text:
What does the subscript of atomic orbital mean?
As everyone knows, the atomic orbital can be classified as $s, p_z, p_x, p_y, d_{z^2},d_{xz},d_{yz},d_{xy},d_{x^2-y^2}$ and so on. I want to know the meaning of $z^2,x^2-y^2$ and so on. Maybe this is a fundamental question, but I'm not familiar with chemistry.

I have some ideas about this:

For example, we consider the $p$ orbital: $p_z$ is symmetric about $z$ axis; $p_x$ is symmetric about $x$ axis; $p_y$ is symmetric about $y$ axis. But what's the meaning of $z^2$ or $x^2 -y^2$?
We consider the symmetry of atomic orbital with group theory. The change of orbital under the symmetry operation is the same as the orbital subscript.

I have deleted my post, if I get correct relevance I will repost. But, you can notice that d_xy is symmetric about x+y axis.

@GODPARTICLE Do you mean x+y axis is x=y axis?

Yes. The vector x+y will be $45^0$ with respect to both x and y axis.

But it does not work for all （╯_╰）

It does for all xy, yz, zx involved subscripts.

I do agree that it doesn't work for other subscripts.

@GODPARTICLE In fact, I think you are in the wrong way. For example, $d_{xy}$ has 3 symmetry axes(if you konw about the point group, the axis should be called $C_2$ exactly).

): May be, but good question. I have something to learn from any of the best answer to your question.

For each azimutal quantum number $l$, the magnetic quantum number $m$ ranges from $-l$ to $+l$. Those give you the number of atomic orbitals "subsripts" you should obtain.
Those are then expressed using spherical harmonics. If you look at the table of spherical harmonics at the $l=1$, you will find that they contain the cartesian axes as $x-y$, $z$ and $x+y$. Therefore it is advisable to form the linear combination out of them, so you get $x$, $y$, $z$. If you go to higher angular momentum, things get more complicated, but there is some tradition what linear combinations to use.
To answer your second question, all this formulas are not some inevitable result you obtain by juggling the numbers. Contrary, it is carefully crafted in a way that the formulas are so simple.

1)What do you mean "you will find that they contain the cartesian axes as $x−y$, $z$ and $x+y$."? 2)I want know why the subscripts are written as $z$, $z^2$, $x^2-y^2$ and so on respectively. 3)You say "it is carefully crafted in a way that the formulas are so simple.", and I'm not clear about "the formulas". Can you explain it more specificly?

Please have a careful look at the table of spherical harmonics at l=1, i.e. p-orbitals in cartesian coordinates and try to understand, what I have written.

I know how to combine spherical to obtain $p_x$, $p_y$...But I do not know why choose x, y, z and so on as subscripts.

The orientation of the orbitals. For $p$, one is oriented in the x direction one is oriented in the y direction and one is one the z direction.

Bach: We chose the orientation for immense convenience. It is great, if you expand some equation and magically 95% of the terms disappear just because you have chosen good coordinate system.
And the labels $p_x$,... just show, which axes and in which power appear in the spherical harmonic expression.

@user2612743 What about $d_{x^2-y^2}$, $d_{z^2}$ and so on?

@ssavec I want to know how the labels show which axes and in which power appear in the spherical harmonic expression.

@Bach: have you followed the simpler example for p-orbitals? And plotted the spherical harmonics in 3D? And formed linear combinations of them?

@ssavec You can explain directly. For example, you explain how $d_{x^2−y^2}$ show which axes and in which power appear in the spherical harmonic expression.

@Bach: I still do not get your point. For example, the $d_{z^2}$ obviously contains z-axis in the power of 2 and is obtained as linear combination: $Y_2^{-2}+Y_2^{+2}+2Y_2^0$ (up to prefactor).
If the problem is in visualization, than you have to use visualization program. At least I am not able to see how the axis in the expression affect the shape of function without producing a graph.

I mean why choose $z^2$ as the subscript? Dose $z^2$ mean someting?

@Bach: I still do not understand. It means it is there in the equation. And this linear combination was chosen to make subsequent calculations much simpler. What type of meaning it should have?

I don't know the subscript's meaning, so I ask this question. I think the subscript is not chosen at random just for definition. For example, when we consider the symmetry of atomic orbital, the subscript make some coincidence.

As you noted, the symmetry is also an important concept, as it governs the magnetic quantum number, so $p_z$ has $m=0$ and the remaining $p_x$ and $p_y$ have $m=\pm 1$. So it is definitely not random. But for $l>1$, so d and higher angular momentum, the exact composition is IMHO not straightforward, but chosen under some constraints.

Late, but OP seemed unhappy with any of the answers and this is a great question I had for years as a chemistry student, suffering what I felt were vague explanations when I just wanted a precise formula.  I wanted to see that subscript as part of the polynomial in the wavefunction.  After all that time, here's my answer:
The orbitals have the same symmetry as the function $\psi(x,y,z) = $ "subscript".
p-orbitals $p_x$, $p_y$, and $p_z$ have the same symmetry as $\psi = x$, $\psi = y$, and $\psi = z$. They have nodes where that function is 0, and they flip sign where that function does, just dampened by a decreasing exponential $e^{-r}$, rather than going off to infinity. Basically $\psi = xe^{-r}$ (Not exactly the solution but very close).
d-orbitals have same symmetry as $\psi = xy$, $\psi = yz$, $\psi = xz$, $\psi = x^2-y²$, ... and the tricky one, $z²$ is short for $z² - (x² + y²)$, which gives you conical nodes and a sign flip with the horizontal radius $\sqrt{x^2 + y²} > z$.
The same holds true for the f-orbitals, with again one of them being a short-hand.  Slightly annoying, I know, but once someone TELLS you, it's ok.
Hope that helps.


text:
Why Gold and Silver react minimally with atmosphere?
Gold and Silver atoms interact with one another to form reasonably strong metallic bonds, but surprisingly they have almost no chemical interaction with atmosphere. How can we explain it based on the atomic structure of Gold and Silver. What is so unique about its chemical structure that is not present in other transition metals that makes them noble (non-reactive)?

Silver, compared to Gold, has a huge chemical interaction with the atmosphere: it tarnishes quite easily.

@mannaia You took advantage of the vague title :D Polishing silverware is a major pain in the lower back ;)

Actually, they do interact with atmosphere. Both gold and silver adsorb oxygen from the atmosphere and can form surface oxide.

Copper, silver, gold. What is the energy of the highest "s" orbital versus the highest "d" orbital?  Gold is a very heavy nucleus.  An "s" orbital has non-zero probability at its nucleus.  Special relativity effects make its "s" electrons more massive, bringing them into the atom.
A silver atom is a soft Lewis acid.  It reacts well with large soft Lewis bases.  Silver quickly tarnishes with sulfide.  Gold is not chemically inert at all to even mild oxidation if the product cation is coordinated by ligands.  Salt water plus air will tarnish gold.  Dilute cyanide plus air will dissolve it (e.g., heap leaching of gold ore).
Thermodynamics proposes, kinetics disposes.  Aluminum is tremendously reactive toward air and water.  Anything that permanently breaches its adherent oxide coating (e.g., abrasion plus amalgamation) sets it off.

Note its the presence of sulphides in sea water (and other sources) that corrode both gold and silver. Simple sodium chloride without oxidisers more powerful than oxygen won't corrode silver or gold.

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text:
How many grams of CH3OH can be made from 30.0g of CO if it all reacts?
Once made by heating wood in the absence of air, methanol ($\ce{CH3OH}$) is now made by reacting carbon monoxide and hydrogen at high pressure:
$\ce{CO_{(g)} + 2 H2_{(g)} -> CH3OH_{(l)}}$
How many grams of $\ce{CH3OH}$ can be made from 30.0g of $\ce{CO}$ if it all reacts?

Hi Brian, try recall the mole concept and show us your efforts we have to follow our homework policy

Moles of CO in equals moles of methanol out (at 100% yield), typically over copper-doped zinc oxide catalyst.  Do the arithmetic.
One gram-mole of a chemical formula is [the magnitude of its atomic mass units] grams.  A nitrogen atom is 14.0067 amu.  A mole of nitrogen gas, $\ce{N2}$, is then 28.0134 grams.

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text:
A molecule absorbs light having a specific wavelength, why doesn't it absorb shorter wavelengths?
I am currently studying computational chemistry and I am trying to understand the UV-VIS absorption of molecules. I know that UV-VIS absorption is electronic excitation and requires a specific energy.
What I don't understand is why a molecule can not absorb a shorter wavelength as it has more energy? A slightly higher energy causes vibrational excitations which is then released through vibrational relaxation (releasing heat). Isn't it the case when a very shorter wavelength is used so that the electrons in the molecule are excited from more lower MOs (lower energy) to the higher ones? Isn't it still absorption? It is supposed to be by definition?

Molecule can absorb a photon that has "red" amount of energy, it can also absorb "green" and "blue" amounts. But why can't it absorb anywhere between red and green or more basically isn't it supposed to absorb anything more than "red"? Just to be clear, by red I mean the energy difference between $n = 1$ and $n = 2$.
Also, I am already aware of the fact that, for example, any energy between $n = 2$ and $n = 3$ can not be emitted fully as some of them will be lost for rotational and vibrational transitions until the energy hits the $n = 2$ state but I am not expecting this to stop a molecule from absorbing that light.

You should quantify what you mean for shorter wavelength...

Lets say the molecule has maximum absorptivity at 400nm and I am sending a light beam having a 300nm wavelength.

You can have (depending on the symmetry) Raman spectroscopy, where light of a much shorter wavelength excites a vibrational change and gets scattered (and shifted in wavelength) as a result.

If I understand this correctly, you're wondering why the system can't absorb a photon that's slightly larger than the n1->n2 gap, then vibrationally relax to n2? I think the answer is that it can happen. I overheard something about electronic Raman scattering the other day, which would be something like what you described, but trying to read the material on it just hurts my head too much for me to delve into it right now.

@chipbuster Yes I am wondering that.

Any molecule has several excited states. Absorbing into the lowest causes its vibrational levels to be excited, as in your picture, leading to lines in the absorption spectrum as determined by Franck-Condon principle. Above this there is another excited state so there is often a gap with little absorption before absorption starts in the next excited state, and so on. Sometimes absorption can overlap so some spectra can be more or less continuous, others have gaps between absorption peaks, e.g.  azulene, chlorophylls.

Classical approach
Absorption occurs when there are "oscillators" that can resonate with the frequency of the electromagnetic wave. In classical weave physics, these frequencies are called natural frequencies.
These oscillators can involve electrons (UV, VIS, XR), bonds (MIR, FIR), nuclei (gamma rays). They are able to absorb electromagnetic waves changing their spins(NMR, ESR) orientation (microwave), configuration (infrared) nuclear configuration  ($\gamma$ rays) and finally electronic distribution (VIS, UV, X-Ray).
Of course, not all the materials have oscillators that absorb in all the spectrum. For wavelength between 10 nm and 100 pm the oscillators, that mainly involve electrons in these regions, have to absorb energy between $10^{7}$ and $10^{9}$ J/mol without being ionized, in this case only the inner electrons of some atoms (almost always metals) absorbs these wavelengths see XAS, so in our organic world seems that most compounds are X-Ray transparent but in fact there are a lot of inorganic compounds that absorb X-Ray.
When radiation is absorbed there are many ways to release his energy(see figure 1). We can divide them into two great categories radiative relaxation and non-radiative relaxation. In fact radiative relaxation is most common when we are dealing with shorter wavelength because with more energy is possible to excite an electron to a higher level and this can sometimes re-emit radiation when it goes from a higher level to the ground level, this as you know, is called fluorescence and is very common when you deal with UV, and X-Ray (see X-Ray Fluorescence spectroscopy). Eventually, there is also non-radiative relaxation process that are the main processes that occur from VIS (note that however is possible a VIS-NIR luminescence for some compounds) to the other spectral region with lower wavelength.

Quantum approach
Compound energy levels are quantized, so compounds can absorb only some discrete energy values. These values are determinate by transitions that respect strictly selection rules. If you know the energy of the transition, from Plank relation:
$$E=\frac{ch}{\lambda}$$
you can see that for every energy value there is a specific wavelength, so this means that compounds can absorb only at specific discrete wavelength. However if you see a UV-VIS spectrum you will see a continuous spectrum because roto-vibrational changes affect the energy levels of the compounds involved.

I have passed all the courses including Quantum Chemistry, Instrumental Analysis, Introduction to Computational Chemistry etc... I am quite familiar with those concepts but I have never been thought about resonance part of the story. So still I don't understand why a resonance is necessary since diffrent waves can destructively and constructively interfere.

Am I missing something very fundamental?

@GunDeniz I don't know, I'm not a physicist and I never done such courses, but how can you explain absorption without resonance? If scattering is due to the constructive and destructive interference how can you distinguish it from absorption? If you notice Hydrogen emission spectrum is the "opposite" of Hydrogen absorption spectrum, this mean that absorption occur at some frequencies, we could say "natural frequencies", that are involved in the process of emission. So normally are allowed only oscillation with these freq. for H2 atoms, can you think that these oscillator can act backward?

I can distunguish because while absorption is the part where the molecule receives energy, emission is another story because after the energy is absorbed the molecule undergoes a lot of things like singlet to singlet or singlet to triplet transitions.

Emission makes perfect sense about a spesific wavelength but absorption doesn't.

@GunDeniz I mean distinguish scattering form absorption ( scattering can't be consider emission, are two different things). The process after the absorption is another story, that is treat in the last part of my answer. The point is how and when the energy can be absorb by the atoms/molecule etc. "Emission makes perfect about a specific wavelength but absorption doesn't.". Why? H2 absorption spectrum seems to contradict this. When I studied spectroscopy I learn about selection rules compounds can absorb only at specific wavelength!

See my update on question.

@GunDeniz I've edited the answer!

Sorry but this does not answer the question. I describes a Jablonski diagram for absorption and emission in solution.

@porphyrin why? The jablonski diagram is only a  part of the answer and it simply clarifies the radiative process the user stated:  A slightly higher energy causes vibrational excitations which is then released through vibrational relaxation(releasing heat) that is not always true. I explained before and after using classical and quantum approach why molecules absorb only to precise wavelength see Plank relation and selection rules.

@G M It does not answer because the question asks about gaps in absorption spectra not fluorescence or non-radiative processes.  (Also in a classic approach discrete energy levels are not possible, these only arise from a quantum approach, so the Jablonski diagram is strictly quantum.)

@porphyrin yes and in fact is what is written in the quantum approach part of my answer...

A cryogenic atom or molecule in vacuum is a narrow bandwidth oscillator (particle in a box).  If you do not hit the right note it will not resonantly electronically ring for you.
If it is polyatomic, warming it will introduce rotational (microwave) then vibrational (infrared) degrees of freedom, broadening lines and introducing new ones.  Put it in a fluid medium and the random jingle jangle of collisions broadens things further.  Put it in a solid and modes may be physically eliminated, e.g., whole molecule rotation.  Water's pale blue color is summed infrared vibronic modes, the only visible color caused by vibration absorption.
The higher the frequency, the more localized the photon.  At very high energies photon collision is literally physical.  Things get collisionally knocked out rather than resonantly pumped up.
A heavy atomic nucleus has density ~$2.3{×10^{14} \ce{g/cm^3}}$.  Add a muon 206.768 times the mass of an electron into U-238.  It decays to orbit at a fair fraction of lightspeed just inside the nuclear surface.  Selection rules for hadrons and leptons are different.  They may not see each other.  Triprismane, $\ce{C6H6}$, sits 90+ kcal/mole above benzene at room temp.  It won't fragment until increased temperature adds another 33 kcal/mole - and it doesn't crack to benzene.  Reaction path is defined by conservation of orbital symmetry.  Tetra-tert-butyl tetrahedrane does not thermally fragment.  The insanely strained molecule is so symmetric that its lowest energy path is synchronous transformation into tetra-tert-butyl cyclobutadiene' at 130°C!

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text:
Structure of NO2 compound?
What is the correct structure for the $\ce{NO2}$ compound (not ion)? I always thought it was like:

since the negative charge would be residing the on the highly electronegative oxygen, but was reading this blogpost by the guys at WolframAlpha, who state that contrary to popular belief, the correct structure is actually:

They explain that it has to do with molecular orbital theory, but don't explain in depth.
Can someone please explain why this occurs?

Though it's hardly a rigorous argument, you can calculate the formal charges on each atom for both proposed structures. In the first, there is a +1 charge on the nitrogen atom, and a -1 charge on the singly-bonded oxygen atom (though really it's -0.5 on both oxygen atoms due to resonance). In the second structure, all atoms have zero formal charge, and so the second structure is expected to be lower in energy and hence to contribute more to the actual picture of bonding in $\ce{NO2}$. A higher level argument would still be desirable.

I would love another one of those research project answers by LordStryker on this one!

Drawings of molecules are only representations to help us understand them, not the actual molecule itself. So both pictures you have there are correct; they both contribute to the actual structure of the $\ce{NO2}$ molecule. I'm guessing WolframAlpha chose the structure they did because it is the structure the most contributes to the actual structure of the molecule, with all formal charges being 0 like Nicolau said, it is the most 'favorable' of possible configurations that the molecule could be in.

Does this mean electron clouds are more likely to stay near the N atom?

According to Electronic Structure of NO2 Studied by Photoelectron and Vacuum-uv Spectroscopy and
Gaussian Orbital Calculations J. Chem. Phys. 53, 705 (1970) :
The highest molecular orbital $4a_1$ is occupied by the 1 unpaired electron.
The electron population of this orbital is (see table VI):
0.53 on the N atom (0.16 2s, 0.37 2pz)
0.24 on each O atom  (0.24 pz)
Experimentally, Oxides and Oxyions of the Non-metals. Part II.
C02- and NO2 of the Chemical Society (1962): 2873-2880 collects values for the partition of unpaired electron density among N and O atomic orbitals.
The most purely experimental values are:
N 2s  0.094
N 2pz 0.364
N 2px 0.054
O 2pz 0.33
and
N 2s 0.103
N 2pz 0.471
O 2pz 0.33
(where the values for O are both O atoms added together)
There is extensive discussion of the various Lewis structures of NO2 from a dimerization point of view (that the opposite spin electrons should be on the atoms that form the dimer bond) starting on page 90 if the 2015 book Bonding in Electron-Rich Molecules: Qualitative Valence-Bond Approach via Increase-valence Structures , and also elsewhere in the book from a monomer point of view.
Overall, it is clear theoretically and experimentally that the unpaired electron is delocalized, having significant density on both N and the two O atoms.  It is wrong to say that a particular one of the Lewis structures is "correct".

How large of a basis set were they able to use? Being 1970, I suspect a rather small one, in which case the electron occupation pattern is likely suspect, due to basis set insufficiency.

@hBy2Py I guess you wouldn't trust the values from this 1962 EPR paper either: http://pubs.rsc.org/en/content/articlepdf/1962/jr/jr9620002873

@hBy2Py There is an experimental value for unpaired electron density at the N nucleus by microwave spectroscopy in http://scitation.aip.org/content/aip/journal/jcp/40/11/10.1063/1.1725010 (1964)

@hBy2Py the 1970 article says "The best-atom-double-zeta (BADZ)
basis set of Basch et al. [reference 26] was used throughout." Reference 26 is http://scitation.aip.org/content/aip/journal/jcp/47/4/10.1063/1.1712069 J. Chem. Phys. 47, 1201 (1967)

I'll have to look around for a recent post-Hartree-Fock study. DFT gives results at odds with the ESR data. It looks likely that $\ce{NO2}$ is a system that DFT is no good for, or that one has to play some tricks in order to get to the correct result.

@hBy2Py here is a modern study: http://pubs.acs.org/doi/abs/10.1021/jp077561y

@hBy2Py and http://pubs.acs.org/doi/abs/10.1021/ja300545e I'll see if there is anything good in there to add to my answer

Ahh, great finds! Yeah, I was fiddling around with some coupled cluster calculations, and actually got an HF solution at the symmetric geometry that exhibited an asymmetric charge distribution, that was of lower energy than the symmetric wavefunction. So, there's definitely some major oddness going on!

Nitrogen dioxide has 17 valence electrons, and is bent with an angle of 134 degrees and bond lengths 0.119 nm. Electron spin resonance experiments place the odd electron on the nitrogen in a $\sigma$ rather than a $\pi$ orbital. Thus formally we can consider the odd electron to be in a sigma bonding orbital (of $a_1$ symmetry species in $\ce{C_2v}$) of $sp^2$ hybrid orbital type. In so far as you want to use these types of diagrams experiment would indicate that the first structure is the most correct.
The ion $\ce{NO2^+}$ is linear, $\ce{NO2}$ is bent and so is $\ce{NO2^-}$  with an angle of 101 degrees. To understand these differences it is necessary to combine the p orbitals on each atom to form various $\pi$ orbitals and then consider how these orbitals change in energy as the molecule bends. To do this is quite easy but working out what happens to the energy on bending is more tricky. However, these have been worked out and we can use Walsh diagrams do do this. All that is necessary is to know the symmetry species of an orbital and fill up the orbitals with electrons.
(To do this is rather off topic for your question but I can add pictures if you are interested.)

Can you link to literature describing the ESR results? A quick UPBE calculation shows all of the spin density divided on the oxygens.

(That is, all three of the Bader, Loewdin, and Mulliken methods put all of the spin density on the oxygens.)

I'm not an epr expert at all so i'm not familiar with the literature. I found the details in 'Stable Radicals: Fundamentals and Applied Aspects of Odd-Electron Compounds' edited by Robin Hicks' chapter  4 by D. Bohle,  section 4.5. There are references there to original papers.

It's an interesting problem. Per my comment on DavePhD's answer, it may be that $\ce{NO2}$ is just a problematic system for DFT to handle.

@ hBy2Py  Yes $\ce{NO2}$ is a tough problem, I'm a bit out of date but I think that its electronic  spectroscopy is not fully sorted out even after many years of study. There are lots of low lying and interacting states that are coupled to one another making life very messy. Presumably low lying states makes calculating things hard too.

Yep, probably it needs a correlated multireference treatment to get it right.

The $^{16}\ce{O}$ nucleus is spin-0, the $^{14}\ce{N}$ nucleus is spin-1.  You can see the real world weighted fractional residence times of the unpaired spin with EPR.  $\pi$-bonding will complicate the spectrum with spin cross-coupling.

$\ce{NO2}$ molecules dimerize to form $\ce{N2O4}.$ The $\ce{N}$ adds to each other through the odd electron at $\ce{N}$ atoms. This may be a strong indication that the first structure of $\ce{NO2}$ has significant contribution.

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text:
Deriving a reduction potential from two other reduction potentials

Given:
\begin{align}
\ce{Co^{3+}(aq) + e-  &-> Co^{2+}(aq)} &   E° &= \pu{+1.82 V} \\
\ce{Co^{2+}(aq) + 2e- &->  Co(s)}      &   E° &= \pu{-0.28 V}
\end{align}
what is the standard reduction potential for $\ce{Co^{3+}(aq) +3e-  -> Co(s)}$?

If I add up these two half-equations to get the desired half-equation, I get $E^\circ = \pu{+1.54 V}$. But the correct answer is $\pu{+ 0.42 V}$!
Am I missing something?

That is because you cannot simply add the electrode potentials algebraically. In both cases, the electrode potentials have to be multiplied by $n$. What you can do instead is use Gibbs free energy change for each reaction which then can be added algebraically.
$$\Delta G^\circ = -nFE^\circ$$
Using this you can get the $\Delta G^\circ$ for each reaction which can be then added algebraically in the same way you add the reactions. The resultant $\Delta G^\circ_\mathrm{net}$, on equating with $-nFE^\circ_\mathrm{net}$, we can get the value of $E^\circ_\mathrm{net}$, which should be your answer.
In this approach, since the value of $n$ for both the reactions are different, the relative weightage of the electrode potentials of each reaction in determining the resultant $E^\circ$ will differ and therefore the final $E^\circ$ will be different than what you get by just adding up the electrode potentials.
For your case:
$$\begin{align}
\ce{Co^3+ + e- &-> Co^2+} & E^\circ_1 &= \pu{+1.82 V} & \Delta G_1^\circ &= \pu{-175.6 kJ mol-1} \tag{1} \\
\ce{Co^2+ + 2e- &-> Co} & E^\circ_2 &= \pu{-0.28 V} & \Delta G_2^\circ &= \pu{+54.0 kJ mol-1} \tag{2} \\
\end{align}$$
$$\begin{align}
\Delta G^\circ_\mathrm{net} &= \Delta G_1^\circ + \Delta G_2^\circ \\
&= \pu{-121.6 kJ mol-1} \\[3pt]
E^\circ_\mathrm{net} &= -\frac{\pu{-121.6 kJ mol-1}}{3F} \\
&= \pu{+0.42 V}
\end{align}$$


text:
Concentration from Absorbance vs Time
I am given a protein, BAPNA, and the initial concentration of the protein. The experiment involves reaction rates of varying protein concentrations. Each sample cuvette is inserted into a spectrometer, 100% transmittance is set, has the enzyme inserted, and then has transmittance measured every 20 s for 600 s.
I understand absorbance is given as: $A = 2 - \log (T\%)$ and that's no issue.
The problem I run into is trying to solve for the concentration or molar extinction constant. I understand that Beer-Lambert law, $A = \epsilon bc$  is used for this purpose. The only given concentration is the initial concentration. Assuming $b = 1$, we can solve for $\epsilon$ or $c$, but having 100% set in the beginning causes $A = 0$, meaning initially $\epsilon = 0$ if $c \neq 0$.
This is where my calculations run into a dead end. I'm not given $\epsilon$ or concentration at any other point, but I'm supposed to be able to calculate the final concentration and maximum reaction rate. I've only a set of transmittance numbers over time, and that initial concentration.
Is there something I am misunderstanding or not seeing?

The only given concentration is the initial concentration [...]

Unless there is a major misunderstanding on my side, you have all the data you need.
Given the initial concentration $c_0$ and knowing the path length $d$ of your cuvette, you measure the extinction $E$ at this point and use - as you have correctly stated in your question - the Lambert-Beer law to calculate the molar extinction coefficient $\epsilon$.
$$E = \epsilon\cdot c\cdot d$$
$$\epsilon = \frac{E}{c\cdot d}$$
Did I mention units? Units are your friend!

Write the Lambert-Beer law with units!
Percent (%) is NOT a proper unit for the initial concentration of your protein.

but I'm supposed to be able to calculate the final concentration and maximum reaction rate

Having determined $\epsilon$, you can now correlate at any point along your reaction the measured extinction with the actual concentration of your sample, including the final concentration.
From the slope of the curve of extinction $E\ $ vs concentration $c$, the rate of your reaction can be obtained.
If the relation is not a straight line, logarithms are your friend.

The problem I run into is that the experimental procedures do not call for a measurement of Extinction before the enzyme is added. Transmittance is set to 100 at that point, giving Extinction = 0. The equation becomes unsolvable at that point.


text:
Why only two atoms share an electron and not three?
In a covalent bond between two atoms, an electron from one of the either atom is shared by overlapping of their orbitals.
So,
Why can't three atoms share an electron and overlap their orbitals?

The normal case for a covalent bond is indeed a 2-centre 2-electron bond.
There are however cases, where 3 centres (= atoms) either share 2 or 4 electrons.

In organic chemistry, the most famous case for an electron-deficient 3-centre 2-electron bond is the 2-norbornyl cation.
In inorganic chemistry, boranes with their banana bonds are the role models.

Hypervalent molecules with 3-centre 4-electron bonds, like $\ce{SF6}$ or the bifluoride anion $\ce{HF2-}$ have been discussed here too.

@AwalGarg Maybe I misunderstood your question then. In a "normal" covalent bond, the two atoms (centres) share two electrons, one valence electron from each partner. They are not sharing one, but two electrons. It appeared to me that you were asking for the "less normal" cases.

@AwalGarg In organic chemistry, I'm not aware of anything more electron-deficient than the non-classical cations (3-centre 2-electrons) and I doubt that something like 5-centre 2-electrons exists even as an extremely short-lived intermediate. On the other hand, what about metallic bonds in a crystal lattice: Is this the case for an $\infty$-centre $\infty$-electrons system?

Ok, then let's take a step back and look at the organic electron deficient cases. In the case of the norbornyl cation, we could imagine that to be like a normal $\ce{C-C}$ bond plus a cationic centre near by that wants to mix in. For a larger number of centres, sharing the same number (2) of electrons, I see two problems (i) lack of binding force: what will hold them together and (ii) geometrical restrictions: how will all these atoms manage sufficient orbital overlap. So, unless proven otherwise, I'd say 3-centres 2-electrons is the end of the road here.

But I'm sure some guys here, well-versed in theoretical chemistry, might want to give some (less pictorial) answers too :)

authors: Αντώνιος Κελεσίδης


text:
Why is it necessary to avoid even traces of moisture from a Grignard reagent?
This fact was written in my text book. I wonder why that should be so.

Grignard reagents are highly reactive and react with any source of proton to give hydrocarbons. Even water, alcohols, amines are sufficiently acidic to convert them to corresponding hydrocarbons.
$\ce{RMgX + H2O -> RH + Mg(OH)X}$
Where, $\ce{R}$ is a hydrocarbon.

Aside from decomposing as described, Grignard formation will not initiate given a trace of water.  There is a bulky lore of getting the things started even under carefully dried conditions.  Consider an oven-dried roundbottom containing absolute ether, Mg turnings, and a small portion of the total amount of  alkyl halide.  It routinely sits there doing nothing.
Crush a turning to expose fresh surface. Add a small iodine crystal to corrode the Mg.  Heat the thing, oh yeah, then scramble for the exotherm.  Making Grignards spawns neurotic behavior like lucky lab coats, flame drying everything, always adding a drop of freshly distilled $\ce{Me3SiCl}$ (always adding a drop of old $\ce{Me3SiCl}$), saving some of the last batch to set off the next.

I have engaged in much of the neurotic behaviour described above. I now just try to use organolithiums for all instances where Grignards are proposed.

The first answer doesn't really answer the question. It merely shows, correctly, what happens to water which is introduced into a Grignard solution already prepared.
The second one hints at it. Water passifies the surface of the Mg (presumably coating it with MgO or Mg(OH)2) making it almost impossible to initiate the reaction. I have found that the techniques used in answer usually work.

The grignard reagent (magnesium halide) has then already formed. The passivation of the magnesium surface will not play a role.


text:
Conservation of angular momentum in electronic transition
My book says the "angular momentum is always conserved in a transition". But if suppose we have a s electron then it's angular momentum is $\hbar\sqrt{l(l+1)}+\hbar\sqrt{s(s+1)}$ $\implies \hbar\frac{\sqrt{3}}{2}$ and suppose it transits to a p electron (which is in accordance with the selection rule as $\Delta l=1$) which has an angular momentum of           $\hbar\sqrt2 + \hbar\frac{\sqrt{3}}{2}$. As the angular momentum of a photon is $\hbar$ then it means (mathematically) that we would have to add $\sqrt2$photons to the atom but that's not possible. Then how is the angular momentum conserved??
Also I can't see why the spin angular momentum of a photon is $\hbar$ as spin angular momentum is $\hbar\sqrt{s(s+1)}$ and as $s=1$ for a photon so it should be $\hbar\sqrt2$.
P.S. the book I am using is Elements of Physical Chemistry by Peter Atkins and Julia de Paula.

I think you have misunderstood some concept. The angular momentum of spin is just $\sqrt{s(s+1)}\hbar$. The spin which we always say is spin projection quantum number in fact not the angular momentum. And I'm not familiar with chemistry, but I know that we can not add up the orbital and spin angular momentum directly because they are not in the same direction. Sometimes we can add up the quantum number. If the spin and orbital angular momentum in opposite direction, we should subtract.

I have some idea about your example. The electron transition is "electron jumps from one energy level to another". So the energy must change. Where does the energy come from? If it is photon, this example has no paradox. But can the energy come from other way? I have no idea.

I had that in the back of my mind because orbital and spin angular momentum are both vectors but I thought it would be better to clarify here.

@ShrayanshJyoti I doesn't understand what you say. In fact, angular momentum in quantum is not a classical angular momentum.

"The classical definition of angular momentum as $\mathbf{L}=\mathbf{r}\times\mathbf{p}$ can be carried over to quantum mechanics, by reinterpreting r as the quantum position operator and p as the quantum momentum operator. L is then an operator, specifically called the orbital angular momentum operator."--Says Wikipedia.

Yes, that's right.


text:
What purpose do chain reactions serve?
I just learned what chain reactions are, but I can't seem to wrap my head around them because it seems utterly useless and doesn't accomplish anything. At least that's how I understood it because the textbook doesn't have any real world applications for it.
So what I am asking is do chain reactions serve any purpose. If so, it would be great if you could support your answer with a real world example or application.

Would you mind to give an example for an utterly useless chain reaction?

Are you joking, open your textbook and see nuclear reactor:). I was just joking, nuclear reactor is one of the major application of chain reaction, it is common in all textbooks.

I imagine you mean to say it's very hard to control the products of typical radical reactions. While true, it is also true that radical mechanisms can operate on molecules which are otherwise almost inert and very hard to derivatize, such as alkanes. C-H bond activation is a major issue in organic chemistry today, yet we've known for over 100 years that radical reactions can easily substitute hydrogen atoms linked to carbon. It just happens that radical C-H activation usually comes at the cost of low chemical selectivity and the formation of many byproducts.

[...] it would be great if you could support your answer with a real world example or application.

Free radical polymerizations of monomers with $\ce{C=C}$ double bonds are not completely useless ;)
Let me give some examples:

Polymethyl methacrylate (PMMA) is obtained by polymerization of methyl methacrylate (MMA). Think acrylic glass.
Styrene can be polymerized to polystyrene (PS). Think Styrodur and food containers.
Vinyl chloride is polymerized to polyvinyl chloride (PVC). Think pipes, insulation for electric cables, flooring.

Nuclear reactors and warheads, addition polymers (all hail Parylenes, too, and linear low density polyethylene, epoxy resins, truck liner beds, synthetic rubbers), genomes into proteomes, fire...to "gray goo."  Now, some footnotes.
http://en.wikipedia.org/wiki/Living_polymerization
CONTROL.  A free radical polymerization or classical Ziegler-Natta zips until it dies.  Big molecular weight spread puts physical properties all over the map.  Living polymerization is an non-terminating polymerization.  The hot end(s) reversibly cap, allowing blocks to be assembled such that component pieces and the whole are hard by identical molecule to molecule.  This is called "low dispersivity."  The polymer soles of your shoes are filled Kraton thermoplastic elastomers  (anionic living polymerization).  They are an identical short block of polystyrene, identical long block of disordered hydrocarbon, identical short block of polystyrene. It flows when hot, the polystyrene end segments being soluble in the melt.  It reversibly crosslinks as it cools, polystyrene domains coming out of solution and crystallizing.  There is no manufacturing waste.  Trimmings and rejects are remelted.  Physical properties are remarkably controllable and absolutely reproducible.  The polymer is inexpensive.
Polyethylene was a miracle, though an uncooperative one.  Dow's Insite catalysts changed the world with linear low density polyethylene.  The molecules now can be dialed into compositional anything, with molecular weight control.  Price/performance plummeted.  A plastic shopping bag weighs about five grams.  Less hair-thickness film has astounding strength. On the other end Dyneema and Spectra ultra-high molecular weight linear polyethylene, especially Plasma crystallized.  Much stronger than Kevlar/weight, does not hydrolyze, and it floats.
Chain reactions, their engineering and control, construct civilization on huge bulk scales.
For extra credit:  Why is there exactly one silicon atom in every molecule of Kraton triblock polymer regardless of molecular weight?  Where is it within the molecule?


text:
Low-volatile liquids that 'ball-up' on siliconoxide or perfluoroalkanes
This is a very simple and short question. I am looking for liquids that are low/non-volatile at 20$^\circ$C (at least substantially less volatile than water) which have a high contact angle with glass and oxidized silicon (in air) or with a coating of perfluoroalkanes (specifically Perfluorodecyltrichlorosilane).
The requirement on the contact angle is say $\theta>70^\circ$, i.e. the liquid should 'ball-up' a bit. The reason I am asking is that I cannot seem to find any (common) liquids that have both the low-volatility and the high contact angle $\theta$. Does anyone know some examples? Or, if no such thing exists, can someone explain what the fundamental reason is that these liquids don't exist?

Here it comes... mercury.  75-wt% Ga 25 wti%, mp. 15.5 C and
http://en.wikipedia.org/wiki/Galinstan
in general wet glass.  Maybe ionic liquids,
J. Am. Chem. Soc. 129(13), 3804 (2007), DOI: 10.1021/ja070169d
Aust. J. Chem. 57(2) 113-155 (2004)
http://www.researchgate.net/publication/233906460_Wetting_of_surfaces_with_ionic_liquids
http://dx.doi.org/10.1016/j.jcis.2008.10.019
http://ura.unisa.edu.au/R/?func=dbin-jump-full&object_id=63927
http://www.ppmp.pwr.wroc.pl/sorpa/file/eee7c6b05fb9497a9a59507ca0346479/download
[DMIM][Cl], 1,3-Dimethylimidazolium chloride, is reported 105.9 contact angle on glass.  That might be a typo - it's a huge number for the class of compound.
http://english.cas.cn/ST/BR/br_progress/201106/t20110622_71442.shtml

Also consider putting one or a pair of really long alkyl groups on the imidazolium to make it greasy toward glass, and with a bis(trifluoromethanesulfonyl)imide counterion.

Sulfate appears to be the internal binding energy opposite (very high) of bis(trifluoromethanesulfonyl)imide anion (very low), giving much larger surface tensions.


text:
Questions about Buffer System (Lab)
I will be doing a lab on Hydrogen Phosphate buffer system tomorrow and I have a few questions about the buffer system. I am not sure if my assumptions are correct.
(1) What would happen to the rate of pH change if 5.0M HCl and NaOH were used instead of 0.50M?
My assumption is that, since the pH of the buffer does not depend on the actual concentration of the buffer, but on the ratio of the two parts, therefore, the rate of pH should not expect to change. Is this correct?
(2) Lets say I prepared a buffer at a different pH, how would the capacity of my buffer system change?
My assumption is that, the buffer capacity would not change upon the change of pH, is this correct?

+1 Hi Jesse, this is how student questions should be, well done! What I don't understand is if you are adding HCl and NaOH together or if you are using a solution of HCl and then of NaOH to test your buffer. How was the original experiment?

Let's consider two buffer solutions of the same kind ($pK_a=4.75)$, but of different concentrations:

$\ce{CH3COOH}\,0.01\,\text{M}$ and $\ce{CH3COO-} 0.01\,\text{M}$
$\ce{CH3COOH}\,0.1\,\text{M}$ and $\ce{CH3COO-} 0.1\,\text{M}$

The starting buffer volume is the same for both and equal to $100\,\text{mL}$.
Now, add a strong base $\ce{NaOH}\,0.05\,\text{M}$, few $\text{mL}$ only: see below $pH$ as a function of $\ce{NaOH}$ added volume:

Have a look how "fast" the $pH$ changes and you will conclude that higher concentrations buffers are more effective. Differences can indeed be appreciated from the graph: they witness the evidence that larger concentrations better "buffer" the addition of the base.
I would suggest you to do some calculations (e.g. for $1\,\text{mL}$ of  $\ce{NaOH}\,0.05\,M$ added) and see whether you get the same values of the graph (hopefully I have done them right...). Remember that the addition of a strong base will provoke a decrease in concentration of the acid, $\ce{CH3COOH}$ and a consequent increase of its conjugated base, $\ce{CH3COO-}$.


text:
Strength identification in conjugate acid-base pairing
I'm a student in highschool AP chemistry. I'm confused about a certain problem in my chem textbook. In summary, I am given a species and asked to identify its basicity as strong, weak, or negligible. I must then give its conjugate acid and identify likewise for its acidity. (in the context of Brønsted-Lowry theory)
I understood all parts of the problem except for $\mbox{O}^{2-}$, to which I assigned the conjugate acid as $\mbox{OH}^{-}$. It was correct thus far. My guess that they were both weak was wrong, however. According to the solutions page, they're both strong!
How is $\mbox{O}^{2-}$ a strong base? How is $\mbox{OH}^{-}$ a strong acid? To my knowledge, neither are found in the canonical strong acid/base lists, and neither seem to satisfy the "completely ionize in water" definition.

$\ce{H2O}$ is a weak acid.  Abstract a proton to give $\ce{HO^{-}}$, a strong base.  Abstract a proton from that to give $\ce{O^{2-}}$, a very strong base.  See a species as a whole, up and down for a property.
Try it with ammonia.  Feel free to wonder why $\ce{Li3N}$ is red.  Look what a very strong base can do,
$\ce{Li3N + 2 H2 → LiNH2 + 2 LiH}$ (reversible at 270°C)

But why is OH- a strong base? What criteria does it meet for strength?

Hydroxide is the strongest base allowed in water. Stronger bases give you hydroxide plus their conjugate acid. For stronger bases than hydroxide use a less acid solvent, e.g.,  liquid ammonia, or dipolar aprotics like DMSO, HMPA, possibly DMF or MeCN.


text:
Why does sweetening milk need less sugar as compared to tea?
If I have a cup of milk and a cup of tea; to get them to a sweetness I like, I usually add less sugar to the milk as compared to the tea.
Why is it that milk needs less sugar to be just as sweet as tea?

You might also like to consider asking this question at http://cooking.stackexchange.com/

It's all in your taste buds. Lactose in milk is a form of sugar. It is naturally a little sweet. Tea is slightly bitter so it might seem you need to add more sugar to make it sweet.

Additionally, uncalibrated mouth (see http://www.iso.org/iso/home/store/catalogue_ics/catalogue_ics_browse.htm?ICS1=67&ICS2=240 ) is very simply misguided by bitterness of the tea. It is difficult to assess the sweetness independently of other tastes. But it is possible after training.

I pretty much agree with "Mistergreen" The Lactose in milk does make it slightly sweet and the Tannic acid in tea makes it bitter and a bit chalky. When making Yogurt, the milk needs to be heated in order to kill the bacteria present in the milk. Lactobacillus bulgaricus and Streptococcus thermophilus are not the only bacteria that can convert lactose into lactic acid although they are the most commonly added bacterias for Yogurt making. it is the conversion of the Lactose to Lactic Acid that makes the milk in Yogurt Tart and Not Sweet Therefore before you start making yogurt, it is necessary to heat the milk so that the only bacteria it contains are the ones that are added. Unbeknownst to many, the other reason to heat the milk to 180-185 deg. F. is to change the structure of the Caseins (proteins) contained in the milk among other protein like whey. If you fail to heat the milk to an adequate temperature for at least several minutes, the Yogurt produced will be thin and can have a granular texture to it. I have made my own Yogurt for over 20 years and it is a little bit of an art form as well as a Chemistry Experiment. I normally use a Greek Yogurt for a started that contains the aforementioned Bacteria as well as well as Lactobacillus Acidophilus; Bifidobacterium Bifidus; Lactobacillus Casei. the only other ingredient in the Yogurt I use for a starter is nonfat pasteurized milk. I do use greek yogurt. The difference is that it has been strained through a fine cloth that feels like lightweight and very soft canvas. There is other cloth like cheese cloth that are not as fine a weave or you can use a very fine chinois, But I have found that the cloth I use makes a much better textured Yogurt. The Greek Yogurt Process is one step longer, but i prefer it due to the reduced Carbohydrates and it doesn't filter all of the Whey out.If I use my own Yogurt as a starter, the resulting product tends to increase as the I use it more and more and then after 6-7 uses it tends and I use fresh starter. I often use fresh starter when I want a milder tasting yogurt that will get a bit more tart as it sits in the refrigerator over a week or so. Thatit. Sorry i tend to ramble, I hope that anyone who like yogurt will try making their own. Just heat milk to 180-185 deg. F. for 3-5 min. Cool to 100 Deg. F pull out a cup or so of the milk and wix 102 Tbs. in it, mix well to get out all lumps. Then set it is in a place that is between 90 deg.F. and 110 Deg F. (I use a gas stove with the light on and a tea candle or two on a plate away from the yogurt containers. sit at Temp for 6-10 hours and cool it in fridge, then strain if you want Greek Yogurt. The more you strain the thicker it will be. Just use a hand electric mixer to blend iot to a creamy state. Namate, Brian


text:
Volume required to dilute solution for a pH change

A $\pu{100 mL}$ solution of $\ce{NaOH}$ has a $\mathrm{pH}$ of $13$. What volume of water in $\pu{mL}$ must be added to change the $\mathrm{pH}$ to $11$?

My steps:

Begin by calculating conc. of $[\ce{OH-}]$ for $\mathrm{pH}~13$.
Find the moles by multiplying concentration and volume ($\pu{0.1 L}$)
Determine new concentration of $[\ce{OH-}]$ for $\mathrm{pH}~11$.
Use $\text{amount}/\text{new concentration}$ to find the total volume.
Use total volume ($\pu{100 mL}$) to find volume needed to be added.

The above doesn't produce the right answer, which is $\pu{9.900 mL}$. Can someone please fix my working out, or provide alternative steps to calculate this?

Is this really just a simplifying approximation.  Using the same logic, if I diluted this to a total volume of 1000 liters, I would have a pH of 9, and diluting it to 100,000 liters would yield a pH of 7.  If I diluted it to a total volume of 1,000,000 liters then the pH would be 6.  I'm sure that no matter how much you dilute it, it would always be basic, or at least never acidic.  So is the formula that you showed just an approximation that works for smaller changes in pH, but doesn't extrapolate for larger changes?

Since the amount of substance stays equal, you can calculate the volume of your final solution $V_\mathrm{end}$, form your start concentration $c_\mathrm{start}(\ce{OH-})$ and your initial volume $V_\mathrm{start}$ and your target concentration $c_\mathrm{end}(\ce{OH-})$. Your working equation should be:
\begin{align}
c_\mathrm{start}(\ce{OH-})\cdot V_\mathrm{start}
&= c_\mathrm{end}(\ce{OH-})\cdot V_\mathrm{end} \\
V_\mathrm{end} &= \frac{c_\mathrm{start}(\ce{OH-})}{c_\mathrm{end}(\ce{OH-})}\cdot V_\mathrm{start}
\end{align}
You can work out the start concentration for the initial pH and the target concentration from your final pH. Given (at $T=298.15~\mathrm{K}$, $p=1~\mathrm{atm}$):
\begin{align}
\ce{pH}_\mathrm{start} &= 13 \\
c_\mathrm{start}(\ce{OH-}) &= 10^{-(14-\ce{pH})}~\mathrm{mol/L} = 0.1~\mathrm{mol/L} \\
V_\mathrm{start} &= 0.1~\mathrm{L} \\
\ce{pH}_\mathrm{end} &=11 \\
c_\mathrm{end}(\ce{OH-}) &=10^{-(14-\ce{pH})}~\mathrm{mol/L} = 0.001~\mathrm{mol/L}
\end{align}
Hence
$$
V_\mathrm{end}= \frac{c_\mathrm{start}(\ce{OH-})}{c_\mathrm{end}(\ce{OH-})}\cdot V_\mathrm{start} = 10~\mathrm{L}
$$
and therefore
$$
\Delta V = V_\mathrm{end}-V_\mathrm{start} = 9.9~\mathrm{L}= 9900~\mathrm{mL}.
$$

instead of 10 L i kept writing 10 mL at the end.. -.-

Ah! You're confused.

Here is a different approach:
$\ce{NaOH}$ is a strong base and will be completely dissociated in water. A solution of $\ce{NaOH}$ with $\mathrm{pH} = 13$ will contain $\pu{0.10 M}$ $\ce{NaOH}$, since
$$[\ce{OH-}] = 0.10 \implies \mathrm{pOH} = 1.0 \implies \mathrm{pH} = 14.0 - 1.0 = 13.0.$$
What will the $\mathrm{pH}$ be of a $\pu{0.0010 M}$ solution of $\ce{NaOH}$?
We estimate the $\mathrm{pH}$ by assuming the water autoprotolysis will not significantly affect the $\mathrm{pH}$.
$$[\ce{OH-}] = \pu{0.0010 M} \implies \mathrm{pOH} = 3.0 \implies \mathrm{pH} = 14.0 - 3.0 = 11.0.$$
So, I suggest you dilute the original solution of $\pu{100 mL}$ $\ce{NaOH}$ $100$ times, i.e. you should add $\pu{9900 mL}$ of water.

Using the definition of $\mathrm{pH}$ (which is a log value) i.e. the difference of $\mathrm{pH}~1$ value is 10 folds difference in real terms.
You have $\pu{100 ml}$ of $\ce{NaOH}$ @ $\mathrm{pH}~13.$ If you dilute by $10$ times, then the new $\mathrm{pH}$ will be $12,$ so you'll need to dilute it another $10$ times to reach $\mathrm{pH}~11.$ If you start with $\pu{100 ml},$ you'll need to end with $10$ litres in total ($100$ folds). You already have your $\pu{100 ml},$ so you'll need to add $\pu{10000 ml} - \pu{100 ml} = \pu{9900 ml}.$
Obviously, if the required final $\mathrm{pH}$ had fractions this simple mental arithmetic will not work.


text:
Calculate the moles of acid and conjugate base needed
Question

Using the following salts listed below, calculate the moles of acid and conjugate base needed to make $V = \pu{100 mL}$ of $c = \pu{0.50 M}$, $pH \in 6.5, 6.7, 6.8$ phosphate buffer solutions. ($\mathrm pK_\mathrm a(\ce{H2PO4}) = 6.64$)
Molas masses of salts:

$\ce{KH2PO4} = 136.09 ~\pu{g/mol}$
$\ce{NaH2PO\cdot H2O} = \pu{137.99 g/mol}$
$\ce{Na2HPO4 \cdot 7H2O} = \pu{268.09 g/mol}$
$\ce{K2HPO4} = \pu{174.2 g/mol}$

I tried using the Henderson-Hasselbalch equation but I am stuck at the point where I have found the base-acid ratio and I need to find the moles of the acid and the moles of the base.
$$
\mathrm pH = \mathrm pK_\mathrm a + \log\left(\frac{[\text{Conjugate base}]}{[\text{Acid}]}\right)\\
n = \pu{100 ml} \times \pu{0.5 M} =\pu{ 0.05 mol}
$$
For pH 6.5:
$$
6.5 = 6.64 + \log\frac{[B]}{[A]}\\
\frac{[B]}{[A]} = 0.72
$$
Here I got the base-acid ratio as 0.72 but what should I do next? I tried looking up for solutions and a person suggests using $A + B = 0.05$ But how do I know what A and B are just by knowing the ratio is 0.72?

You can use substitition, [B] = 0.5 - [A].

One thing has to be explained by the one who put up this question before a tangible answer without assumption can be made. The question mentions 100mL of c = 0.50M. What substance is c representing? It is molarity, but molarity of what chemical? Buffers have two chemicals in them, the weak acid and the salt which has the same anion (negatively charged species) as the weak acid. That anion is called the common ion. So if you say molarity is 0.50M, which of the two species are you referring to, the weak acid (which from your pKa should be H2PO4-) or the salt (which in this case should be HPO4--)

Since it is a buffer solution, the volume is constant ($V =\pu{0.1 L}$), hence
$$
\frac{c(b)}{c(a)} = \frac{n(b)}{n(a)}\overset{calc.}{=}0.72
$$
 and with
$$
n_{\text{tot}}=n(b)+n(a) \overset{\text{def}.}{=}\pu{0.05 mol}\\
n(a)=\frac1{1.72}\cdot\pu{0.05 mol}\\
n(b)=\frac{0.72}{1.72}\cdot\pu{0.05 mol}
$$
Then $$m=M\cdot n$$ and you should be fine.


text:
What is the origin of the colour of azo dyes?
Is it the same reason as to why transition metal complexes have colour?

What do you mean for same reason?

Yes, it is all about the absorption of light at specific wavelength.

Azobenzene, the parent compound has an absorption maximum around $\lambda$= 430 nm in the visible spectrum.
The interesting part is: The absorption can be tuned by substitution of the arenes. This is done before the azo coupling.

Some examples are Allura Red (1), Chrysoine Resorcinol (2), Janus Green B (3) and Direct Blue 1 (4).
The colours of the molecules are chosen to resemble the colour in solution.
In order to make the dyes soluble in water, either sulfonate groups are attached or the dye molecule is just one fat cation.
As already mentioned by GM in another answer, +M substituents elongate the $\pi$ system (auxochromic effect).

Basically it is an application of the particle in a box model.

I think is not right to say that is the same reason whereby transition metal complexes have colour, are involved d-orbitals? However maybe they are both conjugated system and so are similar?

@GM I disagree. The fundamental point isn't about d-orbitals, but absorption in the VIS range, resulting in the complementary colour observable.

@KlausWarzecha  this is of course true, but from this point of view most of the substances has their color for the same reason... And you differentiate it only from color due to interference phenomena... I've interpret the answer at a deeper level: understanding if the color is due to a similar change in electronic distribution... I will ask to the user... Thanks for the clarification!

@Klaus Warzecha  I understand that the azo-dye absorbs a specific frequency of light and we see the complementary colour. However, I am confused about WHAT absorbs the light. For e.g. in transition metals, there is a d-d transition of an electron. I asked my teacher, she told me that it has something to do with the electron cloud of the ring that absorbs a frequency from visible light and goes to a higher energy level- that didn't make sense to me :/

@Eliza Your teacher is right, no d-d transition here. But $\pi$ systems. Aromatic systems (remember your question about the resonance stabilization) and double bonds. More double bonds in conjugation: better. Ethene absorbs at $\lambda$= 160 nm, 1,3-butadiene at 285 nm, trans-stilbene has a maximum at 295 nm. We have transitions from a bonding to an "antibonding" (not occupied) $\pi$ orbital here :) That's the "electron cloud" your teacher mentioned.

@Eliza Admittedly, that's part of the truth ;) In azobenzene, we have the lone pairs on the nitrogen atoms too. These are not isolated (the nitrogen atoms are part of a $\ce{Phenyl-N=N-Phenyl}$) system and can also be excited). But we will not talk about $n\rightarrow\pi^*$ transitions here ;)

@ Klaus Warzecha: This is so interesting :D So, let me get this straight... the $π$ system of the phenyl ring absorbs a frequency in the visible spectrum and promotes to a empty higher energy state $π$ system... we just see the complementary color. So is this why the more the number of phenyl rings in an azo dye, the more intense is the color? (Just curious)

It's not just one phenyl. The aromatic parts are in conjugation through the azo group. It's one $pi$ system. The bigger that gets, the higher is the energy of the highest occupied molecular orbital (HOMO). At the same time, the LUMO (lowest unoccupied molecular orbital) gets lower in energy. Electron excitation between these two determines the colour. In a simple absorption process, the energy absorbed by the molecule would be released in form of heat. Finished. Here, the very intense colour results from light emitted from the excited dyes. We are talking about fluorescence now :)

@Klaus Warzecha : So in $Phenyl−N=N−Phenyl$, the delocalised  $π$ electrons actually merge together to make 1 $π$ system.. This ENTIRE π system absorbs energy from visible spectrum :O However, I am confused with one thing: do we see the complementary color or the one the molecule absorbed... the fluorescence part confused me :/

@Eliza Let's better refer to real compounds ;) Direct Blue 1 definitely is fluorescent. With all the curiousity (which I like), your teacher will have to work real hard for the money :D

What is the origin of colours?
Most of the colours that we perceive are originate by the selective absorption of some spectral bands and the reflection of the others wavelength, some times with the contribution of fluorescence from the absorption at a higher wavelength. If we exclude colour due to interference (e.g. some butterfly wings!) all the other phenomena are linked with the absorption of the photons. However, this absorption could be due to different causes and so we can differentiate between different absorption mechanisms.
Absorption of electromagnetic waves through bonding and anti-bonding $\pi$ and $\sigma$ orbitals

Note: I suppose you are familiar with MO theory

One very spread mechanism of absorption of electromagnetic waves is through electrons in pi and sigma orbitals. The absorption occurs at different wavelengths depending on the compound but however, most of the transitions occur normally in the UV region (400nm is the limit of VIS region). I have made this picture with TikZ from this site you can see that only $n \rightarrow \pi^{*}$ is in the visible region.
How can azo dyes absorb light?
In the case of azo dyes, the Chromophore is $\ce{-N=N}-$. This group permits the absorption of light. However, Azobenzene has its absorption peak in UV region (it only absorbs a little bit of blue).

So how can azo dyes absorb light (of course for light we means 400-700 nm radiation) and so be colourful? One way to do that is to form a bigger conjugated system to lower the energy, in this case delocalizing the electrons across the molecule. For azobenzene you can also add two auxochromes the hydroxyl group ($-OH$) and the amino group ($-NH_{2}$) these groups permit the formation of a charge-transfer complex CT Complex that should be something like this:

Is it the same reason as to why transition metal complexes have colour?
I'm not sure that is right to say that is the same reason whereby transition metal complexes have colour. However, this is a tricky question because we have seen that there are many factors that cause the absorption so of course, a transition metal can be considered a conjugated system as said Dan S, most of the time, however, these interactions are called  ligand-to-metal charge-transfer (LMCT).

I disagree on azobenzene being colourless. Are you certain that you had a look at the UV-VIS spectrum of azobenzene and not silbene, which indeed is colourless?

@KlausWarzecha You are right my professor told me that was colorless but in fact absorbs some blue... Thanks for the comment..

We're talking about "azobenzene" here the whole time, but actually we are having two: trans (the more stable one) and cis. We could even isomerize trans -> cis upon irradiation in the 300 nm region, and cis back to trans thermally or by irradiation around 440 nm. I would have to look up the reference though.

Found it: DOI. Sandra Monti and Pietro Bortolus from Bologna, where Ciamician (!) worked.

@G M: Just one more thing, my teacher told me that the more the number of phenyl rings in an azo dye, the more intense is the colour. If $\ce{-N=N}-$ is what absorbs the frequency of light and we see the complementary colour... how does the number of phenyl rings affect the colour we see?

@Eliza Increasing the number of phenyl increase the delocalization of the electrons (you have a bigger conjugate system, se my edit answer) you should have a shift to lower wavelength absorption but I'm not sure that you will have a more intense color only a different nuance.

Is it the same as transition metal colors? If you define this as "an electron is promoted to a higher energy state" and that photon absorption defines the color then the answer is yes.  If you define it as an absorption by d orbitals then the answer is no.  As others have indicated these dyes can absorb in the visible wavelengths due to the aromatic delocalized pi electrons.  These are electrons that are shared over many atoms.  In a transition metal you are simply promoting electrons in the d shell from one state to another.  The electrons aren't "wandering" over a bunch of other atoms.

On the subject of the comparison between colours of azo dyes vs transition metals, I was going over the below question w a student and the mark scheme threw me. Specifically the idea of "split energy levels". For the same marking point, the other options make more sense, ie: an electron is promoted to a higher energy level with an energy corresponding to a wavelength of visible light. I just thought the idea of the energy levels of the delocalised electrons being split was more of a transition metal explanation than an azo dye explanation. Interested to hear anyone else's thoughts.


text:
How much potassium permanganate is needed to completely oxidize ferrous oxalate?

Amount of substance of $\ce{MnO4-}$ required to oxidize one mole of ferrous oxalate completely in acid medium will be:
(a) 7.5 moles
(b) 0.2 moles
(c) 0.6 moles
(d) 0.4 moles

In acidic medium, 1 mole of $\ce{MnO_4^-}$ accepts 5 moles of electrons. Oxalate is dibasic so will give 2 moles of electrons when oxidized to $\ce{CO2}$.
$\ce{Fe^2+ -> Fe^3+ + e-} $ so total 3 moles of electrons.
$\text{Amount of substance of } \ce{MnO_4^-}=\frac{3}{5}=0.6$
However, the given answer is 0.4 moles indicating that only oxalate is considered to be oxidized.

Does  $\ce{Fe^{2+} -> Fe^{3+} + e^-} $  actually occur in this reaction?
What would be the net reaction when $\ce{KMnO_4}$ is used(not just the ions)?

See here.
The "green salt" is $\ce{K3Fe(C2O4)3.3 H2O}$.
\begin{align}
\ce{MnO4- + 8H+ +5e- &-> Mn^2+ + 4 H2O}\\
\ce{C2O4^2- &-> 2 CO2 + 2e-}\\
\ce{2 MnO4^- + 16 H+ + 5 C2O4^2- &-> 2 Mn^2+ + 8 H2O + 10 CO2 (g)}\\
\end{align}
One expects all the oxalate is oxidized to $\ce{CO2}$ and $\ce{Fe(II)}$ is oxidized to $\ce{Fe(III)}$.


text:
What is the difference between doxycycline HCl and doxycycline hyclate?
I would like to ask what are the main differences between doxycycline HCl and hyclate?
I work in the pharmaceutical/veterinary raw materials industry and some clients are saying that they prefer hyclate because it is more concentrated than HCl, but I have yet to come on any articles confirming this. Is this true?

[...]they prefer hyclate because it crystallizes more readily than HCl salt

The principal component is doxycycline - in both cases. The active component is a metabolite.
Doxycycline is an antibiotic and belongs to the tetracyclines (= four rings).

In order to prepare nice ions that can traverse membranes and pumps by the body easily, pharmaceutics containing amino groups or 'alkaloids' (see the red nitrogen atom in the structure), are often converted to their hydrochloride, citrate, oxalate, bromide etc. salts.
These are salts with the (protonated) pharmaceutical as the cation and chloride as the anion. Salts are water soluble.
Here, the notation for this is doxycycline$\ce{.HCl}$.
The crystallisation is done in a solvent or a mixture of solvents.
When the doxycycline$\ce{.HCl}$ precipitates (=when it forms a solid), molecules of the solvent may get embedded, trapped, or entrained in the crystal structure (but the crystals gets pulverized anyway and re-crystallized).
Note that this is not just a wet solid! There would be little or no absorption in that case.
Exactly this happens when the hyclate is formed during the crystallization of doxycycline hydrochloride in a mixture of water and ethanol.
Hyclate is just an abbreviation for hydrochloride hemiethanolate hemihydrate, which means that each molecule of doxycycline comes with 1 $\ce{.HCl\ }$  and 0.5 $\ce{.CH3CH2OH}$ and 0.5 $\ce{.H2O}$ in the crystal.
Whether it is more concentrated (= more active component in one pill) only depends on the formulation of the respective pharmaceutical companies.

hi thanks for answering! but can you please explain in simpler terms, sorry but i am having trouble understanding your answer.. so if i understand correctly doxycycline hyclate is just the same as hcl?

@user1928162 No, they are not the same. Maybe the answer is more clear now.

I think that crystallisation of pharmaceutically active ingredients is not required for easier indigestion but for better and easier quality control. I’m not sure, though.

I believe the doxycycline hyclate have better absorption than other forms when ingested orally. In the hyclate, the bond between the doxycycline and ethanol breaks in a specific area(middle part if am not mistaken) of the small intestine where there are less Calcium and Magnesium salts, therefore less chelation and more of the compound for which doxycycline was a prodrug available to the capillaries and microvilli to be absorbed along with food and chylomicrons.

hyclate because it is more concentrated than HCl

Looking at on ChEBI it appears that a 1 mole, of "doxycycline hyclate" provides two moles of "doxycycline"1. Compared to "doxycycline HCl" which only provides one 2.
@klaus-dieter already says this above, but I think part of the answer your after is that there is twice the amount of doxycycline when you compare molar quantities of the substances. In reality, the actual strength of drugs is likely to be expressed in mass units, and quite possible based on "doxycycline".
E.g. doxycycline (as Hydrochloride) 100 mg or doxycycline (as hylate) 100 mg


text:
Most common material in the universe
I was told that 'water' was the most common material in the universe and I disputed that claim. I was met with the argument that hydrogen and oxygen were the most common atoms in the world so water was as well. If this is true, then how come we can't find water anywhere in outer space? It must be the collection of 2 hydrogen atoms and 1 oxygen atom that is so rare then?

Try to read this wiki http://en.wikipedia.org/wiki/Abundance_of_the_chemical_elements#Abundance_of_elements_in_the_Universe)(http://en.wikipedia.org/wiki/Abundance_of_the_chemical_elements#Abundance_of_elements_in_the_Universe

It is true that hydrogen is the most common atom in the Universe, and oxygen is one of the most common, too (though a couple orders of magnitude rarer). If that's the case, then it should be apparent that $\ce{H2}$ should be more common than $\ce{H2O}$, since the latter has the same amount of hydrogen but requires one more atom. Indeed, according to Wikipedia, hydrogen gas is the most common molecule in interstellar space, followed by carbon monoxide (presumably because of the exceptionally strong bond in $\ce{CO}$).

The Big Bang started with about 3/4 hydrogen atoms and 1/4 helium atoms.  Add a pinch of lithium and a whisper of boron.  Supernovae since then have released a lot of helium.  Heavier elements are still rare.
"The most common [molecular] material in the universe" must survive photolysis, radiolysis, and reaction.  Water is poor candidate.
http://www2.astro.psu.edu/users/alex/astro497_7.pdf
What is out there.  CO is the marker.
http://physics.nist.gov/cgi-bin/micro/table5/start.pl
Molecules in space, isotopically and by
http://science.gsfc.nasa.gov/691/cosmicice/interstellar.html
http://www.astro.uni-koeln.de/cdms/molecules
atom count


text:
Does potassium aluminum sulfate have any effect on sexual performance/sperm counts or anything relating to the functionality of the sexual organs?
I recently read about potassium aluminum sulfate that it has astringent property and has the ability to constrict body tissues and restrict blood flow. Since, for penis, the erection happens with the flow of the blood being directed to penis' blood vessels, does this effect of potassium aluminum sulfate cause an erection problem or any other effect on male/female sex organ?

Oral ingestion of simple aluminum compounds is not absorbed.  Topical genital application is definitely not recommended.  There is no obvious connection between aluminum and tumescence, one way or another
Soluble aluminum is toxic, likely cumulatively neurotoxic.  It is not absorbed from the GI tract unless complexed.  Aluminum hydroxide antacids as foaming tablets( citric acid plus sodium bicarbonate) gave aluminum citrate that is bioavailable.  Bad idea.  Pickles are made with alum to keep them ever so crunchy. The soluble aluminum is claimed not to be absorbed. Antiperspirant contains aluminum chlorhydroxide or glycinate to stopper sweat glands.  Topical application is not absorbed.  Exposing mucus membranes to soluble aluminum is definitely not recommended.


text:
Why is the bond length of CO+ less than that of CO?
According to molecular orbital theory, the bond order of $\ce{CO}$ is 3. When $\ce{CO+}$ is formed, the bond order decreases to 2.5, and thus the bond length should increase.
However, the bond length of $\ce{CO+}$ is found to be less than $\ce{CO}$. How is that so?

The HOMO of CO has slight anti bonding character. That might be the reason for this obeservation.

Very related question: Why is the bond order of CO+ 3.5?

Unfortunately, the arguments presented by buckminst and Uncle Al aren't completely right. The MO schemes are correct but the HOMO-$\sigma$ orbital ($s_{\sigma}^{*}(5\sigma)$ in buckminst's diagramm, $\sigma_{3}$ in Uncle Al's diagram) is not antibonding but slightly bonding in character because there is some mixing with the $\ce{p}$ atomic orbitals of the right symmetry (in this respect the MO scheme of Uncle Al is better than the one of buckminst because it indicates this mixing).
So, the original problem still persists.
One possible explanation for the bond shortening after ionization is that the ionization leads to a shift of the $\ce{CO}$-electron-polarization (on ionization an electron is lost from the mostly $\ce{C}$-centered HOMO-$\sigma$ orbital and this leads to a formation of a positive partial charge on the $\ce{C}$ atom).
This strengthens the covalence of the $\ce{CO}$-bond and thus reduces the bond length (since the HOMO-$\sigma$ orbital is only slightly bonding there is not much bonding lost by taking away one of its electrons - the lost bonding is outweighed by the gain in covalency).
You can think of this strenghtening of the bond covalence in the following way: Two atomic orbitals can interact better (form stronger bonds) if their energies are close. Without the positive partial charge on $\ce{C}$ the AOs of $\ce{O}$ lie energetically a lot below the AOs of $\ce{C}$ (this is most pronounced for the $\ce{s}$ AOs which form the  HOMO-$\sigma*$ orbital). But with the positive partial charge on $\ce{C}$ the AOs of $\ce{C}$ are shifted down in energy and thus their energies are closer to the related AOs of $\ce{O}$ which leads to a stronger interaction when bonds are formed.

(+1) More insight on this topic from a theoretical point of view: here. The positive charge at the Carbon also causes the strong $\pi$-bonds to be more polarized towards the Carbon, making the bond stronger.

@Martin Thanks for providing the reference. I guess it's not by chance that you give a reference written by a professor from Marburg :)

I just went through your post again (unfortunately I already upvoted on it) and I noticed, that there might be an important argument missing: Upon ionisation, the nuclear charge in relation to the electronic charge is higher. Hence the orbitals have to become more contracted as the fewer electrons are more attracted. So another reason would be plain and simple electrostatics. (Upon ionisation, the former HOMO of CO also drops below the $\pi$ orbitals in CO(+1) - missing spin pairing, lower electron repulsion, and $\sigma$ is more contracted than $\pi$.)

So the other two answers are incorrect?

@Martin I'm not sure whether I understand your comment correctly. Are you saying that my answer is incorrect or that it might only be missing the electrostatic argument you're making? Also, you state "Upon ionisation, the former HOMO of CO also drops below the $\pi$ orbitals in CO(+1)". That is interesting: Have you calculated this yourself or how do you know that? But I fear I'm not able to draw the connection between the higher effective nuclear charge leading to more contracted orbitals and ionized CO having a shorter bond length than normal CO. Could you eleborate on this some more, pls?

@Dissenter In my opinion the other answers are indeed incorrect for the reasons stated in my own answer. Have a look at the article linked by Martin in his topmost comment. It explains things quite neatly and in more depth than I was able to provide.

My comment is complementary to your already correct answer (which I would have upvoted again, if I had not done it already. In retrospect my comment was articulated badly). I have to think a little more about the contraction part and try to feed it with some data. I might post an answer later on. About the HOMO part, I will check that with a different method again, I do not want to jump to conclusions. Since CO is an awesome simple molecule, too complicated for a lot of computations, I want to make that double dead sure, but yes that statement was based on a result of a calculation I did.

@Martin Thanks for the effort. I will certainly look forward to an answer of yours. And if there is another effect at work here that needs to be considered I will gladly add this to my portfolio of things to watch out for when analysing a bonding situation - I'm always very interested in things like that :)

http://www.meta-synthesis.com/webbook/39_diatomics/co.jpg
Measured $\ce{CO}$ bond length is 1.128 Å, $\ce{CO^{+}}$ is 1.115 Å.  Given the following carbon monoxide MO diagram, from where is the electron removed to afford the cation with a shorter bond length?  $\ce{CO}$ has a dipole moment of 0.122 debyes, net negative charge on the carbon. One then suspects the electron is removed from the sigma anti-bonding orbital on carbon.

This is getting exciting!  IR stretch:  CO, 2143 cm^(-1), CO+, 2184 cm^(-1) - definitely a tighter bond.  The cation is more strongly bound, http://131.104.156.23/Lectures/331/331_chapter_6.htm  In LCAO terms, perhaps the sp-hybrid leaves two non-bonding lobes.  The carbon lobe has more than half the electron density. and preferentially ionizes.

There was definitely more to this question than I initially thought. The matter has some aspects that are not very intuitive - the MO diagram of CO makes the answer seem obvious but actually leads to the wrong reasoning (see my answer).

Molecular Orbital theory does NOT predict a longer bond. It correctly predicts a shorter bond. The HOMO for $\ce{CO}$ is a sigma antibonding orbital. When it is ionized, the bond order is increased to 3.5, and the bond length decreases.

@UncleAl it is not. But the orbitals are nice. Look at the shape of $s_{\sigma}*$ orbital and try forming that from only two s-orbitals.


text:
Charge of polyatomic ions doesn't equal the charge of ions it consists of?
I'm taking entry level collage chem. class, and we got to this topic that I don't really understand. I think I understood what an ion is (a atom whose amount of electrons is not equal to amount of protons, so it has positive or negative charge). But then polyatomic ions... Lets say peroxide; consists of two Oxygens, whose ions have charge of -2. Consequently, two of these anions are suppose to have charge of -4, but peroxide has charge of -2. How does this work? If polyatomic ions don't just sum the charge of ions they consist of, how else can I predict and understand their charge?

H-O-O-H.  The bond in the middle is covalent.  Deprotonate to $\ce{{HOO}^{-}}$ (hydroperoxide) then $\ce^{^{-}{OO}^{-}}$, peroxide dianion, $\ce{O2^{2-}}$.  You learn it by doing it, having examples, learning what to look for, and gaining understanding.  Both oxygens are in the -1 oxidation state.
Even good answers here have a legitimate diversity of explanation.  The world is complicated and dirty.  Unlike physicists (quantum gravitation, SUSY, dark matter), we can test our predictions with experiments.  Chemists are not too stinking proud to admit we were wrong, e.g., benzene displacements then palladium catalysis.  That is how we make real world progress, by discovering footnotes.
Introductory inorganic does not have wiggle room.

I would not consider peroxide to be a 'standard' polyatomic oxo anion, as Uncle Al describes.


text:
Why does free chlorine in the stratosphere lose its ozone-depleting potential after about 100,000 reactions?
Free chlorine ($\ce{Cl}$) in the stratosphere can deplete ozone ($\ce{O3}$)  as follows:
$$\ce{Cl + O3 -> ClO + O2}$$
The chlorine atoms can then react with oxygen and return to the beginning of the cycle:
$$\ce{ClO + O -> Cl + O2}$$
The EPA estimates the ozone-depleting potential of a single chlorine atom:

It is estimated that one chlorine atom can destroy over 100,000 ozone molecules before it is removed from the stratosphere.

From the equations above, it seems that each chlorine atom returns to its original form. Why, after about 100,000 cycles, does it lose its ozone-depleting potential?
As a side question, how is this estimated?

Since the stratosphere is nowhere near a closed system, a chlorine atom will eventually leave it. Look at the phrasing again:

It is estimated that one chlorine atom can destroy over 100,000 ozone molecules before it is removed from the stratosphere.

It is not stated, that it looses its potential, it just leaves the region, where there is sufficient ozone concentration.
In a closed system, the catalytic cycle would continue until the ozone concentration is small enough, that the chance of chlorine atoms recombining is bigger.
Thanks to Nicolau's comment I decided to further look into the issue. Also qwersjc gave some valuable insights. For further reading I suggest, the following sources:

http://www.nobelprize.org/nobel_prizes/chemistry/laureates/1995/ (overview)
Nobel Lecture by S. Rowland (Free Access)
Stratospheric Chlorine: a Possible Sink for Ozone. R. S. Stolarski, R. J. Cicerone; Canadian Journal of Chemistry, 1974, 52(8): 1610-1615. (Open access)
Original Paper by M. J. Molina and F. S. Rowland; Nature 249, 810 - 812 (28 June 1974) (Might not be free), reprint on ozone.unep.org

The Ozon Cycle
The ozone cycle is what protects us from quite dangerous UV-B radiation ($M$ is an inert Air molecule like $\ce{N2}$):
$$
\ce{O2 ->[h\nu] O + O} \\
\ce{O + O2 ->[<M>] O3}\\
\ce{O3 ->[h\nu] O + O2}
$$
The Troublemakers
The main reason for chlorine atoms in the stratosphere is man made. Chlorofluorocarbons were very common in many products. These molecules are colourless and water-insoluble. Diffusion taking its course the end up in the stratosphere, where they may undergo dissociation.
$$\ce{CCl3F ->[h\nu] Cl + CCl2F}$$
So the main problem is, that free chlorine radicals are produced in the stratosphere. Also these CFCs have a lifetime in the stratosphere of up to 150 years.
There are some natural sources (of close to insignificant importance) that also produces free Chloride Radicals, i.e. volcanic eruptions. Those may transport Hydrochloride into the stratosphere, where it can react with the Hydroxide radical:
$$\ce{HCl + OH -> Cl + H2O}$$
Removal of Chlorine
There are several steps involved, that lead to the removal of Chlorine from the stratosphere. Basically it will stay there for quite a while and cause significant damage. The stratosphere is very dry and water-soluble compounds, like HCl, will not be washed out quickly. In order to leave the stratosphere, the chlorine atoms or compounds therefore have to be near the troposphere.
Probably the most important reaction is with methane forming Hydrochloride, which may the be washed out by water (causing at least a part of acid rain nevertheless).
$$\ce{Cl + CH4 -> HCl + CH3}$$ The likeliness of this event is about 1000 times smaller than colliding with another ozone.
However eventually, after about two years (a number I picked up on the way), a chlorine will leave the stratosphere. In the troposphere is more water present, so Hydrochloric acid will be formed and washed out.

In our estimates, termination of the Cl--ClO chain results from a downward diffusion of the longer lived species in the chain (HCl, ClO) and eventual removal by tropospheric process. From: Nature 249, 810-812 (28 June 1974).

There are also a number of reactions that decrease the potential of Chlorine radicals.
$$\ce{ClO + NO2 -> ClONO2}\\
\ce{ClO + O2H -> ClOH + O2}\\
...$$
However these also form reservoir molecules that might diffuse back to the stratosphere.
The calculations behind this are rather complex and use a wide variety of empiric data along with complex models.
I think there is still a lot of research going on in this area, considering more molecules like $\ce{CO2}$ and other greenhouse gases, as well as global climate change.
The Nobel lecture (1995) closes with the following remarks and we can only hope that this holds true:

However, with the limitations just expressed, the threat of extensive further stratospheric ozone depletion during the 21st century appears to be under control.

It isn't really true to say that a chlorine leave the upper atmosphere in two years. Some Cl atoms leave very quickly; some very slowly: it is the average atom that determines the overall lifetime and this is what is being expressed by "leaves after two years".

Updated link:https://www.nobelprize.org/prizes/chemistry/1995/rowland/lecture/

https://web.archive.org/web/20060720145739/http://ozone.unep.org/pdf/stratopheric.pdf

According to Wikipedia, Martin is correct; a chlorine molecule, being slightly heavier than oxygen, will eventually be transported out of the ozone layer back to the troposphere (taking approximately 2 years to do so). Additionally, presence of other atoms in the ozone layer have a small chance of destroying the radical chlorine, causing it to form $\ce{HCl}$ or $\ce{ClONO2}$.
So most likely this estimate is based on some math crunching between the time-frame that the chlorine is in the ozone, the probabilities of chance reactions, and the rates/probabilities of these reactions.


text:
Why can't the molecularity be determined from the rate law?
Basically can someone explain why the molecularity cannot be determined from the rate law?

The coefficients only correspond to the rate law coefficients if the reaction is elementary. Elementary reactions are ones where the reagent(s) combine through a mechanism implied by the reaction. These are simple. Many reactions aren't as simple and have multiple steps.
Check out this link, example two. The reaction is actually comprised of two reactions where the slow step determines the rate coefficients.
As you can see, you cannot just read the equation and get the coefficients. One can only determine whether a reaction is elementary (i.e. find its rate law coefficients) by experimentation.

I might add that even if you have the rate law coefficients it is not always possible to infer the elementary reaction steps from it, there might be multiple possible paths that result in the same rate law. For further distinguishing between the paths you have to use in-situ techniques that can monitor intermediates while the reaction is going on


text:
Coconut aldehyde mechanism
How would you draw the mechanism going from non-3-enoic acid to coconut aldehyde (5-pentyloxolan-2-one)? I know I added sulfuric acid and sodium carbonate, which I was confused how they worked into the mechanism. I am certain there is some type of internal reaction to create the five membered ring.
Those are the only chemicals that could, in my opinion, be involved. The others are not needed in the mechanism. Pretty much confused on the first step of what happens when sulfuric acid is added. What does it protonate? How does this lead to an internal reaction and where does that reaction take place? Does the C-C double bond have anything to do with it?
Would the lone pairs on the oxygen readily take the proton? But how would that affect the overall state of the compound in order to form that ring. I'm not seeing how that could be right.

Welcome to Chemistry.SE! Please have a look at our homework policy. Could you elaborate a little more on what your own thoughts are on the matter? Where specifically do you have trouble working out the mechanism? Furthermore it would be helpful if you make sure that you have mentioned chemicals that are used in the reaction - is it only 3-nonenoic acid, sulfuric acid and sodium carbonate or is there something else involved.

I have the impression that you a very close to find the answer yourself!
I will not give a full answer here, but some hints.

[...] I know I added Sulfuric acid and sodium carbonate [...]

Let's assume that you did not add them simultaneously. Then, it seems likely that sulfuric acid does the magic and sodium carbonate is added to neutralize the reaction mixture after your product has been formed.

I am certain there is some type of internal reaction to create the five membered ring.

Good point! Let's draw the starting material and the product.

3-nonenoic acid is a carboxylic acid with a non-conjugated $\ce{C=C}$ double bond. With other words, there are two functional groups.
Now think in the sulfuric acid! It's a strong acid while its anion is a rather weak nucleophile.

"Coconut aldehyde" is  not an aldehyde at all, but a furanone, a five-membered lactone. Think intramolecular ester.

EDIT

Is it conceivable that the $\ce{C=C}$ double bond is protonated?
If so, what could possibly add to the resulting cation?

A good hint would also be to think about the stability of carbocations.

So the job of sulfuric acid is to protonate something. Would the lone pairs of the oxygen take a proton? I'm still missing steps getting to the point where I can have a favorable condition for that internal reaction to occur

It is the double bond that gets protonated, and a secondary carbocation is formed. The carbonyl oxygen of the carboxyl group adds to this cation, forming the five-membered ring. This reaction is favorable because the resulting cation is more stable.


text:
Can an excessive reactant amount result in a longer reaction time?
I am performing an experiment with DTZA and Ferrate. After a point of increasing the DTZA concentration too much the Ferrate takes longer (lower reaction rate) to react with the DTZA. Is this something reasonable or is it due to error?

http://www.ncbi.nlm.nih.gov/pubmed/21780694 Email the authors and ask. Is it a pH issue? solubility?

Let me first say that I don't know this specific reaction, so I am not sure whether my answer holds for it, but let me answer the general question: can a high concentration slow down the reaction rate?
The answer is, yes it can. The mechanism by which this happens is called inhibition and it is often (although not necessarily) connected to reactions on surfaces. I will explain inhibition in terms of a surface reaction here, but keep in mind that it can also happen in solution due to 'crowding', in which there is so much of 1 reactant that it basically 'cannot find' the other reactant.
For surface reactions the story is as follows. You will get a competition between adsorption on the surface and reaction on the surface.
Qualitatively, if you have a large amount of one of the two reactants than many of the surface sites will have this reactant adsorbed to it meaning that some of the reactant on the surface has neighbors that it cannot react with (because it is the same molecule), this can slow down the reaction rate.
Quantitatively, the reaction order will be concentration dependent. A famous example of this is Langmuir-Hinshelwood kinetics
$$r =k C_s^2 \frac{k_1 k_2 C_A C_B}{(1+k_1 C_A+k_2 C_B)^2} $$
where $C_s$ is the concentration of available adsorption sites and $C_A$ and $C_B$ are the concentrations of your reactants.
There are many possible ways in which this reaction rate equation can behave, depending on the values of the rate constants $k$, $k_1$ and $k_2$, but let me single out two cases that will explain a shift from increasing to decreasing reaction rate with increasing concentration:

a very high concentration of $C_A$ will result in $k_1C_A>>>1, k_2C_B$. This allows us to neglect the $1$ and $k_2C_B$ terms in the denominator, turning the rate equation into this:
$$r =k C_s^2 \frac{k_2 C_B}{k_1 C_A} $$
which indeed decreases with increasing concentration of $C_A$.
a very low concentration of $C_A$ will result in $k_1C_A<<<1, k_2C_B$. Then the rate equation becomes:
$$r =k C_s^2 \frac{k_1 k_2 C_A C_B}{(1+k_2 C_B)^2} $$
which has an increasing rate with increasing concentration of $C_A$

So what you see here is that a change of $C_A$ from low to high concentration can change the reaction order in $C_A$ and therefore indeed result in an initial increase of the reaction rate with increasing $C_A$, but later a decrease if you increase it even further. When this crossover occurs depends on the values of the rate constants ($k$, $k_1$ and $k_2$) and the concentration of the other reactant.


text:
Determine which molecule has stronger intermolecular force?
SiH4 and CH4
The only intermolecular force they both have is London Dispersion forces
Strength of LDF is determined by molar mass
molar mass of SiH4 = 32.132
molar mass of CH4 = 48.42
Therefore shouldn't CH4 have a stronger LDF??
The answer says that SiH4 has a greater intermolecular force...why is that?

Only problem is that you have the wrong molar mass for ch4.

Look at the boiling points.  Interaction is based on geometry (surface area) and polarizability, among other things - and the phase: solid, liquid, vapor.  Molecular weight is not diagnostic.  SiH4 is "fluffy" versus methane being "hard."  Compare the melting and boiling points of decane and isodecane  versus adamantane, all being $\ce{C_{10}}$.  Compare the melting and boiling points of n-octane versus 2,2,3,3-tetramethylbutane, both $\ce{C8H_{18}}$.  Complicated.
You might check your MW for methane.  Bells and whistles should have gone off, $\ce{CH4}$ versus $\ce{SiH4}$.

The molar mass of SiH4 is ~32, however the molar mass of CH4 is ~16. This explains the textbook's answer. Uncle Al, however, provides a more detailed answer.


text:
Figuring out the theoretical mass of reactants
I'm working on my chemistry lab report. The initial instructions were to test the hypothesis that the mass of products of a reaction will equal the mass of the reactants.
Basically we where given a equation, our job was to balance and figure out the theoretical mass of reactants, and the theoretical mass of products.
There we're a total of three questions. In the first one I got the following differential values, 0.2 and 0.1. This seemed to be the most logical for me since I knew the law of conservation of mass implied that mass cannot be created nor destroyed but could be transformed. I attributed this small differential value to this transformation theory.
However, the next two questions I attempted still gave a very large differential value (40-120), which rejects my initial hypothesis.
I believe I may have missed something in the second and third question, below are my attempted solutions.
Question 1:
Initial Equation: $\ce{CH4 + O2 -> CO2 + H2O}$
Balanced Equation: $\ce{CH4 + 2O2 -> CO2 + 2H2O}$
$$
\begin{array}{ccccccc}
\ce{CH4} & + & \ce{2O2} & \ce{->} & \ce{CO2} & + & \ce{2H2O}\\
1(12+4) & + & 2(32) & \ce{->} & 1(12+32) & + & 2(2+16)\\
1(16) & + & 2(32) & \ce{->} & 1(44) & + & 2(18)\\
16 & + & 64 & \ce{->} & 44 & + & 36\\
\end{array}
$$
Theoretical Calculation:
$$
\begin{array}{ccc}
\ce{CH4}=\pu{16g} & & \ce{H2O}=\pu{36g}\\
+ & \ce{->} & +\\
\ce{O2}=\pu{32g} & & \ce{CO2}=\pu{44g}\\
\end{array}
$$
Actual Calculation:
$
\ce{CO2}=\pu{43.8g}\\
\ce{H2O}=\pu{35.9g}\\
$
Differential Value(Theoretical Calculation - Actual Calculation):
$
\ce{CO2}=\pu{0.2g}\\
\ce{H2O}=\pu{0.1g}\\
$
Question 2:
Initial Equation: $\ce{C2H6 + O2 -> CO2 + H2O}$
Balanced Equation: $\ce{2C2H6 + 7O2 -> 4CO2 + 6H2O}$
$$
\begin{array}{ccccccc}
\ce{2C2H6} & + & \ce{7O2} & \ce{->} & \ce{4CO2} & + & \ce{6H2O}\\
2(24+6) & + & 7(32) & \ce{->} & 4(12+32) & + & 6(2+16)\\
2(30) & + & 7(32) & \ce{->} & 4(44) & + & 6(18)\\
60 & + & 224 & \ce{->} & 176 & + & 108\\
\end{array}
$$
Theoretical Calculation:
$$
\begin{array}{ccc}
\ce{C2H6}=\pu{60g} & & \ce{H2O}=\pu{108g}\\
+ & \ce{->} & +\\
\ce{O2}=\pu{224g} & & \ce{CO2}=\pu{176g}\\
\end{array}
$$
Actual Calculation:
$
\ce{CO2}=\pu{78.59g}\\
\ce{H2O}=\pu{48.2g}\\
$
Differential Value(Theoretical Calculation - Actual Calculation):
$
\ce{CO2}=\pu{97.41g}\\
\ce{H2O}=\pu{48.2g}\\
$
Question 3:
Initial Equation: $\ce{C3H8 + O2 -> CO2 + H2O}$
Balanced Equation: $\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}$
$$
\begin{array}{ccccccc}
\ce{C3H8} & + & \ce{5O2} & \ce{->} & \ce{3CO2} & + & \ce{4H2O}\\
1(36+8) & + & 5(32) & \ce{->} & 3(12+32) & + & 4(2+16)\\
1(44) & + & 5(32) & \ce{->} & 3(44) & + & 4(18)\\
44 & + & 160 & \ce{->} & 132 & + & 72\\
\end{array}
$$
Theoretical Calculation:
$$
\begin{array}{ccc}
\ce{C3H8}=\pu{44g} & & \ce{H2O}=\pu{72g}\\
+ & \ce{->} & +\\
\ce{O2}=\pu{160g} & & \ce{CO2}=\pu{132g}\\
\end{array}
$$
Actual Calculation:
$
\ce{CO2}=\pu{82.5g}\\
\ce{H2O}=\pu{45.0g}\\
$
Differential Value(Theoretical Calculation - Actual Calculation):
$
\ce{CO2}=\pu{49.5g}\\
\ce{H2O}=\pu{27.0g}\\
$
Not sure if it matters, but we where given this program to enter our calculations to simulate the reactions.

Try to answer to these questions, don't cheat, check the answer passing over the yellow box.
I think this software is confusing you always start from how much reactants you have and then calculate how much products you will have, suppose you have $\pu{100g}$ of oxygen and $\pu{44g}$ of propane.
How many mole of propane and oxygen ($\ce{O2}$) do you have?

 Oxygen moles$=\frac{\pu{100g}}{\pu{32g/mol}}\sim \pu{3mol}$,  Propane moles$=\frac{\pu{44g}}{\pu{44.096g/mol}}\sim \pu{1 mol}$

If this is your balanced equation:
$$\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}$$
How many oxygen molecules do you need to react with one molecule of propane?

 You need 5 oxygen molecules for every propane molecule. So you need 5 moles of oxygen for every mole of propane.

Which is the ratio?

 $\frac{1}{5}$ if you have $x$ moles of oxygen you need $\frac{x}{5}$ moles of propane. e.g. if you have 5 moles of oxygen you need? 1 mole of propane.

But you have only 3 moles of oxygen to get the products, oxygen is a limiting reagent! So how many moles of propane will react:

 $$\pu{\frac{3}{5}mol} =\pu{0.6 mol} $$

So how many moles of the products will you have?

 $3\times0.6$ moles of $\ce{CO2}$ and $4\times0.6$ moles of water

And so how many grams?

 for $\ce{CO2}$, $\pu{1.8mol}\times \pu{44g/mol}$ for water, $\pu{2.4mol}\times \pu{18g/mol}$

$\pu{0.1 gram}$ is ($\pu{E4 kg}$)($\pu{299792458 m/s2}$) = $\pu{8.988E12 J}$ or two kilotons nuclear yield equivalent.  Let's be conservative and say "experimental error."
Mass-energy is conserved because time is homogeneous throughout the universe.  Run this continuous symmetry through Noether's theorems to obtain the First Law of Thermodynamics.  However, time is not locally homogeneous!  We live at the bottom of a gravitational potential well.  GPS atomic clocks run slower in orbit for special relativity (orbital velocity), but faster still for general relativity (higher in the potential well).  Therefore, don't transform "missing mass" except in physics labs.  Their party, their mess.