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Decarboxylation of THCA during CO2 Extraction of Cannabis
We have a strange problem. We are performing CO2 extraction on cannabis and we are seeing considerable decarboxylation of THCA to THC.
We have high quality temperature sensors at the heat exchangers, the extractor vessels, and the separator. We have not observed any temperature spikes that would explain it.
Various parameters have been tried, i.e.
40 - 50C (104-122F) / 115 - 280 bar (1770 - 4050 psi)
We can't find any correlation between the temperatures / pressures and the level of decarboxylation.
The system has been run for 24 hours at 120bar/50C and we are getting approximately equal amounts of THC and THCA out.
The starting materials has been tested to being only slightly decarboxylated. Disturbingly, the spent material is also being decarboxylated.
The same equipment is used to test both the starting and spent material. The results on the starting material have been verified by a 3rd party lab. It seems unlikely to be a testing problem.
We have multiple extractors and we are seeing the same behaviour so it is unlikely to be a sensor issue. The temperature controllers also have their own thermostat that agrees with the extractor sensor. The extractors are all temperature controlled and all temperatures are logged every 30 seconds.
There is a research paper by Rovetto et al. that also observes significant decarboxylation but not to the extent that we are observing it:
https://www.researchgate.net/publication/315417305_Supercritical_carbon_dioxide_extraction_of_cannabinoids_from_Cannabis_sativa_L_plant_material
Is it possible that the normal decarboxylation threshold (e.g. 115C) is lowered when at high pressure?
I understand that THC extracts far more readily than THCA, however looking at the mass balance it seems that selective extraction can not explain it.
If anyone has any insights to share, I would very much appreciate it.
Thanks

+1 for a well written first SE question! Have you checked for any leaks in the system? (humor)

Thanks for the encouragement! I wish I could find anything that might explain it :(

A paper by Iffland et al indicates that decarboxylation occurs at a lower temperature than 110 C. I think in the plant material that THCA predominates, so your own data of an extraction at 120bar/50C indicates conversion at temperatures as low as 50C. So at 50C the decarboxylation seems to be a kinetic problem. A longer time should convert more. // The decarboxylation reaction would seem to be irreversible since you're using CO2.

Also see "Decarboxylation Study of Acidic Cannabinoids: A Novel Approach Using Ultra-High-Performance Supercritical Fluid Chromatography/Photodiode Array-Mass Spectrometry" by Wang et al. They give a formula for the reaction rate. I didn't read the paper carefully enough, but I'm assuming that is for pure THCA. There might be something in the plant material that would stabilize the THCA a bit more.

Thank you MaxW, I have seen 80C as the start of decarboxylation according to a few sources now, appreciate the links to the papers. I wonder though (and my chemistry knowledge is weak) - could this 80C be lowered significantly if reacting under huge pressures?

You're performing supercritical extraction. Is that absolutely necessary? Decarboxylation of CBDA and THCA occurs at room temperature and is obviously accelerated by temperature, THCA being the easier to decarboxylate. Are there any other components that could actually accelerate the process more? e.g. Other acids, bases etc. Also the position of your temperature sensors may not pick up an initial temp spike during CO2 compression as gas is forced out the system and SC CO2 replaces it,  possibly?

Have you also considered that the rate of reaction of THCA to something else e.g CBN is skewing your ratio making you think it's decarboxylation?  What's the mass balance?

@Beerhunter Unfortunately supercritical extraction is necessary to extract cannabinoids, if it is subcritical, it is only efficient at extracting terpenes. I have been thinking about the initial temp spike and that could be a factor - the probe is at the top of the cylindrical extractor and the co2 fills from the bottom. We will be careful about slowing that down. THC is easier to extract than THCA (twice as easy) but unfortunately that doesn't account for it in the mass balance. Thanks very much for your input

Which solvent?: https://youtu.be/n7cP6YWO_nc?t=853 - or no solvent: https://www.youtube.com/watch?v=dxQq9wuykpU ?

If this problem has not already been solved by the author, then here: Supercritical extraction, by nature, breaks the carbon bond in the dissolution phase, which is why you are even seeing your biomass decarbed. It's not a heat thing, it's a function of SCF CO2.


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Can someone intuitively explain the reason for the units of entropy (J/K )?
If entropy is just a measure of the "randomness" or "chaos" of a system, then why does it have specific units. What is the reason behind the units of entropy being J/K?

Entropy is not a measure of the randomness. It is equal to the constant of gases R multiplied by the logarithm of the probability of a given state. It has the same unit as R.

Wikipedia has what seems to me like a good explanation of the relation between the information-theoretic definition of entropy and the thermodynamic definition: https://en.wikipedia.org/wiki/Entropy_in_thermodynamics_and_information_theory

The units of of energy over temperature (e.g. J/K) used for entropy in the thermodynamic definition follow from a historical association with heat transfer under temperature gradients, in other words, the definitions of temperature and entropy are intertwined, with entropy being the more fundamental property. Entropy can be thought of as a potential and temperature (or its inverse, rather) as a generalized force associated with displacements along energy dimensions in the entropy potential. The (inverse) temperature is a measure of the effect of changes in the amount of energy on the entropy of the system, as the following definition suggests:
$$\frac{1}{T}=\left(\frac{\partial S}{\partial U}\right)_V$$
However if you start from a statistical mechanical definition of the entropy,
$$S=k_\mathrm{B}\log\Omega$$
then it is easy to see that a unitless definition might be just as well suited:
$$S_{\mathrm{unitless}}=\log\Omega$$
However, in the absence of the Boltzmann constant you need some other way to relate entropy and energy (e.g. heat transfer) in our conventional system of units. You can of course subsume $k_\mathrm{B}$ into the definition of a new temperature scale:
$$\frac{1}{T_\mathrm{new}}=\left(\frac{\partial S_\mathrm{unitless}}{\partial U}\right)_V$$
where the old and new temperature scales are related as $T_\mathrm{new}=k_\mathrm{B}T$. The new temperature scale then has units of energy.

The "unitless entropy" should have some sort of unit even if it's not mathematically required. Like how an angle is 1 radian, not just 1 (even though we do not use radian units in the context of pure calculation, you still need to specify radians in order to understand what the number 1 actually means).

@user253751 In the "unitless" case the entropy is the logarithm of the number of states, so in a  mathematical sense it does not have units. I wouldn't know what to call the units in this case.

Entropy does have units, and the unit is based on the logarithmic base, e.g.: https://en.wikipedia.org/wiki/Nat_(unit)

...and I guess most of the readers have heard about the unit of entropy if the logarithm is base-2, which is called bit.

Entropy isn't "just" a measure of randomness.  It is the only physical property that gives the universe a temporal direction, i.e., that provides an "arrow of time".
Also, a "measure of randomness" a crude way to characterize entropy.  Rather, the entropy of a system is proportional (through Boltzmann's constant) to the log of the number of possible macroscopically indistinguishable microstates that a system in a given state can sample, weighted by the relative probabilities of those microstates.
And what determines ($k_B$ times the log of) how many microstates a system at a given temperature can sample? It is how much heat we have let flow (reversibly) into the system divided by the temperature at each point.
And that is why the units of entropy are energy/temperature, because it is the integral of energy flow/temperature that determines the number of available states, and it is the number of available states that in turn determine the entropy.
More precisely, the entropy of a system is the integral of the inverse temperature times amount of reversible heat flow needed to bring that system from absolute zero to its current temperature.  I.e.:
$$S(T') =\int_{0}^{T'} dS= \int_{0}^{T'}\frac{\text{đ}q_{rev}}{T}$$

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Tertiary carbanion in more alkylated alkene
Saytzeff rule says that more alkylated alkene is more stable and the reason is hyper conjugation.
Going by this, 2,3 Dimethyl but-2-ene should be most stable alkene due to 12-alpha hydrogen.
But on drawing hyper conjugative structures of the molecule and a less alkylated alkene, say 2-Methyl propene; there was formation of a tertiary carbanion in 2,3 Dimethyl but-2-ene! and not in 2-Methyl propene. Instead in this molecule there is a primary carbanion which is much more stable than tertiary. I know that there will be 12 hyper conjugative structures in 2,3 Dimethyl but-2-ene rather than 6 structures but isn’t a tertiary carbanion poorly stable and make overall molecule less stable than 2-Methyl propene?
Can anyone please prove me wrong?
Thanks in advance

There are two types of description of hyperconjugation: (1) Valence bond approach & (2) Molecular orbital approach. Your question is pertaining to valence bond approach as you are interested in the stability of the canonical forms.
In Valence bond approach, hyperconjugation is said to happen due to conjugation of C—H bond electron pair with adjacent double bond. We call this phenomenon 'hyper'-conjugation because there is no real negative charge that is getting delocalized. Rather, the partial negative charge which is accumulated on the carbon due to electronegativity difference with hydrogen is getting delocalized.
In a real carbanion, the carbon has ~100% share of the negative charge. However, in this case, the carbon has only partial share of the negative charge. To be precise, about 7%, because 7% is the ionic character of a C—H bond (Calculated using Hannay-Smith's Relationship for Calculating Percentage Ionic Character [1]).
In case of conjugation of real carbanion, the ~100% negative charge is getting delocalized, and you can give importance to the stability of canonical forms to detrmine whether that conjugation is relevant. However, when ~7% negative charge is getting delocalized, it is more important to give importance to the number of such conjugations, than the relevance of each conjugation based on stability of canonical forms, because anyway, ~7% negative charge won't cause much destabilization even in a tertiary carbanion!
In Molecular orbital approach, hyperconjugation is said to happen due to overlap of bonding σ oribitals of C—H bond with adjacent antibonding π* orbital (in case of double bond). Owing to energy & orientation mismatch, the overlap is minimal and thus stabilization arising from hyperconjugation becomes significant only when the number of hyperconjugations is greater. So, this explains why 2,3-dimethyl-but-2-ene assumes much greater stabilization from hyperconjugation compared to 2-methyl-propene.

" carbocation, the ~100% negative charge" - negative? You're mixing up too many things IMO.

Corrected! That is carbanion. What else have I mixed?

Also removed reference of hyperconjugation-aided-carbocation-stability for adding clarity to the answer.

Thank you Sujay Ghosh for such a logical explanation!

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Does Magnesium metal react with carbonic acid?
I use rainwater in my house for my water supply. I collect the water in plastic-lined tanks. Some of my piping at home is copper and I have blue stains in my shower because I think the copper from my pipes is being dissolved by the carbonic acid in the rainwater.
I'm planning to hang magnesium rods into my rainwater tanks, in the hope that these rods will react with the carbonic acid instead of my copper pipes. Does that make sense? Would this work?

Then you'll have white residue all over the place. Admittedly, Mg is less harmful than Cu. Then again, are you sure the blue stains were there because of Cu?

You may try. But pure water in contact with air dissolves CO2 extremely quickly. So even with magnesium rods, I'm afraid water will quickly redissolve the CO2 from the air to compensate the losses due to the action of magnesium

Mg could help if it would be in direct contact with Cu (sacrificial cathodic protection) but then it would dissolve pretty quickly.

Distilled water ( rain) plus CO 2 is pretty aggressive to many metals. You could try adding calcium as adding limestone gravel to your storage tank; one source is aquarium gravel ,aragonite/crushed coral. It will add some hardness to the water and lower the acidity . I can't think it would create any problem and is cheap and easy.

I wonder if other items show blue stains: kitchen or bathroom sink, toilet? perhaps detergent items used in the shower (body wash, soap, shampoo, conditioner) might be responsible for increasing the blue color by complexing or precipitating the color. Changing one or more of these items might help. Or you could investigate a cleaning system better adjusted to removing the blue, like one which contains a chelating agent, or is more acid than your rainwater.

Hot water heaters normally come with anodes. Aluminum is cheaper than magnesium so aluminum is more common. Replacement magnesium and aluminum water heater anodes are readily available online. That seems a handy source for either one. Magnesium has a higher potential and is consumed more quickly. In addition, I see some ads say magnesium has health benefits.
Both will react with carbonic acid.

First, thanks for reporting some obvious unexpected chemistry occurring. Others in your community may also have similar metal leaching issues, which especially for children, may be a health concern (see, for example, this article).
My speculation, on what is occurring, is that the iron pipe in your shower is displacing (upon standing over time) some dissolved copper (possibly from created bicarbonate). The likely acidic rainwater may also contain chloride or nitrate salts, which may further complex with the fresh active copper precipitate. Here is a related illustrative source, per an article 'Electrochemical detection of nitrate and nitrite at a copper modified electrode' to quote:

The development of a reagentless electrode system for the determination of nitrate is presented. The approach is based upon the deposition of a macroporous copper deposit which shows marked selectivity for nitrate ion under mildly acidic conditions (pH 3) with a linear range extending from 10 to 200 mM nitrate.

So, do not employ Magnesium, address the iron presence. The coloration means there is some, but in actual amounts, most likely limited copper ion presence along with nitrate.


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Is it possible to determine the composition of a gas mixture from the effusion of a gas through a pinhole?
Typically all the problems which I've seen regarding rate of diffusion involve relating the rate of one gas to the other and from there finding either speed of effusion or diffusion or getting the formula weight. However what sort of procedure should be used when it is needed to find the composition of a mixture?
The problem is as follows:

A certain volume of nitrogen gas $(\ce{N2})$ pass through a pinhole in $\pu{20 s}$. Find the composition of a mixture between $\ce{O2}$ and $\ce{CO2}$ if this mixture is let pass through the same slit and the measured time for this effusion is $\pu{24 s}$ and of Avogadro's conditions.

45.5% $\ce{O2}$
69.3% $\ce{CO2}$
12.4% $\ce{O2}$
69.3% $\ce{O2}$
45.5% $\ce{CO2}$

I'm not very sure if it is possible to find what it is being asked here. It's obvious that the approach here is to use Graham's law of gas diffusion. But the problem arises on what volume should be used to establish an equation from where to find the composition of the mixture.
What I thought was:
$$\frac{v_{\ce{N2}}}{v_\text{mixture}} = \sqrt{\frac{FW_\text{mixture}}{FW_{\ce{N2}}}}$$
The thing here is that the average molecular weight can be used to find the composition of the mixture.
$$\frac{v_{\ce{N2}}}{v_\text{mixture}}=\sqrt{\frac{\chi_{\ce{O2}} \cdot FW_{\ce{O2}} + \chi_{\ce{CO2}} \cdot FW_{\ce{CO2}}}{FW_{\ce{N2}}}}$$
But the thing here is I am still stuck on where to go from here. What should be the right way to go to get the composition of the mixture. Could it be that the volume of the nitrogen is needed?. So far I'm going in circles with this. Regarding the source of this problem it doesn't indicate as far I'm aware since it is a collection of problems from different topics in gases.

Go on! You are nearly at the end. If you square your equation, and replace $y$ by $1-x$, you obtain: $$(24/20)^2 = \frac{32·x + 44(1-x)}{28}$$
You finally obtain: $x = 1/3$ and $y = 2/3$.

@Martin-マーチン♦ It seems that you got wrong the result for the molar fraction. Solving the system yields $x=\frac{23}{75}=0.30667$ and $1-x=0.69333$ which given as a percentage would be $69.3%$ of $CO_{2}$ which according to my book is the right answer.

@Martin-マーチン♦ The reason why I was almost at the end was my confusion regarding if should I assume that the same volume of gas was for the mixture as well?. Because in the problem did not explicitly said this. Other than assuming that by mentioning Avogadro's conditions $22.4\times V$ of the nitrogen gas and $22.4\times V$ of the mixture. Should this rationale be used in all sorts of problems of this kind?.

@Chris I only slightly edited this answer, I don't even really know where the numbers come from or what they mean. Sorry, I cannot help you there.


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The rate of change in pH v.s Temperature
I have somewhat established that there is a linear relationship between change in pH of carbonic fizzy drinks over time. I also realise that the pH increases linearly faster for the same carbonated fizzy drink when the temperature of the drink is changed (increased).
Now the problem is, when I graph the relationship between the rate of pH increase (y-axis) and the temperature (x- axis), I get an exponentially increasing relationship. I do not know if this relationship is true, but even if it was true I do not know how to explain why this is true.
Are there any equations I can manipulate to show a mathematical relationship between rate of pH increase and temperature of the same carbonated fizzy drink?

Check the ratio of $\mu$/kT, since pH can be defined as the chemical potential ($\mu$) divided by thermal energy (kT). Hence pH is proportional to the chemical potential due to $\ce{H+}$ concentration reduced by kT. If, for a given system, the ratio of $\mu$/kT remains constant with temperature, pH is expected to be independent of temperature. Otherwise, pH could change accordingly.

I appreciate the help, but I am not sure how to do this as I am just doing this for a practise experiment with not so much background in chemistry...

You are dealing with temperature variation of carbonic fizzy drinks. So, the situation is somewhat tricky.
Lets approach the problem step by step.

If you raise temperature of neutral water, the rate of decomposition of water will increase. So, the water will contain more protons at higher temperature than at lower temperature, although the neutrality is maintained. Therefore, the pH of neutral water will decrease with temperature. pH of water at 100°C is ~6.126.
Carbonic fizzy drinks contain sodium & potassium bicarbonate salts, which ionize completely in aqueous solution to yield alkali cations & bicarbonate anions. The initial pH is between 3 and 4. (1)
Bicarbonate anion can capture a proton from the system to yield carbonic acid. Carbonic acid is unstable and thus decomposes even at room temperature to produce carbon-dioxide, & thus the bubble! So, even at room temperature gradually the solution turns less acidic, then neutral & finally alkaline. So, here you are not increasing temperature & just letting time to flow, & the pH rises gradually. Therefore, I can expect 'linear relationship between change in pH of carbonic fizzy drinks over time'.
When you raise temperature of carbonic fizzy drinks, the pH increase gets faster because the decomposition of carbonic acid is very much temperature sensitive. It is worthwhile to mention that the rapid pH rise is somewhat countered by the pH lowering due to water's decomposition, but that effect is much less significant as water's decomposition is much less sensitive to temperature. In this case, I can certainly expect deviation from linear behavior (nonlinear or exponential increase) aided by temperature influenced decomposition of carbonic acid.

You may be able to arrive at a mathematical relationship by combining the temperature dependency of carbonic acid & water decomposition.

I somehow amended it. Please check.

I appreciate the feedback, however I do not know exactly how to arrive at a mathematical derivation even though experimentally it seems to show a non linear behaviour as you also mentioned..

To draw a plot it is important to know the initial composition of the carbonic fizzy drink. It usually contains sodium chloride, sodium citrate, sodium bicarbonate, potassium bicarbonate, potassium citrate, potassium sulfate, or disodium phosphate, depending on the product. pH is increasing due to the presence of $Na^+$ & $K^+$ which were added as biocarbonates. After complete decomposition of carbonic acids, charges of those $Na^+$ & $K^+$ ions will be balanced by $OH^-$ ions, and hence you can measure the final pH value.

Now you need to determine the initial concentration of carbonic acid. Remember that carbonic acid is a weak acid, and its concentration is linked with the initial concentration of $HCO_3^-$ through the equilibrium constant. Then find the rate of decomposition of carbonic acid, and how that rate increases with rise of temperature.


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Is a compound that reacts with another to form hydrogen for fueling aircraft possible
Hydrogen is the most abundant element in the universe and has great properties when reacting with oxygen for use as a fuel. Yet here on Earth, its production is prohibitively expensive, its storage without wastage is almost impossible because its molecules are so small no know substance or material can store it safely without leaks. Not to mention safety, as it is highly explosive.
Given its energy density, the above mentioned problems and other inherent problems I did not mention with using hydrogen as a fuel for aircraft, is it possible to come up with a compound or two compounds that can be safely stored in two tanks and have an efficient energy to weight ratio, such that when they react by mixture or process, they produce enough hydrogen to power airliners at the same rate or much more efficiently than with regular fuel?

The real question is almost never "is such-and-such reaction possible", but "where are the money".

Such compounds certainly exist but they are 1- expensive to make relative to hydrogen 2 - heavy, you don't get much hydrogen per kg of compound

Apparently Palladium absorbs easily Hydrogen. But If you succeed in making a compound with 1 atom Palladium + 2 atom Hydrogen, the obtained product would be made of 98% Palladium + 2% Hydrogen in weight. The fuel is only 2% of the mass.

However you put this question, it's either blatant, or too broad, or opinion based. Reversible storage of hydrogen in compounds is like whole branch of chemistry for a long time, and so what? You're not gonna see airliners running like that for a long time, maybe never.

Chemical hydrogen storage materials like hydrazine, borohydrides etc. are considered potential candidates for this purpose (1).
Nanoparticles, nanoalloys & functionalized nanoparticles are extensively trailed as catalysts for efficient on-demand hydrogen generation from those materials. We are, however, yet to reach a stage where the hydrogen generation is rapid enough to be used in applications such as in an aircraft.


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How is mercury in drinking water detected?
A document I've been reading states that:

Both the Maximum Contaminant Level Goal and the Maximum Contaminant Level were set at 2 parts per billion because current technology allows public water suppliers to detect and remove mercury levels that low.

But I know of drinking water test strips that detect 0, 50, 100, ... 1000 ppb and are 1,5-Diphenylcarbazide based.
On a chem supplier catalog I've found that 1,5-Diphenylcarbazide can detect up to 0,000002 g/ml, which is 2000 ppb. But I know from a question on this website that the above reagent can detect up to 250 ppb.
And as per urine and blood concentrations we are looking at 10 ng/ml they detect.
So what is the threshold for 1,5-Diphenylcarbazide test and how they detect those minute amounts anyway?
Is there a DIY method of detecting something like 2 ppb at home?
P.S. I am talking about Hg2+ ions, if it is not clear

I doubt that any water analysis lab would be using  1,5-Diphenylcarbazide strips for analysis. Almost certainly ICAP would be used. // It is also unlikely that most municipal water supplies would have mercury as only $\ce{Hg^{2+}}$.

In wiki, look up “Atomic absorption spectroscopy” and scroll down to “Cold-vapor generation”. Detection limits are down at the parts per billion level. The chemical stuff is generally not as sensitive, especially test strips.

Typo: replace “generation” by “atomization”.


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T1 spin-lattice relaxation times, deviation from exponential relationship
For longitudinal relaxation, one would think that the magnetic moment obeys this simple relationship
$$M_z = M_0\left[1 - 2\exp\left(\frac{-t}{T_1}\right)\right].$$
However there is always a point of inflection when magnetisation is zero. Does anyone knows why?
edit: I deleted the image and added some from my own experiment for mineral oil.
In this scenario, the sample is placed under a large longitudinal z-magnetic field. To measure T1, a 180 degree pulse is applied to flip Mz, then a 90 degree pulse after time t. Mo is the equilibrium longitudinal magnetic moment.
I don't think this is due to the mixture of compounds, because it is observed in all samples I tried e.g glycerine, Fe(NO3)3.

Please cite the source of the image and/or data and define the variables in the equation to avoid ambiguity.

Possibly the data refers to a mixture of compounds whose chemical shift  are overlapping. Look for data on a known pure compound for comparison.

Can you clarify the question, please? I don't see any inflection point there -- your graph is quite clearly concave -- and I don't think I've ever seen an inflection point in the inversion-recovery experiments I've done before (example). Btw, your equation is missing a minus sign inside the exponential.

The inflection point is quite slight, if you look at the ln(Mz) against time plot, there is a  small difference between the gradient below ~0.02s and after. It occurs when the sign of the Mz flips. Maybe because your data started at positive Mz ? @orthocresol (also added a -ve sign)

No, it doesn't start at positive Mz. I don't have an explanation for what you're seeing, unfortunately; I just can't say I've seen it before myself.

It's not clear what you mean. Could you post the equation you are using to generate the log plot, because you must be subtracting a constant before taking the logarithm. Magnetization damping can be an issue and non-exponential behavior is certainly not uncommon for many reasons, instrumental not being the least. But this inflection is as you point out very faint and I cannot think of an explanation.

Also, what kind of setup are you using, lowfield with no spectral resolution or high field, with spectral resolution and picking selected spectral regions?


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Significant Figures for Exponents and Logs
What are the Sig Fig rules for exponents and logs. Ex. if I have 10^-3.83, or log(4.32) how should the answer be represented

In wiki, look up “Propagation of uncertainty” and, for a simpler treatment, “Significant figures”. The trouble with significant figures is that they are attempts to express, in one number, both a best estimate and an uncertainty. This cannot be perfect, but the rules work reasonably well.

You asked about the significant figures in $10^{-3.83}$. The digits in the characteristic don't count, but the digits in the mantissa do. Thus there are only 2 significant figures in $10^{-3.83}$.

The 'rule of thumb' is that the log of the number should have as many digits to the right of the decimal as there are significant digits in the number, e.g. $\log_{10}(0.000012345)=-4.90850$

if I have 10^-3.83

-3.83 is less than -3.82 and more than -3.84. If we raise 10 to those powers, we get:
$$10^{-3.82} = 0.000151$$
$$10^{-3.84} = 0.000144$$
So the answer is 0.00015 or $\pu{1.5e-4}$

if I have [...] log(4.32)

$$\log(4.31) = 0.6344$$
$$\log(4.33) = 0.6364$$
So the answer is 0.64.

What are the Sig Fig rules for exponents and logs?

Once you use exponents and logs a lot, you might also know some calculus to support a better model of error propagation. If not, the rule for logs is that the error is equal to the error of the argument divided by the argument. The rule for exponentials is that the error is equal to the error of the exponent times the value of the exponential. Very few introductory textbooks bother to state these, though.


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What is d-orbital collapse?
When I was perusing the works of Schwarz on atomic structure, I came across the unfamiliar term of d-orbital collapse.
He describes it as a variation in energetic sequence from group 1 to 3 elements such that the order changes from $\ce{4s < 4p < 3d}$ in $\ce{K}$, $\ce{4s < 3d < 4p}$ in $\ce{Ca}$ and $\ce{3d < 4s < 4p}$ for $\ce{Sc}$ and subsequent elements.
He continues to say "It is the result of the interplay of nuclear attraction, of imperfect shielding by the inner-core electrons, and of angular-momentum dependent centrifugal forces. It explains why 4s becomes occupied in $\ce{K}$ and $\ce{Ca}$ before $\ce{3d}$, but $\ce{3d}$ before $\ce{4s}$ in the transition elements."

He states the orbital energies $ε$ are represented by the effective quantum numbers
$$n_\mathrm{eff} = \frac{1}{\left(\frac{\sqrt{|ε|}}{\pu{13.606 eV}}\right)^3}.$$
This seems to imply the orbital energy of $\ce{4s}$ is lower than $\ce{3d}$ for $\ce{K}$ and $\ce{Ca}$ hence the $\ce{4s}$ orbitals are occupied in preference to $\ce{3d}$ in their atoms.
This contradicts various literature which state according to Hartree-Fock calculations the 4s orbital never has a lower energy than $\ce{3d}$, regardless of which atom is being considered. ("Transition Metals and the Aufbau Principle" VPD, J. Chem. Educ. 1994 71 (6), 469; "Why Is the 4s Orbital Filled before the 3d" MS, J. Chem. Educ. 74: 498–503, 1996; "4s is always above 3d! Or, how to tell the orbitals from the wavefunctions" Pilar, J. Chem. Educ. 1978 55 (1), 2; "Transition metal configurations and limitations of the orbital approximation" Scerri, J. Chem. Educ. 1989 66 (6), 481)

Criticisms of a similar graph used in textbooks is that it features out-dated calculations made by R. Latter in the 1950s that seemed to indicate the energy of the 4s orbital was lower than that of 3d. The graph which shows the crossing-over of the energies of the 4s and 3d orbital energies, "is now known to be erroneous". ("A Critique of Atkins' Periodic Kingdom and Some Writings on Electronic Structure" Scerri, Foundations of Chemistry 1: 297–305, 1999) Schwarz also alludes to this in his article: "The graphic is quantitatively correct, in contrast to qualitatively incorrect artwork in many chemical and physico-chemical textbooks.", yet his graph reproduces the same phenomenon.
Does Schwarz's d-orbital collapse actually imply 4s orbitals having lower energy than 3d? Is there any way to renconcile the contrasting information presented?

Prof. Schwarz is apparently active. Scerri is more of a historian and no need to rely too much on J.Chem.Ed. It may be worth contacting the author http://www.tc.chemie.uni-siegen.de/schwarz/. I was lately looking for a historical reference and contacted the author directly. He was kind enough to scan original works and send them to me along with some details.


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How big does an atom need to be to have dispersion forces be greater than other intermolecular forces?
I know as an atom gets bigger the dispersion forces grow with it. But how big does an atom, e.g methanoic acid, to have dispersion forces that outrank dipole-dipole forces?

Methanoic acid is not an atom.

How big should a man be in order to be more effective in sumo wrestling than in karate? Some men certainly are big enough, and some certainly aren't, and you won't be able to pinpoint the exact border. Same thing here.

I think you mean molecules, not atoms. Regardless let's just say you meant molecule and by intermolecular you're really just focusing on Dipole, the answer is- youre comparing apples to oranges.  But you don't look at 1 molecule interacting with another rather a whole bunch of molecules interacting simultaneously.  At some moment when electrons are gravitating toward 1 atom in a molecule and another molecule nearby has dipolarization and the relative distance can cause bonds to break and realign on another molecule.  But I think what you mean is, what kind of atoms in a molecule would be impacted by dispersion force the most? And that's large molecules and also large atoms because large atoms like iodine, can have an induced dipole from a atoms its not bonded to much easier than a tiny atom.  How big? There really isn't any way to answer that question.


text:
Substructure search with RDKit
I'm trying to search for substructures with RDKit.
I would like to search for structure 1 as substructure in other structures. I don't understand why it doesn't find this structure in structures 3 and 5. I assume this is not a bug in RDKit, but more the result of my limited chemistry understanding. Something with aromaticity? Something wrong with the structures?
Can someone explain to me why it doesn't find structure 1 in structures 3 and 5?
How should the search structure look like that I will find all structures 2-5?

from rdkit import Chem

smiles_list = ['C12=CC=CC=C1C3=CC=CC4=C3C2=CC=C4', 'C12=CC=CC=C1C3=CC=C4C5=C(C=CC2=C35)C6=C4C=CC=C6', 'C12=CC=CC=C1C3=CC=C4C5=C(C=CC2=C35)C=C4', 'C12=CC=CC=C1C3=CC=C4C5=C(C=CC2=C35)C6=C4C=C7CCCCC7=C6', 'C12=CC=CC=C1C3=CC=C4C5=C(C=CC2=C35)C6=C4C=C7CC8=CC=CC=C8CC7=C6']

pattern = Chem.MolFromSmiles(smiles_list[0])
for idx,smiles in enumerate(smiles_list):
    m = Chem.MolFromSmiles(smiles)
    print("Structure {}: pattern found {}".format(idx+1,m.HasSubstructMatch(pattern)))

# Result:
# Structure 1: pattern found True
# Structure 2: pattern found True
# Structure 3: pattern found False
# Structure 4: pattern found True
# Structure 5: pattern found False

thank you for pointing to this. I think I found a solution (see my answer), but I'm not sure if this solution might cause some other surprises or unexpected results. Should I discuss this on the mailing list? What do you think?

@GeoffHutchison See above response.

@theozh Just a heads up, if you reply to a comment, you have to tag the person, otherwise, they won't see it. Welcome to Chem.SE! I love your question and will be following it. Very interesting work you're doing.

I just realized you're not new to chem.se, but welcome anyway! You have some seriously cool questions!

The following looks like a solution, unless anybody disproves it. Thank you @Unskilled, for pointing me in the right direction.
If you use structure 1 with smiles_1a: C12=CC=CC=C1C3=CC=CC4=C3C2=CC=C4, you will not find structures 3 and 5.
If you take OpenBabel and convert smiles_1a to SMILES you get: c12ccccc1c1cccc3c1ccc3.
If you take this SMILES string and convert it to mol via Chem.MolFromSmarts() you will find all structures. However, I don't want to use additional external tools.
So, in RDKit, if you convert smiles_1a to mol and this mol back to SMILES again, you get c1ccc2c(c1)-c1cccc3cccc-2c13. If you search with this, you will still not find structures 3 and 5. Probably because of the defined single bonds. However, if you replace - by ~, you get smiles_1b: c1ccc2c(c1)~c1cccc3cccc~2c13. With this, you will find also structures 3 and 5. Happy End, hopefully.
Code: (I also added Benzene to have a non-match)
from rdkit import Chem

smiles_list = ['C12=CC=CC=C1C3=CC=CC4=C3C2=CC=C4', 'C12=CC=CC=C1C3=CC=C4C5=C(C=CC2=C35)C6=C4C=CC=C6', 'C12=CC=CC=C1C3=CC=C4C5=C(C=CC2=C35)C=C4', 'C12=CC=CC=C1C3=CC=C4C5=C(C=CC2=C35)C6=C4C=C7CCCCC7=C6', 'C12=CC=CC=C1C3=CC=C4C5=C(C=CC2=C35)C6=C4C=C7CC8=CC=CC=C8CC7=C6','c1ccccc1']

def search_structure(pattern):
    for idx,smiles in enumerate(smiles_list):
        m = Chem.MolFromSmiles(smiles)
        print("Structure {}: pattern found {}".format(idx+1,m.HasSubstructMatch(pattern)))

smiles_1a  = smiles_list[0]
pattern_1a = Chem.MolFromSmiles(smiles_list[0])
smiles_1b  = Chem.MolToSmiles(pattern_1a).replace('-','~')
pattern_1b = Chem.MolFromSmarts(smiles_1b)

print("\nSMILES 1a: {}".format(smiles_1a))
search_structure(pattern_1a)
print("\nSMILES 1b: {}".format(smiles_1b))
search_structure(pattern_1b)

Result:
SMILES 1a: C12=CC=CC=C1C3=CC=CC4=C3C2=CC=C4
Structure 1: pattern found True
Structure 2: pattern found True
Structure 3: pattern found False
Structure 4: pattern found True
Structure 5: pattern found False
Structure 6: pattern found False

SMILES 1b: c1ccc2c(c1)~c1cccc3cccc~2c13
Structure 1: pattern found True
Structure 2: pattern found True
Structure 3: pattern found True
Structure 4: pattern found True
Structure 5: pattern found True
Structure 6: pattern found False

I'm not sure about why it's not matching but when I carry out substructure matches in rdkit I use SMARTS instead https://www.daylight.com/dayhtml/doc/theory/theory.smarts.html
Here is what I would have done.
from rdkit import Chem

smiles_list = ['C12=CC=CC=C1C3=CC=CC4=C3C2=CC=C4', 'C12=CC=CC=C1C3=CC=C4C5=C(C=CC2=C35)C6=C4C=CC=C6', 'C12=CC=CC=C1C3=CC=C4C5=C(C=CC2=C35)C=C4', 'C12=CC=CC=C1C3=CC=C4C5=C(C=CC2=C35)C6=C4C=C7CCCCC7=C6', 'C12=CC=CC=C1C3=CC=C4C5=C(C=CC2=C35)C6=C4C=C7CC8=CC=CC=C8CC7=C6', 'c1ccccc1']

pattern = Chem.MolFromSmarts('[#6]12~[#6][#6]~[#6][#6]~[#6]1[#6]3~[#6][#6]~[#6][#6]4~[#6]3[#6]2~[#6][#6]~[#6]4')
for idx,smiles in enumerate(smiles_list):
    m = Chem.MolFromSmiles(smiles)
    print("Structure {}: pattern found {}".format(idx+1,m.HasSubstructMatch(pattern)))

and the output
Structure 1: pattern found True
Structure 2: pattern found True
Structure 3: pattern found True
Structure 4: pattern found True
Structure 5: pattern found True
Structure 6: pattern found False

I've added a Benzene smiles to show that it's not just matching everything. The SMARTS pattern I made probably could be better I think. I hope this helps anyway.

Thank you for testing. Interesting. Then I need to check how I can reliably convert SMILES to SMARTS. Why it doesn't work with SMILES? Still a bug in RDKit?

I'm not sure why it doesn't work with SMILES, I've always used the m.HasSubstructMatch method with SMARTS. But learning SMARTS is pretty easy, its similar to SMILES just follow the link in the answer.

I just read that a SMILES is also a valid SMARTS expression. But if I use Chem.MolFromSmarts(smiles), all matches are False.

Yes most SMILES patterns are valid SMARTS but the SMARTS pattern 'C12=CC=CC=C1C3=CC=CC4=C3C2=CC=C4' is not the same as '[#6]12~[#6][#6]~[#6][#6]~[#6]1[#6]3~[#6][#6]~[#6][#6]4~[#6]3[#6]2~[#6][#6]~[#6]4'. The [#6] means any carbon atom aromatic or aliphatic while in the first pattern the C means an aliphatic carbon. Since at some point rdkit will make certain carbons in your molecules aromatic it will mean that it will not match. Also ~ means any bond while = in the first pattern is a double bond rdkit will at some point change some of your molecules bonds to aromatic bonds so will not match.

OK. I see. But how do I get a SMARTS from a SMILES? So far, couldn't find it in RDKit. From SMARTS to Mol and from Mol to SMILES, but not the other way round. OBabel? I wanted to avoid an extra tool. I need to search further.

I don't know if there is a function of sorts to do this but I literally took the SMILES 'C12=CC=CC=C1C3=CC=CC4=C3C2=CC=C4' then replaced the characters C ->  [#6] and = -> ~ to get '[#6]12~[#6][#6]~[#6][#6]~[#6]1[#6]3~[#6][#6]~[#6][#6]4~[#6]3[#6]2~[#6][#6]~[#6]4'.

ah... that easy? A simple search&replace? I hope this will also work for more complex structures, heterocylces, etc... I will check and let you know.

yea for this case but please read through https://www.daylight.com/dayhtml/doc/theory/theory.smarts.html the SMARTS pattern I made is very basic and might match some thing you may not want it to. Once you understand SMARTS you'll be able to make your own patterns that will be specific to the things you want.


text:
Change in inductive effect of a susbstituent when -R group replaces -H atom
There are several examples where the negative inductive effect of a substituent gets increased when a hydrogen atom on that substituent is replaced by an alkyl group.
Some particular examples:
$$\ce{-NR3+} > \ce{-NH3+} \qquad \ce{-OR} > \ce{-OH}$$
I don't understand why it should be that way, as an alkyl group shows more positive inductive effect than a hydrogen atom which should, in turn, reduce the negative inductive effect of the substituent.
The answer here gives the reason that a carbon atom is more electronegative than a hydrogen atom and hence it should increase the −I effect (the answer also talks about bond length but I think it's irrelevant to mention that). But wouldn't the inductive effect shown by the whole group be a better parameter than the electronegativity of its primary atom to judge its electron withdrawing/donating ability?
Then there are also examples where replacing hydrogen atom by an alkyl group reduces the negative inductive effect of the substituent, as with $\ce{-CHO}$ and $\ce{-CRO}$ groups. I would usually explain it using the +I effects of $\ce{-R}$ and $\ce{-H}$ groups, but since that logic doesn't hold for the previous case, there should be some other reason for it.
So my question is: How does replacing an H atom in a substituent by an alkyl group affect the inductive effect of the substituent?
P.S. There is a related question, but it is too specific and does not have an answer to my question.

There are several examples where the negative inductive effect of a substituent gets increased when a hydrogen atom on that substituent is replaced by an alkyl group.
Some particular examples:
\begin{align}
\ce{-NR3^+} &> \ce{-NH3^+}& \ce{-OR} &> \ce{-OH}
\end{align}

About comparing $\ce{-NR3^+}$ and $\ce{-NH3^+}$, at first, the electron-donating inductive effect of the methyl groups (+I) makes us feel that the electron-withdrawing inductive effect trend should be the opposite. Neither, could I find any reliable sources with an explanation. So, I came up with a few of my own.
Part I: The Bent's Rule Interpretation (More Reliable)
To analyse the inductive effect, let's take the groups attached to a hydrocarbon chain and then see.

The $\ce{C-H}$ bond orbitals will have more s-character than in the $\ce{N-C}$ bond orbitals because there is more electron-density in the former due to the greater electronegativity of carbon over hydrogen and nitrogen over carbon, which requires the presence of more, lower-energy s-character in them to lower the system's overall energy. Thus, an implication of this would be the increase of p-character in the three $\ce{N-C_{Me}}$ bond-orbitals and the subsequent decrease of p-character in the $\ce{N-C_{chain}}$. This results in an increase of electronegativity of $\ce{N}$, thus the (-I) trend is:
$$\ce{-NR3^+} > \ce{NH3^+}$$
Part II: I don't know what to call this (Need help with this)
Let's consider the examples of phenol and anisole. It is considered that phenol donates more electron density to the benzene ring than anisole because it exists in equilibrium with the phenoxide ion, which is a far-more electron-donating group, despite anisole's $\ce{-OCH3}$ group being a better electron-donating group.
If we apply this to our case, we can assume that $\ce{-NH3^+}$ may be in equilibrium with $\ce{-NH2}$ which makes it a better (+I) group and poorer (-I) group, while $\ce{-NMe3^+}$ can't deprotonate itself and establish an equilibrium like that, making it a better (-I) group.
(somebody please help here)
Part III: Should we really consider $\ce{-CH3}$ as a (+I) group here? (Just thinking, very reliable*)
It is considered that the inductive effect of $\ce{-CH3}$ is mainly attributed to its hyperconjugative effect. However, hyperconjugation requires an empty p orbital carrying a positive charge into which it donates its $\sigma_{\ce{C-H}}$ bond-electron density. However, nitrogen's orbitals are sp³ hybridized here, with no empty orbital carrying the charge as such. So, I think (not really sure*), we should only consider carbon's greater electronegativity over hydrogen, thus making $\ce{-NMe3^+}$ more inductively electron-withdrawing than $\ce{-NH3^+}$. You can also apply this to $\ce{-OR}$ and $\ce{-OH}$.
* Just saw Jan's answer in the middle of my writing process. That makes this a valid point. Great answer, a must check-out.

Although the last point can provide solution to all the questions that I have asked, it's not a very popular or widely known fact. Even orthocresol, whose answer is just below Jan's answer to the same question, didn't mention anything like that.

I admit it's not well-heard. Just on that account, I mentioned my uncertainty on it. My motive was to provide different ways in which I could think about the situation. Although if you see the first part, you can also use that for answering both your questions. It applies to the OH/OR comparison as well. Edit: If you find the interpretations helpful, an upvote would help :)

I didn't understand the first point. Why should N-C bond have lower electron density? In fact, I read this on SE Chemistry only, that the electron density in NR3+ is more on N and C atoms, while the positive charge is spread on H atoms.

May I have a link to the SE Chem answer that states otherwise?

Sorry, I just made a mistake in my comment. The Bent's Rule suggests in the case of CH₃F that the fluorine will lessen the electron density in the CF bond orbital, thus increasing p character in it, and in turn increasing s-character in the CH bonds where there is more electron density than in CF, where F has pulled it substantially. That leads to somewhat acidic hydrogen due to increased electronegativity in CH bond orbitals. But, you know about Bent's Rule so no worries. :) The proposition that the electron density is more on N and C atoms doesn't seem to be valid because the carbon directly

attached to the positive carrying nitrogen should have less electron density. I do feel that hydrogens at beta positions would have a tendency to protonate (like you said, positive charge on H atoms) to undergo Hoffmann Elimination however I don't feel it would be with the hydrogens on the methyl, which are at alpha.

https://chemistry.stackexchange.com/a/72216/43333

@VijayBhardwaj if we attribute the inductive effect of an alkyl group to its hyperconjugative effect, then how would one explain the higher acidic strength of butan-1-ol than that of tert butyl alcohol?

I have updated your post with chemistry markup. If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details.

@AumkaarPranav the most straightforward explanation here is that tBuO- is poorly solvated because of steric crowding around the oxygen. Less stable conjugate base = weaker acid.

Also, I really don't buy the Bent's rule explanation: there are a few holes in the logic here. Firstly, the shading doesn't make sense, because s- or p-character is a characteristic of orbitals, not bonds; and there are two orbitals to a bond. So each bond should have two colours. The second thing is that the orbitals that C uses to bond to N has no influence on the orbitals that N uses to bond to C (Bent's rule only deals with orbitals on the same atom). So there isn't any "chain of reasoning" that lets you figure out the orbitals that atom Z uses, based on the orbitals that atom X uses.


text:
Calculating exposure time to inactivate coronavirus by UV light
Sorry for a layman question. I'm having difficulties understanding some basic math apparently.
According to a paper, for a coronavirus, the UVGI dose for 90% inactivation is 339–423 μW sec/cm2.
I have a UV lamp with the following specifications:

it's 15 watts
it takes 1h30 min to 90% inactivate coronavirus in a 20 square meters room with a 3-meter ceiling.

But that doesn't make any sense if you believe the paper's number.
This is how I calculate the necessary exposure time:

the surface of the floor, ceiling, and walls in the specified room is 94 square meters (or 940,000 square centimeters.)
15 watts is 15,000,000 microwatts
from the 339–423 μW sec/cm2 in the paper, take the highest number

So since our surface is 940,000 cm2, we multiply 423 * 940,000 and that gives us 397,620,000 μW sec/surface of the room.
Since our lamp is 15,000,000 microwatts, we divide 397 million by 15 million, and we get 26 (watts provided by our lamp) * sec/surface of the room.
So 26 seconds is the necessary time exposure, but it should be 5400 secs as specified in the manual. Is there an error somewhere?

That paper says "airbourne" viruses. And the guy who sells that UV lamp is guaranteed to be a quack. ;)

At 15 watt light bulb only refers to the amount of electricity needed, and not to the number of photons produced whose total energy will be several orders of magnitude less.

Also, aside from actual energy emitted, efficacy depends on the region of the UV spectrum. UV A and B have little or no germicidal effect. UV C is effective... and also damages skin and eyes in the short term, and can cause skin cancer and cataracts over time. Which type of lamp do you have: safe and useless, or effective and unusable in an inhabited area?

The latter kind.

I want stare at or watch even a UV a / b

yes, there are several errors in your calculation.
Germicidal lamps specify an effective power output in UVC. This lamp
https://www.osram.de/ecat/PURITEC%20HNS%20UV-C-UV-C%20lamps%20for%20purification-Ultraviolettstrahler-Industrie-Spezialbeleuchtung/de/de/GPS01_1028570/ZMP_1022255/
consumes 16W of electricity but has a UVC output of 4W. You need to calculate with the 4W value.
Then you need to calculate the intensity of your irradiation on the surface you want to inactivate the virus on. The distance from the lamp to the surface is key, as the 4W UVC power remains contant over the surface of the light cone that shines from it. In the lamp's data sheet it says that this is valid for up to 3m from the lamp. (after 3m it most likely decreases due to other effects.) So you need to calcuate the distance from your lamp, and you need to calculate your powerdistribution over the surface of the sphere/cylinder (depending on your geometry), with the lamp in the center. It does not do to calculate the surface of the room as a whole. Given the irradiation intensity on the surface of that sphere you can now calculate the time needed until the UVC output of the germicidal lamp manages to reach the 90% inactivation dose for a given virus. For corona viruses different values for that dose are given in different papers. Some say 10.3mJ/cm^2 for 99%, some quote 0.6mJ/cm^2 for 90% inactivation. This one gives a good overview: https://www.nature.com/articles/s41598-020-67211-2.pdf
I recommend to choose a simple geometry for calclation of the irradiation power. Living rooms are quite tricky. Spheres are better.

Likely related disinfecting questions are, apparently, daily becoming more of interest, and while not always best presented, are certainly worthy of scientific-based comments.
My suggestion, perhaps staining a virus-infected surface, with say a B-12 based photocatalyst (see, for example, Enhanced photocatalytic activity of a B12-based catalyst co-photosensitized by TiO2 and Ru(ii) towards dechlorination), that releases hydroxide radicals capable of decomposing organics (including DNA) with UV light irradiation (or strong sunlight), is, coming from a non-expert in biocides, a suggested path. However, there are supporting sources (see this 2010 article available for free online: 'Applications of Photocatalytic Disinfection' and this 2015 paper 'Advances in photocatalytic disinfection of bacteria: Development of photocatalysts and mechanisms').
Upon complete decomposition of all present stained organics, the solution becomes colorless (likely measurable).
Interestingly, the article mentions measurable results in 30 minutes with UV irradiation. My experience with sunlight acting on aqueous B-12, and also intensely colored B-complex solutions, demonstrates surprising results (decoloration) in a few hours of sunlight.
In your particular case, add B-12 to a disinfecting alcohol mix (for example, one containing over 50% isopropanol, or especially ethanol for addressing a virus concern, but have at least, 10% water) onto a white surface and apply your UV product (or, say irradiation by a natural sunlight electric bulb) for testing (take periodic pictures). Note, decoloration by itself, is only a chemistry-based presumption of disinfecting effectiveness.
My proposed solution replaces the estimation issue with visible results, at least with respect to decoloration, by applying a doable photocatalytic spray. If your UV lamp is 'safe and useless', perhaps you can convert it into something that is part of a safe and effective protocol.

The effectiveness of the lamp diminishes at a rate of twice the rate of the increase in distance. You have a kill rate o lets say 100 at 1 ft. You will have a kill rate of 50 at 2  ft an a kill rest of 25 at 3 ft you don't realy have any major worries about skin issues with the small lamps, if you are not sleeping nude under them. The ozium is as big a health issue as eye burn linet your exposure to breathing the sweet air produced by the lamp it as effective at killing germs and viruses as the light more effective at killing bacteria and mold. Protect your eyes with a welding eye shield for prolonged exposure and at least a good par of sunglasses for anything more than a minutes of direct eye exposure (looking at the light directly). Protect your pet's also signs of excessive eye exposure is red burning eyes that feel like thay are full of corse sand the same as if you have been watching someone weld without eye protection. If you are a few feet away, you don't have much skin worries with the small low power lamps. UVC will not penetrate your clothes.


text:
Comparing basic strength of acetamidine and 2-imidazoline

My approach:

Conjugate acids of both molecules have the same number of stable resonance structures. So no conclusion from here.
Conjugate acid of 2-imidazoline will have negative inductive on the lone pair donating nitrogen by the other nitrogen atom. Since this effect is absent in acetamidine, acetamidine should be a stronger base.

Is the above approach right? Can ring strain affect the basic strength of 2-imidazoline in any way? Should I consider any other factor too to make the right comparison of the basic strength of the above two molecules?
Related Questions:
Comparing basicity of imidazole and 2-imidazoline
How to predict the basicity of organic amines?


text:
Why doesn't ethanol conduct electricity?
As far I know ethanol is soluble in water in all proportions, and since water is polar it would imply that ethanol is also polar and the only possible polar group in ethanol is the -OH group.
Oxygen is much more electronegative than hydrogen, so oxygen has to attract the electron pair towards itself forming a polar compound.
This means it should have to disassociate into ions in aqueous solution and conduct electricity but it doesn't.
In some other posts of the same site mentions, why ethanol doesn't disassociate much in water, but it doesn't explain why the ethanol is still soluble in water in all proportion even it is not get disassociating much.
Why?

Pure water isn't a very good conductor either...

Really, water itself isn't polar enough to dissociate much; why should ethanol be?

Neutral particles are by definition not charge carriers and can therefore not contribute directly to the conductivity of a solution. You need dissolved charge carriers, in the case of a pure solvent at low applied fields usually ions resulting from self-dissociation of the solvent molecules or some other charge transfer mechanism.

Related: https://chemistry.stackexchange.com/questions/31507/electrical-conductivity-of-organic-solvents

For ethanol, the autoionization constant, at room temperature, is  $\mathrm 1.0 x 10^{-20}$. This is what I have seen in various places on the web, so it needs a serious reference, but it seems reasonable.

So the resistivity of pure ethanol at room temperature should be roughly 18 $\text{G}\Omega\cdot \text{cm}$, which is a thousand times higher than pure water at room temperature. This means a given potential applied to a pair of electrodes will result in currents a thousand times smaller that in water, ceteris paribus.

Why should pure water and pure ethanol dissociate ? Substances get dissociated into ions if they win some stability in this process. NaCl is more stable in the dissociated state than in the covalent. In ionic NaCl, all atoms look like rare gases, and respect the octet rule. In a theoretically covalent structure, Na does not respect the octet rule. So it is less stable.
But ethanol is made of atoms which all respect the octet rule. So why should they dissociate ?

If ethanol doesn't disassociate than why is it soluble in all proportions? Please explain


text:
Why do ionic compounds form in the exact ratios that they do instead of more variable ratios?
I read that e.g. $\ce{Mg^{2+}}$ and $\ce{Cl-}$ come together because of electrostatic force or coulomb force and form an ionic bond. Then why is the formula $\ce{MgCl2}$? Why does one $\ce{Mg^{2+}}$ attract only two chlorides (if Coulomb's law is at work) and form an ionic bond with two chlorides only? When the ions come close and form a bond, they don't distribute the charge i.e. their electrons in such a way that they would have the same potential. They remain ions even after forming the bond. Then why don't multiple $\ce{Mg^{2+}}$ attract multiple chlorides and form something like $\ce{Mg_xCl_y}$? Why does a single molecule of the compound contain one $\ce{Mg^{2+}}$ and two chlorides only?Someone said that,because more clorides would then repel one another so more clorides won't be able to stay with one  magnesium ion in one matrix .if we use the coulomb's law kq1q2÷r^2,we should find that the attraction between mg2+ and cl- would be greater than the repulsion between 2 cl-.I mean if we take q1=-1,q2=2,the force will be greater than if we take q1=q2=-1. Then the attraction force between mg ion  and cl ion should be able to surpass the repulsion between cl ions and one mg should be surrounded by more than 2 clorides in a crystal.

In solid $\ce{MgCl2}$ there are no particular molecules of $\ce{MgCl2}$. The ions are in a solid matrix.

Please guys,be sympathetic,I am just learning the ABCDs of chemistry.don't downvote it if u can't help me understand.

Alternatively, magnesium chloride can be boiled (1412°C). The vapor would then likely consist of either MgCl2 molecules or a mixture of these with oligo-mers having that atomic ratio.

Thing is your premise is wrong - it happens pretty much as you think it should! Also title is wrong at the moment. It's not about proportion, but oligomerization, crystal formation and nature of ionic bond.

If you succeed in maintaining Mg2+ ions and Cl- ions in a proportion different from 2:1, the whole system will not exist more than a couple of nanoseconds. The whole system will be charged, and the excess ions will repel one another, destroying the system.

@Maurice Well, if that were to be a billion atoms with proportion 3:1 then yeah, but if proportion was, say, about 2 000 001:1 000 000 then no.

In some sense, OP is not wrong: $\ce{Mg^{2+}}$ will attract and interact with more than two $\ce{Cl-}$ (and vice versa), in that it will form an "infinite" lattice structure - an ionic crystal (and not molecules, except at extreme conditions that I will ignore here). If you consult any depiction of the structure, you see that every $\ce{Mg^{2+}}$ inside the structure is surrounded (we call it coordinated) by six closest neighbours that are $\ce{Cl-}$.
However, if one were to put only ever more $\ce{Cl-}$ ions around a central $\ce{Mg^{2+}}$, the ions with the same charge would begin repelling each other. To avoid this, $\ce{Mg^{2+}}$ begin grouping themselves around the charge that attracts them. In this way, the crystal forms layer by layer.

'if one were to put only ever more Cl− ions around a central Mg2+, the ions with the same charge would begin repelling each other'.why would they repel so strongly?according to coulomb's law the attraction between mg2+ and cl- should be greater than the repulsion among cl- ions.right?the charge of mg ion is 2,whereas the charge of cl ion is 1 if we use the formula kq1q2÷r^2 , that's what we are going to find.so  Has the explanation behind that been found ,or it is what has been observed?

No. Once you have your charges balanced, you have two Cl-. They repel another Cl- with the same exact force that Mg2+ attracts it.


text:
Different gibbs energy equations
For temperature dependence of Gibbs energy, Why we  Gibbs-Helmholtz equation rather than finding directly in the same way as Gibbs energy dependence on pressure?
$$\frac{\mathrm dG}{\mathrm dp} = V \Rightarrow \mathrm dG = V\,\mathrm dp \quad\text{for solids and liquids}$$
By pressure I mean external pressure applied. Like for gases $\mathrm dG = nRT \ln(p_\mathrm f/p_\mathrm i)$. Why we use this equation $(G/T)/T = H/T^2$ for change in Gibbs energy dependence of temperature. Why not this equation $\mathrm dG = \mathrm dT\times(-S)$?

Please add more detail such as what you mean by "Gibbs energy dependence on pressure". Temperature and pressure are not the same variable, so your question does not make sense.

dG/dP = V => dG = dPV (For Solids and liquids).
By pressure i mean external pressure applied. Like for Gases dG = nRT In(pf/pi). Why we use this equation (G/T)/T = H/T^2 for change in gibbs energy dependence of temperature. Why not this equation dG = dT(-S)

I added your clarification directly  to the post.

See the bottom of the Wikipedia page for the Gibbs-Helmholtz equation for a useful derivation: https://en.wikipedia.org/wiki/Gibbs%E2%80%93Helmholtz_equation

Why we use this equation [and ...] not this equation?

It depends what problem you want to solve.
$\frac{dG}{dT}$ is of interest, for example, if you want to know the maximum work available from a process at different temperatures.
$\frac{d(G/T)}{dT}$ is of interest, for example, if you want to know how the equilibrium constant changes with temperature (because $\Delta G^\circ = -R T \ln K$).


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Does toxicity depend on state of matter?
Does the toxicity of an element depend on the state of matter (liquid, solid, gas)?
Could there be an element, which is totally non-toxic when it is solid, but toxic when it is liquid. Which elements would fall in that category?

State of matter has the major effect how much and which way the substance gets into the body. That means, while toxicity is the same, the effective toxicity differs very much.
Similarly, if a medicine is applied, the total amount of the drug is just a part of the effect. The other part is the form and the way how it is applied.
Generally, toxicity ( or medication effect ) decreases in order:
gas/vapour > inhaled aerosol or infusion liquid > swallowed liquid/solution > solid/dust > solid/pieces
as it follows the effectiveness of the substance transfer to body fluids on molecular level.
Typical example is mercury.
Liquid mercury has relatively low toxicity. If you digest it, the major symptom is diarrhea.
Mercury vapours, that are created by slow evaporation of spilled mercury, are highly toxic, if they are inhaled.
Another case are insoluble salts of poisonous metals. Solutions of barium salts are poisonous, but ingestion of thick milk ( rather fluid mash ) of insoluble barium sulphate is/was used in roentgen examination of digestion tract.


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What is meant by position 5' or 3'
I'm a self-learner that likes to keep up with current research. However, I can't find what is meant by expressions such as 5' or 3' positions on chemical structures. But , in particular, what is meant by the phrase "Cis-elements may be located in 5' or 3' untranslated regions or within introns."

You could have just as well asked about the meaning of "x" in mathematics. It could have been anything. But wait... did you mean it in a somewhat narrower scope? A particular kind of equations, by any chance? A particular kind of chemical compounds? Oligosaccharides, by any chance? A particular kind of those, maybe?

The DNA-RNA tag is a hint, as is introns and cis-elements (regarding the specific question).

About all that be said is that for a particular compound there is a specific way of numbering the atoms.

@ChrisGardner  Putting the keywords RNA/DNA and introns with the above comments together, maybe these two sites may help, at least as a starter, to orient along the strangs: https://en.wikipedia.org/wiki/Untranslated_region and https://en.wikipedia.org/wiki/Directionality_(molecular_biology)

I suggest getting hold of a biochemistry textbook, if you can: any half-decent one will tell you what 3' and 5' are (in the context of nucleic acids) and why they are important.

Here is a link to a more than half-decent biochemistry textbook chapter explaining the numbering: https://www.ncbi.nlm.nih.gov/books/NBK22490/
And the relevant figure from that chapter:

[Berg, Tymoczko, Stryer] Backbones of DNA and RNA. The backbones of these nucleic acids are formed by 3′-to-5′ phosphodiester linkages. A sugar unit is highlighted in red and a phosphate group in blue.

Like any other linear polymer sucu as proteins and DNAs, RNA structure has two ends. Like carboxylic acid end and amine end notation given for proteins, two ends of a RNA (and DNA) call 3'-end and 5'-end (see the diagram below):

Note that these notations are derived from the numbering of the ribose sugar function (see the 5'-end of the tetramer in the diagram).
Answer for your particular question: What is meant by the phrase "Cis-elements may be located in 5' or 3' untranslated regions or within introns"?, see diagram below:

Cis-regulatory elements (CREs) such as promoters, enhances, silencers, etc. are usually in 5' or 3' untranslated regions.


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Hydrochloric acid substitute in an Iron-gall ink solution
Recently I started experimenting with making my own ink to be used in fountain pens.
I was wondering if hydrochloric acid in an iron-gall ink is used to prevent sediment ONLY by lowering the pH of the solution or if it also does something else that I don't know.
The recipes found on Wikipedia call for varying amount of hydrochloric acid to be added [3-5cm^3 in 1000ml solution]; however I am quite sceptical of adding it to my ink because of its corrosive nature. I even spoke with one manufacturer of modern iron-gall inks and they said they do not use said acid in theirs; they also said it shouldn't really be a problem since its in such a small concentration.
However I am still reluctant to use it honestly.
I've made couple of test batches with citric acid as a replacement, i added 2 and 5 grammes of it but after about half a day I noticed some sediment forming on the bottom of the container (which I assume is the iron). The ink works alright although I would like the ink to be more stable and predictable.
Here is the recipe I was following, however I cut the water to 500mls:
11.7 g tannic acid
3.8 g gallic acid C6H2(OH)3COOH
15 g iron(II) sulfate
3 cm3 hydrochloric acid (used to prevent sediment forming)
1 g carbolic acid (phenol) C6H5OH (biocide)
3.5 g china-blue aniline dye (water-soluble)
1000 cm3 distilled water

As you can probably see I am no chemist at all, I do not got much clue about it, but it would be lovely if you could tell me if and what I could replace the hydrochloric acid with in order to keep the solution under pH 3. I hope this makes sense.

Most of replacements to keep the same very low pH would be more, not less corrosive.
As it is low pH, which is corrosive, not HCl presence. Replacements like organic acids would be more corrosive as there would be needed a higher molar amount of such an acid for that, taking more of a base to neutralize that.
Additionally, complexing abilities of most of organic acids would contribute to corrosion even more and could affect forming of iron-tannin complexes.


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How to figure out the most stable conformation of the diastereomer?

Question 1: Which of the following Newman projections represent the most stable confirmation of the diastereomer of the molecule given below?

I guess my main question is looking only at the Newman projection, how does one determine the stereochemistry?
In the molecule to the right, I get for the first carbon, ($\ce{Et, H, Me}$) R (I flipped it after moving the $\ce{H}$ back) and for $\ce{C}$2 ($\ce{Et, Br, Me}$) I get S.

Question 2: Which of the following Newman projections represents a confirmation of the enantiomer of the molecule given below?

In the same vein, for the second question, I thought I should flip all the stereocenters to get the enantiomer since it is a mirror image. But I am not getting the answers the professor provided to us.

A side note: For the question in title, you could just move on with the Newman projections to determine the relative stabilities of various conformers. Large groups which are anti to each other have the least energy possible, and hence are more stable. I don't see why you want to convert the Newman structures to 3d bond line structures for the question and I consider it to be unnecessary for this question.

To answer your professor's questions, there is no need to determine the R/S configuration of the two carbon atoms. We have to check the clockwise/anticlockwise arrangement of the substituents around the carbon atom, but checking specifically whether it is R or S is unnecessary.
Before moving further, let's have a look at the following image so that we are clear on how to visualise Newman Projections:

Since in the options provided in the question the carbon atom having bromine is at the back, we'll have to look at the molecule in the following manner:

Hence the Newman Projection obtained would be this:

Now, in question 1, option A can be rejected as it's an entirely different molecule. Option 4 has the atoms in the same order as above, so it is the exactly same molecule. In option 2 and 3, the arrangement of atoms is same for the back carbon but different for the front carbon. Therefore, without finding R or S configuration, we can say that options 2 and 3 are the diastereomers!
Out of these two structures, option 3 is more stable as the bulky ethyl groups are at the farthest position from each other.
Moving on to question 2:
You are right in flipping the arrangement about both the carbon atoms and you should get the answer that way. Maybe you are doing some mistake in getting the right Newman projection. Look again at the Newman projection I've provided above and then compare the options with it to see which option has different arrangement of substituents on both the carbons.

A table of A-values indicate that $\ce{Br}$ substituent on a long chain organic molecule has relatively less steric effect than that of $\ce{Me}$- or $\ce{Et}$-group ($1.7$ and $\pu{1.75 kcal mol-1}$, respectively compared to $\pu{0.38 kcal mol-1}$ for $\ce{Br}$), and even smaller than that of $\ce{Cl}$ ($\pu{0.43 kcal mol-1}$), probably due to longer $\ce{C-Br}$ bond. Thus, it is safe to assume that both $\ce{Br}$-$\ce{Me}$ and $\ce{Br}$-$\ce{Et}$ gauche interactions are smaller than ether of $\ce{Me}$-$\ce{Me}$ or $\ce{Me}$-$\ce{Et}$ or $\ce{Et}$-$\ce{Et}$ gauche interactions.
Now let's look at the questions in hand:

OP's main question: Looking only at the Newman projection, how does one determine the stereochemistry?

I already gave the comparative gauche interactions by $\ce{Me}$, $\ce{Et}$, and $\ce{Br}$ groups. First, I want to point out that the compound $\bf{1}$ is either written wrong or intentionally put a $\ce{Me}$ group on the back carbon (see the group marked by green circle). I think it should be an $\ce{Et}$ group so that it would relate to the given reference compound, $\mathrm{\bf{I}}$. So, I made the adjustment as of it is an $\ce{Et}$ group.
Since the given compound is (3S,4R), its diastereomers should be (3R,4R) and (3S,4S) as depicted. Therefore, it is necessary to find out that all given compounds are diastereomers before you work on the most stable conformer. As indicated, compounds $\bf{1}$, $\bf{2}$, $\bf{3}$, and $\bf{4}$ are either diastereomer at different conformation. I have put how many gauche interactions in each compound. When roughly calculated, $\bf{1}$ and $\bf{2}$ have almost similar energy. However, $\bf{3}$ has less energy than that of $\bf{1}$ and $\bf{2}$ (e.g., $\bf{3}$ has an extra $\ce{Me}$-$\ce{Br}$ interaction in place of $\ce{Me}$-$\ce{Me}$ interaction in $\bf{1}$). However, if you look at carefully, $\bf{3}$ is exact mirror image of $\bf{4}$, thus have the identical energy (See the diagram above). Therefore, actual answer is both $\bf{3}$ and $\bf{4}$ ($\mathrm{C}$ and $\mathrm{D}$ in the grid).
The easiest method to answer OP's second question is to identify the (R,S) configurations of given answers. Since that of the given compound is (3S,4R), its enantiomer is (3R,4S), which is $\bf{6}$ ($\mathrm{B}$ in the grid; I have changed the original numbering for clarity). Other three options: $\bf{5}$ and $\bf{7}$ are diastereomers of $\mathrm{\bf{I}}$ while $\bf{8}$ is identical to $\mathrm{\bf{I}}$.
Note: I have also put relevant gauche interactions in each compound of second question for the readers' benefit.


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Why does ψ^2 ( square of orbital wave function ψ) give the probability of an electron in a given region?
Today I was introduced to the Orbital Wave Function for electrons.
$\Psi$ is a mathematical function for coordinate of electrons and has no physical meaning.
But $\Psi^2$ gives probability of an electron.
How does a function for coordinate give probability distribution when squared ? How is the $\Psi$ working ?
Please explain me in easier to understand terms with out monster equations.

What's the monstrous equation (Schrödinger's equations)? And state of the particle is completely described by $\Psi$, it got it's meaning.. doesn't $\Psi^2$ gives probability density?

Discussed here in the question and answers: https://physics.stackexchange.com/q/194999

@Zenix yes! I don't really understand it. I suppose it relates the various energies of the particle with its coordinates. Am I right?

https://chemistry.stackexchange.com/questions/92244/what-is-the-difference-between-%cf%88-%cf%88%c2%b2-radial-probability-and-radial-distribut

"Today I was introduced to the Orbital Wave Function" - well that suggests that you already know more then you really need on your level of education, and teacher didn't tell you how it work for good reasons.

@Mithoron Yes that might be true but I have some basic knowledge about it and it's graphs and how to find nodes and all ! But why am I getting down votes? Is there anything wrong?

Not much, but it's modulus of function squared and, well, comments about monster functions are better... in comments not in body of question. Also I feel this may be too broad.

Please don't go in depth Navoneel, you don't really need to know more about than this, if you are preparing for competitive examinations. You can't grasp these things in 10+1 grade... little knowledge is always harmful. Remain focused on goal and have faith in your faculties. Rest depends on you, I will upvote so that you don't get demotivated.

This is the Born interpretation and it is a postulate about how quantum mechanics works. As far as we can tell, it's right. There is no easier way in quantum mechanics. You either look at the equations and get some insight, or you don't.

Well,this depends.. Do you know the basic postulates of quantum mechanics,especially the 3rd one regarding eigenvalues and eigenvectors?(You should obviously know the Hermitian operators postulate and the normalization one)

Ok thanks to everyone in the comments! I realise I'll learn this when am eligible (i.e. in higher grade)


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Why does the combustion of graphite have a positive Delta S?

The following reaction is exothermic and entropically favorable:
$$\ce{C (s) + O2 (g) -> CO2 (g)}$$
My question: Why is this reaction entropically favorable?

Clearly there are more ways to arrange $\ce{C (s)}$ and $\ce{O2 (g)}$ in a container than $\ce{CO2}$.
Sure, the $\ce{CO2}$ has degrees of freedom in it's vibrations, but these shouldn't play a role in determining the entropy of the reaction since the reaction is completely unaffected by the type of vibration $\ce{CO2}$ is doing.

Have you got the actual number for delta S?

@Karl +2.9 J K–1

There is only one way to arrange С (s).

It seems that the gaseous state of CO2 is not considered in your question and in the above comment. As by @Ivan Neretin comment, do not expect, not to say clearly, more entropy in C than in CO2.

@IvanNeretin  "There is only one way to arrange C (s)."  Nope, that's only true at absolute zero.  I.e., if there really were just one way to arrange C(s) at, say, room temperature, then the absolute entropy of C(s) at room temp would be zero, which is clearly not the case.  Rather, at non-zero temperatures, C(s) has access to a host of different sets of vibrational states, each of which counts as a different arrangement, giving C(s) a non-zero entropy.

@dljs  "the reaction is completely unaffected by the type of vibration $\ce{CO2}$ is doing."  As you hopefully understood from Buck Thorne's answer, this is absolutely not the case.  $\Delta S$ for any reaction is $S_{products} - S_{reactants}$, so the entropies of both the reactants and products must be accounted for, regardless of whether their entropies come from translational, rotational, or vibrational sources (or any other sources that change during the reaction).

@theorist Yes! This is precisely where I'm confused. I don't understand at a molecular level why the vibrational entropy dictates anything about the reaction spontaneity. Translational entropy makes sense in affecting the reaction (in my mind), but vibrational and rotational do not. Translational entropy gives us the chances that 2 chemical species will be close enough to react, so the translational entropy should play a big role. Vibrational not so much (in my mind at least). I'm about the post a (bountied?) question on Physics Stack Exchange explaining my (flawed) mental model.

@djis  "close enough to react" would affect the kinetics, not the thermodynamics.  Entropy is a difficult concept. But, very simply put, the entropy is a statistical phenomenon: it is  proportional to the log of the number of possible macroscopically indistinguishable arrangements associated with a given state.  So if you allow a state more possible arrangements, regardless of whether those arrangments are translational, vibrational, or rotational, you increase its probability of occurrence, and thus its entropy.

@theorist I know I was exaggerating more than a little.

There are a number of possible explanations for the entropy difference, which can be computed based on standard free energies and enthalpies of formation as $\pu{3.0 J/molK}$. First, it could be that the entropy change is so small ($TΔS^∘<\pu{1 kJ}$, compare this to $ΔH^∘=\pu{−393.5kJ/mol}$) and measurement too imprecise for this difference to be significant.
However, an alternate conclusion can be drawn from basic expressions for the translational and rotational entropy of a linear molecule in a gas, as derived from statistical mechanics:
$$\begin{align} S_{\text{rot}}/R &\propto \log{\left( I\right)} \\ S_{\text{trans}}/R &\propto 1.5\log{\left( m\right)}\end{align}$$
where m and I are respectively the mass and moment of intertia of the molecule. The moment of intertia of $\ce{CO2}$ is roughly 4 times that of  $\ce{O2}$, therefore the above two contributions suggest that the molar entropy of $\ce{CO2}$ should be about $\pu{15.5 J/molK}$ greater than that of $\ce{O2}$.
Finally, note that vibrational contributions to the entropy of the gases are expected to be negligible based on basic statistical mechanical expressions. Vibrations in graphite contain softer contributions and therefore vibrations would be expected to contribute more to the entropy of the solid than to that of either gas. NIST provides a value of $\approx \pu{5.5 J/molK}$ for the standard entropy of graphite. This doesn't entirely account for the small value of $\Delta_r S^\circ =\pu{3.0 J/molK}$ based on enthalpies and free energies of formation, but brings the estimate a little closer, to $\Delta_r S^\circ \approx \pu{10 J/molK}$.


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Do molecules with bridges through rings exist?
Do molecules with bridges through rings (in a manner illustrated by this) exist?

I sometimes get results like this when doing Energy Minimization on molview.org. For example:

Is this actually a thing?
EDIT: As a slightly more realistic example, consider this:

When a lactone is synthesized, a small proportion of the chains close their rings when the chain is passing through a ring previously made. There are no bridges between the two rings, and they cannot get separated. But, who knows. May be in a later step, some bridges may be synthesized between the two rings.

This earlier Q&A entitled, Interlocked cyclic compounds, may be helpful.

Bad examples - these are non-existent.

https://en.wikipedia.org/wiki/Molecular_knot

How about posting that as an answer @Mithoron (sorry for the tone, I don't know how to make this sound more friendly)

You seem to be asking whether such molecules can exist in general. But are you implicitly asking whether the compound you modeled makes sense? Because what you're seeing is a common artifact of energy minimization that has nothing to do with chemistry. I've seen it referred to as "ring punch" but I can't find much more about it...

Not unless you rewrote question.  Even your "more realistic example" is not realistic at all.

@Calimo Do you have a link to where it is referred to as a ring punch?

Yes, molecules like that can exist (see the answer below). But the specific ones you give as examples cannot as 6-membered rings don't have enough space in the middle for anything to pass through. Draw the space filling version of a cyclohexane of benzene ring to see why not.

@schuelermine My comment doesn't belong to the other answers as it doesn't answer your question, and has nothing to do with any of the given answer. I don't have a pointer on top of my head but you should consider asking this as a separate question.

I'm not sure about the existence of molecules with bridges through rings. However, there are several publications of synthesis of molecules mimicking wheels and axles ([2]rotaxanes; The “[2]” refers to the number of interlocked components) as one shown below (Ref. 1):

(The diagram is from Reference 1)
This specific molecule (8; an “impossible” [2]rotaxane) represents a macro-cycle with a straight-chain molecule with bulky end groups going through its center. The inclusion of two bulky end groups prevents the straight-chain molecule leaving the macro-cycle (mechanically interlocked) as depicted in the diagram (See Ref. 2 for the total synthesis of the molecule).
Note that Ref. 1 also cited articles for the synthesis of [2]catenanes, which contain two interlocked rings (instead of one axle and one macrocycle). Keep in mind that there are some advanced catenanes and rotaxanes that exist (e.g., [3]catenanes and [3]rotaxanes).

(The structures are from Reference 1)
References:

Edward A. Neal, Stephen M. Goldup, "Chemical consequences of mechanical bonding in catenanes and rotaxanes: isomerism, modification, catalysis and molecular machines for synthesis,"  Chem. Commun. 2014, 50(40), 5128-5142 (https://doi.org/10.1039/C3CC47842D).
Jeffrey S. Hannam, Stephen M. Lacy, David A. Leigh, Carlos G. Saiz, Alexandra M. Z. Slawin, Sheila G. Stitchell, "Controlled Submolecular Translational Motion in Synthesis: A Mechanically Interlocking Auxiliary,"  Angew. Chem., Intl. Fd. 2004, 43(25), 3260-3264 (https://doi.org/10.1002/anie.200353606).

The examples look pretty much like the catenane chemistry pushed forward by Sauvage (https://en.wikipedia.org/wiki/Jean-Pierre_Sauvage).  The motif of cyclic polyethers / crown ethers, as well as the guiding «pre-orientation»  of a chain into an already existing ring by weak coordination (including transition metal complexation) often occured in Lehn's publications, too.

Catenanes can also be found in biology: the capsid of the bacteriophage HK97 contains 71 rings that pass through each other to form "protein chainmail". ( https://science.sciencemag.org/content/289/5487/2129.long )

A variation on this theme is Ice VII, in which two cubic ice structures are intertwined with hydrogen bonds from each component structure passing through the hydrogen-bonded rings formed by the other component.  Known to occur naturally on Earth as a high-pressure phase trapped in diamonds, Ice VII is a stepping-stone to the macromolecular and superionic ices believed to exist in some giant planets.

Source

Ring-opening polymerization, in particular using olefin metathesis which keeps growing closed rings in the absence of linear olefins, will produce macrocycles that are intertwined to form a pseudo cross-linked network, globally insoluble polymer. This can be later "un-crosslinked" by resuming the metathesis reaction with linear olefins.


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Determining if lithium acetate is anhydrate or dihydrate
I'm interested in trying a new gel electrophoresis buffer using lithium acetate. Before making a financial commitment, someone was able to provide me with enough lithium acetate from a different lab to try the buffer first. Lithium acetate is produced in both anhydrate and dihydrate form. The problem is that this batch was provided to me in a conical tube without any of the original product info, and the person didn't pay attention to this particular detail when taking it out of the original container. For annoying reasons beyond my control (nothing unethical or illegal), getting this info directly is taking a long time. I want to move forward with my work, and getting the molarity correct for the buffer is important, so I'm looking for options to determine which form it is. From what I can tell, the dihydrate form has a melting point of 54-56C, while the anyhydrate form has a melting point of 283-285C. So I was thinking that throwing it in an oven at 65C should give me my answer. Before trying it, though, is there any reason why this might be a bad idea, or just not work? Is there an easier approach?

You can make a solution with known weight of your acetate and titrate it with standard solution of $\ce{HCl}$. Them calculate molarity and MW.

Run a thermogravimetry.

Or DTA/DSC. Or XRD. IR, if you have one.

Since no one else has formally submitted an answer, here goes. I ended up finally getting confirmation that the original container is labelled as dihydrate, so the original question is moot at this point. With that said, I think Mathew Mahindaratne's comment to do a tritation using a standard solution of HCl should be the answer, in part because it can be done with fairly basic equipment. This is relevant for me because I work primarily with DNA, so I don't have access to a lot of specialized chemistry equipment outside of that scope of work.


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What is the entropy of mixing a gas into the whole atmosphere?
Suppose I release 1 mole of pure argon into the atmosphere. What is the entropy of mixing? The equations I have found seem to only apply to the case of a fixed volume. Should I use an estimate of the volume of the whole atmosphere? Some sort of infinite approximation?

Performing an actual calculation could be quite complicated, even if we use the ideal gas approximation, because the atmosphere has enormous variations in both temperature and density.
To simplify the problem, let's think of the atmosphere as an ideal gas of uniform temperature and density.   Then we need to determine how the entropy of the argon released into the atmosphere increases as a result of this release.
Since we're doing an ideal gas approximation, the argon doesn't "know" about the other atmospheric gases – the only thing that affects its entropy is the increase in volume.  But, when doing an entropy calculation, we do need to account for the fact that the atmosphere already contains 0.93% Ar.   Effectively, the presence of the existing argon removes a fraction of the configurational possibilities, since if we replace argon with argon, we don't change the entropy.   Hence the entropy increase is calculated from the effective configurational volume that remains for the argon relased from the box.
Specifically, for the purpose of calculating entropy, the effective final volume for the argon released from the box is:
$$V_\mathrm f = V_\text {atmosphere} - V_\text{Ar in atmosphere}+ V_\text {Ar in box}\approx V_\text {atmosphere} - V_\text {Ar in atmosphere} = (1- x_\text {Ar})V_\text {atmosphere},$$
where $x_\text {Ar}$ is the mole fraction of argon in the atmosphere, which is 0.93% = 0.0093.
Proceeding, we start with the following general expression for a pure substance (again, the ideal gas approximation allows us to ignore the non-argon molecules):
$$\left(\frac{\partial S}{\partial V}\right)_{\!T} =\frac{\alpha}{\kappa},$$
where $\alpha$ is the coefficient of thermal expansion, and $\kappa$ is the isothermal compressibility.
For an ideal gas, $\alpha = \frac{1}{T}$ and $\kappa = \frac{1}{p}$, hence:
$$\left(\frac{\partial S}{\partial V}\right)_{\!T} =\frac{p}{T}$$
If we additionaly assume the expansion is isothermal (i.e., T of the box is the same as that of the atmosphere--which we've already assumed is of uniform T) we obtain, for an ideal gas:
$$\mathrm dS = \frac{p}{T}\,\mathrm dV = \frac{n R}{V}\,\mathrm dV \Rightarrow \Delta S = n R \int\limits_{V_\mathrm i}^{V_\mathrm f}\frac{\mathrm dV}{V} = nR \ln\frac{V_\mathrm f}{V_\mathrm i}=nR \ln\frac{0.9907 \,V_\text {atmosphere}}{V_\mathrm i}$$
where $V_\mathrm i$ is the initial volume available to the argon, and $V_\mathrm f$ is the final volume available to the argon.
As you can see, releasing a mole of argon into the atmosphere will result in a significant entropy increase, once the argon is fully dispersed.
Note that, in the case where the atmosphere is 100% Ar:
$$V_f = V_\text {atmosphere} - V_\text{Ar in atmosphere}+ V_\text {Ar in box}=V_{\text{Ar in box}} \Rightarrow \frac{V_f}{V_i}=1\Rightarrow \Delta S = 0, \text{ as expected.}$$
We can also calculate the corresponding free energy change using:
$$\Delta G = \Delta H - T\,\Delta S$$
With the ideal gas approximation, there is no enthalpy of mixing, hence:
$$\Delta G = -T\,\Delta S = -nRT \ln\frac{V_\mathrm f}{V_\mathrm i}.$$
If we assume STP ($273.15\ \mathrm K$, $1\ \mathrm{bar}$), and take the volume of the atmosphere to be the area of the earth's surface times, say, 50 kilometers*, then we have:
$$\Delta S = (1 \ \mathrm{mol})\ R\ \ln\frac{2.53 \times 10^{22}\ \mathrm L}{22.7\ \mathrm L} = 0.403\ \frac{\mathrm{kJ}}{\mathrm K}$$
$$\Delta G = - (1 \ \mathrm{mol})\ R\ (273.15\ \mathrm K) \ln\frac{2.53 \times 10^{22}\ \mathrm L}{22.7\ \mathrm L} = -110\ \mathrm{kJ}$$
*This is all arbitrary, since we're unrealistically assuming constant pressure and temperature; but I needed to pick a number for the calculation, so I chose the top of the stratosphere ($50\ \mathrm{km}$).  $94\ \%$ of the atmosphere's gases are found below this altitude, so about $94\ \%$ of the argon would end up there as well.
Then again, because of the logarithm, the choice of atmospheric volume doesn't significantly affect the calculation, as you can see if you compare my answer with Buck Thorn's. Indeed, even if we allow the argon to expand to fill the observable universe ($4 \times 10^{83} L$), this only increases $\Delta S$ and $\Delta G$ by a factor of four! [$\frac{83}{22}\approx 4$.]
Calculation done in Mathematica (in Mathematica, ln is represented by "Log"):

Doesn't the atmosphere's initial concentration of ~1% play a role in approximating this? Intuitively it seems that it should reduce it considerably, for example if nitrogen was being released I would expect a lower number?

@DougMcClean  That's a good point (& I upvoted your comment). It seems there should be some effect. For instance, suppose you had a, say, 10 L box of pure Ar at STP, and were standing inside an atmosphere-sized box of pure Ar at STP.  If you opened the box, what would be $\Delta S$?  The answer is zero.  I.e., you could easily reverse the process by simply closing the box, which would put you back to your initial conditions.  If I have a chance, I'll think about how to calculate the effect of 1% of Ar. In the meantime, I'll edit my question to include the assumption of no Ar in the atmosphere.

I believe what you are referring to is a form of Gibbs' paradox: https://en.wikipedia.org/wiki/Gibbs_paradox

@DougMcClean  I've addressed your question about the effect of the ~1% Ar content of the atmosphere.  As you can see, it makes little difference in the calculation.

To compute the entropy change, assuming the process is isothermal, you need to know the initial volume of the gas, in addition to the approximate final volume. The entropy of mixing in this case is nearly identical to the entropy change for the dispersed argon, since the volume increase for the atmospheric gases is negligible.
If one assumes the atmosphere has a thickness $\mathrm{d}R$ then its volume is $\approx 4\pi R^2 \mathrm{d}R$. The value of $\mathrm{d}R$ to use is unclear since  the atmospheric gas concentration varies due to thermal and gravitational effects. One could use the boundary at which atmospheric effects become noticeable due to the presence of gas, $\mathrm{d}R\approx \pu{120 km}$ and a radius $R\approx \pu{6365 km}$, to estimate the atmospheric volume as $\approx\pu{6.1e10 km^3}$. Given the estimated mass of the atmosphere $\pu{5.15e18 kg}$ this results in an estimated average density of $\approx\pu{0.084 kg/m^3}$, well below the density of air at sea level ($\approx\pu{1.2 kg/m^3}$).
Using the above upper estimate of the atmospheric volume, we now need an estimate of the volume in the argon vessel. If the molar concentration is assumed identical in the atmosphere and in the argon vessel, and using an estimate of $\approx\pu{29 g/mol}$ for the average molecular weight of the atmospheric gas, then the initial volume of argon can be estimated as $\approx\pu{ 0.024 m^3}$ (approximately the ideal molar volume at STP). Using the standard formula for the molar entropy change of an ideal gas under an isothermal expansion gives
$$\begin{align} \Delta _{\text{mix}} S_\text{m} &= R \log\left(\frac{V_\text{f}}{V_\text{i}} \right) \\ &= \pu{ 8.3 J/K } \log \left( \frac{\pu{6.1e10 km^3}}{\pu{ 0.024 m^3}} \right) \\ &=\pu{ 410 J/K }\end{align} $$
Alternately considering the estimated mass of the atmosphere and assuming the final density to be uniformly $\pu{1.2 kg/m^3}$ then the relevant final volume is $\approx\pu{ 4.3e9 km^3}$. Then the standard formula for the entropy change gives
$$\begin{align} \Delta_{\text{mix}} S_\text{m} &= \pu{ 8.3 J/K } \log \left( \frac{\pu{ 4.3e9 km^3}}{\pu{ 0.024 m^3}} \right) \\ &=\pu{ 390 J/K }\end{align} $$
Clearly the very different estimates of the final volume of the atmosphere do not result in a great difference in the final estimated entropy change. You can thank the logarithm for that.

Note a slightly smaller estimate  of the final volume ($\pu{4.1e9 km3}$) is obtained if the final volume is limited to a layer 8 km thick.