text:
Particle size vs dissolving rate of sucrose
I’m required to make a lab with this information only (particle size vs dissolving rate of a solid (molecular covalent compound-sucrose)? so far what I’m thinking is using three different kinds of sugar like table sugar, sugar cubes, and one of them grinding it but the issues is how do you find how they dissolving rate?

It depends on the level of precision /accuracy required. If it is merely for demonstration it suffices to put a small amount of sugar and use a watch. Beware that sugar cubes can give the same results as table sugar, they are porous and the effective interface could be as in common sugar. Better replace them with some hard sugar. Or start from sugar candies and use a candy and differently ground ones.

I have found in Germany sugar crystals that are about 5 mm thick. If you are able to get some, you may drop one of these crystals in a beaker with 50 mL water and an stirring bar. If not take a standard sugar cube. Using a watch, measure how long it takes for the crystal to be dissolved. Now start again with another rotating speed. Now break a new crystal in two or three parts with a hammer. And start again with smaller crystals (to be weighed in advance). Repeat the measurements. Report these values on a graph vs rotating rate, or vs. weight of sugar, or vs. temperature. Now start again with powder, and even with the sugar powder used in bakery which is made of extremely fine grains. This is just ideas.

For one, pure water has a certain refractive index, $n_D = 1.333$ at a temperature of $20\,^\circ{}\mathrm{C}$.  For the other, the presence of a solute alters this optical property, which is used, for example to estimate the sugar content in grapes (see, for example an earlier answer by mine here).
«Calibrating your tools» would now request that you prepare a number of samples with known concentration of sugar from nothing = pure water to the max = an saturated solution; each charaterized by its refractive index.  Note that the relation between the sugar concentration $c$ and the refractive index $n$ recorded need not to be a linear one.
With this information in hand, find a large beaker and a stop watch, and repeatedly measure the time variant refractive index quicly.  The temperture variant approach already was determined, e.g. here.
In terms of sugar samples with different granularity, maybe you find sieves in a group working about soil chemistry / geology or process chemistry because knowledge of particle size distributions is of importance to model mecahnical properties, transfer of heat and matter across pebbles / grains of catalyst, etc. (Sieve analysis is a keyword.)


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Are there any reactions that are endothermic and exothermic?
As it says: are there any reactions that require heat and produce heat simultaneously?

Just to address a possible confusion - as Loong says it is not possible for a reaction to be both endo and exo thermic at the same time, they are precise opposites. However that is not quite what the body of the question says. A reaction might be so slow at room temperature that heating is required to make the kinetics sufficiently fast to observe the reaction. But kinetics and thermodynamics are different things, and this initial heat input to speed up the reaction has little to do with the total amount of energy produced or consumed, which is what Enthalpy measures.

Building on Ian's comment, most reactions require energy to start, and release energy as they proceed. such that after all is said and done, more energy is eventually released than was initially consumed. This means that most reactions are exothermic (thermodynamics), but they still have to be supplied a certain amount of activation energy (kinetics).

"require" is a loaded word that requires clarification here.

An exothermic reaction is a reaction for which the overall standard enthalpy change $\Delta H^\circ$ is negative.
An endothermic reaction is a reaction for which the overall standard enthalpy change $\Delta H^\circ$ is positive.
Clearly, it is impossible that $\Delta H^\circ$ is simultaneously negative and positive.

A possible example (which is real and reflective of a class of compounds, energetics) is the accidentally formed problematic explosive nitrogen trichloride from the chlorination of a warmed ammonium salt solution, like $\ce{NH4Cl}$, where the reaction can rapidly transition from an endothermic to an exothermic process (albeit not precisely simultaneously, around one-thousandth of a second or less). Or, one could argue that equivalently a highly exothermic energetic is endothermic on creation where the reaction moves in the reverse direction.
To quote a source:

Why is $\ce{NCl3}$ so unstable?
The enthalpy of formation of $\ce{NCl3}$ is + 232 kJ mol-1, a fairly large endothermic value, a good sign that a compound is unstable. This refers to the process:
$\ce{½ N2(g) + 3/2 Cl2(g) -> NCl3(l)}$
This means that the enthalpy change for the reverse reaction, the decomposition of $\ce{NCl3}$, is pretty exothermic.
$\ce{NCl3(l) -> ½ N2(g) + 3/2 Cl2(g)}$
The positive entropy change associated with the formation of the gaseous chlorine and nitrogen would also favour the decomposition, as would the very high N-N bond energy (~940 kJ mol-1).

Now, $\ce{NCl3}$ in the presence of organics will explode, however, a dust particle likely does not detonate nitrogen trichloride as I have witnessed its transport in an open test tube while being cooled (please, do not verify given the inherent dangers of this compound). Clearly, the dust results in some exothermic activity, but is insufficient in mass to precipitate an explosion in cooled $\ce{NCl3}$, implying some equilibrium process.

source: Stack Exchange

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Is it necessary that hydrolysis of ester will only take place in excess of water?
I was given a question on chemical kinetics in which the statement was given-
"The Concentration of water in hydrolysis of ester does not change to much extent"
I marked the statement to be false because I thought that the hydrolysis need not take place in excess water, there might be a condition where I could take 1 mole water and 1 mole of ester and still carry on the reaction.
Although it might be that the hydrolysis may not go in forward direction to much extent in such a case, but still reaction would take place.

Water excess helps in equilibrium shifting.

No, it is not necessary to have excess water for reaction to occur. Yet usually what books and sources mean by hydrolysis means that water as a solvent reacts with the compound. As you may know without large excess of water you wouldn't be able to go very far with the reaction (Low Yield due to equilibrium).
This is a reason why ester hydrolysis is also a pseudo first order reaction due to large excess of water which does not change effectively as reaction proceeds. Just know that hydrolysis requires excess of water. That is what is meant by hydrolysis. As for your question in the book, it is useless to spend time on such questions demanding plain answers without reasoning.

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How can ionic bonding in lanthanides occur without valence orbitals available for overlap?
I've recently been taught that the 4f orbitals in lanthanides are "core-like", supposedly meaning they have radius smaller than the 4d orbitals, therefore they are not available on the outside of the atom to form covalent bonds with ligands. The lecturer then went on to say that, therefore, the bonding of lanthanides to ligands is entirely ionic.
It is my understanding that a bond is deemed ionic when there is a large electronegativity difference between the two atoms, i.e. the bonding molecular orbital is near-identical in energy to the atomic orbital of the more electronegative atom, as shown below for sodium fluoride:

However, if this is the case, how can lanthanides form any bonds if their valence orbitals are not available? Surely orbital overlap with the ligand is still required to form a molecular orbital, even if the resulting orbital is ionic in nature?

There's no such thing a entirely ionic bond in chemistry, as that wouldn't be chemical bond, but a two separate charged particles.

Lanthanide atoms also have $5d$ and $6s$ valence orbitals available for bonding, and these can overlap effectively with ligands.  Whether this overlap leads to a bonding orbital encompassing both metal and ligand (covalent bond) or transfer if the electrons to a much low-energy ligand orbital (ionic bond), a bond is formed either way.  The bond may not be entirely ionic but is entirely (or nearly so) derived from the $5d$ and $6s$ orbitals of the lanthanide atom instead of the $4f$ orbitals.
Lanthanides do tend to form complexes with high coordination numbers, exceeding the coordination number of six commonly seen with $d$-block transition metal complexes.  From here:

Lanthanides share many similar characteristics. These characteristics include the following:
...
Adoption of coordination numbers greater than 6 (usually 8-9) in compounds
...
A preference for more electronegative elements (such as O or F) binding
...

The coordination number thereby exceeds the number of separate covalent bonds that can be formed from $5d$ and $6s$ valence orbitals alone, so a limited number of covalent bonds must be de-localized over the many linkages.  This would lead to the bonds defaulting towards a more predominantly ionic mode, requiring the ligand orbitals to be low-energy and associated with highly electronegative ligand atoms (also noted in the above excerpt).

Thank you for your response, and my apologies for the late reply - would you be able to clarify what you mean by delocalised covalent bonds over multiple linkages? I can't think about how that works from a molecular orbital standpoint.

For a simple example, look at 3-center 4-electron bonds.  Note carefully that while the overall three-center 4-electron bond is stronger than an ordinary covalent bond, it's distributed over two linkages and thus each individual linkage is weaker.

Ah yes of course, thank you for your answer!


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Endothermic or Exothermic
Is an endothermic reaction always endothermic at all conditions of pressure, temperature etc.? Similarly for exothermic... I mean is sign of ∆H always the same as sign of ∆H°?
This might seem to be a vague doubt but I just feel that there can be a reaction with trends of energies of formation of reactants and products such that endothermic reaction changes to exothermic or vice versa subject to change in concentrations of substances or external factors like temperature, pressure etc.
This is just a random thought of mine and even an example in this direction would be helpful.
Thanks!

There are no known example of reactions changing from exothermic to endothermic reaction when changing the conditions of pressure or temperature.

Depends on what one considers a reaction, or more precisely is it a single specific reaction or whole process.

@Maurice That's a bold claim.  Do you have anything to support it?

Because $\Delta H$ is temperature-, pressure-, and concentration-dependent, it's certainly theoretically possible for $\Delta H$ for a chemical reaction to change sign in response to changes in any of those intensive properties.
I don't have an example of such a chemical reaction handy.  However, there are direct experimental measurements showing enthalpies of solution that change sign in response to both temperature and concentration changes:
For instance, the enthalpy of solution of polystyrene in decalin has been shown to change from ~ –10 J/g at ~20 C to ~ +6 J/g at ~100 C (the ~'s were needed because I'm eyeballing these from a plot).  Reference:
Tager, A. A., et al. "The temperature and concentration dependence of enthalpy, free energy and entropy of mixing polystyrene solutions with an upper and lower critical temperature of mixing." Polymer Science USSR 12.6 (1970): 1497-1511.
And the enthalpy of solution of $\ce{LiNO_3}$ has been shown to change from negative to positive as the concentration increases.  Reference:
Wolf, G., V. Vacek, and V. Pekarek. "Enthalpies of solution and of crystallization of lithium nitrate and of lithium nitrate trihydrate in water at 25° C." Journal of Solution Chemistry 19.10 (1990): 1029-1039.

Thank you for the examples. Does this also have any useful implications other than this factual information?

It could have practical implications in process chemistry, since the sign and magnitude of $\Delta H$ determine the temperature-dependence of the equilibrium constant.  So it's useful to know that, within certain concentration or temperature ranges, increasing the temperature will favor the forward reaction while, at other concentrations or temperatures, increasing the temperature will favor the reverse reaction.


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Possbile to test percent of Ethyl Alcohol in hand sanitizer?
The percent of Ethyl Alcohol in hand sanitizers slowly drops over time. And manufacturers put an expiration date on the bottles. But I'd to test the percent of Ethyl Alcohol for myself for a new bottle and to see how much over time is lost as it gets close to the expiration data and beyond. Is this possible for a chemistry n00b?

If the alcohol is simply lost via evaporation and diffusion, just measure the weight or the liquid level of the bottle.

Flammability (actually, flash-point): https://ask.metafilter.com/95484/At-what-proof-will-spirits-burn

Note that there equally are sanitizers based on isopropyl alcohol instead of ethanol.


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Shoulders of NMR peaks
I want to ask a question about the shoulders of peaks when separating overlapping peaks.
I have the following spectrum circled with relevant shoulders:

I believe that the spectrum above shows an overlapping triplet and doublet, but I want to know what causes these shoulders in the NMR spectrum. I'm stuck to whether to decide there is further splitting of a small $\ce{J}$ value or to just assume it's a shoulder.
Below shows analysis of this region from Acta Pharm. 54 (2004) 163–176.

What is causing these shoulders? Are they part of the signal or just a result of poor preparation leading to artefacts?

It's more like quadruplet of triplets or other complicated multiplet, you just don't have the resolution to show it properly.

@Mithoron I've updated it with the source that states it is an overlapping triplet of doublet. I'm just confused as to how these shoulders are caused. If the resolution is too low, can I assume it is just a shoulder peak and not a splitting with a smaller $J$ value?

I think it is splitting, but with value so small they ignored it.

A good question. The term 'shoulder' is really an informal description used to describe the appearance of parts of peaks in the NMR spectrum. When a peak tends to deviate from the standard Lorentzian, we might describe this as having a shoulder. Shoulders can come about from typically three origins:

Shimming anomaly. Fairly common, but the key to diagnosing this cause is that it will affect all peaks in the same way (within the natural linewidth of the peak). So, have a look at the TMS peak, or solvent peak, or that internal reference of ethyl acetate that your supervisor suggested was an impurity. They will all have exactly that same lineshape anomaly, with identical shoulders. Always check these small molecule peaks when you run your spectra to determine shim quality. As an aside on this spectrum, these 'shoulders' you are looking at aren't really shoulders - they are split peaks. If this lineshape was genuinely due to a shimming anomaly, that indicates a pretty serious z shim issue, and you've probably got some floating solid in the sample.
Overlapping peaks. NMR spectra are simply the superposition of all nuclei in the sample. If two singlet species have a non-identical chemical shift within their natural linewidths, then they will appear as a shouldered peak. Obviously, the relative populations will determine how they actually appear. This could account for this type of spectrum (but more than likely doesn't), and that is why you would do other experiments such as COSY or HSQC to verify this.
Splitting due to coupling Often it is difficult to fully resolve small couplings, especially those that come from aromatic systems or long range pathways, where the size of the splitting might be about 1Hz or so. This is where an understanding of the expected coupling pattern for the molecule is very helpful. So, take a deep breath in.....

For Salicylamide, as described in your paper, we would expect H6 to be a large doublet (ortho to H5, ~7-8Hz), but also show meta coupling to H4 (~1-3Hz), and so we expect this large doublet to exhibit small splittings. Indeed, that is what you are seeing. On a good shimming day, in better solvents than dmso, we might even expect to see para splitting (<1Hz - don't hold your breath though). For the H5, we expect, again, a large ortho doublet coupling to H6, but also a large ortho doublet coupling to H4 (again, ~7-8Hz). That gives us the reported triplet, which is in fact a dd. But, we also expect to see some meta coupling to H3 (again 1-3Hz). Again, we see this, and each of these of these should be split due to the small meta couplings. (And we'll leave alone the discussion about whether they have reported apparent splittings, as opposed to J couplings, otherwise J56=J65),
And breath out.
So what you are seeing is genuine splitting coming from small couplings. These are often not reported in experimental data, because it is quite difficult (or at least time consuming) to resolve all of these couplings. You can get a better understanding of these small splittings if you use resolution enhancement when you process your data. Use a window function such as a small negative line broadened exponential function, or a gaussian function, and you can often split these lines almost to the baseline.

There are two signals. ddd and dd. However, the ddd contains a very small coupling constant, so the ddd appears in part as a triplet.
The following figure shows the case clearly:

A second order effect can also be argued, to explain the roofing effect.


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Peak at m/z = 39 in mass spectrum of 3-chloropropene
I've noticed that in the mass spectrum of 3-chloropropene, there is a peak at m/z = 39, which to me, indicates the formation of the C3H3 ion.
However, given that 3 bonds would have to be broken to form this ion, how can the frequency of this cleavage be explained?

It would indeed be $\ce{C3H3^+}$ assuming predominant isotopic species.  Although there can be other isomers, the cyclic isomer, the cyclopropenyl cation is strongly stabilized by aromaticity.
As PLD comments, there are really only two steps, not three, to forming this cation.  The allyl cation ($\ce{C3H5^+},m/z=41)$ is first obtained by loss of the chlorine atom plus an electron.  The aromatic cyclopropenyl cation noted above is then obtained by loss of $\ce{H2}$.

That makes sense, but why would it form such a predominant peak given that it seems quite unlikely that such a cleavage would occur?

Probably multiple-bond cleavage does occur, but normally we would not see it.  Usually if something breaks three bonds it immediately gets one or two back.  However, if the ion is stabilized it might last long enough to show up in the detector.

It is a consecutive process: homolytic bond cleavage leads to m/z 41. And for C3H5+, the lowest energy pathway for fragmentation is loss of H2 with the formation of the cyclopropenium ion (m/z 39). It is not so unlikely if one considers that EI can introduce quite a lot of energy within the ionized molecule.


text:
DIY hand sanitiser solution forms spongy substance
I am trying to make my own hand sanitiser since it’s sold out everywhere. The formula I use is 1 part aloe vera gel in 2 parts 100% isopropyl alcohol. The problem is when mixing the two, the solution always forms a small white spongy sticky substance while the rest stays liquid, not gel-like:

I tried two brands of aloe vera gel and got the same result:

Palmer Aloe vera gel: Aloe Vera Gel (Aloe Barbadensis), Polyglycerylmethacrylate, Propylene Glycol, Witch Hazel, Algae Extract (Algae), PPG-2-Isodeceth-12, Carbomer, Tocopheryl Acetate, Triethanolamine, Methyl Gluceth-20, DMDM Hydantoin, Tetrasodium EDTA, Quaternium-15, Fragrance (Parfum), Hexyl Cinnamal, Linalool, Geraniol, Limonene, Yellow 5 (CI 19140), Blue 1 (CI 42090).
Thursday Plantation Aloe Vera Gel: Aloe Vera inner leaf juice equiv. to fresh 974mg/g, Hydroxybenzoates, Diazolidinylurea, Potassium Sorbate and Sodium Benzoate

I am starting to think I got the wrong type of alcohol but the label clearly says pure 100% isopropyl alcohol.
What am I doing wrong?

What you are experiencing is a reason why it is not recommended to do it yourself. See for example https://www.cnet.com/how-to/why-you-shouldnt-make-your-own-hand-sanitizer/

I would always use soap and water when possible, but when it’s not available, there isn’t an alternative?

I tried shaking the bottle to mix and experienced the same issue.

If you don´t put your hands into your eyes or nose, there is no reason why you should disinfect them. The chance of catching the flu via food is negligible.

Components are clearly balling up due to lack of solubility in the 100% IPA.  I would have tried 70% IPA/30% water mix, but I'm with the other comments. Follow the WHO method or not at all.


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Mechanistic explanation of salt lowering temperature of ice slush
As we all know, adding salt to ice water lowers its temperature.
I've read plenty of system-level accounts of energy balances, enthalpies, vapor pressures, phase equilibria and freezing points—I understand that chemists' bread and butter is doing energy arithmetic on thermodynamic systems as a whole, but I would like to gain microscopic-level intuition instead of just being handed some thermodynamic abstractions.
I am looking for an intuitive explanation that details, step by step, at an atomic level, what happens at the interface between the ice and the salt water. For each step, I would like to understand:

What energies (kinetic/temperature vs. latent/chemical) are involved, and where?
How exactly (atomically or subatomically) is one type of energy converted into the other?
Why is each step energetically favourable? Or is it not, and it's reversible? How reversible, statistically?

The explanation should be clear enough that I should be able to intuitively answer the following, without consulting diagrams:

How does the concentration of salt change the evolution of the system (again, at a microscopic level)
How does adding e.g. ethanol instead of salt change things?
Is liquid water required at all? Is salt able to melt and cool completely frozen ice, or is external force required (stirring, driving over it, etc.) to cause some initial melting for anything to happen? (I've read conflicting answers to this in the past.)

Do not refer to thermodynamics as abstraction because it was born as a totally experimental discipline, even as an engineering discipline.

Water loses kinetic energy when it detaches from the surface of the ice
Water molecule in the solid phase have more hydrogen bonds than in the liquid phase. It takes energy to break these bonds, so only molecules with sufficient kinetic energy will detach (maybe because they just collided with a water molecule that transferred some kinetic energy to it). So the molecule that detaches will have a low kinetic energy as it goes into the liquid phase. Further collisions will quickly ensure that over time, this particular molecule has an average kinetic energy, but overall the temperature will drop from that detaching event.
Water gains kinetic energy when it attaches to the surface of the ice
This is just the opposite process, so the opposite happens. As additional hydrogen bonds are formed, the molecule has greater kinetic energy (in the form of vibrational energy). Either the molecule had less than average kinetic energy and now has average energy again, or it had average kinetic energy and will have higher kinetic energy, but quickly dissipate this to other molecules.
When you add salt to the liquid water, less molecules attach to the surface of the ice
Compared to pure water, less collisions of particles in the solution with the surface of the ice will lead to attachment of particles because now, some of these particles are not water molecules but sodium or chloride ions instead (they are not incorporated into the crystal lattice of ice but stay solvated by liquid water). As a consequence, there are less attachment events while the detachment events stay the same. With a net melting of ice, the average kinetic energy of particles decreases (the temperature drops). As a secondary effect, this also affects the rates of attachment and detachment, so once the temperature has sufficiently dropped, the two rates will be the same again.

What energies (kinetic/temperature vs. latent/chemical) are involved, and where?

In my explanation, I talked about kinetic energy (reflecting the temperature) and energy necessary to break hydrogen bonds (could be generalized as intermolecular interactions, and they underlie the observed bulk latent heat).

How exactly (atomically or subatomically) is one type of energy converted into the other

There is no chemistry or nuclear physics happening, so it is sufficient to talk about the molecular level. Kinetic energy is converted into potential energy through inelastic collisions. Kinetic energy is redistributed between molecules through elastic collisions.

Why is each step energetically favourable? Or is it not, and it's reversible? How reversible, statistically?

According to microscopic reversibility, all molecular events are reversible. Processes go in a certain direction because it is more probable than in the other direction (2nd law of thermodynamics). The 2nd law of thermodynamics is inherently linked to large numbers of particles, so you need to look at the forest rather than at individual trees.

How does the concentration of salt change the evolution of the system (again, at a microscopic level).

The higher the concentration of the salt, the lower the concentration of the water in the liquid phase, i.e the fewer events of liquid water attaching to the ice.

How does adding e.g. ethanol instead of salt change things?

1 mmol/L NaCl has twice the dissolved particles compared to 1 mmol/L ethanol. Otherwise, the effect is the same (colligative property).

Is liquid water required at all? Is salt able to melt and cool completely frozen ice, or is external force required (stirring, driving over it, etc.) to cause some initial melting for anything to happen? (I've read conflicting answers to this in the past.)

Ice near the freezing point exhibit surface melting. Once there is a tiny amount of water, more ice will melt. On the other hand, if it is cold enough that even saturated solutions will freeze, a mixture of dry salt and ice will not interact.

Related: https://chemistry.stackexchange.com/a/119142 and
https://chemistry.stackexchange.com/a/116323


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Efficiency of H production from FeOOH catalyst?
Research has recently found that FeOOH is 25 times better at hydrogen production than titanium dioxide.
Today, a new electrolysis plant delivers energy efficiency of around 80%. Is 25x the Honda-Fujishima reaction, using FeOOH higher efficiency than that?
ref:
https://www.tus.ac.jp/en/mediarelations/archive/20200227001.html

I have read the article. I am not convinced. They say that by irradiating "something" with a Hg-Xe lamp, and using rust as catalyst, plus some oxygen, they produce more H2 than with another catalyst (TiO2). Well. Catalytic effect of rust has been studied for years, with or without light, without any important result. They say that adding Oxygen would improve the yield. Why ? On the contrary, Oxygen should have the inverse effect.

Can't any chemist check their finding with some rust, some methanol and a xenon lamp? it sounds like a 1 hour experiement. It would be crazy for any magazine to publish something so easy to verify if it is silly and easy to verify.

Does iron oxyhydroxide really assists in production of monoatomic hydrogen (the title says "H")?


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Problem with determining the mass value of a soil sample for gravimetric calculations
I'm having some trouble with how to determine the representative mass value of a prepared soil sample for calculating its $\ce{Cl-}$ concentration in ppm (mg/kg).
Recently I carried out a gravimetric analysis for determining the concentration of $\ce{Cl-}$ in a soil sample by the addition of $\ce{AgNO3}$. The sample was prepared from 10 g of soil, which were filtered with 50 mL of distilled water. Then, an aliquot of 15 mL was taken from the filtrate. The aliquot was centrifuged in order to separate the solid phase from the liquid phase. The liquid was transfered to a volumetric flask and diluted to 50 ml with distilled water. This final volume was the one used for the analysis. The gravimetric processes were developed normally and at the end it turned out that the produced mass of the AgCl precipitate was 0.003 g and by stoichiometry I calculated its content of $\ce{Cl-}$ (which resulted to be 0.00074204 g).
Since I want the $\ce{Cl-}$ concentration in terms of mg/kg, I expressed the mass of the aformentioned ion in mg but my conflict comes when trying to establish the mass of the used soil sample (in kg) because of the dilutions. As far as I'm concerned, the preparation of the sample could be mathematically arranged as follows:
$$\frac{\pu{10 g}\,\text{soil}(\frac{\pu{1 kg}}{1000\,g})}{\pu{50 ml}\,\text{filtrate}}\cdot\frac{\pu{15 ml}\,\text{aliquot}}{\pu{50 ml}\,\text{dilution}}=\pu{6e-5 kg ml-1}$$
but as you can see, the dimensional analysis doesn't result in just kg. I don't really know where the problem is, but I intuit that either I'm missing something or adding more of something, or I'm just having an erronous mathematical concept about the sample preparation.
It will help me a lot if you can help me to identify my mistake.

First I would say that you cannot trust your measurement, You say that you obtain 3 mg AgCl. 3 mg AgCl is nearly nothing. If you are not able to be more precise, it means that your amount of AgCl is between 2.5 mg and 3.5 mg. The uncertainty is about ± 17%. That is huge.

First I would say that your measurement is not reliable enough. You say that you obtain 3 mg AgCl. 3 mg AgCl is nearly nothing. If you are not able to be more precise, it means that your amount of AgCl is between 2.5 mg and 3.5 mg. The uncertainty is about ± 17%. That is a huge uncertainty.
Anyway your 3 mg AgCl contains 0.074 mg Cl which were in the 50 mL flask. As this solution was made from 15 mL of the first solution, the entire initial solution contains 50/15 times 0.074 mg Cl. This is 0.248 mg Cl. As this amount of Chloride is coming from 10 g soil, it means the the proportion of Cl in the soil is 0.248 mg/0.01 kg soil = 24.8 mg/kg. As the uncertainty on the initial weight (3 mg) is high, you may admit that the content of Cl in the soil is 25 mg/kg. With the uncertainty, it is 25 ± 5 mg/kg.
If I were at your place, I would determine the chloride content with a volumetric method. The result of such a titration would be much more reliable. A precision of ±1% is easily obtained.

Just a little remark; the content in mg would be 0.74 mg Cl-. Now that you mention it, a sample prepared in the same way as in the gravimetric method was analysed by titration. A mean volume of 0.20 ml was spent for completing the precipitation. For both methods a 0.1 mol/L AgNO3 solution was used. Interestingly, when doing the stoichiometry the content of Cl- results in a value of 0.709 mg, which is relatively close to the mass obtained in the gravimetric analysis. Anyway, I appreciate your help and feedback.


text:
Problem calculating relative molecular mass
I am starting be re-learn Chemistry after several years.
The Book I am using has a test question to calculate the relative atomic mass of:
$ (CH_3CO)_2O $
My take
Carbon C, $ A_r = 56 $
Hydrogen H, $ A_r = 1 $
Oxygen O, $ A_r = 16 $
$ M_r = ( 56 + 1 \times 3 + 56 + 16 ) \times 2 + 16 = 131 \times 2 + 16 = 278 $
This is nothing like the answer given in the answer key.  Is there an error in my text book?
If not where am I going wrong?

Edit: Problem was a misprint in my text book.  See correct answer below

Wow! Is relative atomic mass of carbon 56?

Solved.  The problem was a typing error in the text book $ A_r $ for carbon should of course be 12 not 56.  This is given correctly in the periodic table shown on the back cover but was misprinted in the body of chapter 1 of my book.
Thanks to Zenix for pointing this out in comments.
$ M_r $ of $ (CH_3CO)_2O $
Carbon C, $ A_r = 12 $
Hydrogen H, $ A_r = 1 $
Oxygen O, $ A_r = 16 $
$ M_r = ( 12 + 1 \times 3 + 12 + 16 ) \times 2 + 16 = 43 \times 2 + 16 =  102$
This matches the answer given in the answer key.
I have not studied chemistry in over 30 years which explains why 56 as a relative atomic mass for Carbon did not scream out as being wrong.

Hey, $A_r$ for carbon, not copper (in first sentence).


text:
Is irreversible work a path function?
In $P/V$ plots of irreversible processes only initial and final state is defined due to sudden compression or rarefaction then how is work done a path function in irreversible processes.

Why won't work done by the process here be only dependent on initial and final states?

Can someone please tell me why is this classified as a homework question?

For any process, the work done by the gas on the surroundings is $\int{P_\mathrm{ext}\,\mathrm dV},$ where $P_\mathrm{ext}$ is the force per unit area the piston exerts on the gas at the piston face, and also the force per unit area the gas exerts on the piston at the piston face.  For a reversible process, $P_\mathrm{ext}$ is also equal to $P,$ the pressure calculated from the equation of state for the gas (such as the ideal gas law).  But, for an irreversible process it is not determined by the equation of state, due mainly to viscous stresses within the gas.
So, if you can plot $P_\mathrm{ext}$ vs $V$ on the graph, you can correctly calculate the work, but, for an arbitrary irreversible process where $P_\mathrm{ext}$ is not adequately controlled externally, this is not always possible.

I had a doubt whether such graphs are  possible for Irreversible processes where P and V aren't defined properly during the process.Can such a graph be plotted experimentally?

Sure.  Just place a flush mounted pressure transducer in the piston face contacting the gas.  This is a direct measurement of $P_{ext}$.

From Fundamentals of Thermodynamics by Borgnakke and Sonntag [1, p. 94]:

Further consideration of a $P-V$ diagram, such as Fig. 4.6, leads to another important conclusion.
  It is possible to go from state 1 to state 2 along many different quasi-equilibrium paths, such as A, B, or C. Since the area under each curve represents the work for each process, the amount of work done during each process not only is a function of the end states of the process but also depends on the path followed in going from one state to another.
  For this reason, work is called a path function or, in mathematical parlance, $\delta W$ is an inexact differential.

FIGURE 4.6 Various quasi-equilibrium processes between two given states, indicating that work is a path function.

Because work is a path function. It doesn't matter your process is reversible or irreversible. If you just consider the initial and final state, but go from two different paths, you'll have two different answers. For example, in the picture I uploaded, the areas under curve A and B are different. Therefore, work is different. But, as you can see, the final and initial states are the same.
References

Borgnakke, C.; Sonntag, R. E. Fundamentals of Thermodynamics, 7th ed.; Wiley: Hoboken, NJ, 2009. ISBN 978-0-470-04192-5.

As for the text, again, there is very rarely a need for posting a screenshot: Chemistry.SE provides a wide variety of formatting utilities for a precise reproduction of the quoted text. Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.

@andselisk Thank you. It's the first time I'm answering. I will improve and edit it.

Just a doubt how is it possible to graph out the P and V during the process as the Thermodynamic Variables wont be properly defined due to sudden compressions and rarefactions. Can such a graph be made experimentally?

@paria I took a liberty to edit your post accordingly correcting citation, converting images to text and improving grammar; hopefully you find my this useful for your future contributions.

That figure is of a process in quasi-static equilibrium which would mean they are reversible reactions not Irreversible.

@AmeyaM As far as I know, we can determine work from other information we have, such as entropy or internal energy or efficiency.

@AmeyaM Diagrams are based on experience. They measure the volume in a specific pressure. And sometimes when it's not possible, they use extrapolation.

@AmeyaM For sudden compression it's not usual to find the diagram. But for doing so, we make some assumptions. In general, if you have just initial and final state you can't find work from diagram. You should use some properties as I said before.


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How does one refer to the amino acids in a peptide?
Once a part of a peptide an amino acid is no longer an amino acid, so how do we refer to the part of the peptide that was once amino acid x?

A part of the peptide that was previously an amino acid molecule when free can be correctly referred to as an amino acid residue when bound in the peptide.
A source


text:
Why are branched isomers less polarisable?
This fact was given as an explanation as to why branched isomers of a compound have lower boiling points.
How are branched isomers less polarisable and how does this relate to London dispersion forces?

Branched isomers have less surface area exposed, thus have weaker Van der Walls forces of attraction.

I was itching to mark this as "homework", but the answer is not so perfectly clear imo. First it should be defined what "polarisable" means exactly.

Does this answer your question? Why do branched chain compounds have lower boiling points than the corresponding straight chain isomers?

@Karl This is what I don't understand- the definition of 'polarisable'. I was thinking that even if there was an induced dipole in a branched isomer, all the surrounding branches would experience an oppositely induced dipole, so cancelling out the overall induced polarity of the isomer. This would not happen in a straight-chain isomer, because there are no branches to provide an opposite dipole. Is this wrong?

The point is that they might be more or less polarisable but that is not the reason, at least not so easy to be immediately seen. Also in this context as Karl said it is not that clear what is meant... The point is even not the WdW forces to be weaker per se, but their cumulative effect being weaker. Indeed see the answer linked by Mithoron.


text:
How to control current and voltage in electrolysis
I am doing an experiment that aims to determine if the rate of electroplating changes with the temperature. I am using a copper rod, a solution of copper(II) ions and a neutral rod to be coated. I also want to keep every other variable the same, which means the voltage across the rods, the concentration of the solution and the current that goes through the circuit. But the times my professor tried to demonstrate electrolysis, either the current or the voltage was always dropping. How do I prevent this from happening? How can I ensure that the voltage and current remain unchanged for the entire duration of a trial?

One option is to write an equation to account for the change in concentration and then adjust the expression for the reaction rate accordingly.

make sure nothing changes. That includes the surface properties of your "neutral rod".

You might want to control voltage and current density rather than current. See the recent answer by @Poutnik here: https://chemistry.stackexchange.com/a/127775/79678

See here as well: https://chemistry.stackexchange.com/q/7866/79678

Look into the Randles–Sevcik equation to get some ideas about how to decrease the resistance

From an electrical point of view $V = \dfrac{I}{R}$.  So there is nothing you can do from the electrical side.  You can however monitor and change the concentration of the substance to keep $ R $ constant.

Remember that the way electrolysis works - the number of atoms moved will be proportional to the current.  As such, what you're looking to demonstrate is that the integral of the current dt changes with respect to temperature.

As ion mobility, conductivity and speed of electrode reactions are all temperature  dependent, you can either control voltage, either current/current density. The other quantity will be temperature dependent.

You can't control both voltage and current electrically:  Your power supply will hold one or the other constant, but the properties of the electrolytic cell will then determine the other one.
So you need to hold all of the chemical parts of the experiment constant:  Have a large ion bath so that the concentration doesn't change much as the plating takes place.  Stir well to keep things mixed. Have large electrodes (i.e. small current density) so that small surface changes don't dominate.  Etc.

You could get all complicated and adjust the one parameter by varying the solution parameters.  This is not practical for hunting control but if you know that the cell will need dilution to maintain your parameters you could add a dosing pump with water (or saturated electrolyte if the other direction is needed).

Try to change the geometry of your electrodes. For example, increase or decrease the distance between the electrodes. This will modify the internal resistance of the solution. So if you have a generator producing a constant voltage, you may reach the desired currant, simply by changing the distance between the electrodes. Of course the electrodes must be hanging at a bar in such a way that their relative position can be adjusted at will by hand.

If ou use flat electrodes you can probably fine tune the cell conductivity by rotating the electrode.  So you can adjust the current with a powersupply to maintain some reasonable value that will not burn the plating or cause excess gas evolution and then use a simple geared servo motor to hold the cell voltage constant by altering the electrode geometry.

You could create a constant-current source instead of using an off-the-shelf power supply.
There are also off-the-shelf constant-current supplies, so ask your lab manager if that's a resource.
An EE colleague can set this up for you if it's too far outside your domain.  Here are a few articles as starting points.  Total cost should be < $15.
http://www.bristolwatch.com/ccs/index.htm
http://imajeenyus.com/electronics/20160530_adjustable_current_source/index.shtml


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Reduction of Organic compounds
I've encountered in chemistry quite a lot of reducing agents. However one question strikes me quite often and  any research done on the same hasn't been fruitful. This is in context to organic chemistry. Now for example, in reductive ozonolysis, we often use ozone in combination with zinc and water. Zinc's presence ensures reductive results and not oxidative ones. Now, some sources(not books) which I have seen, say that some sulphur compounds would do the same like $\ce{Me2S}$. Also, in case of 1,3- dinitrobenzene,  $\ce{Sn + HCl}$ converts both the $\ce{-NO2}$ groups to $\ce{-NH2}$ groups. Now if we use $\ce{NH4HS}$ or some other nitrogen and sulphur containing ammonia like compounds, we get reduction again, but only for one group out of the two. So my question is, how do we :

Decide that there is going to be reduction
Know the extent of reduction (partial/full)

Can some insight into the inorganic properties of the elements provide an answer like d-orbital availability, lone pair, electronegativity, size/charge ratio? I've tried reasoning but haven't gotten anywhere. Or is it that memorization of the reagents is the only way to know it's function.
Note: This question is only in context to some less commonly used reagents, so as to recognize it's function if we ever see one in problems present in question books.

You already underestimated it, a single reagent can give many products in proportions depending on various conditions. In chemistry, experiment tells what does what and when and conducting it in seemingly identical conditions can give surprising effects.

I will try to explain the difficulty of this problem by using a simpler reaction, the reduction of nitrobenzene, which is already difficult enough. The reduction of dinitrobenzene should produce more different possibilities.
Nitrobenzene can be reduced into at least five different compounds. It depends on the choice of the reducing agent. I am afraid nobody can justify the particularly mechanisms used by a given reducing agent to get to each final result.
With ($\ce{Sn + HCl}$) or with ($\ce{Fe + HCl}$), it gives aniline, $\ce{C6H5NH2}$.
With sodium methoxide ($\ce{NaOCH3}$), it gives azoxybenzene, $\ce{C6H5N(O)NC6H5}$
With zinc, it can produce three different products:

With zinc + $\ce{NH4Cl}$ at elevated temperature, it gives
phenylhydroxylamine, $\ce{C6H5NHOH}$.

With zinc and an alcoholic solution of $\ce{NaOH}$, it gives azobenzene
$\ce{C6H5N=NC6H5}$.

With zinc and an aqueous solution of $\ce{NaOH}$ at $\pu{60 ^\circ C}$, it give hydrazobenzene $\ce{C6H5NH-NHC6H5}$.

Yes. I understand that a little change in the reagents makes a difference. But can we not check the strength of reducing agents in organic chemistry? The real aim is to determine: 1. Whether it is a reducing agent or not and 2. If yes, then is it strong or weak? i.e., will it be able to reduce all possible positions or only one like in case of NH4HS.

You check these looking at the literature and by experiment. Solvent, temperature and additional reagents make so much difference that there is no systematic rule. If you want to be able to predict everything from equations study physics.

@PrarabdhShukla Then you check standard redox potential, but it doesn't tell what and how it can actually reduce.

@Waylander Nice answer. Thanks for the clarification. And yeah, physics isn't all predictable. Classical mechanics is, but quantum isn't . And hence we have chemistry, completely unpredictable (to some extent it is) because we don't where the electron is! So the door for studying physics is closed. But I like both!

Possibly if you went into a lot of data collection, you could try using principal component analysis or PLS, but by the time you got the reduction potential data, solvent properties and other variables collated , you could have done the experiments quicker. All this would be substrate dependent too.


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Cadmium Selenide, Cadmium Sulfide, Cinnabar and Lead Oxide toxicity
I have a question about the toxicity of the pigments Cadmium Selenide (Cadmium Red), Cadmium Sulfide (Cadmium Yellow), Mercury Sulfide (Cinnabar or Vermillion) and Lead Oxide (Lead White).
Not long ago the EU banned lead-white pigments from use, even for art. It is now generally impossible to get hold of lead-white oil paint on the internet. However, the cadmium pigments were spared a ban recently. And it is quite easy to find Cinnabar and Vermillion oil paint online even of European production. For example, the dutch company Royal Talens supplies Vermillion paint online, without even warning that it contains mercury: https://www.royaltalens.com/en/catalog/rembrandt-oil-colour/oil-colour-paint-vermillion-311-40ml-tube/
I am not a chemist but my limited knowledge tells me that mercury is a lot more poisonous than lead. How come Cinnabar wasn't banned and what is the relative toxicity of the four pigments I listed, given that they are suspended in oil?
Thank you very much.

The solubility under relevant conditions is a decisive point. And the concentration of the given pigment. Certainly the places of usage (oil paintings are just one, lead white used to be very widely used). Availability of replacement options was surely taken into account. And then a bit of politics gets mixed in. ;)

@Karl I think that Vermillion would be easily substituted for by various organic red dyes and/or cadmium red. I thought may be lead white was actually more toxic than Cinnabar? So why did the cadmium colors come under attack, are they also more toxic than mercury containing colors? That's something I don't understand. Mainly because I started painting recently and am fascinated by the colors these pigments produce, unlike anything you could ever see in a photo :)

The toxicity of mercury depends a lot on its form and way of application. The metal could reputedly be swallowed without much harm, inhaling it is more critical, and dimethylmercury is extremely poisonous. Cinnabarit is very resistant to acids and bases. You could likely eat the pure pigment, it would just pass through your digestive tract.

@Karl Thanks, do you know much about the cadmium sulfide and cadmium selenide?

How about a bit of literature research on your side first? ;) wikipedia is usually a good start. You´re welcome back if you still have questions. Or you can write an answer yourself here.

@Karl I did read all I could find on wikipedia before coming to this site. Regarding the toxicity of CdS and CdSe it only says vaguely: "Cadmium is a toxic heavy metal and appropriate precautions should be taken when handling it" I could also write that and be 100% correct. The thing you said a bout cinnabar for example, it being resistant to acids and passing through the organism. I would guess that CdS and
CdSe would have similar chemical properties because of those elements' positions in the periodic table right?

https://de.wikipedia.org/wiki/Cadmiumfarben (German is not so hard, give it a try.) The gist is that modern Cd colours are not simply CdS and CdSe, but mixed crystals thereof also containing zinc, and they are even less soluble than the originals. Their are still banned from construction use, only for art.

I found an answer to my question. Typically oil paint manufacturers will write what pigments are used in their paints. The pigments are classified according to the 'color index' or C.I. These pigments may actually have different chemical composition, but nevertheless the color index can be used to search the pigment database here: http://www.artiscreation.com/red.html
For example, PR106 stands for genuine Vermillion, which is true Cinnabar and has a toxicity rating of C according to the table linked above, where A is non-toxic and D is extremely toxic.
Most colors that are marketed as 'cinnabar' are actually composed of various different pigments, most often not actually containing mercury.


text:
How to use Google’s Alpha-fold to predict the structure of a two-protein complex?
Alpha-fold won the CASP13 and CASP14 competitions last year and this year. It used deep learning to predict the secondary structure of a protein given the primary amino acid sequence.
Google has released its code a few weeks ago, and I was wondering how to use Alpha-fold to predict the complex of two proteins.
Naively, I'd concatenate the amino acid sequences of both proteins as if they were a single protein. Then I’d input the neural network with the concatenated primary sequence.
The network will predict the lowest energy conformation  of the concatenated proteins. I am afraid that the artificial backbone connection between the two proteins might lead to a completely wrong complex.
I could do some docking of the two individual proteins, but I want to find the denovo prediction of the two-protein complex.
Edit: here is an Pytroch implementation of Alpha-fold2 on github.

Of course you have to do the docking! Two proteins is not one.

And for docking you need to use a program made for docking not for single protein, just because it's good.

@JonCuster   Do a search on ChemSE for Amber, Charmm, Gromacs, or ChemDraw.  You'll see lots of questions and answers.  In addition, his question, and my answer, has general implication for how you can and can't use biopolymer folding prediction software (wherther the biopolymer is a protein or some other biopolymer, e.g., RNA).

Are you asking how to take code developed to find the MFE (minimum free energy) structure of a single protein, and modify it to instead determine the structure of a two-protein complex?
Or are you asking if you can use this code, as is, to fold a concatenated combination of the two proteins, and in turn use this result as a surrogate for the predicted structure of the two-protein complex?
If you're asking the former, I would recommend contacting one of the code developers to ask them for guidance.  It's possible that they have already developed a prototype option to address this.
If you're asking the latter, that approach is problematic, for two reasons:

Folding the concatenated protein would constrain the last amino acid in the first protein to be next to the first amino acid in the second protein.  That's not a constraint that would make sense a priori for a two-protein complex.

You have to ask yourself if, in forming the two-protein complex, you envision both proteins unfolding and then refolding to do a global energy minimization— since that's what folding the concatenated protein would give you (with the additional constraint mentioned in point #1, above).  If, on the other hand, you would like the proteins to each fold into their respective MFE structures, and then (while essentially retaining those structures) find the MFE configuration for the two-protein complex, then folding the concatenated protein clearly would not give you that.

Does Alpha-fold actually compute free energies, or is its performance based on training on known structures that are not necessarily minimum energy, particularly if experimental i.e. folded in vivo?

From the abstract of the 2019 Nature article outlining the success of AlphaFold:

Here we show that we can train a neural network to make accurate predictions of the distances between pairs of residues, which convey more information about the structure than contact predictions. Using this information, we construct a potential of mean force4 that can accurately describe the shape of a protein. We find that the resulting potential can be optimized by a simple gradient descent algorithm to generate structures without complex sampling procedures

One implication of what is stated above is that AlphaFold provides an algorithm that maps sequence to experimental folds with high accuracy but does not attempt to provide a physically accurate multi-purpose interaction potential. It is a statistically generated potential rather than based on fine tuning terms with a physical interpretation (e.g. angular, torsional, steric, Coulombic). This results among other things in potential predictive limitations imposed by the training sequence domain. Like any model it can guarantee success only if relevant training data (or potential terms) are included in creating the model. The sequence-fold relations relevant to quaternary structure might not overlap well with those relevant to lower order folding. I would argue that a subset of the folding interactions (in particular contacts between distant residues not involved in secondary structure) are more relevant to quaternary structure. It is also not clear that the potential would be useful to perform MD simulations to probe non-equilibrium structures or sample local minima.
Despite that, if those potential roadblocks are not significant I do not see why your approach should not work, particularly if you pick an appropriate linker (which is easily tested).
This answer risks might seem opinion-based but is an interpretation based on what can be quickly inferred from information in the linked publication.


text:
Dissolving 304 Stainless Steel
We installed 20 2" drop in anchors to anchor several signs in place near the entrance of our pool. We have to remove the signs about twice a year and after the end of the past season, I noticed there is a small amount of rust forming. I purchased the 304 because it was not going to be in direct contact with the water and because it was solid, not coated, as opposed to the 314, which was only coated. It seems as though the 304 was also just coated so it is beginning to rust. I need to provide some sort of plan as to how to remove the 304 anchors and replace with 316.
Drop in anchors are not made to come out of concrete as the driving pin expands the bottom into place. I had thought about drilling them out but I am hoping to use that as a last resort because I don't want to damage our new concrete and need to keep the holes in tact. I read that Nitric acid can dissolve stainless steel. Will this work in my situation? Or will that also damage the concrete?

You are not going to dissolve 304 and leave the concrete. Use sandpaper to remove the surface rust as needed : Or mechanically ( cut/grind) the 304 to remove it.

Try this, apply a cheap and effective spray onto the iron surface employing hypochlorous acid (prepared here by mixing chlorine bleach containing dilute aqueous NaOCl with 6% acetic acid in vinegar in excess to the NaOCl). Add a source of copper (coins), or better  high surface area carbon particles in the spray, serving as a cathodic zone, in this spontaneous electrochemical (basically a battery) cell. Source: Equation (7) here.
Note: This battery cell differs mainly in chemistry from the bleach battery cell by employing iron (steel) as the anode which will be the subject of considerable anodic corrosion in place of Aluminum. The cathode is high surface area carbon particles.
Mellor on Hypochlorous Acid in 'A comprehensive treatise on inorganic and theoretical chemistry, Volume 2, however, this is NOT the electrochemical based dissolution suggested above:

Iron filings immediately decompose hypochlorous acid with a brisk effervescence produced by the evolution of chlorine; the iron is partly oxidized and in part dissolved as chloride without the formation of any chlorate. A. J. Balard commented on this: "The greater number of other metallic substances do not decompose hypochlorous acid, and I am yet entirely ignorant of the cause of the peculiar behaviour of iron." P. Grouvelle passed chlorine through water with iron hydroxidelin suspension and a bleaching liquid along with ferric chloride was produced, and he found the liquid retained its bleaching properties after boiling for a quarter of an hour; but A. J. Balard failed to confirm this statement; he could not make ferric hypochlorite either (i) by the action of hypochlorous acid on iron hydroxide, for hypochlorous acid does not dissolve ferric oxide; or (ii) by the action of calcium hypochlorite on ferric sulphate, for calcium sulphate and ferric oxide are produced. Hence, adds A. J. Balard, "ferric hypochlorite cannot exist;" the results by P. Grouvelle are due to the formation of ferric chloride and hypochlorous acid in dil. soln.; when the mixture was heated, a portion of the acid distilled off, and the reaction which occurred with the cold soln. was reversed, for ferric oxide and chlorine were formed. Ferrous oxide is oxidized to ferric oxide by hypochlorous acid.

Problematic fumes so an outdoor reaction is strongly advised. Neutralize solution products with Washing Soda at the end of the reaction.
Here is a particular comment relative to steel being corroded in place of iron, without the benefit of a good high surface area cathode in diluted NaOCl.


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Haemoglobin literature spectra does not match
I am studying haemoglobin absorption spectra in its oxygenated and deoxygenated state and I compiled values from different sources including:

A PhD thesis by van Assendelft [1], using extinction coefficients in (mmol/L)-1/cm

A book compiling almost one century of research on the topic by the same laboratory that helped standardizing the haemoglobin cyanide standard [2]. Extinction coefficients in the same unit as [1], numerical values are very close.

Coefficients compiled by Scott Prahl[3]. The unit is the same but there is a factor 3.7 up to 4.0 between his data and that of Assendelft and Zijlstra.

Coefficients measured by Kolyva[4].

All the published coefficients are defined as base 10 extinction coefficient. That is if I
is the light intensity, they are expressed in Beer-Lambert law as:
$I = I_0 \cdot 10^{-c\cdot \varepsilon \cdot l}$
Where $l$ is the light path length (cm), $c$ the concentration in absorbing compound (mol/L) and $\varepsilon$ the molar extinction coefficient (L/mol/cm).
How comes there is a 3.7-4.0 factor between Prahl and Kolyva data on the one hand and Assendelft and Zijlstra on the other hand?
At first I thought of some $log(10)\approx2.3$ difference, as if some of the author forgot to precise that they used base $e$ extinction coefficient at some point but it does not fit. If anyone has an idea on the topic, the help would be most appreciated.

Further information
The cyanmethaemoglobin method is often used for the determination of the haemoglobin concentration of a solution. In particular, its absorbance at 540nm is known to be 11.00L/mmol/cm[5] which is in agreement with Assendelft and Zijlstra published data.
References

O.W. Assendelft, Spectrophotometry of Haemoglobin Derivatives, 1970. ISBN: 902320560X.
W.G. Zijlstra, A. Buursma and O.W. van Assendelft, Visible and Near Infrared Absorption Spectra of Human and Animal Haemoglobin, 2000. ISBN: 9067643173.
Scott Prahl, Optical Absorption of Hemoglobin, 1999, https://omlc.org/spectra/hemoglobin/
C. Kolyva, Systematic investigation of changes in oxidized cerebral cytochrome c oxidase concentration during frontal lobe activation in healthy adults, 2012. DOI: 10.1364/BOE.3.002550. Coefficients are in MEDIA 1.
F. Meng, Determination of extinction coefficients of human hemoglobin in various redox states, 2017. DOI: 10.1016/j.ab.2017.01.002. Table 1.

In Scott Prahl's article, he discusses that some authors consider equivalents and some use molar values and that this would create a 4x difference in the values. Have you factored this in?

@Tyberius Thanks a lot, I had read it long ago and forgotten it by the time I came to look at those spectra once again. Thank you for the fresh look you had on the issue! I answered the question with your remark :)
(Kind of weirdly, I had the reflex to open the different books and articles I had found here and there because they were "publications" but had not that of rereading Prahl's article which I classified unconsciously as a "blog post", thus less likely to contain valuable information. My bad.)

As stated by Scott Prahl :

One of the most confusing things about looking at hemoglobin (Hb)
  spectra is that the values are typically tabulated in equivalents. The
  term equivalent is used to indicate the amount of hemoglobin which
  contains 1 gm atom of Fe and combines with 1 gm molecule of O2 or CO.
  One equivalent of hemoglobin is assumed to be 64,500/4 or 16,125 gm. A
  concentration of 10$^{-6}$ equivalent is 16.125mg of hemoglobin per cc.

It seems that Assendelft and Zijlstra used the equivalent quantification of haemoglobin, meaning that all their extinction coefficients should be multiplied by 4. This leads to the following figure:

Sorry for not having read again Scott Prahl's description before posting. I read it several months ago, as I was gathering data from different source and forgotten it since...


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Coupling in C-NMR

Given the structure below, sketch the $^{13}$C - spectrum in a $^{13}$C marked sample
The data given is the chemical shifts (left) and the coupling constants (right)

The correct spectrum is given at the end of this question
I don't really understand what is meant by "in a $^{13}$C marked sample" however I assume that it means that I am supposed to look at C instead of H as my core atom. So from this I tried to determine the coupling. If looking at the α-C I assumed that its neighbouring H would split it into a doublet and then the hydrogens of the β-C (3+1 = 4) would split that doublet into an multiplet of eight. Which seems to be right if compared to the correct spectrum. Similarly, the β-C is first split into triplets and then into doublets, which makes 6 peaks.
However, why would C=O cause a peak? And how does it become a doublet?

The compound in question is seemingly a polypeptide. If all carbons are labeled $\ce{^{13}C}$, then you can obtain very good  $\ce{^{13}C}$-NMR spectrum withe a few accusations with minimum noise. That's probably what "$\ce{^{13}C}$ marked sample" means.
Since it is a polypeptide, $\ce{^{13}C=O}$ is attaced to a $\ce{-NH-}$ group from right hand side because it is a part of peptide bond (an amide). The relevant chemical shift for carbonyl carbon of an amide is around $\pu{175 ppm}$, and therefore the peak at that position is justified. However, this carbonyl carbon is adjacent to $\ce{C^\alpha}$. Since $^1J_\ce{^{13}C-^{13}C}$ coupling constant is given as $\pu{50 Hz}$, this peak is split as a doublet with $J$ value of $\pu{50 Hz}$.
Similarly, $\ce{C^\alpha}$ is attached to $\ce{C^\beta}$ and $\ce{C=O}$ groups as well as to a hydrogen. Since all $\ce{^{13}C-^{13}C}$ coupling constants are identical (according to the given values), the $\ce{^{13}C}$ peak of $\ce{C^\alpha}$ would split first to triplet with $^1J_\ce{^{13}C-^{13}C}=\pu{50 Hz}$ because of two carbons. Then, each triplet would split to doublets because of a single hydrogen on $\ce{C^\alpha}$ with $^1J_\ce{^{13}C-^1H}=\pu{125 Hz}$. Thus, the $\ce{C^\alpha}$ peak would appear as dt as shown in the answer.
Keep in mind that $\ce{C^\alpha}$ peak and $\ce{C=O}$ peaks would further split to aditional doublets with  $^1J_\ce{^{13}C-^{15}N}=\pu{15 Hz}$ if present nitrogen is $\ce{^{15}N}$-labeled.
I let OP to figured out what splitting would $\ce{C^\beta}$ peak would show.

Thank you for your answer! I just have one question, why isn't the C=O group also affected by the C$^β$?

$\ce{C=O}$ is not affected by $\ce{C^\beta}$. It is affected by only $\ce{C^\alpha}$. That's why it is only a doublet.

13C NMR detects only the 13
C
isotope of carbon, whose natural abundance is only 1.1%, because the main carbon isotope, 12
C
, is not detectable by NMR since its nucleus has zero spin. (https://en.wikipedia.org/wiki/Carbon-13_nuclear_magnetic_resonance)
You seem to have the number of peaks in the  α-C and the β-C confused. For example, the α-C peak is first split into a triplet by the two neighboring carbons, and that triplet is split into a doublet because of the α-C hydrogen.
Considering the α-C and the β-C are split by neighboring carbons, I believe you can then determine why the C=O peak is a doublet.


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Largest change in the water level

Which would produce the largest change in the $\ce{H2O}$ level when added to water in a $\pu{25 mL}$ graduated cylinder?
(A) $\pu{10.0 g}$ of $\ce{Hg}$ $(d = \pu{13.6 g/mL})$
(B) $\pu{7.42 g}$ of $\ce{Al}$ $(d = \pu{2.70 g/mL})$
(C) $\pu{5.09 g}$ of iron pyrite $(d = \pu{4.9 g/mL})$
(D) $\pu{2.68 g}$ of oak $(d = \pu{0.72 g/mL})$

I approached this question by dividing the mass by density to find the volume change. Thus, I got the answer (D), but the answer is (B).
Can anyone please explain this?

Because it’s density is less than 1, the oak floats.
Therefore it displaced less than its full volume; some of the floating oak is above water.
Since it’s 2.68g of oak, it displaces only the volume equivalent to 2.58g of water.

Because oak has a density less than water (1g/mL), it won't be fully submerged. For substances more dense than water, they will be fully submerged so you don't have to worry about it. See this video to calculate how much of the oak will be submerged.


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Difference between H-NMR with C$^{13}$ and C$^{13}$- NMR
I don't know if I am overcomplicating things but I am having trouble understanding the difference between H-NMR with  C$^{13}$ sample  and  C$^{13}$-NMR.
For example, if I have the molecule below and I am asked to sketch the H-NMR spectrum knowing that the carbons are C$^{13}$ and a C$^{13}$-NMR spectrum, what is the difference? For the C$^{13}$-spectrum I know that I should look at the neighbouring carbons and apply the $n+1$ rule and then look at the neighbouring hydrogens of the carbon of interest. For example, the C$^α$ has two neighbouring carbons so is split into a triplet and it also has one neighbouring hydrogen so those triplets are split into doublets making the C$^α$ show up as a multiplet of six.

However, if it is a H-spectrum with C$^{13}$ what do I look at? The correct spectrum is added below and the chemical shifts are:

So for H$^α$ I should get $2 *5 $ peaks. Which I get from H$^α$ having one carbon neighbour (spliting it into a doublet) and then having four neighbouring hydrogens (spliting each doublet into a multiplet of five). Is that the correct way of doing it? It works for H$^β$ as well but then for the H$^N$ am I only looking at the hydrogens again (so $n+1=1+1=2$ a doublet)?

Do you understand why the $n+1$ rule works and, perhaps more importantly, when it doesn't? The $n+1$ "rule" is extremely inadequate for solving this question.

The H-N has a $^3J$ coupling to C. Ten Hz or so sounds, well, reasonable. The next hydrogen would be a $^4J$, so likely not visible.


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For bond dissociation energies (BDEs) of double and triple bonds, is number shown in tables for that second or third bond? Or all of them?
Simple, 'dumb' question....   Is the kJ/mol number given in tables of bond strengths or bond dissociation energies , when it comes to double and triple bonds, just for the 'last' bond or for both (or all three) together?

The tabulated numbers are for dissociating the bond, i.e., breaking it, not just weakening it.

As for you are interested on this, you should have reasoned about this by considering the values, isn't?

I did reason about this, Alchimista, I just wasn't sure...

They're for the entire bond (so for a triple bond, all 3).
From Wikipedia, it can be defined as

The standard enthalpy change when [a bond, be it single, double, or triple] is cleaved by homolysis to give fragments A and B, which are usually radical species

So you get two fragments as your final state.

The part about homolytic cleavage is also important because for some bonds (like HCl) under the right conditions, heterolytic cleavage is much easier. For one thing, it does not produce radicals.


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galvanic cell vs electrolysis
At school(high school), to calculate the voltage needed for electrolysis, we calculate the voltage produce if the elements were in a galvanic cell, preceded by a minus.
So if a galvanic cell with Cu and Cr would produce 1,14V, we would say that an electrolysis with Cu and Cr will need -1.14V.
But why do we do this and is this correct?
Thank you in advance.

Suppose you have two galvanic cells A and B producing 1.14 V. If you connect them together, with both positive poles together, and both negative poles together, nothing will happen. No current will be produced. Now suppose that one of these two cells, say A, produces the same voltage 1.14 V, and B produces a little less than A, say 1.10 V. A works in a galvanic cell mode and will produce some current going through the second cell B. And this second cell B will work in the electrolysis mode, and the current through the cell B will go in the opposite direction with respect to its behavior in the galvanic cell mode. It looks as if B is going to be recharged.

"Suppose you have two galvanic cells A and B producing 1.14 V. If you connect them together, with both positive poles together, and both negative poles together, nothing will happen. No current will be produced" I think the current will be produced and the voltage= 1.14


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CO2 kills humans. Does it also kill organisms?
CO2 kills humans.
Does it also kill organisms?
Does it also kill viruses?

I'm inhaling it right now and am still alive. The dose makes the poison.

Does an extreme does of CO2 kill viruses?

At what temperature and pressure?

temp: 98.6 --- pressure: 14.696 psi

Viruses are not quite alive in the first place.

Larger concentration of $\ce{CO2}$ directly affects only organisms with metabolism involving $\ce{CO2}$.
Killing is indirect as humans and animals fall into unconsciousness and stop breathing.
$\ce{CO2}$ is used, in mixture with $\ce{N2}$, as protective gas for packed food. It may not necessary kill bacteria, rather stops their growth.
A virus does not have metabolism, so it is not affected this way.
It may be affected indirectly as $\ce{CO2}$ dissolved in water causes mildly acidic environment.
In summary, $\ce{CO2}$ cannot be considered as desinfectant.


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Covalent Bonds and Ionic Bonds
Covalent bond is a strong bond compared to Ionic Bonds but Ionic Compounds have higher melting and boiling points then covalent compounds. Why?

Maybe covalent bonds are stronger than ionic bonds. But ionic bonds are even stronger than that.

You cannot generalise from the bonding type to whether a compound is high or low melting. There are ionic compounds that are liquid at room temperature and there are covalent compounds that have very, very high melting points (like graphite).

When a substance is made of molecules, these molecules are made of atoms being strongly attached to one another. But a given molecule is not attached to its neighbor by strong bonds. On the contrary, they are attached by so-called van der Waals forces, which are rather weak. If you heat such a substance, the van der Waals forces are easy to destroy, and the substance melts easily. In the liquid, the covalent bonds are not modified.
When a substance is made of ions, the ionic bonds may not be as strong as covalent bonds. But they are stronger than van der Waals forces. This is why the ionic substances are melting at higher temperatures than substances made of pure covalent molecules.

In a chemistry site one should be not afraid to say that ionic bonds per sé can be very strong, having energies exceeding that of covalent bonds. In principle and it in vacuum it can be. Nevertheless I simple and clear answer which I upvote.

https://chemistry.stackexchange.com/questions/21774/why-is-covalent-bonding-stronger-than-electrostatic-attraction and especially top answer therein.


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Is it appropriate to call this material a hybrid composite?
The material I've synthesised is made up of two primary parts at a 1:1 ratio:

Part 1: agar, glycerol, water.
Part 2: polyurethane resin.

The agar mixture on its own was too soft, so I added the resin to harden the material. Would this material be classified as a hybrid composite? If not, why?

It is likely a composite. Sinthesys became (unfortunately) a slang to replace "making" "preparation" etc...

There is a slight deviation in the literature as to what the hybrid composites are.
Originally, materials were considered hybrid composites if both inorganic (e.g. clay/silicates as the hydrophilic component) and organic (e.g. organic polymer as the hydrophobic component) constituents were involved in building up the matrix (see e.g. a review [1]).
Another review [2] and the more recent definition from Encyclopedic Dictionary of Polymers [3, p. 373]:

Hybrid Composite n Advanced composite with a combination of different high-strength continuous filaments in the matrix.
  Also, composite in which continuous and staple fibers are used in the same matrix.

provide a more permissive definition, according to which hybrid composite materials are the materials in which a combination of reinforcing layers from two or more types of fibers is utilized.
With this in mind, in order to provide a definitive answer for yourself, you probably want to do surface imaging (e.g. with AFM or SEM) to prove that really are more than one type of fibers present.
If you are planning to publish this data, be prepared that you will be asked to provide rheological data as well.
References

Wahab, M. A.; Kim, I.; Ha, C.-S. Silica- and Silsesquioxane-Containing Polymer Nanohybrids. In Macromolecules Containing Metal and Metal-Like Elements; Abd-El-Aziz, A. S., Carraher, C. E., Pittman, C. U., Zeldin, M., Eds.; John Wiley & Sons, Inc.: Hoboken, NJ, USA, 2005; pp 133–160. DOI: 10.1002/0471712566.ch6.
Perov, B. V.; Khoroshilova, I. P. Hybrid Composite Materials. In Polymer Matrix Composites; Shalin, R. E., Ed.; Soviet Advanced Composites Technology Series; Springer Netherlands: Dordrecht, 1995; pp 269–304. DOI: 10.1007/978-94-011-0515-6_6.
Encyclopedic Dictionary of Polymers; Gooch, J. W., Ed.; Springer New York: New York, NY, 2011. DOI: 10.1007/978-1-4419-6247-8.

Thank you. I couldn't exactly find the correlation or significant differences of this material to the description of a hybrid composite, so your information really helped and is appreciated.


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Mechanism for synthesis of Meldrum's Acid
I have a little bit of problem trying to figure out the mechanism for the synthesis of Meldrum Acid. Meldrum acid is synthesis by a condensation reaction of acetone with malonic acid in acetic anhydride.

First, the carboxyl OH will be acetylated by acetic anhydride. But I can't figure out how the two ester bonds are being formed subsequently. I'm thinking that there's an acid-catalysed aldol condensation somewhere that results in the removing of water. Perhaps by attack of the enol (from acetone) to the acetylated carbon. But I'm not really sure about it either.
Any thoughts?

The first step (reaction with acetic anhydride) creates a mixed anhydride, this is a good electrophile with acetate as a leaving group. This is nucleophilically attacked by the oxygen of acetone (drawn as its enol in the scheme, but it is easier to visualise it as acetone) to give a cationic intermediate which is immediately attacked by the -OH of the neighbouring carboxyl group to give the product.

Thank you, I can understand how this eventually gets me to the product. But is there a reason why the carbonyl oxygen is the nucleophile here? Seems quite unusual for a carbonyl oxygen to attack a carbonyl carbon.

The mixed anhydride is very reactive, comparable to an acyl halide. The carbonyl has sufficient electron density to attack (or consider the enol the attacking species). I would emphasise that the intermediate that results reacts very quickly indeed with the carboxyl OH and that reaction is irreversible, so you only need a tiny percentage reaction for it to move forward.


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Software recommendations for overlaying molecular structures
I am looking for a program to overlay the structures of small molecules as an aid for my students in comparing the results of optimization geometries using different optimization methods.
At this stage, we are thinking of atomic coordinates and a visual comparison. However, if it can be complemented with a quantitative comparison that would a welcomed extra.
We work with the Microsoft Windows versions of Gaussian 09 and GaussView 5.

The compare command in free and open source programs Jmol (offline) or JSmol (online) outputs the root mean square of the pairwise differences (RMSD) as well as the superposed coordinates. For an extended tutorial see Proteiopedia — Jmol/Superposition.

You can do a structural comparison with VMD, using the RMSD trajectory tool. The RMSDTT is a default plugin for VMD (it comes with it, no need to do any fancy installation, so should help students access it).

There's a "pair-fitting" function in pymol that will align molecules given some user-defined reference atoms, and output an RMSD across those points of comparison. You can get pymol educational version for free, so students would be able to install it themselves.

I would say Chimera, no question. It has a good GUI with a lot of tools, including ensemble comparison. I find it both much easier and much prettier than VMD which has been recommended here.

Having used both VMD and Chimera, I'd beg to differ :P. By and large, it's down to user preference.

Oh, it was bliss when I found Chimera compared to VMD :P but yes, obviously it is preference. Hopefully the OP will find one program to their liking.


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How to extract a image (vector graphic) from a surface EPS plot with GaussView?
I am quiet new to this field of chemistry. I work with GaussView and I created a surface plot for my molecule with the cube and surface generator. This works very well. Now I like to save and copy this image of my molecule with its surface. My favorite file type whould be a vector graphic type (eps or svg).
I don not know how. I tried the option with File -> Save Image. When I choose vector graphic as type I just get the structur of my molecule but not the surface (as well as a error message...). When I choose all the other file types I get nothing in the file just a black screen.
Where is my mistake? What do I constantly wrong? I see a lot of these plots in publications... I just want this plot for my PowerPoint...
I hope someone of you got a godd tip for me, thanks!


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Is the benzylic carbocation aromatic?
So I've been trying to understand what exactly aromaticity means, and I've come to know that compounds possessing benzene rings are aromatic, like toluene and phenol. It thus made sense to me that the benzylic carbocation should be aromatic, as it fulfills all conditions of Huckel's rule. However, I found no mention of it being aromatic anywhere. Can anyone tell me if it is indeed aromatic or is only resonance stabilised, and if it is not aromatic, why so? What criteria is it violating?
Sorry if this is an elementary question, I really want to know what aromaticity actually means and am stuck on this.

For this case, aromaticity is most easily understood when comparing the stability to other carbocations. That being said, you'll find that aromaticity, much like a chemical bond, is like a unicorn: everyone knows what it should look like, nobody has ever really observed one or can give a quantitative measure except in a few narrow cases.

Basic criteria is a cyclic molecule with 4n+2 pi electrons,  as described in most undergraduate courses on physical organic chemistry

Sorry, forgot to mention that an alternative form of the benzyl cation is the cycloheptatrienyl cation. Hopefully that helps.


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Chemical Composition of NMC cathode in lithium ion battery
I am working with lithium ion batteries having the cathode material as NMC i.e. LiNixMnyCozO2, wherein x + y + z = 1. The proportion of Nickel, Manganese and Cobalt in the cathode varies, but mostly Nickel is either equal to or greater than the amount of Manganese and Cobalt in the cathode (for example NMC-111, NMC-532, NMC-622, NMC-811 are the commonly reported chemistries in literature).
I tried doing Energy dispersive X-ray spectroscopy on the surface of my battery cathode to find out the proportion of Nickel, Manganese and Cathode. The results I obtained were: Ni - 20%, Mn - 20%, Co - 60%. I repeated the EDXS procedure at ten different locations on the battery cathode and still got the same results with 1% error in the values. This result is far from what I expected and I suspect the reason for this is that I am scanning only the surface of the cathode.
So I have two questions regarding this:
1) Should I expect the EDXS results to be different and correct if I perform it on the cross-section of my battery cathode?
2) What are other cheap techniques which can get me the elemental composition of the cathode? One of my friends suggested me ICP-MS (inductively coupled plasma mass spectrometry), which is way too expensive for me.


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Equilibrium: Effect of change in composition and value of Kp, Kc on changing pressure

Di-hydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction:
$CH_4{_{(g)}}+H_2O_{(g)} \ce{<=>} CO_{(g)}+3H_2{_{(g)}}$
How will the values of $K_p$, $K_c$ and composition of equilibrium mixture be affected by increasing the pressure ?

Now,
1) According to this answer $K_c$ only changes on changing temperature i.e. unaffected by pressure.
2) According to Wikipedia the article "Pressure dependence" says, "...However, the composition at equilibrium will depend appreciably on pressure when:
1)the pressure is changed by compression or expansion of the gaseous reacting system, and
2)the reaction results in the change of the number of moles of gas in the system....."

here wiki is using term $K_p$ to denote equilibrium constant. Now the "composition of equilibrium" means $K_p$ and not $Q_p$ because its taking at equilibrium, so the $K_p$ changes with pressure.
3) I have studied that on increasing pressure the reaction shifts the side with less number of sum of  gaseous atoms in the reaction formulae. I know from this question that composition of equilibrium is different from "the side at which the reaction shifts". So it seems my study isn't useful here in deciding the composition of equilibrium
So my question is how will the values of $K_p$, $K_c$ and composition of equilibrium mixture be affected by increasing the pressure ?
Does one change and the other remains unaffected and isn't the composition of equilibrium directly related to $K_p$ and $K_c$
Are dependencies of $K_p$ and $K_c$ different for change in pressure.

By increasing the pressure, you produce more $CH_4$ and $H_2O$, and the amount of $CO$ and $H_2$ decreases. This is a well known method for producing $CH_4$ (+ $H_2O$) from the so called "water gas", made of a mixture $ CO$ + $H_2$ which is produced by sending water on some hot or burning charcoal :  $C + H_2O$ -> $CO  + H_2$

I think you're looking for $\mathrm{K_p} = \mathrm{K_c}(\mathrm{RT})^{\Delta n}$where $\Delta n$ is the change in the number of moles for the reaction. See for example this explanation.

$K_p$ will not change with pressure, only with temperature. $Q_p$ will instantaneously increase upon applying the pressure for this reaction (why? see below), and will drive the reaction to reactants to re-achieve equilibrium (decreasing $Q_p$) until $Q_p$ = $K_p$ once again. From a LeChatelier approach, you  have more gas molecules in the products, so an increase in pressure will drive the equilibrium composition to the reactants.
As for why $K_p$ won't change with pressure:
$\ln K_p=-\frac{\Delta G^⦵}{RT}$ is at a specified $p^⦵$ (usually 1 bar), so $\left(\frac{\partial {\ln}K_p}{\partial P}\right)_T = 0 $.
And as for why $K_c$ won't change with pressure, see the answer here

What about $K_c$

See https://chemistry.stackexchange.com/questions/49225/why-is-kc-not-affected-by-change-in-pressure (also added to answer).


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Most endothermic room-temperature reaction
In terms of energy per gram of reactants, the evaporation of water at -2.2 kJ/g, is much higher than any chemical endothermic reaction that I have seen demonstrated. At extreme temperatures there is no real limit, but for reactions that occur at a reasonable rate at room-temperature it seems water is hard to beat. And it isn't even a chemical reaction.
I have reason to think water is hard to beat: the entropy of gases is much higher than liquids. The strong hydrogen bond network of water makes the liquid phase lower entropy than non-polar liquids because it structures the liquid. Finally, water is a very light molecule so you get many moles of gas per kg of liquid. Despite this huge entropy gain, the phase-transition proceeds sluggishly because it takes so much dry air to shift the equilibrium forward, if it took much more energy it wouldn't happen at a meaningful rate.
Can any chemical reaction approach this at room temperature? If so, can any non-gas-evolving reaction come close?

Most, largest, biggest type questions over all chemicals aren't particularly interesting. First you end comparing apples and oranges. Second, there is always something else to discover.

Why is evaporation not a chemical reaction? Or do you just assume phase changes have nothing to do with chemistry? Seems odd...

Don't know if these are quite in the spirit of what you are looking for since, with the exception of the first reaction, the entropy change alone is not sufficient to drive them  (i.e., at room temp., their equilibria lie far to the left) (and, for ozone formation, the entropy change is actually unfavorable).  Furthermore, ozone formation and photosynthesis aren't single reactions, but multi-step processes.  But they all do take place at room temperature (or below):
Sublimation of ice (yes, snow can evaporate!):  2.8 kJ/g
Formation of ozone (O$_3$) from O$_2$ (photochemical, occurs in stratosphere): 3.0 kJ/g
Photosynthesis: 7.5 kJ/g
Electrolysis of water: 15.9 kJ/g
The values I've listed are all enthalpy changes (at constant pressure, the heat flow is equal to the change in enthalpy; at constant volume, the heat flow equals the change in energy).  In addition, by convention, the sign of the enthlapy for endothermic reactions is positive, since heat flows into the system, so the enthalpy is increased (not negative, as you've used).

One endothermic low-temperature reaction that gets a lot if use, seasonally, is salting of icy or snowy surfaces.  The sodium chloride combines with the ice to form a liquid solution.  Sodium chloride has little heat of solution either way with liquid water, but here the solid ice has to be converted to the liquid.  Therefore we have a reaction that's roughly as endothermic as the latent heat of water melting (80 calories per gram), but does not require directly warming the temperature of the ice.


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Which mechanism should a reaction between the cyclopentylmethanol and the HBr should follow?
I just thought that the reaction should follow $\mathrm{S_{N}2}$ mechanism this is because the alkyl halide is primary but the book says that the reaction will follow $\mathrm{S_{N}1}$ mechanism resulting in the formation of tertiary carbonation and a ring expansion product. Can anyone please explain the reason behind this ?

see http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch04/ch4-5.html

@jezzo is but the alcohol mentionedis also a primary alcohol that's why I am confused because I also thought that the reaction should follow SN2 mechanism.

It's a tricky example, but the $S_N1$ process is favored because the solvent is protic and the resulting 6-member ring is more stable than a 5-member ring.
If ring expansion weren't possible, I believe your intuition is correct that the $S_N2$ mechanism would dominate. See this discussion. You're not alone!


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Is alkyl bromide or alkyl chloride formed when alcohol reacts with thionyl bromide or thionyl chloride
Would alkyl bromide/alkyl chloride form when an alcohol reacts with thionyl bromide/thionyl chloride?

Did you not mean chloride instead of iodide? Or perhaps other way round... 'cause right now it makes no sense. Even then you should have some reason to think why not.

Thionyl chloride is a standard reagent for the conversion of alkyl alcohols to alkyl chlorides. Wikipedia article.
Thionyl bromide converts alkyl chlorides to alkyl bromides but is less used as it is less widely available and less stable, thus more difficult to store. more details


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Why do certain signals in proton NMR experience extensive splitting despite only having 1 neighbour?
I came across the compound 3-methylpentan-2-ol. Looking at its proton NMR spectra, it seems there is a lot of splitting for the signal at 4 ppm.

I'm assuming that the 4 ppm represents the proton in the hydroxyl group (since it clearly is much more deshielded than the other protons), but this would mean it only has one neighbouring hydrogen and hence should exist as a doublet.
I just wanted to clarify, is there any reason why the 4 ppm peak has so much splitting despite only having one neighbour? Or otherwise, I have also considered, could I be incorrect in assuming the 4 ppm peak represents the hydroxyl group, and if so why?

You are not correct in assuming the proton at 4 (actually closer to 3.6 - it all counts) is the proton on the -OH. So a question back to you - which proton is it?

Why wouldn't the proton at 4 be due to OH, as wouldn't it be the most deshielded due to oxygen? But anyways, to answer your question, would it be the proton attached to the C-OH group?

Peaks from labile protons in NMR  are frequently not observed, and have variable chemical shifts, unless great care is taken in sample preparation. This sample was runin CDCl3, and the -OH peak exchanges with H2O, and comes at ~1.6ppm. So, yes, the peak at 3.6 is from the CH(OH). So, can you account for the amount of splitting now? This is actually a very tricky spectrum if you are starting out with NMR.

This question has a few different parts to it.
Part 1 - what's that peak at 3.6ppm?
Well, it's not the -OH peak.  The peak at 3.6 is from the CH(OH).
Part 2 - where is the -OH peak if that isn't it at 3.6ppm?
Peaks from labile protons are frequently not observed, and have variable chemical shifts, unless great care is taken in sample preparation. This sample was run in CDCl3, and the -OH peak exchanges with H2O, and comes at ~1.6ppm.
Part 3 - can we account for the splitting of the peak at 3.6ppm now we know what it is?

The CH(OH) proton should couple to proton on the adjacent carbon (split to a doublet) and also the protons of the methyl group (split to a quartet). It won't couple to the -OH for the same reason we don't readily observe the -OH; it is rapidly exchanging with water. So, we expect that the peak at 3.6ppm should be a doublet of quartets. Looks more complicated than that to me.
Part 4 - It looks  more complicated that a doublet of quartets to me. What's happening there?
Of course, 3-methylpentan-2-ol has 2 stereocentres, and so we have here a mixture of diastereomers in solution in the sample. Both diastereotopic CH(OH) protons have very similar chemical shifts, and so there are in fact 2 sets of doublets or quartets. It's also why the rest of the spectrum looks such a mess.


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Calculating volume of a vessel using idea gas law

To an evacuated vessel with movable piston under external pressure of $\pu{1 atm},$ $\ce{0.1 mol}$ of $\ce{He}$ and $\pu{1 mol}$ of unknown compound (vapour pressure $\pu{0.68 atm}$ at $\pu{273 K})$ are introduced. Considering ideal gas behavior, calculate the total volume of 2 gases at $\pu{273 K}.$

Why can't I add the amounts of substance for helium and the unknown compound and directly use $PV=nRT?$ Why is the vapour pressure given?

HINT - Some of the unknown compound is in the liquid state.

@MaxW Ok. Now I got it. Since external pressure is greater than saturated vapour pressure, the unkown liquid will be in vapour form and by obtaining Partial pressure of He (1-0.68), I can proceed with ideal gas equation.

Yes, the volume will be the volume for which 0.1 mole of He has a partial pressure of 0.32 atm at 273K. The other 0.68 atm will be the unknown compound.


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Will NH3 gas decompose into N2 and H2 at 250 degrees celcius?
If I heat ammonia gas to 250 degrees celsius, will it decompose into its basic components - N2 and H2? What temperature is required for the reaction to occur?

Only on far higher temperature. Afaik 250 is far not enough for that, maybe if you have some catalyzers, but even so is it not sure. As far I know, the composition of NH3 is exotherm, at least on this temperature. The problem of the creation of ammonia is that the triple bonds of the N2 are very hard to dismantle. This is why ammonia creation requires well-controlled temperature (more than 250), pressure and catalyzers, in a many-step process. But I am not very sure in that.

The equation of the reaction is : $$\ce{N_2(g)  +  3 H_2(g)  <=> 2 NH_3(g)} $$
The equilibrium constant for this reaction have been summarized in the Chemistry Data Book, by J. G. Stark and H.G. Wallace, Jon Murray, London, 2006.  At $\pu{400 °C},$ $K_p$ is $\pu{40.7 atm^{-2}}.$ At $\pu{500 °C},$ $K_p$ is equal to $0.035$. So, at $\pu{250 °C},$ $\ce{NH3}$ is not significantly decomposed.

See Highly efficient decomposition of ammonia using high-entropy alloy catalysts which proposes highly efficient ammonia decomposition using a novel high-entropy alloy (HEA) catalysts made of earth abundant elements, namely, quinary CoMoFeNiCu nanoparticles.
However, the reported temperature is high (going up to 2000–2300 K).
Another source notes:

To date, the most effective catalyst for ammonia decomposition consists of ruthenium particles supported on carbon nanotubes (CNT) due to their high conductivity (6353 molH2 mol−1Ruh−1 at 430 °C) [2, 17]. The low temperature activity can be further improved by the addition of an electron donating promoter such as cesium (7870 molH2 mol−1Ruh−1 at 370 °C) [3, 17, 19, 20].

So, to answer your question, not at the temperature you indicated (250 degrees Celsius).