text:
Raman Scattered (inelastic) photons rescattering outside of capture area
Imagine that you are focusing the raman laser right at the center of the vial sample with water (or any liquid) using 532nm laser.

(from "Photometric Standards for Raman Spectroscopy)
For raman 180 degree backscattering sample geometry, can the inelastic scattering photons upon travel to the spectrometer becomes elastically scattered on way to it by the thick liquid sample such that it is scattered outside the area?
If yes, does it mean it is better not to focus so deep into the sample so the return path would be less (that is focusing the raman laser just after the glass on the liquid sample)? The side effect of this is getting some raman noises from the glass itself.

Yes, of course the Raman scattering can suffer from additional scattering on its way through the sample, just like it will generally be attenuated on its way to the detector by any element in its path (mirrors, lenses, glass surfaces, etc.).
If scattering is a problem or not very much depends on your sample. The ideal sample will not scatter your excitation laser or Raman scattering much, like for instance a clear solution, and then I would say the influence on the measurement is negligable. If you have something opaque, such as an emulsion, additional scattering will be relevant and you would want to measure near the sample surface.
If you have problems with a glass-affected background you could switch to quartz glass or use a confocal Raman setup, if available.
And also depending on your sample, absorption might be something you should consider as well.
By the way: I would probably rather go for a planar glass surface in order to avoid refraction as it occurs on a curved glass surface.
Edit due to comments:
Pure water does neither absorb nor scatter light with a wavelength around 532 nm much. If it would, you would be able to see this with your eyes. Wikipedia claims that a fraction of 10-7 of the incoming light is scattered inelastically. Although this should be only used as a rule of thumb, this tells you that a loss of Raman intensity due to "secondary" Raman scattering is not a practical problem you would usually be able to observe. Also, I would not be worried about elastic scattering although it is much more relevant than inelastic scattering (Wikipedia claims 0.1% to 0.01%).
I do not know your Raman setup in detail. But if you have anything like a confocal aperture (pinhole, slit...) on the path to the detector, refraction on the container/water surface will definitely reduce your signal intensity with increasing penetration depth, especially if you do not measure with an immersion objective lens. This is because your focus is severely distorted especially in length when focussing deeper into the sample, it will become much longer. Your confocal aperture then blocks part of your signal because it originates from a part of the sample which is far from the focal plane. This is descibed in detail for example in this paper, and also in this paper. However, the papers are behind a paywall.

What puzzled me is water is affected too. I used a plastic bottled water commonly used. When I sampled near the container surface, the water OH peak counts per frame at 3400 cm-1 is higher (about 14,000 counts) than if the focus is deep inside the plastic bottled water (only 10,000 counts). So if water can be raman scattered by the 532nm system, it can suffer rescattering on the way back to the spectrometer too?

To continue.I think what I wanted ultimately to know is in the case of water. Is it scattering or absorption that can change the counts from 14,000 to 10,000 using 532nm laser? What do you think based on your experience with water raman?

How deep is deep inside the bottle?

laser focused 20mm into the bottled water versus just 3mm. So is it scattering or absorption that mostly lower the counts. What do you think?

Alexander. I think it's water scattering and not absorption because water doesn't absorb visible wavelength but only IR, agree?

"Wikipedia claims that a fraction of $10^{-6}$ of the incoming light is scattered inelastically". We need here the fraction that is scattered elastically since the discussed mechanism of loss is that a Raman scattered photon is then undergoing a second scattering process. That could in principle be elastic or inelastic, but elastic is several orders of magnitude more likely, so p (further scattering) ≈ p (further elastic scattering). It is still only a tiny fraction of of the Raman scattered intensity.

Thank you, cbeleites, you are right, of course. I added the numbers from Wikipedia for elastic scattering to the answer to make the difference between elastic and inelastic scattering intensities a little clearer.

A few points in addition to @Alexander's answer:

Total scattering (including secondary) losses depend on the optical pathlength (cuvette thickness).

for 10 mm clear water, I'd expect this to be $\ll 1 \%$, so it doesn't explain the 30 % loss you encounter.
Yes, secondary scattering can sometimes be noticeable in Raman setups.  In clear media it plays a role only for long interaction lengths (e.g. collection fibers that have the Rayleigh filter only after the fiber may generate background Raman signal of the fiber)
Practical plausibility check: If your laser intensity (via laser intensity meter) after a flat of same thickness as your cylindrical vial isn't noticeably lower than before, then there isn't much of an intensity loss due to any type of scattering over the whole thickness of the cuvette, and in conseqence also not for the Raman scattered photons travelling towards your detector.

Even for geometries with low confocality, the optical geometry with the round cuvette may be the problem here: if the round cuvette is positioned (slightly) off the optical axis, you get higher reflection losses at the surface (the same is true if a flat cuvette is tilted).

A symptom of this should be that the observed intensity varies a lot when taking out the cuvette and putting it in again.
You can again check laser intensity before and after the cuvette: as long as you still get the whole beam onto the sensor area of the laser power meter, higher losses than though a similar flat cuvette are reflection losses (though both entering and leaving the cuvette)

In addition, if you use a high NA objective, that has a wide angle collection cone.

This is typically desired because you collect Raman signal over a wide solid angle and thus get better intensity.
But unless you manage to get your cuvette positioned so that the center of the cuvette is where the focus point in air would be (i.e. collected light leaves the cuvette at 90° to the cuvette surface), you may get substantial losses due to reflection:

(see Wikipedia on Fresnel equations)
Note that these reflection losses are polarization dependent, so may affect different bands differently depening on their depolarization ratio.
Again, an immersion may help.

In consequence, you need to optimize depth into the cuvette (and possibly cuvette diameter), lateral position (centered) and the angle of the cuvette (wrt. polarization) in order to get maximum intensity.


text:
Which is faster- SN1 or SN2?
We were studying about nucleophilic substitution reactions. My professor said that in general SN1 reactions are faster than SN2 reactions. I am unable to understand that how can we generalise the rate of two completely different reactions. In this case, what I think is that the rate will depend on our reagent, leaving group, solvent, etc and in some cases SN1 will be faster while in some others SN2. Therefore, I'm a bit confused.
I would be happy if anyone could help me with this.

Not always the case: Confusion about the SN1 and SN2 Mechanism; Comparison of the rate of SN1 and SN2 reaction.

Yes, but generally, first order kinetics are faster than second order, and the latter slow down much more as the reaction progresses.

@Karl Is that somewhat related to the fact that in second order reactions, effective collisions are required and this takes time?

[OP] In this case, what I think is that the rate will depend on our reagent, leaving group, solvent, etc and in some cases SN1 will be faster while in some others SN2.

SN2 is a second order reaction, so the rate depends on the concentration of the nucleophile. In cases where both SN1 and SN2 mechanisms are plausible (e.g. secondary alkyl halides), you can "push" the mechanism toward SN1 (low concentration of nucleophile, weak nucleophile, protic solvent) or SN2 (high concentration of nucleophile, strong nucleophile, aprotic polar solvent). So even for a given reaction, you can't make the statement that one will be faster than the other. Here is a publication where they studied this.

[OP] My professor said that in general SN1 reactions are faster than SN2 reactions.

I'm not sure your professor would publish that statement. Maybe they said it in passing or for a very specific case. In the comments, there is a discussion about whether SN1 is faster because it does not need a collision. I don't think that that is argument is tight - you need collisions with the solvent to get over the activation barrier.

[OP] I am unable to understand that how can we generalise the rate of two completely different reactions.

You are right, we can't.

source: Stack Exchange

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

sort: votes


text:
Database for experimental diatomic ionization energies
I am looking for a database, in which the experimental ionization energies (and e.g. heat of formation) of atoms and diatomic molecules are listed as files.
If it comes with a Python package, that would even better.
I found the Pitt Quantum Repository, but there seem to be only theoretical values.

CRC handbook of chemistry and physics lists IEs of elements, atomic and several molecular ions. Also, have a look at Resources for learning Chemistry.

Ah. The good old rubber bible. No chemistry shelf is complete without it. :-)

NIST Atomic Spectra Database has the atom part. Not sure about molecules.
https://physics.nist.gov/PhysRefData/ASD/ionEnergy.html

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


text:
Moving water vapour between flasks
I have a flask of silica gel connected to a flask that contains a bit of water, so when I heat the silica gel the water absorbed gets evaporated and I want it to move to the other flask, but it always condenses on the roof of the first flask, so is there any way to force the vapour to move flasks?

To prevent the condensation, you need to create the conditions under which the vapor does not become saturated. The way to do that is to heat up the whole transfer line (the roof of the first flask and all the tubes that follow) - the higher the temperature, the higher the saturated vapor pressure. Heated transfer lines are often employed in commercial equipment when there's need to get condensible gases from one setup (e.g. thermogravimetric) to the other (e.g. mass spectrometer).

In your flask, the water will always go from the drops of liquid water to the silica, and not in the opposite direction. If you want to remove the water from the silica, the only way to do it is to heat the hydrated silica in the air oder under a vacuum, and quickly close the hot silica when dry.


text:
Effect of dilution on the degree of dissociation of an electrolyte
My book says that on dilution, the degree of dissociation of an electrolyte increases. Consider the reaction:
$$\ce{A(aq) <=> C(aq) + D(aq)}$$
Let $\ce{A}$ be the electrolyte that splits into its ions $\ce{C}$ and $\ce{D}$.
According to my book, on dilution, the concentration of $\ce{C}$ and $\ce{D}$ will decrease. Hence according to the Le Chatelier's principle, the reaction must proceed in forward direction.
Won't the concentration of $\ce{A}$ also decrease? In that sense, shouldn't the reaction proceed both in the forward and backward direction?
Please do let me know where am I going wrong.

Your question is good one .
So let me first correct you your argument "conc. Of both product and reactant will decrease on dilution so the reaction will remain in equilibrium Without any change in no of moles of either product or reactant"
First of all Le Chatliers principle talks about combined conc. Of reactant and product so on dilution combined conc will decrease so to increase the combined conc. Reaction will proceed in direction with greater no of moles. I Hope it will help you.

Related: https://chemistry.stackexchange.com/questions/72609/what-is-the-effect-of-dilution-on-the-degree-of-dissociation-of-a-weak-acid-or-b?rq=1

If our dummy dissociation degree is $$\alpha=\frac{[\text{C}]}{[\text{A}]+[\text{C}]}=\frac{[\text{D}]}{[\text{A}]+[\text{D}]}$$ and the total concentration of dissociated and undissociated species is $$x=[\text{A}]+[\text{D}]=[\text{A}]+[\text{C}]$$ the equilibrium constant, on which le Chatelier's principle is (well, should be) based is $$K=\frac{[\text{C}][\text{D}]}{[\text{A}]}=\frac{\alpha^2x}{1-\alpha}$$ Now, we'd like to dilute (to decrease) the total concentration $x$. Since the equilibrium constant cannot change, $\alpha$ must increase to keep the constant constant (sorry). And all the concentrations decrease upon dilution, which you can easily prove by checking whether derivatives $\frac{\partial[\text{A}]}{\partial x}$, $\frac{\partial[\text{C}]}{\partial x}$ and $\frac{\partial[\text{D}]}{\partial x}$ are positive.
I really hate them easy "intuitive" principles. In my opinion, resorting to the discussion of the equilibrium constants gives you a clearer picture.
And yes, the reaction proceeds in both forward and reverse directions, if only because it is an equilibrium process.

Your book is right, On dilution, the degree of dissociation of an electrolyte increases. But your interpretation is wrong. If you increase the degree of dissociation of A, it means that there is less A, and more B and C in the solution. The degree of dissociation is by definition the fraction of A which are transformed into B and C. This fraction increases by dilution, according to le Chatelier's principle.
Finally, by dilution, all concentrations decrease, whatever their degree of dissociation.

Water is often described as the universal solvent because most things will dissolve in water if you have enough time and water.
For a compound that is only slightly soluble in water adding more water will indeed increase the number of electrolytes formed.
It may help to think about your reaction in this format:
$$\ce{A + H2O <=> C + D}$$
If copious amounts of water are needed to get the material to dissolve thinking of water as a reactant may help you to understand the equilibrium. Of course, adding water still has an effect on the concentration of reactants and products.


text:
Why do we use [H⁺] and [OH⁻] in balancing redox reactions instead of using [H₃O⁺] and [OH⁻]?
For what reason is it universally agreed upon that we use $$\ce{H+ + OH-}$$ ions in balancing water on either side of the reaction, but not the $$\ce{H3O+}$$ ions?

Who said hydronium isn't used? You can use either $\ce{H+}$ or $\ce{H3O+}$ or even $\ce{H9O4+}$ — it just adds complexity in terms of bloated stoichiometry. So, if you don't want to focus on what business does a proton have in aqueous solution, in most cases you get away with $\ce{H+}$ just fine.

@andselisk so basically it's because of the accuracy and complexity of which we want to refer to when doing redox reactions?

"on average, each hydronium ion is solvated by 6 water molecules which are unable to solvate other solute molecules."  https://en.wikipedia.org/wiki/Hydronium#Solvation

@user243851 Yes, I suppose you can put it this way.

It is by far not universally agreed upon. My year 10 chemistry teacher taught me to use $\ce{H3O+}$ in balancing. The result is a larger number of water molecules on the other side of the equation but everything else is the same.
Many chemists will prefer $\ce{H+}$ because it is shorter and simpler, that’s all.

Because what we really need to know here is the concentration of protons. The concentration of water is considered to be constant in most of the calculations involving the water autodissociation. Thus, you may hydrate the said proton in your equation however you want — or not at all — but $K_\mathrm{w} = [\ce{H+}][\ce{OH-}]$ would still be equal to $K_\mathrm{w} = [\ce{H9O4+}][\ce{OH-}]$ and, well, who needs extra complications?


text:
Phosphoryl bromide reaction with benzoic acid
I am trying to find out how $\ce{POBr3}$ reacts with $\ce{-COOH}$ group (benzoic acid to be specific). I found that it performs an elimination reaction with alcohols to give alkenes. I have also seen that keto group was converted to $\ce{-Cl}$ using $\ce{POCl3}.$
But I don't think that $\ce{POCl3}$ will eliminate $\ce{-COOH}$ group — that seems like nonsense. Eliminating using $\ce{-OH}$ from $\ce{-COOH}$ would give me a carbon with five bonds. Exchanging the $\ce{=O}$ from $\ce{-COOH}$ would give me something I have never seen and it doesn't even look good…
So I am sort of out of ideas. I'd appreciate any hint what to do.

I've not see this done this way but my gut feeling is that you would get the acyl bromide. Anyone got any references or other thoughts?

I don't know about POBr3 but PBr3 has been used to brominate carboxylic acid to produce alpha-bromocarboxylic acid. This is the base for HVZ reaction. I cannot say if this happens for benzoic acid since it can be brominated using Br2/FeBr3.

SciFinder does not uncover any published reactions that use a carboxylic acid and $\ce{POBr3}$ or $\ce{POCl3}$. Thus, it seems we will have to speculate about the answer.

@Waylander you are correct :) thank you for your help

Well actually @Waylander and @Nilay Ghosh are both correct in a way. To convert carboxylic acids into acyl chloride, $\ce{PCl5}$ is often used as a chlorinating agent , which yields an acyl chloride and $\ce{POCl3}$ (and $\ce{HCl}$) as a by-product.
This $\ce{POBr3}$ (or likewise $\ce{POCl3}$) is essentially a dehydrating agent and is thus used to dehydrate alcohols to alkenes(1)

or brominate(2) (in some cases) to form alkyl bromides.

But this dehydrating capability might also be used on carboxylic acids to convert them into their anhydrides just like $\ce{P4O10}$, though there's no such special mention of the viability of the reaction progress.
But since you're really asking, I think conversion of the acid into its anhydride could be a possible outcome in presence of a dehydrating agent $\ce{POBr3}$.

So the correct answer (according to my teacher) is that R-COBr is generated from the reaction of POBr3 with R-COOH (benzoic acid)


text:
Is it wrong to allege that lemons and limes are alkaline?
Please ELI5; I've never studied chemistry. I read about  alkalinity v. basicity here and on Sciencing.com.
Lemons and limes are doubtless acidic.

Lime: The pH range of lime is 2.00 to 2.35.
Lemon: The pH range of lemon is 2.00 to 2.60. This makes lemon slightly less acidic than lime.

'The pH levels of citrus fruits can cause damage to tooth enamel. Typically, the lower the pH, the more damage is done to the teeth. Oranges have a higher pH level, at 4.35, and are less damaging to teeth than lemons at 2.75, limes at 2.88.'

So isn't it wrong for Yuri Elkaim BPHE, CK, RHN

and this unidentified author to allege lemons and limes to be alkaline?

https://chemistry.stackexchange.com/questions/105503/sugar-solutions-have-a-neutral-ph-in-themselves-but-it-makes-your-body-acidic

Yeah, I've seen such claims too and they always leave me puzzled. I heard also that it's good for the humors to eat a balance of wet and dry foods.

Whole lotta pseudoscience stuff from people who never took a chemistry course, and great marketing to the crowd who believes that they can "alkalize" their body with the snake oil being sold to them.

You're right to be confused. Limes and lemons are indeed acidic, not alkaline.

Grouping of food items into various classes is a very old idea of Ancient Greek, Alchemists & Ancient Indian School of Medicine (Aryuveda). These ideas are perhaps thousands of years old. From a scientific perspective these classifications don't mean anything today. For example, in those schools of thought, fish has a very "hot" effect on the body but yogurt is "cold". Similarly, citrus fruits are "cold" but dried dates are "hot". Now nobody can define what hot or cold means and what is the criterion?
Similarly, the chart shown in the original posts are all wrong. It is just a modern version of these ancient schools of medicine. I think the major source of confusion in calling limes and lemons is that citrates (salt of citric acid) are indeed alkaline. Many antacid concoctions use solid sodium bicarbonate and citric acid. The moment you add a spoon to water, it starts to fizz vigorously. The pH of such solutions are slightly alkaline. Perhaps whoever made those charts had these ideas in mind.

There might be some confusion between lime as in limestone and lime as in lime tree. Even though the current spelling is the same, the former has indo-european roots while the latter has Arabic roots. One is alkaline and the other acidic.

Very interesting, I did not know that the word lime (fruit) was derived from Arabic. In Urdu/ Hindi/ lemon is lee-moo, very close to Arabic (< Arabic līma). The calcium hydroxide (lime) is from Latin  *lîmo- = Latin līmus mud as per the OED.


text:
Intuition for the overlap integral
I want to ask a question about the overlap integral.
I was studying about the MO diagram of $\ce{H_2^+}$ and we managed to derive an expression which had the following term:
$$\langle \mathrm{1s}_\ce{A}(r_\ce{A}) | \mathrm{1s}_\ce{B}(r_\ce{B}) \rangle$$
and our lecturer later explained that this is known as the overlap integral.

The overlap integral is the magnitude to which the atoms overlap when combining AOs to produce MOs.

The trouble is, I later wanted to find a proof for this idea in order to confirm my understanding of this idea, but could not find anything that could explain this phenomenon from principle.
I did come across this question, but this only explained its main definition and not actually any derivation from basic principle.
How can the overlap integral be explained from basic principle?

I don't quite understand what you're asking. The overlap integral is just defined to be that integral (or even more simply, that integral is given the name of "overlap"); there's no "derivation" involved.

Let $\Phi, \Psi$ be normalized wave functions.
Then $0 \le|\langle\Phi\vert\Psi\rangle| \le 1$, with $\langle\Phi\vert\Phi\rangle = 1$ and $\langle\Phi\vert\Psi\rangle = 0$ for orthogonal $\Phi$ and $\Psi$.
Since we can argue that any wave function completely contains itself and for two orthogonal wave functions none of them contains a part of the other, we can interpret the scalar product $\langle\Phi\vert\Psi\rangle$ as a measure for overlap.


text:
Is the vapor pressure of a liquid dependent on the pressure of the atmosphere in which it is measured?
The figure below shows how vapor pressure of a liquid is measured. Initially the flask is at atmospheric pressure. Then the liquid is dropped in the flask. Once equilibrium is reached, the pressure from the vapor created is recorded. My question is - if the flask were initially at a different pressure - let's say 5 atm - would the vapor pressure of the liquid be different? My thinking is that the additional pressure in the flask would keep more of the liquid molecules in liquid form.

No, it doesn't. Unless it changes the substance. For example turns water into "hot ice". .

Yes, it is,see https://en.wikipedia.org/wiki/Poynting_effect

Yes, in principle it could be different, but the effect would be very small unless the increase in pressure were very large.  To put it in simplified terms:  The increase in ambient pressure compresses the liquid, bringing it to a higher-energy state, causing a higher proportion of the liquid molecules to want to escape from the liquid, and into the gas state, thus causing the vapor pressure to increase.  This assumes ideal gas behavior, i.e., that the increased ambient pressure affects only the liquid, not the gas.
DETAILS:
Let's call your liquid "$X$".  At equilibrium, the chemical potential of $X_{(l)}$ equals the chemical potential of  $X_{(g)}$.  The pressure of $X_{(g)}$ at which that occurs is called the vapor pressure.
Now suppose you increase the ambient pressure at constant temperature. Consider what happens to the chemical potential of the liquid: $d\mu = -S_mdT +V_m dP$, where $\mu$ is the chemical potential, and $S_m$ and $V_m$ are the molar entropy and volume, respectively.  Thus, at constant $T$, $d\mu = V_m dP$.  Hence the chemical potential of $X_{(l)}$ increases.  This effect, by itself, would cause the vapor pressure to increase (to get the chemical potential of $X_{(g)}$ to match the increased chemical potential of $X_{(l)}$).
But what about the chemical potential of $X_{(g)}$?  If the gas behaves ideally, then it is unaffected by the presence of other gases, and thus its chemical potential is unaffected as well.  I.e., as with the liquid, its chemical potential is determined by its pressure.  But unlike the case with the liquid, the pressure of $X_{(g)}$ is independent of the ambient pressure -- it depends only on the partial pressure of $X_{(g)}$ itself.  Thus the only effect to consider would be the increased chemical potential of the liquid, and the vapor pressure would increase.
However #1: At high concentrations, gases tend not to behave ideally.  Thus there are two possibilities: the increased partial pressure of the other gas increases the chemical potential of $X_{(g)}$, or decreases it.  If it's the former, and the effect is sufficiently strong, the vapor pressure will decrease.  If it's the latter, the vapor pressure will increase.
However #2: If the ambient gas is soluble in the liquid, increasing the ambient pressure will increase the concentration of the ambient gas in the liquid, which could reduce the chemical potential of the liquid, thus reducing the vapor pressure!
Again, let me emphasize that these are very small effects.

That math makes sense to me, but does anyone have an intuitive explanation at a molecular level for why vapor pressure increases due to atmospheric pressure.

@Ryan I've added some language towards the end, in the paragraph beginning "So, to put it in simplified terms..."  Let me know if this helps.

yes it does help!

@Ryan: Feel free to accept my answer if you think it addresses your question.


text:
Extraction Microrna From Humans Plasm
My question is: what is the best way to extract micro RNA (miRNA) from humans plasma? Whether trizol is suitable, or should I use commercial kits? Thanks

This seems a question better suited for another SE site, such as biology or medical sciences SE.

See: https://www.researchgate.net/post/What_is_the_best_method_of_microRNA_extraction

For commercial advice: https://www.thermofisher.com/se/en/home/references/ambion-tech-support/rna-isolation/general-articles/ten-ways-to-improve-your-rna-isolation.html

A comparison of approaches: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3871541/


text:
Spectroscopic Determination of an Equilibrium Constant
I got values of $\pu{5.0E-4}$, $\pu{2.5E-4}$, and $183$ for the following problems. However, these seem to be inaccurate and I'm not sure what to do.
Here is the procedure:

Obtain 6 disposable test tubes. Label the test tubes 1 through 6. To each of the $\pu{6 mm}$ test tubes, add $\pu{5.0 mL}$ of $\pu{2.0E–4 M}$ $\ce{KSCN}$. If you make a mistake, you must rinse out and thoroughly dry the test tube before starting over.
To test tube 1, add $\pu{5.0 mL}$ of $\pu{0.20 M}\; \ce{Fe(NO3)3}$ and stir. The resulting solution is your standard solution of $\ce{FeSCN^2+}$.
For the other test tubes, do the following: Measure $\pu{10.0 mL}$ of $\pu{0.20 M}$ $\ce{Fe(NO3)3}$ in a $\pu{25 mL}$ graduated cylinder, fill to $\pu{25.0 mL}$ with distilled water, and stir thoroughly to mix. Measure $\pu{5.0 mL}$ of the resulting $\pu{0.080 M}$ $\ce{Fe^3+}$ into test tube 2.
Discard all but $\pu{10.0 mL}$ of the $\pu{0.080 M}$ $\ce{Fe^3+}$, refill the graduated cylinder with $\pu{10.0 mL}$ of the $\pu{0.080 M}$ $\ce{Fe^3+}$ solution to $\pu{25.0 mL}$ with distilled water, and stir thoroughly. Add $\pu{5.0 mL}$ of the resulting $\pu{0.032 M}$ $\ce{Fe^3+}$ to test tube 3.
Again discard all but $\pu{10.0 mL}$ of the contents of the graduated cylinder, refill to $\pu{25.0 mL}$ with distilled water, stir, and add $\pu{5.0 mL}$ to test tube 4.
Repeat above procedure with test tubes 5 and 6.

Calculate the initial concentrations of $\ce{Fe^3+}$ and $\ce{SCN-}$ for the six solutions to be prepared in this experiment (i.e., the 6 solutions prepared in test tubes 1–6). These are the concentrations just after the two solutions are mixed in the large test tube (thereby diluting each other) but before any $\ce{FeSCN^2+}$ has formed. Assume that the volumes are additive and that the $\ce{Fe(NO3)3}$ and $\ce{KSCN}$ are each completely dissociated. What is the concentration of $\ce{Fe^3+}$ in test tube #5?

$$
\begin{array}{c|cc}
\hline
\text{Test tube} & \text{Initial}~[\ce{Fe^3+}] & \text{Initial}~[\ce{SCN-}] \\
\hline
1 & & \\
2 & & \\
3 & & \\
4 & & \\
5 & & \\
6 & & \\
\hline
\end{array}
$$

For the chart in question 1, what is the molar concentration of $\ce{SCN-}$ in tube #4?
If the equilibrium $\ce{[FeSCN^2+]}$ in test tube 2 was determined to be $\pu{8.3E-5 M}$, calculate the equilibrium $\ce{[Fe^3+]}$.

You must first calculate dilution factors. Assuming the volumes are additive, we can calculate dilution using $M_1V_1=M_2V_2$ equation. Also, assuming $\ce{Fe(NO3)3}$ and $\ce{KSCN}$ are each completely dissociated, we can say:
$$\ce{[Fe(NO3)3] = [Fe^3+]} \text{ and } \ce{[KSCN] = [SCN-]}$$
Thus, initial $\ce{[Fe^3+]}$ and $\ce{[SCN-]}$ are $0.20$ and $\pu{2.0 \times 10^{-4} M}$, respectively.
In test tube one: $\pu{5.0 mL}$ of $\pu{0.20 M} \; \ce{Fe(NO3)3}$ and $\pu{5.0 mL}$ of $\pu{2.0 × 10^{−4} M} \; \ce{KSCN}$ are added together, thus dilution factor is 1:2. Therefore, using $M_1V_1=M_2V_2$ equation you can calculate initial concentrations of $\ce{[Fe^3+]}$ and $\ce{[SCN-]}$ after the addition (but before they react with each other) as follows:
$$\ce{[Fe^3+]} = \pu{\frac{0.2 \times 5.0}{10.0} M} = \pu{0.10 M} \text{ and } \ce{[SCN-]} = \pu{\frac{2.0 × 10^{−4}  \times 5.0}{10.0} M} = \pu{1.0 × 10^{−4} M}$$
As a matter of fact, since the original solution of $\ce{KSCN}$ did not undergo any dilution maneuver before final 1:1 dilution, $\ce{[SCN-]}$ of all 6 test tubes are same, and its value is $\pu{1.0 × 10^{−4} M}$.
On the other hand, the original solution of $\ce{Fe(NO3)3}$ has undergone serial dilution of $\pu{10 mL}$ to $\pu{25 mL}$ (hence dilution factor is $10/25$ or $2/5$). Therefore, if original concentration of each solution is $x$ before add together in test tubes, the final $\ce{[Fe^3+]}_a$ before add with $\ce{[SCN-]}$ is:
$$\ce{[Fe^3+]}_a = \frac{x \times 2.0}{5.0} \: \pu{M}$$
Thus, the final $\ce{[Fe^3+]}_a$ before add with $\ce{[SCN-]}$:
$$\ce{[Fe^3+]}_2 = \pu{\frac{0.20 \times 2.0}{5.0} M}= \pu{0.08 M}, \, \ce{[Fe^3+]}_3 = \pu{\frac{0.08 \times 2.0}{5.0} M}= \pu{0.032 M}, \,  \ce{[Fe^3+]}_4 = \pu{\frac{0.032 \times 2.0}{5.0} M}= \pu{0.0128 M}, \,  \ce{[Fe^3+]}_5 = \pu{\frac{0.0128 \times 2.0}{5.0} M}= \pu{0.00512 M}, \,  \ce{[Fe^3+]}_6 = \pu{\frac{0.00512 \times 2.0}{5.0} M}= \pu{0.002048 M}$$
Finally, those concentrations were further diluted in 1:1 ratio in test tubes, and therefore, initial $\ce{[Fe^3+]}$ are half of those $\ce{[Fe^3+]}_a$ values. I tabulated all of them accordingly in following table, as answer to your question 1:
$$
\begin{array}{c|cc}
\text{Test Tube #} &\text{Initial $\ce{[Fe^3+]}$} &\text{Initial $\ce{[SCN-]}$} \\\hline
1 & \pu{0.10 M} & \pu{1.0 × 10^{−4} M}\\
2 & \pu{0.04 M} & \pu{1.0 × 10^{−4} M}\\
3 & \pu{0.016 M} & \pu{1.0 × 10^{−4} M}\\
4 & \pu{0.0064 M} & \pu{1.0 × 10^{−4} M}\\
5 & \pu{0.00256 M} & \pu{1.0 × 10^{−4} M}\\
6 & \pu{0.001024 M} & \pu{1.0 × 10^{−4} M}\\\hline
\end{array}
$$

There are 7 different values in the table. But you have given one
relevant answer, and it is also incorrect.
Find the answer for $\ce{[SCN-]}$ in test tube 4 from the table, which is
identical in all 6 test tubes.
The concern equilibrium is: $$\ce{Fe^3+ + SCN- <=> FeSCN^2+}$$ Reaction mole ratio of $\ce{[Fe^3+]}$ : $\ce{[SCN-]}$ : $\ce{[FeSCN^2+]}$ is 1 : 1 : 1. If equilibrium concentration of $\ce{[FeSCN^2+]}$ in test tube 2 is $\ce{ 8.3×10^{−5} M}$, the equilibrium concentration of $\ce{[Fe^3+]}$ in test tube 2 is $\ce{ (0.04 - 8.3×10^{−5}) M}$ and that of $\ce{[SCN-]}$ in test tube 2 is $\ce{ (1.0×10^{−4} - 8.3×10^{−5}) M}$. Now, you may able to calculate equilibrium constent for the reaction using following equation:

$$K_\mathrm{eq} =  \frac{\ce{[FeSCN^2+]}}{\ce{[Fe^3+]}\ce{[SCN-]}}$$

@Denise If you find the answer useful, the best way to appreciate the effort that went into producing one would be to upvote and accept the answer.


text:
n-Factor of phosphorous acid
It's very confusing that while $\ce{H2SO4}$ has an $n$-factor of 2, whereas $\ce{H3PO3}$ has $n$-factor 2 instead of 3. Many tried explaining it to me by molecular structures, but I don't know how to draw them. Could you please try explaining without the help of molecular structures?

As long as you refuse to know the thing, there is no way you can know it.

The hint: the structure of $\ce{H3PO4}$ is $\ce{OP(OH)3}$ , but the structure of $\ce{H3PO3}$ is NOT $\ce{P(OH)3}$

Also, note that there is no zero in $\ce{H2SO4}.$ And I'm going to be the third commentator who is going to stress out that you won't be able to go far in chemistry without molecular structures.

What is the $n$ factor?

$n$-Factor is basically a method to find out a relationship between the compound and what it is equivalent to in terms of acidic nature or basic nature. Note that the two hydrogen atoms in $\ce{H2SO4}$ are both attached to oxygen. But in $\ce{H3PO3}$ two of the three hydrogens are attached to oxygen and the other hydrogen is directly attached to phosphorous.
Note that only those hydrogens attached to highly electronegative elements can be donated easily in case of inorganic acids. One could explain better by using the concept of stabler structures with the $\ce{H}$ removed from $\ce{O}$ than from $\ce{P}$ by resonance. But for now know it as a thumb rule.
Therefore, since $\ce{H3PO3}$ has only two $\ce{H}$ attached to $\ce{O},$ its $n$-factor is 2 and not 3.

This is more or less correct (just bring the formatting in order), but the reasoning is based on the structural aspects which OP tries to evade. Sort of catch 22 in action.

@andselisk Sorry but I can't use mathjax a lot. Maybe you can help.

I tried my best. In the meantime, feel free to visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.

Thanks andselisk! Ill try to implement the codes little by little from next posts here. And anyway I don't know how the OP wants to get the answer without even the basic Lewis structure?

No prob, take your time, we are all learning here. I honestly have no clue either how to give a meaningful answer without getting the structure involved.

I am afraid the answer if your teacher or textbook has told that the n-factor of phosphoric acid is 2, they are teaching it the wrong way. The concept of equivalent weight is totally dependent on how the reaction equation is written, therefore there is no fixed equivalent weight. Think of the permanganate ion- how many equivalent weights does it have!
I will give an example of phosphoric acid, and you can figure out the same for your compound. So, if you are neutralizing only one proton, in an acid base titration, the n factor is equal to 1, i.e., the equivalent weight is the same as the molecular weight
$\ce{H3PO4 + NaOH -> NaH2PO4 + H2O}$  n factor is equal to 1 in this case.
However, if you are adding enough base that both are two protons are abstracted.
$\ce{H3PO4 + 2NaOH -> Na2HPO4 + 2H2O}$  n factor is equal to 2 in this case
You added more base:
$\ce{H3PO4 + 3NaOH -> Na3PO4 + 3H2O}$  n factor is equal to 3 in this case.
The main lesson is that equivalent weight is calculated the way you write the equation.

The questions is NOT about phosphorIC acid but about phophorOUS acid, https://en.wikipedia.org/wiki/Phosphorous_acid which is indeed dibasic

@IanBush which doesn’t exist as is it actually phosphonic acid ;)

@Indeed Ian Bush. Any person who can count from 1 to 10 would also be able to differentiate between H3PO4 and H3PO3, I gave this 9th grader (OP) an example to follow rather than solve this homework for H3PO3.

@IanBush https://chemistry.stackexchange.com/questions/85253/phosphorous-acid-vs-phosphonic-acid/85265?r=SearchResults#85265

If the $n$-factor is what I assume it is (the basicity, or ‘how many moles of sodium hydroxide can one mole of acid neutralise?’) then …
It is really hard to explain this without molecular structures. Sum formula are of little help to chemists once you reach more than four or five atoms or if there are two types of non-hydrogen atoms. Two compounds may have the same sum formula but different structures explaining a very different reactivity and properties. Take for example $\ce{C4H10O}$ which, among others, can be diethyl ether, the very low viscosity liquid with a boiling point of $\pu{32 ^\circ C}$ that was once used as an anaesthetic and a replacement for alcohol, or can be tert-butanol, a solid at room temperature except in Japanese labs in summer, with a quite different but nonetheless recognisable odour that has no common uses except sometimes apparently as a paint remover according to Wikipedia. So over time you will have to accustom yourself to molecular structures.
As long as you don’t resort to molecular structures, the only real answer is that even at $\mathrm{pH\ 14}$ the anion will remain $\ce{HPO3-}$ and the final hydrogen will never be lost. The answer to why lies in molecular structures.
You can approach the experimental result in different ways, too. For example, certain methods allow esterification, meaning the attachment of an organic group (e.g. ethyl, $\ce{C2H5}$) to the oxygens replacing the hydrogens there. If you perform esterification reactions with $\ce{H2SO4}$ under appropriate conditions, your final product is diethyl sulphate, $\ce{(C2H5O)2SO2}$ – two hydrogens were replaced, $n$ factor 2. If you perform it with phosphoric acid $\ce{H3PO4}$, your product is triethyl phosphate $\ce{(C2H5O)3PO}$. If you perform it with ‘phosphorous acid’, the product only contains two ethyl groups; the final hydrogen is not removable by this method.
Again, the real answer to why requires structures which you expressed you didn’t want. So all I am going to say is that ‘phosphorous acid’ should actually be termed phosphonic acid because that predicts the correct structure and reactivity.

https://pubchem.ncbi.nlm.nih.gov/compound/Phosphorous-acid#section=Names-and-Identifiers, I think IUPAC accepts phosphorous acid as its name.

This n-factor is taught is taught in Indian high schools (only, I guess). It is an outdated concept of equivalent weights, which was also taught in US/British universities until the 60s. n-factor not only refers to acidity or basicity, it is also used for redox reactions, and it totally depends on how the equation is written. Sadly, it has very little to do with molecular structures. For redox species, one has to account for the number of electrons gained or lost. For acids and bases it is indeed basicity or acidity, for salts it is based on the charge...complicated story of equivalents.

@M.Farooq It helps to look at the correct structure because what IUPAC wants to call phosphorous acid doesn’t actually exist in any meaningful concentrations

Yes, I agree. But knowing the structure of $KMnO_4$ does not help at all in calculating its equivalent weights and their n factors. The point is that I am not against knowing structures but the way equivalent weight is being taught to these students. Now you can imagine why teaching equivalents is not a good idea. It is good for chemical history classes and it has no room in modern science.

@M.Farooq I mean … my point here is … as far as I understand, OP wants to understand why phosphonic (‘phosphorous’) acid is dibasic. That is very easily explained with the corresponding molecular structures plus the knowledge of electronegativity values. It cannot really be explained without structures because there is nothing in the sum formula to explain the different hydrogens. But the redox chemistry of permanganate is an entirely different beast altogether, given that it can reduce to $\ce{Mn^2+, MnO2}$ or $\ce{MnO4^2-}$ depending on conditions.

(You will never get $\ce{Mn^{\pm 0}}$ using regular redox methods though). The key to finding the different possibilities of $\ce{MnO4-}$ is to look at the electronic properties like oxidation state and field split, not structure. So if OP had asked about $\ce{MnO4-}$ that would have prompted an entirely different answer, the core of which is ‘you can’t tell from the structure anyway’.


text:
Is it necessary to add prefixes of halogens to the IUPAC name alphabetically?
If I draw the structures of 2-bromo-4-chloropentane and
4-chloro-2-bromopentane, I get the same result, when I try to draw the structures. So why is it necessary to name prefixes alphabetically? Can we write IUPAC names not alphabetically?

I think it's just a discipline.. nothing much.

With the preferred IUPAC name (PIN), there is exactly one way that you expect to find a compound written.
Assume you want to buy the compound from the chemical supplier of your choice but don’t have a CAS number at hand. What do you do? You search for the compound in the catalogue. In the old days, these were always paper books that you could use as a weapon but nowadays electronic catalogues exist too. Still, if you’re looking for 1-bromo-3-chloropentane but it is entered into the catalogue as 3-chloro-1-bromopentane, you’re going to have a very hard time searching for it.
There are also other reasons why having one defined standard way of naming compounds is strongly preferred over random but clear.

As side note to the OP, for some weeks now, resources for learning chemistry includes a dedicated section «References about Nomenclature», too.

Anyway, it is always better to have only one name for a given substance. Chemists may realize that 2-bromo-4-chloropentane and 4-chloro-2-bromopentane are one and the same substance. But the clerk at the customs, the manager and the caretaker may not be able to do this reasoning.


text:
Force on an electron in a given orbital
I was wondering whether there is any quantum mechanical  equation to describe the force on a given electron in a given orbital of an atom (lets say, for the simplest case, for hydrogen atom). Also if it exists then what is the use of effective nuclear charge?

The force concept exists in the Bohr's model. However, it is more "natural" to talk of angular momentum (recall it is related to force) in the quantum world which have no classical analogs. It is better to keep the ideas of Bohr model separate from QM.

In QM model, there exist the potential in the Schroedinger  equation, that is in direct relation to the electrostatic force.

Your question doesn't make much sense. If you think that your electrons go in circles then get it out of your head expeditiously.

No I don't think of electron moving in a circle.

Also asked on Physics. Please don’t cross post like that.

Well, electron described by any orbital can be found in virtually any place in atom so why you ask about a "given" orbital?

An orbital is not an orbit. Also for hydrogen atom an electron in its lowest energy state resides in 1s orbital.


text:
Cis/trans with alkene substituents on cycloalkanes
My chemistry book is saying that the double bond indicated by the red arrow can generate cis/trans isomers:

I don't understand how this can be so, as you need to use E/Z priority rules in this case. If you go by the rule for cis/trans of finding the longest carbon chain containing the double bond, and seeing if the chain ends up on the opposite or same side of the bond, you find that by going around the ring, both options of cis or trans have the same length.
Am I missing something here, or is this a case where you really should be using E/Z, not cis/trans?

:/ Just put geometric instead of "cis-trans"; yeah it needs E/Z to be named, but it's same kind of isomerism!

This is a terminology problem. The bond clearly has the same kind of isomerism as a simpler system with a double bond but is complicated enough that the naming convention for the simpler system won't work anymore. I think you are right to argue for the E/Z convention here and you could have a productive debate with your teacher on the right way to describe the type of isomerism involved.


text:
Factors affecting hardness of water - EDTA titration
I am thinking of doing a project involving EDTA complexometric titration to determine the hardness of the water. I am not sure if the pH will affect the hardness of the water. If it did, I would control it by changing the buffer solution's pH. If buffer pH does not affect the hardness of the water, are there any other factors that I could potentially explore.
Thanks

related https://chemistry.stackexchange.com/questions/88183/determination-of-hardness-of-water-using-edta

https://chemistry.stackexchange.com/questions/44778/is-water-hardness-dependent-more-on-calcium-or-magnesium https://chemistry.stackexchange.com/questions/57720/how-do-i-determine-the-amount-of-calcium-in-a-substance-using-edta-titration


text:
Can aniline be made with the method described in Dr Stone?
In Dr Stone, the main character makes aniline by rinsing coal tar with HCl, then mixing in ethyl acetate. See the image below. Is a method like this possible? What is occurring here?

It's extraction, not synthesis; and seems reasonable.

Actually he's getting anilinium hydrochloride

@Waylander Sure, but extraction with acetate suggest not mentioned neutralization step.

Believe it or not, aniline was actually isolated from coal tar way back in 1834. A chemist called Friedlieb Runge isolated aniline along with phenol from coal tar. He discovered that aniline produced a beautiful blue color when treated with chloride of lime (calcium hypochlorite). He named it cyanol/kyanol and that was the birth of the first aniline dye but it was made a dye by the exhaustive work of another chemist, William Perkin.
Perkin distilled the components of coal tar like benzene, aniline, toluene, anthracene, phenol, naphthalene and other aromatic hydrocarbons. He first worked around toluene to make allyl toluidine which he then transform to quinine by oxidizing it with a mixture potassium dichromate and sulfuric acid. He was left with a useless reddish-black powder. He then took aniline and repeated the experiment. He was left with a black gooey substance. Disappointed by the results, he took the test tube and washed it off but then came out a beautiful looking purple residue which hung off at the wiping cloth. He named the dye "mauveine". Perkin was so amazed by the result that he started working with other distilled products hoping to make other dyes$\ce{^{[4]}}$ . His discovery revolutionized the dye industry. Perkin proposed the following statement:

Benzene, toluene, and other components of coal tar were colorless
because they absorbed ultraviolet light undetectable by the human eye.
But if those aromatic hydrocarbons were treated with an acid to create aniline or another amine, after some additional steps the
newly synthesized molecules very efficiently absorbed light particles
from specific wavelengths in the visible spectrum.

So, Senku (the main character of Dr. Stone) being well-versed in chemistry knows what he's talking about.
Notes and references

https://trc-leiden.nl/trc-needles/materials/dyes/aniline
https://www.scientificamerican.com/article/toms-river-excerpt-on-aniline-dye/
https://www.cavemanchemistry.com/cavebook/chtar.html
Magentas from toluene, reds from anthracene, pinks from phenol, and indigos from naphthalene.
Dr. Stone is an amazing anime. I recommend fellow readers to check it out.

My recollection is that Hofmann, Perkin's teacher, put him up to the task of synthesizing quinine based solely on formulas. I don't believe he every succeeded in his goal.


text:
Change in enthalpy equation
If the formula for change in enthalpy is ΔH=ΔU+pΔV, and ΔU = q - pΔV then shouldn't the formula be simplified to ΔH = q regardless of whether the volume or pressure is constant? Is this only true when pressure and volume is constant?

The general formula is ΔH=ΔU+Δ(pV)

Just wondering, would the change of enthalpy be equal to q when volume is constant?

If pressure is constant as well, then yes.

The general formula is $\Delta H=\Delta U + \Delta (pV)$.
Since $\Delta U=q_v$ (heat at constant volume), then $\Delta H=q_V+V\Delta p$, so no, heat at constant volume does not equal enthalpy, unless p is constant, which just brings you back to the original relation $\Delta H = q_p$.


text:
Conditions for the reduction of dichromate by manganese(II) to chromium(III)

The unbalanced redox reaction
$$\ce{Mn^2+ + Cr2O7^2- -> Mn^3+ + Cr^3+}$$
occurs in what type of condition?
(a) Acidic
(b) Basic
(c) Neutral
(d) Indeterminable

Guess that would happen in any conditions.

Or perhaps in no conditions at all :D Depends on perspective... Presence of dichromate should be a dead giveaway for what answer is expected, though.

So is the fact that they put $\ce{Mn^{3+}}$.  Would not expect most trivalent ions to remain in solution when it's a_i.

$$\ce{Mn^{2+} + Cr2O7^{2-} -> Mn^{3+} + Cr^{3+}}$$
If you were to balance this reaction, using the appropriate molecules of water and protons, you'd get:
$$\ce{ Mn^{2+} + Cr2O7^{2-} + 14H+ + 5e- -> Mn^{3+} + 2Cr^{3+} + 7H2O }$$
The presence of protons the left side of equation suggests that this reaction has a higher rate when the concentration of protons is higher. That is, it is favored under acidic conditions.

With all due respect please avoid answering homework questions (even a one that is somewhat interesting to some extent, as this one) because this would encourage OP to post more textbook questions without bothering to make an attempt towards solving it (and they did). Also, showing intermediate steps in obtaining final equation would make it a better answer as now for the untrained chemist it probably looks similar to the tutorial for drawing an owl:)


text:
How do these restrictions on the wavefunction logically arise?
My lecturer notes that the following restrictions on the wavefunction (in addition to it being a solution to the Schrodinger equation) are logical:
i) Continuous (mathematical definition of a continuous function)

ii) Single-valued (every x input has precisely one y output)

iii) Have a continuous derivative i.e. can define the gradient at every point
Why are these restrictions necessary?

Hi, if you add some of your own thoughts to why these requirements for the wavefunctions arise, people are more inclined to help you understand in detail why.

Well, the second one is not a restriction at all, since all functions are single-valued. Also, $\psi$ has to be continuous (and moreover, even differentiable) so as to be a solution of a differential equation. As for the third restriction, it is simply not there. For example, a particle in an infinite potential box violates it.

In case (ii) it means that wavefunctions cannot have a form such as $y=\sin^{-1}(x)$ which has repeated values for $|x|\le 1$. (i) means that an infinitesimal change about a point must not cause a big change in its value, such as happens in a step function.

The simple answer is that you will use this function to calculate eigenvalues/expectation values of operators, and you would have difficulty to define those for functions that cannot do I - III. Note, sometimes we use functions that are not smooth at every point (eg particle in a box).

@IvanNeretin the nonphysical infinite potential in that case makes the third condition impossible (or at least only gives the trivial solution), but it should hold under realistic physical conditions.

@Tyberius Well, if we stick to the realistic physical conditions, then pretty much all functions have continuous derivatives.


text:
Decomposition of Metal Nitrates
An entrance JEE Main & Advanced AIEEE 2003 exam question:

Which one of the following nitrates will leave behind a metal on strong heating?
A. Copper nitrate
B. Manganese nitrate
C. Silver nitrate
D. Ferric nitrate

I looked up the answer and it was given to be silver nitrate, and I took it up as something to be remembered. However, when I checked online for the decomposition of other options such as ferric nitrate I found many research papers discussing the same which is not something one asks students who have just studied chemistry up to 12th grade.
Also, is there a way to decide which one will, because if it says strong heating, shouldn't most metals leave their salts and exist as free metals if we go on increasing temperature?

Most metal nitrates decompose on heating to give metal oxides. Decomposition of silver nitrate is different as it yields elemental silver.
$$\ce{2AgNO3 -> 2Ag + O2 + 2NO2}$$
Usually the metals which lie very low in the activity series decompose to give their elemental form.
I don't think it's possible to extract metals from just any nitrate by strong heating. They will yield their respective oxide which won't give the element rather they'll just fuse (melt) just like stable metal oxides like $\ce{Al2O3}$ does.


text:
Distillation of methanol and water - Temperature Change
We tried the synthesis of methyl benzoate in the lab via the Fischer Esterification reaction with excess starting methanol. After 1 hour of refluxing the reactants we immediately set up a simple distillation procedure to remove the excess methanol and here lies my question. By closely monitoring the temperature of the thermometer the mixture started boiling at 70+ oC and continued to rise slowly as expected. At one point the temperature started dropping slowly and after asking the professor she said that the distillation was over. I fail to understand why the temperature dropped. Can someone provide insight as to why it dropped and furthermore what should the thermometer show from start to finish and why? To my knowledge, the solution before the distillation contains methanol, some benzoic acid, methyl benzoate, sulfuric acid and water. Thank you in advance.

The thermometer that records the temperature is in the air high above the solution (usually above a Vigreux column). Your distillation was under ambient conditions, so there was air in contact with everything.
The thermometer can only records a rising temperature if the temperature around it rises. During the distillation, this is due to methanol vapour (around $\pu{70 °C}$) which can also condense on the thermometer to transfer heat.
Once all methanol has been distilled away, there is nothing coming from below that could transfer any additional heat: water, sulphuric acid and benzoic acid all have far too high boiling points and low vapour pressures. Thus, the thermometer is again in contact with the surrounding air and cools down.
While I have described this with air, basically the same thing will happen in a vacuum distillation, too.


text:
Varying sizes in LUMO of acrolein
I want to ask a question about the varying sizes in the LUMO of acrolein during a 1,2- or 1,4-addition.
Today in our introductory lesson to Organic Synthesis with 1,2- and 1,4-addition we were listening to the reason why some nucleophiles may attack one functional group compared to another.
The following picture was discussed:

Now, as shown here, the far orbital on the alkene is far larger than the orbital on the $\ce{C}$ of the $\ce{C=O} $ bond. This implies that for a larger orbital, a nucleophile finds it is easier to attack the alkene than the carbonyl.
This makes sense.
However, when finding an explanation for these varying sizes in orbitals, I couldn't find anything to explain this.
This question was the closest which explained the effects of varying orbital sizes but did not differentiate between $\ce{C1}$ of the alkene and $\ce{C3}$ of the carbonyl.
Why are there two different orbital sizes?

Don't take the picture too literally. I believe the apparent size in fact represents the coefficient with which every AO contributes to the MO.

The coefficients of the LUMO on each atom are determined via mathematical methods. E.g. Applying a perturbation to the orbitals derived from Huckel theory. I believe there is no chemically intuitive way of deriving these coefficients on your own.


text:
Cross-conjugation
Could somebody please explain the concept of “cross-conjugation” (Organic Chemistry) to me? In what kind of conjugated systems is it observed, what are its effects on resonance of the system as a whole and so on? Also, I read this line in my book, possibly with respect to the same topic-

If any group has more than one π bond in conjugation, then only one π bond will take part in delocalisation.
Example:- In But-1-en-3-yne, only one π bond will take part in resonance.

Does this mean, that in every compound that shows cross-conjugation, only one, not all π bonds are actually involved in delocalisation? Or it it just that cross-conjugation nullifies the much-expected stabilising effect of resonance on the concerned organic compound? I am quite confused and my textbook is not really helping.

Cross conjugation is a phenomena due to which (in simple words) resonance from two groups in opposite direction cancel out. It is a destabilising effect.
It is something like this:

These are some examples which will help you understand why this is a destabilising effect:

In the above two examples the stability of carbanion decreases drastically as the two "resonances" cannot happen simultaneously (if it did, we would have 5 bonds with carbon and we don't want that).
Therefore, only one of the conjugated system out of the two present resonate. That implies the effective resonance of one conjugate system is decreased to half (where the molecule is symmetrical).
I hope this helps.

So, we can say that linear extended conjugation has more resonance energy than cross conjugation?


text:
Calibration curves and dilution factors
I am currently taking a course where we are exploring nitrate concentrations in water samples. We are using a UV-Vis Spectrophotometer and have created a calibration curve in the first part of the experiment.
Now in the second part of the experiment we need to measure the absorbance rates and use the calibration curve produced in the first part of the experiment to find the nitrate concentrations.
I expect that I need to multiply the concentration obtained from my calibration curve by the dilution factor used during the determination of the absorption rate, however, I would like to get this confirmed and understand better how this works.

Well, without more detailed information it's difficult to provide detailed assistance. Sounds like your intuition is right, though.

Show the calibration curve and your attempt to solve it. Homework questions get a better response when you share the exact problem.

Thank you both! I checked with my professor and I understood it and did it correctly.


text:
Why is the carboxyl proton in salicylic acid more acidic than the phenol proton?
I know I'm just wrong about this but I'd appreciate some help seeing why.  It seems to me like the phenol proton in salicylic acid should be more acidic than the carboxyl proton.  When I try to draw resonance structures for the conjugate base, it looks to me like the charge gets distributed over two oxygen atoms and the whole ring (if anyone can tell me what software makes the nice pictures in some answers here I'd take it):

Then when I try to draw the conjugate base when the carboxyl group is deprotonated, as far as I can tell only the oxygens take the charge, and trying to distribute over the ring results in invalid resonance structures because you aren't supposed to add formal charges to already charged species.

What am I doing wrong?

More resonance structures does not necessarily mean more stabilization.
Phenolate has a negative charge at an oxygen atom which cannot be shared without "disturbing" the aromaticity of the ring. This is in part because in the resonance structures drawn, there are negative charged carbons within the aromatic ring, which either implies that:
The ring is still aromatic, so there must be a free electron pair in a p orbital (very high in energy).
Or:
The free electron pair is in an sp³ orbital, which means the ring is no longer aromatic.
In the case of the carboxylic acid, the charge is distributed between two atoms more electronegative than carbon and the ring retains its aromaticity.

In this case, when the carboxyl group is deprotoneted and we draw the resonating structures the resonating structures are equivalent. But in case of phenolic deprotonation the resonating structures are not equivalent. So the carboxyl anion will be more stable, although it have less number of resonating structures than phenolic anion. So the carboxyl H will be more acidic.


text:
Spectra from Dipole Moments
If there is a change in dipole moments such as absorption spectroscopy,  there are infrared absorptions (and/or transmittance) corresponding to them. Is the reason change in dipole moments are in infrared frequency simply because of the size of the molecules or scale of wavelength?   Or are the dipole moments producing particular frequencies simply because of their sizes just like in antenna? If not, why are they mostly in IR?

(Note: image from https://en.wikipedia.org/wiki/Electromagnetic_spectrum)

possible duplicate: https://chemistry.stackexchange.com/questions/73565/why-are-some-molecules-unable-to-absorb-infrared-radiation?rq=1

The energy of IR radiation matches the energy of the transitions between vibrational states. So it is a match in energy, not in length.

Could you edit your question to make sure it is clear what you are asking. If you look at the answers and the comments, folks are guessing what you want to ask.

Karsten. You answered it directly, I thought the energy of IR radiation is a match in length. You rightly answered it is a match in energy.

The transition energy, or wavelength if you prefer, are governed by the nature of the molecules involved. Whole body free rotational motion (in the gas phase) gives rise to microwave spectra ( fractions of wavenumbers cm$^{-1}$  or THz frequencies). The frequency is determined by the molecules moment of inertia. At higher energy comes vibrational motion, stretching and bending of chemical bonds, a few hundred to a few thousand cm$^{-1}$. These bond frequencies are approximately given by $\sqrt{k/\mu}$ where $k$ is force constant which relates force to bond extension as in Hook's law, and $\mu$ is the reduced mass. There are 3N-6 types of vibration for N atoms, (3N-5 for linear molecules). At higher frequencies are electronic transitions where and electron gets promoted from a bonding to anti-bonding orbital. Typically these transitions are in the visible and uv regions.
The exact number of transitions is difficult to estimate because each vibrational level has rotational levels superimposed on them and there are also vibrations that are combinations of levels. There are, however, fewer electronic transitions because there are generally few molecular orbitals than types of vibrational modes, and each excited state has its own set of vibrational and rotational levels.
One of a chemist's essential tools, arguably more important that any other spectroscopy, is NMR spectroscopy which is caused by transitions between nuclear spin states when in a magnetic field and occurs at very low frequency, 100-800 MHz range.

I think your own answer here: [https://chemistry.stackexchange.com/questions/73565/why-are-some-molecules-unable-to-absorb-infrared-radiation?rq=1 is much closer to what the OP is attempting to figure out...

Thanks, I was not sure what the OP knew about this so here made general answer.

Your comment was very helpful, porphyrin. I was asking initialy if the IR is due to the match in length. It is not, but a match in energy. But it is true the length of the atom matches the length of x-ray? But how come atom still emit x-ray?

As you can see in the comments, it is the energy between levels that matters. An atom can, under appropriate conditions, emit/absorb photons in the infra red, visible, uv and xray regions. Similarly an atom and a big molecule, such as C60, can have almost identical absorption and emission wavelengths.


text:
Electrophilic base / nucleophilic acid / basic electrophile / acidic nucleophile
What I was understanding is that all bases are nucleophiles and acids are electrophiles means the condition for them is same and all nucleophiles are basic and all bases are nucleophiles and same for acid and electrophiles but my sir said that it isn't completely true . Can someone give me an example of electrophilic base / nucleophilic acid / acidic nucleophile / basic electrophile

https://chemistry.stackexchange.com/questions/20061/whats-the-difference-between-a-nucleophile-and-a-base

All these aren't inherent properties of compounds, but are relative  and depend on situation. Some molecule may act as nucleophile/base/... While it's a common saying that for example H2SO4 is an acid, it's rather that it usually acts as one, but it can act also as base. If one says it is an acid it's rather because of  human need for categorization then inherent property.

https://chemistry.stackexchange.com/questions/20061/whats-the-difference-between-a-nucleophile-and-a-base

$\ce{HSO4-}$ is acidic ($\mathrm pK_\mathrm a = 1.99$) yet also reacts as a nucleophile e.g. in the acid-catalysed hydration of olefins, which according to my mechanism lecture procedes via a sulphate diester intermediate.
If I thought for long enough, I probably could come up with a basic compound that acts as an electrophile in a given situation.
However, having single counterexamples to a general trend isn’t necessarily helpful; it is better to understand the differences in terminology. For that, I would like to direct you to Ben’s answer in the linked question, that clearly highlights the difference between nucleophilicity (a kinetic phenomenon) and basicity (a thermodynamic phenomenon).


text:
Isolating barium(II), copper(II) and zinc(II) from aqueous solution

Q15 Each metallic ion was separated from an aqueous solution containing $\ce{Ba^2+},$ $\ce{Cu^2+},$ and $\ce{Zn^2+}$ by the procedure shown in the following figure. From ①–⑥ in the table below choose the correct combination of ions that predominates in the precipitates a, b and the filtrate c.

For the precipitate a, I know the answer is $\ce{Ba^2+}$ ion because it has a highly negative standard reduction potential than the other ions, so it tends to displace the sulfuric acid to form insoluble barium sulfate.
But I don't know which one can displace $\ce{NaOH}.$ $\ce{Na+}$ ion reduction potential is a lot more negative than both $\ce{Cu^2+}$ and $\ce{Zn^2+}$ ions. I suspect there is something in 'added in excess' part, but I don't know how can it be any different.

Reduction potential is quite irrelevant here. You might want to review what do you know about the amphoteric properties of Zn.

$\ce{NaOH}$ is added in excess to first neutralize the sulfuric acid from the first step and then to make the pH go up beyond 7.

@AegisNine I suggest you look up the properties of Zn(OH)4 dianion

So, this related to properties of transition metal, and not reduction potential, but Ba is not d-block metal, how can I compare it? This is an undergrad admission exam, so is it really necessary to know the specific properties of a compound?

The answer is a=Ba, b=Cu, c=Zn from answer key, but from searching a lot it looks like both Cu and Zn can reach with NaOH, Zn with its amphoteric properties, Cu by firstly forming oxides(?)

Of course both Cu and Zn will react with NaOH. But they will react in different ways.

Once $\ce{Ba^2+}$ is eliminated as poorly soluble barium(II) sulfate, there is a solution of soluble copper(II) and zinc(II) sulfates.
Considering this has been done by careful/gradual addition of sulfuric acid, and the fact that both remaining sulfates undergo hydrolysis, the solution is going to be slightly acidic — this is one reason for sodium hydroxide "added in excess".
Another reason, as both Ivan and Waylander pointed out in the comments, is the difference between reactivity of copper(II) and zinc(II) in alkali medium.
Long story short, you are probably supposed to know the specific properties of the common transition metals cations as the reduction potential isn't really helpful here.
Copper(II)
Copper(II) with $\ce{NaOH}$ first forms colloidal copper(II) hydroxide; the gel is then solidifies to light-blue amorphous substance upon subsequent addition of hydroxide:
$$\ce{CuSO4(aq) + 2 NaOH(aq) -> Cu(OH)2(s) + Na2SO4(aq)}$$
$\ce{Cu(OH)2(s)}$ appears to be precipitate b.
Sodium tetrahydroxocuprate(II) $\ce{Na2[Cu(OH)4]}$ doesn't form here: it's a very moisture-sensitive and in the air quickly turns dark-brown:
$$\ce{Na2[Cu(OH)4](aq) -> CuO(s) + H2O(l) + 2 Na(OH)(aq)}$$
Zinc(II)
Zinc(II), on the other hand, with $\ce{NaOH}$ first forms white precipitate of zinc(II) hydroxide:
$$\ce{\ce{ZnSO4(aq) + 2 NaOH(aq) -> Zn(OH)2(s) + Na2SO4(aq)}}$$
which is then dissolved again forming colorless soluble sodium terahydroxozincate(II) (filtrate c):
$$\ce{Zn(OH)2(s) + 2 NaOH(aq) -> Na2[Zn(OH)4](aq)}$$
Excess of $\ce{NaOH}$ is also necessary to keep $\ce{Na2[Zn(OH)4]}$ in solution: upon diluting (lowering pH, adding water/acid), terahydroxozincate(II) decomposes to zinc(II) hydroxide.
Note: chemical reactions are adapted from [1].
References

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


text:
Is black tea a pH indicator?
Today, I made a cup of lemon tea in a different way than usual. Instead of pouring hot water in to a mug containing lemon juice and black tea (my normal routine), I let the tea steep for a little while before squeezing the lemon. Before adding the lemon, I noticed that the tea was much darker than I am used to. After squeezing the lemon, the tea changed color to a lighter brown and became more translucent.
I could see only a half inch down the side of the cup before squeezing the lemon, but afterwards I could see three inches down, all the way to the bottom. I did not notice any precipitated solids on the bottom of the cup. What happened?

Something in black tea changes its colour from brownish to less brownish at low pH. An indicator, if you want.

The colour of black tea is given by a complex mixture of theaflavins and thearubigins.
These compounds are formed by enzymatic oxidation and condensation of polyphenols of green tea, typically of gallocatechin and epigallocatechin gallate ( EGCG )
Basic dissociated forms of these compounds have strong absorption of visible light due to delocalization of phenolate electrons. This absorption  decreases a lot when phenolates are protonized in acidic environment.
So yes, in some sense, black tea infusion is a $\mathrm{pH}$ indicator, similar to litmus or juice from red beet roots. Many natural colourful compounds have $\mathrm{pH}$ dependent colour, as their molecules undergo acid-basic reactions.
But synthetic indicators are better due stability, sharper and better defined colour transition.


text:
How is energy not stored in bonds?
Consider the paraphrase from Ebbing:

It takes energy to break bonds & energy is released when bonds are formed

Let's look at the first statement  It takes energy to break bonds.

Is this true of spontaneous processes?
If yes, is this true only in a closed-environment sense, where the energy required for a reaction to occur between two molecules is the energy contained within those molecules; i.e. no outside energy is necessary?
If yes to 2), is this energy, outside the kinetic energy of the molecules themselves, potential energy (PE) stored in the bonds?

Then for the second statement, energy is released when bonds are formed.
When the PE of the reactants is greater than the PE of the products, heat is released and we have an exothermic process. Consider the conversion of ethane to ethene. This is an endothermic reaction requiring approximately 220 kJ/mol of energy (bonds broken (2X C-H) - bonds formed (C=C).

As the PE of ethene is greater than that of ethane, why is this not reflected when considering the sum of bond energies in each molecule? For example, for ethane we have 2,812 kJ/mol and for ethene we have 2,246 kJ/mol.
Given the bond energy of C=C (602 kJ/mol) is higher than C-C (346 kJ/mol), why would it be incorrect to assume the energy necessary to drive the reaction is stored in the C=C bond, given the PE of ethene is greater than that of ethane?

Lastly, when teaching cellular respiration, we speak of energy being stored in the phosphate bonds when ADP is converted to ATP and in photosynthesis, of the energy from sunlight being stored in the bonds of glucose.

Is this line of thinking correct, if not why?
If that line of thinking is correct, how do we reconcile this fact with "Energy is released when bonds are formed?
Would it be appropriate to say energy is both stored and released as bonds are formed?

I am a tutor hoping to master this content to best help my students. Please lend a hand to whichever questions you feel most confident in answering.

The planetary model is of course incorrect, but think of the electrons "dropping" into lower orbitals. Being closer to the nucleus the electron "picks up" energy. However the electron can't have that energy in the closer orbital, so the electron has to shed the energy somehow.

In the ethane/ethene case don't forget hydrogen.

Let's start with the specific examples. For the conversion of ethene to ethane, you need a reaction partner to provide the extra hydrogen, e.g. dihydrogen:
$$\ce{H2C=CH2 + H2 -> H3C-CH3}$$
In the reaction, a single bond and the pi-bond of the double bond are broken, and two single bonds are formed. If you look up the four bond dissociation energies (two broken and two formed), you get a fairly good estimate of the reaction enthalpy.
$$\ce{ADP + phosphate -> ATP + H2O}$$
I'm leaving out the charges and the protonation state of phosphate to make things simpler. If we look a bit closer, we can write:
$$\ce{RO3P-OH + H-OPO3R' -> RO3P-OPO3R' + H-OH}$$
So when ATP is formed, we are not only making the phospho-anhydride bond (and a bond within water), but also breaking two bonds.

Lastly, when teaching cellular respiration, we speak of energy being stored in the phosphate bonds when ADP is converted to ATP and in photosynthesis, of the energy from sunlight being stored in the bonds of glucose.

Is this line of thinking correct, if not why?

You could say the energy is stored in the reaction, and when the reaction runs in the other direction (hydrolysis), it gets released again. Similarly, glucose is a source of free energy only if oxygen is available as reaction partner. Breaking glucose into atoms is not a source of free energy.

Let's look at the first statement It takes energy to break bonds.

Is this true of spontaneous processes?

Here is the reaction this refers to:
$$\ce{molecule -> free atoms}$$
The entropy of reaction is expected to be positive (more particles on product side). If the reaction were also exothermic, these molecules would not be stable under any condition (Gibbs energy of reaction would be negative at any temperature). So bonds of molecules that actually exist are expected to have a positive bond dissociation energy.
The other questions
The answers to the rest of the questions follow from the information above.

I like your use of logic in the last paragraph, even though I've yet to encounter a bond with a negative bond dissociation energy :-)

On breaking a bond: dissociation will occur if the energy to overcome the attraction that is holding the atoms together is somehow made available to the bond, such that it can be excited into a higher energy state leading to dissociation. "Spontaneity" would imply access to the excess energy that will drive the process forward. No energy, no dissociation. The energy can come from laser photons with just the right energy to excite the appropriate vibrational mode and trigger dissociation, or, just as well, the energy might be provided by a heat bath that results in energy being transferred through collisions into the molecule and eventually into the vibrational mode that results in dissociation. If you had only two molecules, dissociation would require inter- or intramolecular transfer of energy into the cleaved bond. The molecules would evidently have to be excited, loaded with excess energy that when transferred would raise the vibrational state of the cleaved bond well above the ground state. Ultimately the energy would be stored as potential energy of the now dissociated bond plus kinetic energy from the excess above the required dissociation energy. The heat bath would provide kinetic energy to the products.
Molecular process are reversible, which means bond formation could proceed through precisely the same path, but in reverse. Atoms would come together to form a highly vibrationally excited product that could shed that energy by distributing it to all available degrees of freedom, including eventually into surrounding molecules.
Regarding whether energy can be stored in bonds: yes, it can, in the same way a ball may come to rest in a higher energy dip. However some of these questions might be better dealt with using a thermodynamic treatment that addresses the free energy of the substances in specific states. These questions are answered in other posts.


text:
Hyperconjugation in bridged carbon systems
While studying about hyperconjugation, I came across the following compound which posseses one α-hydrogen (with respect to the $\mathrm{sp^2}$ hybridised carbon atom with an empty $\mathrm{p}$-orbital), however, apparently does not show a hyperconjugation effect.

This led me to draw an ‘assumption’ that an alpha hydrogen on a bridged carbon atom may not get involved in hyperconjugation when placed next to a carbocation, alkene, free radical, or an arene. Is this true? If so, then why?

Careful with these kind of compounds; they usually tend to form nonclassical cations.

@Martin-マーチン What do you mean by “non-classical”?

See What is a non-classical carbocation?

Hyperconjugation is indeed very difficult to practically in cases involving bridgehead protons. The underlying reason is essentially the same that also explains why these protons cannot be enolised if there is a carbonyl group adjacent to them. This is best explained by drawing out the compound in a three-dimensional manner, which I have done for you in the figure below.

Figure 1: approximate three-dimensional representation of the cation in question; cation represented by the p orbital.
If you look at figure 1 closely, you can see that the empty p orbital and the bridgehead hydrogen σ-bond are almost at a $90^\circ$ angle towards each other. For hyperconjugation, you want approximately parallel arrangements, so this obviously fails.
However, you should always be aware that this type of compounds, substituted norbonyl cations, are derived from theprime example species for non-classical cations. In this case, instead of a hyperconjugation involving the $\ce{C-H}$ bond one essentially expects the $\ce{C-C}$ bond of the ‘back half’ of the six-membered ring to provide that hyperconjugation. For a more accurate and detailed explanation, please consult the Wikipedia page or the textbook of your choice.


text:
Size of target substance in IR and Raman Spectroscopy
Usually what is the size (in nanometer) of the sample or target substance in IR and Raman Spectroscopy?
Which one has largest size of sample it can scan, how many molecules or atoms are involved? If you can make the size bigger or scan coverage bigger, then the more intense would be the IR or Raman stroke signal? Or no difference in intensity in the case of Raman even if sample size is increased? Please answer separately for IR and Raman since they are different. One uses IR light, the other laser. I guess IR light needs to be focus at a spot just like a laser too.

Why do you call it Raman "stroke"?

Because most raman spectrometers only measure the raman stroke.. and not the antistrokes.

The correct term is Stokes and Anti-Stokes Scattering (named after a person)

This all depends on the exact geometry of your spectrometer light path, if you measure in transmission or reflection, using ATR, etc. AND of course on your sample concentration.
If you want to ask about the minimal size: As small as you can get your focal point or laser spot. A few dozen microns, maybe less. Local sample heating can be a problem. You can buy scanning IR and FT Raman microscopes.
In transmission, you can measure IR on a cm sized sample, given that is is translucent. With Raman, that would be tricky, because you should focus the scattered light again.
Minimum number of molecules for IR is very low, but that also depends on how strong the absorption line in question is. Raman used to be very insensitive, that has become a lot better with FT Raman.

For raman, is there also measuring in "transmission" or "reflection"? I know there is only stroke shift of the laser. Where is the transmission or reflection in that?

Raman is a scattering effect. You cannot observe the scattering in the beam direction, because it is completely drowned in your laser light.

Isn't it notch filter filtered these rayleigh scattered light so it can be done in the beam direction?

That would make no sense, the scattered light goes basically in all directions (there is an angle dependance i remember), and you want to collect as much of it as possible.

Im focusing on raman (pun unintended). I know raman effect comes from electron polarizability of molecules. If your target sample is bigger (whats the size usually?)  Can it increase the raman intensity (stroke shift, as most raman devices dont measure the anti stroke shifts) or no changes?

I think not. You would have to enlarge every optical part of the instrument. The costs will increase overproportionally, the SNR much less.

So for an  average Czerny-Turner Raman spectrometers. What is the size of the target spot. How many nanometers across (the biggest possible so largest number of sample possible)? And when mixing substance. Can diffusion affect such sizes such as  adding sugar or alochol to water, etc?

Why "nanometers"? We are talking about an optical system here! You should write a new question, in which you explain what this is all about. Mixing substance? Diffusion? This has nothing to do with the original question.

My original question was how many nanometers (biggest possible) is the sample scanned by say an ordinary raman spectrometer. Or is it smaller than a nanometer? Remember my original question was "
Usually what is the size (in nanometer) of the sample or target substance in IR and Raman Spectroscopy?". You haven't answered in the case of raman spectroscopy. I want to know the biggest size for typical ordinary raman spectrometers.

@Jtl Do you know how many orders of magnitude there are between one nanometer and the wavelenght of your Raman laser source, and then how small you can (and should) usually focus a rather powerful coherent light source?

IR instruments often work with beam diameter at sample (for macrochamber in
transmission) of several mm to a cm or so.
The other extreme for "normal" IR instruments is diffraction limited, so in the order of magnitude of 10 μm.
There are FTIR instruments with Focal Plane Array (or Line) detectors that simultaneously collect spectra from many such pixels (e.g. 128 x 128), and optics are available to adjust pixel size to your needs (basically again between several mm FOV and diffraction limited pixel size)
If you need higher lateral resolution, you can e.g. go for near-field/nano-IR or photothermal spectroscopy.

For Raman, the largest spot sizes/measuerment volumes are AFAIK achieved by fiber optic probes. Here, the Kaiser PhAT probe is the one with largest laser diameter I'm aware of (I haven't personally worked with one, though). Other fiber optic probes e.g. this one have lateral diameter of the measurement volume in the range of a few 100 μm. The large spots come at the cost of low NA, and thus low collection efficiency, but this can be partially compensated as the larger measurement volume allows higher total laser power without burning the sample.
Microscopic setups have allow high NA and thus also better collection efficiency, but the sampled volume is correspondingly smaller (lateral diameter somewhere between diffraction-limited and a few μm).
Thus, for Raman you usually have severe undersampling. Several strategies are used to overcome this, such as continuously moving the sample or doing multiple short acquisitions at different positions. Ideally, you should use knowledge about the heterogeneity of your sample to determine a suitable strategy (how many spots, how far apart, excitation power & time per spot)

For gases, both Raman and IR spectroscopy may be done involving very long interaction path lengths (can be several 100 m).


text:
Explanation for why nickel turns green in hydrochloric acid
My daughter has a chemistry test that I am helping her prepare for. One question is

Nickel sulfate, $\ce{NiSO4(aq)}$ is a green solution. Nickel chloride, $\ce{NiCl2(aq)},$ is a yellow solution. If you add nickel, $\ce{Ni(s)},$ to hydrochloric acid, $\ce{HCl(aq)},$ what color solution do you expect?

Wikipedia states that it will turn green but doesn't give an explanation for why. Can anyone fill in the blanks for us? (This is for high school intro to chem so try to keep it at that level.)

Could you add a link to where it says that the solution will turn green? Just for the record. And here's a related question.

I think the question is trying to get the student to think about what salts come from what acids. Since HCl does not produce sulfates, one can rule out the green option based on the information provided by the question.

High school science teacher here, so definitely biased, but I'd contest the notion that the question is badly worded. Notice that it asks what outcome you expect, not what the outcome will be. The latter sets the the student up for rote learning, while the former sets them up for deductive learning, a central concept in science. If the outcome is not as you expected, then that in in itself is interesting, and something we can build on further: why is that?

@AkselA Agreed, the question gives you information to get an expectation - all the teacher is asking is that you know that nickel will form salt with chlorine.

@TomášZato actually, all the teacher is asking is that you know that sulfur and oxygen cannot be spontaneously created, and that what comes out of a reaction can only contain elements which are supplied as input ;)

The question is really badly worded. For starters, let’s look at solutions of nickel(II):

Figure 1: Nickel(II) solutions. From left to right: $\ce{[Ni(NH3)6]^2+}$, $\ce{[Ni(en)3]^2+}$, $\ce{[NiCl4]^2-}$, $\ce{[Ni(H2O)6]^2+}$. Image taken from Wikipedia, where a full list of authors is available.
You can ignore the left two but the rightmost is a standard nickel(II) solution (could be $\ce{NiSO4}$) while the second from the right is a chlorido-complex—basically $\ce{NiCl2}$ but dissolved in hydrochloric acid. I wouldn’t call the chloride solution ‘yellow’, more like ‘yellowish-green’ but your mileage may vary. Next, let’s look at solids.

Figure 2: salts of nickel(II) chloride. Left or top: hexahydrate. Right or bottom: anhydrous. Images taken from Wikipedia (hexahydrate, anhydrous), where a full list of authors is available.
The picture on the right is dry, i.e. completely water-free nickel(II) chloride (anhydrous). On the left, the salt contains tightly bound water molecules which can be removed by further drying although the salt itself would be dry in everyday terms; this is the hexahydrate of nickel(II) chloride.
It might seem easy that the salt on the left gives a greenish solution while the salt on the right gives a yellowish one. Here’s the catch: they should both give a greenish solution, because there is no excess of chloride as was required in the solution in figure 1.
But what about the actual question? Well, dissolving nickel in hydrochloric acid means that there should in fact be excess chloride, meaning that yellow (or yellowish green) would be the correct answer.
tl;dr: That question is a perfect opportunity to complain if it was marked as ‘incorrect’.

Anhydrous nickel chloride is yellow. However most simple divalent salts of nickel are green. This should be good enough for high school. The question is poorly worded because it is ambiguous (hopefully the textbook will improve it). If they discuss metal complexes it can be mentioned that $\ce{[Ni(H2O)6]^2+}$ is green in color.

Jan already gave a great answer explaining the real chemistry behind this, which you and your daughter should absolutely read. This answer also makes the point that the question is poorly worded and could be contested on that basis—that is absolutely true, but unfortunately rather common in high school chemistry courses.
In high school chemistry, one of the things they're trying to drill into young minds is how reactions proceed and which products will form. That's what they're trying to test here: can the student draw a conclusion about which product(s) will form when given the reactants?
Here, the reactants are given as excess aqueous hydrochloric acid and free nickel. Formulaically, that is the following reaction:
$$\ce{HCl(aq) + Ni(s) \rightarrow ?}$$
The student is supposed to be able to figure out that this is a single replacement/substitution reaction, a characteristic reaction of acids with metals, in which the ionic compound nickel (II) chloride will form while hydrogen gas is liberated:
$$\ce{2HCl(aq) + Ni(s) \rightarrow NiCl2 + H2(g)}$$
Once you figure this out, the question is designed to give you all of the information you need to answer the question insofar as it tells you what the colors of two different nickel salts are:

Nickel sulfate, $\ce{NiSO4(aq)}$ is a green solution. Nickel chloride, $\ce{NiCl2(aq)},$ is a yellow solution.

Essentially, you're supposed to decide which of these two salt solutions will be formed by the described reaction. Since the described reaction will yield nickel chloride (not nickel sulfate), the question is expecting you to answer that the solution will be yellow in color.
The question does not expect you to know, be able to figure out, or go research what color the solution will actually be if you do the reaction in the laboratory. Overthinking it may cause you to arrive at a different answer than was expected, which will lead you to getting the question wrong. You can absolutely contest this, but at least now you understand what the question wanted you to say and how they expected you to arrive at that conclusion.

The other answers provide valuable insight into what actually happens and are very helpful in understanding the chemistry behind it.
But to correctly answer the question, a much more basic understanding of chemistry (and how tests work) would suffice.
Let's look at what is given in the Question:

Nickel sulfate, NiSO4(aq) is a green solution. Nickel chloride, NiCl2(aq), is a yellow solution. If you add nickel, Ni(s), to hydrochloric acid, HCl(aq), what color solution do you expect?

So what the teacher gave us is:

two solutions with their names, colors and formulas:

Nickel sulfate, NiSO4(aq), green
Nickel chloride, NiCl2(aq), yellow

two chemicals to be mixed, with their formulas:

Nickel, Ni(s)
hydrochloric acid, HCl(aq)

Now, if we can assume that the student in question knows the notations, she can figure out that the elements going in are Ni, H and Cl. Let's look at the options we were given for output:

One of the two possible output solutions contains Ni and Cl - these are going in, so this solution would be possible to produce.
The other one contains Ni, S and O - Ni is going in, but where would the S and O come from? (Technically the O could come from the air I guess, but even allowing for that the S would still be unaccounted for)

So with basic knowledge of chemistry and the notation used, basic logic can give us the answer. There's no way to produce Nickel sulfate because there's no sulfur or oxygen going in, which means we should expect to get the other one. Which is colored yellow, making "yellow" the answer.

This question does look ambiguous. The question states that Nickel is added to aqueous Hydrochloric acid. The word aqueous might suggest that the complex that the present in the solution is [Ni(H2O)6]2+ and not [NiCl4]2-   Now the thing to be noted is, NiSO4(aq)has the characteristic green color due to the formation of the complex cation [Ni(H2O)6]2+ which is green in color.   And also as mentioned by some of the people who answered above, Nickel chloride in anhydrous form is yellow in color whereas, its hexahydrate form has a distinct green color.
  This is the image of Nickel sulfate in anhydrous and hexahydrate form. As you can see here, the color is due to the H2O ligand forming a complex with Nickel.