text:
Average Oxidation State - Utility?
What is the purpose of calculating an average oxidation state for a certain element in a compound?

Hence the “average oxidation state” of carbon in acetic acid is 0.

And

the oxidation state value for carbon atoms in acetic acid is, however, computed as zero. This latter value represents an average of the values for the two carbon atoms derived by the rigorous assignment rules.

What does this tell me? Why not just stick with individual oxidation states? Is there something the overall oxidation state is supposed to tell me about acetic acid?
The only reason I can think of is that perhaps this average oxidation state of 0 for the carbon atoms tells me that acetic acid has some ionic character. Hence the oxidation state of 0 for carbon (as opposed to +4 for carbon in its elemental form).

Ionic half-reactions method is represented here:
$$\ce{Oxidation : 2S_{2}{O}_{3}^{2-}\rightarrow S_{4}{O}_{6}^{2-} +2 e^{-}}$$
$$\ce{Reduction : I_2 +2e^{-}\rightarrow 2I^-}$$
Then, as the same number of electrons is transfered in both processes, you just sum these equations to get the balanced reaction. If the number of transfered electrons wasn't the same, then you need to find the greatest common denominator and multiply each equation with a number so the number of transfered electrons gets equal to the greatest common denominator. Then you sum the equations.

It helps you when you want to balance a red-ox reaction by the mean of the oxidation number change. For example, if you want to balance the reaction of thiosulfate with iodine to tetrathionate and iodide, you realize that the average oxidation number of sulfur changed from +2 to +2.5. The oxidation half-reaction looks like this:
$$\ce{4S^{+2}-> 4S^{2.5+} +2e-}$$

still, it gives some strange ideas without definitive simplification. So the practice is questionable.

Oxidation number per se is an invented model to simplify some problems and explain some phenomena. The usage of the oxidation number change method or the ionic half-reactions method (which I prefer more) is up to the person.

What's the ionic half-reactions method?


text:
How to obtain pure aniline from a mixture of phenol and aniline?
There is a mixture of aniline and phenol. I need to obtain (just the procedure) pure aniline.
So I think the mixture should heat with Zn powder(to convert phenol to benzene) and then I thought to synthesize aniline from benzene.
I am not sure about this method?
Please help me

This link should help you
https://uk.answers.yahoo.com/question/index?qid=20100910210700AACatO3
Good luck!

Is the objective to remove the phenol from the mixture, or to convert phenol to aniline. It is fairly easy to purify the aniline from the mixture.

I would try to dissolve the mixture in dichloromethane, then extract phenol with concentrated water solution of sodium hydroxide and finally precipitate aniline from ethanol or methanol with water. Even though working with sodium hydroxide requires goggles, you can care less about inhaling its fumes, unlike concentrated HCl. Re-synthesis is usually a very unvise choice even for moderately complex aromatics.

are aniline and phenol soluble to each other?

First dissolve phenol and aniline in diethyl ether and add excess HCl solution .
Then aniline will react with HCl and dissolve in water layer.
Easily remove organic layer.
Take the water layer and add excess NaOH to it and aniline will be precipitated.

Use a solvent such a methylene chloride so aqueous layer need not be removed from funnel. Aniline and phenol are liquid and reasonably if not very soluble in water.

Aniline is a base, phenol an acid. Add the mix to a sep funnel add either dilute HCl or dilute NaOH and mix well. Extract several times with small portions of methylene chloride [or chloroform] [bottom layer] You figure out which you have in the organic layer. Neutralize the aqueous solution and reverse the acidity. Extract the aqueous mess again with methylene chloride you have the other in this batch. Distill each organic batch separately. This should all be done in a hood with appropriate safety precautions. The order depends on circumstances. I would probably prefer that the final aqueous solution be acidic so I would extract the aniline first. The amounts concentrations etc. have to be carefully worked out.

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text:
Why do many transparent plastics turn opaque when deformed?
Many transparent plastic materials, rigid or not, turn opaque (white) when subjected to stress, deformations or when folded (if not rigid) . The effect seems to be irreversible.
What is the cause?
I would have guessed that some internal crystalline structure is broken, but plastic (like glasses) have no crystal structure, no?

I would say that the reason for the white color is that by deformation you introduce a lot of small "cracks" into the material which scatter the incident light such that the deformed areas appear white.

The only commodity polymer that springs to mind which is absolutely noncrystalline is polystyrene.

The process of forming an opaque area when you apply stress to a polymer film is known as crazing.
Polymers are long chains and, when coated, often form tangled, amorphous, randomly aligned polymer regions.  When stress is applied to the film, the stress can be relieved by the polymer chains aligning.  In other words, energy introduced into the film by stressing it can be used to align some polymer chains.
Increasing the alignment in the stressed region means that the density of polymer in this region of the film will be different from the density in non-stressed regions.  This difference in density will cause a difference in light reflection and scattering accounting for the different optical properties between these two regions.
Further, within the stressed area of the film, mechanical stresses are produced between areas that are more and areas that are less aligned.  These stresses can be relieved through the formation of micro-cracks between the regions.  These small cracks also result in different light reflecting and scattering properties of the film.
Finally, as the polymer chains align, they are more likely to crystallize.  The formation of small crystallites within the stressed (aligned) region will also alter the light reflecting and scattering properties of the film.  Often, upon heating or standing at room temperature (for a longer time) the optical changes may reverse as the thermal energy supplied will:

cause the polymer structure to randomize back to its amorphous structure,
allow the polymer to "flow" and fill the micro-cracks, and\or
the crystallites can redissolve.

References: The following links provide more information about crazing and associated phenomena at different levels of complexity:
reference 1
reference 2
reference 3

@Philipp already mentioned microscopic cracks. This is called crazing. When a plastic is stretched, the extra space is filled randomly/irregularly. The gap between two polymer chains will grow, but most chains won't break. Also, if the material is already crystallized locally, the crystals will pull apart.
As the overall structure is non-crystalline, the cracks will have different sizes, and in particular quite a few will have sizes smaller than a micrometer - optical wavelengths, in other words. And with the variety in sizes, you don't get oen dominant color but a mixture of all colors which is of course white.
The other reason for the optical change is crystals forming. Again, these small crystals have different sizes around the wavelengths of visible light, so there's no dominant color from these either.

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text:
What is cheap material that absorbs water in packaging?
I am thinking cheap material that can be used between two layers of carton that absorbs water.
I want to use air but I still think that there may be other materials available that

absorb water
can be easily removed when the outer carton layer is damaged

Possible materials

carbon (well absorbing)
air (cheap and easy)
sand (should be bad)

What is cheap material that absorbs water in packaging?

You could probably also use Sodium Polyacrylate, which is used in diapers to absorb water. This can absorb up to more than 200 times its weight in water. It comes in little white pellets, lab grade is $9.90 per 100 g (absorbs more than 2 gal) from Carolina Science (Sodium Polyacrylate).
More information can be found at https://en.wikipedia.org/wiki/Sodium_polyacrylate

Can you compare it with silica gel? I think it can absorb mre than silica gel but I cannot find a source for the ratio.

Well according to this website silica gel can adsorb about 40% of its weight in water.

Yes, how much then sodium polyacrylate?

About 200 times its weight. (Note: Sodium Polyacrylate expands into a gel-like material while it absorbs water). You might be looking for silica gel as it won't blow the box up if it absorbs too much water.

What you are looking for is most probably silica gel.
http://en.wikipedia.org/wiki/Silica_gel
It is a highly water absorbent material with large surface area. It is often used in packaging as well as in laboratory in desiccators.
You can actually buy it packed or in bulk easily (let me not link explicit offers here). You can regenerate used (old) silica by heating / drying / pumping, so you don't need to buy every single time new one if the old gets less effective.


text:
What are Lattice Points?
What are lattice points exactly? For what purpose are they used in solid state chemistry?

this link might help you

There are different types of lattices. A lattice point is a point in any of these lattices. Lattices are used to describe  highly ordered systems such as crystals and some supersolids. I am unsure whether lattices can only describe periodic systems (see quasicrystals). However, I will be assuming that we are talking about periodic lattices. I will also be assuming that we are dealing with crystals.
Crystal lattice
The crystal lattice is used to describe the lattice of a real crystal. For example, in NaCl, a lattice point in a crystal lattice represents the position of a sodium ion or a chloride ion.
Bravais lattice
Bravais lattices are more mathematical and abstract than crystal lattices. They are pretty much the same as crystal lattices. Unlike the crystal lattice, however, lattice points in the Bravais lattice no longer represent a position of a particular atom. Instead a lattice point represents a position in which an atom can be placed. In other words, a lattice point in a Bravais lattice is a point, which is equal and indistinguishable from any other another point. What matters in a Bravais lattice are not the points themselves, but how they are arranged (i.e. symmetry).
Reciprocal lattice
The reciprocal lattice is the Fourier transform of either the crystal lattice or a Bravais lattice. More often than not, it is used to refer to the Fourier transform of the Bravais lattice. More of that is explained in Physics.SE. However, I personally think that this video explains it better. A lattice point here can represent an atom or a point, depending on the context.
Finally, as mentioned by other posts, these points can be vacant.

When we speak about crystal symmetry or the space group of a crystal do we refer to the Bravais lattice or the crystal (lattice+basis)?

@CoffeeIsLife " I am unsure whether lattices can only describe periodic systems (see quasicrystals)". By definition of lattice (infinite array of points in space where each point has identical surroundings to all others), you cannot define a lattice for a quasi-crystal. Moreover, the concept of 'crystal lattice' is meaningless. What is meaningful is the concept of 'crystal structure', that is obtained by the convolution of the lattice with the motif (consistuted of one or more atoms).

@Anton the symmetry of the crystal structure is defined by the space group type. In fact several space group types (belonging to a same system) exist with a same lattice. For example (in the orthorhombic system) you have Pnma, Pmmm, P222... all these space group have a primitive (P) lattice, but are distinguished by the associated groups of symmetry opoerations.

Lattice Points are the points in a crystal lattice where atoms or ions can be placed. Lattice Points can be vancant. (See Crystal Defects.)

How many Lattice Points are there in each of the Unit Cells?

Its depends on which type of Unit Cell we are talking about.
See this for full refrence of Unit Cells including their types.

lattice points are mathematical objects. In fact, a lattice is an infinite array of points in space where each point has identical surroundings to all others. A lattice is thus a purely abstract mathematical object.
In 3 dimensions there exist the 14 Bravais lattices filling all space. Their importance in solid state chemistry is that every crystal structure results from the convolution of one these Bravais lattice with a basis (a basis is an atom or a group of atoms).

Lattice points are the positions (or coordinates if you wish) where you can place an atom. Therefore you can find either atoms or vacancies in the lattice points of a crystal structure.
I dont quite understand what you mean by "How many Lattice Points are there is each of the Unit Cells?"

The points in a crystal lattice, representing the positions of structural units(atoms,molecules or ions) in the representative crystal are called lattice points .
Each lattice point in the crystal has the same environment as the structural units have in the real crystal.
How many lattice point in the unit cell, depend on the type of crystal .


text:
Arrange these compounds: CO2, CH3OH, RbF, CH3Br in order of increasing boiling points

Arrange these compounds: $\ce{CO2}$, $\ce{CH3OH}$, $\ce{RbF}$, $\ce{CH3Br}$ in order of increasing boiling points.

I think I should consider the forces between them, that is:

$\ce{CO2}$: dispersion forces
$\ce{RbF}$: dispersion and  ionic forces
$\ce{CH3OH}$: Dipole-dipole interactions, Hydrogen bonding and dispersion forces
$\ce{CH3Br}$: Dipole-dipole interactions and dispersion forces

It is obvious that $\ce{CO2}$ is the smallest one and because $\ce{CH3OH}$ stronger than $\ce{CH3Br}$ it will have higher boiling point
But how to arrange the rest? Or how to compare ionic forces with Dipole?

Ionic bonds stronger than dipole dipole attractions.
Ionic bonds are strong electrostatic attractions between, cations and anions.

You know $\ce{CO_2}$ is gaseous at room temperature, so let's put that at the bottom. Methanol  forms hydrogen bonds, so that will be above bromomethane which does not. At last we have rubidium fluoride which is a salt. Salts generally have a very high boiling point (> 1000 °C, much higher than molecular structures) because of the ionic (electrostatic) interaction between the ions, so that one will be at the top. Ionic forces can be seen as extreme dipoles in a certain way, there is a grey area when electronegativity becomes large enough, that it can be seen either as a molecular structure or ionic structure. Consulting online information about the boiling points of these compounds (i.e. just check Wikipedia or some MSDS site) confirms the theory.

The following is the order from lowest boiling point to highest based on the types of forces these compounds have:
$\ce{CO2}$ - dispersion forces
$\ce{CH3Br}$ - dispersion and dipole-dipole
$\ce{CH3OH}$ - dispersion, dipole-dipole, and hydrogen bonding
$\ce{RbF}$ - ionic


text:
Structure of diamond
"diamond structure. Every carbon atom is bonded
to four other carbon atoms forming a three-dimensional lattice of chair conformations."
SENTENCE FROM A BOOK
What does this mean?
Where I am struck is that I knew carbons occupy zinc blende like lattice so, how can this line be true.

Where I am struck is that I knew carbons occupy zinc sulfide like
  lattice so, how can this line be true.

Zinc sulfide can exist in two different crystalline forms, Zincblende (sphalerite) and wurtzite.  Zincblende is face-centered cubic (also known as diamond cubic), each ion is tetracoordinate and has local tetrahedral geometry - just like the carbons in diamond.  The wurtzite structure is hexagonal close-packed with interconnected 6-membered rings.  You may have been thinking of the wurtzite structure, while it is the zincblende structure that is related to diamond.  You can see pictures of the two structures here.

YES I TO KNOW IT IS ZINC BLENDE  STRUCTURE . BUT, THE BOOK SAYS "Every carbon atom is bonded to four other carbon atoms forming a three-dimensional lattice of chair conformations"

If you know the fcc structure, maybe you should be more specific what you cannot understand.

well what is this "chair conformation" related to to diamond structure. I know chair conformation are for hexane . But how does diamond form this structure

Take a look at the link in my answer and look at the picture for the wurtzite structure - do you see the chair arrangement of the carbon atoms?  If not, here is another link with a good picture of the chair arrangement http://link.springer.com/article/10.1007/BF00721087#page-1  When 6 sp3 carbons are placed in a ring, the chair arrangement is a low energy arrangement, no matter whether those carbons are bonded to hydrogen atoms (cyclohexane) or other carbon atoms (diamond).


text:
How can two electrons lie together in an orbital?
Two electron of opposite spin can lie in a single orbital.. But what about the electron-electron repulsion.
Okay! I got that the nuclear charge rather the large Z-effective overcome this repulsion by pulling them together towards the nucleus.
One thing more turned out in mind that it may be the attraction of the two unlike pole of the magnet developed due to opposite spin of the electrons of an orbital, which may be reason for the two opposite spin electron lying in an orbital..
Am I right ?

Preludium
In chemistry we often use some techniques to factorize the (quite complicated) many-particle ground state of atoms and molecules into a series of one-particle states that have to fullfil certain requirements; examples are Hartree-Fock-Theory or Kohn-Sham-Density-Functional-Theory. These one-particle states are what we usually call orbitals or, when you differentiate them according to their spin, spin-orbitals. In the theories that I quoted you then have to solve a series of one-particle Schroedinger-equations which will give you certain discrete energy levels that correspond to bound states of the system. Those energy levels are a fundamental property of the system, i.e. the eigenvalues of its Hamilton operator. You can't have energy levels in between or below. Now, the problem is that the orbitals and their energy levels depend on the number of electrons you have in the system, i.e. considering some species $\ce{A}$ you will find different orbitals at different energies for its anion $\ce{A-}$, neutral species $\ce{A}$ or cation $\ce{A+}$. That is so, because the electrons interact with one another and populating one orbital changes the potential which the other electrons feel.
Main answer
But let's assume for a moment that the orbitals for $\ce{A-}$, $\ce{A}$ and $\ce{A+}$ are the same and only taking into account the electron-electron-repulsion of electrons occupying the same orbital: Starting from $\ce{A+}$ where we have all lower-lying orbitals doubly occupied and one orbital $\psi_n$ with energy $E_n$ singly occupied. Now, if you want add one more electron to this system to get to $\ce{A}$ you have to choose where to put this electron: You can either put it into $\psi_n$ or into some higher-lying orbital, like $\psi_{n+1}$ or $\psi_{n+2}$. If you put it into $\psi_n$, thus pairing the electrons up, you have to pay for that with the electron-electron-repulsion energy between the paired electrons (also known as spin pairing energy) $\Delta E_{\mathrm{pair}}$. So, compaired to $\ce{A+}$ the energy of your new system $\ce{A}_{\text{paired}}$ will be higher by an amount of $E_n + \Delta E_{\mathrm{pair}}$. If you instead put the new electron into $\psi_{n+1}$ you don't have to pair up any electrons, so you don't have to pay $\Delta E_{\mathrm{pair}}$. But the energy $E_{n+1}$ of orbital $\psi_{n+1}$ is higher than $E_n$. So, compaired to $\ce{A+}$ the energy of your new system $\ce{A}_{\text{unpaired}}$ will be higher by an amount of $E_{n+1}$. Now, the question: Which situation is better? For that you have to look at the energy difference between $\ce{A}_{\text{paired}}$ and $\ce{A}_{\text{unpaired}}$. This will be
\begin{equation}
  \Delta E = E(\ce{A}_{\text{paired}}) - E(\ce{A}_{\text{unpaired}}) = (E_n - E_{n+1}) + \Delta E_{\mathrm{pair}} \ .
\end{equation}
If $\Delta E$ is negative then $\ce{A}_{\text{paired}}$ is lower in energy and thus energetically favored, i.e. pairing electrons is favored when $|(E_n - E_{n+1})| > \Delta E_{\mathrm{pair}}$. So, coming back to your original question: Typical values of the pairing energy are between 200 and 300 kJ/mol (or 2-3 eV) - for an example have a look here: there you can see that the pairing energy for carbon and nitrogen lies at about $20000 \, \mathrm{cm}^{-1}$ which is ca. 240 kJ/mol. The seperation between energy levels in atoms is usually much greater as can be seen in the table of ionization potentials given here (the differences between the ionization potentials of $\ce{s}$ and $\ce{p}$ valence orbitals are usually greater than 6 eV). Only in molecules and transition-metal complexes with frontier-orbitals that lie close in energy (or are degenerate) you have so called high-spin configurations where it is favorable not to pair up the electrons in a lower-lying orbital but instead to occupy the higher-lying orbital. So in conclusion, it really does "cost" energy to pair up electrons in a single orbital but it usually would cost a lot more energy to force an electron into a higher-lying orbital instead.
Finally, a word of caution: My answer oversimplifies things as I tried to point out in my preludium. There are other effects at work that might have a big influence in some cases.

You're thinking is correct, is does cost energy to pair electron spins in an orbital (the Pauli energy or exchange energy required to pair spins).  However, the alternative is to place that next electron into a higher energy orbital - that requires energy too.  Usually it takes less energy to pair spins than to place the next electron into a higher energy orbital.  But note cases like $\ce{O2}$ where there are two orbitals at the same energy (degenerate orbitals).  Now when we add the last electron it doesn't cost any extra energy to put that electron into the next available empty orbital because it is at the same energy.  So in cases like this (or where the next orbital is very close in energy to the partially filled orbital) we can put that next electron into a different orbital and leave their spins unpaired.


text:
Mechanism of reduction by elemental sodium
How exactly does sodium form an alkoxide from an alcohol? I understand the general idea - sodium has a low first ionization energy. Sodium can thus give up its electron to something that'll take it. Sodium can therefore somehow cause hydrogen gas to leave.
How exactly does this mechanism work? I'm imagining that because alcohols are weak acids, that some hydrated protons exist in solution, and that sodium can donate an electron to these hydrated protons, which then can pair up with each other and form hydrogen gas. Is this accurate?

When alcohol molecules adsorb on the surface of the sodium metal, an electron can be transferred from sodium to the alcohol O-H bond with simultaneous disproportionation  (homolytic cleavage) of the O-H bond thereby generating the alkoxide anion and a hydrogen radical.  The hydrogen radical (probably still on or near the metal surface) can then go on to dimerize and produce hydrogen gas.

EDIT: Support of Mechanism
The mechanism proposed above involves a single electron transfer (SET) from sodium.  This is based on the fact that:

All dissolving (sodium) metal reactions, such as the Birch reduction of alkynes and aromatics, proceed by the SET mechanism
Other reactions involving sodium metal such as the acyloin condensation, the pinacol coupling, the Bouveault–Blanc reduction, etc., proceed through the SET mechanism
The reaction of sodium and water still proceeds in the gas phase where ionization seems less likely, but electron transfer still viable.

Searching on-line I found that water can solvate electrons just like ammonia can.  The chief difference being one of stability; in ammonia solvated electrons are stable for days, whereas in water they are stable for less than 1 msec.  Given the preceding mechanistic evidence and the fact that solvated electrons can exist in water, it seems reasonable to suggest that the mechanism of alkoxide formation, when sodium is reacted with an alcohol, also involves the SET mechanism.

This is an additional pathway to the aforementioned or the exclusive pathway?

exclusive pathway

Oops, so we were wrong before?

I'm not following your last comment.

Another poster proposed a mechanism that did not involve disproportionation of the alcohol as a result of reaction with sodium.

As far as I am aware, the reaction of solid sodium metal with organic compounds involves a single electron transfer mechanism involving the metal surface.  I applied the same reasoning here.

prooflink please?

Is the other mechanism at least plausible?

Here's an example of the SET mechanism involving sodium  http://en.wikipedia.org/wiki/Bouveault%E2%80%93Blanc_reduction  Also, sodium and water react in the vapor phase - it seems less likely ions would be involved in the primary step in the gas phase.

I believe an SET mechanism is operating, but doesn't that radical anion have nine electrons on oxygen? Is it possible that the first two steps are concerted? The other referenced reactions involve carbonyls with lower lying pi* orbitals. This mechanism must fill in a sigma* orbital if I'm thinking about it correctly.

Good point jerepierre!  I've been looking into the matter further.  In the case of gaseous water, the thinking is that a solvated sodium atom may insert itself into the O-H bond, then upon interaction with a second sodium atom electron transfer to form a $\ce{Na^{+}Na^{-}}$ entity occurs (vibrational timescale) followed by disproportionation to sodium hydroxide and sodium hydride.  I'm wondering if "insertion" might be the same as orbital interaction between the sodium AO and oxygen lone pair orbital to form a lower (then sigma*) energy orbital that can hold the ninth electron.

still can't see any proof. Links to readuction of compounds with double bond are irrelevant by definition, as such compounds have low-level unoccupied orbitals. I downvote the answer as misleading.

@permeakra  Thanks for sharing your view.  I would say that calling my answer "misleading" may be harsh.  I spent a considerable amount of time researching the problem, and like so often in organic chemistry, and particularly with reactions involving metals, the precise details of the mechanism are still being explored.  Nonetheless, I tried to fashion together what I had learned in order to offer a reasonable answer.  I presume you find the alternate answer less misleading and more accurate since you did not downvote it.

The alcohol splits like this:
$$ROH\rightarrow RO^-+H^+$$
since the pka of alcohol is around 15-17 and then a redox reaction happens:
$$2Na+2H^+\rightarrow 2Na^++H_2$$
Since $H_2$ is released it is entropically favourable and drives the reaction.
Better, alkoxide is represented as:
$$RO^-Na^+$$

I think you dropped a charge on the second equation. You're now missing a charge on the right side.

Are you sure ions can't exist in isolation in solution? I'm guessing you're saying there's always some ion pairing going on?

@Dissenter no it is not always true, i was also told that ions do not exist in isolation until i studied ionic equilibrium

Okay! So ions can exist in isolation? Also your second equation is still not balanced for charge.

Also hmm now that I've thought of it this mechanism makes sense since the reaction of sodium with water produces hydrogen gas. The protons came from the auto-ionization of water in that case.

@Dissenter yes you are correct

I think that the sodium when it dissolves in either water or an alcohol will form solvated electrons. These solvated electrons will react with water or the alcohol in reactions such as
e- + H2O --> H + OH-
The hydrogen atoms can then dimerise.
The chemistry of solvated electrons in various media has been studied at great length by pulse radiolysis, this is an experiment similar to flash photoylsis. Instead of a flask of light to start the reaction the pulse radiolysis method normally uses a pulse of high energy electrons from a LINAC.
This article in Chemistry World sheds some light on the reaction of sodium metal with water. https://www.chemistryworld.com/news/tamed-sodium-in-water-comes-out-of-the-blue/1017231.article


text:
Concentration of hypochlorous acid and hypochlorite

Given a total chlorine concentration  $c(\ce{Cl})_\text{tot}= \pu{0.1 M}$, with $\mathrm{pH}(\ce{Cl}) = 8$ and $\mathrm{p}K_\mathrm{a}(\ce{HOCl}) = 7.5$, how would you find the concentration of $\ce{OCl-}$ and $\ce{HOCl}$ disinfectants?

So far I think that I should start this way:
$K_\mathrm{a} (\ce{HOCl}) = 10^{-7.5}=3.16\cdot 10^{-8}$
I wrote up the reactions:
$$\ce{H2O + 2e- + OCl <=> Cl- + 2OH-}$$
Where $\ce{OCl-}$ is the conjugate acid and $\ce{OH-}$ the conjugate base.
$$\ce{HOCl<=> OCl- + H+}$$
The conjugated acid is $\ce{HOCl}$, the conjugated base $\ce{OCl-}$.
Now I use
$$\mathrm{pH}=\mathrm{p}K_\mathrm{a} + \log\left(\frac{[\text{conj base}]}{[\text{conj acid}]}\right)$$
and/or for weak acids
$$K_\mathrm{a}= [\text{products}]/[\text{reactants}]$$
But I'm not sure if this is correct and what values to use.

Note that chlorine is Cl, not "CL."
I'm unclear on what you mean by "total pH of Cl."

@Dissenter Yes oops I meant Cl, and total concentration of Cl. I will change it

edit your question using latex

You really overcomplicate things (as I did initially, because I thought you were talking about the dissociation of $\ce{Cl2}$ gas in water and the subsequent reactions).
Here are the basic pieces of information needed:

Reaction equation
Equilibrium equation
Mass balance equation

Reaction Equation
The reaction is a simple acid-base equilibrium reaction:
$$ \ce{HOCl <=>[K_\text{a}] H+ + OCl-} $$
Equilibrium Equation
The equilibrium equation follows easily from the reaction equation:
$$ K_\text{a} = \frac{[\ce{H+}][\ce{OCl-}]}{[\ce{HOCl}]} $$
Mass Balance Equation
The total amount of chlorine is calculated as follows:
$$ [\ce{Cl}]_\text{tot} = [\ce{HOCl}] + [\ce{OCl-}]$$
From the reaction equation also follows
$$ [\ce{OCl-}] = [\ce{H+}] $$
The Solution
Since the concentration of protons is known, we can easily calculate the concentration of the hypochlorite anion:
$$ [\ce{OCl-}] = [\ce{H+}] = 10^{-\text{pH}} = 10^{-8} $$
Since we know the total concentration, simply reforming the mass balance gives us the concentration (strictly speaking, the activity) of the hypochlorous acid:
$$ [\ce{HOCl}] = [\ce{Cl}]_\text{tot} - [\ce{OCl-}] = 0.1 - 10^{-8} = 0.09999999 \approx 0.1 $$
Slap on some units and we're good to go!
$$ \begin{align}
c(\ce{HOCl}) &\approx 0.1~\mathrm{mol\, L^{-1}} \\
c(\ce{OCl-}) &= 10^{-8} ~\mathrm{mol\, L^{-1}}
\end{align}$$

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text:
Kb of hydride anion
I can't find any data about the Kb value of the hydride anion except that it's huge. Does anyone know what it is, or how I can characterize it?
I know I can use electrochemical means but the only reaction I find with hydride ion is this one:
http://en.wikipedia.org/wiki/Hydride#Hydride_ion

I think I got it figured out using electrochemical means, thanks to Aditya’s suggestion to consult $K_\text{a}$ values rather than $K_\text{b}$ values.
Consider these two half reactions:
$\ce{2e- +H2->2H-}\tag{$E^\circ=-2.25~\mathrm{V}$}$
$\ce{H2 + 2H2O -> 2H3O+ + 2e-}\tag{$E^\circ=0.00~\mathrm{V}$}$
Coupling these two half reactions results in:
$\ce{2H2 +2H2O ->2H3O+ +2H- }\tag{$E^\circ=-2.25~\mathrm{V}$}$
Application of the Nernst equation can help us find an equilibrium constant for this reaction.
$\Delta G^\circ = -nFE^\circ = -(2)(96\,500~\mathrm{C/mol})(-2.25~\mathrm{V}) = +434\,250~\mathrm{J/mol}$
Value makes sense; we’d expect the reaction of hydrogen gas as an acid with water to be highly disfavorable.
$\Delta G^\circ = -RT\ln K=-\left(8.31~\mathrm{J/(mol\cdot K)}\right)(298~\mathrm{K})\ln K=+434\,250~\mathrm{J/mol}$
$K = 6.97464\times10^{-77}$
Now, this $K$ correspond to this equilibrium:
$\ce{2H2 +2H2O ->2H3O+ +2H- }$
So we must take the square root of the found equilibrium constant to generate a value for $K_\text{a}(\ce{H2})= 8.35\times10^{-39}$.
And finally this lines up well with Aditya’s finding that the $\mathrm{p}K_\text{a}$ of $\ce{H2}$ is 35; the −log of the above $K_\text{a}$ value I found is 38. Nice.

Recall that $\mathrm{p}K_\text{b}$ of hydride is actually $14-\mathrm{p}K_\text{a}$ of hydrogen gas. The value I found is $\mathrm{p}K_\text{a}=35$, so $\mathrm{p}K_\text{b}=14-35=-21$.

You mean Kw/Ka of hydrogen gas?

That 35 value also sounds like a pKa value rather than a Ka value. Can't imagine hydrogen gas being that acidic.


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Can anyone explain the reaction mechanisms of these synthesis?
Can anyone explain the reaction mechanism of the following synthesis?
$\ce{2NH4Cl + Ca(OH)2 -> 2NH3 + 2H2O + CaCl2}$
$\ce{(NH4)2SO4 + 2NaOH -> 2NH3 + 2H2O + Na2SO4}$

The ammonium chloride, calcium hydroxide, ammonium sulfate, and sodium hydroxide are all ionic compounds, and when dissolved in water they dissociate into $\ce{NH4+}$, $\ce{Cl-}$, $\ce{Ca^2+}$, $\ce{SO4^2-}$, and $\ce{Na+}$ ions. The $\ce{NH4+}$ ammonium ion can donate a proton to the $\ce{OH-}$ ion, creating $\ce{NH3}$ and $\ce{H2O}$. The other ions don't really do anything, so these spectator ions remain in solution but the charges are still balanced. If you removed the ammonia and water, which is easy with a vacuum, you'd be left with the $\ce{CaCl2}$ and $\ce{Na2SO4}$ salts. These reactions will have some equilibrium, with $\ce{H2O}$ donating a proton back to $\ce{NH3}$ to form $\ce{NH4+}$ and $\ce{OH-}$ again.


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Reaction of alkyl halides with strong unhindered bases
Please look at the image on this website: http://www.physicsforums.com/showthread.php?t=765853
For (b) and (c), what the solution manual says makes no sense to me.
You're reacting a primary alkyl halide with an unhindered base $\ce{EtONa}$. They say that the only product is the alkene. Why do they not consider the $\mathrm{S_N2}$ reaction that WILL take place in larger quantity?
I'm uploading the question too on the other website

I agree with you, I'd expect some $\mathrm{S_N2}$ product too. I'm not sure about the relative quantities of the $\mathrm{E2}$ and $\mathrm{S_N2}$ products though.
I wouldn't flat out say that you'd expect more $\mathrm{S_N2}$ product than $\mathrm{E2}$ product, because although the substrate is unhindered, the solvent is ethanol, which is protic. The solvent interacts with the nucleophile through hydrogen bonding.
This hinders the nucleophile. This slows down the rate of the $\mathrm{S_N2}$ reaction because now the nucleophile is dragging along lots of solvent molecules, and it also increases the bulk of the nucleophile.

How does the solvent being protic hinder the nucleophile?

The solvent can hydrogen bond with the nucleophile.

Ok. Now, does that mean that E2 is favored in a strict sense?

Er that's a hard call to make without actually doing the experiment.

Ok. I just consulted my textbook. It seems as though Sn1 and Sn2 are really impacted by solvent. E2 and E1 are indirectly concerned and as you say need the experiment. Great help! I'll post another question that was racking my brains.

I'm not sure what the effect of protic vs aprotic solvent would be on the product distribution here. However, it is definitely not true there is no solvent effect on basicity. Just as hydrogen bonding decreases nucleophilicity, it decreases basicity.

Wouldn't the protic solvent hinder the nucleophile? Just as F- is a bad nucleophile in water, it is a good nucleophile in a non-protic solvent.

Yes, that's true. Similarly though, hydroxide is a strong base in water, but it's stronger in an aprotic solvent or in the gas phase.


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Synthesis via Sn2/Sn1, E2, E1: Choice of solvent
I uploaded pictures on http://www.physicsforums.com/showthread.php?t=765106
I was looking at the answers in my textbook. They seem to make incongruent choices of solvents.
In e) and j), why choose EtOH and H2O/CH3OH? I would rather choose an aprotic solvent like DMF to better allow Sn2 to occur. Especially with j), there is a lot of competition with E2 because OH- is a strong base and because we are attacking a secondary carbon.
I see this again in k). Secondary carbon attacked: why not add DMF to prevent E2 and get best yield?

I'm not seeing what you're seeing, OP. I don't see any methanol in e, for example.

sorry to poke but k) part is not there

The competition from $\mathrm{E_2}$ in problem (j) is less of an issue than it may at first appear. The $\mathrm{E_2}$ elimination would require a ring flip that places the two substituents on the ring in axial positions (recall that $\mathrm{E_2}$ typically proceeds only when $\beta$-hydrogen and leaving group are anti-periplanar, and in a cyclohexane ring that can only occur if the two are axial). The conformer with both substituents axial is substantially less stable, so $\mathrm{E_2}$ is hindered (though not ruled out) by the energy barrier associated with the ring flip.

Sorry Aditya! I have no idea how to upload pictures. Can someone teach me?

DMF is a last resort, as it is hell to get rid of. \

@GregE. Great comment! Do you have any idea what would happen with e) and k)? (By the way, (k) is in one of the lower comments on the physicsforum site.) Why did they chose protic polar solvents? The question still remains for (j), even though it is less of a problem.

Does DMF not maximize yields however? (I never actually saw or used DMF.)

@Dissenter, yes, in (e) there is no methanol but there is EtOH, which winds up to being a polar protic solvent.

NOTE: SOMETHING IMPORTANT I FORGOT TO ADD: The book says : When alternative possibilities exist for a synthesis, you should be careful to choose the one that gives the better yield.

@yolo123, $\mathrm{E_2}$ is still possible for (j), it's just slower due to the pre-equilibrium for chair conformer interconversion. Maybe with careful temperature control, the ring flip is enough to favor the $\mathrm{SN_2}$ there, but I would expect a mixture even in the best case. The choice of protic solvents is theoretically sub-optimal for the reasons you've mentioned, though it's not really a problem in (e) because there are really no alternate pathways. Solubility is a possible practical reason, as jerepierre suggested in his answer. As for (k), I don't have an account on that site.

Your thinking is not wrong. For (e) DMF is a better choice of solvent - this doesn't mean that MeOH won't work just that DMF is better and is typically used in these reactions. For even better results you can use DMSO. If you dare, as DMSO/cyanide is a very toxic cocktail.
(j) I don't think it is a practical synthetic reaction since it can possibly give you elimination and/or $S_N1$ (ie no inversion of stereochemistry) products.

For reaction (e), given the strength of the nucleophile, the primary benzylic leaving group, and the lack of any protons beta to the leaving group, SN2 is the only mechanism that can operate. The choice of ethanol as solvent likely has much more to do with getting everything into solution than influencing one pathway over another. According to Wikipedia, NaCN is soluble in EtOH but only slightly soluble in DMF.
For reaction (j), I would guess that the water/methanol mixture was chosen to get everything into solution. There might be some concern about hydroxide deprotonating methanol, and then methoxide carrying out the displacement. My answer would probably be to use a water/THF mixture, if you were really worried about that.

In both e and j, you would choose the protic solvent over aprotic solvent, because the protic solvent will stabilize the resulting bromide anion. Sometimes, the stability of the other product should also be concerned, not just the main product.


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Ammonia and hydroxide ion - Lewis basicity
Which is the stronger Lewis base, and why?
I'm thinking that the answer is "it depends." With regard to the hydrogen proton, which hydroxide ion is obviously the stronger Lewis base. However, with regard to other, more complex Lewis bases, the distinction may not be so clear, as hydroxide ion has several lone pairs (three) and these can weaken the stability of Lewis adducts through lone pair/"bonding pair" (the pair that forms the coordinate covalent bond) repulsions. Ammonia only has one lone pair so there isn't an issue of lone pair/"bonding pair" repulsion.

I agree in principle with your premise that "it depends." There's a slew of complicated effects at work here, some related to HSAB and FMO theory (the importance of frontier orbital energies vs. Coulombic interactions), some to solvent effects (reactivity in the gas phase, for example, sometimes differs drastically), and other factors. The nature of the electrophile in question is crucial here: some molecules are highly reactive as nucleophiles with certain substrates, while being comparatively unreactive with others.

I think you should also probably distinguish between kinetic and thermodynamic basicity/nucleophilicity. Reaction rates doesn't necessarily correlate well with equilbrium constants in all cases.

With respect to Brønsted basicity, definitely hydroxide is the stronger base. With respect to Lewis basicity, I think the same scenario holds. As you said, hydroxide has three lone pairs, whereas ammonia has only one lone pair. As far as the energetics of these situations are concerned, the magnitude of lone pair to lone pair repulsion is far greater than that of lone pair to bond repulsion. Thus, it would make sense for hydroxide to grab the proton instead of ammonia.

$pKa$ of conjugate acid($NH_4^+$) of ammonia($NH_3$) is $9.9$ and of conjugate acid($H_2O$) of hydroxide($OH^-$) is $15.7$.
So, hydroxide is a strong base.

I'm talking about Lewis basicity not just Bronsted basicity.

@Dissenter I just treat all acids and bases equally, what's the difference between them?

@Dissenter the molecule doesn't diffrentiate the two, so why would we. I actually know the difference


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Ionic compounds dissolve in nonpolar compounds?
I think ionic compounds dissolve in just about any compound including nonpolar. My reasoning is this:
Neutral electric charge attracts both positive and negative (This I know from doing things with static electricity and putting it on a neutral surface).
Ionic compounds like strong acids and bases disassociate into individual ions, or if it is not dissolved but melted ionic dimers (like an individual Na+Cl- dimer for example).
Do ionic compounds such as NaCl dissolve in just about every solvent from polar to nonpolar to liquid ionic?

First, I think you have some misconceptions about electrical charge. There isn't neutral charge, it's either positive or negative, if it doesn't have either, it has no charge. Why are you seeing static charge interaction with neutral surfaces? It might be polarization, where a charged particle warps the electron cloud around a neutral atom generating partial charges around the neutral atom. Without knowing more about your experiment I can't say more.
Meanwhile I did an experiment of my own. I took 5 20mL scintillation vials, and added about 0.5g of NaCl to each. I then added 5mL of deionized water, methanol, 200 proof denatured ethanol, isopropanol, or acetone to each vial and tried to dissolve the salt. Only water dissolved all of the salt. I can't say there isn't ANY salt dissolved in the other solvents, but clearly the solubility is solvent dependent, and only water is polar enough to dissolve all the salt.

One of the best answers I've read on this site.

I have watched videos about static electricity and they say that if something is electrically neutral it will attract both positive and negative which explains partly why some atoms are more stable when N > P where N is neutrons and P is protons.

and I have done this experiment: I took a balloon and rubbed it in hair. charge exchange happened with the positively charged balloon and negatively charged hair. I then put it on a wall which is electrically neutral from both positive and negative touching it. The balloon was attracted to the wall because of the charge in the balloon and 0 charge in the wall. This proves that neutral attracts both positive and negative.

Neutrons and protons in the nucleus of an atom are not held together by electrostatic forces. Neutrons play a role in nuclear stability, but they do it through the nuclear force, which is described here: http://en.wikipedia.org/wiki/Nuclear_force   I can only conjecture that your wall was not as neutral as you assumed.

I think ionic compounds dissolve in just about any compound including nonpolar. My reasoning is this:

Ionic compounds don't always dissolve in just about any compound. Consider sodium chloride. This isn't soluble in acetone.

Neutral electric charge attracts both positive and negative(This I
  know from doing things with static electricity and putting it on a
  neutral surface).

If something has no net charge (i.e. is neutral) it won't be electrically attracting anything. Only oppositely charged things attract.

Ionic compounds like strong acids and bases disassociate into
  individual ions(Or if it is not dissolved but melted ionic dimers(like
  an individual Na+Cl- dimer for example).

Not all strong acids or bases are ionic. Consider sulfuric acid. The S-O and O-H bonds, granted, have some ionic character, but are generally characterized as covalent. Also consider the halide series of binary acids - HCl, HBr, HI. Strength increases in that order, yet ionic character decreases in that order.
Plus a base isn't necessarily ionic. NaOH is an ionic salt, and yes, it's a strongly basic salt. But the hydroxide ion is also a strong base. In isolation, it's not ionic.
Also you should define what a dimer is before incorrectly applying it. The definition of a dimer according to IUPAC is:

The transformation of a molecular entity $\ce{A}$ to give a molecular entity $\ce{A2}$.

So no, the sodium and chloride ions cannot dimerize in the fashion you suggested - with sodium ion coming together with chloride ion. That's not dimerization. Now, if a chlorine radical joined with another chlorine radical, that would be dimerization, since both starting molecular entities are the same.

Do ionic compounds such as NaCl dissolve in just about every solvent
  from polar to nonpolar to liquid ionic?

No, and this has been illustrated by user137's excellent demonstration. And consider this table of solubility products. A bunch of ionic compounds, yes. But all of them are considered semi-soluble salts because of their low degree of dissociation into their constituent ions (hence the miniscule Ksp values).

The "Ionic compounds like strong acids and bases disassociate into individual ions" doesn't mean all strong acids and bases are ionic but rather that just like how strong acids and bases disassociate into individual ions, ionic compounds disassociate into individual ions(assuming aqueous solution).

Not all ionic compounds dissociate into individual ions in aqueous solution. Consider calcium fluoride.


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How do breaking chemical bonds turn into kinetic energy
I’m trying to understand how breaking chemical bonds turn into kinetic energy. If AB is a chemical which breaks into A and B which zip away from each other, is it because when the bond breaks they suddenly repel each other? Another way I'm thinking of this is what happens if one single molecule of AB alone in a vacuum breaks apart.
I think the point I'm trying to make is the reaction is making the products move faster than when they started. It takes a force to make things move. So it has to be either gravity, electric, magnetic, or one of the nuclear. I'm pretty sure we can rule out nuclear and gravity.

Maybe we can explain it by temperature. If PV = nRT, and the energy from the bond is turned into heat, then n, T and P must increase in the immediate vicinity of the molecule. ( n increases because AB has become A + B ) So to get back to equilibrium the molecules move apart. If I had better evidence I'd make an answer, but this is pure speculation.

Breaking increases the potential energy of the system, making lowers it.  When making bonds the lost potential energy is converted to kinetic energy.  A reaction involves making and breaking bonds.  Perhaps you have come across the infamous ATP example and a flawed text has said that breaking bonds in ATP releases energy.  It does not.  The reaction involves breaking existing bonds and making new bonds for a net reduction in the potential energy of the system.

Breaking chemical bonds does not turn their potential energy into kinetic energy. All bond dissociation enthalpies are positive (for stable bonds that actually exist), which means that breaking them is an endothermic or heat absorbing process, and therefore they absorb kinetic energy (heat is a transfer of molecular kinetic energy).
What you might be thinking of is exothermic reactions, where the enthalpy of the reactants is lower than the products, and so the reaction releases heat, or increases the kinetic energy of the surroundings.
The short answer to your question is "because energy is conserved, it has to go somewhere, and heat and light are the only places it can go."
I'll try to think about a better way to explain it that isn't too far away from reality, but at the same time isn't too complicated. If I can, I'll edit this answer later to include it.

As the other answers have mentioned, a simple A-B bond dissociation reaction is significantly endothermic and does not happen spontaneously. A single molecule may break apart and its pieces may be scattered if energy is introduced in the molecule (for example, a collision with a very energetic target or the absorption of a photon may trigger strong vibrational excitations which end up breaking the molecule apart). It is, however, also possible for an unstable molecule to spontaneously rearrange into a more stable isomer, and part of the released energy may cause the rearranged molecule to shake apart.
In any event it's a case of causing a vibration to stretch atoms strongly enough that the bond between them snaps and the pieces fling away, not unlike a rubber string being pulled until it tears. Perhaps this video may help clarify what's going on, though it doesn't specifically show reactions occurring.

Edit: I somehow had completely misread the question as $\ce{A +B -> AB}$ instead of $\ce{AB -> A +B}$, so I apologize for that. My previous answer isn't actually an answer, but I figure I might as well leave the information below in case anyone is interested:
As far as I understand, most often the chemical potential energy released when two molecules react exothermically turns into vibrational potential energy in the bonds of the product molecule/entity for a small amount of time, and this vibrational energy is quickly dispersed into other molecules by collisions (timescale of a few picoseconds), turning mostly into rotational and translational kinetic energy dispersed over several molecules, as vibrational excitations are "frozen out" at temperatures close to ambient.
If the two molecules meet in an absolute vacuum and can react exothermically but have nowhere to dump the energy, then the molecules actually have a large chance of not reacting at all, even if it's extremely spontaneous! The only way they can meet and stay bound is if, during the short time they are in proximity, a low-probability event may occur where the reaction transition state emits a photon carrying away most of the releasable chemical potential energy, otherwise the molecules will just fly apart again. This is a very important consideration in astrochemistry. You can read more about this specific topic in "The Physics of the Interstellar Medium", 2nd edition, 1997 J.E. Dyson and D.A. Williams (IoP), chapter 3, section 3.4.
Actually, I should clarify that the second paragraph is only really valid for reactions of the type $\ce{A +B->C}$. A reaction of the type $\ce{A +B->C +D}$ can happen more easily as one of the species being formed can also act as a dump for excess energy.

You are thinking about it a little wrong.  The fact that the bond exists means it is in a state of lower energy than when A and B are separated -- therefore you would have to input energy into the system to get the bond to break.  But if you had two species A-B and C and you could form a more stable bond B-C which means it has a lower energy level, then the excess energy E(A-B) - E(B-C) for the reaction:
A-B + C -> A + B-C
could be released as kinetic energy of the A and B-C species moving away from each other.


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Ethene to ethanol: electrophillic additions involving ethyl sulfates
http://www.chemguide.co.uk/mechanisms/eladd/symh2so4.html
I was reading my textbook and this website. I just cannot find the mechanism for what happens when we go from CH3CH2OSO3H to ethanol by the addition of water and heat.
I understand the electrophillic addition part.
But, what happens next? Is there some kind of Sn2 reaction to make HSO4- the leaving group? Then, why does the EtOH2+ molecule give out one of its acidic protons to HSO4- and not to H2O which is a more powerful base?
I hate when my textbook leaves out the details and I obsess on them :(

Yes, I believe you are correct. Water attacks at the sulfate containing carbon by an SN2 mechanism. You are also right that sulfuric acid is a stronger acid than hydronium (or protonated ethanol), which means that in solution the bisulfate anion will predominate. Often people (including textbook authors) will draw a mechanism that goes to neutral products. Most likely water will carry out the final deprotonation (especially given that water is the solvent).
Incidentally, most likely this reaction is self catalyzed, as one sulfate protonates another, activating the sulfate for SN2 reaction by a weak nucleophile.

Great answer. Small questions:
we said HSO4- won't be taking the final proton from protic ethanol. Did you draw it like that just to point out the fact that most textbook authors try to put neutral products?
Then, in that catalyzed mechanism, why does the sulphate part NEED to get a proton before acting as a leaving group?

Yes, I drew the protonation of bisulfate just to show how you would get to the indicated products. In the catalyzed version, the extra proton acts to make the sulfate into a better leaving group. Water displacing the protonated sulfate will be faster than the neutral sulfate. You will see this regularly in organic chemistry: when there is a weak nucleophile (like water and many neutrals) the elecrophile will be activated by protonation.


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Sodium benzoate in soft drinks reacting to produce benzene?
Regarding chemical synergies, I read: if someone ingests various products containing sodium benzoate (for example soft drinks that exist) AND if one has had some quantity of vitamin C there might be 'reactions' that produce benzene, a carcinogen.
Some people have told me you'd have to drink a lot of pop for this to happen (assuming you have had some vitamin C during the day), but how much pop is a lot?
Given one has had a glass of orange juice say, how much pop would one have to drink before a benzene molecule is formed?

I've never heard of this before. Can you support your statement with a claim? This sounds like it should be on Skeptics.

Can sodium benzoate and vitamin C  interact 'badly'?

The reaction taking place is described in article Benzene production from decarboxylation of benzoic acid in the presence of ascorbic acid and a transition-metal catalyst and its importance for food safety recently reviewed in Evaluation of Accelerated UV and Thermal Testing for Benzene Formation in Beverages Containing Benzoate and Ascorbic Acid.
To summarize, the reaction CAN occur, namely at elevated temperatures (>50 °C) and UV-light and in presence of transition metal. Even if the reaction takes place, the amount of benzene formed is either below detection limit, or below limit for drinking water.
For the end user it means, that in extremely unlucky situation, the benzene can form in quantity, which is not considered safe to drink in unlimited amount for the whole life (1 ppb). As such, I would consider the benzene formation as very weak argument against such beverages (mainly if compared with other problems, such as sugar content and associated obesity).

Thank you for the info. Would drinking an average of 4 L. of pop per day make benezene formation more likely?

I have troubles understanding your problem as a valid one. I am not aware of any reasonable diet containing 4L of pop per day. Recommended WATER intake is ~2L per day. If you are about to drink 4L of pop per day, be ready to die from anything else but benzene caused cancer.

I'm not saying this is part of a reasonable diet yet I have seen many young adults buy 2 L. bottles of Coke say seemingly to drink them for the caffeine and sugar buzz. I think a lot of young people and adults use Coke and Pepsi in a way similar to energy drinks. In this obsession to be energized and/or stay awake aren't you worried some may be drinking too much in one day. If a young person drinks a lot of Coke in one day for weeks to stay awake while studying ,say; it may not accumulate enough benzene  to do harm but does this molecule stay in the body for a while?

Please see the toxicology of benzene, very brief: http://www.crios.be/Benzene/toxicology.htm, very long: http://www.atsdr.cdc.gov/toxprofiles/tp.asp?id=40&tid=14
The benzene is really not a problem here.

If sodium benzoate is harmless given many people drink a lot of pop during the day and ingest it in many processed food items; there may be a tiny amount of benzene formed that is within 'safe' standards. Yet how efficient is the human body at getting rid of such molecules as benzene ,even if it exists in the body in tiny amounts? Over time it may accumulate..

Actually, the toxicology of benzene is quite interesting. The problem is not the benzene itself, but benzene oxide, which is formed as the body tries to get rid of benzene. The benzene itself does not accumulate, but mutations caused by benzene oxide do. But really please, read some toxicology, nice short and accurate here: http://www.hpa.org.uk/webc/hpawebfile/hpaweb_c/1194947391801
To say, gallons of coke may contain the same amount of benzene as you inhale during smoking of 1 cigarette.

So from ingesting A LOT of sodium benzoate in pop drinks and many common food products in one month ,say could the human body in trying to get rid of benzene make a fair amount of benzene oxide that causes the accumulating mutations. We as a modern society seem to use a lot of preservatives (that are only for longer 'shelf life' and save a company money). OVER a few years could these mutations accumulate to harmful levels?

Although quite understandable, your position misses important thing, at that is food security. The longer "shelf life" is a reason for no famine in developed world in last century. We are not only able to produce enough food, but also store it efficiently, so that we do not have to throw it away rotten. So if you can, buy fresh food. If not, buy preserved food. If not, death of starvation is inevitable.

I did mention I was talking about modern society not the developed world. Should ANYBODY rich or poor have to ingest preservative laden food. Maybe a company could offer a preservative- free version of all their products for a slightly bigger price. It seems most average North Americans including 'tons' of students buy processed food.  Why does Europe seem to use less preservatives than in North America?

So benzene oxide can cause mutations that accumulate. It forms as the body tries to get rid of benzene. So my question should really apply to Benzene Oxide. If we drink a lot of pop and eat processed food with sodium benzoate and potassium benzoate , over a month would this form a lot of Benzene Oxide in the Body?? ( which accumulates and causes mutations)

Can a 2 litre bottle of Coke say 3 times a week give one enough benezene oxide to cause mutations?


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How do the fields of medicinal chemistry, pharmaceutical chemistry, pharmacy, and pharmacology differ?
I would like to know what the difference between medicinal chemistry, pharmaceutical chemistry,  pharmacy and pharmacology. All of them seem to involve the development of drugs, so I am confusd as to how to distinguish them.
Also, does specialising in medicinal chemistry require a biology background? It seems to lean more towards biology.

I'm in a medicinal chemistry department, the floor below is the pharmaceutics department. We are both located in our university's college of pharmacy. The pharmacology department is under the college of medicine in same building as biochemistry, microbiology, etc. There is little interaction between departments here and that's not good.
But we can break the differences in each discipline down into some order based on drug development. Medicinal chemistry and pharmacology come first, with the discovery of molecules that bind to receptors and other proteins and have biological effect. This used to be based almost entirely on natural products, and those still play a large role, but high throughput screening and computer modeling are becoming more important for discovery of drug candidates. While pharmacology focuses on the effects of drugs and other molecules on cells and the body, medicinal chemistry has traditionally focused on the chemistry behind them. A medicinal chemist might look at a natural product with desirable activity, and ask "how can I make this better?" and apply organic synthesis to produce new versions of that drug, with better activity, reduced toxicity, and longer half-life. Of course the biology cannot be ignored, and has become more important in recent decades. Our department used to be pure synthesis, but now only 2 of our 8 labs are pure synthesis, and 2 almost entirely biological.
Once the medicinal chemists have developed a drug candidate to the point they think it can be used in the real world, it moves on to pharmaceutical chemists. They not only have to refine the synthesis of the drug to produce it efficiently and reliably on large scale, they have to develop the formulation the drug is delivered as. If it's a tablet, they have to determine what to make the bulk of the tablet from, how to press the tablets, how to store them, etc. A large part of pharmaceutics is stability, how long can the product sit on a shelf and not degrade? How does temperature and humidity affect degradation? What does it degrade into and are those products dangerous? Pharmaceutics is also concerned with drug absorption and excretion. They often try to develop "extended release" formulations so that a tablet takes a longer time to break down in your gut, releasing the drug over a longer time and keeping an effective concentration of drug in your blood longer, thus requiring fewer pills and less variation in concentration. Pharmaceutics is a lot of organic chemistry, physical chemistry, analytical chemistry, and quality control.
Pharmacy is the end product of all of these. After a drug is developed and approved the pharmacist is responsible for distribution. While doctors prescribe drugs, they usually have very little training in how the drugs work and interact. The pharmacist needs to know more than just how to sell pills, they must have at least a working knowledge of the pharmacology behind the drug, be aware of potential interactions and side effects, and advise the patient on the best practices for using the drug. They must also be aware of the rules and regulations, and make sure the controlled substances under their care are not stolen or lost or abused. Substance abuse is actually the biggest workplace hazard for pharmacists.
I have some knowledge on these things because I'm in a med chem department, my father is a chemist for a pharmaceutical company, and my brother just got his pharmacy license, so we see all parts of the process.

Wow — my father’s a pharmacist working for a pharmaceutical company, I’m a chemist … but I don’t have a brother. But the similarities are shocking ;)

The difference between the fields is significant, but as all become more interdisciplinary, some simplification are inevitable. So better look what an expert in given field does:

pharmacy - general overview of all drugs, their action and interactions
pharmacology - detailed study of a given drug, how it is distributed in body, which enzymes it affects, how it is transformed and excreted, ...
medicinal chemistry - detailed study of the mechanism of action at molecular level, enzyme variations which affect the efficiency, ...
pharmaceutical chemistry - how to synthesize the drug, more or less specialized organic chemistry


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Why can't light pass through a gold foil, but alpha particles can?
Rutherfords experiment showed that most of the alpha particles passed through a thin gold foil undeflected. So why can't light pass through the foil, which is supposedly thinner than the heavy alpha particles?

Search for "effective skin depth".

Light can pass through a gold foil though, it just has to be thin enough.
Pure gold is a very malleable substance and can be beaten with a hammer into foils of around 100 nm thickness. Sources suggest the gold foil used in the Geiger–Marsden experiment (known more commonly as the Rutherford gold foil experiment) was about 86 nm thick. Somewhere around this region of thickness, gold starts letting light through, as well as alpha particles. It is well known that ~20 nm thick gold (or any other metal) deposited onto a surface by thermal evaporation in a vacuum is essentially invisible and can be used as a transparent conducting surface/electrode. As the foil gets thicker, the gold starts to absorb a noticeable amount of photons, especially in the violet/blue part of the electromagnetic spectrum, while scattering away the red light. The light scattered off a semi-transparent gold film acquires a reddish/yellow tinge, while the light that still does make it through is blueish. When the film reaches 100 nm or so, it becomes practically as opaque as the bulk metal.
In truth, comparing the mean free path or collision cross-sections of photons and alpha particles through a gold sheet is somewhat more complicated than just saying alpha particles should penetrate less because they're massive, so it would be conceivable for a film of appropriate thickness to block photons yet still allow alpha particles through. A very coarse, hand-waving explanation is that photons are absorbed in the electrosphere of a gold atom, while an alpha particle is only significantly pushed away if it bangs almost straight into the nucleus, a region composing approximately $10^{-15}$ of the volume of the atom.

The plasma frequency for gold is in the visible, resulting in the blue transmission and red reflection. Just to add some explanation for this behavior.

This opens opens the question considerably, but light in physics is EM radiation of any wavelength. There are a myriad of interactions between matter and EM radiation.


text:
Environmental Alkylating Agents
I am looking for the names of some environmental alkylating agents (in context of DNA damages) found for example in the fume produced by industry. Most of the papers which I have found is concentrated on cigarette smoke, chemotherapy ect. I found something about methyl chloride, that is produced by plants and fungi, but also industrially. However there is no more details about that industrial part and in general the amount of methyl chloride in the air is rather low. Are there more relevant examples? Also I came across the information that alkylating agents may be found in food, but again there was no details.

Mutagen X is found in chlorinated drinking water and can cause DNA damage by adduct formation: http://www.oehha.ca.gov/prop65/pdf/MX_HID.pdf

A much more significant example of this is methyl bromide, which has historically been used as a soil fumigant. Its use has been slowed phased out due to its toxicity and ozone depletion abilities.

This is interesting in general. Could you please provide some further reading material, a source of any kind.

Here's a ling to an open source review paper on alternatives to methyl bromide: http://dx.doi.org/10.1094/PHYTO.2002.92.12.1337 . As for other sources, many of the good papers on the topic are in journals which require subscription or payment. It you're at a university, I recommend checking there, or at least using Google Scholar( http://scholar.google.com/ ) to search for something like "Methyl Bromide Poisoning".

It would improve your post immensely, if you would [edit] in this information.


text:
Does there exist any relation between metallic property and oxidation potential?
I was having a revision on periodic table I was thinking about different properties I could connect most of I could not connect metallic property and oxidation potential .

The short answer is yes. The reason for this is that metals exhibit metallic properties because their electrons are relatively loosely bound to the atoms. A metallic bond is one in which the valence electrons are free to move throughout the lattice structure of the metal. Here's a schematic diagram from wiki commons:

The idea is that the positively charged metallic "cores" (which includes the nucleus and the more tightly bound inner core electrons) are sort of suspended in a crystalline structure within a "sea" of mobile electrons.
This model explains most of the  metallic properties we observe, from heat and electrical conductance, to malleability and ductility.
Now think about oxidation - in an oxidation-reduction (redox) reaction, one element gains electrons (it is reduced) and the other loses electrons (it is oxidized). Here's an animation, also from wikipedia:

So, for an element to be metallic, its valence electrons need to be relatively "loose" - it can't take too much energy to detach them. For an element to be oxidized, something else has to "want" the electrons more - the energy cost of removing the electron has to be lower then the gain from adding them to the reducing agent.
As a result there is a very strong correlation between the metallic nature of elements and their relative ease of oxidation. There is also a very strong correlation between both of these properties and the trends in ionization energy, electron affinity, and  electronegativity.
Of course, in chemistry, things are never that simple. There lots of details and it is tough to make general statements that are always true. For example, you can say that often times the more "metallic" an element is, the more easily it will be oxidized. However, when you compare elements using a standard electrode potential table (which gives us a way of measuring relative ease of oxidation) you find that it is easier to oxidize calcium ($-3.8 \space \rm{V}$) than it is to oxidize potassium ($-2.9 \space \rm{V}$), even though potassium is closer to the "metallic side" of the periodic table, it has a lower first ionization energy, and it is less electronegative.
This illustrates a limitation of thinking of elements in terms of metals and non-metals: it is really a shorthand way of describing all of the things that affect how tightly those valence electrons are bound to the atom, in comparison to how badly other elements "want" to remove them. With that amount of generalization, we lose too much detail to be able to make accurate predictions in every situation.

"Of course, in chemistry, things are never that simple. There lots of details and it is tough to make general statements that are always true." - I agree with this statement wholeheartedly, and it actually made me reconsider and revise my post to better reflect chemical reality.

Oxidation is defined as an increase in oxidation state, and an increase in oxidation state can come from the loss of an electron. According to IUPAC:

oxidation
1) The complete, net removal of one or more electrons from a
  molecular entity (also called 'de-electronation').
2) An increase in the
  oxidation number of any atom within any substrate.

http://goldbook.iupac.org/O04362.html
Therefore, it might be useful to examine first ionization energies of metals to determine their oxidation potentials.

Consider for example the oxidation of the alkali metals. Sodium has a highly positive oxidation potential.
$\ce{Na -> Na^+ + e^-}$ $E^o=2.71 V$
This highly positive oxidation potential makes intuitive sense; a positive potential indicates a spontaneous reaction. The above reaction is definitely exothermic, thereby satisfying one criterion of spontaneity; solvated sodium ion is more stable than elemental sodium. The above reaction is also likely entropically favorable; solvated ions have more degrees of freedom than a solid.
Finally, note that the first ionization energy of sodium is $496 ~kJ/mole$.

Now, consider the oxidation of potassium metal:
$\ce{K -> K^+ + e^-}$ $E^o=2.93V$
The first ionization energy of potassium is $418 ~kJ/mole$.

As we can see, a lower first ionization energy seems to corresponds with a more positive standard electrode potential.
This makes intuitive sense; the oxidation half-reactions examined above are fundamentally ionization reactions. And the easier it is to strip an electron; the lower the first ionization energy, the greater the standard cell potential.

However, this trend doesn't necessarily hold in all cases. Consider, for example, the oxidation of lithium.
$\ce{Li -> Li^+ + e^-}$ $E^o=3.05V$
Lithium is the smallest alkali metal on the periodic table. Accordingly, it has the highest first ionization energy of all the alkali earth metals ($520 kJ/mole$).
Nonetheless, it still has the highest oxidation potential (greater than three volts!). So it seems that we can only use ionization energies as a rule of thumb.
Here's an example in which ionization potential can work:
Consider fluorine.
Do you think that fluorine would rather lose an electron (be ionized) or gain an electron? Gain is the answer. By gaining an electron fluorine becomes isoelectronic with a noble gas. Sure, fluorine could also lose all seven of its valence electrons to be isoelectronic with helium, but that's obviously a more roundabout path to stability. Given this information, you can probably deduce that fluorine would rather be reduced than oxidized. And this is true; fluorine is a powerful oxidizing agent.


text:
How to Experimentally Quantify the Response of a Receptor to a Ligand?
In medicinal chemistry, there are multiple reasons why one would want to experimentally determine the effect of a particular molecule on a receptor e.g. measuring how 'strongly' that molecule brings about a chemical response (an agonist) or how strongly it prevents a response (an antagonist).
If we were looking at an insulin receptor antagonist (for whatever reason), then it would be easy to quantify the strength of the antagonist by a simple blood test of the test subject (human or animal), to see how much insulin was present.
What happens when we're faced with a scenario inherently difficult to quantify? How do we measure the action of a drug then?
Say we've found a natural neurotransmitter receptor agonist and want to compare this to a synthetic drug of a similar nature. I'd struggle to see how we'd experimentally be able to analyse the amount of neurotransmitter present in the synapses in each of the cases to compare.
Do we then look at less direct, perhaps qualitative methods of analysis? Such as 'state of agitation' or 'how many times the subject's arm twitched in a minute'?

What you're asking about is a wide topic.
You have the basic ideas of agonist and antagonist, but there are also molecules called inverse agonists. Every receptor has some base level of activity, even without an agonist to activate them, or in the presence of an antagonist. Inverse agonists push this activity even lower. There are even "partial agonists" which are molecules that activate receptors, but not as well as the endogenous ligand, so act like competitive inhibitors because they occupy some of the receptors the endogenous ligand would have. In addition, the binding site is important. The endogenous ligand binds to a location called the orthosteric binding site, but other molecules may bind to other locations on the receptor. These are allosteric binding sites. Allosteric binders may activate or inhibit a receptor, but they may also be modulators. Modulators don't directly activate a receptor, but allow a stronger response when another ligand binds at the orthosteric site. Here's a schematic to help show how different types of ligand affect receptor activity.

Anyway, ligands have two important properties, affinity and efficacy. Affinity is how strongly a ligand binds to a receptor, and efficacy is how the ligand affects the receptor's activity.
Before we can dig too deep into agonists or antagonists, we should think about receptor theory. Receptors, like all proteins, have complex 3 dimensional structure, or conformation. This conformation is affected by solvent, pH, temperature, and other factors, but always has some fluctuation. Classical receptor theory used a lock and key model, where an inactive receptor bound a ligand and became active. However, this did not explain the base activity and activity of inverse agonists, so an equilibrium based model was developed. Here, the population of receptors has an ensemble of conformation states, some of these are active. When an agonist binds, it drives the equilibrium of the population toward the active state by stabilizing the active conformation. This diagram shows a simplified example of how drugs affect the equilibrium. Ri is inactive receptor, Ra is active. D is drug. Ra leads to some activity, while Ra-D leads to strong activity.

Affinity is relatively straightforward to measure becuase the same idea can be applied to almost any receptor-ligand system. Usually, a radiolabled version of the drug is added to a known amount of receptors in a binding assay. The goal is to determine a dissociation constant, Kd. Kd is used to quantify how well a ligand binds to a receptor, and smaller numbers are tighter binders.
Look at this reaction: D + R <--> DR with forward rate constant k1, reverse rate constant k2. ( how to get the right arrows here? )
At equilibrium, k1[D][R] = k2[DR]
and $K_d =\frac{[D][R]}{[DR]}$ = $\frac{k_2}{k_1}$
So incubate the radioligand and receptor until some DR complex is formed, then seperate the bound and unbound ligand so you can measure D and DR. The seperation method is important because if it's too slow the DR complex will revert to D and R, creating inaccurate data. But this is commonly done using dialysis, centrifugation, and vacuum filtration. Dialysis works well when dissociation rate is slow. You put receptor in one side and drug in both sides. It's slow, but it's theoritically most accurate becuase you can measure the true steady state by taking an equal volume of solution from each side of the membrane without affecting equilibrium. But you also need a large supply of receptor and drug.
Centrifugation works when you can immobilize the receptor on a large particle, usually you do this by using whole cells. You add an excess of of drug, then spin it down and take off the supernatent. Timing is still an issue and free drug can get caught in the pellet.
Vacuum filtration is usually considered the best because it can seperate the components very quickly, by pulling the solution through a filter and trapping the receptor on the surface. However this requires specialized and expensive equipment.
Once you measure bound and free drug at several concentrations, you can do a double reciprocal plot
Measuring efficacy is different because you need to choose an activity to measure. Idealy it will be objective and easy to quantify. If the receptor is an enzyme, you can measure its enzyme activity. If it activates a gene transcription factor, you can use cells modified to express green fluorescent protien or luciferase when that transcription factor is activated. If it's an ion channel you can measure a cell's membrane potential. If you cannot measure anything else, you can use behavioral assays. There are several assays used to measure mouse activity and try to get at certain aspects of their brain and are often used to study depression and other mental illnesses. These are subjective. Efficacy is quantified by EC50, the concentration at which response is half of maximum response. Smaller EC50s are more efficacious.
This field is much much larger than my answer. For more you should read this article : http://www.ncbi.nlm.nih.gov/pubmed/15063082


text:
2,2,6,6-Tetramethylpiperidine - Brønsted Basicity
I understand that this molecule can only act as a Brønsted base because it is sterically hindered by its multiple methyl groups.

However, isn't the proton bound up in solution? Doesn't that increase the effective size of the proton? How then exactly is a proton expected to reach the site of basicity if it's tied up with multiple water molecules? Or is the size of the hydrated proton still quite small? Or does the proton separate from the water molecules once the proton is sufficiently close to the site of basicity (the lone pair on the nitrogen)?

I don't think there will be steric hinderence to a small cation proton

Right but we know for a fact the proton is solvated quite heavily in water, with many users on this very forum insisting that H3O+ is inadequate to describe the extent of solvation of the proton in water. So wouldn't this solvation bulk up the proton?

did you reseacrched its pKa, it is 11.07, stronger than methylamine, carbonate,phenoxide,ammonia,etc.

pKa of what? Hydronium ion? Also I don't see phenoxide ion as having a particularly high pKa in the first place.

see here, you do no research on google?, it is stronger than unhindered amines.

I still remember you saying "The basicity of amines has to do not only with the degree of substitution with electron-releasing alkyl groups but also factors such as steric hindrance and degree of solvation." here, see my answer you'll recognize it's basicity on this page

But back to the question: does solvation increase the size of the proton; if so, is the proton then sterically hindered?

yes solvation does, it is not correct to comment whether it eould be strecially hindered, but i guess it would be, wait i'm working with 3d models and will tell if i get something

I don't see much hinderance[note that they all have been optimised to stable configurations]
Blue: nitrogen,White: Hydrogen,Grey: Carbon
Side View:

Top View:

The methyls on the left seem eclipsed in the top pic, is that the most stable conformed? Also can another Lewis acid get to the lone pair on nitrogen? I.e would say a metal cation be too big to fit?

@Dissenter I just used optimise geometry function and this is what i got.

Cool. What program?

@Dissenter avogardo


text:
Is tropone aromatic?

Looking at resonance structure 1a, it doesn't seem that tropone is aromatic. However, the resonance structure 1b with $\ce{C+-O-}$ satisfies Hückel's rule for aromaticity. Is it correct to say that tropone is then aromatic?

Tropone, or 2,4,6-cycloheptatrien-1-one, is an aromatic, non-benzenoid hydrocarbon.  If you look at the resonance structures in the drawing below you'll see that structure B

depicts a molecule with a continuous, planar loop of 7 p-orbitals that contain 6 pi-electrons.  Any planar (or near-planar), cyclic system with a continuous loop of p-orbitals is considered aromatic, or stabilized, if it contains 4n+2 pi-electrons and anti-aromatic if it contains 4n pi-electrons.  On this basis, molecules like the cyclopropenyl cation (4n+2, n=0), benzene (n=1), cycloheptatriene (tropylium) cation (n=1), cycooctatetraene dianion (n=2) would be considered aromatic systems.  Correspondingly, the cyclopropenyl anion (4n, n=1), cyclobutadiene (n=1), cyclopentadienyl cation (n=1), cyclooctatetraene (if it were a planar molecule, n=2) are considered destabilized or anti-aromatic molecules.

Does it mean that resonance structure B is far more abundant than A? That is, the real structure of tropone is better represented with structure B than A?

Here is some data.  The dipole moment of tropone is 4.17 D while that of cycloheptanone is 3.04 D; the C=O stretch is at 1638 cm-1, weaker than a typical alpha-beta unsaturated ketone (~1665 cm-1); the tropone ring undergoes Diels-Alder reactions and the carbonyl reacts with 2,4-dinitrophenylhydrazine.  My guess would be that while structure B is a significant contributor to the overall description of tropone, it is not the major contributor.

@ron actually, here I found C13 and H1 data https://www.chem.wisc.edu/areas/reich/nmr/c13-data/cdata.htm. Tropone shows C13 shifts that are, unlike in cyclohexenone, pretty uniform through ring atoms except for carbonyl carbon. The same applies to H1-NMR. Moreover, shifts in H1-nmr may be said to be closer to thats of benzene than to cyclohexanone or dienones, though this is debatable to some extent. So, I would say, that according to NMR data tropone has very strong aromatic character. But this aromaticity does not affect its reactivity much, it seems that stabilization is low.

@Permeakra  Are you suggesting that resonance structure B is the major contributor?  I agree that the nmr chemical shift data is debatable.  However, note that the coupling constants are suggestive of significant bond alternation, "tropone shows larger bond-alternation effects than the aromatic tropylium ion"  http://www.chem.wisc.edu/areas/reich/nmr/05-hmr-05-3j.htm

@ron Afaik, in modern days aromatic ring current is considered as the main sign of aromaticity. Such current creates its own magnetic field, that manifests in specific shifts in NMR data. So, I think that accurate analysis of NMR data should be the method of establishing degree of tropone aromaticity, and I presented my view on the topic. That's not a field of my major interest, though, so I can't provide deep analysis of the subject.

I think the data in these comments would be better off in the answer. After all, it is nice to say that there are resonance structures which are aromatic. But it remains to be shown which resonance structure is more important.

Does tropone undergoes substitution reaction characteristic of unsaturated cyclic aromatic systems.

The best answer to this theme is probably given by Barach in his book
BACHRACH, Steven M. Computational organic chemistry. Annual Reports Section" B"(Organic Chemistry), 2008, 104: 394-426. p 144

there is no single definition of aromaticity but rather a collection of properties associated with aromatic molecules. These properties include the following
Aromatic compounds are stable relative to some nonaromatic reference
Aromatic compounds display magnetic properties related to a ring current. These include unusual chemical shifts, especially protons, large magnetic anisotropies, and large diamagnetic susceptibility exaltations
Aromatic compounds tends to be planar and have equivalent C-C bond lengths,even though formal Lewis structure indicate alternating single and double bonds about the rings.
Even though their Lewis dot structures show C=C bonds, aromatic compounds prefer substitution reaction to addition reactions

According to these definition Tropone can be considered aromatic.
I've found also useful information in  SCHERRER, Robert (ed.). Anti-inflammatory Agents Part II: Chemistry And Pharmacology. Elsevier, 2012.
there you can find that there wasn't agreement on the aromaticity of tropone  because

... investigation of tropone, tropolone, and their derivatives seemed to confirm the aromatic character, which was attributed to a significant contribution from the dipolar resonance structure. ... Bertelli and Andrews (1969) revaluated the question of aromatic character in tropone and related compounds using data derived from dipole measurements ...Shimanouchi and Sasaada in 1970 .. found that the molecule was almost planar.. The authors pointed our that the average C-C lenght (1.407 Angrstrom) agreed with the standard aromatic value... Deewar and Trinajstic (1970) suggest that tese compounds more closely resemble polyenones

Kalsi in his book KALSI, P. S. Spectroscopy of organic compounds. New Age International, 2007. report that tropone represents a series of dipolar structures which constitute the aromatic tropylium system) and also Agrawal in his book AGRAWAL, O. P. Organic chemistry reactions and reagents. 1996.
report that being miscible with water, boils at 113°C, strongly polar ($\mu$ 4.3 D) and lacks of ketonic properties. Since

all these properties may be explained on the basis that tropone is a resonance hybrid of three structures).. the resonance hybryd II of tropone also indicates that the tropone ring contains 6$\pi$-electrons in a cyclic resonating system and thus behave like an aromatic compound. Chemically, tropone behaves partly as an unsaturated compound, viz. it decolourises aqueous potassium permanganate solution, add up chlorine and bromine, and undergoes Diel's-Alder reaction with maleic anhydride. On the other hand it also show benzenoid characteristic as indicated by the formation of 2,4,7-tribromotropone and 2-aminotropone on bromination and amination respectively.

So I guess there is not only one answer but depends on the aspect that you are looking at but in order to summarize
pro

is almost planar with an average C-C length that agree with standard aromatic

miscible in water (hint an interaction with $\pi$ electrons)

lack of chetonic propertiers

stable due to the the fact that is a resonance hybrid
5.it reacts to form 2,4,7-tribromotropone and 2-aminotropone by bromination and amination respectively

cons

dipole measurements
decolorise aqueous potassium permanganate solution
add up chlorine and bromine
undergoes Diel's-Alder reaction with maleic anhydride


text:
Disposing of hazardous waste indicated by numbers on chemicals
I'm doing a book for the African market and the syllabus asks for discussions on proper disposal of chemicals and waste products, particularly in the classroom setting
The authors said something along the lines of:

Chemicals numbered 1 to 4 can be rinsed down the drain with water. Chemicals labeled 5 to 8 must be disposed of with the help of an Environmental Agent or official.

What are the numbers they mention? Is there some numerical system (similar to scheduling on prescription meds) which guides us on how to dispose of certain chemicals?

The numbering clearly is not a part of the international GHS (Globally Harmonized System of Classification and Labelling of Chemicals) labeling scheme.

On a side note, GHS requires

hazard pictograms,
coded hazard statements (Hnnn and EUHnnn)
coded precautionary statements (Pnnn)

but typically does not give any information on the how/where to dispose chemicals.

For classroom settings, the questions are:

How many different waste containers are needed?
How to label them?
Which chemicals go where?

To my knowledge, a global or EC-wide (European Community) scheme does not exist for this and I'd be astonished to find that the African Union has agreed upon a common scheme.
Instead, this is typically regulated on a nation-wise basis only.
To give an example: In Germany, the classification scheme is provided by the Deutsche Gesetzliche Unfallversicherung (mandatory accident insurance) and you may find their suggestions for waste disposal in schools here (PDF in German, Stoffliste zur Regel
„Unterricht in Schulen
mit gefährlichen Stoffen“, 2017).
Consequently, without knowing the (national) origin of the source you are citing in your question, it is very difficult to say which number refers to which compound class.


text:
Bicycle tube inflation with 16g CO2 cartridge
How much pressure will a 16 gram cartridge of $\ce{CO2}$ (carbon dioxide) provide in a bicycle tire? Remembering my high school chemistry (forty years ago) and googling, I came up with this answer:
Figure the volume of the tube (a torus): area of a cross section times the circumference, for a $30\:\mathrm{mm}$ tube on a $70\:\mathrm{cm}$ wheel, I got $1,554\:\mathrm{cm^3}$, or $1.554\:\mathrm{L}$. One mole is $24.47\:\mathrm{L}$. $1.554 / 24.47 = .06353$. One mole of $\ce{CO2}$ is $44\:\mathrm{g}$ (using atomic weights of carbon, 12, and oxygen, 16). $.06353 \cdot 44 = 2.795\:\mathrm{g}$. So the volume of the tube at one atmosphere is $2.795\:\mathrm{g}$ of $\ce{CO2}$. $16 / 2.795 = 5.7\:\mathrm{atm}$. One atmosphere is $14.7\:\mathrm{psi}$. $5.7 \cdot 14.7\:\mathrm{psi} = 84\:\mathrm{psi}$.
Is this correct?

I strongly suggest to search data for density of $\ce{CO2}$ at different pressures. Most real gases do not behave as ideal gases at any moderate or highter pressure.

Are you working at 25 degrees Celsius?  If not, the 24.47 is incorrect...'

Don't forget to subtract 14.7 to convert your absolute pressure answer to the gauge pressure you'll measure after filling your tire...

I did not understand why you said the "volume of tube at one atmosphere is 2.795g..." you divide g/g and result atm ?

Assumptions. The $\ce{CO2}$ will cool down a bit as it expands out of the cartridge (due to the Joule-Thomson effect). But let's assume that you're interested in the pressure in the tire when it warms back up to room temperature, $25~^\circ\mathrm{C}$, so there is essentially a zero temperature change before and after filling. Let's also assume that the full $16~\mathrm{g}$ of $\ce{CO2}$ is contained by the cartridge and tire when they are connected; none of it is vented during filling.
Estimating the volume of the tire and cartridge. A $30 \times 700$ bicycle tire has a $30~\mathrm{mm}$ outside tube diameter and a $70~\mathrm{cm}$ wheel diameter. Let's say the rubber in the tire is $3~\mathrm{mm}$ thick. You then have an inside diameter of about $24~\mathrm{mm}$. If we assume the tube has a circular cross-section, the volume inside the tire when fully inflated is.
$$V = \pi \left(\frac{2.4~\mathrm{cm}}{2}\right)^2\cdot (\pi \cdot 70~\mathrm{cm}) \cdot \frac{1~\mathrm{L}}{1000~\mathrm{cm^3}} = 0.99_5~\mathrm{L}$$
Let's assume the $\ce{CO2}$ cartridge has a volume of $21~\mathrm{cm^3}$. The total volume of the $\ce{CO2}$ in the attached cartridge and tire will then be about $1.0_2~\mathrm{L}$.
If you have $16~\mathrm{g}$ $\ce{CO2}$, and the tire is fully deflated to begin with, the moles of gas is.
$$n = 16~\mathrm{g} \cdot \frac{1~\mathrm{mol}}{44.01~\mathrm{g}} = 0.36_4~\mathrm{mol}$$
Estimating the pressure. Since we're dealing with a gas with significant intermolecular interactions, which is also at pressures above $1~\mathrm{atm}$, let's compare an ideal gas calculation of the pressure
with a van der Waals calculation to see if nonideality makes any difference.
$$\begin{align}
P_{ideal} &= \frac{n R T}{V}\\
          &= \frac{(0.364~\mathrm{mol})(0.082059~\mathrm{L~atm~mol^{-1}~K^{-1}})
              (298~\mathrm{K})}{1.02~\mathrm{L}}\cdot
              \frac{14.696~\mathrm{psi}}{1~\mathrm{atm}}\\
          &\approx 129~\mathrm{psi}\\
P_{vdw}   &= \frac{n R T}{V - n b} - \frac{n^2 a}{V^2}\\
          &= \frac{(0.364~\mathrm{mol})(0.082059~\mathrm{L~atm~mol^{-1}~K^{-1}})
             (298~\mathrm{K})}{1.02~\mathrm{L} - (0.364~\mathrm{mol})
             (0.04267~\mathrm{L~mol^{-1}})}\\
          &~~~~~~ - \frac{(0.364~\mathrm{mol})^2(3.592~\mathrm{L^2~atm~mol^{-2}})}
                    {(1.02~\mathrm{L})^2}\\
          &= 8.43~\mathrm{atm} \cdot \frac{14.696~\mathrm{psi}}{1~\mathrm{atm}}\\
          &\approx 124~\mathrm{psi}\\
\end{align}$$
A tire gauge reads (gas pressure - atmospheric pressure), so you'd read about about $109~\mathrm{psi}$ if you started with a fully deflated tire. If you started with a tire filled with air at $1~\mathrm{atm}$, though, you'd get about $124~\mathrm{psi}$.

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