text:
Why is the dipole moment of acetone higher than ethanol?
It is true that acetone is less polar than ethanol.
I thought the dipole moment is proportional to polarity, and if so, the dipole moment of acetone should be lower than ethanol.
But the dipole moment of acetone is higher than ethanol.
Why is it higher?

Regarding your statement about differences in polarity, did you consider differences in hydrogen bonding capabilities?

I change comment because I see your point better. You are rather asking why ethanol is more polar than acetone instead of why acetone has higher dipole moment. . It is not the same question. But you got the answer.

Dipole moment is just one of aspects of solvent polarity, as is well known in HPLC separation  analytical technique where the following effects take part:
Dipole moment interaction, permanent and induced
The above for the whole molecule and for particular bonds
Electron pair donor
Electron pair acceptor (including $\ce{H+}$ donation)
Depending on solute molecules specifics, the solvent polarity order is variable, as various molecules have different responses to above interactions.
See also Using solvent triangle in HPLC

The bond dipole μ is given by:
$ \mu =\delta \,d$.
The bond dipole is modeled as +δ — δ- with a distance d between the partial charges +δ and δ-. It is a vector, parallel to the bond axis, pointing from minus to plus, as is conventional for electric dipole moment vectors.
In your case, acetone has a higher dipole moment than ethanol because it has a carbonyl group more towards the center and thus acetone has lesser boiling point than ethanol.
The definition is from Wikipedia, : https://en.m.wikipedia.org/wiki/Bond_dipole_moment
EDIT : Alcohol hydrogen bonding is the major effect here, as in the case of water. (by @Poutnik)
Thanks for the edit, @MathewMahindaratne
Think for yourself

Acetone has 3 carbons, ethanol 2 carbons. Major effect is alcohol hydrogen bonding, similar to water.

Who said acetone has higher boiling point than ethanol?

Read carefully.

@ShishirMaharana Note that every question and answer has dedicated edit history visible not just to you. Correcting the mistake which another user helped you to discover and then suggest to "read carefully" isn't polite and constructive at all. If there were a flaw in my answer, I would rather be grateful and seek other means to improve my answer instead.

I know that! I meant read again carefully!

That is why, I first acknowledged Poutnik & then MathewMahindaratne, who made changes in my answer so that it be helpful to our Chemistry SE community!

Thanks for paying attention though @andselisk. I am very grateful to your noble deed. I respect you to the highest degree sir. Thanks to all the advisors who helped me edit this answer! Thanks to Chemistry Stack Exchange! All hail Chemistry Stack Exchange! rofl!!! :)


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The synthesis of 1-phenylprop-1-ene from propiophenone
I would like to know how to make 1-phenylprop-1-ene ($\ce{C6H5-CH=CH-CH3}$) from propiophenone ($\ce{C6H5-CO-CH2-CH3}$)?
I think carbonyl group ($\ce{C=O}$) of propiophenone is electron-withdrawing group because of Oxygen of carbonyl group.
So the carbonyl group of propiophenone will be attacked by nucleophile. And this will be oxidized to make double bonds.
But I don’t know the detail reaction mechanism to make phenylprop-1-ene from propiophenone.
Can anyone help me understand this reaction mechanism?

A detailed reaction mechanism is an orthogonal question to a synthetic route. At the very least, you are confused about what kind of answer you need for this problem. (It's probably not a mechanism, though that may be helpful.)

Your suggested synthetic rought is not going to happen. Thus, can't suggest a mechanism. Yet, the best way to do this is given in Waylander's answer. I think you should accept that answer because it is best way to do this transformation.

I think I should ask a synthetic route, too as you teach me. I didn't understand the exact meaning of reaction mechanism.

This can be done it 2 simple steps. You are not adding to the carbon count so all you need to do is operate on propiophenone.
Step 1 - reduce the carbonyl group to a secondary alcohol. Sodium borohydride in $\ce{MeOH}$ or $\ce{EtOH}$ will do this in high yield and the workup is straightforward.
Step 2 - dehydrate the secondary alcohol to give the double bond. As it is a benzylic alcohol, protonation of the $\ce{-OH}$ group and formation of the benzylic cation is strongly favoured and will give you the alkene you want. Catalytic para-toluenesulfonic acid in refluxing toluene (with water removal by Dean-Stark apparatus) will do this easily.

I can understand it is necessary to make alcohol from carbonyl group with the help of hydride reduction and to make alkene with dehydration reaction. Thank you so much for your great answer!

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Why is it considered acid rain with pH <5.6?
I recently read in a book that rain is considered acid rain if the pH falls below 5.6. However a substance is acidic when the pH is below 7; so why is the boundary for acid rain 5.6?
I was thinking pH between 5.6-7 would be too diluted to have an effect on limestone and other materials, and that's why the pH has to be under 5.6. But I wasn't sure if this is true so I wanted to find out.

Because "normal rain" still is a bit acidic.

Actually, just the carbon dioxide the rain picks up on the way down is enough to have an effect on limestone - that's how limestone is dissolved with no human activity, and that's how all those wonderful caves with stalagmites and stalactites are formed.

Would you consider something at a pH of 6,9 acidic? How about 6,99? What's the point of that if the substance is relatively neutral? Keep in mind pH is logarithmic, a couple of points is a major difference. A line is drawn, it had to be drawn somewhere, and this is where they put it.

@Luaan This, of course, begs the question of how those big limestone masses had managed to form in the first place. :-)

@oakad It's quite interesting, actually - rain water is acidic enough to dissolve limestone, yes. But the deposits form out of seawater, which is basic rather than acidic. And then you have life, of course. Those deposits then either get uplifted, or release their carbon dioxide as they get subducted. If the deposit gets enough rain, it will slowly be dissolved away back into the oceans (building up caves and sinkholes in the meantime).

You are forgetting an important component of the air: carbon dioxide. When it dissolves in pure water (=rain water), it makes it acidic. It is not considered that harmful.
Acid rain has a negative connotation; it is mainly caused by anthropogenic activities. The low pH of acid rain is due to sulfur oxides and nitrogen oxides and it is indeed below 5.7.
IUPACs definition "Rain with pH values < about 5; commonly results from acids formed from pollutants. 'Pure' rain water equilibrated with atmospheric CO2 and naturally occurring acids in relatively clean air usually has a pH>5."

but why is it considered acid rain only below pH 5.7? Why not below 7?

Because natural rain, which is free from all types of anthropogenic pollution has a pH < 7. You need to "distinguish" natural acidity vs. excessive acidity due man's activities. Otherwise what is the point of calling each an every rain as acid rain. Also remember there is little bit of nitric acid formed from lightning which is the plant food. That will not be termed as acid rain either. As I said, acid rain has a negative meaning.

"Why not below 7?"
I think this has to do with how you parse the term "acid rain". You seem to consider it to mean "an acid (anything with a pH < 7) which is raining". As the other answers have explained, acid rain means rain which is more acid than normal rain.

@M.Farooq So why choose 5.7 as a boundary value? Acidic soils having pH < 5 or even 4.5 are not uncommon, so such a rain would not harm them. Of course, such pH values are harmless for humans.

@trolley813, see Loong's answer. A pH of 5.67 is the border pH of CO2 in water. Soil chemistry is a complete different story. Acid rain was not named on the basis of its effect on soil by Angus Smith in 1872. He saw deleterious effects on plants and chlorophyll from "acid gases". His book is available online.

The $\mathrm{pH}$ of pure water (rain as well as distilled water) in equilibrium with the atmosphere ($p_{\ce{CO2}}= 10^{-3.5}\ \mathrm{atm}$) can be calculated as follows.
$$[\ce{H2CO3^*}]=K_\mathrm H\cdot p_{\ce{CO2}}$$
where $[\ce{H2CO3^*}]$ is the total analytical concentration of dissolved $\ce{CO2}$, i.e. $[\ce{H2CO3^*}]=[\ce{CO2(aq)}]+[\ce{H2CO3}]$, and
$K_\mathrm H= 3.39\times10^{-2}\ \mathrm{mol\ l^{-1}\ atm^{-1}}$ is Henry's law constant for $\ce{CO2}$.
$$\begin{align}
\log[\ce{H2CO3^*}]&=\log K_\mathrm H+\log p_{\ce{CO2}}\\
&=-1.5-3.5\\
&=-5.0
\end{align}$$
The commonly used first acid dissociation constant of carbonic acid $\mathrm pK_{\mathrm a1}=6.3$ (at $25\ \mathrm{^\circ C}$) actually is a composite constant that includes both the hydration reaction
$$\ce{H2O + CO2(aq) <=> H2CO3}$$
and the protolysis of true $\ce{H2CO3}$
$$\ce{H2CO3 <=> H+ + HCO3-}$$
For a weak acid
$$\begin{align}
\log[\ce{H+}]&\approx\frac12\left(\log K_\mathrm a+\log[\ce{H2CO3^*}]\right)\\
&=\frac12\left(-6.3-5.0\right)\\
&=-5.65\\
\mathrm{pH}&=5.65
\end{align}$$
Thus, pure rain in equilibrium with the atmosphere has about $\mathrm{pH}=5.65$. Any acid rain with lower $\mathrm{pH}$ would be caused by additional acids.

Under atmospheric pressure, dissolved carbon dioxide can reach an equilibrium state in water that yields a pH of as low as 5.7

slightly related "rain safety" question: Is there any reason to fear personal exposure to rain with a pH of 3.1?

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What is the fate of sulfur in cysteine when it participates in gluconeogenesis?
Amino acids can be broken down to yield pyruvate, which in turn can be used to construct glucose molecule in the process of gluconeogenesis.  What happens to the sulfur atom in the amino acid cysteine when it is processed into pyruvate?

Image source: NCBI Bookshelf

Cysteine is a dispensable amino acid required for synthesis of protein and non-protein compounds. These non-protein compounds include pyruate (hence acetyl coenzyme A), taurine, sulfate, and glutatione (GSH). The fate of the cycteine is hence sketched in following diagram (Ref.1):

As evidence from the diagram, the sulfur atom in cysteine has been converted to sulfate ion ($\ce{SO4^2-}$) through $\ce{HS-}$ ion. This fact has been previously proved experimentally (Ref.2). Accordingly, the first major step in cysteine catabolism in mammalian tissues is the oxidation of cysteine to cysteine sulfinic acid catalyzes by cysteine dioxygenase (CDO). The reaction mechanism of this non-heme iron dioxygenase has been recently reviewed (Ref.3).
References:

G. Courtney-Martin, P. B. Pencharz, “Chapter 19: Sulfur Amino Acids Metabolism From Protein Synthesis to Glutathione,“ In The Molecular Nutrition of Amino Acids and Proteins; Dominique Dardevet, Ed.; Academic Press, Elsevier Inc.: London, United Kingdom, 2016, pp. 265-286 (https://doi.org/10.1016/C2014-0-02227-7).
Catherine L. Weinstein, Rudy H. Haschemeyer, Owen W. Griffith, “In vivo studies of cysteine metabolism. Use of D-cysteinesulfinate, a novel cysteinesulfinate decarboxylase inhibitor, to probe taurine and pyruvate synthesis,” The Journal of Biological Chemistry 1988, 263(32), 16568-16579 (http://www.jbc.org/content/263/32/16568.full.pdf+html).
Crisjoe A. Josepha, Michael J. Maroney, “Cysteine dioxygenase: structure and mechanism,” Chem. Commun. 2007, (32), 3338-3349 (DOI: 10.1039/B702158E).

According to https://www.uniprot.org/uniprot/P25325, 3-mercaptopyruvate sulfurtransferase is also involved in the catabolism of cysteine:

Transfer of a sulfur ion to cyanide or to other thiol compounds. Also has weak rhodanese activity. Detoxifies cyanide and is required for thiosulfate biosynthesis. Acts as an antioxidant. In combination with cysteine aminotransferase (CAT), contributes to the catabolism of cysteine and is an important producer of hydrogen sulfide in the brain, retina and vascular endothelial cells. Hydrogen sulfide H2S is an important synaptic modulator, signaling molecule, smooth muscle contractor and neuroprotectant. Its production by the 3MST/CAT pathway is regulated by calcium ions.


text:
Order of internal pressure resistance to expect from laboratory glassware?
I have a need to use more pressure differential than can be afforded by vacuum vs atmospheric pressure alone, and so I've come to consider using pressurised laboratory glassware for part of my setup.
I've made a little bit of research here on stack-exchange and on google and the general information I've gathered is that the physical shape of laboratory glassware and reaction vessel lends better to resisting external pressure (when we use them with internal vacuum), than to resisting internal pressure, though I'm left wondering what to expect to be doable in terms of pressurising glassware.
I have two pieces of glassware I'm considering using on the pressurised end of my setup: One is a 5,000mL flat bottom flask and the other is a 20,000mL flat bottom flask. Both are borosilicate glass, neither are perfect (they both have one or two tiny air bubbles in the glass at places) but both have shown very good resistance while being used with internal vacuum.
Is it realistic to expect them to be fine while adding an internal pressure up to 3 Bars? (I would put the glassware inside of a big metal cylinder for safety in case of "explosion")
What will be the best Resistance vs Glassware Shape / Glassware Volume / Temperature / etc. ?
Does anyone have any tips on the topic?

Glass in general is relatively bad at withstanding internal pressure, and glass joints in particular are even worse.

@IvanNeretin That's interesting. Instinctively I would have worried the least about the joints as they seem to be the most resistant parts of the vessels. Do you mean the part where they connect to the body of the vessel?

No, I mean the part where two vessels meet. When you have vacuum inside, they are pressed together with great force, which is good. But internal pressure seeks to tear them apart.

What level of internal pressure (in Bars) would you think each of my two flasks could withstand with less than 10% risk of breaking at those joints?               10% is high but I'm fine operating with 1/10 chance it will break, as long as it's not much more than that. If it breaks at the joint and the flask is surrounded by a glass cylinder at worst the joints and connection will fly/jump upwards. It would be better than the body breaking as there's less chance I'd have to deal with a liquid mess. I may even be able to salvage the solution and filter out possible glass shards.

Or do you mean the joints not breaking but dissociating (as the vessel dissociating from the piece of glassware connected to one if its joints without either breaking)? I've seen metal clips available, if they are any better at keeping joints together?

I am pretty sure you won’t find anyone stating their glassware will take anything greater than 1atm. That puts it into pressure vessel territory.

In the worst case I could but the flask at the bottom of a swimming pool, with 5 meters depth that would allow me to put 0.5 bar of additional pressure inside, all things kept equal. I'd just have to figure out the +5 meter long glass joints out of it... :S  Maybe not worth the effort for just +0.5 bar.

*put the flask at the bottom

Yes I thought of assemblies that disconnect without breaking. Metal clips are not strong enough to counter that. If they were made stronger, they themselves would break the glass when without pressure.

Most chemists who need to do reactions at pressure use specialist vessels designed to do so not standard glassware which very much isn't designed to do so. Large volume pressure vessels will usually be made of metal (which is far better under pressure than glass). Or, perhaps, internally glass coated metal vessels). There are some specialist thick-walled glass vessels designed to cope with some modest pressure, but they will usually be fitted with specialist pressure seals: standard glass joints don't work under internal pressure.

Oh, and a good generalisation on shape is that the more pressure the "rounder" the shape needs to be. Vessels with flat portions or sharp edges will be very poor under pressure.

@matt_black Thanks! I had actually been wondering whether a cylinder or a spherical container would take most pressure. Somewhat instinctively, it was clear to me that both would hold more pressure than an erlenmeyer but I failed to rationalise about the angle.

Put the flask in a metal container, with a large (at least three times the volume of the flask, for 3 atm pressure) bladder or sylphon (metal bellows) in the supply to the inner flask. Pressurize the outer metal container with air or inert gas.
The absolute pressure in the reaction vessel should be the same as the outer vessel (unless the reactants outgas rapidly), but there would be no differential pressure difference, ergo, no additional stress on the glass.
Or just conduct the whole experiment in a hyperbaric chamber... not a wimpy 0.1 atm plastic bubble, but something like this 10 atm unit, with margin for error. ;-)

This sounds like the way to go. I'll just have to find the equipment (not the hyperbaric chamber though, I can't afford that). Do you know of something affordable with connections going out of the chamber or do I have to be able to put my entire setup inside of it? (part of my setup will be under vacuum, so the chamber would put additional stress on it, though it seems glass resists better that way around)

The chamber looks really cool ^^ I want one! I can't afford it, but just out of  curiosity do you have any idea of the order of pricing of such a chamber? Makes carrying out pressurised stuff so much more simple.

Sorry... the atmosphere in marketing is toxic, and they're under pressure not to release that info ;-)

:D You made me laugh


text:
Same Activation Energy for Forward and Backward Reactions
While reading about equilibrium, I read that the equilibrium constant, $K_\mathrm{c}$, changes with temperature due to the forward and backward reactions having different activation energies. Is it possible for a reaction to have the same activation energy for the forward and backward reactions, such that $K_\mathrm{c}$ does not change when temperature changes? If so, please provide an example.

The van't Hoff equation describes how the equilibrium constant changes with temperature:
$$\ln \frac{K_2}{K_1} = −\Delta H_R^\circ (\frac{1}{T_2} − \frac{1}{T_1})$$
So if the reaction enthalpy is zero for a given temperature interval, the equilibrium constant will not change. This also means that the activation energy in either direction is the same, at least if entropy is not considered (the typical diagram explaining activation energy is for a single molecule with potential energy as the y-axis, enthalpy as the difference in energy between reactant and product, and activation energies as the difference between reactant or product and the activated complex).

If so, please provide an example.

Take the transfer of protons across a membrane. Equilibrium is reached when the concentrations on either side are equal. A change in temperature will not change the equilibrium constant (which is 1, with a reaction enthalpy of zero).
Another example would be a reaction that turns one enantiomer into the other, e.g. D-alanine to L-alanine. In the absence of chiral co-solvents or solvents or other unusual circumstances, again K = 1 and enthalpy = zero over a range of temperatures.

Nice answer (+1)! I was thinking of something like hydroxide ion taking a proton from water, yielding water and a hydroxide ion, but that is too rinky dink!

@EdV If you want to do a self-exchange reaction such as that one, you have to used isotopic labeling to keep reactant and product apart.

Sure, but if you simply do not care, then both sides are the same and we know it must happen, as can be proven with isotopic labeling. But this is just too rinky dink, in any event!


text:
Remineralization of ro water
Hello we are trying to prepare a solvent based solution of different kinds of minerals which can be added to reverse osmosis permeate water so that some of the essential minerals can be added back to the water. Do anyone knows which solvent we should use?

For future reference: be more specific with your questions. This one, with «different kinds of minerals», «some of the essential minerals», does not qualifies for such.

You really want to add a solvent back into drinking water???

I think water is the best solvent, assuming you want to remineralise the RO water for drinking it. Your solution might become a slurry depending on the required volume. Have you considered simply using a solid mix of these minerals? A solid mix would allow for a more homogeneous mix of different minerals than a low volume liquid.


text:
Why does hybridization produce a more stable structure?
I am aware of the fact that hybridization does not ALWAYS produce a stable structure (example: PH3)
However, my question is, what makes hybrid orbitals so effective in forming stable bonds? Although exceptions like PH3 do exist, what makes hybridization the best option for stability most of the time?
Is it due to the fact that the unequal lobes formed ensure greater overlapping and hence more stability?

It seems that the A to this Q https://chemistry.stackexchange.com/q/2769/48509 is what you are looking for

Atomic orbitals only exist in single atoms. If you want to make a bond with it, you need to have most of the electron density between the two aroms. That's why the shape of the MO is different from AOs, and "hybridisation" is only a way to describe the shape as a mixture of AO's shapes.


text:
Cathode ray discharge tube
I am learning about 'cathode ray discharge tube'. I know when electrons are given out from cathode, they travel to anode, because anode being positively charged attract those electrons and continue to accelerate those electrons. When the electron hit any of the gas atom in tube, it helps to knock out electron from it and therefore, electron knocked also start moving towards anode. But i think that one stage would come when there are no electrons left with gas atoms those can be knocked out and when electron hits the gas ions, it would just slow down the electrons(emitted by cathode) and eventually leading to no ray at the end.

Electrons do recombine with positive ions rather quickly. Somewhat similar scenario to the one you've described occurs in cathode dark space. Note though that stripping off all electrons (and often even more than just a few) requires tremendous energies and extreme conditions, e.g. deep inside stellar matter.

@andselisk. or just a very good vacuum, with just a few hydrogen atoms in the CRT, perhaps <10^-14 torr. Not many atoms to ionize...

A CRT is nowhere near 1E-14 torr. In the 1E-5 or 1E-6 perhaps if lucky.

Why it involves knocking out and leading to chain. Why not the electrons travel directly from cathode to anode?

The behaviour of electrons in a cathode ray tube depends on the amount of gas in the tube. When there is lots of gas there are lots of collisions and what you observe is very different from when the tube is a hard vacuum and most electrons go straight from cathode to anode.

For a cathode ray tube (old-style TV), you want the electrons traveling collision-free in a straight line from the cathode to the screen (anode), where they generate light hitting a phosphorescent layer. For a neon light, you want the electron hitting gas particles so that the entire tube glows.
This is further explained here: Why does the cathode ray tube only start glowing at low pressures?

Then for cathode ray tube don't we need any gas? As the path of electrons would remain straight only if they don't hit other gas atoms. Suppose after hitting too they remained in the straight path, then what's the use of gas here as the electrons can directly reach from one end (cathode) to the other end (anode).

Cathode rays are electrons emitted from the cathode. The discharge is from gas collisions. In a TV tube, there's no gas involved, and the electrons zoom past the anode to hit the screen (they are then collected by other anodes). If they are in a tube with gas, they can hit gas molecules and light up the tube as they head toward the anode (no screen).


text:
Infinite Increase in Entropy when Energy added to Absolute Zero
My textbook states the following:

If a system were at absolute zero, an additional small amount of heat energy would lead to an infinite increase in entropy. Such a state is impossible. Absolute zero can never be achieved.

It also provides the equation:
$$ \Delta S_\text{surroundings} = \frac{-\Delta H_\text{system}}{T}$$ Where $T$ is given in kelvin.
From the statement, will the entropy of the system or surroundings increase; from what I can deduce, I would say the entropy increase would be within the system as the surroundings is losing energy, but I am not sure.
The second part of my question is, why would this lead to an infinite increase in entropy? Please provide a comprehensible mathematical explanation and also, preferably, an analogy.
Textbook: Pearson Baccalaureate: Higher Level Chemistry, 2nd Edition. By Catrin Brown and Mike Ford Pages: 254-255

The textbook is referring to the entropy change of the system.  While the textbook is correct that absolute zero can never be attained, its statement that the entropy change is infinite is wrong. The authors' rationale for thinking it is infinite likely stems from a misinterpretation of the definition of entropy change:
$dS = \frac{\text{đ}q_{rev}}{T}$
where $\text{đ}q_{rev}$ is the reversible heat flow into the system.
Let's start with a system at $0 K$, and then warm it to a temperature $T = T'$.  Then:
$\Delta S =\int_{0}^{T'} dS= \int_{0}^{T'}\frac{\text{đ}q_{rev}}{T}$
Let's consider what happens when we flow the first, infinitesimal amount of heat, $\text{đ}q$, into the system.  And to make the calculation easy, let's assume the heat flow is reversible (it doesn't have to be; if it weren't, we'd just need to find a reversible process that gets us to the same final state).  As soon as we do this, the temperature is no longer zero! And once the temperature rises above zero, the singularity disappears and we no longer have a concern about infinite entropy change.
But, you may ask, what about the mathematics right at the very beginning, when the temperature is indeed zero?  Here, to properly calculate the entropy change, we need to use a limit, recognizing that both $T$ and $\text{đ}q$ are approaching zero.  The easiest way to understand this is to consider the Debye $T^3$ law, which models the constant-volume heat capacity of solids as they approach absolute zero:
$C_v = C_v(T) = k T^3$, where $k$ is a constant, and where I've written "$C_v(T)$" as an explicit indicator that $C_v$ is temperature-dependent.
Since, for a constant-volume reversible process, $\text{đ}q_{rev} = C_v(T) dT$, we have:
$\Delta S= \int_{0}^{T'}\frac{\text{đ}q_{rev}}{T} = \int_{0}^{T'}\frac{C_v(T)}{T} dT = \int_{0}^{T'}\frac{k T^3}{T} dT=\int_{0}^{T'}k T^2 dT$
I.e., in evaluating what happens to $dS = \frac{\text{đ}q_{rev}}{T}$ as $T \rightarrow  0$, it is necessary to consider what is happening to both the numerator and the denominator. The textbook authors' mistake was in not understanding this.  Clearly, once we find a reasonable functional form for $\text{đ}q$ in the limit as $T \rightarrow  0$, the singularity disappears.
Think of it this way (borrowing from one of my comments): At constant volume, $dS=\frac{C_v}{T} dT$. As you approach absolute zero, it's not just $T$ that's going to $0$; $C_v$ is going to $0$ as well. And since $C_v$ is going to $0$ faster than $T$ is going to $0$, $\frac{C_v}{T}$ goes to $0$ rather than infinity (for the same reason that $\frac{x^3}{x}$ goes to $0$ rather than infinity as $x$ goes to $0$). I.e., the mistake would be in assuming that $C_v$  is a constant; it's not.
Here's another way we can understand that the textbook's statement is wrong:  Thermodynamics allows us, in principle, to determine absolute entropies for any substance.  The absolute entropy is given by integrating the entropy change from absolute zero to whatever temperature the substance is at:
$\text {Absolute entropy at } T' \equiv S(T') =\Delta S_{0 \rightarrow T'}= \int_{0}^{T'}\frac{\text{đ}q_{rev}}{T}$
If the entropy change between $0 K$ and any higher temperature were infinite, the absolute entropies of all real substances would likewise also be infinite! [Unless the substance were at absolute zero, which is unatainable.]
Here's a link showing how standard molar entropies might be calculated (I don't know if the procedure described here is what is used to determine the official CODATA values):
https://www2.stetson.edu/~wgrubbs/datadriven/entropyaluminumoxide/entropyal2o3wtg.html.
Essentially, it mentions using the Debye extrapolation up to ~$15 K$, and then using a differential scanning calorimeter to determine the reversible heat flow between $15 K$ and $298.15 K$

Your answer seems great, but the mathematics and explanations are way out of my scope of my comprehension. From what I can understand though, your explanation is what I was looking for. However, I will have to revisit it in a few months, once I have learned integral calculus.

It helps to understand integral calculus, but you can get the essence of my argument without it if you can understand limits. Think of it this way: At constant volume,  dS=Cv/T dT.  As you approach absolute zero, it's not just T that's going to $0$; Cv is going to $0$ as well.  And since Cv is going to $0$ faster than T is going to $0$,  Cv/T goes to $0$ rather than infinity (for the same reason that x^3/x goes to $0$ rather than infinity as x goes to zero).  I.e., the mistake would be in assuming that Cv is a constant; it's not.

I've never seen a differentiation operator denoted with a dyet (đ). Is there a certain meaning attached to this symbol besides using it for derivation? If you would like to use an upright operator, you probably want to use \mathrm d or \mathrm{d} instead.

@andselisk  As I'm sure you know, q and w are not state functions; rather, they are path-dependent. Hence their differential forms are inexact differentials, rather than exact differentials.  The dyet is used to indicate that (though typically it is an angled slash rather than a horizontal bar; don't know how to make that with LaTeX).  Since they are inexact differentials, any integration with respect to them must be a path integral.   While not all physical chemistry/thermodynamics texts use this notation, several well-known ones do, including Callen, Reif, Castellan, and Engel & Reid.

@theorist A-ha, I see, thank you for the explanation. It's just that the symbol looks a bit off due to the MathJax font support. I guess you may leave it as is, or adapt the representation from "d-crossbar" to "d-hat" with \bar{\mathrm d} $(\bar{\mathrm d}).$ Nice answer!

@theorist That makes more sense to me now and thank you for your eloquent answers!

@andselisk On a side note - some texts (such as Rock's Chemical Thermodynamics) use lower-case delta ($\delta$) for inexact differentials. I find that easier to use in MathJax, although it seems it is sometimes confused with partial derivative notation ($\partial$) by students unfamiliar with inexact differentials.


text:
Does equality of molarity or molality determine constancy of boiling (or freezing) points of different solutions?
I encountered a question among papers for JEE (2005) that demands completing a statement:

Equimolar solutions in the same solvent have _______

There are four options and the correct answers are

same boiling point

and

same freezing point

I don't think this is correct as the change in boiling and freezing point depends on molality and not molarity. Hence, according to the options, the one saying different boiling and freezing points seems to be correct to me.
However, such exams very rarely have mistakes which are not corrected later. Please clarify.

Is this in a question block about ideal solutions? Because at higher concentrations, both answers would even be wrong if you ignored the "lality"/"larity" mistake.

@Karl it's about ideal solutions. I forgot to mention it

Equimolar solutions are also equimolal !

@Maurice not at all! Molarity is a volumetric property, therefore depends on density.

@Buck Thorn. Equimolar solutions do not have the same molality. But compared to one another, their molalities are equal.

@Maurice I don't understand your statement, the second sentence seems inconsistent with the first. In any case, for dilute solutions (within the scope of the problem) the units are interchangeable.

At low concentrations for which the ideal bp elevation/fp depression expressions usually apply, molarity is linearly proportional to molality, therefore the statements are equivalent.
This webpage explains nicely why low concentrations are important:

Raoult's law only works for low concentration solutions. Why? Well, in order for our approximation to work, the interactions between the solute and solvent molecules must be nearly identical. If the interactions are stronger, then the heat of vaporization of the solvent will change and thus our whole approximation falls apart. Since we know that intermolecular forces vary greatly between molecules, the affect of those forces must be kept at a minimum by keeping the solute concentration real low.

In any case (as explained by additional statements in the abovementioned website, which discusses dissociation of ionic solutes), the statement made in the answer to your exam would not be true generally even if molality instead of molarity was mentioned.
The point of the question seems to be that a student should associate a universal response of the solvent that is independent of solute with colligative properties as described using Raoult's law, which is behavior most often approximated by dilute solutions (the statement that Raoult's law only applies to dilute solutions happens to be false, btw), and that in this realm of application molarity is approx linearly proportional to molality.

Hm, the question never says that the cocentration is low.

@Karl No, some details seem implicit - or at the very least, the point seems to be that a student should associate a universal response that is independent of solute with colligative properties as described using Raoult's law, which as my answer explains, is behavior most often approximated by dilute solutions (the statement that Raoult's law only applies to dilute solutions happens to be false, but then the point is not to write an essay on solution chemistry, just to clarify the confusion), and that in this realm of application molarity is approx linearly proportional to molality.

@BuckThorn Very well said


text:
What is the most convenient way to prepare ferrous oxide (FeO) in the laboratory?
The Wikipedia page for ferrous oxide states that $\ce{FeO}$ can be prepared by the thermal decomposition of iron(II) oxalate, with the following reaction:
$$\ce{FeC2O4 → FeO + CO2 + CO}$$
And that the procedure is conducted under an inert atmosphere to avoid the formation of ferric oxide.
Sicius' Eisengruppe: Elemente der achten Nebengruppe: Eine Reise durch das Periodensystem states that $\ce{FeC2O4}$ should "slowly" be heated in a vacuum and then rapidly cooled down.
Both of these resources, however, omit critical information as to the preparation process. These questions remain:

At which temperature does this reaction really start occurring, and what is the ideal temperature range to carry out this thermal decomposition?
How "fast" should it be cooled down? And what are appropriate methods to do so?

I'd be very thankful for any additional information on the topic.

If you were an undergrad student, this would qualify as "homework": literature research. Are you? ;-)

In appears that the synthetic procedure for preparation of iron(II) oxide from iron(II) oxalate has been first described by Günther et al. [1] and subsequently summarized in Brauer's Handbook of Preparative Inorganic Chemistry [2, p. 1497]:

Iron (II) Oxide
I.
$$\ce{\underset{143.8}{FeC2O4} = \underset{71.8}{FeO} + \underset{28.0}{CO} + \underset{44.0}{CO2}}$$
Thermal decomposition of $\ce{FeC2O4}$ yields pure $\ce{FeO}$ only under specific conditions.
  The decomposition is carried out in a quartz vessel (Fig. 332) whose lower section is kept at $\pu{850 °C}$ by means of an electric furnace.
  The joint is surrounded by a water-cooled lead coil or a rubber hose.
  The nascent gases should be removed as quickly as possible; for this reason, the reactor is connected to two parallel mercury pumps and a good forepump; the gas is carried into two liquid-nitrogen-cooled traps containing activated charcoal.

  Fig. 332. Preparation of iron (II) oxide.
The starting $\ce{FeC2O4}$ $(\pu{0.5 - 0.8 g.})$ is placed in the small bulb above the quartz vessel, and the water of crystallization is completely vaporized by heating in vacuum for 12 hours at $\pu{200 °C}.$
  The bulb is turned in the joint, and the $\ce{FeC2O4}$ drops into the heated lower section of the reactor where it is rapidly decomposed to $\ce{FeO},$ $\ce{CO}$ and $\ce{CO2}$ (the decomposition is complete in about 20 seconds). The product $\ce{FeO}$ is retained by a quartz wool plug, which must be loose enough to prevent a buildup of pressure during the decomposition.
The furnace is now removed and the hot quartz tube is chilled as rapidly as possible in cold water, since $\ce{FeO}$ is unstable in the range of $\pu{300-560 °C}$ and decomposes according to:
$$\ce{4 FeO = Fe3O4 + Fe}$$
(this decomposition proceeds most rapidly at about $\pu{480 °C},$ but
  ceases below $\pu{300 °C}).$
  The above procedure yields a jet-black product, readily soluble in dilute acids; it is rapidly oxidized in air, but does not ignite.

Alternatively, iron(III) oxide can be reduced by iron to iron(II) oxide [2, p. 1498]:

II. The preparation from stoichiometric quantities of commercial $\ce{Fe2O3}$ and reduced iron can also be recommended.
  The mixture and a few drops of water are sealed into a preevacuated quartz tube, heated for about three days at $\pu{900 °C},$ and quenched in cold water.

$$\ce{Fe + Fe2O3 ->[\pu{900 °C}] 3 FeO}$$

Iron(II) oxide can also be produced from iron(III) oxide reduction by carbon monoxide at the temperature range between $\pu{500 °C}$ and $\pu{600 °C}$ (adapted from [3, p. 415]):
$$
\begin{align}
\ce{3 Fe2O3 + CO &->[\pu{400 °C}] 2 Fe3O4 + CO2} \\
\ce{Fe2O3 + CO &->[\pu{500-600 °C}] 2 FeO + CO2} \\
\ce{Fe2O3 + 3 CO &->[\pu{700 °C}] 2 Fe + 3 CO2}
\end{align}
$$
as well as by thermal decomposition of magnetite [3, p. 415]:
$$\ce{2 Fe3O4 ->[>\pu{1538 °C}] 6 FeO + O2}$$
References

Günther, P. L.; Rehaag, H. Über Die Thermische Zersetzung von Oxalaten II. Mitteilung. Darstellung von Reinem Ferrooxyd. Z. Anorg. Allg. Chem. 1939, 243 (1), 60–68. https://doi.org/10/cqzgk8. (in German)
Handbook of Preparative Inorganic Chemistry, 2nd ed.; Brauer, G., Ed.; Academic Press: New York; London, 1965; Vol. 2. (Archive.org)
R. A. Lidin, V. A. Molochko, and L. L. Andreeva, Reactivity of Inorganic Substances, 3rd ed.; Khimia: Moscow, 2000. (in Russian)

The paper linked below indicates that the proper temperature is north of  535°C. This decomposition should be carried out with great caution though as some of the products might enflame, or even initiate a thermite reaction if carried out in the presence of Aluminum (as in the experimental setup used for the paper hereafter).

Source: https://pubs.rsc.org/en/content/articlelanding/2006/jm/b514565a#!divAbstract

The CO would give one pause, unless this was being done with proper facilities and safety equipment.

Agreed! Maybe with a way to vacuum extract the gases from the container once the decomposition has proceeded


text:
How can some enzymes work faster than the diffusion rates of the reactants allow?
Quoting Wikipedia: "Some enzymes operate with kinetics which are faster than diffusion rates, which would seem to be impossible." Which are those enzymes and how can they be so fast?
One example is catalase in which peroxide is catalyzed faster than the diffusion rate of peroxide allows. Lionel Milgrom discusses this case in Water Journal no. 7.

It doesn't work faster than diffusion. It's diffusion-limited. However, if diffusion rates were faster, the enzyme would also work faster. The bottom line is that you are rate-limited by diffusion in solution.

The conceptually easiest case is that of a positively charged active site with a negatively charged substrate. The substrate (i.e. reactant) enters the active site with kinetics that are faster than diffusion because there is a long-range electrostatic interaction.
Catalase works near the diffusion limit. The document the OP cites is a hypothesis, without experimental data. It suggests that catalase acts at a distance to catalyze reactions outside of its active site. I would not worry about this hypothesis until there is some experimental evidence supporting it.
Wikipedia has a list of 9 examples of diffusion limited enzymes:

Acetylcholinesterase
β-lactamase
Catalase
Carbonic anhydrase
Carbon monoxide dehydrogenase
Cytochrome c peroxidase
Fumarase
Superoxide dismutase
Triosephosphate isomerase

Some of them have neutral substrates, so the electrostatic attraction would not apply to them. There are other hypothetical mechanisms, though, like weak binding sites on the surface of the enzyme that increase the local concentration of substrate, for example.

I think enzyme kinetics in the case of the OP would refer to overall turnover rate, not just to binding rate. So whatever the mechanism of association, what's fast is the combination of binding, reaction in the active site, and release.

@BuckThorn If it is diffusion-limited, the binding step is limiting. If it is faster than the theoretical diffusion limit, the puzzling thing is how it can bind faster than that limit. It is amazing in any case that practically every binding event leads to a reaction, very different from what happens in the absence of a catalyst.

Regarding enzymes seemingly beating the diffusion limit,
there is an additional factor to consider, and that is stirring!
Can enzymes stir a solution?
The answer at first sight is no, but then enzymes are large, complex and may have inward and outward channels. In cases were the active site design is such, the substrate may be forced to enter from one side to form a productive complex, be converted and leave on a different side, thus producing a flow. If the reaction free energy is favorable and if oriented properly, that flow would literally impulse the protein molecule in some direction and could stir the solution nearby.
In catalases, the active site design creates a bottleneck and the site seems designed to filter out water at the inward channel, leave peroxide in, and the reaction's free energy change is large and negative (Dominguez et al., 2010 10.1016/j.abb.2010.05.017). In theory, this may create a net flow and propel the protein in some direction.
The effect however, does not seem to have been tested experimentally.

Are there means to detect such motion in the small volume that they would occur in?

May be difficult to see it in a single-molecule experiment. However, you could place a small drop of concentrated enzyme in an agar plate, containing a small concentration of hydrogen peroxide. Then compare the rate of protein diffusion in the plate, against, the diffusion of a protein of similar size, with no catalase activity (i.e. BSA).  It might be possible to see a systematic reproducible difference, because the enzyme would consume the local peroxide and a gradient would be created. The difficult part is to measure protein diffusion.

May be difficult to see it in a single-molecule experiment. However, you could place a small drop of concentrated enzyme in an agar plate, containing a small concentration of hydrogen peroxide. Then compare the rate of protein diffusion in the plate, against, the diffusion of a protein of similar size, with no catalase activity (i.e. BSA). It might be possible to see a systematic reproducible difference, because the enzyme would consume the local peroxide and a gradient would be created.
The difficult part is to measure protein diffusion. You could stain the agar with coomasie blue, but that would mean a lot a agar replicates. A fluorescent catalase may help, but the label would have to withstand peroxidation (most probes would loose their flourescence). That is the part I have not figured out how to tackle-Any ideas?


text:
Long term composition of matter
If I have an igneous rock shown by geologic study to be 10,000 years old, approximately what percentage of the atoms (or protons, etc) in the rock might be the same ones that were in it 10,000 years ago?  Is it reasonable to even think of a "particular", individual atom?  Is it the "formation" of particles that persists over time, or is it the actual "matter"?

It might be worth reading up about radioactive decay.

The radioactive atoms in an igneous rock would be a very very small fraction of all of the atoms. So 99.99+% of the atoms would be the same as when the rock was formed.

Short answer: 10 000 years in geological scale is a very very short time, so practically nothing happens and 100% of atoms are the same. A little longer answer: There are two main mechanisms that can change individual atoms that pops in my mind. One is radioactive decay (see other answers), the other is a chemical exchange of an atom to a similar or same atom. This later mechanism leads to the transformation of rocks or the formation of fossils. Both are very slow (see the short answer), and both are very much dependent on the chemical composition of the rock and the environment.

"Is it reasonable to even think of a "particular" individual atom?" No, the atoms are unlikely to change position in typical solids unless subjected to extreme conditions of temperature and pressure, or to chemical processes such as dissolution. "Is it the "formation" of particles that persists over time, or is it the actual "matter"" This is a very unclear question!

To answer your question, we can use the idea of radioactive decay;
According to the IB Physics Book from Oxford University, radioactive decay is "a naturally occurring process in which the nucleus of an unstable atom will spontaneously change into a different nuclear configuration by the emission of combinations of alpha particles, beta particles, and gamma radiation. Alpha particles are simply a Helium atom while beta particles consist of either electrons or positrons (the antiparticle of the electron).
Radioactive decay is a random process and therefore it is impossible to predict which exact nucleus may decay next. However, with a large sample of nuclei, say of your igneous rock, it is highly probable that in a given amount of time, a predictable amount of nuclei will decay, even if we cannot predict which ones exactly. Hence we have the idea of half-life, which is the amount of time it takes for the half of the initial nuclei of a sample to decay.

approximately what percentage of the atoms (or protons, etc) in the rock might be the same ones that were in it 10,000 years ago?

It depends on what the igneous rock consists of; if it is made out of ununoctium-294, an infinitesimal amount of it will remain, as the half-life of it is 5 milliseconds. On the other hand, if the rock consists of Uranium-238, after 10 thousand years the composition of the atomic composition of the rock will have barely changed, as Uranium-238 has a half-life of 4500000000 years. So to answer your question, you would need to look at the half-life of your igneous rock. For example, if your igneous rock has a half life of 10 thousand years, then only 50% of the nuclei would be what they were 10 thousand years ago. However, it would also depend on how much of the rock is radioactive, as MaxW mentioned.

Is it reasonable to even think of a "particular", individual atom?

As already mentioned, it is difficult to predict which particular nucleus will decay next, therefore, looking at the substance as whole facilitates predicting how much will decay in a certain period of time.

Is it the "formation" of particles that persists over time, or is it the actual "matter"?

I do not quite understand this question. Nonetheless, I would say that it is the matter that persists over time, as the matter consists of these particles. In the photo provided below, you can see that the nuclei are decaying, but they are not decaying row for row, yet more in a random manner. Hence, I would not consider it to be the formation of particles that consists over time, but the matter as a whole.

I studied radioactivity in physics and not in chemistry, even though I would rationally assume that it is actually chemistry. I am mentioning this as you might have more luck looking at radioactivity in the physics stack if the chemistry stack is not too helpful. Hope I could help! :)

Thanks for your comments -- the question came up in reference to the fact that atoms being indistinguishable,  any number of them could be replaced and the object would be the same.


text:
Cycloaddition vs Electrocyclic Pathway
Is there a molecular orbital/Woodward-Hoffmann explanation as to why Diels-Alder reactants don't commonly undergo electrocyclization instead of cycloaddition?

The Woodward-Hoffmann rules only tell you whether a reaction is thermally allowed or not, they do not give any indication of the relative rates of multiple allowed reactions.
I guess you're asking why the 4π component in a [4+2]-cycloaddition doesnt undergo a conrotatory 4π electrocyclic ring-closure? This is plausible, but I suppose even if it did happen the cyclobutene product is highly strained so the equilibrium would lie in favour of the ring-opened diene anyway.


text:
How is serine ionized in serine proteases?
I read that serine isn't strongly acidic so how is it that in serine proteases it is able to have its pKa lowered by the catalytic triad?

Could you edit to show a bit more what you already found out. For example, what is the pKa of serine in water, and what is it in the context of a serine protease? Also, how low do you think the pKa has to be in order for it to be a good catalyst?


text:
Why do impacting electrons have a low probability of causing ionization?
In a section on ionization efficiency and ionization cross section, my mass spectrometry textbook, Mass Spectrometry by Jürgen Gross [1, pp. 38–39], says the following:

The ionization energy represents the absolute minimum energy required for ionization of the neutral concerned. This means in turn that in order to effect ionization, the impacting electrons need to carry at least this amount of energy. If this energy were then to be quantitatively transferred during the collision, ionization would take place. Obviously, such an event is of rather low probability and therefore, the ionization efficiency is close to zero with electrons carrying just the IE of the pertinent neutral. However, a slight increase in electron energy brings about a steady increase in ionization efficiency.

Why is such an event of "rather low probability"? It seems to me that if the impacting electrons are carrying at least the necessary ionization energy, as the author suggests, then shouldn't ionization occur (be a certainty or, at least, a high probability event)?
I would greatly appreciate it if people could please take the time to clarify this.
References

Gross, J. H. Mass Spectrometry: A Textbook, 3rd ed.; Springer International Publishing: Cham, Switzerland, 2017. ISBN 978-3-319-54397-0.

The electrons in the atoms of the molecule have a small cross-section, hence most electrons will just pass through the molecule without causing ionization.

Ionization has a low probability of occurring if the energy of the incoming electron is exactly equal to the ionization energy. Using a classical analogy, if  ionization occurred the incoming electron would be left with no energy, ie, stationary. The incoming, now stationary, electron would not scatter away but rather remain trapped by the nuclear attraction, such that no net ionization would have occurred. It would also imply (classically) a colinear collision with the departing electron, a seemingly unlikely event. On the other hand, if the incoming electron has a significant energy surplus, such that it has sufficient energy to escape the nuclear attraction (scatter), then ionization is more likely to occur.
There are obviously more accurate non-classical ways to describe the process, but the analogy illustrates the problem. There are also intermediate scenarios, such as those exploited in Auger spectroscopy.

In a quantum context, there just aren't that many final scattering states to choose from, so the overlap integral is really small (basically the same thing - the collision has to be just right or you get nothing).


text:
Neutralize the smell of alcohol in liquid chalk
i'm trying to mix my own liquid chalk, which is a simple mixture of chalk and a high percentage alcohol in a ~2 part chalk 1 part alcohol relation.
Unfortunately the smell is rather intensive and I would like to find a way to neutralize or cover it up.
Any suggestions?
I do not have a chemical backround.

Chemically, chalk is CaCO3. But blackboard chalk is mostly CaSO4.2H2O. If you want to mark something, the aim is to lay a coating of a certain (dry) thickness so as to be visible.
One way to reduce the odor is to reduce the amount of alcohol used by 1) making the paste richer in the chalk by a) just adding more chalk (but that might increase the viscosity too much), so b) grind the paste with a mortar and pestle (this may thin out the paste by causing a better variation in particle size and packing), or c) add some of whichever "chalk" you are not using (i.e., if you are using CaCO3, add some blackboard chalk and grind).
Or, 2) dilute the alcohol with a little water. You probably use alcohol to get a fast dry time, so may not be able to use much water. There are other solvents that have less odor (like heptane), but have other objections. Isopropanol has a higher boiling point (less odor, slower drying, but not much slower) than ethanol (but you are probably not using ethanol).
Or, 3) add a little TiO2. This will dramatically increase the visibility of thinner films, allowing application of much less paste and reducing the amount of alcohol exposure.
Or, 4) add a dispersant to reduce viscosity. Assuming you do not have access to many chemicals, I would suggest a little trisodium phosphate. Or a tiny amount of dishwashing detergent (be careful not to stir too vigorously so as to incorporate air bubbles); this may help wet particles or displace tiny air in pockets between particles. Then you can add more solid.
5) Chemical "masks" are available that claim to cover up obnoxious odors. In my experience they do not cover, but only add extra odor. But adding some delightful fragrance could distract from the alcohol odor to a more pleasing experience. Commercial additives I have used are coconut and cherry - both quite nice. You don't need much. Some food flavorings have enough odor to be useful here. Some people find acetone to be more pleasant than alcohol - a little acetone could change the odor to be more acceptable.
6) Slowing the evaporation by adding wax or high molecular weight alcohols (octadecanol) might reduce the odor while extending the dry time, if that is acceptable. Octadecanol is used to reduce the evaporation rate of water, and may work with alcohol; wax is often used with solvents.
Other than that, I can't think of anything.

Im using it to dry the skin on my hands for rock climbing. It sticks well and as far as i know leaves fewer traces than just dry chalk or Magnesia. So I'll try Option1, 4 and 5. Thanks!

Aaakkk! Na3PO4 is pretty alkaline, might dry your skin too much. Washing soda (sodium carbonate) is less alkaline, but would still be a skin dryer. Surfactant residue might become slippery if your hands sweat. Have you tried rosin like ball players use? Why not a paste of water and chalk - then wave your hands around a bit?


text:
Why do carbonyl groups lower antibonding orbital energies of conjugated systems?
In the chapter on conjugation in Organic Chemistry by Clayden, Greeves, Warren and Wothers, there's a brief discussion of pigments with conjugated systems that includes the following paragraph (first edition, p. 171; emphasis mine):

Transitions from bonding to antibonding π orbitals are called π → π* transitions. A much smaller energy gap is available if we use electrons in a nonbonding orbital as the electrons start off much higher in energy and can be promoted to low-lying antibonding π orbitals. We call these transitions n → π*, where the ‘n’ stands for nonbonding. It is easy to find coloured compounds throughout the whole range of wavelengths by this means. The colour of blue jeans comes from the pigment indigo. The two nitrogen atoms provide the lone pairs that can be excited into the π* orbitals of the rest of the molecule. These are low in energy because of the two carbonyl groups. Yellow light is absorbed by this pigment and indigo-blue light transmitted.

The bolded statement is not elaborated on or, as far as I can tell, addressed directly in the rest of the text. I took "these" to mean "π* orbitals" rather than "lone pairs"; my best attempt at deducing an explanation is something like this:

The bonding orbitals in the carbonyl group are polarized towards the O atom, exposing a positive charge on C and shifting electron density in the π* orbitals toward C. These two effects combined mean that any electrons in π* experience a strong attraction to the carbonyl group, lowering the overall energy in π*.

Is this roughly correct?


text:
1H-NMR of vanillin details in Nilered video
The image below is from a Nilered video How to extract Vanillin from Vanilla Sugar, where he displays the $\ce{^1H}$-$\mathrm{NMR}$ of the final product created:

The labels A, F, & E make sense to me, but I'm confused about D and C - I feel they should be swapped. To my understanding, B and D share a more similar chemical environment (though not identical), specifically they are near a $\ce{C-O}$ bond, moving their chemical shift to left. Also, the NMR peaks labelled D display a doublet, suggesting it is next to a single proton. Though B, C, and D are all next to a single proton, C is the odd one out.
Is this a case of me missing something, or is the picture mislabeled?

Assignment looks correct to me; don't forget the effect of the carbonyl group. Not quite sure what you're trying to say with C being the 'odd one out'. It's hard to see any multiplet structure in the screenshot, especially when two of the protons are overlapping with each other. But ideally, C should be a doublet of doublets as it is coupled to both B and D. On the other hand, B and D should both be doublets, and the D doublet should have a larger separation (coupling constant) than the B doublet.

Thanks for the answer. A few follow up questions then: 1) why is C coupled to both B & D (and not vice versa)? Does it have something to do with the benzene ring? You mention the carbonyl group - but what effect does that have? It appears far away from D/C/B, no?

Hm, thinking more about your carbonyl comment: is it because it is a meta-directing group and will change the environment of D. Likewise, -OH is meta-directing and again will change the environment of D.
I still don't understand why C should be a doublet of doublets.

You may want to go over this with a teacher if you have one. There are several problems with your comments. (1) B and D are both coupled to C, that's exactly why they appear as doublets. (2) Whether a group is o/p- or m-directing is not of any consequence here. They affect all protons on the ring, albeit not to the same extent (it depends on the group). (3) OH is most certainly not m-directing.

Right, I meant ortho-para for -OH. AFAIU, coupling occurs between protons on neighbouring carbons, but B is not neighbouring to C. I'm back to square one: what effect does the carbonyl have on the NMR?

Hm, I think I need to know more about coupling in benzene rings. This should help any future readers: https://www.sas.upenn.edu/~genette/NMRcoupling.pdf

There are spectra catalogues about IR and NMR spectroscopy (e.g., Sadtler, Aldrich) and compilations like AIST's spectra collection about organic compounds which may guide your training / self study.  If you are aware that each model-based predictions may err to some degree, you may consider predictions by software, e.g., advanced versions of sketchers like ChemSketch, ChemDraw, Chemdoodle or on-line services like nmrshiftdb2, too.

"Normal" coupling is vicinal coupling across 3 bonds such as $\ce{H-C-C-H}$. For aromatics, ortho-coupling is of this type. But one can also get other couplings such as long range coupling, e.g., meta-coupling across 4 bonds. Long range coupling tends to be smaller than vicinal coupling (smaller $J$, the coupling constant). Thus C/D coupling is stronger than B/C coupling. And remember coupling must work both ways C to D, and D to C. The problem here is that B and C are overlapped so tough to see their multiplicities. C has higher shift than D due to position relative to the substituents.

[...] but I'm confused about D and C - I feel they should be swapped. To my understanding, B and D share a more similar chemical environment (though not identical), specifically they are near a C-O bond, moving their chemical shift left. Also, the NMR peaks labelled D display a doublet, suggesting it is next to a single proton. Though B, C and D are all next to a single proton, C is the odd one out.

First, in this statement, your assignment of "[...] B, C and D are all next to a single proton," is not correct. Clearly, B is not next to a single proton:

Now, after said that, your confusion is due to your ignorance of the electron withdrawing nature of aldehyde group $(\ce{CHO})$. That group is ortho to both $\ce{H_B}$ and $\ce{H_C}$  while meta to $\ce{H_D}$. Also, it is para to the electron donating $\ce{OH}$ group. Because they are para to each other, their effect may be optimized on each other. Regardless, $\ce{H_D}$ experiences significant electron density due to the electron donating $(\ce{OH})$ group at ortho position, while $\ce{H_C}$ (and $\ce{H_B}$) experiences significantly lesser electron density. When compare electron donating ability of $\ce{OH}$ group and $\ce{OCH3}$ group, it is safe to say that $\ce{OH}$ group has been the superior (compare $\sigma_{o,p}$ values of $\ce{OH}$ and $\ce{OCH3}$ groups). Since both $\ce{H_B}$ and $\ce{H_C}$ would experience significant electron withdrawing from $(\ce{CHO})$ group ortho to them in addition to experiencing aforementioned electron density from $(\ce{OCH3})$ group. Keep in mind that since $(\ce{CHO})$ group is meta to $\ce{H_D}$, it makes less effective on deshielding. Thus, based on these facts along, one can say safely that  $\ce{H_B}$ and $\ce{H_D}$ as well as $\ce{H_C}$ and $\ce{H_D}$ are significantly different chemically and magnetically. Therefore, their chemical shifts should be vastly different as assigned.
When compare the effects of substituents on $\ce{H_B}$ and $\ce{H_C}$, one can also say that they are very similar magnetically due to the effects of meta $\ce{CHO}$ group and ortho, para $\ce{OCH3}$ group (ortho to $\ce{H_B}$ and para to $\ce{H_C}$). The only difference is the $+I$ effect of $\ce{OCH3}$ group on each proton (with pata being the minimal). However, seemingly that difference is not been very significant in this case (again, compare $\sigma_{o}$ and $\sigma_{p}$ values of $\ce{OCH3}$ group for comparison). Thus, one can say that the chemical shifts of $\ce{H_B}$ and $\ce{H_C}$ should be very similar as assigned.
Note: The resonance of $\ce{H_D}$ should be a clear doublet since it does not have any meta-protons to have long-range couplings (usually $\pu{1-2 ppm}$). Yet, $\ce{H_B}$ and $\ce{H_C}$ are meta to each other and show multiplicity.


text:
Cloud-based molecular modeling software
I’m an IT Acquisitions and Licensing specialist researching well-supported cloud software alternatives to Fujitsu’s molecular modeling SCiGRESS. I am coming in with a tech background, but little to none in chemistry, so I’m seeking the expertise of users of chemistry software.
While my online research has yielded software with (purportedly) similar functionality to SCiGRESS, most of these seem to be of the traditional download and install varieties, with most search engine results for “molecular modeling cloud software” and “computational chemistry cloud software” detailing the benefits of performance in the cloud as an alternative to expensive, dedicated hardware rather than software that runs in the cloud to which one can subscribe. The exception to this trend is ChemCompute, but the combination of GAMESS and TINKER do not seem to fulfill all of the functionality of SCiGRESS.
As such, I’m putting the question forward to you chemistry professionals: are there any molecular modeling and/or computational chemistry software that have similar functionality to SCiGRESS, but being implemented as well-supported cloud software rather than requiring packaging, installation, and management?
Edit (19/07/30): The usage is to be in a two-year community-college environment, so lower undergraduate uses will be the norm. Still awaiting reply from professors to determine their exact needs for this software.
Edit (19/08/01): I've received a reply from one of the coordinators that they will poll all users and get back to me, but to expect a wait of at least two weeks.

The brief overview about Scrigress you indicate lists a wide range of computations possible, it would be helpful to know which of these are interesting for you / the clients you care for.  Consequently, assuming it could be around material's properties I would indicate Materials Project, https://materialsproject.org/.

I have reached out to the clients to clarify their primary use cases for the SCiGRESS software, and will edit my post with that information once they respond. That may take some time, however, since replies are notoriously slow during the summer.

Andrew Gilbert offers QChem access via IQmol (http://iqmol.org/) that can be accessed anywhere but it uses a small local program to interface with.  Time limits are short given the tutorial aspect of the application. QChem supports a wide variety of computational methods (http://www.q-chem.com/qchem-website/features4.htm).  While MM is not supported, QChem interfaces with CHARMM, a MM program that has a version that is free.  Not sure if this is what you're looking for.

This is a tough nut I guess. Back when this question was asked, [mattermodeling.se] did not exist. Now it is too old to migrate there. I see no harm in asking for an updated opinion there.

I have been developing a cloud platform for computational chemistry: https://calcus.cloud/ You can give it a try for free without registering.
The code is open-source (https://github.com/cyllab/CalcUS) and can be used as standalone as well. We deployed the platform to the cloud to make it more accessible and convenient, just like you describe.
If you are interested, get in contact with me (email on the website). I am open to adding new features to meet your needs.


text:
I am having difficulties with analysing steroids in GC/MS
Is there anyone here that has already analysed estrone and 17beta-estradiol on GC/MS (HP5-MS column and MSTFA as derivatizing agent)?
Is it necessary to make a mixture of other reagents with MSTFA or can I just use MSTFA in the process of derivatization? How long does that process take to happen?


text:
Which of the Following is Optically Active?

Which of the Following is Optically Active?

I understand that for a compound to be optically active, it needs to be chiral and not have any planes of symmetry. I know 1 is symmetric, but I am confused about 2 and 3.

The dotted line to HO means it is further away while the bold line to OH or HO means it is closer to you, both respective to the pentagon or hexagon. Perhaps this might help.

In the same way that you might show 1 is symmetric, do you think you could show either 2 or 3 to be symmetric?

I am quoting from IUPAC GOLD book$^1$ .

Meso-compound : A term for the achiral member(s) of a set of diastereoisomers which also includes one or more chiral members. For example:

There is a plane of symmetry in the examples given in IUPAC GOLD book (shown in colour) .

Molecule $\ce{3}$ also has a plane of symmetry as shown in the figure below. This makes it achiral.

Molecule $\ce{2}$ does not possess plane of symmetry , and therefore chiral.
In conclusion , molecule $\ce{3}$ is achiral and molecule $\ce{2}$ is chiral.
Refereneces
$^1$ : IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific Publications, Oxford (1997). Online version (2019-) created by S. J. Chalk. ISBN 0-9678550-9-8. https://doi.org/10.1351/goldbook.


text:
What is the maximum amount of medicine that could accumulate in the body if the compound has a half-life of 24 hours?

If a patient is prescribed 25 mg per day of a compound that has a half-life of roughly 24 hours, what is the maximum accumulated amount of the medicine that would build in the patients body?

1 + 1/2 + 1/4 + 1/8 + ... = 2

1st day: 25 mg
2nd day: (12.5 + 25) mg
3rd day: (6.25 + 12.5 + 25) mg
4th day: (3.125 + 6.25 + 12.5 + 25) mg
The amount get closer and closer to 50 mg but never exceeds it.

Out of curiosity, is there a mathematical formula that would represent this idea? Something like $$ \sum_{k=25mg}^mx_.5k$$

@doremi the formula you would want in general would be $$25\sum_{i=1}^{k} (\frac{1}{2})^{i-1}$$ where $k$ would be the number of days you take the drug. As the answers have stated, as $k\to \infty$, the sum approaches 2 and after multiplying by daily dose you get 50mg.


text:
Which is more basic: 1,2‐dihydropyridine or 1,4‐dihydropyridine?

Which compound is more basic of the two?

My thinking: In compound A, the lone pair of nitrogen are in resonance with two other double bonds. The resonance chain here seems to be longer as compared to in compound B. However, in compound B, the lone pairs of nitrogen seem to be more 'busy' because of the two double bonds on its either side,when compared to compound A (@YUSUFHASAN provided this second logic).
So, which of the two compounds should be more basic? (The main idea is that, the lesser 'busy' lone pair electrons of nitrogen means more 'free' electrons to pick up $\ce{H^+},$ which is what bases do.)
Do note that this question is supposed to be answered qualitatively. (By this I mean, not looking at tabular data from research papers or books!)

In compound $\ce{A}$ , lone pair on $\ce{N}$ is in conjugation with two $\ce pi$ bonds (as shown below). Therefore this lone pair is delocalized (resonance structures in the figure below) to a greater extent.
In  compound $\ce{B}$ , lone pair on $\ce{N}$ is in conjugation with a single $\ce pi$ bond , while the other pi bond is not in conjugation or is cross conjugated.
Since lone pair on $\ce{N}$ in compound $\ce{B}$ is less delocalized then in compound $\ce{A}$ , compound $\ce{B}$ is more basic then compound $\ce{A}$.

So what you're saying is that, cross-conjugation is of not much importance when it comes to delocalization of the lone pair electrons? So whenever there are same number of electrons, and if they are in regular conjugation (as in my compound A) the delocalization will always be more than a cross conjugated system (as in my compound B)? I would love to draw some conclusions from your answer :)

General tech note: bright green color is usually not a good choice for presentation, especially combined with the red color. The majority of occurrences of color blindness is related to green color, and, especially when combined with the red color, this makes reading such illustrations for people suffering from deuteranopia quite troublesome. If possible, avoid using any color and stick to black-and-white images: this is a good tone for any scientific/technical publication. Also, on the nitpicking side, try to use font of the same size (and typeface) on the pictures and in the text.

Although the OP said that he wanted to do this qualitatively... Yet I would be grateful if you could pull up some pKa values for both these compounds above to support your rationale and prove mine wrong

@ChakravarthyKalyan could you please repsond to my comment above? Thank you! I would love to have a conclusion drawn :-)

@McSuperbX1 , I am currently in work area , which forbades me to answer. I will respond , when I am back to my PC .

@McSuperbX1 , in 1,4-dihydropyridine (compound B) , part of the structure is cross-conjugated .This double bond may not be part of lone pair delocalisation on N . Therefore compound B may be more basic. This conclusion is based on qualitative treatment. I have tried for Pka values and am unable to find them. I welcome from anyone for Pka values.

@ YUSUF HASAN ,I have tried various sources and  am unable to come up with Pka values. That said , the answer is based on qualitative treatment .I certainly welcome any one eith relevent Pka values .


text:
How many calcium ions is needed to displace hydronium ions in an strong acid ion exchange resin?
If 1 mole of Ca2+ can displace 2 moles of Na+ in an ion exchange resin, can it do the same for H+? Will Ca2+ ion displace 2 moles of H+ in a strong acid resin?

Yes and no.
Yes, as  $\ce{1 Ca^2+}$ displaces $\ce{2 H+}$.
No, as you need to involve more then $\ce{1 Ca^2+}$ to displace $\ce{2 H+}$ for exchange comparable with ionex capacity, as some  $\ce{Ca^2+}$ passes through before $\ce{H+}$ is washed out.

I see. So to completely wash out the H+ ion in the solution, Ca2+ needed to be in excess? Am I right?

Yes, you are. But not in so big excess as  opposite, i.e. washing out calcium. Strong catexes prefer calcium to hydrogen, in contrary to weak catexes..

Thank you! This helped me a lot on my exam.

It should not. Using SE to solve exam tasks is a bad way to use it.

We have a policy which states that you should show your thoughts, effort and attempts to understand underlaying principles. It'll make us certain that we aren't doing your homework for you, and that the Q/A is beneficial for broad audience.
Basically any question with the wording your question has is considered homework; it needn't be literally one. It can be self-study questions, puzzles, worked examples etc.

This was not a homework, our professor taught us how to do calculations for Ion exchange resins. However, he did not showed us how to calculate for the stoichiometry of divalent cations with monovalent cations. I am reading materials online and saw an example that has H+ embedded on the resin instead of Na+ so I used SE to know if the stoichiomrtry between Ca and Na also applies for H

"not literally". The key part is you have not provided any own reasoning, what is a no no in Chem SE.     Why should the stoichiometry be different between free and bound functional groups ?

Because prior in posting my question here, I asked quora first and one response to my question is that the stoichiometry of Ca2+ ion to H+ is 1:1 and that confused me a lot. So I went to SE to see if my thoughts are incorrect or not.


text:
Reversibility of a reaction
I was doing some questions on equilbrium. I got stuck in this particular question.

Identify which of the following reactions are reversible.
$$
\begin{align}
\ce{Fe + S &-> FeS }\label{rxn:1}\tag{1}\\
\ce{NaCl + AgNO3 &-> AgCl + NaNO3}\label{rxn:2}\tag{2}\\
\ce{H2SO4 + Ba(OH)2 &-> BaSO4 + 2 H2O}\label{rxn:3}\tag{3}\\
\ce{H2 +I2 &-> 2HI }\label{rxn:4}\tag{4}\
\end{align}
$$

I know that \eqref{rxn:2} and \eqref{rxn:3} cannot be reversible because one of the products is precipitated.
I could not reach a conclusion between \eqref{rxn:1} and \eqref{rxn:4} (although I know that \eqref{rxn:4} is reversible).
Is there a way to determine without actual experiments whether a reaction is reversible or not?

Strictly speaking, the answer depends on conditions and, as a consequence, states of aggregation.
Reaction (1) can be reversible if iron(II) sulfide is subsequently heated above $\pu{700 °C}$ in vacuum to prevent oxidation of the elements [1, p. 422]:
$$\ce{FeS(s) ->[\pu{700 °C}][vac] Fe(s) + S(g)}\tag{1a}$$
Reactions \eqref{rxn:2a} and \eqref{rxn:3a} are indeed irreversible if carried in aqueous solution — as you already mentioned, a poorly soluble salt is formed in both cases:
$$
\begin{align}
\ce{NaCl(aq) + AgNO3(aq) &-> AgCl(s) + NaNO3(aq)}\label{rxn:2a}\tag{2a}\\
\ce{H2SO4(aq) + Ba(OH)2(aq) &-> BaSO4(s) + 2H2O(l)}\label{rxn:3a}\tag{3a}
\end{align}
$$
Reaction \eqref{rxn:4a} is a canonical example of a reversible reaction used to study equilibrium in a gas phase:
$$\ce{H2(g) + I2(g) <=> 2 HI(g)}\label{rxn:4a}\tag{4a}$$
I'd say it's impossible to tell in advance what reaction is reversible or not without conducting an actual experiment or at least estimating free Gibbs energy.
References

R. A. Lidin, V. A. Molochko, and L. L. Andreeva, Reactivity of Inorganic Substances, 3rd ed.; Khimia: Moscow, 2000. (in Russian)


text:
What does the formula (CH2O)106(NH3)16H3PO4 represent?
The molecular formula $\ce{(CH2O)106(NH3)16H3PO4}$  appears in the wikipedia entry for "anoxic bodies of water". A quick search suggests that it represents the product of photosynthesis, or the generic starting point for breakdown of organic matter in the nitrogen cycle, in particular the process of denitrification of "organic matter". On page 271 of Ref. (1) it is suggested  that it represents the structure of "marine plankton". This formula seems relatively widespread in use. Exactly what "organic matter" is this supposed to represent?  What is its origin (who first proposed and perpetuated its use)?
References

Derek E. G. Briggs (Ed.), Peter R. Crowther (Ed.). Palaeobiology II (1st Ed.). Wiley-Blackwell (March 5, 2001).

It seems the original work denoting elemental ratio C:N:P = 106:16:1 (aka the Redfield ratio) as the most common among marine organics is Fleming, R. H., 1940. The composition of plankton and units for reporting populations and production. Proc. Pacif. Sci. Congr., 1940, Vol. 3, pp. 535-540. Unfortunately, I wasn't unable to find a full-text document yet.

Indeed from the context isn't a molecular formula but elemental ratios, so while the body of text poses a sensible question, "molecular formula" shall be removed from the title.

@Alchimista $\ce{CH2O}$,  $\ce{NH3}$,  and $\ce{H3PO4}$, are molecular formulas, suggesting they are building blocks of a larger structure. There must be a good reason why this "molecular-like formula" (if you prefer) and not an elemental ratio was reported.

https://en.m.wikipedia.org/wiki/Redfield_ratio. See also respiration and nitrogen cycle.

@Alchimista Thanks! Yes, I saw the wikipedia page after following up on andselisk's helpful comment. I understand now that the formula was originally a ratio. If I'd known that it would have been unecessary to pose my question. And I still haven't read the explanation for why the ratio has been recast as a molecular formula, but have some leads.

Why not? That is a composition of matter, even a formula but not a molecular one.  You can do it. It can be done in Hill system or like that,  grouping likely to reflects some chemistry behind. For instance full combustion / respiration end in a block like CO2.

@Alchimista The point of the question is to figure out exactly where the formula comes from. Now that the source, a ratio, is clear, next step is to figure out why this particular formula was posited. Perhaps not extremely complicated, but, more complicated than combustion, methinks.

Even sulfur is also involved in the formula: https://biology.stackexchange.com/questions/96884/what-is-the-representation-for-sulfur-in-organic-matter. It's a good question. Probably, you can raise a bounty so the biochem expects here can have a go ;)


text:
Is the cathode and anode on the cell notation of an Electrolytic cell on the same side as if how you will write it for a Galvanic cell?
I was reading an online material then I came upon a cell notation for an electrolytic cell (I determined that it was an electrolytic cell because of the absence of a salt bridge). Usually, I will proceed on reading but then I realized, since cathode and anode is switched for an electrolytic cell, which side of the cell will be the cathode and the anode?
The example given was:

$$\ce{Ag | AgCl(s), Cl- (\pu{0.200 M}), Cd^2+ (\pu{0.00500 M}) | Cd}$$

This could be simply a Ag|AgCl electrode, a Cd electrode and a solution that is 0.200 M in chloride ion, 0.005 M in cadmium ion and, say, 0.190 M in sodium ions. All in one beaker, so no salt bridge. The context in what you were reading matters.

So, is the Ag l AgCl side the anode?

This is where the Nernst equation get used: you calculate the two reduction potentials and the most positive one is the spontaneous reduction. Then the other one is the oxidation side and it is conventionally on the left.

A good mnemonic to remember which is the anode or the cathode is to remember that Right side represents Reduction in a cell diagram. This is fixed by those who invented cell notation.. Also, by definition, reduction occurs at the cathode. Whether it is a spontaneous cell or not, is another question.
Now as EdV said, calculate the cell potential as written i.e., consider cadmium half cell as the cathode, and the other silver electrode as the anode. If the Ecell=Ecathode-Eanode turns out to be positive, everything is fine, all that means that the cell, as written, is a spontaneous cell.
If Ecell is negative, then the cell is non-spontaneous.


text:
Should High Vacuum oil smell?
I recently have purchased a high vacuum pump and am currently repairing it, the internals have a strong smell of oil. I was going to do an oil change anyway but should I clean all the old oil. High vacuum oil shouldn't have a smell because that would indicate out-gassing? Is that right?

New oil has very little odor. Used oil can definitely carry odors depending on what has run through the pump and the condition of the pump (overheating is bad).

If it smells then it contains either burned oil or lots of  whatever has been pumped through it.  Definitely clean out as much oil as you can, use solvents only if you must, then when completely dry, and, if you can,  bake the whole darn thing for ages to get rid of any remaining volatile stuff.

I wouldn't worry too much. Clean out what you can, refill, and let the pump run with the ballast valve open (and perhaps with a small leak in your apparatus) for a while.

Oh, somehow the first edit didnt save properly, it is a turbomolecular pump.


text:
Modelling the Oregonator
Assuming this is the correct place for a post like this, I need to write a piece of code to show the oscillations in the Oregonator. I have successfully done so for the Brusselator, but have been unable to find the appropriate rate constants and starting concentrations for the Oregonator.
Any help or information on a better place to post this would be appreciated.

see Field & Noyes, J. Chem. Phys. v60,p1877,1974 for the original.  Scott,  ' Oscillations Waves and Chaos in Chemical Kinetics' (publ OUP) gives the scheme, rate constants and reduced variables to use in this scheme.

I am not able to access these papers or find a copy of this book online. If possible, could you please provide me with the starting concentrations and the necessary rate constants. Thanks.

yes but much later on today

I believe this is the paper [PDF].


text:
Sources of carbon monoxide in dwellings
I am doing some research on indoor environmental quality and come across a number of spikes in CO in a dwelling.
This is a dwelling in central London with no gas-fired equipment: electric cooker and heating from a centralised district heating system.
Could this be from cooking? And/or people themselves?

I'd say both would be possible, but for a megalopolis like London the major source is likely car exhaust. Having a comparable density with air, CO tends to accumulate in a poorly ventilated space over time. Also, it might be worth to check CO pollution levels in your region from time to time; probably there is a wildfire nearby.

@andselisk Light or heavy gases only have buoyancy if they occur in high concentrations, and CO certainly does not accumulate, unless you have a source in the room.

Humans do not give off carbon monoxide, unless they are suffering an acute poisioning. If your sensor isn't an actual mass spectrometer, I would suspect it's confused by sth else, or indeed an external source, like someone smoking on the toilet. :-D

The sensors we are using for the research are high grade research instruments manufactured by Eltek AQ110. The research participants do not smoke and live on the sixth floor of a tall building not facing any road. https://www.google.com/url?sa=t&source=web&rct=j&url=http://www.eltekdataloggers.co.uk/pdf/user_instructions/TU1082_AQ110A_from_serial_no_32655.pdf&ved=2ahUKEwjS57DTvN3jAhWKa8AKHQRSAToQFjAYegQIAhAB&usg=AOvVaw3JqeoPsXZudzhJTS-WK1rP

It's still not a mass spectrometer and could be fooled. And people who "don't smoke" tend to do it in secret. Have you put two identical sensors in the same room an checked if they spike at the same time? Correlated the occurence with the presence of your participants? You don't say, which makes me suspect, bloody suspicious scientist that I am, that you haven't. :-)

Check at ground level at the same time to help pinpoint the source: car exhaust, old gas lines underground (producer gas contains substantial amounts of CO), a neighbor, insider or outside, using a charcoal grill...

The dwelling is occupied during all spikes. The participants live in a flat with a joined kitchen and living space and have a sensor in both the kitchen and living rooms, which show the same spikes, but with a higher concentration in the kitchen. 'General' CO levels cycle in a range of 0.2-1.6mg/m3. 5no spikes in CO (all on different days), 3.5@22:00, 7.6@21:00, 6.1@14:00, 4.4@02:00, 8.4@21:00. The spikes are a clear sharp increase in concentration followed by a sharp decay in CO levels.


text:
Number of aromatic rings in triphenylene
My question is how to find the number of aromatic rings in triphenylene:

My professor said that there are 3 aromatic rings, after which he added that the carbons of the middle rings are only supporting ortho-fusion of rings and hence the center ring is not to be considered.
But I am not satisfied. Why not consider the center ring since the electron after all delocalizes in the entire system giving me 4 as the answer? Someone, please clear my confusion out here.

Clar's rule

Basically, you're right and your prof is wrong. The middle ring is not that aromatic (alternating long and short bonds), but it's flat, and certainly the center of one big pi-system, not just holding together three individuals.


text:
Which of the following reactions doesn't represent an oxidation-reduction

Which of the following reactions doesn't represent an oxidation-reduction?
$$
  \begin{align*}
    \ce{2 P + 5 HClO + 3 H2O &-> 2 H3PO4 + 5 HCl} \\
    \ce{Zn + 2 HCl &-> ZnCl2 + H2} \\
    \ce{Mg + CuSO4 &-> MgSO4 + Cu} \\
    \ce{NaOH + HNO3 &-> NaNO3 + H2O}
  \end{align*}
$$

I want to find which of those reactions is not an oxidation-reduction reaction, but I'm not sure how to do this and how to find the oxidation numbers, should I memorize them?
Is there some rules like when bases and acids react this can't be an oxidation-reduction reaction — the last reaction implies this — so I can know if the reaction is an oxidation-reduction reaction or not when I look at it without searching for every oxidation number?

Only the last one doesn't. This case can be solved trivially without memorizing anything as in the first three reactions an element is a reactant, and the fourth reaction is a neutralization reaction. But in general, yes, you need to memorize typical oxidation numbers of elements and charges of common anions and cations.

@andselisk so if there is an element in the reactants, does this means that it must by an oxidation-reduction reaction?

If there is an element in the reactants and it is not an element in the products then yes, oxidation/reduction has occurred.

If there is an element in the reactants and it is not an element in the products then yes, oxidation or reduction has occurred.
Acid-base reactions are pretty much non-redox reactions. But it is always handy to learn the oxidations numbers.
Extra sources
Do non-redox reactions exist?

"Well acid-base reactions are pretty much non-redox reactions".
Do you mean that almost all acid-base reactions are non-redox or are you emphasizing that all acid-base reactions are non-redox reactions ?

Well if you mean by acids, the species which give H+ ion in solution and by base, thes species which give OH- ion in solution, yes. I haven't known anything yet. However, it doesnot account for lewis base and lewis acids.

And i would add up to @Waylander. "Vice-versa"

I have just suggested an edit to contain both to accept it as an answer.


text:
Red Squares in RDKit?
I'm newish to RDKit and couldn't find an answer to this online.  The following is a minimal representation of how I drew a molecule
from rdkit import Chem
from rdkit.Chem import Draw as Draw
from rdkit.Chem.rdMolDescriptors import CalcMolFormula

# .. read mol_a from file... then
print(repr(mol_a))
<rdkit.Chem.rdchem.Mol object at 0x7fd5c5494dc0>

print(CalcMolFormula(mol_a))
C4H11N

Draw.MolToImageFile(mol_a, 'pic.png')

Here is what I get:

What are those red boxes?
I'm running Ubuntu 16.04, Python 3.7.3
I had three semesters of college chemistry about a decade ago.
Thanks!

In your 3d data some atoms have the same coordinates.
If all coordinates are singular like here
from rdkit.Chem import AllChem as Chem
from rdkit.Chem.Draw import IPythonConsole
from rdkit.Chem import Draw

mb1 ='''
     RDKit          3D

 16 15  0  0  0  0  0  0  0  0999 V2000
   -1.4420   -0.8874   -0.7989 C   0  0  0  0  0  0  0  0  0  0  0  0
   -0.3758    0.0725   -0.4996 N   0  0  0  0  0  0  0  0  0  0  0  0
   -0.9154    1.1745    0.2995 C   0  0  0  0  0  0  0  0  0  0  0  0
    0.7376   -0.6096    0.1788 C   0  0  0  0  0  0  0  0  0  0  0  0
    1.9820    0.2624    0.3084 C   0  0  0  0  0  0  0  0  0  0  0  0
   -1.0693   -1.6891   -1.4463 H   0  0  0  0  0  0  0  0  0  0  0  0
   -1.8570   -1.3405    0.1088 H   0  0  0  0  0  0  0  0  0  0  0  0
   -2.2571   -0.4005   -1.3461 H   0  0  0  0  0  0  0  0  0  0  0  0
   -0.1723    1.9659    0.4378 H   0  0  0  0  0  0  0  0  0  0  0  0
   -1.2580    0.8409    1.2857 H   0  0  0  0  0  0  0  0  0  0  0  0
   -1.7617    1.6439   -0.2155 H   0  0  0  0  0  0  0  0  0  0  0  0
    0.4357   -0.9719    1.1698 H   0  0  0  0  0  0  0  0  0  0  0  0
    1.0275   -1.4907   -0.4080 H   0  0  0  0  0  0  0  0  0  0  0  0
    1.8280    1.0937    1.0028 H   0  0  0  0  0  0  0  0  0  0  0  0
    2.2816    0.6703   -0.6627 H   0  0  0  0  0  0  0  0  0  0  0  0
    2.8161   -0.3344    0.6923 H   0  0  0  0  0  0  0  0  0  0  0  0
  1  2  1  0
  2  3  1  0
  2  4  1  0
  4  5  1  0
  1  6  1  0
  1  7  1  0
  1  8  1  0
  3  9  1  0
  3 10  1  0
  3 11  1  0
  4 12  1  0
  4 13  1  0
  5 14  1  0
  5 15  1  0
  5 16  1  0
M  END'''

You get a pic without the red box.
mol1 = Chem.MolFromMolBlock(mb1, removeHs=False)
Draw.MolToImage(mol1)

If I give the H on position 15 the coordinates from the C from position 5
mb2 ='''
     RDKit          3D

 16 15  0  0  0  0  0  0  0  0999 V2000
   -1.4420   -0.8874   -0.7989 C   0  0  0  0  0  0  0  0  0  0  0  0
   -0.3758    0.0725   -0.4996 N   0  0  0  0  0  0  0  0  0  0  0  0
   -0.9154    1.1745    0.2995 C   0  0  0  0  0  0  0  0  0  0  0  0
    0.7376   -0.6096    0.1788 C   0  0  0  0  0  0  0  0  0  0  0  0
    1.9820    0.2624    0.3084 C   0  0  0  0  0  0  0  0  0  0  0  0
   -1.0693   -1.6891   -1.4463 H   0  0  0  0  0  0  0  0  0  0  0  0
   -1.8570   -1.3405    0.1088 H   0  0  0  0  0  0  0  0  0  0  0  0
   -2.2571   -0.4005   -1.3461 H   0  0  0  0  0  0  0  0  0  0  0  0
   -0.1723    1.9659    0.4378 H   0  0  0  0  0  0  0  0  0  0  0  0
   -1.2580    0.8409    1.2857 H   0  0  0  0  0  0  0  0  0  0  0  0
   -1.7617    1.6439   -0.2155 H   0  0  0  0  0  0  0  0  0  0  0  0
    0.4357   -0.9719    1.1698 H   0  0  0  0  0  0  0  0  0  0  0  0
    1.0275   -1.4907   -0.4080 H   0  0  0  0  0  0  0  0  0  0  0  0
    1.8280    1.0937    1.0028 H   0  0  0  0  0  0  0  0  0  0  0  0
    1.9820    0.2624    0.3084 H   0  0  0  0  0  0  0  0  0  0  0  0
    2.8161   -0.3344    0.6923 H   0  0  0  0  0  0  0  0  0  0  0  0
  1  2  1  0
  2  3  1  0
  2  4  1  0
  4  5  1  0
  1  6  1  0
  1  7  1  0
  1  8  1  0
  3  9  1  0
  3 10  1  0
  3 11  1  0
  4 12  1  0
  4 13  1  0
  5 14  1  0
  5 15  1  0
  5 16  1  0
M  END'''

you get the red box


text:
OpenBabel: how to get correct SMILES from Gaussian output?
I would like to get the molecular structure as SMILES from Gaussian output files. OpenBabel seems to be the tool made for such tasks.
However, the structure is not always correct.
One example:

OpenBabel conversion Gaussian09 to SMILES (left):
N1(c2ccccc2)[C]2C=CC=CC2=[N]=C1c1ccccc1

SMILES generated by some drawing program (right):
C12=CC=CC=C1N(C3=CC=CC=C3)C(C4=CC=CC=C4)=N2

Is there a way (maybe a special option in OpenBabel which I missed) to directly get a correct output or to do some "postprocessing" to correct the SMILES afterward for example with RDKit or others?
Update:
One way to increase the yield of correct output seems to be the following.
If you convert the Gaussian output to InChI first, for the above example you'll get:
InChI=1S/C19H14N2/c1-3-9-15(10-4-1)19-20-17-13-7-8-14-18(17)21(19)16-11-5-2-6-12-16/h1-14H

Then, if you convert this InChI to SMILES, you'll get:
c1ccc(cc1)c1nc2ccccc2n1c1ccccc1

Which results in the right picture above.
However, even with this workaround some compounds, e.g. with ...C#N... triple bonds are sometimes still incorrect.
I placed a bug report on GitHub. Let's see...
Any strategies, tips, and tricks to reliably and directly get a correct SMILES output are appreciated.

The left structure simply is wrong; that should probably be a bug report for open Babel. Do you have other examples? Also the smiles generated for the right side not apply aromaticity, so I wouldn't regard this as correct either.

Often structures containing Benzimidazoles are wrong... I will check a bug report for OpenBabel.

@Martin - マーチン♦, by the way, do you maybe know what exact section OpenBabel is extracting from the Gaussian output file (start/end keywords or headers)?. I found that if you convert first to InChI and then afterwards convert  these InChI to SMILES gives a higher "yield" of correct SMILES structures (but still not 100%).

I don't know, but I assume it's extracting the molecular structure, then probably forms internal coordinates, flattens the molecule, form a graph, translates it to smiles, canonicalise it. InChI just is probably less ambiguous. As for why it might yield 'correct (er)' results, I'm guessing the algorithm is different and more robust. In any case, a canonical smiles should have a 1:1 correspondence to InChI. But still, I'm guessing here... And open Babel is way too complex for me to understand or look into.

Do feel free to submit your findings as an answer! :)

@MelanieShebel well, sorry, no response at all on GitHub so far after 1.5 years... I have to test with the latest version...

@theozh I'm sorry to hear that.

@theozh - At the moment, perceiving occurs via a set of heuristics, which certainly can make mistakes. It's impossible to debug either your post here or the bug report without a test file. We've made many changes to bond perception in both 3.0 releases and 3.1 releases. Without something to test, I can't say whether your particular cases are fixed or not.