text:
Diazotisation of aniline derivatives
I want to know if N-methyl aniline $(\ce{PhNHCH3})$ can undergo diazotisation under suitable conditions and in the presence of $\ce{NaNO2}/\ce{HCl}.$
The mechanism involves loss of two $\ce{H+}$ during the formation of the diazo compound, but if we have one $\ce{CH3}$ and one $\ce{H}$ instead of two $\ce{H}$ (as we have for aniline), will the $\ce{CH3+}$ act as a good leaving group? Is diazotisation of $\ce{PhNHCH3}$ feasible?

$\ce{PhNHCH3}$ is N-methylaniline, not N-methylphenylaniline (which is $\ce{Ph2NCH3}$). Check the names/formulas and fix them accordingly.

Sorry for the inconvenience.

Secondary amines react with nitrous acid to give N-nitroso compounds reference here
N-Methylaniline will give N-Methyl-N-nitrosoaniline as shown below

Which further gives Liebermann nitroso reaction...

Indeed  https://chemistry.stackexchange.com/questions/118764/what-is-liebermann-nitroso-test


text:
Determining the torsion energy profile of butadiene by molecular dynamics
Disclaimer: this question is related to my previous question, although it is much more detailed
Consider the following rotation of butadiene:

I want to determine the relative energy of each dihedral angle by molecular dynamics (MD).
My initial idea was to simulate butadiene in the NVT ensemble and use the relative frequencies of each dihedral angle to determine the relative energy. In the NVT ensemble, the probability of each microstate $P_i$ is:
$$P_i = \frac{\mathrm e^{-E/RT}}{Z}$$
Thus, I performed MD simulations in the NVT ensemble and recorded the dihedral angle periodically. With $N$ total dihedral angles, I made a histogram of the dihedral angles (e.g. a bin every 5 degrees). If we define the number of angles contained in a given bin as $n_i$, I am using the following equation:
$$n_i = \frac{\mathrm e^{-E/RT}}{N}$$
This equation can be rearranged to isolate the approximate energy associated with that dihedral angle:
$$-RT \ln(n_i N)= E$$
However, the curve obtained does not match the distribution I obtained by a simple relaxed scan of the dihedral:

Where is the problem in my method?

I would change the tag thermodynamics for statistical mechanics.

@CharlieCrown you can ping editors directly (i.e. @orthocresol), so that I get notified ;)

@orthocresol yes but sometimes when you play with the bull, you get the horns

I really only want to make a comment, but it is too long...
Your equation $n_i = \exp^{-E_i/RT}/N$ does not make sense to me.
I agree with
\begin{equation}
P_i = \frac{\exp ^{-E_i/RT}}{Z}
\end{equation}
where $Z = \sum \exp ^{-E_i/RT}$ is the configurational partition function.
I can also accept that using a histogram, you could say $P_i = n_i/N$ where $n_i$ is number of samples in that bin, and N is total samples.
I do not see however how
\begin{equation}
n_i N = \exp ^{-E_i/RT}
\end{equation}
This does not make sense to me. Perhaps you could swap $P_i$ for $n_i/N$ so that you have
\begin{equation}
P_i =\frac{n_i}{N} = \frac{\exp ^{-E_i/RT}}{Z}
\end{equation}
which could lead to
\begin{equation}
n_i = \frac{N}{Z}\exp ^{-E_i/RT}
\end{equation}
Which could be rearranged to get
\begin{equation}
E_i =-RT \ln \left(\frac{n_i }{N}Z\right)
\end{equation}
If you are comparing to a reference energy then you could get
\begin{eqnarray}
E_i - E_{\rm ref} = \Delta E_{\rm i,ref} &=&-RT \ln \left(\frac{n_i }{N}Z\right) +RT \ln \left(\frac{n_{\rm ref} }{N}Z\right) \\
&=& -RT\ln \left(\frac{n_i }{n_{\rm ref}}\right)
\end{eqnarray}
It would be interesting to see if this works? I don't have the raw data, so I cannot try it (I also don't have the time, only the interest).

You are right, I made a leap at that part, thanks for pointing it out. However, it does not appear to cause the mismatch. Since N and Z and in the logarithm, they will subtract/add a constant to the energies. If I calculate the relative energies afterwards, the result is the same.

@Raphaël The probability of one conformer to another would not need the partition function, but that is a special case. Overall, just by making things relative, you still have a Z, just a simpler Z made up of a sum of $\Delta E$'s rather than a sum of $E_i$'s. i.e, if $n_i = N \frac{\exp^{-E_i/RT}}{Z}$ then you could say $\frac{n_i}{n_j}=\exp^{-\Delta E_{\rm ij}/RT}$, but that cancellation of both N and Z is because it is a ratio of 2 specific configurations.

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sort: votes


text:
Why is formic acid stronger than acetic acid
Even when acetate ion is stabilized due to resonance why is it a weaker acid than formic acid. It is true that acetic acid has +I effect due to methyl group present on it but resonance is a stronger effect than inductive effect. Then why is it so?

The resonance is identical in both ions.

In comparing acidic strength between molecules firstly make its conjugate base , by -H+. Remember the fact Stability of conjugate base is directly proportional to acidic strength of its corresponding molecule.  Also while comparing acidic strength check factors that are uncommon to all molecules and increase stability (check electronic effect and also steric hindrance should be kept in mind) .electron withdrawing group increases acidic strength and electron donating group decreases acidic strength. Here, after making conjugate base we have, formate ion and acetate ion. Uncommon part is CH3 bonded and H bonded to acetate and formate.Resonance in C double bonded to O bonded to
O- is common to both. Ch3 is electron donating thus decreases acidic strength. H bonded to C is reference and 0 inductive effect is taken for it.Thus, Acetic acid is weaker than Formic acid.

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text:
Quantization of particle on a sphere
While studying particle on a sphere, I read that boundary condition on $\theta$ and $\phi$ results in quantization.
I understand boundary condition for $\phi$ part which varies from 0 to 2$\pi$. As $\psi(0)$ should be equal to $\psi(2\pi)$ otherwise in standing wave the wave function can have more than one value and can interfere destructively.
But I am not able to figure out boundary condition for $\theta$ part which varies from 0 to $\pi$. In books,the result is directly quoted as "Boundary condition in $\theta$ part resulted in quantization which restricts the quantization in $\phi$ part". Quantization in $\theta$ part is called as azimuthal quantum number$(l)$ and $\phi$ part is called magnetic quantum number($m$) which varies from $-l$ to $l$.
Please tell what would be the boundary condition for $\theta$ part. I am not able to figure out and visualize it.

The idea of boundary conditions is nice and intuitive, but this is one case in which intuition cannot really help us.
In general when you are solving the Schrödinger equation in spherical coordinates, you can separate the radial and angular parts (unless you have some pathological potential energy term, I think). See Griffiths' Introduction to Quantum Mechanics (2nd ed.), Chpt 4 for details. It doesn't matter whether you're solving the H atom, a particle on a sphere, or indeed a free particle: in all cases the potential energy has no angular dependence in the potential energy, so the angular component is the same problem.
When you solve the angular bit, you can (again) separate out the two angles, so you end up with two differential equations: one in $\theta$ and one in $\phi$. The $\phi$ bit is easy enough:
$$\frac{1}{g}\frac{\mathrm d^2 g}{\mathrm d\phi^2} = -m^2$$
where $m$ is unconstrained for now, and $g$ is a function of only $\phi$. This is easy to solve and applying the "common-sense" boundary condition that $g(\phi) = g(\phi + 2\pi)$, you can find that $g = \exp(\mathrm i m\phi)$ with $m$ an integer. So far so good.
The $\theta$ bit is substantially more difficult, but this differential equation is so important that it has been studied for centuries and has its own name, the Legendre equation:
$$(1 - x^2)\frac{\mathrm d^2\! f}{\mathrm dx^2} - 2x\frac{\mathrm df}{\mathrm dx} + \left[l(l + 1) - \frac{m}{1 - x^2}\right]f  = 0$$
where $f$ is again a function of only $x$, and we have made the substitution $x = \cos\theta$. In our case, it turns out that unless $(l, m)$ obey certain constraints (i.e. the now-familiar $l = 0, 1, 2\ldots$ and $m = -l, \ldots, +l$), then the solutions $f(x) = f(\cos\theta)$ go to $\pm\infty$ at $x = \pm 1$, i.e. $\theta = 0$ or $\theta = \pi$. That's not acceptable for a well-behaved wavefunction, so we have to reject all the other cases.
So, you could say that the boundary condition on $\theta$ is that the wavefunction  must be welll-behaved (i.e. not diverge to $\pm\infty$) at $\theta = 0$ and $\theta = \pi$. But that is not something we could say without delving into the maths. A priori, we could certainly say that it is important for the wavefunction to be well-behaved at all points. But we cannot pick out the two particularly crucial points $\theta = 0$ and $\theta = \pi$ just using common sense, or "by observation".
More details can be found in Griffiths (there is a fair bit of mathematical detail in there), or any other book that covers differential equations, really.
Bonus: if you multiply the solutions $f(\theta)$ with $g(\phi)$, then normalise the entire thing, you get what are known as the spherical harmonics, which are also extremely important functions in their own right; that's where the shapes of the atomic orbitals come from.


text:
Does an increase in pressure always causes increase in molar concentration of each gaseous substance involved?
Question
Statement :
"An increase in pressure (caused by decrease im volume) at equilibrium results in increase in molar concentration of each gaseous susbstance  involved"
This statement is given true, so i tried this reasoning:
Since decrease in volume(V) causes increase in pressure and hence increase in no. of moles(n), so n/V increase  hence molarity increases.
Is this reasoning correct? Please give the real reason for the statement.

Your reasoning is correct as long as there is a decrease in volume. Note that pressure can also increase with temperature and the volume can stay constant. In this case, the molarity will not change. Molarity is not a very useful unit for expressing gas concentrations in equilibria because you have to quote pressure, volume, temperature and of course moles to quote molarity.

I think the no of moles will remain constant the pressure and volume of a gas alternatively changes(they are inversely proportional when temp and mike's are constant).thus due to the decrease in volume and the no of moles remaining constant increases.
Apply the equation P=1/3(c(vrms)^2) now v RMS and m are constant in a particular temperature thus if  pressure increases c also increases.


text:
Why is ionization energy of indium less than gallium?
In group 13 we observe an irregular trend in ionization energy: B > Tl > Ga > Al > In.
Gallium has a filled 3d subshell, but indium has a filled 4d and 3d subshell. Thus it should have more poor shielding along with more ionization energy, but that is not the case why?

See: https://chemistry.stackexchange.com/questions/99323/why-is-the-electronegativity-of-indium-greater-than-that-of-thallium

The theoretical values of ionisation potential are pretty close to each other.in fact ionisation potential of gallium is almost equal to that of aluminium.Now due to MORE shielding in indium but lack of lanthanoid contraction the affect balances out.As the atomic radii and shielding affect simultaneously takes place from your viewpoint you can claim that the increase in the atomic radii predominates over the zeff.thus indium will have lesser ionisation energy.


text:
Ideal Gas equation and its Assumptions
The two assumptions are that the gases are points of mass that move, they have no volume and that there is no interaction between other molecules.
But even the points of mass can collide with each other so wouldn't that be considered interaction between molecules?

The second assumption takes care of that.

There are probably a ton of other answers to similar posts but ironically enough my own answer to the above linked post was the first one I spotted that addressed the OPs question.

Part of the assumptions set of an ideal gas is that there are no attractive or repulsive forces between them.
Also, part of the assumption is that the volume of molecules obtain a negligible amount when compared to the overall volume of the gas.

I don't think collisions between the ideal gas molecules are prohibited. They are just elastic. This is one of the postulates of kinetic theory of gases from which the ideal gas law can be derived.

@M.Farooq Yes! You are clearly correct. I will remove. Thank you.

There is no ideal gas in reality. If you consider the molecules as point masses, they have no volume by definition of a point. If point masses, or a better word, gas "particles" collide, their collision is perfectly elastic i.e., there is no loss of energy by either particle. When you go to advanced classes you will study other gas laws which do not rely on point masses but include molecular volume and molecular interactions. If you are interested see Van der Waals gas equation.

IUPAC only defines an ideal gas phenomenologically:

Gas which obeys the equation of state pV=nRT...

https://goldbook.iupac.org/terms/view/I02935
There's good reason for that, since there is no universally-agreed-upon microscopic definition of an ideal gas.
The definition I prefer is that an ideal gas consists of point masses that don't interact with each other, but only (elastically) with the walls of the container. [A monatomic ideal gas consists of point masses; a diatomic ideal gas particle consists of two point masses joined by a massless 2D connecter.]
An alternate definition is the one to which you alluded (and which can be found on, e.g., https://en.wikipedia.org/wiki/Ideal_gas ), which is that an ideal gas consists of point masses whose only interaction consists of purely elastic collisions: they extert no attractive or repulsive forces, except for the hard-sphere repulsive potential responsible for the elastic collision.
The problem with the latter definition is that point masses have zero volume, and thus zero scattering cross-section.  I.e., if they have no volume, they will never collide.


text:
Order of reactivity of alkyl halides in Sn1 reactions
Problem:: Predict the order of reactivity of the following compounds in SN1 reactions .
i) $\ce{PhCH2Br}$
ii) $\ce{Ph2CHBr}$
iii) $\ce{PhMeCHBr}$
iv) $\ce{Ph2MeCBr}$
Answer::
SN1:iv>ii>iii>i
Reason:: SN1resonance due to phenyl groups.
I didn't understood why it doesn't consider loss of planarity as two or three phenyls groups can't arrange themselves in a plane around one carbon atom because of ortho hydrogen clash with each other which will no allow
Effective resonance
Please provide some examples or explanation to it
Here is an example of loss of planarity in a compound with 3 phenyl groups


text:
How to calculate Wiener index for aromatic and cyclic organic compounds?
I have been searching for literature on calculating the Wiener index for aromatic and cyclic organic compounds. For acyclic aliphatic compounds the methodology is straightforward obtained from obtaining the distance matrix of molecules . However, for aromatic and cyclic compounds is there a general methodology in obtaining the Wiener index?


text:
Can methanol act as a ligand?
As methanol is both a lewis acid and a lewis base, the lewis base part suggests that it can act as a ligand as many bases are. So, is methanol a ligand? And what is the reason behind it?

To me, the presence of lone pairs is enough of a reason.

If water can act as a ligand, why can't methanol?

There are several complexes where methanol acts as ligand. Some examples are:

methanol coordinated vanadium(V) chalcogenido complexes: $\ce{V(Chal)Cl2(OMe)(HOMe)2}$ where $\ce{Chal = O, S, Se}$
bis(μ-methoxo)dichromium(III) complexes, $\ce{[(L^{Se})2Cr2(μ-OCH3)2(CH3OH)2]}$ and $\ce{[(L^{Se})2Cr2(μ-OCH3)2(CH3OH)(CH3O)]}$, where $\ce{L^{Se}}$ represents the dianion of 2,2‘-selenobis(4,6-di-tert-butylphenol)
di-µ-chlorido-bis[dichloridobis(methanol-κO)iridium(III)] dihydrate $\ce{[Ir2Cl6(CH3OH)4]2H2O}$
$\ce{M(CH3OH)3SO4}$ where $\ce{M = Ni, Co, Zn}$

Reference
Google: Methanol ligand complex. First 4 search results.


text:
Why do things want to lower their energy?
Like if we consider a container of gas molecules as a system, Why does it want to lower its energy?
I have read that always more spontaneous events are favored. So Gibb's Free energy should have a negative value i.e ΔG=ΔH-TΔS<0. So Entropy Contributes towards more spontaneity. But Randomness contributes towards more entropy and having more Internal energy causes more random motion.
So overall more energy should contribute towards more Spontaneity which is favored in nature. But we observe that things tend to lower their energy. Please Explain.
Edit: Well I guess I might have found the answer -
When a system lowers its energy, ΔH becomes negative ( due to it being exothermic) so at moderate temperatures, the entropy contributes to lower spontaneity but ΔH is more negative so ΔG becomes more negative(more spontaneous). While at high temperatures, endothermic processes are favored (ΔH positive)
Thermodynamics is so beautiful!

It is quite difficult to answer such a question. First no physical system want something. It does. Also associating randomness to entropy is a difficult exercise. Something moving faster is not more disordered, just think of fixed frame shots. While the answer by Maurice deserve downvotes - one is mine - it is true that what we observe lead us to conclusion. Thermodynamics goes this way. In this respect it is correct that you won't be much surprised to listen that heat (energy flow) goes from a warmer to a colder body, or that a body in a gravity field falls to lower its potential E.

I hope someone has the skill and patience to help you getting the right track

We have discovered a property of something we call matter which we call energy, and that follows the rules that it does. Beyond that lies philosophy.

This is one of those questions that will take you to the frontiers of physics and out into the unknown in short order. Basically, it is because on a macroscopic scale time has a direction. As to why...

They don't want to, they are forced to.  The Universe wants to raise its entropy, which requires extracting heat from things, which lowers the energy of said things.

Maybe my shortest answer ever:
Energy is loosely a measure of how disturbed a system it is from its natural, grounded, non-excited state, and it is from here that the "less energy, more stable" thing comes from. Your question reverse-engineers that.

It is an observation coming from the study of the gravitation. If you allow an object situated on a table to move somewhat, it may happen that the object is at the edge of the table, and fall to the ground. The contrary never happens spontaneously.


text:
Did both temperature of nernst equation same
I saw the nernst equation
I have a doubt.Did the temparature at which Standard potential is measured is also the temparature at which electrode potential is measured.
I mean Did the temperature (T) in the equation should always taken as the standard temperature at which Standard potential is measured.
See the below link
https://photos.app.goo.gl/EPxTHcPZ4Tynwvqc6
https://photos.app.goo.gl/cAbVFVHexCPgPXE88
In the above given Links it was written
V(metal)-V(solution)=kT/ze×ln{C(reference)/C(metal)} +kT/ze×ln{C(solution)/C(reference)}
Here kT/ze×ln{Preference/C solution}=V electrode

The factor $\ce{2.3 RT/zF}$ is valid at any temperature, not necessarily at  25°C. If you considering working at $25$°C, and only in this case, this factor can be simplified into $\ce{0.0591/z}$

@Maurice 0.0591 V/z

Look at the Wikipedia webpage Standard electrode potential (data page). The standard potentials are taken at 25°C. Since the half-cells are for real chemical reactions, the potential of the half-cell can be corrected for temperature using the Nernst equation.


text:
1-propanamine to cyclopropane?
I was solving some questions on amines when I happened to see the below question:

propan-1-amine + ($\ce{NaNO2 + HCl}$)= a mixture of products
  (originally, the mixture in brackets on the reactant side was just
  written as $\ce{HNO2}$)
The mixture contains:
a.    1-Propanol
b.    2-Propanol
c.    Cyclopropane
d.    All of these

The answer happened to be "d. All of these". I could understand the formation of 1-propanol and 2-propanol but not cyclopropane. The answer given in the textbook wasn't clear and I couldn't understand. It said that the carbocation intermediate had something to do with it which I couldn't follow.
Can someone explain what is behind all this?

Diazotisation of primary amines gives the diazonium salt which decomposes to the primary cation and $\ce{N2}$. From this, in aq. $\ce{HCl}$, you directly get the primary alcohol, primary chloride and primary nitro by immediate trapping of the primary cation by ions in solution and the terminal alkene by 1,2 elimination.
A cation shift to give the secondary cation also occurs to give secondary alcohol and secondary chloride.
The cyclopropane arises from a 1,3-elimination of the primary cation.

pics from reference 1
References

Amines with Nitrous Acid - chem.libretexts.org
Nitrosation, diazotisation, and deamination by J. H. Ridd  Q. Rev. Chem. Soc., 1961, 15, 418-441 DOI: https://doi.org/10.1039/QR9611500418

Reaction equations added


text:
In acid/base chemistry, are there amphoteric substances that undergo something that resembles disproportionation in redox chemistry
Some substances disproportionate. This means a species with an intermediate oxidation state yields two species with higher and lower oxidation state. For example:
$$\ce{Hg2Cl2 -> Hg + HgCl2}$$
My question is if there are substance that have the analogous acid/base behavior, where an intermediate protonation state yields the conjugate acid and the conjugate base. You could write it like this:
$$\ce{2AH -> A- + AH2+}\tag{1}$$
An example would be water:
$$\ce{2H2O <- OH- + H3O+}$$
What I am looking for, however, is a case where the mixture of conjugate acid and the conjugate base (of the same amphoteric starting material) are the major species, i.e. a reaction of the type (1) with an equilibrium constant larger than 1. This would mean that the pKa values are in the opposite order than usual, i.e. it is "easier" to lose the second proton than it is to lose the first. For disproportionation, there is a similar feature of the reduction potentials (it is easier to accept the second electron than it is to accept the first).
Are there any such examples?
Update: To clarify my question, I am looking for substances that skip a protonation state, i.e. pKa2 < pKa1. If you could isolate the intermediate protonation state, it would yield the higher and lower protonation state. This is similar to species that skip an oxidation number (or have the corresponding species at very low concentration) because the reduction potentials favor disproportion of the (hypothetical or barely detectable) species with intermediate oxidation state.

So, you mean if autodissociation constant can be more then 1? In theory, yes, in practice, no. BTW it would be nice if you edited it for clarity.

Complete hydrolysis might be an example, though admittedly somewhat far-fetched.

I don't think that such substance should exist in nature, though may be created in-situ. As clearly by your statement "...with equilibrium constant larger than 1..." should mean that equilibrium is shifted towards products side. But as you want them to be acid and base themselves, they will react with each other again resulting in the reactant, unlike products of disproportionation. So actually, this shouldn't be true. But observation and experiments is the final truth. Edit: Sorry, didn't see Mithoron had said the same thing, realised now.

@Mithoron How about dinitrogen pentoxide? At least in the solid phase, it seems the autodissociation constant is much greater than 1. Admittedly, such cases are very rare, and this case is also a Lewis acid-base, no protons involved. Also, Karsten, I know of a single example where a compound has a second proton which is more acidic than the first: the hydrated vanadyl complex. Does this get vaguely in the direction which you're thinking?

@NicolauSakerNeto Yes, that's exactly it. Thanks to your wikipedia reference, I now have a name for it: an amphoteric substance that does not follow "Pauling's first rule". I wonder what the graph for the speciation looks like, and how this affects buffering capacity. Anyway, I accept your answer, even if it is a comment.

Glad I could be of help. I'll see if I can write up a formal answer later. A small erratum, the correct name for the cation is pervanadyl. I do vaguely recall seeing some example on the web a long time ago, where speciation curves were calculated for the unusual case where pKa2 < pKa1. Hopefully I can pull that up too.

@NicolauSakerNeto I heard about amide complex with second proton more acidic than the first, but this isn't autodissociation. If that's all the same for OP then yeah, there are some cases like that. There was even similar question already, afair.

@Mithoron Yes, all the same to me. It was more a coincident that my examples all look like autodissociation. I could have said it differently too: some species "skip" oxidation states, but there are very few species (like pervanadyl) that "skip" protonation states (said in another way, there are two electron carriers but I'm not aware of many two proton carriers). Thanks for helping me to clarify my question.

As I mentioned in the comments, I know of one case where a polyprotic acid has a second proton which is more acidic than the first: the aqueous pervanadyl complex, $\ce{[VO_2(H_2O)_4]^+}$. According to Wikipedia,
$$
\begin{array}{rcl}
\ce{[VO_2(H_2O)_4]^+ &<=>& H3VO4 + H+ + 2 H2O} &  \mathrm{pK_{a_1}} = 4.2\\
\ce{H3VO4\ \  &<=>& H2VO4^- + H+} &  \mathrm{pK_{a_2}} = 2.60\\
\ce{H2VO4- &<=>& HVO4^2- + H+} &  \mathrm{pK_{a_3}} = 7.92\\
\ce{HVO4^2- &<=>& VO4^3- + H+} &  \mathrm{pK_{a_4}} = 13.27
\end{array}
$$
Though it certainly feels quite odd that $\mathrm{pK_{a_2}<pK_{a_1}}$, I suspect this feature is not absurdly uncommon in larger/more complex systems, where structural reorganisation upon deprotonation can be severe. Perhaps what makes the pervanadyl complex unusual is its comparative simplicity.
However, Wikipedia provides no source for those acidity constants. I looked around for an article discussing the speciation of vanadium ions in aqueous solution, and discovered this nice reference:

Huang, J.-H.; Huang, F.; Evans, L.; Glasauer, S. Vanadium: Global
  (Bio)Geochemistry. Chemical Geology 2015, 417, 68–89.
  DOI:10.1016/j.chemgeo.2015.09.019

The relevant portion in this article is section 4.1. (Aqueous speciation chemistry of vanadium). The authors use thermodynamic data obtained elsewhere to establish the following equilibria:
$$
\begin{array}{rcl}
\ce{[VO_2(H_2O)_4]^+ &<=>& H3VO4 + H+ + 2 H2O} &  \mathrm{pK_{a_1}} = 3.67\\
\ce{H3VO4\ \  &<=>& H2VO4^- + H+} &  \mathrm{pK_{a_2}} = 3.40\\
\ce{H2VO4- &<=>& HVO4^2- + H+} &  \mathrm{pK_{a_3}} = 8.06\\
\ce{HVO4^2- &<=>& VO4^3- + H+} &  \mathrm{pK_{a_4}} = 13.28
\end{array}
$$
As you can see, the first two constants are rather different from the values quoted on Wikipedia, and now they've become difficult to distinguish. However, clearly this is still an unusual case. They take these data and then calculate speciation curves to obtain the following graph:

Evidently, the speciation curves don't really look anything out of the ordinary. It's likely this is partly due to the proximity of the values of $\mathrm{pK_{a_1}}$ and $\mathrm{pK_{a_2}}$ used, and also because the solution is highly dilute. It would be nice to know what happens in more concentrated solutions, but unfortunately we get something like this:

Like some other transition metals, vanadium has a tendency to oligomerise in solution, forming complex mixtures of polyoxometalates. Therefore, it's basically impossible to get a good visualisation of the effects of $\mathrm{pK_{a_2}<pK_{a_1}}$.
Fortunately, we can forget the chemistry and just study how the mathematics behaves. For simplicity, let's consider a diprotic acid $\ce{H2A}$, where we choose the values of $\mathrm{pK_{a_1}}$ and $\mathrm{pK_{a_2}}$. Also for simplicity, let us safely ignore the autodissociation of water by working with relatively concentrated solutions. Therefore, the mole fractions of the species $\ce{H2A}$, $\ce{HA-}$ and $\ce{A^2-}$ will be:
$$X_{H_2A} = \frac{[H^+]^2}{[H^+]^2 + K_{a_1}[H^+] + K_{a_1}K_{a_2}}$$
$$X_{HA^-} = \frac{K_{a_1}[H^+]}{[H^+]^2 + K_{a_1}[H^+] + K_{a_1}K_{a_2}}$$
$$X_{A^{2-}} = \frac{K_{a_1}K_{a_2}}{[H^+]^2 + K_{a_1}[H^+] + K_{a_1}K_{a_2}}$$
Let's keep $\mathrm{pK_{a_1}}=5$ and decrease $\mathrm{pK_{a_2}}$. Here are the individual speciation curves for $\mathrm{pK_{a_2}=10}$ to $\mathrm{pK_{a_2}=-1}$, in decrements of 1 unit of $\mathrm{pK}$ (click on graphs for a larger version).

As you can see, there is no sudden catastrophe as the second proton becomes more acidic than the first. As the second ionisation becomes more favourable, it smoothly and continuously suppresses the formation of $\ce{HA-}$ . Something nice does happen at the crossover point, where $\mathrm{pK_{a_1} = pK_{a_2}}$, namely all species are present in the same concentration when $\mathrm{pH = pK}$. However, this not surprising if you look at the equations above.
For good measure, here are all the plots once again, now with a logarithmic y-axis. It's nice to see that even with the suppression of $\ce{HA-}$, it still exists as a trace in the background, and still behaves smoothly.

And to finish things off, now here are graphs showing the mole fraction of each unique species as $\mathrm{pK_{a_2}}$ varies from 10 to -6, in decrements of 1 unit of $\mathrm{pK}$. I rather like the subtle way the curves for the fully protonated and fully ionised species change, as the intermediate ion stops being a significant contributor in the equilibria.

All you need to do is heat most metal hydroxides; they decompose according to reactions such as the following:
Magnesium hydroxide: $\ce{Mg(OH)2 -> MgO + H2O}~ (\pu{350^\circ C})$
Calcium hydroxide: $\ce{Ca(OH)2 -> CaO + H2O}~ (\pu{580^\circ C})$
and even with one of the alkali metals: $\ce{2LiOH -> Li2O + H2O}~(\pu{924^\circ C})$
The lithium hydroxide decomposition temperature is close to that of calcium carbonate.
In all these cases the protic disproportionation
$\ce{2OH^- -> O^{2-} + H2O}$
is driven by a combination of two factors: the more favorable lattice energy of the oxide and the entropy of the escaping water, which is converted to steam. Only the hydroxides of sodium and heavier alkali metals, formed from the most strongly basic anhydrides and with less favorable lattice energy of the oxides, appear to form gas under atmospheric pressure without decomposing.

@orthocresol subtle question. Why do I need to use $\circ$ as a superscipt when I have degree symbol on my keyboard? I've noticed that edit on several of my answers.

Hmm, I think degree symbol is fine: $\pu{350 °C}$ $\pu{350 °C}$

Personally, it's just a 'that's how I've always done it' thing, so when I edit, typing  ^\circ C comes out quite naturally. :) And that probably stems from the fact that LaTeX (or pdflatex) does not support Unicode characters out of the box. But certainly not anything you're doing wrong.

OK thanks. Also tell me more about your discovery of transmuting magnesium to calcium :-). (Apparent copy/paste error.)

Ah just some backyard nuclear fusion... my bad, thanks for spotting it!


text:
Are there acid/base processes in aqueous solution that are so slow that they are kinetically controlled?
This question came up in figuring out what the fundamental differences are between redox chemistry and acid/base chemistry (if any).
In redox chemistry, strong oxidants and strong reductants can co-exist in solution because they react slowly with each other (or not at all). An example from biology is a low NADH/NAD+ ratio in the same compartment as a high NADPH/NADP+ ratio.
Are there substances that act as acid or base that are far from equilibrium in aqueous solution for a long (minute, hours) time period? Or are all acid/base reactions in aqueous solution fast and reach equilibrium within seconds?

Tautomerisation can be that slow.

The cleavage of esters in aqueous solution can be slow.

@MaxW Yes, but I would argue that the slow steps are not the acid/base ones but the carbon-oxygen bonds forming and breaking.

I guess the Grotthus mechanism is winning every time. At least in those media where you could actually observe proton movement and acid base equilibration. But the key word to the comment is still guess.

Would proton transport into the cleft of a protein count?

Since water is a protic solvent, the predominant form of aqueous acid-base reactions involve proton transfer and that mechanism tends to be quite fast.
But not always.
Most readers are familiar with the use of phenolphthalein as an indicator for titrating acid solutions with a strong base such as sodium hydroxide. When the acid is neutralized, the next increment of sodium hydroxide deprotonates the phenolphthalein molecules and they rapidly turn pink/fuchsia. We proceed to record the data, unaware that with too much additional sodium hydroxide the phenolphthalein  color may fade.
That is the slow reaction, for instead of a relatively rapid proton transfer this excess stage is a much slower nucleophilic addition of hydroxide ions to the deprotonated phenolphthalein. From Wikipedia:

The lactone form (H2In) is colorless between strongly acidic and slightly basic conditions. The doubly deprotonated (In2-) phenolate form (the anion form of phenol) gives the familiar pink color. In strongly basic solutions, phenolphthalein is converted to its In(OH)3− form, and its pink color undergoes a rather slow fading reaction[1] and becomes completely colorless when pH is greater than 13.

How slow is "rather slow" is indicated in the educational article by González-Arjona et al.[2]. The table below, taken from this reference, indicates that the second-order rate constant for the nucleophilic reaction is about $0.02\text{ mol}^{-1}\text{ s}^{-1}$.

At pH $13$ we multiply by $0.1\text{ M}$ hydroxide ion concentration to get a pseudo-first order constant of $0.002\text{ s}^{-1}$, so the time constant here is about $500$ seconds — certainly long enough for kinetic control and convenient for laboratory kinetic measurements.
So if you want to use phenolphthalein indicator for a chemically inspired "It's a girl!" reveal, sodium hydroxide is not the best base to set off the color reveal. A small amount of milk of magnesia in water has the optimal pH to make the pink color but not set off the slow fading reaction.
Reference

Kunimoto, Ko-Ki (February 2001). "Molecular structure and vibrational spectra of phenolphthalein and its dianion". Spectrochimica Acta Part A. 57 (2): 265–271. Bibcode:2001AcSpA..57..265K. doi:10.1016/S1386-1425(00)00371-1. PMID 11206560.

González-Arjona, D. & Domínguez, Manuel & López-Pérez, German & Mulder, Willem. (2019). "Primary Kinetic Salt Effect on Fading of Phenolphthalein in Strong Alkaline Media: Experimental Design for a Single Lab Session". The Chemical Educator. 24. 126-132.

(+1) Marvelous answer! I bumped into this one during a lecture demo I was doing!

I remember we were discussing this earlier. Lime can be too strong/soluble a base too if you want to keep the color, hence my recommendation of magnesia.

At the pH of a phenolphthalein end point ~8-8.5 the decolorization is absorption of CO2  or possibly ionization of a weak acid present not further reaction of the indictor.

Still a magnesia-based system would last longer due to the lower solubikity of Mg(OH)2 vs Ca(OH)2. Next time we cover the beaker after the reaction is over.


text:
Are all dynamic equilibrium reactions with neutralization, must also be buffers?
Reactions like:
$\ce{H2CO3_{(aq)} + H2O_{(l)} <--> H3O+_{(aq)} + HCO3-_{(aq)}}$
or
$\ce{C5H5N_{(aq)} + H2O_{(l)} <--> C5H5NH+_{(aq)} + OH-_{(aq)}}$
Does that mean that the last stage of reaction of a buffer system must be of a weak acid/base?

With a strong acid or base, the equilibrium will be shifted all the way to the right, hence no buffer.

Yes, because a buffer is basically a weak acid or weak base as far as I know. These hydronium ions are also like weak acids on top of carbonic acid which is also a weak acid. Carbonic acid is a weak acid. It has a pH of 5 or 6.


text:
Validity of graph-based molecular representation in ML
I was reading about molecular graph based generative models, which attempt to generate new molecules by training on a subset of a molecular dataset, such as QM9. In these works, there are 2 types of constraints that define the validity of the output:

The sum of the order of the bonds that are incident to an atom may
not exceed its valency.
The output should be a single connected graph.

While we can go further and do DFT calculations to see if these are also stable structures, I was wondering if there are any other simple constraints that we can impose to check the validity of the output molecular graph, without the need for DFT-type calculations.

The idea of running DFT calculations implies that you have you graph together with some tentative geometry, not just connectivity. Is that so?

@IvanNeretin I can theoretically have, but I'm not. Essentially, I am training my model with graphs that only have the atom type information. As coordinates are also available in the training dataset, I can add them as additional features for each atom (that is something that I'm thinking of, but haven't found a potential benefit for that, yet).

@Blade coordinates could be useful, along with atom type, to determine reasonable distances between atoms. You could have graph of two hydrogen atoms that satisfies your constraints, but if the atoms are 1000 angstroms apart, you can quickly determine that the graph doesn't represent a stable molecule.

That's what I was thinking about, too. Bond length checks are almost as fast as the checks you already have in place, while DFT is many orders of magnitude slower.

While that is true, that is defining a new output, and a new check for that output (maybe I shouldn't have mentioned DFT, as my output didn't have enough information to do those computation in the first place). Without any new outputs, based on knowing the type of atom and type of the bonds only, is there anything else other than the two mentioned above, that helps us to say that this molecule is an invalid molecule?

In terms of additional constraints, you will also want to have a minimum valency for each atom (C-H doesn't make sense as a whole molecule) and account for multiple bonds (quadruple bonded carbon doesn't makes sense, but quadruple bonded tungsten would). Beyond that, I can't think of any other obvious graphical constraints. You might be able to make some pseudogeometric considerations of things like a max/min ring size.

Hmm, that's (ring) actually a good point, specially for larger molecules like ZINC dataset.

As an aside, based on your previous questions, you might be interested in the new Materials Modeling SE. It might better address some of your questions that are more on the simulation/computation side.


text:
Difference between organic and inorganic chemistry
This question may seem vague at a first glance, and even lead to the thought: "come on, just go Google it!"
But it's a real question because it does not make enough sense to me (even after Googling it, a lot).
I was just presented to the words, but not to their definition.
Like: "Hey, now we will speak about organic nomenclature".
At some point, books and videos start talking about inorganic and organic as if it was a common sense thing.
A frequent rule I found while trying to understand the difference:

Organic: has carbon
Inorganic: does not have carbon

But then, some inorganic compounds do have carbon too, and there may even be some  compounds that some call organic, and others call inorganic, like $CO_2$.
As I have felt it, in my learnings so far, it's like inorganic chemistry is the default chemistry and organic chemistry goes a step beyond.
But I don't quite grasp the difference.
What is the real semantics behind the word "organic"?
For example, we humans are made of loads of water, and that's a pretty organic thing to me. But then, water is inorganic.
Diamonds are the carbon top of the cake, and do not transmit the idea of being an "organic" thing.
Another very confusing thing are polymers, chanins of loads of carbons with other elements, in many shapes and textures. To me, a piece of "plastic" is not a very organic thing, but indeed, they are!
That brings the semantics into an even more confusing level.
And of course, there must be historical reasons for those chosen words.
Could someone please point out where this distinction comes from and why it is important?
With all my respect to science and the people who made chemistry a useful thing. This question is not about critics, it's about not knowing the facts, so of course I am the ignorant here.
Related and useful: What is the definition of organic compounds?

Organic chemistry refers to study of compounds that are derived from nature and their derivatives, and was thought to be impossible to create by us. When Woehler did so, still such compounds were still continued to be called 'organic' to continue the 'tradition'.

@B.Anshuman That is a very interesting point. Thank you a lot.

For future reference, use \ce{}   for writing chemical formulas. Please take a visit tour page, and a page on meta, on how to edit your question (I am too lazy to search for the links).

Related: https://chemistry.stackexchange.com/questions/102586/does-the-field-of-organic-chemistry-cover-all-group-14-elements-or-only-carbon/102587

@Mithoron It's very helpful information, too. Thank you. But I think it's useful to keep this question and not mark it as duplicate because it's actually a different question with a different approach.

It's a standard duplicate target, and here's seems to be one answer too much already  https://chemistry.stackexchange.com/questions/14246/what-does-organic-non-organic-molecule-means-exactly

IUPAC is the International Union of Pure and Applied Chemistry, they make recommendations on the nomenclature. IUPAC mentions that the difference between organic and inorganic is not distinct. To quote "The boundaries between ‘organic’ and ‘inorganic’ compounds are blurred." in Brief Guide to the Nomenclature of Inorganic Chemistry R. M. Hartshorn, K.-H. Hellwich, A. Yerin.
Since the terminology of organic vs. inorganic is all human classification, it is not a binary system 0 or 1. What we can say now is that traditionally, all organic compounds do contain carbon. It can come from natural sources or purely synthetic. There is no such restriction. Plastic is an organic compound because it contains a lot of carbon chains.
Note that this word organic, as used in chemistry, has nothing to with the buzz word used in marketing of organic food, organic fruits, organically grown stuff. The word organic comes from French organique designating the jugular vein, hence related to organs or living beings.
It is a myth that Wohler started the distinction of organic vs. inorganic. There is a nice French article. The 16-paged paper is titled “Les origines de la chimie organique au-delà du mythe fondateur”  (The origins of organic chemistry beyond the founding myth), Comptes Rendus Chimie, 2012, 15, 553–568.
Read the table at the end, and the abstract.
It clearly shows that Berzelius coined the term organic chemistry between 1807-1813.
Use Google Translate or DeepL to translate it.

This is a beautiful precious answer, thanks a lot for it. I will read it again a few times. Link to the paper you mention: https://www.researchgate.net/publication/257674546_Les_origines_de_la_chimie_organique_au-dela_du_mythe_fondateur

Plastic is an interesting edge case. It is, by definition, organic chemistry. But most of its properties are studied as part of inorganic chemistry... And many other polymer type materials sneak their way into the inorganic sphere because we have found they can have inorganic type qualities. OLEDs, graphene, LCDs... The line is not easily drawn these days. (But I'll still get angry if people assume me for a filthy organic chemist!) j/k...

Historically, organic chemistry was the chemistry done by living organism. It was (back in time) supposed to recquire some sort of divine element/intervention to be done, and therefore could not be done artificially (outside of a living organism). Since the discovery of artificial synthesis of urea in 1828 by Friedrich Wöhler, and the many more that followed, the definition of organic chemistry changed a lot.
Today, organic chemistry designate the chemistry of covalent bond. Those are mostly done by carbon and hydrogen atoms, and a few other hetero-atoms mostly oxygen, nitrogen and halogen. The way of reasoning in organic chemistry is to have a carbon skeleton, on which you will put functional group (which often will include hetero-atom).
Despite not having carbon atom, you can consider the water molecule as an organic molecule (bonds are covalent). Some organic reaction can produce or consume water molecule, or need to be done in water. Same goes for $NH_3$.
Inorganic chemistry designate the chemistry of ionic and coordination bonds. It mostly focus on metal atoms. For a chemist, a metal atom is an atom that can lose up to 4 electrons (it is different from the physics definition, which defines metal by their macroscopic properties). A simple example of ionic bond is NaCl.
Coordination bond involve metal and organic molecule, therefore do mind that both fields are not exclusive.
You can do inorganic chemistry with organic molecule (chelation of metal ions by ethylenediamine for example), or organic chemistry with metals (synthesis of crown ether using the template effect of a metal).
Plastics are organics polymers, salts can be seen as metal polymers (but nobody does). You can create polymers that use both organics molecules and metals, in which case you will do both organic and inorganic chemistry! (keywords for example would be "self-assembly coordination polymers", unfortunately all examples I can find come from publication behind paywall)
Finally, to go back to living organism, we know today that metal ions play a huge part in living organism. Bioinorganic chemistry is the field that study reactions of metal inside living organism. The early anti-cancer drug cis-palladium is the textbook example for this field.

Amon, If you could access and read the article Les origines de la chimie organique au-delà du mythe fondateur The origins of organic chemistry beyond the founding myth"https://www.sciencedirect.com/science/article/pii/S1631074812000331, you will see that it is a myth that Wohler did a magical urea synthesis and ushered a new era in organic chemistry. It is in French, but machine translation of it is pretty good, that what I read last year for some other pupose.

This is a very hepful answer too, Amon. And I will keep researching on all the facts that you expressed here. I will keep M. Farooq's as the accepted answer because it actually answers the question, as well as your answer. But I appreciate them both and the real answer is a combination of both. Thanks a lot.

@M.Farooq Sorry, I was still writing my answer when you post yours. Regarding your comment, of course organic chemistry didn't appear out of thin air and it took time for the "vitalism" to disappear. Synthesis of urea is very often considered to be the starting point of this change of paradigm. Making a full history of the beginning of organic chemistry, albeit very interesting, is not the subject of the question.


text:
Water color additive that will maintain color when water is a thin film
I'm producing thin sheets of liquid (50 to 100 microns) from a nozzle and trying to photograph them.  I think it would help if I could color the water. However, typically as water becomes more of a thin film it tends to lose color.  I'm wondering if anyone has any thoughts on this.  I think white would be a particularly good color.   Food coloring containing titanium dioxide might work.  I know that just using milk will work, but it's a mess to deal with because I am dealing with sprays traveling at a minimum 5 meters per second.

Use fluorescein or other fluorescent dye, along with an effective UV filter (AKA haze or skylight filter) on the lens and an ultraviolet source, such as a xenon strobe (speed light) or UV flashlight. A UV-transmission filter on the strobe can increase contrast, and can be made cheaply, but might not be needed. Because fluorescence makes the fluid become a source of light, it is quite obvious against a dark background. You can even use most laundry detergents, which contain blue-fluorescing dyes -- though foaming might be a nuisance.
See this image of fluorescent soap bubbles or this one. Soap bubble film may be just a few hundred nm thick, yet show clearly with fluorescence. Also watch a demo of strobe illumination of water droplets using sodium fluorescein for contrast. In the video, notice, in particular, that the stream tends to break into pairs of droplets, one perhaps a millimeter in diameter, and the sibling, just a few microns across, but all clearly visible!
Choose a near-UV light source and fluorescent dye for safety, since shorter wavelengths are more damaging to eyes and skin. A search for UV flashlight returned items from US$10 and up, some of which might be suitable. Most were in the 395 to 365 nm spectrum, relatively innocuous.

I should have mentioned that I am taking pictures in the dark using a microsecond flash so that I can freeze the water.  I don't know if a flash would work in conjunction with a black light. I'll have to do some internet research to see if anyone has tried that.

Xenon strobes (speed lights) provide ample UV... no extra black light needed, then.

Oh, well that would be great.  Thirty years ago, I used coumaric acid and it produced a nice green fluorescence.  But I don't remember whether it was ortho, para or meta.

My apologies.  I should have read your first response more carefully.  You provided this information before as well as a lot of reference info.  Thanks so much. If this works, it should be really cool!

@BobD, you didn't make any error reading the original... after you pointed out the use of strobes, I edited the answer. Best wishes for the photos;  perhaps post one as part of your question, if the answer was of help.

This is a detailed and helpful answer that should get upvoted, so here is mine. The only thing I would add is that fluorescent dyes can be messy if spills or sprays occur: I know this from working with dye lasers.

Methylene Blue produces a very intense colour at quite low concentrations. It's also very cheap from places like eBay.It is also relatively non toxic

"Methylene Blue produces a very intense colour at quite low concentrations. It's also very cheap from places like eBay.It is also relatively non toxic"  Thanks.  I'm also looking in to the "ouzo effect".

@BobD. Maybe try oil in water emulsion of metal cutting fluid?


text:
Long Term Storage of D-Cell Battery
I plan to store D-cell batteries in a winter home.  The shelf-life is finite: is there any reason not to freeze or refridgerate the batteries to extend the shelf life?

Freezing destroys the seal, so the battery dies rapidly after thawing. As for chilling, I've found no difference in shelf life (perhaps a negative effect from the dampness in the fridge).

D cells? That´s a bit 1990ies, isn´t it? ;)

There are different types; do you mean the old carbon/zinc s?

Shelf life of name brand batteries these days is a surprising number of years (Duracell, for example, claims 10 years). That said, I use a shelf in the fridge but that is as much because there is then one place in the house for them to be as anything else.

@DrMoishePippik Freezing might destroy seals. Might as well not, depending on the type and brand. And the air in a fridge is surely not damp, but usually very dry. Things get damp in a fridge if you open the door too often in summer.

He says winter home. So I'd assume that he must leave minimal heat on to avoid freezing all sorts of things in the home; plumbing, drywall, can goods, and so on.


text:
Do alkylhydroxylamines form N-nitrosamines?
Usual secondary amines can form N-nitrosamines with nitrosating agents, e.g., $\ce{KNO2}$, and are therefore forbidden in metalworking fluids.
But can alkylhydroxylamines form N-nitrosamines with nitrosating agents in metalworking fluids?
I assume the basicity of the N-atom in alkylhydroxylamines is lowered by the inductive effect and the mesomeric effect. But is this enough to prevent formation of N-nitrosamines?

Yes they can in acidic solutions according to the abstract of this paper$^{[1]}$:

O-Alkylhydroxylamines react with nitrous acid with a rate law $v = k_2\ce{[R^1NH2^+OR^2][HNO2]}$; the mechanism involves electrophilic nitrosation at the nitrogen atom of the free base. At acid concentrations below $\pu{2M}$-per-chloric acid, hydroxylamine and N-methylhydroxylamine react with nitrous acid by an acid-catalysed mechanism with a rate law $v = k_3\ce{[RNH2^+OH][HNO2]}h_\circ$; this involves initial nitrosation at the oxygen atom of the conjugate acid, followed by migration of the nitrosonium ion to the nitrogen. At higher acidities the O-nitrosation can be reversed, and the migration of the nitrosonium ion becomes the rate-determining stage.

Reference

T. D. B. Morgan, G. Stedman, M. N. Hughes, "Kinetics and mechanism of the reaction between nitrous acid and hydroxylamine," J. Chem. Soc. B 1968, 344-349 (DOI: https://doi.org/10.1039/J29680000344).


text:
What are some Qualitative test/s that you can use for benzophenone?
I have determined one of my unknowns to be benzophenone and in a lab setting (but that isn't possible because of COVID-19) I would conducted the 2,4 DNP test which should yield a positive result. However, I am a bit stuck on other tests that would make it very definitive that a ketone is present.
I know if I use the following test, I would get the following results:

The Tollen's test- it would be negative because an aldehyde isn't present
The iodoform test- would be negative because the ketone on benzophenone would be negative
The Chromic Acid test- would be negative because Aldehydes are oxidized by chromic acid, ketones are not.

But are there any other tests that can be used?
I have seen the previous post on this question dated back to 4 years ago and it didn't really help.

It would give negative Schiff's test, Benedict and Fehling's test.

Take an mp of sublimed material (benzophenone, mp 48.5oC). Reduce with NaBH4 to benzhydrol (mp. 69oC).  Dissolve unknown in anhydrous THF under nitrogen. Add sodium. Blue sodium benzophenone ketyl forms.

Not really, I had seen that post.


text:
Is there any elemental molecule which does not react well with anything but oxygen?
For example, phosphor reacts with oxygen but it also easily reacts with many halogens, metals, etc.
Is there any elemental molecule that doesn't react well with almost anything but oxygen?

No, there isn't. Asking similar question about fluorine would make more sense.


text:
Why is atomic orbital one electron wave function? Why distance from centre is proportional to angular wave function?
Consider the following statements:

An atomic orbital is one electron wave function $\psi(r,\theta,\varphi)$ obtained from the solution to the Schrödinger equation.

There two electrons in an atomic orbital, so shouldn't atomic orbital be two electron wave function? Or since we get three eigenvalues from Schrödinger's equation, not four, so atomic orbital is one electron wave function?

In angular wave function plots, the distance from the centre is proportional to numerical values of $\Theta\Phi$ in that direction and is not the distance from centre of the nucleus.

Distance is from which centre if not from centre of nucleus? Also why distance from centre is proportional to $\Theta\Phi$?
Lastly, I guess, it is impossible to plot the variation of $\psi$ with $r,\theta,\varphi$, because it would require 4 axis (or four-dimension graph). Am I right?

@Martin - マーチン can you please answer the second part?

OK, let's address these one by one.

Yes, it should be a two-electron wave function, i.e., the one that depends on 6 spatial coordinates. Just imagine the amount of trouble required to simply write it down, let alone produce a meaningful plot or do anything else with it. So we go with two one-electron functions instead (see Hartree-Fock method). That's an approximation, of course, but it would suffice for now.
Yes, in angular wave function plots, it is the numerical value of $\Theta\Phi$ that is actually plotted by means of the distance from the origin. There is no physical reasons behind this. It is not a physics question at all. It is a question about drawings made by humans on pieces of paper. We may plot whatever we like against whatever else we like. If you find the type of plot which gives you better intuition of what the spherical harmonics look like, you are welcome to use those. (Hint: people have been using these for quite a while now.)
The "not the distance" part of the quote is cryptic to me. I can only guess it is a botched attempt to point out that the plot is not related at all to the radial part of the wave function, and hence to the distance from the electron to the origin.
Yes, that would require a four-dimensional graph, which we can't do on paper. Wait, we can't even do a three-dimensional graph on paper. That's why we resort to cross-sections and other weird plots in the first place.

So it goes.

For 3. I meant about visual graphics also... Are there visual graphics showing variation with respect to all three? I am waiting, if someone would tell more about 2..

What is visual graphics? Well, I guess you can build a 3D graph on screen, turn it this way and that, and so get a relatively good idea of its shape. Too bad that's still not enough. Like you said, we'd need a 4D picture, which just can't be done.

There two electrons in an atomic orbital, so shouldn't atomic orbital be two electron wave function?

No, it should not. The thing is that the actual one-electron wave function is the so-called spin orbital which is a function of not only spatial coordinates of a single electron $(r,\theta,\phi)$ but also of its spin coordinate ($m_s$). Spin orbitals are constructed from spatial orbitals by using the so-called "spin up" and "spin down" functions defined as follows:
\begin{equation}
    \alpha(m_{s}) =
    \begin{cases}
        1, & m_{s} = +1/2 \\
        0, & m_{s} = -1/2
    \end{cases} \, ,
    \quad \quad \quad
    \beta(m_{s}) =
    \begin{cases}
        0, & m_{s} = +1/2 \\
        1, & m_{s} = -1/2
    \end{cases} \, ,
\end{equation}
and in general chemistry it is assumed that spin orbitals always come in pairs - from each spatial orbital $\chi$ two spin orbitals are constructed:
\begin{equation} \label{eq:rhf_spin_orbitals}
    \begin{aligned}
        \psi_{\alpha}(r,\theta,\phi,m_s) &= \chi(r,\theta,\phi) \alpha(m_s) \, , \\
        \psi_{\beta}(r,\theta,\phi,m_s)   &= \chi(r,\theta,\phi) \beta(m_s)  \, .
    \end{aligned}
\end{equation}
Thus, both electrons that occupy the same spatial orbital (say, atomic one) are described by the wave functions (spin orbitals) that share exact same spatial part and this spatial part (spatial orbital) still is a one-electron function in a sense that it depends on spatial coordinates of a single electron only. Note that this apparent paradox arises because calling spatial orbitals wave functions is strictly speaking incorrect - the spin orbitals are the actual wave functions, not spatial ones.

Distance is from which centre if not from centre of nucleus? Also why
  distance from centre is proportional to ΘΦ?

This is tough one without the context. Can share more text with us or reference a book you are apparently reading?

Lastly, I guess, it is impossible to plot the variation of  with
  ,,, because it would require 4 axis (or four-dimension graph).
  Am I right?
  This is obviously true. Being limited with plane sheets of paper we can only draw two-dimensional graphs.


text:
Deviation from ideal gas law curves
So I was reading through my textbook on deviations from an ideal gas, and they had plotted a curve of experimental PV/nRT against the pressure.

They talk a lot about why the volume is really smaller than what the ideal gas predicts, and so on, and thats why as the pressure increases, the deviation from the ideal gas law increases. They further accredit the imbalances to IMFs, and how when the molecules are in closer contact, they tend to form stronger and more IMFs. This all makes sense. This would imply that as the pressure increases or as the temperature decreases, the deviation from the ideal gas would be increasingly over 1. However, when looking at the curve they present, it isn't a clear direct path from no deviation to much deviation. For example, when looking at Methane, CH4, at 200 atm it has a certain negative magnitude, yet at 400 atm, it is back to acting like a ideal gas.
What causes this section of the curve[P=0 to P=400] of Methane that doesn't follow the general trend of increasing pressure --> more deviation from ideal gas?

There are 2 deviation terms of the opposite sign.
The negative deviation term relates to the attractive forces between molecules, decreasing pressure,and is directly related to condensation ability of real gases.
The positive deviation term  relates to non-negligible own volume of molecules, compared to the total gas volume, leading to higher than ideal pressure, as with the higher own volume, the frequency of collisions with wall is higher.
These 2 effects sum to the overall deviation and in some circumstances may cancel each other.
For "permanent gasses" like hydrogen, helium, neon, nitrogen, the negative term is small and there is more or less steady increase of positive deviation with pressure.
For gases that are easy to turn into liquid ( see critical temperature and pressure ) like methane or carbon dioxide, the negative term is dominant for lower pressures and temperatures. If they are high enough, the dominance is overtaken by the positive term.
A lot of insight can be gained from van der Waals equation Wikipedia page

Why is the volume the positive deviation? Cause by account by the non-negligible volume, v should decrease, and since PV/nRT is directly correlated to a decrease in V, shouldn't it also decrease(in respect to the ideal gas).

The same volume, temperature and molar amount, but bigger molecules, leads to more frequent collisions with the walls, and therefore to higher pressure.


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Wouldn't a better term for smoking Crystal meth be condensing or vaporizing?
There isn't actually any chemical reaction taking place.  The meth is heated, first it melts, then it vaporizes, but in the process it doesn't mix with other chemicals.  It just merely changes state.  Smoke, as I understand it is a chemical reaction.  it combines with oxygen and releases smoke and ash, the end product is 2 entirely different chemicals.
Not that it really matters, people are probably going to go on calling it smoke anyway, and not that I condone illicit drug use in any way, but was just curious, technically is it smoke or vapor that is inhaled?

I don't know what takes place when crystal meth is "smoked".  But, yes, if it's the case that, in "smoking" crystal meth, all users are doing is converting it (and the other substances with which it is mixed) to the vapor state (rather than combusting them) then vaping, rather than smoking, would be more technically correct.


text:
How to solve this dimensional analysis/stoichometry problem
This problem has been giving me a headache for the last hour on my homework. Does anyone have any pointers in the right direction?

Finally, on January 28th, 2014, in Germany, a barn containing 90 cows exploded when a stray spark caught the methane trapped in the building on fire. Assume that the 90 cows were producing 250 L of methane per day and that there was a 15 hour collection in the barn when it ignited. What mass of water was generated in the explosion?

This problem is specifically about combustion, and I'm assuming I'd have to multiply the combustion equation for methane by 90 somehow. I'm also assuming I'd have to divide 250 / 15 to get the correct liters of methane produced at this time.

Most of the information is redundant. The key information is 250 L of methane/ 24 h. See how much methane would be there in 15 h. Write an equation on the combustion of methane. What are the products? How many moles of water are generated?

This was just the right think I needed to get the gears going. Thank you!

LOL - If ever there was a problem needing reference to a spherical cow then this is it.// Assume 550 kg/cow, that a cow is 70% water, and that the explosion vaporized the cows.

There is no 'correct' answer here as it is, indeed, a combustion problem. As such, per a source, "Methane gas has an LFL of 5.0%. If the atmosphere has less than 5.0% methane, an explosion cannot occur even if a source of ignition is present". The latter implies that 5% or more of the volume of the barn criteria was met. So, depending on the volume of the barn, the answer will decidedly vary!  I am also assuming no loss of gas in the sealed barn (also a problematic assumption even given the volume).

Here is the reference link, for those interested. https://en.wikipedia.org/wiki/Flammability_limit . Also, the range between Lower and Upper limits is narrow, between 5,0% to 14.3% with a spark, so if the cows produce too much gas per unit volume, the cows are just fine also.

This is based on a real accident, https://www.bbc.com/news/world-europe-25922514. However, there was a jet of flame, with slight damage to the roof, not an explosion. One cow suffered burns, and no cows were vaporized. It is even unclear whether the methane originated from the cows or indirectly from manure stored in the barn. It is possible that the question implies that each cow is producing 250 L of methane per day, not all 90 together.

Figured it out, thanks to @M. Farooq's help.
I first did some cross multiplication of 250/24 times X/15 to find out that in 15 hours, there would be 156.25 L of methane. This is equal to 86.56 grams of methane. Using that, I found out there were 5.393 moles of methane, and multiplying that by 2 (because for every one methane it makes 2 waters) I found that there would be 10.79 mol of water when the reaction was done. This, times the molar mass of water, gives me 194.4 grams of water after the explosion.

Good job on the problem.

To be precise, I would qualify your answer as 'at most'. A good question in that my comments on LFL, and also of interest is the range between UFL and LFL, may be educational to many. What is especially terrifying is the wide range for H2 (4% to 75%). So, recharging your fuel cell car battery in your garage/house and flipping the light switch in the morning, could be a real bang.


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Writing Ionic Equation
According to the rules of writing ionic equation, only compounds in aqueous state can be split or dissociated into their ions, whereas elements or compounds in solid, liquid or gas state remain as they are. But what about molten sodium chloride which is NaCl(l) in liquid state, isn't it dissociated into ions when melted? So can we split it into its respective ions when writing ionic equations?

Whose rules? Such a claim should have a specific reference and an explicit quote. I dislike a reference to vague and unknown authority.

@MaxW http://www.occc.edu/kmbailey/Chem1115Tutorials/Net_Ionic_Eqns.htm I got the information from here, perhaps I phrased it in a wrong way.

I think you mostly misunderstood. The webpage starts off with "When aqueous solutions of sodium phosphate and calcium chloride are mixed together..." So that is the context.

@MaxW so imagine when molten sodium chloride is one of the reactants which is NaCl (l) right? Can we separate it into its ions when writing the ionic equations? As 3rd point in the link said, "Bring down all compounds with (s), (l) or (g) unchanged."

How does molten sodium chloride have anything to do with the context of an aqueous solution of sodium phosphate and calcium chloride? You must be careful not to extrapolate statements beyond their context. // I'd agree that the webpage could use some editing. It is always nice to have several people read something to reduce the possibility of misconceptions in reading.

I mean molten sodium chloride in another chemical reaction, not referring to the one in the link. But seems like we don't use molten sodium chloride for chemical reactions. Please correct me...

Molten sodium chloride is definitely used for reactions, it just wasn't referenced within the context of the discussion on the webpage that you pointed to as the source of your confusion. If I'm reading a book on inorganic qualitative analysis and it says that "The $K_{sp}$ of silver sulfide is $6.31\times10^{−50}$." the context is an aqueous solution, and the statement doesn't imply anything about what happens with $\ce{Ag2S}$ in molten sodium chloride.

The statement, "only compounds in aqueous state can be split or dissociated into their ions," is false in a number of ways.

As you state, molten salts are ionic, and there are room-temperature ionic liquids, such as ethylammonium nitrate , $\ce{(C2H5)NH3+·NO3-}$.
In addition to water, there are other ionic solvents, such as nitromethane, $\ce{CH3NO2}$. Salts in these solvents break up to form ions.
Further, not everything dissolved in water is ionized. Sugar (sucrose), for example, is not ionized.

http://www.occc.edu/kmbailey/Chem1115Tutorials/Net_Ionic_Eqns.htm I got the information from here.

Perhaps I phrased it in a misleading way. Sorry.

Yes Sodium and chlorine seperate into their ions. NaCl can turn into sodium ion and chlorine ion when molten. You could seperate it into respective ions if you want to write ionic equations.


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Open source for atomic spectra raw data
Is anyone aware of a open source atomic spectra database? In particular I am looking for high resolution Na atomic emission which can be downloaded in csv or Excel. I am aware of NIST but I could not find a atomic spectrum. Secondly do they allow re-use of their figures? Thanks.

Any US government publications can be distributed and copies at will. One should certainly be ethically honest and acknowledge the source, but there is no fee for use.

It is a tricky situation in a publication. I want to publish a spectrum. Sometimes, the copyright is held by another publisher. Many IR spectra don't belong to NIST but they have been digitized from very old commercial catalogues.

Do I need reprint permission to use material from a NIST publication? --- "In general, publications of the National Institute of Standards and Technology, as publications of the Federal government, are in the public domain and not subject to copyright in the United States. Permission to reprint or copy from them is therefore not required. The original source should be credited. ..." // It in general, I'd think that webpages fall under unlimited use.

Okay, thanks. That sounds encouraging.

Does this answer your question? Open resource for spectroscopic raw data

No Cody, I had asked that question. I am asking for a downloadable spectrum of Na.

Based on your previous question and your comment above in lines of «NIST's policy to use published data may be fine» I question if your literature search included a visit at NIST's Handbook of Basic Atomic Spectroscopic Data.
In the case of sodium, recorded lines for either air or vacuum are listed next to the data about persistent lines, and analogue data about the single ionized form.  While not yet in a .csv format, copy-paste into an editor is eased because you may toggle between the output to be table-like formatted form, or ASCII.
Addition:
An alternative could be to digitize emission spectra like the following:

(reference, final spectrum about the hot high pressure sodium lamp)

Thanks Buttonwood, I was looking for an entire real "spectrum", as the output from a detector, so that one could look at the peak shapes.

@M.Farooq Please have a look if an emission spectrum like the added one suits your interest better.  Equally, I created a chat room for this at https://chat.stackexchange.com/rooms/111693/apropos-sodium


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carbon dioxide and its behaviour in acid base balance
In human body acid base balance, why and how does the dissolved carbon dioxide increases the hydrogen ion concentration with increasing in its partial pressure..? At the same time bicarbonate ion is staying in same concentration while hydrogen ion are increasing?

The disolved carbon dioxide increases the acid content of the human body. This happens when a carbon dioxide molecule bumps into a water molecule. Carbon dioxide makes water break into hydrogen and hydroxide ion. Hydroxide merges with the carbon dioxide molecule creating bicarbonate ion. Sometimes the hydrogen ion can become part of increasing acidity. Sometimes the hydrogen ion combines with the bicarbonate molecule and make carbonic acid. Carbonic acid could turn into carbonate ion and two hydrogen ions. This is what makes the human body acid base balance. The amount of bicarbonate is staying the same because the bicarbonate that does get converted into carbonic acid or carbonate ion an equal amount of bicarbonate is created from water and carbon dioxide. This just results in a huge amount of hydrogen ions.

Where can I get that in literature? @Roghan Arun

you can get the link here: http://ion.chem.usu.edu/~sbialkow/Classes/3650/Carbonate/Carbonic%20Acid.html

If you don't mind ....Do u know Stewart approach to acid base balance in human body


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Gibbs free energy where T, V, and P change simultaneously
So $\Delta G = \Delta H - T\Delta S - S\Delta T$.
Is there a way to express $\Delta G$ for a process where T, V, and P change simultaneously only in terms of the initial and final T, V, P? Assume this process is for an ideal gas.
Since $G$ is a state function I can calculate it separately for an isothermal process and an isochoric process, but in that case $\Delta G$ for the isochoric process has dependence on the entropy of one of the states.

To 'change simultaneously' means we could also be looking at instantaneous change, which was previously addressed on this forum here https://chemistry.stackexchange.com/questions/50569/what-is-the-difference-between-%CE%94g-and-%CE%94rg .

Your original equation is incorrect.  It should read $\Delta G=\Delta H -\Delta (TS)$, and $\Delta (TS)\neq T\Delta S+S\Delta T$ unless the changes are infinitesimal.
That said, for an ideal gas, irrespective of path, $$\Delta S=S_f-S_i=C_p\ln{(T_f/T_i)}-R\ln{(P_f/P_i)}=C_v\ln{(T_f/T_i)}+R\ln{(V_f/V_i)}$$and$$\Delta H=C_p(T_f-T_i)$$So, $$\Delta G=G_f-G_i=C_p\Delta T-\frac{(T_f+T_i)}{2}\Delta S-\frac{(S_f+S_i)}{2}\Delta T$$Clearly the 3rd term on the rhs depends on either $S_i$, $S_f$, or both, so $\Delta G$ depends on these.  It then all comes down to how you assign your reference state values for G, H, and S, such that $G_{ref}=H_{ref}-T_{ref}S_{ref}$.


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The IUPAC name of a multifunctional carboxylic acid compound
What is more preferred name when ester is present as junior functional group bis-carboxylic acid? (refer the image attached):

to count carbon of ester and consider two branches on it (one oxo and other like methoxy).
to consider whole ester group as branch.

(According to me answer by first method: 4-(formyloxy)-4-oxobutanoic acid and by second: 3-(formyloxycarbonyl)propanoic acid) please correct me if I'm wrong.

There is no ester.

See https://chemistry.stackexchange.com/questions/59536/iupac-naming-for-compounds-containing-carboxylic-acid-and-anhydride-together


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Clausius–Clapeyron equation: shape of phase diagrams makes no sense
I am trying to model the melting point of a substance at varying pressures (ranging from very small to very very large). All I am trying to do is make an equation that relates melting temperature to pressure, so $T(P)$ is some function. To do this, I am trying to use the Clausius–Clapeyron equation (CC), which states that
$$\frac{\mathrm dP}{\mathrm dT} = \frac{L}{TΔV}.$$
In other words, the slope of the equilibrium line on the phase diagram should decrease as temperature is increased.
However, this is not the case; the curve of the equilibrium line is exponential and the slope $\mathrm dP/\mathrm dT$ increases as $T$ increases. Integrating CC we arrive at a logarithmic function, which again is not what empirical measurements reflect.
As I see it, the empirical results and the equation that is supposed to describe them are mutually exclusive. There is no way to arrive at an exponential curve from a slope that varies with $1/x.$ The CC equation and phase diagrams cannot be both be true at once and it is driving me mad.
Why is this the case? Is the CC equation valid at all because it seems to be totally false? What function do I use to model melting points at different temperatures?
The results that are so stupefying are these:

The shape of the curve is exponential. But, the supposed derivative is $1/T$, in which case the slope of each curve (red and blue here) should flatten as $T$ increases, but it steepens. Also, integrating that supposed derivative gives us $\ln (T)$ which is definitely not the shape of the phase diagram. This discrepancy is true for both the liquid/solid and liquid/gas curves.
I hope this clarifies the question!

СС usually works well enough for me. What are your empirical results that contradict it so outrageously?

Clausius-Clapeyron does not apply to the fusion process. It applies to the vaporization process !

I object. CC applies to all phase transitions.

In fusion, is the volume of liquid larger or smaller than the volume of ice?  If smaller, then $\Delta V$ is negative.

@IvanNeretin At least as far as I was taught, the CC equation is derived from the Clapeyron equation (which is exact) on the assumption that the volume of the vapor is much greater than that of the condensed phase, and that the vapor behaves as an ideal gas. The OPs equation is what is commonly called the Clapeyron equation.

@BuckThorn You seem to be referring to a specific particular case when the volume change itself changes with pressure in a certain way, as if all of it were caused by the gas, and the gas behaved as ideal.

@IvanNeretin Again, this is what I was taught, and I would have to look into a textbook to confirm, but my understanding is that the CC equation is an approximation applicable when a vapor phase is in equilibrium with a condensed phase, the latter having negligible volume. So yes, these approximations allow you to ignore changes in the condensed phase with T or p.

The equation in this post does not make such assumptions, it is general. So this whole argument is really of little importance.

You should present actual experimental data rather than a cartoon from the Wikipedia.

Good point! For all we know, the green curve might just as well be a scribble made by a drunken artist arrested for indecent behavior, while two policemen were dragging him away by the feet.

It looks like your fundamental misunderstanding is that you are assuming that L and deltaV are both independent of P and T. Hopefully porphyrin’s answer clears that up for you.

This is a summary of the equations to use to calculate phase transitions.
The Clapeyron equation $\displaystyle p_2-p_1=\frac{\Delta H}{\Delta V}\ln\left( \frac{T_2}{T_1} \right)$ is used for a solid-liquid transition. The changes in enthalpy and volume relate therefore to changes occurring in fusion.
The Clausius-Clapeyron equation describes solid-vapour and liquid-vapour changes because the final volume is far greater than the initial one, and is $\displaystyle \frac{dp}{dT}=p\frac{\Delta H}{RT^2}$ where $\Delta H$ the enthalpy change at the liquid–vapour or sublimation transition. Integrating this last equation from pressure $p_1 \to p_2$ and temperature $T_1 \to T_2$ gives $\displaystyle \ln\left(\frac{p_2}{p_1} \right) = -\frac{\Delta_{vap}H}{R}\left( \frac{1}{T_2}-\frac{1}{T_1} \right) $.
The change in the volume during fusion is $\displaystyle \Delta_{fus}V = m\left(\frac{1}{d_l}-\frac{1}{d_s} \right)$ where $m$ is the molar mass and $d_l$ and $d_s$  the densities of the liquid and solid. The pressure variation for the solid to liquid (melting or fusion) change is
$$\displaystyle p_2=p_1+\frac{\Delta_{fus}H}{\Delta_{fus}V}\ln\left(\frac{T_2}{T_1} \right)$$
and for evaporation and sublimation
$$\displaystyle p_2=p_1\exp\left( -\frac{\Delta_{vap}H}{R}\left(\frac{1}{T_2}-\frac{1}{T_1} \right) \right)$$
with the appropriate $\Delta H$. This is $\Delta_{vap}H$ for evaporation and $\Delta_{fus}H + \Delta_{vap}H$ for sublimation. Sublimation is treated as two steps merged into one; melting and instantaneous evaporation.
The $p$ vs $T$ plot for benzene is shown in the figure. Notice how this is different to how these are generally shown. This is sometimes due to the fact that log pressure is sometimes plotted but this is not always shown on the figure. Notice also how the solid liquid line is effectively vertical.

The data used is
R = 8.314  #( J/mol/K)
dens_sol = 981.0  #( kg/m^3)
dens_liq = 879.0  #( kg/m^3)
mol_mass = 78.0/1000.0 #( kg/mol)
DH_vap   = 30.8*1000   #( J/mol)
DH_fus   = 10.6*1000   #( J/mol)
p3 = 36.0/760*101325 #( triple point pressure Pa)
T3 = 5.5 + 273.16    #( triple point temperature K)
DV_fus = mol_mass*(1/dens_liq-1/dens_sol)                          # delta volume fusion
Python/numpy  functions for the pressure are:
p_liq_vap= lambda T: p3*np.exp( (DH_vap/R)*(1/T3-1/T))
p_sol_vap= lambda T: p3*np.exp( ((DH_fus + DH_vap)/R ) * (1/T3-1/T) )
p_sol_liq= lambda T: p3 + DH_fus/DV_fus*(np.log(T) - np.log(T3))


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What is the difference between bond disassociation energy and lattice energy?
Bond disassociation energy refers to amount of energy required to break bonds of one mole of atom of particular type whereas lattice energy is the energy required to separate a crystal lattice containing one mole ions into their respective gaseous states.
What i can understand from the above information is that bond dissociation energy has to be a broader classification and lattice energy must be a subset of it.
Is it true or am i wrong. Please correct me, i am a bit confused.
Any help would be appreciated.

Why are you speaking of ionic compounds, to begin with? Both terms are much broader than that.

What do you mean?

I mean just that. Bond disassociation energy is not limited to ionic compounds. Neither is lattice energy.

if you are a senior secondary student (class 11 or 12) like me, then my point may mean some sense-bond disassociation enthalpy is the energy required to break a bond between two atoms of the same molecule while lattice enthalpy is the energy required to overcome the intermolecular forces of attraction between the molecules in a crystalline system ( it may be an ionic or covalent or molecular crystal as @Ivan Neretin quoted.


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Measuring instruments and significant figures
To start with, I am familiar with the rules that we have for sig-figs.
I have two chemistry books by different authors (One by McMurry and the other one by Nivaldo J Tro).
I was doing some casual readig, I am familiar with sig-figs (or that's what I thought).
According to the first book, let's say I measure the thickness of the cover of a hardbound version of a book with a ruler that is accurate up to the nearest millimeter, and I find that it is 3±1 mm thick, or 0.3±0.1 cm thick. The book says I can claim just one significant figures when I use a ruler because it is accurate up to 0.1 cm, or 1 mm. The book clearly wants to say that I can't claim more than one sig fig if I use the ruler. For example, I can't say that the cover is about 3.2 mm thick, or 3.4 mm thick because the limitations of the ruler, it is not accurate up to 0.1 cm, or 1 mm.
On the other hand, if I use an instrument that is accurate up to 0.01 mm (Vernier caliper, for example), I find that the book is 3.25±0.01 mm thick. It means I can claim three sig figs if I use a Vernier caliper because it is accurate up to 0.01 mm. Out of these three sig figs (3 2 5), 3 and 2 are certain digits and 5 is uncertain.
The other book, however, says that if I use a ruler (accurate up to the nearest mm) to measure the cover of the book, I can be sure the the cover is at least 3 mm thick, and then I can mentally divide the little space for a mm on the ruler into 10 parts and I can extrapolate that the cover is about 3.2 mm thick (or 3.3 mm, whatever). Now I can say that the cover is about 3.2 mm thick, out of which 3 is the certain digit and 2 is uncertain. That's what the book says.
In other words, there are two sig-figs according to the 2nd book, when we used a ruler to measure the thickness of the cover of a hardbound book. On the other hand, the first book says that we can claim only one sig-fig if we use the same ruler that is accurate up to the nearest mm.
I am confused, which one is correct? From what I understand about significant figures, I would think that the first book is correct in saying that we can't have two sig-figs if we use a ruler that can accurately measure up to the nearest mm. But the second book says that we can have two sig-figs.
Please help me with it

@Poutnik In large samples, where error of standard deviation estimation is small, there can be overlap of 2 digits. Isn't it why we should say it is 3±1 mm, and not 3.2±0.1 mm? Unless I did not understand what you said. Can you please elaborate?

Is this a site about measuring books now?

@Mithoron not books, just covers

If you're an old timer like me you used a slide rule and interpolating between lines was common. You also have to (well at least you should...) interpolate when using a burette. You'd also run into that problem when reading a mercury barometer which would probably have a vernier scale which would help greatly. So the second book is right.

Last significant digit is for the most of common measurements the first unsure digit. The eye+ruler can estimate $\pu{0.1  mm}$ , so it is $\pu{ 3.2 +- 0.1 mm}$, not $\pu{3 +- 1 mm}$.
There is big difference between e.g some 1st generation digital voltmeter, displaying integer values of voltage, and analogue voltmeter with integer voltage marks, where you can estimate the position between marks. The 1st book speaks about the former, the 2nd book about the latter. The common sense must tell you you know better than the thickness is within $\pu{ 2-4 mm}$.
In large samples, where error of standard deviation estimation is small, there can be overlap of 2 digits. That applies e.g. to massive and very accurate parallel measumurent of physical values. Like $1.12345678 \pm 0.00000034$,  by convention written as $1.12345678(34)$.

Thanks. So in some situations, I need to see if I can extrapolate another digit if it gives me more useful (accurate) information?

It is done always when discrete marks are plotted along  continuous scale.


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What is an exothermic reaction in terms of bond breaking and bond formation?
Nitrogen + Hydrogen -> Ammonia
This reaction has a $\Delta H$ of -93kJ
This reaction is exothermic. Explain what it means in terms of bond breaking and bond formation
I know an exothermic reaction produced heat to the surrounding.
Energy graph for exothermic reactions

It shows that reactants have higher energy. Enthalpy change (∆H) is probably negative because when we calculate we say ΔH = Hproducts – Hreactants then we get a negative number.
When I google I find that exothermic reactions are bond formation reactions. How is an exorthermic reaction bond formation. Does it have any ting to do with enthalpy change? Do we also have bond breaking in exothermic reactions.

Well, it means just that: we break some bonds and form some stronger bonds.

∆ always designates current - previous, so in case of a reaction: property of product - property of reactant (when you are making products). You have it the other way around for ∆H.

@Mithoron enthaly wasn't my main question it just popped up while I was typing

@IvanNeretin I think this has something to do with what i am looking for

Maybe take a look at this one: https://chemistry.stackexchange.com/questions/117923/i-cant-find-reaction-enthalpy-%e2%88%86h-for-water-formation-reaction/117926#117926

@KarstenTheis so I read in your answer you said Breaking bonds costs (positive sign), making bonds gains (negative sign), now can I say exorthermic reactions have higher bond formation energies than bond breaking energies

Yes, Bret. I do think @IvanNeretin said it more elegantly in the very first comment.

Exothermic reactions are those reactions in which heat is evolved. If we are breaking a bond we need to energy to break it so those are endothermic reactions. But when a bond gets formed energy gets liberated those are known as exothermic reactions.
In the formation of ammonia three nitrogen - hydrogen bonds are formed which results in the form of loss of energy.
In the formation of bonds energy is released as they are getting stabilised due to bond formation so atoms used to release energy. The amount of energy released is dependent on various factors like electronegativity, electron affinity and various other factors.
This is a general trend. I don't know if there are exceptions as well please tell me if there are so so even I can know.


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Treating styrene in the atmosphere
I was actually inspired to ask this question in certain circumstances. In the outskirts of my city, there happened to be a styrene gas leak early morning today which killed thirteen and hospitalized many people.
Since styrene is an aromatic compound, it is suspected to be carcinogenic in nature but I think it is inert towards other atmospheric gases as well as many compounds in non-laboratory conditions. The temperature here is 31 °C, so it can stay for a large time unaffected at the same area without much spreading. I think so because it has a high molecular weight(104 g/mol).
My question is: "is there a way to treat this styrene in the atmosphere without causing much more toxicity in that area?" (For information regarding its toxicity, refer to this safety data sheet). Of course, I understand that styrene is broken down within 1-2 days in the atmosphere naturally but is there any other way to do it quicker?

Regional government, those who are responsible for safekeeping people's life must do the needful. How will you be able to help? Not sure if it's a medical advice question, I hope others will answer..

@Zenix I just want to know how to treat such compounds in such circumstances

Please make the initials capital, 'i' as I, put proper space before opening bracket, and if you can't, accept others edit proposals. Removing newspaper article won't change the question motive much.

Good for you for trying to ameliorate a problem, whether theoretically, or to suggest remedies for individuals or the area as a whole.

@JamesGaidis thank you for your encouraging comment. I 'd just like to know, how to treat styrene in the atmosphere.

Spraying another compound to neutralize styrene (which is difficult) or whatever it was is probably not helpful, as it takes too long to spread and other potentially harmful materials are introduced, as one would need to apply the substance in excess.
Having readily installed outlets for water curtains at the plant might be helpful. The quickest emergency measure would have been if every household in the area was equipped with sufficient amounts of respirators and suitable absorbers ("filters") and training the people how to use them correctly (and teaching them that the absorbers have a limited capacity) might give enough time to escape. This would be a good and rather cheap provision for the future. Otherwise ventilation to dilute the materials is the only thing one can do, I'm afraid.

Good idea, but provided that the solubility of styrene is low in water (just 0.03%) how do you think it is useful to a significant extent? Of course, spraying water was what is being done, but i don't understand how it is useful. Is there any other way which employs lesser water as it's summer and water crises may arise.

Spraying water was just a suggestion to aid in the precipitation of the substances that have been released. Wondering if it was only styrene, as most of it should have been escaped as liquid, because of its high boiling point (nevertheless, it smells awful). In any case, spraying water should reduce the explosion hazard that is going along with the liberation of large amounts of combustible chemicals. Water crisis or not, anything that helps in saving the health of the people who are living around the plant should be considered beneficial. It would be done for a short time only.

Could you tell what sorts of reagents are useful to counter styrene gas. Styrene was heated in the plant for polymerisation, yet even other compounds are suspected to have leaked as just styrene alone cannot spread in a radius of nearly 5 km

I wonder if UV light could be used to activate the styrene and cause it to oxidize faster. UV is being used to clean subway trains in NY city (coronavirus). Could the water sprays be laced with H2O2 (just a little) to give some radical attack?

That is a clever idea. I would hypothesize that the concentration of the styrene gas would be too low to induce polymerization; however, the presence of the aqueous peroxide may convert the styrene to the 1-phenylethanol.


text:
Question related to Hydration of Alkynes
There will be Markovnikov addition of water and hence —OH will be added to the carbon with triple bond having less number of hydrogen and —H will be added to the one having more number of hydrogen. Then the adduct will tautomerise to form the product corresponding to option A but the answer given is option B. Where am I going wrong?

I think it is due to the greater intensity of negative inductive effect at beta carbon wrt ketone. And as there is no possibility of conjugation to take place,  inductive effect should dominate.

If indeed B is the answer, think about the carbonyl oxygen reacting with the mercurinium ion. [This is a terrible question. It should read,"Why is B formed and not A or why does B predominate over A?"]

BTW, there is water in this reaction.

The mercuration of acetylene 1 leads to mercurinium ion 2.3 Participation of the carbonyl oxygen via a five-membered transition state with hydration affords 3. Protonation of this intermediate leads to aldehyde 4, which undergoes exchange of mercury for hydrogen giving the product 5. This mechanism is based on the observations of Stork and Borch1,2 who observed that hydration of 6 involves carbonyl oxygen participation via a six-membered transition state affording a 1,5-diketone. In the case of 7, a five-membered transition state provides a 1,4-diketone.

1) G. Stork and R. Borch, J. Am. Chem. Soc., 1964, 86, 935.
2) G. Stork and R. Borch, J. Am. Chem. Soc., 1964, 86, 936.
3) For 5-endo-dig cyclizations, see: M. A. Sajid, et al., Molecular Diversity, 2020, 24, 295.

(+1) So, Cl got no major role in product formation?

No! I suspect it may be a madeup problem or unpublished. A Chem. Abst. search did not find the reaction with or without chlorine.

Due to -I effect from the carbonyl group and the halogen, not far enough for it to be negligible, the stability of the primary vinylic carbocation is counter-intuitively more than that of the secondary vinylic carbocation. Markonikov rule is applicable in absence of permanent electron shift effects, like resonance and inductive effects. The Markonikov rule, under these restrictions, translates into the stability of the carbocation, decided by the positive inductive effect of neighbouring carbons. As a clearer example, consider the hydration of acrolein,
The reaction predominantly yields the primary alcohol, as opposed to the product predicted by the markonikov rule, which again is not applicable in presence of permanent electronic effects. Hope that helps!