text:
What happens to boiled (BMIM)PF6?
[BMIM]PF6, or 1-butyl-3-methylimidazolium hexafluorophosphate, is commonly used as an ionic liquid. As has been shown by the accepted answer to this question, sodium chloride that has been made to boil actually exists either as a monomer (Molecule of NaCl), or a dimer, rather than ions, in the gas phase. What happens to 1-butyl-3-methylimidazolium hexafluorophosphate, or any other ionic liquid, that has been boiled? Would it, like NaCl, also exist as a monomer? What would happen to an ionic liquid that has been placed into a vacuum chamber from which air had been removed?

Ionic liquids have very special properties, the most important of those of interest is that they are liquids at low temperature with extremely low vapour pressure. This is in fact the main reason they have caused excitement among chemists. Quoting (1) from a report comparing the volatility of various ILs:

An extremely low vapor pressure (e.g., ca. 100 pPa at 298 K for [C4mim][PF6] 1 compared with 3 kPa at 298 K for H2O 2) is one of the extraordinary properties of room temperature ionic liquids (RTILs), i.e., molten salts with melting points below 100◦C.  As a consequence, RTILs such as [C4mim][PF6] at 298 K are liquids which do not evaporate significantly even under ultrahigh vacuum (UHV) conditions (i.e., for a pressure range 100 nPa...100 pPa [3]), which offers the possibility to use RTILs, e.g., as substitutes for volatile organic solvents [4, 5].

When an IL vaporizes it does so primarily as a neutral species, just like in the case of sodium chloride in the linked post, because charge separation (separation of oppositely charged ions) consumes a lot of energy.
One challenge in studying the vapor of ILs is that they have very high boiling points (typically >500K, see Ref 2) and degrade at higher T. By measuring the vapor pressure over broad T ranges it has been possible however to estimate the normal boiling point of some ILs by extrapolation, as explained in Ref. (3):

Moreover, due to the extremely broad temperature range, the estimation of the boiling temperature of [EMIm][NTf2] from thefit of the Clarke–Glew-equation can be considered as the first reliable boiling point of the archetypical ionic liquid obtained from the experimental vapor pressures measured in the most possible close proximity to the normal boiling temperature.

References

http://arxiv.org/abs/1006.2090v1
F. Mozaffari, Journal of Molecular Liquids 209 (2015) 657–661
Ahrenberg et al., Phys. Chem. Chem. Phys., 2016,18, 21381-21390

Very interesting, I hadn't seen a measured vapor pressure of an ionic liquid before. Now if only I can find the vapor pressure of a polymer melt...

@nicolau it does not have to be ionic. Check out what happens when we try to render the triple-point pressure of gallium.

Ionic liquids are not magical liquids. They are just typical organic compounds with certain properties. These compounds generated a storm in the tea cup in the late 1990s and 2000s, but they are as toxic as other solvents. Certainly they have niche applications especially as a gas chromatography stationary phase, where they can be used at high temperature (up to 350 centigrade!).
Now you can guess based on the GC application, that ionic liquids cannot be distilled or vaporized. They start to decompose by complex path ways above these temperatures. We can easily see that in a gas chromatograph. Surprisingly, not much has been studied as what happens to ILs at high temperatures.
Only one recent study exists Dicationic ionic liquid thermal decomposition pathways. There are complex decomposition products in the absence of oxygen. Like any organic compound, they will burn in the presence of air. NaCl is another story. It is not an organic compound. In short do not compare NaCl with ionic liquids.


text:
Determining Oxidation Number
In NH3, what's the oxidation number of H? Should it be +3 or +1? I'm a little confused what to use.

What was your reason for ruling out all numbers except +1 and +3?

The oxidation number of hydrogen is mostly $1$. Unless it is in elemental state, i.e. as $\ce{H2}$, here the oxidation number of $\ce{N}$ is $-3$ and hydrogen is $+1$.
Hydrogen can also take up $-1$ depending on the atom it is attached to. What you really need to lookout for is the electronegativity of the atom attached to it.

OS of H isn't always +1, it is only in case when H is linked with a non-metal. NaH, MgH2 are some examples of -1 OS of H.

source: Stack Exchange

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

sort: votes


text:
Sodalime method for multiple functional groups

what exactly happens? I get a little confused when it comes to multi functional groups. can some one explain soda lime as simple as they can.
I apologize if its a really simple question but I'm a high school student.
some good weblinks would be very helpful.

Hint : ketones don't undergo hydrolysis, but esters go. What is the product?

so, from the hint ive concluded that ester will undergo nucleophilic substitution to give carboxylic acid which in turn will undergo soda lime, and become  alkane. is that it (●'◡'●)

Soda lime is usually used to induce decarboxylation https://www.chemguide.co.uk/organicprops/acids/decarbox.html

Almost, as I couldn't call it alkane cuz it has keto group :)

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text:
Approximate Hamiltonian in variational method (Example of Helium atom)
As one know, the exact Helium Hamiltonian
\begin{equation}
        \hat H =
        \left( -\frac{\hbar}{2m_e}\Delta_1 - \frac{Ze^2}{r_1} \right)
        +
        \left( -\frac{\hbar}{2m_e}\Delta_2 - \frac{Ze^2}{r_2} \right) +
        \frac{e^2}{r_{12}}
    \end{equation}
Now, we can use variational method for determining approximate value of energy of ground state with $\Phi = \frac{\zeta^3}{\pi}e^{-\zeta(r_1 + r_2)}$ as
\begin{equation}
        E_{0_{\min}} = -\zeta_{\min}^2 = -\left(Z - 5/16\right)^2 = -\left(Z - \sigma\right)^2.
    \end{equation}
Can we say that we actually found the Hamiltonian for a given function $\Phi(r_1,r_2;\zeta_{\min})$ in form
\begin{equation}
        \hat H_0 =
        \left( -\frac{\hbar}{2m_e}\Delta_1 - \frac{(Z-\sigma)e^2}{r_1} \right)
        +
        \left( -\frac{\hbar}{2m_e}\Delta_2 - \frac{(Z-\sigma)e^2}{r_2} \right)
    \end{equation}
? I.e. is the such $\Phi(r_1,r_2;\zeta_{\min})$ eigenfunction of $\hat{H}_0$?


text:
An equilibrium question involving mixing equal volumes of HCl and H2SO4
Question: Calculate the concentrations of H3O(+) and SO4(2-) in a solution prepared by mixing equal volumes of 0.2M HCl and 0.6M H2SO4. (Ka1 for H2SO4 is very large, and Ka2 for H2SO4 is 1.2E-2)
Here is my thought process:

Since the Ka1 for H2SO4 is very large, it will completely dissociate into HSO4- and H3O+. I then used Ka2 to find the equilibrium concentration of HSO4-, H3O+, and SO4(2-) before mixing it with HCl.

When the solutions of equal volumes were mixed, the concentration of each species in both solutions were halved. The concentration of H3O+ is now the sum of the halved H3O+ concentrations from each solution. Since the concentrations have changed, I calculated Qc from the new concentrations and determined that the dissociation equation proceeds to the right.

Lastly, I used Ka2 again to calculate the equilibrium concentrations in the mixed solution.

The final answer I got was [H3O+]=0.40856M and [SO4(2-)]=0.00856M.
I do not have the answer key for this question, so I would greatly appreciate it if you could help verify my thought process/answer. Thank you!

calc of part 1 is irrelevant. Please show calculation, or st least step up for part 3.

Mixing equal volumes of 0.2M $\ce{HCl}$ and 0.6M $\ce{H2SO4}$ would yield a solution with nominal concentration of 0.1M $\ce{HCl}$ and 0.3M $\ce{H2SO4}$.
Also given that $K_\mathrm{a2}$ for $\ce{H2SO4}$ is $1.2\times10^{-2}$.

RANT- I have come to absolutlye hate problems that don't use significant figures consistently. Is the initial HCl concentration really supposed to be 0.1 molar, or perhaps 0.10 molar?
Since the $K_\mathrm{a2}$ is given to two significant figures, I'll assume two significant figures for the concentrations.

Now the $\ce{HCl}$ will ionize completely, and the $\ce{Cl^-}$ anion can assumed to be a spectator ion.
Now the first ionization of $\ce{H2SO4}$ will ionize completely, and the second only partially.
Thus if $x=\ce{[H+]}$ from the ionization of $\ce{HSO4-}$, then the total $\ce{[H+]} = 0.4+x$
That means at equilibrium $\ce{HSO4-} = 0.3 -x$, and $\ce{SO4^{2-}} = x$
So:
\begin{align}
K_\mathrm{a2} &= \dfrac{\ce{[H+][SO4^{2-}]}}{\ce{[HSO4-]}} \\
1.2\times10^{-2} &= \dfrac{(0.40+x)(x)}{0.30-x} \\
0 &= x^2 + 0.412x - 3.6\times10^{-3} \\
\end{align}
Thus $x=0.008560$
$\ce{[H+]} = 0.40 + 0.008560 = 0.41$
$\ce{[HSO4-]} = 0.30 - 0.008560 = 0.29$
$\ce{[SO4^{2-}]} = 0.008560 = 0.0086$


text:
Yellowing/Browning from Mixing Ascorbic Acid with Calcium Carbonate and Glycerol
I slowly get some yellowing / browning upon mixing ascorbic acid with calcium carbonate and glycerol and letting it sit at room temperature over time.
At first I thought it was ascorbic acid oxidating but it seems to happen even when kept in a full sealed container.
It's very humid where I'm at so I expect there is water in there too and suspect it accelerates the reaction as I get small amounts of a brown liquid in there as well.
Anyone knows what I'm dealing with?

What is the reason of mixing the acid and the carbonate in the first place ?

What is the purity (source) of the different compounds?

@Poutnik: The ascorbic acid has a dual effect, so it is also for CaCO3, the reason for the combination is that upon adding water there is a mild "foaming" effect, if it can be called so.

I would keep it as a binary mixture.

@Buck_Thorn The CaCO3 is fairly clean (>99%), so is the Ascorbic Acid (meets USP standards), and the Glycerine (>99.5%).


text:
Why is carbon tetrachloride a liquid at room temperature?
From what I can understand, $\ce{CCl4}$ has only London dispersion forces between its molecules. However, it is still liquid at room temperature. $\ce{CCl4}$ has just five atoms, making it a relatively small molecule, so how come its intermolecular forces are strong enough for it to be a liquid at room temperature and atmospheric pressure?

It may have only 5 atoms but 4 of them chlorine so it is quite heavy.

I'm not sure why it is a surprise. Pentane, with half the molar mass and also non-polar, is a liquid. There are far more factors than just polarity to explain why things are liquids.


text:
Reactants used for manufacturing of mustard gas

A compound 'X' which is used in manufacture of mustard gas reacts with (ozone) $\ce{NH4Cl}$ to form 'Y' and elemental sulfur ($\ce{S}$). The compound 'Y' reacts with $\ce{NaOH}$ and $\ce{Ag}$ in separate reactions to liberate 'M' and 'K'. The compound 'M' is heated with $\ce{Cu2O}$ to give product 'K'. The value of $\frac{M_K}{\text{Atomicity}_Y}$ is:

The answer given is $3.50$.
Mustard gas is $\ce{Cl-CH2-CH2-S-CH2-CH2-Cl}$ and I know that it can be produced by the reaction between $\ce{S2Cl2}$ or $\ce{SCl2}$ with ethene, however, none of the reactants satisfy the reactions given in the question. What could be the reactants?

I think the question statement is wrong, as N isn't present in any of the reactants. myPAT adv?

@Rahul Verma I am PCB student, but yes, one PCM student sent me this.

I think M is NH3, and so NH4Cl may be part of the next statement. Since NaOH and Ag with ammonium chloride gives a reaction that feels similar to tollen's test.

Welcome to Chemistry! A screenshot or picture of an exercise is not searchable. Please consider rewriting it, so that it can be of help for future visitors.

I think i got the answer, not sure (about one part) so i am not posting an answer. First reaction is S2Cl2 reacting with Ammonium chloride (Not sure of role of O3) to give S4N4 and then S4N4 reacts with Ag to give Ag2S and N2 and S4N4 reacts with OH- to give NH3. Ammonia then reacts with cupric oxide to give nitrogen. So K is dinitrogen and Y is S4N4 and so atomicity would be 8 and mass of K is 28. So answer is 3.5. please Correct if there is a mistake somewhere.

@Ouch X reacts with ozone to form Y, sulfur and amm. Chloride. So, X must be a compound containing sulfur, nitrogen, hydrogen and chlorine atoms which seems to be impossible. The closest I could find is trithioazyl chloride, S3N3Cl3.

@SafdarFaisal Nice attempt. I would say the question is flawed. Something is not correct. Also bad denotion is used. K can also mean potassium.

@NilayGhosh they said 'K' is a product. so that is fine.. https://onlinelibrary.wiley.com/doi/abs/10.1002/9780470132401.ch26
If somebody has access to this, this was the closest i could find for the reaction to involve Ozone, but is behind a paywall.

@Ouch don't take it too seriously. There were formatting issues. and there were some issues with the question itself. So the downvotes were for the issues with the actual question more than yours.

@Safdar Faisal thanks, your comment makes sense. Don't hesitate to post it as an answer..

@Ouch I cannot tell you why some users have down-voted, I can only guess. The initial ones were probably due to the image in the question, that is for many of our users a trigger. Then often, if the exercise question is longer than the attempt of making sense of the question, it won't be well received. From the amount of comments here, I'd say you'll get quite a bit of help anyway.

@Martin - マーチン Thanks for the support. Your and Nilay Ghosh's link (your initiated discussion itself) helped me. I am sorry for my previous comment, I got a little bit frustrated in beginning, I will gradually learn about site working. Thanks

Thanks for your understanding and engagement. I see your question is taking up some momentum now. I understand that it can be frustrating, especially for new users, when your question is down-voted. I am glad you have taken the time to improve your question and I am positive you'll get an answer. Good readability will always help a question.

In the given question, it is given that 'X' is a compound used in the production of Mustard Gas $\ce{(Cl-CH2CH2)2S}$. Therefore 'X' may be $\ce{S2Cl2}$ or $\ce{SCl2}$ since they react with ethene to give $\ce{(Cl-CH2CH2)2S}$.
Since it is given that X reacts with $\ce{NH4Cl}$, this shows that X is $\ce{S2Cl2}$, since
$\ce{6S2Cl2 + 4NH4Cl -> S4N4 + S8 + 16HCl}$
So, we get the compound Y is $\ce{S4N4}$
Now onto the next statement which says that 'Y' reacts with $\ce{NaOH}$ to release 'M' and reacts with $\ce{Ag}$ to release 'K'.
$\ce{S4N4 + 6OH- + 9H2O -> S2O3^2- + 2S3O6^2- +8NH3}$
$\ce{S4N4 + 8Ag -> 4Ag2S +2N2}$
Therefore, we get that 'M' and 'K' are $\ce{NH3}$ and $\ce{N2}$.The last statement states that 'M' reacts with $\ce{Cu2O}$ to give 'K'. Seeing this we confirm that 'M' is $\ce{NH3}$ and 'K' is $\ce{N2}$
$\ce{NH3 + CuO -> N2 + Cu + H2O}$
Therefore, since Mass of 'K' ($\ce{N2}$) is 28 amu and atomicity of 'Y' ($\ce{S4N4}$) is eight. And so the answer is $\frac{28}{8}$ = 3.5


text:
Why use 1.008 g/mol instead of 1.0079 g/mol calculating the number of moles in 75.0 g H_2 molecules?
Why are figures removed from the calculation for values that are known in more detail? E.g. if you want to calculate the number of moles in 75.0 g H2 molecules. Why should one here use
for the relative atomic mass/molar mass of H 1,008 u or g/mol instead of 1,0079 u or g/mol?
What is the purpose of this?
In addition to that, should one round interim results like the substance amount, the molar mass or should one just round the final result?
Many thanks in advance

Welcome to Chemistry! Please note that the proper term for "number of moles" is amount of substance. The former would be the same as referring to the mass as "number of kilograms".

Think about the error of input data and error propagation.

(I deleted the last comment, I thought that 1.0079g/mol was the molar mass of H). The molar mass of H lies between [1.00784:1.00811] g/mol, so why use 1.008 g/mol?

H molar mas has relative error circa 1e-4. the given mass has rel.error circa 1.3e-3. n=m/V has relative error circa sqrt((1e-4)^2 + (1.3e-3)^2)=sqrt(1e-8 + 1.8e-6)=circa 1.3e-3. Error of M has minimal impact on the error of n,in contrary to the error of m.

Similar illustrative example: if you have expression a . b, and if you know 3 valid digits of a, it is useless to use e.g. 5 or more valid digits of b, even if they are known.( unless computed, in such case it does not matter ). But the result error is determined by the more errorneous parameter.

In my experience, mostly cause we don't care.
In my lab if I'm measuring 3mL - 12mL of a liquid I use a syringe that only has lines on the syringe every 0.2mL. So if I were to be measuring, say, 9mL, I would only have about 2 sig figs. Also, to be honest, I sometimes don't really care if I'm off by a mL. So in doing the math I'd round 1.0079 to 1 because it's easier and the extra precision will be lost when I measure out the chemical anyway.
As for when to round. You should always keep one or two extra sig figs during the calculation and round at the end.

Molecular mass of hydrogen
For $\ce{H2}$ there are 2.016 g/mol (4 significant figures)
The IUPAC value [1.00784, 1.00811] indicates that there is significant isotopic variation in the atomic mass of hydrogen due to the abundances of $\ce{^1H}$ and $\ce{^2H}$ (comparatively there is too little $\ce{^3H}$ to matter). So for hydrogen in some arbitrary compound it would be expected that  $1.00784 <= m_\mathrm{H} <= 1.00811$.

If you look at the individual isotopes the atomic mass is known to much greater precision.
\begin{array}{|c|c|c|} \hline
Nuclide &   Isotopic\ mass & abundance \\ \hline
\ce{^1H} & 1.00782503224(9) & 0.999885(70)\\ \hline
\ce{^2H} & 2.01410177811(12) & 0.000115(70)\\ \hline
\ce{^3H} & 3.01604928199(23) & trace \\ \hline
\end{array}
Rounding
With a modern calculator try to avoid intermediate calculations as much as possible to avoid rounding errors.
If you do calculate intermediate results which can't be stored in the calculator, then I'd write down two extra digits for the intermediate result.
Round only the final calculation to the proper number of significant figures.

You write "molecular mass", but it looks like you report the value of molar mass (in g/mol). These are not the same thing, especially in the context of the question.


text:
What’s the difference between?
In my exam paper, I got this question:

Give the name of the fourth member of the homologous series of carboxylic acids.

My answer was butanoic acid, but the answer was butyric acid. What’s the difference?

I have impression you have not really bothered yourself even with quick search. Why should then others put any effort to answering, if the ready to be read answers are easy to get ? Hint:  Ethanoic vs acetic acid.

Now you’re saying ethanoic acid and acetic acid are the same thing ??

@Googagaman. what we’re saying is a quick search would show you the answer to that

Oh.............

@Goo gaga man. Many long known compounds have older legacy empirical name and a newer systematic name recommended by IUPAC. Homologous acid serie is the typical example. Searching for one usually provides the other as well. In the lab or applied chemistry, empiric names are often preferred in "real life" of chemists, as they have tradition, even if harder to remember, as they cannot be ( usually ) derived.

Butyric acid also known under the systematic name butanoic acid is a carboxylic acid with the structural formula $\ce{CH3CH2CH2COOH}$

Source
So butanoic acid and butyric acid are the same substance.


text:
Can you purify a metal oxide with a laser?
Is there any metal oxide (like aluminum oxide, titanium dioxide etc.), that can be put into a vacuum and hit with a heat source, like a laser and be converted into pure metal with a slightly weaker vacuum having oxygen in it?
From my understanding pure metals tend to oxidize at higher temperatures, however, since "nature hates a vacuum", is it possible to use heat to do the opposite process by providing enough energy to break it's chemical bonds and then the oxygen joins the vacuum leaving the metal behind?
If this is impossible. Is there another way to use a laser to purify metals? If heat doesn't do the trick even in a unique environment, perhaps is it possible to focus ionizing radiation to essentially perform electrolysis on it?
Again, this might not work. So one last idea. For example, aluminum oxide and titanium dioxide are dielectrics. If you were to hit one side with an electron beam from a cathode ray tube to negatively charge one surface and use ionizing radiation on the other side could you then create a chemical reaction to purify the metal?

Even if it the energy does split the bonds with the oxygen you'd have some equilibrium with part of the liberated oxygen reacting back with the metal. You could influence on that by increasing the size of the vacuum relative to that of the amount of metal, but it might not be a very efficient production process. Normal vacuum pumps are not designed to operate with extremely high temperatures at the inlet so I assume you couldn't be continuously maintaining the vacuum by that means.

Maybe have a look at Laser ablation to see what tends to happen when lasers zap solids. On a trivial note, you could certainly use an appropriate laser, e.g., at 405 nm, to thermally decompose HgO.

There is a well known analysis technique in materials chemistry called ICP-MS (Inductively Coupled Plasma Mass Spectrometry) where you have a sample, and using an inductive electric field, the plasma of the sample containing ions of the elements present is created. This is then passed through a mass spectrometer to know the elemental composition of a material synthesised. It is one of the best known techniques for elemental analysis.
However, the idea of using ICP-MS like technique for purification of metal oxides is novel to me. Instead of induction to introduce electric field and rip apart the electrons from the atoms, you can achieve the same using a laser. The rest idea is same.
References:

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


text:
How is bond length of C-O and C-N same?

In NCERT and many other books it is given that bond length of $\ce{C-O}$ and $\ce{C-N}$ is same. But, how can it be possible as size of oxygen is smaller than that of nitrogen so bond length of $\ce{C-N}$ should have been more but it is not. similarly in $\ce{C-O}$ and $\ce{N-O}$, bond length of $\ce{N-O}$ should be smaller as nitrogen is smaller than carbon. but in books it is just reverse of that.
Bond order of all these molecules is same so bond length should depend on size of the bonding atoms but is not so in many cases. So what is the reason behind this.
is there any other factor that is responsible for such result?

this is not really true. it varies based on the bond in a given compound. Some examples are given here https://en.wikipedia.org/wiki/Carbon%E2%80%93nitrogen_bond#:~:text=Similar%20to%20carbon%E2%80%93carbon%20bonds,triple%20bonds%20as%20in%20nitriles.

CO and NO bond lengths are not the same, and any source trying to state otherwise is most likely either trying to make an approximation or it's just textbook erratum

The bond lengths given in your example tables are average bond lengths. That means, the actual bond length in given compound can be larger or smaller than the given value. Keep in mind that bond lengths are not just proportional to sizes of atoms involved making them. As pointed in the other answer, they are determine by other factors as well, which is a broad subject. One such factor is chemical structure of a compound. For example, let's compare chemical bonds in oxazole nucleus in oxazole derivatives with at least 2-substitutions (Ref.1-3):

$$
\begin{array}{c|ccc}
\text{Bonds} & \text{Bond length in $\bf{I}$} & \text{Bond length in $\bf{II}$} & \text{Bond length in $\bf{III}$}  \\
\hline
\ce{O_{(1)}-C_{(2)}} & \pu{1.370 \mathring{A}} & \pu{1.356 \mathring{A}} & \pu{1.356 \mathring{A}} \\
\ce{C_{(2)}-N_{(3)}} & \pu{1.299 \mathring{A}} & \pu{1.294 \mathring{A}} & \pu{1.297 \mathring{A}} \\
\ce{N_{(3)}-C_{(4)}} & \pu{1.382 \mathring{A}} & \pu{1.410 \mathring{A}} & \pu{1.405 \mathring{A}} \\
\ce{C_{(4)}-C_{(5)}} & \pu{1.333 \mathring{A}} & \pu{1.310 \mathring{A}} & \pu{1.332 \mathring{A}} \\
\ce{C_{(5)}-O_{(1)}} & \pu{1.375 \mathring{A}} & \pu{1.399 \mathring{A}} & \pu{1.402 \mathring{A}} \\
\hline
\end{array}
$$
These data demonstrate how bond lengths in oxazole ring deffere by its substitutions and attached ring systems. Even two $\ce{C^\mathrm{sp^2}-O^\mathrm{sp^3}}$ bonds in the same ring gives two different values (c.f., $\ce{O_{(1)}-C_{(2)}}$ and $\ce{C_{(5)}-O_{(1)}}$ values of each compound) because of substitution differences.
References:

Boon-Chuan Yip, Hoong-Kun Fun, Siang-Guan Teoh, Omar Bin Shawkataly, "Structure of 2-(1-naphthyl)-5-phenyl-1,3-oxazole ($\alpha$-NPO)," Acta Cryst. C 1993, C49, 1532-1534 (https://doi.org/10.1107/S0108270193001192).
A. Albinati, M. G. Marcon, P. Traldi, P. Cavoli, "The structure of 2-amino-1,3-oxazole," Acta Cryst. B 1981, B37, 2090-2092 (https://doi.org/10.1107/S0567740881008078).
P. Luger, G. Griss, R. Hurnaus, G. Trummlitz, "The $\alpha_2$-adrenoceptor agonists B-HT 920, B-HT 922, and B-HT 958, a comparative X-ray and molecular-mechanics study," Acta Cryst. B 1986, B42, 478-490 (https://doi.org/10.1107/S0108768186097859).

Bond length is not just proportional to atom size. Determine by other things too. One other determining factor of bond length is electronegativity. Bigger the difference in electronegativity the tighter the bond. I made up a table and stuff to illustrate this point but, as it so happens, the electronegativities didn't explain the data super well either.
I think it's good to keep in mind (with this and also the rest of chem) these are trends and predictors. If you want to get reasonably close to real life you gotta use schrodinger's equation.
(Here's a table on atomic radii which also show that there's many different ways to measure the radius of an atom. One might line up better than another.)


text:
Aluminium oxide phase diagram
I have been looking for the phase diagram of alumina or Aluminium oxide, could it be that there is not something like that?
thank you

There is really not much to it.  Basically you have a metal and an oxide phase, and never the twain shall mix significantly even after melting the oxide above 2000°C [1].  The suboxide that is sometimes mentioned, $\ce{Al2O}$, is not indicated here.

From Ref. 1
Reference
1.
S. Das, "The Al-O-Ti System", J. Phase Equilibria 23(6) (2002), 525-536.

https://www.sciencedirect.com/science/article/pii/036459169290005I?via%3Dihub is another reference - it also shows no other binary phases, although it does show an ionic liquid of Al2O3 composition with a miscibility gap to the Al liquid.

This diagram shows the ionic liquid as well, its saturation line is the short segment on the upper right.


text:
Can there still exist equilibrium with a very high equilibrium constant?
Problem

Consider the reaction translated by:
$$\ce{2 CO(g) + O2(g) <=> 2 CO2(g)} \quad  K_c = 2 \times 10^{22}~\text{at } \pu{1000 K}$$
The order of magnitude of the equilibrium constant, $K_c$, at the temperature considered, indicates that
(A) the reaction has a very high yield;
(B) both reagents were completely consumed.

Answer

 (B)

My thoughts
In my point of view, the higher the equilibrium constant, the more extensive the direct reaction is. Thus the conversion of reagents into products of the reaction is "bigger", so the yield of the direct reaction has to be very high. With that, my answer was (A).

Can we assume both reagents were completely consumed just by analysing the magnitude of the equilibrium constant or can still exist equilibrium even though the magnitude of the constant is so elevated?
Having in mind your chemistry knowledge, what answer do you think it's the "most correct"?

It really is a coin flip. Note that to get (B) you'd have to start off with $\ce{CO2}$. There is no other way to get an exactly stoichiometric mixture of $\ce{CO}$ and $\ce{O2}$. But I'd say that (A) is always correct and therefore the better answer.

@MaxW thanks for answering. I'm sorry but I don't understand why I would have to start off with CO2. Could you explain, please?

It is a poor question. Raise your objection politely once and let it go. All in all it isn't worth getting into a brouhaha with your teacher over this.

I'm having dificulties understanding why couldn't I start with CO and O2 and getting these two to completly be consumed... Can you help me please?

In order for "both reagents" to be "completly consumed", you'd have to have exactly a 2:1 ration of carbon monoxide to oxygen molecules. Let's say that there was exactly one mole of  carbon monoxide. You'd have to have exactly two moles of oxygen. There isn't any way to really do that. But if you started with pure $\ce{CO2}$, then due to the reverse reaction the carbon monoxide and the oxygen would be in exactly a 2:1 ratio.


text:
Priority order of attack of grignard reagents on functional groups

Recently I was solving carbonyl compounds and I came across this. Is there some priority order, or logic behind why alcohol reacts first with Grignard reagent and then ketone. Whereas ester could've reacted with 2 moles of RMgX but didn't.

What is the difference between Grignard reagent reacts with either $\ce{R-OH}$ or $\ce{R-CO-R'}$? There is a pretty good reason for that.

so  the first one seems like a base so acid-base reaction? And the second one will undergo nucleophilic addition ? so acid-base before substitution . is that it?(●'◡'●)

Looks good but is the structure of R shown?

@shreya: You are correct, when acidic hydrogen is present it react first such as tin the case of given 1st reaction to give $a$.

For the compound given in the question, the priority order of Grignard reagent attack follows,
$\ce{R-OH > R-CO-R' > R-COOR'}$
Reason:

Acid-base neutralisation are very fast.

Carbonyl carbon is more nucleophilic than carboxylic carbon, as the latter one is stabilized by resonance.

Therefore, first alcohol is attacked and hence P is (a), then the carbonyl part is attacked and hence Q is (b).
Finally, one more mole of RMgX gets used up on the carboxyl part, hence R is not (d).


text:
What does K° signify in this reaction?

What does the K° signify in the above reaction. Is it just a typo, or is it supposed to symbolize something?
EDIT: This screenshot is from the solutions manual. In the original textbook, the author has used $K ^ 0$.

How do you write the oxidation of a pure element (metal or non-metal)?

Gauri agrawal, Please tell us the full reference from where you saw this?

@M.Farooq This was practice problem 11.14 in Solomon's Organic Chemistry

@gauriagrawal, Thanks. Which edition?

@M.Farooq It is the 11th edition

K° here means elemental potassium (zero oxidation state). The author wanted to avoid the confusion of K with the equilibrium constant. There are more quantities than we have symbols, so letters and symbols get re-used. It is called a pigeon hole effect.
Now, that the OP has posted the reference, Solomons, Organic Chemistry, Ed 11, Problem 11.14 Solution manual, it is clear that the circle is a typo for zero. In the 11 and the 12th edition, the symbol stands corrected to proper numerical "zero" in the main textbook (pg 510). It is not unusual to find typos in solution manuals and textbooks. We are all humans-after all.

Thank you for finding the source, and for another proof that this is not a good textbook:) I scrolled through the solution manual looking for the use of uppercase circle and I wasn't disappointed. They also have $\ce{Mg^\circ},$ as well as three different notations for Celsius scale: $\mathrm{5~^\circ C},$ $\mathrm{5^\circ~C}$ and $\mathrm{5~^\circ~C}.$ And I was clearly mistaken about $\mathrm\LaTeX,$ the main textbook was prepared in InDesign, which could explain why editors missed the inconsistency in symbols.

I have invariably noticed that the solution manuals rarely have the proper formatting. The authors and editors don't want to spend money on it.

Exactly, they either type the solution manuals themselves in MS Word as best as they can and then publish at the university's typography, or outsource their handwritten papers to the freelancers or a typography on the other end of the world. And this is a problem.

It is possible that author's intent was to denote that the reaction with potassium metal is carried out under ambient conditions.
One could use a superscript circle to refer to the standard state of a pure substance, and even though only the pressure is standardized, the notation of standard state is often utilized for denoting a temperature of 25 °C.
This is non-conventional use of a standard state notation and should not be encouraged.
Why it's probably not a notation for a zerovalent metal?
First, a zerovalency is denoted with, well, zero, not a circle.
Second, it appears that the image for the reaction scheme was produced using a $\mathrm\LaTeX$ compiler, and usually those who are using such typesetting system know the difference between zero “0” and a circle “○”.
One does not occasionally type K^\circ $(\ce{K^\circ})$ in place of K^0 $(\ce{K^0}).$
Why it's probably not an equilibrium constant?
Balance aside, it could be a standard equilibrium constant $K^\circ,$ but since the symbol is upright $(\ce{K^\circ})$ and not slanted/italicized, it eliminates this possibility.
There is no need for extra symbols to distinguish between the symbols for an element and for a variable.
Same goes, e.g. for phosphorus $\ce{P}$ and power $P,$ or nitrogen $\ce{N}$ and number of particles $N.$

Dear andselisk, nobody is as perfect in notation as you. People are sloppy and we should not over interpret that notation for *elemental potassium" to mean anything else. The author meant to denote that circle as zero, rather, he should have used a numerical zero. Although it is all unnecessary. He could written state symbol with potassium or he should written the full name.

@M.Farooq The notations are the part of math which is the language of science. If one wouldn't speak it, or would speak incorrectly, they will not be able to communicate. Unfortunately, I don't have any information about how literate the author is, so I cannot operate with phrases like "author wanted" or "author meant". There must be "probably" in between unless you know more than me about the author. Nobody's perfect, but I try to assume the the best in people. Am I having difficulties because of this? Sure, but it's better to overestimate a person rather than to underestimate.

@andseliak, I have asked for a reference. It is quite likely, this is from a test preparation book or some school textbook for Indian students.

@andselisk - I'm hesitant to join this debate, but I wonder how you determined that the image appears to be compiled from LaTeX? Many people still use chemistry-specific drawing programs like ChemDraw for reaction diagrams, and I don't know of a way to tell from the final image how it was created. In a drawing program, inserting a degree symbol is a quick way to make something that looks somewhat like a superscript 0. I'm not condoning that, but I agree with M Farooq that an intended meaning of 0 oxidation state seems most likely. That said, I appreciate you raising an alternate possibility.

@Andrew I'm not 100% sure about LaTeX, but the longer bond length, alignment about the arrow (uniform spacing below and above) and thinner arrow and this arrowhead look a lot like defaults of chemfig. The serif font looks a like CM. GUI programs like ChemDraw or MarvinSketch use different template by default (shorter bonds, thicker double bonds etc.), and stick with Arial or other sans-serif fonts.

@Andrew I don't think there is much to debate about. Here, our answers are just opinions (equally plausible, I think, despite M.Farooq's self-confident manner of expression) and only the author of the text can clarify what the original intention were.

@andselisk, I got the screenshot from the solutions manual, in the original book it shows as $K ^ 0$. So now it makes sense. Thanks!

@gauriagrawal No prob, thank you for reporting the source. I guess I'll leave my answer as is, it still answers the original question and IMO worth every downvote:)


text:
Limitations of Henry's law
Consider the diagram

Henry's law is a gas law that states that the amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid. The proportionality factor is called Henry's law constant.

Suppose the weights on the piston are so increased such that the pressure of the gas becomes numerically equal to Henry's law constant.
Then Henry's law gives that the mole fraction of the gas in solution is equal to 1.
Mole fraction of solute in solvent being equal to 1 implied that there is absolutely no solvent.
But in the situation described by the diagram, the solvent molecules are always present.
How is this so? Perhaps there is a limitation to Henry's law which has not been mentioned?

That's right, Henry's law ceases to work before you get to pressures that high.

(Source: chem.libretexts.org)
In the image, we see that for a "real gas" (indicated by non-ideal solution), Henry's law is applicable for $\ce{\chi} << 1$.
As, $$\ce{p = K_H \cdot \chi}$$ hence the pressure should be low.
Limitations of Henry's law:

The pressure should be low and the temperature should be high i.e the gas should behave like an ideal gas.

The gas should not undergo compound formation with the solvent or association or dissociation in the solvent.

If you think of Raoult's law as a special case of Henry's law (since in both cases the vapor pressure of a substance is proportional to its mole fraction in solution) then there isn't really a problem.
The case you describe where the solute concentration in solution spans the range $\chi=0-1$ and observes the equation
$$\chi=\frac{p}{K}$$
where $K$ is equal to $p*$, the vapor pressure of the solute in the pure liquid state, is the case of a substance exhibiting ideal Raoult's law behavior over its entire concentration range.
However, if $K_H$ is not equal to $p*$ then Raoult's law is not observed over that concentration range, and this is then a more general case of Henry's law with an upper concentration limit on the ideal behavior of the solution. Where that limit is somewhat arbitrary, it occurs when the solute deviates sufficiently from the extrapolated behavior under dilute conditions. In fact, the best way to understand Henry's law is as a limiting law applicable strictly for infinitely dilute solutions:
$$K_H = \lim_{\chi\rightarrow0}\frac{p}{\chi}$$
that is, the constant $K_H$ is defined as the slope of the curve of p versus $\chi$ as the concentration goes to zero.
Where the deviation from linear behavior is observed and the law is violated is in general an empirical question: we apply Henry's law when it works well enough, and not beyond that. The ideal behavior in both Raoult's and Henry's laws means, as explained in another answer, that where the rules apply the behavior of the solute in solution does not depend on its concentration. Note however that even an associating or dissociating compound is expected to display Henry's law behavior at sufficiently low concentrations.


text:
Solubility of Amides
I was looking into how the solubility of amides compares to other compounds e.g. acids and amines, and came across these two statements on the wikipedia page for amides:
https://en.wikipedia.org/wiki/Amide#Hydrogen_bonding_and_solubility
"The presence of a C=O dipole and, to a lesser extent a N–C dipole, allows amides to act as H-bond acceptors. In primary and secondary amides, the presence of N–H dipoles allows amides to function as H-bond donors as well."
"Typically amides are less soluble than comparable amines and carboxylic acids since these compounds can both donate and accept hydrogen bonds."
I understand that amides can act as both H bond donors and acceptors, so I'm wondering how the second statement can make sense?
Additionally, would it be correct to say amides have greater solubility in water since they can form more H bonds in total compared to amines and acids?

Amides have electron delocalisation, so the net charge on an individual atom is less

Amines can be better soluble because of their basicity - acid-base reactions are primary  enhancer of solubility.

Urea ($\ce{H2N-CO-NH2}$) is the quintessential amide. It is extremely soluble in water, and you could say it is a hydrogen donor and acceptor in water.
However, if you replace one of the $\ce{-NH2}$ groups with an alkane or other organic group, solubility is reduced, depending on the size of the group.
Amides generally have higher boiling points and melting points than related alcohols and acids - because of their unique ability to be internal hydrogen donors and acceptors. This donor-acceptor capability makes their solid state more stable and less able to be solvated to a greater degree in water. You could think of the internal bonds being like a dimerization, where the organic chain effect is doubled, while the double $\ce{-CO-NH2}$ interaction gains little benefit from solution in water.


text:
Why does the internal energy increase when ice melts to give water?
Ice melts at 0 °C to give water at 0 °C. The internal energy apparently increases. Thermal energy is transferred to ice causing this to occur. Since the temperature doesn’t change, all the thermal energy is used to increase the potential energy(is this correct?).
Now I know that water is MORE dense than ice. Considering that, shouldn’t the potential energy DECREASE when ice melts? Consequently, shouldn’t the internal energy decrease?

Potential energy refers in this case primarily to the energy associated with intermolecular hydrogen bonds between water molecules. The total strength of H-bonds for each water molecule is on average weaker in the liquid. The density is a function of the geometry of the solid or liquid required to accommodate the bonds. Water is odd in that it acquires a spacious bonding arrangement in ice, but this does not mean that bonding is any weaker for that matter. On the contrary, you might say that the geometry is special in order to accommodate those strong intermolecular bonds.
To complement a point made in another answer: the change in internal energy  during a thermodynamic tranformation is the sum of the heat transferred to and work done on the system. Note that work done by a system (here the water) decreases its internal energy. In the case of ice melting under atmospheric pressure, the volume contraction means that the internal energy increases due to work done on the system by the surroundings. This increase in internal energy is on top of the gain due to a transfer of heat from the surroundings.
Note however that the free energy is unchanged during the process (if carried out reversibly), or, equivalently, the joint entropy of system and surroundings is constant. So the transfer of heat to the system (which reduces the entropy of the surroundings) is compensated by an increase in the entropy of water with melting.

Buck Thorn has already addressed your second question, so allow me to address your first:

Since the temperature doesn’t change, all the thermal energy is used to increase the potential energy(is this correct?).

Not in this case.
Temperature is not a measure of kinetic energy.  It is, more precisely, a measure of the kinetic energy per available degree of freedom. See my answer here:  What exactly is temperature?
Intuitively, liquid water has more available degrees of freedom (rotational and translational) than ice.  Thus, at the same temperature, liquid water will have more kinetic energy than solid water.
Think of it this way: As we change solid water into liquid water, we need to flow more thermal energy into it to "fill up" those additional kinetic degrees of freedom, to keep it at the same temperature. Consequently, the thermal energy used to melt ice increases both its potential energy and its kinetic energy.

The internal energy of a substance is directly linked to the random motion of its constituent particle. Water as ice is a crystalline structure and a firm solid however when it melts the water molecule absorbs heat and oscillates more rapidly, hence the H-bonds between layers of water break, leading to more furious motion of water molecules, increasing its internal energy, you can also associate it with entropy of system while phase change.

Density of a substance doesn't guarantee less entropy to the system . A system can be more dense yet having less firm structure . Density of ice is less due to the interstitial voids inside it and not due to weak bonding


text:
What the difference in data if calibrate pH-meter with 2 and 3 solutions?
pH-meter (12 bucks one) can be calibrated with 2 different solutions (let's say 4.0 and 6.86), but as an option 3 can be used (4.0, 6.86 and 9.18). If I use 2 instead of 3, what error I'm going to get.
For instance I need to determine if liquid higher or lower than 5.5.

You are the only one who can answer this question, because you have the instrument in your hands and can compare. Of course a third calibration point outsinde the range you are interested in should be quite useless, don´t your think?

@Karl I agree with the gist of your comment, but the data at pH 5.5 is intermediate to the two extreme calibration points in the 3-point calibration.

The instrument might have internal electronic settings that are adjusted when it encounters voltage input in different ranges. This is a question of electronics as much as of math and chemistry. It would be interesting to have a link to the product in question.

For pH around 5.5, 2-point calibration is fine. For alkaline solutions, the answer to the question is the measured pH of the 3rd calibration buffer.

@Karl 9.18 is in the kit for calibration.

@Buck Thorn you must be right, 9.18 is in the kit for calibration, I think they know what they doing. A typical yellow pH-meter that you can find on Aliexpress or Amazon.

@BuckThorn 5.5 is very much in the middle of 4.0 and 6.9. A third point well in the alcalic range cannot improve accuracy here. Of course it shouldn´t hurt also, but I have not seen the firmware source nor any calibration curve of that instrument. ;-)

If the pH meter’s response is highly linear, then a two point calibration would seem adequate: the calibration is forced to be linear. If the meter’s response is slightly non-linear, as might be expected for a very inexpensive pH meter, then a three point calibration might help somewhat: the calibration might be piecewise linear.

You cannot "predict" any error by thinking that if we calibrate with two buffers the error will be x, and if we calibrate with three buffers the error will be y.
Ideally, in each case, the pH of the sample should be identical. Good quality pH meters tell you how close they are to the theoretical slope in terms of percentage. It should be close to 100%.
For example, at 25°C the theoretical slope is ~ -59 mV/pH. If your calibrate with two buffers and  is close to -58 mV/pH, the slope percentage is 98.3%. It is good to go. As electrode age, the percentages goes down.
As EdV said, three points might be better for the $12 pH meter.

Better 3 even if I know that pH of solution is somewhere between pH of those 2 calibration powders, is that correct?

It all depends on the slope. If the slope is close to the theoretical slope, 2 buffers are fine. A $12 pH meter will not show the slopes.


text:
Can we determine pH of liquid at home without pH-meter or litmus paper?
Are there alternative ways to determine $\mathrm{pH}$ of a liquid without using $\mathrm{pH}$-meter or litmus paper? In case if we didn't cook the liquid ourselves and don't know concentration of acid in it.
I want to see if it's below or above 5.5, the closer it's going to be to 5.5 the harder it would be to say if it's 5.4 or 5.5 or 5.6 using visual methods.
If yes, how precise it's going to be if compare with those two, what accuracy a method has?

Aquarists routinely use the sets to determine water pH. A solution of pH indicator mixture is dropped into water, resulting color then compared with color of the set of test tubes or paper. I guess accuracy can be within 0.2.

If you search the internet, you can find many suggestions for common food items that can be used as pH indicator dyes. For example, this list: https://www.thoughtco.com/home-and-garden-ph-indicators-601971

A single roll of 5m universal indicator paper costs less than five Euros. How precise do you want it?

@Karl to be able to say if it's higher or above 5.5.

@Karl ...so visual methods wouldn't make it.

https://de.wikipedia.org/wiki/Bromthymolblau

@Karl I mean looking at a color I couldn't say if it's 5.2 or 5.8 for example.

BTB has transition 6.0-7.6, AFAIK.

Maybe methyl red, from the chart here: https://en.wikipedia.org/wiki/PH_indicator

Yes you will. Unless your solution has it´s own colour, methyl red or BTB are well suited to make that distinction. With universal indicator, a difference of 0.6 is at the limit. You still haven´t said what kind of precision you need. Which leads me to conclude you don´t know exactly. (I understand that´s hard to find out, no sweat!) Please update your question if you can be more specific.

Take four test tubes with sample, add bromothymol blue in two of them, methyl red in the rest. Add a few drops of HCl to one of the BTB tubes, and a few drops of NaOH in one of those with methyl red.
One of the tubes will have a noticable change of colour.

I want to see if it's below or above 5.5, the closer it's going to be to 5.5 the harder it would be to say if it's 5.4 or 5.5 or 5.6 looking at colors, this method seems couldn't provide precision of 0.1 in that case. But still looks great, thanks.

If you buy a precision pH meter, and it shows you 5.50, what will you do then? That´s like buying a slide rule and asking wether the 50cm mark is more or less than 50cm from zero. It sure is one or the other, but nobody would care!

Then I'll assume it's 5.50 and not less or more :)

By using a indicator, you cannot achieve $\mathrm{pH} \ \pm 0.1$ accuracy. To get that accuracy, as Karl suggested, you must buy a precision $\mathrm{pH}$ meter (you still has to calibrate it before use). The best indicator to get fairly accurate reading within the $\mathrm{pH}$ you are interested is bromothymol blue indicator. The following is a display of color change according to increasing $\mathrm{pH}$ values:

The color starts to change from yellow to green when $\mathrm{pH}$ reaches the value of 5. At $\mathrm{pH} \approx 6.4$, it is more green than yellow. You can make a method to determine how close it is to $\mathrm{pH \ 6.4}$ such as counting the $\ce{NaOH}$ drops added until the yellowish solution becomes light green. Other than that, there is no other way you can reach needed accuracy without a $\mathrm{pH}$ meter.


text:
Why doesn't but-2-yne show acidic character?
I found this in several sources and also found this while solving a few questions that but-2-yne does not show acidic character (or atleast not a significant  amount of acidity) compared to terminal alkynes
This could easily be attributed to the fact that but-2-yne doesnt have a hydrogen attached to a sp hybridised carbon.
However I felt that if the methyl group attached to the sp hybridised carbon was to donate a proton,  the resulting anion would be resonance stabilized,  because there has to exist a pi bond in the plane of the lone pair on the terminal carbon.
I couldn't see why this resonance shouldn't be possible,  as pi electrons are spread out over 3 atoms(though in different planes) , instead of all of them being concentrated between 2 atoms in a triple bond.
But I couldn't find an explanation anywhere for this.  Please help

Your analysis is quite correct. The only hitch is that that CH group is so weakly acidic that even a bit of resonance stabilisation can't make it significantly acidic. Look up the pKa of toluene, for example: that has one whole aromatic ring to delocalise the negative charge around, and its pKa is still larger than 40.

But most of the sources said that terminal alkynes were more acidic than but-2-yne. But by resonance it should atleast be more acidic than the regular terminal alkynes right? (Marginally but still comparable)

However removing the proton is difficult in the first place. You are having to remove it from a sp3 carbon which doesn't stabilise the the negative charge very well in the first place

@SafdarFaisal But doesn't resonance compensate that?

Well, resonance is not this one factor which automatically trumps everything else. Resonance can be very stabilising (look at pKa of acetone); or not very much at all (look at pKa of toluene), or indeed everything in between.

@orthocresol I get your point. But why does that make it incomparable with terminal alkynes? Is stabilisation by s character here more prominent than resonance?

@VamsiKrishna, I'd encourage you to look at it the other way around: instead of trying to use the model to explain the facts, use the facts to refine your model. Our understanding of resonance, inductive effects, etc. and our understanding of how influential they are, are derived only by careful observation of a wide range of chemical properties. So, this is a great example from which you can deduce that yes, in this particular case, the hybridisation argument is actually far more important than resonance.

@orthocresol Thanks for your time! I can understand it now!


text:
Dependency of concentration terms on temperature
I wanted to know why molarity is temperature-dependent whereas molality is temperature-independent.

Density is temperature dependent, but mass is not. So a volume of liquid solvent changes with temperature whereas a mass of solvent does not.

So then what is the temperature at which densities are calculated?

Any temperature. Generally, densities are measured at chosen temperatures and then the dependence is approximated by an empirical function ( as it is smooth enough ) and eventually tabelated.

The glassware, e.g., volumetric flash, pipette, etc., are calibrated by the manufacturer and the temperature is typically either on the glassware or in accompanying documentation. Commonly the temperature is 20 degrees C.

@Shreyansh Kuntal Your last question should rather be: "So then what is the temperature at which volumes are measured? ", as this is the one Ed V has answered.

This happens because molarity is number of moles per volume of solution and molality is number of moles per mass of the solvent. Now, volume is temperature dependent property since most substances expand with increase in temp and vice versa. Hence the ratio of moles to volume change with temperature. But mass is a temperature independent property hence it does not change with temperature unless the substance or the solvent is getting evaporated constantly due to the conditions in which it is kept. Hence molarity is temperature dependent whereas molality is not.


text:
Importance of halogens and nitrogen in degree of unsaturation
Why is oxygen (or other elements) isn't present in the formulae for calculation of the value of degree of unsaturation $(\mathrm{DU})$ as it usually more common than nitrogen and halogens.
And whats so special about nitrogen and halogens to be in the formulae
$$\mathrm{DU} = C + 1 - \frac{H + X - N}{2},$$
where $C,$ $H,$ $N,$ $X$ are the numbers of carbon, hydrogen, nitrogen and halogen atoms present in the compound, respectively?

Try to derive DU=f(C,H,X,N,O) for various heterogenous componds. How does DU change, if -CH2- is replaced by -CO-, or  C-H by C-OH, or C-C by C-O-C ?  Do the same for other elements.

Forget the formula! Try the methods here under Degree of Unsaturation. http://ursula.chem.yale.edu/~chem220/chem220js/StudyAids.html#Unsaturation

Hint:Think about how many bonds that atom can form and make sure you so not change the DBI

The degree of unsaturation is defined as the index of hydrogen deficiency (IHD) that determines the total number of rings and π bonds. It means the removal of two hydrogen atoms from a molecule is equal to one added $\mathrm{DU}.$
$$\text{rings} + \pi~\text{bonds} = C - \frac{H}{2} - \frac{X}{2} + \frac{N}{2} + 1$$
If you add a halogen to a molecule, you need to remove a hydrogen atom (decreased half-one $\mathrm{DU}).$ If you do the same for a nitrogen atom, you need to remove a hydrogen atom from the molecule and add two others to the nitrogen (for amines as a saturated substituent), so you have one more hydrogen (increased half-one $\mathrm{DU}).$ As a result, the number of halogen and nitrogen atoms must be considered in the equation, but about other atoms such as oxygen $(\ce{C-OH})$ and sulfur $(\ce{C-SH}),$ there is no change in the total number of hydrogen atoms when adding them as saturated substituents to a molecule.
In the case of the unsaturated substituents such as imines and carbonyls, it's clear that the total number of hydrogen atoms decreases, so the $\mathrm{DU}$ increases.

Please note that Chemistry.SE has a native support for MathJax, so there is no need to insert equations as images. Feel free to visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.

So here is the method I learned years ago without memorizing an awkward formula. The concept is to reduce the molecular formula to a hydrocarbon and compare it to the most saturated hydrocarbon bearing the same number of carbon atoms.  Consider the infamous free-base, hydroxychloroquine, $\ce{C18H26ClN3O}$.

The following atom replacements may be done in any order. Ignore divalent atoms ($\ce{O,S}$). The formula is now $\ce{C18H26ClN3}$. Replace all halogens with an equal number of hydrogens $\Rightarrow \ce{C18H27N3}$. Substitute $\ce{CH}$ for each nitrogen or trivalent atom $\Rightarrow \ce{C21H30}$. The most saturated $\ce{C21}$ hydrocarbon is $\ce{C21H44}$. Now do the math: $\ce{C21H44}$ - $\ce{C21H30}$ = $\ce{H14}$. Now divide 14 by 2. The Degree of Unsaturation is 7.
As to the question as to why oxygen is dropped from the equation, if the oxygen in hydroxychloroquine weren't there, the degree of unsaturation would still be the same. For example, the Degrees of Unsaturation of tetrahydrofuran and cyclobutane are the same, D.U. = 1.

The question was about why you would not need to consider Oxygen. Not on how to find DU,

Thats a nice way for DU calculation.

If you think so, you might want to consider accepting it. ;)


text:
How to control the optimization step size in Gaussian?
I would like to perform a geometry optimization for a weakly bound methane dimer structure with Gaussian 16. During this process, the Newton-Raphson step should be controlled such that its length should not exceed 0.01 bohr. For this purpose, I created the following input (strictly for testing purposes):
%nprocs=120
%chk=test
%mem=100GB
#hf/sto-3g opt=(trustupdate,nrscale,maxstep=1,maxcycles=100,tight)

test

0 1
C        0.000000000      0.000000000     -1.821797542
H        0.000000000      0.000000000     -2.931818538
H       -0.906328348      0.523268916     -1.451790544
H        0.906328348      0.523268916     -1.451790544
H        0.000000000     -1.046537831     -1.451790544
C        0.000000000      0.000000000      1.821797542
H        0.000000000      0.000000000      2.931818538
H       -0.906328348     -0.523268916      1.451790544
H        0.906328348     -0.523268916      1.451790544
H        0.000000000      1.046537831      1.451790544

Here maxstep=1 should restrict the NR step below 0.01 bohr. Nevertheless, in the 9th step, I received the following report on convergence:
         Item               Value     Threshold  Converged?
 Maximum Force            0.001795     0.000015     NO
 RMS     Force            0.000287     0.000010     NO
 Maximum Displacement     0.079485     0.000060     NO
 RMS     Displacement     0.024807     0.000040     NO

It seems to me that the 0.01 bohr criterion is not fulfilled for the optimization displacement. Is there any way to really constrain the step length during optimization?

Gaussian by default uses a dynamic trust radius for optimizations to a minimum (you also ensured this by including the trustupdate option). When you specify a maxstep this is only guaranteed to be the initial trust radius. In fact, the second step will ensure the trust radius is at least 0.05.
You can see the trust radius for a given step (if you have #p in the route section) by searching for "Trust test". In that same line, it will say "DXMaxT set to 5.00D-2", with the value printed being the trust radius used for this step.
To prevent these updates, simply replace TrustUpdate with NoTrustUpdate. Keep in mind that they likely force the radius up to 0.05 for a reason and your optimizations may fail or take much longer without these updates. You should only use a step this small if you think you have a geometry very close to the minimum or a very messy potential energy surface that you want to step through cautiously.


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spatial degeneracy in the states of atoms and molecules
I would like to know if the exact ground state of a closed-shell atom or molecule can have a spatial degeneracy. If we ignore relativistics effects, the Hamiltonian of the system $H$ commutes with the Spin  operators $S^{2}$ and $S_{z}$, so, to each eigenstate of $H$, there are another $2S + 1$ eigenstates that have the same energy. In the case of a closed-shell system, $S = 0$ and there are no spin degeneracy. However, my doubt is about spatial degeneracy, that is, if it can exist two states $\Psi$ and $\Psi'$ with the same energy eigenvalue that have the same spin. (in this case, $H$ is the full Hamiltonian, that includes electron-electron interaction)

The nearest to your requirement would seem to be a particle on a ring where $\psi(\theta)=e^{\pm ni\theta}/\sqrt{2\pi}$ and where $n=0,\pm1,\pm2\cdots$ and when $n>0$


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Laboratory setup for the distillation of sulfuric acid
Purification of Laboratory Chemicals, 5th Edition, by Armarego, reads (p. 53):

Reagents such as water, ammonia, hydrochloric acid, nitric acid, perchloric acid, and sulfuric acid can be purified via distillation (preferably under reduced pressure and particularly with perchloric acid) using an all-glass still.

as well as (p. 479):

Sulfuric acid, and also 30% fuming $\ce{H2SO4}$, can be distilled in an all-Pyrex system, optionally from potassium persulfate. Also purified by fractional crystn of the monohydrate from the liquid. Dehydrates and attacks skin—wash immediately with $\ce{H2O}$.

I've never seen sulfuric acid fume (only spread with steam when heated), and Wikipedia states a boiling point of $\pu{337 ^\circ C}$, along the fact that it slowly decomposes to $\ce{SO3 + H2O}$ above $\pu{330 ^\circ C}$, so I'm wondering, mostly from a theoretical standpoint:

How far lower can the boiling temperature effectively get under reasonably strong vacuum?

Does the vacuum not also bring down the temperature level at which sulfuric acid degrades?

What would be an ideal laboratory (not industrial!) setup and process for the safe distillation of sulfuric acid in fair enough amounts for laboratory use (say producing in a batch the equivalent of $\pu{5L}$ at 30% concentration)?

Nota Bene: I know it's a better idea to buy sulfuric acid at the desired level of purity, but that is not the question.

Just on the purification of sulphuric acid, well the one he gets has a dark green tinge, and what is in the sulphuric acid is anticorrosive inhibitors, etc, and they are all organic in nature, and the simple heating up a beaker of sulphuric acid, and before it gets to hot, slowly add 100 ml of 3% grocery store bought food grade hydrogen peroxide, and stir whilst slowly heating, well before long the acid will be crystal clear, any fumes are the skeleton structures of the organic additives, and concentrating it back to 98% sulphuric acid is either density determined, temperature, to insure water has been removed hence both will get a much purer, crystal clear sulphuric acid. Anything organic will be consumed by the peroxide and what doesn't will turn into a gaseous product that evaporates of. The product described has served his associates and himself, for many years.

@Babs_Lawrenson: This information can be useful for some, but I'm not sure this really answers any of the central aspects of my question.

This doesn't answer the question (which is about distillation and purification not the destruction of additives which still leaves a solution that has to be purified). And it is fucking dangerous even with weak peroxide.


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How do you dispose of K2SO4 and MnSO4?
I want to do an experiment by showing how generated oxygen can affect an ember.
I saw on the video that adding peroxide and potassium permanganate will do this.
$\ce{2 KMnO4 + 3 H2SO4 + 5H2O2 = 2 MnSO4 + K2SO4 + 5O2 + 8H2O}$
If I were to do this at home, how should I dispose of the solution afterwards?
Can I put it down the sink or is it considered a hazardous material and appropriate disposal measures need to be taken?

Not particularly hazardous.

Using H2O2 and a catalyst (e.g. potato) instead would avoid most of the reactants, which are hazardous if sufficiently concentrated. We can't answer questions about personal safety on this site.

Thank you for your comment. Are you suggesting that I could get a similar reaction (lots of generated oxygen) using a pulverized potato instead of the potassium permanganate?  What would be the reaction formula?

Check out MSDS of potassium sulfate and manganese sulfate

@Tolure, Potato (or horseradish) are catalysts, not the source of oxygen.

For most of us, putting anything down the drain is not a plan. At home, small amounts, perhaps, but your sewer authority may have a problem with it.

The problem is the Manganese ion presence (see, for example, this article) as Manganese sulfate is fairly soluble.
Suggested remedy: Add inexpensive Washing Soda (Na2CO3) to the solution as:
$\ce{MnSO4 (aq) + Na2CO3 (aq) -> Na2SO4 (aq) + MnCO3 (s)}$
where the created Manganese carbonate is, according to Wikipedia, to quote:

It is a pale pink, water-insoluble solid.

As discussed in the comments, always check the Material Safety Data Sheet (MSDS) for any compound if you are unsure of proper handling or disposal methods.
Whether or not something may be poured down the drain is a function of local municipal by-laws rather than actual toxicity or any other dangers, however, so check the municipals bylaws for your area.
That said, the products of the reaction are quite soluble in water and are used both in fertilizers and in food suppliments for both humans and livestock and are not prone to causing corrosion problems with most materials of construction used in sewage systems. So you are unlikely to blow up or clog your drains by flushing it down the sink.


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Relative reactivity of alkyl and aryl halides in nucleophilic-substitution reactions
In case of alkyl halides, the order for relative rates of reactions is $\ce{RI > RBr > RCl}$.
The reason is clear that $\ce{I-}$ is a better lg because of negative charge stabilisation and also $\ce{C-I}$ bond is weaker than $\ce{C-Br}$ bond which in turn is weaker than $\ce{C-Cl}$ because of overlap of orbitals.
But why is it the reverse order in case of aryl halides, $\ce{ArF > ArCl > ArBr > ArI}$?
Why is it so even though $\ce{C-F}$ bond strength is higher and $\ce{F}$ is also not able to stabilise negative charge?

Two different mechanisms with different features have different relevant factors.

There are three mechanisms to discuss here: $S_N2$, $S_N1$ for $RX$ and nucleophilic aromatic substitution by addition-elimination for $ArX$:
$S_N2$ and $S_N1$:
(Image from https://www.quora.com/Why-steric-hindrance-doesnt-affect-Sn1-reaction)

$S_N2$ is concerted (one step), so this one step is rate determining, or the slowest of the mechanism. In $S_N1$, formation of the carbocation is rate determining. In both rate determining steps, the leaving group is being kicked off. This means the easier it is for the $C-X$ bond to break, the faster the reaction goes. The bond enthalpy decreases as size of halogen increases, giving the order you stated above.
Aromatic addition-elimination:
(Image from https://www.masterorganicchemistry.com/2018/08/20/nucleophilic-aromatic-substitution-nas/#seven)

In this case, step 1 is rate determining, which does not involve the leaving group departing. The halogen, however, can speed up the reaction by stabilising the negative charge formed during this step. The stronger the inductive effect of the halogen - which outweighs the +M mesomeric effect - the more it can draw the negative charge from the ring, stabilising it. The bond strength is irrelevant here, as the step during which it is cleaved is faster than the previous step regardless, and is always 'waiting around' for the first step. This gives the reverse order to the previous one.


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Comparison between hot conc. KMnO4 and cold KMnO4

Recently while solving carbonyl compounds I came across this. What all can cold KMnO4 do?
Can it oxidize aldehydes too, along with oxidizing alkene to vicinal diols?
(I need a thorough comparison in hot conc. and cold`s properties.)

(hot conc KMnO4)looks a bit like oxidative ozonolysis

I found all these but I cant seem to find specific conditions for oxidation of aldehydes.

i got to know that aldehyde remains intact in cold KMnO4(●'◡'●)

Check here and here.

Hot acidic KMnO4 is pretty strong oxidizing agent than cold basic KMnO4. It can oxidize and decompose pretty much anything. It can be explained by looking at the reduction potentials for MnO4- .You'll see that reduction potentials of MnO4- is different in acid and alkali, and thus the reaction pathway is different. In alkaline solution, the reaction is generally messy because the reaction stops at Mn(IV) leaving a brown mess of manganese dioxide.


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What is the density of sodium trimethylsilylpropanesulfonate (DSS)?
I tried to look for the density of sodium trimethylsilylpropanesulfonate (DSS) online but could not find anything. Perhaps someone here measured it/knows where it may be found?

Did you read the Wiki entry carefully? It says it is a white solid.

Don't solid materials have a density?

Yes, solids do have densities, but there are plenty of types of "densities" for solids, e.g. bulk density, single crystal density, packing density etc.

The reported crystallographic density for sodium 3-(trimethylsilyl)-1-propane-sulfonic acid monohydrate is $\pu{1.32 g cm^-3}$ [1].

The title compound which is used as reference for some NMR spectra in $\ce{D2O},$ was obtained casualy during an attempt to crystallize a cephalosporin of a serie we are interesting for […].
The X-ray analysis was undertaken because the expected cephalosporin and the two independent molecules of the title compound which were in fact present in the unit cell have nearly identical molecular weights.
The compound crystallizes from water in the triclinic space group $P\bar{1}$ with unit cell dimensions $a = 5.914(3),$ $b = 10.816(7),$ $c = \pu{18.869(9) Å},$ $α = 97.96(5),$ $β = 95.46(4),$ $γ = 89.76(5)^\circ,$ $V = \pu{1190(1) Å^3}.$
The calculated density is $\pu{1.32 Mg m^-3}$ for $Z = 4.$

Figure 1. Unit cell of sodium trimethylsilylpropanesulfonate monohydrate (CSD: BEHNIL).
Reference

Tinant, B.; Coene, B.; Declercq, J. P.; Germain, G.; Meerssche, M. van. X-Ray Crystal Structure of 3-(Trimethylsilyl)-1-Propane Sulfonic Acid Sodium Salt Hydrate $\ce{C6H15O3SSi^-Na^+ · H2O}.$ Bulletin des Sociétés Chimiques Belges 1982, 91 (1), 95–96. DOI: 10.1002/bscb.19820910116.


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Multiple ways of balancing a reaction
The reaction XeF4 + H2O gives Xe + O2 + HF + XeO3 can be balanced in more than one way and I can not understand why. Please help.

Check here and the other answers there as well.

Actually, if the cited reaction produces transient radical species, I would postulate that it is possible that the reaction can NOT be balanced, with respect to measured observed products, in a  single reaction equation. Examples, the action of HNO3 on Copper metal. Unclear if this is the intent of the question, a clue would be at what level is the science being studied.

Please consider giving an upvote to the most helpful of the posted answers either at the link I gave or at one of the linked duplicates. It encourages people to put some thought and time into crafting answers that are factually correct, relevant, understandable and likely to be of benefit to those, in future, who encounter the question and accepted answer. It is a small reward for those who volunteer their considerable time, effort and experience to aid others and they might well look favorably at future questions from the same person. Thanks for considering this!

There is a quicker way to solve this equation, without any algebra. It is to imagine that, first $\ce{XeF_4}$ is hydrolyzed by $\ce{H_2O}$ forming $\ce{HF}$ and a hypothetical xenon oxide without any change of oxidation number. And secondly this hypothetical oxide $\ce{XeO_2}$gets disproportionized into $\ce{Xe}$ and $\ce{O_2}$, or into $\ce{Xe + XeO_3}$ $$\ce{XeF_4 + 2 H_2O -> XeO_2 + 4 HF}  .......\ (1)$$ $$\ce{XeO_2 -> Xe + O_2} ....................(2) $$ or $$\ce{3 XeO_2 ->  Xe + 2 XeO_3}    \  ..............(3) $$
The simplest solution is adding ($2$) and ($3$), and then adding $4$ times ($1$). $\ce{XeO_2}$ disappears, and it gives :
$$\ce{4 XeF_4 + 8 H_2O -> 16  HF + 2 Xe + 2 XeO_3 + O_2}$$
But there is an infinite number of other solutions.
Multiplying ($2$) by x, ($3$) by y, and ($1$) by x + 3y, then adding the obtained equations, you get the general solution of the equation :
$$\ce{(x + 3y) XeF_4 + 2(x + 3y) H_2O -> 4(x + 3y) HF + (x + y) Xe + 2y XeO_3 + x O_2}$$
For example, choosing x = $1$ and y = $10$, one gets $$\ce{31 XeF_4 + 62 H_2O -> 124 HF + 11  Xe + 20 XeO_3 +  O_2}$$

Per https://books.google.com/books?id=oAlxIVRzrN8C&pg=PA737&lpg=PA737&dq=XeF4+H2O&source=bl&ots=P0M6dIppJb&sig=ACfU3U31bSkmRk5yiV7_p-76hTrJ6nGdBg&hl=en&sa=X&ved=2ahUKEwjm0sGz28PqAhXuUN8KHV0aC2MQ6AEwC3oECAoQAQ#v=onepage&q=XeF4%20H2O&f=false  also possible F2, no O2. Adding H2O forms HF,  and O2 or O3, see https://www.chemistryscl.com/reactions/fluorine+water-reaction/index.php#:~:text=Reaction%20%7C%20F2%20%2B%20H2O-,Fluorine%20and%20Water%20Reaction%20%7C%20F2%20%2B%20H2O,compared%20to%20other%20halogens%20do.&text=Excess%20amount%20of%20fluorine%20gas,fluoride%20and%20ozone%20as%20products.

Maurice, per my last two references, the reaction system is complex with varying products depending on the amount of water present. Also possible XeOF2. Your cited reactions thus appear to be largely speculations unless you have a source,

@ Ajkoer. You are right. My method is a speculation. I have told it : it is hypothetical.  It has the advantage of being able to quickly solve the problem. And of course you could propose XeOF2 as intermediate.


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Multiple hydride shifts
Can we have multiple hydride or methanide shifts? Specially in the case where after first shift it becomes less stable and after the second shift it becomes more stable.
For example, the first step of acid catalysed dehydration of 3-methylbutan-1-ol. The step in which H+ is added to the hydroxy group and it departs leaving a carbocation.

Could you be more specific and provide an example.

@user55119 Please check now, I have also stated an example.

@RahulVerma What about the example above?                  I'm telling to check the example written in the question. Can the carbonation have multiple Hydride shifts?

hydration or dehydration?

@SafdarFaisal sorry dehydration i will correct it

@SafdarFaisal please check now.

In this given scenario, after the first hydride shift, the abstraction of H takes place from C3. so no shift takes place. Even if it did shift it would give an carbocation with the same stability

@SafdarFaisal OK cool, but what if i had 3-methylpentan-1-ol. Then the most stable structure would be formed after the second Hydride shift.

One more thing to be considered is the fact that rearrangement only takes place if the new carbocation is more stable so your first statement is flawed

Let us continue this discussion in chat.

@SafdarFaisal OK got it.

This may help you https://chemistry.stackexchange.com/questions/24604/what-is-the-maximum-extent-of-a-hydride-shift-to-form-the-most-stable-carbocatio

This answer is related to your question https://chemistry.stackexchange.com/a/27930/95066

Yours is a primary alcohol and it has $\beta$ hydrogen atoms also. Thus, it will undergo $\pu E2$ mechanism. In this mechanism removal of protonated hydroxyl group and $\beta ~\pu H$ happen at once (in a single step). Therefore, there will be no scope of hydride shifting.
You may compare this with the dehydration of neo-pentyl alcohol.


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Is there a chemical reaction without a chemical change?
Chemical transformation alters the chemical properties of a material, which includes reactivity with various chemical species.  The new specials thus formed have different characteristics from the initials.
The chemical reaction breaks bonds, forms bonds, or both.  Basically, the atoms within the species reorganize into a new combination, producing a new chemical species with different chemical properties.
Can a chemical reaction happen without a transformation?  There is a chemical balance that pH and chemical properties do not change ...
If there is a chemical reaction in which there is no transformation, what example can you give me?

$$\ce{OH- + H2O -> H2O + OH-}$$

Isomerization of ethylene

trombho, it will be just a waste of time to continue thinking about this "problem". It is a play of words. A chemical reaction at equilibrium appears to be static yet bonds are being broken and re-formed. What is a chemical change?

Sure, there are great many such reactions, the simplest probably being $\ce{H^. + H2 -> H2 + H^.}$. Not that they matter much, though.

The H+H2 (and D+H2 and H+D2) reaction has been extensively studied both theoretically and experimentally in low pressure vacuum using molecular beams. The rotational and vibrational energy in the products fits with quantum calculations. Search for 'reaction dynamics and 'reactive scattering'.

@porfirina Minimal changes, so whenever a chemical reaction occurs, a transformation will inevitably happen?  I am confused by this, because I had read books that these words are different from the other

What you are looking for describes the state of thermodynamic equilibrium. Many examples you've been presented are reactions of the form $$\ce{A^* +B<->B^* +A}$$ which do not "transform" the properties of the system as a whole. The reactions permute the arrangement of atoms into chemically identical arrangements. From a thermodynamic standpoint they are equivalent initial and final states.
If you look at a collection of molecules in aggregate you might identify other reactions called isomerizations, including tautomerizations, that involve rapid changes in the structure of a molecule between various structural forms. In a given sample you will encounter all such isomers in equilibrium concentrations, and if you focus on just one single molecule you will see it sample all of the possible isomers, provided you wait long enough. During that time, if the system is in equilibrium, the properties of the sample as a whole will not change and the relative proportions of the different isomers will be preserved.

A clear distinction can be observed with a hypothetical example of assuming a dynamic chemical equilibrium between 2 species A and B in a mixture. A converts into B and with same rate B converts into A. A→B and B→A occurs simultaneously. If rate of forward reaction is same as that of backward reaction, it is known as dynamic equilibrium. So 2 chemical reactions are taking place everytime but no visible change appears. No chemical change occurs in the mixture as the chemical properties such as pH of mixture experience no change but reaction is taking place everytime.

I think there may be a chemical reaction without a chemical change, I am still not convinced that it is not (at the same time I am confused)


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Nitro-substitution

Here I expected (b) to happen via nucleophilic substitution, but contrary to that (a) is the one occuring. What exactly is happening here?

I’m voting to close this question because it's a bunch of crappy pictures. Up your game. ChemSE is, without a doubt, the best resource for all things chemistry-related. Play the game, get rewarded.

The reaction given in options (a) and (b) are called formation of nitrate esters. The following nitrate ester is called "ethyl nitrate",

Firstly, $\ce{HNO3}$ and $\ce{H2SO4}$ reacts to form the $\ce{E+}$. As we know, the former is relatively weaker acid than the latter, therefore $\ce{HNO3}$ donates $\ce{OH-}$ and neutralisation takes place,
$$\ce{H2SO4 + HNO3 -> H2O + HSO4- + NO2+}$$
So, here electrophile is $\ce{NO2+}$, which gets attacked by lone pairs of alcoholic oxygen and finally gives "nitrate ester" as shown correctly in (a).
(Here is a video explanation of the mechanism by Sal, KhanAcademy)
So, according to me, I guess (b) and (c) are not according to the reaction, hence they should be correct.

Further, if anyone is interested in (c), they should have a look at the answer in Why does NaBH4 reduce double bonds conjugated to carbonyl groups, while LiAlH4 does not?


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Meaning of "MS1 and MS2 resolution" and "MS1 and MS2: Unit" in a tandem mass-spectrometry parameter list
I'm translating a Certificate of Suitability that very briefly describes procedures used to measure several impurities in a drug substance. There is a short description of the parameters of a Gas Chromatography-MS/MS procedure to detect an impurity called NDBA.
I don't understand the final line here:

What is the meaning of Resolution and especially of Unit? I can get by with "resolution" by simply translating it into Russian without knowing what it means, but there is no suitable Russian word to translate "Unit" - I must use either the Russian words for "Unit operation" or for "Equipment unit" or for "Measurement unit", and I have no idea what this Unit means in the original text.
I tried reading up on MS/MS but I haven't understood it well enough to guess the meaning of this line.

Can you ask the source who provided the method? Looks like their internal jargon. There is something unit resolution but not sure what unit means.

@M.Farooq - I'll try but I doubt that I'll get any answer, because I now work not for a single company but for a translation bureau, which gets each document from a different client, and the client may be quite distant from the original authors of the document. I'll just mark this as "impossible to translate".

@CowperKetlle, I think most probably it is equipment unit. It does not make sense in English either.

The term "unit" probably means the size of the isolation window used in the QQQ.  The term "unit" means the nominal isolation window is 1 Da (i.e. one mass "unit") wide.
The "resolution" is a standard term in mass spectrometry, and here they are saying it is about 1 Da.


text:
Stereoisomers of metaldehyde and paraldehyde

(1) In case of paraldehyde I see 3 similar chiral carbons. So I divided it into 2 subcases
-two methyls above the plane and one below the plane
-all three above the plane
so, that makes it 2. When I looked this up on Wikipedia I found this,

can,(1) and (4) not be obtained by simply ring flipping, doesn't that make them conformers? Do I need to count them separately in my count of stereoisomers?

this according to me should have 4 stereosiomer(not including conformers) can someone confirm this answer? Are there any more im missing out?

Looking good! Shows you what Wikipedia is worth. Conformers of paraldehyde do not count as stereoisomers.

Your analysis for metaldehyde is correct. There are four achiral stereoisomers. The example cited here is a bit different. To understand the use of R, S, r and s, see this discussion on the inositols.

Shreya, you might be interested in this: http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/isomers/RS14272/metaldehyde.html


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What is peak identification?
I am sorry if this question is too vague, but what is peak identification? I am working with a chemistry professor to program an autosampler and he says he would like to "program the robot for a peak identification feature that he would use on his project." Can someone please explain in detail what peak identification is?
Many thanks in advance for any help

One can discuss this topic if it is related to analytical chemistry but you have to add more details to your question. What are you doing, what type of measurements are you making. Above all, did you ask your professor if are not sure about it, what does he want and what is that "project"?

Autosampler (in analytical chem work) has nothing to do with peak identification.

i didnt know what tags to put. i just put analytical chem. what tags should i put

Sorry, I don't know what you are doing. You have to describe your work and the nature of measurements so that others can suggest a better tag. Yes peak detection is indeed part of analytical chemistry.

Wait, I think I am understanding. Is peak identification being able to identify peaks in a certain experiment? For example, I remember I was doing an experiment in biology where we drew a graph of the number of cells in the sample. There was a certain peak of the number of cells over a period of time. In this case, do you think that the professor wants to program something that identifies those peaks and records them?

Haversine, You really have to ask your professor because he can tell you his requirements. I really have no idea what are you doing and you seem to be unclear as well. My friend has a very good website on all you would ever need to know on signal processing. Here is the link to Peak Finding. https://terpconnect.umd.edu/~toh/spectrum/PeakFindingandMeasurement.htm

Haversine should at least know and tell us if the peak is obtained in mass spectrometery or in gas chromatography or in another technique. Without this information, we cannot help him.

Autosampler - is a piece of equipment that aspirates the sample from a vial and introduces it into a chromatographer or a detector. It has nothing to do with peaks or their identification - that comes later.
After the sample is injected into the detector and the measurements are taken you get back data describing physical properties of the sample. If you inject directly into the detector you can plot it as signal intensity vs. physical property, here's an example of spectra from Mass Spectrometer:

You may need to find peaks on such graph in order I don't know.. to identify how many impurities there are.
But often before introducing the sample directly to the detector chemists would run chromatography - it'd separate different components of the mixture and make them elute at different time. Now we'd measure physical properties not once - but do it at some rate. If we measure (and sum up or average) the signal at different time we may get peaks when the compound is eluted from chromatographer:

Chemists would then have to define the boundaries of the peaks and integrate their area in order to find how many substances there are in the mixture, in some cases calculate how much of each there are. Software can automate some of this work - identify the peak boundaries for you. This may be what your professor is asking for. Or maybe he wants you to write algorithms determining what substance is represented by this or that peak.
Note that this problem is very challenging and not universally solved - finding peaks can be problematic because:

different chromatograms will have different amount of high-frequency noise
there could be a baseline drift (low-frequency noise)
peaks often are skewed
compounds can elute very closely to each other, so you get merged peaks (rider peaks, shoulders, round peaks)

Due to these reasons you may not always be able to find the peaks reliably e.g.

So for autodetection you may want to go over several algorithms. And the person who uses them may have to understand and tune the parameters for it to work successfully on a particular chromatogram.