text:
Difference of dipole moments of dichloromethane and trichloromethane?
Why is the dipole moment of $\ce{CH2Cl2}$ ($1.60 ~\mathrm{D}$) greater than that of $\ce{CHCl3}$ ($1.08~\mathrm{D}$)?
Based on my knowledge of vectors, I feel it should be the other way around.

Just to add some quantification to Ben Norris's answer. Consider each $\ce{C-Cl}$ bond, which has a bond dipole moment of magnitude $A$. The contribution from $\ce{C-H}$ is neglected here to simplify the calculations.
Now consider dichloromethane. The resultant of the two $\ce{C-Cl}$ dipoles will have a magnitude of $$2A\cos\frac\theta2$$
where $\theta\approx109.5^\circ$, the angle between the vectors from the ends of tetrahedron to the centre. This turns out to be $1.154A$.
For tricholoromethane, the three $\ce{C-Cl}$ dipoles will be at an angle $\theta$ such that $\cos\theta=1/3$ with a straight line symmetrically passing through its centre. Thus the resultant in this case will be $$3A\cos\theta$$
which will be $1A$, lesser than in the case of dichloromethane.
To visualise how can you get those angles and cosines, see this page.

Since you know about vectors, then the piece you are missing is probably the proper three-dimensional structure of the molecules.

Trichloromethane (chloroform)

Notice that the third chlorine atom in trichloromethane is pointing away from the other two. More precisely, the three chlorine atoms are at the bottom three points of a tetrahedron. Thus the $x$ and $y$ components of the individual $\ce{C-Cl}$ bond dipoles cancel, leaving only the $z$ component.

Dichloromethane

For dichloromethane, the $y$ components fully cancel, but the $x$ components do not.

But how do you know that the $x$ components in Dicholoromethane sum up to be greater than the $z$ component in Chloroform?


text:
How many grams of each substance will be present at equilibrium in the reaction of ethanol and ethanoic acid?

In the following reaction, $K_c=4.0$:
  $$\ce{C2H5OH + CH3COOH <=> CH3COOC2H5 + H2O}$$
A reaction is allowed to occur in a mixture of $\pu{17.2 g}$ $\ce{C2H5OH}$, $\pu{23.8 g}$ $\ce{CH3COOH}$, $\pu{48.6 g}$ $\ce{CH3COOC2H5}$, and $\pu{71.2 g}$ $\ce{H2O}$
(a) In what direction will a net change occur?
(b) How many grams of each substance will be present at equilibrium?

This is what I did for part (a):
$$Q_c = \frac{[\ce{CH3COOC2H5}][\ce{H2O}]}{[\ce{C2H5OH}][\ce{CH3COOH}]}$$
First, get the concentration of $\ce{C2H5OH}$, $\ce{CH3COOH}$, $\ce{CH3COOC2H5}$, and $\ce{H2O}$ using the given amount of each substance.
$$\begin{align}
[\ce{C2H5OH}]
  &= \pu{17.2 g} \times \frac{\pu{1 mol}}{\pu{46.1 g}} &
  &= \pu{0.373 mol} \\
[\ce{CH3COOH}]
  &= \pu{23.8 g} \times \frac{\pu{1mol}}{\pu{60.1 g}} &
  &= \pu{0.396mol} \\
[\ce{CH3COOC2H5}]
  &= \pu{48.6 g} \times \frac{\pu{1 mol}}{\pu{88.1 g}} &
  &= \pu{0.552mol} \\
[\ce{H2O}]
  &= \pu{71.2 g} \times \frac{\pu{1 mol}}{\pu{18.0 g}} &
  &= \pu{3.96 mol} \\
\end{align}$$
Assuming the overall volume is $\pu{1 L}$, then
$$Q_c = \frac{(\pu{0.552 M})(\pu{3.96 M})}{(\pu{0.373 M})(\pu{0.396 M})}
      = 14.8 $$
Since $K_c$ is $4.0$, and $Q_c$ is $14.84$, $Q_c$ is larger than $K_c$ and the direction of the reaction is to the left, towards the reactants.
I know how to solve for (a) but I have no idea how to get the amount of each substance at equilibrium for (b). The answer for (b) should be $\pu{68 g}$. I tried doing the ICE chart, but I did not get $\ce{68 g}$. Do I use the ICE chart for this part of the question or which method should I be using?

You say raction evolves to the left. Calculate by which amount concentrations vary until equilibrium is established.

Do not understand why for each substance the mass should have the same value (68 g)...

@mannaia Yep, it's only for water. The question should have asked specifically how many grams are produced when the ester is produced from the dehydration reaction.

I'll not give the correct answer yet, but the sum of grams of each substance in the final equations is different from the initial, attention to the mass conservation!!!
At 25C and 1 ATM, the correct Kc for this reaction is 4.634, so what is the correct answer?

First of all welcome to Chemistry SE. You're right there are mistakes in above answer, another member has further pointed it out. Second, no mention is given in the original question that working conditions are standard. So $K_{c}$ can have other values than the standard one.

@mannaia Yep, I assumed this was occurring at $25^\circ\text{C}$ and 1 ATM. I resolved the calculation errors. Alternatively, we can use the van't Hoff isochore equation if the temperature is changed.

Knowing that $K_c=4.0$, equilibrium is achieved when $Q_c = K_c$.
Let's set up an ICE table and write down our equilibrium expression like you did:
Reactants:
$$\ce{C2H5OH + CH3COOH <=> CH3COOC2H5 + H2O}$$
\begin{array} {|c|c|c|c|c|}
\hline
\text{Initial conc.} & \pu{0.373 mol} & \pu{0.396 mol} & \pu{0.552 mol} & \pu{3.96 mol}\\
\hline
\text{Change conc.} & -x & -x & +x & +x\\
\hline
\text{End conc.} & \pu{0.373 mol} - x & \pu{0.396 mol} - x & \pu{0.552 mol} + x & \pu{3.96 mol} + x \\
\hline
\end{array}
Assuming that our reaction is at equilibrium:
$$ K_c = \ce{\frac{[CH3COOC2H5][H2O]}{[C2H5OH][CH3COOH]}}$$
We plug in our values for our change in equilibrium:
$$ K_c = 4.0 = {\frac{[0.552 + x][3.96 + x]}{[0.373 - x][0.396 - x]}}$$
Solving for $x$, which is the change in concentration:
\begin{align}
4 &= \frac{x^2+4.512 x+2.18592}{x^2-0.769 x+0.147708}\\
0 &= 3x^2-7.588 x-1.59509
\end{align}
Using the quadratic equation, or thankfully polysolv on modern calculators, we get two values for $x$:
\begin{align}
x_1 &= 2.72449 \\
x_2 &= -0.195154
\end{align}
The negative zero is chosen, since the reaction will evolve to the left (being $Q_c>K_c$), thus increasing the concentration of the reactants. This will only be accounted for, if one takes the negative zero. If one chooses the other zero, one will go to the wrong opposite direction (towards a diminution of reactants concentration). In the particular case of the present exercise, one would even result in a negative concentration. Therefore:
\begin{align}
\ce{[CH3COOH]}
  &= \pu{0.373 mol} - (\pu{-0.195154 mol}) &
  &= 0.568154\\
\ce{[C2H5OH]}
  &= \pu{0.396 mol} - (\pu{-0.195154 mol}) &
  &= 0.591154\\
\ce{[CH3COOC2H5]}
  &= \pu{0.552 mol} + (\pu{-0.195154 mol}) &
  &= 0.356846\\
\ce{[H2O]}
  &= \pu{3.96 mol} + (\pu{-0.195154 mol}) &
  &= 3.764846
\end{align}
Therefore:
\begin{align}
\pu{0.568154 mol} \ \ce{CH3COOH} \times \frac{\pu{60.1 g}}{\pu{1 mol} \ \ce{CH3OOH}} &= \pu{34.1 g}\\
\pu{0.591154 mol} \ \ce{C2H5OH} \times \frac{\pu{46.1 g}}{\pu{1 mol} \ \ce{C2H5OH}} &= \pu{27.3 g}\\
\pu{0.356846 mol} \ \ce{CH3COOC2H5} \times \frac{\pu{88.1 g}}{\pu{1 mol} \ \ce{CH3COOC2H5}} &= \pu{31.4 g}\\
\pu{3.764846 mol} \ \ce{H2O} \times \frac{\pu{18.0 g}}{\pu{1 mol} \ \ce{H2O}} &= \pu{67.8 g}
\end{align}
Initial product sum is $\pu{160.8 g}$.
Equilibrium product sum is $\pu{160.6 g}$. $\pu{0.2 g}$ is likely due to truncation error in calculations.
This problem is really cruel... whoever assigned it... :/

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text:
How to treat Magnesium powder and ribbon
I am a master student of nuclear physics. Due to some measurements I'd like to make, I have to use Magnesium as a beam target. Searching through the net to see it's properties I found that it can be easily oxidized. In addition I've read that it's highly flammable.
I have in hand two forms of Magnesium:powder and ribbon. In order to make the targets I have two alternatives.

Use the powder to make highly pressured pellets. I thought of flushing a bottle with Argon or Nitrogen, making the pellet as fast as possible using this procedure(which involves acetone) in the atmosphere and then after the pellet is formed put it in the Argon or Nitrogen bottle.
Use the ribbon as is. The only problem is that it not wide enough, so I will have to make a spiral of this ribbon around something(I don't yet know what the "something" will be, so any idea is welcome!). The ribbon I have has a width of 3mm but I want a width of 6mm. So I thought of spinning the ribbon around an item, to form a spiral.

The goal is to make the target as safely as possible and avoid Magnesium oxidizing, because I don't want to have Oxygen in my target. So my question has three parts.

How to safely treat magnesium powder/ribbon so as to avoid any possible accident and preventing it from oxidizing?
Which of the above would be the most effective way to make the target?
How to store the target until the experiment will be held?

Note that after the target is made, it will go to an evaporator, in order to evaporate some gold. Any advice, help or direction is more than welcome!

Magnesium powder and ribbon are relatively stable in air at normal conditions, probably because of thin protective layer of oxide. However, ribbon can burn in air with blinding light and powder gives self-burning (i.e. not needing external oxygen sources like air) mixtures with many common substances, some of them explosive. Moreover, burning magnesium may continue to burn in $\ce{CO_2}$ atmosphere, under quartz sand and especially under water (producing hydrogen, that can form explosive mixture). The only common material that can safely extinguish burning magnesium is iron powder. Surprising enough, being stable in cold water for short time, magnesium readily react with methanol (and probably ethanol), chlorinated hydrocarbons and hot water.
The good thing is that magnesium ribbon is not easy to lit up, and powder is almost impossible to lit up when on surface of some massive object (common gas lighter flame lit magnesium ribbon after 15-60 seconds of heating)
Magnesium must not be heated on air or under nitrogen. Period. Technical grade argon probably contain significant amount of nitrogen, so it should not be used either.
Magnesium theoretically can be melted under eutectic mixture of $\ce{NaCl}$-$\ce{KCl}$-$\ce{CaCl_2}$, but by doing so without experienced supervision you ask troubles. Still, theoretically speaking, if you find an iron (not silicate, porcelain or corundum) crucible you probably can melt magnesium in it under mentioned mixture of dry salts, let it crystallize and then carve it out.
However, magnesium alloy sheets are available on the market, so maybe you can find something and cut needed form from commercially available sheets. Try this first before doing anything requiring heating magnesium. I have doubts, that you will be able to press good tablets from magnesium powder, so if commercially available sheets and maybe some magnesium compounds (oxide, sulfide and others) won't not work for you, you probably will end with melting magnesium. I hope you won't need it, it is dangerous.

Thank you very much for your answer! I also want to avoid melting Magnesium, because I don't have equipment to do so, let alone safety. The ribbon I have has a width of 3mm but I want a width of 6mm. So I thought of spinning the ribbon around an item, to form a spiral. Can this be achieved?

Probably it won't keep form. However, magnesium is relatively soft, so if you need relatively small widening, may be cold forging or press on ribbon will work?

I need a widening but I also need the thickness to remain as is.

Let me ask. What elements are tolerable in the target?

That's not trivial to answer! It certainly must not have Oxygen or near elements(in atomic number). It will  also have evaporated gold on it. Which elements do you have in mind?

Ouch. Well, if it will be treated with high temperatures, I'm out of options. Bad news: some oxygen on the surface is almost absolutely inevitable, you will have to grind the surface under pure argon to get rid of it and it will reform on air almost immediately.

I am not sure if I will have to face high temperatures. I would say I will use it on room temperature and in high vacuum($10^{-6};Torr$). Note that the ribbon I have is enclosed in an Argon envelope and it is still shinny.

You probably can evaporate magnesium onto desired surfaces, but making a thick film this way is probably economically inefficient. In case you have shiny magnesium ribbon, you probably can safely wrap it around iron rod, if it suits your need.

That could work, but it would require a lot of time.

Which elements did you had in mind and what for? I would make a simulation to see if these elements will affect the measurement.

basically, you can take magnesium powder and bind it with something, dissolved in proper solvent, with subsequent evaporation of the solvent. For example, polystyrene in benzene distilled over metallic sodium. This way you will add about 0.5-5% in mass of binder, but will be able to make almost any form from it. I used polystyrene in benzene to bind various magnesium-containing pyrotechnic compositions and it worked just find, but well, it is introduction of carbon (atomic number 6, while for oxygen it is 8 - probably too close). Some salts, in diethyl esters should work too.

let us continue this discussion in chat

You can buy magnesium fabrications any way you like.  It does surface air oxidize, but only divided metal is hazardous - powder, shavings, chips - unless you grind or saw on it.  Magnesium burns under air, nitrogen, carbon dioxide, and water; and sand.  Near anything containing oxygen or halogen will feed it.  An argon flood should kill it.  Also look at Metal-X extinguisher ratings.
If surface oxidation is a problem, ion mill it clean in vacuum, then deposit a micron of Parylene. Parylene C may be reactive toward naked magnesium.  Parylene as such should work... but ask the coating vendors or test it.
http://en.wikipedia.org/wiki/Parylene
http://www.paryleneengineering.com/why_use_parylene.html
http://scscoatings.com/what_is_parylene/parylene_properties.aspx
https://www.google.com/patents/US5156919

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text:
Why is the alpha particle called a particle when it is made of four particles?
We know the alpha particle is the nucleus of helium. It contains four subatomic particles - two protons and two neutrons. The protons and neutrons are further made of particles called up and down quarks. So why call an alpha particle a particle. And similarly, why do we say electrons, protons, neutrons etc. are particles, even though, they are made of particles?
I can understand the case of electrons, protons, and neutrons as when they were discovered and named, quarks were not discovered. But the helium nucleus is a different case.
I reckon talking about an English word's meaning changes in the frame of reference. We do not call a quark an object, although, an object could be any real or virtual thing. So in chemistry, I don't think calling an alpha particle a particle is a good practice. I do understand there are dust particles and all but we mention them only while talking about the macroscopic world. Not so in the case of chemistry. I hope you get my point.

also known as 'alpha radiation', just the terminology  that stuck I think - will be interesting to see an answer to this

@Amaterasu I am also eager to get an answer to this question. Yes, alpha radiation does relate somehow (I don't know exactly how..). If you think this is interesting, please up-vote the question and any subsequent interesting answer.

I think electrons are elementary particles. It still isn't known whether they actually have some volume or not.

@AwalGarg done and done and done - always a good thing to learn something new

@Amaterasu good thing. The answers below are really intriguing.

A helium nucleus is considered a single particle.

@AwalGarg wouldn't you mind to adjust the title of your question ? A pair of four particles yields 8 particles....

@mannaia thnx for pointing that out

Actually, the alpha particle contains two protons and two neutrons. It is what is emitted from the nucleus under alpha decay, giving off what was first classified as alpha radiation by Rutherford at the end of the 19th century (very late 1800's).
In 1932 the neutron was discovered. Only then did it become clear that the nuclei of atoms contain more than just some positively charged particles (which we know as protons).
So there is this historic reason for the naming of the alpha particle.
"Particle"
The word particle does not only pertain to elementary particles (and in [quantum] chemistry, we stop at the level of nuclei as particles, everything that goes deeper is nuclear physics). There are smoke particles, dust particles, etc.
There is also the wave-particle duality. In principle, since we can observe this "alpha radiation" from radioactive processes, there must be a particle that belongs to the wave. With electromagnetic radiation, this is the photon, and in alpha radiation it is the alpha particle.

Thnx for pointing out the mistake. Updated the question - my bad. I reckon talking about an English word's meaning changes in the frame of reference. We do not call a quark an object, although, an object could be any real or virtual thing. So in chemistry, I don't think calling an alpha particle a particle is a good practice. I do understand there are dust particles and all but we mention them only while talking about the macroscopic world. Not so in the case of chemistry. I hope you get my point. Still, grateful for your answer.

Well, maybe then this question is more suited for English.SE than Chemistry.SE. As I understand, your question is: "Why do we still use 'alpha particle' in the light of the evidence that it is made up of four elementary particles?"

lol. I don't know if there is someone at English.SE who would know something about this. There may be of-course. But may be they would migrate it back here. Are you at English.SE?

I'm not participating at English.SE, it just struck me that the question you're trying to ask has more to do with the usage of English than actual chemistry.

To expand on the wave-particle duality comment, there are commercial surface analysis instruments that operate by observing the diffraction of helium from a surface. It's about the largest particle that can be reliably observed interfering in a 2-slit type experiment (ignoring things like Bose-Einstein condensation and superfluidity).

@craigim Actually, some rather remarkably large molecules have been shown to exhibit diffraction pattens in double-slit experiments, as expected of a wave. A while ago, the experiment was performed successfully with $\ce{C_{60}}$, and recently with molecules about 50% larger. Last year another type of quantum interference experiment worked with a molecule containing over 800 atoms!

Note that calling something a particle does not imply calling it a fundamental particle. Anything can be a particle (a particle of dust, a particle of sand, a particle of cheese, etc). However, only non-subdivisible particles (particles with no further inner structure) are called fundamental (electrons, quarks, neutrinos, photons, etc).
We used to not know what radiation actually was, so we classified them broadly into three groups according to the way their trajectories bent in an electric or magnetic field, and the three main types found were given an effectively placeholder name (alpha, beta, gamma). The names stuck, even though we now know their real identities.

Thnx for your answer. I am not saying that particle only implies for fundamental particle. But in the microscopic world of chemistry, a particle, in my view, should not be made up of particles. Do we call an atom a particle? No. Because it does not satisfy the condition of being a particle, i.e. its not only divisible, but in-fact, made of further small particles which are complete entities themselves. I hope you get my point. Still, grateful for your answer.

I think it's completely arbitrary whether you want to call them particles or not. We don't usually call things particles in chemistry because in common use it has a very general meaning when used informally (of course physics usually regards particles as specific entities). In chemistry, we don't call atoms particles because we have a more specific way to refer to them (atoms!). We do sometimes use the term particle in chemistry, such as in nanoparticles, which are composed of many atoms themselves.

Hmm, that nano-particles' thing is really interesting.

The alpha particle is more "particle-like" than many other nuclei because it is so tightly bound together, thereby making it much more difficult to break into its constituent parts. The stability of an alpha particle is a result of the number of neutrons and protons in the nucleus. Nuclei that have complete nuclear shells have a larger binding energy per nucleon, giving rise to so-called magic numbers of nucleons (2, 8, 20, 28, 50, 82, 126) that result in particularly tightly bound nuclei. In an alpha particle, the two protons form one complete nuclear shell and the two neutrons form a second complete nuclear shell. This makes the alpha particle doubly magic. The stability and simplicity of the alpha particle makes it energetically favorable to be produced from the fusion of lighter particles and to be emitted in the radioactive decay of heavy nuclei.

Particle denotes a logical partition of components of a system in a manner which retains an accurate model (on some particular scale) of a physical system while greatly simplifying analysis of the system.
Your example of the dust particle is analogous to the alpha. To someone studying dust interactions, they might create a useful abstraction by ignoring the microstructure of the dust particle and just regard the particle as a single unit. In the case of analysis of radiation transport of alpha particles in a shielding application, the internal structure of the particle is useful and important, but only to the extent that it affects the external microscopic properties of the particle, such as its reaction cross sections, charge, and kinetic energy.
If for almost all intents and purposes nothing is lost, and much is gained, by treating some system as a particle, then it is useful to do so.

Particles can be elementary or composite. Composite particles are those made of simpler particles. Elementary particles are not made of anything else and are the building blocks of nature.

Even molecules are often considered particles: List of composite particles

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text:
Why is the melting point of p-dichlorobenzene higher than those of o-dichlorobenzene and m-dichlorobenzene?
I was going through alkyl and aryl halides and noted the following statement in my textbook:

p-dichlorobenzene has higher melting point and solubility than those of o-dichlorobenzene and m-dichlorobenzene.

I thought over it but could not come to a conclusion. Why is the above assertion true?

o-dichlorobenzene has a higher density than p-dichlorobenzene, and a small dipole moment. That it has a lower melting and boiling point really baffles me. I'm looking forward to a good answer to this one.

Generally, the melting point of para-isomer is quite higher than that of ortho- or meta-isomers. This is due to the fact that it has a symmetrical structure and therefore, its molecules can easily pack closely in the crystal lattice. As a result, intermolecular forces of attraction are stronger and therefore, greater energy is required to break its lattice and its melts at higher temperature. This question is most likely to be answered easily.
If you consider solubility. These are insoluble in water but soluble in organic solvents. Generally, para-isomers are highly soluble in organic solvents than ortho-isomers. This is interesting question.
Every compound (more specifically solids) does not dissolve in a given compound. In general, a liquid dissolves in a liquid if the intermolecular interactions are similar. This is in accordance with the rule "like dissolves like," ionic or polar compounds dissolve more readily in polar solvents. Non-polar compounds (covalent or organic) are soluble in non-polar solvents.
If you consider p-dichlorobenzene, it has zero dipole moment and thus is more non-polar than o-dichlorobenzene which has dipole moment of about $\pu{2.54 D}$. So, non-polar p-dichlorobenzene dissolves more readily than o-dichlorobenzene in non-polar (organic) solvents.

What do you mean by "it has a symmetrical structure" because all those have an axe of symetry.. ?

In the Carbon plane, the 1,4 isomer is symmetrical along one additional axis than the 1, 2 and 1, 3 isomers because it is symmetrical across a line that includes the two Chlorine atoms and also symmetrical along  a line that is perpendicular to the first line, and bisecting the molecule.

Oh wow this is interesting.

I had some comments to make, but they're too large to actually fit into the comments, so I'm making another answer. I'll take the opportunity to give a bit of background, which may be going a little too far for you right now, but which may come in useful in the future. I'm sorry all my answers tend to be overly lengthy and dry!
There are many factors that combine to create a substance's observed physical properties. When talking about solubility, melting points, and boiling points, it's essential to analyse the intermolecular forces between molecules.
First let's start with solubility. After defining the concept of polarity for a substance, (based on the strength of the electric dipole present in a molecule) one quickly discovers that, in general, "like dissolves like." This is a very approximate rule of thumb that relies on a qualitative description of intermolecular forces. For a substance $A$ to dissolve in another substance $B$, then it is necessary that intermolecular $A\dotsb A$ forces and $B\dotsb B$ forces be broken in order to create $A\dotsb B$ forces, and this process must be favourable. We say "like dissolves like" because if $A$ and $B$ are similar, then the $A$-$B$ interactions formed  are likely about as strong as $A\dotsb A$ and $B\dotsb B$, and so the molecules of each substance will tend to mingle well. If $A$ and $B$ are very different, then it's less likely that $A\dotsb B$ interactions will predominate over $A\dotsb A$ and $B\dotsb B$, and hence the substances stay apart.
Intramolecular interactions are all about electrostatic interactions at long range. From Coloumb's law of electrostatics, it's clear that there are stronger attractions and repulsions the higher the charges involved. Even in neutral molecules, there can often be parts of the molecules with a slightly concentration of negative charge and another part with a slightly concentration of positive charge. We call these oppositely-charged areas in a same object poles, and when there are two oppositely charged areas connected, they form a dipole. The dipole is characterized by something called the dipole moment, which is a vector (a quantity that has direction as well as magnitude). Depending on how spatially concentrated the excess negative and positive charges are, the dipole is said to vary in strength. Weak dipoles will have little charge separation, while strong dipoles have lots of charge separation. The strength of the dipole is what dictates a substance's polarity, which is a quantification of just how polar it is.
As I mentioned, most intermolecular forces are electrostatic in nature. Smaller charges and dipoles tend to cause weaker intermolecular interactions, and larger charges and dipoles tend to cause stronger intermolecular interactions. If you mix substances $A$ and $B$, both with weak dipoles (low polarity) or both with strong dipoles (high polarity), then $A\dotsb B$ interactions are similar to $A\dotsb A$ and $B\dotsb B$, so they tend to dissolve. If, however, you mix $A$ and $B$ such that $A$ has low polarity and $B$ has high polarity, then the electrostatic forces between $B\dotsb B$ will be stronger than $B\dotsb A$, and there is no drive for $B$ to mix with $A$.
All of this talk is to justify why polarity is something you'd expect to be important in determining whether two substances mix. In your specific case, you're studying mixing water (a high polarity substance) with dichlorobenzenes. It is possible to approximately calculate the dipoles present in molecules by comparing the placement of atoms in the molecule and the atoms' electronegativities (electronegativity quantifies how badly an element wants electrons). For example, carbon and hydrogen atoms have similar electronegativities, so when they bond, they form approximately non-polar bonds. When chlorine and carbon bond, the former has a significantly higher electronegativity, so it will tug on the bond's electrons a bit, such that the chlorine acquires a small negative charge, and the carbon acquires a small positive charge, and we say the bond thus formed is polar.
In dichlorobenzenes (DCBs), this means there are dipole moments in the $\ce{C-Cl}$ bonds. In ortho-DCB, the dipoles moments of each bond are pointing in a similar direction (two equal vectors with a $60^\circ$ angle between them), so their effect adds up and the substance has some polarity. In meta-DCB, the dipoles are kind of facing away from each other (two equal vectors with a $120^\circ$ angle between them), so their effects cancel about a bit and the molecule is only left with a slight polarity. For para-DCB, the vectors are exactly opposite and equal (two equal vectors with a $180^\circ$ angle between them), so they cancel out entirely; this means para-DCB is a non-polar molecule, even though it has some polar bonds.
Since water is a highly polar substance, if we assume that the dichlorobenzenes with higher polarity will be more soluble, then we expect that the solubilities increase in the order p-DCB < m-DCB < o-DCB. Experimental data show that their solubilities in water are approximately $\pu{80 mg/L}$, $\pu{125 mg/L}$, and $\pu{156 mg/L}$ for p-DCB, m-DCB, and o-DCB, respectively. Just what we expected! Note that even o-DCB, the most polar of the three, has a very low solubility in water. This is because the polar regions in the DCB molecules are quite small relative to the non-polar regions.
So we got through solubility. Now with the same background in intermolecular forces, explaining the boiling point behaviour is relatively simple. When a substance is heated close to its boiling point, you are giving the molecules almost enough energy to break the intermolecular $A\dotsb A$ forces. The stronger the $A\dotsb A$ forces, the more energy is needed to overcome them. All other things being equal, $A\dotsb A$ interactions are stronger for high-polarity molecules than low-polarity molecules, as there are bigger charges and bigger electrostatic interaction in high-polarity molecules.
Again, by looking at polarity alone, we expect the boiling point trend to be p-DCB < m-DCB < o-DCB. Experimental data shows that the boiling points are $\pu{174 ^\circ C}$, $\pu{173 ^\circ C}$, and $\pu{180 ^\circ C}$ for p-DCB, m-DCB, and o-DCB, respectively. The expected and experimental trends are not exactly the same. This is likely because the polarity difference between p-DCB and m-DCB is small enough that other factors need to be taken into account.
Explaining melting points is quite a bit harder than boiling points. Several things other than polarity can strongly influence melting temperatures. I have a previous post on this matter with some general information. In the case of the dichlorobenzenes, it seems that molecular symmetry and solid packing is the dominant effect. Chem.SE user Uncle Al has provided a list of compounds showing how strong this factor can be.

I think this reasoning will suffice (not sure though):
p-Dichlorobenzene is more symmetrical than o-and m-isomers. For this reason, it fits more closely than o-and m-isomers in the crystal lattice. Therefore, more energy is required to break the crystal lattice of p-dichlorobenzene. As a result, p-dichlorobenzene has a higher melting point than o-and m-isomers.
Any thoughts?

I believe half right.  You may want to modify answer.  Symmetry does not necessarily provide "closer" packing.  In this case there is not closer packing because p has the lowest density of the three.  Symmetry in this case though allows the molecule to have hydrogen atoms on both sides of the ring that can interact with delocalized electrons on perpendicular rings without the Halides pushing them away.  Having two halides on the same side only leaves hydrogen free to interact with delocalized electrons on one side.

The p-isomer being nore symmetrical fits closely in the crystal lattice and thus has a stronger inter-molecular forces of attraction than tose of o- and  m-isomers.Since during melting or dissolution, the crystal lattice breaks. therefore, a larger amount of energy is needed to melt or dissolve the p-isomer than the corresponding o- and m-isomer. In other words,melting point of the p- isomer is higher and its solubility lower than the corresponding o- and m-isomer.

The p- isomer is more linear than the o- and m- isomer, hence it has a larger surface area for inter-molecular attraction.

Why does it have a larger surface area? The atoms are the same and occupy the same area...

I think it's technically true that the surface area available for interactions is smaller in the ortho isomer (and arguably the meta isomer); the region between the two chlorine atoms is too small, so no molecule or atom can come close to interact with that region. However, I don't think that is the best answer.


text:
Reaction of phenol with Zinc dust
I have seen this reaction many times:

$\text {Phenol}+\text{Zn}\ce {->}\text{Benzene}+\text{ZnO} $

But what is the mechanism?! Tried a lot but couldn't cook up one...
Thanks for any help...

How about a reference. I see folks using it on SE but I have not found it in the primary literature.

The net result is $\ce{C6H5-OH + Zn -> ZnO + C6H5-H}$, but under the reaction conditions (zinc DUST and strong heating; b.p. of phenol is $\pu{182 ^\circ C}$), I suspect the reaction mechanism involves the zinc surface rather than individual $\ce{Zn}$ atoms or $\ce{Zn^2+}$ ions. And the zinc surface will be a complex oxide. I have read that the reaction is not clean and produces several byproducts, typical of high temperature reactions.
The acidity of phenol is not great enough to dissolve zinc (at least, not easily), but it could adsorb on the zinc/zinc oxide surface. First step: probably a phenolation of an oxygen to phenolate and hydroxide:
$$\ce{Zn-Zn-ZnO-Zn-Zn- (surface) -> Zn-Zn-Zn(OH, OC6H5)-Zn-Zn}$$
At high temperature, the $\ce{H}$ on the hydroxyl could reach the phenyl group in a five-membered ring and go off as benzene ($\ce{C6H5-H}$), leaving $\ce{Zn-Zn-ZnO -ZnO-Zn}$.
Magnesium and zinc have some similarities (ionic radius, oxide coat), but magnesium is significantly more reactive. I suspect that magnesium in phenol would react at lower temperatures and give magnesium phenolate and hydrogen: $\ce{Mg + 2 C6H5OH -> Mg(OC6H5)2 + H2}$, no benzene. In this case, the acidity of phenol would be enough to dissolve the magnesium; fine powder (pyrophoric) would not be needed.

@andselisk: I get multiple SE requests to confirm your rejection of a proposed edit (which I do confirm). The message says the matter has been handled, and yet asks for my agreement.  I can't seem to make the matter go away. Am I missing something?


text:
Why aren't there 8 double bonds in octasulfur (S8)?
I know that the $\ce{S8}$ molecule contains 16 pairs of lone electrons. My teacher always says that we must try to reduce the no of lone pairs, so, had there been double bonds between 8 atoms of sulfur then the no of lone pairs would be reduced. Then why aren't there any double bonds?

An $\ce{S}$ atom in a molecule must have 2 bonds to be neutral and in the $\ce{S8}$ ring it needs both of these bonds to continue the chain/ring. If you introduce some double bonded S atoms into the ring each of these will have 3 bonds and will thus be positively charged. So, while reducing the number of free electron pairs you increase the charge of the compound which would be wrong because $\ce{S8}$ isn't charged. You can only use free electron pairs to form double bonds if you have some electron deficient atoms in the neighborhood which can "receive" them.


text:
Difference in the functioning of alkaline and acidified KMnO4
What is the difference in the functioning of alkaline and acidified $\ce{KMnO4}$ as a reagent in organic chemistry?
E.g., in the oxidation of a primary alcohol to carboxylic acid, we use acidified $\ce{KMnO4}$ and not alkaline $\ce{KMnO4}$. Why?

Permanganate has the highest oxidizing power in acidic solution and is reduced down to Mn(II). I think I answered that here

I can't see how that answers my question... What about the synthesis of the carboxylic acid?

@Apurv As you want to oxidize your primary alcohol the reaction proceeds better if you use a stronger oxidizing agent. So doing this under acidic conditions seems logical as $\ce{KMnO4}$ is a stronger oxidizing agent in acid than in base.

Mostly for economical reasons. Potassium permanganate is quite costly, and in acidic solution it is reduced to Mn(II), while in extremely basic solution it is reduced to Mn(VI). And in neutral solutions it is usually reduced to Mn(IV) forming hard to wash away $MnO_2$ deposits.

According to wikipedia: "Normally, these oxidations are performed under strong alkaline conditions using a ca. 1N NaOH or KOH solution, because this promotes a greater oxidation speed and selectivity. In substrates sensitive to strong base, the reaction can be carried out at a lower pH – or even under acidic conditions — at the cost of a greatly decreased reaction velocity. " (see http://en.wikipedia.org/wiki/Oxidation_of_primary_alcohols_to_carboxylic_acids)

@permeakra Your comment is quite close to being an answer :)

@LievenB Two different aspects have to be considered here: 1. Oxidizing power of the permanganate: higher at lower pH and 2. ease of the oxidation of the organic species: easier when deprotonated. Even carboxylates (!) can be oxidized with suitable methods ;)

yeah - sorry about that, didn't thoroughly think of those aspects ;)

Permanganate as oxidizing agent works most efficiently in acidic solution, because it is reduced to the greatest extent in this medium, from oxidation state +VII in $\ce{MnO4-}$ to +II in $\ce{Mn^2+}$.
$$\ce{8H+ + MnO4- + 5e- ->~ Mn^2+ + 4H2O}$$
Therefore, the number of electrons transferred from oxidized species per mole $\ce{KMnO4}$ (5 electrons) is the largest, and the least amount of $\ce{KMnO4}$ is needed for the reaction with a certain amount of alcohol. This makes oxidation in acidic medium the best choice for economic reasons.
The oxidation in neutral medium is less efficient, as manganese(VII) is only reduced to manganese(IV), yielding insoluble $\ce{MnO2}$ as a byproduct which needs to be removed from the reaction mixture during purification of the desired product (carboxylic acid).
$$\ce{2H2O + MnO4- + 3e- ->~ MnO2\downarrow + 4OH-}$$
Oxidation in strongly alkaline medium is least efficient in terms of transferred electrons per mole permanganate, as the latter is only reduced to manganate(VI).
$$\ce{MnO4- +e- ->~ MnO4^2-}$$
However, when the reaction is done under these conditions, a higher reaction speed and selectivity can be achieved, presumably because the alcohol will react more readily with $\ce{MnO4-}$ when deprotonated. Nevertheless, oxidation in acidic medium can be preferred for reasons stated above and substrates which are sensitive to strong bases.

It is worthy to mention that in acidic env. the true oxidizing agent is often $\ce{Mn^3+}$, as oxidation by $\ce{MnO4-}$ is often initially slow, until some $\ce{Mn^2+}$  is created and eventually oxidized to $\ce{Mn^3+}$ See e.g. manganometric titration of oxalic acid, when there are intentionally added some $\ce{Mn^2+}$  ions to catalyze the oxidation.

"Permanganate as oxidizing agent works most efficiently in acidic solution, because it is reduced to the greatest extent in this medium" - Is there are a reason behind this?


text:
How can I make an acidic pen to burn paper on writing on it?
If I have a pen with refillable ink, can I just pour in concentrated Hydrochloric acid or concentrated sulfuric acid in the ink chamber to make it an acidic pen?
Would it be better to use the acid of suitable molarity instead of being concentrated?
Then if I write with it on a paper, will it burn?
Is this a good idea to play with?
Is it dangerous in any way?
Is it illegal to make such things for experimenting and fun?
Is there any other safe method to achieve the same?
What are the odds that the poured acid will burn the ink chamber itself and start leaking?
Can it be used in any constructive way? Like cutting many sheets of paper at once. Or using it in industries suitably to make smooth cuts in materials that can be burned with the used acid.
Will it be fun to you?
Thanks to everyone.

Edit:
I would like to add a few details.
Assume that the acid is not concentrated but of suitable molarity to just cut the paper. I mean, to pierce through it and burning its edges to just make a nice cut. Like a glowing incense stick makes when made to touch a paper.
Also, the nip of the pen and the ink chamber are strongly inert; enough to survive the concentrated acid's attack and prevent leaking. Like, the chamber can be made of glass (in which acids are usually kept) and the nip can be made of galvanized/electroplated iron.
Or if there is still more, we can just make this into a thought experiment.

Nice idea! Would never try it though... Because to cut through paper would mean to use conc. acid, which means taking the risk of burning the pen or yourself! Would like to see interesting answers for this question...

What exactly do you mean by "burn" paper? You're not thinking the paper will combust on contact with the ink, are you?

@KyleStrand I mean, to pierce through it and burning its edges to just make a nice cut. Like a glowing incense stick makes when made to touch a paper.

You can pour anything into your pen. Is it a good idea? Most likely not. Is it dangerous? You bet! Is it a good idea to play with? NO! Whether it is legal or not to create such an object, you'll have to check with legislation in your country.
If you want paper to ignite on touch with this pen, you would need some truly strong acid (and probably undiluted). I'm certain that if this substance has the capability to burn paper, it would also corrode away the feed and nib of the pen. In the worst case, it will even eat through the ink container and leaking everywhere. In any way, this is no viable option to cut paper in a controlled fashion.
Is there a safer method? Well, there would be laser-assisted cutting of paper, but the good ol' knife or saw would still be cheaper.

Are you sure? What about Uncle Al's answer?

The question explicitly mentions acids, and as such I've limited my answer to them. I personally think that filling your pen with an oxidation agent is just as bad, and I'm not sure if the cellulase solution actually would work.

Soluble cellulose derivatives enzymatically cleave as fast as solutions are mixed.  Bulk cellulose is slower. The is no point in using chemistry with residues when lasers are so fast and exact, e.g., laser greeting cards.

In principle, a solution of white phosphorus in carbon disulfide ($\ce{CS2}$) can be used as an "ink" to set paper on fire. This video demonstrates the effect.
Is it a good idea? NO!
While the disgusting smell of carbonyl sulfide ($\ce{COS}$) often found as an impurity is just a nuisance, the solvent itself is neutotoxic. But it gets worse:
White phosphorus may cause awful deep third degree burns on your body!
So, unless you want to qualify for the Darwin Award, don't mess around with this!

Thanks for the video link. I had initially thought of using a white phosphorous sprinkler, but then, I realized it would be too much clumsy. I don't want a Darwin Award. I was just trying to make some James Bond kinda guddies...

http://en.wikipedia.org/wiki/Cellulase
Load the pen with a cellulase solution.
http://en.wikipedia.org/wiki/N-Methylmorpholine_N-oxide
http://pubs.acs.org/isbn/9780841200067
Load the pen with a cellulose solvent.
Paper is bulk cut with blades or lasers.


text:
rate determination by 1H NMR spectroscopy
Currently I investigate a reaction where a starting material A reacts with an excess of thiophenol B towards products C and D.  As shown by recurrent TLC analysis of the reaction, the formation of D is slower than the one of C, while A and B are still detectable.
Perhaps the reactions are consecutive, i.e.
$\ce{A + B ->[k_1] C ->[k_2] D}$
or they are simultaneous, i.e.
$\ce{D <-[k_2] A + B ->[k_1] C}$.
A, B, C, and D are distinguishable by 1H NMR spectroscopy; NMR data of the purified compounds A, B, C, and D were  recorded successfully.
Based on a recurrent 1H NMR spectroscopy, I would like to use the integrations of the signals to determine the rate constants; perhaps there is evidence for the first case (similar to Bodenstein principle).  The advancement of the thioetherifications alter the multiplicities of the signals observed, and simultaneously indicator signals vanish.
Question:  Prior to get work on my data, where may I find a tutorial of reactions with known starting material(s)/product(s), determined rate constant(s) and NMR data-used-to-determine-these-rate-constants?

So you want to measure the kinetic of the reaction with NMR, measuring 1D spectra continuously while the reaction happens in the NMR tube (or you run a regular reaction and take samples for NMR), or? From your question you seem to know what you want to do, what is the part you're unsure about? Do you want to know how to set up the NMR experiment, how to extract the rates or how to analyze the rates and determine the mechanism?

Md Scientist is right, this is very straight forward, provided you are not generating any persistent free radicals during the reaction.  Go to Journal of Chemical Education and poke around for a simple one if you need practice.

As mentioned in the comments, J. Chem. Ed is a good resource for NMR kinetics experiments. I have had success with the experiment published by J. Orvis in J. Chem. Ed 80(7), 2003, 803 link.  This publication describes the synthesis of trans-dichlorotetraamminecobalt(III) and its subsequent reaction with water to produce cis-aquachlorotetramminecobalt(III).  There are several advantages to this experiment as a model study:

The synthesis is straightforward and does not require materials unlikely to be already in the laboratory, although it is critical that the directions are followed precisely.
The aquation reaction occurs within about one hour - slow enough to easily monitor via NMR but not so slow as to take up precious instrument time
The reaction rate is temperature sensitive, which makes it a good model for extracting thermodynamic properties from the reaction

My undergraduate students are able to complete the synthesis about 50% of the time, and the yields using the published procedure are about 400 mg, so one successful synthesis is suitable for a number of NMR experiments.


text:
Phase stability of alcohols
Tert-butyl alcohol seems unusual among alcohols in that its melting point is high (25°C) while its boiling point is also still low (82°C).  I am looking for more materials with phase-unstable liquid regions like this so I'm curious what makes the liquid phase so unstable relative to the solid and gas phases?
Do the methyl groups align and act like alkanes to stabilize the solid but the molecule is still small enough to have a low boiling point?
To my larger point, any advice on characteristics of materials that exhibit low liquid phase stability?

Highly symmetric molecules tightly fit into crystal lattices, give high melting points for their composition.  Their small surface areas minimize London dispersion forces and van der Walls forces, giving high volatility.  They tend to sublime.  Compare mp and bp data for cyclohexane and benzene versus their perfluorinated derivatives. Less exotic: n-octane, iso-octane, 2,2,4-trimethylpentane, 2,2,3,3-tetramethylbutane, and cubane.
http://en.wikipedia.org/wiki/Neopentane
n-Pentane and isopentane versus neopentane.
Chromium, molybdenum, and tungsetn metal versus their respective hexacarbonyls.  $\ce{-SiMe3}$ and $\ce{-CF3}$ plus symmetry confer remarkable volatility.  $\ce{I(CF3)7}$ melts and boils around 0 C.  That is molecular weight 609.95 - and it has no static molecular structure (Bartell mechanism).
http://www.patentgenius.com/patent/3992424.html
US Patent 3992424


text:
What is the formula for finding the average atomic mass of an element based on the abundance of its isotopes?
In my class my teacher showed us how to find the average atomic mass of an element with a method, but he didn't really state a formula one could use. I came  up with a formula of my own, and from what I observed it works:
$$
\frac{i_1x + i_2y}{100} = A,
$$
where $A$ is the atomic mass,
$i_1$ is the first isotope's atomic weight,
$i_2$ is the second isotope's atomic weight,
$x$ and $y$ are the percentages of the isotopes, respectively,
and they add up to 100, i.e. $x+y = 100$.
Therefore: $$
\frac{i_1x + i_2(100-x)}{100} = A
$$
I'm just wondering, is there an official formula, or is this method it?

Have you studied weighted averages before? It's exactly that type of mathematics.

In my class we just did three questions. My teacher would give us an element such as Rubidium, the isotope's atomic mass, and the percentage of one isotope, then we would have to find the atomic mass. Or he would give us enough information to find another variable in my equation. (When I see my equation I'm not saying I made it, it's probably out there somewhere, I just don't know where)

The average relative atomic mass of an element comprised of $n$ isotopes with relative atomic masses $A_i$ and relative fractional abundances $p_i$ is given by:
$$
A = p_1 A_1 + p_2 A_2 + \dots + p_n A_n = \sum\limits_{i=1}^n p_i A_i
$$
For example carbon:
\begin{array}{lrrr}
\text{Isotope} &
\text{Isotopic Mass $A$} &
\text{Abundance $p$} &
A\times p\\\hline
\ce{^{12}C}:   & \pu{12.000000 u} & 0.98892 &= \pu{11.867 u}\\
\ce{^{13}C}:   & \pu{13.003354 u} & 0.01108 &= \pu{00.144 u}
\end{array}
Since we have $p_1A_1$ and $p_2A_2$, we add those together to find $A$, therefore
 the chemical relative atomic mass of carbon is
$$A = \pu{00.144 u} + \pu{11.867 u} = \pu{12.0011 u}.$$

Once you know how it works, you never want to do that by hand again. For example for bromine you can use this python code:
isotopes = [79, 81]
abundances = [50.69, 49.31]

weighted = map(lambda x, y: x * y, isotopes, abundances)
# multiply isotope mass and abundance

weighted = sum(weighted)/100
# sum them up and divide by 100

print weighted
>>>> 79.9862


text:
Heterogeneous Equilibrium Without Gas
We know that for homogeneous mixtures the equilibrium constant is determined using both concentration of reactants and of products. For heterogeneous equilibrium involving a gas and a liquid or a solid, we only use the concentration of the gas because the concentration of liquids and gases are constants, and can thus be omitted from the equation. But what of the heterogeneous equilibrium between a solid and a liquid? Both are incompressible, so what do you do?
Thanks!

You're right, they don't really have concentrations or pressure to work with, which would lead to 0/0 craziness.
I think we can go back a step and look at how the equilibrium constant is defined. When you derive the equilibrium constant you don't use pressure or concentrations. You actually use the substances' activities which we can relate to change in Standard Gibbs Free Energy for the reaction (which is easy to find with tabulated data):

For a full derivation refer to Wikipedia's article on the equilibrium constant.

Ok, so I haven't learned about Gibbs Energy yet. Is it safe to assume that for my exams we won't be tested on this?

I suppose...? Haha the equation is commonplace in high school chemistry but the connection wasn't made in my coursework until physical chemistry. (college upper div chemistry)


text:
Extraction of nickel from its ores
How is nickel mined and extracted from its ores?
What are the appropriate word and chemical equations for this process?

Nickel can be extracted directly from its ore by reduction by hydrogen or carbon monoxide at elevated temperature at $600\ \mathrm{^\circ C}$ to $650\ \mathrm{^\circ C}$:
$$\ce{NiO + H2 -> Ni + H2O}$$
$$\ce{NiO + CO -> Ni + CO2}$$
Also, it can be extracted by treatment with dilute sulfuric acid following by precipitation:
$$\ce{NiO + 2H+ -> Ni^2+ + H2O}$$


text:
Conjugate acids and bases
Consider the reaction
$\ce{NH3 + H2O -> NH4+ + OH-}$
Here, $\ce{NH3}$ is a weak base, and thus $\ce{H2O}$ becomes a weak acid. Thus, shouldn't its conjugate be a strong acid? However, $\ce{NH4+}$ is a weak acid, instead of the strong conjugate acid it should be. Why is this?

I think you need to compare the strength of acid and base using the $pK_{a}$ values, it is not quantitative enough to just say 'strong' and 'weak'.

Please elaborate? I thought that if you had a weak acid or base it would produce a strong conjugate? Is that not the case?

Consider the $pK_b$ of the base $\ce{NH3}$ and the $pK_a$ of the acid $\ce{NH4+}$.
$pK_b$ = $4.8$
$pK_a$ = $9.2$
$\ce{NH3}$ is a stronger base than what $\ce{NH4+}$ is as an acid, being $4.8$ smaller than $9.2$: let's compare $pH$ for equal concentrations, say $0.1 M$.
For $\ce{NH3}$, you get $11.1$, while for $\ce{NH4+}$, you get $5.1$.
$11.1$ is about 100 times more "distant" from neutral $pH$ than $5.1$, if we move from the $pH$ scale to the concentration scale. That's what also stems from the comparison of the $pK$ values.
To simplify the problem, take the $pK_a$ for the acid/base couple, e.g. $\ce{NH4+/NH3}$, that is equal to $9.2$. If that value is larger than $7$, then, within that couple, the base will be stronger than its conjugated acid. The other way round, if $pK_a$ is lower than $7$.


text:
Why does the equilibrium constant change on scaling the stoichiometric coefficients of a reaction?
I read in my textbook that if we multiply a chemical reaction by some factor(let's say $b$) its new equilibrium constant becomes $K^b$.But I don't understand why this happens..
What is the difference between the reaction $\ce{A} + \ce{B} = \ce{C}$ and $b \ce{A} + b \ce{B} = b \ce{C}$ ?
Aren't the two reactions same? So why does the equilibrium constant change?

Related question: https://chemistry.stackexchange.com/questions/6641/why-does-equilibrium-constant-change-when-the-reaction-is-multiplied-by-a-number

Have a look at this answer of mine.
There you can find the derivation of the formula that defines the equilibrium constant
\begin{equation}
\log \underbrace{\prod_i [a_{i}]^{\nu_{i}}}_{= \, K} = -\frac{\Delta G^{0}}{RT} \qquad \Rightarrow \qquad \log K = -\frac{\Delta G^{0}}{RT}
\end{equation}
of a general reaction
\begin{equation}
 \ce{\nu_{1}A + \nu_{2}B + ... + \nu_{$x$}C <=> \nu_{$x+1$}D + \nu_{$x+2$}E + ... +\nu_{$n$}F}
\end{equation}
and it is also shown how this definition relates to the maybe more familiar definition
\begin{equation}
 K_{c} = \prod_i [c_{i}]^{\nu_{i}} \ .
\end{equation}
Now, the reason why the equilibrium constants of the general reaction above and the same reaction multiplied with a constant factor $\alpha$:
\begin{equation}
 \ce{\alpha\nu_{1}A + \alpha\nu_{2}B + ... + \alpha\nu_{$x$}C <=> \alpha\nu_{$x+1$}D + \alpha\nu_{$x+2$}E + ... + \alpha\nu_{$n$}F}
\end{equation}
are not the same but are related to each other via:
\begin{equation}
 K_{\alpha} = K_{1}^{\alpha}
\end{equation}
comes from the definition of $K$.
That there must be a difference is also plausible: The defining formula for $K$ relates it to the standard Gibbs free energy of reaction $\Delta G^{0}$ which in turn is defined to be the amount of Gibbs free energy that is produced per formula conversion (meaning that it is defined for an extend of reaction $\xi = 1 \, \text{mol}$).
So, it is the amount of Gibbs free energy produced by the reaction of $\nu_{1}\, \text{mol}$ of $\text{A}$ and $\nu_{2}\, \text{mol}$ of $\text{B}$ etc. to $\nu_{x+1} \, \text{mol}$ of $\text{D}$ and $\nu_{x+2}\, \text{mol}$ $\text{E}$ etc.
But if you multiply the reaction by $\alpha$ the standard Gibbs free energy of reaction is the Gibbs free energy produced by the reaction of $\alpha\nu_{1}\, \text{mol}$ of $\text{A}$ and $\alpha\nu_{2}\, \text{mol}$ of $\text{B}$ etc. to $\alpha\nu_{x+1} \, \text{mol}$ of $\text{D}$ and $\alpha\nu_{x+2}\, \text{mol}$ $\text{E}$ etc. and so there should be $\alpha$ times as much Gibbs free energy as before:
\begin{equation}
 \Delta G^{0}_{\alpha} = \alpha \Delta G^{0}_{1} \ .
\end{equation}
And this is indeed the case when you have $K_{\alpha} = K_{1}^{\alpha}$ because according to the logarithm laws
\begin{equation}
RT \log K_{1}^{\alpha} = RT \alpha \log K_{1} \ .
\end{equation}

Let's take the simplest reaction equation $$\ce {A <=> B}$$
In this reaction, your stoichiometric coefficients are simply $$1A:1B$$
Coefficients are the factor you will need to raise your ratio to when you calculate your equilibrium constant. For this reaction, the equilibrium equation would be:
$$K^1_{eq} = \frac{[B]^1}{[A]^1}$$
Now, to make things easier to see, lets say your concentrations are:
$$A=10 : B=5$$
This would mean that
$$K^1_{eq}=\frac{10}{5} = 2$$
Now, let's multiply our equation by a coefficient say, 2:
$$\ce{2A<=>2B}$$
Now when you go to calculate your equilibrium constant, your exponents for a will be 2 instead of 1:
$$K^2_{eq} = \frac{[B]^2}{[A]^2} $$
Replace A and B with your concentration since they haven't changed, only their stoichiometric coefficients have:
$$K^2_{eq}=\frac{10^2}{5^2} = \frac{100}{25}=4$$
Compare this new K value, call it K2 to your first K value, call it K1:
$$K_2=4 , K_1=2$$
Remembering that K2 is the square of K1, prove to yourself the reaction is the same by saying:
$$K_2=K_1\iff K_1^2=K_2 \iff K_2^\frac{1}{2} = K_1$$ which just shows that
$$4^\frac{1}{2} = 2 $$
$$2=2$$
Your rate hasn't necessarily increased, the kinetics of the reaction are just the same, it's just that an equilibrium constant serves as a way to analyze how many moles of your product you are gaining compared to how many you are losing. If you add more moles of your reactant, your product amount has to increase commensurately.
If you want to prove it to yourself, do the same thing but
$$K^3_{eq} = \frac{[B]^3}{[A]^3}$$ You should still get a $$K=2$$ if you use $$B=10 , A=5$$


text:
Why does $\text{Cr}^{3+}$ not have the same electron configuration as $\text{Sc}$?
To be explicit, $\text{Cr}^{3+}$ and $\text{Sc}$ have the same number of electrons, the only difference is the nuclear charge.
This is not an isolated anomaly, it seems: the book I'm using claim that with all fourth period transition metal cations, the $4 \text{s}$ electrons are removed first. This indicates that the $4 \text{s}$ electrons are at a higher energy than $3\text{d}$.
My question: why is this the case, given that the $4\text{s}$ electrons are 'added' before the $3\text{d}$ electrons as one moves rightwards across the fourth period, indicating that they are of a  lower energy? I cannot reconcile this apparent contradiction.

The electron configuration of chromium is : $\ce{[Ar] 4s^1 3d^5 }$ (chromium is one of those examples of special electron configurations : the electron configuration is not $\ce{[Ar] 4s^2 3d^4 }$ as many people might suspect, an electron is moved to the 3d orbital because this configuration is more stable  --> that is the reason why chromium has this special electron configuration)
So the $\ce{Cr^3+}$ ion loses an electron from $\ce{4s}$ and loses 2 electrons from $\ce{3d}$ to give it an electron configuration of $\ce{[Ar] 3d^3 }$
Scandium has an electron configuration of $\ce{[Ar] 4s^2 3d^1  }$ and isn't one of those exceptions.
These are the reasons why the chromium(+III) ion doesn't have the same electron configuration as scandium, despite the fact that they have equal amounts of electrons.
As for why the 4s orbital is emptied first:

In all the chemistry of the transition elements, the 4s orbital
  behaves as the outermost, highest energy orbital. The reversed order
  of the 3d and 4s orbitals only seems to apply to building the atom up
  in the first place. (Using the Aufbau principle) In all other respects, the 4s electrons are always
  the electrons you need to think about first.

(this link is interesting for you in case you wanted a more detailed explanation about the 4s / 3d problem :http://www.chemguide.co.uk/atoms/properties/3d4sproblem.html )
EDIT : Chromium, just like all other elements, tries to obtain more stable configurations by losing certain amounts of electrons. Now, in the case of $\ce{Cr^3+}$ half/fully filled subshells can't be achieved ( you get $\ce{[Ar] 3d^3}$ ) but that doesn't necessarily mean that it is not stable. But chromium does have other ions in which it does have more stable electron configurations. Take a look at the $\ce{Cr^6+}$ ion : here the ion has an electron configuration that is equal to the one of argon, which does have completely filled valence subshells. ($\ce{[Ne] 3s^2 3p^6}$)
So ions can't always achieve these more stable half/fully filled subshells, but that doesn't necessarily mean that they are not stable.

I was under the impression that the cause of the stability that gives rise to the electron movingg to the 3d orbital in Chromium was that all the 3d subshells would become half-filled if an electron moved from 4s to 3d. However, this is not the case in $\text{Cr}^{3+}$ so I don't see how this logic applies (I'm assuming my dilemma arises from the argument in my first sentence being flawed).

I've edited my question - I hope that is the answer that you were looking for :)


text:
calculating Kp with partial pressures
This is a question I solved in class:

At 35 degrees, $\ce{N_2O_4}$ is 15% dissociated in a flask at a pressure of 1.2 atm. Calculate
  $K_p$  Equation:
  $$\ce{N_2O_4(g)<=>2NO_2 (g)}$$

This was how the answer went...
Partial pressure of $\ce{N_2O_4} = 0.85\times 1.2$
Partial pressure of $\ce{NO_2} = 0.1\times 1.2$
then $K_p$ was calculated using the standard formula
My question:
The part I cannot agree to with my teacher is that how is the partial pressure of $NO_2$ equal to 0.15x1.2. If I begin with 1 mole of $\ce{N_2O_4}$, if 15% dissociated at 1.2 atm, I would end up with 0.3 moles of $\ce{NO_2}$... the the partial pressure would be ${0.3\over 1.15}$x$1.2$... this does not go with the general answer stated above

I would say, you are right: the partial pressure of $\ce{N2O4}$ should be $\frac{0.85}{1.15} \cdot 1.2 , \text{atm}$ and the partial pressure of $\ce{NO2}$ should be $\frac{0.3}{1.15} \cdot 1.2 , \text{atm}$.

I do agree with your point and Philipp explanation. Getting more formal, your $K_p$ at 35°C is:
$$ K^\circ_p(308.15\,\mathrm{K}) = \frac{\left(\frac{p_\mathrm{tot}}{p_\circ}\right)^2\chi^2_\mathrm{NO_2}}{\left(\frac{p_\mathrm{tot}}{p_\circ}\right)\chi_\mathrm{N_2O_4}} $$
Where $p_\mathrm{tot}$ is the pressure in your flask and $p_\circ$ the standard pressure. The $\chi$ quantities stand for molar ratio. For the reaction given, those ratios are:
$$ \chi_\mathrm{NO_2} = \frac{n_{NO_2,0} + 2\xi}{n_{N_2O_4,0} + n_{NO_2,0} + \xi} $$
and:
$$ \chi_\mathrm{N_2O_4} = \frac{n_{N_2O_4,0}  -\xi}{n_{N_2O_4,0} + n_{NO_2,0} + \xi} $$
Those identities come from amount balance, where $\xi$ is the molar coordinate of the reaction and zero-indexed quantities amount of matter when starting experience.
Assuming that initially there was only $\mathrm{N_2O_4}$, and you are using the following definition of dissociated ratio:
$$ \tau = \frac{\xi}{n_\mathrm{N_2O_4},0} = 0.15$$
You will obtain by substitution, molar ratio values:
$$ \chi_\mathrm{N_2O_4} = \frac{0.85}{1.15},  \chi_\mathrm{NO_2} = \frac{0.3}{1.15}$$
Then you inject those values in the $K^\circ_p$ expression above. Do not forget to express pressures in bars - this is a common mistake, you must remeber that $K_p$ are dimensionless. Therefore, $p_\mathrm{tot} = 1.2 \times 1.01325 \,\mathrm{bar}$ and $p_\circ = 1.0 \, \mathrm{bar}$. This is also an approximation because we are assimilating pressure as activities. But in your case you can consider fugacities close to unity and assimilate your gas mixture as an ideal gas.
Yes, you were right about molar ratio. Do not forget that equilibrium constants are dimensionless: they are ratio of activities!


text:
How would one compare the magnitude of covalent character between SnCl4 and SnF2 using Fajan's Rules?
It is easy to compare two ionic compounds when one of the ions is same. However, how do we compare two compounds if one of the ions is the same element but just has different charge and the other ion is completely different?

Here the change of anion and the increase of oxidation, both act towards producing a similar change in the ionic character of the compound. Can you tell why aren't you able to apply Fajan's rules to solve this problem? It appears a straightforward application to me.

@SatwikPasani I am having difficulty applying the rules. Could you please apply the rules and solve this problem? Thanks!

@SatwikPasani SnF2 is apparently not a straightforward case. Studies employing Mossbauer spectroscopy have shown that SnF2 is covalently bonded.

Fajan's rules for the prediction of bond character can be summarized in two points:

Large cation with low positive charge + small anion $\rightarrow$ ionic
Small cation with high positive charge + large anion $\rightarrow$ covalent

In $\ce{SnF2}$, we have a large cation with low positive charge. $\ce{Sn^2+}$ is larger than $\ce{Sn^4+}$ because it has 2 more electrons in the $5s$ orbital. The anion $\ce{F^-}$ is smaller than its heavier group homolog $\ce{Cl^-}$. For this combination large cation + small anion, the bond character is predicted to be more ionic and less covalent.
On the other hand, $\ce{Sn^4+}$ is a small cation with high positive charge, and $\ce{Cl^-}$ is the large anion, as already stated. Therefore, according to the rules, $\ce{SnCl4}$ is the tin halide with more covalent bonds, and indeed, tin(IV) chloride consists of tetrahedral $\ce{SnCl4}$ molecules in the solid, liquid and gaseous state.
However, the predictions made with these rules do not always agree with the real bonding situation, and $\ce{SnF2}$ is an example. There is experimental evidence for covalent $\ce{Sn-F}$ bonds in tin(II) fluoride (references: 1, 2).

Complementary the HSAB Theory can be applied.

SnF2 is a bent molecule with a a stereo-active lone pair in a hybrid orbital, so it must be covalent. I have this one professor who has spent his whole life studying tin chemistry and tin crystallography and he assures me that SnF2 is covalent.
"With small and highly electronegative non-metals (F, Cl, O, S), bonding is alwayscovalent, and as a result, the tin orbitals are hybridized, usually sp3, sp3d or sp3d2,to give an electron pair distribution that is pseudotetrahedral SnX3E, pseudotrigonal bipyramidal SnX4E or pseudooctahedral SnX5E, respectively" -Georges Denes & Abdualhafed Muntasar


text:
Do soap molecules form liposomes?
I know soap molecules form micelles. Do they also form liposomes in water? I need the reasons too.

Liposomes are a lipid bilayer separating an aqueous internal compartment from the aqueous phase. Micelles are closed lipid monolayers with a fatty acid core and polar surface, or a polar core with fatty acids on the surface (inverted micelle, Aerosol-OT).  A soap would then form a micelle but not a lipsoome for its ionic water-soluble head.
A non-ionic surfactant, e.g., diblock short PEO-PPO, could stretch the definitions.

So basically that ionic ends stops the formation of liposome?

So it seems.  Ask your prof.  Some definitions are matters of fashion or convenience.  Carbonium ions (-onium - highest valency) became carbenium ions after Magic Acid.  The nomenclature is still all over the map.

Amphiphiles spontaneously form a variety of self-assemblies in water, such as
micelles, vesicles, and bilayers. The simplest form of self-assembly for single chain amphiphiles are micelles.
The concept of packing parameter proposed by J. Israelachvili allows a simple
and intuitive insight into the self-assembled aggregates of amphiphiles in water
The packing parameter $P$ is defined as:
$$P = \frac v{ l \times a_0}$$
where $v$ and $l$ are the volume and length of the surfactant hydrophobic tail, respectively, and $a_0$ is the optimal surface area per amphiphile.
Let me briefly explain the concept of optimal surface area per amphiphile ($a_0$). It is based on the principle of opposing forces; the hydrophobic interactions act to pack the soap molecules closer together to avoid exposing their hydrophobic tails to water, while the repulsive energy between the charged headgroups of the soap molecules increases as molecules are pushed closer together. The surface energy of the self-assembly is minimized when the amphiphilic molecules occupy the optimal area.
Single-tailed amphiphiles (like your soap molecules) have a large head-group area and have a packing parameter $P < \frac13$. Thus, they form spherical micelles. Ionic surfactants such as sodium dodecyl sulphate (SDS/NaDS) generally forms spherical micelles in solution at low salt concentrations. Single-tailed amphiphiles with a small head-group area ($\frac13 < P < \frac12$) will form elongated or cylindrical micelles. Non-ionic surfactants and ionic surfactants such as
SDS and CTAB in high salt are known to form elongated micelles. Double-tailed
amphiphiles with a large head-group area (phosphatidylcholine, phosphatidylserine, phosphatidylinositol etc) have a packing parameter approaching unity and tend to self-assemble into planar bilayers and vesicles/liposomes.
The figure here will explain this schematically:


text:
Learning chemistry with software?
I'm interested in learning organic chemistry, and am following a text (Organic Chemistry by Clayden) and an online course. I also have access to ChemBio3D.
I feel as though the way chemistry is taught conventionally is more difficult than it needs to be. Some online courses utilize industry software when teaching things (such as using MatLab in engineering), which makes visualizing and understanding a lot of the theories and concepts taught in books much more intuitive, since you can replicate the theory directly into a program.
Does a similar course exist somewhere for chemistry? I'm not talking about the little java applets that are available everywhere that mimic certain reactions or visualize concepts, I mean a course that utilizes a powerful software suite such as ChemBio in order to teach organic chemistry? Something that will teach both organic chemistry at a proper undergrad level and a good piece of industry software at the same time?

So what do you want exactly? You want a software "which makes visualizing and understanding a lot of the theories and concepts" but not " little java applets ...  that visualize concepts"? What is powerful about ChemBio3D you want to use? It is mostly for drawing, property prediction, nothing about concepts..

Do your chair parade.  See what the Kahn Academy has to offer.  Have more than one texbook.  If organic is not the best thing in the world pursued clothed, consider other options.  Employment at the bench will be under management that has other goals.  Grant funding considers discovery to be unacceptable risk.  Remember what they called the first designed degenerate Cope rearrangement - BULLVALENE.  It worked.  So did its two-step synthesis using no reagents.
LCAO for hybridization, geometry, resonance, and 1,3-shifts.  Nomenclature and chirality.  Woodward-Hoffmann comes later.  Carbocations, carbanions, radicals.  Carbonyl condensations.  Hard and soft acids and bases, Lewis and Brønsted acids and bases.  Reductants and oxidants; reagents in general.  Name reactions (that you already know as pedagogy).  Practice looking into stereograms.
NMR, IR, UV-Vis (not so much now), mass spec with database and software.  You should have some glassblowing under your belt - Schlenkware;  T-, butt, and ring seals; making an ampoule.  Own your own didymium glasses and Kennametal cutter.  Burn a pencil for its "lead."
Synthesis.  Work backwards via the synthon approach.  We used to read the new Aldrich catalog, and Fieser and Fieser, but both have has gotten out of hand.  Spend a weekend/month skimming the main journals - JACS, Angew. Chemie, JOC, Tetrahedron - and what interests you.  Never fill a reaction flask more than half full, because the universe hates you.  COTTON lab coats, because they don't melt onto your flesh.  Don't do anything stupid twice.  When in doubt - palladium.
Buy HyperChem Lite, oh yes indeed.  Remember to add hydrogens, assign atom types, then minimize well below the recommended cutoff for 10 times the recommended iterations, at least.  Note how the software is boggled when minimizing high symmetry cage ketones
If it isn't fun, you might be a theorist.  There is no shame in that - but you don't get to make crystals.


text:
Determine the equilibrium constant for this reaction
The question is:

Ions $\ce{B}$ and $\ce{C}$ react to form complex $\ce{BC}$. If $\pu{15.0 mL}$ of $\pu{1.00 M}$ $\ce{B}$ is combined with $\pu{15.0 mL}$ of $\pu{1.00 M}$ $\ce{C}$, and our initial concentration of BC is $0$, yet the end in equilibrium for complex $\ce{BC}$ is $\pu{0.00450 mol}$ determine the equilibrium constant for this reaction.

Firstly, I converted the total volume $\pu{15.0 mL + 15.0 mL = 30 mL}$ to L and I got $\pu{0.03L}$
Secondly, I used the given $\pu{0.00450 mol}$ and divided it by $\pu{0.03 L}$ and I got $\pu{0.15 M}$ for BC
Thirdly, I did the ICE chart

After that, I used the quadratic formula to solve for $x$ and calculate the concentration for each substance, as well as getting the $K_\mathrm{c}$ at equilibrium

I got $1.53$ for $K_\mathrm{c}$ but I am not sure if the whole process is correctly done. Please point out my mistakes

You made a slight error in calculations for B and C. While those concentrations are correct, that is 1 M at 15 mL. When doing RICE tables and equilibrium calculations, we can use concentrations, however I prefer to convert concentration into moles.
We are trying to find $K_c$, and we know that equal moles of the reactant react to give us 0.0045 mol of BC.
$$\ce{B + C <=> BC}$$
\begin{array} {|c|c|c|c|}
\hline
\text{Initial conc.} & \text{0.015 mol} & \text{0.015 mol} & \text{0 mol}\\
\hline
\text{Change conc.} & -x & -x & +x\\
\hline
\text{End conc.} & \text{0.015 mol} - x & \text{0.015 mol} - x &  0 + x \\
\hline
\end{array}
We know that our change in equilibrium is 0.00450 mol, as that is the end amount of product that we have in our equilibrium calculations.
Accounting for the concentrations, we divide our number of moles by the total volume to get mol/L. 0.5 mol/L, 0.5 mol/L, .15 mol/L.
Assuming that our reaction is at equilibrium:
$$K_c = \frac{[\text{0} + \text{.15 mol/L}]}{[\text{0.5 mol/L} - \text{.15 mol/L}][\text{0.5 mol/L} - \text{.15 mol/L}]} $$
$$K_c = \frac{[\text{.15 mol/L}]}{\text{[0.35 mol/L]}^2} $$
$$ K_c = 1.22 $$
Thanks @Nicolau for pointing out the error in the question and reviewing my work.

Thank you for showing the work in how to do this question and pointing out the fact that concentrations can also be used in this case. However, I am confused in how to get 0.015 mol for B and C. Did you divide 0.00450 mol by 0.03L to get 0.015 mol? Moreover, do we need to use the quadratic formula to answer this question? Based on your work, it seems like we do not need to.

@Jesse Dear Jesse, for B and C, I divided 1 M by 15 mL to get the number of moles. However, at the end, I needed to convert once again back to concentration, so I divided by the total volume 30 mL and multiplied by 1000 to get mol/L. We do not need the quadratic formula, since we are trying to find K. We know what the end of concentration is in our RICE table, the lower right hand corner. Since we know by reaction stoichiometry that it increased by .15 mol, therefore x is .15 and we plug that into our equilibrium expression.

Dear Mr. Kwak, thank you once again for the clear explanation! Since 1 M = 1 mol over 1000 mL, and we have 15 mL, so we multiply 1 mol over 1000 mL by 15 mL and we are left with 0.015 mol. Is this correct? I finally get it, thank you! :)

@Jesse Exactly. And since we are in moles at the end, we have to multiply by our total volume to get back our concentrations. And thus, we plug those concentrations into our equilibrium expression. It was my pleasure. :)


text:
What is the term for the amount of volume change at a phase boundary?
Apologies for the extremely basic question, but I would like to look up the amount by which the volume (or equivalently, density) of various substances changes when they freeze. However, I don't know the correct term for this, and Google isn't helping much.
To clarify, I'm asking about the vertical line segment on the plot below (from Wikipedia). If the $y$ axis was internal energy then the height of this line segment would be called "latent heat of fusion", and I could easily look it up for any given substance. I would like to know the equivalent term for the discontinuous volume (or density) change, in whichever units are most standard.

I realise that unlike latent heat, this will depend quite a bit on the pressure. However, I would hope that it would be tabulated for atmospheric pressure at least.

Phase transition can be distinguished by analysing their chemical potential, such classification has been introduced by Ehrenfest. For a given transition, lets call it $\phi$ transition:
$$ \ce{X_{\alpha} \rightleftharpoons X_{\beta}} $$
You can assess transition molar specific volume in the following way:
$$ \Delta_{\phi}V = v_{\beta} - v_{\alpha} = \left(\frac{\partial \mu_{\beta}}{\partial p}\right)_{T} - \left(\frac{\partial \mu_{\alpha}}{\partial p}\right)_{T}$$
Thus, by modelling your potential in term of pressure and temperature, you can assess molar volume change of a phase transition. This is the key to solve your question. In real life application, we use diagrams and tables to assess it. But building a model is always valuable for comprehension.
In a similar way, molar entropy is given by:
$$ \Delta_{\phi}S = s_{\beta} -s_{\alpha} = \left(\frac{\partial \mu_{\beta}}{\partial T}\right)_{p} - \left(\frac{\partial \mu_{\alpha}}{\partial T}\right)_{p} = \frac{\Delta_{\phi}H}{T_\phi} $$
Because $\Delta_{\phi}V$ and $\Delta_{\phi}H$ are not null for first order (in term of first derivative) transition, slope difference is not null and therefore potential slopes are different both side of phase transition when Gibbs parameters vary ($p$ and $T$). Thus there are discontinuities in specific volume, enthalpy and entropy when a first order transition occurs.
Ehrenfest did a classification of phase transition based on this criterion which led to: first order, second order (smooth, no slope discontinuity) and $\lambda$ (sharper) transitions regarding how such chemical potential discontinuity happens.
This discontinuity also impact heat capacity $c_p$. In case of water, whose phase transitions can be assumed as a first order, when a phase transition occurs, discontinuity leads to a virtually infinite heat capacity, explaining why fusion or vaporisation happens at constant temperature (when pressure is held constant). This is why we need an extra term called latent heat (temperature independent) which contrast with sensible heat (temperature dependent).
You will find detailed explanation (models, diagram and tables) of those phenomenons in Atkins, Physical Chemistry, Chapters 2-7.

I greatly appreciate the effort you've gone to to give a detailed answer. Unfortunately I know all this stuff already! All I'm asking for is the name of $\Delta_\phi V$, so that I can search online for tables of its values.

You will rarely encounter tables for $\Delta_\phi V$ but specific volume $v_\gamma$ (molar or mass) for a given phase $\gamma$ are tabulated or plotted. Look for an Handbook of Chemistry or Thermodynamic Handbook for Air or Water. NIST Chemistry Handbook has numerous substance and phase transition data.


text:
Why do the trends in reactivity not apply for francium?
Why is francium not included in the reactivity series?
Why is potassium considered more reactive than francium?
I know that reactivity increases down the group, but why does it not apply here?

Francium is radioactive. All isotopes of francium are radioactive. The most stable isotope of francium has a half life of only 22 minutes. From the Wikipedia article I linked:

Bulk francium has never been viewed. ... However preparing such a sample is impossible, since the extreme heat of decay (the half-life of its longest-lived isotope is only 22 minutes) would immediately vaporize any viewable quantity of the element.

This extreme radioactivity makes it hard to determine the chemical properties of francium. Some are known. For example, francium's ionization energy is 392.8 kJ/mol (cesium is 375.7 kJ/mol). Since francium has a higher ionization potential than cesium, which it does because of relativistic effects, francium should be less active than cesium. However, since observable quantities are difficult (nearly impossible) to come by. It is hard to say for sure.

I know that reactivity increases down the group, but why does it not
  apply here?

Francium has a $\ce{[Rn] 7s^1}$ electron configuration. It is a relatively heavy element with atomic number $Z=87$. In such heavy elements, relativistic effects become significant and impact the reactivity and other physical characteristics of the element. In francium, the inner $\ce{s}$ electrons travel at roughly $\frac{[Z]}{[137]}$ or 64% the speed of light.  These relativistic speeds lead to a significant contraction in the radius of all s orbitals as explained here. As the $\ce{7s}$ orbital contracts, the $\ce{s}$ electron

moves closer to the nucleus,
is lowered in energy (because it is closer to the nucleus),
becomes more shielded by other electrons (because it is closer to the nucleus),

and, therefore, is less likely to be involved in chemical reactions.  Consequently, the reactivity of francium is significantly less than anticipated.

It is predicted that francium is less reactive than caesium due to the speed at which the outer electron spins, adding strength to the bond between the atom's nucleus and the outer electron. While I am unable to fully comprehend quantum physics, a script of a 'Chemistry in its Element' podcast at the Royal Society of Chemistry, covers the subject about 2/3 of the way down:
Francium (via the Internet Archive).


text:
When a thin lead sheet is hammered, what happens to internal energy
When a thin lead sheet is hammered, what happens to internal energy, does it increase, decrease or remains the same or first increases and then decreases?

What are your thoughts or predictions?

When you hammer a sheet of lead, you are adding energy in form of work into it, increasing it's internal energy. If this experiment were to be done in complete vacuum, it'd retain most of the energy, however if this experiment were to be on air, it'd lose a lot of energy in the form of sound.

when a thin lead sheet is hammered, its internal energy is increased. this is because when we are hammering it, we are providing energy to the system which results in the increase of the energy of the system. :-)

It gets hotter then cools down, net change 0.  A very narrow lead sheet formed into a spring would be another story because it would lose potential energy due to a change in shape.

Change in shape still is important as surface atoms have higher energy.


text:
Standard reference for thermodynamic properties for octane
where can I find thermodynamics properties for octane at saturated Temp.$T_s$?
I need this properties or where can i find all ?
$$liquid-density=\rho_l=?$$
$$vapor-density=\rho_v=?$$
$$latent-heat-of-evaporation=h_fg=?$$
$$liquid-kinematic-viscosity=\mu_l=? $$
$$liquid-conductivity=k_l=?$$

When you ask for something like that, please always define all you symbols. For one thing, the symbols are not universal and furthermore, you can't search for a symbol.

sorry I edited it ...

Have you checked the NIST database, as well as a standard MSDS for Octane?

how can i check it ?

Check the infobox on the wikipedia page. If this is not "standard" enough for your purposes, why not check the sources of the infobox itself?

The NIST Chemistry Webbook maintained by the National Institute of Standards and Technology is a freely searchable database of chemical and physical data on a vast array of substances.
Everything you want for octane can be found there.


text:
Organic chemistry Reaction mechanism
Which one is more stable, alkyl carbocations or aryl carbocations and why?

What are your own thoughts on the matter. Homework question should show some efforts from you before we help you, so that we can see where you have problems and guide you along the way. Especially as this question is a quite basic one which gets addressed in every introductory organic chemistry textbook.

If you want it nice and stable, you want charge delocalization, like in allylic and benzylic cations. Note that more aryl residues mean more delocalization, additional +M substituents provide further stabilization.

Take the cations of malachite green (1) and crystal violet (2) as an example and draw some resonance structures.
Compare these with an alkyl cation or an aryl cation.

Steric hindrance might kinetically stabilize a tris-alkyl carbenium ion, such as (tert-Bu)$\ce{_3C^+}$.  Delocalized aromatic systems leave it in the dust - tropylium.


text:
Combustion of thiazole
I am learning Chemistry from a book Heterocyclic compounds and I have bumped onto a following image with given tasks:

What is the balanced reaction?
What is the RMM of thiazole?
What is the fuel-air mass ratio?
What is the oxygen depletion?
What is the yield of CO2?
What is the yield of H2O?
What is the yield of N2?
For ideal combustion, what is the yield of carbon-monoxide?
Now, from this I can see that aformentioned product is Thiazole( checked on wiki ) and that products of combustion are carbon dioxide, water, hydrogen sulphide (H2S) and nitrogen. Now how can I solve this matter having in mind following data that air is consisted of 21% O2 and 79% of N2.

First I have to ask about the premise of your question- why are you teaching yourself chemistry from a book on heterocyclic compounds?  The rest of your question suggests you've skipped the basics and jumped to fairly advanced stuff.  There are many good introductory chemistry that will walk you through problems similar to this one that will be a far easier starting point than heterocyclic compounds.
Assuming your general question is "how do I do calculations on oxidation reactions?", you fill find whole chapters and even books on the topic.  I would highly recommend starting with something like "What is the balanced reaction of combusting propane (C3H8) in pure oxygen?"  You will need this kind of basis to be able to work up problems like the one you ask about with many possible products and spectator species.
This will get you started: http://en.wikipedia.org/wiki/Combustion#Chemical_equations with answer to the first question.  In general though, pick up a used introduction to chemistry book up off Amazon for $10 and start with those walk throughs.
Hope this helps answer your real question and please let me know if I misunderstood and you had a more specific question about this particular problem.
EDIT: For the yield questions, you can use the the balanced equation and molecular weights to come up with a yield.
You'll have something like:
$$yield=stoichiometricRatio*\frac{MW_{product}}{MW_{thiazole}}$$
$$\frac{g_{product}}{g_{thiozole}}=\frac{mol_{product}}{mol_{thiozole}}*\frac{g_{product}}{mol_{product}}*\frac{mol_{thiazole}}{g_{thiazole}}$$
If they give you a mass (or mass flow rate) of thiazole, you could further convert this to a mass (or mass flow rate) of products.

You have a valid point. I have answered on all Q. except on the following ones:5,6,7. I just don't know how to find out yield of some compound?

I've updated the answer to include a more detailed explanation of calculating yields.  The question is a little vague on what it is looking for but this is how I would interpret it.


text:
How does malachite react with acid?
I performed an experiment where I did this:

Add one spatula of malachite to a test tube, add dilute hydrochloric acid to the malachite until the test tube is one third full. Note observations and identify gas produced.

The green solid turned to a blue solution and there was an effervescence of hydrogen. However, I now realise that copper doesn't react with acid and malachite is a copper carbonate hydroxide mineral...
I researched online and found that

It dissolves in sulfuric acid to give Copper sulfate, Carbon dioxide
  and Water.
  http://wiki.answers.com/Q/What_happens_when_malachite_is_dissolved_in_sulphuric_acid?classic=true

So how did I get that malachite reacted with acid and did produce hydrogen with a squeaky pop test?
Thanks in advance

P.S: Please could you include the formula to explain? Thanks again ;)

No, pure malachite $\ce{Cu2CO3(OH)2}$ usually doesn't produce hydrogen because it is a copper carbonate. Copper is already in a higher oxidation state (+2) so it can't reduce hydronium to hydrogen. The right reaction is the following:
$$\ce{Cu2CO3(OH)2 +2H2SO4 -> 2CuSO4 + CO2 ^ + 3H2O}$$
However, if there are traces of metallic copper some hydrogen can be produced but not with sulfuric acid because:
$$\ce{Cu + 2H2SO4 -> CuSO4 + 2H2O + SO2 }$$
Nor with $HCl$ (copper doesn't react with $HCl$)  but only with nitric acid or aqua regia you can produce hydrogen from the metallic form. So, in fact, the gas you see is carbon dioxide.

@TimTimmy the up arrow $\ce{ ^ }$ indicates that the compound become a gas and can escape from the solution the down arrow $\ce{ v }$ indicates that the compound precipitate. Have a look at the bottom here for latex reference and here for a general introduction to the symbols used in chemistry.

Just one last issue - in your last equation, copper reacts with sulphuric acid - why? I know that it is lower than hydrogen in the reactivity series, so how? Thank you very much for your time!

@TimTimmy yes in fact if you use sulphuric acid you normally add some hydrogen peroxide as oxidant agent however hot concentrate sulphuric acid is a good oxidant agent so it can dissolve copper. However is not very effective (see here for reference).

Last thing hydrogen is not involved in this reaction so  the effective oxidant agent in concentrate sulphuric acid I think is $SO_4^{-}$ that is reduce to $SO_2$, this is the point.

@TimTimmy I've found this and this take a look maybe you could find them helpful. You are welcome!Thank you!