text:
Enthalpy of a reaction changes with temperature, does this mean bond strengths change with temperature?
During chemical reactions, the bonds between atoms break or form to either absorb or release energy. The result is a change to the potential energy of the system. The heat absorbed or released from a system under constant pressure is known as enthalpy.
Mathematically, we can think of the enthalpy of reaction as the difference between the potential energy from the product bonds and the potential energy of the reactant bonds.
But we know that enthalpy of a reaction depends on the temperature according to Kirchoff's equations. So does this mean that bond energies also change with temperature.
Also can someone explain to me at the molecular level why does enthalpy of the reaction change with temperature?
And since the enthalpy of a reaction depend on the constant pressure at which reaction is carried out, does this mean bond strengths also depend on pressure?

No: bond strengths as such do not change with temperature. The potential energy surface holding the atoms together, i.e. all the bonds, is constant for a given molecule.
However, all molecules have vibrational and rotational energy levels and molecules also have bulk kinetic energy due to thermal motion in a gas or liquid. As the temperature increases the bulk thermal energy increases, i.e. molecules move faster, but also more vibrational and rotational energy levels become excited, which means that the vibrational and rotational quantum numbers increase. This means that less extra energy is needed to break a bond if the temperature is already high, compared to that needed at low temperature, simply because the bond is already has a lot of internal (vibrational/rotational) energy.
Not all molecules have exactly the same energy but instead there is an exponential distribution of energies (the Boltzmann distribution) where the higher the energy is the fewer molecules are there that have this energy.
The subject of  Statistical Mechanics describes these molecular effects and also forms the foundation for Classical Thermodynamics, i.e. what we usually call 'Thermodynamics', and which was developed before atomic/molecular theory.

I think the answer is already in the question. You're talking about enthalpy, so we're measuring change in energy under constant pressure. This means that the system is able to do work as a result of volume changes. Volumes do change as a function of temperature, and different substances have different coefficients of thermal expansion, so there is a enthalpy differential when this is taken into account.

But i have been taught that we can write enthalpy of a reaction by using only bond energies of reactants and products. I think this method only calculates ∆U of the reaction and ignores the work done. Then how is this correct?

Well, that's not correct because enthalpy and energy are not the same thing. You can estimate roughly for constant temperature, which is sketchy for very endothermic or exothermic reactions...

Then what are the other factors that lead to enthalpy change other than making and breaking of bonds? Where is the other part of enthalpy change comes from?

It's not just bonds. Bonds just happen to be the easiest piece to account for. Any kind of steric de-stabilization or electronic effect that stabilizes will alter $\Delta U$ of a reaction. But if you consider a reaction that involves a gas, the pressure-volume work is heavily dependent on temperature...

Do i need to go into these many details ?

I don't know. What are you trying to do? Your question was in regards to whether enthalpy is a function of temperature.

Apart from bonds, steric factors etc, the kinetic energy of molecules also changes ( although temperature remains same but number of moles may change ). Are we neglecting that also? I think it is being covered in the expansion work done, as it is equal to P∆V = ∆nRT. So the total kinetic energy of the system also changes and that is somehow equivalent to the expansion work  done in changing volume? And this is the thing that we are not taking into account if we think enthalpy is only due to bond breaking/forming?

I think that about covers it. Nonbonding interactions and pressure-volume work. Maybe someone else has something that I missed...

I'd like to address the one part of this question not treated in the answers:

Also can someone explain to me at the molecular level why does enthalpy of the reaction change with temperature?

This is a really interesting question I'd not considered before.
As Zhe mentioned in his answer, $\Delta H_{rxn}$ includes a $\Delta (pV)_{rxn}$ term.  At constant $p$, this becomes $p \Delta V_{rxn}$, and, with gas-phase reactions, $\Delta V_{rxn} \propto T$.
But this is an added macroscopic effect that exists in addition to the direct temperature effect on enthalpy at the molecular level.  Thus, to simplify things, and get more directly at your question, I'd suggest reframing it to look at the T-dependence not of enthalpy, but of the internal energy, at constant volume (which removes the added pV term).
Let's start phenomenologically, with the thermodynamics:
$$\left( \frac{\partial \Delta U_{rxn}}{\partial T}\right)_V = \Delta C_{V\, rxn}$$
Thus what's causing  $\Delta U_{V\,rxn}$ to change with temperature is the difference between the heat capacity of the reactants and products.
How can we understand this at a molecular scale? As we increase the temperature, thermal energy will flow into the reactants and products. This includes energy flowing into their bonds, thus putting their bonds at a higher-energy state, making them easier to break.
If both reactants and products have the same constant-V heat capacity, this effect will be the same for both, and there would be no net change in $\Delta U_{V\,rxn}$.  But suppose the reactants have a higher constant_V heat capacity than the products $(\Delta C_{V\, rxn} < 0)$. In this case the reactants are raised in energy more than the products (their bonds become easier to break), causing $\Delta U_{V\,rxn}$ to become more favorable (more negative):
$$\left( \frac{\partial \Delta U_{rxn}}{\partial T}\right)_V = \Delta C_{V\, rxn} < 0$$

The connection between quantum mechanics and classical thermodynamics is not straightforward. Or perhaps I should say it isn't simple. Quantum Mechanics, as you may know, isn't an easy subject. Richard Feynman famously said:"... no one understands quantum mechanics." So, looking too hard for intuitive understanding of the way the world works isn't likely to be productive. As has been widely discussed since it was invented/discovered in 1925, you can't use your intuition to understand it, you have to use the math. Does temperature affect bond strength? Temperature is a measure of molecular motion. Will motion affect bond energy? It depends, if the two atoms involved in the bond have sufficiently different momentum, the bond will break. So as you raise temperature of a material, you get closer to the point of disintegration. How about pressure? Well, you may know that water ice has various crystal structures depending on T and P? You can't explain that if you assume there's no change in energy and entropy of the system at different P or T. But chemicals are generally assumed to be at their LOWEST energy state, and compressing them (ignoring the PV work) will result in more repulsion between electrons, right? So you'll have to overcome that repulsive energy hence compression adds to potential energy.

This does not really answer the question at hand at all.  Please don't use answers for commentary.


text:
Is the Ebullioscopic constant temperature-dependent?
In tables, the ebullioscopic constant is usually written as a single value.
But, when the value is not available, it can be calculated by applying the formula: $$K_{\mathrm b} = RT_{\mathrm b}^2/\Delta h_\text{vap}$$
I was wondering: since in this last definition the ebullioscopic constant does depend on temperature, being the value of $\Delta h_\text{vap}$ temperature-dependent (its value is different if calculated at room temperature or at the boiling point @1 atm) what is, in real life, the best approximation of the behaviour of $K_b$?
In other words, does it behave like a constant or its value changes at two different temperatures? (eg: boiling point elevation for a solution @atmospheric pressure vs boiling point elevation for the same solution @reduced pressure)

I'd say that Kb can be described as constant in the same sense as the enthalpy of vaporization can be considered constant; it's approximately constant over a small T range or when T << Tcritical. In other words, for Kb to be exact at a given temperature, the latent heat of vaporization would have to be corrected for temperature differences.

Short answer
As long as external pressure is constant, it does not make sense to discuss $K_b$'s temperature dependence. If external pressure is changed, boiling point of solvent changes and hence also the ebullioscopic constant.
Longer version
Condition for equilibrium (skip if familiar)
Boiling is a reversible equilibrium process. By the second law of thermodynamics,
$$\mathrm{d}S_\text{universe}=0.$$
Assuming constant pressure,
$$\mathrm{d}S_\text{universe}= \mathrm{d}S-\frac{\mathrm{d}H}{T}.$$
Provided additional constant temperature,
$$-T\mathrm{d}S_\text{universe}= \mathrm{d}H - T\mathrm{d}S = \mathrm{d}G_\text{P,T}.$$
We also know that $G=G(P,T,n_1,\ldots, n_k).$ Thus
$$\mathrm{dG}=V\mathrm{d}P-S\mathrm{d}T + \sum_{i=1}^{k}\mu_i\mathrm{d}n_i,$$
and therefore
$$\mathrm{d}G_\text{P,T} = \sum_{i=1}^{k}\mu_i\mathrm{d}n_i=0.$$
Currently, we have two phases: a gas and a liquid. Boiling is a dynamic equilibrium, the differentials are non-zero, $\mathrm{d}n_1 = -\mathrm{d}n_2$, and so
$$\mathrm{d}G_\text{P,T} = 0 \iff G_\text{liquid} = G_\text{gas} \iff \mu_\text{liquid} = \mu_\text{gas}.$$
Clausius$-$Clapeyron equation (skip if familiar)
Suppose a closed system with initial equilibrium, constant temperature and pressure. We saw that
$$G_\text{liquid} = G_\text{gas}.$$
Now change the temperature. To achieve a new equilibrium in a closed system, the pressure would also need to change. At this new equilibrium, $G'_\text{liquid} = G'_\text{gas}$, which is why
$$\mathrm{d}G_\text{liquid} = \mathrm{d}G_\text{gas},$$
or equivalently,
$$V_\text{liquid}\mathrm{d}P - S_\text{liquid}\mathrm{d}T = V_\text{gas}\mathrm{d}P - S_\text{gas}\mathrm{d}T.$$
Now
$$\frac{\mathrm{d}P}{\mathrm{d}T}=\frac{S_\text{gas}-S_\text{liquid}}{V_\text{gas}-V_\text{liquid}}.$$
Note that the RHS depends again only on the initial and final conditions of phase equilibrium. Thus we are allowed to use $\Delta S = \Delta H/T$ to arrive at
$$\frac{\mathrm{d}P}{\mathrm{d}T}=\frac{\Delta H}{T(V_\text{gas}-V_\text{liquid})}.$$
Because $V_\text{gas} >> V_\text{liquid}$, and due to the ideal gas law
$$\frac{1}{P}\frac{\mathrm{d}P}{\mathrm{d}T}\approx\frac{\Delta H}{RT^2}.$$
Assuming $\Delta H$ is not a function of temperature,
$$\ln \frac{P_2}{P_1} \approx -\frac{\Delta H}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right).\tag{Clausius–Clapeyron}$$
NOTE. There is a small but important detail to keep in mind here. The assumption that $\Delta H$ is independent of temperature will only make our equations more inaccurate but it will not affect our answer. See the definition further below.
Boiling point elevation
We have
$$\ln \frac{P^\text{s}}{P^\circ} \approx -\frac{\Delta H_{vap}(T^\circ_\text{b})}{R}\left(\frac{1}{T^\circ_\text{b}} - \frac{1}{T^\text{s}_\text{b}}\right)\tag1$$
where

$P^\text{s}$ – vapor pressure of solution at initial boiling temperature $T^\circ_\text{b}$,
$P^\circ$ – saturation pressure of solvent, by definition of boiling, $P^\circ = P_\text{ext}$,
$\Delta H_{vap}(T^\circ_\text{b})$ – entalphy of vaporisation at initial boiling temperature $T^\circ_\text{b}$,
$T^\text{s}_\text{b}$ – new boiling temperature at some external pressure $P_\text{ext}$,
$T^\circ_\text{b}$ – boiling temperature of solute at the same external pressure $P_\text{ext}$.

From Raoult's law
$$P^\text{s} \approx P^\circ X_\text{solvent} = P^\circ(1 - X_\text{solute}),$$
and thus from $(1)$
$$\ln (1 - X_\text{solute}) \approx -\frac{\Delta H_{vap}(T^\circ_\text{b})}{R}\left(\frac{1}{T^\circ_\text{b}} - \frac{1}{T^\text{s}_\text{b}}\right).$$
From first-order Maclaurin series, and since boiling point elevation is usually small for our equations to be a good approximation,
$$\Delta T_\text{b} \approx \frac{R(T^\circ_\text{b})^2}{\Delta H_{vap}(T^\circ_\text{b})}X_\text{solute}.$$
Granted molar mass $M$ is in units $\mathrm{g/mol}$, and molality $c_m$ has units of $\mathrm{mol\ /\ 1\ kg\ solvent}$, then
$$\Delta T_\text{b} \approx \underbrace{\frac{R(T^\circ_\text{b})^2}{\Delta H_{vap}(T^\circ_\text{b})}\frac{M_{\text{solvent}}}{1000}}_{K_\text{b}}c_m \equiv K_\text{b}c_m$$
or
$$K_\text{b} \equiv \frac{R(T^\circ_\text{b})^2}{\Delta H_{vap}(T^\circ_\text{b})}\frac{M_{\text{solvent}}}{1000}. \tag2$$
It does not matter whether our previous equations are mere approximations, this is now a definition of the ebullioscopic constant. It is evident that it is in no manner dependent on external temperature. It is only dependent on the boiling temperature of a pure solvent at some well-defined external pressure $P_\text{ext}$. To change this temperature, we would have to change the external pressure, or pick a different solvent altogether.
If the external pressure is transformed, $T^\circ_\text{b}$ will not remain the same. There is no reason to believe that $(T^\circ_\text{b})^2$ at this new pressure $P'_\text{ext}$ will cancel with the change in vaporisation entalphy. And thus, the ebullioscopic constant is, in fact, not constant.
Conclusion

If external pressure remains the same, it makes no sense to talk about temperature dependence. This is because the ebullioscopic constant is defined as $$K_\text{b} \equiv \frac{R(T^\circ_\text{b})^2}{\Delta H_{vap}(T^\circ_\text{b})}\frac{M_{\text{solvent}}}{1000}$$
which is dependent only on the original boiling temperature of the pure solvent at some well-defined external pressure.
If external pressure is changed, the pure solvent's boiling point $T^\circ_\text{b}$ will change to a new value. There is no cause to believe that the square of the boiling temperature will cancel with the altered entalphy of vaporisation. Hence, ebullioscopic constant will change, and is, in this sense, dependent on temperature. The exact function will vary with the solvent at hand but is most likely approximated by polynomials with empirical coefficients if need be.

I skiped all your answer because I was familiar with it all. Shall I downvote it ?

@Hexacoordinate-C Haha. One cannot comprehend the answers by those who have more rep. That's not how the system works

In short: depending on the solvent in question, the temperature of ebullioscopic determination and the accuracy you aim for, the temperature dependence of the ebullioscopic constant may be negligible.
Alexander Apelblat published with Springer a book Citric Acid (ISBN 978-3-319-11232-9), partially visible in the book view by Google Canada.  In chapter 2, entitled "Properties of Citric Acid and Its Solutions", he provides an example for water's ebullioscopic constant and its temperature dependence. Page 28 starts with this:

The page in question may be accessed for example from here, then jumping to chapter 2 / page 13, and manual browsing to page 28.  (As a side note, the preview does not allow to see page 29.)

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text:
What are the decay products of HTO?

Tritium, symbol $\ce{T}$, is hydrogen with a mass of $3\ \mathrm{amu}$. It is radioactive and undergoes $\beta$ decay. Which of the following could be the only products after a quantity of $\ce{HTO}$ undergoes decay?

$\ce{HeOH}$
$\ce{H2O}$, $\ce{O2}$ and $\ce{He}$
$\ce{H2O}$, $\ce{H2}$ and $\ce{He}$

The answer is 2 only, however I am finding it difficult to understand how I could have worked this out. Tritium undergoes beta decay, so: $$\ce{^3_1H+ -> ^3_2He^{2+} + e- + \bar {\nu}_e}$$
I had assumed it wouldn't be possible for the electron emitted in beta decay to become an orbital electron. I have seen the following question: What does HTO decay into?, but it doesn't answer my specific query. I could state that the helium ions produced gain 2 electrons from the oxygen ions to become helium, with these oxygen atoms then reacting to form $\ce{O2}$ and some of the hydrogen combining with oxygen to form water, leading to answer 2, but this feels somewhat arbitrary. What approach could be taken to solve this problem?

You walk to a local market with 50$$$ in your pocket, then you check again and find nothing (0$$$). Were you robbed? You didn't notice anything suspicious. Well, supposedly, you could compare 0 and 50 and do the arithmetic, but this feels somewhat arbitrary... No, it is not arbitrary at all. This is the conservation law, arguably the most reliable thing in chemistry. There is no going around it.

I don’t consider your working to be arbitrary at all. Let’s first consider the β decay including the associated $\ce{OH}$ part of $\ce{HTO}$:
$$\ce{_1^3THO -> _2^3He^2+ + OH- + e-^ + \bar{\nu_e}}\tag{1}$$
As you suggested, the electron created in equation (1) won’t stick around, it is emitted into the environment, leaving us with a very reactive $\ce{He^2+}$ ion. Obviously, that is a strong oxidant as it wants to get reduced itself. The closest electron sources are $\ce{OH-}$ and $\ce{HTO}$, so we can consider a number of other reactions. However, it makes sense to organise it in a simple redox reaction. The reduced component is obviously going to be $\ce{He^2+}$ which only leaves $\ce{HTO}$ being the oxidised component — and since both hydrogen and tritium are in the $\mathrm{+I}$ oxidation state in $\ce{HTO}$, it must be oxygen which is oxidised to either peroxide or elemental oxygen. Indeed, peroxides are not stable and typically disproportionate into oxygen and water anyway, so we might as well just consider the formation of oxygen. Here goes:
$$\textbf{Reduction:}\\
\begin{gather}\ce{He^2+ + 2 e- -> He^{\pm 0}}\tag{2}\end{gather}$$
$$\textbf{Oxidation:}\\
\begin{align}\ce{2HTO^{-II} &-> O2^{\pm0} + 4 e-}\tag{3.1}\\
\ce{2 HTO & -> O2 + 4 e- + 2 H+ + 2 T+}\tag{3.2}\end{align}$$
$$\textbf{Redox:}\\
\begin{gather}\ce{2 He^2+ + 2 HTO -> 2 He + O2 + 2 H+ + 2 T+}\tag{4}\end{gather}$$
Equation (3) and (4) would be similar if I had chosen to oxidise only to $\ce{H2O2}$ and then would have been followed by an equation (5) to show the disproportionation of hydrogen peroxide:
$$\ce{2 H2O2 -> O2 + 2 H2O}\tag{5}$$
In the end, you will arrive more or less at the same substances.
Note that to end up with hydrogen gas, you would have needed a reductive environment capable of reducing $\ce{H^{+I}}$ to $\ce{H2^{\pm0}}$ — not the case if you have a severly electron-desiring ion like $\ce{He^2+}$ around.

The reactions are impressive looking but mostly bogus.

@MaxW Are they now? On what basis? Suggest better ones that lead to the final products? Or suggest entirely different final products altogether?

First off HTO doesn't exist as a pure compound. H and T would exchange so that you's have a mixture of $\ce{H20}$, $\ce{HTO}$, and $\ce{T2O}$. The electron from the decay is energetic enough to cause all sorts of weird ionizations. $\ce{H2O^{2-}}$ could only exist for a fleeting moment until it would grive an electron to anothyer water molecule to form two $\ce{H2O^-}$.

@MaxW Where do you see $\ce{H2O^2-}$? I also decided to completely ignore the hydrogen/tritium exchanges, since on average $\ce{HTO}$ should be the most prevalent compound by statistics. You are probably correct about the electron, though.

The answer really can't be deduced by an understanding of the various chemical reactions, but I think we can all agree that answer (1) is obviously wrong. So the answer must be either (2) or (3).
Since the decay of tritium is consuming a species of hydrogen the solution will be hydrogen deficient, so (3) is out.
Thus (2) is the answer and it is more from logical deduction rather than an understanding of all of the chemical equations.

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text:
Plotting a Langmuir adsorption isotherm
We have,
$$\theta = \frac{Kp_\mathrm a}{1+Kp_\mathrm a}$$
where, the fractional coverage $\theta$ is defined as the number of sites occupied by adsorbate $A$ over the total number of sites. $p_\mathrm a$ is the partial pressure of the adsorbate and $K$ is the equilibrium constant.
$$\theta = \frac{[A_{\text{ad}}]}{[S_{\text{total}}]}$$
I need to plot the following data to verify the monolayer adsorption of $\ce{CO}$ on 1 g of $\ce{Pt}$ powder.
$$\begin{array}{cc}
\hline
\text{Equilibrium Pressure (P)/mmHg} & \text{Adsoprtion Amount(V)/ mmHg} \\ \hline
 2.50 & 0.507\\
5.00 & 0.92\\
10.0 & 1.51\\
15.0 & 1.97\\
20.00 & 2.32\\
30.00 & 2.75\\
40.00 & 3.13\\ \hline
\end{array}$$
The question gives me the following hint

Because adsorption amount $V$ is proportional to $\theta$, $\theta$ can be expressed as $\frac{V}{V_\mathrm m}$ where $V_\mathrm m$ is the adsorption amount at $\theta = 1$

I generated the following graph in Mathematica, simply by plotting $V$ vs $P$

I don't quite understand what the hint wants me to do? Does that require that I normalise the data fashion, i.e by dividing by a particular $V_\mathrm m$, so instead of having $V$ on the y-axis, I have $\theta$ varying from 0 to 1.
Is $V_\mathrm m$ something I obtain from the data/graph by calculating, or is it a tabulated value that I can look up in table somewhere.

If you plot 1/V vs 1/p, you should get a straight line.  How would you interpret the slope and how would you interpret the intercept?

@ChesterMiller I am guessing $V_m$ is the volume of the monolayer? I have generated the appropriate plot, and have attempted to linearise the equation. Is this what you meant?

Self Answer
Additionally, I plotted $\frac{1}{V}$ and $\frac{1}{p}$, as suggested by @Chester Miller in the comments below, and indeed I get a straight line.

A linear model fit was performed and the equation $4.39274 x+0.214246$ fits the data with an $R^2 = 0.999929 $. From my limited knowledge on this topic (I haven't formally studied this subject), the intercept should give me $\frac{1}{\text{adsorption capacity}}$, and the slope should give some indication of the affinity of the adsorption sites?
So going back to the initial equation, $$\theta = \frac{V}{V_\mathrm m} = \frac{Kp_\mathrm a}{1+Kp_\mathrm a} $$
And now considering $$\frac{V_\mathrm m}{V} = \frac{1+Kp_\mathrm a}{Kp_\mathrm a} = \frac{1}{Kp_\mathrm a} + 1$$
and dividing throughout by $V_\mathrm m$ we get,
$$\frac{1}{V} = \frac{1}{V_\mathrm m} + \frac{1}{V_\mathrm mKp_\mathrm a}$$
And we get the appropriate values for $V_\mathrm m$ and $K$ from the slope and intercept of the $1/V$ vs $1/p$ plot, and then obtaining $\theta$ and plotting vs. $p$ is a trivial exercise.
Anyway, using the calculated parameters, I once again plotted $V$ vs $p$ and fit a curve of the form $\frac{V_\mathrm mKp_\mathrm a}{1+Kp_\mathrm a}$

I have a question regarding your answer. I have a similar research question, where I want to calculate K and fit the same equation as you did. As a first step, I used your data to test if I can reproduce your results. Unfortunately, I already have problems with your data. The last step (fitting of the curve to the data) does not work with your data in my calculations. Can you please explain in more detail, how you get K from the plot 1/V with 1/p to plot VmKp/1+Kp?


text:
Temperatures dependence of aromaticity
Does aromaticity depend on temperature?  I recently read about a compound (tropone) which changes its aromatic nature with temperature. Is it plausible? If yes then why
does it happen?

Are you asking if the (impossible) compound in the photo you provided meets that criteria? As of right now I'm not really sure what kind of explanation you want.

@ringo Maybe I'm running in low-brain mode due to a cold, but the compound looks like tropone to me. The corresponding resonance structure would be a tropylium oxide. The latter is aromatic.

@KlausWarzecha I didn't see that it was a seven membered ring... Maybe I'm the one running in low-brain mode after all...

@ringo No worries,  I counted more than once. Go figure :D

This compound is non aromatic at room temperature but at higher temperature it is aromatic.

Also called tropone and is aromatic. The energy needed to disrupt the electronic energy involved in aromaticity will be huge, the molecule would probably decompose in trying to do so.


text:
DOSY-NMR: How to calculate the volume of the molecule from the diffusion coefficient?
DOSY-NMR allows for the determination of the diffusion coefficient of the analyte molecule. How do I calculate the volume/radius of the molecule from the diffusion coefficient?

You can try and use the Stokes-Einstein equation for the diffusion coefficient $$D=\frac{k_\mathrm{B}T}{\zeta}$$ where the friction term is $$\zeta = 6\pi\eta r$$ and $k_\mathrm{B}$ is the Boltzmann constant, $\eta$ the solvent viscosity $(1\ \mathrm{cP} = 10^{-3}\ \mathrm{kg~m^{-1}~s^{-1}} = 10^{-3}\ \mathrm{Pa~s})$ and r the molecule's radius . You only have one number in diffusion constant so you can only get radius or volume assuming the molecule is a sphere.
The model assumes that the molecules are larger than the solvent molecules, i.e. solvent is effectively continuous not molecular. There are slightly different expressions for the friction, 4 instead of 6 for example depending on whether 'slip' or 'stick' limits are assumed.

Are there other ways to calculate the volume without assuming the molecules to be a sphere. Can I calculate the diffusion constants using e.g. MM or DFT?

@ketbra  What exactly are you after? Don't you know what substance you are measureing?

@Karl I know the substance but whant to show wheather its a monomer/dimer/tetramer and want to argue with the size of the molecule.

If you only have one measurement then you can only assume a sphere, you do not have enough information to assume a prolate or oblate spheroid. You should be able to distinguish monomer etc just from the diffusion coef as the molecules probably coil up or cluster in some way . You can certainly get size & shape from calculation. If you could measure rotational diffusion, say by fluorescence anisotropy, then this gives a better idea of shape.

@ketbra You can surely differentiate betwen monomers, dimers and trimers spectroscopically. After that the diffusion cofficients are distinctive.

@Karl That's a funny comment. Sure the diffusion coefficients will be different. The question is not how to distinguish them, but how to identify them: giving only the single measured diffusion constant, deciding on whether its a monomer, dimer or other oligomer. E.g. the diffusion constant is 50 is it a monomer or an oligomer?

@porphyrin "If you only have one measurement then you can only assume a sphere" OK, then I will just perform the NMR measurement twice thanks ; )
"You should be able to distinguish monomer etc just from the diffusion coef" OK, but how exactly?

DOSY is diffusion ordered spectroscopy. What is that stuff that you cannot identify monomer, dimer and oligomer out of the spectrum? Or is this a hypothetical question? Are you measuring a pure substance? Mixture?

@ketbra, ha-ha: very good comment :) As to monomer/dimer etc presumably you know roughly the size of molecule you study, thus compare  diffusion coeff with literature values to others of similar sizes, then look at you data to see how much bigger/smaller  your values are. At least this will give you a start. Also your data should show different types of molecules dispersed in value, perhaps grouped also and that should help identify monomer , dimer etc.

@porphyrin I see : ) I just thought that there maybe is a program similar to Mobcal for collision cross section to predict diffusion coefficient but maybe I will just write my own then ; )


text:
Loaded and unloaded solutions
What is the difference between a loaded solution and an unloaded solution? I was reading this article and I didn't understand these parts:

The reaction kinetics of $\ce{CO2}$ with loaded aqueous MEA solution...
$\ce{CO2}$ loading from $\mathrm{0-0.4~mole}~ \ce{CO2/ mole amine}$ for $30$ weight percent MEA solution...
kinetic constants are developed taking into account loading, temperature ...

Unloaded solutions are solutions without absorbed $\ce{CO2}$ and loaded solutions contain absorbed $\ce{CO2}$.  Please see the this link.

This is an answer, so not deleteable.


text:
Why H2S2 is hydrogen disulfide and not hydrogen persulfide?
Context: I was checking this question and in the answer, the compound $\ce{FeS2}$ was named "iron disulphide" (it was previously named "iron persulfide" but later changed after responding to the comments).
The structure of "hydrogen disulfide" is quite analogous to hydrogen peroxide i.e. containing the sulfur-sulfur linkage bond ($\ce{-S-S -}$) similar to peroxide bond. The bond angles of both compounds are also similar. So, why name it "disulfide" instead of "persulfide"?
In a group, compound are named in a similar manner in order to follow a certain fashion/trend. For e.g. methane, silane, germane (group 14). So, why $\ce{H2O2}$ is hydrogen peroxide and $\ce{H2S2}$ is hydrogen disulfide? Why isn't it named hydrogen persulfide as per analogy?
Related:
Why is H2O2 named hydrogen peroxide?

Hydrogen persulfide is one of the names of $\ce{H2S2}$.

Hydrogen peroxide can be alternatively called dihydrogen dioxide (note the prefix -di). As such, $\ce{FeS_2}$ could be called either iron(II) persulfde or iron(II) disulfide. Structurally speaking, though, in both cases the per-option is more appropriate.

"Hydrogen persulfide" is technically a very crude mixture of various hydrogen polysulfides ($\ce{H2S_n ; n\geq2}$). It has long sulfur chains and is formulated as $\ce{H-(S)_{n}-H}$. To avoid confusion, each members of the mixture is named according to number of sulfur atoms. So, $\ce{H2S2}$ is hydrogen disulfide.
More information on the preparation and properties of hydrogen persulfide can be found here: http://sulphur.atomistry.com/hydrogen_polysulphides.html

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text:
Verification of variants of the Ebullioscopic formula
Wikipedia cites the following formula for the ebullioscopic constant $K_{\mathrm{b}}$ regarding the elevation in boiling point of a volatile solvent upon addition of a non volatile solute (https://en.wikipedia.org/wiki/Ebullioscopic_constant):
(All symbols carry their usual meaning, unless specified otherwise)

$$K_{\mathrm{b}} = \frac{RT^2_{\mathrm{b}}}{\Delta H_{\mathrm{vap}}}$$

But my textbook cites another formula :

$$K_{\mathrm{b}} = \frac{RT^2_{\mathrm{b}}}{L_{\mathrm{v}}}$$

Where $L_{\mathrm{v}}$ is the latent heat of vaporization. Since there exists the equation $$\Delta H_{\mathrm{vap}} = {mL_{\mathrm{v}}}$$
my textbook further modified its formula to:

$$K_{\mathrm{b}} = \frac{RT^2_{\mathrm{b}}m}{1000\Delta H_{\mathrm{vap}}}$$

Where $m$ is the mass of solvent in grams.
Now my question is, which one's correct? The only difference between the equation on Wikipedia and the one in my textbook is the mass term $m$. I can't seem to find any reliable references on this :/. So a good reference would be appreciated.

I am familar with the version with $\Delta_{\text{vap}}H$, and in the derivation somewhere I remember that the Gibbs-Helmholtz equation is invoked and that is the origin of this term. Moreover, $\Delta_{\text{vap}}H$ and $L_v$ have the same physical meaning, and only the dimensions are different ($\mathrm{kJ\ mol^{-1}}$ vs.$\mathrm{kJ \  kg^{-1}}$. The multiplicative factors that result from the conversion between the two, can just be absorbed into the constant itself, believe. So guessing the tabulated values might differ based on definition. Plus, I think it is an empirical parameter anyway.

This answer expands upon my first comment:

I am familiar with the version with $\Delta_{\text{vap}}H$, and in the derivation somewhere I remember that the Gibbs-Helmholtz equation is invoked and that is the origin of this term.

I have tried to perform the derivation I alluded to, and it is given:
Assuming a binary mixture, A denotes the solvent and B denotes the solute. Superscript (*) implies pure solvent without solute (e.g $T^*$ is boiling point of pure solvent). Also, I assume the enthalpy of vaporisation doesn't change over the temperature range we are interested in.At equilibrium the chemical potential obeys:
$$\mu^*_A(g) = \mu^*_A(l) + RT\ln(\chi_A)$$
$$ \ln(\chi_A) = \frac{\mu^*_A(g)-\mu^*_A(l)}{RT} = \frac{\Delta_{\text{vap}}G}{RT}$$
Differentiating with respect to $T$ and invoking the Gibbs-Helmholtz equation
$$\frac{\mathrm{d}\ln(\chi_A)}{\mathrm{d}T} = \frac{\mathrm{d}(\Delta_{\text{vap}}G/T)}{R\mathrm{d}T} = -\frac{\Delta_{\text{vap}}H}{RT^2}$$
Integrating from $\chi_A = 1 \ \text{to an arbitrary value}$
$$\int_{0}^{\ln(\chi_A} \mathrm{d}\ln(\chi_A) = \frac{-1}{R}\int_{T^*}^{T}\frac{\Delta_{\text{vap}}H}{T^2}\mathrm{d}T$$
$$\ln(\chi_A) = \ln(1-\chi_B) = \frac{\Delta_{\text{vap}}H}{R}\left( \frac{1}{T}-\frac{1}{T^*}\right)$$
For dilute solutions, $\chi_B < 1$ so the approximate expression
$$\chi_B =\frac{\Delta_{\text{vap}}H}{R}\left( \frac{1}{T^*}-\frac{1}{T}\right)$$
Since, $T \approx T^*$
$$\left( \frac{1}{T^*}-\frac{1}{T}\right) = \frac{T-T^*}{TT^*} \approx \frac{\Delta T_b}{T^{*2}}$$
Now, we can rearrange to get the desired expression,
$$ \Delta T_b = \overbrace{\left(\frac{RT^{*2}}{\Delta_{\text{vap}}H}\right)}^{K_b} \chi_B$$

Moreover, $\Delta_{\text{vap}}H$ and $L_v$ have the same physical meaning, and only the dimensions are different ($\mathrm{kJ \ mol^{−1}}$ vs. $\mathrm{kJ\ kg^{−1}}$  The multiplicative factors that result from the conversion between the two, can just be absorbed into the constant itself,I believe. So I am guessing the tabulated values might differ based on definition. Plus, I think it is an empirical parameter anyway.

Note, that the relation given above is in terms of molar fraction however it is easier to work with molalities when you are doing experiments. The same relation can thus be reformulated
$$\Delta T_b = \tilde{K_b}b$$
and this $\tilde{K_b}$ has the definition which involves $L_v$, I believe. (just look at the dimension in both the expressions and you'll see).

The formatting's beautiful, seriously! Thank you! And yeah, they have the same dimensions ^_^

Wikipedia has the same formula as your textbook .
The formula from Wikipedia
$${K_b} = \frac{MRT^2_{\mathrm{b}}}{\Delta H_{\mathrm{vap}}}$$
Simply use the relation $\Delta H_{vap} =ML_{vap}$ and you will arrive at the equation in your textbook.

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


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How can I estimate the triple point of a substance given the densities of its liquid and solid phases and its enthalpies of fusion and vaporization?
I am trying to find the triple point of H2O given only the densities of water and ice, the enthalpy of vaporization, and the enthalpy of fusion.  I know that I should be able to use some combination of the Clapeyron and Clausius-Clapeyron equations to yield a system of two equations in two unknowns, after which I can use a computational engine to solve the system.  However, I am having trouble putting this knowledge to use.
A search on stackexchange brought me to the following question, which provides some answers, but confuses me in new ways:
Triple Point Equation
The answer given shows me that I am on the right track, but I am not sure what the lowercase p with the subscripts of sol-liq and liq-vap represents, and while the answerer points out that T1 and P1 are the triple point temperature and pressure, I am not certain of what the T with no subscript is meant to be.  My best guesses so far is that the lowercase p is vapor pressure and that the T with no subscript is the standard temperature for the given phase transition.

With reference to the other post, you can make a P vs T phase diagram. The $p_{sol-liq}, p_{liq-vap} $  are the pressure curves in the phase diagram that separate solid-liquid and liquid-vapour (gas) phases . T and P are the temp (in kelvin)  and pressure (bar) you use to plot the graph. The sublimation curve is not given but use the first equation with values for $\Delta H$ indicated underneath it. If you plot this you will easily see the triple point.

I think that you will need some other bit of information to fix the triple point. You can solve the equations with lots of different pairs of values of the triple point. The easiest would be to use the boiling point then your calculated  liquid/vapour curve is fixed to pass through this point.


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Why is KOH less soluble than NaOH in alcohols?
Alkali metal hydroxides solubility in water increases down the group. Because the mechanism of dissolution is the same in alcohols, why the anomaly?

On a molar basis, though not by a large amount, potassium hydroxide is also less soluble than sodium hydroxide in water at 25°C.  Check out the WP data tables for each of these compounds.

Sometimes answers are wrong, but in this case, the question is wrong. Sodium hydroxide is 4 times more soluble in water and in methanol than potassium hydroxide.

Your question has multiple aspects to be discussed:

Upon your observation; Solutes follow Like dissolve like principle to decide the extent to which they will get dissolved into solvent. It means more ionic solute will dissolve more in a polar solvent than a non-polar solvent and vice-versa.

KOH has a greater ionic character than NaOH and hence needs a more polar solvent to get dissolved as compared to the equivalent amount of NaOH needs to be. Since alcohols are very less polar, they may break bindings of NaOH effectively but are not able to break those of KOH to that extent. As a result KOH is less soluble in alcohols. This is also evident from the fact that KOH is more soluble in water as compared to NaOH because water is a polar solvent and hence it finds KOH more perfect match than NaOH.

As we all know solubility is a temperature-dependent property and does not possesses same slopes for all solutes. Hence you may get a lower solubility for KOH at some temperature.

By the word "Alcohols" we do not have a specific solvent and hence different  solute-solvent pairs are possible and do not know exactly which alcohol you're discussing as some halogenated alcohol have greater affinity for NaOH.

Because of smaller size of sodium ion, hydration ease is another factor you should not ignore while concerning this.

Finally there are several other factors which leads to the greater solubility which include charge density, bond stability, percent ionic character etc. studying which may or may not give results contrary to your question.

This is overly simplistic. There are a lot of factors that determine solvation enthalpies, and I think it's a disservice to use a simple rule like "like dissolves like." Because of the larger size of K+, the solvation spheres are larger, so overall, you're getting less stabilization. On the other hand, you do have to pay for this entropically, as the ordering in the solvation spheres leads to a more negative entropy change for solvating a sodium ion. So, it is possible that at a different temperature, KOH is more soluble. (All data is in water.)


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Systematic IUPAC names for beta-glucose and lactic acid
What are the IUPAC systematic name for $\beta$-glucose and lactic acid (as in the monomers used in polymerization for PLA)?
For lactic acid, I found this:

2-hydroxypropanoic acid

Is this correct?
For $\beta$-glucose, I have not find anything conclusive so far. I did find on this site (Carbohydrates) that "carbohydrates have been given non-systematic names". Is this true, and does that imply that "$\beta$-glucose" is simply $\beta$-glucose?

For your second compound, Google IUPAC name lactic acid for confirmation (yes, you're correct).

A compound that can be isolated from a natural source is commonly given a trivial name. These trivial names are usually related to the biological origin of the material. When the full structure is known, a systematic name may be generated in accordance with the IUPAC recommendations for nomenclature of organic chemistry; however, this name may be too cumbersome (for example for repeated use in the text of scientific publication).
According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the preferred IUPAC name (PIN) for lactic acid is the systematic name 2-hydroxypropanoic acid. Nevertheless, the name lactic acid is retained for use in general nomenclature.

P-65.1.1.2.3 The names citric acid, lactic acid, glyceric acid, pyruvic acid, and tartaric acid, related to natural products, are also retained; no substitution is recommended, but the formation of salts and esters is allowed.

For carbohydrates such as glucose, the current IUPAC recommendations do not identify a PIN. The systematic name for the β-pyranose form of ᴅ-glucose is (2R,3R,4S,5S,6R)-6-(hydroxymethyl)oxane-2,3,4,5-tetrol. Since this name may be too cumbersome, a semisystematic name may be formed based on the retained name ᴅ-glucose as follows.

P-102.3.4 Cyclic forms of monosaccharides
Most monosaccharides exist as cyclic hemiacetals or hemiketals. Two aspects of the internal cyclisation must be examined: first, the size of the ring, and secondly, the conformation of the newly created chirality center.
P-102.3.4.1 Ring size
Out of the various possible heterocyclic ring sizes resulting from hemiacetal or hemiketal formation, those with five and six members, including an oxygen atom, prevail and are discussed in this Section. Their names are based on those of the parent heterocycles furan and pyran, respectively. Names are formed by including the terms ‘furan’ and ‘pyran’ before the ending ‘ose’ in the name of a sugar. (…)

Thus, the retained name ᴅ-glucose is changed to ᴅ-glucopyranose to indicate the cyclic form having a six-membered ring.

P-102.3.4.2 Anomeric forms; the stereodescriptors ‘α’ and ‘β’
P-102.3.4.2.1 In the cyclic form, the configuration of the newly created chirality center ‘C-1’ must be expressed. This center is called the ‘anomeric center’. The two stereosiomers are called ‘anomers’; they are designated by the stereodescriptors ‘α’ and ‘β’ according to the configurational relationship between the anomeric center and the so called ‘reference center’.
P-102.3.4.2.2 Configurations ‘α’ and ‘β’ for monosaccharides
The anomeric reference center in a monosaccharide having a retained name is the configurational atom as defined in P-102.3.3. (…)
  The anomeric stereodescriptor ‘α’ or ‘β’, followed by a hyphen, is placed immediately before the configurational stereodescriptor ‘ᴅ’ or ‘ʟ’ of the carbohydrate name.

Therefore, the resulting name for the β-pyranose form of ᴅ-glucose is β-ᴅ-glucopyranose.


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Surface Tension and Evaporation
Why does acetone evaporate faster than ethanol yet still has a higher surface tension?
I would expect it to have a lower surface tension... I got the data from surface-tension.de.

Surface tension and evaporation are not as linked with each other as posed by your question. They are two distinct characteristics of liquids which generally follow direct relationship but not always. There are several other factors which you should not ignore:

Surface tension is the inherent force of liquid described by its tendency to minimize its surface area. A liquid with high molecular attraction takes account for a higher surface tension. It's temperature dependent and does not have similar slopes for all liquids. The case you provided is for standard temperature i.e. 20 C so you may get a contrary result at some other temperature. Compare: http://www.ddbst.com/en/EED/PCP/SFT_C11.php and http://www.ddbst.com/en/EED/PCP/SFT_C4.php2.
Evaporation usually deals with heat capacity, thermal energy required to overcome inter-molecular attractions and other thermal properties. Usually higher surface tension means energy required to overcome inter-molecular attractions is higher hence it may also have a higher value for the liquid having higher surface tension but it is not necessary. So liquids resembling each other in thermal properties usually form intersecting surface tension curves. i.e. one have a larger value till intersecting temperature and later on second has more. Ethanol and propanone is also such borderline case to be compared.


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How to extract thymol from oregano leaves?
This is my second graders in highschool project of science report.
I try to a make natural hand sanitizer using thymol from oregano leaves as  the antiseptic ingredient instead of alcohol.
I want to extract using a maceration method.

What solvent should I use?
How to separate thymol from the solvent?

This sounds like a nice project and I'm tempted to answer in too much detail, but since this is homework […]. Here are some hints:

Find information about thymol. This is crucial!

What is the structure?
In which solvents is it soluble?
Is thymol soluble in water?

Is it soluble in acidic or alkaline solution?

What about the boiling point and the melting point?

Plant leaves have a cell structure in which your thymol is somehow bound and hidden. In order to break of the structure, you may want to

dry and powderize the leaves or
cut the fresh leaves into pieces and grind them in a mortar with sea sand

Oregano leaves are green. You know that they contain chlorophyll. There's probably a lot of other compounds too, not just thymol.
Supposed that you treat the leaves as described in 2. and start to extract using ethanol in a Soxhlet extractor, you'll end up with a nice cocktail. Do you need to separate thymol from the other extracted compounds? If so, how will you proceed? Knowing the properties of thymol will help you here!

Remember that the easiest work is the one that you can avoid ;-)
Think in analogies. Thymol is a monoterpene. A lot of other members from this family have a typical smell and are used as fragrances. There must be an established, mild procedure to obtain these compounds from plant material.

 Yes there is! Search for steam distillation :)

If there is an established prodecure that avoids column chromatography or other purification techniques, maybe somebody has already used it for thymol and wrote about it.
Back in the days, our professors used to say with a smile:

A week in the lab saves you an hour in the library

Replace library with internet and have another search for a suitable and easy procedure already published.

Update
In a follow-up comment, you asked:

Then, how much ethanol should I use?

My suggestion is: Avoid subsequent laborious workup and do not use any ethanol at all!
Read Introducing Organic Chemistry Students to Natural Product Isolation Using Steam Distillation and Liquid Phase Extraction of Thymol, Camphor, and Citral, Monoterpenes Sharing a Unified Biosynthetic Precursor by Katherine A. McLain, Kenneth A. Miller, and William R. Collins, J. Chem. Educ., 2015, 92, 1226-1228. Follow the given procedure, which was deliberately designed to be performed in a students' lab:

One, 3 h laboratory period is required to complete the experiment. Dried plant material (10 g) or fresh plant material (20 g) is placed in a high-speed blender with distilled water (200 mL) and blended to a fine
  slurry. The slurry is transferred to a round-bottom flask (500 mL) with a distillation apparatus attached and a stir bar, and it
  is placed directly on a hot plate turned to high heat. A beaker
  (200 mL) is used as the receiving vessel to collect
  approximately 100 mL of distillate. The bioactive molecule is
  extracted from the distillate with an appropriate organic solvent
  (30 mL). The organic phase is dried with magnesium sulfate,
  and the organic solvent is removed on a hot plate at medium
  heat.

For further details, have a look at the Supporting Info.


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What is the hybridisation of carbon in transition state in SN2 mechanism?
Normally, a carbon atom is $\ce{sp^3}$ hybridized, whereas in a transition state five bonds are present, then what's the hybridisation of carbon?

http://www.chemgapedia.de/vsengine/vlu/vsc/en/ch/12/oc/vlu_organik/substitution/sn_2/sn_2.vlu/Page/vsc/en/ch/12/oc/substitution/sn_2/uebergangszustand/uebergangszustand_sn2.vscml.html

Related: How are SN2 transition states stabilised by adjacent double bonds and carbonyl groups?

In transition state of SN2 mechanism, the carbon atom is $\ce{sp^2}$ hybridised due to its planar structure. At this point, carbon almost acquires 'pentavalency' with three full bonds and two 'partial' bonds, and a planar complex is formed, which is $\ce{sp^2}$ hybridised.


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Why is sugar syrup so sticky?
On the festival of Diwali, Indians exchange sweets and share their love. I did just that this very morning, but things don't always go the right way. I dropped a box of sweets and the sugary syrup spread all over the floor. Now while cleaning up the floor, a question popped up in my mind: Why is sugar syrup is so sticky? Is there any chemical phenomenon responsible for its stickiness?
Since I'm a high school student and not a chemist, I can't seem to think of any possible explanation. I shall be thankful if someone could direct me the right way here.

Probably due to a phenomenon called association. It happens a lot for non polar solutes (sugar) in polar solvents (water). Normally it arises out of something called Hydrogen bonding . Also steric factors could also come to play when you're dealing with highly concentrated solutions.

I know about  h-bonding and yes it is associated with the phenomena association.But how can we relate association and sticky effect.

Fortunately, it was a small box and just a minor disaster. Happy Diwali to you and your loved ones!

MinuteEarth - Why Is Syrup Sticky? https://www.youtube.com/watch?v=WLgrY8x7Q_A

The "sugar" you refer to is sucrose, which is a disaccharide (a fancy term we use for a type of sugar, composed of two other bits of sugar called monosaccharides; which in this case, are glucose and fructose). I felt I ought to have brought up the sucrose specification, since when you talk chemistry, "sugar" can refer to a whole lot of things.
Anyways have a look at this diagram:

Observe carefully, and you'll see that there are 8 -OH groups (hydroxyl groups) in a molecule of sucrose. Now why did I direct your attention to these particular groups?  It's because they're capable of taking part in a fascinating phenomenon called hydrogen bonding (I believe you already know a bit about it, so I won't go into details).
Now when you add a solute to water, that is capable of H-bonding, it sorta causes all the water molecules and solute molecules to 'clump' together. The greater the amount of H-bonding the solute displays with water, the greater the tendency for all those interaction to put up some resistance to the flow of water. This ability to 'resist flow', that is inherent to every real fluid (to different degrees of course), is what you call viscosity.
Now 8 hydroxyl groups, can result in some pretty extensive H-bonding. To give you a rough idea as to how much that contributes to the viscosity of the sugar solution, think of glycerine (glycerol). Rub some of it between your fingers, it's pretty viscous right? Now the thing is, a glycerol molecule has only 3 hydroxyl groups. So imagine how much more viscous/sticky the sugar solution ought to be, when it's got nearly 3 times as many hydroxyl groups! [No, I'm not saying H-bonding is the only factor you should consider]. Of course, the more concentrated the sugar solution, the more the sucrose molecules and hence more the H-bonding; so it'll get stickier and stickier.
Now another thing you ought to consider is the geometry of the sucrose molecule. Now it isn't as important as the H-bonding bit, but it's still noteworthy. What you must know is that the sucrose molecule is not a planar molecule. It isn't some 2-dimensional structure. It's got all sorts of groups sticking out above and below the plane of the carbon rings. Now at low concentrations this isn't really an awfully important thing, but the story changes when you're dealing with concentrated solutions. That's when the 'jagged' physique of the sucrose molecule comes into play. It serves to compound the effect of H-bonding and further obstructs flow.
Want an analogy for that bit on the importance of the molecule's geometry?
Imagine you've got this large sack of coins. Dump them onto the floor and they spread out so well, that you wouldn't be wrong if you said they flowed. But suppose you had nothing better to do, and decided to glue a couple of screws on both faces of each coin. Put all your coins back in the sack, and proceed to dump out the whole lot. You can easily picture how well this 'modification' obstructs the flow of the coins, and probably now, you can appreciate the importance of molecular geometry in this regard better.

"Is there any chemical phenomenon responsible for its stickiness?" You mean compared to a supernatural one? Almost everything that matter does in the everyday world has a chemical basis. (Radioactivity, weight and movement are examples of physics which most would agree aren't "chemical".)
Sugar is a polar organic compound. Actually, "sugar" is the name for several related polar organic compounds. "Table sugar" (in the USA) is the common name for sucrose. Sucrose can be chemically broken apart into glucose and fructose, both of which are also sugars, and are also sold commercially. The three dietary sugars are glucose, fructose, and galactose - and you'll note that sucrose isn't included although it is the most common! (Since the body naturally breaks it down into fructose and glucose, it's "included by implication").
The reason sugar is sticky is because of "hydrogen bonding" (H-bonding). In high school, you may have learned that there are two types of chemical solids (compounds): ionic and covalent. All chemistry is "caused" by electric charges. In ionic compounds, the charges behave without directional preference - charge density is spherically uniform while in covalent compounds charge density is confined and is usually shared between "bonded" atoms. (I am ignoring metal bonding here, which arguably is a third type of solid). Despite its name, H-bonding is not usually included when we're talking about chemical bonding (sometimes it is, usually it is not). H-bonding is weaker than the weakest covalent bond. Water is an example of a material which gets many of its physical properties because of its H-bonding. The size of rain drops, the ability of insects to walk on its surface, and its melting and boiling points all are greatly influenced by the H-bonding that naturally occurs in it. By far the most common H-bond is the one between an -OH group (part of a large molecule) and an -O- "group" (although we'd usually say O atom). Often you will see H-bonds drawn as -OH...O- or something similar. They are much stronger than the typical interactions between separate molecules (and, as I said, weaker than the bond interactions between covalently bonded atoms) (of course, these things aren't black/white yes/no questions, there is a scale of strength...)
So, sugar! If you take a look at the structure of sucrose (see Wikipedia), you'll see that there are 8 -OH groups on each molecule and that there are 3 more O 'groups' (a total of 11 O atoms capable of forming H bonds). That's a whole lot! So in a concentrated solution of sucrose, you'll get a lot of H-bonding which acts to link the fairly small sucrose molecules into chains. In fact, one experiment you can do is to heat a concentrated solution of sugar and see if you can make strings or threads of it by putting a spoon into it and pulling it quickly out. Cotton candy is popular at fairs in the USA and is based on this (although has been improved by trial and error changes to the process and recipe).
Plastics are polymers which exist as chains. Most glues also are polymers. Our skin is composed of polymers. As a matter of fact our skin is composed of polar polymers. And sugar, also being polar, will tend to "stick" to our skin. But not only are they polar, both are capable of H-bonding. So, you have two different chemical phenomena: concentrated sugar solutions have a cohesive strength (it sticks together) and it has an adhesive strength (to polar surfaces). These are what make it sticky.
One thing I've not explained is what I mean by "polar". It has to do with electric charge. In the context of molecules, when the negative charge density is centered at the same point as the positive charge density, a molecule is nonpolar. When the centers don't coincide, the molecule is polar (or if the positive charge is on an entirely different atom than the negative charge AND its a whole number, then we'd call that an ionic molecule) In other words, polar molecules are molecules which have fractional separation of negative and positive charges. It is a bit more complicated than that: In CO2 there is more negative charge "on" the O atoms than on the central C atom. But because the molecule is linear (the 3 atoms form a straight line in space) the center of negative charge is on the central C atom, despite its being slightly positively charged. NO2 is a slightly 'bent' molecule, it's not straight so although the negative charge on both O atoms is equal, they don't "cancel out" and the molecule is polar. So, we have a "spectrum" of polarity from something like H2 which is non-polar (and no charge separation) to CO2 which is also non-polar but with charge separation to NO2 which with a central bond angle of 134.3° is polar, but only "slightly" to HCl (in the gaseous state) which is very polar (but still covalent) to H(+)Cl(-) in solution which is ionic.
Polar molecules will interact more with other polar molecules than they will with non-polar molecules (everything else being equal). Nonpolar molecules will interact with other non-polar molecules only slightly (due to "dipole-dipole interactions"). They'd interact more with polar molecules (due to "induced dipole interactions") and once you get to polar-polar molecular interactions, you're talking significant effects on some properties. H-bonding is a very strong type of polar-polar interaction; and it is very specific to the H atom.


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Are alkenes with bulky groups attached (like the one shown in the picture) chiral or achiral?

Is the compound in the picture chiral or achiral?

I think it should be achiral for the following reasons:

There is free rotation. I know, but for every configuration produced by free rotation there is an exactly opposite configuration.
Also at certain point of time all the groups can come to the same plane.

This question is from my textbook exercise. And the answer says it is chiral. But I've no idea why. Also is it optically active or inactive?
EDIT:
@K_P gave a very useful article related to these type of compounds:

Is there free rotation around the bond between the phenyl ring and the double bond?

I think so @K_P tell me if I am wrong

related http://chemistry.stackexchange.com/questions/37794/when-can-a-molecule-be-considered-freely-rotating-at-room-temperature

Have you sudied atropisomerism? It is a very important topic and your exercise is an example of this

also http://chemistry.stackexchange.com/questions/24901/do-molecules-with-axial-chirality-have-stereogenic-units

@K_P  I read it now. But how to understand whether the group is bulky enough for atropisomerism ?

Your picture is not right, you need to draw the methyl groups (and the bromine atom) on the benzene ring so it will be more clear why there is hindered rotation. If this is useful at all, I was not sure if there is atropisomerism, I assumed there was, because of the answer in your textbook and then did some googling that confirms it: article

Related: http://chemistry.stackexchange.com/a/35515/14857

Is the compound in the picture chiral or achiral?

The answer in practice is: it depends. You say that there would be free rotation along the phenyl–vinyl bond. However, every adjacent carbon atom — the 2- and 6-positions of the phenyl ring and the 1-position of the double bond — are substituted. For free rotation to occur, these substituents must pass each other. Since they are all bulky, though, free rotation is very effectively hindered. A planar rotamer with full π conjugation between the phenyl ring and the double bond is not populated at any extend which can be shown by the NMR shifts of the corresponding carbon atoms (and could also be shown if the methyl group on the double bond were a proton, because it’s shift would be vastly different from one in styryl conjugation).
Under low enough temperatures, the compound you have been given will indeed form to separable enantiomers due to the bromine atom which destroys the molecule’s symmetry. These are examples of axial chirality; if the chlorine atom is above the phenyl ring’s plane, it is (aR), if it is below, the compound is (aS). The two enantiomers can be separated, e.g. by chiral HPLC, and they will show optical activity.
At higher temperatures, thermal energy will be enough to overcome the rotation barrier and an optically active single enantiomer will racemise. This temperature is typically above room temperature for a system such as your’s but well below the temperature required for alkene (E/Z) isomerisation.
This concept is called atropisomerism and is also the reason why BINOL and BINAP are chiral.

Do you know of any web resource which has listed down all known atropisomers ?

@S007 All known is wa~y too broad; I don’t think that even all bisphenyl-based atropisomers are listed anywhere.  Unless of course you’re talking about different conceptual building blocks but then again I still don’t know. (But at least the list would be manageable.)


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Additives to allow wax to melt in the microwave?
I'm working with microcrystalline wax that I would like to be able to microwave to heat up. I've found that wax doesn't seem to microwave, with the most common answer I've come across for why being that it doesn't contain any water.
Adding water doesn't seem like a valid solution in this instance, since even if I could add it to a wax mixture, it would eventually dry out anyway. Therefore, I'm looking for an additive that, once mixed in with the melted wax, will allow me to microwave it after that point. The mix ratio isn't all that important as long as it's kept below 30 percent or so, although this would depend on the additive.
In searching for a potential solution, I came across a concept called the "dielectric constant". Unless I'm misunderstanding, the higher this number the greater the capacity for heating in a microwave. Looking up numbers, this seems to make sense as the DK of paraffin (the closest material I could find to microcrystalline) is around 2, whereas water is around 80.
A potential material I found was titanium dioxide. I'm getting different numbers, but it seems to have a DK somewhere in the 100-300 range. Before going out and buying some, is it feasible that I could use this as a filler for my wax that would allow it to be microwaved?
If not, or if I'm on a completely wrong track with the dielectric constant concept, are there are other suggestions on how I could solve this problem?

Why cannot add water just before warming? Cheap, easy, easy to remove unlike TiO2

Well I am looking for a filler for my wax regardless, so if TiO2 meets both my needs (filler and microwavable) then it's no issue to add at all. I'm not even sure how I could add water. I could always use a water-bath / bain-marie, but that detracts from my need for convenience, which is the purpose of making this microwavable in the first place.

I would try graphite powder. It is conductive so it should absorb microwaves.

I think you are onto something with the idea of thinking about the dielectric constant, though not quite the right thing. Polar molecules and salts can greatly increase microwave absorption (salty foods often microwave unevenly as the surface absorbs so much of the energy)...

Whether you need to include the salt or material in the wax is unclear. It might be possible to create a vessel that will warm enough to melt wax if made of the right materials. Something simpler than the carbon-lined alumina kilns used to get 1,000 °C from microwave heating might work.

Microwave heating works best on polar molecules, such as water and alcohols, as it then causes the molecules to rotate or "jiggle" violently. Waxes are nonpolar, so heat poorly, but if your application can use fatty acids in the mixture, such as palmitic or stearic acid, you might try mixing some into the microcrystalline wax. The dielectric constant is not significant in your application.
However, the longer the fatty acid molecule, the less effective it is in absorbing microwave radiation and converting it to heat, and on the other hand, the shorter molecules have strong odors and do not mix well with wax... do not use butyric acid for that reason. The suggestion of using an absorbent container, by @Li Zhi, seems to be more effective.

I actually have some stearic acid I could use, but I may need to use this wax with a latex mould, and from what I've read latex and stearic don't work well together. Alternatively, I also have access to lanolin which is a long-chain fatty acid, but I can't find any information about whether it could be used with latex safely or not. I'm assuming not since it's also considered an acid. Thanks for the help though, I may end up just going with lanolin or stearin and just try to avoid latex moulds.

Would triglycerides not work then @DrMoishe Pippik?  Coconut has fairly short chains.  Also what about glycerin, or some kind of ester or vitamin E.

Powdered sand?  I am sure silicon dioxide heats up without water

No, SiO2 is the main ingredient in glass, which does not heat much in a microwave.

Are you sure.  I remember reading about a fire started in a microwave from sand heated in a paper towel.  Also, I have ceramic dishes that get hot with nothing in them.  It could be due to metallic content though, iron in the sand and cobalt maybe in the ceramic.

I just tried and I got Silicate sand to heat up pretty well.  Also found this:  http://physics.stackexchange.com/questions/136219/will-a-microwave-heat-sand  @DrMoishePippik

So my bone china bowls get hot but my ceramic ones don't.

It's a matter of what absorbs energy best: if you have a glass bowl or mug of water and equivalent container filled with sand in the microwave at the same time, the water will absorb most and the glass and sand stay cool. If there's nothing else but sand, or even Teflon, that will absorb the radiation. See http://www.rfcafe.com/references/electrical/dielectric-constants-strengths.htm for absorption coefficients. N.B. high-lead glass and glass with metallic ornamentation behave differently.

You are, imho, on the completely wrong track. Microwave ovens are "tuned" to be particularly effective in heating (transferring energy to) water. For the simple reason that our foods almost universally contain it (but try to heat dried beans in a microwave, or even cooking oil). Wax is, of course, similar to cooking oil and won't be efficiently heated in that way.It is possible that you can find some additives which heat the MC wax efficiently without substantially changing its properties. I've no idea why you'd want to go this route, primarily because efficiency will be proportional to additive content, and at high enough concentrations to be effective, you're not going to have MC wax anymore...unless you get really really lucky (snowball's chance in Hell, I'd say). There are a multitude of microwave dishes that will themselves heat up in a microwave. Use one of them.

Actually I may test this out but I seem to remember silicates and sand heating up in the m w.  Silicon or powdered sand?

Actually microwave energy is also strongly absorbed by conductive additives and salty materials. They are not as fine tuned to water as you might think.


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Dry water using graphite
Dry water or "powdered water" is essentially a bunch of extremely small water droplets surrounded by $\ce{SiO2}$.
Some time ago, I was messing around and mixed some powdered graphite with water. (The graphite was obtained from a pencil, so it was not pure. Also I mixed the two substances with a stick, not with a high-speed blender.) When a droplet of this was splashed onto the table, it could be blown around with air and even held on the hand without breaking, until it was squished.
The important part of this is, that that droplet could be placed back into the original mixture of water and graphite powder, and it would roll around on the surface.
So does anyone know if it's possible to use powdered graphite rather than silica particles to make dry water? If is is, what properties are different from the ordinary powdered water?

Silica is highly hydrophilic, graphite is not quite like  that.

@Ivan But the 'dry water' uses hydrophobic silica particles (surface modified). I think that nanographite would be a suitable substitute.

I may end up doing something similar to this for a school project. If I decide to mix the graphite with water instead of ink, I will measure its conductance. I could post my results here, maybe.

Graphite for pencils is mixed with clay, which is largely composed of silicates. I don't know whether pure graphite powder would display the same behavior you describe.


text:
Understanding molecule infrared absorption
My chemistry course requires us to know which molecules absorb infrared, and under which vibrational modes that they do. My professor provides us the following rule, which is very confusing for me and I still don't know how to start.

Rules for IR absorption: (1) The frequency of the IR radiation
  matches the frequency of the vibrational mode. AND (2) There is a net
  change in the symmetry of electrical charges during the vibrational
  mode.

So, say we have a molecule such as CF2Cl2. This absorbs infrared under all vibrational modes: symmetric stretching, asymmetric stretching, and bending. How do you get to that answer?
Thank you very much.

All forms of  radiation match the difference in two energy levels, electronic, vibrational, rotational & nmr etc. For  IR transitions we refer to molecular vibrations. The molecule must have a permanent dipole or produce a transient dipole as a result of vibrating in one of its normal modes.
So homonuclear diatomics have no IR spectrum, they of course have energy levels, but the radiation cannot 'couple' to these and so be absorbed. Heteronuclear molecule have a dipole and so also have IR spectra. The classical idea is that the electrons oscillate as the nuclei move in a vibration, this oscillating charge can interact with electromagnetic radiation of the correct energy and cause a transition from a lower energy level to a higher one.
Molecules such as benzene vibrate in many ($3N-6$) normal modes, some such as asymmetric stretches produce transient dipoles and have IR transitions. Others do not and so have no transitions. What determines the are called selection rules.
The only secure way is to look at the symmetry of the molecules using a point group table, its very easy when you know how, but a bit daunting at first sight. Have a look at this answer Understanding group theory easily and quickly for more details.
(The molecule you mention belongs to the $\ce{C_{2v}}$ point group, see molecule-viewer.com for a 3D model with symmetry elements.)

Thank you for your input, but some of it is hard to understand as I'm only a beginner at chemistry. We will not be needed to know group theory, nor have we discussed selection rules or anything like that. I just need to be able to tell whether a given molecule will absorb IR, and under which vibrational modes. How can I do that?

Then the simplest 'rule of thumb' until you learn more is to see if there is any change in the molecules dipole as the molecules bonds vibrate, if there is then there is an IR absorption.  This is presumably the 'net change in symmetry' in your question.  All heteronuclear diatomics have IR absorption due to vibration, none of the homonuclear diatomics do.


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In a combustion reaction, why is carbon in methane oxidised even though it starts and finishes with 8 electrons?

In $\ce{CH4}$, carbon has 8 electrons due to 4 covalent bonds.
In $\ce{CO2}$, carbon has 8 electrons due to 2 double bonds.

In both cases, carbon has 8 electrons. So why is carbon oxidised? Doesn't oxidation occur when electrons are lost?
The crux of my question is how to treat covalent bonds in redox reactions.
I know we can say carbon has gained oxygen so is oxidised, but I am after an answer in terms of the electron definition of reduction and oxidation.

(Image Source)

Oxygen is more electronegative than hydrogen and carbon , so it attracts the electron cloud more.

So then is every single reaction a redox reaction because one of the elements in a molecule (unless its a diatomic molecule with the same atoms)  will be more electronegative than the other and will have the electrons slightly towards it?

Actually you need to check the oxidation numbers for identifying redox reactions. In methane carbon has -4 oxidation number but in carbon dioxide it is +4.  Increase in oxidation number means oxidation.

To determine oxidation states and thus the outcome of redox reactions, polar covalent (even if they are only very slightly polar) are treated as fully separated ionic bonds.
In methane, carbon is more electronegative than hydrogen ($2.5 > 2.1$), so when setting up oxidation states, we treat methane as if it were composed of $\ce{C^4-}$ and $\ce{4 H+}$ — leading to the oxidation states of $\mathrm{-IV}$ on carbon and $\mathrm{+I}$ on hydrogen.
Likewise, oxygen is more electronegative than carbon ($2.5 < 3.5$), so in carbon dioxide we assume the electrons to be completely transferred over to oxygen; as if it were made up of $\ce{C^4+}$ and $\ce{2 O^2-}$ leading to the oxidation states of $\mathrm{+IV}$ and $\mathrm{-II}$, respectively.
This is an entirely formal treatment. However, there is some experimental basis behind it, namely that in methane carbon has a slightly higher electron density and a slightly negative partial charge while in carbon dioxide it is the oxygens that have a partial negative charge. The on-paper treatment is, however, an extremely gross exaggeration.
For ionic compounds, of course you observe complete electron transfer from one species to another signified in the ion charge, which may seem as if the concept is much more ‘true’ to ionic compounds.


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How to find the molar mass of a compound given mass, vol, temp, and pressure?

A $2.1\ \mathrm g$ sample of a liquid vaporizes and exerts $120\ \mathrm{mmHg}$ pressure at $1.5\ \mathrm L$ volume and $80\ \mathrm{^\circ C}$ temperature. Find the molar mass of the gas.

I first get the equation $PV=nRT$ and replace $n$ with $m/M$ since $M=m/n$
Then the equation becomes this $$PV=\frac{mRT}{M} \Rightarrow M=\frac{mRT}{PV}$$
$$\begin{align}
M&=\frac{mRT}{PV}\\[6pt]
\Rightarrow M&=\frac{(2.1\ \mathrm g)\times(62.364\ \mathrm{L\ mmHg/(mol\ K}))\times(80+273.13)\ \mathrm K}{(120\ \mathrm{mmHg})(1.5\ \mathrm L)}\\[6pt]
\Rightarrow M&=257\ \mathrm{g/mol}
\end{align}$$
But the answer key says $150\ \mathrm{g/mol}$. What am I doing wrong?

Your method seems okay. Just check calculations. If it is correct, then your answer key is wrong .

The calculation is correct, too.

Sometimes (too often) answer keys are wrong. In this case we've now triple checked the calculation, and you, Jan (as implied in his comment) and myself have all come up with a molar mass of $\pu{257 g/mol}$. At this point, we have to say the answer key is wrong and you are correct.
Your answer is $\pu{1.71}$ times that given by the key, and neither that nor any other comparison appears to correlate to any obvious error on their part. It simply appears to be a misplaced or incorrectly entered answer in the key, end of story and move on. And good job getting the correct answer ;)


text:
How to develop a retrosynthesis plan for 2-[(E)-2-(3-bromophenyl)ethenyl]-1,3-benzoxazole?
I need to retro-synthetically design a valid synthesis plan (in two steps, from cheap starting materials available over at Sigma Aldrich) for a target compound, pictured below.

How would I go about doing this?

Extempore, this should be available by the reaction of 3-bromocinnamaldehyde with 2-aminophenol. Samarium triflate as a catalyst might be a good idea.

In any case, I'd start the retrosynthesis at the benzoxazole moiety.

Thank you very much! In the meantime I tried to develop another strategy as well, I've done some research and found another way to synthesize this via reaxys. If I may, I do have a follow up question regarding this: would this reaction go the same way if I were to react the benzoxazole without the halogen (F)? If not, is there a way to remove it from the target molecule? Here's a link to the . Once again, thank you very much for the assistance.

How about a more classic Wittig reaction between 3-Bromobenzaldehyde and 2-(chloromethyl)-1,3-benzoxazole?


text:
Is vegetable shortening a fully hydrogenated product or is it partially hydrogenated?
Have the oils used to make vegetable shortening been fully hydrogenated or have they been only partially hydrogenated?

What's the label say?

As @ Todd Minehardt states, it depends... coconut oil, for example, is naturally 86% saturated.


text:
How is the specific heat of an alloy obtained?
I was wondering:
How is the specific heat of an alloy obtained?
My first attempt is to assume a thermal equilibrium between a mechanical solution and an homogeneous solution of two metals, A and B.
\begin{align*}
    Q_{A+B}&=Q_A+Q_B\\
    n_{A+B}C_{A+B}(T_{eq}-T_{A+B})&=n_{A}C_{A}(T_{eq}-T_A)+n_{B}C_{B}(T_{eq}-T_B)\\
    C_{A+B}(T_{eq}-T_{A+B})&=\frac{n_{A}}{n_{A+B}}C_{A}(T_{eq}-T_A)+\frac{n_{B}}{n_{A+B}}C_{B}(T_{eq}-T_B)\\
    C_{A+B}(T_{eq}-T_{A+B})&=X_AC_{A}(T_{eq}-T_A)+X_BC_{B}(T_{eq}-T_B)\\
    C_{A+B}&=\frac{X_AC_{A}(T_{eq}-T_A)+X_BC_{B}(T_{eq}-T_B)}{(T_{eq}-T_{A+B})}
\end{align*}
Where Q=heat, C=specific heat, n=moles, X=molar fraction and T=temperature.
Is it correct?
Is there another way of doing this?
I have previously searched and this are some relevant things on Chem.SE what I have found:

What is this equation for?
Properties of steel and aluminum alloys

I did not find a solution or an approximation in any of those links.
My aim is to prepare a class and teach my students how to determine the specific heat of any alloy, if possible, without going to a lab. (There is no basic infrastructure for special labs).

I suggest reading some journals: http://journals.aps.org/pr/abstract/10.1103/PhysRev.77.669

https://books.google.co.kr/books?hl=en&lr=&id=LIPsUaTqUXUC&oi=fnd&pg=PR11&dq=specific+heat+alloy+theoretical&ots=wzTKnNrLQL&sig=n_l6xdd6J_xTj_45-SOxc7eh-4s#v=onepage&q=specific%20heat%20alloy%20theoretical&f=false

http://journals.aps.org/prb/abstract/10.1103/PhysRevB.23.5628

The enthalpies and calorific capacities of certain mixtures are tabulated in standard references. If you do not have such data, you can use the following approximation:
Rule 1: For a mixture of gases or liquids, calculate the total change in enthalpy as the sum of its changes in the pure components of the mixture. In fact, the enthalpy changes associated with component mixing are neglected, which is an excellent approximation for gas mixtures and similar liquid mixtures such as pentane and hexane, but is bad for different liquids, such as nitric acid and Water.
Rule 2: For highly diluted solutions of solids or gases in liquids, neglect the change in enthalpy of the solute. This approach is better, as the solution is more diluted.
The calculation of the enthalpy changes for heating or cooling a known composition mixture can often be simplified by estimating the calorific capacity of the mixture as follows:
  \begin{equation}
    C_{p,m}(T) = \sum_i y_i\cdot C_{p,i}(T)
\end{equation}
Where:
$C_{p,m}(T)$ Calorific capacity of the mixture
$y_i$ Mass or mole fraction of the i-th component
$C_{p,i}(T)$ Calorific capacity of the i-th component

Reference:
Richard Mark Felder and Ronald W. Rousseau. Elementary Principles of Chemical Processes. John Wiley & Sons Inc.; 3rd Revised edition. 702 pages.


text:
Formation of pi-allyl Pd cation complex

(from pg. 1334 of Clayden Organic Chemistry 1st ed.)
In the above reaction catalyzed by Pd, I'm not exactly sure what's going on at the second step. It says in the book that Pd is using its electrons to push out X, and the electrons are returned to Pd when Nu comes and pushes out Pd.
My understanding is that in the first intermediate, the double bond's 2 pi electrons are complexed to Pd. When X leaves, the 2 pi electrons (that are now spread out between 3 C's) are still complexed to Pd. By the drawing of the mechanism, it seems to me that Pd has also used 2 of its electrons to bond with the 3 C's. So in the second intermediate, are the Pd and the allyl cation sharing a total of 4 electrons? If so, shouldn't there be a double bond, not a single bond, from Pd to the allyl cation?

This is a classic oxidative addition. You're kind of outside the classic world of single and double bonds. You have 4 electrons spread of 3 carbon atoms that link stabilize a bonding interaction with the palladium. Two of those electrons came from the palladium, so we think of it as having been oxidized, hence the name of the process.
See reference for general $\mathrm{S}_{N}2$ type oxidative addition. The only difference from your example is that the electrophile has a double bond that serves as an additional ligand for the palladium.

So when organometallic complexes are drawn, a bond from the metal to a ligand doesn't necessarily mean they only share 2 electrons right? They can share any amount of electrons?

That depends on the type of ligand. There are X ligands and L ligands. For X ligands, it's shared. For L ligands, they exclusively come from the ligand. Allyl is a XL ligand when it's in the $\eta_{3}$ mode where all three carbon atoms are coordinated. In this case, the X portion means a covalent-ish bond with the palladium. In reality, it's probably fairly delocalized.


text:
What are the effects of storing drinking water in a copper container?
I recently learned that Ayurveda recommends storing drinking water in copper containers for positive health benefits.
Upon my simple experimentation, I've noticed the following effects:
Water Type : activated carbon filtered drinking water
Duration of storage: 8 hours
The taste of the water changed, it tasted more metallic. I think this is because of the increase of dissolved copper in the water.
If I'm drinking 4-5 liters of water per day, would that be still under the safe limit of 2 mg/day? ref 2 Note: I don't have any apparatus to measure the copper content in the water.
I've researched a bit and found that it does have some positive effects on killing some microbes such as bacteria ref 1. Note: I'm not able to deduce any conclusion about the increase in copper content from this paper even though the author said it is "within the permissible limits."
I would like to know more about the studies related to this claim and I'm really interested to know about the side effects.

As you state, copper storage containers have long been used to prevent growth of pathogens in potable water.
As for the danger of copper toxicity, it depends on the pH and water temperature. Between pH 8 and 11, about 1 ppm of copper dissolves, so if you're consuming ~5 l/day or about 5,000 g/day of water, and if the water has had sufficient time to become saturated, then you'd get ~5 mg/day of copper, exceeding suggested safe limits for long-term exposure.
That said, I think that, if the water pH is between 8 and 11, and if it is refrigerated (e.g. 4 C), and if it has been standing quiescently for no more than eight hours, it would not have time to be saturated with copper, so it probably would be safe to drink if that were the only source of copper in your diet.
Given all those mealy-mouthed caveats, then, you might at least check the pH of the tap water, and use a different container if it's acid. You would not want to be like the drinkers from Yudono spring, unless you're really into Heavy Metal.


text:
Bond lengths: 1-D particle-in-a-box for 1,3-butadiene
When estimating the length for a one-dimensional particle-in-a-box model for a hydrocarbon (such as 1,3-butadiene), do we include the bond lengths between the (terminal) carbon and hydrogen atoms at the ends of the chain?

No, they are not in part of the double bond conjugation and so the $\pi$ electrons are not delocalised into these bonds (nor onto and CH bond). Only use delocalised electrons in particle in a box.

As porphyrin said in his comment, the $\ce{C-H}$ bond is not part of the delocalized $\pi$ system that makes up the one dimensional box. This is partially because hydrogen does not have a p orbital to contribute to the system. The $\pi$ system that creates the box is the result of overlap between p orbitals found on adjacent carbon atoms, meaning that you would be able to approximate the length of the box by summing up the bond lengths between all adjacent carbons in the delocalized system. While resonance means that there are not actually alternating double and single bonds across the $\pi$ system, all of the $C-C$ bonds will be the same length, which should be close to the average of the bond lengths for the most stable resonance contributor.


text:
Isothermal paradox(kind of)
I have recently read that for an isothermal expansion ∆u =0 which implies that work done =heat released as temperature  remains constant ....but also I read that q=m×c×∆T which in this case should be zero as ∆T=0,,,then this implies work done = 0 ..but this is not true
Where am I committing mistake

In freshman physics, we were taught that $q=mC\Delta T$, but in all the cases we considered, no work was being done.  When we got to thermodynamics, we learned that q represents energy in transit, and is a function of process path.  We also learned that, when work is being done, q is not equal to $mC\Delta T$.  Instead, we learned about the internal energy of a system U, which is expressed as $$\Delta U=q-w$$  When no work is being done, $$\Delta U=q=mC\Delta T$$All this misunderstanding is resolved and made consistent in thermodynamics when we change the definition of heat capacity slightly, defining it strictly in terms of U, rather than in terms of q:$$C_v=\left(\frac{\partial U}{\partial T}\right)_V$$Then, whether or not work is being done, we get the correct answer for the heat transferred and the change in internal energy.  The advantage of this definition is that C is properly recognized as a physical property of the material being processed rather than a function of the process path.


text:
Effect of impurities on melting point
This is a 10th-grade question.

The melting point of a substance decreases with increase in presence of impurities in it. The melting point of ice decreases from 0 °C to -22 °C on mixing salt in it in proper proportion. That is why salt is added to make freezing mixtures to keep ice creams frozen.

I want to ask that suppose ice melts at 0 °C and we are now adding salt to lower melting point but now when we have lowered the melting point the ice will melt more quickly, then what is the purpose of mixing salt?

In order for the ice to melt it needs heat.  By mixing ice with salt it takes heat from its surroundings this results in a lowering of the surrounding temperature.  If you have an ice bath the temperature is zero deg C until all the ice melts but if you had salt and mix there is a fall in temperature.  Try it at home ...

Mixing salt will result in liquid water that can stay liquid below $0^oC$.  This is dependent on the core of the ice being below zero at the beginning of the process because the heat has to flow from higher to lower temp material.  In other words if you add salt to ice that is homogeneously right at the freezing point you won't get liquid water below zero because there is no where for the heat to flow.

Note that impurities in a solid will affect the melting point, but the freezing point depression effect (a colligative effect) has to do with impurities in the liquid phase, not in the solid. The paragraph is therefore a tad confusing.

The presence of impurities in a substance lowers the melting point but increases the boiling point of the substance.

The purpose of mixing salt to ice is to get a simple way to obtain a temperature lower the 0°C without a fridge.

In the old fashioned process, we used buttermilk sugar and a bit of vanilla. We put them into a churn (which is just a small bucket with a mixing blade in it so you could hand mix the contents) and then added ice around the churn (outside the churn) to cool it down. And then you mixed and mixed and mixed and mixed. But there was a problem. As the ice melted, the water it created was at $\pu{32°F}$ and even though the ice was colder (a commercial freezer is often about $\pu{15°F}$ (but a home freezer is warmer)) the water kept the churn at around (usually above) $\pu{32°F}$ ($\pu{0°C}$). And ice cream (being an "impure" water mixture) is a liquid, not solid, at $\pu{32°F}$. So, this wouldn't work. Enter salt. If you added salt to the ice, then the ice would melt well below $\pu{32°F}$ and be able to cool the churn to the freezing point of the ice cream mixture (I think its about $\pu{25 °F}$, but my memory shouldn't be trusted). So, the salt and ice form a salt water mixture which can be well below $\pu{32°F}$, and so can cool to below $\pu{32°F}$. If you were to add salt to the recipe, then you'd have to cool it even MORE to get it to freeze (although a little salt makes it taste better, imho). So, salt is added to the cooling mixture and (supposedly) not to the recipe's ingredients. I guess I should point out that you need to consider walking over hot rocks vs walking in say an inch of water at the same temperature. Which would burn you faster? There's no question that your feet would have more contact with the water than the solid rocks, so your feet would be more damaged by the water (water at $180$ or $\pu{200 °F}$ WILL burn you). My point is you actually WANT the ice to melt into a salt solution, because that will better cool the churn - as long as there's enough salt so that the liquid is cold enough.

The presence of impurities in a substance results in a lower melting point due to a process called melting point depression. Melting point depression is the reason why adding salt to frozen streets helps to melt the ice.
Melting point depression occurs due to the nature of a material's solid state. Most solids, such as ice, form as crystalline lattices of repeating ions or molecules. This lattice is held together by intermolecular forces that create a strong, stable structure. The solid must be heated to a certain temperature in order for this structure to be disrupted and for the solid to melt. However, the presence of impurities weakens the lattice, making it less stable. As a result, the compound melts at a lower temperature.
When working with organic compounds in a lab, the purity of the compound can be partially determined through the use of a precise measurement of the melting point. If the melting point is within the scientifically accepted range of the material's melting temperature, then the material is presumed to be pure. However, if the observed melting temperatures are outside of this range, then the compound is not pure.
One interesting effect of this process is that sea water freezes at a lower temperature than pure water. Pure water freezes at $\pu{32°F}$ while sea water freezes at $\pu{28.4 °F}$.
I hope you get it. For any problem, leave comment.

In regards to why you do it.
Lets say you live in an environment where the temperature in the winter is -10 °C. In this case pure water at this temperature will be in the solid phase and exist as ice because the surrounding temperature is lower than the melting/freezing point. For driving and general transport, ice on the roads/walkways, etc. is a hazard so this is a concern. One way to deal with it is to physically remove the ice but this is very difficult.
Instead think about what happens if you add an impurity to the water and it reduces the melting point way down to -22 °C as was given in your textbook. Now the temperature at which the water freezes is much lower than the surrounding temperature so the water will still exist as a liquid form even when the temperature is lower than the normal freezing point. Thus ice problem when driving is solved (aside from the environmental concern of all the heavy salts).

I want to ask that suppose ice melts at 0 °C and we are now adding salt to lower melting point but now when we have lowered the melting point the ice will melt more quickly, then what is the purpose of mixing salt?

You want the temperature of the liquid water to be lower. The process of melting the ice cools down the ice/salt water mixture. This is analogous to evaporation cooling down our bodies when our sweat evaporates.

[comments by Buck Thorn] Note that impurities in a solid will affect the melting point, but the freezing point depression effect (a colligative effect) has to do with impurities in the liquid phase, not in the solid. The paragraph is therefore a tad confusing.

For the ice cream making process, we start with pure ice and have the impurity in the liquid. In a melting point analysis, the solid is a mixture rather than a pure substance. For both cases, the change in melting point has to do with homogeneous mixing in the liquid state, though, according to Lisa Nichols:

An impure solid is typically heterogeneous on the microscopic level, with pure regions of each component distributed through the bulk solid much like granite. When an impure solid is warmed, microscopic melting first occurs in a pure region by the component with the lower melting point (compound A in Figure 6.7a). This microscopic melting is not visible to the eye.

The preliminary melting of compound A in Figure 6.7a forms tiny pools of liquid that begin to dissolve compound B from the bulk solid. As compound B is dissolved into the melt (causing it to become more impure), the freezing point of this mixture is depressed. Compound B will continue to dissolve in the melt, until it reaches the eutectic composition (point a in Figure 6.7b), and the system will continue to melt at this composition until the entirety of the minor component (the impurity) is dissolved. Once the minor component is completely dissolved, further melting continues of the bulk component. This increases the purity of the melt, so the melting temperature increases somewhat. The system follows the melting line in Figure 6.7b either to the left or right of the eutectic temperature (depending on which side of the eutectic point is started), adjusting its melting temperature as the bulk component increases its concentration in the melt. This continues until the entire sample is melted.

Source: https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/06%3A_Miscellaneous_Techniques/6.01%3A_Melting_Point/6.1C%3A__Melting_Point_Theory

@BuckThorn I tried to address your comments to the OPs question in my answer.


text:
Is necessary for carbon to complete its octet?
I have seen a lot of carbon compounds; all of them have a complete octet for example HCN, CO2, etc. What about more exotic ones?

It's not necessary, but highly preferable. Any carbon species without an octet is nearly always very reactive.

@orthocresol can you give some examples???

http://www.sigmaaldrich.com/catalog/product/aldrich/164577?lang=en&region=US    So stable, you can buy it.

Which is why I absolutely had to add the qualifier "nearly always"!

I changed second sentence because it didn't making sense. Feel free to edit if you don't like it now, but be careful with grammar, spelling etc.

No. Carbenes and carbocations do not have complete octet. But that makes them very very reactive.

There's a number of well-known persistent radicals and salts that contain carbon atoms that do not have complete octets. Examples are

Gomberg's (triphenylmethyl) radical and derivatives,
Triphenylmethyl hexafluorophosphate,
compounds that contain tropylium cations (e.g. $\ce{[C7H7]+[BF4]-}$), or
componds that contain the 2-norbornyl cation (e.g. $\ce{[C7H11]+[Al2Br7]- \cdot CH2Br2}$).

No. $\ce{CH5+}$ has been observed, and $\ce{CH3}\,^\bullet$ is a common chemical reagent. ($\ce{CH5+}$ isn't stable, and $\ce{CH3}\,^\bullet$ is only 'stable' under special conditions, but never-the-less, both do exist.)
In the vast majority of organic chemistry you'll be exposed to, assuming carbon is tetravalent is an excellent starting place. The octet rule is best for the Groups 13-18 and for the 2nd period especially with the 3rd period elements usually obeying it.

In CH5+ carbon still has octet on contrary to for example CH3+

Furthermore, stable carbon-centred radicals exist (triphenylmethyl radical) as do stable carbocations (triphenylmethyl cation). And let’s not forget NHC’s.


text:
Largest molecule that isn't polymer?
What is the largest molecule (synthetic or natural) that isn't a polymer -- let's say, bigger than 3000 dalton?

Do covalent solids count? In that case, imagine the largest possible diamond, quartz crystal, etc.

Welcome to Chemistry.SE. I am worried that this question will not have a definitive answer. What I mean is that everyone who answers will provide the largest molecule they can think of. Then someone will discover a bigger molecule and invalidate all the existing answers.

I wouldn't expect a definitive answer, since the notions involved are not really well-defined. That being said, maitotoxin sounds like a good example.

Are biomolecules such as proteins or DNA considered to be polymers?

@aventurin - yes.

An old German book about chemical records (Quadbeck-Seeger, H.-J., Ed. Chemie Rekorde; Wiley-VCH: Weinheim, 1997) gives the Cullinan Diamond as the largest found natural molecule and a synthetic diamond of 38.4 carat as the largest made artificial molecule.