text:
How to make approximation of rotational partition function of diatomic linear molecules?
Using the rigid rotor approximation to the level energies, and such other appropriate assumptions, we can approximate rotational partition function, $Q_{\mathrm{rot}}$, of linear molecules as follows:
\begin{eqnarray}
  Q_{\mathrm{rot}} & = & \sum_{J=0}^{\infty}{(2J+1)\exp\left(-\frac{E_J}{kT}\right)}\\
  & \simeq & \sum_{J=0}^{\infty}{(2J+1)\exp\left(-\frac{hB_0J(J+1)}{kT}\right)} \\
\end{eqnarray}
In the book, Microwave Molecular Spectra (Gordy & Cook, 1984), Chapter 3, Equation 3.64, the authors made approximation as:
$$Q_{rot}\simeq\frac{kT}{hB_0}+\frac{1}{3}+\frac{1}{15}\left(\frac{hB_0}{kT}\right)+\frac{4}{315}\left(\frac{hB_0}{kT}\right)^2$$
Also, there is another approximation made by McDowell, R. S. 1988, Journal of Chemical Physics, 88, 356:
$$Q_{rot}\simeq\frac{kT}{hB_0}\exp\left(\frac{hB_0}{3kT}\right)$$

And we can see from the picture that $Q_{\mathrm{rot}}\simeq\frac{kT}{hB_0}+\frac{1}{3}$ is also a good approximation. Then I can guess how to derive McDowell's approximation, but I cannot derive Gordy & Cook's approximation.
What I have got are as follows:
Since $$n\exp(-x(n+1)n)+\int_{(n+1)n}^{(m+1)m}{\exp(-xy)}\,\mathrm dy+(m+1)\exp(-x(m+1)m) \\
\leq\sum_{J=n}^{m}{(2J+1)\exp(-x(J+1)J)} \\
\leq\sum_{i=n^2}^{m^2+2m}{\exp(-xi)}$$ hence, $$\int_{0}^{\infty}{\exp(-xy)}\,\mathrm dy=\frac{1}{x} \\
\leq\sum_{J=0}^{\infty}{(2J+1)\exp(-x(J+1)J)} \\
\leq\sum_{i=0}^{\infty}{\exp(-xi)}=\frac{1}{1-\exp(-x)}$$
i.e., $$\frac{kT}{hB_0}<Q_{\mathrm{rot}}<\frac{1}{1-\exp\left(-\frac{hB_0}{kT}\right)}$$
While the latter one approximates $\frac{1}{1-\exp\left(-\frac{hB_0}{kT}\right)}\approx \frac{kT}{hB_0}+\frac{1}{2}+\frac{1}{12}\frac{hB_0}{kT}-\frac{1}{720}\left(\frac{hB_0}{kT}\right)^3$ when $T$ is rather high.
Is there anyone knowing how to derive Gordy & Cook's approximation? Please tell me and I appreciate it very much.
Besides, another little question, are such approximations still useful as nowadays computers are greatly powerful?

This question appears to be off-topic because it is better suited for Physics.SE

I only just discovered the question so I couldn't answer it earlier. I've provided an answer which hopefully explains everything.

@Martin, still thank you and the answer has come:)

Sorry for the late answer. I just discovered this question.
Preludium
Firstly, I think you might have an error or some non-standard notation in your formula for $Q_{\mathrm{rot}}$. The rotational energy levels are given by
\begin{align}
  E_{\mathrm{rot}} = \frac{\hbar^2}{2 I} J ( J + 1 ) \ ,
\end{align}
where $I$ is the moment of inertia.
Then the rotational constant $B_{0} = \frac{\hbar^2}{2 I}$ (in units of energy) is introduced, so
\begin{align}
  E_{\mathrm{rot}} = B_{0} J ( J + 1 ) \ .
\end{align}
Thus
\begin{align}
  Q_{\mathrm{rot}} = \sum_{J=0}^{\infty}{(2J+1)\exp\left(-\frac{B_0 J(J+1)}{kT}\right)} \ ,
\end{align}
whereas your exponent contains another $h$ which should not be there. At least that's the standard notation I am used to. For convenience I'll also introduce the rotational temperature $\Theta_{\mathrm{rot}} = \frac{B_{0}}{k}$, so that
\begin{align}
  Q_{\mathrm{rot}} = \sum_{J=0}^{\infty}{(2J+1)\exp\left(-\frac{\Theta_{\mathrm{rot}} J(J+1)}{T}\right)} \ .
\end{align}
Main Answer
Ok, now that that's sorted out, let's start the real work. It is actually a pretty standard approximation that is used here: They use the Euler-Maclaurin formula
\begin{align}
  \sum_{n = a}^{b} f(n) &= \int_{a}^{b} f(n) \, \mathrm{d} n + \frac{1}{2} \bigl( f(b) + f(a) \bigr) + \sum_{i=1}^{\infty} (-1)^{i} \frac{b_{2i}}{(2i)!} \Bigl( f^{(2i - 1)}(a) - f^{(2i - 1)}(b) \Bigr)
\end{align}
where $f^{(k)}(a)$ is the $k$th derivative of $f$ evaluated at $a$. The $b_{2i}$s are the absolute values of the even Bernoulli numbers, $b_{2} = \frac{1}{6}$, $b_{4} = \frac{1}{30}$, $b_{6} = \frac{1}{42}$, $...$.
It provides a way to approximate the summation of a function over a discrete variable by an analogous integration over a continuous variable.
In the special case at hand,
\begin{align}
  f(J) = (2J+1)\exp\left(-\frac{\Theta_{\mathrm{rot}} J(J+1)}{T}\right) \ ,
\end{align}
where the summation limits are $a=0$ and $b=\infty$ and where $f(\infty) = f^{\prime}(\infty) = \ldots = 0$ this formula boils down to
\begin{align}
  Q_{\mathrm{rot}} = \sum_{J = 0}^{\infty} f(J) &= \int_{0}^{\infty} f(J) \, \mathrm{d} J + \frac{1}{2} f(0) + \sum_{i=1}^{\infty} (-1)^{i} \frac{b_{2i}}{(2i)!} f^{(2i - 1)}(0)
\end{align}
The first term is what is usually called the classical or the high-temperature limit. It can be easily calculated by making the substitution $K = J(J + 1)$ $\Rightarrow$ $\mathrm{d}J = \frac{\mathrm{d}K}{2J + 1}$. So
\begin{align}
  \int_{0}^{\infty} f(J) \, \mathrm{d} J &= \int_{0}^{\infty} (2J+1)\exp\left(-\frac{\Theta_{\mathrm{rot}} J(J+1)}{T}\right) \, \mathrm{d} J \\
  &= \int_{0}^{\infty} \exp\left(-\frac{\Theta_{\mathrm{rot}} K}{T}\right) \, \mathrm{d} K \\
  &= \frac{T}{\Theta_{\mathrm{rot}}}
\end{align}
Now, only the derivatives $f^{(k)}(0)$ need to be sorted out. Those that contain terms below the third order in $\frac{\Theta_{\mathrm{rot}}}{T}$ can be calculated to be:
\begin{align}
  f(0) &= 1 \\
  f^{(1)}(0) &= 2 - \frac{\Theta_{\mathrm{rot}}}{T} \\
  f^{(3)}(0) &= - 12 \frac{\Theta_{\mathrm{rot}}}{T} + 12 \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{2} - \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{3} \\
  f^{(5)}(0) &= 120 \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{2} - 180 \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{3} + 30 \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{4} - \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{5} \ .
\end{align}
Doing this by hand is very tedious. Best you use something like Mathematica. Plugging all this into the Euler-MacLaurin formula and dropping all terms above second order in $\frac{\Theta_{\mathrm{rot}}}{T}$ you get exactly the result of Gordy & Cook
\begin{align}
  Q_{\mathrm{rot}} &\approx \frac{T}{\Theta_{\mathrm{rot}}} + \frac{1}{2} \cdot 1 + (-1) \frac{1}{6 \cdot 2!} \left( 2 - \frac{\Theta_{\mathrm{rot}}}{T} \right) \\
 &\quad + (-1)^{2} \frac{1}{30 \cdot 4!} \left( - 12 \frac{\Theta_{\mathrm{rot}}}{T} + 12 \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{2} - \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{3} \right) \\
  &\quad + (-1)^{3} \frac{1}{42 \cdot 6!} \left(120 \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{2} - 180 \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{3} + 30 \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{4} - \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{5} \right) \\
  &= \frac{T}{\Theta_{\mathrm{rot}}} + \frac{1}{2} - \frac{1}{6} + \frac{1}{12} \frac{\Theta_{\mathrm{rot}}}{T} - \frac{1}{60} \frac{\Theta_{\mathrm{rot}}}{T} \\
  &\quad + \frac{1}{60} \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{2} - \frac{1}{252} \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{2} + \mathcal{O}\left( \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{3} \right) \\
  &= \frac{T}{\Theta_{\mathrm{rot}}} + \frac{1}{3} + \frac{1}{15} \frac{\Theta_{\mathrm{rot}}}{T} + \frac{4}{315} \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{2} + \mathcal{O}\left( \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{3} \right)
\end{align}
As a side note: If you have a homonuclear diatomic molecule you need to introduce the so called symmetry number $\sigma$ which represents the number of indistinguishable orientations a molecule can have. It originates from symmetry constrains: for homonuclear diatomic molecules there is a two-fold axis of symmetry perpendicular to the internuclear axis which gives rise to two indistinguishable orientations, so the above formula for $Q_{\mathrm{rot}}$ overcounts the number of available quantum states
by a factor of two because of the inherent indistinguishability of the nuclear pair. Thus, $Q_{\mathrm{rot}}$ needs to be divided by $\sigma = 2$. For heteronuclear diatomic molecules you simply have $\sigma = 1$.
With the introduction of $\sigma$ the equation for $Q_{\mathrm{rot}}$ becomes:
\begin{align}
  Q_{\mathrm{rot}} &\approx \frac{1}{\sigma} \left( \frac{T}{\Theta_{\mathrm{rot}}} + \frac{1}{3} + \frac{1}{15} \frac{\Theta_{\mathrm{rot}}}{T} + \frac{4}{315} \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{2} + \mathcal{O}\left( \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{3} \right) \right) \\
  &= \frac{T}{\sigma \Theta_{\mathrm{rot}}} \left(1  + \frac{1}{3} \frac{\Theta_{\mathrm{rot}}}{T} + \frac{1}{15} \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{2} + \frac{4}{315} \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{3} + \mathcal{O}\left( \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{4} \right) \right)
\end{align}

+1, this is a really exemplary answer. I just wanted to provide a link to an article that I found useful when studying the connection between the Euler-Maclaurin formula and Bernoulli numbers in the context of studying the calculus of finite differences.

@GregE. Thanks for the kind comment and the link. Looks interesting.

@Philipp, thank you. I really appreciate it. As for $h$, is it due to c.g.s units? C.g.s units are still commonly used in astronomy, and it might still need a long time for the conversion to S.I. units.

@puresky Glad to hear that my answer is helpful. Unfortunately, I'm not familiar enough with c.g.s units to tell you whether they are the reason for the additional $h$. Might well be but I can't say for sure. This would be a question better suited for Physics.SE I think. By the way, I added some information about homonuclear diatomic molecules and the symmetry number that needs to be introduced when you want to calculate the partition sum for them at the end of my answer.


text:
Are ther known compounds of Argon? What is their molecular geometry and hybridisation?
Can Argon hybridize orbitals and/or form covalent-like/ionic-like compounds?
Is there any study of that? I would be happy to read concrete references.
What kind of molecular geometries for argon compounds are known or hypothesized at the current time?
Is there some peculiarity to apply something like VSEPR or molecular orbital theory to theoretical or experimentally known argon compounds?

Argon is a quite unreactive element, however very few compounds of it are known.
The most popular, and currently the only neutral compound, is argon hydroflouride, $\ce{HArF}$, which was reported by L. Khriachtchev et.al..[1]  In the same issue there is also a comment on the matter by my former supervisor G. Frenking.[2]
Another interesting read on the subject is the prediction of the metastable $\ce{HHeF}$, by M. W. Wong.[3] It generally treats all compounds of the $\ce{HNgF}$ family.
All calculations done so far indicate, that the molecule is linear and only kinetically stable. The highly accurate CCSD(T)/aug-cc-pV5Z level predicts, that the $\ce{HArF}$ molecule is $5.87~\mathrm{eV}$ higher in energy than its fragments $\ce{HF}$ and $\ce{Ar}$. The minimum barrier on this level of theory is a mere $0.35~\mathrm{eV}$ along the stretching coordinate of the hydrogen argon bond. The bending is predicted to be higher in energy. The reported bond distances for $\ce{H-Ar}$ and $\ce{Ar-F}$ are $133~\mathrm{pm}$ and $197~\mathrm{pm}$, respectively.
An NBO analysis on the MP2/aug-cc-pVTZ level of theory shows a strong ionic character of the molecule, it is reported, that there is a strong $\ce{(HAr)+F- }$ charge transfer in the electronic structure of the molecule.
Wong performed accurate calculations on the QCISD/6-311+G** level of theory and combined it with a Quantum Theory of Atoms in Molecules approach. He calculated a positive value of the Laplacian of the electron density $\nabla^2\rho_\mathrm{b}$ for the bond critical point of the argon fluoride bond, which indicates a mainly electrostatic interaction. The contrary is the case for the argon hydrogen bond, where the Laplacian is found to be negative, i.e. charge concentration, and therefore indicating a dominantly covalent interaction.
The following picture is a snapshot of this behaviour at the BP86/cc-pVTZ level of theory.[4]

Solid lines are regions of charge depletion, $\nabla^2\rho_\mathrm{b}>0$, while dashed lines are regions of charge concentration $\nabla^2\rho_\mathrm{b}<0$. Red lines are zero flux surfaces, where $\nabla\rho_\mathrm{b}=0$. And red spheres indicate bond critical points.
This molecule is quite interesting for many reasons, however, since it is only kinetically stable it has to be treated with fullest force of quantum chemical calculations. The calculations I did on a low level DFT method are not sufficient at all on its own, as has been investigated by J. Panek et.al..[5]]
Hybridisation is in a wider sense the mixing of s and p orbitals;[6] and upon further examination of the electron density this will be observed in this molecule. Most likely for the hydrogen argon covalent bond. As hybridisation is just a mathematical transformation of the original atomic orbitals there is no real interpretative value in this case. For example, the NBO analysis at the BP86/cc-pVTZ level of theory give only a negligible 3% s contribution of argon to the $\ce{H-Ar}$ bond.
The VSEPR theory is only a observational concept for well-behaved molecules. It basically lacks a solid scientific basis and should not be used any more, but especially in cases like this, it must not be used.

Literature and Notes

Leonid Khriachtchev, Mika Pettersson, Nino Runeberg, Jan Lundell, and Markku Räsänen, Nature, 2000, 406 874-876.
Gernot Frenking, Nature, 2000, 406, 836-837.
Ming Wah Wong, J. Am. Chem. Soc., 2000, 122 (26), 6289-6290.
To avoid copyright issues I performed a quick calculation to illustrate the above raised point.
Jarosław Panek, Zdzisław Latajka, and Jan Lundell, Phys. Chem. Chem. Phys., 2002, 4, 2504-2510.
Hybridisation is a concept that dates back to the beginnings of Valence Bond (VB) Theory. It was originally implemented in the strict fashion of creating $s^{1/(n+1)}p^{n/(n+1)}$ or $sp^n$ hybrid orbitals for $n = \{1,2,3\}$. H. Bent later went on suggesting fractional hybridisation schemes, see Henry A. Bent, Chem. Rev., 1961, 61 (3), 275–311. Nowadays the term s-p-mixing in Molecular Orbital theory is almost synonymous with the VB theory term hybridisation.

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text:
How to calculate the pH of a buffered solution with Henderson Hasselbalch?
Lets say you have a buffered solution containing $0.50~\mathrm{M}$ acetic acid ($K_a = 1.8 \cdot 10^{-5}$) and $0.50~\mathrm{M}$ sodium acetate, and you want to calculate the $\mathrm{p}\ce{H}$ of this solution.
This can be done in two ways: using the standard equilibrium chart (initial, change, equilibrium), or using the Henderson Hasselbalch equation.
My question is why do these two give the same result.
The Henderson Hasselbalch equation is derived starting with the equilibrium equation:
$K_a = \frac{\ce{[H+][CH3COO^{-}]}}{\ce{[CH3COOH]}}$ and then the $\ce{[H+]}$ is found, and then the $\log_{10}$ is taken. However, doesn't this assume that the system is already at equilibrium and that the concentrations of $\ce{[CH3COO^{-}]}$ and $\ce{[CH3COOH]}$ are the concentrations at equilibrium?
However, that is clearly not the case as shown by the ICE chart. However, both give the correct pH. Why is this?

The main equilibrium in such a solution is the following:
$\ce{CH3COOH +CH3COO^- <=>CH3COOH + CH3COO^-}$.
This should be your ICE chart's equation too. I suspect that the equation you have for your ICE table is the following:
$\ce{CH3COOH +H_2O <=>CH3COO^- + H_3^+O}$
You can certainly solve the second equation with an ICE table by plugging in the initial concentration of the conjugate base, $\ce{CH3COO^-}$.

However, doesn't this assume that the system is already at equilibrium
  and that the concentrations of $\ce{[CH3COO-]}$ and $\ce{[CH3COOH]}$
  are the concentrations at equilibrium?

Yes. The concentrations (or more accurately, activities) used in the H-H equation are equilibrium concentrations.
In the case of a 0.5 M acetic acid/acetate ion buffer, we can, for most purposes, conclude that the initial molarity of the main system components - acetic acid and acetate ion will remain constant. This is because the main acid-base reaction in the system is the following:
$\ce{CH3COOH +CH3COO^- <=>CH3COOH + CH3COO^-}$.
As we can see, the above equation is symmetric. There is no net change in system composition resulting from the above equation and the given molarities.
We also have these equilibria:
$\ce{CH3COOH + H2O- <=>H_3^+O + CH3COO^-}$
$\ce{H2O + CH3COO^- <=>CH3COOH + HO^-}$
But both acetic acid and its conjugate base, acetate ion, are weak. So these won't do much to change the system composition. Plus these aren't the favored reaction in the system; that would be the one above (the one between the strongest acid and strongest base in the system).

If you wanted to solve this system analytically - i.e. you want an even more accurate measure of the system's pH, then you would have to employ the law of conservation of mass and invoke the concept of solution electro neutrality.
Mass conservation:
$1.0 M$ $\ce{=[CH3COOH] + [CH3COO^{-}]}$
Solution electroneutrality:
$\ce{[HO^-{}] + [CH3COO^{-}] = [H_3^+O]}$.
These two statements, coupled with $\ce{K_a(CH3COOH)}$ and $\ce{K_b(CH3COO^{-})}$, and $\ce{K_w}$, give you a system of equations in which you can solve for $\ce{[H_3^+O]}$.

I think I got it. So this can also be used for the calculating the pH when titrating a weak acid with a strong base (or weak base with strong acid), right? Because once the reagent acid reacts with the original base, it will produce the conjugate acid and what is left is now a buffer, whose pH can be determined with this, right?

However, it cannot be employed for calculating pH when titrating a strong acid with a strong base because in that case the equilibrium concentrations will be completely different from the initial ones, right?

You can definitely use the HH equation to find the pH of a titration between a weak acid and a strong base. Just make sure you're using equilibrium concentrations; strong bases will completely abstract protons from the acid. Do not use initial concentrations when dealing with anything strong.

If you are titrating a strong acid with a strong base I would just take the easy way out by finding how much of the strong acid and strong base are consumed when they react with each other and then just take the negative log of the hydronium or hydroxide ion concentration. Note that this method only works for non-very dilute solutions.

@DissenterI try to solve this system analytically according to the following equation :
$$
[\ce{H+}]^3
+([\ce{CH3COONa}]0 + 1)[\ce{H+}]^2
-(K\mathrm{w}+[\ce{CH3COOH}]0K\mathrm{a})[\ce{H+}] -K_\mathrm{w}=0

$$Is the equation correct?thank you.

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text:
What happens when a solution of sodium chlorite is acidified?
I'm told that when sodium chlorite $\ce{(NaClO2)}$ is mixed with aqueous acid, chlorine dioxide ($\ce{ClO2}$) is produced and remains mostly dissolved in water.
I would like to know how the strength of the acid (concentration of acid and/or the type of acid) affects this reaction. I would also like to know what happens (afterwards) to the $\ce{ClO2}$ over time. For example, if undisturbed, does the $\ce{ClO2}$ enter the air, convert into another compound, or something else?

When acid is added to sodium chlorite in solution (note: $\ce{ClO2-}$ is the chlorite ion), chlorous acid ($\ce{HClO2}$) is transiently formed which then goes on to decompose. However, you need hydrochloric acid, not just any acid, because chloride ion ($\ce{Cl-}$) itself acts as a catalyst for the decomposition. What happens is that chlorous acid disproportionates; that is, some of it gets oxidized while the rest gets reduced.
Specifically, out of five $\ce{Cl}$ atoms in chlorous acid (oxidation state +3), one will end up as chloride in the oxidation state $-1$, and four will end up as $\ce{ClO2}$ in the oxidation state +4:
$$\ce{5HClO2 -> 4ClO2 + Cl- + H+ + 2H2O}$$
Pretty much any concentration will do. Calculating the proportions is easy if you know a bit of stoichiometry: you need 5 moles of chlorite for 4 moles of hydrochloric acid. The reaction proceeds to completion, and you get 4 moles of $\ce{ClO2}$.
The $\ce{ClO2}$ remains in solution and is a good bleaching agent; see Wikipedia. It can undergo further disproportionation but if the solutions are kept in the dark, this process is slow.

sorry, I messed that one up real bad.


text:
Why are solutions electrically neutral?
I keep reading that solutions are electrically neutral, but I've never read an explanation. I've read after the fact justifications, such as one problem which asked me to calculate the force of repulsion between two beakers with a tiny, tiny bit of excess charge (it ended up being over 10,000 pounds).
So why are solutions electrically neutral? Not interested in after the fact explanations like the above example.

What do you mean by "after the fact explanations," exactly? Does an explanation based on Coulombic repulsion/attraction qualify? Or the accumulation of potential energy resulting from buildup of like-charged particles and/or separation of oppositely charged ones (and hence, the work done by, e.g., an E-field necessary to produce and/or maintain such an arrangement)?

it will be more useful if you give why you think they aren't or how can electrical neutrality of solution be disturbed ; because we never added charge externally, we just added salts, for an example thus sustaining it to be neutral.

Possible duplicate (http://physics.stackexchange.com/questions/62181/why-do-aqueous-solutions-always-have-to-be-electrically-neutral) or(http://chemistry.stackexchange.com/questions/4805/why-do-aqueous-solutions-always-have-to-be-electrically-neutral)

Take a look at this question, too.

@GregE. I mean that the Coulombic repulsion/attraction problem I read in my textbook was like saying humans aren't 100 feet tall because ceilings aren't 100 feet high. In other words it seemed to have cause and effect reversed.

I think there is a misunderstanding here about the difference between explanations and observations.
The fact that two beakers of aqueous solution do not have a measurable force between them is an observation that proves the solution has no (large) charge. Calculating a hypothetical force due to a charged solution and finding that it is much larger than what we observe is not an after-the-fact justification, it is a falsification of a hypothesis by an empirical observation.
The explanation is that no matter what the ionic compound that you started with was, it was electrically neutral. Therefore, the solution must still be neutral.

Are macroscopic solid ionic compounds really neutral though, or is that an idealisation? I don't think it's possible to make an argument that any real object must be exactly electrically neutral, just that it has to be neutral to within a good approximation.

So why must the ionic compound we started with be neutral? Again, I understand the stability arguments - i.e. isolating ions means we'd also be isolating electrons. So I guess our not being able to isolate electrons is an explanation rather than an observation. Thanks!

@NicolauSakerNeto I agree, we can "charge" glass by rubbing it with a silk cloth. Can we similarly add charge to a solution?

@Nicolau - let's put it this way: They are as neutral as any macroscopic compound can be.

It is more acurate to say that they aren't very far from being electrically neutral. If you take a charged sphere and drop it into a beaker of some electrolyte sitting on an insulating counter, the result is a charged solution, albeit slightly.
Also, in spite of your amusing ceiling analogy (it made me chuckle), the argument about electrostatic attraction and repulsion does justify why solutions (and solids) can't be too far from being electrically neutral.  The flaw in the analogy is that humans don't need ceilings to exist; but you can' t have two charged beakers without a force between them. If each has a charge of 1 C -- which is a mere $10^{-5}$ times the charge of a mole of electrons -- and they're sitting 1m apart the force between them is about a million tons -- clearly this situation won't last very long!

Because the solute is neutral when its solid so it must me neutral when it's aqueous.


text:
Finding the initial concentration
The solubility product of $\ce{AgBr}$ is $7.7 \cdot 10^{-13}\:\mathrm{mol^2/L^2}$.
What was the initial concentration of $\ce{AgNO3}$ solution, if the precipitation of $\ce{AgBr}$ appears after the addition of $20\:\mathrm{mL}$ of a $0.001$ molar solution of $\ce{NaBr}$ to $500\:\mathrm{mL}$ of the $\ce{AgNO3}$ solution.
I got the solution as $0.054\:\mathrm{M}$. I'm confused with the procedure. This is what I have done.

Precipitate occurs at $K_{sp}=Q$ and $Q=[\ce{Ag+}][\ce{Br-}]$
$[\ce{Ag+}] = [\text{(Vol of $\ce{AgNO3}$)}\cdot\text{Molarity} ] / \text{Total Volume of the mixture}$
Similarly for $\ce{Br-}$
$[\ce{Ag+}][\ce{Br-}] = 2\cdot10^5$

The answer I got it $0.054\:\mathrm{M}$ ? Is that correct ?

Would be correct if you used the number at step 4 -- the number you correctly wrote in step 1! Where did $2\cdot 10^5$ come from?

This is a titration problem for quantitatively determining the concentration of a solution.
What reaction does occur?

 $\ce{AgNO3 (aq) + NaBr (aq) <=> AgBr v + Na+ (aq) + NO3- (aq)}$
 or essentially $\ce{Ag+ + Br- <=> AgBr v}$

Why is the solubility product important?

 The solubility product tells you about the extent of the reaction. In this particular case it tells you, that you reached equilibrium between ions in solution and the precipitated salt. It tells you exactly the product of concentrations in a saturated solution.

What can you tell about the equilibrium state at the point when the first precipitate falls?

 The solubility product is matched, hence $\ce{[Ag^+][Br^-]}<K_s=7.7\cdot10^{-13}~\mathrm{\left(\frac{mol}{L}\right)}^2$

What is the amount of bromine ions added to the solution?

 $n(\ce{Br^-}) = V(\ce{NaBr})\cdot c(\ce{NaBr}) = 0.020~\mathrm{mL}\cdot 0.001~\mathrm{\frac{mol}{L}} = 2\cdot10^{-5}~\mathrm{mol}$

What can you tell about the concentrations in the final mixture?
First, what is the concentration of bromide ions in this mixture?

 $V_0(\ce{AgNO3}) = 0.5~\mathrm{L}$, $V(\ce{NaBr}) = 0.02~\mathrm{L}$, $V_t = 0.52~\mathrm{L}$
 $c_t(\ce{Br^-}) = \frac{n(\ce{Br^-}}{V_t} \approx 3.8\cdot10^{-5}~\mathrm{\frac{mol}{L}}$

Second, what can you tell about the concentration of silver ions in the final mixture?

 $c(\ce{Ag^+}) = \frac{K_s}{c(\ce{Br^-})} = \frac{K_s}{\frac{V(\ce{NaBr})}{V_t}\cdot c(\ce{NaBr})} = \frac{K_s\cdot V_t}{V(\ce{NaBr})\cdot c(\ce{NaBr})} \approx 2\cdot10^{-7}~\mathrm{\frac{mol}{L}}$

What is the number of moles of silver ions in the final mixture?

 $n(\ce{Ag^+}) = c(\ce{Ag^+})\cdot V_t = \frac{K_s\cdot V_t^2}{V(\ce{NaBr})\cdot c(\ce{NaBr})} \approx 1\cdot10^{-7}~\mathrm{mol}$

What is the initial concentration of the silver nitrate solution?

 $c_0(\ce{AgNO3}) = \frac{n(\ce{Ag^+})}{V_0(\ce{AgNO3})} = \frac{K_s\cdot V_t^2}{V_0(\ce{AgNO3})\cdot V(\ce{NaBr})\cdot c(\ce{NaBr})} = 2.08\cdot10^{-8}~\mathrm{\frac{mol}{L}}$

Martin - Are you sure that it's not just $\ce{[Ag+]}[\ce{Br-}]=7.7\cdot 10^{-13}$ (mol/L)$^2$ at the moment the solution starts to get cloudy?

@SilvioLevy I am very sure that is true. I understood the question in the way that the concentration of silver nitrate is looked for, before the sodium bromide is added to this solution.

Yes, the question asks for the concentration before the addition of NaBr, but  what I'm talking about the concentrations at the moment the solution turns.  Why is $[\ce{Ag+}]=[\ce{Br^-}]$?  To put it another way: your answer does not use the solubility product. If the molarities are the same at your "equivalence point", how does $\sim$0.00004 molar of bromide coexist in solution with $\sim$0.00004 molar of silver ion, just before the solution turns?  That would mean $[\ce{Ag+}][\ce{Br-}]=1.6\cdot 10^{-9}\gg 7.7\cdot 10^{-13}$. (See also my reply to the comment you added to another answer.)

@SilvioLevy You are right, I was thinking about a titration with Mohr's method (there is no english wiki on this), but there you add an indicator to ensure you reached the equivalent point, which is not true in this case. I have to rework the answer or delete it all together.

Right, in Mohr's method $\ce{Ag^+}$ is the titrant and there is indeed an equivalence point (longer after precipitation has started) that is detected by the turning of an indicator that reacts with excess $\ce{Ag^+}$.  Good memories! And I hope you do rework your solution and don't delete it: I love the way the answers appear when you hover over the empty boxes! Also your answer is the only one so far to call attention to the 0.5L versus 0.52L subtlety.

@SilvioLevy Thanks again for pointing out the mistakes. Please check my edit, I think it should be correct now.

Perfect answer now, but I've made a suggestion for clarity, by editing the 3rd answer directly.  I think you have the reputation to see it and approve it you think it helps.

The key is to get the concentration of bromide ions and use that value in the Solubility Equation as defined in step 1 to get $\ce{[Ag^+]}$:
$K_{sp} = [Br^-][Ag^+]$
The analysis and procedure is fine, except the product in step 4 it's a little bit big. Check the algebra reorder there. The answer I get is $2\cdot 10^{-8}~\mathrm{M}$. I would comment but I'm new in Chemistry Beta and can't do that. Hope that helps,

The way you posted your calculation is confusing. You should be clear on what you want in your statement.
First, find the number of moles of $Br^-$,

$\#\ moles\ Br^-=0.020L\cdot 0.001 M$
$\#\ moles\ Br^-=2\cdot 10^{-5} moles$

Now find the concentration of $Ag^+$ in the 520 mL solution,

$K_{sp}=[Ag^+][Br^-]$
$[Ag^+]=\frac {K_{sp}}{[Br^-]}$
$[Ag^+]=\frac {7.7⋅10^{−13}mol^2/L^2}{\frac {2\cdot 10^{-5} moles}{0.520L}}$
$[Ag^+]=\frac {7.7⋅10^{−13}mol^2/L^2}{3.84\cdot 10^{-5} moles/L}$
$[Ag^+]=2.00\cdot 10^{-8} moles/L$

Now find the concentration of $AgNO_3$ of the initial solution

$[Ag^+]=2.00\cdot 10^{-8} moles/L\cdot \frac {0.520 L}{0.500 L}$
$[Ag^+]=2.10\cdot 10^{-8} moles/L$

So the concentration of $AgNO_3$ of the initial solution is $2.10\cdot 10^{-8} moles/L$.

This answer is basically right, but it fails to take into account that the volume of solution grew from 0.5L to 0.52L.  @LDC3, maybe you can fix it and then whoever downvoted it will reconsider?

@SilvioLevy The question states "What was the initial concentration of $AgNO_3$ solution?" I just didn't make that statement at the end.

@SilvioLevy I see what you're saying. I made a mistake in calculating the silver concentration.


text:
Homoassociation and Acid Strength
I understand that the strength of HF differs significantly depending on the concentration. I understand the basic theory behind this - homoassociation - and stabilization of the highly negative charge dense fluoride anion through hydrogen "bonds"/intermolecular attractions with undissociated solute $\ce{HF}$ molecules.
However, I am unclear on the Wikipedia explanation of homoassociation:
http://en.wikipedia.org/wiki/Homoassociation
Regarding the equations - I cannot tell if the authors meant to refer to pure, liquid $\ce{HF}$ or a solution of HF. I'm thinking that the authors are referring to liquid $\ce{HF}$ because auto-ionization of HF is mentioned. I'm thinking that auto-ionization would only be significant in liquid $\ce{HF}$. Or in a concentrated solution.
Also, regarding the second equation - the authors make it look like the fluoride anion is forming an actual, intramolecular, bond with the $\ce{HF}$ molecule. I think the authors are implying the formation of an intermolecular hydrogen bond, correct?
On the other hand the bifluoride anion seems to be a real molecular entity (http://en.wikipedia.org/wiki/Bifluoride).

This only muddies the water further. I.e. which of the following best represents how the fluoride anion is stabilized, causing concentrated solutions of $\ce{HF}$ to ionize further than one would expect from $\ce{K_a}$ values?
$\ce{HF}$ ... $\ce{F^-}$ (hydrogen bonding)
or
$\ce{HF_2^-}$ (formation of bifluoride anion)
Also, any further links to resources about homoassociation would be appreciated.
ETA: Also, can't the fluoride anion form hydrogen bonds with water? The fact that concentrated HF is stronger than dilute HF than what mere $\ce{Ka}$ values suggest implies that H-bonding with water isn't as effective as H-bonding with HF molecules. Why? Is water less polar? On the other hand water can form a maximum of two hydrogen bonds with the fluoride anion, while a HF molecule can only form a maximum of one hydrogen bond with a fluoride anion.

I think this Wikipedia article on hydrogen fluoride will answer your questions.  It discusses the acidity of $\ce{HF}$ from dilute to concentrated aqueous solution, and then pure $\ce{HF}$.  They present four equilibria to describe these situations
\begin{aligned}
\ce{H2O + HF &~<=> [H3O^+ {.} F^{-}]&&(I)}\\
\ce{[H3O^+ {.} F^{-}] &~<=> H3O^+ + F^{-}&&(II)}\\
\ce{[H3O^+ {.} F^{-}] + HF &~<=> H3O^+ + HF2^{-}&&(III)}\\
\ce{3HF &~<=> H2F^+ + HF2^{-}&&(IV)}\\
\end{aligned}
In dilute solution equilibrium (I) lies far to the right and the second equibrium (II) far to the left.  So even though the $\ce{HF}$ is extensively dissociated, the resultant hydrogen bonded ion pair is very stable and reluctant to undergo significant dissociation.  Hence the solution is weakly acidic.  As the concentration of $\ce{HF}$ is increased homodissociation comes into play [equilibrium (III)] producing the hydrogen difluoride ion as a discrete entity which is stabilized through intermolecular hydrogen bonding.  The increase in $\ce{H3O+}$ makes the resultant solution very acidic.  In pure $\ce{HF}$, autoionization occurs producing an exteremly acidic solution.  Page 4 of this link shows a graph of HF acidity from near zero to 50% concentration.

Why would H2F- be written as a molecule if in fact it is comprised of hydrogen bonds?

I'm not sure of how you are using the phrase "comprised of hydrogen bonds".  Methane is comprised of hydrogen bonds (to carbon). $\ce{H2F^{-}}$ is an ion with real $\ce{H-F}$ bonds.  The hydrogens in these ions then form intermolecular hydrogen bonds with other other $\ce{H2F^{-}}$ molecules.

I see. I thought the hydrogen bonds were formed between HF and F-. The Wiki article is supremely unclear as it starts out saying: "In acid–base chemistry, homoassociation (an IUPAC term)[1] is an association between a base and its conjugate acid through a hydrogen bond." It appeared to me that by base, the article means F-, and that by conjugate acid, the article meant HF. Not H2F(-), as that would be the fluoride anion's second conjugate acid.

"It appeared to me that by base, the article means F-, and that by conjugate acid, the article meant HF"  That's close, but in this case the F- is (was) locked up in the $\ce{[H3O^+ {.} F^{-}]}$ ion pair

@Martin, why didn't you consider hydrogen bonding between the fluoride anion and water? Why isn't this significant? It appears to me that this should be significant given the relative concentration of water to hydronium ion. Also ron - with regard to your earlier comment about H-bonds in methane - I meant H-bonding as in the intermolecular association force (and sometimes intramolecular force).


text:
Systematic name of OCS (COS)
What is the systematic name of carbonyl sulfide? My professor commented that the systematic name was "very odd" so he didn't bother to mention it. Is it really that odd? And what is it, actually?
I found the name "Thioxomethanone". Is that really odd? From my limited knowledge, I've seen the root "thiol" before in describing something with a sulfur in it. So at least the root doesn't seem weird. I also see "oxo." There's an O in OCS.

well, "thioxomethanone" name surely is decipherable, like e.g. "benzene-1-ylwater" or "norcresol" for phenol :) but is it even organic compound?

Wikipedia lists the IUPAC name as "carbon oxide sulfide."  Odd? Sounds kind of logical to me.

Is IUPAC the same as systematic name?

Yes, it's the only system I'm aware of for naming compounds.

http://youtu.be/QBJk_L0NmMg?t=6m1s

I think this is what he came across: http://www.chemspider.com/Chemical-Structure.9644.html

please be noted that the Wikipedia article was improved to contain referenced IUPAC names (see my answer)

I'm not sure if there's definitive answer to not-yet-asked question, whether this compound can be considered organic or inorganic compound.
While organic nomenclature books do not mention it (acording to the current rules it might be called sulfanylidenemethanone, which is somewhat absurd), official IUPAC inorganic chemistry nomenclature (Red Book 2005, table IX, p. 292) lists two names for this compound:

carbonyl sulfide
oxidosulfidocarbon

Your mentioned (organic) systematic name thioxomethanone is just a variant of already mentioned sulfanylidenemethanone, as S= group's preferred name ("preselected prefix") is now sulfanylidene-* instead of thioxo- (rule P-15.5.3.3)

*) see analogy H2C=/H3C− methylidene-/methyl-, and
S=/HS− sulfanylidene-/sulfanyl- (older name: mercapto-)

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text:
Is there a possible effect of mechanical vibration on the stability of organic molecules?
Is there a possible theoretical effect of mechanical vibration on the stability of
organic molecules.  E.g. some medicines degrade if they are not stored in a fridge, is there a theoretical foundation that suggests that low frequency vibrations might have the same effect?

What do you mean by "low frequency vibrations"?  The term "extemely low frequency" is used to describe earthquake type frequencies (3-30 Hz, give or take), whereas "low frequency" is used to describe frequencies in the 30kHz - 300 kHz range.  If you mean "low frequency" as defined in the previous sentence, then yes, that frequency range of vibration is known to have chemical effects.  The process involves the formation and collapse of bubbles (cavitation) in the solution.  High temperatures and pressures can be generated during the bubble formation and collapse process.  The typical lab apparatus can homogenize samples, produce catalytic effects by cleaning metal surfaces or increasing phase transfer, disrupt cells, and break some (weak?) chemical bonds.  This reference reports,"Ligand-metal bond cleavage in transition metal complexes to give coordinatively unsaturated species or modified complexes as well as complete strip off of ligands to produce amorphous metals disrupture of the solvent structure altering the solvation of reactants Sonolysis of molecules (homolytic fragmentation to radicals, rupture of polymers, generation of excited states, cell disrupture)."


text:
Separation of Binary mixture?
I will be given a binary mixture containing two unknown compounds(the mixture could be either liquid or solid- I don't know yet). The plan of the experiment is to identify the two compounds. So my primary concern is to separate the binary mixture and purify the individual compounds(then the subsequent classification tests/NMR/IR analysis should be straightforward).  I am thinking of doing an acid-base extraction but don't know how to since I don't know any physical properties of the compounds. Any suggestions would be appreciated.

I've tried to give a very general answer to your question, but it's hard to make specific suggestions without more information. Any further details you can provide will enhance the quality of responses.

I am doing this experiment for a OChem lab at school. Unfortunately this is all I know about the experiment(we don't even get a list of possible choices) and my instructor only allows for IR once we have done the classification tests(on the separated compounds). So I guess I will be trying random solvents.

chromatography on silicagel in  organic solvents is routinely used to separate organic compounds. It has its limitations, but its worth a shot in case nothing else pops up

N.B.: the advice below mostly applies to organic compounds. If you're dealing with inorganic salts, or solutions thereof, the approach will be very different.
For a solid mixture, I would take an IR first. That should allow you to rule out potentially useless chemical tests by identifying the functional groups present in your mixture without wasting any of the sample. This would also give you some indication whether acid/base extraction is likely to be of any use. After that, it seems logical to test solubility in a wide range of solvents and at different temperatures. If you can find one that dissolves your mixture only partially at low temperature, but completely at high temperature, then you can likely separate the components immediately by recrystallization. Judicious use of a centrifuge can help if the crystals form a fine suspension.
In the case of a liquid mixture, I would again take an IR initially. In some cases, one component of the mixture can be chemically separated, and IR results would again be useful in determining the viability of that approach. Ketones and aldehydes, for example, will react to form semicarbazones that can be recrystallized and then identified by melting point. As far as physical methods of separation, standard are distillation and fractional freezing.


text:
Nitric acid vs. chloric acid strength

Here the central chlorine atom is best described as $\ce{sp^3}$ hybridized. This makes sense; the molecule is pyramidal. Also, due to the molecule's pyramidal geometry, this rules out the possibility of effective resonance due to less overlap between the  $\ce{sp^3}$ lobes, which are angled, as opposed to the parallel pi lobes.

This explains quite well the difference in acidity between nitric acid and chloric acid; nitric acid is stronger than chloric acid. At first it seems difficult to differentiate between the acidities of both; the nitrate anion also has three peripheral oxygen atoms, which provide a similar (negative) inductive effect by delocalizing electron density.

With regard to electronegativity, the chlorate anion seems like it should be more slightly stable than the nitrate anion, implying that chloric acid is stronger than nitric acid; chlorine is more electronegative than nitrogen (if only slightly so - difference is 0.12 on the Pauling scale).
Oxidation state analysis suggests that the two should be similar in acidity; both the chlorine and the nitrogen have oxidation states of +5.
So it seems that the best rationalization of why nitric acid is stronger than chloric acid is the "better" (overlapping pi orbitals) resonance form of the nitrate anion.
With respect to the above rationalization,
1) How, exactly, do parallel pi orbitals enable "better" resonance? I get the principle of maximum overlap. If I had to guess, I would say that more overlap enables greater distribution of electrons among the relevant orbitals ... is there a more concise and scientific way of putting this? Is this even correct?
2) Are parallel pi orbitals necessary for resonance? I got the impression it is from organic class. Or are parallel pi orbitals only necessary for a special form of resonance - conjugation? Is the picture on the bottom referring to conjugation?
3) Are there any other ways to rationalize the difference in strength between these two acids?

I have no definitive answer, but two things I'd note: (1) don't ignore solvation; nitrate is smaller than chlorate, which likely means better charge stabilization by solvent molecules (especially protic), (2) resonance is an abstraction, not a real phenomenon; it's a convenient device that allows electron delocalization to be represented within VB theory. Conjugation refers to systems of overlapping $p$ AOs that combine to form alternating $\pi$-bonds. This requires planar geometry, and probably isn't fully applicable to chlorate. MO theory is likely necessary for an adequate explanation.

How does size exactly affect charge stabilization? I'm not understanding the underlying mechanism.

A larger molecule requires a larger shell of solvent molecules to solvate it. In the case of an anion in a protic solvent, solvation is stabilization of the negative charge by intermolecular bonding with the $\delta^{+}$ of hydrogen atoms. This need for a larger shell may have an undesirable enthalpic effect (typically if the charge is localized and inaccessible due to bulky groups, probably not the case here), as well as an undesirable entropic one, because the shell of solvent molecules is highly ordered. Hence, the change in free energy associated with solvation becomes less favorable.

Ah, I see, thank you! Good considerations, and ones I'm sure my prof would appreciate. Also to be clear, you mean that steric factors can cause the solvation energy to be less exo/possibly endothermic?

Also is water a protic solvent? Finally I'm trying to think through how the mechanism behind the enthalpy part works. The entropy part makes sense though.

Yes, when I refer to the enthalpic effect, I do mean that steric factors may cause the solvation to be less exothermic than it would be otherwise. And yes, water is a protic solvent. The term refers to any substance capable of hydrogen bonding, the relevance being that hydrogen bonding is particularly good at stabilizing negative charge.

Okay I think I have an idea of the enthalpic factors; I think of frustrated Lewis acids and bases. And ah, good old H bonding; brings me back to another question of mine regarding the strength of HF. And that brings me to nitrous acid which can hydrogen bond within each molecule. Probably enthalpically favorable but not entropically unfavorable to tie oneself up like that in the case of HONO.

Oops double negative, I meant entropically unfavorable to h bond intramolecularly

@GregE. You are correct in your assumption, that the MO scheme for nitrate is not applicable for chlorate. Conjugation is simply not possible for the latter.

A few comments, 1) "a neighboring nonbonding pair (in a p orbital)", it doesn't need to be a p orbital, for example it could be an $\ce{sp^3}$ orbital; 2) "they must be separated by one bond", to include the possibility of homoconjugation perhaps "separated by at least one bond" might be more accurate, 3) "the p orbitals must be parallel", IIRC the overlap fall-off is a function of $\ce{cos^2\Theta}$ so you still have >90% overlap with a 15 degree twist between 2 p orbitals

The two ions you are comparing to each other are very different. In $\ce{ClO3-}$ there are two more electrons than in $\ce{NO3-}$. This is also the reason why the former pyramidalizes, while the latter does not.
In nitrate the negative charge can be delocalised over the whole molecule, since there is one molecular orbital with $\pi$ symmetry. One basic requirement for these orbitals is, that they are antisymmetric with respect to a mirror plane, that includes the bonding axis.
This MO scheme would also be thinkable for the chlorate ion, but here you have two more electrons, so you have to fill an antibonding orbital. The charge can no longer be delocalised as well in this molecule. The conclusion is, that the symmetry of the whole molecule is lowered and the $\pi$ orbitals vanish. A lone pair is always more stable in an orbital with high $\ce{s}$ contribution, which leaves only the $\ce{p}$ orbitals of chlorine for bonding with oxygen.
Resonance is a model which is necessary to describe bonding situations, that cannot be described by one Lewis structure alone. It is therefore more a rule of thumb than any evidence. $\pi$ bonds (or orbitals) are not necessary for valid resonance structures. They are however necessary for ([negative] hyper-)conjugation.

First, we must see $pK_a's$, too see which factors which we must account:
$$\begin{array}{c|c|}
&\ce{HNO3}&-1.4\\
&\ce{HClO3}&-1\\
\end{array}$$
The strength of a Bronsted-Lowry acid depends upon the extent to which the proton can be separated from it and transferred to a base.Removing the proton involves breaking a bond tp the proton, and it involves making the conjugate base more electrically negative.
Relationship between structure and acidity can be seen through factors:

Hybridisation
Inductive Effects(Electronegativity)

There is also an effect of the solvent on acidity.
Now let's compare both acids:
$$\begin{array}{c|c|c}
&acid&EN&\text{Hybridisation}\\
&\ce{HNO3}&N\equiv3.0^*&sp^2&\\
&\ce{HClO3}&Cl\equiv3.0^*&sp^3&\\
\end{array}$$
$^*$may vary a little
So Hybridisation suggests N to be more acidic as %s character is more and thus the negative charge on O will be better stabilised; But Solvation is different due to size. Both have almost same electronegativity.In water essentialy both ionize to 100%.So you must here only cosider solvation to be major factor.

Parallel pi-orbitals are necessary for resonance; consider pyridine where there is no resonance of N's lone pair as N is $sp^2$ which makes its lone pair in the plane of benzene(type) ring and thus this leaves a lone pair of electrons available on nitrogen.

But the chlorate anion can resonance despite its lack of p orbitals.

@Dissenter Actually, it can't. The order of Cl-O bonds is 1. And nitrogen in nitric acid and nitrate-anion participates in exactly 4 bonds

"The chlorate ion cannot be satisfactorily represented by just one Lewis structure, since all the Cl-O bonds are the same length (1.49 Å in potassium chlorate[2])."
http://upload.wikimedia.org/wikipedia/commons/thumb/d/d2/Chlorate-resonance-2D.png/1920px-Chlorate-resonance-2D.png

I think you mean that parallel pi-orbitals are necessary for conjugation.

yes necesarry...

But not for resonance, correct?

yes necessary for conjugation


text:
Hydrogen atom energy level
The energy level of a hydrogen atom in its ground state is $-13.6\ \mathrm{eV}$.
How is this value calculated and how can we calculate the same for different orbits?

From the Bohr theory:
$$\large E=-\frac{Z^2e^4m}{8\epsilon_0^2n^2h^2}=-13.6\times\frac{Z^2}{n^2}\ \mathrm{eV}$$
where:
$$\begin{array}{c|c|}
& Z&\text{atomic number}&-\\
& e&\text{charge of electron}&1.60 \times 10^{-19}\ \mathrm{C}\\
& m&\text{mass of electron}&9.10 \times 10^{-31}\ \mathrm{kg}\\
& \epsilon_0&\text{electric permittivity}& 8.85 \times 10^{−12}\ \mathrm{F/m}\\
& n&\text{principal quantum number}&-\\
& h&\text{Planck's constant}&6.62 \times 10^{-34}\ \mathrm{m^2\ kg/s}\\
\end{array}$$
You can put values of $Z$ and $n$ for any shell of a hydrogenic species like $\ce{H}$, $\ce{He+}$, $\ce{Li^2+}$, $\ce{Be^3+}$, $\ce{B^4+}$…

The equation gives the answer in Joules. To convert to electron volts use $\pu{ 1 eV = 1.602~10^{-19} J}$. To show that the equation is in Joules you need to know that the Farad has units of Coulomb/Volt $\pu{F=C/V}$ and $\pu{V=J/C}$ and $\pu{J= kgm^2s^{-2}}$. The  $13.6 $ is the Rydberg constant in eV.


text:
Does the distance increase as the polarity increases?
According to the dipole moment's definition, as polarity increases, the distance between the centres of positive and negative charges increases. In the ionic bond, as the polarity is maximum, the distance between the two charged centres is maximum. Is this concept correct? If not, what is the relationship between distance and polarity?

The dipole moment $\vec\mu$ is
$$\vec\mu = q\cdot\vec d,$$
where $q$ is the magnitude of charge on any particle and $\vec d$ is the distance between them.
This must imply that if charge increases or distance between the charges increases, dipole moment increases. The larger the dipole moment, the greater the polarity.

Actually when the more electronegative elements pulls electrons, the partial charge, $q\le e$, increases and in case of $100\,\%$ polar compound, $q=e,$ whereas in $100\,\%$ covalent compounds $q = 0.$
So, you can see dipole moment and polarity is maximum for ionic compounds. But if two atoms have greater bond length, surely dipole moment and polarity increases.

If suppose a compound can exist in covalent and go to ionic form[totally hypothetical]in which one is more electronegative,then as dipole moment is max. in ionic compound,is this because there is whole charge than partial charge or as it goes from covalent to ionic,the distance increases or both? Thanks again.

can't say about the distance, as in real there is no change of covalent to ionic, so changing nature will change distance, can't say.

I'm not sure if the guideline actually works out in empirical observations. There are at least two factors affecting bond length - atomic radii - and polarity of the bond. A more polar bond will have a less even distribution of electrons. The electron density will generally be closer to the more electronegative atom. This isolation of electrons may have the effect of "stretching" the bond.
Consider the following two molecules, both of which are linear and polar:

From polarity considerations alone, we might expect the $\ce{H-F}$ bond to be longer than the $\ce{H-Cl}$ bond because fluorine is more electronegative than chlorine. However, chlorine is a bigger atom than fluorine by nature of having additional sublevels of electrons.
Also, consider the below data: bond length actually increases with decreasing polarity (quantified through dipole moments).
Fluoromethane Bond length 136pm, Bond strength 108 kcal/mol, Dipole moment (D) 1.85
Bromomethane Bond length 178pm, Bond strength 84 kcal/mol, Dipole moment (D) 1.87
Chloromethane Bond length 193pm, Bond strength 70 kcal/mol, Dipole moment (D) 1.81
Iodomethane Bond length 214pm, Bond strength 56 kcal/mol, Dipole moment (D) 1.62
The unfortunate thing about your guideline is that I thought that it would work at least across periods - in which atomic size stays relatively constant. However, the facts do not fit the case. Consider $\ce{H-N}$ and $\ce{H-O}$. Oxygen is more electronegative than nitrogen; we'd expect the $\ce{H-O}$ bond to be more polar, and thus longer than the $\ce{H-N}$, at least according to your guideline. The opposite is actually true; the $\ce{H-O}$ bond is actually shorter than the $\ce{H-N}$ bond.

So what is the role of distance in measuring the degree of polarity.According to u , polarity is inversely proportional to the distance?Can u plz explain?

I didn't say that. I think that you should consider bond length on a case by case basis as there are several factors.

Can u give me the actual answer:what is the role of $d$ in measuuring the polarity ?

How does $d$ affect the dipole moment?

d is a measure of the dipole moment.

What is the difference between dipole moment and polarity?

The bond length decreases as polarity increases because in a polar bond, one side is partially negative and one side is partially positive, so these two sides attract even more, get closer together, and the bond length decreases. So, more polarity means more attraction (or attractive pull), which brings the atoms closer together, hence decreasing the bond length.
That is why when there is more difference in electronegativity between atoms, their bond length is shorter; when there is more difference in electronegativity, there is more polarity, and shorter bond length.

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text:
Can a clay brick waste in a soil be a source of minerals for plants?
Clay brick is made by heating them so their clay structure changes. What i understand is ultimately all materials have one or another base minerals in it particularly here this brick still it remains in some other form. So after process of heat, those minerals or nutrients can remain in such condition a plant can absorb? Can a brick dust be a good source of nutrients for plants to grow?

In general clay is not a good source of nutrients. Clay is a class of hydrated alumina-silicates containing small amounts of metals. The alumina and silica are not plant nutrients, but some of the metals can be, in particular: potassium, magnesium, calcium, and iron.
Of these, only potassium is needed in large amounts by plants, and conversely, is fairly uncommon in clays. Calcium and iron are much more common, but are needed by plants in very small quantities.
For any mineral nutrient in any case I can think of, other sources would be better than clay.
Therefore, in the specific case where there was a mineral deficiency in the soil, and you didn't have another option available, clay could provide that nutrient.
However, clay can help plants in another manner. This article describes the ability of clay soil to attract and hold positive ions (metals that can be absorbed by plants are positive ions) due to the high surface area and negative surface charge of the clay particles. Since leaching can be a problem in nutrient-poor soils, having a soil component that "holds on" to those nutrients can be helpful.
Whether or not brick dust would be helpful to plant growth would depend on the specific case, in particular:

The type of plant(s)
The composition of the rest of the soil
The type of clay used to make the brick
The climate conditions (amount of rainfall, etc.)


text:
I heard aluminium doesn't produce fumes when burnt but the opposite happened, how?
One day I took aluminium foil(sanitary brand,  don't know if you've  heard it)  then I casually burned it over  a gas lighter then it burned and gave off Grey fumes and the foil became a dull Grey where it had burnt.  I mistakenly may have inhaled a bit of fumes.  I had a slight pain in my right nostril the whole day. How is that?  Is that dangerous?

It doesn't. Consumer aluminium foil often is a thin film on thicker polymer film, it is much more durable this way. The most common polymer for films is likely polyethylene, and when heated it produces nasty mix of pyrolysis and oxidation products. While not extremely toxic, they are definitely irritating and unhealthy.


text:
Number of hyperconjugation structures
Is the number of hyperconjugation structures for a hydrocarbon with number of 'alpha' hydrogens 'a' equal to 'a' or 'a+1'?

Let's use the n-propyl carbocation as an example.  It has two hydrogens alpha to the cationic carbon and two resonance structures involving hyperconjugation of those alpha hydrogens can be drawn.

So total number of structures are 3 or 2 as itself is also a structure?

Total number of resonance structures is 3

What about the analogous structure using the C-methyl bond instead of the C-H bond?

Yes, resonance structures incolving C-C hyperconjugation are also valid resonance structures.  In this case, C-C hyperconjugation probably contributes less to the overall picture of the molecule than C-H hyperconjugation. I didn't include such structures since the OP was asking about the alpha-hydrogens.


text:
Steam distillation of açaí powder
I have a steam distillation setup with the flask for boiling the açaí powder with distilled water. Are you able to extract the essential oil from the boiling material?

Wiki on açaí oil
Açaí  oil is obtained by cold pressing.  Steam distillation may be able to extract compounds like vanillic acid but may not get the anthocyanins or the triglycerides.  Many of the health benefits are from the healthy triglycerides and the anthocyanins.

Does the cold pressing have to be done with the berries only or can you do it with the acai powder?


text:
Effects of orbital overlap and electronegativity on polarity of binary halogen acids
I have discerned these factors affecting orbital overlap:

Symmetry
Type of orbital
Nodes along the internuclear axis
$n$ quantum number of the atomic orbitals - the number of shells of electrons
Size of the atoms
Geometry of molecule
Angle of overlap

With respect to the binary halogen acids, I still have the following questions:
I can understand how one might argue that the binary halogen acids become more polar as one goes down group 7, in that electron density isn't very well shared; overlap is poor because of the small 1s orbital of hydrogen and the increasingly larger hybridized orbitals of the halogens. But this is half of what constitutes ionic character.
I.e. what about the electron on hydrogen? As the halogens get bigger, electronegativity decreases. How can the electron from hydrogen be isolated toward the halogen as EN decreases?

Is your question referring to HX or H+ X- (reactant or products)?

It's referring to reactants.

Why do you think the electron from hydrogen is isolated toward the halogen as EN decreases?

I don't see how the hydrogen electron is isolated toward the halogen as EN decreases.

OK, it's not isolated toward the halogen as EN decreases.  What's your question?

Why then would one describe the halogen binary acids as increasing in ionic character?


text:
Why is the ionization energy of sulfur anion more than that of oxygen anion?

Which of the following are in the correct order of their ionization energies? (multi-answer question)

$\ce{O > S > S- >O-}$
$\ce{F > F- > Cl- > Cl}$
$\ce{O > O- > S- > S}$
$\ce{F > Cl > Cl- > F-}$

The answers are 1 and 4.
Why is ionization energy of $\ce{S-}$ greater than $\ce{O-}$? The sulfur ion has an extra shell than an oxygen ion, so the atomic size of sulfur should be greater than oxygen. Therefore, the ionization energy of an oxygen ion should be more than a sulfur ion because the electrons in oxygen are closer to the nucleus when compared to sulfur. But why is the reverse true?

The ionisation energy of an anion is the same as the electron affinity of the element. $$\text{IE}(\ce{A-}) = \text{EA}(\ce{A})$$
Your reasoning is not wrong and one might expect exactly this behaviour. Another point for this reasoning is the substantial lower electronegativity of sulfur ($2.6$) versus oxygen ($3.4$).  However, experimental findings contradict this reasoning.
$$\begin{array}{lll}
&&/\mathrm{kJ\cdot{}mol^{-1}}\\\hline
\text{IE}\ce{(O)}  &        &=1314\\
\text{IE}\ce{(S)}  &        &=1000\\
\text{IE}\ce{(S− )} &=\text{EA}\ce{(S)} &=200\\
\text{IE}\ce{(O− )} &=\text{EA}\ce{(O)} &=141\\\hline
\end{array}$$
Sources: Wikipedia articles on ionisation energies, electron affinities
Basically it all boils down to electron-electron repulsion. Oxygen ($152~\mathrm{pm}$) is much smaller than sulfur ($180~\mathrm{pm}$) (van der Waals radius, source). This goes hand in hand with the smaller orbitals (lower main quantum number for the valence shell)  The electrons can therefore only occupy lesser space and repulse each other more. This is mainly the reason, why sulfur has a greater electron affinity than oxygen.
The same reasoning may be applied to fluorine and chlorine.

Write the electronic configuration for all atoms:
$$\begin{array}{lll}\hline
\text{atom}&\text{before ionization}&\text{after ionization}\\\hline
\ce{O}    &1s^22s^22p^4             &1s^22s^22p^3(^*)\\\hline
\ce{O-}   &1s^22s^22p^5             &1s^22s^22p^4\\\hline
\ce{S}    &1s^22s^22p^63s^23p^4     &1s^22s^22p^63s^23p^3(^*)\\\hline
\ce{S-}   &1s^22s^22p^63s^23p^5     &1s^22s^22p^63s^23p^4\\\hline
\ce{F}    &1s^22s^22p^5             &1s^22s^22p^4\\\hline
\ce{F-}   &1s^22s^22p^6(^*)         &1s^22s^22p^5\\\hline
\ce{Cl}   &1s^22s^22p^63s^23p^5     &1s^22s^22p^63s^23p^4 \\\hline
\ce{Cl-}  &1s^22s^22p^63s^23p^6(^*) &1s^22s^22p^63s^23p^5 \\\hline
\end{array}$$
$(^*)$:Stable Configurations
Since $\ce{O^-}$ and $\ce{S^-}$ go to stable configurations  $\ce{S,O>S^- ,O^- }$.
Since size of $\ce{S}$ is larger than $\ce{O}$ $\implies$ $\ce{S < O } $.
Assuming IE of $S^-$ reverse of EA of $S$ and $O^-$ reverse of EA of O, $O^-<s^-$ because S has large size to accomodate electron and less repulsion than small O.
[My incorrect order $\ce{S<O < S- < O- }$ was actually $\ce{S<O < S+ < O+ }$]
Thus, $\ce{O^- < S^- < S < O }$.
Apply same concept to others; Remember:

a)Nuclear Charge
b)Shielding
c)Atomic radii
Penetration Effect[s>p>d>f]
d)Stability of fully and halfly filled orbitals.
Successive Ionisations are always harder starting from neutral compound.$IE_1\ll IE_2\lll IE_3$

As a rule of Thumb:

$IE\propto \frac{(a)(d)}{(b)(c)} $, very roughly

Thats excatly what I did but the answer key says O > S > S- > O-  Is there something to do with electron-electron repulsion?

What is this data, are there any units given? What is the source? What are $a,b,c,d$?

The data sheets contradict your reasoning: $\ce{IE(O)=1314}$, $\ce{IE(S)=1000}$, $\ce{IE(S- )=EA(S)=200}$, $\ce{IE(O- )=EA(O)=141}$, all in $\mathrm{kJ/mol}$. IE, EA

@Martin please elaborate as to waht is contrdicting and what you meant by EA?

This data is from wikipedia, all in KJ/mol ; a,b,c,d are points marked just above

@Yashas no, nothing more than this

The answer key says O > S > S- > O- but you are saying O- > S- > O > S

maybe the answer key is wrong, cause I based my reasoning upon data

The answer key is correct, please see my previous comment. Your answer is incorrect. (And next time please provide the source in your answer) Somehow the EA link got messed up: http://en.wikipedia.org/wiki/Electron_affinity_%28data_page%29

@Martin I'm sorry, accidently i saw $ie_2$ of the atoms.


text:
Is carbon dioxide more acidic than carbon monoxide?

Which of the given compounds is most acidic? $\ce{CO}$ or $\ce{CO2}$ ?

How do I solve this question, and how do I explain it?
I took the oxidation state of carbon in both. The oxidation state in $\ce{CO2}$ is higher, therefore it is more electronegative, so therefore $\ce{CO2}$ may be more acidic.
For $\ce{CO }$ I don't know about the charge so that I can use it to understand if it would give electron or accept electrons, so that I can apply Lewis acid/base theory.

Carbon dioxide is a much stronger Lewis acid than carbon monoxide. In carbon dioxide, the carbon has a much higher oxidation state (formally, $+4$ vs. the $+2$ of carbon monoxide), and the two oxygen atoms heavily withdraw electron density from the carbon atom, leaving a very large partial positive charge ($\delta^{+}$). This is rigorously confirmed by electrostatic potential maps and calculated orbital coefficients. As a consequence, the electron-deficient carbon of $\ce{CO2}$ is highly electrophilic, and reacts with a variety of nucleophiles (e.g., Grignard reagents, organolithium reagents, etc.).
Carbon monoxide, on the other hand, actually contains substantial electron density at the carbon atom (in spite of the bonds being generally polarized towards the oxygen atom), which possesses a fairly localized and active lone pair. Consequently, a few notable exceptions aside, carbon monoxide mostly functions as a Lewis base.

Formal charge analysis reinforces the fact that carbon monoxide is often a Lewis base; the carbon portion has a lone pair of electrons and a negative formal charge. Also see the case of the cyanide ion, in which the carbon is the more nucleophilic of the two constituent atoms.


text:
Solid salts of hydronium ion
Are there any solids with hydronium ion in its crystal structure?
If so, what are they?
If not, why not? I understand there are hydroxide-containing solids and gels such as aluminium hydroxide but why not hydronium ions? Does this have to do with the positive charge of hydronium ion? I could see the positive charge of hydronium ion excluding the possibility of forming any salts/solids with the group 1 and 2 metals.

There is a section in this Wikipedia link entitled, "Solid hydronium salts."  It says that any acid with an ionization constant of 10^9 or higher may form stable hydronium ion salts.  Examples they give include hydronium perchlorate and carborane-based hydronium ion salts.

Why 10^9? What's the reason for this Ka value?

They note it as a general rule of thumb, based on experimental observation.  They cite HCL with an ionization constant of 10^7 not forming a stable hydronium ion salt with water at any HCL:water ratio at RT.

What do you mean salt with water?

I see, monohydrates. Why must hydronium ion form nonohydrate salts? Bc of the water bound up in hydronium?

1)not forming a stable hydronium ion salt when mixed with water, 2) monohydrates probably form do to space limitations in the crystal

Are there any solids with hydronium ion in its crystal structure? If so, what are they?

There are crystal structures of acid monohydrates where the acid is deprotonated and the crystal water is protonated, i.e. the hydronium salt has formed. In addition to the hydronium perchlorate (DOI: 10.1107/s0108270103010461) mentioned in ron's answer, there are structures of hydronium tosylate (Acta Cryst B 27(7), 1293-1298, (1971)) and a crown ether complexed to hydronium with perchlorate as counter ion (http://scripts.iucr.org/cgi-bin/paper?S1600536810048622) shown below.
In the animated GIF, the hydronium ions are shown as space-filling while the crown ethers and the perchlorate ions are shown as ball-and-stick. One hydrogen atoms in each hydronium ion hydrogen bonds with perchlorate. If there were a proton transfer between the hydrogen bonding partners, the structure would be perchloric acid monohydrate, but the crystal structure shows that the hydrogen atom is closer to the oxygen of the hydronium than that of the perchlorate, so calling it a hydronium salt it appropriate.

Here is another view showing the hydrogen bond (1.7 Å distance vs 1.2 Å H-O bond length), with the hydronium ions as ball-and-stick and the remainder of the structure in all-bond representation.

What a distorted view!  The oxygen atoms in the hydronium ions  like five times the diameter of the others.  Hard to read the structure at first.  Authors should have been more consistent IMO.

@OscarLanzi I made another one where you can see bond lengths and the hydrogen bond interaction. The coordinates (https://doi.org/10.1107/S1600536810048622/hb5736sup1.cif), the software (https://chemapps.stolaf.edu/jmol/jsmol/simple.htm) and the script (http://proteopedia.org/wiki/images/7/76/Wobble.spt) are all available to the public, and there are many choices to be made.


text:
Activities of Solutes - especially acids and bases
What is the concentration of hydrochloric acid needed to have a solution with the activity of the hydrogen proton = 1? I read 1.18 M somewhere but I am unable to find a corroborating source. Also, how can the activity of the hydrogen proton = 1 when it doesn't exist by itself, suggesting that its activity should = 0 (since it can't act as an independent particle)?
Also how would one get to this result? Does one use a high-grade pH meter? Don't analytical-level pH meters measure activities rather than concentrations of hydronium ion?

For almost the last 100 years, the definition of pH has been -log of hydrogen ion activity.  Prior to 1920, the definition was -log of hydrogen ion concentration.
Hydrogen ion activity or concentration does not refer to bare protons, but solvated protons, as $\ce{H_3O^+, H_9O_4+}$ etc.
See Determination of Hydrogen Single Ion Activity Coefficients in Aqueous HCl Solutions at 25°C for authoritative data.

Would solvation of hydrogen ion decrease its activity?

not sure I understand what you are asking.  For gas phase HCl the hydrogen ion activity and concentration are many orders of magnitude lower than for the aqueous phase, so in that sense solvation greatly increases activity.  See the following table where "GP" means gas phase: http://www.ut.ee/ams/acidity-of-strong-acids-pka-in-water-and-organic-solvents-3/


text:
Why does acetyl have negative charge?

"Acetylation (adding an acetyl group) and phosphorylation (adding a phosphate group) make the histones more negatively charged because acetyl and phosphoryl groups are negative. They are "glass is half empty" molecules. By making histones more negatively charged, their grip on DNA will be much looser because DNA is also negatively charged. Similar charges (negative and negative) repel one another".

I know oxygen and carbon have different electronegativities but that generates a partial negative charge and a partial positive charge. So why do we consider the negative charge prevails over the positive charge in order to histones "repel" DNA?

The acetyl group itself does not have any significant charge. But the residue that gets acetylated is lysine in this case, which has a positive charge. The acetylation results in a neutral amide, removing the positive charge.

From Wikimedia Commons
I'd add that the effect of histone acetylation is likely more subtle than just the charge effect. This kind of posttranslational modification often affects the protein structure in some way, or is recognized by other proteins interacting with that site.


text:
Refluxing with tilted condenser
Can you reflux without any problems/dangers with a tilted condenser (e.g. because you don't have enough space in your fumehood)? I would say there isn't a problem, because both the surface area of the refluxed liquid and the length of the condenser (and thus the length of the trajectory of the water flowing through the condenser) remain unchanged. Is this thought correct?
This is my current setup

Also you should use countercurrent exchange right?

As long as everything is clamped and balanced - so that there isn't excessive weight on the joint connecting the flask to the condenser - the system should be OK.  After all, many rotary evaporators have the flask plus condenser tilted at an angle.

Yes, I just realized that. Would it work for a Allihn condenser too? Thanks Ron.

I recommend to estimate the thickness of the glass in the condenser here.

@Jori  I would think so, as long as it is not "packed" with anything.  And since you're not rotating your flask like a rotary evaporator does, you can clamp it and the condenser to alleviate stress on the connecting joint.

@ron "packed"? Also note that I clamped the condenser and the flask, but I guess you mean a broader clamp (like the one used to hold the condenser to the stand).

@Jori  1) Sometimes columns are "packed" with glass helices or other high surface area materials to increase the fractionation efficiency of the column.  You wouldn't want to do that with a tilted column as it might cause flooding of the column (to much liquid in the column unable to drain back to the flask quick enough).  2) Looks OK, now I see it, couldn't see the second clamp before.

@ron A rotary evaporator is NOT a reflux reaction vessel, for heavens sake! And a condensor/cooler of ANY SORT is not a column!

I can think about three problems in the case
1) Glass is fragile, so if any significant load falls onto flask's or condenser's necks, it is a risk. Thus, the condenser should be light and have separate support, and the flask should have a second support for bottom or have a very small load (say, if it would be more than 100 g, I would hesitate.)
2) even round-bottom flasks does not necessarily have a perfectly spherical bottom, so fast stirring is absolutely out of question, and even gentle stirring for medium loads is not good.
3) If highly tilted, Allihn condenser may trap significant amount of liquid in its spheres.  Dimroth and Inland Revenue condenser may trap some (little) amount of solvent on their coils too. Dimroth condenser MAY be used for refluxing even if should not ideally, and for solvents with high BP (say, toluene) should be OK. In fact, in case if violent reaction may occur it is PREFERABLE to use condenser with low internals resistance to straight float so the stream of boiling reaction mixture runned through the condenser in did not lift and smash it or teared the flask apart.
Actually, if no gas exhaust is expected, solvent, reactants and products do not  have low bp and high toxicity I would prefer to reflux outside of the fume hood rather then tilted. It is OK. Even if some gas exhaust is expected, I would shrug of nonviolent non-toxic gas exhaust.

You mean that the flask would break from the condenser at the joint? I could hardly imagine that, if you keep reactants weight under 1kg, especially if the flask also rests on the bottom of a water bath. Adding an extra support for the condenser could be an option, although again, does it really gain that much weight from the condensing liquid?
???
Can't this be compensated with a little bit of extra reactants in situ?
Gas exhaust is toxic (cyclohexylamine and perhaps nitrogen dioxide)

Well, the tilted flask/condenser is a lever. It has MUCH larger load near joint than it looks imaginable and given glass fragility... I wouldn't take risks. 2) Anchor for a magnetic stirrer, especially for heavy loads and fast stir rates should rotate at the axis of symmetry. Otherwise it may break the flask.  3) It probably can,  but I would prefer Liebig condenser in this setup. And post experiment details if possible.

Yes I will. If everything works out fine and I'm not killed in an explosion or anything, you'll be the first to hear! Also note that I clamped the condenser and the flask, but I guess you mean a broader clamp (like the one used to hold the condenser to the stand). "Anchor for a magnetic stirrer, especially for heavy loads and fast stir rates should rotate at the axis of symmetry. Otherwise it may break the flask" you mean because of the extra side ward pressure on the joint due to centripetal force? (should also be balanced out by an extra clamp I guess)

I mean, the condenser must have support in the middle. Otherwise, load near the joint may become too much. In fact, it is common to fic large condenser at the middle even in vertical setups. ||| Nope, if uncentered, it may either cause vibration at best case or break from its position and start jumping in the flask. At higher RPMs the risk is much higher, at lower RPMs are almost negligible. ]

Add that the fluid runs back into the vessel at the wall, where it starts to boil before reaching the reaction mixture,.

Support notwithstanding, those water hoses indicate the device is a Graham condenser, in which case it should always be vertical. Perhaps a Dimroth condenser would be better employed here.


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Why is standard reduction potential an intensive property?
My book says that

Since the number of electrons lost must equal the number gained, the half-reactions must be multiplied by integers as necessary to achieve the balanced equation. However, the value of e° is not changed when a half-reaction is multiplied by an integer. Since a standard reduction potential is an intensive property (it does not depend on how many times the reaction occurs), the potential is not multiplied by the integer required to balance the cell reaction.

However, I do not understand why the reduction potential is unaffected by the number of mols. If you have a greater number of mols of the metal, doesn't that allow for a greater potential difference?

Look at the unit volt. 1 volt = 1 joule per coulomb. If the number of moles is doubled, the coulombs are doubled, as is the number of joules. Think of the volt as the driving force behind an individual electron in an electrochemical system.

Good answer.  This question is just another example of how physics is dropping the ball by continuing to use outdated units in which coulomb and mole exist as separate units, just reporting electrical potential in J per mole and current in mol per s, etc.

Basically, Imagine that you have a waterfall that is 100 meters high, and 50 meters long, this scenario  makes reduction potential an intensive property: now let's say that you have two of the said waterfall--you will have in total (when put side by side) a larger waterfall that is 100 meters high, and 100 meters long. You do not need to multiply it by two because the your "potential" (usually determined by height in physics) is unchanged.

I'll take a shot at this, so don't insta-hate me. :)
The best way I think I can show you this through is an example.
Say we have 2 apples in one bin and 3 oranges in another bin. Assuming the $\ce{E}$ is the difference of the apples and the oranges you get $1.00$.
Even if we doubled the apples and the oranges the difference would be the same. In other words no matter the quantity the voltage will be equivalent.
Now, you might be thinking: what if we just double the apples.
Well it doesn't matter because we can consider apples and oranges as reactants to a apple-orange pie. So the more apples we have: it would just be excess.

Take a look at this. Explains it beautifully.


text:
Simulating enzymatic change rates of Carbonic anhydrase
I'm trying to simulate effects of carbonic anhydrase on cellular pH as caused by changes in bicarbonate. I assume a bicarbonate homeostatic mechanisms which replenishes intracellular concentrations from
$$\ce{CO2 + H2O <=> H+ + HCO3- }$$
One paper I’ve found states the assumptions and rates, but doesn’t parameterize the differential (difference) relation between the substrates and products. The paper merely states that given the time scales I’m considering, it’s ok to assume $p(\ce{CO2})$ is constant, since carbon dioxide’s diffusion is fast in aqueous media. Secondly, the article states that

The equilibrium constant of the reaction was $10^{-6.35}\ \mathrm{M}$ and the rate constant for $\ce{CO2}$ hydration was $10^6\ \mathrm{s^{-1}}$.

How do should I think of this reaction, and more importantly how do I simulate it?


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How is the charge of covalently bonded atoms determined?
My question is about what it is - at the subatomic level - that determines the charge on atoms which participate in covalent bonding in reality. I understand that formal charge is just that - it's a formality, which gives human beings a very useful tool for modelling reality and making predictions. I, though, would like to know what is responsible for the charge we find in covalently bonded molecules.
Let me explain. If we picture an ionic bond as a highly polar covalent bond, we can see that in an atom like NaCl we have the state of affairs where the highly electronegative chlorine atom has "pulled away" an electron from the sodium atom.

(Orbitals omitted for ease of drawing)
Ultimately this means that Na, with its 11 protons has just 10 electrons, which gives it a positive charge, while Cl, with its 17 protons, now has 18 electrons, giving it a negative charge. Electrostatic forces between the two nuclei result in the bond.
Of course, it might be more accurate to say that the Na nucleus simply ceases to be significantly affected by its former 3s1 electron. As a result the 11 protons in the Na nucleus, which result in a +11 charge, are countered by only 10 electrons, giving a +1 charge overall. Meanwhile, the 17 protons of the Cl nucleus (+17 charge) experience the negative charge of 18 electrons (-18) giving a -1 charge overall.
Let's compare this with the situation we find in the carbonate ion. (Link) For the purposes of my question I won't worry too much about resonance, I don't mind too much whether we have the accurate -2/3 charge, but let's keep it simple and just look at one of the contributing structures.

(Orbitals again omitted for ease of drawing. Orange indicates an additional electron from some other source. Oxygen has only eight electrons ordinarily, but one will be gained when forming this ion)
In this case we have covalent bonding. In the case of the singly bonded oxygens, we have 8 protons and 10 electrons. This is two more negatively charged electrons than positively charged protons. This, though, is not ionic bonding, it is covalent bonding and the polarisation is not such that the electron contributed by carbon dissociates completely from the Carbon. Instead it is "shared" between the two. Nevertheless, the "sharing" is not equal and oxygen attracts the electrons more strongly than the carbon, so we have a -1 charge.
My question, though, is - why is it that the charge works out like this? Why is it that when you have oxygen (8 protons) (charge of +8) and a filled outer shell (8 electrons plus buried inner core gives -10 (charge of -1 on each electron)) you have a charge of -1 not -2, is it simply the covalent bond, is it simply the fact that the two electrons in the sigma bond are sufficiently far enough away in reality?
But then why should it be that the doubly bonded oxygen, which also has ten electrons surrounding it, has no charge at all? Meanwhile, the carbon atom in the  middle is surrounded by 10 electrons also, whereas it has only 6 protons and yet carries no charge at all.
Naively the answer seems obvious. The bonds are covalent and the electrons are "shared." So, in the double bond (which, of course, has a different nature to the single bond) one must assume that the electrons are positioned in such away that they are balanced by the protons of the oxygen nuclei. There is therefore no charge on this atom.
In the case of the Carbon atom we surely again understand that the right balance of charges and distance from the nucleus is arrived at.
Overall we have 32 electrons and 30 protons so the overall -2 charge on the whole atom makes sense.
The ultimate question I've been trying to lead up to is what is it that dictates the distribution of charge. What is the difference - at the subatomic level - between two ionically bonded atoms, a covalently bonded molecule like the CO32- ion which carries a charge, and a covalently bonded molecule like H2O - which is also polar (hence we can have hydrogen bonding) but isn't usually assigned a charge. On what basis do we assign charge?

To confuse you more: there is no universal agreed definition of what is 'atomic charge'. Several definitions exists, mostly based on location of electronic occupation of the atoms and electrostatic field of the molecule. The problem is that it is impossible to put a definitive border between two atoms.

As you point out, formal charge is a "bookkeeping" method of assigning electric charges to atoms in a way that lets us easily approximate where the electron density is.
Chemistry is made up of layers of successively less detailed approximations. The concept of "formal charge" is near the top of the stack - very little detail is involved. It is only slightly more detailed than Lewis structures, in that we are at least trying to acknowledge that covalent bonds don't really share electrons equally.
With that in mind, we will look at the next higher and next lower level of detail, to see why formal charges are useful.
Whenever we draw a Lewis structure, we imply that each line is made up of two electrons that share a molecular orbital that exists somewhere in the space between the two atoms that it connects. Without knowing any better, we would assume that each electron can be wherever it "needs" to be to fill the valence shell of an atom. So, in the case of $\ce{CO3^2-}$, the carbon has four bonds and no lone pairs, giving it a full octet. The oxygen atoms have either three lone pairs and a single bond, or two lone pairs and a double bond, again giving each a full octet.

Note that we counted each electron twice for the purpose of checking the valence configuration. It's as if we are saying "each atom can borrow electrons from the other whenever it needs them." We are effectively ignoring any tendency of atoms to pull electrons towards each other - they float between atoms as needed, as if maintaining a full octet were the most important job in the world.
Next we look at the formal charges. The rules for determining formal charge say that we count the number of valence electrons an atom normally has (without bonds), then subtract the electrons we assign to it. We assign to an atom all of the electrons that are in lone pairs, plus one half of the bonding electrons. In other words, instead of saying that each atom gets the electrons whenever it wants them, we are saying "each atom in the bond divides the electrons equally between themselves." When you do this, you find that some atoms have "extra" electrons (more than they would normally have in the valence shell) and some have too few - thus, a formal charge.

This starts to give us a slightly more correct picture of how electrons are distributed inside of a molecule, but it doesn't really tell the whole story. Notice that we didn't care about what type of atom it was when we split up the electrons. We only looked at that to find the "natural" number of electrons it would have.
The next layer of detail involves electronegativity. In any covalent bond between two different elements, one element "likes" electrons more. The reasons for this are complex, and you have to go down another couple of layers to get to the reason. For now, just know that it is related to quantum mechanics.
If we know the relative electronegativities of elements, we can start to account for this tendency as well. We do this when assigning oxidation numbers. With oxidation numbers, you assign to each atom all of its loan pair electrons. Then, for each bond, you give all electrons in the bond to the more electronegative atom. Then we subtract the total from the number of valence electrons the atom normally has.

Notice what we are doing. First we said "all electrons are shared equally all the time." Then we said "Divide the electrons equally." Finally we said "the most electronegative atom gets all of the electrons."
None of these is actually true, but each gives a quick approximation that is useful for a particular type of problem. To get a real picture of what is happening to the electron density, you have to do detailed quantum mechanical calculations. They are really hard to do! We typically use computers to solve them. But, if we know what the limitations are, we can use quick approximations to get an idea of where the electrons are at that is good enough to, for example, make reasonable predictions in organic chemistry.
To summarize:

Valence count - assume all electrons are shared equally
Formal charge - assign one half of bonding electrons to each atom
Oxidation number - assign all bonding electrons to the most electronegative atom


text:
How to determine the molecular formula of a compound given its molecular mass and mass percentage contribution of the involved elements?
Hydralazine is a drug used to treat heart disease. It is $59.99\%$ carbon, $5.03\%$ hydrogen, and $34.98\%$ nitrogen and has a molecular mass of $160.178$.
What is the molecular formula for hydralazine?
I did:
\begin{align}
59.99~\ce{C}/12.011&=4.994\\
5.03~\ce{H}/1.01&=4.9801\\
34.98~\ce{N}/14.007&=2.4973.
\end{align}
Then divided all $3$ by $2.4973$, and got 2 carbon, 2 hydrogen, and 1 nitrogen atoms i.e. $\ce{C_2H_2N}$. The online homework said that the stoichiometry was wrong. I don't know what I did wrong.

Can you show your calculations that led you to that answer?  That way someone can show you where you went wrong.

Good, but does the molecular weight of $\ce{C2H2N}$ add up to 160.178?  If not, what do you need to multiply by to get there?

The percentages are based on the molar mass. You have to multiply the percentages with the molar mass and divide by the mass of the atoms.
\begin{align}
\ce{C}:&& \frac{0.5999 \cdot 160.178}{12.011} &\approx 8\\
\ce{H}:&& \frac{0.0503 \cdot 160.178}{1.01} &\approx 8\\
\ce{N}:&& \frac{0.3498 \cdot 160.178}{14.007} &\approx 4\\
       && \implies \ce{C8H8N4}\\
\end{align}


text:
Calculating the electronegativity of an element from bond dissociation energies

Calculate the electronegativity of fluorine from the following data:
$$\begin{align}
\text{BDE}(\ce{H-H}) &= \pu{104.2 kcal/mol} \\
\text{BDE}(\ce{F-F}) &= \pu{36.6 kcal/mol} \\
\text{BDE}(\ce{H-F}) &= \pu{134.6 kcal/mol} \\
\chi_\ce{H} &= 2.1
\end{align}$$

I thought that electronegativity should be related to effective nuclear charge, as it is a measure of how strongly atoms "pull" on bond pairs.
Is there a formula to calculate the electronegativity from bond dissociation energies, and what is the rationale behind it?

Considering the data you have been given, the equation you need is the one that defines the Pauling electronegativities:
$$\left|\chi_\ce{A} - \chi_\ce{B}\right|=\left(\text{eV}\right)^{-1/2}\sqrt{E_\mathrm d(\ce{A-B})-\frac{E_\mathrm d(\ce{A-A})+E_\mathrm d{\ce{(B-B)}}}{2}}$$
In the equation, $\left(\text{eV}\right)^{-1/2}$ is in the equation to negate the units of the bond dissociation energies. This means that you need to convert all your bond dissociation energies from $\pu{kcal/mol}$ into $\pu{eV}$. You can use the BDEs in units of $\pu{kcal/mol}$, but then the formula will require a constant in front to adjust for the difference between units.
$\pu{1 eV}$ is equal to $\pu{1.602 \times 10^-19 J}$, so you need to be able to convert from $\pu{kcal/mol}$ to $\pu{J/mol}$, then divide through by Avogadro's constant $N_\mathrm A = \pu{6.022 \times 10^19 mol-1}$. There is a handy website which can help with this.
$$\begin{align}
E_\mathrm d(\ce{H-H}) &= \pu{4.519 eV} \\
E_\mathrm d(\ce{F-F}) &= \pu{1.59 eV} \\
E_\mathrm d(\ce{H-F}) &= \pu{5.837 eV}
\end{align}$$
Plugging the information you have been given, we find that
$$\begin{align}
\left|2.1 - \chi_\ce{F}\right| &= \sqrt{5.837 - \frac{4.519+1.59}{2}} \\
&= 1.668
\end{align}$$
This gives you two possibilities for $\chi_\ce{F}$, $0.4$ or $3.8$, depending on whether the quantity $(2.1 - \chi_\ce{F})$ is positive or negative. This is where we need to use some knowledge of chemistry: fluorine is more electronegative than hydrogen, so we should have $\chi_\ce{F} > \chi_\ce{H}$, and consequently we take the solution $\chi_\ce{F} = 3.8$.

Why should bond energies be related to electronegativity? The idea that Pauling had when he devised this equation was that in a covalent bond between two of the same atom, there is no electronegativity difference and hence no ionic contribution to the bond strength. Hence, the "average covalent contribution" of each atom to a bond could be measured by the mean of the two bond strengths, i.e. $[E_\mathrm d(\ce{A-A}) + E_\mathrm d(\ce{B-B}) ]/2$.
However, when there is a difference in electronegativity, the bond strength has an ionic contribution. The larger the difference in electronegativity $\left|\chi_\ce{A} - \chi_\ce{B}\right|$, the larger the ionic contribution. This ionic contribution could be roughly measured by subtracting the covalent contribution, $[E_\mathrm d(\ce{A-A}) + E_\mathrm d(\ce{B-B}) ]/2$, from the $\ce{A-B}$ bond strength.


text:
Transition metal order of filling electron
There is a specific order for filling electron for transition metals, but I don't know if it applies to other elements too?
Some thing to take note of in this picture is the order of 3d and 4s. My belief is: 3d has a higher energy level than 4s, and but 4s is still the outermost shell.
The reason that transition metal would form colorful ions is that there are empty or half filled d orbitals, so when excited electrons drop back to the d orbitals, they emit energy in form of light. I am a bit confused about this.( it is from a book) what electron is it referring to? Is it the d electron or electron of lower energy level orbitals, as for instance 3p?

This is the order of electrons filling for all elements but typically this diagram is only used for transition elements because when writing electron configurations for any other element its simple and intuitive but transition metals are somewhat irregular so a helpful diagram like that is useful.
Its transition metal compounds or complexes that are coloured not the metal on its own. Its coloured because when it forms compounds or complexes the orbitals in the 3d subshell aren't the same energy as they would be if it wasn't part of a compound or complex. The electrons make transitions between these different energy 3d orbitals when they absorb photons of visible light. The colour of the compound will be the complementary colour of the colour absorbed.
The 3d orbital is typically seen as the higher energy orbital but when a transition metal is ionized the 4s electrons are taken first.

So which orbital has the valence electrons?

Then what is the change in electron configuration when a transition metal forms compound with another element? Why it produces complementary colors? Aren't they supposed to release the amount of energy it absorbs?

Do the photons light come from the forming of the compound?

In response to your comments on my last answer: the term valence electrons isn't really used for compounds - especially ionic compounds. The valence electrons of the element will be the 4s electrons or electron in the case of Chromium and Copper. It's not so much a change in electron configuration its more about the energy or the orbitals. Each 4s orbital will have a different energy level. The electrons make transitions between these orbitals - this is why it is key that the transition metal ion has a partially filled d sub-shell (so there's room for the electrons to go to and from). It's not about the colour it emits, its about the colour it absorbs when visible light is passed through it. A good website to explain is "http://www.chemguide.co.uk/inorganic/complexions/colour.html". It talks about complexes but its the same idea for compounds.

By the virtue of your answer, if it does not absorb any light at all, doesn't it emit any light? So if this is performed in total darkness, we will see no color, right?

It does absorb light - the light that passes through it. You're right about total darkness. Take a look at the link in my last answer. It has some very useful diagrams.