text:
Comparing basicity of imidazole and 2-imidazoline

I recently sat for an exam in which I was asked to compare the basicity of the above compounds. I thought imidazole would be more basic because it is aromatic but according to the answer key 2-imidazoline is more basic. The reason provided is that the lone pair on imidazole is a part of aromatic system and hence it's not basic.
Shouldn't the other lone pair which is on nitrogen double-bonded to carbon be basic? After all it's not involved in resonance due to being out of plane of the pi-system. What is the real reason behind higher basicity of 2-imidazoline (if it is really more basic)?

Are you sure you read the question correctly? Are we just talking about basicity or the basicity of the nitrogen with the hydrogen on it?

The question asks to compare the basicity of the compunds. It doesn't mention any particular atom/element. I have just checked it again :)

The sp2 carbon should be more electron withdrawing, so I am inclined to agree with your answer without more context.

related https://chemistry.stackexchange.com/questions/18906/in-which-direction-does-imidazole-affect-the-ph https://chemistry.stackexchange.com/questions/5999/is-there-a-difference-between-imidazolidinone-and-imidazolidone

@Zhe thanks for putting this important concept on sp2 carbon

I was unable to find experimental $\mathrm{p}K_\mathrm{a}$ values of imidazoline. However, the tools one can use to calculate and thereby predict $\mathrm{p}K_\mathrm{a}$ values are pretty good nowadays as can be seen comparing the predicted and experimental values for imidazole.
$$\textbf{Table 1: }\text{predicted and experimental }\mathrm{p}K_\mathrm{a}\text{ values of imidazole and imidazoline}$$
$$\begin{array}{lcccc}\hline
\text{compound} & & \mathrm{p}K_\mathrm{a,1} & \mathrm{p}K_\mathrm{a,2} & \text{ref}\\ \hline
\text{imidazole} & \text{(e)} & \phantom{0}7.13 & 12.70 & [1]\\
& \text{(p)} & \phantom{0}6.97 & 13.40 & [2]\\
\text{imidazoline} & \text{(p)} & 10.17 & \text{N/A} & [2]\\ \hline\end{array}$$
Note: $\mathrm{p}K_\mathrm{a,1}$ refers to the $\mathrm{p}K_\mathrm{a}$ values of the protonated heterocycle. The higher this value, the higher the basicity of the compound in question.
Obviously, the answer set’s statement that imidazoline is more basic is correct; the question remains why that is the case.
Both compounds obviously have the same protonated form, namely it is the imino nitrogen that bears a proton and thus a formal positive charge. Both of these protonated forms are $C_\mathrm{2v}$ symmetric. At first sight, the electronics of the two are pretty much equal. Yet obviously, imidazoline’s nitrogen must experience a greater electron density for some reason.
The key difference is the size of the π system which spans five atoms in imidazole but only three in imidazoline. At first sight, it seems like imidazole should be more basic since the positive charge can be delocalised onto five atoms compared to only three in imidazoline. However, this explanation fails to acknowledge that all π electrons are delocalised across more atoms in imidazole’s case.
With the π system being larger and spanning more atoms, we expect less electron density to be concentrated on each of the two nitrogens. On the other hand, the much smaller π system of imidazoline means that a greater electron density can be found on both nitrogen atoms at any time.
Further effects, such as the greater electronegativity of $\mathrm{sp^2}$ carbon atoms or the $+I$ effect of $\ce{C-H}$ bonds should be neglegible.

References:
[1]: The Evans $\mathrm{p}K_\mathrm{a}$ table
[2]: Calculated by ChemAxon’s Chemicalize service.

I agree with your assessment, but I think the focus is really on the lone pair on the "pyrrole-type" nitrogen (i.e. the singly bonded N). This lone pair isn't the one getting protonated, but it is crucial to stabilisation of the conjugate acid via resonance. In imidazole it is more delocalised (for reasons you stated - more specifically because that nitrogen releases electrons into the π system - if it withdrew electrons then there would be greater electron density on it) and hence contributes to stabilisation less.

The answer is right and the explaination given is also perfectly right and you're interpretation of which nitrogen is basic is also right. :)
To understand the answer, you need to first understand how nitrogen works here. After the double bonded nitrogen in either of these compounds is protonated it acquires a positive charge which is stabilized by the lone pair of the non double bond nitrogen which is in conjugation with it.
Now in imadozole this lone pair (non double bond N) is also part of the aromatic sextet. Hence it's availablity to stabilize the other nitrogen in the conjugate acid is greatly diminished. This is not the case in imadoziline where the line pair on N is localised (not part of aromaticity).
So what the answer key refers to as lone pair in aromaticity is the lone pair on the non basic nitrogen (non double bond N).
pKa of imadozole= 7.1
pKa of imadziline=11

Following this discussion which touched on concepts regarding factors influencing the stability of bases (and leaving groups-though off topic for this question though :)), and incorporating @Zhe's comment above: In order to determine strength of these bases, there are things worth noting in this case:
When evaluating the basicity of a nitrogen-containing organic functional group, the central question we need to take note of is: how reactive (and thus how basic) is the lone pair on the nitrogen? In other words, how much does that lone pair want to break away from the nitrogen nucleus and form a new bond with a hydrogen?
However the corresponding structure is conserved in both compounds and hence cannot be further used to segregate these compounds with regards to basicity
(Note: I have put this statement so that it clears doubts)
In this case the possible factors include:

Aromaticity and resonance
Rings do not necessarily need to be 6-membered in order to have six π electrons. In this case imidizole, has aromatic 5-membered rings with six π electrons (Hückel rule).

In both of these molecules one nitrogen atom is $\mathrm{sp^2}$-hybridized. In imidazole, one lone pair occupies a $\mathrm{2p}_z$ orbital and is part of the aromatic sextet, while the second occupies one of the $\mathrm{sp^2}$ orbitals and is not part of the sextet.  However, imidazoline has no sextet (no extra 2 electrons to complete aromaticity)
This means that for imidazole, these electrons are very stable right where they are (in the aromatic system), and are much less available for bonding to a proton (and if they do pick up a proton, the aromatic system is destroyed).  For these reasons, imidazole is not strongly basic
Resonance effects involving aromatic structures can have a dramatic influence on acidity and basicity. The delocalization of electron density has a stabilizing effect, and the greater area over which the delocalization is possible, the greater the stabilisation
The situation is slightly different in imidazoline, as it has one lacking π bond and the system is not aromatic and thus no greater room for stabilisation due to delocalisation and hence more basic.
As a side note: The less the stability, the more the basic.
Carbon bond hybridisation
Another factor contributing to increased basicity of imidazoline lies in the carbon bond hydridisation. This phenomenon can be explained using orbital theory and the inductive effect: the $\mathrm{sp^2}$ orbitals of carbon bond in position 2 is one-part s and two parts p, meaning that it has about 67% s character.
On the other hand, this carbon (in same position on imidazoline) has $\mathrm{sp^3}$ orbitals, conversely, are only 25% s character (one-part s, three parts p).
Because the s atomic orbital holds electrons in a spherical shape, closer to the nucleus than a p orbital, $\mathrm{sp^2}$ hybridisation implies greater electronegative than $\mathrm{sp^3}$ hybridisation. Recall the inductive effect: more electronegative atoms absorb electron density more easily, and thus are less basic (and more acidic).

Remarks, the explanation you noted as the reason for the answer is valid as explained above, and your understanding also is correct, however is it was not enough to explain difference in basicity as other options needed to be taken into consideration (as discussed above) Thus we have sone picture on why imidazoline was chosen as more basic than imidazole.

The aromaticity and resonance explanation misses a key point. In both compounds, there are two different nitrogens. In both cases, the nitrogen with the sp2 lone pair is more basic than the one with the 2p lone pair. The explanation is correct that protonating at the nitrogen with 2p lone pair breaks the aromaticity of imidazole. However, a similar effect occurs in imidazoline: the 2p lone pair can be delocalized to the other nitrogen, and this is interrupted by protonation at this lone pair.

Thinking about it on the product side, in both cases, protonating at the sp2 lone pair allows for distributing the positive charge over both nitrogens. Since this occurs for both imidazole and imidazoline, I don't think the fact that one is aromatic and the other isn't has any substantial effect on the basicity difference.

As Rishon D'souza's answer and orthocresol's comment above state, the conjugate acid of imidazoline is more stabilized because the lone pair of the other nitrogen is not as delocalised as in the case of imidazole. Just adding an image representing the two resonance structures of imidazoline's conjugate acid for further clarity:

An easy way of justifying the fact that imidazole is less likely to protonate i.e. less basic than 2-imidazoline:
imidazole: Already aromatic
2-imidazoline: Aliphatic amidine
In the conjugate acid of imidazole there is considerable loss of aromatic character because now the pyrrole like Nitrogen atom also gets positive formal charge on it in some of the resonating structures. Such N atom now hesitates more to delocalise its electrons over entire ring. This causes higher barrier to protonation in imidazole.
No such loss in aromatic character in 2-imidazoline because it is aliphatic. Hence, low barrier to protonation in 2-imidazoline.


text:
Is graphene a combustible material?
Various sources such as https://news.ycombinator.com/item?id=7679154 speculate that graphene will combust at higher temperatures.
Is this true? Is the flash point/ temperature of combustion for graphene known?

Graphene is not a material at all, to begin with.

I would hardly consider that link a "source"...

The Wikipedia article for graphene links to a Journal Phys. Chem abstract claiming that it combusts around 350 C. I have no access to the full article with the specific details.

Graphene is a name for rather diverse materials (size, purity, concentration of defects etc can be different, occasionally stabilizers) so actual behavior may vary.

Graphene combusts at 620 K. See http://pubs.acs.org/doi/abs/10.1021/jp410044v.
It is postulated that it burns at this temperature "because of a large interlayer spacing of graphene sheets (i.e., 5.1 Å)".

As @airhuff pointed out, the Wikipedia article of graphene says that graphene combusts at 350℃. This statement is taken from another link which states  as to why graphene combusts:

it(graphene) has a high specific surface area of ca. 1168 m2 g–1 and
  starts to burn at 350 °C because of a large interlayer spacing of graphene sheets (i.e., 5.1 Å).

Graphene is made from carbon atoms, just like diamond and graphite.
Just like diamond and graphite, it will burn in oxygen.

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text:
Does simmering sparkling water change/decrease its mineral composition significantly?
I know that heating sparkling water will cause it to lose $\ce{CO2}$.
Does this have any other effects on the water that remains, such as reducing its mineral content?  I understand when non-carbonated mineral water evaporates, its minerals stay behind and create a greater mineral concentration in the water that's left. Does sparkling water behave in the same way, or do other reactions occur that cause the water to change in additional ways?

OP asks about a fairly basic concept, but that's no reason to be snide, @Karl.  It's not out of the question that the carbonate may introduce some different reactivity as the water is heated.  Actually, the solubility of many carbonates decreases with increasing temperature, now that I think about it -- I may have to revise my answer....

No, simmering sparkling water should have negligible effect on its mineral composition.
You're exactly right that simmering it will drive off the $\ce{CO2}$. This $\ce{CO2}$ leaves as only/exactly $\ce{CO2}$ molecules, and any minerals that might have been associated with the carbonate/bicarbonate $(\ce{CO3^{2-}}/\ce{HCO3^-})$ will remain behind in the water.  Most such minerals will be present in dilute enough of concentrations that they will remain dissolved in the no-longer-sparkling water, as long as only a relatively small amount of the water itself is boiled away.

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text:
How to define the initial guess for the electron density?
This is a technical question. If I have  a one-dimensional box of length L, and an electron density existing in the box and integrating to N electrons. And if I want to determine the electron density that minimizes the energy of the system, assuming the following:

electron density is zero at the boundary
a fixed external potential of any kind and form
the Thomas-Fermi kinetic energy functional is used. $v_{TF}(x)=C^{TF}*\rho^{2/3}(x)$
absence of electron-electron interaction, $v_{HXC}(x)=0$

the question is related to the initial guess for the electron density. I've been trying the following definition for the initial $\rho(x)$
$g = \frac{l}{(gp-1.0)}$
$\rho(x)= \frac{N}{(l-g)}$
where
$g$ is the grid gap,
$gp$ is the number of grid points,
$l$ the length of the box, and
$N$ the number of electrons.
with this density, the minimized density does not preserved the number of electrons. Do you guys have any suggestion to define a suitable initial guess?
Also I am including a figure where the density at each grid point is plotted. I assumed two electrons and 200 grid points.

Thank you

Assumming an iterative process: I admit I never tried it, but I would think that after the first iteration, the number of electrons should be correct, and that any problems in that direction point to a problem with the iterative process, not the initial guess. In my DFT experience, I have come across qualitatively wrong solutions due to a bad guess, but never wrong electron numbers.

It's not clear to me what system are you modelling. No electron-electron interaction includes no electrostatics? What about exchange interactions which are inherent to fermions?

This is a a simple training problem where you can applied any potential. For simplicity both exchange and coulomb are assumed to be zero.
Also I added the resulting density at each grid point, it is no clear for me why the plot gets that concavity. Or maybe is correct but I am not seeing it. Thanks for your comments.

Have you tried various values of l? It seems to me that the density in one dimension is trying to escape the boundaries for being in a confined box.

Is your plot of your initial guess? Then your definition of $\rho(x)$ looks strange to me:
$$\rho(x) = \frac{N}{l - g}\text{.}$$
If $N$, $l$ and $g$ are constants, where is the dependency on $x$? Your graph must be wrong. Or is it your final density that's plotted there?

Besides, can you show us your functional? It should contain a Lagrange term for conservation of electrons, e.g.:
$$\min_\rho C^{TF} \rho^{\frac{2}{3}}{\left (x \right )} \\text{subject to} \int_{0}^{l} \rho{\left (x \right )}, dx = N$$
is equivalent to
$$\min_\rho C^{TF} \rho^{\frac{2}{3}}{\left (x \right )} - \lambda \left(\int_{0}^{l} \rho{\left (x \right )}, dx - N\right)\text{,}$$
where $\lambda$ is a new parameter to optimise.


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How do the numbers on the pH scale compare?
How do the numbers on the pH scale compare?
Example – is a pH of 4 twice as strong as a pH of 2? Remember the pH scale is not linear! Also, "is 4 twice of 2" isn't my question. I just want to know the way the numbers compare to each other.

They don't. There is no such thing as "twice as strong" among acids.

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

Ever heard about logarithms?

Perhaps it's better for you to ask your own question in your own words, instead of (what seems to be) copy-pasting somebody else's question here.

I'm voting to close this question as off-topic because it is something that is typically taught in schools the minute the pH scale is introduced.

$\mathrm{pH} = -\mathrm{log}[\mathrm{H}^+]$
i.e. the pH stands for the negative log (base 10) of the concentration of H$^+$.
this means in a solution of HCl (0.1 mol dm$^{-3}$ ) where the HCl is fully dissociated, i.e. $\ce{HCl -> H+ + Cl-}$ the concentration of H$^+$ is 0.1 mol dm$^{-3}$, so $\mathrm{pH} = -\mathrm{log}(0.1) = 1$
if the concentration of $\ce{H+}$ is 10 times greater, i.e. 1 mol dm$^{-3}$, so $\mathrm{pH} = -\mathrm{log}(1) = 0$

The pH scale tells you how acid is a solution, or more formally the concentration of hydronium ion ($\ce{H3O+}$) you have in the solution. Remember that
$pH = - \log([\ce{H3O+}])$
therefore the concentration of $[\ce{H3O+}]$ in a solution with $pH = 4$ will be two orders of magnitude lower than the one with $pH = 2$.
If you do the $Antilog$ of -2 and -4 you will have $\pu{1E-2}$ and $\pu{1E-4}$ respectively.
So the higher the pH the less concentration of $[\ce{H3O+}]$ and less acid (or more alkaline).

For future reference, you may find https://chemistry.meta.stackexchange.com/questions/86/how-can-i-format-math-chemistry-expressions-here/88 and https://mhchem.github.io/MathJax-mhchem/ helpful for formatting questions and answers on ChemSE (and, partially, LaTeX, too).

Oh, and in addition to @Buttonwood, don't abuse MathJax for emphasizing text. Use *italic* for italic and **bold** for bold.

The pH is a measure of:
 $$\mathrm{pH} = -\mathrm{log}[\mathrm{H}^+]$$
It's a logarithm, for being more exactly it's a common logarithm (his base is 10).
In other worlds, this means that each pH is 10 times less "powerful" or acid that the last (it's one order of magnitude lesser).
For example:

$\mathrm{pH}=\mathrm{4}$ is 10 times more acid than $\mathrm{pH}=\mathrm{5}$, or it's one order of magnitude (x10 times) higher.

It can be some strange to think, but, higher numbers are not more acids, it's the opposite, higher numbers are less acids and lower numbers are more acids.
I hope this could help.

Why the down-vote?

Base 10 logs are called "common logs". Base $e$ logs are the "natural logs".

@PrittBalagopal Ups, sorry, I'll fix that.

And I shall remove my downvote :)


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What is the systematic name of ammonium bisoxalato cuprate (II)?
What is the systematic name of ammonium bisoxalato cuprate(II)?
I already tried $\ce{(NH4)2[Cu(C2O4)2]^{2+}}$ but i am uncertain if that answer is correct.

The systematic name of ammonium bisoxalatocuprate(II) is ammonium bisoxalatocuprate(II). Where is your question?

Did you mean to ask "what is the chemical formula of ammonium bisoxalatocuprate(II)", yes? If so, the compound is a salt and should be chargeless.
$$\ce{(NH4)2[Cu(C2O4)]}$$
This is because oxalate is -2 charged and there are two of them making a total of -4, copper is +2 charged, and 2 ammoniums make a total of +2 charge. All of these charges cancel.


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Terms related to the structure of atom
In my chemistry textbook, the chapter "Structure of Atom" deals with the old Bohr's model and the new Quantum mechanical model. But when I read the contents of the new model of atom, there were terms like energy levels, shells being repeated. I failed to understand the context. Moreover, Bohr's model has been discarded so why do we still talk about energy levels/ orbits in an atom?
Basically, I want a satisfactory explanation of these terms: energy levels, orbits, sub orbits, shells, sub-shells, orbitals and principal quantum number in accordance with the Quantum mechanical model of the atom.
I tried to understand this from two other books but there too the distinction was vague.
I also read the contents of:
Is there a difference between energy levels and electron shells?
https://www.khanacademy.org/science/chemistry/electronic-structure-of-atoms#orbitals-and-electrons
PS: Please provide an answer keeping in mind that the questioner is a Grade 11 student.

There are no orbits, suborbits, and subshells in QM.

What is the problem with the common QM definition of shell, subshell etc?

So, does the book not provide definitions of those terms? Or are you confused about those terms?

https://chemistry.stackexchange.com/questions/32860/quantum-mechanical-model-of-atom-and-the-quantum-numbers https://chemistry.stackexchange.com/questions/18466/difference-between-shells-subshells-and-orbitals

@Greg I can't tell if that comment is directed at my comment or the OP.

Possible duplicate of Difference between shells, subshells and orbitals

@JonCuster both


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Why does the Michaelis-Menten constant remain constant in the event of non-competitive enzyme inhibition?
A while back, I "answered" the question: Why does the Michaelis-Menten constant decrease in the presence of non-competitive inhibitor?
At the time of posting my answer, both the OP and I were under the impression that non-competitive inhibition reduces the Michaelis-Menten constant ($K_\mathrm{M}$).
However, the Wikipedia article on "Enzyme inhibitors" states that $K_\mathrm{M}$ remains constant in the event of non-competitive inhibition. (second emphasis mine)

In non-competitive inhibition, the binding of the inhibitor to the enzyme reduces its activity but does not affect the binding of substrate. As a result, the extent of inhibition depends only on the concentration of the inhibitor. $V_{max}$ will decrease due to the inability for the reaction to proceed as efficiently, but $K_m$ will remain the same as the actual binding of the substrate, by definition, will still function properly.

But this goes against what I thought earlier (see my answer to the linked question). Wikipedia defines $K_\mathrm{M}$ as:

The Michaelis constant $K_\mathrm{M}$ is the substrate concentration at which the reaction rate is half of $V_\mathrm{max}$. Here, $V_\mathrm{max}$ represents the maximum rate achieved by the system, at saturating substrate concentration.

Wikipedia attributes the invariance of $K_\mathrm{M}$ in non-competitive inhibition to the fact that the actual binding of substrate to an active site is unaffected. Now I agree that the actual binding of the substrate is unaffected here, but how does that have any effect on $K_\mathrm{M}$?
When $V_\mathrm{max}$ decreases during non-competitive inhibition (the only point I and Wikipedia concur on), shouldn't the $K_\mathrm{M}$ decrease correspondingly (based on the definition of $K_\mathrm{M}$)? Is the Wikipedia article incorrect in this regard?

It depends on whether you define relationships based on the uninhibited values and then with inhibition, some of these are held constant. For example $\mathrm{V}{\mathrm{max}}$ here changes, but maybe $\mathrm{K}{M}$ is based on the uninhibited value of $\mathrm{V}_{\mathrm{max}}$?

This is a question that is relatively easy to answer mathematically.
To keep the algebra a little simpler, let me use a simplified Michaelis-Menten mechanism, where we assume that the enzyme-substrate complex (ES) and the free enzyme are in equilibrium.  (We could use the less restrictive pseudo-steady state hypothesis on (ES) and obtain the same result for non-competitive inhibition, at the expense of more complicated math.)
$$\ce{E + S <=>[K_M] (ES) ->[k3] E + P}$$
Non-competitive inhibition means that the inhibitor doesn't care at all whether the substrate is bound to the enzyme.  Either on (ES) or on (E), it will bind with affinity $\mathrm K_{\mathrm i}$.  We assume that (ESI) does not react, at all, to form products.  So the only other two reactions we need are:
\begin{align}
\ce{E + I &<=>[K_i] (EI)}\\
\ce{(ES) + I &<=>[K_i] (ESI)}
\end{align}
Now to get the rate equation, we first perform a mass balance on enzyme:
$E_0 = E + ES + EI + ESI$
We use the equilibria we've assumed above to get:
\begin{align}
(ES) &= \frac{(E)(S)}{K_M}\\
(EI) &= \frac{(E)(I)}{K_i}\\
(ESI) &= \frac{(E)(S)(I)}{K_M K_i}
\end{align}
Thus
\begin{align}
E_0 &= E\Big(1 + \frac{(S)}{K_m} + \frac{(I)}{K_i} + \frac{(S)(I)}{K_M K_i} \Big)\Big)\\
E &= \frac{E_0}{\Big(1 + \frac{(S)}{K_m} + \frac{(I)}{K_i} + \frac{(S)(I)}{K_M K_i} \Big)}
\end{align}
The rate equation we are after is just the formation of product, $\frac{dP}{dt}=k_3 (ES)$.
$$k_3 (ES) = k_3 \frac{(E)(S)}{K_M}$$
Now we know $(E)$ from our enzyme mass balance, so
\begin{align}
\frac{\mathrm{d}P}{\mathrm{d}t} &= k_3 E_0 \frac{(S)}{K_M \Big(1 + \frac{(S)}{K_m} + \frac{(I)}{K_i} + \frac{(S)(I)}{K_M K_i} \Big)}\\
&=k_3 E_0 \frac{S}{K_M + S + \frac{I}{K_i}\left(K_M + (S)\right)}\\
&=k_3 E_0 \frac{S}{(K_M + S)\cdot (1 +\frac{(I)}{K_i})}\\
&=\frac{k_3 E_0}{(1 +\frac{I}{K_i})} \frac{S}{(K_M + S)}
\end{align}
The only thing different about this expression is that the "$V_{max}$" part of the expression, which would have been $k_3 E_0$ for a simple reaction with no inhibitor, is now $\frac{k_3 E_0}{1 + I/K_i}$.  In other words, $V_{max}$ is the only parameter changed by the inhibition mechanism we have assumed.
The apparent $K_M$ of the non-competitively inhibited reaction is still $K_M$, because if you plug in $S = K_M$, you get a rate that is half of the maximum (of the inhibited reaction).

Wikipedia attributes the invariance of  $K_M$ in non-competitive inhibition to the fact that the actual binding of substrate to an active site is unaffected. Now I agree that the actual binding of the substrate is unaffected here...but how does that have any effect on $K_M$?

You can think of $K_M$ as an apparent equilibrium constant for the binding of substrate to free enzyme.  So if something doesn't affect substrate binding, it probably won't affect $K_M$.

So, when $V_{max}$ decreases during non-competitive inhibition (the only point I and Wikipedia concur on), shouldn't the $K_M$ decrease correspondingly (based on the definition of $K_M$? Is the Wikipedia article incorrect in this regard?

No, Wikipedia is correct.  Enzymologists have adopted very confusing terminology for no good reason, but non-competitive inhibition is not the same uncompetitive inhibition.  In uncompetitive inhibition, both apparent $K_M$ and apparent $V_{max}$ are changed.
If I gave you two vials, A and B, of the same enzyme, and one of them had been pre-mixed with a high concentration of a non-competitive inhibitor, and I asked you "what's the $K_M$ you measure for each of these two different enzyme preparations?,"   then you'd get the same number for each vial.  If 5 mM of substrate is the concentration at which enzyme from vial A reaches half of its maximum rate, then you'll find that 5 mM of substrate is also the concentration where enzyme from vial B reaches half of its maximum rate.

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text:
How to dissolve slime jelly?
Suppose I have a small kid that had two cans of slime jelly (a Jelly made of water, glycerin, thickeners, dispersant, flavorings and pigments).
And that two cans were inside a backpack, and somehow opened and spread in the bottom. Involving some dolls, small toy cars, entrancing in some holes. Even after trying physical processes not all jelly could be removed.
Adding water and soap did not difference, as jelly is still there, and even the two colors haven't mixed.
How can I dissolve those jelly, preferably leaving the nylon backpack, the metal cars and the plastic dolls without significant damage?

I like the surreptitious  "Suppose I have a small kid that had two cans of slime jelly..." :D

I'd put everything (perhaps not the toy cars) in the washing machine, use a long program without heating, and let nature take it's course. No detergents, just water and mechanical agitation.

@paracetamol oh, I would end my question stating that "bonus points" would be making that kid happy again with her backpack. But I don't have enough reputation to offer a bounty... so I forgot to end it that way.

It may help to have the particular brand of slime jelly. The thing that's causing you the issue is the thickener/polymer being used, not the water/glycerin/pigments. Knowing the exact chemical makeup of the polymer doing the gelling would assist greatly in figuring out how to get rid of it. -- I suggest mentioning the brand, because that's our only lead into the identity of the polymer, if it's not listed explicitly on the container.

@R.M. It's a small brand from Brazil, manufactured by Asca Toys (www.ascatoys.com). It's called "Amoeba". There's no information in the package/site besides that the patent involves cellulose. Now I'm trying with some bleaching product, those that don't use chlorine.


text:
Does the mechanism of the Schmidt reaction involve a nitrene as an intermediate?
Per the mechanism described here the reaction does not involve nitrene as an intermediate.(EDIT-1---I initially thought the written mechanism described so, but as Zhe points out,I was wrong in my observation.)
But, can anybody shed some light on why the reaction does not involve nitrene; though it possibly can?

The mechanism is similar to that of a Curtius, for which there have been a lot more mechanistic studies. The reaction is thought to be more concerted, avoiding the intermediate nitrene (which although proposed, has never been observed).

I don't think the linked mechanism is quite correct

Nitrenes are higher energy species; why make one when you can have a concerted migration that presumably proceeds via a lower energy pathway?

Nowhere in that link is there a nitrene formed. All steps are concerted, that is, you see that alkyl shift happens at the same time as dinitrogen fragmentation.
So indeed, nitrenes are not intermediates. I suspect that mechanistic studies have been done. You might consider looking at this paper.


text:
Weighing by difference
When preparing a known mass of something (e.g. when producing a standard solution) it is often that in my school lab that we first use a weighing boat to measure the mass and then once transferred into a beaker we reweigh the boat and calculate the mass transferred.

Do we this even though we zero the scale anyway?

Now to my knowledge would this not increase the uncertainty as instead of just using the beaker on the scale and adding the mass directly which would lead to one uncertainty we instead weigh it twice which would thus have 2x the uncertainty in the scale.

Could somebody explain where my reasoning is wrong and why this method is more accurate?

Beakers are heavy. Lots of balances can't handle that much weight. Also what if your beaker had stuff in it already and you added to too much?

Also, the weight of the container may change, e.g. because two similar pieces of filter paper may weigh slightly different amounts, or the same beaker, though used again, could have a drop of water or precipitate clinging to it after washing, or a small chip removed.

Would the problem proposed by DrMoishe not be solved by zeroing the apparatus since we are only using it once?

Although there are several sources of weighing error that would be minimized by using the weighing by difference method (as discussed in the comments), I want to illustrate what I see as the limiting problem, that of the relative uncertainty of the measurement, which roughly scales (no pun intended, really) with the mass being measured.
Lets use a bit of an exaggerated situation for illustration. In particular, we'll say that we have one balance that can measure down to $\pu{0.001g}$, and can handle up to $\pu{100g}$ total mass. In reality you would likely have a high-mass balance for weighing directly into the beaker, or a low-mass balance for using the weighing boat. We are just using one really good balance to simplify this illustration, but the concept is the same either way.
Lets say our beaker has a mass of $\pu{50g}$, our weighing boat is $\pu{1g}$, and we are to measure out $\pu{1g}$ of material.
The beaker plus the material to be measured will then have a measured mass of ~$\pu{51.00 +/- 0.05g}$. This means that the relative uncertainty in the measured mass of the material is no better than $\pu{5\%}$($\pu{0.05g}$ out of $\pu{1g}$).
The weighing boat plus the material to be measured out will have a measured mass of ~$\pu{2.000 +/- 0.002g}$. This means that the relative uncertainty in the measured mass of the material is $\pu{0.2\%}$ ($\pu{0.002g}$ out of $\pu{1g}$).
So, even with two measurements rather than one, using the weighing boat method results in much less total uncertainty in the measured mass of your material as compared to weighing a small quantity of material directly into a large beaker.

This assumes that the balance error scales linearly as the total mass. Quite reasonable though you may want to justify.

Where do the uncertainty values come from? does the uncertainty in the mass increase the uncertainty in the measurement?

Most chemists in the lab I have seen so far (caution: sample distribution may be skewed) usually do not reweigh whatever they weighed into after transferring into the reaction vessel or beaker. For most substances, this is justifyable since you can see if significant bits remain in the boat or weighing device you used. Also, typically the mass remaining in the boat will be in the dimension of $\pu{0.1 mg}$ or less and thus not already in the same dimension of the scale’s uncertainty. In rare cases, especially when I see lots of a substance remaining, I will reweigh the boat or weighing vessel, though.
As to the question of weighing into a boat or the beaker this is more complicated. For one, the uncertainty usually increases roughly linearly with the weight of the sample measured. Many scales in the lab I’m working in now have five sub-gram digits but once the total weight is larger than eighty grams or so, the final digit is hidden. The smaller and lighter the vessel you use, the less it affects the total weight on the scales and thus also the less inaccurate your weighing is.
Weighing directly into the beaker or flask can also have practical disadvantages. If — what is often the case in synthesis — you need to weigh more than one compound, you should keep them separate in the weighing step since you will probably have to add and remove to get to the mass you want to have. You naturally cannot remove a pure sample from a mixture (‘so which of the white crystals am I taking out now?’). Furthermore, these reaction vessels typically allow for a somewhat easy addition of solids into them, but make it very hard to remove a bit of solid from them, e.g. because you can’t access the solids with a spatula through the opening. These problems come on top of the scaling uncertainty, making weighing directly into a flask more and more unsuitable.


text:
What are some materials that expand a lot proportionally to heat?
There's never any source on the internet that can simply give me a list of materials that have the kinds of properties I'm looking for. I'm wondering what materials, like for a solid block of something, in general expand either a lot or at least very quickly when exposed to heat. I also can't remember what property this is exactly, but I think it's the coefficient of thermal expansion.
Am I looking for something with a high coefficient or low coefficient to achieve this?

Dry ice would vaporize and expand quickly and it's practical to use. Liquids such as hydrazine expand very quick but are dangerous to use. I'm not of the co efficient

Dry ice is an interesting idea, but I'm looking for something more stable like at room temperature.

Anything that is stable at room temperature most likely will not expand at a fast rate (violently) like what your describing. Hence the stability factor. Most materials will vaporize and expand at some rate when heat is applied tho. Some may require a   lot of heat , some only a bit.

Right, but I'm telling you right now, I'm looking for "solid materials," something with which to construct something else. Last time I checked, you can't build a machine out of glycerin or ethanol.

Oh you want to build a machine! I thought you were reffering to solid phase (solid,liquid,gas) not specifically solid enough to build a machine. While there are plenty of solids that will expand quickly  when heat is applied I'm not sure about a material solid enough to build a machine . There must be something that acts like that tho maybe a resin,plastic or fibreglass with explosive properties that could be molded

You are looking for thermal expansion and its coefficient.  Of course such tables, and diagrams exist

(source)
as it is of relevance in daily live.  Thermometers based on liquid mercury, or alcohol, depend on this.  Bridges take this into account in expansion joints, too:

(source)
Joining two materials of different expansion coefficient equally is of relevance for the design of bimetallic stripes, found not only in thermometers and mechanical clocks:

(source)
So in reply to your "there is no source available" ...

For solids you are looking for the linear thermal expansion coefficient $α_L$:
$$α_L = \frac{\mathrm{d}L}{L\cdot \mathrm{d}T}$$
where $L$ is length and can be uniform for isotropic materials (amorphous compounds, some polymers) and varied for anisotropic materials (crystals).
For crystalline materials $L$ is often given for each axis $a, b, c$ of a crystal's unit cell resulting in direction-dependent thermal expansion.
Obviously, the higher the $α_L$, the higher the value of expansion $(\mathrm{d}L/L)$ per temperature change $\mathrm{d}T$, so the expansion of the solid is directly proportional to $α_L$.
The Engineering ToolBox provides a decent list of thermal expansion coefficients for common solids at NTP.
Looking for the materials with high thermal expansion coefficients, you probably have to look at polymers such as ethylene ethyl acrylate $(α_L = \pu{205e-6 K-1})$, ethylene vinyl acetate $(α_L = \pu{180e-6 K-1})$ or polyethylene $(α_L = \pu{108e-6 K-1}~\text{to}~\pu{200e-6 K-1})$.
Paraffin wax also has high $α_L = \pu{106e-6 K-1}~\text{to}~\pu{480e-6 K-1}$ (depends on exact composition) and expands drastically when reaches its melting point.
Among crystalline matter, the current record is held by silver(I) hexacyanocobaltate(III), $\ce{Ag3[Co(CN)6]}$ [1].
Due to mobile layer of silver atoms in between $\ce{Co(CN)6}$ sheets the crystal lattice can fold in one direction and stretch in another.
Such high anisotropy results in $α_a = \pu{130e-6 K-1}~\text{to}~\pu{150e-6 K-1}$ and $α_c = \pu{-130e-6 K-1}~\text{to}~\pu{-120e-6 K-1}$.
This is the record values for both negative and positive thermal expansion in crystals.
References

Goodwin, A. L.; Calleja, M.; Conterio, M. J.; Dove, M. T.; Evans, J. S. O.; Keen, D. A.; Peters, L.; Tucker, M. G. Colossal Positive and Negative Thermal Expansion in the Framework Material $\ce{Ag3[Co(CN)6]}$. Science 2008, 319 (5864), 794–797. https://doi.org/10.1126/science.1151442.

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


text:
Why doesn't chocolate ever feel cold?
I noticed that every time I put chocolate in the freezer, then I touch it or eat it, it somehow doesn't feel cold at all.
I figured that it must be that it heats up extraordinarily fast from body heat, because if it had a super high coefficient of heat then it would be just as hard to warm it up as it would appear to get cold.
What property would give chocolate that capacity? I've never seen any other substance be put in a freezer and somehow not feel cold.

I like to freeze my choc to make it last longer - it certainly does feel colder than if refrigerated & takes much longer to melt (my main reason to freeze it!)

One possibility is that chocolate has a lower thermal conductivity than water-based foods, which make up much of the food in our freezers.
If you open your freezer and touch a cardboard food container, then touch a metal ice tray, which one feels colder? They are both the same temperature, but the metal tray feels colder because it conducts heat better than the cardboard box, so the heat gets transferred out of your hand quicker and it feels colder.
Water is a relatively good conductor of heat and is present in large quantities in many foods. Chocolate is largely made up of fatty organics, which do not conduct heat as well. It is thus very possible that chocolate has a lower thermal conductivity than many frozen foods and therefore doesn't feel as cold to the touch in comparison.

that's what I was thinking...the chocolate does not not feel cold; it does not feel cold enough quick enough.

So somehow it doesn't conduct heat well, yet it melts exceptionally easily? Those two seem mutually exclusive.

@RayOfHope With "it melts exceptionally easily" reads like referring to a low (relatively to what?) melting point.  Comparing water and copper, for example, with water with the lower melting point ($0$ vs. $\pu{1084 ^\circ{}C}$, but with copper the better heat conductivity ($401$ vs. $\pu{0.591W/(m·K)}$ (source: https://en.wikipedia.org/wiki/List_of_thermal_conductivities) is not supporting an easy causal relation between the two.  Nor that the two mutually exclude their presence, either.

@RayOfHope , as Buttonwood pointed out, even though it may have a slow transfer of heat, the melting point is low so it doesn't take much heat to get there. Just another way of looking at it.

I think this is a psychological effect rather than a physical one. Room temperature chocolate is pretty close to its melting point (about 30 $^\circ$C), so it probably noticeably softens pretty quickly when it comes out of the freezer and absorbs your body heat, which may trick your perception into thinking it is actually heating up faster than it really is. In fact, the opposite is true; as a substance melts, even though it absorbs energy, the temperature remains constant. The energy is used to re-arrange the molecules.
Edit: Upon thinking about it a little more, due to chocolate being relatively soft, perhaps when it starts to melt due to your body heat, you "squish" it and allow your body heat to penetrate deeper into the material, thus heating it evenly. Compare with ice which is much harder; your heat doesn't get into the center until the outside is melted.


text:
How to calculate Lennard-Jones potential with quantum mechanical methods
I want to know the procedure to calculate the Lennard-Jones potential for a metal-halogen pair (specifically vanadium-chlorine). Is it possible to calculate using any QM packages like Mopac, NWChem, or Gaussian?
I am specifically looking for values of A and B in the 12-6 potential.

Scan the bond length around its equilibrium and fit the energies afterwards.

@pH13-YetanotherPhilipp That won't work, since it will incorporate all interactions up to the method level (electrostatics, polarization, exchange repulsion, dispersion, and so on). The LJ potential should only be used to approximate exchange repulsion and dispersion.

@pentavalentcarbon Do you have any link/source to provide further information on that? I never heard of such a limitation.

@pH13-YetanotherPhilipp You can tell just based on the mathematical form. The Coulomb potential goes like $1/r$, and since the gross features of interactions between atoms/molecules tend to be dominated by charge-charge interactions rather than charge-dipole/higher-order ones, I wouldn't expect $(1/r^{12})-(1/r^{6})$ to handle this well. For anything charged a Coulomb term would be necessary. https://en.wikipedia.org/wiki/Interatomic_potential

Yes, this is technically possible. A basic tutorial for this is in the excellent Psi4Numpy project, which I'll reproduce here with minor modifications. Their example fits the counterpoise-corrected MP2/aug-cc-pVDZ total interaction energy of the helium dimer.
from __future__ import print_function

import psi4

import numpy as np

import matplotlib as mpl
mpl.use('Agg')
import matplotlib.pyplot as plt

he_dimer = """
He
--
He 1 **R**
"""

distances = [2.875, 3.0, 3.125, 3.25, 3.375, 3.5, 3.75, 4.0, 4.5, 5.0, 6.0, 7.0]
energies = []
for d in distances:
    # Build a new molecule at each separation
    mol = psi4.geometry(he_dimer.replace('**R**', str(d)))

    # Compute the Counterpoise-Corrected interaction energy
    en = psi4.energy('MP2/aug-cc-pVDZ', molecule=mol, bsse_type='cp')

    # Place in a reasonable unit, Wavenumbers in this case
    en *= 219474.6

    # Append the value to our list
    energies.append(en)

print("Finished computing the potential!")

# Fit data in least-squares way to a -12, -6 polynomial
powers = [-12, -6]
x = np.power(np.array(distances).reshape(-1, 1), powers)
coeffs = np.linalg.lstsq(x, energies)[0]

# Build list of points
fpoints = np.linspace(2, 7, 50).reshape(-1, 1)
fdata = np.power(fpoints, powers)

fit_energies = np.dot(fdata, coeffs)

fig, ax = plt.subplots()
ax.set_xlim((2, 7))  # X limits
ax.set_ylim((-7, 2))  # Y limits
ax.scatter(distances, energies)  # Scatter plot of the distances/energies
ax.plot(fpoints, fit_energies)  # Fit data
ax.plot([0,10], [0,0], 'k-')  # Make a line at 0
ax.set_xlabel(r'intertomic separation ($\AA{}$)')
ax.set_ylabel(r'interaction energy ($\mathrm{cm^{-1}}$)')
fig.savefig('1b_molecule.pdf', bbox_inches='tight')

Once the set of energies is calculated, a least-squares polynomial fit is performed, giving
$$
f(x) = 6677721.45419193x^{-12} - 11394.79882998x^{-6}.
$$
This isn't a good (efficient or general) way of evaluating the fit function, but illustrates the placement of correct signs in the above code:
def f(z):
    return coeffs[0]*z**powers[0] + coeffs[1]*z**powers[1]

From the above we can see that the first and second coefficients in the polynomial fit correspond directly to $A$ and $B$, respectively. In this example, the energies are in wavenumbers ($\pu{cm^{-1}}$). Most quantum packages print in atomic units (Hartrees) and dynamics packages print in kJ/mol, which is fine, but be aware that in the above fit the units are incorporated into $A$ and $B$. Here is the resulting fit compared against the individual interaction energy calculations:

This was not in the question, but I think it's important to ask if this is the correct approach and investigate further. Notice how the fit is qualitatively ok near the minimum, but the shape in the dissociation region is qualitatively wrong, and the potential simply isn't attractive enough in the limit of infinite separation. This could arise in three ways:

The curve fitting isn't working properly, or there's a bug in the code.
The functional form of the chosen interatomic potential is incorrect.
The (Boys-Bernardi) counterpoise correction, which is known to overcompensate for BSSE, is giving strange results at large separation.

Assume that point 1 isn't an issue (if there's a bug anywhere, please comment). Point 2 can be investigated by also fitting against the Buckingham and shifted Morse potentials, which should display better short-range and long-range behavior:
\begin{align}
V_{\text{Buckingham}}(r;A,B,C) &= A e^{-Br} - \frac{C}{r^{6}} \\
V_{\text{Morse}}(r;A,B,C,D) &= A \left( 1 - e^{-B(r-C)} \right)^{2} + D
\end{align}

Note that for these energies, an unshifted Morse potential (no $D$ parameter) fails miserably. See here for more information about fitting Morse-type potentials. The Buckingham and Morse fits require a more general (arbitrary) curve fitting routine, and the script that made this plot can be found as an HTML comment in the post source code below.

I find it interesting that the Buckingham potential gives a worse fit than the Lennard-Jones one, even though it is supposed to reproduce the repulsive wall better with $e^{-r}$ rather than $\frac{1}{r^{6}}$. The Morse fit is remarkably good, which makes sense considering there are 4 free parameters rather than 3 (Buckingham) or 2 (Lennard-Jones). Before going any further, a check on point 3 by setting bsse_type=nocp:

This reveals that the counterpoise correction is definitely interfering with the quality of the fit, and the functional forms of the Lennard-Jones and Buckingham potential appear to not describe dissociation properly, though extended to infinite separation, the Morse potential is the one that is qualitatively incorrect. This can be attributed to using a single functional form to describe the total interaction or binding energy, and not the different components. I suspect that the apparent poor fit is due to a non-zero Coulomb interaction, rather than van der Waals-type interactions which at least the Lennard-Jones potential is meant to be used for.
To test this, here are interaction energy calculations at two levels of sophistication, both based on symmetry-adapted perturbation theory (SAPT). This is still the helium dimer, at 7 angstroms separation, with the aug-cc-pVDZ basis set and the monomer-centered basis approximation. All units are kcal/mol.
$$
\small
\begin{array}{lrr}
\hline
\text{Component}  & \text{SAPT0} & \text{SAPT2+3(CCD)}\delta_{\text{MP2}} \\
\hline
\text{Electrostatics} & -0.01796877 & -0.01777334 \\
\text{Exchange}       &  0.00000000 &  0.00000000 \\
\text{Induction}      &  0.01587019 &  0.01508578 \\
\text{Dispersion}     & -0.00013235 & -0.00016773 \\ \hline
\text{Electrostatics + Induction} & -0.00209858 & -0.00268756 \\ \hline
\text{Total}          & -0.00223093 & -0.00285529 \\
\hline
\end{array}
$$
At both levels of SAPT, the non-dispersive part of the interaction energy accounts for 94% of the total interaction! Is this true at 3 angstroms separation?
$$
\small
\begin{array}{lrr}
\hline
\text{Component}  & \text{SAPT0} & \text{SAPT2+3(CCD)}\delta_{\text{MP2}} \\
\hline
\text{Electrostatics} & -0.04852183 & -0.04836334 \\
\text{Exchange}       & 0.02045965 & 0.02219227 \\
\text{Induction}      & 0.04202349 & 0.04081412 \\
\text{Dispersion}     & -0.02228767 & -0.02843936 \\ \hline
\text{Electrostatics + Induction} & -0.00649834 & -0.00754922 \\ \hline
\text{Total}          & -0.00832637 & -0.01379630 \\
\hline
\end{array}
$$
The non-vdW interactions now account for 78% and 55% of the total interaction energy for each SAPT flavor, respectively. Although induction (also called polarization) is repulsive, I am including it in the Coulomb-type interaction since it is not purely quantum mechanical in nature like the exchange term. It is the exchange term, not induction, that is closer to the charge cloud penetration picture. Even so, the interaction between two helium atoms is not purely based on dispersion, and requires fitting another nonbonded term for electrostatics, the Coulomb term:
$$
V_{\text{Coulomb}}(\vec{r}_{i},\vec{r}_{j},q_{i},q_{j}) = \frac{1}{4\pi\epsilon_0} \frac{q_{i}q_{j}}{|\vec{r}_{i} - \vec{r}_{j}|}
$$
Our goal is to separate out the Coulomb-like terms from the interaction energy, so that $A$ will describe only the exchange contribution and $B$ will describe only the dispersion contribution; electrostatics and induction will be handled separately in the Coulomb term. This will be done by fitting only to the combination of exchange and dispersion, which are taken from SAPT calculations. MP2 binding energies are not CP-corrected.

As you can see, using SAPT energies provides a much better fit than naively using the total interaction or binding energies. I'm not sure why the total SAPT energy works so well, but it may be due to the absolute magnitudes at each point being so small.
The importance of only fitting exchange and dispersion is obvious for $\ce{Na^+---Cl^-}$ (using def2-SVP for the basis).

This is an unfair example, since the dispersion interaction is dwarfed by exchange, and it is an ionically-bound molecule. As a final example, consider $\ce{V(III)---Cl^-}$ (using def2-SV(P) for the basis, vanadium as a quintet).

I had some trouble converging many of the calculations, presumably due to a non-optimal spin state at long range. For this reason (spin-state crossing), Lennard-Jones type models are poor for metal-x interactions. The problem appears to be similar to the sodium chloride, but the exchange plus dispersion fit looks good.
If you want to reproduce any of the work, rather than fish through the source for inputs and scripts, everything is in a GitHub repository. More references for SAPT can be found here, along with the Psi4 manual. Note that I didn't do any literature searching here, but I expect people have used energy decomposition approaches in the past as part of force field design. Hopefully this serves as a good starting point.

This is just awesome!

All of the credit should go to the Psi4 developers; if you're a Python hacker, they have the best quantum package out there.

On second thought, I suspect that the real reason the LJ fits for most of the curves didn't work is due to a bad guess in scipy.optimize.curve_fit. This is worth investigating and would invalidate some of my argument.

Thank you soo much for this answer but I really couldn't understand how to use it. May be because I don't know how to use phyton. Is there any video tutorial on a molecule?

@BarisVvolf I really suggest learning Python. For intermolecular potentials, I recommend searching for that phrase or "Lennard-Jones"; the Wikipedia article and related ones are very good. Increasing the level of detail, Anthony Stone's "The Theory of Intermolecular Forces" is excellent. For curve fitting, YouTube has plenty. For running Psi4 from Python, YouTube! For this specific example, I could be convinced to make a video...

Thank you soo much again @pentavalentcarbon I will try your recommandations.


text:
Will a solid object lose or gain atoms on standing over time (without being acted upon by external influences)?
Suppose I'm in my room with my keyboard, my desktop and my desk. Are these objects losing or gaining atoms every second? Or are they forever stuck with the same number of particles until you scratch, break or pulverize them?
Or suppose, for simplicity's sake, I have a pure gold sphere in my room. Neglecting the presence of dust or oxide layers...would it be losing or gaining atoms over time on account of simply standing untouched in one place?

That depends on how you define those objects (spatially). Take your desk for example...how would you go about defining it? Does it just consist of wood? Or does that layer of paint on it count as a part of the desk? Do the particles of dust that cling to the paint also qualify as being part of the the desk? Do you consider the gas molecules that are adsorbed ("stuck") to the surface of the paint as a part of the desk? Moreover... where does that layer of paint event begin? Where does it end? I hope you see what I mean here... ;)

Ahahah that's a beatiful insight, you're right I'm too vague. What if I edit the question and write that I have a pure glass sphere, or gold sphere on my desk, no painting, only dust of course but it doesn't count....I'm tryng to know if a solid object of the same element or of the same solid compound will eventually lose some of its aotms without major events happening, just interacting with air...Will this be a more answerable question? ;)

I'll be editing your question to retain its original essence...it would probably help future users. Nice question by the way :D

Of significant relevance are the international prototype kilogram (IPK) and its copies. These ingots of metal were designed to define the kilogram, and thus require their mass to be as stable as possible. Even with the tremendous amount of effort in their fabrication, storage, handling and measurement, they still seem to be varying in mass over time. No one really knows why for sure.

If you have a sphere of pure gold in your room, and you advertise this fact, don't be surprised if it vanishes as soon as your back is turned. You can't fight physics! :-)

Short answer: Yes they do.

Firstly, you need to arrive at a proper definition for an "object". Why you ask? Because it would otherwise present philosophical problems which become apparent when we begin to work with those objects.

... what is an object? Philosophers are always saying, “Well, just take a chair for example.” The moment they say that, you know that they do not know what they are talking about any more. What is a chair? Well, a chair is a certain thing over there … certain?, how certain? The atoms are evaporating from it from time to time—not many atoms, but a few—dirt falls on it and gets dissolved in the paint; so to define a chair precisely, to say exactly which atoms are chair, and which atoms are air, or which atoms are dirt, or which atoms are paint that belongs to the chair is impossible. So the mass of a chair can be defined only approximately. In the same way, to define the mass of a single object is impossible, because there are not any single, left-alone objects in the world—every object is a mixture of a lot of things, so we can deal with it only as a series of approximations and idealizations.

If you can't define an object (more specifically: If you can't define its location in space) exactly...you cannot know its mass exactly. If you do not have its exact mass, you cannot make an exact/complete description (be it physical or chemical) of the body. [In fact, our ability to provide accurate descriptions of bodies is because we make good approximations]
I get what you are trying to ask in your question, and it is per se, a reasonable question; but the example (your room) is (unfortunately) difficult to deal with...the single glaring issue being the inability to define an object in an absolute sense (I'm not blaming you, in fact, no one can define any real-life object exactly :P)
Just picking out some other holes in your "room" example: Is you room sealed (can it be considered a closed system)? Is your room evacuated (I mean a "Vacuum")? Are the different "objects" present in the room in contact with each other? Are there foreign particles present on the walls, ceiling and floor? You can see where this is going...
You go on to mention using a "pure glass/gold sphere" in order to circumvent some of the difficulties of defining an "object". Unfortunately, 100% purity is inconceivable (the main issue being the dust/oxides that are present at the surface). To neglect the presence of these layers would be to forgo reality...and the question (using the "pure" spheres) would be reduced to a purely theoretical proposition.

Trying to address you question, I'll do this in two ways: first I will deal with an idealized case and secondly, I shall deal with the abomination you call your "room" (introducing a few constraints of course, to keep things simpler) :D

First: The (highly) idealized version of your example:
You have a really large ("infinitely" large, if you will, since this is a thought-experiment) sealed region of space, and this "chamber" (as we shall now call it) shall be totally isolated from its surroundings: No energy transfer, no mass transfer, no (external) gravitational field, no (external) electric or magnetic fields...no nothing. The walls of the chamber have no substance and are infinitesimally thin. The inside of the chamber is maintained at an arbitrary temperature of 25°C.
Suppose your chamber has a small amount of a mono-atomic gas (take neon for example) in it. Now you introduce an "object" in the chamber instantaneously. The "object" is a 1000 atom x 1000 atom x 1000 atom cube of absolutely pure, solid gold (and we shall assume there's no oxide layer on the surface).
Now what happens?
Simply put; the neon atoms will impinge on the gold cube and some may end up "sticking" to it. The gold atoms will slowly (like reeaaally slowly) "diffuse" out of the original block and co-exist with the neon atoms as a gas (you can think of this as the "evaporation" of the solid gold cube). You'll observe that even though neon diffuses into the gold cube, and the gold atoms diffuse into the neon gas...over a period of time there is a net diffusion of gold atoms out of the block. Keep watching, and you'll find that you have a gaseous mixture of neon and gold, with no solid gold left to speak of. Handwaving the kinetics involved here, this is an example of the Second law of Thermodynamics at play. The entropy (or the "randomness") of the system goes on increasing over time, and will continue to do so until an equilibrium condition has been achieved. {You might also want to read about the Heat Death of the Universe  <--- Interesting thought}.
Even if you didn't have the neon in the chamber (I put it in there to "simulate" the air in your room), the cube would still "evaporate" into a gas, albeit, at a slightly faster rate.

Second: Your room (with a few restrictions):
I shall proceed to define your room (albeit with a few constraints). You room is, also sealed so that no mass can be exchanged between the inside of your room and the surroundings. Everything inside your room is still subject to external gravitational, electric and magnetic fields. The temperature inside your room is 25°C. You room is not dust-free either. Your room is now set.
Start the clock...and what happens?
The results are similar (albeit, a scaled down version) to the idealized case. Every "object" in your room loses atoms/molecules and gains some too. Like before, there is a net diffusion of particles out of the objects in your room...entropy increases yet again. Leave it there for a few hundred million years, and you'll find that the objects in your room will have changed significantly (remember, the time frame was in hundreds of millions  of years).
As @Ruslan pointed out in the comments: Unlike the ideal case where the gold atoms can evaporate from the cube and continue to spread ad infinitum...your furniture and computer don't have that kind of luxury (their ability to spread out in space is limited by the walls of your cramped room) and hence won't evaporate completely (but there is evaporation nonetheless).

To understand that last bit ("incomplete" evaporation in your room, but "complete" evaporation in the idealized chamber), it will help to consider an analogy with the evaporation of water:
Here's an excellent piece (The Feynman Lectures on Physics, Volume 1, Chapter 1) you must read through before we continue:

What happens at the surface of the water? We shall now make the picture more complicated—and more realistic—by imagining that the surface is in air. The figure shows the surface of water in air.

We see the water molecules as before, forming a body of liquid water, but now we also see the surface of the water. Above the surface we find a number of things: First of all there are water molecules, as in steam. This is water vapor, which is always found above liquid water. (There is an equilibrium between the steam vapor and the water which will be described later.) In addition we find some other molecules—here two oxygen atoms stuck together by themselves, forming an oxygen molecule, there two nitrogen atoms also stuck together to make a nitrogen molecule. Air consists almost entirely of nitrogen, oxygen, some water vapor, and lesser amounts of carbon dioxide, argon, and other things. So above the water surface is the air, a gas, containing some water vapor.

Now what is happening in this picture? The molecules in the water are always jiggling around. From time to time, one on the surface happens to be hit a little harder than usual, and gets knocked away. It is hard to see that happening in the picture because it is a still picture. But we can imagine that one molecule near the surface has just been hit and is flying out, or perhaps another one has been hit and is flying out. Thus, molecule by molecule, the water disappears—it evaporates. But if we close the vessel above, after a while we shall find a large number of molecules of water amongst the air molecules. From time to time, one of these vapor molecules comes flying down to the water and gets stuck again.
So we see that what looks like a dead, uninteresting thing—a glass of water with a cover, that has been sitting there for perhaps twenty years—really contains a dynamic and interesting phenomenon which is going on all the time. To our eyes, our crude eyes, nothing is changing, but if we could see it a billion times magnified, we would see that from its own point of view it is always changing: molecules are leaving the surface, molecules are coming back.
  Why do we see no change? Because just as many molecules are leaving as are coming back! In the long run “nothing happens.” If we then take the top of the vessel off and blow the moist air away, replacing it with dry air, then the number of molecules leaving is just the same as it was before, because this depends on the jiggling of the water, but the number coming back is greatly reduced because there are so many fewer water molecules above the water. Therefore there are more going out than coming in, and the water evaporates.

Now back to where I left off; why do we see partial evaporation in your room, but total evaporation in the "infinite" chamber?
Based on the excerpt, I'll try to develop an analogy for what's going on in 1) Your room and 2) The "infinite" chamber I mentioned first.
1) Analogy for what's going on in your room:
You fill a perfectly dry bottle halfway up with water and screw on the cap. The water molecules constantly evaporate from the surface and molecules of oxygen, nitrogen, etc from the air inside the bottle constantly impinge on the water surface and may get dissolved into the water as a result. However, their ability of the water molecules to spread out in space (increasing entropy) is limited by the bottle's boundaries. Though we have an initial net evaporation of water...we're soon faced with an equilibrium state (described in the excerpt) where there is no net evaporation of water. You're still left with water that is mostly in its original (liquid) state.
It's pretty much for reasons similar to this, that the object in your room won't completely evaporate. There simply isn't enough space for the particles to "comfortably" move out in to.
2) Analogy for what's going on in the idealized case:
You pour a glass of water on the side-walk. Unlike the case of the bottle, water molecules can evaporate and expand till infinity (well, it's "infinite"  relative to the the space you had in the bottle).
It's pretty much for reasons similar to this that the gold cube in the idealized chamber will completely evaporate.There is enough space for the particles to "comfortably" occupy outside of the solid that they were initially constrained to.

So yes... over time, all the "objects" in your room lose and gain particles over time. Ultimately however, the "object" you once knew will no longer be the same >:)

There cannot be a single, simple answer to this as the answer depends on the chemistry of the objset
Whether a particular object gains or loses mass over time depends on the chemistry of the object and its environment. Chemistry matters, and there are a lot of different substances with different chemical properties, so there simply won't be a single general answer.
Even very chemically inert substances in a sealed environment–like the platinum-iridium standard objects used to define the kilogram mentioned in a comment–don't keep absolutely the same mass over time, though changes are so small they require special equipment to notice them.
Were the object made from osmium, the effect would be bigger as oxygen reacts on the surface to produce OsO4 which is volatile (it is also dangerous and smelly which is why the atomic name is derived from the greek for "smell").
Other elements might gain mass from reacting with air or its impurities to form solids on their surface. Were the object made of potassium, it might get heavier even in dry air as various oxides would form on the surface and stick to it.
Real-world compounds (with a far, far wider variety of properties) would show an even bigger range of behaviours.  Some plastics might lose mass as as the plasticisers used to make them might slowly evaporate. Other substances might gain significant mass by reacting with the atmosphere.
The main point is that generalising about the behaviour of an object is futile as there are so many possible substances to make the object from that a wide variety of totally different behaviours is possible.

authors: Bob Jarvis - Слава Україні


text:
e1 reaction products - cis/trans
The question below ... that goes through an E1 reaction ..
I have came out with the different possible products that can be formed .
I learnt that E1 reaction forms a cis and a trans product (ignore E,Z,etc for simplicity).
I'm struggling to find which of the products are cis/trans.
Product 2 has obviously no cis/trans. Its between 1 and 2 . Thanks !

I'll give you two words: steric hindrance. Use it to get the answer.

@PrittBalagopal steric hindrance explains why the reaction may go via E1 - it doesn't rationalise the nature of the product formed. Indeed, E1eliminations show little dependance upon the base used  The reaction usually occurs in the complete absence of a base or the presence of only a weak base (Wikipedia).

@NotEvans I think he's talking about allylic strain, etc. in the product.

Your issue is that cis- and trans- are better used to describe alkenes that are 1-,2-disubstituted. Otherwise, you should prefer to use (E) and (Z) notation instead.

E-Z notation should be used here. But to answer your question, the longer chain off the double bond takes the priority. In this case, it is the ethyl group. So product 1 is trans while product 3 is cis.


text:
What is the underlying chemistry behind the triboelectric effect?
I remember briefly studying the triboelectric effect in Physics but as I now approach the domain of Chemistry, I'd like to know what processes are involved (at the molecular level) when the the triboelectric effect comes into play. What variables (if any) would have have to be considered when conducting an analysis of the triboelectric effect?

If you already know the "physics" behind the triboelectric effect, then ipso facto, you know the "chemistry" involved as well (it's the same). Physics and Chemistry are human descriptions of the essentially the same thing: our universe. Chemistry looks at the finer details of things...Physics is just the "big-picture" :P

ahah you're probably right, I studied it  in physics but my lecturer  just told us that some "materials" if rubbed with others tend to gain or lose electrons, I just wanted to dig a little bit deeper. I made some reaserches but I've found no detailed explanation even wikipedia says " The triboelectric effect is not very predictable, and only broad generalizations can be made"

You would've probably heard of the commonly given example, rubbing a glass rod with silk. The glass rod becomes positively charged, and the silk becomes negatively charged.
But why? That's your question I believe, and for that we need to zoom in a little further, until we can see the molecular structure.
Glass, is a polymer mainly composed of silicates. It is a network polymer, with a bunch of $\ce{Si-O}$ groups, as evident in it's structure:

(source: mechanicalengineeringblog.com)
You can observe that some of the oxygens are negatively charged, and are countered by sodium ions.
Now, lets observe silk,
Silk, like most natural fibers, are composed of proteins. These proteins exist as $\alpha$-helices. Sericin is the outermost protein in the silk structure, so I'll show a bit of it's structure,

Now, let's get to the rubbing part!
What happens when you rub them?
You're agitating the two molecules and bringing them in contact. Loose bonds can be broken off. Consider the $\ce{N-H}$ bond in the Sericin structure. It can be easily removed as a $\ce{H+}$ ion, since the resulting negative charge can be shared with the nearby oxygen in the adjacent carbonyl group. This leaves silk with a net negative charge.
This $\ce{H+}$ can now bind to the negatively charged oxygen atoms on glass, giving the glass rod, a postive charge.

I shall end my answer with a note:
Whenever in doubt, look deeper into the structure.


text:
What is a hydrogen bond acceptor and donor?
I recently read about hydrogen bonding. One thing about which I am really confused is how to distinguish between hydrogen bond acceptors and donors.
I looked it up on SE and found this post, which said that water can form four hydrogen bonds as it has 2 hydrogen bond acceptors (the H atoms) and two H bond donors (the two lone pair of electrons on O atom).
When I looked it up on Wikipedia, it said:

In a hydrogen bond, the electronegative atom not covalently attached to the hydrogen is named proton acceptor, whereas the one covalently bound to the hydrogen is named the proton donor.

And both of them seem to be contradictory. Which is correct?

Dissenter's answer which you linked appears to be wrong (as of now). I have left a comment.
The conventional way in which we talk about hydrogen bond acceptors/donors is: in a hydrogen bond $\ce{X-H...Y}$, the hydrogen bond donor is $\ce{X-H}$ and the hydrogen bond acceptor is $\ce{Y}$. (See for example Chem LibreTexts, or this other question on hydrogen bonds.)
Of course, water is capable of acting as both an acceptor and as a donor:

I wouldn't call it a "proton donor" like Wikipedia does, though, as that suggests some kind of acid-base reaction occurring, like
$$\ce{\color{red}{H2O} + \color{blue}{H2O} <=> \color{red}{OH-} + \color{blue}{H3O+}}$$
In this reaction the red water molecule is acting as a proton donor (Brønsted acid) and the blue water molecule is acting as a proton acceptor (Brønsted base). But these terms should only be used in the context of a proton transfer, and not hydrogen bonding.

I would say it is a bit of a misnomer, saying that something is accepted or donated. A hydrogen bond is an attraction between a very positively polarised proton and a well defined lone pair, from N, O or F. When this attraction is broken the molecules separate, no protons are donated or received (overall, of course proton exchange can and does occur)


text:
Formation of α,β-unsaturated ketones
In Claisen-Schmidt condensation I've used $\ce{NaOH}$ catalyst. Can I use $\ce{KOH}$ instead of $\ce{NaOH}$?

In principle yes, though it depends on what solvent you are using. Its the OH which is important for the reaction not the counterion, though the counterion can affect the solubility.

Solvent is ethanol.

Then there should be no problem


text:
Conceptual confusion about NMR
I am quoting Clayden et al 2nd edition:

The sample is irridiated with a short pulse of radio-frequency energy. This disturbs the equilibrium balance between the two energy levels: some nuclei absorb the energy and are promoted to a higher energy level.

When the pulse finishes, the radiation given out as the nuclei fall back down to the lower energy level is detected using what is basically a sophisticated radio receiver.

Now, my confusion is about the equilibrium of nuclei in higher & lower energy level , after the introduction of radio wave.
Are the nuclei in higher level steady i.e. not radiating and coming back to the lower level constantly?
Or, are they radiating and constantly re-energized by the radio wave till the radio wave is on(still keeping a equilibrium constant between nuclei in lower and higher energy level)?

Do you understand the fundamental way in which modern Fourier transform NMR works generally, or is the whole concept new to you?

It's a new concept for me @NotEvans

I'll try to be brief. Suppose a spin exists in the state
$$\require{begingroup} \begingroup \newcommand{ket}[1]{\left|#1\right>} \ket{\psi} = c_\alpha\ket{\alpha} + c_\beta\ket{\beta}$$
where the coefficients $c_\alpha, c_\beta$ are time-dependent. When a $90^\circ$ RF pulse, with strength $B_1$, is applied about the $x$-axis, the Hamiltonian of the system becomes (in the rotating frame)
$$\hat{H} = \omega_1 \hat{I}_{\!x}; \quad \omega_1 = -\frac{B_1}{\gamma}$$
If you plug this into the time-dependent Schrodinger equation
$$\mathrm{i}\hbar \frac{\mathrm{d}\ket{\psi}}{\mathrm{d}t} = \hat{H}\ket{\psi}$$
you will get two coupled differential equations (the proof is detailed in most QM texts, as far as I know)
$$\begin{align}
\frac{\mathrm dc_\alpha}{\mathrm{d}t} &= \frac{1}{2}\mathrm{i}\omega_1c_\beta \\
\frac{\mathrm dc_\beta}{\mathrm{d}t} &= \frac{1}{2}\mathrm{i}\omega_1c_\alpha
\end{align}$$
These can be solved using standard methods to get
$$\begin{align}
c_\alpha(t) &= c_\alpha(0)\cos\left(\frac{\omega_1 t}{2}\right) - \mathrm{i}c_\beta(0)\sin\left(\frac{\omega_1 t}{2}\right) \\
c_\beta(t) &= c_\beta(0)\cos\left(\frac{\omega_1 t}{2}\right) - \mathrm{i}c_\alpha(0)\sin\left(\frac{\omega_1 t}{2}\right)
\end{align}$$
For a $90^\circ$ pulse, the flip angle $\omega_1 t = \pi/2$, and this simplifies to
$$\begin{align}
c_\alpha(t) &= \frac{1}{\sqrt{2}}(c_\alpha(0) - \mathrm{i}c_\beta(0)) \\
c_\beta(t) &= \frac{1}{\sqrt{2}}(c_\beta(0) - \mathrm{i}c_\alpha(0))
\end{align}$$

Now let's talk about a spin that happens to be initially 100% up spin (which is the "lower energy state" for a nucleus with positive gyromagnetic ratio), such that
$$\ket{\psi(0)} = \ket{\alpha}; \quad c_\alpha(0) = 1, c_\beta(0) = 0.$$
After applying a $90^\circ$ pulse, the wavefunction will be changed into
$$\ket{\psi(t)} = \frac{1}{\sqrt{2}}\ket{\alpha} - \frac{\mathrm{i}}{\sqrt{2}}\ket{\beta}$$
i.e. equal amounts of up and down character, in the sense that the mod-squares of the coefficients are the same. This is the so-called "excitation".
Likewise one can verify that if $\ket{\psi(0)} = \ket{\beta}$, then the final wavefunction after the RF pulse will be
$$\ket{\psi(t)} = -\frac{\mathrm{i}}{\sqrt{2}}\ket{\alpha} + \frac{1}{\sqrt{2}}\ket{\beta}$$
i.e. equal amounts of up and down character, as well. So if a spin is initially in the "higher-energy state" $\ket{\beta}$, the pulse will actually cause it to return towards the lower-energy state!

In fact there is no stipulation that the spins have to be initially either up or down. This misconception is very widely propagated in introductory accounts of NMR. In truth the initial coefficients $c_\alpha, c_\beta$ can be anything. There is no restriction on them such that they can only be $(1,0)$ or $(0,1)$. In general the spins in a sample will have all sorts of values of $c_\alpha$ and $c_\beta$. Some of them will be leaning more towards $\alpha$ than they do towards $\beta$, and vice versa. This is a direct consequence of the small energy gap between the two levels - there isn't a very large driving force for the spins to fall into the lower level.
Nevertheless, we can say that because of the external magnetic field, in an ensemble of spins (i.e. many spins), the values of $|c_\alpha|^2$ tend to be larger than the values of $|c_\beta|^2$. Formally, we would write
$$\overline{|c_\alpha|^2} > \overline{|c_\beta|^2}$$
where the bar indicates an ensemble average. These two quantities are called the populations of states $\alpha$ and $\beta$. If you are interested, the idea of an ensemble average is accounted for very neatly by density matrices.
If you work through the relevant maths, you would find that after the $90^\circ$ pulse, you would get
$$\overline{|c_\alpha|^2} = \overline{|c_\beta|^2}$$
i.e. equal populations of the two spin states. Therefore, there is a net conversion of the spins in the system, from $\ket{\alpha}$ to $\ket{\beta}$, and this is the "excitation" that textbooks like to speak of.
However, we can only speak of the populations in terms of an ensemble average. The so-called "excitation" only means that as a whole, the system loses $\alpha$-character ($\overline{|c_\alpha|^2}$ decreases) and gains $\beta$-character ($\overline{|c_\beta|^2}$ increases).
That tells us nothing about each individual spin. As shown in the previous section, it is perfectly possible to have an individual spin lose $\beta$-character ($c_\beta$ decreases) and gain $\alpha$-character ($c_\alpha$ increases).

As an addendum: after this excitation, the main form by which $\overline{|c_\alpha|^2}$ and $\overline{|c_\beta|^2}$ return to their equilibrium values is not by spontaneous emission of a photon (the rate of spontaneous emission is proportional to $\nu^3$ and is hence incredibly small) but rather spin-lattice relaxation, which is a non-radiative process.

The nuclei absorb the rf radiation and the spin population in the upper level increases at the expense of that in the lower level. There is now a chance that the upper levels will radiate away their extra energy and so return to equilibrium. However, in an nmr experiment the excitation pulses used in most if not all nmr machines are very short, just a few microseconds (and so spectrally wide) but the natural radiative lifetime of the excited levels is some seconds. Thus it is possible to excite the nuclear spins and build up a relatively large population before any significant decay occurs.
(This is one of the reasons why nmr works so well considering the tiny population difference that is produced because the energy levels are so close in energy and so all levels are significantly thermally populated.  The long lifetime of upper levels means that it is possible to coherently build up a significant macroscopic magnetisation in the sample and this what is detected.)


text:
Why is it wrong to use the concept of hybridization for transition metal complexes?
I have asked a lot of questions on coordination chemistry here before and I have gone through a lot others here as well. Students, including me, attempt to answer those questions using the concept of hybridization because that's what we are taught in class and of course it's easier and more intuitive than crystal field theory/molecular orbital theory. But almost all of the times that I attempted to use the concept of hybridization to explain bonding, somebody comes up and tells that it's wrong.
How do you determine the hybridisation state of a coordinate complex?
This is a link to one such question and the first thing that the person who answered it says: "Again, I feel a bit like a broken record. You should not use hybridization to describe transition metal complexes."
I need to know:

Why is it wrong? Is it wrong because it's oversimplified?
Why does it work well while explaining bonding in other compounds?
What goes wrong in the case of transition metals?

Tetrahedral complexes
Let's consider, for example, a tetrahedral $\ce{Ni(II)}$ complex ($\mathrm{d^8}$), like $\ce{[NiCl4]^2-}$. According to hybridisation theory, the central nickel ion has $\mathrm{sp^3}$ hybridisation, the four $\mathrm{sp^3}$-type orbitals are filled by electrons from the chloride ligands, and the $\mathrm{3d}$ orbitals are not involved in bonding.

Already there are several problems with this interpretation. The most obvious is that the $\mathrm{3d}$ orbitals are very much involved in (covalent) bonding: a cursory glance at a MO diagram will show that this is the case. If they were not involved in bonding at all, they should remain degenerate, which is obviously untrue; and even if you bring in crystal field theory (CFT) to say that there is an ionic interaction, it is still not sufficient.
If accuracy is desired, the complex can only really be described by a full MO diagram. One might ask why we should believe the MO diagram over the hybridisation picture. The answer is that there is a wealth of experimental evidence, especially electronic spectroscopy ($\mathrm{d-d^*}$ transitions being the most obvious example), and magnetic properties, that is in accordance with the MO picture and not the hybridisation one. It is simply impossible to explain many of these phenomena using this $\mathrm{sp^3}$ model.
Lastly, hybridisation alone cannot explain whether a complex should be tetrahedral ($\ce{[NiCl4]^2-}$) or square planar ($\ce{[Ni(CN)4]^2-}$, or $\ce{[PtCl4]^2-}$). Generally the effect of the ligand, for example, is explained using the spectrochemical series. However, hybridisation cannot account for the position of ligands in the spectrochemical series! To do so you would need to bring in MO theory.

Octahedral complexes
Moving on to $\ce{Ni(II)}$ octahedral complexes, like $\ce{[Ni(H2O)6]^2+}$, the typical explanation is that there is $\mathrm{sp^3d^2}$ hybridisation. But all the $\mathrm{3d}$ orbitals are already populated, so where do the two $\mathrm{d}$ orbitals come from? The $\mathrm{4d}$ set, I suppose.

The points raised above for tetrahedral case above still apply here. However, here we have something even more criminal: the involvement of $\mathrm{4d}$ orbitals in bonding. This is simply not plausible, as these orbitals are energetically inaccessible. On top of that, it is unrealistic to expect that electrons will be donated into the $\mathrm{4d}$ orbitals when there are vacant holes in the $\mathrm{3d}$ orbitals.
For octahedral complexes where there is the possibility for high- and low-spin forms (e.g., $\mathrm{d^5}$ $\ce{Fe^3+}$ complexes), hybridisation theory becomes even more misleading:

Hybridisation theory implies that there is a fundamental difference in the orbitals involved in metal-ligand bonding for the high- and low-spin complexes. However, this is simply not true (again, an MO diagram will illustrate this point). And the notion of $\mathrm{4d}$ orbitals being involved in bonding is no more realistic than it was in the last case, which is to say, utterly unrealistic. In this situation, one also has the added issue that hybridisation theory provides no way of predicting whether a complex is high- or low-spin, as this again depends on the spectrochemical series.

Summary
Hybridisation theory, when applied to transition metals, is both incorrect and inadequate.
It is incorrect in the sense that it uses completely implausible ideas ($\mathrm{3d}$ metals using $\mathrm{4d}$ orbitals in bonding) as a basis for describing the metal complexes. That alone should cast doubt on the entire idea of using hybridisation for the $\mathrm{3d}$ transition metals.
However, it is also inadequate in that it does not explain the rich chemistry of the transition metals and their complexes, be it their geometries, spectra, reactivities, or magnetic properties. This prevents it from being useful even as a predictive model.

What about other chemical species?
You mentioned that hybridisation works well for "other compounds." That is really not always the case, though. For simple compounds like water, etc. there are already issues associated with the standard VSEPR/hybridisation theory. Superficially, the $\mathrm{sp^3}$ hybridisation of oxygen is consistent with the observed bent structure, but that's just about all that can be explained. The photoelectron spectrum of water shows very clearly that the two lone pairs on oxygen are inequivalent, and the MO diagram of water backs this up. Apart from that, hybridisation has absolutely no way of explaining the structures of boranes; Wade's rules do a much better job with the delocalised bonding.
And these are just Period 2 elements - when you go into the chemistry of the heavier elements, hybridisation generally becomes less and less useful a concept. For example, hypervalency is a huge problem: $\ce{SF6}$ is claimed to be $\mathrm{sp^3d^2}$ hybridised, but in fact $\mathrm{d}$-orbital involvement in bonding is negligible. On the other hand, non-hypervalent compounds, such as $\ce{H2S}$, are probably best described as unhybridised - what happened to the theory that worked so well for $\ce{H2O}$? It just isn't applicable here, for reasons beyond the scope of this post.
There is probably one scenario in which it is really useful, and that is when describing organic compounds. The reason for this is because tetravalent carbon tends to conform to the simple categories of $\mathrm{sp}^n$ $(n \in \{1, 2, 3\})$; we don't have the same teething issues with $\mathrm{d}$-orbitals that have been discussed above. But there are caveats. For example, it is important to recognise that it is not atoms that are hybridised, but rather orbitals: for example, each carbon in cyclopropane uses $\mathrm{sp^5}$ orbitals for the $\ce{C-C}$ bonds and $\mathrm{sp^2}$ orbitals for the $\ce{C-H}$ bonds.
The bottom line is that every model that we use in chemistry has a range of validity, and we should be careful not to use a model in a context where it is not valid. Hybridisation theory is not valid in the context of transition metal complexes, and should not be used as a means of explaining their structure, bonding, and properties.

In case of Ni octahedral complex what do you mean that the 4d orbitals are "energetically inaccessible"? Is the energy gap between the 4s/4p and 4d orbitals that high so that they cannot form hybrid orbitals?

No, it is not really to do with hybridisation. Broadly speaking, if you want two orbitals to form a strong bonding interaction, you need two criteria: 1) strong overlap 2) small difference in energy. The 4d orbitals are high in energy and consequently interact very poorly with ligand-based orbitals; that's what is meant by "energetically inaccessible".

@orthocresol Why they don't interact with 4s though? Why metal must interact from the 3d orbital even if they are closer to nucleus than 4s or 4p?

To say that tetrahedral metal complexes are sp3 is a false straw man. By group theory, sd3 is equally valid, and it is clear from the energy levels that this is a better (though still not perfect) representation. This is partly why Td geometry is commonly seen with transition metals with low d electron counts. I don't think hybridization is a good concept from metal complexes, but done correctly it's not as incompatible as you are presenting it.

@Andrew, that's a fair point, and the group theoretical argument makes sense. I'm sorry I don't really have the time right now, but I'm happy if you want to edit my post!

How I wish the blind curriculum designers of high school chemistry in India (and maybe other places too - I don't know) would see this post! And something more we are taught - that even in CF theory, after splitting into t2g and eg levels, the ligands' electron pairs still populate hybrid orbitals of type sp3, dsp2 or sp3d2, whatever applicable. CFT supposedly continues using the notion of hybridization...

For the most part, coordination bonds involve d electrons from the transition metal interacting with some other electrons on a separate molecule. Hybrid orbitals are constructed from a single atom's orbitals mixing. If you do hybridization for the metal, then you are mixing its s, p, and d orbitals together in attempt to explain geometry. But ultimately, it is incorrect as really only the d orbitals interact. That's why people get upset when you use hybridization theory for coordination bonds.

Any typical MO diagram for a ML6 complex will involve the s and p orbitals mixing with appropriate ligand group orbitals, though. See e.g. http://4.bp.blogspot.com/-JhBU8xnwX-A/U2rN0dBSAwI/AAAAAAAAAlA/JKBmxPKeBrY/s1600/Octahedral_MO_diagram.jpg

Hybridization theory states that methane's four sigma bonds are equal (sp3).
However, if you apply group theory to the four sigma bonds (use the character table for Td point-group), the irreducible representation, if memory serves, = A1 + T2 (A1=single and T2 = a triple degeneracy).
H.T. is a nice tool to help understand introductory molecular geometry and/or o-chem, but, as it is superceded by M.O. theory, it has absolutely no basis in reality.
There is a very good description of a square planar complex above.  The simple answer to your question:  Almost every d8 transition metal complex is square-planar.
Regarding the rest (this also depends upon the d electron count and is a generalization ), Oh is more favorable than Td, and Td is more favorable than SP.  Again, take this with a grain of salt, but it is a decent starting point.  Consider the #of ligands and the # of occupied co-ordination sites on the TM center resulting from a L occupying a single co-ordination site--then remember that if a ligand can occupy multiple CN sites, it typically will do so.  Regarding the above 'favorability generalisation', remember that lower CN complexes are less stable and, thus, more open to attack because of unoccupied CN sites--think about a medieval town vs a walled medieval town;  more L or more occupied co-ordination sites= bigger wall (our TM friend wants to have the largest wall allowed by the system).  More often than not, your textbook's and/or prof's examples will rely upon a strict adherence to the 18 electron rule, so keep this in mind when trying to determine whether or not the addition of a MM bond and/or consideration of a dimer is apt (see below).
You should also consider whether or not a M-M bond is favorable (this is typically encountered with odd # d TM complexes--this is not an absolute) because then you would need to consider whether or not dimerization has occured (remember though that poly-dentate ligands may also result in the formation of multiple metal-center complexes that lack MM bonds).  H.T. completely fails here--this is why you should learn how to determine a molecule's point-group and ignore H.T.


text:
How does hydrogen bonding affect the boiling points of chemical compounds?
My main question is why is the boiling point of methanol so much different from that of water?
I understand that both compound are able to develop hydrogen bonding, and obviously water can develop one more hydrogen bond than methanol.
But on the other hand:

Due to presence of an electron-releasing alkyl group, the oxygen atom of an alcohol molecule is less-electron withdrawing on the remaining hydrogen atom. Therefore the electron deficiency of the hydrogen atom is less than that of a water molecule resulting in the formation of a weaker hydrogen bond.

What does this actually mean?
Both hydrogen, in methanol and water, are connected to oxygen, which has the same electronegativity and takes electrons off the hydrogen making it have a partial positive charge.
Why does a "less electron withdrawing oxygen atom" lead to a weaker hydrogen bond?
I thought that a less electron withdrawing oxygen atom would mean that the H is more negatively charged, meaning it would be stronger for hydrogen bonding, since F, O, N want more electrons as they are more electronegative?

There is an answer to a related but similar question here https://chemistry.stackexchange.com/questions/72132/why-alcohol-is-volatile-although-it-has-hydrogen-bonds/72150

Source: This is from page 223 of Understanding Advanced Organic and Analytical Chemistry: The Learner's Approach, World Scientific Publishing Company, Sep 29, 2016 by Kim Seng Chan and Jeanne Tan.

A hydrogen bond is a particularly strong dipole:dipole interaction with some covalent character. If we ignore the covalent character for a moment, we can say the higher the partial positive charge on the hydrogen, the stronger the hydrogen bond (keeping the O-H bond distance constant). If you relate "electron deficient" to positive partial charge, it might help you to make sense of the textbook answer.
You write:

Both hydrogen, in methanol and water, are connected to oxygen, which has the same electronegativity and takes electrons off the hydrogen making it have a partial positive charge.

Yes, that is the primary reason compounds like water, methanol and - say - hydrogen peroxide and acetic acid are hydrogen bond donors. But the chemistry beyond the OH group also plays a role. To illustrate, just compare the acidity of the proton connected to the oxygen in these four cases. They are very different. In a similar manner, the hydrogen bonding will be different (not as different, though).

For a study about the effect of substituents on pKa and hydrogen bond strength in aqueous and non-aqueous solutions, see this paper

See. First of all, your question is so much confusing that no one may get what you are saying or what are you confused of.
 I would simply answer directly to you main question.
Consider the following points:
1. In methanol, the oxygen atom is directly attached to the methyl group. And we know that all alkyl group are electron releasing group.

This means that the methyl group attached to the Oxygen will increase the electron density on the oxygen atom, thereby decreasing the polarity of the O-H bond. (That is the partial positive charge on hydrogen atom will be lowered due to the presence of the methyl group)
Whereas in a water molecule, no methyl group is directly attached to the oxygen atom, so the polarity of the O-H bond will be greater as compared to that in methanol.(this means that the partial positive charge on hydrogen atom attached the oxygen atom will be greater as compared to that in methanol).

Hence in case of a water molecule, the hydrogen atom will be easily to removed.
This is simply an answer that I wanted to give.
Get knowledge from this and I hope that this will be helpful.
Cheer up chemistry!!