, “Density of Perchloric Acid Solutions,” J. Am. Chem. Soc. 1941, 63(3), 874-875 (https://doi.org/10.1021/ja01848a509).
Langhorne H. Brickwedde, “Properties of Aqueous Solutions of Perchloric Acid,”  Journal of Research of the National Bureau of Standards 1949, 42, 309-329 (PDF).

Great data showing that volumes are not additive! // It is a big dangerous to extrapolate a fitted curve beyond the bounds of the data. I'd agree that the 70% number is good to 2 decimal places. Extrapolating to 72.5% gives 1.713 vs 1.768 with equation 2. However extrapolating to pure perchloric acid is too much.  // I looked for the density of the pure acid and couldn't find it...

density of 70% acid is 1.6644 @ 25 C which agrees with an earlier study // "Properties of Aqueous Solutions of Perchloric Acid"; Langhorne H . Brickwedde; Journal of Research  of the National Bureau of Standards Volume 4 2, March 1949, page 309

75.0% acid = 1.731 @ 25 C // Pub Chem - Other Experimental Properities // data from 10th edition Merck Index

"Perchloric acid; authentic, scientific and practical considerations relative to its use as an important research and routine analytical chemical reagent with special reference to rapidity, accuracy and economy"; Data compiled by G. Frederick Smith; Columbus, G. Frederick Smith Chemical Co., 1940, 4th ed., page 60  // table for 25 C for 75% to 65% by 0.25% to 5 decimal places. // 75% =1.73175; 74%=1.71824; 73%=1.70743; 72%=1.69122; 71%=1.67771; 70%=1.66420; 65%=1.59665

100% Perchloric acid 1.7676 @ 20 C // "PERCHLORIC  ACID  AND  PERCHLORATE" by ALFRED  A.   SCHILT,    G.  FREDERICK   SMITH  CHEMICAL   COMPANY,   867  McKinley   Avenue   Columbus,   Ohio  43223  LCCN   79-63068   ©   COPYRIGHT  1979, table 2, page 14 which references A.  A.  Zinov'ev   and   N.  A.  Shchirova,  ZNK,  5,  1299  (1960).

same source as above in Table 5  lists azetrope as 72.40% perchloric acid boiling at 203 C.

100% Perchloric acid 1.764 @ 22 C // Proceedings of the American Pharmaceutical Association at the Annual Meeting, Volume 49, "Non-Hydrated Perchloric Acid - Preparations and Properties", D. Vorlaender and R. V. Schilling, page 762 // data on page 763

an early study to which I lack access // "Untersuchungen über das System: Überchlorsäure und Wasser", H. J. van Wyk, Zeitschrift für anorganische Chemie Volume48, Issue1, 19 Dezember 1905, Pages 1-52,  https://doi.org/10.1002/zaac.19050480102

MaxW: You have done a lot of research. Thanks for all valuable data.

@MaxW: I have corrected my post according to your comments and publications.

This isn't an answer but a plot of the density data from:

100% Perchloric acid 1.7676 @ 20 C // "PERCHLORIC ACID AND PERCHLORATE" by ALFRED A. SCHILT, G. FREDERICK SMITH CHEMICAL COMPANY, 867 McKinley Avenue Columbus, Ohio 43223 LCCN 79-63068 © COPYRIGHT 1979, table 2, page 14 which references A. A. Zinov'ev and N. A. Shchirova, ZNK, 5, 1299 (1960).

The density curves exhibits a maximum at about 80% perchloric acid. Also note the danger of extrapolating curves beyond the fitted data range.


text:
Are all ions stable?
My science teacher says that all ions are stable, which seems wrong to me.
For example $\ce{He^{-1}}$ is an unstable ion of helium, and although it might not exist in the real world, it is still technically an ion of helium. Is it right?

Of course there are plenty of unstable ions. But maybe your teacher wanted to say something different. He might have thought of a given group of ions, for a given application...

I have a feeling the point was all ionised atoms are stable by themselves. As you can see, it's not true for anions.

Even calling a positively charged ion 'stable' ignores the fact that it is quite happy to grab spare electrons...

+1 for not just swallowing a sweeping statement without thinking through its limitations. // $\ce{He+}$ isn't "stable" either. // The statement would be true for ionic salts. So in NaCl, the sodium and chlorine atoms don't flip-flop between ions and uncharged atoms.

In a way, everything that exists is stable.

I will point out that the fact that the negative ion of helium is metastable for long enough is important for tandem accelerators - one can do Rutherford Backscattering with helium since the metastable lifetime is long enough to get to the terminal to then be stripped to being a positive ion. So, indeed, the negative ion of helium 'exists' at various times around the world.

Stable relative to what? You need some reference point to compare energies. And there better be the same amount of matter present (no comparing energies of neutral atoms and ions).

Your teacher made an awfully strange comment if not given within a greater context. Presumably it was meant that ions of a soluble salt are stable in solution (they don't tend to recombine to form the solid).

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text:
Gas / Gas mixture dynamic viscosity temperature and pressure dependance
I’m trying to get an accurate interpretation of certain gases ($CO_2$, $N_2$, $NO$, etc.) dynamic viscosity [Pa.s] as a function of temperature and pressure (afterwards determine the dynamic viscosity of the whole gas mixture).
I have found more than one equation and I’m struggling to choose a proper one which can be interpreted in MS Excel. For example, it’s not that easy to find all the coefficient for each equation (unfortunately, same goes for the gas mixture dynamic viscosity: there is Sutherland-Thiesen (found on the internet), Herning–Zipperer (found in the literature) equation, etc. …).
Thanks for any help.

See Transport Phenomena by Bird, Stewart, and Lightfoot, Chapter 1.  If the gases and mixtures can be approximated as ideal gases, then their viscosities are independent of pressure.

@ChetMiller Thanks, I'll look at it. Overall, do you know anything "Excel" based that would allow to define equation as $\eta = f(T,p)$?

@ChetMiller I have one more question, I have calculated all the substances dynamic viscosity using Lennard-Jones parameters from the book you've suggested, but what if the gas mixture contains water vapour $\rightarrow$ related to the moist gas: $H_2O$: 10 % vol. The book says that calculating water viscosity (I suppose $H_2O$ can't be considered gas even in the form vapour mixture, because 10% vol. is just a water) at different temperature gives not very accurate viscosity estimation. Do you have any suggestions, please? Thanks.

I don't understand what you are asking.  What does "because 10% vol. is just a water" mean?  Are you dealing with an ideal gas mixture?  What is the pressure?

@ChetMiller No, it's exhaust gas mixture: contains $CO_2, O_2, CO, NO, SO_2, N_2$ and $H_2O$. The content of H2O in the mixture is 10% vol. I have computed the viscosities of the gases using the Lennard-Jones parameters, but they can't be used for $H_2O$ the other chapter shows how to calculate viscosity of water, but says it's not very accurate. The pressure changes, that's why I need to make $\eta = f(T,p)$ - pressure ranging from 100 kPa to 300 kPa.

That's a pretty low pressure.  I would still treat that as an ideal gas, and neglect the pressure dependence.  If the water is only 10% by volume, it can't have much of an effect on the viscosity of the mixture.  I would use the results in the other chapter to get the viscosity of  pure water for applying the mixture rule.  Understand that, in the end, this is not going to give you a super-accurate estimate of the mixture viscosity, but it is about the best you're going to do.

@ChetMiller OK, I see. Thanks, I'll do as you say if the difference is negligible (I have been also struggling finding a proper formula to calculate water density as a function of pressure and temperature: $\rho_{H_2O} = f(p,T)$. I guess that simple equation of state won't do the trick...) I can use the problem 1A.4: Estimation of liquid viscosity as an example. What exactly do you mean by the mixture rule? I thought it's sum of each substance's viscosity multiplied by their content in the mixture. Thanks for your help (Please, note that I'm not chemist, so some of questions may seem stupid)!

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text:
Resonance energy in benzene
Quoting from my textbook:

Although stabilized by resonance compared with other carbocations, its resonance energy is much less than that of the starting benzene ring.

Here is more context :

Isn't that good? Isn't it more stable if it had less energy? Shouldn't it stay that way?
I feel like I'm missing something very obvious but I can't pinpoint it.

Maybe try suggesting an alternative next time :)

It is potentially badly worded. Resonance energy is a negative quantity: it is how much more stable something is compared to the alternative (benzene is more stable than the theoretical cyclohexatriene with three double and three single bonds). But benzenonium ions have less resonance energy than benzene but more than other carbocations so have higher energy than benzene but lower energy than typical carbocations.

I would not even say that is poorly worded.


text:
What are the GDB-13 criteria for "synthetically accessible organic molecules"?
I'm trying to understand the QM datasets (QM7, QM8, QM9), however, in the description of QM7 the data is described as follows:

[The] QM7 dataset, which is a subset of GDB-13 (a database of nearly 1 billion stable and synthetically accessible organic molecules) containing up to 7 heavy atoms C, N, O, and S. The 3D Cartesian coordinates of the most stable conformations and their atomization energies were determined using ab-initio density functional theory (PBE0/tier2 basis set)

What exactly does the phrase "stable and synthetically accessible organic molecules" mean? If I had to guess I would have said it means they can be synthesized in a lab, but I didn't want to guess.

@DarienSchettler Sorry for being a bit curt, but I do not understand why someone with "little to no background" would look into such datasets. Going deep into applied theoretical chemistry with no idea about chemistry? So please explain what exactly you find unclear about the description, and then we can discuss it, no problem. It also helps if you give a bit of a context to your question, so you get an answer you can actually understand. We have zero idea about your background.

See my answer below... this is all I wanted. As to your point, I don’t feel it’s necessary to provide that level of background knowledge for this question. That being said... I’m a deep learning engineer working with relevant parties on Tensorflow Quantum development in industry. I have two degrees, one in Systems Engineering and an Honours B.Sc in Biological Sciences (Math Specialty). I am attempting to further my understanding of data sets that may contain ‘quantum data’ and happened across the GDP13 subsets. I’m just doing my best to learn. Please consider this question closed. Thanks.

What you have provided is a definition of all the terms involved. This is fine, but there remains the (rather interesting) question of exactly how this is defined for a molecule in silico, especially because your question asks specifically about the GDB-13 database. Of course, one could evaluate "synthetic accessibility" by trying to make it in the lab: if you succeed, then it is synthetically available, and if not, then it isn't. Likewise, one could measure "stability" by (potentially) making it and studying its thermodynamic properties.
However, for a database of 1 billion compounds, this is obviously unfeasible. So the designers of the database have to resort to using some heuristics, or rules, to determine whether a given molecular structure is "synthetically accessible" and "stable". The rules are outlined in the original paper in which the GDB-13 database was published,[1] as well as the first GDB paper from slightly earlier.[2] Both are very interesting papers and well worth a read if you are interested in cheminformatics in general.

Because all organic molecules have a carbon chain or ring as their backbone, the first step is to construct all possible frameworks of carbons and hydrogens using graph theory (up to a maximum number of carbons, of course). Here there is not yet any consideration as to what other atoms may possibly exist in a typical organic molecule. However, there are some computer-generated frameworks which are almost certainly impossible to make because of strain or other considerations (such as Bredt's rule). On top of that, it appears that they have intentionally removed three- and four-membered rings, which are stable but would have dominated the database, thus making it highly unrepresentative:[2]

The vast majority of these graphs (99.8%) contained three‐ and four‐membered rings and was excluded to avoid generating a database consisting almost exclusively of such small rings.

The next step is to introduce heteroatoms, which can be done fairly straightforwardly because in organic chemistry atoms have very specific bonding patterns. Thus, for example, because carbon forms four bonds and nitrogen three, you could replace a $\ce{CH2}$ group in a molecule with $\ce{NH}$. This is not trivial to do in the lab, but is very easy for a computer:[2]

[...] all possible atom‐type combinations by introducing carbon, nitrogen, oxygen, and fluorine (as a model halogen) at each node

although for GDB-13, they have ignored fluorines:[1]

We also eliminated fluorine because it was rarely found and never considered in our group for synthesis in virtual-screening guided drug discovery applications of GDB-11.

and also added chlorines (replacing $\ce{OH}$ groups in molecules which have them) and sulfurs (replacing $\ce{O}$ atoms).

The problem with doing that exhaustively is that too many heteroatoms tends to make a molecule very unstable. For example, directly joining heteroatoms via single bonds is (generally) a good recipe for making explosive compounds. So, essentially all such molecules were removed. The authors found that most of this work could be very quickly automated by using an even simpler heuristic: simply removing any compound with a high heteroatom:carbon ratio.[1]

Because most of the rejected molecules contained multiple heteroatoms, we reasoned that it might be possible to accelerate the database computation using a very fast “element-ratio” filter. Analysis of databases of known compounds suggested cutoff values of (N + O)/C < 1.0, N/C < 0.571, and O/C < 0.666

and specifically disallowed cases, which have to be filtered out in another step, included:[1]

The following functional groups are discarded as too unstable to be considered: hemiacetals, hemi-aminals, aminals, acyclic imines, non-aromatic enols, orthoesters and analogs, carbamic acids, non-aromatic enamines (except acylated enamines and vinylogous enamines), beta-keto-carboxylic acids and beta imino-carboxylic acids, and all compounds containing both a primary or secondary amines and an aldehyde or ketone.

This is also helpful for preparing a database of "druglike" molecules because molecules with too many heteroatoms are very polar, which makes it almost impossible to diffuse across cell membranes (see e.g. https://en.wikipedia.org/wiki/Lipinski%27s_rule_of_five).

What you get at the end is probably not likely to be 100% "synthetically accessible". However, the filters applied mean that if you were to pick a random molecule from the database, there is an exceptionally good chance that you could make it in the lab if you so desired. Quoting from the authors one final time:[2]

The database construction strategy chosen also ensures that the majority of GDB, although presently unknown, should be synthetically accessible.

References

Blum, L. C.; Reymond, J. 970 Million Druglike Small Molecules for Virtual Screening in the Chemical Universe Database GDB-13. J. Am. Chem. Soc. 2009, 131 (25), 8732–8733. DOI: 10.1021/ja902302h.
Fink, T.; Bruggesser, H.; Reymond, J. Virtual Exploration of the Small-Molecule Chemical Universe below 160 Daltons. Angew. Chem. Int. Ed. 2005, 44 (10), 1504–1508. DOI: 10.1002/anie.200462457.

While I appreciate the effort and it is quite helpful in general... A definition of the terms was sufficient as it communicates what the authors meant. My initial question prior to editing was What is a Synthetically Accessible Organic Molecule. You are correct in clarifying exactly HOW (heuristically) the creators of the GDB-13 database would have attempted to impose this definition. You are also correct in answering your own question of what are GDB-13 criteria for a Synthetically Accessible Organic Molecule. As for the answer to my initial question though... most of this is superfluous.

That being said, thanks for your time and effort. I found your answer educational. It's clear you're an expert and I appreciate you taking the time to try and help/clarify. Hopefully, in addition to educating me, your answer can help someone in the future as well.

I see what you mean. Well, maybe I have overinterpreted it, then; and if that's the case, then all you needed was that "synthetically accessible" = "stuff the authors thought would be sensible to make in the lab". The thing is that "synthetically accessible" can have slightly different meanings depending on the context. So I thought (think) that the setting that you provided was (is) important. For example, maybe you're looking at a database of known reactions where it would mean "things that actually have been made". But as you say, maybe somebody else can use the extra info in future.

After spending the afternoon googling around I've written a simple answer to my own question (essentially a basic definition of terms). For a more complete answer see the accepted answer by orthocresol♦:
Synthetic Accessibility - Refers to ease of synthesis. I.e how difficult a compound is to make (synthesize) in a lab.
https://jcheminf.biomedcentral.com/articles/10.1186/1758-2946-1-8
Organic Molecule - An organic molecule is a molecule that contains carbon atoms (generally bonded to other carbon atoms as well as hydrogen atoms). Although carbon is present in all organic compounds, other elements such as Hydrogen, Oxygen, Nitrogen, Sulphur, and Phosphorus, are also common in these molecules.
https://www.siyavula.com/read/science/grade-12/organic-molecules/04-organic-molecules-01
Stable Molecule - in general this is more difficult to ascertain as an absolute. Generally the idea would be that a molecule that does not decompose (ie is persistent) in most environments may be considered stable. For more details the following link contains quite a bit of information.
https://www.reddit.com/r/askscience/comments/li6nb/what_makes_molecules_stable/?utm_source=share&utm_medium=ios_app&utm_name=iossmf
———
So, In short.
The phrase “stable and synthetically accessible organic molecules”, refers to molecules that contain Carbon (generally bonded to other atoms and Hydrogen), can be created with relative ease in a laboratory, and are relatively persistent in most environments (ie Are not prone to spontaneous decomposition to lower energy states).


text:
How can should one separate and purify benzophenone from salicylic acid?
I am working with two solids and I know that they are  benzophenone (s)(ketone) and salicylic acid (s) (carboxylic acid). I am curious to know what is/are the best method of extraction and purification?
In other words, would extraction be the only method that these two can be separated because one is neutral and the other is an acid?
Also, what kind of qualitative test/s can you use for salicylic acid? I mean I understand that the acid/base test is most commonly used but can you use it again to test for -COOH functional group?
Apologizes for the questions but not being able to physically see this makes this 10X hard basing this off of hypotheticals.

Are you familiar with solvent-solvent extraction?

Google the water solubulities of the two compounds

The partition coefficient (https://en.wikipedia.org/wiki/Partition_coefficient) is used to describe the relative solubility of a compound in two immiscible solvents (e.g., organic:aqueous). This is the principle upon which the sovent-solvent extraction is based. This is where you would use a sep funnel (https://en.wikipedia.org/wiki/Separatory_funnel).

An additional thought... You might also consider taking advantage of the fact that salicylic acid is an acid and would react with sodium bicarbonate. Looking up partition coefficients (logP) in PubChem gives logP = -1.43 in octanol:water for sodium salicylate and a computed logP = 3.4 for benzophenone.


text:
How to switch Mercury-vapor lamp between UVA and UVC bands?
I have Mercury-vapor lamp emitting UV light. I am looking for the nice solution to cut off spectral lines for UVA or UVC, to make it one day to generate ozone and another day to use it for disinfection as UVC band lamp. I mean using precoated tubes and put mercury vapor lamp inside or using coated glass as a shield.
Related: What property allows ozone to absorb uv light

This comes under no free lunch. I collect fluorescent rocks. To change between UV bands I have to switch bulbs and filters on the lamp.

From your text it looks like you suppose UVA (320-400 nm) generates ozone. It does not. UVC does.( <280 nm)

I’m voting to close this question because it's not about chemistry.

@Mithoron
"I’m voting to close this question because it's not about chemistry. – Mithoron
Asking alike question in Physics I was asked to ask it in Chemistry, so life is not easy.
But answer given by Nicolau Saker Neto is really excellent, deserving 100 Points, so don't close this question so fast

@MaxW
"I collect fluorescent rocks. "
Ok,  I have already visited your website for fluorescent rocks.
" I have to switch bulbs and filters on the lamp. "
Ok, but I test 150W, 250W, 400W high pressure, low pressure sodium, mercury lamps, generating a lot of heat, so the only filters that can work in high temperature are industry made fluorescent coatings for the glass or the special glass sorts, which can itself absorb some UV light band, " If you are interested in blocking UVC and letting UVA through, a borosilicate glass (e.g. Pyrex(TM)) tube would be sufficient" by Donovans

TL;DR - You need two different lamps, one that has a phosphor to absorb the 254 nm light, and one that doesn't. Alternatively, you can use one lamp, and use a filter to select the wavelengths.
Mercury vapour lamps emit a significant amount of both UV-C light (around 254 nm) and UV-A light (around 365 nm).  This includes fluorescent bulbs used for lighting a room. The ultraviolet emission is a fundamental property of the element, not some kind of engineering accident. Light at these wavelengths is harmful, so you don't want it flying all over the place. The solution? Coat the inside of the lamp with a substance that absorbs the UV light. In some cases (but not all), the glass of the lightbulb itself can absorb the UV light.
Except now you're turning lots of luminous energy into heat, which is wasteful. A smarter solution is to coat the lamp with a phosphor, which absorbs the UV-C and UV-A light and reemits the energy at longer wavelengths. Depending on the phosphor, the reemitted light will have a different spectrum.
If you look at a 254 nm UV-C bulb and a 365 nm UV-A bulb (while they're turned off..), you'll notice that the 254 nm lamp is completely see-through, whereas the 365 nm lamp often has a white powdery coating the inside. This powder is a phosphor, which is chosen to absorb the 254 nm UV-C light, while letting the 365 nm UV-A light through (as well as other lower energy visible wavelengths). Some 365 nm UV-A lamps also have a coating to absorb the any visible light generated, such that relativity pure invisible 365 nm light shines through. This is popularly called a "black light".
Note that this does NOT imply every mercury lamp with a phosphor coating blocks the UV-C light it generates.

The ultraviolet emission is a fundamental property of the element, not some kind of engineering accident. This is not quite right. The intensity profile of the emission spectrum of a discharge tube varies with both pressure and temperature. You can see this easily with the naked eye when looking at many discharge tubes, such as hydrogen, because the color appears different in different parts of the tube.

@NicolauSakerNeto
thank you for your excellent answer, deserving 100 Points
I accepted your answer immediately since another guy tried to close my question, not giving me a chance to name the winner.
I have some comments, which follow soon, since exactly some sodium and mercury bulbs come with "A smarter solution is to coat the lamp with a phosphor"
I have to check if my sodium  see-through lamp emits UV, generating ozone, checking ozone's smell.
I must check sodium, mercury lamp spectral lines in UV band. removing outer phosphor coated bulb.
I can smell ozone in case of mercury

Ok, since phosphor coating is a complicated thing, all I can do is to remove phosphor coated bulbs from sodium and mercury lamps and use CD/DVD as difraction grating to study spectral lines, as discussed in https://en.wikipedia.org/wiki/Diffraction_grating
https://en.wikipedia.org/wiki/File:Fluorescent_lamp_spectrum.jpg
Study on fluorescent lamps
https://en.wikipedia.org/wiki/Fluorescent_lamp#Phosphors_and_the_spectrum_of_emitted_light

"Characteristic spectral power distributions (SPDs) for an incandescent lamp (left) and a fluorescent lamp (right).
https://en.wikipedia.org/wiki/Color_temperature

Light and Ozone Production
Ozone is generated not by UVC or UVA but by Vacuum UV (VUV) in the 100-200nm wavelength range 1.
Low pressure mercury lamps generate strong VUV (184nm) and UVC (254nm) light, and medium pressure lamps generate light at various frequencies in the range of 200nm-600nm 2.
When it's said that UVC light is disinfecting or germicidal, it works through the absorption of light energy by DNA which then breaks down the molecules.  Ozone (O3) is effective at disinfection because it is a reactive species of oxygen (causing faster oxidation than O2) 3.  The germicidal effectiveness of light by DNA disruption peaks at around 270nm, but while the 254nm light produced by mercury lamps is not ideal, it is still about 75% as effective as 270nm light.  The 180nm light band produced by mercury lamps is about 1% as effective as 270nm through that mechanism  4.
UVA light can produce germicidal effects, but in this study, it took 3 minutes for UVC to produce total irradiation and 60 minutes for UVA.
While medium pressure mercury vapor type lamps, including the HID type and flourescent tube type, produce UVC, the borosilicate or silicate glass they are generally made from absorbs >99% of the UVC light, protecting humans from the effects of those wavelengths 6.  To use the lamp for germicidal purposes, it must be made from fused quartz or fused silica glass that allows UVC to pass 7.
From personal experience, medium pressure mercury vapor lamps for germicidal applications do not produce significant ozone.  It's obvious if you're in the vicinity of a lamp that can produce ozone as you will smell it quickly.
Filtering Light
If your mercury lamp is producing VUV and UVC, it might be difficult to block UVC while not blocking VUV, but if you just want the ozone, cover the light, blocking all the light, but provide airflow between the lamp and the shade to move the ozone where you want it to go.  If you are interested in blocking UVC and letting UVA through, a borosilicate glass (e.g. Pyrex(TM)) tube would be sufficient.
Blocking UVA and letting through UVC has you blocking longer wavelengths and passing shorter wavelengths.  This is done with UV passing, visible light blocking filters that create black lights.  I don't know of a similar material that blocks UVA while passing UVC.  A material like this would also probably have some safety concerns as it would likely block much of the visible light too, yielding an invisible sunburn/eyeburn causing machine.
Phosphors, as used in flourescent lights, generally absorb shorter wavelength light and use that energy to emit longer wavelength light known as the Stokes shift.  Although some materials can absorb a longer wavelength and emit a shorter wavelength, I don't know if that is something that has been commercially exploited 8.  There are phosphors that can absorb VUV and emit 290nm, germicidal, wavelength light, and ones that can absorb VUV and UVC to emit UVA wavelengths.

closing high quality questions, followed by high quality answers is never smart


text:
In the following reaction is E2 or Sn2 favorable?
Question

Doubt
It is quite obvious that carbonyl compound is going to convert into alkane, but what confuses me is whether Sn2 or E2 will occur at bromide position
Although there is heating( which I believe favour E2) and a strong base, the carbon is 1° which would favor SN2 more than E2.
Please help.

I think you will get an SN2 at the alkyl bromide to give the substituted hydrazine, this may go on to do a second substitution with further alkyl bromide. You certainly will not get b. a could be the product if by manipulating the hydrazine addition conditions you can get selectivity for the reaction at the ketone. I consider this unlikely. Remember that the intermediate hydrazide can also react with the bromide.

But the answer is given B?

This is simply wrong. The person setting this question has assumed that the hydrazine will be completely selective for the ketone over the alkyl bromide and that the strong basic conditions will also cause elimination of the Br- to give the styrene. Selectivity seems most unlikely give the typical forcing conditions used in W-K reductions
Hydrazine is a powerful nucleophile and reacts with alkyl bromides under mild conditions often giving polyalkylation reference here.
I consider it most unlikely there will be any selectivity unless confronted by literature example.


text:
chlorination of aromatic compounds
I was reading some notes on Electrophilic Substitution and I came across the topic chlorination of Toluene. I'm wondering why we cannot substitute chlorine in the meta position of toluene?

We do get meta-chlorotoluene when toluene is chlorinated by EAS, but only in minor amounts.  This paper describes a process designed to selectively produce the ortho isomer over the para one, but in addition to some para-chlorotoluene still being formed the abstract reports 4.0% meta-chlorotoluene.  There are also small amounts of dichlorinated products and even benzyl chloride.
This Youtube video describes a synthesis route from benzene that favors meta-chlorotoluene.

Nitrate the benzene.
Apply Friedel-Crafts methylation which gives substitution meta to the nitro group.  The remaining steps ultimately exchange the nitro group for a chlorine atom.
Reduce the nitro group to amino using tin plus acid.
Convert the amino group to a diazonium ion with nitrite plus acid.
React the diazonium ion with cuprous chloride plus HCl.

Meta-chlorotoluene is available at 98% purity from Sigma-Aldrich.


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Calculating the pH of a sodium phosphate solution
I have the following question in a chemistry homework:

Calculate the $\mathrm{pH}$ of a solution made by dissolving $\pu{113 g}$ of sodium phosphate in enough water to make a solution with a final volume of $\pu{475 mL}$. Find the equilibrium concentration of all species present in this solution.

It shouldn't be that hard of a problem, but my teacher really doesn't explain well at a distance (with the COVID-19 situation). I have found that it dissociates into $\ce{NaOH}$ and $\ce{H3PO4}$ when in water:
$$\ce{Na3PO4(s) + 3H2O(l) → 3NaOH(aq) + H3PO4(aq)}$$
But I don't really know what to do next...
Thanks in advance!

You do not think it can be strongly acidic and strongly alkalic at the same time, do you ? Google 3 equilibrium dissociation reactions of phosphoric acid.

No of course not, but if $Na_3PO_4$ dissociates and just the $PO_4$ ion reacts with water, it could make $HPO_4$, $H_2PO_4$ or $H_3PO_4$, right?

Yes, right, but you need to quantify that. Technote: If you use mhchem MathJax extension, you can Write formulas as $\ce{H3PO4}$ what will look like $\ce{H3PO4}$

See phosphoric acid section acidic properties.

Sorry, it is not an easy problem. If she really means "find the equilibrium concentration of all species present in this solution."

It can be approximated as calculation of pH for a PO4^3h base with pKb = 14 - pKa3, and from that pH to calculate ratios of phosphate forms, as the pKax are far enough from each other.

113 g of solute in 475 mL of solutions is far from a dilute, near-ideal solution. Ionic strength will surely not be negligible.


text:
What is the product of following iodoform reaction?
Question

Attempt
I think the product A would be as I have attempted below, but I am getting the answer as A but book says D.

Where am I wrong? Please help.

if there was excess I2 present would it add across the double bond of the iodoform product that you have drawn?

But still even if I2 adds I+ and OH- across double bond, how would we get d?

The intermediate hydroxy acid gets oxidised to keto acid by I2 perhaps?

Alpha keto acids readily decarboxylate https://pubs.acs.org/doi/full/10.1021/acs.orglett.6b01964

Can you give a mechanism? Because although i understand how C-OH will convert to C=O and help in decarboxylation, I dont understand what will happen to I that attacked C=C in first place?

Form the hydroxy iodide by addition to the double bond then OH opening the iodonium giving the tertiary iodide/alpha hydroxy acid. I2 oxidises the hydroxy to the keto acid which decarboxylates. OH- eliminates I- to form double bond.


text:
What would be the product of following elimination reaction?
Question

Doubt
The answer is given B. But why cant it be C after all formation of cyclopropane would increase angle strain?

Where are the most acidic hydrogens?

At adjacent site of C=O. But how come compound B is stable . I mean in E2 first Cl- should leave and after that OH- abstracts proton, but it seems here than it is following E1cb?

[3] rings are reasonably stable, lots of examples of them.

The first thing you notice is an alcoholic base reacting with a molecule. The base looks for the most acidic hydrogen in the molecule. Since the no.of moles of $alc.KOH$ is not mentioned, we go for both the acidic hydrogens at the adjacent positions to the carbonyl functional group. Once the acidic hydrogens are abstracted, it's a simple elimination of $Cl$ that will happen.

I agree with the comment made by @Waylander as there are plenty of examples with three-membered rings. Well, (c) is not possible but you can say (a) is a possibility. Of course (a) forms too by a simple elimination reaction but the most probable one is the abstraction of acidic hydrogen in the molecule. The stability of [3]rings can be brought about by banana bonds.


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catalytic activity of Pd and Pt
what phenomenon takes place during catalytic activity of Pd and Pt during hydrogenation ?
in some books it is given as absorption,in others adsorption .some say that it is occlusion .
please help me

Occlusion is same as "surface adsorption". And catalytic hydrogenation is surface phenomenon...

but Sir, during formation of hydrides by d block elements ,hydrogen occupies intersticial spaces ? (i think so)

@DVS but the molecules that are being hydrogenated usually can't access the interstitial spaces through a standard surface.


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How can I dissolve a precipitate when stuck to the side of a boiling flask
I’ve been experimenting with producing Iron Acetates and I’ve run into a snag. Here’s my process so far:

I started by adding white vinegar and ~50% Hydrogen Peroxide to a reaction vessel with steel wool sponges and allowing it to react for a month. After that I ran the solution through a filter multiple times until it was fairly clear to remove most of the unreacted solids. After that I tested an aliquot of the fluid by placing some leather in it to see if the leather turned black. It did quite quickly. After that I placed it in a boiling flask and heated it with a flame directly (because I don’t have a big enough beaker or other apparatus to indirectly heat it) and it turned orange. I’m guessing this happened because the elevated temperature allowed it to react with atmospheric oxygen. I continued heating till the solvent was gone leaving a red-brown-black solid on the side of the boiling flask.

I want to dissolve the solid so I added ~100mL of ~50% Ethanol (Smirnoff blue label vodka) to the flask to dissolve it and left it overnight, but it’s barely dissolved at all. I thought about heating it to create pure ethanol vapor which would do the trick but I don’t have a condenser to condense the ethanol again. Is there a better way of dissolving it?

I gather the acetate should dissolve in ethanol? The solid precipitate is surely iron oxide.

Possibly, but the acetate is supposed to be able to crystallize so presumably the ethanol just isn’t pure enough to break apart the bonds between the molecules very quickly in order to dissolve them. At least at low temperatures

This is not acetate, this is rust.

Try dissolving it with vinegar.

Also common rust removal chemicals contain nitric acid.

@NilayGhosh, I feel you are mixing two issues with three different suggestions, acetic acid, citric acid and then nitrice acid for rust removal in three different comments. There is a difference between removing surface stains versus removing rust from metallic objects. Rust stains on surfaces are invariably treated by oxalic acid because it forms a complex, not because it is an acid. Similarly, powerful reducing agents are also used in removing surface rust stains such as on glass, marble, wood etc.

contd.  Removing rust from metallic objects is another story, it involves the use of strong acid(s) but never vinegar. It is too weak to remove rust quickly.

@M. Farooq thanks for that and that does seem quite useful to know that oxalic acid forms a complex with iron oxide (although anyone who spends time working with nanoparticles would probably know that). I appreciate the advice, however that’s kinda useless to me because oxalic acid is a reagent I don’t have. The same is true of nitric acid. I’m trying to get both of them but if anyone has suggestions I can pull off with more common chemicals I’d love to hear them.

This is all rust which has precipitated on the walls and it is very hard to remove it. You might have to use oxalic acid to remove rust stains on glass. Oxalic acid plays a dual role, it dissolves the rust like an acid, but it also complexes the iron, which prevents further hydrolysis and re-settling.

Citric acid also works.

Ughh I hate that this is right. You mentioned hydrolysis and I checked and ya, that’s what happened. The iron acetate broke down when the vinegar solution was heated causing the iron acetate to turn into iron oxide. That explains the rather sudden change to orange

Time to post the follow up question I guess

Your precipitate is very likely iron oxides. First make sure you get the remains of the hydrogen peroxide out completely (e.g. by flushing several times with a little distilled water), then use an acid to dissolve the oxide. Possibly acetic acid (vinegar) or citric acid and a little heating will do the job. If not, try 20% sulphuric acid or 10% hydrochloric acid.
BTW reacting possible organic impurities on steel wool etc. with rather concentrated hydrogen peroxide might end up in peroxide formation and serious explosions, especially if one is heating the setup. Something I'd never do in my own lab without lots of safety precautions (a setup as small as possible, fume hood, perspex shield, etc.) and definitely not in my kitchen. Having glass shards picked out of one's face and eyes is no fun.

Thanks for the reply! I tried adding a citric acid solution to the flask and heating till just below boiling and the solution turned a dark orange color and some of the residue looked like it started to dissolve. However, adding leather to that solution has yet to change its color so I assume you were right about the residue being at least partly iron oxide, but there should still be some iron acetate left which has yet to dissolve. Could the acetate have denatured or something?


text:
Looking for an old (1950s) "chemical experiments" book
Not sure if this is precisely on-topic, but I'll risk some reputation and give it a try.
I got a chemistry set as a kid in the early 1970s, and the subject really captured my interest. I grabbed every chemistry book I could find from our libraries. They ranged from college textbooks to "Chemical Magic" collections of demonstrations. These latter tended toward pyrotechnics, and were pretty appalling safety-wise.
There were several favorites, but there was one book in particular that stuck with me, found in a rural Virginia county library around 1976 or so. It featured adolescents in coats and ties (and maybe one in blouse and skirt) demonstrating experiments, but the experiments were a couple of levels beyond what I found elsewhere.
The ones I definitely remember:

Removing stains by soaking clothes in potassium permanganate solution, then removing the manganese-dioxide residual stain with sodium bisulfite
Preparing white phosphorus by heating red phosphorus in a test tube, and condensing the vapors onto a cold test tube nested inside it
Preparing lithium metal by melting a lithium salt in a crucible and electrolyzing it
Synthesizing Bakelite from phenol and formaldehyde

There were probably a dozen or so others. I think one involved generating chlorine gas from an aqueous solution in the bottom of a beaker, then dropping in pellets of calcium carbide to watch the acetylene spontaneously combust.
I've searched using Google, but can't seem to get past The Golden Book of Chemistry Experiments -- yes, that's another treasured memory from the time, but it's not the one I have in mind. (For one thing, the one I'm asking about had no color plates at all.)
Another near miss that's quite familiar to me is Leonard A. Ford's Chemical Magic (I remember the 1959 original). I don't have it in hand, but Google's preview let me read far enough to see familiar material -- including pyrotechnic mixes using potassium chlorate instead of perchlorate, which got me in no small amount of trouble some years later. The book I'm searching for was more oriented toward science, rather than showmanship.
If the book I've described sounds familiar, please post an answer with any additional information you find -- title and publisher would be great, but additional experiment descriptions would also be helpful. If you have a favorite book from that period that doesn't seem to match this description, I'd be interested in hearing about it as well.
Finally, if this is a bad question, I welcome guidance on turning it into a good one, or why it's not worth pursuing. Thanks!

The replacement to The Golden Book of Chemistry Experiments  is Illustrated Guide to Home Chemistry Experiments (I think).

I came across Morgan, A. P. Things a boy can do with electrochemistry. 1940 541.37 See also pages in the following books: Morgan, A. P. Things a boy can do with chemistry. Does this title match your memory?

I think you are talking about old series of Chem Craft book. Google has images of boys with a tie.

@M.Farooq The Chemcraft books are familiar to me, but the one I'm looking for is not from that series. (Chemcraft books usually have a prominent line on the cover about "no dangerous poisons or explosives", and that definitely rules out the experiments I'm remembering!)

The book I sought, and a compilation of similar books
I've found the book! It is Chemical Magic, by Kenneth M. Swezey, McGraw-Hill, 1956. All the experiments I remember are there. There are quite a few others that I'd done, but forgotten finding in this particular book.
I found it listed in this brilliant compilation of home-chemistry and "chemical magic" books, compiled by William B. Jensen. He lists many other books that I remember from childhood.
Thanks to those who offered help, and I invite anyone interested to take a look through these books. I do miss the days when it was conceivable for a schoolchild to work with "dangerous materials" (never mind "drug precursors")...

William B. Jensen is an amazing chemistry historian and perhaps the most knowledgeable today. He published couple of my historical questions in the Journal of Chemical Education as well. Too bad that he discontinued his columns- Ask the Historian.

There are 2 books in German which inspired me to become Chemist. I had them in the late 70s.
"Chemische Experiemnte, die gelingen" (Chemical experiments which succeed) and "Organische Chemie im Probierglas" (Organic chemistry in a test tube). Both published originally in 1939 by Hermann Römpp and Hermann Raaf. Kosmos Verlag (Kosmos Publishers). No idea if there ever were English translations, but you might still find them in antiquary bookshops.

Thanks, these both look interesting! I don't think I encountered them as a child; I don't see any evidence of English translations, and my high-school German classes were still a few years in the future.


text:
What are practical examples of simultaneous measuring of quantities?
I am a maths teacher and I'm looking for practical examples of measuring quantities simultaneously, so that mathematically one needs to construct and solve a system of equations. I thought that in the world of chemistry there could well be many possible such problems, but alas I have no experience of chemistry, hence I thought to ask here.
I searched online and found two possible situations.

Given an alloy of say two metals, we can use the combined weight as one equation and the combined volume as a second equation and hence we can find the amount of each metal in the mixture. What I like about this is the simplicity whereas I assume it is much more difficult to extract the metals separately.
Given a mixture of two gases, we can see how much light of one frequency is absorbed and again for a second frequency which allows the amount of both gases to be simultaneously found.

What other such situations are there is chemistry; when would one actually need to do such simultaneous calculations?

Careful: The volume of an alloy is not just the sum of the volumes of the constituents.

You could use an emulsion as an example. Or a foam.

For the math, do the students know just algebra or do they know some calculus too? // gas/light notion - as light goes through gas the intensity decreases. Say 1 inch of travel cuts light down by 50%, 2 inches yields 25% and so on. One would really need to understand the basics of calculus to understand how the intensity/concentration relationship works.

I'd assume that the students have had no chemistry and that the problems should be formulated in a way that presumes no knowledge of chemistry?

@Max That's right. They won't know calculus at that point, basic algebra. My idea was to give a little statement of where the equations come from and then they need to solve them.

You may also give the weight of a sample made of an alloy $\ce{Zn-Mg}$. Then you dip it into some diluted $\ce{HCl}$. Both metals will react simultaneously according to $$\ce{Zn  +  2 HCl  ->  H_2 +  ZnCl_2}$$ and $$\ce{Mg  +  2 HCl  ->  H_2 +  MgCl_2}$$ This will produce an important amount of $\ce{H_2}$  gas. You can measure its volume. So you have two measured data, the original mass and the volume of gas. It is enough for calculating the proportion of $\ce{Zn}$ and $\ce{Mg}$ in the original sample.
Example : Take a mixture of $0.1$ mol $\ce{Zn}$ + $0.3$ mol $\ce{Mg}$. Of course the molar masses of $\ce{Zn}$ and $\ce{Mg}$ are respectively $65.39$ g/mol and $24.32$ g/mol. The total mass is : $6.54$ g + $7.29$ g =  $13.83$ g. Suppose the temperature and the pressure of the gas is such that $1$ mole gas occupies $24$ L. So $0.1$ mol $\ce{Zn}$ will produce $0.1$ mol $\ce{H_2}$ which occupies  $2.4$ L gas. Then $0.3$ mol $\ce{Mg}$ will produce $0.3$ mol $\ce{H_2}$, or $7.2$ L gas. The total volume of $\ce{H_2}$ is $2.4$ L + $7.2$ L = $9.6$ L.
If now you simply give the following data to your student : Total mass = $13.83$ g, and total volume of $\ce{H_2}$ = $9.6$ L, they should discover two equations with two unknowns. Of course you must tell them that $1$ mole gas occupies $24$ L. If x is the number of mole of Zn, and y the number of moles of Mg, these equations are :  $$\ce{x·65.39 + y·24.32 = 13.83}$$ $$\ce{x + y = 9.6/2.4}$$ When these equations are solved, the student still has to calculate the mass of the two metals.
You may of course change the choice of the unknowns, and the nature of the metals.

I think that if you state in the problem that 65.4 grams of pure zinc gives 24 liters of gas and that 24.3 grams of pure magnesium gives 24 liters of gas then you untangle the chemistry and reduce the problem to a math problem.

@MaxW. Yes. I have reduced the problem to a math problem. I did it on purpose, because Geoff required a math problem, not a chemistry problem !

I was thinking let x= grams Cu, y = grams Mg. Cu-> 3.70 L H2/(g Cu), Mg->0.988 L H2/(g Mg) $$x + y = 13.833$$ $$x \times 3.70 + y\times0.988 = 9.6$$ That avoids any discussion of moles.

@MaxW. Well ! You should know that Copper does not react with Copper !! And does not produce any H2.

Dah... I don't know how I got Cu in my mind. Thanks for the correction. The numbers in my comment above are for Zn.

In chemistry the Principle of Atom Conservation says that there must be the same number of like atoms on both sides of the equation. Balance the following chemical reaction by determining the appropriate values for $a, b, c, d$ and $e$ using Diophantine linear equations:
$$\ce{aS + bHNO3 <=> cH2SO4 + dNO2 + eH2O}$$
$$\text{i.e. Sulfur + Nitric acid}\ce{<=>} \text{Sulfuric acid + Nitrogen dioxide + water}$$

For example, the same number of like atoms on both sides of the equation would mean 6 oxygen atoms on the left and 6 oxygen atoms on the right.
The subscripts on the atoms indicate the number of that kind of atom in the molecule. Thus $\ce{H2SO4}$ contains 2 Hydrogen atoms(H), 1 sulfur atom(S), and 4 oxygen atoms(O).
Standard form dictates that the greatest common multiple of the coefficients must be 1.

HINT For hydrogen(H): $b = 2*c +2*e$

A nastier equation would be:
$$\ce{aK2Cr2O7 + bH2SO4 + cSO2 -> dCr2(SO4)3 + eK2SO4 + fH2O}$$

If you can put three different answers to the same question, then you should write any.

@Mithoron - I changed all the answers to community wiki so I wouldn't be unethically be getting 2 extra reputation points.  Happy? // I only separated the problems so that any discussion about a problem type would be on that problem. I wasn't trying to do anything nefarious.

My point is rather that it's obviously too broad. In such cases sometimes there's equally broad answer given, but adding multiple examples just isn't doing it.

@Mithoron - I found the question intriguing. - Can you write some chemistry problems that use simultaneous equations that can be solved with no real understanding of chemistry? Maurice's Cu/Mg problem is really nice if you fiddle with it just a bit.

@MaxW. I never proposed a problem with Cu and Mg. Copper will not react with HCl. No ! I proposed a mixture Mg-Zn. These two metals do react with HCl.

@Maurice - Thanks for the correction. Sorry for my confusion.

How about this?
Acetic acid, $\ce{CH3COOH}$, ionizes in water according to the following chemical equation to produce a hydronium ion, $\ce{H3O+}$, and an acetate anion, $\ce{CH3COO-}$.
$$\ce{CH3COOH + H2O <=> H3O+ + CH3COO-}$$
The mathematical equation for the equilibrium is:
$$1.8\cdot10^{-5} =\dfrac{\ce{[H3O+][CH3COO-]}}{\ce{[CH3COOH]}} $$
However water itself also ionizes according to the chemical equation:
$$\ce{ 2H2O <=> H3O+ + OH-}$$
The mathematical equation for that equilibrium is:
$$\ce{1\cdot10^{-14} = [H3O+][OH-]}$$
where the brackets, [], indicate concentration.
If acetic acid is initially added at a concentration of 0.1, and $\ce{[H+] = [OH-] = 1\cdot10^{-7}}$ what is the concentration of $\ce{H3O+}$ when the system reaches equilbrium?
HINTS:
(1) At equilibrium $$0.1 = \ce{[CH3COOH] + [CH3COO-]}$$
(2) Charges must balance in the solution so:
$$\ce{[H3O+] = [CH3COO-] + [OH-]}$$

I offered the hints and use a nebulous term "concentration" to try to take the required chemistry knowledge out of the problem. // The solution to this will require solving a quadratic equation.

I think a useful problem with the properties you describe is finding how a acid with several ionized forms changes with pH in the medium. Recently I explained how to do it for $\ce{CO2}$ here. Look How to calculate bicarbonate and carbonate from total alkalinity. The general solution scales nicely with complexity of the system:
ACID TYPE       DIFFERENT FORMS                 GENERAL SOLUTION                                                                EXAMPLE
Monoprotic      2 (undissociated, +1)           Fraction, degree 1 polynomial at denominator, one of its terms at numerator     Acetic acid
Diprotic        3 (undissociated, +1, +2)       Fraction, degree 2 polynomial at denominator, one of its terms at numerator     Carbonic Acid
Triprotic       4 (undissociated, +1, +2, +3)   Fraction, degree 3 polynomial at denominator, one of its terms at numerator     Phosphoric acid
...             ...                             ...                                                                             ...

Plenty of aplications too. For example, explains why ocean acidification affects calcifying organisms.

Copper, $\ce{Cu}$, is oxidized by concentrated nitric acid, $\ce{HNO3}$, to produce $\ce{Cu^{2+}}$ ions. The nitric acid is reduced to nitrogen dioxide, $\ce{NO2}$, or nitric oxide, $\ce{NO}$, according to the following chemical equations.
$$\ce{Cu  +  4HNO3 ->  Cu(NO3)2  +  2NO2(gas)  +  2H2O}\tag{1}$$
i.e. When 1 atom of copper is dissolved then 2 molecules of $\ce{NO2}$ are formed.
$$\ce{3Cu  +  8HNO3  ->  3Cu(NO3)2  +  2NO(gas)  +  4H2O}\tag{2}$$
i.e. When 3 atoms of copper are dissolved then 2 molecules of $\ce{NO}$ are formed.
When the reaction was complete by volume 63% of the gas was $\ce{NO2}$ and 37% of the gas was $\ce{NO}$, thus the ratio $\ce{NO2:NO :: 63:37}$. What percentage of copper dissolved by the first reaction?

Is the answer here 36.2% ?

@Geoff- yes 36.2


text:
The H NMR analysis of the hydrogens in salicylic acid
I am currently working with an unknown and from the IR spectrum I was able to figure out that the compound I am working with is salicylic acid.
I am looking at the H NMR, but I am a bit confused on the deshielding, integration, and upsheilding of the hydrogens that are near the $\ce{OH}$ and $\ce{COOH}$ groups on the compound. Can someone explain this to me:
Attached is a picture

The resolution of the picture is painfully low.

Apologizes but this is the only picture I received...

"Working with an unknown" means you got the IR and NMR spectra as a homework to identify, right? ;)

Since you have identified your unknown as salicylic acid (2-hydroxybenzoic acid), I'd help you understand how easily understand the substituents of aromatic nucleus induce shielding-deshielding effects. The following diagram illustrate the $\mathrm{^1H}$-$\mathrm{NMR}$ assignments of salicylic aromatic protons:

A lot of experts do not like explaining shielding-deshielding effects using electron densities around the sought atom. I agree with them, but a student of your capacity could easily related to that explanation. A good rule of thumb is o,p-directing electron donating groups (coursing positive mesomeric or resonance effect) usually shield the relevant protons at o,p-positions (inducing $\delta^-$-charge at the positions). Similarly, m-directing electron withdrawing groups (coursing negative mesomeric or resonance effect) usually deshield the relevant protons at o,p-positions (inducing $\delta^+$-charge at the positions). Accordingly, $\ce{OH}$-group shields $\ce{H_a}$ and $\ce{H_c}$ while $\ce{COOH}$-group deshields $\ce{H_b}$ and $\ce{H_d}$.
Thus, it is safe to say that the most up-field aromatic signals (sheilded) at $\delta \ 6.97$ (apparent $triplet$) and  $\delta \ 7.03$ (apparent $doublet$) are due to $\ce{H_a}$ and $\ce{H_c}$. The assignments of $\delta \ 6.97$ is belong to $\ce{H_c}$ (two vicinal protons) and  $\delta \ 7.03$ is belong to $\ce{H_a}$ (one vicinal proton) are solely due to the number of neighboring (vicinal) protons to each. Keep in mind that chemical shifts of these two signals can be moved to up- or down-field positions depending on the using locking solvent. For example, the locking solvent used to get present $\mathrm{^1H}$-$\mathrm{NMR}$ spectrum here is apparently $\ce{CDCl3}$ (based on solvent residue at $\delta \ 7.28$). However, two signals have been moved to more up-field positions (compared to that in $\ce{CDCl3}$) when the locking solvent is changed to $d_6$-$\ce{DMSO}$ (Ref.1). In that spectrum, two signals at $\delta \ 6.92$ ($t,\: ^3\!\!J=\pu{7.4 Hz}$) and  $\delta \ 6.96$ ($d,\: ^3\!\!J=\pu{7.5 Hz}$) are assigned for $\ce{H_c}$ and $\ce{H_a}$, respectively.
Similarly, the most down-field aromatic signals (desheilded) at $\delta \ 7.55$ (apparent $triplet \ of \ doublet$) and  $\delta \ 7.96$ (apparent $doublet \ of \ doublet$) are due to $\ce{H_b}$ and $\ce{H_d}$ (Note: The splitting of each signal with a small coupling constant is due to meta-or $\omega$-coupling generated by the presence of proton at meta to each; This phenomenon is generally called $^4\!\!J$-coupling). The assignments of $\delta \ 7.55$ is belong to $\ce{H_b}$ (two vicinal protons) and  $\delta \ 7.96$ is belong to $\ce{H_d}$ (one vicinal proton) are also due to the number of neighboring (vicinal) protons to each. Similar to $\ce{H_a}$ and $\ce{H_c}$, these two signals have also been moved to more some what up-field positions (compared to that in $\ce{CDCl3}$) when the locking solvent is changed to $d_6$-$\ce{DMSO}$ (Ref.1). In that spectrum, two signals at $\delta \ 7.52$ ($t,\: ^3\!\!J=\pu{7.7 Hz}$) and  $\delta \ 7.82$ ($d,\: ^3\!\!J=\pu{7.7 Hz}$) are assigned for $\ce{H_b}$ and $\ce{H_d}$, respectively.
References:

Milena Jadrijević-Mladar Takač, Dražen Vikić Topić, “FT-IR and NMR spectroscopic studies of salicylic acid derivatives. II. Comparison of 2-hydroxy- and 2,4- and 2,5-dihydroxy derivatives,” Acta Pharmaceutica 2004, 54(3), 177-191 (PDF).


text:
Nickel salts from hypochlorites
(JUST A HEADS UP I’m 15 years old and new to chemistry) So I was recently looking around the house looking for products that I can extract chemicals from and I found a bottle of cleaning Vinegar ( 5% acetic acid), so I decided to make some acetate compounds, but when I first started I thought it would be a simple 1 step synthesis. I was wrong. I soon realized you must have hydrogen peroxide and acetic acid to make zink and copper Acetate.
So I tried making nickel acetate, with no luck. So then I tried to make nickel chloride knowing this new hydrogen peroxide step. I took a pure nickel sheet and put it in a beaker then added calcium hypochlorite and hydrogen peroxide. And it worked! The solution turned light green over a few hours, and it had a gas being produced. And also a black Precipitate which was most likely nickel oxide (I filtered the precipitate off before taking the picture)
But just to check myself I tried to find this process online, to no avail. So now I’m wondering if I did something wrong and got the wrong product or no one does it through this process. If you can check my process to make sure it’s possible and I didn’t waste my time I would Appreciate it. Thanks in advance.

The solution is actually green. I had some UV lights off to the side for a different project that made it look blue. But I turned those off. And I also replaced the nickel strip for a nickel coin to be 100% shure it was infact nickel and not zink or some other Metal.

A US nickel is 25% nickel and 75% copper.

I’ll go find some more nickel strips

Your result is surprising, for the following reasons : 1) Nickel compounds and ions are mainly green. And you say your solution is blue. 2) When treated by hypochlorite ions, Nickel will produce a black precipitate of $\ce{Ni_2O_3}$, which is not what you got . 3) You obtain a solution without any precipitate, although the hypochlorite solution is always basic, which must produce a Nickel Hydroxyde precipitate. 4) The only possibility to have a blue solution with Nickel, is adding ammonia $\ce{NH_3}$, which produces a blue Nickel-Ammonia complex in solution. So my question : Are you sure that you have taken pure Nickel in the beginning ? Other question : With this method you will obtain a lot of calcium chloride with a small amount of nickel (or copper ?) compound.


text:
Silver nitrate mirror on 3d printed sla resin
For a project, I need to develop special array of reflectors.
Usually this is done through a process of plastic injection and vapor deposition, however the prototyping cost is very high.
Another solution I would like to explore is to 3d print the array of reflector with SLA printer like Form 3 and then deposit silver through the process of silver nitrate on solution.
However, I wonder if that could work and if the bond would be good enough, if the resin part would need prior treatment.
Another method could be by depositing a conductive layer and then do electroplating.
I'm not a chemist and don't really know where to start looking.
Thanks

Will a 3D-printed surface be smooth enough for your application? For your "array of reflectors" are you able to attach glass substrates?

Can you identify the specific resin that you will use? The chemistry of the resin may facilitate direct deposition of the mirror coating.

There are ways to smooth the surface by mechanical, or by over coating it. We don't need telescope grade optics though.

There are ways to bond silver to surfaces by chemical means without electroplating. I can describe possible approaches as an answer below.

Just an idea, but similarly to metal-loaded PLA filaments for Fused Deposition Modeling (FDM) 3-D printers (look at Filamet, by The Virtual Foundry), there may be some SLA resins working in a similar way (example). The principle is that you first print and then put your 3D printed model in an oven / kiln for sintering. The obtained surface can then be polished for a mirror-like aspect. The advantage is that you have a metallic surface at the end, which can thus be electro-plated with various reflective coatings...

My experience with glass reduced-silver mirrors is that bonding is poor, and that unprotected Ag rapidly degrades in air (a matter of days, in some cases). The bond to organic surfaces, I believe, might be worse.
If you have access to a fairly good vacuum pump (less than 10-4 Torr), it's not hard to aluminize. Watch a demo online.
You might also request a quote from an aluminization service, such as those in this list. In 2010, a 200 mm diameter mirror was said to cost about US$70 to aluminize. This British service has an online price list, starting at £20.

I was looking into that, although seems to be fairly hard with a simple setup, and very low yield and is also time taking and difficult to scale. Panels are fairly large, at least 150x150mm but I will look more at commercials service. For oxidation of course it is known and need a protective coating atop.

A traditional way of applying a silver coating to glass is through a chemical reaction using a reagent known as Tollens' Reagent. This reagent is prepared from silver nitrate as the source of silver.
Depending upon the surface chemistry of your cured SLA polymer, it may be possible to coat the polymer directly, as one would coat glass. There may also be a coating that will adhere to the SLA polymer and serve as a surface primer to which the silver from the Tollens' Reagent will adhere.
My recommendation would be first to identify the specific SLA resin that you will be using. This information should be available on the container, its packaging, or a Material Safety Data Sheet (MSDS).
Some research, such as an internet search, may provide additional information once the polymer chemistry is known. You might also find some helpful information from the SLA resin or printer manufacturer in the form of "Application Notes" that address applying surface coatings to SLA 3D-printed components.
I recommend performing a small-scale test to see what the Tollens' Reagent reaction does on the surface of a 3D-printed part that is made from the same resin that you will use. You might also experiment with different resins. Several resins may work equally well for forming your part while one provides a better surface for applying a mirror coating.
It will be necessary to apply a protective coating to the silvered surface, once prepared, to prevent oxidation.


text:
Does solute concentration vary significantly with depth?
If I dissolve glucose in an ordinary beaker filled with water, are the concentrations of glucose at the top and bottom significantly different? What if we use NaCl as the solute? Is there a molecule large enough to ``sink'' to the bottom, and if so would it still be classified as a solute? Likewise, are there solutes which are lighter than H2O, so they float upwards?

theres a famous experiment where you dissolve salt without mixing in an aquarium to obtain a gradient of refractive index, and so bending of light rays passing through. Im not sure but i guess as the salt dissolves stronger concentration falls to the bottom

here: https://youtu.be/PAK_1C-Zqo0

(1)If I dissolve glucose in an ordinary beaker filled with water, are the concentrations of glucose at the top and bottom significantly different?
(2) What if we use NaCl as the solute?

1 & 2 are basically the same question. The difference in gravity between the top of the beaker and the bottom of the beaker is real but insignificant. So due to gravity alone there won't be a concentration gradient.
However there are solutions which can be centrifuged which will create a concentration gradient. Essentially the centrifuge creates a large gravity gradient. Close to center gives lower gravity, far from center gives higher gravity. Once you stop the centrifuge however the concentration gradient is not at equilibrium. So over time the concentration gradient would disappear.

(3) Is there a molecule large enough to ``sink'' to the bottom, and if so would it still be classified as a solute?

The above discussion about a centrifuge confirms the basic notation, but there is an unstable gradient, not a sinking per se.
Obviously if you drop a marble in a glass of water it sinks. But the marble isn't dissolved in the water.

(4) Likewise, are there solutes which are lighter than H2O, so they float upwards?

This is a bit more interesting. Open a plastic bottle of soda and put a balloon over the neck of the bottle.  Carbon dioxide was dissolved in the soda under pressure. When the soda is opened, the pressure is released and the excess carbon dioxide will "float up" out of the soda and blowup the balloon. Since the gas phase is less dense than the liquid phase (soda) it is on top. The gas doesn't really "float" on the soda however. It is also important to realize that once equilibrium is established the carbon dioxide remaining in the soda is homogeneously distributed throughout the soda.
A different sort of twist are soap molecules which will "float" to the surface, not because they are lighter, but because of the surface tension of water. The famous soap/pepper experiment illustrates this.
Another sort of twist here is a bartender who makes fancy cocktails by floating one layer of a liquor over another due to the higher layers being less dense. The segregation of the layers is unstable however. One good swirl with a straw and the layers mix.

For 3 and 4 there is what is called a phase diagram. For example it shows if two liquids will mix, or if there will be two phases where one liquid would float on the other.
Think of gasoline and water. There is a gasoline layer floating on the water so there are two liquid phases because there is an interface. However a very tiny amount of gasoline will be dissolved in the water and a very tiny amount of water will be dissolved in the gasoline. In this case the gasoline that is dissolved in the water is distributed homogeneously throughout the water. Likewise the water dissolved in the gasoline is distributed throughout the gasoline.

I used gasoline as a conceptual example. Gasoline however isn't a pure compound but rather a mixture. So for those that want to be picky substitute n-hexane for gasoline.

Good point about the presence or absence of intermolecular interactions. However, e.g. dissolved oxygen will not rise or sink in a lake due to difference in mass or density. Also, the barometric formula does not rely on changes in gravity in different layers of the atmosphere - it is just a Boltzmann distribution.

"But since the gas is less dense than the liquid it will flow out of the liquid." This is potentially really misleading. Carbon dioxide gas (or oxygen gas or nitrogen gas) is less dense than water, but is carbon dioxide (or oxygen or nitrogen )  dissolved in water less dense than water?

@KarstenTheis - Thanks for pointing out the problem with fix the gas/liquid example. I've tried to fix it. // I thought about using the atmosphere as an example of a density gradient due to gravity, but there is also a huge temperature difference.

@BuckThorn - Thanks for pointing out the problem. I definitely wrote that part badly. I've tried to fix it. // The density of carbonated water is interesting. A study found that it is. I didn't study the part on partial molar volumes carefully, but I'd expect that a carbon dioxide molecule tends to replace a water molecule. Since CO2 is heavier than H2O the density would go up. But that would also mean that the volume of the solution would increase when carbon dioxide is injected too.

Yes, there is a gravity gradient within a centrifuge tube—the centrifugal acceleration increases with the distance from the axis of rotation. However, the reason a centrifuge creates a density-based concentration gradient isn't that there's a gravity gradient. Rather, it's that the centrifugal acceleration is so high. I.e. what matters is how high the acceleration is, not that it varies. Indeed, you'd get more of a concentration gradient with an acceleration that was a constant, say, 10^5 g from top to bottom, than you would if it were 10^5 g at the bottom of the tube and <10^5 g at the top.

In order to "see" the effects you are interested in you will have to forget the beaker. So we need to think at a very large scale. So instead of beaker, consider the Earth and its atmosphere. Air is a very nice example of a solution which is homogeneous mixture of nitrogen, oxygen, carbon dioxide, water in gas phase, noble gases (He, Ne, Kr, Rn, etc.) and some rare gases (H2 etc) . You might wish to consider nitrogen as the solvent and the rest as solutes. Guess why there is no hydrogen in the atmosphere and why helium is so rare in the atmosphere? Helium is continuously being produced by radioactive decay in rocks. Because these gases are so light that they "float upward" i.e., Earth's gravitational pull  is not strong enough on these small gases and they escape into the universe. This is why helium is so expensive. So this partially fulfills your molecules floating upward example.
Now comings to making some heavier molecules settle downwards in a solution. Take the example of air again and carbon dioxide. You can actually pour carbon dioxide over candles and they can be extinguished.

As you would have guessed now, you either need very large scale system or very powerful gravitational field in order see the effects you wish to see. There are devices are ultra-centrifuges, which can achieve the separation of heavier molecules from the lighter molecules by differences in their weight. Earth's gravitational pull is not strong enough where you can observe dissolved glucose or NaCl settling to the bottom of a beaker. You can only see these effects with much larger molecules, so called macromolecules.

There are two issues that need to be carefully separated: density of samples with different composition (aquarium example from OP and CO2 example here), and separation from a homogeneous mixture (hydrogen example).

H/He is an interesting example, but the composition of the atmosphere is much more complicated of course. http://wordpress.mrreid.org/2014/08/01/the-composition-of-earths-atmosphere-with-elevation/

This is an interesting question -- clearly marbles sink and molecules dissolve, and we don't commonly encounter objects between those length scales. It is helpful to look at colloids here, which are small particles of between 1nm and 1000nm, which means they're in-between the realm of molecules and of macroscopic objects. Since we can observe them with an ordinary microscope, they're often used as "super atoms".
You can get suspensions of colloids in water (they form when you add water to aniseed flavoured spirits, for example). In a very simple case, a colloid particle has two features affecting its energy: gravity, $mgh$, and thermal energy $k_\text{B} T$. Some particles will randomly have more thermal energy,
and thereby be able to float higher.
We can calculate the balance of these two energies with a Boltzmann distribution, which says the concentration of colloid particles at a height $ h $ compared to the concentration of colloid particles at the bottom is given by
$$ \frac{\eta(h)}{\eta(0)} = \exp\left(-\frac{mgz}{k_\text{B} T}\right)$$
Famously, Jean Baptise Perrin used this equation to calculate a value of $k_\text{B} T$ in 1916, which helped him earn the Nobel Prize in 1926.
We can put some numbers into this equation now. What height of beaker would we need to see 10% fewer particles at the top than at the bottom?

For a virus of mass $1\times 10^{-17}\text{kg}$, we'd need a beaker that was 0.04mm tall to see 10% less at the top than the bottom.
For a protein of mass $ 5\times 10^{-18}\text{kg}$, we'd need a beaker that was 8mm tall to see 10% less at the top than the bottom.
For a glucose molecule of mass $ 2.98\times 10^{-25} \text{kg}$, we'd need a beaker that was 149 metres tall to see 10% less at the top and the bottom.
For sodium chloride, we'd need a beaker 1170m tall to see 10% less $\text{Na}^+$ at the top than at the bottom!

Putting that last point in perspective, if we had a 10cm tall beaker full of salt in water, we'd find that the concentration of $\text{Na}^+$ at
the top was $99.99991\%$ of the concentration at the bottom -- it's such a small difference that we can ignore it. For glucose, the top of a 10cm beaker would have $99.9929\%$ of the concentration at the top compare to the bottom.

This would be true in vacuum, I think. However, none of these particles would be in the gas phase, they would all form solids (at least at room temperature and for an appreciable number of particles). In solution, you have to consider not the mass, but the difference in density (buoyancy). To sediment large proteins in solution, you need an ultracentrifuge (where particles experience on the order of 200,000 x g acceleration).


text:
Which one is the stronger nucleophile? CH3CH2O or CH3CO2

Not really sure what the principle would be. Maybe it has something to do with the resonance capability of the two Os in CH3CO2

"Not really sure what the principle would be. Maybe it has something to do with the resonance capability of the two Os in CH3CO2". What is the consequence for the negative charge in acetate vs ethanolate?

Since a nucleophile is a species that is donating a pair of electrons, it’s reasonable to expect that its ability to donate electrons will increase as it becomes more electron rich.
As electron density increases, nucleophilicity increases.
In our first option(CH3CO2-), resonance is occurring between the two oxygen atoms, decreasing the electron density on oxygen atom.
Nothing like this occurs in our second option(CH3CH2O-) which means it has more electron density than the first compound, hence second compound is more nucleophilic.


text:
Confusion on when to use mass of entire solution when performing q=mc∆T
I am studying for the AP Chemistry exam in May 2020. While studying, I encountered a doubt.
In Problem 1 (the question on the top), the educator saw that the problem was referring to the solution as a whole (the problem states ∆Hsoln), and used 100 g + 10 g = 110g grams for the mass in Q = mc∆T and then, on the next part, did ∆H = -Q/n to find the enthalpy of solution.
However, in Problem 2 (the question on the bottom), the educator uses the total mass of the solution (200 grams) even if the problem refers to the reaction (the problem states ∆Hrxn) and not the solution. When I saw the question, I thought I had to convert the 1 mol Ag+ and 1 mol Na+ to grams, and add those masses for (m) in Q=mc∆T since the problem referred to the reaction and not the solution.
As a result, I am confused as to when exactly I should use the mass of the entire solution instead of the mass of the reactants in my Q = mc∆T and ∆H = ±Q/n calculations.

In short you are saying that mass of water has not been taken into account in the second question while it has been in the first. - is that what you are trying to say?

Are you saying because one is a dissolution process and the other is an acid base reaction, you thought they should be treated differently? But it is always the entire solution (i.e. mostly water) that changes temperature.

Karsten, an acquaintance recently showed me a similar first year undergraduate problem and he was in a very good university. It was hopeless to see how this was being taught to him (online classes have made it worse). In some online solutions to such problems, 10 g was being considered and in other online solutions, 10 g salt was ignored. This exact crazy problem is circulating here and there. My opinion was that it must be included in the calculation. Someone should address this problem in detail here so that the confusion ends.

M. Farooq, in my experience as an educator, there are often mistakes in solutions manuals. The total mass of the solution must be used. Another way to verify this is that heat capacities tend to be given per mass solution, not mass solvent

I am confused as to when exactly I should use the mass of the entire solution instead of the mass of the reactants in my Q = mc∆T and ∆H = ±Q/n calculations.

Ideally, you always take the mass of the entire solution, and the heat capacity of the product mixture. The two questions tell you what assumptions to make with respect to the heat capacity. There are other assumptions you also have to make, even if they are not stated.

[top question] ... assuming that the specific heat capacity of the solution is $\pu{4.18 J/(g ^{\circ}C)}$.

The instructions are straightforward. You know the mass of the product mixture (conservation of mass) and the assumed specific heat capacity. This is a dissolution process, indicated by the subscript "soln", i.e. $\Delta H_\mathrm{soln}$. The first sentence in the problem description should read "enthalpy of dissolution" instead of "enthalpy of solution" to be less confusing.

[bottom question] Assume the density of the solution is $\pu{1 g/mL}$ and the and the specific heat capacity is the same as that of water.

Here, solution refers to the product mixture. You have to make an additional assumption, namely that the volume of the product mixture is equal to the sum of the volumes of the reactants. For the given situation, you are not making a big systematic error assuming that. With this assumption, you can calculate $Q$.
What $Q$ are we calculating?
The reactants are at a certain starting temperature, and the product mixture are at a different temperature. In a thought experiment, you could heat up or cool down the product mixture back to the starting temperature. That is the magnitude $Q$ you are calculating (opposite sign, though). Because the entire mixture is "too" hot or cold, the mass (and specific heat capacity) of the entire product mixture needs to be considered.
Approximations in other problem sets
In other problems, you are instructed to consider the mass of water instead of the mass of the entire product mixture. This is a fine approximation for diluted solutions. In rare cases, problems such as this give you a more realistic value for the specific heat capacity of the product mixture instead of asking you to use the value for water. Again, using the value for pure water is fine for diluted solutions.

Just a quick question though; why do we usually use the heat capacity of the solution after mixing, and the mass after mixing? Basically that would be like a thought experiment "dissolving" the salt without enthalpy change and then adding the heat with the salt solutions' heat capacity and mass after mixing

Is it because enthalpy is a state function and we can model it after this thought?

@Mäßige Yes. It's a cheap setup. Instead, you could have the reaction vessel in a calorimeter heat bath, and either keep the bath temperature constant (isothermal calorimetry) or almost constant (traditional large heat bath). In the first case, the reaction mixture would have the same temperature before and after (enthalpy is reflected in cooling or heating necessary to keep bath at constant temperature), and in the second case, the temperature change is tiny (and the enthalpy would almost be equal to the heat transferred to the bath because it has the higher thermal mass).

In mcΔT, you must use the total mass of the system that is changing temperature. In this case, that is the solution.
The same thing is occurring in the second problem. The temperature of the solution changes, and therefore the solution mass is the mass you should use.
It wouldn't make sense if the solute changes temperature but the solution it is dissolved in doesn't change temperature, does it? I find it always helps me to think about the physical process that is happening for questions like this. Always identify what your system of interest is, and the initial and final states of the system. Doing so can clear up a lot of these problems, and will also help you get through more difficult problems if you ever take a thermodynamics course.


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Para effects on substituents in aromatic compounds
I wanted to know how to compare the following compounds on basis of acidic strength.

I have to find the acidic strength of following compounds with $\mathrm{X}=\ce{CH3, \ OCH3,}$ and $\ce{NO2}$.
For ease of explaining, denote the first compunds as (A), the second as (B), and the third as (C)
Due to steric inhibition of Resonance (SIR) effect, compound (A) will be the most acidic, as its anion will be the least stable.
Now for ordering the other two, I know that there are two things affecting acidic character:

Bond strength; the weaker the bond, the higher its acidic strength
Charge on hydrogen; the more positive charge their is on the hydrogen, the more its acidic strength.

So now if we take $\mathrm{X}=\ce{CH3}$, we should have that $\mathrm{(A)>(C)>(B)}$; as $\ce{CH3}$ is an electron donating group and it would make the hydrogen less positive, and inductive effect is distance dependent. This answer is incorrect as per my textbook, which says $\mathrm{(A)>(B)>(C)}$.
If we take $\mathrm{X}=\ce{OCH3}$ which is an electron withdrawing group, we should have $\mathrm{(A)>(B)>(C)}$. This answer is correct.
If we take $\mathrm{X}=\ce{NO2}$, which is also an electron withdrawing, we should have $\mathrm{(A)>(B)>(C)}$. But my textbook says the answer is $\mathrm{(A)>(C)>(B)}$.
I wanted to know where am I going wrong, and if there is some other effect I am not taking into account.
Note: my earlier question was closed due to the similar question: Ortho-effect in substituted aromatic acids and bases.
But that question only deals with the ortho effects of substituents, not the para.

Actually, where are your structures (A), (B), and (C)?

I have "Solomons, Fryhle, Snyder Organic Chemistry" open in front of me, and it seems there are a few difference between the book's order and the order given by you.
Acidic strength order for

$\mathrm{X}=\ce{CH3}$: m-dimethylbenzene >  p-dimethylbenzene >  o-dimethylbenzene
$\mathrm{X}=\ce{OCH3}$: m-methoxyphenol > o-m-methoxyphenol > p-m-methoxyphenol
$\mathrm{X}=\ce{NO2}$: o-nitrotoluene > p-nitrotoluene > m-nitrotoluene

$\ce{CH3}$ shows +I and hyperconjugation, $\ce{OCH3}$ shows +M and -I and $ce{NO2}$ shows -R and -I effect.
The dominating orders of effects is: mesomeric>hyperconjugation>inductive.
This and the answer by Bitthal Maheshwari should explain the order.
For the halotoluenes but it came out m-halotoluene > o-halotoluene > p-halotoluene for me but I do not have the data so I cannot confirm it. Perhaps someone else could do it.

At meta position the M effect is not considered. The reason behind is that during resonance of a six membered cyclic compound charges never reaches at meta position that why it is so try to proof it by drawing resonance structure and if you didn't get us then tell me in the comments below.

I still don't get why there would be no resonance, could you please elaborate?

There would be resonance but when you draw resonance hybrid(expecting you know how to draw it) of benzene you will observe that the slight(+ or -) charges will be at ortho and para position not at meta position so that is the reason that resonance effect is not observed at meta position.


text:
Calculate amount of enzyme needed for in vitro phosphorylation?
I have 100 micrograms of STAT1 and its molecular weight is 90 kDa. I calculated the number of moles to be 1 nmoles. The specific activity of JAK1 (the kinase) is 110 nmoles of phosphate per minute per mg of total protein at 30°C and the activity was determined at a final concentration of 0.83 micrograms/milliliters. What amout of JAK1 do I need?
I tried many ways but I don't get how to link the data.
Thank you.

How many phosphorylation sites does STAT1 have? How long to you want to wait for the reaction to be done?

Only one site Tyr701, I'm not sure about the amount of time of the reaction.

"I have 100 micrograms of STAT1 and its molecular weight is 90 kDa. I calculated the number of moles to be 1 nmoles. "
Calculate again.

It would take 1 minute to phosphorylate 0.11 nmoles of substrate if you add a microgram of enzyme (in a reaction volume of 1 mL). If you have 1 nmole substrate instead, it would take about 10 minutes with the same amount of enzyme.

Thank you! How did you get to the 1 minute___0.11 nmole relation?


text:
A reaction of Hydrogen Iodide and Ether
The problem is:

1,2,3-trimethoxy-cyclohexane reacts with $x$ moles of $\ce{HI}$ to give $y$ moles of $\ce{CH3I}$ and iodo-cyclohexane as product. Then find the value of $\frac{x+1}{y+1}$.

I am not getting that how is iodohexane formed and also I am not able to think of any mechanism for this reaction.

It would be more accurate to say iodocyclohexanes as there are 3 possible iodinated cyclohexane products. As for the mechanism, HI is a strong acid where will it protonate?

I used the same language as was used the question.

The question is poorly worded

Okay,it might be.

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$\ce{HI}$ attacks the molecule: $\ce{H+}$ attacks the oxygen atom and as a result, the oxygen-cyclohexyl bond cleaves in SN1 fashion and $\ce{I-}$ gets substituted on the ring. Result: three $\ce{HI}$ molecules are consumed, three $\ce{CH3OH}$ molecules are formed, and 1,2,3-triiodocyclohexane is formed.
First, 1,2,3-triiodocyclohexane becomes 6-iodocyclohexene in a manner similar to glycerol's reaction with excess $\ce{HI}$ as described here (or via the Internet Archive). Second, each methanol molecule formed in the above reaction consumes one mole each of $\ce{HI}$ to form $\ce{CH3I}$. Result: three $\ce{HI}$ molecules are consumed, three $\ce{CH3I}$ molecules are formed, 6-iodocyclohexene is formed.
Reaction of 6-iodocyclohexene to form iodo cyclohexane:
The former reacts with a molecule of $\ce{HI}$ to form 1,2-diiodocyclohexane. It undergoes a reaction similar to the second part of step 2 to give cyclohexene. Then, cyclohexene reacts with another molecule of $\ce{HI}$ to form iodo cyclohexane. Result: two molecules of $\ce{HI}$ are consumed and iodocyclohexane is formed.
Finally: The number of molecules of $\ce{HI}$ consumed  is $x = 8$, the number of molecules of $\ce{CH3I}$ formed is $y = 3$. Hence, $\frac{x+1}{y+1} = \frac{9}{4} = 2.25$. The mechanism is similar to the reaction of glycerol with excess of $\ce{HI}$.