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Ratio between heat tranfer coefficient and thermal conductivity
Consider the Robin boundary condition for the diffusion/heat equation $u_t=a(t)u_{xx}+f(x,t)$:
$$-k(t)u_x(0,t)=h(t)u(0,t)$$
or
$$u_x(0,t)+\frac{h(t)}{k(t)}u(0,t)=0$$
where $k(t)$ thermal conductivity and $h(t)$ heat tranfer coefficient.
My Question: Is it possible that the ratio $h(t)/k(t)$ to be constant? Could anyone please help me? I have really no idea.

İ want to solve the equation and if this ratio is constant i can continue. I just want to know that whether it has a physical meaning or not

From dimensional analysis it should be clear that the ratio is time independent...

@J.LS thanks for answering. Can you give a specific reference so that I can look at it.

The heat transfer coefficient and the thermal conductivity can often be treated as constants in practical heat transfer problems.  However, the heat transfer coefficient can change if the exterior convective flow conditions change.

Thanks for answering. Do you know a physical/ chemical process in which such boundary condition occurs and the ratio in problem can be taken as constant.


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What types of chemicals/ionic liquid could perform this type of separation?
Ionic Liquids separation of oil and sand.

In the video from PennState, Aron Lupinsky tells that he is working in the group of Prof. Painter.
Combining these information, a cursory search furnishes the article
Recovery of Bitumen from Low-Grade Oil Sands Using Ionic Liquids by Phillip Williams, Aron Lupinsky and Paul Painter, published in
Energy Fuels, 2010, 24, 2172–2173
(DOI).
On the first page of the article, it is mentioned that a mixture of toluene and 1-butyl-2,3-dimethylimidazolinium tetrafluoroborate, $\ce{[bmmim][BF4]}$, was used for the extraction.

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Predicting product of a Hofmann elimination with a cyclic amine
What is the final product of this reaction?

Here's what I have:

Does this look correct? I'm concerned because I'm used to seeing the nitrogen removed from the product (in the cases of primary amines).

With excess of methylamine, I wouldn't exclude additional methylation, yielding a trimethylanilinium iodide.

However, the reaction should stop there since a further hydrogen abstraction from a $\beta$-position isn't possible.

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Why does alcohol crack acrylic (plexiglass)?
I have seen videos and discussions about alcohol cracking acrylic, and waterblock manufacturers reminding users to not use alcohol in cooling systems with plexiglass acrylic, but why is this so?
The question proposed as duplicate only asked for the effect, but here I already know the effect. I am asking for the reason for the effect, which is not answered in the answer to the 'duplicate' question.

possible duplicate of Ethyl alcohols use on polycarbonate and acrylics

I don't think this is a duplicate.  The other question didn't ask about mechanism.  There was no "why" component in the earlier question but there is here.

Thanks Mithoron for finding a related question. Thanks Curt for helping me to explain. Yup I am asking for why it happens, not what will happen when alcohol is applied to plexi.

Try cleaning a pair of Oakley sunglasses, they all but explode. It happens very quick!

According to chemical resistance of plexiglass acrylic alcohol is not as aggressive as you mention with plexiglass acrylic.
But, we have to signal that acrylics is a big family of polymers (thermoplastics or thermosets) derived from acrylic acid, methacrylic acid or other related compounds. Plexiglass is only a member of this family.
Any way, in response to your question why  alcohol might cause severe crazing  and cracking?
It's simply because alcohol, as a good solvent, has the ability to attack the surface of acrylics and dissolve the polymer chains.

Yes, acrylics are a wide family, but the second part of this explanation is so oversimplified it become plainly wrong.

If a solvent causes an amorphous polymer to craze, this is because the polymer is principally soluble in it. It diffuses into the surface, but without instantly dissolving it, as the polymer chains are still stuck in the glassy network of the substrate. The polymer swells (like a fruit gum in water), but the volume increase can only happen perpendicular to the surface.
The polymer chains thereby end up oriented towards the surface, weakening the structure. When the solvent dries off again, the chains adhere to each other again, but cannot reentangle in time. The surface has become thicker now, but as it is still the same amount of polymer, the additional volume is made up by the cracks and crazes appearing in it.

I believe the name for this phenomenon is environmental stress cracking. We blew more than one acrylic apparatus (thankfully not terribly expensive!) by cleaning with ethanol before we realized what was going on.

The given explanation cannot be correct, because I've watched (and listened to) an acrylic glass cracking and splitting and developing leaks after pouring in a splash of bourbon.  No drying out is involved.  It may be simply swelling of the plastic as ethanol penetrates; this could cause cracks if the bulk of the plastic is too rigid to stretch, and can't accommodate the swelling.

An experimental observation deserves positive consideration, not a downvote, although this answer could have been worded a little more gently: "Perhaps the above answer is not the whole explanation." I've seen the effect with polystyrene and some solvents, but never with acrylics - I must be more careful. I wonder if the acrylics that fail are substandard in some way.

Definitively makes some sense. Also the solvent would relieve orientation, leading to strong swelling on one direction, and very little or none in the other.

@JamesGaidis I wouldnt say "substandard". Its just another type, which might be cheaper, or have some other redeeming properties.

Indeed. I don't think drying has anything to do with the process. I submerged a plexiglass character display in pure IPA and cracks started forming right away. When I took it out, the surface felt like rubber. After a short while it became rock hard again

A factor I haven't seen mentioned is which processes the acrylic part was subjected to during manufacturing. If the acrylic was flame polished during post processing (common in PC water cooling parts to achieve desired aesthetics), it can significantly weaken the bonds near the surface of the material and even cause microscopic crazing to occur.
When alcohol or a similar solvent is then applied, it will rapidly dissolve the weakened bonds and cause crazing, seep into the cracks, and dissolve the polymer within the cracks. The polymer will then swell, move upwards in the channel, and potentially crack the material upon drying. This is the process described in @Karl's answer.
Anecdotally, I have used near-pure isopropyl alcohol to clean the residue off many lasercut acrylic sheets without issue, and have never encountered crazing or cracking over the lifetime of those parts. However, many people report immediate cracking when cleaning milled acrylic water blocks with isopropyl alcohol. It seems likely that flame polishing is largely to blame for the part's poor alcohol resistance.

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In regards to the addition of hydrogen to alkynes, does the amount of moles of Hydrogen matter?
Is there any difference between having the addition of 1 mole of hydrogen to 2-pentyne with a platinum catalyst compared to the addition of 2 moles of hydrogen(H2). I was under the impression that both would result in an alkane yet I have 2 problems of that nature.
would one only result in a alkene, though I thought that only happens with the poisoned catalyst.

Your question is not entirely clear, but I think that you ask whether or not an excess of hydrogen gas would interfere with the reduction of an alkyne to an alkane.
This is not the case. A lot of the time hydrogen is just drawn into the reaction flask directly from a large reservoir (20L $\ce{H2}$ gas cylinder under pressure or a simple balloon). When the pressure in the reservoir does not change anymore (in the case of the balloon when it does not shrink anymore) it means that the reaction is finished, i.e. no more hydrogen can be added.
In general over-reduction cannot be prevented by limiting the amount of hydrogen present. It will just result in an awful mixture of products (alkanes/alkenes/alkynes). A poisoned catalyst is required to prevent this (e.g. Lindlar's catalyst).


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Suggest a method, other than spectrophotometry, for determining the reaction rate
For the hydrolysis of $\ce{\mathit{trans\hyphen}[Co(en)2Cl2]Cl}$ (and cis) we measured reaction rates via spectrophotometry. Plots of change in absorbance were used to determine the rate law and the rate for the reaction.
The lab report asks:

Suggest another method, other than spectrophotometry, for determining the reaction rate measured in this experiment.

Knowing little about measuring rates, I said one could measure the enthalpy and entropy of the reaction at a given temperature, and use those to calculate the rate constant. This is based off of the equation
$$\ln\frac{k}{T}=-\frac{\Delta H_\text{act}}{RT}+\ln\frac{k_\mathrm{B}}{h}+\frac{\Delta S_\text{act}}{R}$$
but, my professor said thermodynamics couldn’t be used to characterize kinetics. Also, reflecting on my answer, I didn’t really specify a technique.
What’s another technique that could be applied here?

Spectroscopy is often the best method, since it allows for (almost) continuous monitoring of the concentration.
If that is not wanted, ask yourself:

What is released in the hydrolysis of the complex?

Chloride would be a good guess.

Can I take samples and perform a wet chemical analysis to determine the concentration of new species?

Looking at the graph in your other related question, the answer is yes. The hydrolysis is slow enough.

Which methods can I possibly use to determine the concentration $\ce{Cl-}$ in solution?

An argentometric titration, based on the low solubility of $\ce{AgCl}$ is a good idea. What about the Mohr titration, using $\ce{K2CrO4}$ as an indicator?


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Justify assuming first-order kinetics for a complex
For the hydrolysis of $trans-\ce{[Co(en)2Cl2]Cl}$ (and cis) we measured reaction rates via spectrophotometry. Plots of change in absorbance $(\ln(A-A_\infty))$ were used to determine the rate law and the rate for the reaction. Using my data, I created the first-order plot seen below:

The question is for the report was:

Justify assuming first-order kinetics

My explanation was that first-order reactions can be expressed as $$[\ce{A}]=[\ce{A}]_0\operatorname{e}^{-kT}$$
and because the concentration decreases exponentially over time, we can assume its first order. Other rate laws I looked up didn’t show an exponential function, so this seemed like sound reasoning, but I was told I was wrong.
What would be the justification for thinking this is first-order?

As you have explained, the reaction is a hydrolysis:
$$\ce{[Co(en)2Cl2]+ + H2O -> [Co(en)2(H2O)Cl]^2+ + Cl-}$$
Therefore, you may expect a second-order rate equation:
$$r = -\frac{\mathrm{d}c}{\mathrm{d}t} = k_1 \cdot \left[\ce{[Co(en)2Cl2]+}\right] \cdot \left[\ce{H2O}\right]$$
However, you are using a dilute aqueous solution. Compared to the concentration of the complex, the concentration of water is very large. Thus, the concentration of water remains approximately constant and it can be included in the rate constant, obtaining a pseudo-first-order rate equation:
$$r = -\frac{\mathrm{d}c}{\mathrm{d}t} = k_2 \cdot \left[\ce{[Co(en)2Cl2]+}\right]$$

I like this answer, but doesn't that imply then that the reaction rate is actually second order (which is actually not the case)?

@JohnSnow That is correct; the overall reaction equation implies a second-order rate equation (even if it effectively is pseudo-first-order). However, the detailed mechanism may be more complicated. If the first step (removal of one chlorido ligand) is very slow (i.e. rate-determining), the reaction actually has a first-order rate equation (not just pseudo-first-order).

Finally talked to my professor and this was the correct answer!

What would be the justification for thinking this is first-order?

The KISS principle :)
Before you start to model complex kinetics, begin with a simple model. If it fits, consider your job done.
Good enough is good enough!
(Imagine that you perform therapeutic drug monitoring. It's not your job to elucidate the exact metabolism of a drug here, you just want to be sure that the patients receive the right dose.)

This might be an oversimplification, but my reasoning for assuming that it's first order would be because the plot of the log of the change in absorbance versus time gave a straight line.
The log of the concentration vs time will normally indicate a first order equation (given a straight line) but using the Beer-Lambert law:
A = ecl (where A is absorbance, e is the extinction coefficient, c is the concentration and l is the path length)
and assuming that the path length (l) is 1, you can see that the absorbance is proportional to the concentration. So plotting the graph of absorbance is analogous to plotting the graph of concentration, in that you should get a similar straight line and the result for the reaction constant k.

I feel like thats similar to my answer of it decreasing exponentially?

Pretty much. I feel like you may have been told you were wrong because they wanted you to use the graph to justify your answer, which is what I tried to do above. I did a similar experiment in freshman chemistry and that's the sort of thing they were looking for me to say. I think it was poor form of them to tell you that you were wrong and not to give you proper feedback on your answer.

To be fair, I didn't approach the professor about it. I think I'll talk to him next class and see what the appropriate answer would be.

You were told you are wrong, but you are right. This is clearly a first order reaction, just because of the reasoning you exposed. But I will try to present you a simple mind exercise to realize why in an heuristic way (your argument is the more accurate one in my opinion, but many chemist need an inaccurate representation to feel that the concept is understood).
I bet that it is just a misunderstand of reaction order concept. In the case of elemental reactions in gas phase, it will be (just for statistical reasons) the sum of reactive molecules involved. BUT, here we are talking about a reaction in aq. phase, where one of the reactives is not another solute. So, thinking this hydrolysis just as molecule $A$ plus only one water molecule is not accurate.
The water is making a field around each molecule of your reactive, so represent (in your head) it like your molecule surrounded a electric field is more accurate that only thinking in two molecules. Thinking so, the first order for this reaction arises naturally.


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How do cloves break down plastic?
When my family buys powdered cloves in bulk, we put them in plastic until we get home - which takes at very most an hour. As soon as we get home, we transfer them into glass because we have had issues with them eating through the plastic cups until they are soft and have holes in them - this takes maybe a week.
I do not know exactly what type of plastic the cups are made out of; if this helps, they have the recyclable number 6 on them.
So, my question is, what is it in the cloves that is causing them to eat through the plastic at such a high rate? Are they really acidic? Or is it perhaps due to their oils?

I manage a large commissary kitchen and we are noticing our powdered clove is doing the same thing to EVERY kind of plastic we use to store it in. it arrives in a 1lb clear plastic bag and we've tried storing it in polystyrene containers as well as #5 hard plastic containers like cambro/lexan. The thinner, softer polystyrene breaks down rapidly and looks as if it is being melted slowly over time. When stored in the thicker/heftier containers we are noticing that something similar is happening - the once clear plastic begins to turn a milky white and will forever smell like cloves no matter how

Most of the odor of cloves comes from volatile molecules, especially eugenol, found in the cloves.  Eugenol is a water-immiscible, oily liquid at ambient temperatures.
Plastic with resin identifiction code #6 is polystyrene.
Polystyrene is not compatible with most organic solvents, because it tends to be softened, swelled, or even completely dissolved by them.
Thus I suspect that the problem is that oil in the cloves is attacking the plastic that you use.  You could avoid the problem (mostly) by using a plastic with resin code #2 or #5 (i.e. high-density polyethylene or polypropylene) instead of #6.  But, you are probably right that glass would be better than any plastic for storing your cloves.


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How do I create azeotrope mixture?
I want to decrease the boiling point of a compound. So I need to create an azeotrope mixture from two or three compounds. Where can I find out how make it?

Welcome to chemistry.SE! What is the compound you want to decrease the boiling point of?

Thank you. I want to decrease the boiling point of methyl jasmonate.

Boiling point can be decrased by lowering pressure over liquide. If you insist on the azeotrop preparation, find azeotrop diagram such as:
or data table such as: wiki and determine the composition that must be achieved to obtain the azeotropic mixture. Then mix it accordingly.

If you cannot find a phase diagram like the one above for methyl jasmonate, you can use the principle that a mixture of two dissimilar substances is likely to have a lower boiling point. You might try water plus methyl jasmonate, for example - it should boil slightly below 100C.  Purifying it this way is called steam distillation. Also, another way to lower the boiling point is to lower the ambient pressure - i.e. do a vacuum distillation.


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Colorimetric test to measure dissolved oxygen in salt water
Are there any simple colorimetric tests for measuring the dissolved oxygen in salt water, which do not involve titration like in Winkler method ?  i.e. Are there any chemical reagents, which when added to a salt water sample, would cause a change in the colour of the solution , giving us the measure of DO ?
Regards.

Unfortunatly I don't think there is an easy colorimetric technique available.
There are reagents that are strongly colored and oxygen-sensitive.  One exampleis reduced methylviologen radical.  That species can be prepared by mixing methylviologen with sodium dithionite in aqueous solution.  The one-electron reduced species is intensely blue.  It is rapidly oxidized by oxygen, leading to a loss in color.  In theory, you could use the degree of this color loss to assay DO.
The problem in real life with applying this reagent to DO determination is that oxygen in the air will also react, very quickly.  So you would have to put your seawater sample under an inert atmosphere (and also prepare the reduced MV radical solution under an inert atmosphere).  That is very inconvenient, and could create problems where your seawater sample outgases oxygen into the anaerobic environment before or during measurement.


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How to calculate vibrational spectra of organic molecules?
I watched a ted talk (Luca Turin: The science of scent) and was wondering if anyone had any examples of how to calculate vibrational spectra of molecules (used as fragrances)?  Would love to figure out how he computed these charts.

Going off this definition: A molecule is formed when two or more atoms join together chemically. A compound is a molecule that contains at least two different elements. All compounds are molecules but not all molecules are compounds. I would like to start with the easiest which looks like molecules and go towards compounds.

The usual approach is:

Find the equilibrium geometry (within the Born-Oppenheimer app.).
Expand the energy in Taylor series till second order term. The first order will be 0 because of the minimum energy condition.
Include this expression in the Hamiltonian.
Generate a coordinate transformation to get normal coordinates.
Relate second order derivative of the potential energy with the force constant of Hook law (Using second order term we can identify it with a harmonic oscillator). Then get the harmonic frequencies.
The software employed should give also the intensity for each frequency.
Plot intensity vs. frequency/energy/wave length. You'll get only "points". To transform them to a continuous function you can use a Lorentzian like function.

Details can be found in most computational quantum chemistry books.

You can build molecules and calculate vibrational frequencies with Molcalc.  Molcalc does have a molecular size limit, so for bigger molecules you must download some programs.  I have written about computing vibrations with the GAMESS program.


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What does "dep" mean?
I'm looking at some data with substrate concentration on the $x$-axis and $k_\text{dep}$ on the $y$ axis. What does $k_\text{dep}$ mean? I understand that it must be a rate of some kind but what is the significance of the 'dep' part?
The data I'm looking at can be found in Fig. 9 in this paper (PDF):

Essentially this is a dose response curve for a substrate and enzyme. I suspect $k_\text{dep}$ is simply the rate but why $k_\text{dep}$ and not just $k$?

Material deposition on a surface? I'm just guessing too!

I looked at the paper.  It uses a lowercase k with a subscript "dep" for the symbol you are asking about.  You can use those symbols in your questions -- see here for more info.  This time, I'll edit your question to give you an example of how to do this.

The "dep" probably stands for "depletion": read the paper you linked to for more context.  Often times, subscripts on rate constants are used to refer to which reaction in a system of multiple reactions.  You'll see $k_1$ and $k_2$ for rate constants of "reaction 1" and "reaction 2", etc.

The "dep" probably stands for "depletion": read the paper you linked to for more context. Often times, subscripts on rate constants are used to refer to which reaction in a system of multiple reactions. You'll see $k_1$ and $k_2$ for rate constants of "reaction 1" and "reaction 2", etc.
I suspect in the paper, they were measuring the depletion of a substrate, and not what the substrate was converted to.  So they wouldn't know if they were measuring the rate of a single reaction $\ce{A->B}$, or the combined rates of multiple reactions such as $\ce{A->B}$ and $\ce{A->C}$.  Thus (probably) the desire to indicate $k_{depletion}$ instead of just $k$.

yes, the Fig. 9 caption reads "...The
depletion of 4-oxo-atRA (A) and 18-OH-atRA (B) as a function of time
was measured at six different concentrations of each substrate and
the depletion rates were determined (insets in both panels). All
depletion experiments were conducted in duplicate. The depletion
constants as a function of substrate concentrations were used..."


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Converting methanol into methane
I know that methane to methanol is a common practice, but is it possible to convert methanol back into methane, and would it be a relatively easy process to achieve?

Methanation is a process whereby synthesis gas is converted to methane.
$\ce{CO + 3H2 -> CH4 + H2O}$
Methanol, like nearly all organic molecules, can be gasified to form synthesis gas at elevated temperatures.
$$\ce{CH3OH -> CO + 2 H2}$$
(Note: the gasification reaction is actually a complex balance of many types of reactions.  Often times mixtures of $\ce{CO}$, $\ce{H2}$, $\ce{H2O}$, and $\ce{CO2}$ are obtained.  The reaction above is the ideal case where complete conversion is achieved.)
Thus a process for converting methanol to methane could proceed via gasification followed by methanation.
However, since methanol is made commercially from methane by essentially the reverse process, this route is unlikely to be economically viable!

You wrote that this is unlikely to be financially profitable (economically viable). However, it could be very ecological or environmentally friendly. A farmer could produce methanol by fermenting wood chips or other organic material. After distillation, gasification, and methanation, the resulting methane gas could be used to heat the farmer's home in winter. The combustion of methane gas in the presence of oxygen does not form black particulates (Soot and smoke). Air pollution would be minimal. Burning fossil fuels moves solid carbon underground into the air.

Continuing from my earlier comment, growing corn, or trees, and fermenting corn or wood pulp would pull some carbon out of the air, store the carbon in the corn or wood pulp, and then put than carbon back into the air when the methane gas is burned ina furnace to heat a home in winter. Some carbon in corn comes from the soil, but less than 100% of carbon in corn comes from the soil. When we burn fossil fuels, 100% of the carbon come from underground. We don't want to move carbon from underground into the atmosphere too quickly. The greenhouse effect would cause some plants to die.

Human use plants to filter air, water, and as a source of food. Destabilizing the climate and weather patterns by moving large amounts of carbon from underground into the atmosphere is inadvisable. Perhaps we can ferment food crops into methanol and ethanol. After gasification, the methan or Ethan or other hydrocarbons can be used in older traditional natural gas burning furnaces to heat a person's home in winter. The alchohils need not be transported in tanker trucks which burn diesel fuel if you own a small farm. Alchohol can be produced by fermentation at the same site as where it is burned

That's all true.  But it's worth noting that the partial oxidation of methane, that is, converting methane to methanol, has been a "holy grail" of chemical engineering for many decades.  That is, an oxidative, (comparatively) low-temperature, low-pressure process to achieve $\ce{CH4 + 1/2 O2 -> CH3OH}$ efficiently & economically, has been a goal of industrial chemistry for a long, long time.  That's because methanol is easy to transport and store, and people are willing to pay more for that.  And because natural gas is very, very cheap.  So methanol -> methane isn't likely to be practical...

Some methanogenic miroorganisms isolated from human feces are known to digest $\ce{H2}$ and $\ce{CH3OH}$ to produce $\ce{CH4}$.


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Bonding and Phase Changes?
When water boils into water vapor, its temperature (average kinetic energy) does not change because the inputed energy is used to break its bonds. But how does that work?
Also, in a liquid, aren’t intermolecular bonds broken and reformed constantly anyway? Why does a phase change require a time interval for breaking bonds?

Temperature involves kinetic energy; however, bonds and intermolecular forces (IMF's) involve potential energy.  The potential energy of the system increases when water vaporizes, but kinetic energy does not change.
In a liquid, for each IMF broken, another is made, resulting in a net change of zero.
A phase change requires a time interval because it takes time to add energy to the system.

+1. Indeed the answer lies within the word "potential".

Wait, the potential energy of the system increases when bonds are being broken?

Yes.  When two objects that are attracted to each other by a force are moved farther apart, potential energy increases.

The trivial reason that the phase change liquid to vapour takes time is that this can only occur at the surface of a liquid, but I suspect that this is not what you mean when you talk about a ‘time interval’ to break bonds. In fact time does not come into it. I try to explain what happens below. I've added some background also in case you are not familiar with it; if you are you need only read the paras around the figure.
Boiling occurs when the vapour pressure of the liquid equals the external pressure, normally 1 atm (or 1 bar; almost the same thing). To boil a liquid, energy has to be continuously suppled as molecules are escaping as vapour and taking energy away with them in doing so. (Holding a rapidly evaporating solvent feels cold). At the boiling point the temperature remains constant until all the liquid has evaporated.
The average energy a liquid or gas has at temperature T is $3k_BT/2$. Trouton’s rule states that the latent heat (enthalpy) of vaporisation is related to the boiling temperature $T_B$ as $L_{vap} /T_B \approx 80 \pu {J K^{-1}mol^{-1}}$. This borne out by many liquids, with a range of $\pu{70 - 90 J K^{-1}mol^{-1}}$; those with values greater than this range are hydrogen bonded, e.g. water, ethanol, and if less than this range, form dimers in the vapour, e.g. acetic acid. From this value the cohesive energy can be estimated and comes to $\approx 9 k_BT$.  By cohesive energy is meant all the pairwise interactions between one molecule and its neighbours. Thus molecules will condense to form liquids when their cohesive energy exceeds $\approx 9 k_BT$. Thus to reverse the process, which is most quickly done by boiling, this much energy has to be supplied per molecule.
Energy is present in a molecule as both kinetic (translation plus vibrations/  & rotations ) and potential(bonds). In addition, when in a liquid there is always inter-molecular potential energy, e.g. via pairwise dispersion forces, which are always present between any atom or molecule, and dipole and induced dipole forces etc, which are generally called van-der-Waals forces. These form the cohesive energy. (We shall assume that chemical bonds are so strong that boiling does not change them which is usually the case. This makes hydrogen bonds inter-molecular bonds).
As you mention in your question inter-molecular bonds are made and broken all the time and this because there is a distribution of energies. The Boltzmann eqn. $\ce{exp}(-E/(k_BT))$ gives the chance of having energy E. This distribution of energies is also why it is possible for a liquid to evaporate at room temperature, albeit slowly compared to boiling, as a few molecules / second will randomly gain enough energy to escape the liquid.
The nature of these intermolecular forces is important. At infinite separation the intermolecular potential energy is zero, as molecules approach one another the potential energy falls (an attractive interaction) and reaches a minimum value, say, at 0.5 nm or some similar small separation which is somewhat larger than a chemical bond length. At yet smaller separation there is  strong repulsion between the electrons on either molecule and the potential energy rises very rapidly. In practice ‘infinite separation’ can mean just 3 or 4 times the minimum, equilibrium separation. The shape of the inter-molecular potential typically has the form given by a Lennard-Jones or Morse equation. See the figure for a sketch, which is not to scale.

The minimum energy is negative. Note that if the intermolecular potential were harmonic in form,  then liquids could never evaporate as the harmonic potential energy extends to infinity.
As the average energy increases the cohesive (intermolecular) potential energy increases also until it reaches zero and only kinetic (internal and translational, rotational) is left. The molecules can now escape the solution.
(Finally, note that a molecule has the same average kinetic energy in the liquid or vapour at the same temperature and is $3k_BT/2$. In the liquid the motion is restricted to a smaller space around the potential energy minimum than it is in the gas phase.)


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How to calculate molecular volume from the wavefunction?
Very short question but maybe not a short answer !   As the probability of presence of electrons in a specific spatial position is defined as square of the wave function, I am wondering if there is a technique for molecular volume calculation based on the wavefunction.
Is there such a technique?
Update:  I could run the volume keyword in Gaussian.  This is the output for methane , but I can not find the volume in it !

Usually molecular volume is computed simply as the volume inside a contour of a particular electron density. The choice of the contour is basically up to you, Gaussian, for instance, uses the value of $0.001 , \mathrm{electrons}/\mathrm{Bohr}^3$.

@Wildcat thanks for the reply. I have not access to Gaussian ( I am in Colorado state ! ) .  Can you give a suggestion for GAMESS or using third parties as Avogadro over GAMESS output ?

MoCalc according to its description can handle Gamess output and calculate "molecular surface area and volume", so I suggest you give it a try.

@Wildcat I installed MoCalc but can not find where I can calculate that. I just opened the GAMESS output but there is no option for volume calculation. Can you tell me how to find it please?

Molecular volume can be defined different ways (one may say, rather arbitrary), so I think it is an important question what you would like to use the result / what kind of experimental data you want to compare to. Also, whatever definition you use it is sensitive to geometry, therefore you may need a thermal average...

@Aug, forget about MoCalc. It could not do what you want.

The way to determine the volume of an atom in a molecule (Therefore, the molecular volume) is by doing a quantum calculation to obtain the wave function and, then, integrate it.
The way to obtain the wave function from GAMESS is with the AIMPAC keyword. Look at page 2-18 of the GAMESS manual. :D
You can obtain the volume of your desire molecule by following these steps:

Do a quantum calculation with the desire method and a very good wave function.
In the input file, you have to specify to the software of your preference to print out the wave function file (in gaussian, you can do it with output=wfn, more directions: Directions to print wave function from gaussian).
After the quantum calculation is done, you have to integrate the wave function with a specialized software (see at external links of QTAIM software).
After integrating the wave function, you will get the integrated values of several properties, one of them is the volume of each atom in the desire molecule. (See page 16 of The Quantum Theory of Atoms in Molecules).
Finally, sum up all the atomic volumes to get the molecular one.

Hope this answer would be useful for your purposes

And even for the title of your question :D

I think that QTAIM is too much of an (computational) effort to obtain the molecular volume. You could calculate the van der Waals volume of a molecule from knowing the positions (and the vdW radii) of the involved elements alone. I also think that the van der Waals volume is not equal to the volume you would obtain from a cutoff value of the electron density.

Related to vdW is true that QTAIM does not obtaine it, I will edited my post to be purer. In QTAIM, an atom is based purely on the electronic charge density and uses what are called zero flux surfaces to divide atoms. On this surface, the charge density is a minimum perpendicular to the surface, which is a minimum between atoms and this is a natural place to separate atoms from each other. Therefore, Based on this partition, it defines several electronic properties: volume per atom. This point of view can be the correct way to obtain the molecular volume from the wave function by summing up

@Aug no bounty :(

QTAIM is meant to divide the volume among atoms in a molecule.  For division of the atom with the external space, it uses the total density criteria--meaning that the total volume really isn't an AIM property;  rather, its the volume in a fixed density contour, so my nature is exactly the same as what Gaussian does by default.  But this certainly gives you a number.

the volume seems to be already in the output:
$\text{Molar volume}=305.994~\mathrm{bohr^3/mol}\ (27.304~\mathrm{cm^3/mol})$
But of course, this is defined with the volume inside a density contour of $0.001~\mathrm{e/bohr^3}$.  That may be different from what you measure, because volume is not a quantity with well defined meaning, and what you find in an experiment depends on how you measure it.
Mathematically any atom or molecule have a volume of infinity since the electron spreads to the entire space, but this is clearly not the physically meaning of volume as we usually know it.

Yes but I really can't understand it. As I ran it for hydrogen atom it gives 111 bohr^3/mol. As I know bohr is radius of hydrogen and its volume can not be 111 bohr^3/mol. On the other hand it is divided by mol. Should I divide it by Avogadro number ?

As I said, the definition of volume itself is somewhat ambiguous.  Bohr is the expectation value of the radius.  The radius of hydrogen atom is 1 born only in the Bohr model.  In the real quantum mechanical treatment of hydrogen atom, the wave function stretches to infinity and takes the entire space, so volume depends on where you define the probability density to be "too small" to be considered almost zero.  In Gaussian this value is set to be 0.001, which is quite small so i can imagine that would cause the radius to be much larger than the Bohr radius.

In fact, even the expectation value of distance between electron and the nuclei is larger than the Bohr radius in a hydrogen atom.  On the ground state this expectation value is 1.5 times Bohr radius, and that is the distance you expect to find the electron on average.  Gaussian defines radius differently, namely the distance where you can confidently say the electron will almost never go beyond.

The "distance to expect to find the electron on average" can be really promising.  Is the a technique to convert that 0.001 e/bohr into this average?  Such a conversion can be the definite answer to the question .

For hydrogen atom, you basically just do the integral and calculate the expectation value through <ϕ0|r|ϕ0>.   For molecules that becomes a vague definition because there is no clear center, and a molecule isn't exactly a ball like hydrogen atom.   I do not think this is available in QC packages.


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Fragmentation of phenyl radical cations in mass spectrometry
What are the main fragmentation pathways and products of a phenyl radical cation in mass spectrometry ? I notice sometimes that there's a m/z = 51 peak after the m/z = 77 for phenyl radical cation; I assume due to loss of acetylene.

Can you give more details?  What type of ionization?  What is the source of the phenyl radicals?  What types of collision cells etc. are used in the mass spectrometry you're thinking of?

Phenyl radical source: benzylic acid; I'm not familiar with the experimental method.

Without more experimental details, it isn't 100% clear that the small peak is exclusively derived from the phenyl radical cation.  It also isn't clear that observed the "phenyl" radical cation is actually cyclic!  It could be another isomer of $\ce{C6H5^{+.}}$.  With benzilic acid as the ion source, it is possible that the small peak contains an oxygen atom, although searching for possible formulas with appropriate "exact" masses makes me think this isn't true (see below)
The fragmentation spectrum of benzoic acid shows similar peaks.  The "exact" mass of the observed ion 51.0237 Daltons, within a milliDalton of the mass of $\ce{C4H3+}$.   The negative mode fragmentation spectrum of benzoic acid shows a corresponding ion at 49.0084, corresponding to $\ce{C4H1-}$ within a mDa.
I think the existence of something like $\ce{C4H1-}$ in negative mode supports Klaus's answer of cyclic structure because it's hard to see how else such an "unsaturated" cluster of carbon atoms can form.  Perhaps a tetrahedrane-type structure is even possible?
The easiest way to get more information on this ion's structure would be to fragment a variety of partially deuterated benzene rings.
Update: I found a paper talking about fragmentation of fluorobenzene, and they argue the "phenyl" cation fragment derived from that molecule may not be cyclic after all:

In combining these approaches, several mechanistic variants can
  be excluded, while a new, hitherto not considered pathway
  involving the formation of acyclic $\ce{C6H5+}$ is proposed.

I don't know if the same thing happens during the fragmentation of benzilic acid, but if so, then I wonder if the observed peaks could be protonated butadiyne (positive mode) or deprotonated butadiyne (negative mode).

What about $\ce{phenyl^{*+} -> Dewar\ benzene^{*+} -> cyclobutadiene^{+*} + C2H2}$?

That's definitely a plausible pathway, thanks !


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Detect Polyethylene
How can we detect polyethylene from a distance?
The method should be mobile as I am thinking of employing it a mobile robot.
I have no idea about it.
Any suggestion or external link would be welcomed.

It looks problematic at best. There are many types of PE; also it would be very hard to discern if it's not PP for example.

Cross posted to Robotics.

There is a growing literature on using near-infrared (NIR) hyperspectral imaging (HSI) to detect a variety of materials, including plastics.  Here's one example that focuses on resolving polyethylene from the very similar polypropylene using NIR HSI.

Are they avialable in the market?

When I google for hyperspectral imager, many different companies selling products come up.  Resonon, Bayspec, Imec, etc. etc.  Depending on the setup, I'd expect a very basic camera to be $15000 to $30000 or more.

authors: 京都府立大学機能分子合成化学研究室


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Claisen Condensation to Ketone
Can someone explain (and preferably draw) the intermediates of this reaction?

I understand that it is a Claisen Reaction, and likely a condensation due to the heat, but I am not sure why there isn't an ester in the final product.

The Claisen reaction is a type of acyl substitution ... as opposed to acyl addition.

Thanks for that, although that doesn't quite fill in the blanks for me. Could you draw the intermediates?

The Claisen is step 1. Step 2 might appear just to be a workup, but several things actually happen when the product from step 1 is heated in acidic aqueous conditions. Start by drawing the Claisen product (as you say, it should be an ester), and then think about what could happen when reacted with water. If you include those thoughts in your question, someone might take you the rest of the way.

@UofTStudent - I meant the Claisen condensation is a type of acyl substitution ... and try thinking of the differences between substitution and addition ... that will help.

Step 1 is a Claisen ester condensation with loss of ethoxide from the ester acting as the carbonyl part. The product is the $\beta$-ketoester (2).

Upon acidification and heating, the ester is cleaved, the resulting $\beta$-ketoacid decarboxylates to yield the symmetrical ketone.

Can someone explain (and preferably draw) the intermediates of this reaction?

Ethoxide - a base comparable in strength to hydroxide ion - abstracts a proton from the alpha position of the ester. This forms an enolate ion. Note that while protons alpha to a carbonyl center are mildly acidic - they are still not very acidic. Definitely not acidic enough for ethoxide to completely deprotonate. So we have both enolate anion and untouched, still-protonated ester.
Enolate anions are nucleophilic at both their oxygens and carbons. However, we mostly consider the reactions of enolates at their negatively charged carbons rather than their oxygens as we are interested in forming C-C bonds (as opposed to O-C bonds). Also, forming O-C bonds isn't particularly favorable versus C-C bonds - you'll have to study a bit of hard-soft acid-base theory to understand that.
Given that we have enolate anions in the system and untouched esters, which have electrophilic carbonyl carbons, we can make carbon-carbon bonds. The enolate anions can attack the carbonyl carbons, forming a tetrahedral intermediate. This intermediate collapses and ejects ethoxide ion.


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Why might two different chemical structures smell the same?
From what I understand, it is the functional group attached to a structure that determines the odor it will give off, for the most part. Which explains why two similar molecules, except for their functional groups, would smell different. So, would that make it so that two different molecules with the same functional groups would smell the same?

The molecular determinants of odor perception is a disputed area in fragrance biochemistry.  There are two basic theories: the "shape" hypothesis and the "vibrations" hypothesis.  Luca Turin is the foremost exponent of the more controversial and less widespread "vibrations" hypothesis, but he does a good job of explaining both theories in his TED talk.
According to the vibrational theory, since different functional groups have different vibrational modes, they should smell quite different.  An example from Luca's talk is a complex alcohol vs. the corresponding thiol.
According to the shape theory, molecules which have similar but not identical functional groups should smell similarly, unless the protein receptors in our nose that sense odors have very exquisite shape sensitivity, enough to perceive the very slight difference in the shape of thiol vs. an alcohol in a large otherwise identical molecule.

Why might two different chemical structures smell the same?

According to one theory, they would smell the same if they had similar vibrational (i.e. infrared) spectra.  In 2011, PNAS published is a very intriguing result that showed this was true for fruit flies in the case of some odorants.  They changed the vibrations by isotopic substitutions, a method that should leave essentially unchanged molecule's shape and its chemical reactivity.

So, would that make it so that two different molecules with the same functional groups would smell the same?

In my mind this is true if you believe the vibrational theory as well.  But it is harder to understand why it is true if you believe the shape theory.
Of course, the real answer may be a combination of both theories.  I'm a tepid supporter of the vibration theory myself, but I would be very interested in a chem.se answer from another expert who is less impressed with Dr. Turin's theories than me.

The big problem with the vibrational theory is that it seems to be just an inference from isotopologues and such and I think there could be other explanations for the observations (which are not crystal clear in themselves) that may not include vibrational modes. Until someone can examine the relevant receptors and propose a detailed mechanism, I'm pretty skeptical.


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Color of pheomelanin?
Do we know what is the wavelength of the color of pheomelanin?

Pheomelanin has a very broad absorption spectrum in the visible, with monotonically increasing absorption as the wavelength decreases until a maximum is reached at 300 nm.
I think that this would correspond to a dark red/brown colour.
Source: Phys. Chem. Chem. Phys. 2011, 13 (20), 9119–9127


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Why is the deposition of silver more uniform from [Ag(Cl)₂]⁻ (aq) than from Ag⁺?
What determines the uniformity? Just a fun fact I am wondering about, because I saw cyanide solutions used in electroplating.

The uniformity of a deposition depends on more parameters, but in this case the difference is in electromigration effect. $\ce{Ag+}$ ions are atrracted to the electrode surface, but the complex $\ce{[Ag(Cl)2]-}$ is repulsed. Cyanides form  complexes as well: $\ce{[Ag(CN)2]-}$ or $\ce{[Ag(Cl)3]2-}$ and they are also  repulsed.
Silver electrodepostion form $\ce{AgNO3}$ ($\ce{Ag+}$ in solution) is without additives nearly impossible as dendrites are created. Fukunaka, Yamamoto, Kondo, Ionic Mass Transfer Associated with Dendrite Growth of Electrodeposited Silver in AgNO3 Solution, 1989 ECS.
The metal deposition from a complex molecules is described in Wilkinson, P. Understanding Gold Plating. Gold Bull 1986, 19, 75 – 81

Some chemicals, that are usually added to electrolyte, affect the deposition because they create a film as they are adsorbed at  inner Helmholz plane. (E.g. acidic Cu deposition with alcohol addition Popescu et al, Materials Chemistry and Physics 72 (2001) 332–336.)


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What is the probability of us being "orderly"?
At first this question might get the following response, "What are you talking about?". But really all I am kind of asking is:

What is the probability of us humans being disorderly? Or in other
  words, I am kind of talking about the Second Law of
  Thermodynamics. But just to be clear, in how many ways can we humans
  be "ordered" or "arranged" to still be human and living.

This is a different question than what you usually see on Chemistry SE, but I just want to know what is the probability of being human?
Thanks

Have you read about the related concept of Boltzmann brains? Maybe some research into the term will turn up some probability, though I would not be surprised if it were on the order of $1/10^{10^{10}}$ for a human configuration to assemble randomly from a mixture of the atoms contained in an average person.

I'm not certain that this question is appropriate for this site. However, there is an awfully large range of configurations that are human. You can have physical defects, genetic anomalies, mutations, and you are still human. That's not even mentioning all the more typical human variations like height. But strictly mathematically, there are probably many more non-viable configurations of your cells than viable, so you might say that the probability is low.
BUT! Your invocation of the Second Law seems wrong to me. This only applies in a closed system, and that does not apply to the creation of a new human. Lots of energy is expended in that creation, which is not at all in defiance of the Second Law because we are not in a closed system (e.g. we ingest food and convert to energy).

Additionally, we humans are well-ordered entropy engines operating far from equilibrium and kept there by kinetic feedback loops. The traditional understanding of entropy as a measure of "disorder" breakdown far from equilibrium as an overabundance of "order" begins to look chaotic.


text:
Gibbs Free Energy of Solutes and Solutions
Using the equation $$\ce{CaSO4 (s) <=> Ca^2+ (aq) + SO4^2- (aq)},$$ and $\Delta H/\Delta S$ values, calculate $\Delta G$ at $50~^\circ\mathrm{C}$ when the solution is saturated with $\ce{Ca^2+}$ and $\ce{SO4^2-}$.
My Attempt
$$\Delta G = \Delta G^\circ + RT \ln Q$$
I calculated $\Delta G^\circ$ at $50\ \mathrm{^\circ C}$, But I am confused as to how we know what $Q$ is if there is no reactant then does $Q$ not become $\ce{[Ca^2+][SO4^2- ]/0}$, but this is impossible.

without reactant how can we get product? Any example?

No I do not think so.

This problem might be much easier than you think.  If the solution is "saturated", the dissolution process is at equilibrium.   What do you know about $\Delta G$ for processes at equilibrium?

For the reaction
$$\ce{CaSO4 (s) <=> Ca^2+ (aq) + SO4^{2-} (aq)},$$
the appropriate equilibrium constant using activities can be set up as
$$K_\mathrm{eq}=\frac{a(\ce{Ca^2+})\,a(\ce{SO4^2-})}{a(\ce{CaSO4})}.$$
The activity $a$ of a pure substances in condensed phases is approximately one, since their behaviour can be often treated ideally, i.e. the chemical potential $\mu$ at the considered conditions is approximately the same as the chemical potential at standard conditions $\mu^\circ$. This derives directly from the definition of the relative activity: $$a = \exp\left\{\frac{\mu-\mu^\circ}{\mathcal{R}T}\right\}$$
For substances where the solubility is very low, concentrations of the ions are very low, so in a further approximation we can assume that in these solutions the activity coefficient is approximately one, therefore, we can write concentrations instead of activities and obtain for the equilibrium constant the expression, that is also known a solubility product.
$$K_\mathrm{eq} \approx K_\mathrm{sp} = c(\ce{Ca^2+})\,c(\ce{SO4^2-})$$
However, this is completely irrelevant to the question itself. Since the solution is saturated, it also means, that the solid is in equilibrium with the solution, which means that the kinetics of dissolution and precipitation are equal of magnitude. The Gibbs energy has a very distinct value for these equations.

 chemical equilibrium
 Reversible processes [processes which may be made to proceed in the forward or reverse direction by the (infinitesimal) change of one variable], ultimately reach a point where the rates in both directions are identical, so that the system gives the appearance of having a static composition at which the Gibbs energy,  $G$, is a minimum. At equilibrium the sum of the chemical potentials of the reactants equals that of the products, so that:
 $$\Delta G_\mathrm{r} = \Delta G_\mathrm{r}^\circ + \mathcal{R}T\,\ln\,K = 0\\\Delta G_\mathrm{r}^\circ = - \mathcal{R}T\,\ln\,K\\$$
 The equilibrium constant, $K$, is given by the mass-law effect.
 Source: IUPAC goldbook

Our solution is at equilibrium so $\ce {Q} = {K}$. That is, $\ce{K_{eq}} = {K_{sp}}$.
$$\Delta G = \Delta G^\circ + RT \ln K$$
At equilibrium, $\Delta G = 0$.
So, $$\Delta G^\circ = - RT \ln K$$ Now, you can use the value of $\ce{K_{sp}}$ and the other known values given to determine $\Delta G^\circ$.

Your issue is that you are falsely considering the $ \ce{CaSO4}$ to be part of the system. The system under consideration is the solution, which the solid is not a part of. This is the idea of the solubility product.
Your reaction quotient is therefore: $Q= \ce{[Ca^{2+}]·[{SO4}^{2-}]}$

I believe this is wrong. In a saturated solution, there always will be/ has to be solid. This is of course part of the system, as it is in equilibrium with it. That is what the chemical equation is all about.

@Martin while I do agree that the solid is part of the overall equilibrium system, the solid is not part of the solution. You cannot measure the concentration of something that is not dissolved in solution.

While it is certainly correct, that you cannot measure the concentration of something, that is not in solution, neglecting the conditions, that lead to the equilibrium is wrong. And the explanation of the solubility product therefore is wrong.

You can leave this solid phase alone. Formally it's part of Nernst equation of equilibrium and for the formality's sake you can put it there, BUT its activity is ONE, so it really doesn't influence the description of the equilibrium.


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How much ozone is produced by ionization of air and how turn ozone into oxygen
I want to make a plasma speaker. I'm worried about the amount of ozone being produced by the ionization of the air and it possibly not being a safe amount. Even if it doesn't produce a dangerous amount, I don't like the smell of ozone. I'd like this all to be continuous. I am very unfamiliar with the chemistry here but I have read that UV light, around 250nm, can cause $\ce{O_3 \to O_2 + O}$ and also $\ce{O_2 \to 2O}$
But also without any help it seems that $\ce{2O \to O_2}$ and also $\ce{O + O_2 \to O_3}$. So it seems that if I shined UV light of the right frequency on the spark I'd make a lot of $\ce{O}$ and perhaps that would cause more $\ce{O_2}$ than $\ce{O_3}$ but I'm not really sure.
I was also thinking that it might oxidize some sort of metal filter really well and I could just keep replacing those but I wasn't sure. I could also heat the filter slightly to speed up the reaction if necessary.
Even if the UV light idea doesn't work out very well I still want a way to do this. Is the amount of ozone produced by a continuous electrical current though air harmful? Is there a way of getting rid of the ozone to make some odorless non-harmful chemical?

We've banned CFC's precisely because they're very effective at the reaction you want. But they're banned so it's not really an answer.

Even if the UV light idea doesn't work out very well I still want a way to do this

With some photochemical background, I suggest to forget about running such a UV lamp

Remember that your plasma will be formed in air. For every molecule of ozone formed, there are much more harm- and odourless $\ce{O2}$ molecules around. If you cleave $\ce{O3}$

\[\ce{O3 + O2->[\mathrm{h}\nu][\lambda\ \lt\ \mathrm{300\ nm}] [O2 + O* + O2] -> O2 + O3}\]
the resulting oxygen atom will be captured by the next $\ce{O2}$ to form ozone again.
The overall process will not reduce the number of ozone molecules, unless $\ce{O*}$ reacts with $\ce{N2}$, which might lead to some other unpleasant species ;)
Note that these processes are fortunately effective in the ozone to protect us from short wave UV radiation, which brings us to the next aspect of the setup.

In order to photolyze ozone, you'll neeed a light source that emits UV at $\lambda\lt300\,\mathrm{nm}$. It's not that these light sources would not be available, but this UV range is not that of a tanning booth lamp! Remember that the ozone layer protects our DNA from running havoc! The least you can expect from short-wave irradtion is the sun burn of your life! Consequently, you will have to shield the whole setup properly!

In summary, I absolutely like the idea of building a plasma speaker yourself and there even seem to be some DIY kits around, but I suggest to forget about the UV part, which can be much more dangerous than the ozone formed.
UPDATE
Burak Ulgut brought up a very interesting point. Back in the days, plasma speakers using other gases (or gas mixtures) were tried, and even went commerical.
This site in German describes a helium driven plasma speaker system :)

Thanks! This is an excellent response. Im definitely not doing this now!

I would say it would be much easier to try to keep the area oxygen free. Atomic oxygen (i.e. $\ce{O}$) is a highly reactive radical. This would react either with another $\ce{O}$, or an $\ce{O_2}$. It is hard to avoid this unless you can get rid of the $\ce{O_2}$.
Not sure what kind of a geometry you are thinking of, but purging the volume with $\ce{N_2}$ gas might be an easy way to achieve this.

The mention of an alternative gas is an interesting point!

You were wondering about a metal filter. That's a viable idea, but not all metals are created equal. Iron indeed would require frequent replacements, it rusts rapidly when in contact with ozone.
Copper is a fairly noble metal (oxidizes slowly), still reasonably cheap, and it works as a catalyst to break down ozone. The oxygen radicals on the copper surface often recombine to form O2, instead of forming Cu2O/CuO. (Copper(I) oxide and copper (II) oxide).
There are more copper-oxygen reaction products that can form. The initial reaction forms CuO3 (basically the ozon molecule sticking to one copper atom) and CuO4 (when ozon sticks to an existing CuO molecule). Another observed reaction product is Cu2O3. With all those combinations possible, a lot of pathways for chemical reactions open up.
In comparison, a free Ozone molecule can't simply "lose" one of its three oxygen atoms. Creating a free oxygen atom costs energy (that's why ozone is stable), so you need two ozone molecules to react with each other. That in turn means the decomposition rate is quadratic in the ozone concentration, not linear. So with low ozone concentrations, the reaction rate is very low.

Could you expand your answer a little more, in its current state it is more a comment than an answer. For example, in which state has copper to be present to have the described feature. How would this reaction take place? What would the turn over be for such a reaction? How could you force the ozone to even come in contact with the copper?

Your right about the UV light.  A UV light at 254 nm will efficiently destroy ozone.  Therefore if this were on, and in the vicinity of the ozone production, you should keep the overall ozone production relatively low.  Just make sure your UV light is focused at 254 nm as lower wavelengths of UV light do a great job producing ozone:
http://www.oxidationtech.com/ozone/ozone-production.html#answer2
For metals, aluminum oxide would do a good job destroying ozone.  It is commonly used for ozone destruction.
The other easy bet is a carbon filter.  Carbon breaks down ozone quickly to CO and CO2.  The only downside is that the carbon will become consumed over time and the carbon filter would need to be replaced.


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Why can you not make a compound with a quaternary alpha carbon using malonic ester?
So my textbook tells me that using a malonic ester to make a quaternary alpha carbon is impossible (I get that this might be hyperbole). However, it doesn't explain why and I don't understand why.
My thought was that after the first addition the tertiary compound would be too bulky to act as a nucleophile. However in the next section, an ethyl acetate is treated with LDA and acts as a nucleophlile 3 times to create a quaternary alpha carbon.
I also considered that you'd be making a tertiary carbon anion to make a second addition to a malonic ester's alpha carbon. However you also make a tertiary carbon anion using LDA on ethyl acetate and then alkyl halides.
I've added a screenshot of my textbook. Skip to the part in parenthesis if you want to read it.

It might be useful to add which book you use and on what page the pictured excerpt is from, so that it is a proper reference. I guess it is Vollhardt & Schore.

Perhaps the author was trying to say that by using just the malonic ester synthesis you can't prepare a compound with 3 (potentially different) alkyl groups on the alpha carbon.  You can only have 2 (potentially different) alkyl groups on the central carbon because the 2 other positions are occupied by the ester groups.  Of course using two additional steps you can 1) decarboxylate the malonic ester and then 2) introduce the third alkyl group via alkylation.  But using only the malonic ester synthesis you can't get 3 alkyl groups on the alpha carbon.

Double alkylations of the $\ce{CH2}$ centre of malonic esters do work!
Have a look at this marvel by R. P. Mariella and R. Raube, published in Org. Synth., 1953, 33, 23

EDIT 1
The use of sodium metal in a properly dried alkanol (in the case of ethanol, refluxing over magnesium ribbon is a good choice) was (and still is) an excellent and cheap method to generate sodium alkoxide in situ:
\[\ce{2Na + 2ROH -> 2NaOR + H2}\]
I don't know if nowadays commerically available sodium ethoxide and/or sodium hydride were that much used back in the days when the procedure was published. Fact is that both these can rot away in humid air when somebody leaves the bottles and cans open for too long, and this happens frequently in university labs.
With sodium metal, you're on the safe side: a freshly cut piece of metal is always good to go.

EDIT 2
For the example given above, no further direct alkylation at that position is possible, and this is exactly, what ron pointed out in his
comment to the question:

Perhaps the author was trying to say that by using just the malonic ester synthesis you can't prepare a compound with 3 (potentially different) alkyl groups on the alpha carbon.

He also suggested the viable solution, which I had outlined in another question on this site. Upon acidification and heating, $\beta$-ketoesters and 1,3-diesters undergo decarboxylation, which makes them great synthons to introduce $\ce{-CH2CO-CH3}$ and $\ce{-CH2COOH}$, respectively.
Converting a 1,3-diester to the corresponding carboxylic acid this way means that the new structure now again has an $\alpha$-proton that can be abstracted. The resulting enolate can be alkylated.
In the case of a carboxylic acid, this is usually done in a solvent like tetrahydrofurane or 1,4-dioxane, using two bases. For the deprotonation of the acid, $\ce{NaH}$ is sufficient, while the abstraction of the $\alpha$-proton, a strong base like $\ce{LDA}$ is used. (Exactly this is outlined in the textbook for the alkylation of an ester).

Some very neat regiochemistry there. Why was $\ce{Na}$ used? I think perhaps $\ce{NaOH}$ and certainly $\ce{NaH}$ would be sufficient Or is $\ce{NaH}$ produced in situ from the reaction of sodium with the solvent (ethanol)?

Also this does not really totally answer the question (sure it does from our perspective, but it would be good to know why the authors wrote what they wrote, assuming they don't write nonsense).

Traditional malonic ester synthesis post-processing includes hot acidic hydrolysis, which should lead to substituted malonic acid. However, such acids are prone to decarboxylation, leading to tertiary carbon.
In case you deviate from the tradition, there is no problem with getting quaternary  carbon (though general bulkiness still applies and may prevent some synthesises)


text:
How to find the mass of a solute given a mass fraction?

How many grams of $\ce{CuSO4 * 5 H2O}$ are needed to prepare a $20\%$ (w/w%) solution of $\ce{CuSO4}$?

Welcome to Chemistry.SE. Please note that the homework policy on this site requires you to show some personal effort.

here's what i come up with so far ..
MA($\ce{CuSO4*5H2O}$)=(65.5+32+64+90)=251.5 g/mol
=>20% of ($\ce{CuSO4*5H2O}$)=50g !?
i don't know i don't have any clue
@KlausWarzecha

Yes! Using the molecular weights is the right approach! From the weight percentage in the solution, you can calculate the mass of $\ce{CuSO4}$ in 1 litre. With the molecular weight of $\ce{CuSO4}$, you get the molar concentration. It's the same number of moles you have to add to the water. Now you just have to convert that to a mass of $\ce{CuSO4*5H2O}$ again, using the same relation between moles, mass and molecular weight. Feel free to give the answer to your question yourself, that's perfectly ok! If you do, please describe how you did to. That will help the next with a similar problem :)

In a $20\%$ solution of $\ce{CuSO4}$ we have $\pu{20 g}$ of $\ce{Cu}$ and $\pu{80 g}$ of $\ce{H2O}$. Then
in every $\pu{259.6 g}$ $\ce{CuSO4 * 5 H2O}$ we have $\pu{159.6 g}$ $\ce{CuSO4}$,
so for every $x~\pu{g}$ $\ce{CuSO4 * 5 H2O}$ we have $\pu{20 g}$ $\ce{CuSO4}$
$$x = \pu{31.87 g}$$

I think there is a very minor mistake in the molar mass of the pentahydrate, which should be 249.6 g/mol. Therefore, your final result is a little bit off. Anyway, the important thing here is to understand the concept. My suggestion is: Always keep the units in your calculations! These are a good measure to see where you have to divide or multiply.


text:
Relation between molar mass and van der Waals force
In the book that I'm reading, it is mentioned that higher molar mass results in stronger van der Waals force but didn't continue why.
I wonder why is that so? Molar mass actually depends on the number of protons and neutrons, how can it have an effect on the van der Waals force?

Just to make sure, by Van der Waals forces do you mean the induced dipole - induced dipole force? Since it's wrong to think that way for all of them.

@MARamezani Yes. it is said "The van the waals forces" so I think it include induced dipole-induced dipole too. could it be wrong?

From chemwiki.Ucdavis:

The more electrons a molecule contains, the higher its ability to become polar. Polarizability increases in the periodic table from the top of a group to the bottom and from right to left within periods. This is because the higher the molecular mass, the more electrons an atom has. With more electrons, the outer electrons are easily displaced because the inner electrons shield the nucleus' positive charge from the outer electrons which would normally keep them close to the nucleus.

This seems totally legitimate. However, (as chemwiki subtly points out) the main factor in work here is the volume occupied by electrons. Polarizability increase is mainly due to a volume increase 1. So, all you need to take into consideration is whether a molecule is bigger than the other in terms of comparing the stronger London dispersion force.

Simply put, the ability of a molecule to get polarized is called polarizability. See Wikipedia for more info.

Thanks. So it is something like that:
higher molar mass >> larger shell >> more Polarity  >> Higher van der waals forces , right?

I would say "more polarizability". "Polarity" is a different concept. And this statement is generally true.

Somewhat related to: Hydrogen and Helium ideal gas behavior in room temperature
The induced dipole-dipole interactions operate through the polarizability of the molecules. Generally speaking heavier atoms will have more electrons and they will have some of those in the outer, higher radius shells. These higher radius elements will be more polarizable, hence stronger Van der Waals forces.
Though I can see where the statement comes from, I doubt it is a completely true statement. I would think one can have higher molar mass compounds that are less polarizable(though I cannot come up with an easy example right now)

Depends on what we call "true". I rather call it a statement teachers teach to students in order to make the comparison easier for them.


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Balancing Redox Equations - Half Reactions
How do I balance this using the half-reaction method: $$\ce{Zn(s) + HCl(aq) -> Zn^{2+} (aq) + H_2 (g)}$$
I would first split it up in its ionic components:
$\ce{Zn(s) + H^+ + Cl^- -> Zn^{2+} (aq) + H_2 (g)}$.
Then we have first the oxidation-reaction: $\ce{Zn(s) -> Zn^{2+} (aq)}$. Balancing this with electrons gives $\ce{Zn(s) -> Zn^{2+} + 2e^-}$.
The reduction-reaction is $\ce{H^+ + Cl^- -> H_2 (g)}$. I need to balance this with $\ce{Cl}$? So that would get me $\ce{H^+ + Cl^- -> H_2 (g) + Cl}$. Then adding hydrogen to the left side with $\ce{H^+}$ gives $\ce{2H^+ + Cl^- -> H_2 (g) + Cl}$.
However, I feel like I made a mistake somewhere. Any help please?

Try to blanace electrons, like you did: Zn(s)→Zn2+ +2e−, where these two electrons are going to?

Keep the chlorine ionic and don't react it with anything. Cl- should stay Cl-.

I don't know where the electrons are going. I'm confused because in the equation $H^+ + Cl^- \rightarrow H_2 (g)$ there is a $Cl^-$ ion the left, but then it dissapears? Where does it go then?

The mistake is right at the beginning in the initial equation. The chlorine has disappeared. So you need to rethink this equation and the subsequent reduction equation.

I recently answered a question like this. It's not the top rated answer, though I feel it may be helpful: http://math.stackexchange.com/questions/624/balance-chemical-equations-without-trial-and-error/1223348#1223348

You are right for the half reaction for oxidation is:
$$\ce{Zn(s) \rightarrow Zn^{2+} + 2e^-}$$
Now, you have to write properly the  half reaction for reduction:
 $$\ce{2H^+(\mathrm{aq}) + 2e^- \rightarrow H_2(g)}$$
This means, that the electrons given by zinc are taken by ion hydrogen.
So, if we add the two half reactions:
$$\ce{Zn(s) + 2H^+(\mathrm{aq}) \rightarrow Zn^{2+}(\mathrm{aq}) +  H_2 (g)}$$
Notice, that ion chloride doesn't change its oxidation state. Its role is to keep the medium neutral electrically. So, we'll add two ions chloride to the two members of the equation:
$$\ce{Zn(s) + 2(H^+_{\mathrm{aq}} + Cl^{-}_{\mathrm{aq}}) \rightarrow (Zn^{2+}_{\mathrm{aq}} +2 Cl^{-}_{\mathrm{aq}}) + H_2(g)}$$
Remember, redox reaction are reactions of electrons exchange between an oxidant and an reducer.


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What are the meanings of dotted and wavy lines in structural formulas?
So I came across this diagram,the other day:

As  a chemistry student, I am well-versed with the dashed and wedged lines, but I was wondering what the wavy and the dotted line represent?

See also Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008) and Graphical Representation of Stereochemical Configuration (IUPAC Recommendations 2006).

I was wondering what the wavy and the dotted line represent?

A dashed line indicates that the bond is extending behind the plane of the drawing surface
A bold-wedged line indicates that the bond is protruding out from the plane of the drawing surface
A solid line indicates that the bond exists in the plane of the drawing surface.
A wavy line indicates that the stereochemistry of the bond is unknown.
A dotted line indicates that the bond is not a full bond, it is only a partial bond as in a hydrogen bond or a partially formed or broken bond in a transition state. Also, as Mrigank Pawagi notes in the comments, the dotted line can be used along with the solid line to denote bonds that have partial double bond character due to resonance.
An arrow indicates a dative bond in which both electrons in the bond originate from one atom (not common)

As a side note, I would guide the OP to their second virtual friend.

I think dotted lines are more often used for intermolecular attractions and dashed lines for partial bonds.

A wavy line could also mean isomerises rapidly, e.g. a methyl group between two carbonyls.

The Dotted line is also used to denote bonds in resonance hybrids.

The arror is not uncommon at all, look at n-oxides

The wavy line usually represents either that that part of the molecule is unimportant or that the stereochemistry at that specific position is unimportant (or unknown).
The dotted line usually represents intermolecular interactions, like hydrogen bonds. Dashed lines refer mostly to partial bonds such as in the transition state.

authors: Κωνσταντίνος Πανταζής


text:
Are all lattices Bravais lattices?
Can all lattices be described as one of the fourteen Bravais lattices?
Is the hexagonal close packed structure also one of the fourteen Bravais lattices?

From Ashcroft and Mermin's Solid State Physics:

A fundamental concept in the description of any crystalline solid is that of the Bravais lattice, which specifies the periodic array in which the repeated units of the crystal are arranged. The units themselves may be single atoms, groups of atoms, molecules, ions, etc., but the Bravais lattice summarizes only the geometry of the underlying periodic structure, regardless of what the actual units may be.

So, a real crystal structure is defined by the Bravais lattice, and the unit placed at each point on the Bravais lattice.  Now, it turns out that the hexagonal close packed structure (hcp) is not, in fact, a Bravais lattice.  Hcp is based on the simple hexagonal Bravais lattice, but with two atoms as the basis (one could also call it two interpenetrating simple hexagonal Bravais lattices).  The bottom line is the hcp is not itself a Bravais lattice (see further details in Ashcroft and Mermin). I'll skip any discussion of point groups and space groups, since those are just further refinements of what it means to combine the symmetry of a Bravais lattice with the symmetry of the repeating unit.
There is a hierarchy of symmetry - 7 crystal systems, 14 Bravais lattices, 32 crystallographic point groups, and 230 space groups. For hcp, the point is that it can be represented as a simple hexagonal Bravais lattice with a two-atom unit, so having hcp as another "Bravais lattice" would totally change the definition of what a Bravais lattice is - the fundamental symmetry is simple hexagonal.
The general idea is to start with something you can repeat in a specified way to fill space.  For example, starting with a cube, it is pretty easy to see how to stack them up to fill space.  If, instead of being cubes they are stretched along one axis to get a rectangular solid, you can still fill space, but it will look different depending on what direction you are looking.
So, we get the seven crystal systems: Cubic (with simple cubic, body centered, and face centered cubic Bravais lattices); Tetragonal, which is a cube stretched along one side (with simple tetragonal and centered tetragonal Bravais lattices); Orthorhombic, a unit with perpendicular faces but unequal lengths on each side (with 4 Bravais lattices); Monoclinic, which tilts an orthorhombic unit to make one angle non-90 degrees (2 Bravais lattices); Triclinic, which has no two faces perpendicular to each other (1 Bravais lattice); Trigonal, which is obtained by taking a cube and stretching it along a body diagonal (1 Bravais lattice), and Hexagonal, which has one Bravais lattice.
You get those 7 crystal systems and 14 Bravais lattices by being able to stack the appropriate units to fill space.  Now, while this concept is pretty fundamental, there are further symmetries imposed on the crystal by what you stack up.  Lets consider something fairly simple, simple cubic.  If you start with a big pile of white cubes, the resulting object you put together will look the same regardless of which direction you look at it, or how you rotate it around.  BUT - if you start with a pile of white cubes with one face painted red, you just changed things. To conform with the Bravais lattice, you would stack them so that all the red faces pointed in the same direction.  You would still get a big cube made of the smaller cubes.  But now it would look different depending on what direction you looked at it - you have reduced the symmetry of the crystal.
For simple hexagonal, one starts with a hexagonal right prism (hexagon base, some height perpendicular to the base).  Stacking those one on top of each other results in a simple hexagonal Bravais lattice.  Lets make the hexagonal prisms out of clear acrylic, but place a marble at the center of the prism.  Now when you have stacked them up you would see a simple hexagonal lattice of marbles, or heck, lets call them atoms.  Now, exactly where you put the marble in the clear prism doesn't matter - you would still see the exact same arrangement of the marbles in space.  OK, lets place the first model in the center of the base of the prism, and place a second marble halfway up one of the edges.  When you stack them up, you will now get an arrangement of marbles in space that looks like the hexagonal close packed structure.  But you stacked the prisms following the rules of the simple hexagonal Bravais lattice.  Placing that second marble in the Bravais prism generated a two atom basis.  It clearly is different from simple hexagonal, but the reason is not how you stacked the prisms, it is what the prism contains.  The underlying Bravais lattice does not change, but the symmetry of the resulting crystal is changed by the basis unit used.
Now, use different colored marbles in the two positions, and it changes once again.  Throw three marbles in in different places in the prism and it changes yet again.
So, the Bravais lattice occupies a fundamental level in understanding the crystal structure - it is the rules for stacking things up (and only the rules for stacking things).  The objects you stack up may have their own properties, meaning that two things stacked using the same Bravais lattice may in the end look very different.  This combination of the lattice and the unit are what create the 230 space groups.

Can't all possible crystal structures be defined as one of the 14 lattices? If not - like hcp cannot be - why weren't more than 14 lattices used so that every possible lattice was covered in these?

All possible lattices are covered by the 230 space groups that arise from combining the 14 Bravais lattices and all possible symmetries of the unit you place on the Bravais lattice.  There is a hierarchy of symmetry - 7 crystal systems, 14 Bravais lattices, 32 crystallographic point groups, and 230 space groups.  For hcp, the point is that it can be represented as a simple hexagonal Bravais lattice with a two-atom unit, so having hcp as another "Bravais lattice" would totally change the definition of what a Bravais lattice is - the fundamental symmetry is simple hexagonal.

I'm sorry - I don't quite know what space groups, crystal systems, crystallographic point groups, or two atom units are. If you could tell me of a source where I could find this explained in a brief and not too mathematical or detailed manner (just high school level) I'd be really grateful.

But why do we have only 230 space groups? Like in your example I can have any combination of marbles inside the hexagonal prism which will be stacked periodically to cover the whole space. Won't this lead to infinite number of space groups?


text:
Spectral reflectance identification of types of iron rust
Based on the Wikipedia page about rust, there are red and green types of rust that occur on iron.  Red rust is the reaction with oxygen and water and the green with chlorine (in oxygen poor environments).
I understand the chemical reactions that forms these types of rust, but in the Wikipedia article, it states that:

Several forms of rust are distinguishable both visually and by spectroscopy

(bolding mine). Visually, it'd be pretty obvious (one is red, one is green); spectroscopy-wise, I would imagine it to be a similar story - but what aspect of the types of rust cause this difference - hence, the question:
Are the spectral reflectence features that distinguish between the red and green rust types based on what the iron reacts with (oxygen/water or chlorine), or is it based on the type of iron present (Fe2+ vs Fe3+)?
Full disclosure, I have a similar question on Open Data.SE here, but are asking for an explanation in terms of spectral reflectivity.

If available, the best spectroscopic technique for iron is usually Mossbauer.


text:
Electroplating a spoon (electrolytic cell) using a electrochemical voltaic cell
I drew a rough picture of the two cells, and marked each position with a distinct letter:
The whole reason I am asking this question is that I am bit confused and unsure about some of the chemistry and physics concepts. It will be extremely helpful if you guys can tell me if what I did is correct and why.
What I think is correct: to make this whole thing work.

Connect alligator clip x to position A, right on the zinc anode not the wire.

And connect clip y to B, on the copper cathode plate.

From what I read, the electron is forced from Ag plate, anode, through the wire (with two alligator clips on each end), then reduces $Ag_{+}$ ions to $Ag$ metal on the spoon cathode. But from the picture, it looks more like all the electrons came from the zinc anode of the voltaic cell and all the electrons from the $Ag$ anode of the electrolytic cell go to $Cu$ cathode of the voltaic cell.
Some physics aspect of this: if x clip was to be connected to position C and y clip to position D, is it true that there will be no potential difference?(unless the wire itself has a high resistance)

Let's start with the last question: The points A and C have the same potential, because we  generally neglect the resistance of the wire. The same argument is valid for points B and D. But according to your scheme, you have a short circuit (as C and D are connected).
Now, to electroplate the spoon, you want the silver to be oxidized at the silver electrode. So, silver electrode is the anode, and must be connected to pole (+) of the voltaic cell (point B):
$$\ce{Ag -> Ag+ + e- }$$
The spoon is the electrode where ion silver will be reduced. So, it's the cathode and must be connected to pole (-) of of the voltaic cell (point A). The electrons enter the cathode and reduced the silver ion according to the equation:
$$\ce{Ag+ + e- -> Ag}$$
So, you're right!

What is short cutting? Do you mean short circuit (actually it doesn't make too much sense)? What do mean exactly by "short cutting"?

I mean short circuit and this occurs because  the positive and negative terminals of the battery are connected with a wire. In fact,due to the low resistance in the connection, a high current exists, causing the cell to deliver a large amount of energy in a short time.

If AC have the same potential CD should also do.

Electrolysis is always carried out using direct current because here the electrodes have definite polarity. Generally low voltage and high current are preferred. Towards A.C the behavior of the solution depends mainly on the electrodes. For most metallic electrodes namely Cu, Zn the solution behaves much like resistance and the energy is wasted in heating of the solution.


text:
Rate constants for water exchange
Are rate constants for water exchange of transition metal aqua-complexes tabulated anywhere ? I haven't been able to find a resource which is anywhere near comprehensive.

I think this part of Wikipedia article touches this

Yes, see Fig. 1 and tables VI-VIII of  SOLVENT EXCHANGE ON METAL IONS Advances in Inorganic Chemistry volume 54 pages 1-69.
Values vary over almost 20 orders of magnitude.
To make this not a link-only answer, roughly, per second exchange rates are:
$\ce{Ir^3+}$  $10^{-10}$
$\ce{Rh^3+}$  $2.2 \times 10^{-9}$
$\ce{Cr^3+}$ $10^{-6}$
$\ce{Ru^3+}$ $10^{-5}$
$\ce{Pt^2+}$ $10^{-3}$
$\ce{Ru^2+}$ $10^{-2}$
$\ce{V^2+}$, $\ce{Fe^3+}$ $10^2$
$\ce{Pd^2+}$, $\ce{V^3+}$ $10^3$
$\ce{Ni^2+}$ $10^4$
$\ce{Ti^3+}$ $10^5$
$\ce{Co^2+}$ $3.2 \times 10^6$
$\ce{Fe^2+}$ $4.4 \times 10^6$
$\ce{Mn^2+}$,  $2.1 \times 10^7$
$\ce{Zn^2+}$ great uncertainty $5 \times 10^7 - 10^{10}$
$\ce{Cd^2+}$ $10^8$
$\ce{Cr^2+}$, $\ce{Hg^2+}$ $10^9$
$\ce{Cu^2+}$ $5.7 \times 10^9$


text:
Unit cell of CdCl2
I understand that $\ce{CdCl2}$ has a rhombohedral unit cell as shown in this book. However I often find it being referred to as a "CCP analogue" of $\ce{CdI2}$. Presumably it is the chlorides which form a cubic lattice; how would you draw the cubic unit cell ?

Cubic close packing is another (confusing) name for face-centered cubic.  So, to visualize these structures, you start with a hexagonal 2D layer of the chlorine or iodine.  Now start stacking those layers, ABAB... for hcp (iodine), or ABCABC... for fcc (chlorine).  Finally, stick the Cd in between alternate layers of the chlorine/iodine. The net result for CdCl2, is that the unit cell is indeed rhombohedral, since there is one Cd layer for every two Cl layers, so it is stretched in that direction.

Why does the ABCABC stacking correspond to a FCC lattice though ? I can't see the right orientation in a 3d structure model (http://www.chemtube3d.com/solidstate/_CdCl2%28final%29.htm).

Orient the live 3D model so you are looking perpendicular to a face on the long side of the indicated rhombohedron. Now the green layers (at least on my browser) are the close packed planes of chlorine. They repeat in a pattern of A-B-C..  Now, go find a picture of an fcc metal.  You will note that if you look along the body diagonal you see an atom on the bottom point, the atoms on the faces of the bottom sides, the atoms on the faces of the top sides, and finally come back to the same position with the top vertex atom.  These are the close packed planes stacked up as A-B-C-A.

I can see the alternation but not where the fcc unit cells of Cl are...

Again, compare with a simple fcc metal such as copper. Turn the traditional unit cell so that the body diagonal is pointing up.  Take the bottom vertex as an atom on a hexagonal plane.  The atoms at the center of the lower faces are also a hexagonal pattern, and are the B position.  Finally, the atoms at the center of the upper faces are yet another hexagonal plane which line up with neither A or B, so they are C.  The atom at the top vertex is back to A, and it repeats from there.  For FCC, the stacking is along the 111 direction.

Horray, I can see it now ! Thank you so much.

You are quite welcome.  I think that the standard unit cell representation, and calling it 'cubic', gives us a bias that is hard to overcome.


text:
What is the pH of CsBr?

What is the pH of CsBr?

3.8
7
9
12

According to me,
$\ce{Cs+ + H2O <<=> CsOH + H+}$   (because $\ce{CsOH}$ is a strong base, the forward reaction should not occur, right?)
$\ce{Br- + H2O <<=> HBr + OH-}$ (because $\ce{HBr}$ is an acid, the forward reaction should also not occur, right?)
So the answer should be $\mathrm{pH} = 7$.

There's always equilibrium constant for such reactions, but your reasoning is otherwise OK

If there was an equilibrium constant present, how would this change/affect the answer?

Both CsOH and HBr are soluble strong (base/acid) which helps explain the relative lack of effect. In aqueous solution anyway, the effect would be very small

The reactions you have shown could in fact occur (in the forward direction) to a slight degree in a solution of cesium bromide in water. But I would expect them to occur to a lesser degree than the autoprotolysis of water (formation of hydronium and hydroxide ions from two water molecules), so their effect on pH would be insignificant, and the pH would effectively be neutral.


text:
To what state would copper wiring revert if abandoned for centuries?
Would copper wires from power transmission or telephone lines acquire a patina but remain largely intact, or oxidize into dust after a century or more without replacement or maintenance?
Obviously environmental factors would be enormous; ideally I'd like to know what the metal's state would be in ideal (underground?), typical (temperate zone), and worst-case (near salt water?) scenarios. Clearly some circumstances allow copper objects to survive largely intact for several thousand years, since such artifacts have been found by archeologists, but does this require very exceptional circumstances?
This question is inspired by an answer on the Worldbuilding site, which claims that a post-apocalyptic society would have a difficult time developing metallurgy as the rusted remains of human civilization would be more difficult to smelt than the naturally-occurring ores which were relatively abundant during the Stone Age. Hopefully by asking about this copper wires specifically, I have an appropriate question for this site. If so, I may ask another question about iron and steel.
I did not have much luck in finding relevant tags. Any assistance there is appreciated!

Copper sheathing for roofs can last a very long time, indeed acquiring a nice patina.  It might not be a great roof after a century, but it will still be there.

True. The Clarksville Customs House is over 100 years old, and the roof has survived tornadoes. However, copper wiring is presumably much more widespread. Though I understand that most power lines are largely aluminum these days.

Let me talk first about how iron corrodes (as in rusts) and then compare that to copper. The simple answer is that when iron corrodes, the iron oxide (iron which has chemically bonded with oxygen, which it loves to do) that forms on the surface of the pure iron underneath, does not seal the surface against further intrusion of more oxygen. When copper corrodes, the copper oxide does seal the surface protecting underlying copper against further corrosion. I don't know if iron oxide is simply more porous to oxygen than copper oxide is but I do know the following. Iron can chemically combine with oxygen in multiple ratios. When you examine a highly corroded iron surface you'll notice that the first layer of rust is very red. If you brush or maybe lightly sand that red layer of rust away you'll notice a black layer underneath it. That's still rust. If you sand your way through that you'll get to the pure iron underneath which has more of a dark silver color. The red rust has more oxygen in it than the black rust but can pass some of that oxygen to the black rust turning it red. Similarly, the black rust can pass some of its oxygen on to the pure iron underneath it. That process continues until oxygen can leach all the way through the iron. Chemically, iron has more than one oxidation state. Oxidized copper atoms are much more reluctant to release their oxygen to adjacent pure copper and act as a barrier to the intrusion of more oxygen.

Right, iron is notorious that way. I heard that it's because the oxide has a greater volume, and thus the outer layer expands and separates from the inner layer, exposing it. Regarding copper, though, will the layer of oxide protect the inner part indefinitely, as long as there is no mechanical action to expose fresh metal? Will it survive for years, decades, centuries? What if it's exposed to rain, or standing water?


text:
Is time a measurement of Entropy?
Is time a measurement of entropy? Is it because time is increasing forever?
Well after reading about the Second Law of Thermodynamics and working with Entropy, I began to realize that if time can be interpreted as a measure of Entropy, since it is always increasing.

Since almost everything decays with time, I thought that time could be
  a measure of Entropy. Since it is achieving the lowest energy state
  possible.

Thanks

Could you elaborate and improve your question? Right now it looks like you thought too much about II law of thermodynamics ;)

Yep, you seem to be confused... Is it connected with the most common problem with thermodynamics, like: http://chemistry.stackexchange.com/questions/24669/does-adsorption-violate-thermodynamics ?

yeah, It is kind of measure of entropy

Related fun fact: http://numb3rs.wolfram.com/511/

Time and entropy has not much to do with each other.
From practical point of view: Equilibrium systems can hang around for infinite time, and their entropy does not change at all - so time would be an impractical measure.
From conceptual point of view: Entropy comes from statistical behavior of the system of many bodies. Time is unrelated to statistics; it exist even in single body systems.

Most systems that are said to be in perpetual equilibrium (say a pendulum which ignores any friction) would go for ever. However, in reality it does loose it’s energy as “time” goes by. Causing its entropy to increase until it remains stationary. Similarly the entropy of the whole universe is going in the same direction, and this is just a statistical phenomena (the second law).

@RSM This is not true. You mix up thermodynamic and mechanical systems in an incorrect way. A given volume of gas will not just cool down due to some friction.


text:
Why is the melting point of TiCl4 lower than TiCl3
I was wondering why the melting point of $\ce{TiCl4}$ was below that of $\ce{TiCl3}$
The melting point of the former is  $\ce{-25 Celsius}$ whereas the latter is $\ce{425 Celsius}$ .
Thanks!

-The high melting point of  $\ce{TiCl3}$
is mainly due to the short bond length $\ce{Ti-Ti}$ , which infer
 a strong metallic character to these bonds between titanium ions in  $\ce{TiCl3}$.
-While in $\ce{TiCl4}$, the main intermolecular forces are of van der Waals type.

What leads to this short bond length? Is it less shielding by Cl or greater charge, which would lead to a smaller radius per unit cell?

Most metal chlorides are polymers, wherein the chloride atoms bridge between the metals. TiCl3 (crystallographic form beta) structure consists of chains of TiCl6 octahedra that share opposite faces such that the closest Ti—Ti contact is 2.91 Å. Even, for the other crystallographic forms of TiCl3 (alpha, gamma and delta) where the distance Ti-Ti  becomes bigger, the metallic character of the bond is dominant.

Oh I see! TiCl3 has a polymer/ crystalline structure but TiCl4 has a molecular structure. That makes sense! Thank you for clarifying!


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Computational Chemistry (Ab Initio), what should I study?
I'm interested in ab initio computational chemistry. I'm trying to figure out what I should study. I know that I should study physical chemistry, mathematics, and programming.
However, I am not sure in what I should be focusing on. It seems to be a fairly new field and books on the topic are rather specific.
Can anyone working in computational chemistry research provide me with a list of (math, computational chemistry, physical chemistry) textbooks they studied or any additional resources to get me started?

Computational chemistry is not a new field, but yes it can be difficult to find the right approach to it, because it is so diversified.  And that means to some degree what you need depends strongly on what you want to study.  Most of the time, you're basically solving the Schrodinger equation all kinds of different ways.
It will be better if you can specify what level of knowledge you do have at this time. Assume that you've just got into college, then there's a list of things you should study:
Basic theory

Math.

You need linear algebra and mathematical analysis.  Any good old textbook will do.
What you use the most everyday would be linear algebra, so you better be very very good at it, along with a good expertise in vectors, matrices and in some cases tensors.  There may also be some functional analysis involved but for most people it really just boil down to linear algebra.
Mathematical analysis is basically just differentiation, integration, and all kinds of tricks related to it.

Quantum mechanics.

You some of it from P-Chem II, but if you take it from the physics department it will be more helpful.  I recommend the Cohen-Tannoudji textbook.  Make sure you start from the beginning, go through every chapter to the very end.

Quantum chemistry.

With a background you can now proceed to understand the conventional methods of quantum chemistry.  For this I would recommend Modern Quantum Chemistry by Szabo and Ostlund.
Courses to take:
In addition to what was already mentioned above, many universities will also have quantum chemistry and computational chemistry courses. I also recommend you to take group theory and calculus of variation.
Programming skills
And that prepares you for the theoretical part.  Whether you need comprehensive programming skill depends on what you want to do.  Some may make do with existing programs, some may need to write there own.  But even if you use an existing program, it requires no less amount of understanding of the quantum chemical theories you're using.
For programming skills, you will certainly benefit by taking numerical methods course, and mastery over one programming language. Take statistics and machine learning if you have time. Any one will do, it doesn't matter. For what we do, its straight forward to switch to a different language whenever you want.  Computational chemists are not usually very big on the algorithms themselves. Most of the times, you just take something right out of a numerical analysis text book or program package and adapt a little bit.  It is much more about the underlying physics, i.e., what form of wave function you choose, what approximation you make, etc.
The most important thing about programming, I feel, is hands on experience.  If you can understand quantum mechanics, programming is not gonna be hard.
Chemistry
As for other branches of chemistry, you don't have to be the expert, but you definitely need a good understanding because those are the problems you're trying to solve.  You need to understand very well the experimental procedure you're trying to study.