text:
How can we get the pure mass of an unclean metal?
So I'm facing a small chemistry problem that I can't find a way to solve. I'm not looking for answers, I'm just looking for how can I approach this problem.
The problem is as follows :

We put 5 g of non-pure zinc (with some impurities) into a 200 ml solution containing hydronium ions. At the end, we get some gases and we are left with 0.05 mol of hydronium ions in the solution.
Find the pure mass of zinc in that non-pure 5 g of zinc.

I really don't have any idea how to approach this problem. I have written and balanced the chemical equation:
$$\ce{Zn + 2H^+ = Zn^{2+} + H2}$$
I've arrived at the above equation from redox half-reactions. From here I don't know where to go.
Generally I would use the degree of purity formula:
$$P = \frac{C\cdot M}{10\cdot d}$$
where $P$ is degree of purity, $C$ is concentration, $M$ is the molar mass, and $d$ is the density, but this one is used for gases and liquids only. Besides, we've never studied how to compute the density of a metal.

There is not enough information to solve the problem as you have described it. From the nature of the problem it seems that some amount of acid was used to dissolve the zinc. So two moles of acid dissolved one mole of zinc. Your problem statement does indicate how much acid was left, but not how much you started with. It would also seem that whatever the impurities were, that they did not dissolve in acid.

The only info I have about the initial state of the acid is it's volume : 200ml, and that at the end we have 0.05 moles of that acid .

Again, there is not enough information to solve the problem as you have described it. It could just be a bad problem, or you might not have translated it properly. I don't know which.

Okay, it might be the problem that is bad or missing some info, but let's say I have the molar quantity of the acid at initial state how can I proceed in this case ? (please write an answer with this assumption so that I can mark it as solved if it helped me) .

It depends on the "impurities": if they were alkali metals (e.g. Na or K) or other metal more reactive than zinc, you have no way of knowing the proportions of impurity to Zn.

It would make sense to know the amount of gas (=hydrogen) evolved.


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IUPAC nomenclature benzene ring and functional group
What will be the IUPAC name of the benzene ring compound with SO3H and COOH as substituents?

According Section P-41 (seniority order for classes) and in more detail Section P-42 (seniority order for acids) of the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), carboxylic acids $(\ce{-COOH})$ are senior to sulfonic acids $(\ce{-SO2-OH})$. Therefore, a compound that contains both functional groups is named as carboxylic acid. In case of benzenecarboxylic acid $(\ce{C6H5-COOH})$, the retained name ‘benzoic acid’ is the preferred IUPAC name (PIN).
Benzoic acid is substituted using the usual principles of substitutive nomenclature.  In particular, a sulfonic acid group is expressed in substitutive nomenclature as follows.

P-65.3.2 Substitutive nomenclature, prefix mode for sulfonic, sulfinic, etc. acids
P-65.3.2.1 When another group is also present that has seniority for citation as principal group (see P-41, P-42, P-43), or when all groups cannot be expressed as suffixes, organic oxoacids of sulfur, selenium or tellurium are named by adding to the name of the parent compound the appropriate prefix given in Table 6.2. (…)

According to Table 6.2 in Subsection P-65.3.0, the preferred prefix used to denote a sulfonic acid group is ‘sulfo’. Therefore, a benzene ring with a carboxylic acid group and a sulfonic acid group is named as ‘sulfobenzoic acid’, for example: 4-sulfobenzoic acid.

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text:
Clausius-Clapeyron - Slope of the best-fit-line
I'm using the equation $\frac{y_2-y_1}{x_2-x_1}$ to find the slope of the best fit line for this problem but I can't get the figure shown in the book (-3773x K).  I've been at this a while and I really need to know so I can move on.  What am I doing wrong?  I've looked at half a dozen general chemistry texts and none of them make this step explicit.

What values for $x_1$, $x_2$, $y_1$ and $y_2$ did you get from the diagram?

I used the values .09531-6.25570/.005-.00333= -6.16030/0.00167= -3689

Then you have done it right and your result is probably within the error bounds. You can use linear regression if you want to treat the problem mathematically more rigorously.

Thank you both for your input.  I can not tell you how much of a relief it is to get some clarification on this problem.  As I move on I will get a better understanding of what you mean by 'error bounds'.  Linear regression and programming will have to wait, but I will remember your remarks.

I still have some questions about this kind of graph.  First, on the x-axis, why do we begin with the inverse of 300K?  This creates a negative slope.  Why don't we plot the inverse of 200K as our first point?  Second, and this may be beyond my mathematical understanding, what is it about the natural log that allows us to make a linear graph of pressure as a function of temperature?  Why don't we see disproportional jumps in the natural log of pressure in relation to temperature ultimately producing a curve?

Using your formula, $\mathrm{slope}=\frac{y_2-y_1}{x_2-x_1}$, and eyeballing the most extreme points on the chart, $\mathrm{slope} = \frac{0-6.2}{0.005 - 0.0033}$, I get a slope of 3647 K, pretty close to the "official" answer.
If you want to get more fancy, you could use a software program to fit this data.
my_data <- data.frame(vapor_pressure = c(1.1, 6, 28, 95, 293, 521),
                      temperature = c(200, 220, 240, 260, 280, 300))

# linear regression on transformed variables
my_data$ln_vp <- log(my_data$vapor_pressure)
my_data$inverse_T <- 1/(my_data$temperature)
linear_fit <- lm(data = my_data, formula = ln_vp ~ inverse_T)
linear_fit$coefficients['inverse_T']

inverse_T: -3787.46829114867

This answer is even closer to the "official" answer.  The remaining difference is probably due to rounding errors or differences in fitting algorithms used.  It only affects the answer by 0.3%.

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In w=-p∆v why there is a minus sign before p
In the 1st law of Thermodynamics $\Delta U=q+w$, $w=-p\Delta v$ why there is a minus sign before $p$?

Because simple rearrangement maths.

How can you illustrate

In very simple terms, when considering work you're talking about the energetic interactions within the system. You'll notice that, when an expansion takes place, ΔV has a positive sign as the final volume is greater than the initial volume. With the outer minus sign, the whole value adopts becomes negative. Negative values of energetic magnitudes are associated with the liberation of energy, considering it a loss of it. The work was done by the system. In the inverse case, during a compression, the final volume is smaller than the initial volume, therefore the ΔV is negative and the outcome of W is positive. This informs us of an absorption of energy, caused by the pressure the system was subject of. Now, the work was done on the system. Hope this helped.

@SeemantShekhar Please do not post thank you comments. If you find an answer helpful consider up voting/accepting it.


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Water hydrogen bond
In my book it states each water molecule can hydrogen bond to 4 other water molecules. However, it does not state why this is? Why is it 4 molecules and not 3? Is this related to the electron structure of water?

Two hydrogen atoms and two lone pairs on the oxygen.

Look at the water molecule structure:

Each hydrogen bound to the oxygen participates in the sigma bond with their single electron. The oxygen remains with two unshared pairs of electrons, thus, a remarkable dipolar moment is created and there's an heterogenous distribution of the electronic density in the molecule in this particular atom. In contrast, the hydrogens are stripped of electronic density in the covalent bond. When another water molecule approaches, the highly negative charges of its oxygen will undoubtedly have an attraction with each hydrogen; the sharing of one of these lone pairs with one of the hydrogens will cause what he know as hydrogen bonding. This far, we have two molecules stuck to the hydrogens.
Moving back to our original water molecule, we still have the oxygen with two lone pairs. In a similar fashion, each pair can interact with another hydrogen, therefore, we have our four adjacent molecules, as the following picture depicts:

In a discrete water molecule, there are two hydrogen atoms and one oxygen atom. Two molecules of water can form a hydrogen bond between them; the simplest case, when only two molecules are present, is called the water dimer and is often used as a model system. When more molecules are present, as is the case with liquid water, more bonds are possible because the oxygen of one water molecule has two lone pairs of electrons, each of which can form a hydrogen bond with a hydrogen on another water molecule. This can repeat such that every water molecule is H-bonded with up to four other molecules, as shown in the figure (two through its two lone pairs, and two through its two hydrogen atoms).

This is from the wikipedia article on hydrogen bonds.
Hydrogen Bond
This explains the phenomenon with simplicity.

Reference material should be used to illuminate an answer you have written yourself, not make up the entire answer.  Please either work this quote into some explanatory prose (and link to the original source) or remove this answer.


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Could anyone suggest some catchy chemistry experiments or demonstrations, to be performed in front of a young audience?
I'm wanting to volunteer at some of our local children centers, and do some experiments to interest them in Science/Chemistry. One cool, flashy experiment is to light my hand on fire - and not have to go to the hospital after.
The general scheme I've found for it (I've not tried it - and the instructions aren't .edu or other '95% correct' sites, so if you know of any corrections for this please chime in) is:
50 - 50 EtOH and Water, plus a small amount of salt for color. Dip your hands in, light them on fire, the EtOH burns coldly enough that the water doesn't evaporate and your hand stays safe and insulated.
Of course, I'll test it on small scale on paper in lab first.
However, I'm initially hesitant to play with fire around children. For obvious reasons.
Does anyone have anything similarly flashy or attention grabbing that they've used for a similar purpose with children? Obtaining restricted materials isn't an issue, but low-cost is preferable.
Note: I know this stackexchange frowns upon questions that don't have a specific answer, with the rules stating that discussion type questions are non-preferred - but this question isn't for the meta site, and I trust this community of educated scientists to have better advice on experiment safety than say, the educator stackexchange, so forgive me if this doesn't fit the general site mold.

I would advise against doing this demonstration altogether, especially around young children, simply because some of them might think it is a good idea to try it out for themselves.

Yes I agree. I quite dislike the idea @getafix, its maybe a good party trick but thats it.

There are many, many experiments with a "wow" factor out there that you can use, and require more adult-like skills to acquire materials (like, ordering stuff on the Internet with a credit card) than your audience will possess.  One of them is the "wine into water" trick (Google it). Another is a demonstration of the Tyndall effect (see this, for instance). The list goes on and on. I suggest you choose experiments where (1) the materials are not easy to obtain; or (2) the materials will, upon reacting, yield something that's not toxic/harmful/etc.

(a) luminol expt looks good as you drop in reactants and get flashes of light, (b) fluorescent dyes with uv lamp, can quench away with iodide. (c) iodine clock, (d) far more adventurous :H2/Cl2 light initiated explosion in a plastic pop bottle. Use a security lamp+ screen. v loud bang, rubber bung in bottle hits ceiling (keep in dark before hand)(e) H2/air explosion + match, H2 from cylinder enters by rubber tube into bottom of old coffee tin. Lid with hole spins off with bang after H2 turned off  and H2/O2 left burning in air. Done both of these many times but need proper lab for prep.

To talk to kids about chemistry I don't think you just need some exotic experiments.
I visited the Sterling Hill Mining Museum and one of founders had a neat gimmick that got kids attention. He asked them to name some product that had nothing to do with mining. He'd then rattle off a number of things that are mined to make that product. So for a truck the oil and gas are from an oil well. The steel comes from an iron mine and the copper in the electrical wires comes from a copper mine. So have the kids try to name a product that doesn't use chemistry.
Another thing that has always bugged me is the perception of the general public about "chemicals." It seems that the public believes that only things created in some chemical processing plant are chemicals. Well table salt is a chemical. Sugar is a chemical. A strawberry is a bag of water with a zillion chemicals in it. Take a vitamin tablet? The tablet is full of chemicals that our bodies need. Our bodies run on chemistry. So chemicals are all around us.

Heck, even water is a chemical.

A lot will depend on the age group of your audience. By "children centers", I'm getting the impression this is pre-school age kids, or at least young kids under 10. You need to set your content to match your audience. Their attention will be grabbed by very simple things, if done well. Everything is new to this age group, so there is no need to impress young minds with whirling-flames-of-death type stuff. They will be equally impressed with bicarb+vinegar volcanoes, especially if you throw in a generous dollop of washing detergent and food dye to make colourful bubbles. Some general things to consider:

you have to set extremely high standards of safety. You're not running a magician's festival. That means safety glasses and clean lab coats for all 'experiments'. Young minds will be easily impressioned by the way you do things, not just what you do.
Very safe experiments. Stuff you find in the kitchen is usually about as hazardous as you want to get with chemicals. Remember, it is all new to them, so that's ok.
Kids will want to touch and get involved and try to repeat what they see. Where possible, design an experiment that they can participate in. It will be a much more memorable experience if they get to 'do' rather than 'see'. At least think about getting one helper (they will love having to put on safety glasses and a labcoat) to add some drops of vinegar to your bicarb volcano, for instance.

That said, here are some examples of age-appropriate experiments/demonstrations to consider:

bicarb/vinegar volanco. Works well with several drops of food dye and washing detergent so you create colourful bubbles. (http://www.preschoolinspirations.com/2014/05/28/easy-baking-soda-and-vinegar-volcano-eruption-for-kids/)
pH indicator from red cabbage. Most kids won't have met a red cabbage before, I bet. (http://chemistry.about.com/od/acidsbase1/a/red-cabbage-ph-indicator.htm) This works really well done on a large scale, and allows helpers to pour indicator into your solutions of general household acid/base stuff.
invisible ink with lemon juice or milk or some pre-prepared ones using phenolphthalein/bicarb. Washing detergent whitening agents under a black light work great also. One thing I have done successfully with 7-10 year olds is the place some large cardboard sheets around the room that I have drawn large pictures using clothes whitening agents (check they glow under black light first). They are there all the time you are talking and doing other things, and then when you turn off the lights, and turn on the black light....oooh aaah. Scorpions glow under black light also (we find them all around where we live, so are easy to source).
coke-mentos fountain is another old favourite that is great for older kids. This is one that is all about nucleation,so we experiment with different types of materials and drinks. Rough sandstone and iron slag works best in normal coke, but if done in the dark, tonic water under a black light gives a great effect. (messy though)
static electricity is another great one. Bend water and make the teacher's hair stand on end.
making a rubber egg from a normal egg soaked in vinegar. This is a 'here's one we prepared 3 days ago' experiment.
clean an old coin (or half a coin is better) in coke. Again this takes a while, so set it up at the start of your session and come back to it.
there are some great ideas off this webiste that I have used before (http://www.csiro.au/en/Education/DIY-science/Chemistry). The cornflour slime is great, and kids can have a go for themselves, and don;t care when they get the ratios wrong.
Anything with magnets...(perhaps not chemistry though)

+1. Supersaturated sodium acetate is another good one to go on the list.

I very much agree with the safety focus. I think the experimenter should specifically mention safety. For young kids relate it to safety in the kitchen which would be familiar.

1) Know your surroundings.  Do you have running water?  What can you dispose of onsite, either in a sink or trash?  Will you have a cart? Where will the demo materials go when they are not being use?  How well will the audience be able to see the demo?  How loud can you be?  Is ventilation an issue?
2) Mentally and physically rehearse and troubleshoot logistics.  A demo might seem great at home or on paper, but when you do it, or mentally rehearse it while visualizing your audience and surroundings you may realize that there are logistical issues that you didn't think of.  They can be as simple as not having a place to set something.  REALLY rehearse onsite if possible.  Trying to perform steps of a demo and at the same time to teach requires practice.
3) Proceed slowly and thoughfully.  It will seem slower to you than to the audience.  Slow down to do "little" things like put caps on bottles before proceeding with the next step of the demo, or move liquid containing vessels to where they can't be knocked over.
4) Have a fairly simple, straightforward learning objective for each demo.  Have the kids repeat it and explain what it means.  Don't try to overload a given demo with trivial points that may detract from the ability of your target audience to learn a key point.  That doesn't mean that you can't make 2-3 points during a demo.  For primary grades you might want 1-2 key points.
5) Have a method of illustrating things on a molecular level. You should either have a white board, projector OR models that you can move.  Models can be as simple as large circles of construction paper or card stock with letters written on them.  For example, you can have a circle covered in aluminum foil to represent aluminum or other metal atoms, and others to represent two atoms in a compound.  Neon colors work well.
You are responsible for assessing the safety and feasibility of the demo.  Generally, demos with flammables, corrosives, and the potential for pressure to build up may REQUIRE safety glasses, shields, ventilation, eyewashes.  You also need to know if you will be producing a waste material that requires special disposal.
Examples:
I personally am comfortable making hydrogen gas from calcium metal and ice water in a thick walled jar, but you end up with a vat of strong base.
I am OK with filling a balloon with $\ce{H2}$ in certain settings, and igniting it, but it will be loud, and could trip a fire alarm, and might violate local ordinances, and there are subtle issues with the procedure that you only appreciate with experience.  The whoosh bottle with alcohol is similar though you need to decide if you want to use materials that your audience may have easy access to.
You can combine Lead Nitrate with Sodium Iodide to make a bright yellow precipitate, but you now have lead based paint to dispose of.
Here are some possible demos.

Make a large density column in a cylindrical vase using corn syrup, water, oil and alcohol.  You can add food coloring to the water, corn syrup and alcohol (red, green, blue).  You can also have things for kids to come up and drop into it and talk about how far down they will stop.
Use hydrogen peroxide and yeast enzymes to make oxygen gas in an inverted, submerged bottle.  Then place a taper inside and see how bright it gets.  You can talk about how enzymes can break apart hydrogen peroxide, or how the candle burns faster with more oxygen, and that the air we breath is not mostly oxygen.
Solutions of $\ce{KSCN (clear) + FeCl3 (yellow)}$ will combine to make a blood red/brown product.  Color change can be a sign of a chemical reaction. You can dip a strip of paper towel in to pull up some of the solution.
PVA will polymerize in the presence of sodium borate to make slime.
$\ce{CuCl2}$ can be dissolved to make a solution which will react with aluminum cans or foil.  You can show that salts contain metals as the copper gets replaced by the aluminum.  $\ce{CuCl2}$ is caustic and toxic but you will end up with aluminum chloride in the solution.  (There are health concerns with aluminum chloride.  You could use zinc.)
If you drop alka seltzer in vinegar with a pH indicator you can watch it neutralize acid.  Again, a large vase may be cheapest, but will crack easily if you tip it over and also produce sharper shards than lab grade glassware.
You can show flame colors from salts with different metals.  https://www.youtube.com/watch?v=jJvS4uc4TbU  It looks like he may have dissolved salts in a mostly alcohol and lit the alcohol.  This method would not require a gas burner.
You can also have kids to paper chromatography using coffee filters and magic markers.  All they need is a filter, a few markers and a cup of water.
There are chemical hand warmer reactions where kids can mix chemicals in a plastic bag and feel it get warmer.
Something as simple as having 4 solutions is large flasks, 1 M HCl, vinegar, water and 1 M NaOH and placing a few drops of universal indicator in each one.

Firstly let me say that what you're suggesting is a horrible idea. The fact that you understand the kids will try to copy you makes this even worse. Do not do it!

The most impressive chemistry demonstration I've ever seen (in a classroom setting) was one in which a ~15L round bottom flask was used to demonstrate the luminol-peroxide reaction. It may not be as impressive if the light level can't be sufficiently low.
There are several reactions in which a solution goes through a series of color changes.
The production of nylon at the interface of two liquids is pretty impressive for older grade/middle schoolers.


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Predict which hydrogen is more acidic in the following organic compounds?

Between 2-chloro-2-phenylpropionic acid and hydrogen phthalate, which is the more acidic compound?
I'd say the $\ce{H}$ on the second one is more acidic as the $\ce{-COOH}$ group is directly attached to the aromatic group. Is that correct?

Two negative charges in the same molecule is destabilizing.

In the second molecule, the charge on the carboxylate ion is not stabilised through resonance, so we can rule out any resonance stabilisation. Moreover, the fact that the benzene ring is in fact electron donating within the $\pi$ system doesn't help either.
More importantly, however, (as it was pointed out to me by @orthocresol and @hippalelectryon in chat) one can expect intramolecular hydrogen bonding.
Moreover, after the second deprotonation, the second molecule would have two negative charges in close proximity to each other, and this, again, is obviously destabilising.
A combination of these factors would make the second molecule a weaker acid.
In the first case, however, the inductive electron withdrawing effect would increase the acidity, (and the benzene ring is far enough to not have an effect).
Thus, I would say it is the first compound that is more acidic.

@Hippalectryon used a software to compute a pKa of 3.87 for the first one, and Wikipedia records 5.51 for the second one, which confirms my claim.

You could include the computed value for the first one and the measured value for the second one...

@user34388 good idea! feel free to edit the answer and add that in :)

As a very, very general rule, the two $\mathrm{p}K_\mathrm{a}$ values of a diacid will differ by a value of at least $1$ (likely more). Take for example phosphoric acid and aspartic acid whose $\mathrm{p}K_\mathrm{a}$ values are the following:
$$\begin{array}{ccccc}\hline
\text{acid} & \mathrm{p}K_\mathrm{a,1} & \mathrm{p}K_\mathrm{a,2} & \mathrm{p}K_\mathrm{a,3} & \text{Source}\\ \hline
\ce{H3PO4} & 2.148 & 7.198 & 12.319 & [1]\\
\text{aspartic acid} & 1.99 & 3.90 & 9.90 & [2] \\ \hline \end{array} $$
From that alone we would expect the monoacid to have a higher $\mathrm{p}K_\mathrm{a}$ than the diacid. And there are more contributing factors, as partly noted in getafix’s answer:

Chloroacetic acid already has a lower $\mathrm{p}K_\mathrm{a}$ value than benzoic acid ($2.9$ versus $4.2$) implying that the molecule on the left (which is pretty similar to chloroacetic acid) has a lower one than the right-hand one (closer to benzoic acid).
A chlorine atom is electron-withdrawing ($-I$). A non-conjugating phenyl ring would also be slightly electron-withdrawing but chances are that the two carboxylic acid groups don’t deviate too strongly from planarity making the general effect more of a $+M$ one.
A (very strong) cyclic hydrogen bond can form between the two acid groups on the right-hand molecule leading to a strong stabilisation of the monoanion.
The two anions would even be in very close proximity, which isn’t good (they are better separated in aspartic acid).

Sources:
[1]: https://en.wikipedia.org/wiki/Phosphoric_acid
[2]: http://academics.keene.edu/rblatchly/Chem220/hand/npaa/aawpka.htm

@Lydia Hernandez. 2-chloro-2-phenylpropionic acid is certainly more acidic than hydrogen phthalate. The reason is very simple, A -COOH group has a well defined conjugated system of its own. It is because of this conjugated system the acidic behavior of this group is defined. On the other hand whenever a -COOH group is attached to the benzene ring directly it (benzene ring) disturbs the conjugation of -COOH group, thus the leaving of H becomes difficult. This disturbance is known as "cross Conjugation". You can observe the same with the help of following figures -

You can verify the answer via pH or pKa values.


text:
Why are basis sets needed?
I am not sure whether this question is even reasonable, but here it goes. We are taught about the different types of basis sets (extended, minimal, double-zeta, plane wave), but I do not think it is clear as to why they are needed. After all, it is possible to do computational chemistry without a basis set (see James R. Chelikowsky, N. Troullier, and Y. Saad, Phys. Rev. Lett. 1994, 72, 1240).
From what I understand basis sets are needed because we use LCAO. We find a set of functions (a basis set) that resembles atomic orbitals. Is this true? Or is the picture more complicated?

Finite-difference method, as mentioned in the reference of the question, is in-general exponentially complicated for non-linear polyatomic system. Perhaps the best result so far is still water molecule by Toru Shiozaki..

Spatial orbitals $\phi_i$ in modern electronic structure calculations are indeed typically expressed as a linear combination of a finite number of basis functions $\chi_k$,
\begin{equation}
    \phi_i(1) = \sum\limits_{k=1}^{m} c_{ki} \chi_k(1) \, .
\end{equation}
In the early days, atomic orbitals were built out of basis functions, while molecular orbitals were built out of atomic orbitals, which is where the name of the approach, linear combination of atomic orbitals (LCAO) originated.
Today both atomic and molecular orbitals are built out of basis functions, and while basis functions for molecular calculations are still typically centered on atoms, they are usually differ from the exact atomic orbitals due to approximations and simplifications.
Besides, basis functions centered on bonds or lone pairs of electrons, or even plane waves, are also used as basis functions.
Nevertheless, the approach is still commonly referred to as linear combination of atomic orbitals.

Now, to the very question of why do we do things this way?
On the one hand, the LCAO technique made its way into quantum chemistry as just another example of a widely used approach of reducing a complicated mathematical problem to the well-researched domain of linear algebra. To my knowledge, this was proposed first by Roothaan.1 However, and as it was mentioned by Roothaan from the beginning, the LCAO approach in present day quantum chemistry is also attractive from the general chemistry point of view: it is tempting to construct molecular orbitals in modern electronic structure theory from their atomic counterparts as it was done by Hund, Mulliken and others already in the early days of quantum theory2 and as it is though in high-school general chemistry courses today.

1) Roothaan, C.C.J. "New developments in molecular orbital theory." Reviews of modern physics 23.2 (1951): 69. DOI: 10.1103/RevModPhys.23.69
2) Pauling and Wilson in Introduction to Quantum Mechanics with Applications to Chemistry refer to the following works in that respect (p. 346):

F. Hund, Z. f. Phys. 51, 759 (1928); 73, 1 (1931); etc.; R. S.
  Mulliken, Phys.  Rev. 32, 186, 761 (1928); 41, 49 (1932); etc.; M.
  Dunkel, Z. f. phys. Chem. B7, 81; 10, 434 (1930); E. Hückel ,Z.f.
  Phys. 60, 423 (1930); etc.

Hund's papers are unfortunately in German, but Mulliken's ones are quite an interesting read. Especially the second one,
Mulliken, Robert S. "Electronic structures of polyatomic molecules and valence. II. General considerations." Physical Review 41.1 (1932): 49. DOI: 10.1103/PhysRev.41.49
which (again, to my knowledge) introduced the very term "molecular orbital".

The idea of using basis (in general sense, not just basis set) in electronic structure theory is due to the fact that we can't solve the wave function of a system analytically except those with only one electron (H, H2+,..). So, we need to introcude some approximations to find the wave function.
There are actually two levels of approximation:
1. Many-electron (or N-electron) approximation:

The wave function of N electrons (or a state of N electrons) can be approximately a Slater determinant (SD) or a linear combination of SD. A SD is an
antisymmetric product of one-electron basis functions. These
one-electron basis are called molecular orbitals or MO or orbital. An orbital is nothing but a state of one electron.
In sum, a many-electron wavefunction is constructed using
orbitals which are one-electron states. That's what the name implies: many-electron (or N-electron) approximation

2. One-electron approximation:

Now, the question is: we know that we need the orbitals to construct
the wave function but how do we find those MOs or
orbitals? Here, we introduce another approximation. We use a set
of functions (called basis set) to construct the MOs. MOs are now
constructed by a linear combination of those functions (or LCAO which
is already explained very well by the above answer). The coefficients
used in that construction are called MO coefficients, which are
varied to minimize the energy of the corresponding many-electron
wavefunction (the system of interest). Since one-electron state (MO) is constructed by some functions (basis set), this is called one-electron approximation. I don't like the term "atomic orbital" since they are just some mathematical functions. They are not even a wave function of one single atom. They can be "any function" providing it is convenient to work with. For example Gaussian function is often used since it is easier to calculate the integrals in comparison with Slater function. Plane wave is often used in periodic systems since the periodicity is conveniently imposed.
In fact, we can use an infinite number of those functions to
construct MOs (or orbitals), in that case, the calculation reaches
the complete basis set (CBS) limit. In practice, we use a finite
number of basis set, and many people came up with many types of basis
set with different type of functions used (Gaussian, Slater, Muffin,
numerical, or plane wave..) and with different sizes (minimum, double zeta,
triple zeta,...)

In summary:
In ab initio calculations, one has to provide these information:

Geometry of the systems (atom coordinates)
The total charge of the system
Spin state (singlet, doublet, triplet, ...)
A many-electron approximation (one SD, e.g. HF, DFT, MP2, CCSD OR
many SDs like multiconfigurational methods, e.g. CI, MCSCF). This is
also preferred as the level of correlation since different ways of
constructing the wave function will recover different amounts of
correlation (coulomb interaction between electrons). Methods using
one SD like HF cannot capture the real interaction between electrons.
A one-electron approximation (a basis set).

The entire information above is all we need to calculate electronic structure of a system, and this is known as a quantum chemical model defined by John Pople

This answer, I hope,  complements those earlier ones and give some little explanation as to how the basis set calculation works.
As pointed out already it is not necessary to use a basis set to solve the Schroedinger equation because for simple systems, such as a harmonic or Morse potential, analytical i.e. algebraic  solutions are available. Numerical methods, such as the Numerov and Shooting methods, can be used for more complicated potentials but soon become intractable.
Generating a solution using a basis set is often computationally the best method, and sometimes the only method that can be used. This method involves solving the Schroedinger equation without direct integration.
The idea is to choose a known set of functions which when added together in the correct proportions will generate the wavefunction for the target problem. This is the same idea as using a fourier series  to generate a target function; the more terms that are included in the series the better the fit to the target function becomes.  An example of a fourier series is to form a polynomial using many sine/cosine waves of different amplitude and frequency.
In forming a basis set we choose a set of orthogonal functions, and numerically solve the matrix Schroedinger equation using as many of these functions as needed for the accuracy required. Normalised orthogonal functions have the property that $\int \phi^*_n\phi_mdx=\delta_{nm}$ where $\delta_{nm}$ is the Knonecker delta, if $n=m$ then $\delta_{nm}=1$ otherwise it is  zero.
As an example we may use the sine-wave solutions to the particle in a box (PIB) problem to solve a harmonic well potential with a central 'hill' or barrier. If the PIB equation is written as $H\phi = E\phi$, the eigenvalues (energies) are
$\epsilon_n=\int\phi^*_n~H~\phi_ndx$.
The new problem we want to solve has the equation
$$ H^{’} \psi = E^{’} \psi  \tag {1} $$
which has a different Hamiltonian to reflect the different potential energy. The new wavefunction for the $n^{th}$ level is assumed to have the form
$$\psi_n=c_{1,n}\phi_1 + c_{2,n}\phi_2 \cdots c_{N,n}\phi_N$$
where the coefficients c are to be found. (The calculation follows that in J.Chem. Educ. 2012, v89, p1152). Perhaps twenty or a hundred terms in the basis set will be needed to obtain an accurate answer both for the eigenvalues (energies) and eigenvectors (wavefunctions).  If $N \rightarrow \infty$ then the energy and wavefunctions would be exact.
The PIB basis set used here is a good starting point in our example  as we know that the wavefunctions for a harmonic oscillator oscillate somewhat in a sinusoidal manner and the barrier harmonic well should be similar.  If we used exponential functions many, many more would be needed to achieve the same result as these just do not have the same shape as the wavefunction. However, exponential functions would be  good to use in describing an atoms wavefunctions as they have a shape similar to atomic orbital radial functions.
The energy for level m with the new Hamiltonian $H^{’}$ is found by left multiplying  eqn. 1  by $\phi^*$
$$\int \phi^*_mH^{’} \psi_ndx =E^{’}_n\int \phi^*_m\psi_ndx$$
and substituting for $\psi$ gives
$$\int \phi^*_mH^{’}(c_{1,n}\phi_1 + c_{2,n}\phi_2 \cdots)dx=E^{’}_n\int \phi^*_m(c_{1,n}\phi_1 + c_{2,n}\phi_2 \cdots)dx$$
On expanding this out and using the orthonormality the equation reduces to a matrix equation
$$ \bf HC=E^{’}C$$
where $\bf H$ is a square matrix of elements
$$H_{m,n} = \int \phi^*_mH^{’}\phi_ndx$$
and $\bf C$ is a column matrix of coefficients. The matrix equation is $\bf(H-E^{’}I)C=0$ where $\bf I$ is the unit diagonal matrix and $\bf 0$ a null matrix containing only zeros. Solving this equation involves evaluating many simple integrals $H_{m,n}$ in the secular determinant and then diagonalising to find the characteristic equation which is a polynomial in $E^{’}$ and hence the energies.


text:
Why is it possible to form bromoalkanes but not iodoalkanes from free radical substitution?
For a free radical substitution (FRS) reaction between iodine $\ce{I2}$ with ethane and bromine $\ce{Br2}$ with ethane $\ce{CH3CH3}$:
$E_\mathrm d(\ce{H-I})=299\ \mathrm{kJ\ mol^{-1}}$
$E_\mathrm d(\ce{H-Br})=366\ \mathrm{kJ\ mol^{-1}}$
$E_\mathrm d(\ce{C-H})=410\ \mathrm{kJ\ mol^{-1}}$
I understand that the reaction for iodoethane to form from FRS is energetically unfavourable, since the energy released from the formation of a $\ce{H-I}$ bond (from reaction between an iodine radical and an ethane molecule) is only $299\ \mathrm{kJ\ mol^{-1}}$, which is insufficient to compensate the energy needed to break the $\ce{C-H}$ bond to even form the first ethyl radical in the first place for further propagation of the reaction.
However, it is to my knowledge that bromoethane can form from FRS, despite the formation of a $\ce{H-Br}$ bond releasing only $366\ \mathrm{kJ\ mol^{-1}}$ of energy, which is still lower than the $410\ \mathrm{kJ\ mol^{-1}}$ of energy needed to break the $\ce{C-H}$ bond. This doesn't really make sense if the justification for iodoethane is correct.
I would appreciate if anyone could explain to me the reason why bromoethane can form from FRS despite being apparently energetically unfavourable too, thanks!

This earlier answer may prove helpful.

Mechanistic Summary

Initiation: Involves homolytic cleavage of the $\ce{X-X}$ bond to generate $\ce{X^.}$ radicals. Initiation is endothermic, and the energy requirement is usually met by thermal treatment or photoirradiation.
Propagation I: Abstraction of a $\ce{H}$ atom by the $\ce{X^.}$ radical, to produce a $\ce{R-H2C^.}$ radical. This transformation usually takes place via a $[\ce{R-H2C...H...X}]^\ddagger$ transition state. Also, this process is endothermic
Propagation II: This stage involves the abstraction of $\ce{X}$ atom from $\ce{X-X}$ by the the $\ce{\ce{R-C^.}}$ to prduce $\ce{R-H2C-X}$. This process occusrs via a transition state that looks so something like this: $[\ce{R-H2C^....X...X}]^\ddagger$. This process, if it is sufficiently exothermic, compensates for the previous endothermic step and the reaction proceeds.

Termination: Involves radical-radical recombination.

Now let us look at the energetics of the reaction. The first propagation step is exothermic for fluorine and this explains its extreme reactivity. For the subsequent halogens, it becomes increasingly endothermic from chlorine to iodine. This trend can be explained on the basis of decreasing bond strengths of $\ce{H-X}$ bond. (Data given in the table above)
Now let me compare to extremes, fluorination of methane and iodination of methane. As one can observe, the first propagation step for fluorination is not only exothermic, but it also has a small activation barrier.
Thus in the transition state, $[\ce{H3C...H...F}]^\ddagger$, the $\ce{H-CH3}$ distance is only slightly greater than that in $\ce{CH4}$, and $\ce{H-CH3}$ bond is weaker than the $\ce{H-F}$ bond. Thus only a slight shift of the hydrogen atom towards the fluorine radical is sufficient to break the carbon-hydrogen bond and form the fluorine-hydrogen bond. This kind of a transition state is called an early transition state, and is typical of fast,exothermic processes.
In contrast, iodination displays a late transition state which is typical of slow,endothermic processes.The activation barrier for the first propagation step is relatively high, in fact it is almost the same as the endothermicity of the reaction. The transition state is not reached till the the carbon-hydrogen bond is almost fully broken, and the hydrogen-iodine bond is fully formed.
These rules about transition states are called Hammond's Postulates, and I encourage you to read about them.

Now, indeed the reason why iodination doesn't proceed is that even though final step is exothermic it doesn't release sufficient energy to compensate for the extremely endothermic first step. Thus, the net reaction is endothermic.
However, it is not the exothermicity of $\ce{H-X}$ bond formation that drives the halogenation reaction, rather it is the energy released upon $\ce{-C-X}$ bond formation (in the second propagation step) that ensures reaction spontaneity. (this can be seen in the data presented in the table)
This is why bromination, or for that matter chlorination can take place.
My apologies, the question asked about iodethane and bromoethane, however, the examples I took up was that of iodomethane (not the first time I have misread a question in my rush to post an answer). However, the ideas presented here still apply
References: Data table, and reaction profile taken from Organic Chemistry: Structure and Function by Vollhardt&Schore.


text:
Is methyl orange acidic or basic?
It has been saying that if adding too much indicator in solution, the ph value of solution would be changed. From this, it can be saying that indicators have a pH value. Methyl orange is the indicator I used in the experiment. "My question is whether methyl orange is basic or acidic so that it affects the pH value of solutions."

In an acidic solution, the protonation methyl orange takes place, as shown below

The deprotanted form depicted on the left is basic, and is yellow in colour. The red form is acidic and is depicted on the right.
In general, a lot of acid base-indicators rely on the following mechanism of action
$$\ce{In^-}+  \ce{<=>[\ce{H^+}][\ce{OH^-}]} \ce{HIn}$$
At low pH, i.e high $\ce{H^+}$ concentration the $\ce{HIn}$ form is dominant since the equilibrium lies to the right, however, at higher pH $\ce{HIn}$ (which is weakly acidic) dissasociates to give $\ce{In^-}$ (its conjugate base).
Herein, $\ce{HIn}$ (the acid), and $\ce{In^-}$ (its corresponding conjugate base) have different colours. The colour changes are due to changes in absorption, due to change in the length of the $\pi$ conjugated system (as can be seen in the case of methyl orange pictured above).

It depends on what form of it you have. The yellow form is basic, the red form is acidic.
This change is  done by the yellow one picking up a proton to form the red form.

In a solution which is getting less acidic, methyl orange will move from a red colour to orange and then at last to yellow with the reverse taking place for a solution increasing in acidity.
Therefore, in an acid, it is reddish and in alkali, it is yellow
Source:

http://chemguide.co.uk/


text:
Why do electrolytic concentrations affect voltage in electrochemical cells?
I know that the voltage of a cell is calculated via the Nernst Equation as $$E_\text{cell}=E^\circ-\frac{RT}{nF}\ln Q_\mathrm r$$ where $Q_\mathrm r$ represents $\frac{\text{concentration at the anode}}{\text{concentration at the cathode}}$.
However I don't understand how these concentrations actually affect the potential difference. Could anybody explain the physical processes that mean changing the concentrations affect the voltage of the cell?

Potential is work to move charge and this work is done by the chemical energy stored in the reactants, so if the concentration of reactants increases then there are more reactants which will do more work so this implies that potential increases with comcentration.

Yes but when the concentration at the anode decreases the total cell voltage increases. To me this is counterintuitive and this is would I wanted to understand.

@AlexJones the concentration at anode which in the formula is probably of oxidised product which will not contribute to potential constructively and further helps in decreasing the potential.


text:
Which platonic solids have been built as carbon skeletons?
There are 5 platonic solids: the tetrahedron, the octahedron, the cube, the dodecahedron and the icosahedron.
How many of these have been realised as carbon skeletons by chemical synthesis?

Is this what you mean? https://en.wikipedia.org/wiki/Platonic_hydrocarbon

@orthocresol Yes. But knowing a bit about how they were made and how stable they are would also be interesting.

Why restrict ourselves to carbon? Icosahedral sceletons are quite commonly made of boron. As for octahedra, there are metal clusters with that shape.

I love the variety of shapes which have been made with hydrocarbons, just exploring geometry.  We've made prismanes, fenestranes, propellanes, pagodane, and the list goes on and on and on!

http://onlinelibrary.wiley.com/doi/10.1002/anie.197805201/abstract

The previous answer and comments have already listed out the known platonic hydrocarbons, however, since the OP had expressed an interest in the synthesis of these platonic hydrocarbons in one of his comments, I have provided some commentary to that effect along with links to the original papers that first reported these compounds. Also, of interest would be this paper
Tetrahedrane
Technically speaking, Tetrahedrane ($\ce{C4H4}$) still remains a hypothetical compound, as an unsubstituted version hasn't been synthesised yet.

Two stable derivatives, whose synthesis has been achieved include tetra-tert-butyltetrahedrane (prepared in 1978 by Günther Maier) and tetra(trimethylsilyl)tetrahedrane (reported in 2002).
Tetra-tert-butyltetrahedrane
The original paper discussing the synthesis can be found here. The synthesis has been summarised below. The authors rationalise the stability of this compound as follows:

If  all  the  corners of the tetrahedrane unit are occupied by bulky groups then on lengthening  one bond of the  ring  skeleton , two  tert-butyl groups  tending  to increase  their  mutual  separation  will  be forced  back  together  by  the  other ones  ("corset  effect"); a tetrahedral  structure permits  the  four  substituents  to adopt a regular spherical distribution and hence a maximum mutual separation.

In simpler terms, trying to break a bond forces the bulky substituents together increasing strain.

Tetrakis(trimethylsilyl)tetrahedrane

Again, the synthetic route is given below. The original paper is also by Günther Maier.
Reaction with methyllithium has yielded the stable tetrahedranyllithium derivative. These coupling reactions with this lithium compound have allowed access to more derivatives, for example the dimer in the last step of the reaction scheme given above.
This compound is far more stable than the tert-butyl analogue. On account of the silicon–carbon bond being longer than a carbon–carbon bond, the corset effect is reduced. $\sigma$-donor character of the trimethylsilyl explains the increased stabilisation of the tetrahedrane.
The tetrahedrane skeleton is made up of banana bonds, and hence the carbon atoms are high in s-orbital character
Cubane

Cubane was first synthetically realised by Eaton and Cole in 1964. The synthetic scheme from the original paper is given below:

Note: In the original synthesis, the starting material for this, 2-bromocyclopentadienone, was obtained from 2-cyclopentenone via allylic bromination with N-bromosuccinimide in carbon tetrachloride followed by addition of molecular bromine to the alkene gives a 2,3,4-tribromocyclopentanone. After treatment of this compound with diethylamine in diethyl ether, which causes elimination of two equivalents of hydrogen bromide, we get the desired diene product.
Again, the cubane is kinetically stable because of lack of readily accessible decomposition paths.
Dodecahedrane

Dodecahedrane was synthesised by Leo Paquette in 1982, and later by Prinzbach (via pagodane) in 1987.
The pagodane synthetic route has been given below.

Paquette's 1982 organic synthesis takes about 29 steps, and starts from cyclopentadiene, and for the sake of brevity, I have not presented it here.

There are the three painfully obvious platonic hydrocarbons: tetrahedrane, cubane, dodecahedrane.
The latter one also exists as a fullerene, i.e. $\ce{C20}$, which is also believed to be the smallest fullerene.
From here on it will be very difficult to come up with the other platonic structures.
The octahedral structure of $\ce{C6H6}$ probably cannot exist dues to the fact that the carbons would have five bonds (four to carbon and one to hydrogen) and would quickly rearrange to a trigonal prism. However, there is tricyclo[1.1.1.11,3]hexane, $\ce{C6H8}$, which has a distorted octahedral carbon cluster.
The icosahedral version would require a penta-coordinated carbon (or hexa-coordinated if we consider hydrocarbons), which is one of the rarest to come across in the first place. However, you can have a carbon skeleton with icosahedral symmetry. The famous Buckminsterfullerene (ChemSpider), $\ce{C60}$, is a truncated icosahedron, which is an Archimedean solid.

"believed to be the smallest fullerene" - believed, or considered?

@JanDvorak probably should just be "is" (depending on which definition of a fullerene you use)

@JanDvorak "believe" because I cannot find compelling evidence that this is a fact. There have been a number of carbon clusters proposed to be stable, $\ce{C_2n}, n < 10$, which would fit the broadened description of fullerenes (Gold book: The term has been broadened to include any closed cage structure consisting entirely of three-coordinate carbon atoms.). See R. O. Jones and G. Seifert
Phys. Rev. Lett. 1997, 79, 443.


text:
Reduction of a ketone in the presence of an aldehyde
An aldehyde group can easily be reduced to an alcohol using the mild reducing agent $\ce{NaBH4}$ at low temperature in the presence of a ketone group. This chemoselectivity is due to the fact that an aldehyde is less sterically hindered compared to a ketone, which makes it react faster than the ketone.
Is it possible, however, to chemoselectively reduce a ketone in the presence of an aldehyde using $\ce{NaBH4}$ as reducing agent?

The best strategy is probably to chemoselectively protect the aldehyde before the borohydride reduction.  Protecting groups are derivatives of certain moieties in a molecule that are far less reactive under most conditions, but which can be easily removed under very specific conditions to regenerate the original moiety.
For carbonyl compounds -- this includes both aldehydes and ketones -- the most common protecting groups are all introduced by ketalization/acetalization.   Common examples are dimethyl acetals, 1,3-dioxolanes, and 1,3-dioxanes.
However, in general, aldehydes are far more prone to acetalization than ketones are to ketalization.  That offers a means to chemoselectively protect aldehydes.  Acetals such as 1,3-dioxanes are stable to borohydride reductions, but, as you note in your question, ketones are sensitive.
Poking around reveals several potential strategies for selective chemoprotection of an aldehyde.  Here are two examples.

Tetrabutylammonium Tribromide (TBATB) as An Efficient Generator of HBr for an Efficient Chemoselective Reagent for Acetalization of Carbonyl Compounds:

This convenient, mild, chemoselective method allows acetalization of an aldehyde in the presence of ketone...

Chemoselective Protection of Aldehydes in the Presence of Ketones Using RuPVP Complex as a Heterogeneous Catalyst

Putting it all together, a recipe for selective borohydride reduction of a ketone in the presence of an aldehyde would be something like:

Protect the aldehyde.  The best choice is probably as a 1,3-dioxane acetal by reacting your compound with 4 equivalents of 1,3-propanediol, 1 equivalent of triethyl orthoformate, and 0.01 equivalent of tetrabutylammonium tribromide.  Under suitable conditions, the ratio of protected aldehyde (desired) to protected ketone (undesired) should be >50.
Borohydride reduction of the aldehyde-protected compound.  Only the unprotected ketone will react.
Deprotection.  Acetals can be removed in acidic conditions.  A number of catalysts that will effect deprotection without requiring acid have also been developed.

An easy and convenient method to achieve this is the use of the Luche reduction.[1,2]
The ‘organic chemist’s bible’, Strategic Applications of Named Reactions in Organic Synthesis by L. Kürti and B. Czakó notes on the page of the Luche reduction (p. 268):

8) the combination of $\ce{CeCl3/NaBH4}$ is excellent for the chemoselective reduction of ketones in the presence of aldehydes, since under these conditions aldehydes undergo rapid acetalization, which prevents their reduction.

Indeed, when a $1:1$ mixture of hexanal and cyclohexanone is subjected to Luche conditions with $1.5~\mathrm{eq}$ of $\ce{NaBH4}$, a quantitative yield of cyclohexanol and a nigh-quantitative ($98~\%$) of hexanal are recovered. Without the addition of $\ce{CeCl3}$, $1.5~\mathrm{eq}$ reduce cyclohexanone quantitatively and $49~\%$ of hexan-1-ol are also recovered.[1]

Reference and note:
[1]: J. L. Luche, A. L. Gemal, J. Am. Chem. Soc. 1979, 101, 5848. DOI: 10.1021/ja00513a075.
[2]: The article [1] is only two pages short. The first page is displayed as an ‘abstract’ on pubs.acs.org (‘in lieu of an abstract’, ibid.). Thus, opening the abstract of S. Nagase, T. Fueno, K. Morokama, J. Am. Chem. Soc. 1979, 101, 5849 (DOI: 10.1021/ja00513a076) can serve as a cheap hack to see the entire article.


text:
Why is this cycloalkane name giving the lower number to methyls over ethyl?
This compound is given the name 5-ethyl-1,1-dimethylcyclooctane in a video I watched, and also in ChemDraw:

My question is: Why is it called that instead of 1-ethyl-5,5-dimethylcyclooctane?
Two sources taught me that, first, we ignore prefixes like di and tri (but not iso), and, secondly, the substituent that comes first alphabetically gets the lower carbon number, or we count the carbons from it. And since, in this case, ethyl comes before methyl alphabetically, I would've thought that it gets the 1, not the methyl? Why is this not the case here?

In the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the order of citation of substituent prefixes and the numbering of the corresponding locants are laid down in two different rules.
Simple substituent groups that are named by means of prefixes (such as ‘methyl’ and ‘ethyl’) are arranged alphabetically.

P-14.5 ALPHANUMERICAL ORDER
Alphanumerical order has been commonly called ‘alphabetical order’. As these ordering principles do involve ordering both letters and numbers, in a strict sense, it is best called ‘alphanumerical order’ in order to convey the message that both letters and numbers are involved
Alphanumerical order is used to establish the order of citation of detachable substituent prefixes (not the detachable saturation prefixes, hydro and dehydro), and the numbering of a chain, ring, or ring system when a choice is possible.
(…)
P-14.5.1 Simple prefixes (i.e., those describing atoms and unsubstituted substituents) are arranged alphabetically; multiplicative prefixes, if necessary, are then inserted and do not alter the alphabetical order already established.

Therefore, the correct alphanumerical order for the compound given in the question is x-ethyl-y,y-dimethylcyclooctane (not x,x-dimethyl-y-ethylcyclooctane).
The locants x and y are used to indicate positions of the parent structure at which modifications represented by suffixes occur. The relevant rules concerning the numbering of locants for substituent prefixes are:

P-14.3.5 Lowest set of locants
The lowest set of locants is defined as the set that, when compared term by term with other locant sets, each cited in order of increasing value, has the lowest term at the first point of difference; (…)

and

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

Note that Rule (f) takes precedence over Rule (g).
In accordance with Rule (f), the compound given in the question is named as 5-ethyl-1,1-dimethylcyclooctane, not 1-ethyl-5,5-dimethylcyclooctane since the locant set ‘1,1,5’ is lower than ‘1,5,5’. And thus, Rule (g) is not required in this case.

The concise, focused answer to my specific question, which I found before reading any answers here, is as follows:
The wrong name I suggested, 1-ethyl-5,5-dimethylcyclooctane, has the locant set: 1,5,5. The correct name, 5-ethyl-1,1-dimethylcyclooctane, has the locant set: 1,1,5. According to the IUPAC “lowest set of locants” rule, which has nothing to do with the sum of locants, but the “first point of difference”, we can see that the first point of difference between the locant sets in question is between the second locants: 5 and 1; 1 is lower, so its set wins, or is the correct one to use. That's why the correct name of the compound in question is 5-ethyl-1,1-dimethylcyclooctane.


text:
A book like Feynman's Lectures to Physics, which deals with Chemistry and provides insights
I have been reading The Feynman Lectures, and it provides a lot of basic knowledge in physics. I would like to know if there exists a book for chemistry like this too.

You might find this helpful: http://chemistry.stackexchange.com/questions/37303/resources-for-learning-chemistry

About 20 years ago, I have got the German translation of Atoms, Electrons and Change by Peter W. Atkins. This short and simple book was inspired by the lectures on chemistry given by Michael Faraday (such as The Chemical History of a Candle).

Atkins does have several other chemistry books written for the layman as well. https://en.wikipedia.org/wiki/Peter_Atkins#General_readers


text:
Why does oxidation of the electrode take place in a solution?
What makes the electrode atoms start to ionize? Perhaps because of the solutes in the solvent?
So probably if you place it in distilled water it will not ionize, right?
Follow up question: If the solutes in the water are the reason of ionizing the electrode, Do they rip off the electrons from the electrodes atoms? (Most of websites illustrate that both electrons (on the electrode) and ions (in solution) contribute to the voltage of the electrode).
So why do I use electrode potentials if I have lets say a bowl made purely of $\ce{Zn}$ atoms and it contains lets say Copper ions. Even though I only have positive ions of copper. Does the copper ions generate the same voltage by their own without electrons?

Sorry to be so catty but the overall line of questioning indicates a profound lack of understanding. I can't explain all about of redox reactions in a few paragraphs to let me try to answer some of the questions in a general way and hope that those answers will provide some focus.

What makes the electrode atoms start to ionize? Perhaps because of the solutes in the solvent?

Every redox reaction has both a reduction reaction and an oxidation reaction.
I'm going to assume two electrodes in the solution. Thus in a redox reaction one electrode will have an oxidation reaction and the other will have a reduction reaction. The two electrodes will be connected via a wire through which electrons flow. Without the wire no reaction will occur.

So probably if you place it in distilled water it will not ionize, right?

There are two aspects to the answer to this question.
First distilled water doesn't conduct well. So even with our two electrodes in the distilled water we wouldn't see much current flow through the wire connecting the two electrodes. In this case it isn't the wire limiting current flow but the counter flow of ions through the distilled water.
Second chemists chose "electrodes" that will be inert in the solution being used. For example if a sodium metal "electrode" was dipped into water then a violent reaction would occur spontaneously on the contact surface between the electrode and the water. The products would be hydrogen gas and sodium hydroxide. So a sodium metal electrode in an aqueous solution just won't work.

Follow up question: If the solutes in the water are the reason of ionizing the electrode, Do they rip off the electrons from the electrodes atoms? (Most of websites illustrate that both electrons (on the electrode) and ions (in solution) contribute to the voltage of the electrode).

An overall redox reaction can be written as two half-cell reactions. The gist is that we can use electrons to balance the formula. This makes it easy to balance an overall redox reaction since we just have to balance the electrons.
So for an oxidation half-cell for a zinc electrode we have:
$\ce{Zn -> Zn^{2+} + 2e^-}$
For a reduction half-cell for a copper electrode we have
$\ce{Cu^{2+} + 2e^- -> Cu}$
Since we have two electrons in both half cells we can just add the equations as is to get:
$\ce{Cu^{2+} + Zn -> Cu + Zn^{2+}}$
Now there is a quirk in this particular reaction. If the copper ions in solution, which are to be reduced, come in contact with the Zn electrode then the copper ions would plate on the Zn electrode. So a membrane of some sort must be used as a barrier which keeps the copper ions away from the Zn electrode but still allows anions to flow through the barrier. Of course if we used ordinary glass as the barrier then neither anions nor cations could flow through the barrier and the battery formed by the two half-cells wouldn't work.

So why do I use electrode potentials if I have lets say a bowl made purely of Zn atoms and it contains lets say Copper ions. Even though I only have positive ions of copper. Does the copper ions generate the same voltage by their own without electrons?

Well a bowl can't be pure Zn and have Cu too.
So let me go to a hypothetical situation. I haven't checked Cu/Zn alloys (which would be possible but complicates the general idea here). So let's just assume a Cu/Zn alloy. When metals or alloys become solid they tend to form grains which are interlocked together by their 3D surfaces rather than one homogeneous material. The grains will vary in chemical composition. So the different grains have different electromotive potentials because of their compositional differences.  Thus corrosion will occur. Some grains will dissolve and other grains will plate the dissolved ions back to metallic atoms.
Now let me clarify that corrosion is a very complicated subject for which I have made a gross generality. The actual corrosion reactions will depend on the metals in contact, and the exact composition of the solution with which they are in contact.

._., I didn't mean a galvanic cell. I just meant a half cell by it self. If you put an electrode in it, it starts to create ions and create a potential.
http://www.chemguide.co.uk/physical/redoxeqia/introduction.html

That is an entirely different question. Can I assume that your confusion is with the first three figures shown for the magnesium electrode?

Yes let me break the questions apart:
Suppose you have a half cell and You place Mg electrode inside. Explain why:

Why does Mg atoms form ions in the water? Where do the electrons get the energy from to escape Mg atoms pull?
2)I have a problem with electrons being left on the electrode and not carried away ( if I say by a solvent).
I will post a picture of What I want

Waiting for your picture, but let me make some comments about this. (1) This opens a can of worms. For all practical purposes you can ignore such dissolution at what I perceive your understanding level.  (2) In order to measure any sort of electrode potential then you'd need two electrodes in the solution. (3) There are two phases here. One is solid the other liquid. The phase boundary is sort of like them being separated by a "force field"  from the Starship Enterprise. ;-) So the ions or electrons have to somehow "jump" that barrier.

I would like that can of worms for a better understanding .-.,
http://i.imgur.com/yZDoOQl.png
So at the top you can see a electrode disassociating in water forming copper ions and leaving electrons on the electrode(according to the website and you still didnt explain why :C) So from a physics point of view both electrons and ions contribute to the field generated between the ions and electrons thus should have a bigger voltage than (at the  bottom) just copper ions their voltage should be a bit smaller than in the electrode. But we still interpret the copper ions and zin ions as E-pote


text:
Would magnesium and calcium hydroxide undergo a single replacement reaction?
Can calcium hydroxide react with magnesium (if given sufficient activation energy) to produce magnesium hydroxide and calcium, in the same way that NaOH + $\ce{Mg} \ce{->} \ce{Mg(OH)2} + \ce{Na}$ ?
Or, is calcium's electronegativity (or some other factor) prohibitive of this?

The key part of the question is given sufficient energy.  The difference in electronegativity between Ca and Mg is ~0.3 eV. The relationship of eV:K is about 1:12,000. Given that, you should be able to calculate a temperature at which the reaction $\ce{Ca(OH)2 + Mg -> Mg(OH)2 + Ca}$ is favored over the reverse.
However, temperature is based on the random motion of particles, so the energy distribution follows a Gaussian curve. Some Mg atoms are sure to have enough energy to knock loose some Ca atoms at room temperature.
Furthermore, you can force the suggested reaction by making use of relative solubilities. Find a solvent (not water, since magnesium and calcium are far more electronegative than hydrogen, and would decompose water to form hydroxides of both metals) in which $\ce{Ca(OH)2}$ is more soluble than $\ce{Mg(OH)2}$. Precipitation will "lock up" the $\ce{Mg(OH)2}$ as it forms, little by little, forcing the reaction to proceed even though it is endothermic.
Of course, you could also force this to proceed in molten salts by driving it with an electric current.


text:
Is there an analytical chemistry alternative to the Scoville scale for the hotness of chilli?
The traditional way of measuring the perceived heat of a chilli is the Scoville scale which involves panel testing of diluted extracts taken from chilli. But panels are very variable and the results appear to be a little unreliable.
But we know the class of chemicals that creates the sensation of heat: capsaicinoids. Can't we just measure those directly?
Is there a better analytical method for measuring the heat of a chilli?

Matt did find a method that uses HPLC on the peppers alone, not in a dish.
So given that the capsaicinoids is a class of compounds, each different capsaicinoid would have a different perceived hotness. So a chemical test would measure the concentration of each type capsaicinoid and then use a weighted average over the human perceived hotness for each member of the capsaicinoid class.

Apparently not everyone agrees. See this on fivethirtyeight.


text:
Is it true that calcium carbonate is more soluble in NaCl than deionized water?
I have two mixtures: $\ce{CaCO3}$ in NaCl and $\ce{CaCO3}$ in deionized water (DI). Is it possible to get more $\ce{Ca^2+}$ ions in NaCl solution than DI?

NaCl? You mean salt water?

You claim to have a solid solution of Calcium carbonate in solid (or molten? you didn't specify temperature!) sodium chloride. I doubt it.
Because $\ce{CaCl2}$ is more soluble in water than $\ce{CaCO3}$, addition of $\ce{Cl^-}$ ions would be expected to increase the solubility of $\ce{CaCO3}$ in water. In other words, "yes, it is possible." One crude way to think about it is that the "average" environment for the $\ce{Ca^{2+}}$ ions is more "friendly" in a mix of $\ce{CO3^{2-}}$ and $\ce{Cl^-}$ ions than in a mixture of $\ce{CO3^{2-}}$ions alone. The effect is called the "Common Ion" effect. Look it up on Wikipedia.

It depends... I'm going to assume no carbonate in either the pure water or the NaCl solution.
Calcium carbonate is fairly insoluble. The $\rm K_{sp}$ from Wikipedia is $3.3×10^{−9}$.
Now we typically just use the concentrations of $\ce{Ca^{2+}}$ and $\ce{CO3^{2-}}$ when using the formula, but it is really the activities that should be used not the concentrations.
In a weak solution of NaCl then the $\ce{Na^{+}}$ cations and $\ce{Cl^{-}}$ anions won't effect the activity of the $\ce{CO3^{2-}}$ so more calcium carbonate won't dissolve.
In a strong NaCl solution, just because of the ionic strength the activities of $\ce{Ca^{2+}}$ and $\ce{CO3^{2-}}$ will be lowered. In addition there will be clusters/complexes such as $\ce{NaCO3^{-}}$ forming which lower the activity of the $\ce{CO3^{2-}}$ species even more. This will cause a little more calcium carbonate to dissolve than it would in pure water.
I'll digress and point out that the solubility product itself really isn't a constant but depends on the ionic strength of the solution.
But, off the top of my head, I can't remember if there is a way to calculate [$\ce{Ca^{2+}}$] as a function of [NaCl]. Such a calculation would have to consider both the activities of $\ce{Ca^{2+}}$ and $\ce{CO3^{2-}}$ and the change in the solubility product itself.


text:
Temperature and moles of gas given volume, pressure, enthalpy and constant pressure molar heat capacity
I've been assigned this question as homework and have been attempting it for hours but cannot seem to get anywhere. Any help would be greatly appreciated!

An airbag of volume $40\ \mathrm L$ is inflated adiabatically through the following reaction:
$$\ce{NaN3(s) -> Na(s) + 3/2N2(g)}$$
One mole of $\ce{NaN3}$ decomposed produces an enthalpy of $-23.1\ \mathrm{kJ}$. $60\ \%$ of this energy is deposited in the gaseous product; assume the decomposition is instantaneous. $C_{\mathrm m,p}(\ce{N2})=29.12\ \ \mathrm{J\ mol^{-1}\ K^{-1}}$. When fully inflated, the pressure of the $\ce{N2}$ gas inside the airbag is $2.5\ \mathrm{atm}$.
Estimate the mass of $\ce{NaN3}$ needed to operate the airbag and state any assumptions or approximations being made.

So far, I'm only sure that I've done the following correctly:
$H=U+PV=q+w+PV$ where $q=0$ since the process is adiabatic, so $H=w+PV$
For the expansion of $\ce{N2}$ gas:
$H=0.6n(23.1\ \mathrm{kJ})=n(13.86\ \mathrm{kJ})$ where $n$ is moles of $\ce{NaN3}$
$n(13.86\ \mathrm{kJ})=U+PV$ so $U=n(13.86\ \mathrm{kJ})-(2.5\ \mathrm{atm})(40\ \mathrm L)=n(13.86\ \mathrm{kJ}-RT)$
$q=0$, $P=2.5\ \mathrm{atm}$, $V=40\ \mathrm L$ and $C_{\mathrm m,p}(\ce{N2})= 29.12\ \mathrm{J\ mol^{-1}\ K^{-1}}$
I'm not sure what else to do from here. I'd really appreciate any advice or if anyone could point me in the right direction. Thank you!

Hint: A general outline of what I think is the idea behind this problem/ This is a quick back of the envelope estimate:
$$\ce{NaN3(s) -> Na(s) + 3/2N2(g)}$$
The reaction given above takes place, and releases:
$$\Delta H = -23.1\ \mathrm{kJ} $$
60% of this heat is transferred to gaseous nitrogen, and since the process is adiabatic no additional heat enters/leaves the system. All of this energy goes into expansion work
$$Q = \frac{60}{100} \times 23.1 \ \mathrm{kJ} = 13.86\ \mathrm{kJ} $$
Temperature of $\ce{N2(g)}$ (assuming initially it is at $298\ \mathrm K$) can be found using (first, assuming one mole of azide decomposes releasing 3/2 moles of nitrogen)
$$Q = nC_{\mathrm m,p} \Delta T$$
$$T_{\text{final}} = \frac{Q}{nC_{\mathrm m,p}} + T_{\text{initial}} = 615.3\ \mathrm{K}$$
Volume occupied by 3/2 moles of nitrogen at the above temperature and given pressure is
$$V = \frac{nRT}{P} = 30.3\ \mathrm{L}$$
You need another $9.7 \ \mathrm{L}$ of gas to completely fill the airbag, I am sure you can complete the calculation. I'll come back and post a full, detailed answer later.

Complete Answer
So, we need a volume of $ 40 \ \mathrm{L}$, at an unspecified temperature.
$$\frac{PV}{n_{\ce{N2}}R} = T_\text{final}$$
Additionally, $$T_{\text{final}} = \frac{Q}{n_{\ce{N2}}C_{\mathrm m,p}} + T_{\text{initial}}$$
I am assuming since the process is adiabatic no additionally heat enters/leaves the system. The way I see it, the energy generated during this decomposition would heat the nitrogen produced to some (as of yet unknown temperature), and would be used for doing expansion work. I assume everything is initially at $298 \ \mathrm{K}$
$$\frac{PV}{n_{\ce{N2}}R} = \frac{Q}{n_{\ce{N2}}C_{\mathrm m,p}} + T_{\text{initial}}$$
From the reaction stoichiometry, $n_{\ce{N2}} = \frac{2}{3}n_{\ce{NaN3}}$
Thus, the expression we are interested in is
$$\frac{PV}{(2/3)n_{\ce{NaN3}}R} = \frac{Q}{(2/3)n_{\ce{NaN3}}C_{\mathrm m,p}} + T_{\text{initial}}$$
Now, it is just a matter of plugging in appropriate values, and solving for $n$
I used,

$ P = 2.5 \ \mathrm{atm}$
$ V = 40 \  \mathrm{L}$
$ R = 0.0821 \ \mathrm{L \ atm \ mol^{-1} \ K^{-1}}$
$ Q = 0.6 \times 23.1 \times 1000 \times n_{\ce{NaN3}} \ \mathrm{kJ}$
$C_{\mathrm m,p} = 29.12 \ \mathrm{J\ mol^{−1}\ K^{−1}} $

Answer given below:

 Solving all of this I get no. of mol of azide $\approx 1.33401 \ \mathrm{mol}$ and, no. of mol of nitrogen produced $ \approx 2.001015 \ \mathrm{mol}$. Additionally, $T_{\text{final}} = 615.307 \ \mathrm{K}$ Plugging all of this back into the ideal gas equation and solving for volume at the given pressure gives me $40.4\ \mathrm{L}$

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text:
Why does the energy required to break a bond decrease as the number of bonds increase?
I noticed that the energy to break a single C bond is $348\ \mathrm{kJ}$, to break a double C bond would be $612\ \mathrm{kJ}$, and to break a triple C bond would be $837\ \mathrm{kJ}$. How come the energy required per bond breakage decreases as the number of bonds decrease?
It seems that for double bonds, the energy required to break a bond is $(612/2) = 306\ \mathrm{kJ}$. Why is this so?
Thank you.

It guess its obvious why it takes more energy to break a double bond that a single one, ( more electrons in a double bond for example) but are you asking why the energy to break a double bond is not twice that of a single?

I thought the bond-breaking enthalpy is different for different compounds...

"for double bonds, the energy required to break a bond ..." double bond is one bond...

You are making the false assumption that a double bond is composed of two single bonds, when in fact it is not so.  The two bonds in a double bond are not identical.  In simplistic terms a double bond is composed of a sigma bond and a pi bond.  Sigma bonds are the strongest type of covalent bond and therefore require more energy to break than a pi bond.  This is why a carbon-carbon double bond doesn't require twice the energy to break as a carbon-carbon single bond.  Other factors come into play, but thats the answer in simplistic terms.

There is no simple answer other than "because chemistry and stuff". The bonds are different; one is $\sigma$, another is $\pi$. To make everything worse, think of nitrogen, which is the other way around: triple bond $\ce{N\equiv N}$ is stronger than three ordinary bonds $\ce{N-N}$.

@Keaton I'd like to encourage you to transform your comment into an answer. While it is indeed a very simple explanation and Ivan is also correct, it does provide a good starting point for the OP (and you might gain some reputation).

You are considering all the bonds equivalent and that's where you are making mistake.The energy required to break second bond will depend upon several factors such as repulsion, configuration, also note that there are different types of bond formed (sigma and pi bonds) and we know that energy required  to break a sigma bond is greater than that of pi bond.

I would be interested to know why a sigma bond takes more energy to break than a pi bond.

The reason to that is that a sigma bond is formed by the head on overlap of orbitals,whereas a pi bond is formed by sideways overlap,resulting in a less stronger bond


text:
What is the reasoning for higher pH values in this scene?
In this article regarding low-boiling point amine-catalyzed biodiesel transesterification reactions, it is stated that,

"From Table 1, we can also see that, for the isomers, the
  branched amines always showed higher yields of methyl acetate
  than the normal amines. For example, the yields of methyl
  acetate for n-dipropylamine, n-butylamine, and n-propylamine
  were 25.3%, 46.1%, and 48.2%, respectively, while those for
  diisopropylamine, tert-butylamine, and isopropylamine were
  42.1%, 49.6%, and 50.7%, respectively. This was probably due
  to the high pH values of the branched amines than the normal
  amines."

Why is this the case? Why do high pH values correlate to higher yields? Does the geometry (branched or linear) of the amine play a role in this?

Branched amines are more basic than their linear counterparts because inductive effect of alkanes places more electron density in the adjacent carbon atom. The higher the electron density near/at the amine group, the higher its pH will be.

Also, steric factors also make the amine more planar, so the lone pair orbital is going to have less mixing/stabilization from the orbitals of the other substituents.


text:
How does an anode metal come into solution if it cannot react with an electrolyte?
From what I understand, an anode material is chosen so that it does not react with the anode electrolyte solution.
I was just wondering, then, that when you have a galvanic cell which has a Mg metal anode submerged in sodium chloride solution, how the magnesium metal gives two of its electrons and come into solution (as magnesium ion) even though it does not react with (and cannot react with) sodium chloride?
What I mean is, in a typical single displacement reaction, there is a metal that reacts with the solution and comes into solution. For example, with $\ce{Zn}$ and copper sulfate, the zinc replaces the copper and comes into solution as zinc ion. However, magnesium doesn't react with sodium chloride (because the activity series predicts it can't). How then can we say that we have magnesium ion in solution?
Is it that the metal is more stable in its ion state and thus readily gives up its electrons?
Traditionally, the anion is the one responsible for giving electrons: the chloride ion in the sodium chloride solution should come into contact with magnesium anode and give its electrons, which the magnesium metal should then carry. And thus the reaction should be
$$
\ce{ 2Cl- \!(aq) -> Cl2\!(g) + 2e-}
$$
and not
$$
\ce{Mg\!(s) -> Mg^2+\!(aq) + 2 e-}
$$
However, my books don't say this: they say that magnesium gives up two of its electrons and enters the solution.
What am I understanding wrong?

For understanding the reason, you must have knowledge of electrode potentials and electrochemistry

RE: From what I understand, anode material is chosen so that it does not react with the anode electrolyte solution. You don't want an electrode which would spontaneously react by itself in the solution. For instance a sodium electrode in a galvanic cell with a water medium would decompose in the water to sodium hydroxide and hydrogen gas.

@MaxW Magnesium anodes are used quite frequently in electrochemistry. They do react with the electrolyte, but quite slowly; slowly enough that they are perfectly safe to use, and last sufficiently long to be of practical value. See, e.g., here and here.

I was just wondering then that when you have a galvanic cell which has a $\ce{Mg}$ metal anode submerged in sodium chloride solution, how the magnesium metal gives two of its electrons and come into solution (as magnesium ion) even though it does not react with (and cannot react with) sodium chloride?

Your misunderstanding stems from assuming that the electrode reaction has to be the same sort of single-displacement redox reaction that would occur with no electrodes present at all.
At a dissolving magnesium anode$^\dagger$, the only reaction that needs to occur is the one you wrote involving magnesium:
$$
\ce{Mg\!(s) -> Mg^2+\!(aq) + 2e-}
$$
The potential applied to the electrode pair is what enables the above reaction to occur without a paired reduction reaction at the anode surface.  Instead of being taken up in another chemical reaction, the electrons are drawn into the anode and on into the connected electrical circuit. Those electrons are then "pushed" into the cathodic reaction, whatever it might be.

From what I understand, anode material is chosen so that it does not react with the anode electrolyte solution.

This is true, but does not have to strictly hold. MaxW is right: $\ce{Mg}$ does react directly with water. But, the reaction is slow enough that it's perfectly safe, and that one can conduct useful electrochemical operations even while it's going on (e.g., see here and here).

Is it that the metal is more stable in its ion state and thus readily gives up its electrons?

Despite what one reads from a table of standard reduction potentials or a metal reactivity series, in practical electrochemistry, the above is true for most metals, except for the noblest of the activity series. In my experience, it's usually a lot easier to dissolve metals than to electrodeposit them.  You may have to use additional chemistry to make it happen (e.g., including $\ce{HF}$$^\ddagger$ to strip the passivating oxide from $\ce{Ti}$, $\ce{Nb}$, $\ce{Ta}$, etc.), but once the metals are oxidized/dissolved they tend to stay that way.

 $^\dagger$ Operating at $100\%$ faradaic efficiency. Other reactions that might occur in this system, depending on the applied potential, include the chloride oxidation reaction you noted and water electrolysis to oxygen gas.
 $^\ddagger$ $\ce{HF}$ is very dangerous, do not use at home!

can you please elaborate on the sentence "The potential applied to the electrode pair is what enables the above reaction to occur without a paired reduction reaction at the anode surface."

@user510 The potential applied to the two electrodes allows the reduction and oxidation reactions to occur separated in space. Thus, in theory, in this system it is possible for the only reaction at the anode to be magnesium dissolution; the corresponding reduction reaction will take place at the cathode, instead of also at the anode. Without the electrodes and the potential, the reduction reaction would have to occur at the surface of the magnesium.

if the potential applied to the electrode pair is what enables the above reaction to occur, then why is there resistance in battery even with weak electrolytes (i.e. I could use any metal and it would give up its electron).

Also, why is it the potential that allows magnesium to give its electron?

@user510 if the potential applied to the electrode pair is what enables the above reaction to occur, then why is there resistance in battery even with weak electrolytes (i.e. I could use any metal and it would give up its electron). I'm sorry, this question doesn't make sense to me.

@user510 Also, why is it the potential that allows magnesium to give its electron? See, e.g., this question and its answer.

this answer is whether the reaction is spontaneous, but I am not asking that

Let us continue this discussion in chat.

Hmmm. Electrochemistry can definitely be confusing. Context is quite important. Anyone who makes a broad claim that anodes don't react is just plain wrong. It depends on the set-up, including the applied current (and/or voltage). Anyone who makes the statement that in measuring electrolytic reactions both electrodes should be inert is going to be right a lot (that is, more often than someone will be able to find exceptions).
Magnesium is a VERY reactive metal! The only reason it doesn't explode into flame upon contact with water is that the coating of MgO slows down the reaction. Anyone who thinks putting a piece of Mg metal into a salt solution is a good idea, needs to be very well aware of the hazards.
When thinking about electrochemical reactions, you have to be aware of all of the possible oxidations and reductions that can go on. Running current through pure water can generate H2, O2, as well as several other compounds (in trace amounts, usually). As you probably know, the concentration of ions in pure water is very low [H] = [OH] = 1E-7. Which means that the amount of current (work) you can do is quite limited. So, NaCl (or some other electrolyte) is added. This complicates things since potential reactions (is that punny or what?) include Na(+)→Na° (not going to happen), Cl(-)→Cl2 ( one way to make chlorine gas), and things like Cl(-) + H2O→ OCl(-)+... (and a bunch more). Only if you know what the reaction potentials are for all of the possibilities can you make good predictions. Some electrode materials can act as catalysts (Pt electrode, for instance). These lower the EP necessary while remaining "inert".
It's been so long since I did any electrochemistry (decades) that I don't recall which electrode is the "anode" and which is the "cathode"...doesn't it depend on the type of cell involved? I've forgotten. I do remember that Leo says Ger (Leo is another name for lion, and Grrr is the English for the sound a threatened (or threatening) animal makes.) LEO = Loss of Electrons is Oxidation. GER = Gain of Electrons is Reduction. You have written one oxidation and one reduction. In a driven electrolytic cell, they will NOT occur at the same electrode!!!!!!!!

Magnesium will react with water, but it is nearly as reactive as sodium. Even if you cleaned off the typical oxide coating it still won't react violently but very slowly. Any such reaction though means that magnesium would be a poor choice for a battery that used any water in the electrolyte.

@MazW do you mean magnesium submerged in aqueous solutions is bad?


text:
What is the solution and the solvent in 80% solution of ethanol in water?
I was wondering whether the solvent is water or ethanol, I think it must be ethanol because it is present in larger quantity; however, on the other hand water is the universal solvent so shouldn't it be the solvent.

The solvent is ethanol. Since both ethanol,water and solution are in the same phase and ethanol is in greater proportion, it will be called as the solvent

Of course the notion depends on context too. If I were studying a bunch of alcohols at different concentrations in water, then I would probably refer to water always as the solvent and whatever alcohol as the solute. Imagine trying to create a table and flipping from water being the solvent to ethanol. It would be really confusing.

You're severely misinterpreting the meaning of the phrase "universal solvent". All it means is that water can dissolve a lot of things, i.e. it can be a solvent for many chemicals. (And on top of that, that phrase is nearly always used in a biological context, because water is rubbish at dissolving most organic compounds.) It does not mean that water always takes on the role of the solvent.

@orthocresol - Geologists and mineralogists  "sometimes" use water too. It seems that in spite of the human preference for ethanol, mother earth likes water.

Magical thinking, such as thinking that because a certain chemical has a certain name, that (absent a systematic nomenclature system which requires the name to correspond to certain properties) it will magically exert some unique influence, is almost always incompatible with science. There is no requirement that a green house must be green, nor that the "universal solvent" (both water and time are often called that) means everything dissolves in it. Logical analysis requires the recognition that "a rose by any other name will smell as sweet". That is, what something is called may or may not help to characterize it.
Your question shows that you haven't discriminated between an 80% solution by mass fraction, by volume fraction, or by mole fraction - all of which are different compositions, in general. The more pointed question is if a mixture is exactly 50% of one, and 50% of the other is it a solution? And if so, which is the solvent and which is the solute? Note that the mass fraction is a conserved quantity (it doesn't change), but the volume fraction may not be a linear function of mass % (often abbreviated v/v and w/w, respectively (w is for "weight" even though chemists should know better) There are a variety of conventions which determine which of two liquids is the solvent and which is the solute. In dilute solution, the choice is obvious, but as concentration rises above 10%, there isn't necessarily one answer correct answer. Consider a solid dissolved in water at 80%. We'd still call that an aqueous solution. Even 80% sulfuric acid (not that "pure" sulfuric acid is a stable compound, but it should be considered a liquid (at STP) for most purposes) is generally considered an aqueous solution. Note that both of these solutions are quite un-water-like, compared to solutions where the solute is present at 0.1 molar or less (and is of fairly small molecular mass, say <1000 g/mole (but there's no definite cut-off)). Chemistry, more so than physics, is a pragmatic science. To give a good answer to your question, you'd have to explain the context you wanted to use "solvent" and "solute" in describing your mixture. The general rule of thumb is that the solvent is the continuous phase, but that answer has some problems (which I'll not go into here). A 1:1 mixture by moles would have 18g of water for every 46 g of ethanol or would be about 72 weight % ethanol. (EtOH/Water mixtures do not conserve volume, so that adding 28 ml of water to 72 ml of EtOH would result in less than 100 ml. But volume % is generally meant to imply volume of the starting materials rather than the volume each compound occupies in the mixture). So an 80% solution would mean that there's slightly more EtOH (by volume, much more by mass and slightly more by moles) than water. There are three perfectly valid answers: 1) EtOH is the solvent. 2)Each is dissolved in the other. 3)Water is the solvent. 4)It's better called a mixture than a solution, (even better a "homogeneous mixture"). Oh wait, did I say 3? LOL! To reiterate, as the concentration increases, it becomes less and less USEFUL to characterize one liquid (in a binary system) as the solvent and the other as the solute. So the answer is moot.

I think you have very good points, but it could do with a bit of paragraphing. By the way, to make a new paragraph, you need to press Enter twice.

To identify a solvent:
1) If the components are in different phases, then component having same phase as that of solution is considered as solvent.
2) If both the components are in same phase, component having low b.p. is considered as solvent.

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text:
Internal Energy vs Enthalpy
I struggle to see the difference between internal energy and enthalpy, or at least the need for differentiation between the two. I understand that enthalpy, supposedly unlike internal energy, accounts for the expansion work required for the system to exist, but it seems that that would only be necessarily stated if one had failed to account for the negative contribution to work in the first place.
Could enthalpy, as $$H = U + pV$$ and thus, $$H = q + w + pV$$ not be written as equally validly as, $$H = q + w_1 + w_2$$ i.e. $$H = U \ ?$$
It would seem that this definition of enthalpy implies that internal energy accounts only for work done to the system, disregarding work done by it.

In general, $H \neq U$ (most definitely). I don't quite follow your train of thought, could you please elaborate.

I can certainly try. So, from what I understand internal energy is derived from contributions of work and heat. Enthalpy is then taught as a more useful property because it also accounts for the expansion work done on the surroundings by the system in order for it to exist (i.e. gases "pushing back the atmosphere" to make room for themselves). This would obviously require a transfer of energy from the system to its surroundings and could thus be considered a negative contribution to the work making up the internal energy. Should this negative not just be counted as a change in U itself?

You have really bad misconceptions... Firstly $U \neq q + w$.. heat and work are ways of changing the internal energy, viz. $\Delta U = q + w$.. they are not the "two components of internal energy". Secondly, $w \neq pV$, but rather $w = -p\Delta V$... And lastly, just get this definition of enthalpy out of your head. It doesn't help at all to try and assign a meaning to the total enthalpy of a system.

Please, go and look up a textbook, or ask a teacher. I recommend Levine's Physical Chemistry. The point of using enthalpy, is because enthalpy changes are equal to heat supplied at constant pressure. Therefore the change in enthalpy of a system is something that one can directly measure, e.g. using calorimetry: $q = mc\Delta T$. On the other hand, internal energy changes are equal to heat supplied at constant volume. The issue: most chemical rxns that we do are carried out at constant P, not V. Hence, $\Delta H$ more useful than $\Delta U$, not $H$ more useful than $U$!

That equation only holds at const-p. Not that it matters, I just wanted to get that out of the way. 2) So, you have $\Delta U = q + w = q - p\Delta V$ (definition of p-V work) which means $\Delta H = q - p\Delta V + p\Delta V = q$ That term $+p\Delta V$ is not a further component of work, in fact it cancels out the work term arising from $\Delta U$. As for the reasoning for why $H$ is defined in that way, it is not because of some magical significance of the total enthalpy (...)

(...) but rather because defining it that way gives it certain special properties, of which the most important is $\Delta H = q_p$. Levine puts it this way. http://i.stack.imgur.com/BIXjn.png

Trying to figure out a good intuitive physical interpretation of enthalpy has caused thermodynamics student over the last couple of centuries to waste huge amount of their valuable time (in my judgment).  The really fundamental physical quantity is internal energy U, not enthalpy, and its physical interpretation is simple and straightforward.  I regard enthalpy as just a convenient function to work with in solving many types of physical problems.  If the function enthalpy had never been invented, we would still be able to solve these problems, but the results would not be as concise.

Your derivation is obviously flawed, and unfortunately you reasoning doesn't make much sense to me either.
You assume $U = q + w$ which is clearly not true; similarly $ \mathrm{d}w = -p \int \mathrm{d}V$ (and not $w = pV$ as your derivation seems to suggest).
Your reasoning/intuition about enthalpy seems ill founded as well, sorry. Perhaps, it's simplest to think of enthalpy as heat exchanged at constant pressure. Why? Because it turns out to be an extremely useful and convenient state function if we define it that way. Since most reactions, experiments take place at constant pressure (and calorimetric determination of heat exchanges is not particularly hard).
I did a derivation (of sorts) for enthalpy using Legendre Transforms, a while ago for another question. The question has since been closed, which is why I have reproduced the derivation below in case you are interested in it. Perhaps, it will help you understand why enthalpy is defined the way it is.

First a brief note on the mathematics of Legendre Transforms
Consider a function in two independent variables: $ f(x,y)$
We right the differential of this function as follows:
$df(x,y) = \Big( \frac{\partial f}{\partial x}\Big)_ydx + \Big( \frac{\partial f}{\partial y}\Big)_xdy$
Now, I shall define $u := \Big( \frac{\partial f}{\partial x}\Big) $ and $w := \Big( \frac{\partial f}{\partial y}\Big)$
so, $df = udx + wdy$ [equation 1]
Here, u and x (and similarly w and y) are called conjugate variables. (i.e $ux$ and $wy$ will have the same units as $f$)
[Think of it this way: In thermodynamics, PV, TS have the same units as H, U,G,A]
Now, our goal is transform our original function $f(x,y)$ to a function $g(x,w)$. We wish to replace y, with it's conjugate variable and the function $f $ and $g$ will have the same units.
so, we take the differential of the product $wy$ : $d(wy) = ydw + wdy$
We subtract this from [equation 1] and get $d(f-wy) = udx - ydw $
and can deduce that our Legendre Transformed function $ g(x,w) := f-wy$
Now, we can easily apply this method to Internal Energy (U) to obtain Enthalpy (H)
We have $U(S,V)$ and $dU(S,V) = \Big( \frac{\partial U}{\partial S}\Big)_VdS + \Big( \frac{\partial U}{\partial V}\Big)_SdV $
Now, here we seek $H(S,P) \equiv g$
a)  $f \equiv U$
b) $S \equiv x$ is the variable we hold fixed
c) $V \equiv y$ is the variable we switch
d)  $\Big( \frac{\partial U}{\partial S}\Big)_V := T \equiv u $(the conjugate of the variable we are holding in place
e) $\Big( \frac{\partial U}{\partial V}\Big)_S := -P \equiv w$ (the conjugate of the variable we change
$ g =f-wy $
so, $H(S,P) = U-(V)(-P)$
Or, $H(S,P) = U + PV$
From here, you can proceed to write down: $dH = dU + PdV + Vdp$
(or for macroscopic changes: $\Delta H = \Delta U + \Delta (PV) $)
We impose the constant pressure condition to get
$\Delta H = \Delta U + P\Delta V $


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Do Pigments, Dyes (with plastics) or Plastic Itself Exist that Appear White but are NIR transparent?
I am looking to make a NIR (Near Infrared) window for a project I am working on but need it to appear white rather than the typical Reddish Black NIR window found on most remotes.   Does a dye or pigment exist that makes PMMA or PC (open to other injection moldable thermoplastics!) look white but allows NIR (specifically 940nm) to pass with low loss?

What about the plastics used to encapsulate LED NIR emitters? These are usually, I think, epoxy resins. In fact, these days many IR-based remote controls will use a naked LED with no other window so do you really need an extra window?

interesting.  Most I have seen are either a blackish appearance or clear.  I will look around for white and report back my findings.