text:
Do MO's span a bond or the whole molecule?
I'm reading Organic Chemistry by David Klein (2nd Ed), and when discussing the causes of torsional strain, he writes this

Based on recent quantum mechanical calculations, it is now believed that the staggered conformation possesses a favorable interaction between an occupied, bonding MO and an unoccupied, antibonding MO. This interaction lowers the energy of the staggered conformation.

This favorable interaction is only present in the staggered conformation. When the C—C bond is rotated (going from a staggered to an eclipsed conformation), the favorable overlap above is temporarily disrupted, causing an increase in energy.

I've searched online and most sources provide a similar explanation and diagram. My issue is mainly with the diagram: why are the MOs illustrated as though they span only the C—H bonds? I thought that MOs were supposed to span the entire molecule, rather than being localized along/parallel to the internuclear axes of two nuclei. Are these MOs correct? Do MOs cover the entire molecule or do they each exist between two bonding atoms?

You at the sticky point of a lot of the hand waving that goes on in chemistry. You're holding a water molecule in your hand. Quantum theory says an electron from that molecule can be found on the other side of Mars. Pragmatically most of time the electron will be near the water molecule. So think of the shape of the orbitals drawn as the 95% probability boundaries. Does that help?

@Max While this is quite accurate, I think the problem at hand is a little more complex. This is a quite common mix-up between localised bond models, e.g. valence bond theory as depicted in the diagram, and delocalised bond models like molecular orbital theory.

The unfortunate answer is: both, and it is rather complicated. It can be noted at this point, however, that between two atoms connected by covalent (or even ionic bond) there are MOs.
Molecular Orbitals, the Wave Function, and the (Electron) Density
In quantum chemistry, we can define:

(spin) MO: a one-electron wave-function. Contains one electron, can be combined to with another spin MO to form what we usually think of as an MO, that is, doubly occupied.
Wave Function (WF): Typically a product of MOs, at least in the simpler approaches (Hartree-Fock [HF], common Density Functional Theory). The product is anti-symmetrized such that it obeys the Pauli principle. These products of MOs can be linearly combined to yield more accurate, complicated and expensive approaches.
Density: The expectation value of a WF when applying the density operator. Out of the three, this is the only true observable, for instance by X-ray diffraction. If you have a open shell molecule, you may be able to observe a spin-density, for instance by Electron Spin Resonance spectroscopy.

What's the point of these definitions?
They illustrate that a MO and the observable density have a few steps in between. Let's assume that someone gives us a set of MOs which yield the correct density of a molecule (as well as other correct observables). This is already a tall order, by the way. It is a mathematical fact that one can "rotate" the MOs, that is, create new MOs from linear combinations of old ones and still yield the correct density. So the set of "correct" MOs is not unique by the reasonable measure that it should give us the correct density.
OK, so there is more than one set of "correct" MOs. So what?
For one molecule/method/basis set combination, many ways of "choosing" your MOs are possible and lead to very different pictures, even when choosing reasonable contour values (say, 85%) for the visualization. An incomplete list is:

canonical MOs: these are eigenfunctions of the Fock operator, which is central to HF, but also of relevance for post-HF methods. They also preserve meaningful orbital energies (which relate to ionization potentials), but they are usually completely delocalized over the whole molecule.
localized MOs: "rotations" of the MOs have been performed to yield orbitals of (hopefully) minimal spatial extent. They are more easily interpreted, but the mathematics involved is iterative, meaning that minor changes in molecular geometry could lead to rather different local MOs. There is also more than one scheme to do this, and the results can be fairly different.
natural MOs: suppose you have performed the very accurate/very expensive calculation of full Configuration Interaction (full CI) for your molecule. The resulting orbitals are then eigenfunctions of the true Hamiltonian (in the given basis set). You will probably never encounter these due to the sheer cost of the calculation - and they may not look all that different from canonical orbitals.

And then there is the small issue of virtual/unoccupied orbitals
They do not enter the density (duh), but are often relevant to discussions of mechanisms, such as initial point of attack or stabilization. In my opinion, there is basically no way to pick a good scheme for them - canonical orbitals sort-of yield electron affinities.
Then all hope is lost. There's no way to choose.
Well, for occupied orbitals, there kinda is. One should compare orbitals (of one computational level) with each other across molecules to get to useful results. The textbooks usually do not give you this opportunity because you have no idea where those orbitals come from.
Back to OP's question: yes, canonical orbitals would extend over basically the whole molecule. You are being shown, in all likelyhood, orbitals that have been localized somehow. They are used as illustrations by the authors and that's usually not a bad idea. Take 'em with a grain of salt.

Very good explanation. One minor clarification: The wave function is a linear combination of products of MO as a pure product of MO (Hartree product) cannot be anti-symmetrised.

In addition to TAR86's good explanation I add an example to visualise the different molecular orbitals.
A few general remarks up front though.
Quite a few textbooks (especially in organic chemistry) use the term quantum mechanical calculations (or variations thereof) in a (somewhat) sloppy fashion. They offer (generalised) phenomenological explanations based on such calculations. This is in principle good, as we are moving away from a purely Lewis-structure based understanding of bonding and man-made concepts as ring/ torsional/ angle strain, ±I/M, resonance, etc. (the list is almost infinite) towards a more generalised, complete, overarching understanding, but it can also lead to ambiguity. Sometimes (read: often) books forget to mention the limitations of such explanations and the justification of such models is left as an exercise of critical thinking to the reader.
As the average chemist progresses from basic to advanced understanding of bonding it can get confusing since some books use similar terms for different things. One of the common things is arguing with localised orbital schemes, predefined hybrid orbitals and the likes, which allow little to no flexibility. While these arguments are also deeply rooted in theoretical chemistry, i.e. Valence Bond theory, the source of the computation is most times in terms of Molecular Orbital theory. Rather than performing the extensive (and quantitative) calculations applying the former, the more common Molecular Orbital approaches (Linear combination of atomic orbitals to form molecular orbitals; or LCAO=MO) are used, such as Hartree-Fock, post-HF and Density Functional Approximations. The latter are so well developed that it won't be long that we can run simple calculations on the fly on our phones (The future is here ...).
When you start learning quantum chemical calculations you might be confused at first, because a lot of such tempting model explanations don't seem to fit any more.
Orbitals. First and probably most important is to realise that the shapes we recognise as orbital depictions are not real in any sense. These are visualisations of mathematical functions at a certain cut-off value. This is necessary as a wave function (in principle) only vanishes at infinity. In this sense, orbitals obviously always span the entire molecule. But this is in general not the reason why we say molecular orbitals span the entire molecule; this is a general feature of Molecular Orbital theory (as opposed to Valence Bond theory).
Molecular Orbitals Interpretation. I have used DF-BP86/def2-SVP to run a quick geometry optimisation of staggered ethane with Gaussian09. The approach to obtain molecular orbitals on this level of theory is similar to Hartree-Fock as in that it uses one Slater-Determinant (not a product of wave functions!) as the trial function only.
The results are thus similar to the canonical orbitals you would obtain from a HF solution. (For all intents and purposes in this example we regard them as equal.)
The general feature of this approach is that all atomic orbitals contribute in some fashion to the set of molecular orbitals. The coefficients to the atomic orbitals are optimised in a way that the expectation value of the energy of the trial function is minimal. Therefore in many cases they look like the following picture.

In the graphic above, the orange/blue orbitals are occupied and the red/yellow ones are virtual; core orbitals omitted. Because of symmetry we have some degenerate orbitals in the mix. Strictly speaking only the occupied orbitals are actual solutions to the pseudo-Eigenvalue problem, but we can use these to generate virtual (high energy) solutions, too. While these orbitals offer a general understanding, and often reproduce ionisation potentials and electron affinities, it is difficult to interpret them with the conventional understanding (of Lewis structures) that the organic chemists love so much.
Localised Orbitals and the Valence Bond Interpretation. There are numerous localisation schemes for canonical orbitals out there. Different approaches use different criteria. A common approach is to use linear combinations of the canonical orbitals to produce a different set of molecular orbitals with a compacter spatial extend, more of the electron density is contained in a smaller volume. The general aim is to produce orbitals which represent bonds where we expect them. Through this approach we loose some of the properties, especially those which are energy related.
One of these approaches in Natural Bond Orbital theory, which aims to provide a description as close as possible to Lewis structures. I have performed such a localisation with NBO 6.0 and it resulted in a much more familiar orbital picture:

(I have pulled the σ(*)-C-C bond orbital below the sp3-C-H bonds. In the output, below, these are between the triads.)
One of the main features of such localised orbitals are fractional occupation numbers. These then allow some more interpretations like the ones you have encountered, i.e. interactions between (mostly) occupied and (mostly) unoccupied orbitals. In this scheme your picture results in the following representation:

The above picture could be calculated quantitatively with a valence bond orbital approach, which uses hybrid orbitals and molecular conformations as an input. This usually needs considerably more calculation time as it is in theory already more extensive and more expansive. However, qualitatively it would probably reproduce the above picture (just more accurately), and often offers only for extraordinary bonding situations considerably more insight. As such, MO theory is complementary to VB theory; at their respective complete treatments they are equal.

Addendum. The calculation was run with Gaussian09 rev. D01 with tight convergence criteria and loose symmetry recognition. Convergence into a local minimum has not been checked. The optimised structure is in Cartesian coordinates in angstrom:

scf (DF-BP86/def2SVP) done:   -79.763139
 C    -0.111323     0.000000    -1.496408
 H    -0.111323     1.032093    -1.906398
 H     0.782495    -0.516047    -1.906398
 H    -1.005142    -0.516046    -1.906398
 C    -0.111323     0.000000     0.035289
 H    -1.005142     0.516047     0.445279
 H    -0.111324    -1.032093     0.445279
 H     0.782496     0.516046     0.445279

The displayed canonical orbitals have the following contributions (Display threshold 5%):

 Alpha occ 3 OE=-0.672 is C5-s=0.34 C1-s=0.34
 Alpha occ 4 OE=-0.544 is C1-s=0.19 C5-s=0.19 H2-s=0.07 H7-s=0.07 H4-s=0.07 H8-s=0.07
                          H3-s=0.07 H6-s=0.07 C5-p=0.07 C1-p=0.07
 Alpha occ 5 OE=-0.384 is C1-p=0.30 C5-p=0.30 H7-s=0.13 H2-s=0.13
 Alpha occ 6 OE=-0.384 is C1-p=0.30 C5-p=0.30 H3-s=0.10 H4-s=0.10 H8-s=0.10 H6-s=0.10
 Alpha occ 7 OE=-0.329 is C1-p=0.40 C5-p=0.40
 Alpha occ 8 OE=-0.298 is C5-p=0.21 C1-p=0.21 H7-s=0.19 H2-s=0.19
 Alpha occ 9 OE=-0.298 is C1-p=0.21 C5-p=0.21 H6-s=0.14 H4-s=0.14 H8-s=0.14 H3-s=0.14
 Alpha vir 10 OE=0.048 is C5-s=-0.26 C1-s=-0.26 H7-s=0.24 H2-s=0.24 H3-s=0.24
                          H8-s=0.24 H6-s=0.24 H4-s=0.24
 Alpha vir 11 OE=0.090 is H2-s=0.22 H7-s=0.22 H3-s=0.22 H4-s=0.22 H8-s=0.22
                          H6-s=0.22 C1-s=-0.22 C5-s=-0.22
 Alpha vir 12 OE=0.100 is H3-s=0.24 H8-s=0.24 H4-s=0.24 H6-s=0.24
 Alpha vir 13 OE=0.100 is H7-s=0.32 H2-s=0.32 H6-s=0.08 H8-s=0.08 H4-s=0.08 H3-s=0.08
 Alpha vir 14 OE=0.134 is H8-s=0.22 H6-s=0.22 H4-s=0.22 H3-s=0.22 C5-p=0.05 C1-p=0.05
 Alpha vir 15 OE=0.134 is H2-s=0.29 H7-s=0.29 H3-s=0.07 H6-s=0.07 H4-s=0.07 H8-s=0.07
                          C1-p=0.05 C5-p=0.05
 Alpha vir 16 OE=0.195 is C5-p=0.43 C1-p=0.43

The displayed NBO have the following contributions (cleaned):

     (Occupancy)   Bond orbital / Coefficients / Hybrids
 ------------------ Lewis ------------------------------------------------------
   3. (1.98857) BD ( 1) C  1- H  2
               ( 61.06%)   0.7814* C  1 s( 23.30%)p 3.29( 76.67%)d 0.00(  0.04%)
               ( 38.94%)   0.6240* H  2 s( 99.91%)p 0.00(  0.09%)
   4. (1.98858) BD ( 1) C  1- H  3
               ( 61.06%)   0.7814* C  1 s( 23.30%)p 3.29( 76.66%)d 0.00(  0.04%)
               ( 38.94%)   0.6240* H  3 s( 99.92%)p 0.00(  0.08%)
   5. (1.98858) BD ( 1) C  1- H  4
               ( 61.06%)   0.7814* C  1 s( 23.30%)p 3.29( 76.66%)d 0.00(  0.04%)
               ( 38.94%)   0.6240* H  4 s( 99.92%)p 0.00(  0.08%)
   6. (1.99640) BD ( 1) C  1- C  5
               ( 50.00%)   0.7071* C  1 s( 30.13%)p 2.32( 69.83%)d 0.00(  0.04%)
               ( 50.00%)   0.7071* C  5 s( 30.13%)p 2.32( 69.83%)d 0.00(  0.04%)
   7. (1.98858) BD ( 1) C  5- H  6
               ( 61.06%)   0.7814* C  5 s( 23.30%)p 3.29( 76.66%)d 0.00(  0.04%)
               ( 38.94%)   0.6240* H  6 s( 99.92%)p 0.00(  0.08%)
   8. (1.98857) BD ( 1) C  5- H  7
               ( 61.06%)   0.7814* C  5 s( 23.30%)p 3.29( 76.67%)d 0.00(  0.04%)
               ( 38.94%)   0.6240* H  7 s( 99.91%)p 0.00(  0.09%)
   9. (1.98858) BD ( 1) C  5- H  8
               ( 61.06%)   0.7814* C  5 s( 23.30%)p 3.29( 76.66%)d 0.00(  0.04%)
               ( 38.94%)   0.6240* H  8 s( 99.92%)p 0.00(  0.08%)
 ---------------- non-Lewis ----------------------------------------------------
  10. (0.00902) BD*( 1) C  1- H  2
               ( 38.94%)   0.6240* C  1 s( 23.30%)p 3.29( 76.67%)d 0.00(  0.04%)
               ( 61.06%)  -0.7814* H  2 s( 99.91%)p 0.00(  0.09%)
  11. (0.00902) BD*( 1) C  1- H  3
               ( 38.94%)   0.6240* C  1 s( 23.30%)p 3.29( 76.66%)d 0.00(  0.04%)
               ( 61.06%)  -0.7814* H  3 s( 99.92%)p 0.00(  0.08%)
  12. (0.00902) BD*( 1) C  1- H  4
               ( 38.94%)   0.6240* C  1 s( 23.30%)p 3.29( 76.66%)d 0.00(  0.04%)
               ( 61.06%)  -0.7814* H  4 s( 99.92%)p 0.00(  0.08%)
  13. (0.00098) BD*( 1) C  1- C  5
               ( 50.00%)   0.7071* C  1 s( 30.13%)p 2.32( 69.83%)d 0.00(  0.04%)
               ( 50.00%)  -0.7071* C  5 s( 30.13%)p 2.32( 69.83%)d 0.00(  0.04%)
  14. (0.00902) BD*( 1) C  5- H  6
               ( 38.94%)   0.6240* C  5 s( 23.30%)p 3.29( 76.66%)d 0.00(  0.04%)
               ( 61.06%)  -0.7814* H  6 s( 99.92%)p 0.00(  0.08%)
  15. (0.00902) BD*( 1) C  5- H  7
               ( 38.94%)   0.6240* C  5 s( 23.30%)p 3.29( 76.67%)d 0.00(  0.04%)
               ( 61.06%)  -0.7814* H  7 s( 99.91%)p 0.00(  0.09%)
  16. (0.00902) BD*( 1) C  5- H  8
               ( 38.94%)   0.6240* C  5 s( 23.30%)p 3.29( 76.66%)d 0.00(  0.04%)
               ( 61.06%)  -0.7814* H  8 s( 99.92%)p 0.00(  0.08%)


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Can cationic species form micelles?
Most of the books dive into formation of micelles by soaps. They go on talking about alkyl sulphate and carboxylates, but never do they mention anything about cations. My textbook says that at critical micelle concentration, molecules of colloid aggregate to form micelles.
Can things like $\ce{RNH4+}$ or $\ce{R4N+}$ form micelles, why/why not?

Sure, there are cationic surfactants as well.

Cationic micelles similar to your examples can and do exist.  For example cetyl trimethylammonium bromide (Growth of Cationic Micelles in the Presence of Organic Additives, P. A. Hassan and J. V. Yakhmi, Langmuir, 2000, 16 (18), pp 7187–7191, DOI: 10.1021/la000517o).
Additionally, there are cationic micelle detergents (Ionization of cationic micelles: Effect of the detergent structure, Journal of Colloid and Interface Science, Volume 78, Issue 2, December 1980, Pages 330-337).
The bottom line is that there are a variety of use cases for cationic micelles, in detergents and otherwise.

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Light: Waves and Particles
I have three questions:
a. How do pictorial representations of electromagnetic waves translate to real life? They're shown as having perpendicular fields in phase,  but how can the strength of a given field,  be it magnetic or electrical,  be perpendicular to the other.  Basically,  how does this wave actually propagate through space?
b.  Einstein theorized that the energy of a single photon increases with raised frequency. Frequency? What charicteristic of a photon makes it periodic so as to possess a frequency?  Does its amount of energy fluctuate, or does this refer to the number photons per second,  etc?
c. How do photons propogate through space,  this is almost  a parellel question to my inquiry in point a.

This is of course a fascinating question for most of us.  Unfortunately, unless someone way smarter than me knows how to address this in a couple pages, the question is probably too broad for this forum.  My only technical advice is to just forget about the wave-particle "duality" thing; waves are waves, particles are particles and photons are photons.  That photons exhibit both wave and particle like properties just contradicts what we can experience on our macro scale lives, so just don't look too hard for analogies to macro scale phenomena like waves and massive particles. Big help, huh ;)

There are answers to most of these questions on Physics StackExchange. They do not seem to come to an agreement as to what a photon is, however, so it is not a trivial question.  Experimentally we find that in some experiments it behaves like a particle, in others like a wave. In question (b) The energy of the photon is directly proportional to its frequency; $E=h\nu$ where h is Planck's constant and $\nu$ frequency. (see photoelectric effect experiments for proof of energy vs. number of photons).

Your question must be move their http://physics.stackexchange.com/ because chemistry not study how light is and travel in the space.

@Hexacoordinate-C Disagree on that. Asking what a photon is and how it interacts with matter is a perfectly valid chemical question.

@Zhe : Disagree too, this is the role of the Quantum Electrodynamic https://en.wikipedia.org/wiki/Quantum_electrodynamics "In essence, it describes how light and matter interact and is the first theory where full agreement between quantum mechanics and special relativity is achieved."

@Hexacoordinate-C If that's your view point, then we should be shunting all questions about spectroscopy to physics.SE... :)

@Zhe does this question ask about spectroscopy ?

No, but spectroscopy is just light interacting with matter. I feel like your claim would suggest that spectroscopy should not be asked in chem.SE because it is the realm of QED?

I'm not opposed to moving this specific question to physics, but I think your metric for determining what should be moved does not match mine. I'm just arguing that a basic understanding of a photon and what is does is pretty important in many areas of chemistry. Totally fine if you disagree, but this comment thread is getting a bit long, so I'm just going to leave it at that. Happy to discuss more in chat.

I'm voting to close this question as off-topic because it would be better suited to physics.se.

The problem here is that you're asking the wrong question.
I believe you are confused by the concept of particle-wave duality, which doesn't really exist. The electromagnetic wave is a macroscopic phenomenon that arises because quantum mechanical particles in aggregate may reproduce certain wave-like effects due to the nature of the wave function.
Basically, a photon is just a particle. The energy of this particle is an innate property. At the macroscopic scale, this may manifest as a "frequency" or "color," but neither or those properties is required to explain any physical phenomena.
The photon travels through space like any other particle. You don't ask how an electron travels through space, or if you do and get stuck, then you'll also have the same issue here...

The whole answer isn't that simple. First most !@#$%^&*( diagrams show a photon as an infinite sine wave which is totally wrong. A particular photon would have a wave pattern more like a harmonically damped oscillator.

This response doesnt help me cuz it only answers one question and you wrote it like your talking to a Chem professor who knows what a harmonically damped oscillator is,  or what an infinite sine wave is.

@MaxW I've not seen anyone show a photon as an infinite wave. Generally, the photon is shown with a spread in frequencies centered around a point (this then allows us not to violate the uncertainty principle). (example) Again, this is a particle but must satisfy the usual laws with regard to quantum mechanics.

Zhe - Infinite sine wave. Look at the Wikipedia article for instance... I see it all the time. https://en.wikipedia.org/wiki/Electromagnetic_radiation

I'm starting to feel like you didn't actually read my answer... :) The electromagnetic wave is not a photon. You only get that in the limit of a large number of photons.

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text:
Why can a sigma bond rotate?
My book says

Because it (the sigma bond) lies along the line joining the two carbon atoms, the sigma bond can rotate about this axis.

But that doesn't make sense to me. So what if the sigma bond lies along the bond axis? How does that matter?

The sigma bond has symmetry akin to that of a cylinder. The energy is the same no matter what the rotation angle, thus rotation is possible. The energy does, however, depend greatly on separation of atoms as the balance of charges (protons/electrons) is upset. In a pi bond the symmetry is different and energy is needed to remove the p to p electron overlap on the two atoms, so free rotation is not possible.

Sigma  bonds  are  defined as  having their  electron  density  along the  bond  axis, while  pi  bonds  have  their  electron  density  above and  below  the  bond  axis. What this mean is that pi bonds cannot rotate the same way as sigma bonds since rotation would break the pi bond interaction. See the picture below for clarification.

If you're wondering why it rotates then the more  freely  that  a  bond  rotates, the  more  favorable the entropy.  In  general, the  more  kinds  of  motions  and  the  more  unconstrained  those motions are, the  more  favorable the  entropy.

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text:
Is this an aldehyde?
$\ce{CH3CH2CH2CHO}$
I think it is an aldehyde but I saw somewhere that it might be an alcohol...
I believe that it looks like this. Is that correct?

that's an aldehyde

In an alcohol, the alcohol group would be indicated as OH in a properly written formula.

yeah that's an aldehyde

That's an aldehyde as it contains ($\ce{-CHO}$) group.
Its not an alcohol because in alcohol the carbon and oxygen bond in a single (sigma) bond whereas in aldehyde it is a double bond.
$\ce{CH3CH2CH2CHO}$ cant be a alcohol because as you can see that the terminal carbon is attached to only 3 groups ($\ce{C}$, $\ce{H}$ and $\ce{O}$) which means that it is still missing 1 bond to complete its octet. Therefore this carbon makes one more bond, a double bond, with the oxygen atom to complete its octet and the oxygen also attains a complete octet by this.


text:
How fast a surface oxide layer is formed on bulk chromium?
If a chromium piece is cut in half, how long would it take for an oxide layer to form on the new surface (room temperature)? i.e. is it a matter of seconds/minutes/hours/days/weeks?
The closest study I found is by G. Thurner and P.H. Holloway (1992), but I can't really tell from their results what would happen at ambient conditions:

(I found several publications about high temperature oxidation, but none for surface oxidation at ambient conditions)

First, a definition of the Langmuir unit, used in the graph in the question, which is not often run across except by vacuum (particularly UHV) folks. One Langmuir is the exposure of a surface to a gas pressure of $10^{-6}$ Torr for one second. It roughly corresponds to one monolayer equivalent of gas hitting the surface. At normal atmospheric pressure, with 20% oxygen in air, 1 Langmuir of oxygen hits the surface in about 6.6 nanoseconds (yes, nanoseconds).
Another paper that could be of interest is C. Palacio et al., "AES, XPS and EELS study of the initial oxidation of polycrystalline chromium", Surface Science 182(1-2) 41-55 (1987). They see chemisorption of oxygen in the first 1-2 L, nucleation and lateral growth of oxide islands up to 10 L, and continued growth of an oxide film above 10 L.
Overall, there is not a huge difference between the two papers. By ~10 L of exposure to oxygen there is a continuous oxide on the surface at 300K. Following from the numbers in the first paragraph, that occurs in less than 100 nanoseconds. Basically, it oxidizes as it is cut.

Great expansion! How did you get 6.6 nanoseconds? I'm getting 0.16 nanoseconds

I went with 760 Torr times $10^6$ times 20%, getting 152,000,000 L in one second (for oxygen). Then, one L happens in (1/152000000) of a second.

ohhh it's pressure multiplied by time. makes sense now.

Right. So 1 Langmuir could be a microTorr for 1 second or a nanoTorr for 1000 seconds.

In learning about different steels, I've picked up some information that can hopefully help you.
I have been told by metal vendors and fabricators that if you have some pure chromium, it will passivate (form a chromium oxide layer) in ambient conditions. This layer is only a few molecules thick so I would hazard a guess that within a day you should achieve >90% thickness of this chromium oxide layer.


text:
Heat transfer with surroundings: why considered reversible?
I do not understand how the fact that the surroundings are large makes heat transfer with them reversible? I understand that the temperature of the surroundings is not going to change in the limit, but what about processes in which there is a definite temperature gradient with the system, such as a highly exothermic reaction?
I was sure that this question had already been asked somewhere on Stack, but I can't find anything similar – if you know of one, I would really appreciate if you could refer me to it.

The surroundings are typically modelled in thermodynamics as an ideal (reversible) reservoir with infinite heat capacity and negligible temperature gradients (i.e., infinite thermal conductivity).  The assumption is that the temperature gradients are confined to the system, rather than the surroundings, and the entropy generation is confined to the system.  Of course, in real life, this is only an approximation.

Thank you! So for example, if you burn something, do you consider the burning material and the shell of hot air it generates as the "system", and put the boundary between the system and the surroundings where the temperature difference between the air inside the shell and the air outside it becomes infinitesimal?

Sorry, how do I reference your name? @Chester Miller and any variations aren't recognised.

Apparently it did.

No.  The temperature difference at the boundary is always infinitesimal.  Temperature is continuous at a boundary.  I'm not talking about an interface with air outside the shell, but with an ideal reservoir (e.g. a liquid) featuring an infinite heat capacity and infinite thermal conductivity.

Thanks, but I'm still confused. So you're assuming that no temperature gradients with the surroundings can exist in principle, e.g. when something is burning, any heat released is instantly absorbed by the surroundings with infinite thermal conductivity? In this case I don't see how this can be a good assumption?

Or does our definition of the surroundings changes as the surroundings absorb heat (i.e. from a thin shell of surroundings approximately at the same temp as the system, to infinity?)

In most cases in thermodynamics books, heat transfer to the surroundings is regarded as taking place using an ideal constant temperature bath/reservoir.

Thanks! But when applied to real systems, what corrections need to be made and how to make them? The assumption of the reversible heat transfer with the surroundings seems to be at the very core of thermodynamics, so you would probably need to correct for almost every step? Perhaps you know any theoretical treatments of situations where heat gradients with the surroundings are significant?

You would have to treat the surroundings as a separate system and you would have to solve the transient partial differential momentum balance and heat transfer equations for the entire process for both the system and surroundings.  This would include viscous effects and heat conduction/convection effects.  In most cases, it is not possible to apply equilibrium thermodynamics to an irreversible process.


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Direction of gas molecules in kinetic model of gases?
So when learning about the kinetic model of gases, my teacher said that in a volume $A*v_{x}*\Delta t$, exactly half the molecules will have a velocity component of $v_x$ and the other half will have a velocity component of $-v_x$. But I think they are wrong!
Because there should be some small portion of gas molecules that will have a velocity component of: $v_x = 0$
Am I correct? So is the term for the average number of collisions with the wall during the time interval $\Delta t$, just an approximation?:
$(0.5)*A*v_{x}*\Delta t*n*N_a/V$
This is clear with an example. If there are three molecules in a container, one moving left on the x axis, one moving right on the x axis, and one moving directly up the y axis,  then on average, only 1/3 of the molecules is moving towards the right wall.

You have pointed out the problem of how to handle the infinitesimal in probability theory. So yes a molecule could traveling in the $yz$ plane and have 0 $x$ velocity. However the probability of the motion being in exactly the $yz$ plane can be made as small as you like.
Let's make the problem one dimensional and choose a random angle between 0 and 359.999... degrees. So the question is what is the probability that the chosen angle will be for a vector on the x-axis?

To a degree if I chose 0 or 180 then the vector will be on the x-axis. Odds 2/360
To a tenth of a degree if I chose 0.0 or 180.0 then the vector will be on the x-axis. Odds 2/3600
To a hundredth of a degree if I chose 0.00 or 180.00 then the vector will be on the x-axis. Odds 2/36000

and so on...
So what probability should be assigned to a randomly chosen angle being exactly along the x-axis?

Are you basically arguing that the x-velocity component is best described by a continuous distribution and therefore it having an exact value of zero would be zero?

@Nova - I do like that you had your thinking cap on. You really need to look for boundary conditions on the various chemistry factoids that get thrown around.


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What exactly is going on in this reaction: sodium bicarbonate (supersaturated aqueous), isopropyl alcohol, and acetic acid (vinegar)?
I observed this reaction by playing around with vinegar and baking soda and I do not completely understand it.
Everyone knows that baking soda and vinegar go crazy when mixed together and the solution starts creating carbonic acid that decays into carbon dioxide thus forming bubbles, but I never saw anyone adding vinegar to a supersaturated solution of baking soda and water. So, I decided to try it myself.
To no surprise, the reaction goes very similarly to how it would with just solid sodium bicarbonate. However, I then decided to try and see what rubbing alcohol would do. So, I grabbed the 92% rubbing alcohol and added it to the mixture before adding the vinegar.
You could see the alcohol mix slowly with the supersaturated sodium bicarbonate solution. However, when the vinegar was added, well I do not really know.
I added the vinegar to the solution really slowly and some minor bubbling occurred on the top. However, the solution never started to boil up as it did before. Instead, it just sort of sat there at which point, I decided to just take it to the sink and pour it out and try some new stuff, but I accidentally gave the glass a slight bump and all of the sudden, the reaction took place and proceed very rapidly shooting the bubbles out of the volumetric glassware I was using and getting everywhere. I have done this reaction several more times and it is very repeatable, although, I do not know how the addition of alcohol makes the reaction not occur until the solution is disturbed.
Volumes used are 150mls supersaturated solutions of baking soda in water, 20mls 92% isopropyl alcohol, and 40mls of vinegar.
Any help with understanding this reaction would be greatly appreciated. The only thing I have come up with so far is that the alcohol is changing the water's ability to dissolve the baking soda and that when a slight jolt is added to the system the baking soda is dislodged out of the solution long enough for the vinegar to attack it, but that is just a guess.

Seems like you had two different layers that didn't mix immediately. If you cannot see this because the refractive indices are too close, add some food coloring to one of the liquids.

I'm 99% sure that you are using the term supersaturated wrong.

Is alcohol les dense than saturated carbonate solution and is vinegar poorly soluble in alcohol. Just guessing, maybe you could try to mix vigorously the soda and the alcohol and slowly add vinegar just after mixing.

You have created a kind of isopropanol based "Champagne", which is known it may or may not burst at opening, depending on how fine it is treated.

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


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charged particles
Why are charged particles influenced by magnetic fields? Is it the charicteristic charge itself that the field interracts with (positive or negative) or is there something else about the proton or electron that is inherent within it that interracts with the field.

Electrical current (moving charges) and magnetic fields are related by Maxwell's equations. Think of a coil electromagnet. Now if you use a resister instead of battery and you then move the magnet in and out of the coil you can create an AC voltage.

I asked why,  not how. Electrons make up current,  and current can be induced by magnetism... Thus electrons are influenced by magnetism as well... Why?

Please read my question above more carefully

Maxwell's equations explain the why...

The whole reason I got on this forum and asked for answers was to get answers,  not referrals.

I'm voting to close this question as off-topic because it is about the properties of elementary particles, and would be better suited for Physics.SE.

Sorry you're disappointed. I just know that I couldn't begin to explain Maxwell's equations in a few paragraphs. The theory is a truly elegant bit of physics.

Charged particles like electron, if at rest, produce only electric field but if they start to move with a certain velocity then they produce a magnetic field in addition to the electric field.
Charged particles are not influenced by a external magnetic field if they are at rest.
But when they move they produce their own magnetic field and eventually, a magnetic force appears on the charge due to their interaction with the external magnetic field.


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How to remove sulfuric acid from sample
I have problem with my experiment. My experiment is acid hydrolysis of ethyl benzoate. The acid that I used is sulfuric acid. My problem is when I get the precipitate  (impure benzoic acid) I have to filter it from the solution (of course) and I need to remove the sulfuric acid from my sample or else my filter paper will not dry. But the major problem is sulfuric acid takes longer time to dry. It could takes even month. I've tried leave it for a month and it still wet. My lecturer asks me to rinse my sample + filter paper with cold water (distilled water). And when I did my sample which is the precipitate is LESS. Maybe because I used large amount of cold water. Is there are suggestions or methods to remove the sulfuric acid?

Hello and welcome to Chemistry.SE!  At this point I suggest you take the short [tour] of the site and then familiarize yourself with the homework policy by reading through this discussion.  Good luck!

How do you know that you had less precipitate after the cold water rinse? Just by observation?  How much less did it seem to be?

Also, do you have a feel for how much benzoic acid is there or what your theoretical yield should be?

If you have access to some benzoic acid, weigh some and try rinsing with about 3 molar HCl. If the HCl is ice cold even better.

Acid hydrolysis requires a few drops of sulfuric acid. This is easily removed by washing with a small amount of cold water... What exactly are you struggling with?

Perhaps after hydrolysis, add NaOH to bring the pH up so the benzoic acid is converted to a benzoate salt, then add barium chloride to precipitate all the sulfate, the filter, then collect the filtrate to obtain a sulfate-free solution of sodium benzoate, then acidify to convert to benzoic acid, causing precipitation where you can collect the benzoic acid ppt from under a sulfate-free solution?

Maybe extract with ethanol or DEE and then recrystalise.

Sulfuric acid is an excellent dessicant and is also very non-volatile itself (it boils at 337 °C).  In other words it would likely still have been there in yet another month.
Your instructor was correct to suggest that you rinse the sulfuric acid with some cold water.  Benzoic acid does have some non-negligible solubility in water (1.7 g/L at 0 °C).  Unfortunately, it is even more soluble in most organic solvents so cold water may well be your best bet.  The solubility in 25 °C water is 3.4 g/L, twice the solubility that it has at 0 °C, so definitely keep it cold.
Additionally, use as little water as necessary to get the sulfuric acid out.  If you are working with a gram or so of benzoic acid, I would start with a dropwise addition of about 10 mL. If it’s still gooey, or still appears wet the next day, you need to repeat the rinse. It’s best if you can use a vaccum filter system of course, and spread the material thin and even before rinsing.

I have converted your MathJax to html in this instance:  ° for degree Celsius; because it loads faster. Otherwise you might want to use MathJax for the whole expression, like $\pu{0 ^\circ C}$, $\pu{0 ^\circ C}$. What you have used might break on various occasions.

Yea, I was just going over your edits, much thanks. The  ° also doesn't have the silly italicized look of the way I did it. I've head that "it might break" statement before and am just curious...is that a browser-specific thing, or server side or SEO stuff?

MathJax is complicated in itself. It renders client based, and depending on your own configuration it uses installed or web fonts. Basically it is quite unpredictable how it looks. And then we have the apps... something, somewhere is certainly going to break... So far the worst I have experienced are potential line breaks between MJ and text when using mixed styles, i.e. CO$_{2}$ could have the two on the next line. Better wrap the whole thing to be safe, it also looks better (or at least less intrusive). [It was worse before graduation.]


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What is the product of a reaction between an acyl chloride and a secondary alcohol?
For the following two reactants, I want to predict the major product.

I would have thought the second reactant $\ce{CH3CH(OH)CH3}$ splits into $\ce{CH3CHCH+}$ and $\ce{OH-}$ , and the $\ce{OH-}$ would replace the $\ce{Cl-}$ in the first reactant $\ce{Phenyl-COCl}$, giving B.  My main reason for thinking this would occur is that I always see e.g. water form $\ce{H+ and OH-}$.
However, the answer is C. The $\ce{CH3CH(OH)CH3}$ gives up only its  $\ce{H+}$, and the rest of it joins $\ce{Phenyl-COCl}$ to form C.
How do I predict which atom(s) will leave/join? And for this particular question, how do I know the bond in the 2nd reactant cleaves at O-H not at C-OH?

I don't understand what you're asking. The answer is C. Why or what are you asking about D?

Sorry! Typo. I am asking why the H+ leaves the CH3CH(OH)CH3, leaving the CH3CH(O)CH3- to swap with the Cl- (giving C). Rather than CH3CH(OH)CH splitting into CH3CHCH+ and OH-, and the OH- replacing the Cl- in the first reactant.

In other words, why does the 2nd reactant split at the O-H bond and not the C-OH bond. Thank you. Let me know if my question still doesn't make sense and I will try and re-word it.

It's a bit tricky because there's probably a reagent that's missing. You might add DMAP or triethylamine to this reaction which would further activate the acyl chloride. But the alcohol oxygen is the most nucleophilic moiety in this reaction. The most electrophilic moiety is the carbonyl group, especially since it's activated by the electron withdrawing chlorine.
Thus we would predict that a standard acylation (a carbonyl reaction) will take place where we add a nucleophile to the carbon of a carbonyl group. After which, reformation of the carbonyl group will result in kicking out a chloride anion. Finally, a very fast proton transfer will take place that breaks the O-H bond you are referring to. Notice that this is really an after thought and not really the key point of this reaction.
Take a look at the mechanism for the reaction in your textbook (or here).


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How does DMSO act as a carrier for transdermal transport of large molecules?
DMSO.org claims the following:

"[DMSO] is more successful ferrying some drugs, such as morphine
  sulfate, penicillin, steroids, and cortisone, than others, such as
  insulin."

All of these drugs that it is said to carry are much larger than a single DMSO molecule.
So what is the general mechanism by which DMSO can bring these relatively high-molecular weight compounds into your body transdermally, where they would not do so otherwise?  Does it simply “wrap them up” in a large complex, or does it possibly just complex the portions of the other compound that prevent it from crossing transdermally?  What is it about insulin for example that prevents it from being carried through the skin?
A quick internet search will give pages of documents basically saying “this is what it does”  but I’ve never found a remotely technical explanation of how this works, particularly for large molecules.  That’s what I’m asking here.
NOTE 1: I have never used DMSO medicinally and currently have no intentions of doing so.
NOTE 2: I hope this is not considered a duplicate of this question as the answer is specific to the transport of small molecules.

Quote taken from the review (1) (and which is also duplicated in Jacob's 2015 "canonical DMSO" book (2, p. 16)):

There is some evidence to suggest that DMSO can increase diffusion
  through the
  stratum corneum by disruption of the barrier function. This probably occurs through aprotic interactions with intercellular lipids and may
  also include reversible distortion of lipid head groups that produce a
  more permeable packing arrangement.  DMSO may also play a role in
  partitioning as well by forming solvent microenvironments within the
  tissue that can effectively extract solute from vehicle. Finally, DMSO
  can have a profound solubilizing effect on less soluble agents in a
  variety of vehicles, increasing penetration simply by delivering a
  higher concentration to the membrane barrier.

alongside with extensive molecular modelling by Gurtovenko in 2007 (3) are both suggesting quick pore formation and expansion alongside with the thinning of cellular membranes:

The reported membrane thinning and, in particular, pore formation
  provide a credible molecular-level explanation of how DMSO promotes
  permeation of molecules, particularly those of hydrophilic nature,
  through lipid membranes as observed experimentally.

From what I understand, DMSO paves the way for bigger molecules, such as steroids, making inner skin layers and underlying tissue reversibly accessible, sometimes even on nanoscale level. For the precise mechanism I think it is worth reading through the entire paper (3).

(1) Capriotti, K.; Capriotti, J. A. The Journal of clinical and aesthetic dermatology 2012, 5 (9), 24.
(2) Jacob, S. W.; Jack, C. Dimethyl sulfoxide (DMSO) in trauma and disease; CRC Press, 2015.
(3) Gurtovenko, A. A.; Anwar, J. The Journal of Physical Chemistry B 2007, 111 (35), 10453–10460. DOI 10.1021/jp073113e

There's a lot of scientific jargon here that is inaccessible for the typical English speaker. How is that helpful? When you boil it down key phrases are "aprotic interactions with ...lipids, "forming solvent microenvironments", "profound solubilizing effect on less soluble agents", "membrane thinning", "pore formation". My sense is that so many possibilities means ideas being thrown about without people really knowing the answer. If so, why use jargon to cover that reality? In laymans terms DMSO is acting as a solvent for the molecules making up the skin, disrupting their function. Am I right?

@TonyM If you didn't notice, these are citations from papers, you think authors "cover reality"? Andselisk said it's "making inner skin layers and underlying tissue reversibly accessible" which is quite clear. If you don't understand the jargon then learn more about chemistry and refrain from such comments in the future.

@Mithoron Although I've a lot to learn, I do have a PhD in chemistry have published in and been a reviewer for top journals. Nevertheless, I did not present myself as an expert but instead asked, "Am I right?"  When I wrote "cover that reality" I was making reference to the posts here describing the articles rather than the articles themselves. The authors of the original papers should report their observations and explain them in as clear a way as possible; similarly, those trying to explain what they wrote should be also be as clear as possible. Why ask me to refrain from saying that?


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How can hypochlorous acid produce the hydroxide ion?
Hypochlorous acid ($\ce{HOCl})$ is an acid so it should produce $\ce{H+}$. However in reactions with alkenes, it produces the hydroxide ion like bases do. How is this possible? What is the mechanism?

Come to think of it, water gives $\ce{H+}$ like acids do. But at the same time it gives $\ce{OH-}$.

If you think about it, alkenes react with $\ce{Br-Br}$ to form a bromonium ion, and $\ce{Br-}$, right? Now, if you hypothetically replaced one of the $\ce{Br}$ with $\ce{OH}$, you'd get a bromonium ion and $\ce{OH-}$. $\ce{Cl-OH}$ is not all that different. You get a chloronium ion and $\ce{OH-}$.

Not a cation with $\ce{OH}$ plus a chloride ion?  How do we know which end is positively polarized?

you've made a nice comparison orthocresol but I wanted to know about the mechanism of the reaction

Oscar Lanzi it is HOCl gives Cl^+ and OH^-


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Does the frequency of incident light change in Beer-Lambert Law?
The Beer-Lambert law is meant to describe the interaction between a solute and incident light.
But, does the frequency of the light here change due to the interaction with molecules?
I think there is a change in the intensity of light, not in the frequency. But this seems counter-intuitive.
Is there no change, or is it so small that it's just neglected?

Mostly, it doesn't change at all. Some tiny portion does (see Raman scattering).

In absorption the photon is destroyed and it energy taken up by the molecule, so thats that. Any photons not absorbed thus lead to a reduction in the intensity of light transmitted. As @Ivan Neretin writes, in Raman Scattering there is a change in the photon's frequency, but in this case no photon is absorbed just scattered, i.e. an excited state is not produced.

But what happens to the absorbed energy of the molecules?

All the absorbed energy is returned. Some fraction of this energy is re-emitted as fluorescence and/or  phosphorescence. Another fraction may cause a bond to be broken, but if not any remaining ends up as heat. In femtosecond time scale experiments it is possible to see the solvent warm up and then cool as the energy diffuses away from the molecule.

Related: http://chemistry.stackexchange.com/questions/809

The Beer-Lambert law was never intended to apply to complicated processes like inelastic scattering. The law is best used when you can just treat the problem as a decrease in the number of photons, and don't have to start thinking about altering or re-radiating those photons.


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metal oxide analysis
I am looking in to analytical techniques for a future project. I may have a mixture of metals, i.e. an alloy, which I am thinking XRF spectroscopy would be the best quantitative technique. I could even purchase a handheld XRF analyser.
However, I believe that the XRF technology does not detect light elements such as oxygen. Therefore, if I have a mixture of metal oxides in a sample; what is the best quantitative analytical technique for me to identify each oxide?
I have had a look at XPS but I am confused about how I can work out which oxide I have in my sample. For example, if my sample contains Lead oxide and Copper oxide. How do I know whether it is Cu(II) oxide or Cu(I) oxide?

XRF is good for metals but samples must be "homogeneous," at least to a reasonable degree. X-ray Diffraction (XRD) can identify specific oxides, but it isn't a good quantitative technique. I've not done any XRF work since the a handheld XRF analyzers became commercially available. However I double they have the resolution to analyze a Cr/Fe/Ni alloy quantitatively. If you're willing to supply more information on the matrix perhaps  we can provide more help.

If the oxides are separate grains or compounds (a mixture, not a solid solution), you can find this with SEM microprobe (a geology technique).  For tricky new mixtures, may be better than powder x-ray diffraction (find small amounts of new compound, than resolve stoichiometry and do XRD on phase pure samples.  It will not give you the oxidation level though.
You could just do visual examination.  $\ce{PbO}$ is yellow and $\ce{PbO2}$ is black-brown.  $\ce{CuO}$ is black and $\ce{Cu2O}$ is red.  You can use a metallography microscope (high powered optical) to look at individual grains.  [May also be able to see the crystal habit.]
In general, people working with oxides know the expected oxidation state of the materials they have processed given the conditions.  E.g. cerium will always be $+4$, not $+2$ under atmospheric ceramic processing conditions.  Not always possible, but often the case.
If you could narrow down a little more what you are trying to do, might be easier to answer.  What substances, processing conditions and what info do you need on the end result (and why, really)?
ICP or AA are techniques helpful for trace metal analysis but you won't get oxidation state.
sorry, I can't give a more precise answer, but really your question is very broad.
If you can't discuss your project details on the net, consider to talk to the director of a research university chemistry department shared analysis lab.  I really don't think there is a single answer to your question and it will depend more on what you are looking at and how precise an answer you need, which technique is best.

Welcome to chemistry.SE! Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.

I changed "powder xrays" to "powder x-ray diffraction", hopefully it's OK with you. The first is rather slangy and could be confusing for non-specialists.


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Is the absorbent substance in nappies biodegradable?
I am looking to use the absorbent material found in nappies to improve the water retention of the soil in my garden.
To this end, I have two questions that I have not found definitive answers too.

Are disposable nappies made with Sodium or Potassium Polyacrylate?
Is Sodium/Potassium Polyacrylate biodegradable?

This class of polymer is also referred to as superabsorbent polymer(SAP).

Disposable diapers/nappies are likely to be the sodium form. Sodium forms are typically cheaper. Further, the potassium form of the salt is not as soft and would be less suitable for cloth-like wearables.
Polyacrylate SAP does not biodegrade in soil very fast.
For the main chain polymer:

...if at all, 0.12-0.24 % per 6 months.

Refs:
1: A manufacturer of agricultural/hygiene polyacrylate SAP
2: Wilske, B.; Bai, M.; Lindenstruth, B.; Bach, M.; Rezaie, Z.; Frede, H.-G.; Breuer, L. Biodegradability of a polyacrylate superabsorbent in agricultural soil. Environmental Science and Pollution Research 2013, 21 (16), 9453–9460.Springer
Side note: All diapers with an absorbent layer are SAP, either polyacrylate or polysaccharide (rarely). Polysaccharide is the more expensive / less effective 'green' alternative, and if your product contains this, it will tout it on the label.
 (link limit reached)
babygearlab.com/a/11113/What-Is-Inside-Those-Disposable-Diapers
https://blog.bolandbol.com/product-reviews/diapers-review/
http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.608.22&rep=rep1&type=pdf

This is little more than a link-only answer. Since links to the www could theoretically decay quickly, this site’s policy is to include (i.e. quote, using >  at the beginning of lines) the essential information here but leave the link as a citation. Please [edit] to do so or this posting may be eligible for removal.

Sorry about that. New to the process.

Should be sorted now.

To keep it simple: there are as many different polymers as suppliers of the superabsorbent polymer used in diapers.https://en.wikipedia.org/wiki/Superabsorbent_polymer#Current_synthesis.
So, answer to #1. Some are, some aren't. As far as #2, define what you mean by "biodegradable". Take a diaper, put it into a composting bin, and turn it over every week for a million years, and I'd expect little to remain. It is likely based on your question that you are unaware that some diapers are marketed to be "biodegradable" and some aren't. The reality is that the term has no standard meaning - it's marketing hype. However, it's also true that some polymers will be inherently much more prone to decomposition than others and that some of these are incorporated into some diapers.
Your best bet is to call (or visit) the customer service of the specific diaper manufacturer you are interested about. Note that the nih link provided in a different answer doesn't mention any use of the sodium salt for any major brand.  Pampers (P&G), I think, dominates the market and sells 9 different disposable diaper brands. I didn't see any mention on their site about biodegradability, which was an important controversy some generations back. Since even "highly biodegradable" materials can take centuries to decompose in landfills, the question is a bit meaningless. (If you want biodegradable, use toilet paper (not seriously, since TP is designed to fall apart rapidly, too rapidly for practical use in a wearable product).)


text:
How exactly does the electrolysis of water work?
When we pass electricity through water it breaks into hydrogen and oxygen gas.
$$\ce{2H2O -> 2H2 + O2}$$
What I couldn't understand is how the electrical energy is converted to breaking the bonds of water molecule. I've searched a lot but couldn't find any satisfactory answers on how the energy conversion happens.
Few thoughts :

The fast moving electrons from the battery hit the water molecules breaking them into hydrogen and oxygen ions.
I think above idea is incorrect as pure water is an insulator and blocks passage of electrons. Then it might have to do with the electric field that exists through out the water between the anode and cathode. Does this electric field somehow transfer energy to the water molecules and split them ?
Above idea makes sense to me as we can think of each $\ce{H2O}$ molecule as a dipole $\ce{2H+-O^2-}$, so it can interact with electric field.
I feel all above ideas are silly and there is some sophisticated way to reason this which I don't know yet.

Electrolysis of pure water is very difficult, but adding only a small amount of ions make the process easily achieved. In most places, there are enough minerals in the water  that the ionic strength or conductivity of the water is  great enough for electrolysis to effectively occur without needing to add additional ions to the water
Water dissociates into H+ ions and $\ce{OH–}$ ions; the $\ce{H+}$ ions are attracted to the negative electrode (the cathode) and are converted (reduced) to a hydrogen atom ($\ce{H}$) (i.e. $\ce{e- + H+ -> H}$). This is highly unstable and immediately reacts with another hydrogen atom to produce $\ce{H2}$, molecular hydrogen gas.
At the other electrode (the anode), oxidation occurs.  The $\ce{OH-}$ ions are attracted to the positive electrode where they are oxidized to form oxygen gas ($\ce{O2}$) and hydrogen ions ($\ce{H+}$).  However, if chloride is present, it will oxidize (instead of the $\ce{OH–}$ ions) and form chlorine gas, which will then react with the water to form hypochlorous acid.
The cathode reaction is: $\ce{2H+ + 2e- (cathode) →H2 (g)}$
The  anode  reaction  is:  $\ce{2OH– → 4e-(anode) +O2 + 2H+}$
The overall reaction is: $\ce{2H2O → 2H2 (g) + O2 (g)}$

Thank you. I know the reactions and gases that form at each terminal. But I still don't understand how the electrical energy is converted into breaking the bonds. For example, in an incandescent light bulb, the fast moving electrons hit the atoms in filament and make them vibrate. (converting electrical energy into vibrations of atoms). I'm wondering if  something like this happens in electrolysis too..

The electical energy is used to ionise the water into H+ and OH-  and under the electric field they move towards cathode and anode respectively and react there.                    For eg : H+ ion moves towards cathode under electric field and reacts with electrons present on the negatively charged cathode to form hydrogen gas.                    Also, for pure water a huge potential is required to ionise water,so some salts are added to promote flow of current and ionise the water molecules

I think I can help with this question.  Electrolysis is different than the incandenscence effect of light bulbs.
Water electrolysis requires an added chemical potential (a battery, say).  This causes one electrode to be more prone to having excess electrons and behave as a cathode, reducing water to H2:
M + 2e- --> M(2-)
M(2-) + 2 H2O --> MH2 + 2 OH-
MH2 --> M + H2
The metal M acts only as a catalyst (could be Pt for instance).  Note that M here is not a single atom, but a surface of many atoms, and MH2 represents a state in which there are two more-or-less covalent M-H bonds right beside each other.  A quick electron bond rearrangement is all that is needed in the last step, e.g. imagine a box shape for HMMH and merely moving the bonds around: H-M M-H making M-M and H-H. No "hitting" or vibration/heat effect; just chemical potential at work.
The other electrode is, due to the battery, more prone to having electrons drawn away from it, behaving like an anode, oxidizing water to O2:
M --> M(4+) + 4e-
M(4+) + 2 H2O --> MO2 + 4 H+
MO2 --> M + O2
The quick electron bond rearrangement here can be imagined as O=M M=O making O=O and M=M (though O=O has a more complicated electronic description than a simple double bond).
--Allan in Regina


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Are there any dangers while handling cobalt dichloride?
I use a desiccant named Drierite to remove water from air in experiments. It is composed of $98\%\ \ce{CaSO4}$ and $2\%\ \ce{CoCl2}$. Every once in a while the Drierite gets saturated with water vapor and it must be dried out so it can be used again.
According to the manufacturer you put the chemical in a lab oven, set to $230\ ^\circ\mathrm{C}$ until it turns blue. The cobalt dichloride $\ce{CoCl2}$ is an indicator that turns blue when dry and pink/purple when saturated. One of my coworkers noted that cobalt dichloride is a carcinogen, and now he wants me to stop heating it in the lab oven because of the vapor it is giving off that we are breathing.

Could cobalt dichloride give off anything that would cause a person harm espcially when heated to $230\ ^\circ\mathrm{C}$?

I have looked at the safety data sheet but it is not very helpful. It does say that cobalt dicloride is a carcinogen but not how or when it is transferred.

Related: How dangerous is cobalt(II) chloride used in silica gel balls?

@Loong I read this post and safety precautions are in place to keep the dust under control. My and my coworkers concern is if there is any airborne danger because the lab oven vents directly into the lab.


text:
Which is anode and which is cathode?
A maybe (hopefully) simple question about the denotations of "anode" and "cathode". The below image is a schematic of a polymer solar cell (Source (WBM)). (The figure text is quoted as well for the sake of context).
My question is only about the naming of electrodes. Isn't it correct that

the cathode is the reducing electrode, which supplies electrons, while
the anode is the oxidizing electrode, which accepts electrons?

Why would electrons (black spheres) move towards the negatively charged cathode, while positive holes (white spheres) move towards the positively charged electron-absorbing anode? Wouldn't the opposite make sense?
Is it a mistake on the drawing or should I re-think my understanding of the terms "anode" and "cathode"?

Figure 1. The working principle of the solar cell. Light enters the cell through the transparent anode, and is absorbed in the bulk heterojunction layer through generation of excitons (1). The excitons diffuse in the bulk heterojunction until they either recombine or reach a donor-acceptor interface, where they separate into electrons (black) and holes (white) (2). The electrons and holes will then move to the respective anode and cathode, through the donor and acceptor material phase (3).

Edit
If the figure is correct and I am the one being confused, then there are several other cases that seem to teach false. E.g. the following "PANIC"-remembering-rule to remember, which is cathode and which is anode (from this source, slide 8):

"*Negative is cathode", it says, and we are back to my question in the top: why do the electrons (black spheres) move towards the negative cathode in that schematic?
Many more diagrams agreeing with this latter one are found from a simple google search.

In traditional textbooks, the flow of electricity is shown going the opposite way (than it actually functions) because the guy who came up with the concept couldn't tell which was positive and which was negative- so he guessed. Out of respect for his contributions for science (and the fact that the actual direction has little effect on the math) the scientific community kept it that way, and many people still teach it backwards.

Negative and positive are not linked to anode and cathode - it depends whether the reaction goes toward equilibrium (voltaic cell) or is forced by an external potential to go against it (electrolytic cell). For details, see my answer.

Because there is no oxidation or reduction in a photovoltaic cell, it depends on conventions which are different depending on the type of device: https://sc.edu/study/colleges_schools/chemistry_and_biochemistry/internal/research_groups/andrew_b_greytak/docs/which_is_the_anode_abg.pdf

current flows in the opposite direction of "bump" electrons (if such a thing as an 'electron' even exists! ref: Einstein, Tesla, et al), seems to move. So, if it seems backwards, then think of it like the speed of wind passing your car. What is moving? Your car or the wind? Or is it both?

There is no completed electronic circuit in an electrochemical cell
In an electrochemical cell, the anode is the source of electrons to the external circuit and the cathode is the sink. The circuit of charge transport gets completed by ions traveling inside the cell. A solar cell is different from an electrochemical cell in that their is no net chemical reaction. In the solar cell, electrons flow in a closed circuit - round and round in the external circuit and through the device.
Designation of anode and cathode
So labeling the anode and the cathode relies on an analogy between a voltaic cell and a photovoltaic cell as a source of electrical work. It makes sense to use the direction of electron flow in the external circuit to define anode and cathode (electrons flow from anode to cathode in the external circuit). In the voltaic cell, there is no electron flow inside the cell (there is ion flow instead to balance charges). In the photovoltaic cell, electrons flow from junction to anode and holes flow from junction to cathode (or you could say electrons flow from cathode to junction).
Unfortunately, anode and cathode are named using different conventions depending on the type of device, see this overview (and beware that the current I sometimes goes in the same direction as the electrons and sometimes not, again depending on conventions).
Negative and positive electrode
The (+) and (-) designation is confusing even just for electrochemical cells. While the designation of anode and cathode is consistent for voltaic and electrolytic cells (i.e. using and charging a battery), the designation of (+) and (-) switches, so it is uncoupled from the direction the electrons flow through the external wire.
Direction of electron flow
For the photovoltaic cell, maybe this picture helps: Before light hits the cell, anode and cathode are neither negative nor positive. Once light hits the cell, the anode becomes negative because electrons are moving toward it from the junction, and the cathode becomes positive because electrons are jumping from it into holes coming from the junction. If you then attach an external consumer of electrical work, you can predict the direction of electron flow through the external circuit.

Regardless of the polarity, the electrode where oxidation takes place is the called the anode and therefore reduction must take place at the cathode. The electron flow from the anode to cathode is as shown in your top picture.
By way of an example, in an electrochemical cell, suppose that two beakers are connected by a salt bridge. In one beaker is a strip of zinc metal immersed in a $\ce{Zn(NO3)2}$ solution and in the other a strip of silver in $\ce{AgNO3}$ solution. The two metals are then connected by a wire and a current will flow. (The salt bridge supplies a return path so that the solutions remain electrically neutral). The redox of the Zn electrode is $-0.763$ V and that of the Ag $+0.799$ V. This means that electrons will flow from the zinc to the silver electrode. The zinc is oxidised to $\ce{Zn^{2+}}$ and the electrons are released into the metal and flow to the silver electrode through the wire. The zinc electrode is the anode and the silver the cathode.

The anode is the electrode, where substances are losing electrons and are oxidated.
The cathode is the electrode, where substances are gaining electrons and are reduced.
The tricky part for the memorising is, anodes and cathodes flip the position, when the current is reversed, depending on if

the cell is in the mode of electrolysis
the cell is in the mode of providing current.

The convention for the primary and secondary (rechargable) cells is to take the latter case as the power source.
My mnemotechnic trick is to remember, that
At anode is occuring anabasis (*) of electrons, upwards from the electrode, i.e. oxidation.
It means the anode is

the  positive  electrode at electrolysis,(=there is attached the + wire from the external power source, taking  electrons away. It does not  necessarily mean it has more positive potential than the cathode),
the negative electrode in power cell ( the minus electrode of the cell,  providing electrons).

At the cathode is occuring katabasis of electrons, downwards to the electrode, i.e. reduction.
It means the cathode is

the negative electrode at electrolysis, accepting electrons from the external power source.
the positive electrode for power cells. drawing electrons from the circuit.

(*) Anabasis by Xenophon, some 370 BC, "The journey upwards", about dramatic return of Greek mercenaries from Persia to Greece.

You are confused about the definition of reduction and oxidation.

Reduction is the gain of electrons.
Oxidation is the loss of electrons.

Which means, the cathode is gaining electrons, while the anode is losing electrons. The illustration is absolutely correct.

Thank you for the answer. My description of which "gains" and "loses" might not be clear. The actual question is, if the cathode indeed is the negative electrode? So the question is still: Why would electrons move towards the cathode, which is negative? I have added an extra diagram of simple electrolysis to explain this point a bit clearer - (negative) anions move towards the (positive) anode in order to deliver the excess electron. (Positive) cations move towards the (negative) cathode to receive electrons; that is what I meant by the cathode "supplying electrons".

If this is a galvanic cell, then cathode is positive electrode, hence electrons will move toward cathode. If this is a electrolyser, then yeah, cathode, is negative. But this "negative" is different from the negative in cell case. This "negative" means, the negative end of electricity supply is connected to cathode, and cathode receiving electron from that.

There is no reduction or oxidation in a photovoltaic cell. The material that loses electrons in one direction gains electrons coming from the other side.

Cathode is the electrode where Reduction takes place and Anode is the electrode where oxidation takes place.
I use the mnemonic  OIL RIG
Oxidation -> loss of electrons (or increase in oxidation state)
Reduction - > Gain of electrons (or decrease in oxidation state)

a better one : LEO roars GER : loss of electrons is oxidation and gain of electrons is reduction.

The above picture depicts the inside of a power device i.e. the solar cell. That can be considered as a battery. In the below figure the positive charges go through the metal and reach the cathode of the battery. Meanwhile internally in the battery the negative charges move towards the cathode to cancel out the positive charges coming from the metal electrodes.
In summary, externally positive charges move to the cathode and internally negative charges move toward the cathode. Therefore all the pictures depicted are correct.

The Cathode is the positive electrode; the anode is the negative electrode.
During discharge, positive ions flow from anode to cathode.  This makes the cathode positively charged.  Free electrons at the anode flow externally through the load to the cathode.   During charge, an external voltage source has its positive electrode connected to the cathode and its negative electrode connected to the anode.  The negative electrode of the external voltage source supplies electrons to the anode.  Excess positive ions at the cathode are repelled to the anode but move more slowly than electrons move through wires.

You write "The Cathode is the positive electrode; the anode is the negative electrode." <-- that's only true in a galvanic cell / battery.  In an electrolytic cell, the cathode is negative and the anode is positive.  So.. A Galvanic (cell's) cathode is positive, a galvanic anode is negative.  An electrolytic (cell's) cathode is negative, an electrolytic anode is positive.


text:
Why do alkenes react with oxidising agents?
My book says :

The high electron density associated with the double bond means that
  alkenes are attacked by oxidising agents.

And I dont understand how the high electron density in the double bond causes that.

As per my knowledge, oxidation is loss of electron(s) by any species and an oxidising agent is an acceptor of electron(s). So, high electron density on alkenes make them likely to be attacked by oxidising agents.

What you are saying surely does make sense to me but the full sentence is "The high electron density associated with the double bond means that the alkenes are attacked by both electrophiles and oxidising agents." This implies that electrophiles and oxidising agents are two different things but your definition of an oxidising agent applies to both electrophiles and oxidising agents so if they are the same thing why would the book mention them in that way and make them appear as two different things ?

@Sam19KY First of all you didn't mentioned about electrophiles. And by the way, any electron deficient species always searches for electron rich species.

yeah i didnt mention the electrophiles bcz i understood that part ( i know that positive species are attracted to electrons) but now i had to bring it up because of what i said in the 1st comment


text:
Covalency in bonding across the periodic table
Bonding between main group elements is obviously almost entirely covalent and it is often easy to see these covalent bonds as 2c-2e bonds localised between two atoms. The bonding in borane, boron clusters and hypervalent compounds like SF6 introduce a more delocalised view of covalency as the bonding can be rationalised in an MO diagram. Nevertheless, covalent bonding can still be thought of as electrons occupying a molecular orbital which forms when orbitals from two or more atoms overlap (this orbital mixing requires good energy, size and symmetry match).
However, I find it hard to rationalise covalency and trends in it when considering other compounds and I was wondering if there is a good, simple(ish) way of rationalising (even if not predicting) the presence of covalency in other compounds.
People say if you have a polarising cation and a polarisable anion like in AgCl for example, you get covalency. I suppose this also makes sense because that's essentially saying that the atomic orbitals need to be of a similar energy to mix. However I am going to run through three examples that I struggle to rationalise or have questions regarding it.
MeLi
Li is clearly quite polarising (high charge density) and must therefore polarise C and there is some overlap of orbitals = COVALENCY. However, it is tetrameric in solution. Now compare it to MeMgBr; this is monomeric in solution (and in organic chemistry is softer) indicating more covalency - makes sense because of the increased polarisability of Mg relative to Li. However, is the tetramer vs monomer as a result of the difference in covalency, if so why? Is it fair to say that moving towards a more monomeric molecular structure is indicative of covalency (eg BeCl2 - chain vs MgCl2 - layered lattice vs CaCl2 - rutile)?
Suboxide
Group 1 suboxides are possible (eg Rb9O2) when Rb is exposed to limited oxygen. This forms a cluster with delocalised electrons. It seems bizarre to me because this is one of the least polarising cations going - there are no suboxides higher up the group despite it being easier to oxidise these metals lower down the group. The only vaguely analogous example is the FeS clusters seen in biology with highly delocalised electrons but the electronegativities make more sense here.
Zn
Zn is a soft Lewis acid. Why? It easily loses two electrons and then is left with a filled 3d shell. Surely that would mean it is disinclined to covalently bond and should be relatively similar to Ca?

Surely you mean the zinc ion, not zinc metal right?


text:
Reaction of ethylene glycol with PI₃ (P + I₂)
We know that $\ce{2P + 3I2 -> 2PI3}$
Ethylene glycol reacts with $\ce{PCl3}$ to give ethylene dichloride, but why does it gives ethylene when it reacts with $\ce{P + I2}$?
I don't really know why it happens so as it is quite contradictory because in case of primary alcohols, corresponding iodoalkane is formed.

User @tato_mazo already pointed out the reason that due to two iodine atoms being adjacent to each other, ethylene diiodide is sterically crowded and decomposes to produce ethylene and iodine. My answer basically points out the actual reaction involved and some properties of ethylene diiodide.
$$\ce{OHCH2CH2OH + PI3 -> [CH2I-ICH2] -> CH2=CH2 + I2}$$
Ethylene diiodide is an unstable compound which is made by mixing ethene and iodide (damp) under light or slight heating crystallizing into solid, colorless needle like structure (m.p. 75 °C). It again decomposes into corresponding reactants on stronger heating and occasionally at room temperature, though slow. It is only be sublimed or preserved colorless under ethene gas.
References

Ahluwalia, V. K. Textbook Of Organic Chemistry; Ane Books Pvt. Ltd, 2010; Vol. 2. ISBN 978-93-80156-02-6.
Strecker, A. Short Text Book of Organic Chemistry, 1881.

Two Iodine groups adjacent to each other. It is sort of sterically crowded and hence they leave producing ethylene.

Sterics is not the issue. Look here.


text:
Why does tetrachlorocuprate(II) form in aqueous solution even though water should be the stronger ligand?
In aqueous solution, $\ce{Cu^{2+}}$ forms the $\ce{[Cu(H2O)6]^2+}$ complex. Given that water is a stronger ligand than $\ce{Cl-}$, though, why does why does the  $\ce{[CuCl4]^2-}$ complex form upon addition of chloride ions?

Upon addition of more chloride ions to what exactly? o_o

To that same aqueous solution in which [Cu(H2O)6]2+ ions are present.

[Water ligands] being stronger than $\ce{Cl-}$ ligands will form a $\ce{[Cu(H2O)6]^2+}$ complex.

This is incorrect. Water ligands may be stronger field ligands in the spectrochemical series but that does not in any way affect the strength of the ligand–metal bond or the kinetics and thermodynamics of certain complexes, their formation and their breakdown.
Which ligand forms bonds of which strength to which metal is only very generally determinable; and in general you would always denote binding affinities in a $1:1$ manner: ligand $\ce{X}$ forms strong bonds to metal $\ce{Y}$. So sulphide ions will form strong dative bonds to mercury(II) ions but not-so-strong bonds to titanium(IV) ions. Likewise, palladium(II) if presented with both phosphane and amino ligands will preferentially bind to phosphanes while copper(I) will preferentially bind to amino ligands.
The thermodynamic energy differences between different complexes are often relatively small and typically the kinetic barrier for ligand exchange is low, too. Thus, in many cases which complex is formed depends strongly on the concentrations of individual ions. The mere fact that complexes without aqua ligands can and are formed shows that water is not that good a ligand after all. In fact, due to the positive charge of the central metal having an anionic ligand bind is often preferred due to simple electrostatic interactions.
To answer the actual question you posed: $\ce{[CuCl4]^2-}$ will form because it is the predominant species under a given chloride ion concentration. This can be backrationalised by looking at formation constants $K_1$ to $K_4$. However, aside from quantum chemical calculations there is no way to properly rationalise it a priori.


text:
Organic hydrocarbons
Why new double bond or triple bond creates to the center always? When pentan-2-ol reacts with $\ce{Al2O3}$ ∆ the double bond creates to the 2nd carbon. Can't it create on the 1st carbon?

https://en.wikipedia.org/wiki/Zaitsev's_rule

It can and will form a double bond to the 1st carbon, however it will much more likely form one to the second carbon.
You will get a mixture of both 1-pentene and both cis- and trans-2-pentene.  The major component will be the trans-2-pentene because that is the most stable compound.  Remember, more substituents on a double bond mean the bond is more stabilized for a couple reasons.  Steric interferences mean that trans- is more stable as well.
Since the product is more stable, the transition state to form that product is more stable and therefore easiest to form, generally.

You can eliminate pentan-2-ol to both pent-1-ene and pent-2-ene (and to both stereoisomers of the latter).
To a certain degree, the reagents you use determine which product you get. If you turn the alcohol into a leaving group and then add a strong, sterically demanding base, the pent-1-ene content will increase. If you add a small, mild base, you will observe more pent-2-ene. Careful tuning of reaction conditions often allows for these differences in reactivity.


text:
Equivalent weight in case of disproportionation reaction
We all know that
$\text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{n-factor}}$
However consider a reaction like:
$\ce{Br2 + OH- -> Br- + BrO3- + H2O}$
where oxidation numbers of $\ce{Br}$ are $0, -1, +5$ respectively.

What should here be the equivalent weight of $\ce{Br2}$?

PS: I understand that I should also show what I've attempted. But, this is just a question that came up into my head
and I've made it as descriptive as possible. I don't really see anything else that I must add to make this question complete.
Thank you!
UPDATE: I just came across a formula for calculating the n-factor for a dispropportionation reaction $n_f=\frac{n_1\cdot n_2}{n_1+n_2}$ (sources, 1 and 2), for $n_1$ the magnitude of n-factor in oxidation reaction and $n_2$ the magnitude of n-factor in reduction reaction. Can anyone prove how to arrive at this formula?

@IvanNeretin Thanks! But isn't there only one reaction here that we're seeing? Or is it the fact that you're considering oxidation and reduction reactions separately, as if they're the basic parts of the given reaction?

True, it's one reaction, so what? Like I said, either use the idea of equivalent the two-pronged way, or consider it unusable in this case.

@Gaurang Tandon... To your update...The result can be derived by writing the two half reactions, expressing their equivalent weight as M/n1 and M/n2 and equating their sum with M/N, where N is the resultant n factor.  Thus  $N=\frac{n_1\cdot n_2}{n_1+n_2}$ .........Yet I do not understand why n1 and n2 aren't simply added and divided by two.......because n factor is the no of electrons which one mole of the compound reacts with or supplies in a redox reaction.

Equivalent mass in disproportionation = equivalent mass in oxidation half+equivalent mass in reduction half

The $n$-factor of a compound in a reaction gives the number of moles of electron lost or gained by it (during oxidation and reduction respectively) per mole of the given compound. In a disproportionation reaction, the compound undergoing both the oxidation and reduction is the same.
It is important to understand that the $n$-factor depends not just on the change in oxidation state of the element but also the number of moles of the element undergoing that change.
This is the reason why we cannot simply take the average of the two $n$-factors as the $n$-factor of the disproportionate. It is necessary to also consider how much of the reactant is getting oxidised and how much of it is getting reduced, something along the lines of a weighted mean.
It is quite simple to derive the expression you have found for the $n$-factor of a disproportionated compound. Consider the following general reaction:
$$\ce{nA -> xB + $(n - x)$C}$$
Where $x$ moles of $\ce{A}$ (for which $n$-factor is to be calculated) oxidises to $\ce{B}$ and the rest of it reduces to give $\ce{C}$. Let the $n$-factors for the oxidation and reduction respectively be $n_1$ and $n_2$.  Which means $n_1 x$ moles of electrons go into oxidation, and $(n - x)n_2$ moles into reduction.
We have 2 equations: one by conservation of electrons,
$$\begin{align*}
n_1 x &= n_2(n - x) \\[7pt]
\implies x &= \frac{n_2\cdot n}{n_1 + n_2}
\end{align*}$$
and the other from the definition of $n$-factor:
$$\begin{align*}
n_f &= \frac{\text{moles of electrons transferred}}{\text{moles of reactant}} \\
&= \frac{n_1 x}{n} \\
&= \frac{n_1 n_2}{n_1 + n_2}
\end{align*}$$
An interesting observation here is that upon simplifying the obtained equation, you get the formula for the equivalent resistance of resistors connected parallelly in an electric circuit. The correspondence between them is intuitive.
$$ \frac{1}{n_f} = \frac{1}{n_1} + \frac{1}{n_2} $$

It looks like you are partially answering the question, and at the same attempting to ask a new question. Stack Exchange sites are sticking with strict Q&A format meaning a question should receive a short, but complete and a well-defined answer without discussion or explicitly raising new questions. If you have a new question, please ask a new question instead of attaching one to your answer.

You may consider this formalism inapplicable here altogether. Alternatively, you may assume you have two different kinds of bromine (one that gets oxidized and another that gets reduced), each with its own factor.
True, in this way, the same bromine molecule would have two different equivalent weights in the same reaction. There's nothing strange or unusual about that. After all, the same molecule surely can have different equivalent weights in different reactions.


text:
Which of the following para-substituted benzyl alcohols reacts fastest with HBr?
I had a question in a recent test which asked me to pick the compound that would react fastest with $\ce{HBr}$ with the options being:

p-nitrobenzyl alcohol
p-chlorobenzyl alcohol
benzyl alcohol
p-methoxybenzyl alcohol

The answer was p-methoxybenzyl alcohol.
Now what I don't understand is that $\ce{HBr}$ would produce nucleophilic $\ce{Br-}$, which would be looking to attack a region of electron deficiency. In that case option (1) p-nitrobenzyl alcohol would the right answer because the nitro group is an electron withdrawing group, thus making the ring more electron deficient.
Could someone please explain as to why this logic fails?

Inductive effects will play much more of a role in electrophilic aromatic substitution. Here, positive charge on a benzylic position is being delocalized to the aromatic ring, and the question is asking which group will best stabilize this positive charge (via resonance):

Between $\ce{-H}$ (neutral), $\ce{-NO2}$ (which would destabilize the positive charge), $\ce{-Cl}$ (poor at stabilizing positive charge), and $\ce{-OCH3}$ (good at stabilizing positive charge), $\ce{-OCH3}$ is the clear winner.

So are you telling that a Br+ would attack in this case rather than a Br-. So what will the final product of this reaction be then?

No, $\ce{Br-}$ is present in solution and does attack the benzylic position. This is an $\mathrm{S_N1}$ reaction.

So what is attacking the reactant then?

$\ce{H+}$ from $\ce{HBr}$ attacks first, $\ce{H2O}$ departs, and is replaced by $\ce{Br-}$.

So, would the H+ replace the OH- from the CH2OH group to form a p-methoxy toluene?

It does not and cannot replace it. It activates the alcohol oxygen to leave by forming water as a leaving group, which departs, then bromide attacks the same position to form 1-bromomethyl-4-methoxybenzene.


text:
Defining the Turbomole named BP86 functional in Gaussian
According to Turbomole, the BP86 density functional has the following definition:
$$
\mathrm{Exchange = LDA + B88\\
Correlation = LDA(VWN(V)) + P86}
$$
What is the equivalent definition in Gaussian using IOp keywords?

From the Gaussian 09 manual:

User-Defined Models. Gaussian 09 can use any model of the general form:
      P2EXHF + P1(P4EXSlater + P3ΔExnon-local) + P6EClocal + P5ΔECnon-local
The only available local exchange method is Slater (S), which should be used when only local exchange is desired. Any combinable non-local exchange functional and combinable correlation functional may be used (as listed previously).
The values of the six parameters are specified with various non-standard options to the program:

IOp(3/76=mmmmmnnnnn) sets P1 to mmmmm/10000 and P2 to nnnnn/10000. P1 is usually set to either 1.0 or 0.0, depending on whether an exchange functional is desired or not, and any scaling is accomplished using P3 and P4.
IOp(3/77=mmmmmnnnnn) sets P3 to mmmmm/10000 and P4 to nnnnn/10000.
IOp(3/78=mmmmmnnnnn) sets P5 to mmmmm/10000 and P6 to nnnnn/10000.

For example, IOp(3/76=1000005000) sets P1 to 1.0 and P2 to 0.5. Note that all values must be expressed using five digits, adding any necessary leading zeros.
Here is a route section specifying the functional corresponding to the B3LYP keyword:
#P BLYP IOp(3/76=1000002000) IOp(3/77=0720008000) IOp(3/78=0810010000)
The output file displays the values that are in use:
IExCor=  402 DFT=T Ex=B+HF Corr=LYP ExCW=0 ScaHFX=  0.200000
ScaDFX=  0.800000  0.720000  1.000000  0.810000

where the value of ScaHFX is P2, and the sequence of values given for ScaDFX are P4, P3, P6 and P5.

For BP86 specified normally, this looks like:
IExCor=  404 DFT=T Ex=B Corr=P86 ExCW=0 ScaHFX=  0.000000
ScaDFX=  1.000000  1.000000  1.000000  1.000000

Those 3 IOps don't completely define the functional, however, IOp(3/74) is also needed. See the IExCor field, which corresponds to Becke 1988 exchange (400) combined with Perdew 81 + Perdew 86 correlation (4).

Hi,thanks . defination of BP86 functional in log file of Gaussian  is the same what you write. but i want to create the combination of Turbomole named BP86 functinal in gaussian with IOP keywords

I hate to be the guy that says I told zahra so. Consider this answer: http://chemistry.stackexchange.com/questions/39411/definition-of-the-b3lyp-functional-in-common-qc-programs/65113#65113 . This should clear up the confusion about the displayed notation in TM. Bottom line: TM and Gaussian are in agreement.

IOP(3/74=418) and ScaDFX=  1.000000  1.000000  1.000000  1.000000

The turbomole definition of BP86 (with VWN5 instead of VWN3) is accessible within Gaussian as BVP86. Note that you'd have to redefine Dispersion corrections for this functional.


text:
Simple Distillation of Unknowns
Can an identification experiment (using acetonitrile, isopropyl, and cyclohexane) using simple distillation and boiling point be improved by using a digital thermometer?
Can simple distillation be improved in other ways, other than a more accurate thermometer? That's all I've come up with.

In P-Chem lab I remember using a thermometer that I think was good to 0.01 C. The other odd thing is that you had to increase or decrease the Hg in the bulb by "breaking off" some of the Hg at the top of the thermometer. So the thermometer had a range of maybe 10 C. In the end it would read differences well but it was uncalibrated. So you had to then calibrate it to get absolute temperature readings. Oh the good old dinosaur days... https://en.wikipedia.org/wiki/Beckmann_thermometer

There are a zillion ways to tell these compounds apart. What can you use? Wet chemistry, NMR, IR, GC-mass spec, ect ect ect...

@MaxW I've used those methods before. But I'm trying to improve the simple distillation process itself. Experiment was pure liquid BP identification, miscibility test, then distillation to tell azeotrope and non-azeotrope mixtures apart. What makes simple distillation apparatus/process itself error-prone so that aspect can be fixed? I didn't see much room for instrument error other than calibration

It is really hard to get "absolutely" pure anything. So 99% may be ACS reagent grade  but the BP would be off by a degree or so.

@MaxW True. Is distillation an issue if mixture components are fairly close in bp? Fractional would be the way to go.

Is the 1% of "impurity" 1 compound or 5?

Impurity is likely iso-propanol as it was used in both mixtures @MaxW

Ok, back to the start on all of this....
If you were looking for pure compounds or azeotropes what I'd want is a setup where a fractional distillation column was used with a very precise thermometer and a balance measuring the weight of the distillate. You could then nicely plot temperature vs weight.
Assuming that you just want to detect the pure compounds and/or azeotropes and not collect them.
Part of the problem here is that it takes some time to get the whole fractional distillation column up to temperature. So the temperature of the first drops is suspect.
If you're using a heating mantle you might be willing to boil the pot dry. But you wouldn't want to do that with a flame. So you might lose a bit on the back end of the distillation too.
The fractional distillation column doesn't have an infinite number of plates so the transitions from one compound/azeotrope to the next will be a bit "fuzzy." So if you were trying to collect "pure" compound/azeotrope fractions you'd avoid collecting the distillate from the transitions in those "pure" compound/azeotrope fractions.
Does this help?
EDIT
I guess I'd like two thermometers. One over the pot and one at top of the fractional distillation column.
Some sort of computer controlled temperature setting for the pot would be nice too.

Realized fractional distillation couldn't be used in this case. It won't help with a single pure compound, and isn't useful with the  azeotrope. So simple distillation would have to be tweaked in some other way @MaxW

Correct - it isn't useful if the sample is "pure" azeotrope.

So for three compounds you could have 3 different pure compounds, 3 different binary azeotropes, and 1 ternary azeotrope.

Only cyclohexane and isopropanol form an azeotrope @MaxW   Any suggestions on distillation tweaks?

So for the three particular compounds you could have 3 different compounds, and 1 azeotrope as distillates. But could you get a mixture of acetonitrile and cyclohexane (I'd expect them to be miscibile) as your unknown?


text:
Oxidation States?
Metals in low Oxidation states are usually reducing agents. Is there any example of a metal in zero Oxidation state acting as an oxidizing agent.

As with pretty much every "is there ______?" in chemistry, the answer is yes, there is.
Of course, using the pure metal itself as an oxidising agent would be really tough. However, if you use certain tricks to stabilise the negative oxidation state, it is possible. For example, π-acceptor ligands such as $\ce{CO}$ withdraw electron density from the metal. So, $\ce{M[(CO)_n]^m-}$ complexes are stable for certain choices of $\ce{M}$, $n$, and $m$. For more details see: 18 Electron Rule For Determining the Stability of Transition Metal Complexes
Consider $\ce{[V(CO)6]}$, a 17-electron complex. It would be quite happy to accept another electron to become $[\ce{V(CO)6}]^-$. This reduction can be achieved with sodium metal. So, here vanadium(0) is acting as an oxidising agent; it is itself reduced to vanadium(−1).
These examples are not incredibly common, but it can happen.

If metal is in 0 oxidation state would get oxidized, then it would have to achieve a negative oxidation state since it would have to itself get reduced. Since it is a metal, it would have a really low reduction potential. If the reducing agent has a really high reduction potential, then I think it might be possible ( you will have to check the reduction potential values for that )


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What results from the reaction of 5-bromopentan-1-ol and NaOH?
In our chemistry class, we noted three molecules that result from the reaction:

pentane-(1,5)-diol

This one is simple to understand, simply a nucleophilic substitution between $\ce{Br^-}$ and $\ce{OH^-}$.

(1,5)-epoxypentane

This is the main result, resulting from the intramolecular esterification of the two $\ce{OH^-}$ groups of the previous molecule. This is wrong, please see the explanation
by  ringo

4-penten-1-ol

This one, I don't understand. How is this a possible result? Could anyone help me understand? Thank you for your time.

The three reactions are:
An $\mathrm{S_N2}$:

Deprotonation followed by an intramolecular Williamson ether synthesis ($\mathrm{S_N2}$):

An $\mathrm{E2}$:

Thank you for the illustration, that made it much easier to understand. Would you mind me asking what software do you use to draw the equations?

ChemDraw. If you're at a university, they likely have a subscription and can get you a copy for free.


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What is the value of the gas constant derived from the kinetic model of gases?
I used this resource (http://digitalcommons.unl.edu/cgi/viewcontent.cgi?article=1151&context=physicskatz) to derive the ideal gas law from the kinetic model of gases:
$$PV=n(2/3)N_oT$$
with T being equal to $(1/2)mv_{average}^2$.
And now I'm super confused. Does this mean that the universal gas constant R is simply equal to (2/3)*avogadro's number?

http://www.feynmanlectures.caltech.edu/I_39.html

$T$ here does not refer to temperature but rather the average kinetic energy of a molecule.
I can see why there is some confusion. Sadly we only have so many symbols and kinetic energy also happens to be commonly denoted with $T$.
To relate the average kinetic energy to the temperature, you need to bring in equipartition theory
$$\frac{1}{2}m\langle v^2\rangle = \frac{3}{2}kT$$
(assuming three translational degrees of freedom only). If you substitute that into your above equation
$$pV = \frac{2}{3}nN_\mathrm{A}\cdot m\langle v^2\rangle$$
then you will find that $pV = n(N_\mathrm{A}k_\mathrm{B})T \Longrightarrow R = N_\mathrm{A}k_\mathrm{B}$, hopefully a familiar expression. The question of how the value of $R$ is derived then becomes a question of how equipartition is derived - and that deserves another question itself.

As to the equipartition theorem a derivation is given here http://chemistry.stackexchange.com/questions/61255/derivation-of-mean-kinetic-energy/61332#61332 which while it looks very hard mathematically it  is quite easy to follow.