text:
Propagation of uncertainty
I was taught that when adding/subtracting measurements, add the absolute uncertainties, and when multiplying/dividing measurements, add the relative uncertainties.
I have a problem in a lab where I need to find the difference between two measurements that are very close together, and are each very precise; however, when I propagate the absolute uncertainties, I get a huge relative uncertainty for the final result (upwards of 50%). Here it is:
(0.050 ± 5%) - (0.044 ± 3%) = ?
= (0.050 - 0.044) ± ((0.050 * 5/100) + (0.044 * 3/100))
= 0.006 ± 0.00382
This ± 0.00382 corresponds to 63.7% relative uncertainty; am I doing something wrong? Both of the original measurements were pretty accurate (<12% RU). In this case can I just add their relative uncertainties instead of converting to absolute, which is very large in comparison with the difference between the measurements?

There is nothing wrong with the ±0.004 uncertainty you obtained. The relative uncertainty of the difference between close results is supposed to be far larger than that of the original measurements. One way of seeing why is stretching your scenario a bit: suppose you got 0.048 rather than 0.044 as the result of the second measurement. The difference, 0.002, would be small enough to be meaningless, as the second result would encroach into the uncertainty range of the first measurement. Your example is not so extreme; still, the large relative uncertainty is justified.
(Aside: an analogous effect can affect computer calculations done with limited precision. Suggestively enough, it is referred to as catastrophic cancellation).

Thank you @duplode ! Is there any way that this can be overcome?

I don't think so, apart from e.g. switching to a different experimental strategy to avoid needing differences, or using more precise instrumentation.


text:
Why does chlorine have a higher electron affinity than fluorine?
Since fluorine has its valence electrons in the n=2 energy level, and since chlorine has its valence electrons in the n=3 energy level, one would initially expect that an electron rushing towards fluorine would release more energy, as it would land in the n=2 energy level, whereas in chlorine, the electron would land only in the n=3 energy level, and would then not release as much energy.  Thus, one would expect fluorine to have a greater electron affinity than chlorine.  However, why is it that chlorine has a higher electron affinity (349 kJ/mol)  than fluorine (328.165 kJ/mol)?

Fluorine, though higher than chlorine in the periodic table, has  a very small atomic size. This makes the fluoride anion so formed unstable (highly reactive) due to a very high charge/mass ratio. Also, fluorine has no d-orbitals, which limits its atomic size. As a result, fluorine has an electron affinity less than that of chlorine.
See this.

I don't think d orbitals contribute to atomic size, since they're always only populated in an internal shell. Calculated atomic radii decrease continuously across a period, and they decrease a little slower in the d-block because the inner d orbitals shield the valence s/p electrons from the nuclear charge better than adding electrons to the outer shell. If anything, the presence of a populated d subshell actually helps contract the atom, allowing it to reach a higher effective nuclear charge than expected in a period with no d subshell.

....but chlorine doesn't have any d-orbitals either?

@user3932000 Chlorine have vacant d-orbitals. Since empty orbitals are omitted when writing electronic configuration, it is written as $1s^2 2s^2 2p^6 3s^2 3p^5$. Complete electronic configuration would be $1s^2 2s^2 2p^6 3s^2 3p^5 3d^0$.

@ashu You could say the same for fluorine and say fluorine also has vacant d-orbitals, since its configuration would then be 1s2 2s2 2p5 3s0 3p0 4s0 3d0.

@user3932000 Fluorine doesn't have a 3rd (M) electron shell hence no d-orbitals.

@ashu But if we're considering vacant orbitals, all atoms have an infinite number of vacant orbitals, with energy levels from 1≤n<∞?

We're considering vacant orbitals in partially filled electron shells, not empty electron shells.

@ashu There is absolutely no reason to consider vacant d orbitals in partly filled electron shells. You don’t consider f orbitals for bromine either. In fact, while this answer does start off well I have to give it a $-1$ for invoking a concept that does not help and is introduced incorrectly.

The electron being gained by fluorine would be taken in to a much smaller 2p orbital and requires more electron coupling energy than that of much larger 3p orbital of chlorine. Therefore, energy released during the electron gaining process of fluorine is less than that of chlorine.

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text:
Why does copper not react with nitrogen?
Please describe why copper only reacts with oxygen and not nitrogen in the air. Isn't nitrogen a negative ion, and copper positive, so wouldn't that mean that copper and nitrogen would want to form bonds?

Sketching with broad strokes: the first key point is that oxygen and nitrogen are major components of the atmosphere in the form of $\ce{O2}$ and $\ce{N2}$ molecules, and not as ions. $\ce{N2}$ is a very stable compound because the two nitrogen atoms in its molecules are united by a particularly strong triple bond. $\ce{O2}$ is overall more reactive than $\ce{N2}$, largely as a reflex of the bond between the atoms not being as strong. Specifically concerning oxidation reactions (of which those you suggest are typical examples), oxygen atoms having more affinity for electrons than nitrogen atoms contributes towards making $\ce{O2}$ being a better oxidiser than $\ce{N2}$.

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text:
How can we write the wave function in quantum mechanics?
In quantum mechanics, while writing the wave functions we can take a product of the spatial and the spin parts.But what does it means? What is the meaning of the spatial wave function and the spin wave function?

Can you please provide the expressions for the wavefunction?

Welcome to Chemistry.SE! May I ask what background you have in Quantum Mechanics? Are you familiar with Hilbert spaces and Tensor products? Your state of knowledge will determine the kind of answer you get, since this question could require some rather involved mathematical concepts in order to answer it completely.

I have quantum mechanics course in my master science ..... I don't that much Hilbert spaces and Tensor products......

The wavefunction contains all the information about the system of interest.  This is a basic premise of quantum mechanics.
Within the Born-Oppenheimer approximation, we 'index' all the values required to describe where an electron is.  This includes the spatial coordinates, ($\textbf{r}$), and the spin coordinate, $\omega$.  Electrons are characterized by their spin ($\uparrow$ vs. $\downarrow$).  Another way to think about it is this.  The quantum numbers are used to describe everything we need to know about 'where' an electron is ($n$, $l$, $m_l$, $m_s$).  The spatial coordinates (e.g. Cartesian coordinates) take care of the first 3 quantum numbers.  We need the fourth coordinate to characterize $m_s$.


text:
What's the convention for the symbols of enthalpies of formation, reaction, etc.?
I frequently encounter two ways of writing enthalpies of some process, e.g. for the enthalpy of formation:

$\Delta_{r} G$ (or $\Delta_{R} G$)
$\Delta G_{r}$ (or $\Delta G_{R}$)

On the IUPAC Gold Book webpage they used the first notation (although the use of upper- and lower-case descriptors varies, see here and here). Is this the accepted way of doing it (and the other one is simply a relic) or might this be a matter of preference. And how about the upper- or lower-case descriptor: Is there any convention for that?

I think you have already answered your question… IUPAC's usage should be a pretty good indicator of the proper notation.
There is, however, an even better reference than the Gold Book: IUPAC's Green Book, or Quantities, Units and Symbols in Physical Chemistry (full text here), on section 2.11.1 Other symbols and conventions in chemical thermodynamics (page 59 and following):

(iii) Examples of the use of the symbol ∆
  The symbol ∆ denotes a change in an extensive thermodynamic quantity for a process. The addition of a subscript to the ∆ denotes a change in the property.

It says later:

[∆vapH] can also be written ∆Hvap, but this usage is not recommended

The reason is that ∆Hvap could be understand as a change in Hvap, which it is not. It a change of H during a process, so the subscript should be on the ∆.

Thanks for making me aware of the Green Book. I shouldn't have stopped looking after finding the Gold Book.


text:
What units do I use for energy in the Boltzmann distribution?
I am a mathematics student doing a course in thermodynamics and am working on the basics now. The question I am asked is:

Suppose two states differ in energy by one electronvolt. What can be said about the ratio of the population at $T = \pu{300 K}$?

Now it is clear that I will use the formula:
$$\frac{N_i}{N_j} = \exp\left(-\frac{E_i-E_j}{kT}\right)$$
where $k = \pu{1.381 \times 10^−23 J K-1}$. Now if let $E_i$ be the higher energy state, then am I right in saying that $E_i-E_j = \pu{1 eV}$, or should I convert this to some other unit?

Any quantity that is used as a power has to be dimensionless (that is, a pure number with no units). This means that $kT$ has to have the same units as $E_i-E_j$, so the terms in the exponent cancel. If you're measuring temperature in $\pu{K}$ and Boltzmann's constant in $\pu{J/K}$, then the units of energy you have to use are joules, $\pu{J}$.

Alternatuvely, you can convert $k_B$ to using $\mathrm{eV \cdot K^{-1}}$.

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text:
How are non-Stoichiometric compounds determined?
For solids where the compound contains defects, the stoichiometric equation deviates from the "pure" one. How are these compounds determined? I am looking in my chemistry book but I do not see anything on it. Are there some resources on this?
UPDATE:
After doing some thinking, I considered the Auger Electron Spectroscopy technique used to study the composition of surfaces in materials science. These , I believe, are typically used to study the amount of defects in materials. I do not know how this would be applied to the situation I described above though (as this is only applicable to surfaces). I will continue searching.

You can measure non-stoichiometry by determining how many atoms are present.  There are a number of ways to do this.
You can dissolve your unknown material in an acid and use a reagent to precipitate away one element but not another.  By keeping careful track of the number of moles of reagent you used, you can know how many atoms were precipitated and thus determine the stoichiometry of the original sample.  Once upon a time, this approach was used to demonstrate the fact that some chemicals ARE stoichiometric!
Today, another commonly used method is to count the number of X-rays emitted by a sample in an electron microscope.  Each atom emits X-rays with different amounts of energy so you can determine not only what atoms are present, but how many relative to other atoms.
For example, you can measure iron sulfide which stoichiometrically should have one Fe atom for each S.  Fe emits X-rays with about 2.5 times the energy of S, so they are easy to tell apart.  You then place a known stoichiometric iron sulfide (for example, using chemistry as above) in the microscope and measure how many Fe and S X-rays you get.  Let's say, hypothetically, you get one X-ray from Fe for each X-ray from S (just to keep the math simple).  Then you place a non-stoichiometric iron sulfide in and suddenly you get about 0.9 Fe X-rays for each S.  Presto, you know you have about 9 Fe atoms for each 10 S.
Note: For simplicity sake, I left out all the intricacies of the experimental method which would have to be considered in a real measurement.  If you are interested you can google: Energy Dispersive Spectroscopy (EDS), wavelength dispersive spectroscopy (WDS), and wet chemical analysis.


text:
How to calculate molality, molarity and mole fraction given density (g/mole)?
Calculate the molality, molarity and mole fraction of a $\ce{CuSO4}$
in a 17% m/m aqueous solution (density of the solution is 1.367 g/mole)
I did .17 * 1.367 to find grams of $\ce{CuSO4}$
and .83 * 1.367 to find grams of $\ce{H2O}$
Then used to them calculate:
molarity = 0.00146 M
molality = 1.287 m
Mole fraction = 0.023
I'm wondering if I'm doing it right?

First let me correct you, m/m means mass/mass % I think you assumed it to be mole/mole.
Assume you have 100g of solution in which Cu is 17g and solvent is 83 gm. 17 gm copper means 0.268 moles of Cu.
Now, density = Total weight/Total moles = 100/(0.268 + x) = 1.367 where x is moles of solvent, from here you will get x = 72.8849
Thus, mole fraction of Cu is 0.268/(0.268+72.8849) = 0.00366
Molality = Moles of solute/Weight of Solvent(in Kg) = 0.268/0.083 = 3.229 m
Since, density in terms of g/l is not given I don't think it's possible to calculate molarity.


text:
How can you tell how much actually reacted in an acid base reaction
If you facilitate a weak acid base reaction. i.e. $\ce{NaOH + H2SO4}$. All of it may not react and the solutions will not neutralize. How can you determine how much actually reacted?

Welcome to Chemsitry.SE. There is a way to determine the extent of reaction of weak acids and weak bases using equilibrium calculations. However, your acid/base pair of choice is a strong acid $(\ce{H2SO4})$ and a strong base $(\ce{NaOH})$. Assuming you have stoichiometric equivalence between the two, this reaction will go to completion.

@BenNorris that would also be true with a weak acid and a strong base. One should also consider that $\ce{H2SO4}$ is not that strong on the second ionization.

@fp I think you are right. you should answer the question writing it is not a good praxis use comment to answer the question ..

The procedure is called titration. It consists in adding a pH indicator (that changes color depending on pH) to the solution and adding measured quantities of base or acid until the indicator changes color and determining how much reacted from the amount used.
An alternative is to use a pH meter instead of an indicator specially in automated analysis.

As noted by Ben in the comment, $\ce{NaOH}$ and $\ce{H_2SO_4}$ are strong base and acid, respectively, so they will react till one is completely used up. Those react in 1:2 molar ratio, so it would be easiest to determine if you have more moles of one or another, and it will simply react, using one mole of  $\ce{H_2SO_2}$ per two moles of $\ce{NaOH}$, until there is only one left. If there is one left, you will have a non-neutral solution in the end.
EDIT: Obvious, obvious mistake... Thanks for pointing it!


text:
Is there an online database of acid and basic dissociation constants?
Is there an online database of acid and basic dissociation constants, all in one place, free to use?

Is this possibly what you are looking for? => http://www.csun.edu/~hcchm003/321/Ka.pdf and http://www2.ucdsb.on.ca/tiss/stretton/database/equilibrium_acids.htm and http://genchem1.chem.okstate.edu/1314F13/Database/ABKs.pdf .... ?

Here is a very large table of pKas, measured in DMSO which allows the determination of a much broader range of acidities than measurements in water. Note however that acidities in water are not necessarily the same as acidities in DMSO, as the latter is a polar non-protic solvent. For example, $\ce{H2O}$ is approximately $10^{15}$ times more acidic in water than in DMSO (and equivalently, $\ce{OH-}$ is approximately $10^{15}$ times more basic in DMSO than in water).

To add to my comment above, I have since realized that the same site also has an extensive section for acidities measured in water, here.

The Evans' and Bordwell $pK_a$ tables provide an excellent source of data for water and DMSO based measurements, particularly for organic compounds.

There are a number of resources available that provide dissociation constants for acids and bases.  The following are examples:
UC Davis Chemwiki pages E1: Acid Dissociation Constants at 25°C and E2. Base Dissociation Constants at 25°C provide many examples.  These are based on information taken from CRC Handbook of Chemistry and Physics, 85th Edition.
The WikiBook Chemical Principles/Solution Equilibria: Acids and Bases provides a reasonable database of dissociation constants of acids and bases (Tables 5.3 and 5.4), as well as some further information about how the values are calculated.
The California State University webpage p-functions for common acids and bases, also at 25C, of a series of acids and bases.


text:
Why does Ruthenium only have one electron in its 5s orbital in the neutral oxidation state?
So recently I have undertaken some research with some organometallic chemistry involved, specifically regarding that of ruthenium-aryl compounds. Trying to understand the chemistry of the central metal ion, I noticed something weird a couple of weeks ago, from Wikipedia: Ruthenium only has one electron in that orbital ($4d^75s^1$), rather than the two one would expect a transition metal to have, going with a naive Aufbau across the periodic table.
I am aware, I think, in the 4th period period transition metals, of a tendency to promote their 4s electrons to the 3d orbitals, when possible, so that they have either "half-filled" (5 electrons) or completely filled (10 electron) shells.  These are "magic numbers" for transition metals, apparently; just another one of those heuristic things they teach you in inorganic chemistry that I wish I understood more about. But alas, ruthenium has an asymmetric 7 electrons in its 4d orbital, blasted thing.
Does anyone know anything? I tried using Google Scholar and a few databases, but when I type in "orbital" pretty much all I see is molecular orbital calculations specific to particular molecules. I'd really love to see papers pointing at some faint ideas I may be dimly aware about...

Look at this explanation: http://www.madsci.org/posts/archives/2005-07/1120542929.Ch.r.html

The above link is about how to obtain electronic configuration from spectroscopy experiment, but not why the electronic configuration should be. One may wonder, if the ground state of Ru is $^5$D$_4$, then the configuration would be 4d$^6$5s$^2$, like Fe/Os. Then the question is why the ground state of Ru is $^5$F$_5$? I have a pure guess about the reason... see next comment (the link is too long)

From Oxtoby et al, Principles of Modern Chemistry, 7th edi p219 and Levine, Quantum Chemistry, 5th edi, p314, the orbital energies of Sc and the elements beyond, 3d<4s. The electrostatic repulsion makes energy of 3d$^3$ higher than 3d$^1$4s$^2$.
Just my pure guess, Ru is bigger than Fe, electrostatic repulsion is smaller, so 5s$^1$. However, the 6s$^2$ electrons in Os are extra stable due to relativistic effect, so again 6s$^2$.
Quantitative calculation will be highly appreciate for justifying this hand-waving argument

Just to add slightly to this answer, it isn't so much that the d orbitals in $\ce{Ru}$ are large, but rather that the d orbitals in $\ce{Fe}$ are anomalously small. This happens because the 3d subshell is the first of its kind, and as such its orbital wavefunctions have no radial node (the same happens with 1s, 2p, 4f, 5g, etc). This creates unusually high interorbital and intraorbital electron repulsion. Also, in $\ce{Os}$, not only is the 6s subshell relativistically stabilized, but the 5d subshell is slightly relativistically destabilized, favouring the 6s² ground state even more.

@Billare I'm sorry, but I can't remember my sources too well. I know Pekka Pyykkö is an author of several excellent investigations into relativistic effects in chemistry, including this fascinating review from just a year ago. He also dealt with lots of computational chemistry in general, so he may have also discussed the small orbital anomaly somewhere. Another reference I managed to quickly find on Google was this.

@Billare Aha, here is one of the main sources I came across regarding compact orbitals. It really is surprising how the lack of radial node for some wavefunctions explains all sorts of interesting behaviour, such as the high stability of $\pi$ bonds in the second period nonmetals, the reluctance for fourth period transition metals to display high oxidation numbers and their relatively weak bonding, and the striking chemical similarity among the lanthanides.


text:
Bonds between metals and non-metals
My teacher in college says that bonds between metals and nonmetals are ionic.
$ \ce{Metal - Metal} $ $\Rightarrow$ Metalic bond
$\ce{Non metal - Non metal}$ $\Rightarrow$ Covalent bond
I have to write about $\ce{CuCl2}$, and found in Wikipedia that if you subtract electronegativity you get what bond it is:
$$\chi (\ce{Cl}) = 3.16, \quad \chi (\ce{Cu}) = 1.90$$
$$\chi = 3.16 - 1.90 = 1.26$$

$\chi < 0.4 \Rightarrow$ covalent non-polar
$0.4 < \chi < 1.7 \Rightarrow$ covalent polar
$\chi > 1.7 \Rightarrow$ ionic

So it should be really a covalent bond. Which one is correct?

Cl and Cu are almost placed at the same side of the periodic table, so they are non metals. Hence as expected by your teacher and wikipedia, they will form covalent bond. Yes, I agree that Cu also acts as metal, as it may be transitional element. Anyway, bonds are just the names we give, if there is full transfer of electrons, we say it as ionic, if there is less transfer of electrons, we say it as polar covalent, if transfer of electrons is negligible or shared, we say it as non polar covalent. In reality no bond is ideally true, so it becomes convention to accept any of them which dominates.

@CURIE Copper is most definitely a metal, by any standard. You should not doubt it in order to fit an observation to a mere rule of thumb.

Sorry, if my comment is misleading or if it is wrongly predicted.

You should be very careful about extrapolating a physical meaning from comparing electronegativities because EN is essentially an arbitrary value where everything is defined relative to one another. EN is useful for nothing more than giving a reference for how that element is likely to interact with electrons.

Related, with excellent answers: Metal Compounds that bond covalently

https://chemistry.stackexchange.com/questions/6923/bonds-between-metals-and-non-metals/6939#6939

You should be careful with simple associations such as "metal + non-metal = ionic bond". These tend to throw out the idea of understanding the chemistry involved in favour of rote memorization. Note for example that mixing caesium metal with gold will produce a salt instead of an alloy, caesium auride ($\ce{Cs^+ Au^{-}}$). Mixing barium metal and platinum can also produce salts, though their structures are somewhat more complex. One can also argue that there is significant ionic character in solid xenon difluoride, even though both atoms are non-metals.
The idea of using electronegativity to determine covalent/ionic character is also meant as a helpful guide, not as a strict rule with black-and-white limits. Firstly, all bonds have both ionic and covalent character; both concepts are an oversimplification, and in reality it is more correct to say that a bond has a certain contribution from each type of bonding. This means there is a smooth transition from compounds with mostly ionic character and those with mostly covalent character. Also, the inequalities you mention rely on Pauling electronegativities. Electronegativity is surprisingly still a hotly debated topic, as we continue to search more general, more fundamental and more precise ways of defining it. Pauling electronegativities are based on empirical thermodynamic data regarding bond energies after applying a certain equation that was "picked", not derived from scratch. The values are particularly poorly defined for transition elements, such as the $\ce{Cu}$ in your problem. You get some not-so-easy to explain situations, like $\ce{HF}$ as a gas that is a borderline ionic compound.
Finally, in light of these comments, the answer to your question is that bonding in $\ce{CuCl_2}$ (I'm pretty sure that's what you actually meant to write) has intermediate characteristics between a purely ionic and a polar covalent bond, with similar contributions (though pinpointing which is highest sounds like an exercise in futility). A good way to study it more in depth is to analyze Fajans' rules. After a little self-calibration, you can get a good feel for the degree of ionicity and covalency of a compound. Some further but less certain evidence (lots of caveats!) for the intermediate character of $\ce{CuCl_2}$ can be found by looking at the substances' melting and boiling points ($\pu{498°C}$ and $\pu{993°C}$ [decomposition], respectively, according to Wikipedia). They are both quite high compared to substances with polar covalent bonds (dimethylformamide boils at around $\pu{150°C}$), but rather low compared to substances with very ionic bonds ($\ce{NaCl}$ boils over $\pu{1400°C}$).

we aspect that the bond formed between Fe and Cl is ionic because Fe is metal and Cl is  non metal but charged on Fe is +3 and on Cl is-1 so  polarisation of chlorine ion by Fe+3 ion takes place but also with presence of  d orbital in valence  shell of Fe+3 ion polarisation also takes place  to maximum extent  and covalent charachter  arises in  ionic compound FeCl3.

This does not answer the question of what type of bond should be considered in copper(II) chloride.


text:
Why does the reaction quotient use the products (multiplications) of reactants and products, rather than their respective sums?
As you may know, the reaction quotient $Q_c$ is defined by the equation
$$
Q_c = \frac{[C]^\gamma [D]^\delta}{[A]^\alpha [B]^\beta}
$$
for the chemical reaction
$$
\alpha A + \beta B \rightarrow \gamma C + \delta D.
$$
This is something I've been struggling to understand at an intuitive level. Why is $Q_c$ the ratio of the product of the reaction's product concentrations and the product of the reaction's reactant concentrations (sorry for the confusing terminology; I can't think of a more precise way to phrase that), rather than the ratio of the sums of concentrations, like this:
$$
Q_c' = \frac{\gamma[C]+\delta[D]}{\alpha[A]+\beta[B]}
$$
This second, incorrect equation makes more sense to me because concentrations of solutes add up. Is the ratio defined as such so that it is directly proportional to the concentrations of each product and inversely proportional to the concentrations of each reactant? Is there some relation to kinetics that might give me a deeper insight into the meaning of this quotient?

See here for a mathematical derivation of the relationship between the change in free energy in a reaction and the concentrations of the species (or partial pressures, in that case). It turns out that you add and subtract logarithms of the concentrations, and using a property of logarithms it is possible to condense all concentrations into a ratio that we call the reaction quotient.

@NicolauSakerNeto - if you could summarize/explain that derivation it would make a good answer.

@BenNorris Physical chemistry isn't my forte, so I would rather someone else with a better background describe it, if possible.

Related question: Why is the equilibrium constant the way it is?

The derivation cited can help you understand it mathematically. But the answer also turns out to be satisfactory intuitively. Let me try to explain how.
The term $RT\ln Q_c$ is derived form the entropy part or $\Delta G=\Delta H-T\Delta S$, and in a slightly abstract way, represents entropy of mixing of the products and the reactants. This is why reactions which are spontaneous initially, as concentrations change, reach to the equilibrium state before complete conversion of reactants into products. If there was no entropy of mixing, a reaction which had negative free energy change(under any set of conditions) would continue to have a negative change (hence, spontaneous) until all of the reactants is converted to products, which will be the state of maximum entropy. But since there is entropy of mixing, the state of maximum entropy is reached before all the reactants are converted into products (because somewhere in between the entropy of mixing due to having a mixture of reactants and products outweighs the negative $\Delta G$ for the reactants to convert into the products) , and the concentrations at which this maximum entropy state can be reached is precisely what the equilibrium constant tells us.
To calculate the entropy, we need to realize the fact that certain components of the reaction may have a greater contribution in deciding the equilibrium constant than others because, at the maximum entropy state, the components are present according to their stoichiometric relations and hence one might be in a well-defined excess over other. The relative contributions of each species to the entropy is logarithmic, that is, the increase in entropy is a logarithmic function of concentration (related slightly in a abstract manner to $S=k_B\ln \Omega$). Therefore, the contribution to the free energy and consequently in deciding the $K_c$ is logarithmic in the concentrations. The stoichiometric coefficientts are multiplied to the logarithmic contributions of each(which appear as the power to which the concentrations are raised when brought inside the logarithm) and the relative contributions are simply added, but since the addition of logarithms means multiplications of its arguments, we get the familiar $RT\ln Q_c$ term which gives the mentioned definitions of reaction quotient and the equilibrium constant.
I have used reaction quotient and equilibrium constant interchangeably but the RQ represents an unfulfilled equilibrium constant and hence represents the same drive towards maximum entropy as $K_c$ (which is the state of maximum entropy itself).

Thanks for the answer, this is significantly more thorough than anything I could have managed.

@NicolauSakerNeto I'm pretty sure equilibrium is not actually a maximum-entropy state, but is in fact the state where the entropic and enthalpic contributions to the free energy change are in balance.

@Brian Really? Then what would happen in a simple closed universe with only one equilibrating reaction? Would it reach equilibrium or would there be some other universal maximum entropy state? This question has a related answer, and it rests on the fact that the second law of thermodynamics is equivalent to saying "a system with a fixed internal energy self-adjusts its free parameters in such a way that the entropy is maximized," though no mathematical proof of this was given (outside the scope).

@NicolauSakerNeto And this is why, to this day, thermodynamics consistently mystifies me, since it seems impossible for equilibrium to be a maximum-entropy state unless $\Delta S_\mathrm{rxn} = 0$ for every reaction (which, itself, seems paradoxical). (Otherwise, in a closed system it seems like every equilibrium would proceed to completion in the direction of increasing $S$.) There must be some factor in there that I don't understand or am not considering, because many people agree that I'm wrong. :-)

@Brian After mulling over it for a while, I think we may have been talking about different things. We need to differentiate between the entropy of the system and the universe. I think you're right, the entropy of a chemical reaction at equilibrium need not have the highest entropy possible for the system, but the universe containing a reaction at equilibrium and its surroundings will be at the maximum of entropy. The enthalpy change of the system in Gibbs' free energy equation is a surrogate for the entropy change of the surroundings.

In short, minimizing a system's G is equivalent to maximizing universal S, but not equivalent to maximizing system S. For a chemical reaction at equilibrium, G is minimal as suggested by Higgs' post. See also this derivation on Wikipedia. Otherwise, I'm mostly out of my league here.

@NicolauSakerNeto I sorta grok that, I think... that, say, the heat released by a negative $\Delta H_\mathrm{rxn}$ translates to an increase in the entropy of the surroundings?

Is there some relation to kinetics that might give me a deeper insight into the meaning of this quotient?

Yes! The way I handle equilibrium reactions treats them as a "competition" between a particular reaction and its exact reverse:
$$
\ce{\alpha A + \beta B ->[k_1] \gamma C + \delta D} \tag{1}
$$
$$
\ce{\gamma C + \delta D ->[k_{-1}] \alpha A + \beta B} \tag{2}
$$
Written as an equilibrium reaction, of course, it looks like this:
$$
\ce{\alpha A + \beta B <=>[K_1] \gamma C + \delta D}\tag{3}
$$
So, the question then becomes how to relate the two kinetic reactions (1) and (2) to the equilibrium reaction (3). Assuming both Reactions (1) and (2) are elementary reactions as they're written, the two rate expressions become:
$$
r_1 = k_1 \left[A\right]^\alpha \left[B\right]^\beta \tag{4}
$$
$$
r_{-1} = k_{-1}\left[C\right]^\gamma \left[D\right]^\delta \tag{5}
$$
By definition, at equilibrium the rates of the forward and reverse reactions are equal. So, combining (4) and (5):
$$
k_1 \left[A\right]^\alpha \left[B\right]^\beta = r_1 =
r_{-1} = k_{-1}\left[C\right]^\gamma \left[D\right]^\delta \tag{6}
$$
Now, dividing (6) through by $k_{-1}$ and by $\left[A\right]^\alpha\left[B\right]^\beta$ gives (almost) the familiar equilibrium constant expression:
$$
{k_1 \over k_{-1}} =
{\left[C\right]^\gamma\left[D\right]^\delta \over
   \left[A\right]^\alpha\left[B\right]^\beta } \tag{7}
$$
If we define $K_1 \equiv {k_1 \over k_{-1}}$ (as we should!), then (7) is exactly the standard form of an equilibrium constant expression.
It is (I think?) straightforward to see from here why the reaction quotient $Q_C$ is also defined this way, and why if $Q_1 < K_1$ the reaction will proceed forward toward equilibrium, while if $Q_1 > K_1$ the reaction will proceed in the reverse toward equilibrium.

This doesn't answer why they are exponents.

@AdityaRoychowdhury The exponents enter at the "Assuming both Reactions (1) and (2) are elementary reactions as they're written..." part, due to the use of the principle of mass action in defining the rates of elementary reactions.


text:
Understanding Moseley’s law from the Rydberg-type equation?
To explain the characteristic X-ray emission peaks for various elemental targets, a formula was developed which was similar in construct to the Rydberg equation for H-atom as derived by the Bohr-model.
$$\tilde\nu=R\left(\frac{1}{n_\mathrm{f}^2}-\frac{1}{n_1^2}\right)\left(Z-\sigma\right)^2$$
where $\tilde\nu$ is the wavenumber, $\sigma=1$ for the $K_\alpha$ line and $7.4$ for the $L_\alpha$ line. My question is:
Although Bohr model is wrong in light of the modern quantum mechanical concept of atomic structure, why does the energy for transition (and X-ray emission) of complicated atoms with complicated screening effect be represented by a simple adjustment of the $Z$ factor in the Rydberg equation with the value of $R$ used in case of H atom? Why isn’t there a need of complicated corrections (instead of simple subtraction of an experimentally obtained $\sigma$) to predict the X-ray wavelength?

Though the Bohr model is oversimplified, the Rydberg equation itself is a good approximation (I wonder if it can be derived using the more accurate orbital model). Either way the $K$ and $L$ shell electrons are very close to the nucleus and maybe they don't suffer as much from complicated electron-electron interactions (or rather they likely become less important relative to electron-nuclei attraction), so apparently they can be well approximated by just considering a screening effect which changes the effective nuclear charge.

I guess the equation is used as more of a fitting curve for experimental data points rather than an accurate theoretical equation, especially if the value of $\sigma$ is allowed to vary slightly.

Also perhaps part of the reason the Rydberg equation does not work well with transitions between valence electrons is that in the valence shell there is a significant energy difference between subshells. For core electrons, I think the degeneracy is close to restored, so it becomes less inaccurate to work with only the principal quantum number.

@NicolauSakerNeto I guess all these comments could be posted as a good answer. Though a bit speculative, They effectively answer my main concern.

I appreciate it, but I would rather abstain. I hope someone can come forth with a more direct answer. My comments were mostly musings to get some ideas flowing. I don't know the explanation for certain myself.

The concept to understand here is that most theories in physics and chemistry are obtained by observing then theorizing, this results in the theories and models being made in such a way that they can explain most (if not all) observable phenomenon.
So, when Bohr made his model of the atom, it had to explain the the phenomenon of "X-ray emission" otherwise it wouldn't be accepted, because, X-ray emission is an observed phenomenon of atoms. Following on, it was discovered that the Bohr-model couldn't explain all observable phenomenon as equipment got better and more things could be seen (notably the splitting of spectral lines in magnetic fields). So any theory that followed on to explain this splitting of spectral lines had to explain X-ray emission as well, without disturbing the results of Bohr's theory.
A suitable analogy would be the discovery of special relativity by Einstein, whatever discovered had to explain why Newton's Law had stuck around for hundred's of years without people realizing it was wrong. (And it did, the correction at velocities we know are less than one part in a million and that's why we didn't think Newton was wrong)
So the deeper question is, even a one part in a million error is still an error and so we should know that the theory is wrong even early on... this is not the case as most instrumental errors on their own are much larger in this regard (at least in Newton's time), and the same applies here in that older instruments weren't usually sensitive enough to notice the change brought on by quantum mechanics on X-ray emission to that of Moseley's rough equation. (Though now it is usually possible to detect differences to a good degree.)
So to the final part of your question, there is a horrendous equation waiting if you want to plug in exact values to find the exact wavenumber, but most like the instrument barely notices a difference, so in a sense, it doesn't matter which theory we use. (It's very well possible a hundred years down the road, another genius comes up with an even smaller correction that can be observed only with futuristic devices)


text:
Evaporating the water of a solution of only carbonate ions and water gives?
I have a solution with only $\ce{CO_3^{2-}}$ ions and water.
What would happen with the carbonate ions if I were to evaporate the water of that solution (in a vacuum)?

I'm pretty sure that you also have some counter ion in there

but what if the ion was selectively let through by a semi penetrable membrane into this solution?

You can't selectively remove one type of charge and sustain the difference indefinitely. Because the electrostatic  force constant is so huge, without work being done in the form of an opposing electric field, significant net charges would quickly be neutralized by tearing out the opposite charge from whatever they can, far before getting to macroscopically measurable amount of excess ions. Leaving the system in a vacuum is not enough as the charge would jump through empty space.

so would it be possible to do or don't the laws of physics/chemistry allow it to happen?

There is nothing that says solutions must be exactly electrically neutral, just that they can't have too much excess charge. It's a problem of quantifying the limit. After some research, it seems that you theoretically could prepare a solution with excess $\ce{CO^{2-}_3}$, though the excess would have to be quite small (apparently well under $1\ \mu M$). I'm not exactly sure what would happen if you removed the water entirely in a vacuum. Maybe the ions would become gaseous and float around until they hit a surface.

Here are a few sources: 1 2 3

I would love to add some quantum chemical bone to the stew we are trying to cook here.
Let us assume that it is possible to have a solution of carbonate ions, like already discussed by Nicolau in the comments. Then Rauru is also right that you will have to consider several equilibria:
\begin{aligned}\ce{
(1) && 2H2O &<=> H3+O + {}^{-}OH \\
(2) && CO3^{2-} + H2O &<=> HCO3- + {}^{-}OH\\
(3) && HCO3- + H2O &<=> H2CO3 + {}^{-}OH
}\end{aligned}
Now $(3)$ seems a little far fetched, so we will ignore it.
When you do now remove water from the equilibrium through (gentle) evaporation, you will have a mixture of only $\ce{CO3^{2-} + HCO3- + {}^{-}OH}$. Further assuming this is possible you will always have some collisions within this ensemble. Statistical it is therefore always possible to also remove $\ce{HCO3- + {}^{-}OH}$ via $(2)$ from the system. Also single entities of carbonate ions will repel each other as states by G M. So basically it all boils down to on question: Is carbonate stable in vacuum? What happens to carbonate if you put it into vacuum?
Now we have some quite good approximations to these particular conditions (gas phase and 0K), but I do not want to go into detail. I did a calculation on the carbonate with Gaussian09 and as a cheap shot BP86/cc-pVTZ level of theory. The only thing I want to show is the highest occupied molecular orbital.

What can clearly see is that this is no bonding orbital. There are lone pair contributions from all Oxygen atoms (due to imposed symmetry constraints). In addition to the high negative charge, electrons will not be stabilized by the field of the nuclei. Also, Orbitals only vanish at infinity, giving electrons a non-zero probability to actually leave the compound. Furthermore, there are two electrons in this orbital, so the spins of these are paired, which also costs energy.
My suspicion therefore is that the carbonate at least loses one electron. To where? To anywhere. (Also mentioned by Nicolau in the comments.) I do not have any actual proof for this, as it quite hard to find people actually doing experiments and/ or calculations on that topic. I remember that one question of a Professor of mine in the first semester was: "What happens to sulphate if you shoot it into space?" The answer was given without any further explanation and I never bothered to ask: "It looses an electron."

Note:
Another point that should always be considered when dealing with gas phase reactions is, that when ions have enough kinetic energy they might react in the strangest ways. One of them might be
$$\ce{2CO3^{2-} -> 2CO2- + O2^{2-}}.$$
Also collision with any particle may cause electrons to be transferred.

Theoretically you should have an explosion. Because every $\ce{CO_3^{2-}}$ ions exerts a force against the other. You should be able to calculate it with coulomb equation:
$$|F|=K_e\frac{q_{\ce{CO_3^{2-}}}^2}{r^2}$$
Of course you have to consider the force exerts from every particle to each others, summing every force field assuming a certain          distance: a very hard calculation.
By the way your assumption are quite hard to reach in great scale as say Raur Ferro. But a practical application of electrostatic forces that could be enlightening is Electrospray ionization (ESI), a technique used in mass spectrometry to produce ions.
 In this technique the solvent from an areosols that hold the analyte is forced to evaporate (with various expedients) from a charged droplet until the drop becomes unstable due the charges. At a certain point the droplet undergoes to what is called Coulomb fission, so the original droplet explodes creating many smaller droplets. I know this is not the same thing but this can show you how electrostatic forces can act. I think however that this question is more related to physic than chemistry.

As a theoretical experiment, you must think on different aquieus equilibria:
$\ce{H2O <=> H+ + OH-}$
$\ce{CO3^{2-} <=> HCO3- <=> H2CO3}$
$\ce{H2CO3}$ will be formed with water's $\ce{H+}$, and in vacuum, possibly will decompose to water and $\ce{CO2}$.
At the end, may be a mixture of $\ce{OH-}$ and $\ce{CO3^{2-}}$ will remain.


text:
Changes in defects in a material due to heat treatment
Currently i am doing some research on optical properties of  a oxide material and for that purpose i have done some photoluminescence study on it. Because this is an ongoing research i cant divulge much information here. I have used "$Ex$" excitation wavelength for getting emission spectra for my oxide sample. I have prepared two types of oxide samples one is heat treated and the other not heat treated. After taking emission spectra data for both the samples I got peaks similar to this
Samples               Peak wavelength (nm)

Non Heat treated          a, b

Heat treated              b,c,d,e

*a,b,c,d & e are wavelength where emission peaks detected
Both samples (heat treated and non heat treated) has same chemical composition which i confirmed using powder XRD. As you can see from the data the heat treated sample has more emission peaks comparing to non heated samples. I can understand from this data that heat treatment to the sample causes increased defects in the sample. what i thought was that after heat treating a crystalline material; the defect in that material should decrease and not the other way round? Is my rationale wrong. Why heat treated sample has more defects?

Can you be more specific on the XRD powder analysis? The same means really the same? If so, there should be no difference in other properties.
The heating can reduce defects (disappear of a), but induce some different polymorphs, for example (c,d,e).
Therefore I would first not call it a defect. If the defect is so common, that you can detect it in spectroscopy, it is not longer a defect, but something quite regular.

@ssavec yes XRD has same chemical composition but more crystalline and the crystalline size also got increased after heating. I have gone through lots of the papers and the peaks a,b,c,d and e are reported in my oxide mateiral but not together. sometimes ab together or be,adb..etc

It seems, that your oxide can have different crystal polymorphs and probably also other non-crystalline phases.
You synthesize your material as mixture of those, where the composition depends on the preparation process.
Upon heating and slow cooling you probably recrystallize the species to more thermodynamically stable polymorph (new peaks c,d,e) on the expense of less stable one (a).
It can also depend on the speed of cooling, so you could obtain some metastable phases, etc...
Even then, treat your material always as an unkonwn mixture of uknown number of polymorphs and let the spectra tell you in which way the composition shifted.


text:
Can we prepare alkyl chloride or alkyl bromide by Finkelstein reaction?

My book says:
  Finkelstein reaction is particularly useful for preparing iodoalkanes. The iodoalkanes are obtained by heating chloro or bromoalkanes with a concentrated solution of sodium iodide in dry acetone.
  $$\ce{R-X + NaI ->[\ce{acetone,~reflux}] R-I + NaX}\\
(\ce{X}=\ce{Cl,Br};\ \ce{R}=\text{alkyl group})$$
    NOTE: Refluxing is a process of heating a liquid in a flask provided with a condenser. The vapours of the liquid which rise up get condensed back into the flask.
    Sodium chloride and sodium bromide being less soluble in acetone get precipitated from the solution and can be removed by filtration. This also prevents the backward reaction to occur according to Le Chatelier's principle. The reaction gives the best results with primary halides.

MY QUERY

Can we prepare alkyl chloride or alkyl bromide by Finkelstein reaction?
As we can see, first line reads that Finkelstein reaction is particularly useful for preparing iodoalkanes. The lines followed by it signify that $\ce{NaCl}$ and $\ce{NaBr}$ are less soluble in acetone. If we want to prepare alkyl chloride or alkyl bromide, we will be needed to dissolve $\ce{NaCl}$ or $\ce{NaBr}$ respectively in acetone, which is not so possible. Is it true that, we can't prepare alkyl chloride or alkyl bromide by Finkelstein reaction because of the above reasons?

The thing is that you have to start with some compound $\ce{R-X}$ where $\ce{X}$ is a good leaving group. The quality of the leaving group is often thermodynamically defined, and in the Finkelstein reaction that you described, the energy motor for the whole thing is the insolubility of the salt that is produced.
I don't see what favourable starting material you could use for this in order to prepare other haloalkanes, especially since there are other ways for doing that.
That being said, it would in principle be possible to do
$$\ce{R-X + NaY -> R-Y + NaX v}$$
as long as the solubility of $\ce{NaY}$ is larger than the solubility of $\ce{NaX}$.


text:
Mathematical basis of why enthalpy of mixing is 0 for ideal gas
For this case, let's say there are two identical chambers connected by a small tube, constant pressure and temperature. There's a closed valve that separates the two chambers. Within the chambers there are two ideal gases (doesn't matter which, I think; they're ideal gases). I release the valve and the contents of the two chambers mix.
I understand that heat of mixing for ideal gases is 0 conceptually, because the molecules don't have interaction forces between them so introducing the gases to each other doesn't do much with respect to this. But what is the mathematical basis for heat of mixing being 0 for an ideal gas?
quick edit: above might not be clear. I mean that dH=TdS+VdP. In isothermal, isobaric conditions, dP could be zero, but dS isn't necessarily, that'd imply adiabatic conditions. So how is dH=0 for mixing an ideal gas?
Does the above line of thought apply for mixing? If so, do we take the terms before or after the volume increases? (in this case, final volume = 2*initial volume) I feel this is a frame of reference issue, one that comes up a lot for me in thermo.

$H = U + PV$, and $U = n c_V R T$… from which you can calculate $\Delta H$ and check that it is zero

A strategy that often works is to look at the initial and the final states and then compare them. A good place to start is usually the Gibbs free energy of the system (with the intention to use $\Delta_{\mathrm{mix}}H = \Delta_{\mathrm{mix}}G + T \Delta_{\mathrm{mix}}S$ later): for isothermal and isobaric conditions it is
\begin{equation}
  G = \sum_{i} n_{i} \mu_{i} + \mathrm{constant} = n_{\mathrm{tot}} \sum_{i} \underbrace{x_{i}}_{= \, \frac{n_{i}}{n_{\mathrm{tot}}}} \mu_{i} +  \mathrm{constant} \hspace{4em} (1)
\end{equation}
with $n_{i}$, $\mu_{i}$, and $x_{i} = \frac{n_{i}}{n_{\mathrm{tot}}}$ being the amount, the chemical potential, and the mole fraction of the $i$th component in the system, respectively, and $n_{\mathrm{tot}} = \sum_{i} n_{i}$ being the total amount of gas in the system.
So, now we need an expression for the chemical potential of an ideal gas.
It can be shown that this is given by
\begin{equation}
  \mu_{i} = \mu_{i}^{0} + R T \ln\Bigl(\frac{p_{i}}{p^{0}}\Bigr) \hspace{18em} (2)
\end{equation}
where $\mu_{i}^{0}$ is the chemical potential of the $i$th component in a given standard state, $R$ is the universal gas constant, $T$ is the temperatur, $p_{i}$ is the partial pressure of the $i$th component, and $p^{0}$ is the standard pressure (i.e. the pressure connected to the standard state).
For ideal gases we can apply Raoult's law stating that
\begin{equation}
  p_{i} = x_{i} p_{\mathrm{tot}}
\end{equation}
where $p_{\mathrm{tot}}$ is the total pressure of the system.
Using this with equation (2) we get
\begin{equation}
  \mu_{i} = \mu_{i}^{0} + R T \ln\Bigl(\frac{x_{i} p_{\mathrm{tot}}}{p^{0}}\Bigr) = \mu_{i}^{0} + R T \ln x_{i} + R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr) \hspace{6em} (3)
\end{equation}
Furthermore, it is known that the partial pressures in a system must add up to the total pressure, i.e.
\begin{equation}
  p_{\mathrm{tot}} = \sum_{i} p_{i}
\end{equation}
Ok, now that the stage is set we simply have to apply equations (1) and (3) to the initial and the final state of the mixing process.
At the initial state the chambers are seperated (whereby chamber $i$ contains only one sort of gas labeled by $i$) and so these subsystems are completely independent from each other.
Thus, the Gibbs free energy of the total system of chambers is simply the sum of the Gibbs free energies of each chamber, whereby the gas in each of the chambers feels the ambient/total pressure $p_{\mathrm{tot}}$ in the system, since $x_{i} = 1$ (each chamber/independent subsystem holds just one kind of gas).
According to equation (1) the Gibbs free energy of the $i$th chamber is then given by
\begin{align}
  G_{i} &= n_{i} \mu_{i} = n_{i} \biggl( \mu_{i}^{0} + R T \underbrace{\ln 1}_{= \, 0} + R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr) \biggr) \\
        &= n_{i} \biggl( \mu_{i}^{0} + R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr) \biggr)
\end{align}
and the Gibbs free energy of the total system at the initial state is
\begin{align}
  G_{\mathrm{ini}} &= \sum_{i} G_{i} \\
                 &= \sum_{i} \underbrace{n_{i}}_{= \, x_{i} n_{\mathrm{tot}}} \biggl( \mu_{i}^{0} + R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr) \biggr) \\
                 &= n_{\mathrm{tot}} \biggl( \sum_{i} x_{i} \mu_{i}^{0} + \sum_{i} x_{i} \underbrace{R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr)}_{= \, \mathrm{constant}} \biggr) \\
                 &= n_{\mathrm{tot}} \biggl( \underbrace{\sum_{i} x_{i} \mu_{i}^{0}}_{= \, G_{\mathrm{ini}}^{0}} + R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr) \underbrace{\sum_{i} x_{i}}_{= \, 1} \biggr) \\
                 &= n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr) \biggr)
\end{align}
At the final state the chambers are connected and the gases mix, so the chambers are not independent subsystems anymore and the Gibbs free energy cannot be broken down into independent parts like before.
So now, the complete system of connected chambers must be treated via equation (1)
\begin{align}
  G_{\mathrm{final}} &= \sum_{i} \underbrace{n_{i}}_{= \, x_{i} n_{\mathrm{tot}}} \biggl( \mu_{i}^{0} + R T \ln x_{i} + R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr) \biggr) \\
                   &= n_{\mathrm{tot}} \biggl( \sum_{i} x_{i} \mu_{i}^{0} + \sum_{i} R T x_{i} \ln x_{i} + \sum_{i} x_{i} \underbrace{R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr)}_{= \, \mathrm{constant}} \biggr) \\
                   &= n_{\mathrm{tot}} \biggl( \underbrace{\sum_{i} x_{i} \mu_{i}^{0}}_{= \, G_{\mathrm{ini}}^{0}} + R T \sum_{i} x_{i} \ln x_{i} + R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr) \underbrace{\sum_{i} x_{i}}_{= \, 1} \biggr) \\
                   &= n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + R T \sum_{i} x_{i} \ln x_{i} + R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr) \biggr)
\end{align}
Now, that we have the Gibbs free energies of the initial and final state we simply need to subtract the former from the latter one in order to obtain the change in Gibbs free energy during the mixing process:
\begin{align}
  \Delta_{\mathrm{mix}}G &= G_{\mathrm{final}} - G_{\mathrm{ini}} \\
                       &= n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + R T \sum_{i} x_{i} \ln x_{i} + R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr) \biggr)  - n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr) \biggr)  \\
                       &= n_{\mathrm{tot}} R T \sum_{i} x_{i} \ln x_{i} \hspace{22em} (4)
\end{align}
Differentiating equation (4) with respect to the temperature we get the entropy of mixing since $\frac{\partial G}{\partial T} \big|_{p, \{ n_{i} \}} = - S$
\begin{equation}
  \Delta_{\mathrm{mix}} S = - \frac{\partial \Delta_{\mathrm{mix}}G}{\partial T} \bigg|_{p, \{ n_{i} \}} = - n_{\mathrm{tot}} R \sum_{i} x_{i} \ln x_{i}
\end{equation}
Finally, you get the desired result by using the famous equation $G = H - TS$
\begin{align}
  \Delta_{\mathrm{mix}}H &= \Delta_{\mathrm{mix}}G + T \Delta_{\mathrm{mix}}S \\
                        &= n_{\mathrm{tot}} R T \sum_{i} x_{i} \ln x_{i} - T n_{\mathrm{tot}} R \sum_{i} x_{i} \ln x_{i} \\
                        &= 0
\end{align}

@rch I'm glad that it helped. But I now realize that I forgot to address your "quick edit" and why using $dH = TdS + Vdp$ doesn't work. I think the problem might simply be that you've left out the chemical potential term. You'd rather have to use $dH = TdS + Vdp + \sum_{i} \mu_{i} dn_{i}$ where, as you said, $dp = 0$ for isobaric conditions. I haven't tried out that road but that should also get you to the right result eventually.


text:
Why do branched chain compounds have lower boiling points than the corresponding straight chain isomers?

The branched chain compounds have lower boiling points than the corresponding straight chain isomers. For example,

$\ce{CH_3CH_2CH_2CH_2CH_3}$ -
  No branching-Pentane (n-pentane) ($\mathrm{b.p.}=309~\mathrm{K}$)
$\ce{CH_3CH(CH_3)CH_2CH_3}$ -
  One branching-2-Methylbutane (Iso-pentane) ($\mathrm{b.p.}=301~\mathrm{K}$)
$\ce{C(CH_3)_4}$ - Two branches-2,2-Dimethylpropane (Neo-pentane) ($\mathrm{b.p.}=282.5\mathrm{K~}$)

This is due to the fact that branching of the chain makes the molecule more compact and thereby decreases the surface area. Therefore, the intermolecular attractive forces which depend upon the surface area, also become small in magnitude on account of branching. Consequently, the boiling points of the branched chain alkanes are less than the straight chain isomers.

The above extract from my book, mentions clearly that branching makes the molecule more compact and thereby decreases the surface area. Even for isomeric alkyl halides, the boiling points decrease with branching. The reason is said to be because of the decrease in surface area, same as explained before. How does decrease in the surface area make the intermolecular forces small in magnitude?

Actually the answers all miss the main point. The answer is that overall the attractive forces are dependent on the surface-to-volume ratio. https://en.wikipedia.org/wiki/Surface-area-to-volume_ratio

The simple explanation is that weak intermolecular forces (the forces that make something condense to a liquid when things are cold enough) depend on the surface area (as well as many other things). But in the case of relatively similar non-polar isomers (where weak intermolecular forces are the dominant forces), the larger surface area will lead to the large force and hence the highest boilling point.
2,2 dimethyl propane is a compact, almost spherical molecule. Pentane is long and "floppier" so will experience more forces between molecules.
The underlying, but slightly simplified, explanation of this is that intermolecular forces (often called van der Waal's forces) depend on attractions caused by quantum fluctuations in the surface electrons of the molecule. These lead to short-lived dipole moments that can also induce dipoles in neighboring molecules to which the original dipole is attracted. The larger the surface area, the more opportunity for such dipoles to exist and therefore a stronger force.

Straight chain molecules have more places along its length where they can be attracted to other molecules, so there are more chances of London Dispersion Forces to be developed. Hence, they have stronger intermolecular forces as compared to the branched chain molecules which have a compact shape, therefore fewer spaces where they can be attracted to other molecules.

The London / van der Waals forces are weak forces but their effect depends both on the magnitude of the forces themselves and the distance over which they act. Linear molecules can align and lie parallel to each other, giving small distances of separation. This makes attraction produced by the same forces/moments greater.

Straight chain compounds have large size and hence have large polarizability and have strong  London dispersion forces hence high boiling points while branched compounds have compact structure and hence have low polarizability and have low boiling points.

Could you elaborate on this? Currently your answers adds nothing of value to the question.

Straight chain compounds have higher boiling point then branched chain because in straight chain, molecules are strongly entangled with each other (like noodles)and have more contact with other molecules so strong force is required to remove such molecules consequently straight chain compounds have higher boiling point than branched compounds


text:
Crystal Field Explanation of Colors of Coordination Compounds
Crystal Field Theory explains colors of Coordination compounds as follows : A d-orbital splits into multiple orbitals, the process being called crystal field splitting. When white light falls on the compound, an electron makes a transition into a higher state thus absorbing a particular wavelength of light. The rest of the light is reflected. This reflected light is responsible for the color of the compound.
Problem: Suppose, all electrons have now transit into a higher state. If more light falls on the compound, there should be no more transitions as all the electrons are already in the highest state. This means the compounds should lose color after some time in the light.
Question: How is the light absorbed again and again? If we suppose that the electron falls back to lower state, that would emit the same wavelength it absorbed. Even in this case, how do we explain color?
P.S. Also, if they can absorb only a particular wavelength(given quantisation), that should not bring about a color change. There must be a range of wavelengths(however narrow) being absorbed. What is the reason for that?

The electrons go back and forth between ground and excited states http://en.wikipedia.org/wiki/Einstein_coefficient

Your assumption that the electron emits a photon when transitioning back to the ground state is not entirely correct.
The excited electron can also transition back by transferring the excitation energy to heat energy. The compound then warms up ever so slightly and the electron is buzzing away happily in the lower $d$ orbitals.
As to your P.S.-ed question, I'll quote Wikipedia:

The main factors that cause broadening of the spectral line into an absorption band of a molecular solid are the distributions of vibrational and rotational energies of the molecules in the sample (and also those of their excited states). [...] In gas phase spectroscopy, the fine structure afforded by these factors can be discerned, but in solution-state spectroscopy, the differences in molecular micro environments further broaden the structure to give smooth bands.


text:
Why are NaCl and NaBr not soluble, whereas NaI is soluble in acetone (in the Finkelstein reaction)?
My book states that:

Finkelstein reaction is particularly useful for preparing iodoalkanes. The iodoalkanes are obtained by heating chloro or bromoalkanes with a concentrated solution of sodium iodide in dry acetone.
  $$\ce{R-X~+~NaI->[\ce{acetone,~reflux}]R-I~+NaX}$$
  $(\ce{X=Cl,~Br;~R=alkyl ~group})$
    Sodium chloride and sodium bromide being less soluble in acetone get precipitated from the solution and can be removed by filtration. This also prevents the backward reaction to occur according to Le Chatelier's principle. The reaction gives the best results with primary halides.

My question:
Why are $\ce{NaCl}$ and $\ce{NaBr}$ not soluble, whereas $\ce{NaI}$ is soluble in acetone?
All we can observe is that, sodium iodide has greater molecular weight than sodium chloride or sodium bromide. But, how do this account for solubility of sodium iodide?

This answer by Henry Rzepa is great: http://www.ch.imperial.ac.uk/rzepa/blog/?p=439

The solubility of a salt in a solvent is for a good part determined by the difference of stability in the dissolved state (solvated ions) vs. the solid state (ions in crystal lattice). The different solubility of the sodium halides in acetone can be explained by the crystal structures.
With iodide, the solvate $\ce{NaI \cdot 3(CH3)2CO}$ can be formed, in which each $\ce{Na+}$ is coordinated by 6 acetone ligands via oxygen lone pairs, and the $\ce{I-}$ ions fill in the gaps between these octahedral units, being surrounded by the methyl groups of the acetones. $\ce{I-}$ has a larger ionic radius compared to $\ce{Cl-}$ and $\ce{Br-}$ and thus fills the gap completely, thereby minimizing empty space in the lattice and its distance to $\ce{Na+}$, which is, however, still larger than in solvent-free sodium iodide. This larger distance between the ions decreases the lattice energy, which in turn makes the compound readily soluble in acetone.
$\ce{Cl-}$ and $\ce{Br-}$ are too small to fit in these gaps, and as the cation-anion distance becomes too large, the only way to minimize the lattice energy is the formation of a solvent-free structure $\ce{NaX}$ (X = $\ce{Cl-}$, $\ce{Br-}$). When the lattice energy is larger than the energy released by solvation of the ions, namely that of $\ce{Na+}$ by complexation with acetone, the compound becomes insoluble and precipitates from the solution (source).

This is in accordance with Fajan's rule of polarization. It states that covalent character of an ionic compund increases with increase in size of the anion. Here, Iodine anion is larger in size(ionic radius) than chlorine or bromine anion. This makes NaI attain some covalent properties. Therefore, it easily dissolves with dry acetone( which is an organic solvent and covalent character allows compunds to be soluble in Organic solvents).


text:
Can we prepare alkyl Fluoride by Finkelstein reaction?
My book states that:

Finkelstein reaction is particularly useful for preparing iodoalkanes. The iodoalkanes are obtained by heating chloro or bromoalkanes with a concentrated solution of sodium iodide in dry acetone.
  $$\ce{R-X~+~NaI->[\ce{acetone,~reflux}]R-I~+NaX}$$
  $(\ce{X=Cl,~Br;~R=alkyl ~group})$
  Sodium chloride and sodium bromide being less soluble in acetone get precipitated from the solution and can be removed by filtration. This also prevents the backward reaction to occur according to Le Chartelier's principle. The reaction gives the best results with primary halides.

My question:
Can we prepare alkyl fluoride (not aryl fluoride) by Finkelstein reaction?
My book mentions about another reaction called Swarts reaction, through which alkyl fluoride is prepared. In Swarts reaction, alkyl fluorides are prepared by heating alkyl chlorides or bromides with salts such as mercurous fluoride ($\ce{Hg2F2}$), silver fluoride ($\ce{AgF}$), cobalt fluoride ($\ce{CoF3}$) or antimony trifluoride ($\ce{SbF3}$). Why can't we prepare alkyl fluoride by treating sodium fluoride with alkyl halide (Finkelstein reaction)?

Fluoride ions are not good at $\mathrm{S}_{\mathrm{N}}2$ reactions since they are bad nucleophiles.

In the future I recommend consulting March's Advanced Organic Chemistry before asking such questions. It is a priceless reference work for organic synthesis methods. The following text is result of brief consultation with the book.
The equilibrium shift in Finkelstein reaction is governed by relative nucleofility of halogenide-ions. In general, in polar solvents iodine is the strongest nucleophile, because it is poorly solvated, while in nonpolar solvents chlorine is the strongest solvent as it is smaller.
Fluoride in non-polar solvents should be the strongest nucleophile of all halogenide-ions, but it so strongly solvates in protonic solvents that it almost do not migrate in organic phase even if phase-transfer catalists are present. Strong basicity of fluoride does not help either as beta-eliminations becomes significant side-reaction.
This article claims to develop practically meaningful procedure of clean fluoro-de-halogenations.

As far as the Swarts reaction is concerned: There are indeed reports using $\ce{SbF3}$ alone.
Most of the time, however, the active component in the Swarts reaction for the fluorination through halogene exchange is not an $\ce{Sb(III)}$ fluoride, but a mixed $\ce{Sb(V)}$ halide, such as $\ce{SbF3Cl2}$. This may be prepared by reaction of $\ce{SbF3}$ with chlorine (either previously or in situ), or with $\ce{SbCl5}$ (typically in situ).
Anyway, I'd give the reaction of an alkyl iodide with sodium fluoride in dry DMF a try. Alternatively, I'd start from a tosylate.

Extending Finkelstein reaction's starting materials to pseudohalides reportedly works


text:
What does the Sm5Ge4 orthorhombic structure look like?
I see tons of papers referencing this structure but none show what it looks like.

You can look at the $\text{Sm}_5 \text{Ge}_4$ crystal structure at Materials Project here.

In Crystal structure of Sm5Ge4, the structure is described as orthorhombic with $a = 7.75\,\mathring{A}$, $b = 14.94\,\mathring{A}$ and  $c = 7.84\,\mathring{A}$, having four $\ce{Sm5Ge4}$ units per unit cell. The authors further denote:

The structure is conveniently described in terms of three layers which occur along the $b$ axis. Layer $G$ contains $\ce{Ge}$ atoms only; layer $S$ $\ce{Sm}$ atoms only Layer $C$ is a combination of $G$ and $S$, but with the roles of $\ce{Ge}$ and $\ce{Sm}$ interchanged. Layer $C$ serves as a mirror plane in $Pnma$, but only as a quasi-mirror plane in $Pn2_1a$. [...]

(Personally, I find it a bit irritating that this article from 1967, which seemingly is the original reference, is not directly cited on the Materials Project page.)


text:
Why does a wood fire create benzene?
According to this article called
The Fireplace Delusion
a wood-fire will create benzene. This surprised me and I can't find an article online that explains the chemical reaction behind that.

I don't think there is a specific mechanism that produces benzene. Plants naturally have a small amount of aromatic compounds (as we do), and it's very plausible that under high heat the compounds break down into smaller, especially stable compounds, of which benzene is one due to its aromaticity. Furthermore, organic matter combustion is an incredibly complex phenomenon, and its exact products vary with flame temperature, exposure to oxygen, humidity, starting composition of the fuel, etc. It is entirely possible for a poorly oxygenated fire to produce small amounts of hydrocarbons.

Benzene may be produced as an intermediate, but benzene is quite flammable, so it seems unlikely that it lasts very long or escapes the chimney.

@BenNorris Wikipedia states that benzene has a flash point of about -11°C, indeed quite low, but an autoignition temperature of almost 500°C, which is surprisingly high. Though it's very variable, it would seem that fireplaces can generate up to a few tenths of a gram of benzene per hour. Even a charcoal fire can apparently sustain a concentration of around 10 ppm directly over the fuel, so it must be being produced faster than it burns.

Burning organic materials create a whole host of nasties. Dioxins, for example.

Dioxins carcinogenity is not well established.

This is actually just one part of a much more general phenomena, which is the release of large, polycyclic aromatic compounds due to pyrolysis or combustion of organic matter. It is not specific to wood. So, to understand why these compounds form in a wood fire, or any other kind of fire, we must first understand what wood is, and then we must understand how wood burns.
Wood is biomass - that is, matter composed primarily of carbons, hydrogens, and oxygens. We can generalize here, and talk about any biomass - wood, dead or dry grass, leaves, animal carcasses, fibers, sugar, fat, and even things like coal (which is fossilized biomass with special properties). All of these things contain cellulose or similar molecules, which consists of long chains of aromatic rings (the wikipedia article on cellulose is a good place to start, and any textbook on biochemistry will give much more detail). In the context of combustion, this is called "volatile organic matter." Volatile just means that if you heat it up, it burns, gives off heat, and turns into gas.
Living things also have non-volatile and non-organic matter. As an example, people require some trace amounts of metals for biochemical functions, so humans contain some percentage of mineral matter. In the context of combustion, this material is often called "ash." If you have a fuel like wood, this ash is what's left behind after the wood burns, and composes much of what's released into the air in the form of smoke. Some fuels can have large percentages of ash (as an example, when organic matter fossilizes and forms coal, a large amount of rock or dirt interspersed with the organic material can lead to coals with very high ash content).
When you heat up organic matter, what you're doing is dumping a large amount of thermal energy into the molecules composing that matter, and after a certain point (depends on the fuel, but we're talking about 400 or 500 C) the large chains that compose organic matter begin to break apart. If this happens in the presence of oxygen, the material will also begin to oxidize, and the process is called combustion. If it happens in the absence of oxygen, the material will gasify. As the temperature gets hotter and hotter, a soup of big organic macromolecules is formed. (When I say big, I mean, BIG - hundreds of carbon, hydrogen, and oxygen atoms, and thousands of side chains and functional groups). It is in this macromolecular soup that you can find benzene and its many, many, many cousins. (These molecules are too complex to have names, and are so large that they can't be characterized exactly - instead, chemists can only statistically describe the molecules, using techniques like nuclear magnetic resonance (NMR), mass spectrometers (MBMS), or infrared spectrometers (FTIR).) This process of solid matter turning into volatile gases is called pyrolysis.
Big polycyclic molecules form when there is not enough oxygen (if there were enough oxygen, all the carbon and hydrogen would form carbon dioxide and water). They are huge and hydrophobic, and agglomerate to form soot and tar. When you see sticky brown films on cigarettes or smokers' hands or in pipes, this is caused by these macromolecules.
If you looked very closely at a wood fire, and were able to see the actual flame-wood interface, what you would see is jets of hot, flammable gases shooting out of the pores of the wood. It is these gases that ignite - not the wood itself. (Of course, as the wood heats up, and the organic matter is burned, the mineral matter is left behind, and that becomes extremely hot but does not vaporize, so you still have radiant, red-hot material.)
You can still find benzene, even in a hot fire, because flames are not perfectly mixed, and you can get pockets of cold gas interspersed in a cloud of hot gas. If it is cold enough, benzene (which can be very stable) will not be decomposed.
So the next time you watch a campfire, or smoke a cigarette, or set your dinner on fire in the oven, you should be able to better understand what you're seeing - or know where to look.

Wow. Thanks. So they're bad for our health, then?

@noelicus Yes, fire from combustion of organic matter is technically bad for our health if we are chronically exposed to it, but that doesn't stop anyone from having barbecues and heating homes. To quantify how bad it is (as in how much lifespan/quality of life is lost) would be a very difficult task. If you must be completely wary of it, then fires from the combustion of propane/butane are significantly less toxic, and even better would be heat generated from electrical heaters or heat pumps.

Yes! But things that are bad for our health can be bad for different reasons/on different scales. Big polycyclic molecules can oxidize or reduce things- like proteins or DNA (and DNA damage/mutations can lead to cancer). Chemicals like NOx and SOx can form acid when mixed with water- like the water in your lungs. Larger particulate matter, even inert mineral matter, can also be hazardous. When ash gets hot, it turns into molten metal. Tar can coat your lungs and interfere with oxygen exchange. Byproducts of any combustion process basically trash your health...

@NicolauSakerNeto combustion of light olefins like propane and butane aren't necessarily safer/better, because they can form all of these compounds in the absence of oxygen (that's why lighters have yellow flames - the big soot molecules heat up and glow). They can also have crap like sulfur. If your flame has excess oxygen (lean), its very hot and blue and will have less crap in it... But there's still lots of invisible crap.


text:
Deviation from Raoult's law because of adhesive and cohesive forces
A solution made of two or more compounds have adhesive and cohesive forces. Cohesive forces are the attractive forces between the same molecules. Adhesive forces are the attractive forces between different molecules of the solution.
A real solution shows positive deviation from Raoult's law when the cohesive forces are stronger then adhesive forces. It shows negative deviation when the adhesive forces are stronger than cohesive forces.
Why do stronger cohesive forces in real solution make positive deviation and stronger adhesive forces make negative deviation from an ideal solution?

Foreword
This topic is treated in most books and sites so I've tried to explain it to you in a more unusual and friendly way to grasp the concept but maybe the canonicals explanations are more suited for exams and serious conversation.
First step idealize and simplify the Raoult law
You can see Raoult law as the most basic attempt to predict the partial vapour pressure of a multi-components solution. If $A$ is your first component with higher vapour pressure an $B$ is the second component with lower vapour pressure, it seems quite obvious that if you mix an equal amount of $A$ and $B$ you have a vapour pressure equal to the mean of the two.
Back to reality
In fact, the reality is more complex and you have to take into account the force between the molecules of the two components. If you have strong adhesive forces the molecules of the first component near the molecules of the second keep themself more stuck together. So this creates a sort of "net" that doesn't allow the molecules to "escape" from the solution and so the vapour pressure (that in fact quantify the ratio of molecules that "escape" from the bulk) is lower and finally  you have a negative deviation.
But what if the interaction between the molecules of the same component is greater than the interaction between the two components? As you state, this is the case of strong cohesive forces. In this case you have to imagine that the components act as if they were separate.  Molecules of the component A lower the positive interaction between the molecules of the component B interposing them self and avoiding the formation of bonds between molecules of component B and finally increasing the vapour pressure, a positive deviation!
Deep into fantasy
So if you think the molecules as humans exiting from an elevator, humans A are more claustrophobic and so tend to exit more rapidly than humans B when the doors open. If they interact positively with humans B they will keep calm when exiting. If humans B don't find funny humans A they preserve their panic attitude and at the same time separate B humans from each other and rushing out from the elevator pull humans B out more rapidly too!

Nice explanation!

great explanation...

@G M lovely explanation

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


text:
Do chaotropic and kosmotropic affects occur in polar non-aqueous media?
For example can I introduce salts that have a chaotropic or kosmotropic affect on ethanol? Or is this phenomenon restricted to water? If this phenomenon does occur in other polar solvents, does it follow the Hofmeister series as well? Are there any papers out there on this phenomenon in polar non-aqueous solvents?

As far as I understand, chaotropism and kosmotropism are merely fancy words to describe whether a solute has $\Delta S_{solvation}>0$ or $\Delta S_{solvation}<0$, respectively, so I'm pretty sure it extends to any solvent. Surely each solvent has their own series of ordering/disordering solutes, given how different the intermolecular interactions can be.

Chaotropic and kosmotropic effects of solutes can occur in any solvent in which interactions between solvent molecules lead to a more ordered structure, like a hydrogen bonding network in water. Such a network will also be present in alcohols like ethanol, although to a lesser degree, because the alcohol molecules have only one proton which can act as a hydrogen bond donor. The Hofmeister (or lyotropic) series for alcohols might still be similar to that for water (not identical), but it will be more different with increasing structural difference between the solvent molecule and water. In fact, the Hofmeister series for water is still not fully understood on a molecular basis, as it is described in this post, and this makes it difficult to predict Hofmeister series for other polar, non-aqueous solvents.


text:
How does shock trigger nucleation of gases dissolved in a liquid?
When you drop a bottle of soda, the dissolved $\ce{CO_2}$ starts nucleating. Opening it before it's 'settled' will trigger more nucleation and usually makes a mess.
Why does the shock trigger nucleation?
Secondly, is it the same mechanism in play when you shake a bottle? And what about the physical state changes as it 'settles'? i.e., why does the $\ce{CO_2}$ rapidly nucleate if you open the bottle right after you drop it, but it doesn't if you wait a bit?

I don't know the full answer, but I do know that tapping (or dropping) a beer bottle or soda bottle can cause cavitation (http://www.youtube.com/watch?v=OeunRrfvyAU) which creates nucleation sites for the $\ce{CO2}$. This will not happen when you shake the bottle, so there has to be an additional mechanism, but I don't know what

Additionally try this: https://www.google.nl/search?q=what+makes+soda+fizz+when+you+shake+it

@Michiel When you shake the bottle or can, you mix the air inside with the liquid, creating a multitude of tiny air bubbles which will quickly nucleate the supersaturated $\ce{CO2}$ out of solution. No matter how hard you shake a completely filled bottle, nothing happens (though be careful about leaving a tiny bit of air inside and having it nucleate gas while the bottle is practically full, that creates a lot of pressure). Some more information can be found here.

How is nucleation triggered?
There are, as I see it, three major pathways for nucleation of supercritical carbon dioxide in water.
Bottles: Cavitation
As @Michiel points out correctly in the comments, tapping a glass bottle may cause cavitation, which is then the initial nucleation site for the (runaway) bubble formation. However, I think it is noteworthy that not every type of tapping leads to this; It is when you tap on top of an open bottle with a hard object (such as another bottle) that the gas formation is strongest. This is because the shock waves propagating through the glass walls of the bottle meet up at the bottom of the bottle for a huge impulse into the liquid, forming a larger cavity than other methods (don't do this at parties, people will hate you and their beer will be stale). The relative difference of impulse can be immediately grasped from the fact that the tapping bottle will not start foaming, while the tapped bottle most definitely will.
Cans: Deformation
There are however alternative means of containing beverages, for example aluminium cans. Due to the form of the cans, cavitation formation is highly unlikely. More likely, deformation of the can upon dropping will provide the nucleation sites for the carbon dioxide to form bubbles; think jagged edges on the inside, reactive defect sites in the lattice, and a possibly broken oxide layer of the aluminium, exposing highly reactive elemental aluminium for a short time.
Shaking: Air Dissolution
As @Nikolau correctly pointed out in the comments, shaking a bottle will lead to small bubbles of air getting trapped in the liquid and providing nucleation sites for the gas. Nothing magical about that, but I gave it a heading anyway.

Onwards to the second question you had:
Waiting: Why no foam?
The reason that waiting leads to less foam and spewing around of your favourite carbonated beverage is twofold.

As explained in the linked question on Physics.SE, the bubbles will rise to the top of the liquid level. As such, releasing the pressure will cause further nucleation primarily at the surface and not within the liquid, leading to less liquid propagation by gas evolution. In plain English: This means you don't get soaked by the beverage as it is propelled out of the opening, because the gas evolution happens at the surface and "forces" the liquid to stay in place.

The pressure inside the container rises as more gas is formed. If you wait, a new equilibrium state will be reached by redissolving some of the carbon dioxide that nucleated out of the solution, subsequently lowering the pressure again. (I don't think that the pressure goes back to the initial level within a finite amount of time, but I may be wrong here. If someone could clarify this I would be grateful.)
As there are now less nucleation sites, combined with the fact that they rise to the top (see point 1) foam production is severely hindered.


text:
How do I calculate the normality of NaBrO3?

Find the mass of $\ce{NaBrO3}$ required to prepare $\pu{150 ml}$ of $\pu{0.75 N}$ solution based on the reaction
  $$\ce{BrO3- + 6 H+ + 6e- -> Br- + 3 H2O}$$

My analysis:
$\pu{150 ml}$ solution of $\pu{0.75 N}$ means $0.1125$ equivalents of $\ce{NaBrO3}$ are present in the solution. Since, according to the given equation, n-factor (of equivalence) of $\ce{NaBrO3}$ is $6$. So $0.1125$ equivalents means $0.01875$ moles of $\ce{NaBrO3}$. The molecular mass of $\ce{NaBrO3}$ is $151$. Thus, I get the mass as $\pu{2.83 g}$ approximately. Am I correct?

Since the book says the answer should be $\pu{1.42 g}$

To me your answer is correct. It seems that the book result have a '2' dividing the final result. Could you post all the text of the book exercise? Two possibilities, the result from the book is wrong (could be) or there is a mistake in one of the values (75 ml instead 150, 0.375 N instead 0.75 N...).

The lesson to learn here is, of course: don't use normality. It's based on gram-equvialents, an old ambiguous concept from the 1800s when we still figuring out substance formulae. It's about time to bury it.

@NicolauSakerNeto yes I think is quite useless...

The question is exactly as given here. So, probably, the result given in the book is wrong.

@TejasAdsul I've try to solve the problem I have your same result...I think Rauru is right


text:
Magnetic moment of coordination complexes?
To understand the commonly quoted magnetic values of coordination complexes (central ion) we use
$$m_l=\sqrt{n(n+2)} \text{BM where BM}=\frac{e\hbar}{2m_e}\text{JT}^{-1}$$
$n$=number of unpaired electrons.
How did we derive this equation?
I know the orbital angular momentum of electron is given by $\sqrt{l(l+1)}\hbar$, spin angular momentum is given by $\sqrt{s(s+1)}\hbar$ and the gyromagnetic ratio for electron is $$\vec \mu_B=\frac{-e}{2m_e}\vec L$$. There is also the "total" angular momentum given by $n_0\hbar$.
I fail to see how these magnetic moments and angular momentums are
 related, and how they can be added or related to get the corresponding
 values for an entire atom or a complex?.
I have also heard that the Lande g-factor is relevant here.
Key:-
$n_0$=principal quantum number, $l$=azimuthal quantum number,$s$=spin (half for an electron), $\hbar=\frac{h}{2\pi}$, $S=\Sigma s+1$(spin multiplicity), $\vec L$=angular momentum, $\vec \mu_B$=Magnetic moment.

$n/2=s$, $\sqrt{ (n/2)(n/2+1)} = 1/2 \sqrt{ n(n+2)}$...


text:
Can a battery powered by iron and air really provide a feasible power source for automobiles?
I'm referring to the Shuttle Battery recently announced (read more on it here) that suggests that:

The energy density of the device is up to 7,700 Wh/L. It is more than 5 times
  higher than state-of-the-art (SOA) Li-ion (550Wh/L), and comparable to fossil
  fuel (10,000 Wh/L).

It goes on to say that:

It is rechargeable because it is based on a simple reversible chemical reaction, 3Fe + 4H2O = 4H2 + Fe3O4 and 2H2 + O2 = 2H2O. The proof of concept cell demonstrated zero capacity loss after 200 cycles of discharge and charge.

Now the question is: Is this even remotely possible? Or is there a red flag here somewhere that I can't seem to pick up?

Yes, technically $\frac{7700}{550} = 14$ is greater than 5, but I wouldn't normally describe 14 times as "more than 5 times".

@user1915639 I didn't notice that! :)

This is certainly possible.  In fact, there are a large number of metal-air batteries being developed all over the world.
It is, however, strange that such high energy density is achieved, because iron ion itself would not offer very high energy density.  My guess is that likely they did it by having an extremely high matter density.  So this battery may be small in size for its capacity, but at the same time it may also be much heavier than a conventional battery.
That could be why they brand it as a shuttle power supply(well, energy density isn't really relevant for a bus anyways).  A very heavy battery on a hand held device is never good.   Cost is the more important factor there, but it is hard to tell at the moment.  0 loss over 200 cycles seems like a fairy tale, but I will give them the benefit of the doubt at the moment.
This battery will also likely to have a considerable loss of charge when not in use.   Again, this might not be a huge problem for buses, where you keep driving non-stop during days then kept it in the charging station overnight.
Anyways, don't expect this to be on your iPhone 10 or your laptop.  Even if it works as presented, it might not be that ideal for portable electronic devices, where people worry much more about weight and stand-by time than if it occasionally explodes, sadly.

Makes sense - i was thinking about its feasibility for electric vehicles like Tesla cars, for example, where the battery pack is quite often the heaviest component - and mileage isn't great. Thank you for the answer

Current Li-ion cells have $\ce{LiCoO4}$ or $\ce{LiFePO4}$ cathodes where the $\ce{Co}$ or $\ce{Fe}$ are only oxidised or reduced by 1 electron per atom, so if they have managed to make a cell with iron giving up $\frac{8}{3}$ electrons per atom they will get a higher capacity. I find it hard to believe they have got this reaction to happen reversibly and at high current density (i.e. high power/weight ratio) at room temperature, though - if you put iron in water at room temperature it doesn't react very quickly, and hydrogen-oxygen fuel cells also can't get high power densities at moderate temperatures.

Good point. I noticed they used the term hermetical sealing, which I would guess means none of the components escape into the air.


text:
Imiscible fluids for demonstration
I have been considering building a fluid dynamics demonstration.  Basically, it would be a thin plexiglass box, with some turbulence inducing elements attached to the back face, and a pump to move the fluid around.
Can anybody recommend a pair of fluids that will:

not mix/dissolve permanently
can be dyed different colors, while respecting (1)
won't react with plexiglas (or, alternatively, polycarbonate)
isn't toxic or particularly flammable
can be pumped with a "very" low power pump at temperatures suitable for humans (say, 60-100F)

The point of the demonstration is to create a "sustainable" chaotic system.  I might even point an HD camera at it, and extract random numbers from the video stream (maybe).

Have a look at this paper
Summarizing:

put food coloring to 35% isopropyl alcohol in water.
add ammonium sulfate

What happens ? Once ammonium sulfate added and the solution mixed, the ammonium sulfate dissolves, pushing the isopropyl alcohol out of the aqueous solution (and the food coloring as well).
A separate layer thus forms, that is clearly visible thanks to the food coloring.
The weak point of the whole story is that isopropyl alcohol is flammable...
PS: I would have preferred to reply to @Michiel's answer and @Nicolau Saker Neto's comment: but I do not have enough reputation to do that...

You could use water and some low viscosity oil like olive oil or sunflower oil. Water can be dyed easily with food-coloring that you can buy at the local pharmacy or sometimes grocery store. Both substances are eatable and fairly inert (on the timescale of days at least).
Only a very small part of the oil will dissolve in the water and after saturation there is no problem any more. After thorough stirring the demixing process might be a bit slow since both oils have a density around 920 $\ce{kg/m^3}$ which is not that far from water, but you will have to try whether you can for example run the experiment by day and let the whole thing demix at night. If it doesn't properly demix you could consider going for a lower density oil.

Another way to possibly force quicker demixing is to increase the disparity between solvent polarities, which can be done easily enough by dissolving a highly polar substance in water, such as a salt. Sodium chloride is possibly the best idea, though it is slightly corrosive towards some metals. Sugar might work also, though it's less polar and would increase the viscosity of the water