text:
How to construct NMR spectra from chemical shift tensors?
If I know the chemical shift tensor (from ab initio methods), $\sigma$, for each atom in my structure, is it possible to construct an NMR spectra for my system? If yes, how and what mathematical implementations do I need?

Yes, it is possible. It is done often. The solution depends on the dynamics and orientation(s) of the spins in the ensemble (as will be shown bellow). I will assume you want to calculate the spectrum in frequency space.
$$
\omega_{cs} = \sigma_{zz}^{LF}(-\gamma \vec{B}_{0})
$$
We can hide a few variables and rewrite this in a much more useful way as:
$$
\omega_{CS} = (\vec{b}^t_0 \overset {\leftrightarrow}{\omega}\vec{b}_0)^{LF} = (\vec{b}^t_0 \overset {\leftrightarrow}{\omega}\vec{b}_0)^{PAS} =(\vec{b}^t_0 \overset {\leftrightarrow}{\omega}\vec{b}_0)^{RF}
$$
where,
$$
\vec{b_0} = \frac{\vec{B_0}}{B_0}.
$$
The above allows us to calculate the chemical shift in many reference frames. The frame where $\omega$ is diagonal (the PAS) is often convenient.
$$
\omega_{CS} = \omega_{xx}^{PAS}(\vec{b}^{PAS}_{0,x})^2 + \omega_{yy}^{PAS}(\vec{b}^{PAS}_{0,x})^2 + \omega_{zz}^{PAS}(\vec{b}^{PAS}_{0,x})^2
$$
We can define $\theta$ as the angle between $Z^{PAS}$ and $b_0$ and $\phi$ as the angle between $X^{PAS}$ and the x-component of the $b_0$ field. This lets us write:
$$
\omega_{CS} = \omega_{xx}^{PAS} \cos^2\phi \sin^2\theta + \omega_{yy}\sin^2\phi \sin^2\theta+ \omega_{zz}^{PAS}\cos^2\theta
$$
The above is the main equation you will need to calculate a chemical shift spectrum in frequency space. Numerically you will sum over all the angles in your ensemble. If your sample is a powder you could use angles sampled from a Gaussian distribution on a sphere for example. Typically people then apply convolution with a Lorentzian function (to account for relaxation and smooth the numerical artifacts). If your sample is a solution and the molecular reorientation time is fast compared to the inverse of the anisotropy, everything but the isotropic component will be averaged away. To put this in context (and to get to some possibly familiar equations) it will be useful to introduce some convention. In the Haeberlen-convention:
$$|\delta_{zz} - \delta_{iso}| \ge |\delta_{xx} - \delta_{iso}| \ge |\delta_{yy} - \delta_{iso}|$$
$$\delta_{iso} = \frac{1}{3}(\delta_{xx} + \delta_{yy}+ \delta_{zz})$$
$$\delta = \delta_{zz} =\delta_{iso}$$
$$\eta = (\delta_{yy}-\delta_{xx})/\delta$$
Where $\delta$ is called the anisotropy and $\eta$ is called asymmetry. Graphically:
Using this convention and some geometry you can get to:
$$
\omega _{CS} = \frac{\delta}{2} (3\cos ^2\theta -1 + \eta \sin^2 \theta \cos 2\phi) + \delta_{iso}
$$
Also, ab initio chemical shift calculations normally are reported in shielding values, you will probably want to convert to shift values using a sutable reference.
Reference:
Multidimensional Solid-State NMR and Polymers, by Klaus Schmidt-Rohr and Hans Wolfgang Spiess, chapter 2

Great answer. Indeed some software include sets of pre-calculated reference shifts (e.g., TMS) computed with common methods and basis sets. Otherwise you'll need to run a particular reference compound to report shifts that people will understand.


text:
Why is silicon oxide giant covalent structure?
Why is silicon oxide giant covalent structure? Why it doesn't bond like carbon dioxide?

Carbon dioxide is a molecule with double bonds.
Elements of 3+ rows are not happy with double bonds, they prefer single bonds, even in compounds like $\ce{H2SO4}$ 'double bonds' are actually dative single bonds.
The reason for it is that pi-orbitals for 3+ rows are very diffuse, making overall overlapping with p-orbitals of 2-rows extremely ineffective
For this reason, $\ce{SiO2}$ forms structure with only singular Si-O bonds. It happens so that for it the only way to form one is to make 3-d infinite structure.
For similar reason $\ce{N2O5}$ is exactly like this in gas phase while $\ce{P2O5}$ forms tetrahedral $\ce{P4O10}$  and $\ce{SO3}$ loves to form trimers and chains.

Well, there's carbon disulfide.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

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

sort: votes


text:
What is the coordination number for BeCl2?
I am confused, should it be 2 or 4?
If I consider $\ce{BeCl2}$ as a chain structure the coordination number is 4, and in the gas phase it is 2.

Coordination number is determined by compound. It is normal for an element to have several different CN. Preferable CN can change for an element depending on the atom's neighbors. For example, Fe has preferrable CN=6 for fluorine anions, but CN=4 for chloride anions. When heated or under extreme pressure preferable CN may change as well. For example, $SiO_2$ has high-pressure modification with 6-coordinated silicon and 3-coordinated oxygen, while at normal pressure they are 4- and 2-coordinated respectively.
In case of $Be$ it prefers, just like all elements of the 2nd row, to have CN=4, but in many cases is forced to live with smaller numbers. BTW, in gas phase, $BeCl_2$ exists in equilibrium between monomer with Be's CN=2 and dimer with Be's CN=3.
In your case CN for Be in cristalline $BeCl_2$ is 4, while in gas phase it is 3 and 2 for dimere and monomer respectively. There is no 'one CN for Be atom right for all Be compounds.'

Beryllium is the most electronegative of the Group 1 and Group 2 elements (excluding hydrogen) and so rather than lose 2 electrons (and become $\ce{Be^{+2}}$ - a lot of charge on such a small atom) it can often bond covalently.  Just like with carbon, the covalent bonding will involve mixing of beryllium's $\ce{2s}$ and $\ce{2p}$ orbitals.  The coordination number for Beryllium can vary depending upon how the orbitals mix and the what hybridization results.  In the gas phase the $\ce{BeCl2}$ monomer has a linear structure and the coordination number is 2.  This is because the molecule is $\ce{sp}$ hybridized with two empty $\ce{p}$ orbitals.  $\ce{sp}$ hybridization with two empty $\ce{p}$ orbitals requires bonding to two groups (coordination number = 2) in a linear arrangement. In the solid phase, where other $\ce{BeCl2}$ molecules are close, the chlorine atom from one $\ce{BeCl2}$ molecule can also form a dative bond to an adjacent beryllium atom.  In this case the molecules can rehybridize to $\ce{sp^3}$ and form 4 bonds (coordination number = 4) arranged along the vertices of a tetrahedron.  In this case the formerly empty $\ce{p}$ orbitals have become $\ce{sp^3}$ orbitals and share the lone pair of electrons contributed from a chlorine atom on an adjacent $\ce{BeCl2}$ molecule.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
How is square planar geometry possible in metal complexes?
I came across this statement and I don't understand what it meant by it:

"Square planar geometry is favoured by ligands that can form pi bonds by accepting electrons from metal atoms or ions."

How is this possible? I mean a ligand, itself, donates electrons. How is it supposed to be accepting electrons?

Maybe I'll break your world apart, but back-donation is actually pretty common. There are actually systems where it dominates, like antimony/naphthalene adducts. Back-donation is what participates to extreme stability of transition metal carbonyls. The key is, that not only transition metal has vacant orbitals, but the ligand have too, and in some cases even low-energy ones, like $CO$. Such orbitals can accept electron density from filled orbitals of the central atom, say $d_{z^2}$ in square planar complexes. Orbitals for back-donation are usually $\pi^*$ orbitals, and the interaction commonly found in phosphines, alkenes and arenes, especially if the have substituents with significant $-I$ effect

is it due to achieve greater stability or what?

@DSinghvi Extend your question please

What does it mean to extend question @permeakra

@DSinghvi Ergh... specify please the sentence of my post you have trouble with.

why does it accept electron ?

@DSinghvi the simplest level to think about it is "because the ligand have low-level vacant orbital". To get a good grip on the topic you should become familiar with concepts http://en.wikipedia.org/wiki/Molecular_orbital_theory and http://en.wikipedia.org/wiki/Molecular_orbital_diagram . Next, try this article http://en.wikipedia.org/wiki/Pi_backbonding


text:
Questions about combustion and isotopic abundance

the combustion of 0.214g of sulfur provided exactly 0.428g of sulfur dioxide. After a second try, 0.782g of SO2 was obtained. What mass of sulfur was used?
The combustion of lithium forms lithium oxide. Combustion involving a 1.451g sample of oxygen results in obtaining 2.710g of Li2O. What mass of lithium was involved in this reaction?
Natural argon has three isotopes: Ar-36 (35.967u), Ar-38 (37.962u), and Ar-40 (39.962u). if the isotopic abundance of Ar-36 is 0.006%, find the isotopic abundance of the other two

After pulling my hair out:

I still have no idea how to do this question.
I balanced the equation $4Li + O_{2} -> 2Li_{2}O$ then found the moles of both oxygen and lithium oxide, which was 0.090688 mols and 0.090333 mols. Used the smaller number of moles, set up the ratios $\frac{4Li}{2Li_{2}O} = \frac{x}{0.090333mols}$
x = 0.1807 mols, which equals to 1.2647g of Li
Is this correct?
35.967(0.00006) + 37.962(x) + 39.962(1-x) = 39.948
0.00215802 + 37.962x + 39.962 - 39.962x = 39.948
-2x = -0.01615802
x = 0.00807901
which mean Ar-38 is 0.8%, but the answer clearly says it is 0.588%

StackExchange should not do your homework for you, as that would be cheating.

@EricBrown First, this is not homework, they are just extra practices for the exam. Second, I don't know how to do this and there is nothing wrong with asking for help. There are more questions that goes in this format, I only asked one of each type of question so that I can apply the same method on the other questions. If I am intended to have people do my homework for me, I would have post all of them up.

Great.  I'm sure that you can understand why I made my previous statement, as there are two camps of folks who make answers on SE -- the very few who will answer homework questions, and those who will with some demonstration of effort by the person who asks the questions.

So 'demonstration of effort' means something like... showing what I got up to and let people correct my mistakes? If so, then sure I'll do that.

I checked a couple of references and they say the natural abundance of Ar-36 is 0.334%; Ar-38 is around 0.06%

Yes the answers at the back says  Ar-38 is 0.588% and Ar-40 is 99.406% But I have trouble getting these numbers.

I don't get those numbers either.  In the first question, write the equation for the burning (oxidation) of sulfur in oxygen and you'll see that 1 mole of sulfur generates 1 mole of sulfur dioxide.  Convert the masses of sulfur and sulfur dioxide to moles and you'll see that 0.007 moles of sulfur produced 0.007 moles of sulfur dioxide.  Then they did a second experiment and obtained 0.782 gm sulfur dioxide.  How many moles is that; so then how many moles of sulfur did they start with, and how many grams of sulfur is that?

whoops,  wouldn't 3 involve (1-.00006-x)?

BTW, in problem 3, because you are dealing with such widely different isotope natural abundances (e.g Ar-38 = 0.006% while Ar-40 = 99.6%) it is critical to use isotope masses out to at least 6 decimal places.  Otherwise your answer will be way off.

omg that's right I completely ignored the whole 0.00006 thing. Should be 1-0.00006-x. And I tried to keep a lot of decimal places too but the problem is that, usually during the exam, at the top of the page it would say something like 2 marks for keeping all significant figures. Sometimes there isn't any choice of how many decimal places I can keep. Anyway, thanks for the help.

$\frac{0.214g}{0.428g} = \frac {x}{0.782g}$
Your answer is ok, but on an exam you want to solve quickly.
2.710g - 1.451g = ?

wasn't that easier?
3.

35.967(0.00006) + 37.962(x) + 39.962(1-x) = 39.948

This is slightly wrong.
0.00006, x, 0.99994-x


text:
Does the carboxyl group count as a branched chain
Do carboxylic acids (or any compound with a carbonyl group) count as a branch chained organic compound due to the C=O? Or does it need a carbon - carbon branch?

In organic chemistry, the terminology of branched vs. unbranched is generally used to refer to the structure of the hydrocarbon skeleton of a molecule. Hence, heteroatomic functional groups are usually not relevant. The structure in your question is therefore unbranched. For an authoritative reference, see the relevant sections of the IUPAC blue book.


text:
What is true about this equation, at equilibrium, and why?
${H}_2O(s)\leftrightharpoons{H}_2O(l)$
A) the mass of ice must equal to the mass of water
B) the mass water and the mass of ice each remain constant

At equilibrium, the amount of reactant being converted to product is equal to the amount of product being converted to reactant.  Therefor, the concentration of both the reactant and product are no longer changing and the ratio of these no longer changing concentrations can be represented as a fixed number or constant - the equilibrium constant.  $$\mathrm{K_{eq}=\frac{[Product]}{[Reactant]}}$$ where the brackets denote concentrations.  In the case at hand, a liquid-solid equilibrium, the expression becomes (thanks to DavePhD for pointing this out, see his UC Davis link in the comments if you'd like more info.) $$\ce{K_{eq}=[H2O~ (l)]}$$  This equation tells us that at equilibrium, the concentration of liquid water is a constant, it is not changing.  If the concentration of liquid water is not changing, then the concentration of ice cannot be changing.  This does not say that the concentration (or mass, since concentration is related to mass divided by molecular weight per unit volume) of liquid water must equal the concentration or mass of ice; it just says that the concentrations or masses of liquid water and ice must be constant.  Hence "B" is the correct choice.

Does that definition of equillibrium constant apply to a two phase system?  Whether I have a ton of ice and a kg of liquid water or a ton of liquid water and a kg of ice, they are still in equillibrium as long as they are at the freezing point.

Yes the definition still applies, but in a two-phase system not all parts of the system are in equilibrium.  In your example, only the ice that is in contact with the water could eventually be in equilibrium with the water.

Isn't that the formula for density?

you mean where I said, "mass divided by molecular weight per unit volume"? density is just mass per unit volume

ron, "A heterogeneous equilibrium is more complicated than homogeneous examples discussed above and has species present in more than one phase. Writing an expression for K for a heterogeneous equilibrium is similar to homogeneous reactions, with the important difference that you don't include any term for a solid in the equilibrium expression." http://chemwiki.ucdavis.edu/Physical_Chemistry/Equilibria/Chemical_Equilibria/The_Equilibrium_Constant/Calculating_An_Equilibrium_Concentration_From_An_Equilibrium_Constant/Writing_Equilibrium_Constant_Expressions_involving_solids_and_liquids

Isn't concentration either molarity and morality?

@DavePhD  Thanks for the reference, I've edited my answer accordingly.

@user1.618  "Morality"?  Freudian slip :), but I understand.  Yes, concentration, for example molarity, is (gms/MW)/volume, as in moles per liter

So which term does your description fit into?

Mass divided by molecular weight per unit volume does not match with any of those two terms?!

Mass/Molecular Weight = moles, right?  Moles/volume in liters of solution = molarity, right?  What am I missing?

If the ice and liquid water are in equillibrium, and no heat is added to or removed from the system, B) the mass of water and the mass of ice each remain constant


text:
Why is Technetium unstable?
Technetium is probably the most awkward element of the periodic table for me. It seems to me that Technetium is an exception, in the sense that it doesn't have any stable isotopes, despite having a relatively low atomic number when compared to other unstable elements.
So I thought, maybe there is something special about it. Some exceptional property that causes it to be so unstable with such a low atomic number.
Please tell me. What is it that makes Technetium so unstable?

duplicate of http://physics.stackexchange.com/questions/40960/why-is-technetium-unstable

If you google "why is technetium unstable" you'll see that this question has been around for a while.  Over on physics.stackexchange, it is discussed in Why is technetium unstable?.
I've read some of what others have posted on this topic and the short answer appears to be: because we have such a poor understanding of the atomic nucleus, there is no answer to this question at this time.  Again, the link above would be a good place to start.  One of the answers (3 upvotes) claims to be an answer.  The selected answer (10 upvotes) says there is no answer.


text:
Esterification and dehydration synthesis
I know there are many more terms out there like polymerization, condensation, etc. But the important thing is whether they can be differentiated or not.
In Esterification, which I just learned, needs to have a dehydrating agent when involving bad leaving group, such as $\ce{-OH}$.They are typically Brønsted-Lowry acids. Correct me if I'm wrong.
How does heterolysis work in the case of $\ce{CH3COOC2H5}$. [the reverse of the esterification of ethanoic acid and ethanol]

You must mention the medium in which Esterification/Hydrolysis is happening : Acidic/Basic.

The reverse of esterification is called "hydrolysis."  The hydrolysis reaction can be performed under acidic or basic conditions.  This link shows the mechanisms involved for both the acidic and basic hydrolysis of an ester.  Each step in the reaction is explained in detail and accompanied by a picture.


text:
Why are many silver salts insoluble?
For example, silver chloride, silver sulfide, and silver oxide are three exceedingly stable silver salts in water solution.
Why is it that silver forms such stable salts?
Does this have to do with charge density of the component ions? This doesn't seem to hold water because we know that silver oxide is much more soluble than silver sulfide, and the oxide anion is obviously a lot smaller than the sulfide ion, so the oxide ion should have a much higher negative charge density.
So simple charge density rationalizations go out the window. I've Googled, I've looked in books, etc., but I can't find an explanation.
On the other hand, can a Lewis acid/base model be used to rationalize the stability of the silver compounds? For example, the sulfur, by nature of being less electronegative than oxygen, is a better Lewis base/electron pair donor/nucleophile; the sulfide ion is more willing to "donate" its electrons. Also the sulfide ion should be bigger and has empty 3d valence orbitals which may or may not be accessible to a great extent. Nonetheless, metals, by nature of possessing a large number of valence d-electrons, may be able to increase the stability of the sulfide ion by populating its d-valence.
This Lewis acid/base model seems to get a lot of mileage. I also know that silver is a good Lewis acid, as are many metal cations, because of the poor shielding afforded by the d electrons. Also, the silver cation has an empty 5s orbital. Thus, we can form a sulfide-to-cation sigma coordinate covalent bond.
We may also form a pi-type coordinate covalent bond as the far-away d-electron density "drifts" to empty 3d valence orbitals on the sulfide anion, making the sulfide anion now more nucleophilic than ever.
Let me know if my rationalizations are incorrect or if there are better explanations of silver salt stability.

This is just an unproven guess on my part.  Whether compounds are soluble or not depends on 1) how strong the bonding (including lattice) forces are that hold the compound together (e.g. how much energy do we have to put in to liberate the ions, and 2) how much energy is recovered through solvation of the resulting ions.  I don't see anything odd about the ionic radius of the silver ion so I'm guessing that the energy of solvation of the resultant ions would be fairly normal; so it is probably not factor "2" that makes most silver salts insoluble.  Here is a link to an article that suggests that bonding in silver compounds may have a covalent component, it says, for silver salts "the structure has some contribution from covalent bonding. This appears to be bonding the structure together more tightly, as in a giant covalent structure."  The words "more tightly" catch my eye.  This seems to suggest that factor "1" may be in play in that silver salts may have increased bond\lattice energies.  If true, this would help explain the insolubility of many silver salts.  Finally I'll add that some of the soluble silver salts ($\ce{AgF, AgNO3, Ag(CH3CO2)}$) seem to involve anions that would be expected to be strongly hydrated.  The extra energy generated through solvation would further assist in solubilizing these silver salts.

Interesting comment about the covalent nature of the bonding; would this have anything to do the d-electron density "shifting" from the silver to the sulfur? Regarding solvation effects, I know that the fluoride ion would be solvated by water quite nicely, but what about the oxide or even sulfide ions? Both are much, much stronger bases than the fluoride ion in water solution.

I think atomic radii may be more important than basicity.  A higher charge density (as would result on a smaller ion) is probably an important factor in terms of solvation energy.  This argument is used to explain decreasing solubility down the halide salt series.  Dunno what role  d-electrons play in the sulfur case.

Perhaps we can go back to Bronsted acidity/basicity; consider the H+ ion, O(2-), and S(2-). The H+ ion can only form a sigma bond. S(2-) is a stronger base than O(2-). This implies that S(2-) donating an electron pair is more thermodynamically favorable than O(2-) donating an electron pair. Perhaps this can account for why Ag2S is much more stable than Ag2O. As you mention, there is some covalent character. Perhaps sigma bond strength is one factor.

@Dissenter Partial covalent bonding exists even in CsI, it's part usually is several tens of percents even in strongly ionic compounds. As for silver, the atom has high elecronegativity, making formation of $Ag^+$ very unfavorable.

What is Csl? Also silver has an EN of 2.0, not exactly high.

@Dissenter, $\ce{CsI}$ = caesium iodide. I think the point is even highly electropositive elements exhibit some covalent bonding. As for the EN of $\ce{Ag}$, 2.0 is high; only a handful of trans. metals surpass it and, ceteris paribus, that may favor covalent interactions. Finally, I'm fairly sure sulfide is not a stronger base than oxide; the greater polarizability of sulfur allows it to stabilize a negative charge more effectively. Hence, $\ce{H2S}$ is a stronger acid than $\ce{H2O}$. As for $\ce{Ag2O}$ vs. $\ce{Ag2S}$, I would personally look to HSAB theory, even though it may be handwavy.


text:
Meaning of depiction of atomic orbitals
There was a section about atomic orbitals in my organic chemistry textbook that I did not quite understand. First the author explained a bit about the Schrödinger equation, which I understood to be (vaguely, they did not go to any detail, even omitting the formula all together) a equation that describes the likeliness of positions of an electron in 3d space. $\Psi$ (i.e. the Schrödinger equation) can be solved for different quantum numbers that respectively represent the different orbitals.
This made me wonder what a picture (like below) of an orbital actually depicts, as it are chances that are described by $\Psi$, not boundaries. My best guess was that such picture represents an area that has a high chance, say 90% or more, of containing the electron (how they determine the shape, I don't know, but my best guess again here is that they only take those parts of the real orbital that add a significant percentage to the total chance of the electron actually being in the depicted orbital).
Are all these assumptions right?

Without going into a long discussion about what the wave function is, I will try to briefly answer your questions:
1) For atoms, the (approximate) solution to $\Psi$ is the product of two functions:

the radial components (distance from the nucleus)
the angular/spherical components (behavior going around the nucleus, like latitude and longitude)

2) Since the radial components of each component extend quite far, decaying to zero only at infinity, it would be very difficult to make a plot of the 3D object on a 2D surface that does not cover up the features that would be nearer the nucleus (further from eye), without using varying degrees of transparency.
The spherical component is usually solved analytically using complex numbers (i.e. a+b*i) but they can be transformed into real values by linear combinations. So, instead of two "channels" for real/imaginary, we can use only one "channel" (real) for each function, 1s, 3dxy, etc.  Often, these are colored so that if the sign of the real function is positive, it is e.g. red, otherwise, e.g. blue.
A second feature of these types of graphics is called the "isosurface"--which means, plot only if the value of the function is equal to a chosen value.  Sometimes the chosen value is taken to be "90% of the corresponding electron density contribution from this atomic orbital is enclosed" (which, b.t.w. is $\Psi^* \Psi$ or simply $\Psi^2$.  Though $\Psi$ may be real/complex/negative/positive valued (hence need all sorts of colors and "channels" to represent it) the electron density is positive and real.  This is very important, as complex quantities are not observable, but are rather mathematical constructs (vehicles) for arriving at things that are actually observable, such as the electron density.
But practically speaking, in order to see certain features like the doughnut in $dz^2$, people play a little fast and loose with precisely what the value of the isosurface is.
The shading on the surfaces that you have provided could be misinterpreted, because the colors do not correspond in any way to the value of the function being plotted, but rather seem to be an aid to the volumetric rendering of an isosurface that has already been set to either the absolute value or the square of the atomic orbital.
Finally, note that in your center drawing, which adds all of the individual atomic orbitals, that you are beginning to form a sphere.  It had better, as atoms are, in fact, spherical!${}^{\dagger}$  The formal name of the functions are spherical harmonics, and the set of e.g. $p_x$, $p_y$, $p_z$ or the five $d$'s taken together form a spherical shape.
${}^{\dagger}$ assuming the nuclear quadrupole is zero

In some texts I have seen one lobe shaded and one lobe unshaded; according to these texts the shading differentiates between the two signs of the wave function (positive and negative). Not sure how much that helps when just about everything else regarding the wave function is ignored.

@Dissenter When you plot a function is polar or spherical coordinates, as $r = f(\theta)$, you can not distinguish f being positive from f being negative.  Consider the $d_{z^2}$ orbital as $3cos^2\theta-1$, the negative values are the donut part, the remainder is positive.

Right; spherical orbitals only have one lobe.

What I understand of orbitals is simply that ELECTRONS can not be pin-pointed accurately, because they are in continous motion and that too not in a fixed circular path (as Bohr had depicted), so it is virtually impossible to know where an electron is in an atom. So scientist found a way to counter this problem by providing another model - THE QUANTUM MECHANICAL MODEL of the atom, where instead of electrons residing in orbits, it was stated that electrons have a possibility that it may reside in a particular region - what we call as orbitals. It is quite intresting to note that the probality of finding an electron never becomes 0, which means an atom in USA can have its electrons in Australia (atleast in theory).


text:
Calculating Avogadro's Number
For a school project, I am in a group tasked with calculating Avogadro's Number in multiple manners.  Other than the Millikan Oil Drop Experiment, we know of no ways to compute it.  A Google search yielded no methods that could be of use (either complex beyond the high school level, or through using a mole of materials, which defeats the point of deriving the number).  What methods can we (intelligent high schoolers on a limited budget) use to compute Avogadro's Number?

You can use Faraday's electrolysis method. Find the amount of electricity in couloumbs required for the electrodeposition of one equivalent of any substance, and then divide the amount by the charge on one electron. This gives you the avogadro's number

Thanks!  Can you please explain in more detail or link me to a website that explains in more detail?

Placing a drop of oleic acid on the surface of water contained in a pan is a relatively straightforward experiment that has been widely used to determine Avogadro's number.  Here are two links that provide good background on a) how to run the experiment, b) how to analyze the data, and 3) the science behind the experiment.  The two links are just variations on the same experiment, but if you read them both it will really enhance your understanding of the experiment.
link 1
link 2

Here is a crude method of performing the experiment:-

Take two copper electrodes, weigh them, and dip them in $\ce{CuSO4}$ solution
Use a constant current source or a fixed-EMF battery.
Connect an ammeter in series with the battery and the electrolytic setup
Note down the time of closing the switch. Pass a current for a fixed time and also note down the current magnitude.
After say some time $t$ of passing the current $I$, open the switch and remove the cathode and weigh it.
Say the increase in mass of the cathode is $m$. The equivalent weight of copper is $63.54/2$, hence the equivalents of copper deposited is $2m/63.5$.
The amount of electricity passed is $It$. This must be equal to one $2m/63.5$ equivalents of electrons. Since the charge on one electron is $\approx 1.6\times 10^{-19}=e$, the number of electrons that passed is $It/e%$. Therefore the number of electrons in one equivalent or one mole, (which is the avogadro's number) will be $(It/e)/(2m/63.5)$.

You will have to make appropriate arrangements for accurate electrolysis to ensure correct weight measurement.

One issue with this approach is that it requires the charge of the electron which is determined by the Millikan Oil Drop Experiment.


text:
Why does diamond conduct heat better than graphite?
Our teachers made us accept (without any explanation) that diamond conducts heat better than graphite. What is the reason behind this (alleged) fact?

Diamond is one of the best thermal conductors known, in fact diamond is a better thermal conductor than many metals (thermal conductivity (W/m-K): aluminum=237, copper=401, diamond=895).  The carbon atoms in diamond are $\ce{sp^3}$ hybridized and every carbon is bonded to 4 other carbon atoms located at the vertices of a tetrahedron.  Hence the bonding in diamond is a uniform, continuous 3-dimensional network of $\ce{C-C}$ single (sigma) bonds.  Graphite on the other hand is formed from $\ce{sp^2}$ hybridized carbon atoms that form a continuous 2-dimensional sigma and pi bonding network.  This 2-dimensional network forms sheets of graphite, but there is little connection between the sheets, in fact, the sheet-sheet separation is a whopping ~3.4 angstroms.  This might lead us to suspect that heat conduction in the 2-dimensional sheet of graphite would be superior to diamond, but that heat conduction between graphite sheets would be very low.  This is, in fact, an accurate description of thermal conduction in graphite.  Thermal conductivity parallel to the graphite sheets=1950, but thermal conduction perpendicular to the sheet=5.7.  Therefor, when we consider thermal conduction over all possible directions (anisotropic) diamond would be superior to graphite.

it is an even better thermal conductor than silver?

Yes, even better than silver which has a thermal conductivity of 429 W/m-K.

Great answer, but I'd like to provoke: How come is graphite so bad at conducting heat, but considerably good at conducting elctricity, while diamond is considered an insulator. In other words, why is it the opposite, since it metals this two properties are generally proportional?

@Molx  That would make a nice question!  I think basically thermal and electrical conduction differ in mechanism.  Thermal conduction involves phonons and lattice vibrations, while electrical conduction involves electrons and band gaps.  Not too surprising that different substances respond differently.

Ron's answer is great, but I'd just like to touch on the mechanisms behind thermal conductivity so we can rationalize the differences between the behaviour of diamond, graphite, and metals:
There are two ways in which heat is transmitted through solids: phonons and electronic conductivity. The latter occurs in electrically conductive solids, where conduction band electrons are free to move throughout the structure, carrying thermal energy along with them. This is a significant component to the thermal conductivity of metals and explains some of the in-plane conductivity of graphite. Electrically conductive solids tend to also be good thermal conductors for this reason, but this mode of conduction is not accessible to diamond.
Phonons are very complicated, but conceptually, phonons are waves of vibrations in the lattice of a solid. (quantized vibrational modes, actually, but not important for this) Vibrational energy in one part of a solid will be transmitted to other parts of the solid as phonons. What limits how quickly heat is transferred by this mechanism is phonon scattering. Phonons can scatter on many different things: crystal defects, impurities, crystal boundaries, even other phonons. This is why crystalline substances (like diamond) tend to have higher thermal conductivity than amorphous ones (like glass or polymers) as the phonons scatter more when they don't have a nice ordered crystal structure to travel through.
The reason why diamond, in particular, is an especially good thermal conductor even compared to other well-ordered crystals boils down to two factors: the mass of the carbon atoms and the strength of the bonds connecting them. I don't want to get too much into the details, but briefly, much as a harmonic oscillator has a higher frequency with a stronger spring (the strength of the C-C bonds) and a lighter mass (the C atoms), diamond can support higher energy phonons. This means that at a given temperature, fewer phonons near the material's limit will be present in diamond than a material with a smaller limit. The details aren't important, but when phonons near the limit interact, they scatter in such a way that the phonons don't travel through the structure. Because diamond's limit is so high, the probability of this type of scattering is lower.
Phonons and electrons travel very well along graphite's graphene sheets, but poorly between them, due to weak inter-layer interactions and the large distance between layers, explaining the anisotropy of its thermal and electronic conductivities.
Metals exhibit both modes of thermal conduction in varying degrees of importance and conduct isotropically since they lack the layered structure of graphite.

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


text:
Variation of Arrhenius equation with concentration
In the Arrhenius equation do Arrhenius constant and activation energy depend on concentration ? (Their units do)

To my knowledge the activation energy does not depend on concentration while the preexponential factor will depend on the concentration. The preexponential factor is a measure for the probability of two particles colliding with the proper orientation to facilitate the reaction so a higher concentration of particles will lead to a higher preexponential factor.

The Arrhenius equation gives you a rate constant $k$.
For a second order reaction, $k$ has units of $1/Ms$.
For a first order reaction, $k$ has units of $1/s$.
The Arrhenius prefactor $A$ has the same units as $k$.
The rate, as opposed to the rate constant will depend upon concentration. For the first order reaction, the rate constant is multiplied by one concentration to get the rate, and for a second order reaction the rate constant is multiplied by two concentrations to get the rate.
If $k$ and $A$ do depend upon concentration, that would imply that the reaction is not accurately modeled by the rate constant (for example not really first order when you are modeling as first order or not really second order when you are modeling as second order).
As far as activation energy $E_a$, the equation can be written in terms of energy per mole divided by $RT$ or energy per molecule divided by $k_BT$ (where $k_B$ is Boltzmann constant).  $E_a$ does not depend upon concentration.

You said than activation energy is per mole doesn't that imply it relies on concentration

no, it means that $E_a$ is the amount of energy needed for 1 mole, regardless of how many moles there are in an experiment and regardless of how many moles per liter (concentration) are present.

in other words, the fact that it is energy per mole, means that is does NOT depend upon the number of moles.  Energy is an extensive variable and depends upon amount of a substance, "energy per mole" is an intensive variable and does not depend upon the amount of substance.


text:
How is Melatonin Synthesised?
I should like to know roughly how the hormone melotonin is synthetically made for the pharmaceutical industry (i.e. not how it is made in living beings). I can't seem to find any reference to its manufacture on the web, it is seen to have a fairly simple structure on its Wiki page and biologically it is synthesised from the forerunner amino acid tryptophan. So I believe it is likely to be readily synthesised for use in pharmaceuticals without the use of animal products, but I cannot confirm this.
Specifically, a close relative of mine has been prescribed melotonin and I want to confirm that this substance is not made from animal products, as I live in a country where the authorities in their great wisdom have deregulated the food and meat processing industries. Astoundingly, they did this at exactly the same time that a UK royal commission recommended re-regulation of the same to prevent contamination of food and pharmaceuticals by the Creutzfeld-Jacob prion.

I don't wish to be argumentative, but even if someone can provide details of a fully synthetic industrial production route, what guarantee does that provide that the manufacturer you're specifically dealing with uses that route?

@GregE. None, just gathering evidence to get me one step from a state of complete to partial ignorance: an answer of how it is generally done changes conditional probabilities

Fair enough. You may find these useful: 1, 2. The first is an overview of some synthetic strategies, the second is an EMA report on Circadin (a sustained-release melatonin formulation) which describes a synthesis starting from 5-methoxytryptamine. To my (admittedly limited) knowledge, amino acids (and relatives) are rarely produced by extraction methods these days, but rather by enzymatic and semi-synthetic routes.

There's also another document pertaining to Circadin, this one from the Australian Dept. of Health, describing the same synthetic method. If you do a search on tryptamine production, you'll find bacterial and enzymatic routes described, which (to my knowledge) is in line with trends for production of these sorts of compounds. I haven't found a source describing the production of the 5-methoxy derivative specifically, but the synthesis from tryptamine would be a simple one, and I find it unlikely that extraction methods would be used.

@GregE. I think whenever you like, you might post an answer to this....

First, if it is a supplement from some natural product store, it is most probably not made by a chemist, and probably not even subjected to quality control and might not even contain melatonin. Second, if it is a prescribed it is made according to the Good Manufacturing Practice mandatory for all prescription drugs, the source will be indistinguishable and thus the effect. Third, since synthetic drugs are always made from an advanced intermediate found in nature or refined petroleum products which might be some dead dinosaurs there is a strong probability that it will be made from an animal.

I do not know the exact process in industry, but these are the steps that I thought of (someone likely thought of this before me, it is a very simple synthesis).  The main synthesis of the ring is done in Fischer indole conditions.  Therefore I broke up the molecule into a compound known as 4-acetamidobutanal and p-hydrazinoanisole.  The first compound is an amide that is available from the reaction of 4-aminobutanal and acetyl chloride or acetic anhydride in the presence of a base.  The latter compound is available from the reaction of p-anisidine with sodium nitrite and hydrochloric acid to form the diazonium chloride, and then reduction with sodium sulfite and sodium hydroxide.  This is just an example of a synthesis.  The overall synthesis shouldn't be more than 5 or 6 steps.
I found an article highlighting the syntheses, you can take a look.  Apparently the synthesis that I described above is very similar to the most effective synthesis, a one-pot synthesis done in very similar conditions and having similar starting materials.

authors: محمد سعيد عبده العبيدي


text:
Effect of optical activity
I know that optical activity means polarisation of light but what is so special in that that it changes the properties of the substances significantly

Not sure I understand your question, could you add a little more to it?

Actually, the differences in molecular polarisability indicate differences in the 3-dimensional orientation of the molecule, which is significant enough to alter its properties.

I mean that how can rotation of light change the basic properties of substances

The basic properties of substance determine optical activity not viceversa.

TL;DR version: Chirality is a basic property of a molecule, while optical activity is a bulk physical property possessed by samples of chiral compounds. Specific rotation is an intrinsic property of chiral compounds and a measure of optical activity under standardized conditions. Optical rotation is not responsible for the disparate behavior of enantiomers of biologically active compounds.
Chirality is an intrinsic property of a molecule (or, indeed, of certain geometric structures, more generally). A chiral molecule is one that is not superposable on its mirror image molecule (its enantiomer), meaning that no series of rotations or translations will make the two geometrically identical in three-dimensional space. One of the consequences of chirality is that chiral molecules in bulk will measurably rotate plane-polarized light by a specific amount and in a specific direction. (The distinction here is crucial, since any molecule[s] will in general cause some rotation, but the random distribution and spatial orientation of the large number of molecules in any given sample will produce no observable net rotation if the molecules are achiral. The rotation caused by one molecule is likely to be negated by another identical molecule in an opposite spatial orientation.) The key point is that chirality is an intrinsic property of a molecule, while optical activity is a bulk physical property of said molecules gathered en masse. Specific rotation, on the other hand, can be viewed as an intrinsic property, since it is defined for a specific sample under an exact set of conditions (light wavelength, path length, concentration [for samples in solution], density [for pure liquid samples], temperature).
Optical activity is not the cause of any changes in the properties of chiral compounds, nor is it responsible for the sometimes wildly different biological effects of certain enantiomers (which I presume may be what you're actually referring to, forgive me if I'm wrong). The enantiomers of chiral molecules can often differ widely in their biological effects due to the basic nature of proteins, which are the constituent biological molecules that most if not all biologically active substances target in one way or another. Proteins are constructed from amino acids, which often themselves contain multiple chiral centers, and these proteins go on to fold into very specific and complex conformations. Enzyme active sites and cell receptors (among various other biological structures) are mostly comprised of proteins with very specific three-dimensional structures. As such, they interact significantly only with molecules that have the appropriate shape. Hence, enantiomers of chiral substances, which differ in their three-dimensional shape, will exhibit variable degrees and types of interactions with proteins, and can hence impact biological targets in disparate ways. This is largely what makes chirality so important to chemists, and partly explains why so much of the most respected (not to mention economically important) cutting-edge work recently (at least in organic chemistry) has revolved around enantioselective or enantiospecific synthesis techniques.

Optical activity is the rotation of the plane of polarized light. It does not change the substances at all; rather, it is a property of the substance itself.
It's special because compounds that rotate the plane of light can either rotate it to the right (+) or to the left (-) and that's measurable! Enantiomers can be differentiated in this way, since they will rotate the light by exactly the same amount but in opposite directions.


text:
Rate law of the iodine clock reaction
I am dealing with the iodine clock reaction, which has the half-reactions:
$$\begin{alignat}{2}\ce{H2O2 (aq) + 2I- (aq) &-> I2 (s) + 2 OH- (aq)} &&\quad (\text{slow}) \\ \ce{I2  (s) + 2S2O3^2- (aq) &-> 2I- (aq) + S4O6^2- (aq)} &&\quad  (\text{fast}) \end{alignat}$$
Now, I have to find the rate law. What I did was 4 solutions – one with standard amounts, another with double the $\ce{I-}$, one with double the $\ce{H2O2}$, and one with double the $\ce{S2O3^2-}$. I recorded the amount of time for the ppt ($\ce{I2}$) to appear. Now I have to calculate the rate law.
The thing is, I am not quite sure how. Usually you take the "slow" step and set up an equation based on that, like $\text{rate}=\ce{[H2O2]$^x$ [I- ]$^y$}$. However, the reaction is not just dependent on those chemicals, it's also dependent on the $\ce{S2O3^2-}$. When it was doubled, the reaction was significantly slower (for obvious reasons).
How do I set up the rate law for this reaction? Or, if I can't, what else can I do with my data?

IMO, the only reaction that you are studying is the first reaction:  the conversion of $I^-$ to $I_2$ by hydrogen peroxide.  The thiosulphate reaction is simply a dramatic way to indicate when a particular amount of iodine has been produced.
The reaction was not slower when you doubled the amount of thiosulphate.  Twice as much iodine needed to be produced (and reconverted to iodide ion) before the thiosulphate was consumed and the first permanent iodine changed the starch indicator blue.

Although, doesn't the fact that $\ce{[I^{-}]}$ stays constant as a result of the fast reaction mean that it has an effect (not necessarily slower) on the reaction rate?


text:
Why is the exo product formed in the Diels–Alder reaction between furan and maleic anhydride?
For most Diels–Alder reactions, the major product is endo because there are favourable interactions between the newly forming pi bond and the electron withdrawing groups of the dienophile.
Why is the reaction between furan and maleic anhydride an exception? From what I understand, the carbonyl groups of the anhydride are electron withdrawing.

Great question!  It turns out that the rate of formation of the "expected" endo product is actually ~500 times faster than the rate of formation of the exo product. However, the Diels–Alder is a reversible reaction. In this case, the exo product is thermodynamically favored over the endo product by about $\pu{1.9 kcal/mol}$.
So, even though the predicted endo product is the kinetically favored product (and formed first), thermodynamics eventually takes over and the more stable exo adduct is what you wind up with. More details are given in J. Org. Chem. 1978, 43 (3), 518. What a nice example of competing kinetic and thermodynamic pathways.

Additionally, if this reaction is done "neat" (without any additional solvent), the exo product is less soluble (even at the higher temperature). As the exo product crystallizes, Le Châtelier drags the equilibrium toward exo.


text:
Behavior of NaCl in water with electrodes
If I had some NaCl dissolved in water, it is conductive, so if I were to put two electrodes in the container (powered), what is causing the flow of electrons?
Do the NaCl ionic bonds break and the ionic atoms collect at the oppositely charged electrode (how I'd assume electroplating works) with all of the atoms themselves moving to carry current, or are electrons loose in a "sea" like a metallic bond, with some of the atoms being pulled along with the flow?

Good question. I'd want to know the answer too. I remember in an exam question my professor asked about the reaction of equivalent concentrations of a strong acid and a strong base; he asked what interesting property could we expect from the water even though it is neutral. The answer was solution conductivity (there are still counterions left in the system). However, we never went over the nature of conductivity and how exactly these electrolytes convey electrons. I know that for there to be current there must be mobile electrons or mobile ions, but what exactly do these ions/electrolytes do?

$\ce{NaCl}$ once in water dissociates, producing $\ce{Na+}$ and $\ce{Cl-}$ ions. Note: this process is possible only because immediately after dissociation ions are solvated by water, i.e. water molecules bind to ions

This is releasing energy and increasing effective size of ion, decreasing ion charge density. Once relatively free ions are produced, they can move between electrodes, transferring charge between electrodes. On electrodes ions may discharge taking/giving up electrons, and this is what usually happens with $\ce{Cl-}$ ion. However, if this is not an option, as the expected product is no stable enough in water, water molecule may react instead, producing another ion and either hydrogen or oxygen. This is what happens with $\ce{Na+}$: $\ce{Na}$ is too active to be stable in contact with water, so instead of sodium, hydrogen and $\ce{{}^{-}OH}$ are produced instead, resulting in  net reaction (assuming cold solution) $\ce{2NaCl + 2 H2O <=> 2 NaOH + Cl2 + H2}$ with chlorine and hydrogen released on different electrodes.


text:
Scattering of light
How does the path of light become visible due to scattering of light?
According to me it only indicates the presence of colloidal particles, but my book gives the example of a cinema hall projector and says that you could see the path of light due to its scattering. Nevertheless, it's possible to see the lights path without the scattering.

Light can only be seen when it enters your eye, so it somehow has to find a way between the light source and your cornea. Since light travels in more or less straight lines, it somehow has to change direction if you aren't looking into the source of said light directly.
This process is called scattering. It's what makes the path of light visible, because at every step on this path, the light gets scattered. This of course leads to a loss of intensity at the other end of the light path compared to a non-scattering medium.

There are many types of scattering what you are talking about is generically called Tyndall effect. You can "see" the path of the light because some of the photons from the source are scattered in others direction so so they can reach your eyes. But in fact what you see are the photons that don't travel in straight trajectory, so not the beam itself but the path of the beam due to the photons that change direction due the effects of particles suspended in the medium.
Actually you can't see the light path without these effects...


text:
Why does increasing number of salt bridges increase voltage of electrochemical battery?
In an experiment, I set up a cell with lead nitrate (w/ lead electrode) and zinc sulfate (w/ zinc electrode), with a salt bridge containing potassium nitrate.
I observed that by increasing the number of salt bridges, the voltage measured increased. However, given what I know about electrochemical cells, I can't work out why. I initially thought it would have no effect on the voltage/potential difference.

I suspect this has to do with the kinetics of ion transfer between the two cells through the aqueous salt bridge. If the bridge is too thin, then maybe it could acting as a choke. Was the initial bridge very thin? Did each additional bridge cause the same increase in voltage, or did it asymptotically approach a value?

@NicolauSakerNeto: The relationship was linear, i.e. voltage as function of salt bridges N was of the form $V(N)=aN+b$. The salt bridge I used was filter paper soaked in the salt.

I think your bridge was sub optimal. What voltage was you able to reach?

@Jori: With 5 layers of filter paper, around 350 mV. Can you explain why increasing the number of alt bridges increased the voltage?

350 mV?! You should be able to get around 630 mV. Like Nicolau Saker Neto said, it seems that your bridge isn't working properly. I don't know about the ability of Potassium nitrate to conduct electron transfer, perhaps that could be the problem. Try soaking your bridge in Potassium iodide (table salt can also be used, albeit less optimal). What martial is your bridge made of?

@Jori: As I said, it's filter paper (student grade) soaked in potassium nitrate. I don't really have the opportunity to re-do the experiment, it was done in a school lab.

Hmm, ok. Then I'll do it for you. Any particular reason for using zinc sulfate, does it make it easier for the zinc to come off? I normally just use concentrated sulfuric acid to do the electron transfer.

The oxidation process either produces positive ions or removes negative ions from the solution at the anode (or it may change one ion to a more positive one), and the reduction process either removes positive ions or produces negative ions in the solution at the cathode.  This produces electrically charged solutions, and very quickly stops the process before a measureable number of electrons are transferred.
There must be a path for the ions to move between the two solutions in order for electrons to flow continuously through the wire.  This produces an "ion current" within the battery with cations (positively - charged ions) moving from anode to cathode, and anions (negatively - charged ions) moving from the cathode toward the anode.

Now, more the no. of salt bridges, more will be the ion current. And from the standard ohms law equation, $$V=IR$$ , V being directly proportional to I (Current), will increase if the ion current increases.

Only up to a point, you will never beat the equilibrium voltage.

The voltage you measure between the terminals of a voltaic cell will depend on two factors:

The intrinsic maximum voltage $(V_\mathrm{max} = E_\mathrm{cell})$ that the cell could produce, depending on the $E^o_{red}$ of each half cell, the ion concentrations and the temperature. This is calculated from the Nernst equation:
$$E_\mathrm{cell} = E^⦵_\mathrm{cell} - \frac{RT}{zF}\ln{Q_r}$$

Resistive terms that come from within the cell, including overpotentials at the electrodes and other factors that influence the internal resistance of the cell.

Overall, the observed voltage $(V_L)$ measured at the terminals of the cell will be lower than the theoretical $V_\mathrm{max}$ because the internal resistance of the cell involves a voltage drop, in the same way that you would see an external voltage drop across a resistor in series in the circuit.
$V_L$ can be calculated in the following way:
$$V_L = V_\mathrm{max} - V_I$$
where $V_I$ is the internal voltage drop from the internal resistance of the cell. This drop in voltage depends on the internal resistance of the cell ($R_I$) and the total current $(I)$ running through the cell (and the circuit, see farside.ph.utexas.edu):
$$V_I = I \cdot R_I$$
Your question seems to revolve around how the salt bridge affects the internal resistance of the cell. It would make sense that the more salt bridges you add, the less internal resistance would result, so the smaller the voltage drop and the larger the measured voltage from the cell.
It is possible to think of the salt bridge, which involves the flow of ions, as being analogous to a wire, which involves the flow of electrons. The resistance of a wire $(R)$ is dependent on it cross-sectional area (see http://hyperphysics.phy-astr.gsu.edu):
$$R = \frac{ρL}{A}$$
where $ρ$ is resistivity; $L$ is length; $A$ is cross sectional area.
So it follows that the same may be true for a salt bridge. By adding more salt bridges, you are increasing the cross-sectional area, decreasing the internal resistance and increasing the observed voltage, which would approach the theoretical maximum voltage as the internal resistance approaches zero.


text:
Can toluene be oxidized to benzoic acid using Pyridinium Chlorochromate(PCC)?
My question stands:
Is it possible to oxidize toluene to benzoic acid using PCC?

PCC is generally regarded as a milder oxidizing agent, used to effect selective oxidation of alcohols to aldehydes (from $1^{\circ}$ alcohols) or ketones (from $2^{\circ}$ alcohols). While over-oxidation to a carboxylic acid from a primary alcohol is definitely possible, PCC should not react with a simple unfunctionalized benzylic carbon.
However, oxidation at the benzylic position is easily achieved with more aggressive oxidizing agents. In particular, chromic acid (generated in situ from chromium salts in aqueous sulfuric acid, as in the Jones oxidation) as well as aqueous potassium permanganate will fully oxidize virtually any benzylic carbon all the way to the carboxylic acid, providing the benzylic carbon has at least one hydrogen. Of course, the rest of the chain is completely cleaved in the process.
Edit: Out of curiosity, I consulted my copy of March's Advanced Organic Chemistry to determine if any other common reagents were able to effect the same benzylic oxidation. Nitric acid is listed along with the two methods I described above. Naturally, a survey of various more exotic methods is also given. Intriguing was the use of sodium hypochlorite in acetonitrile, or NBS in aqueous sodium hydroxide under photochemical conditions, for the specific complete oxidation of aryl methyl groups.


text:
How to calculate the number of waters of hydration?
I have three questions about waters of hydration, the first two of which I'm really stuck on:

A sample of cobalt(II) nitrate hydrate was heated to remove all the water of hydration. The hydrate was found to be $65.96\%$ oxygen. Calculate the number of water molecules associated with each formula unit of cobalt(II) nitrate.
Epsom salts is $\ce{MgSO4.xH2O}$. The hydrate was found to contain $71.4\%$ oxygen. Calculate the  value of $x$.
Zinc nitrate $\ce{Zn(NO3)2.xH2O}$ contains $21.98\%$ zinc by mass. What is the value of $x$?

I have absolutely no idea for the first two questions.
For the third:

Assume $\pu{100 g}$ of material
$\pu{21.98 g}$ of Zn = $\pu{0.3284 mol}$ of $\ce{Zn}$

I couldn't get any further than this.

Please add what you have attempted towards solving the problem into the body of your question.  For more information, see the site's homework policy for how to ask homework questions. Thanks!

All of the questions involve the same reasoning, so let's look at question #2.
What is the % oxygen in $\ce{MgSO4}$ when there is no water of hydration (x=0)?  It would be given by $$\mathrm{\%~ oxygen~ =~ \frac{(4 \cdot atomic~ wgt~ O)}{(atomic~wgt~ Mg + atomic~ wgt~S + (4 \cdot atomic~ wgt~ O))}}\\\mathrm{=~\frac{(4 \cdot 16)}{(24_{.}31+32+(4 \cdot 16))}}\\=53.2\%$$  How would the equation change if we added one $\ce{H2O}$ of hydration to the molecular formula? How about if we added 2 or 3?

I realize the percent of oxygen would go up by about 3.5% each time I add in an extra H2O. It is possible to do trial and error until I get to 71.4% but there has to be a more efficient way of doing this right?

sure, set up an equation like above but insert "x" times the atomic weights for the contributions from the H and O in the water


text:
Why is a buffer solution most effective when concentrations of salt and acid are the same?
Why does a buffer work most effectively when the concentration of salt and acid is 1:1?

Not all salts are basic. Example: sodium hydrogen sulfate (acidic salt). NaCl - salt which exerts no effect on water solution pH.
Are we talking about conjugate pairs of acids/bases or just acids and bases in general?

@Dissenter: I am referring to acidic buffers e.g. ethanoic acid and sodium ethanoate

It would be nice if you could add some more information, so that we know where exactly you are stuck. (Please edit it into the question.) For example, do you care about a specific buffer?

@Martin, it is valid for acidic buffers, I guess.

@Aditya This website is no place for guessing - questions and answers should be opinion free. And I have no Idea what your comment is targeting. I asked to improve the question, as it is worded poorly, broadly and in a way, I do not want to do my own research.

@Martin I guessed that he was asking for acidic buffers and include acidic buffer in his question, not a specific buffer as you said.

@Aditya Well, these things should be clear from the questions. And that was all what I wanted to know. Btw. This is not the place to discuss how this site works - we have a chat for that.

Using the Henderson-Hasselbalch-Equation:
$$\mathrm{p}\ce{H}=\mathrm{p}{K_a}+\log\frac{\left[\ce{salt}\right]}{\left[\ce{acid}\right]}$$
and buffer capacity:
$$\phi=\frac{\text{number of moles of acid/base added to 1L solution}}{\Delta\mathrm{p}\ce{H}}$$
where:
$$\Delta \mathrm{p}\ce{H} = \left|\log\frac{\left[\ce{salt}\right]}{\left[\ce{acid}\right]}\right|$$
it is minimum when $$[\ce{salt}]=[\ce{acid}]$$

It will take an equal amount of acid or base to change the pH of the buffer the same amount. If the buffer system was established with 1:2, it would be the same as a 1:1 system after adding an equivalent of a base (or acid).
Added
Support: Just after equation 1 on this page:

Since a buffer is most effective when about half is in the acid form (Buffer H+) and half in the base form (Buffer)...
The buffer concentration must be sufficient to maintain the pH within acceptable limits with the changes in proton concentration expected to occur.

Would those people who gave me a -1 explain, please.

Just look at a titration plot of a weak acid or weak base.   When you are half way to the equivalence point the rate of change (slope) of pH vs mL titrant is near zero.  That point is when you are at 1:1 for salt:acid.


text:
Meaning of 1,1,1,2 in 1,1,1,2-Tetrafluoroethane & similar
What is the meaning of the 1,1,1,2 part in the compound 1,1,1,2-Tetrafluoroethane, a refrigerant commonly known as R-134a, some info:
Formula: $\ce{CH_2FCF_3}$
Structure:

Reference: http://en.wikipedia.org/wiki/1,1,1,2-Tetrafluoroethane

The numbers indicate which carbons the substituents - the fluorine atoms - are on. The root of the molecular name is "tetrafluoro," indicating the presence of four fluorine substituents. So now where are they? That's what the numbers tell us; the "1,1,1" part indicates that there are three fluorine substituents on the first carbon, and "2" tells us there is one more fluorine substituent on the second carbon.
Which is the first carbon and which is the second carbon? The numbering scheme should be selected such that it gives the lowest number at the first point of difference.
This way, we can write "1,1,1,2" to indicate the presence of the four fluorine substituents rather than "1,2,2,2."

Thanks for putting it in a way that I, a HS student, can understand. Also, is there a specific name for this "format"?

Actually, IUPAC rules do not specify that the sum of the locants has to be minimized, but rather that the numbering scheme should be selected such that it gives the lowest number at the first point of difference. For example, 1,5,5- is chosen over, say, 2,2,6-. Q.v., the relevant section, rule 2.2, from the IUPAC Blue Book.

Oops, good point!

This "format" is known as IUPAC nomenclature, @baharini

1,1,1,2 in tetrafluro ethane[HFC 134a] is a position of fluorine in ethane.

This is ethane structure in which 3 hydrogen of first carbon (1,1,1) and 1 hydrogen of second carbon (2) will be replaced by fluorine

1,1,1,2 tetrafluro ethane is used instead of chlorofluorocarbon [CFC] as it is less hazardous to environment as compare to CFC


text:
Will ice cream freeze faster with frozen brine or adding salt to ice?
I've been making some ice cream without a ice cream maker, using salt, ice, mixer and steel canister since I can't get a ice cream machine around here and its fun. It works pretty good but it seems slower than a simple ice cream machine that uses a freezer bowl. And the freezer wall volume seems smaller than the amount of ice I use.
Will using a frozen brine (if I can even freeze it) or liquid cool faster than
using frozen ice and then adding salt to melt it and lower the temperature? whats the difference? Is temperature all the matter? (liquid brine at -10c vs ice at 0)
any other factors that can contribute to the time it takes to freeze the mixture?
Any idea what coolant those bowls use? (seems to freeze at freezer temps).

A very good question, I think it may depend in part on whether you can add more salt to the ice at ice-cream making time than you could freeze at chilling time.  If you have surplus Urea or Ammonium nitrate fertiliser you could dissolve them into chilled water to make it freezing, adding them to ice may work even better and then use it to fertilise your plants.

5+ years on I went to write a simplistic answer based on the assertion that you gain 4.2J/cc of coolth using frozen brine for every degree that the brine melts below 0 degrees C. BUT This interesting 1966 conference paper convinced me that I need to know more. https://eprints.lib.hokudai.ac.jp/dspace/bitstream/2115/20328/1/1_p599-610.pdf

If the brine melts at the same temperature you reach with the ice and salt, then it should take the same time to freeze the ice cream since the freezing time depends on heat transfer. The benefit of frozen brine is that there is more heat capacity in the frozen brine than in ice-salt cooled water, so the ice-salt water may become warmer than the frozen brine.
It may be that there is poor heat transfer from the cream to the ice so it takes longer than you think. Also, if you are the one turning the crank, time seems to stretch out since your muscles get tired and yet the ice cream is still not frozen.

Intuitively I think that the frozen brine will have further benefit of having the salt cold.  However you need to test or research the amount of salt you can add to water and still have your freezer reach the freezing point.


text:
Oxidizing 2-Methylpropan-2-ol using potassium dichromate. Why is heating required?
In Chemistry class, we had to combine $\ce{K2Cr2O7}$ with 2-Methylpropan-2-ol and heat it for the oxidation to occur.
But why do we have to heat the mixture when, for example, butanol, can be oxidized via use of the same reagent but without heating it?

Tertiary alcohols (such as 2-methyl-2-propanol, in your specific case) cannot be directly oxidized by $\ce{Cr(VI)}$ salts like $\ce{K2Cr2O7}$ because they have no $\alpha$-hydrogen. On the other hand, a primary alcohol (such as 1-butanol) will be readily oxidized, even under mild conditions and with reagents much gentler than $\ce{K2Cr2O7}$.
With heating and sufficiently acidic conditions, a tertiary alcohol can be eliminated to form an alkene, which can then be oxidatively cleaved to yield two molecules with carbonyl groups. In the case of 2-methyl-2-propanol, one would presumably obtain acetone and either formaldehyde or formic acid, depending on how far the oxidation is permitted to go. (Though, under conditions vigorous enough to effect the oxidative cleavage of the alkene, it's highly unlikely the aldehyde could be isolated. I would also note that carbon dioxide is a product of oxidative cleavage of terminal alkenes when $\ce{KMnO4}$ is the oxidizer, but whether this occurs with $\ce{Cr(VI)}$ salts is unknown to me.) This is really the only reaction I can imagine to have plausibly occurred, if any.
To the best of my knowledge, $\ce{K2Cr2O7}$ is not very efficient for this purpose, and the procedure itself strikes me as pretty odd if the oxidative cleavage described above is the intended reaction. It may be that no reaction was intended to happen, since oxidation reactions of this type can be used as chemical tests to differentiate alcohols by degree of substitution.


text:
Skin permeability, for all chemicals or just some?
On skin permeability, according to Wikipedia:

Human skin has a low permeability; that is, most foreign substances are
unable to penetrate and diffuse through the skin. Skin's outermost
layer, the stratum corneum, is an effective barrier to most inorganic
nanosized particles."

However there is little to be found on exactly what chemicals can permeate the unbroken skin.
This paper(see ref.) states the following:

Many chemicals can cross the unbroken skin.

However, the highly lipophilic nature of most solvents can also result
in dermal uptake when deposited on the skin.

Is Wikipedia's statement incorrect? What chemicals can permeate the skin, which ones cannot, and why?
Ref.: Semple S, Dermal exposure to chemicals in the workplace: just how important is skin absorption? Occupational and Environmental Medicine 2004; 61:376-382, DOI: 10.1136/oem.2003.010645

Just a warning that this is a pretty broad question. Would you be satisfied with some examples rather than some overall theory?

Ideally a simplification of the overall theory, but examples would also be interesting.

"Skin's outermost layer, the stratum corneum, is an effective barrier to most inorganic nanosized particles."
But skin is quite permeable to many organic compounds. DMSO is one example, and it is warned that DMSO is a good solvent for other compounds that could be more toxic than DMSO.
Another example: we all (should) know that mercury is toxic. However, it's not going to hurt you much if you roll it around in your hand for a brief period (don't breathe in, though!). But organic mercury compounds are terribly toxic and skin permeable, including the infamous dimethyl mercury which killed a famous researcher of toxins.

What are some examples of organic compounds that cannot permeate through the skin?

There are many phsico-chemical properties of the penetrant that determine the skin permeability of the substance. I cite this list from Kanerva, Lasse, ed. Handbook of occupational dermatology. Springer, 2000.

Polarity
Volatility
Solubility
Concentration
Molecular weight and particle size
Electrostatic charge

Of course it also depends on the site of application and how is the skin at the moment of the application however:

in general, non polar, lipid-soluble substances traverse skin more readily than do ionic species. Substances that penetrate skin easily include lipd-soluble endogenous substances (hormones, vitamins D and K) and a number of xeonobiotic compounds. Common examples of these are  phenol. nicotine, and strychnine. Some military poisons, such as the nerve gas sarin permeate the skin very readily.

Manahan, Stanley E. Toxicological chemistry and biochemistry. CRC Press, 2002. p.121


text:
Memorising exceptions for soluble salts
My daughter has a chemistry exam on Tuesday. She is doing quite well with her revision except for the following table which she is having difficulty remembering.
Does anyone know of an easy way to remember this? Or perhaps if someone could explain why these exceptions are insoluble that will help her instead?

This would appear to be for an introductory general chemistry class. If that's the case, it's quite likely that the explanations for why these solubility trends are observed will be confusing, since they really require at least qualitative understanding of more advanced concepts (thermodynamics, physical chemistry, etc.). Unfortunately, chemistry at a basic level usually comprises a lot of memorization without proper motivation for same.

As a mnemonic strategy, I would advise trying to find some associations, use grouping, and prioritize. Notice, for example, that sodium and potassium are both alkali metals, group IA (that is, they're grouped in the same column on the periodic table). Barium and calcium are both alkali earth metals, group IIA. Notice also that the exceptions to the rule of insolubility of carbonates and hydroxides overlap with rule #1 (i.e., if rule #1 is simply given priority, the exceptions to rule #5 need not be explicitly memorized as such).

Greg, she's a clever girl and remembers things much better if she knows the reason for it, simple memorisation doesn't work well for her at all.

@PeterMorris If it helps, Barium Sulphate is given as a 'meal' to patients about to undergo an X Ray of their gut as the Ba2SO4 is more likely to absorb the X Rays and hence allow the gut to show up on the X Ray... Since it is toxic it is important that it is not soluble i.e. does not break up into ions and get absorbed by the body. Perhaps the use of other such examples could help ?

Ammonium hydroxide listed among the salts.  That hurts.

Understanding why it happens unfortunately will not help her to remember. This is the classic case where you have to find a dirty way to remember them. Maybe you can make some practical experiment I remember most of them with a practical application of the salts.
There are a lot of online resource regarding Mnemonic Tricks to remember solubility rules, these are two video on youtubes.
Memonic tricks
Song!
Quoting the song:

This is solubility it's so crazy, don't be a fool and follow these rules!


text:
How much carbonation is in one liter of commercial fountain Coca-Cola
From what I understand generally Coke is carbonated to roughly 6.2 g per liter in bottles and cans, or 3.1 volumes of CO2.  I was watching this video because I was curious about how fountain soda machines worked.
https://www.youtube.com/watch?v=RRHNt67xbMs
By the end of the video they suggest that the soda is carbonated to around 5 volumes of carbonation.  Which as I understand would be somewhere in the realm of 11 g of CO2 per liter, almost twice that of the normal soda.  I imagine there are a couple other factors involved, like they are dispensing more CO2 than is actually dissolved in the soda as a consequence of trying to instantly carbonate.  According to everything I know about carbonation I know from beer making they are carbonating almost 20 times higher than a forced beer carbonation technique in home brewing.  Can anyone help me better understand this?  I would think that the carbonation level difference between bottled and fountain soda would be something highly controlled by beverage companies for the sake of product consistency.
I did some additional reading and found that the correct maintenance levels on soda fountains with chilled water to roughly 40 degrees F is 75 psi if that helps.  For 3.1 volumes of carbonation as I understand equilibrium would be just like 5 psig.  I know I missing something big.
I can only assume that the rate of carbonation loss is inversely exponential as the carbonation approaches closer to equilibrium for the environmental pressure.  Would 5 volumes rapidly decarbonate to 2-3 volumes but not get much lower very quickly?

If the soda is not distributed, the rate of loss of $\ce{CO2}$ would be an exponential decay. So changing from 5 volumes to 3 volumes is faster than from 3 to 1 volumes.
A can of soda with 3 volumes of $\ce{CO2}$ has about 15 psi of extra pressure inside the can. If it was increased to 25 psi (5 volumes, like the fountain sodas), the can would need to be thicker to ensure it doesn't rupture, which would increase the cost of the can. There is a balance of the amount of $\ce{CO2}$ and the cost for the drink.

this is in reference to fountain soda though, not canned soda, and the pressure of co2 in fountain soda machines is 76-105 psi, and the discovery channel video suggests its dispensed at 5 volumes.  so im confused.

@user6794 Which do you think tastes better, can sodas or fountain sodas? I prefer fountain sodas. If the formula is the same for fountain and can sodas, the only difference is in the amount of $CO_2$. I was explaining why they don't have more $CO_2$ in can drinks.

Thats my question tho, what is the carbonation level of fountain drinks.

@user6794 Didn't the video say it was 5 volumes. I know can sodas are about 3 volumes.

Does that make sense?  Will that 5 volumes not rapidly reduce down to 3 volumes from dispensing to drinking time?  Or is the rate of decarbonation still pretty similar to 3 volumes  at 5 volumes

@user6794 Dispensing a fountain drink with 5 volumes may drop it down to 3 volumes. Opening a can with 3 volumes may drop it to 2.5 volumes. Pouring it out of the can may drop it below 2 volumes. I would say that you will always have more $\ce{CO2}$ from a fountain drink than from a can (with the same time from dispensing it).

The fountains over carbonate because the beverage is open for possibly a long time before being enjoyed and the idea is to have a happy customer. It is chilled but warming and approaching equilibrium with air (close to 0% CO2) as the dispensing ends.
A lot of the CO2 is already leaving at the time of dispensing which causes agitation and assists in the final mixing of the syrup.  They have to operate at high pressures to achieve the high carbonation prior to dispensing in a continuous process with low dwell time for the gas and water.
Those who drink beer often are displeased with bubbles, they prefer the product to hold a stable head of foam rather than have continuously rising bubbles, using CO2 concentrations as provided by a soda fountain would cause pretty much any beer product to foam out of the cup as the bubbles will be too large and numerous.


text:
Reduce sizing of molecule - Oxygen
Can the size of molecule oxygen reduce smaller? If yes, how is that possible? Is it related to proton and electron surrounding the nucleus?

Surprisingly enough, the answer is yes, though it would be extremely difficult in practice to reduce the size of a molecule as large as $\ce{O_2}$ by the method I will describe here.
First, I must be clear that "shrinking" as seen in many science-fiction stories is completely impossible. Size is not some arbitrarily determined parameter which you can change with a ray gun!
Though the concept of size is a little fuzzy at atomic/subatomic scales, we know it is the valence electrons who are responsible for determining the distance scales in atoms and molecules (since they're the farthest from the nuclei). After solving the Schrödinger equation for a system, one finds that the electrons are concentrated in regions around the nuclei called orbitals. A meaningful definition of size for these orbitals implies that they are around 100 picometres or more across, conveniently described as twice the Bohr radius.
Usually, this would be the limit for any atom or molecule. However, nature has a rather curious quirk. We have discovered that all atoms in the Universe involved in chemistry are made of electrons, protons and neutrons (the latter two being themselves made of different combinations of up quarks and down quarks). However, there are more subatomic particles out there, and there is nothing to stop (exotic) atoms being made out of these other subatomic particles, as long as they are combined in the right way.
Physics doesn't yet understand why, but the fundamental subatomic particles in the Universe come in three generations. That is, there are groups of particles which behave exactly the same (have the same spin, electric charge, among other parameters), but differ only in their mass. In particular, electrons are just one type in a group of three types of particles called leptons. The other types are called muons and taus (or tauons). Again, all leptons share exactly the same properties, except for mass; the muon is approximately 200 times heavier than the electron, while the tau is around 3500 times heavier. We don't naturally see atoms containing these heavier cousins of the electron because they decay very quickly and are hence much rarer. Even though the muon is one of the most stable subatomic particles, its mean lifetime is a mere $2.2\ \mu s$, and the tau is much more unstable still. In the case of the muon, however, it is long-lived enough to produce it artificially and get some chemistry done.
If one were to mix muons with normal atomic nuclei in an appropriate chamber, then for a fleeting moment it would be possible to make atoms and molecules where all the electrons are replaced with muons. The resulting atoms/molecules would have muon orbitals instead of electron orbitals. By solving the Schrödinger equation for muons instead of electrons (changing $m_e$ for $m_{\mu}\approx200\ m_e$), the Bohr radius becomes around 200 times smaller and as a direct consequence the muon orbitals are smaller by approximately the same factor. Thus, an oxygen atom where all electrons have been replaced by muons will be about 1/200th the size, but will display almost exactly the same chemistry; two such exotic atoms would be expected to combine exactly as normal oxygen atoms do, creating an $\ce{O_2}$ molecule with sixteen muons instead of electrons, with an expected bond length of around $0.6\ pm$ instead of the usual $121\ pm$.
This may all seem theoretical, but muonic atoms have already been made in the laboratory before. Given the difficulties in producing and manipulating muons in their short lifespan, the only atoms produced as of yet seem to be muonic hydrogen (where the single electron is replaced by a muon) and quite recently muonic helium where only one of two electrons was replaced with a muon. In the latter case, the mix of electrons and muons makes the atom behave chemically quite differently from regular helium or doubly-muonic helium.

If you could eliminate the Pauli exclusion principle then the size of all atoms and molecules would collapse.    Because electrons cannot have the same four quantum numbers in an atom, $n$, $l$, $m_l$, and $m_s$, the electrons in multi-electron atoms and molecules are forced to occupy the higher energy and larger radius orbitals.   More electrons: bigger atom.
You can read more about Pauli exclusion principle here:
http://www.grandinetti.org/Teaching/Chem121/Lectures/QuantumNumbers
and atomic radii trends here:
http://www.grandinetti.org/Teaching/Chem121/Lectures/MultiElectronAtoms
and here:
http://www.grandinetti.org/Teaching/Chem121/Lectures/AtomicRadii

Meaning that less electron will result to smaller radius orbitals. And what is the practical method to do that? Can the Far Infra Red energy make it done?

There is none.  You can't violate the Pauli exclusion principle any more than you can defy gravity.  Sorry.

This is going to sound simplistic relative to the previous 2 answers, but what about putting a crystal of solid oxygen into an anvil at a temperature below its solidification point, and compressing it. Phase diagrams (e.g. http://commons.wikimedia.org/wiki/File:Phase_diagram_of_solid_oxygen.svg) show that oxygen exists in different crystal forms at different temperatures and pressures. It becomes superconducting at low temperatures and high pressures.


text:
Charges in colloidal dispersion
How does charge come on dispersion medium in colloidal sols ? (It is said that in electro-osmosis flow of dispersion medium takes because it is also electrically charged, but why ? )

Have you read about Zeta Potential?


text:
Halogens and the free energy of activation in a resonant benzene system
If we look at the right-hand-side table here: http://en.wikipedia.org/wiki/Hammett_equation, we notice something particular.  It appears that the free energy of activation (for example, measured in terms of the acidicity of a carboxyl group on a benzene ring), and with respect to substituents in the para- position on the benzene ring, increases as one moves down column 17 on the periodic table.
Why is this happening?  Shouldn't we expect the opposite to be true?  Consider that, as we travel down column 17 corresponding to the halogens, electronegativity decreases and the probability of resonance contributions via pi-bonding should increase.

Resonance between a carbon and a halogen atom decreases as we move down the series $\ce{F, Cl, Br, I}$.  This is because overlap is most effective between orbitals of the same size.  Therefor, carbon-fluorine overlap, both involving $\ce{2p-2p}$ orbitals should be much stronger than carbon $\ce{2p}$ orbital overlap with the $\ce{3p, 4p, etc_{.}}$ orbitals in the larger halogens.  Consequently carbon-halogen resonance becomes less significant as we move down the halogen series.  If we now consider that 1) the resonance and inductive effects for the para-halogens are in opposition to each other (tend to counteract each other) and 2) the resonance effect decreases as we move down the halogen series, then the order of halogens in the Table you mentioned makes sense.

Thanks, this seems to make sense to me.  However, I don't understand your last point.  How does (1) help to explain the table ordering?  It's exactly contrary to what you would expect from inductive effects.  This just seems to mean that resonance effects dominate for halogens (with the one exception of Iodine in the meta- position, which could be due to stereochemical complications)?

The values for the meta-substituents should be caused primarily by inductive effects.  Looking at the Table, we see that they are all very close, within 0.06 units of one another.  We would expect the para-substituents to display a similarly close grouping for the inductive effect, just diminished somewhat due to greater through-bond distance.  If we overlay a decreasing resonance effect as we move down the halogen group, on to a fairly constant inductive effect, we get the observed pattern for the para-substituents.

Sorry, I've accepted your answer, and it is helpful (I was away from my computer for awhile).  However, just to follow up, my point is that the inductive effect shouldn't be approximately constant, right?  Fluorine has an electronegativity of 3.98, chlorine 3.16, bromine 2.96, iodine 2.66, astatine 2.2, and so forth.  So the trend for inductive contributions runs contrary to the trend for resonance contributions.

The meta inductive effect runs (roughly) counter to the electronegativity scale. You'd expect the inductive and resonance effects to be in opposition.  The resonance effect works to pump substituent electrons in, while the inductive effect would take electrons out.


text:
How does oxygen fight odors? (Deodorant ad)
I saw an ad on tv last night for Mitchum deodorant that showed big text on the screen that said "OXYGEN eliminates all odors!" or something similar. Their website indeed claims "Our exclusive formula releases pure oxygen, a powerful odor fighter...".
How does oxygen fight odors, and how can this aspect of the deodorant be more effective than the abundance of pure O$_2$ in the air?

In my estimation, this is basically marketing nonsense. If you examine the list of ingredients, the primary active agent is invariably some aluminum or aluminum-zirconium salt, which act as anti-perspirants primarily by the mechanical actions of shrinking pores (due to their astringency) and clogging ducts in the apocrine glands.
Listed among the inactive ingredients is hydrogen peroxide, $\ce{H2O2}$. At sufficient concentrations, $\ce{H2O2}$ can be a potent oxidizing agent, and probably the source of the oxygen the ad refers to. It might indeed be able to neutralize some odors by oxidizing the molecules which cause them, thereby transforming those molecules into odorless ones. However, as it's listed among the inactive ingredients, it's almost certainly present in very low concentration. This is to be expected, since prolonged skin contact will invariably cause chemical burns unless the mixture is very dilute. Furthermore, much of the odor of sweat is the result of metabolic byproducts from skin-dwelling bacteria, chief among them certain short-chain carboxylic acids that are already maximally oxidized. The only malodorous substances in sweat that can be neutralized by oxidation, as far as I can tell, are certain amines and thiols. That said, I doubt the concentration of $\ce{H2O2}$ is sufficient to be of any use here. $\ce{H2O2}$ does also have anti-septic properties, so perhaps its purpose is bacteriostatic, but there I'm just speculating. Ultimately, I'm uncertain of its purpose in the formulation, but I'm extremely dubious that it helps substantially in eliminating odor.


text:
How can i detect Formalin in food easily
I am trying to detect formalin in different types of food. Formalin $(\ce{CH2(OH)2})$is an aqueous solution of formaldehyde, which is $\ce{CH2O}$ or $\ce{H2CO}$. I used many chemicals to do this, but every time I fail. Please help me to do this. I got some results in my test by using NFPA 704 and ChEMBL, but that is not enough to save all. I would like to make a simple and easy food formalin detector that can save all the people. Please help me or give me some direction.

Perhaps you are unable to detect formalin because it is not there. Formalin/formaldehyde has a very powerful and distinct smell that is extremely unpleasant. It can be detected at fairly low concentrations by your nose, which is perhaps the best sensor available for formaldehyde.

Also, what you you mean by "test by using NFPA 704, ChEMBL"? NFPA 704 is the code for the NFPA standard that gives us the fire diamond. ChEMBL is a database of properties of drug-like molecules for bioinformatics purposes.

This might help you

Thanks to all... I am new in stackexchange. And I like Chemistry. I want to help everyone with my knowledge. If you know the formula or you have any article that I can get some idea please give me. We will very appreciate to you.

@Freddy - Would you be willing to expand/explain/summarize your link as an answer?

Simple commercially available tests contain formaldehyde dehydrogenase. This enzyme selectively catalyzes the chemical reaction of formaldehyde:
$$\ce{CH2O + NAD+ + H2O -> HCOO- + NADH + H+}$$
The enzymatic reaction is typically indicated by reduction of a tetrazolium salt to a coloured formazan dye.


text:
Empirical and Molecular Formulae
I'm not a chemist, and nor do I study chemistry so please try to be gentle.
Glucose has the molecular formula $\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6$ and the empirical formula $\mathrm{C}\mathrm{H}_{2}\mathrm{O}$.
Starting from the empirical formula and working the other way, am I guaranteed that $\mathrm{C}_n\mathrm{H}_{2n}\mathrm{O}_n$, where $n$ is a positive whole number, are all going to be well-define molecular formulae?
Some examples are Formaldehyde ($n=1$), Acetic Acid ($n=2$) and Ribose ($n=5$). As a starting point, does $n=3$ and $n=4$ make sense, i.e. $\mathrm{C}_3\mathrm{H}_{6}\mathrm{O}_3$ and $\mathrm{C}_4\mathrm{H}_{8}\mathrm{O}_4$ respectively?

Yes they make sense but the same molecular formula can represent different compounds, for the first case $\ce{C3H6O3}$ may rappresent:
Glyceraldehyde:
or lactic acid (and its optical isomers):
$\ce{C4H8O4}$ leads to even more possibilities see the three different tetrose

Generally these compounds may be created when the sum of the formal charges of each atom in the molecular formula is equal to zero. Oxygen, hydrogen normally have formal charge of -2 and -1 carbon +4 so:
$$-2\times n + (-1 \times n \times 2) + (+4 \times n)=-4\times n +4\times n=0$$

Thank you for your answer. I really do appreciate your help. If we have an empirical formula, say $\mathrm{X}m\mathrm{Y}n$, then does $\mathrm{X}{km}\mathrm{Y}{kn}$ always exist? (Here $X$ and $Y$ are elements and $k,m,n$ are positive integers.)

@FlybyNight You are welcome. Not always, it depends on different factors I think you can find useful this question about criss-cross method.

Hmmm... This is beyond me. I'm sorry. I have the following empirical formulae: $\mathrm{Li}_2\mathrm{O}$, $\mathrm{NH}_3$, $\mathrm{Al}_2\mathrm{O}_3$ and $\mathrm{H_3}\mathrm{PO}_4$. Which multiples of the subscripts are possible?

No in not applicable to these compounds. $LiO_2$ and $Al_2O_3$ are ionic compounds so you have a lattice not a molecules hence there's no need to choose a larger unit LiO2 can describe all the compounds: Li2O4 etc etc. I know that exist $\ce{N2H6^{++}}$ but I don't know about N2H6. \ce{H6P2O8} doesn't exit up to my knowledge. Have a look at Homologous series for an example of the behavior you showed in your question .


text:
Behaviour of water at exactly 0 °C
If I had a beaker of water (i.e. many molecules), at exactly 0 °C and at atmospheric pressure, what phase would be encountered?
And would the answer to this also apply to other substances at their freezing/melting point, and out of interest, would the same effect be observed at their other "phase intercepts" too?

It depends on how you achieved water at 0 °C. If you cooled water down to 0 °C, then it will remain water, unless you remove more heat, at which point it will start to freeze. Water needs to be at 0 °C to freeze.
If you warm ice up from below freezing to 0 °C, then it will remain ice, unless you add more heat to melt the ice. Ice needs to be at 0 °C to melt.
It will be the same for all substances.


text:
From liquid to solid by beating
I understand how liquids can become solids by decreasing temperature. To my understanding this is because the molecules slow down and are able to form bonds.
However, I'm having problems understanding how we can create a solid out of liquid by beating it, such as the case with cream becoming butter if you beat it long enough.
What is going on in such a process?

Butter is at least 80% milk fat, 16% water and about 3% milk solids (such as proteins and sugars). The milk fat is normally a solid at room temperature. To make butter, cream is separated from the milk and pasteurized. The cream is then churned (beaten) to separate out buttermilk (mostly water) from the milk fat (butter). So, when butter is made, a solid is separated from a solution (actually an emulsion).
The closest thing I can think of where a liquid is transformed into a solid is beaten egg whites. In this case, air is mixed into the liquid to form an emulsion. It appears solid (but is still a liquid) since the liquid surrounds many air bubbles (which don't pop since they are very small). When you cook the beaten egg whites, then the protein in the egg white are denatured and become a solid.

There's also Non-Newtonian Shear Thickening Fluids.  However I cannot personally explain how beating them turns them solid (temporarily).  But as best I understand, nobody is 100% sure so I'm not too embarrassed.
http://en.wikipedia.org/wiki/Shear_thickening
http://en.wikipedia.org/wiki/Non-Newtonian_fluid
However LDC3s explaination of butter is likely a more accurate answer to your question, and its not actually a change of state from liquid to solid as a result of impact force, but rather water being removed from a material that is its own solid.
Think of taking the water out of salt water.

Indeed, shear thickening fluids turn solid (or rather, behave like a solid) when force is applied. However, this is indeed only temporarily, therefore it cannot explain how butter permanently solidifies by beating.

By the way, if you are interested in understanding more about shear thickening. There is an excellent review paper here: http://arxiv.org/pdf/1307.0269.pdf