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Alloys for inkless metal pen
What alloys can be used to make a metal pencil like this one?

"Silverpoint" nibs have a long history in art and significantly predate modern drawing tools such as the lead pencil (made of graphite and clay but possibly called lead as lead metal was one possible material in metal pens). Silverpoint writing and drawing doesn't smudge or fade and many examples have survived almost unchanged over hundreds of years.
There is a useful site with a good history of the original artistic uses of silverpoint pens here. According to this site a wide variety of metals will work. Historically which was used depended on status:

Wealthy and royal persons who needed to keep records (business inventories and so forth) employed these scribes, and the stylus material served as notice to all of the status of the employer - gold and silver for the upper classes, lead for somewhat lesser beings.

It seems tht many metals will work. The dictionary of Art is quoted as saying:

Metalpoint. Drawing instrument (the forerunner of the pencil) made from a small, pointed metal tip, usually of lead, silver, copper or gold, encased in a wooden holder.

The current preferred metal for artistic use seems to be fairly pure silver.

Best results are obtained from a thicker point that is "pure" silver, or ".999 fine." This is much softer than sterling, so it will deposit more metal on the drawing surface

The silverpoint.com web site sells ready made points of both silver and gold.
It isn't obvious what metal or alloy is used in the pens mentioned in the Wired story, but it isn't silver. It apparently contains at least some lead as the pencils come with warnings not to let children chew them!

Thanks for giving the names of this technique. I still wonder what they use in the commercial pens, according to the picture it seems to work very well. I tried silver, and it leaves a rather faint trace and needs a lot of force. I worked with bismuth crystals and remember that it was leaving traces everywhere. Could be a suitable material, but I think it oxidizes too quickly.

I actually make metalpoint pencils and use pure lead for the writing material. It's obviously a bit toxic (because lead), though I'm experimenting with making custom lead alloys (possibly with a percentage of bismuth, as I have some laying around) to reduce the overall lead content. In terms of utility, however, pure lead leaves a reasonably dark and easy-to-make mark.

It could be similar to "Lead or Aluminum Metalpoint". For instance, I use the 2.4 mm and 1/8 inch "soft aluminum" rods obtained in the US from Hobby Lobby in a pack with two of both sizes (about $4 USD). The smaller rods are placed in my drafting technical pencil holder (Koh-I-Noor, 2 mm still works). I use an old wooden brush handle drilled to 1/8 inch diameter hole, for the larger rods. A strong gray/black line is obtained when using "Rite-in-the-Rain All-Weather Sketchbook" with no further "silverpoint" ground treatment. My "Lead solder' (Home Depot/Lowes) stylus works on regular paper with none of the usual titanium white ground treatment.


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Does potential difference or current drive electrolysis?
Would a high voltage ( of the order of a few score kV ) drive electrolysis?
Or, does it require a large current and low voltage? Alternatively, does electrolysis require both large voltage & large current?

For any electrochemcial reaction, electrons are transfer from the oxidant to the reductant. By definition, any flow of charge per unit time period is an electric current.
However, in the case of electrolysis, the electron transfer is not spontaneous. An external energy source is required for the reaction to take place. The energy (work) provided per unit charge is called the voltage.
For any electrolysis reaction to occur therefore requires a supply of electrical energy which means both a voltage and electric current are needed.
Specifically, the amount of work (energy) resulting from a voltage, $V$ and a current, $I$ applied for a time $t$ is given by:
Work produced by electrical energy, $W = V\cdot I\cdot t$
For a given electrolytic reaction to proceed, the required minimum voltage which must be applied is determined from the Gibb's function,
$E^\circ_\text{cell}=\frac{-\Delta G^\circ}{nF}$
Where $E^\circ_\text{cell}$ is the minimum voltage needed for the electrolysis reaction to occur, $\Delta G^\circ$ is the change in Gibb's free energy under standard conditions, $n$ is the number of electrons transferred and $F$ is Faraday's constant (96,485 coulombs per mole).
For example, if pure water is placed in an 'electrolytic cell' with two non-reactive electrodes (eg: platinum), electrons forced into the 'cell' by an electrical source (eg: battery) will react with water molecules at the cathode forcing them to lyze (split) into hydrogen ions and hydroxide ions.
$\ce{H2O(aq) ->[\text{elect}] H+(aq) + OH- (aq)}$
At the surface of the 'anode', hydroxide ion will be oxidised (dontate electrons) to form oxygen gas.
$\ce{4OH- (aq) -> 2H2O + O2 + 4e- }$ (anode half-cell reaction)
Meanwhile, at the surface of the 'cathode', hydrogen ions will accept electrons to form hydrogen gas.
$\ce{2H+(aq) + 2e- -> H2(g)}$ (cathode half-cell reaction)
The overall cell reaction is therefore:
$\ce{2H2O(aq) -> 2H2(g) + O2(g)}$ ; $\Delta G^\circ=237.2\ \mathrm{kJ/mol}$
The positive Gibb's function for this reaction indicates that the reaction will not occur sponteneously but requires an external energy source (eg: electrical energy), which is reflected in the negative cell potential.
The minimum (theoretical) voltage required is:
$E^\circ_\text{cell}=\frac{-\Delta G^\circ}{nF}=\frac{-237.2 \times 10^{3}}{2\times 96485}=-1.23\ \mathrm{V}$
However, in practice, a slightly larger voltage of 1.48 V is required since the  enthalpy (heating) of the products results in slightly lower efficiency which is manifested as an overpotential of about 0.25 V.
Once this critical voltage level is exceeded, the electrolysis reaction proceeds at a rate determined largely by the the current, since the current represents the rate at which charge is delivered to the system. Basically, the higher the current, the more molecules will react (electrolyze) and the more products (hydrogen gas) will form per unit time.

Current is, by definition, a flow of charged particles, so it doesn't drive anything.  Electrical potential is a store of energy (you can think of a voltage source as an ideal "battery" of sorts).
One volt can produce $\pu{1 A}$ of current through a $\pu{1 \Omega}$ resistor.  One ampere of current is equal to a coulomb per second.  One coulomb is $6.24150965(16)×10^{18}$ electrons.  Were you to use enough voltage that oxidation at the anode released enough electrons to produce a massive current, you'd likely deplete your anode material before you ever achieved it.

High voltage and low current is more efficient but must be oscillated at specific frequencies to be effective low cost electrolysis. it is the oscillation that breaks the bond of hydrogen and oxygen not the current or voltage. Current can be below .5 amp at millions of volts and still be non life threatening like a stun gun. But you need to experiment with the frequency of oscillation to find the most efficient electrolysis.

Welcome to Chemistry.SE! To acquaint yourself with this page, take the tour and visit the help center. Furthermore this tutorial shows you how math and chemical formulae can be nicely formatted on this site.

Do you have a reference for this? I have never heard anyone using high voltage for the electrolysis of water.

@JackWayne Roberts: "it is the oscillation that breaks the bond of hydrogen and oxygen not the current or voltage" Could you explain this? As far as I can tell, direct current electrolyses water just fine and has no oscillations. If you are just saying that an oscillating power source is more efficient then could you explain why that is?

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Writing the chemical equation given molarity
If I am given the molarity of two substances and I know how they react, how can I write the balanced chemical reaction? i.e. experimentally I found:
$\pu{1.0 M}$ $\ce{NH_3(aq)}$ mixed with $\pu{0.1 M}$ $\ce{BaCl_2(aq)}$  yielded no precipitate when $\pu{1ml}$ of each were mixed together. My unbalanced chemical equation so far looks like:
$$\ce{N^{-3}H3+(aq) + (1/10)Ba^{+2}Cl2-(aq) -> N^{-3}Ba^{+2} + Cl^{-2}H+}$$
My logic being: since the solutions are administered in equal volumes, their molarity determines the molar coefficients. Right off the bat I can tell that $\ce{NH}$ is not soluble, but the other compound I think is, which would contradict my observations. Either way, I am incredibly lost.

Have you considered the possibility that your reaction is wrong.  where, for example, is the barium on the product side.  I'll give you one hint here:  barium chloride dissociates in water to barium and chloride ions.  Ammonia reacts with water to produce some ammonium ion and the same amount of hydroxide ion.  However, barium hydroxide is totally ionized.
Can you take it from there?

I've fixed the mistake, but I'm still not sure what to do. Do I use solubility rules now? Should I even bother balancing the chemical equation?

Hint: List the possible products and determine which ones are insoluble.

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Acetic acid freezing distilation
How can acetic acid be distilled by freezing when its freezing point is above the freezing point of water.
Examples on the web show the water portion of the vinegar as ice and the acid as a liquid but Wikipedia shows that freezing point of acetic acid is 16.6°C.

All the sources I read confirm that acetic acid crystallizes before water upon freezing, and is separated that way. Can you include links to back your assertion (“examples on the web”)?

http://www.alchemywebsite.com/distillation_of_vinegar.html

The procedure you are describing sounds like fractional freezing. Acetic acid is frequently purified by this method, hence 100% acetic acid is often called "glacial" acetic acid.

As an example of how this works, let's consider a mixture of acetic acid and water.

Acetic acid has a freezing point of 16.6 oC. Water has a freezing point of 0  o. They form a eutectic at -26.7 oC (although I do not know the composition). The phase diagram probably looks something like (but is not) the following.

Phase Legend:

A = Liquid solution
B = Solid solution of acetic acid in water PLUS liquid solution
C = Solid solution of water in acetic acid PLUS liquid solution
D = Solid solution of acetic acid in water
E = Solid solution of water in acetic acid
F = Solid solution of acetic acid in water PLUS Solid solution of water in acetic acid

A mixture of acetic acid and water that is approximately 72% acetic acid is cooled until it begins to freeze (solid blue vertical line), which happens around 2 oC. The phase changes from A (liquid solution) to C (Solid solution of water in acetic acid PLUS liquid solution). The solid solution is enriched in acetic acid compared to the original liquid solution. The composition of the remaining supernatant liquid solution is enriched in water compared to the original liquid solution. The composition of the solid solution can be determined by drawing a horizontal (red) line to the phase boundary between C and E and reading down to 91% acetic acid. The composition of the remaining supernatant liquid solution can be determined by drawing a horizontal (green) line to the phase boundary between A and C and reading down to 63% acetic.
The supernatant phase is removed and discarded, and the solid phase is allowed to melt. The process is repeated. Each successive freezing yields a new solid solution that is enriched in acetic acid, until the purity is as high as you want it.
Note, that since water and acetic acid have a eutectic point, if we has started on the left side of the eutectic point, the phases would have been reverse. The solid would be enriched in water and the supernatant would have been enriched in acetic acid. The supernatant would be remove and kept, and the solid would be discarded. The supernatant would then be cooled and fractionally frozen again to further enrich the supernatant in acetic acid until the eutectic point is reached (at which point the composition of the supernatant is equivalent to the composition of the solid phase). If you want to isolate acetic acid from a solution that is mostly water, I would distil the solution until the composition is mostly acetic acid, and then fractionally freeze it.

This is a great answer, just one thing - every reference I've ever read regarding the phrase 'glacial acetic acid' indicates that the name comes from the ice-like appearance of the crystals rather than the mode of separation.

The eutectic point, is it at 50% for sure? I tried to look it up, but couldn't find it.

A bit late, but according to this link: http://www.ddbst.com/en/EED/SLE/SLE%20Acetic%20acid%3BWater.php (which links to to primary source), the eutectic point is -27°C / 71.6% (mol/mol) / 88.3% (kg/kg).


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How do I type a simple chemical equation in Microsoft Word?
How do I type a simple chemical equation in Microsoft Word? I can do subscripts, but long arrows are more difficult. I can't get them to align with the text.
Also, I can't figure out how to put a delta above the arrow for heat. I have tried the Chemistry add-on from Microsoft, but that does not seem to help with equations.

If you are using MS Word 2007 or newer, use the equation feature. It is designed for math but works okay for chemistry.
Go to the insert tab. (For shortcut you can press Alt+= sight together)

Click on the equation button on the far right.

Type in your equation. Use the buttons in the ribbon to do superscripts and subscripts. Alternatively you can use _ for subscript and ^ for superscript.  The default is to have letters italicized (as variables), so you will want to fix that.

There are also shortcut commands to render most the common things you want. For example, underscore _ creates a subscript and a caret ^ creates a superscript Shortcut for typing subscript and superscript in MS Word 2007|2010|2013|2016 and office 365  . You have access to a wide range of arrows from a pull-down menu, but -> will give you a simple right arrow (although it is not very long). This feature on Word will also accept some (but not all) tex commands for formatting equations.
To get a long arrow, click on the operator but and choose the arrow with the word "yields" written over it under common operator structures. For up arrow and down arrow showing gas liberation and precipitation use \uparrow or \downarrow followed by space Shortcut for typing arrows of chemical equation in Word 2007 and above.

Click on the word "yields" and replace it with as many spaces as you need to create an arrow of whatever length you want. Shortcut for other types of arrows is.

Finally, finish your equation.

If you need to type above or below arrow just type "\above(text above arrow goes here)[space]".Similarly tying below arrow just type \below(test below goes here)[space]". How to type chemical equation and arrows in Word 2007 and above.
For older versions of MS Word, go to the insert menu and click on the equation, which launches the Equation Editor Program (you can also find this program on your computer by searching for eqnedt.exe), which gives you the same ability to create equations.

Even though your question has already been answered (and this is not an alternative answer), but if you're open to it, switching from Word to LaTeX with the chemmacros package (PDF) will benefit you greatly in the long run.

Agreed. In the end, if you plan to do a lot of chemistry (and/or math), learning to use LaTex will be worth the effort.

If you're new to LaTeX, why not try an online compiler such as WriteLaTeX.com? We've added an example chemistry presentation to help new users get started (I'm one of the developers at writeLaTeX - any feedback is much appreciated, thanks). Hope this helps!

I found the wikibook to be a good place to start; it covers about anything one needs and does not require prior knowledge.

link has died...

The answer mentioned above is correct but there is also a shortcut built in which is math autocorrect. It is much like LaTeX. By default it's inactive but you can activate it and is really helpful if you want to write big equations. For example, if you want to type $\ce{H2}$H then you just have to type H_2.
Many more options are available. For example, if you want a superscript character, the caret (^) sign will be converted as superscript. Many such shortcuts are covered in these videos or you can simply search for How to insert mathematical equation (like LaTeX) in Ms-Office: Tips and tricks on YouTube.
These methods will be especially helpful if you are a fast at typing. Moreover, it will save your time which is lost while switching between keyboard and mouse and searching proper option in Word.

Frequently the best solution is to draw the equation in a chemical drawing software (there are several freeware options) and paste it in as a graphic in Word.

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Will gaseous ionic compounds be free moving ions?
I knew while learning about electrolysis that if the ionic compound is molten it becomes free moving ions.
If that is the case, what will happen if I continued heating till it reaches the boiling point so that the ionic compound evaporates?
Will it still be free moving ions?
Also, shouldn't the result be more efficient at electrolysis than in the liquid state due to increased mobility of ions? If not, why?

The phase you describe is a plasma.

I've incorporated your newer question into this one, they are nearly asking the same thing.

Usually not. Boiling point rarely exceeds 4-5 thousand kelvin. A typical ionic bound energy is about 5 eV. 1 eV is roughly 11 thousand kelvin, so ions in low-temperature vapors exist as molecules. When the temperature becomes enough to break ionic molecules, it is enough to strip one or two electrons from atoms, so high-temperature vapors will be plasma with electrons and positive ions.

To complete the previous answers: according to this article http://www.ncbi.nlm.nih.gov/pubmed/17579381 NaCl exists in the gas phase both as a monomer (NaCl), about 73% of atoms at 943K, and as a dimer where the four atoms are arranged in a wobbly square (the remaining 27%).  Traces of Na${}_2$ and Cl${}_2$ are also observed.  I expect that as the temperature goes up the proportion of the dimer drops.
Since the boiling point of NaCl is  1,686 K at 1atm the study must have been conducted at low pressure or with the NaCl vapor mixed in with other gases.
The article also says (GED = gas-phase electron diffraction) :
The determination of the structure of sodium chloride vapor
by GED dates back as far as 1937, when it was studied by
heating salt to around 1000 K. As a consequence of the visual
analysis methods in use at the time, only the predominant and
expected monomer species was detected. It was not until the
1950s that the existence of associated species such as dimers
(Na2Cl2) were discovered using mass spectrometry. Microwave
spectroscopy has provided detailed information on the structure
of the monomer, but the lack of a permanent dipole moment
makes this impossible for the dimer.

I believe that gaseous ionic compounds such as NaCl are in fact diatomic NaCl.  One can't create a plasma by simply boiling ordinary compounds.  The forces between ions are way too strong.

In liquid state, NaCl become free moving ions but the overall system is stable because of the attractions between the oppositely charged ions.
I believe that gaseous NaCl would exists as very polar covalent or ionic molecules with their overall affect being electrically neutral (no plasma).
This is a purely intuited answer and should be taken with a grain of salt
as I could find no numbers to back it up.
If someone could change this into a comment for me I would be much obliged.

It is actually individual Na+Cl- units in the liquid and gas states. What you are thinking of is dissolved NaCl and that would be Na+ and Cl- separate and thus free sodium and chloride ions.


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Can quantum entanglement affect the chemistry of molecules?
I recently heard a talk from a physicist about experiments on simple two-atom molecules in vacuum that showed that certain behaviour of the molecules was explained by quantum entanglement. Now this experiment was about creating holes in the inner electron shells of the molecule, so pretty much in the realm of physics and not chemistry. It was also performed in vacuum, and the lecturer himself said that in solution such an effect would likely be destroyed by frequent collisions in the solvent.
I was wondering if there are any examples of chemical behaviour of molecules that can only be explained by quantum entanglement. Does it ever play a role on the scale of chemistry, or is this effect just not relevant on anything beyond the small moleculs in vacuum stage physicists tend to use?

Entanglement is actually a fundamental property of quantum mechanics, so basically nothing can be explained without it. It's only a question of the point of view whether you're talking about entanglement or just superposition in a specific application.

"Quantum Entanglement" is a modern day buzzword for what we have known since the advent of quantum mechanics. Yes, it can and does affect chemistry because it's just quantum mechanics.  Electrons exist over all space, and so if you blew a hole in one atom, an electron on the moon would certainly know about it.

I've not really read anything on it, but a short while ago research on the role of entanglement in excitonic transfer in the photosynthetic light harvesting apparatus caused quite a stir.
Here is an ArXiV preprint of a paper on entanglement in an antenna protein from C. tepidum. It was published in Nature Physics. The authors note the extraordinary result of entanglement being observed in an inherently quite 'noisy' system.

I've hear that superatomic systems are theorized to involved strange quantum behaviour (compared with other molecular systems).
Non-classical behaviour is also believed to be the root of unusual carbon / graphene behaviour, such as in C60, and other higher-order materials / chemistry.

Quantum behavior is essentially non-classical (per definition). Everything with spin and chemical bond is non-classical. So it would be more meaningful if you can be more specific about what you mean “non-classical behavior”

Quantum entanglement does affect chemical reactions. As of now, there have been a couple of incidents that possibly provide evidence of it. The reason being these experiments are very hard and complicated. But people usually call it quantum interference. Recently our research group at Purdue published an article that classifies entanglement in chemical reactions (Scattering experiment). The method that was employed was generalization of Bell's inequality to include continuous mesurements such as scattering angle. (https://arxiv.org/abs/1904.01141)


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Why do lithium and sodium corrode so easily?
I want to know what is the phenomenon and explanation behind this corrosion. What is the reaction?

Could you improve your question by including what you know and don't know about these metals and about corrosion? It would help use generate a better answer.

$$\ce{2Li + 2H2O -> 2LiOH + H2}$$
$$\ce{2LiOH + CO2 -> Li2CO3 + H2O}$$
$$\ce{4Li + O2 -> 2Li2O}$$
$$\ce{Li2O + CO2 -> Li2CO3}$$
$$\ce{6Li + N2 -> 2Li3N}$$
Lithium and sodium are reacting with the gases in the air; a destructive physical process.

A better answer would also include why these metals (and other alkali metals) do these reactions and why other metals (iron, tungsten, cobalt, etc) don't undergo these reactions (or do so much more slowly).


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Why can dissolution reactions (e.g. dissolution of calcium chloride) be exothermic?
Dissolution reactions are normally endothermic, and crystallization reactions are normally exothermic, but there are some exceptions -- such as the dissolution of calcium chloride. Why is this so?

Could you add a bit about what your level of understanding of the topic is? And what you've already looked into for this answer?

Many dissolutions (especially of acids and bases into water) are exothermic. I agree with @Manishearth. Could you tell us more about what you know on this topic?

The endothermicity (taking up heat) or exothermicity (giving off heat) depends on a balance of factors including the entropy change, the degree with which the ions become associated with water, and the possibility of actual chemical reactions.
The most notorious example is sulfuric acid, which reacts so strongly with water that it can actually boil the water, and spatter the acid, which is why one always adds the acid to the water(water as solvent) and precautionary measures (gloves and proper clothing) are needed to handle the reaction.


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Boron-Boron bonds
I am reading up on boron in my notes and there is a certain paragraph that I do not understand

The tendency of $\ce{B}$ to form $\ce{B-B}$ bonds is less pronounced than that of
  carbon, its periodic table neighbor on the right, to form $\ce{C-C}$ bond,
  but it is much greater than that of the nearest element in the group,
  $\ce{Al}$, to form $\ce{Al-Al}$ bonds

Anyone can explained to me why is $\ce{B-B}$ bonds can be formed in the first place?

Why wouldn't they? — In order to ask a good answerable question, please give more information about what you know already, and what you don't understand. Otherwise, we don't know what helps you or not. For example, my direct answer to your question would be “why not? have you looked at the molecular orbital diagram for homonuclear diatomics?”… but I have no way to guess if that would actually help you or not.

I agree with @F'x. A boron-boron bond will form because it is lower in energy than the two atoms separately (definition of a bond in all bonding theories). A better question is why the B-B bond is less common than the C-C bond, but more common than the Al-Al bond.

I am not sure boron-boron bonding is inherently less favored thsn carbon-carbon bonding.  The variety and complexity of known boron-based clusters seems suggestive.  We may not have studied boron-boron bonding widely because there does not seem to be a lot of boron-based life around.

As mentioned in the comments, why not? A boron-boron bond will form as long as there is an increase in stability of the system. Forming a bond requires energy, but there is a lot of energy released once it gets formed. The feasibility of a bond depends which energy is greater.
One example of where a boron-boron bond occurs is in Aluminium diboride / Magnesium diboride

Here, you have graphite-like "sheets" of boron (pink), with aluminium layers between them.
This will probably be quite stable, due to delocalization. The singly-filled aluminium $p$-orbitals will overlap with the empty $p$ orbitals of Boron, and will delocalize across the graphite-like sheet (I cannot confirm this, though). This will lead to a large increase in stability.
A few more compounds with $\ce{B-B}$ bonds are Lanthanum hexaboride and various Yttrium borides. These are probably very stable due to the crystal structure (though delocalization may be a factor as well)

It's even possible to stabilize the B_{2} dimer with carbene ligands; this complex has an isolated B-B triple bond. See http://www.sciencemag.org/content/336/6087/1420.

See this answer for a discussion of magnesium diboride.  The boron layers (and three-dimensional extended networks in other electropositive-metal borides) contain electron-accepting, bonding orbitals that drive ionic metal-boron bonding -- even with an electronegativity difference less than 1.0.


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Negative Kelvin Temperature
I remember my Physical Chemistry Professor saying that very tiny negative Kelvin temperatures have been achieved on the quantum level. Is this true?

Yes.  But it is a bit of a joke of a sort.  In any system with energy levels, the lower energy levels never have a lower population than the higher ones.  In fact there is an equilibrium arrangement of the populations of such levels characterized by the temperature.  The higher the temperature, the more evenly populated the levels are.  There is even a formula for calculating the populations in this case.
The energy levels of some systems depends on a magnetic field.  You can let the populations of the energy levels come to equilibrium at some temperature $T$ and then quickly invert the magnetic field.  What was a low energy level now becomes a high one and vice versa.  You have an upside down distribution of populations.
Soon enough that upside down population decays to a right side up population.
But while the distribution is "upside down" one finds that it is given by the same formula used for an ordinary distribution, but with a negative temperature.

Here is my rather easy explanation for negative energies, using the mentioned equation: http://chemistry.stackexchange.com/q/36885/19160

I just found this.


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What is in silico energy optimization?
While doing computational chemistry for both organic and inorganic compounds, an energy optimization of that compound is done before starting a study.
For example, if we do drug docking studies using aspirin, first we would make aspirin's chemical structure, and then we would optimize it. So what does "energy optimization" mean in the context of organic and inorganic compounds.
Is energy optimization different for organic and inorganic materials (e.g.,aspirin or gold nanoparticles, respectively)?

The answer to what the headline says is "computational chemistry" or any calculation that uses a computer (hence in silico). But the body of the question asks something very different. Perhaps a much more specific title would help.

Generally, energy optimization is done as iterative process with following steps repeating

calculate forces of nuclei
move nuclei according forces known so final structure were less tense.

Most focus in science is focused on how first step is made, as choice of the second step does not affect final structure except pathological cases.
The main difference is usually tied to computational capabilities of 'silica' available. Complexity of quantum mechanical computations usually grows as $O(n^3)$, $O(n^4)$ or even $O(e^n)$ where n is number of atoms of same type in system. So, the bigger system is, the 'worse' approximation is used. Biological systems are usually so large, that any sort of quantum mechanical computation is not an option at all, so "molecular mechanics/dynamics" approach is used, with empiric formulas of forces based on relative positions of atoms


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Hybridization sp3 and sp2
When we calculate the electrons and orbital numbers in $\ce{CH4}$ $sp^3$ hybridization we have taken 3 electrons from $2p$ of $\ce{C}$ and 1 electron from $2s$ of $\ce{C}$ but in $\ce{C2H4}$ we have $2sp^2$ (2 electrons from $2p$ and 1 from $2s$).

Why do $\ce{C}$'s atomic orbitals make $2sp^2$ in $\ce{C2H4}$ and $sp^3$ in $\ce{CH4}$?
How we associate the numbers of electrons from $2p$ and $2s$?

How much do you know about VSEPR theory, especially about sigma and pi bonds? What do you mean by $2sp^2$?

I know how to explain molecules in VSEPR and know sigma is first bond for C-C pi is used to form second bond. Sigma is from sp3 hybrid molecular orbital and pi formed in sp2. (these what i know about pi and sigma) 2sp2 is used in my book "Solomons and Fryhle Organic Chemistry" when explainin the excited state and hybridization energy diagram.

This can be explained in terms of orbital overlap and therefore bond strength.
Consider a single carbon in an ethene molecule with two sp3 hybridised carbon atoms. The orbitals on the carbon repel each other and arrange themselves into a tetrahedron. Two of the four sp3 hybrids are used to form sigma bonds with hydrogen atoms. The carbon is then left to share its remaining two sp3 hybridized orbitals with its neighbouring carbon. Draw this system and you can see that two sigma bonds can be formed, though they are "bent" as the sp3 orbitals on the carbon are at a 109.5 degree angle to one another.
Now consider a single sp2 hybridised carbon in an ethene molecules with two sp2 hybridised carbon atoms. Again two orbitals are used to bond with hydrogen. This time we have an sp2 orbital pointing directly at the neighbouring carbon with which to form a strong sigma bond. The remaining p orbital which is orthogonal to the C-C sp bond can also forms a C-C pi bonding interaction.
Draw these orbital systems and compare.

The sp2 hybridised system constructs a strong linear sigma bond.
The sp2 hybridised system better shields the nuclei of the bonding atoms.
The sp2 hybridised system provides more orbital overlap, leading to stronger bonding and lower free energy.

So, in ethylene we got two "2sp2" due to two carbon atoms constructing pi bonds.

The fast and simple answer to part 1. of your question is that the orbitals which result in the strongest bonding will arise.

I agree with Horba's answer.  However the usual case is that the bond structure of a given molecule is already known and then the hybridization is chosen to fit the observed structure.
Consider $\ce{CH4}$ and $\ce{C2H4}$.  It is clear that in methane the central carbon has four bonds and hence will need hybridization that gives four bonds.  Thus we get $sp^3$.  Ethene, on the other hand, has two carbons each of which needs three bonds.  So we get $sp^2$.
In other words, hybridization is not a predictive scheme, it is an explanatory scheme.

Ethene, on the other hand, has two carbons each of which needs three bonds. -- this is prone to misinterpretation--technically, a double bond is two bonds.. Could you clarify that?


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Name of a proton bonded to a hydroxyl carbon
What do you call the proton in a CH-OH configuration?
When discussing an NMR spectrum, I can talk about the hydroxyl proton, but it seems clumsy to refer to "the proton on the hydroxyl carbon".

I would call it the $\alpha$-proton to the $\ce{-OH}$. The alpha proton is the proton bonded to the alpha carbon (i.e., the hydroxyl carbon).


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Why are some salt solutions coloured?
I notice that salt solutions of $\ce{NaCl}$ and $\ce{KCl}$ are colourless while those of $\ce{CuSO4}$ and $\ce{FeSO4}$ are coloured.
I got as far as figuring that it has to do with the transition metal ions, but I can't explain why the salt solution of $\ce{ZnSO4}$ is colourless even though zinc itself is one of the transition metals.

You're right--it's got to do with them being transition metals (usually). Transition metal ions form coordination complexes. Their empty $d$ orbitals accept lone pairs from other molecules (called "ligands") and form larger molecules (though we don't call them that--we call them "complexes"). When put in water, the ligand is $\ce{H2O}$, and you get complexes like $\ce{Cu[(H2O)6]^{2+}}$ (this is what you get when you add water to Copper Sulphate). Thus, metals with no empty $d$ orbitals of comparable energy become colourless.
OK, but where do coordination compounds get their color from? Well, $d$ orbitals look like this:

Even though the last one looks different, these are all identical in many respects, the most important being that they have the same energy. So if an electron jumps from one orbital to another, it on't make any difference. In fact, since this is a quantum mechanical system, it (sort of) doesn't even make sense to talk about electrons jumping from here to there.
All this changes when you have some ligands around. In coordination complexes, the $d$ orbitals are hybridised with the $s$ and $p$ orbitals of the next or previous shell. If we're talking about octahedral complexes, all the orbital lobes have to be aligned along the coordinate axes (since the $p$ orbitals are). This can be said to come from the fact that we're "mixing" these orbitals to get orthogonal hybrid orbitals, but that may not be that good an analogy. Now, the ligands approach along the coordinate axes since their final position will be in hybrid orbitals along the axes. So, they approach the $d_{x^2-y^2}$ and $d_{z^2}$ orbitals (as well as the $s$ and $p$ orbitals). The repulsion generated makes these orbitals of higher energy than the other three $d$ orbitals. This is called "d-orbital splitting" {*}

Now, electrons may "jump" between the $d$ orbitals (the energy gap isn't so high that the jump is too improbable). Due to the energy gap, some energy is absorbed/released in the form of a photon and it just happens to fall within the visible spectrum (at least, for most transition metals, it does). This jump is called a $d-d$ transition.
Using this, we can easily show why $\ce{Zn}$ salts are colourless. In $\ce{Zn}$, the $3d$ orbitals are full, so they're useless from the ligand point of view. The $4d$ orbitals are, on the other hand, empty. Ideal for forming a coordination complex with the $4s$ and $4p$, but there won't be any $d-d$ transitions as there are no electrons in $4d$ to jump around. (Remember, this is before the ligand bonds)
*For tetrahedral complexes, the $d$ orbitals take no active part in the bonding, and the ligands approach between the axes. This leads to an opposite gap between the top three(known as $t_{2g}$) and bottom two($e_g$) orbitals, with the $t_{2g}$ orbitals at a higher energy. So we still get $d-d$ transitions and color, though IIRC many tetrahedral compounds do not have a visible color(the energy gap is about half that of the corresponding ocahedral gap--and may not give a photon with a visible frequency)

I would like to add: the key for a substance to be coloured, is the excitation of electrons, because while it goes back from the excited state to the ground one, it releases photons, which we can see as any of the colours. (Niels Bohr).
This is also the reason why glass is colourless. The valence electrons of silicates in glass can't 'afford' the energy of light, so they don't get excited, while adding any transition metal in glass, they will get excited and therefore the glass will be coloured. The relation between absorbance of a molecule and the wavelength of light gives us the spectrum and it is pretty much important to know in order to understand the colours of any given substance.

$\ce{ZnSO4}$ is colorless because it has completely filled d orbitals while $\ce{FeSO4}$ is green because its d orbitals are incompletely filled.


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In a Koppers-Totzek reactor, why do I get CO and H₂ instead of CO₂ and H₂O?
A Koppers-Totzek reactor (there seems to be only a German Wikipedia page) is an entrained flow gasifier that blows coal dust, oxygen and water (as steam, of course) into a burning chamber where the coal burns at about 1600 °C. The output contains around 60% $\ce{CO}$ and 33% $\ce{H2}$, plus some $\ce{CO2}$ and (very little) $\ce{CH4}$.
When burning coal, I would expect $\ce{CO2}$ and $\ce{H2O}$ as output, is the reactor simply providing too little oxygen, or is there something about the process that produces the output observed?

I believe the question may apply equally to other gasification processes.

The reaction occurs in two stages: in first $\ce{C}$ oxidized to $\ce{CO2}$ and then $\ce{CO2}$ is reduced to $\ce{CO}$.
The process of having carbon react with oxygen and water vapor is called steam-oxygen conversion. It is used, because in reality, due to the high reaction temperatures, the allothermic principle is hard to provide.
$$\ce{C + O2->CO2}$$
$$\ce{C + 2H2O->CO2 + 2H2}$$
$$\ce{C + H2O->CO2 + H2}$$
$$\ce{C + CO2->2CO}$$
The equilibrium concentrations of $\ce{O2}$ and $\ce{H2O}$ in the gasification products at all temperatures are negligible. Steam-oxygen gasification products are a mixture of $\ce{CO2}$, $\ce{CO}$ and $\ce{H2}$.
Before carrying out the reaction of the shift
$$\ce{CO + H2O->CO2 + H2}$$ from a mixture of $\ce{CO2}$, $\ce{CO}$ and $\ce{H2}$ must be separated $\ce{CO2}$. To this must be expended work division (again burn some $\ce{C}$).

OK, so the key is the reduction of the CO2, then? And, if I understand you correctly, the rate at which that occurs is a temperature-dependent equilibrium?

Yes, the main thing is the reduction of CO2. The result depends on the temperature. Need to find a ratio of steam and oxygen when the temperature is 1800 K. And at high temperatures resins and other macromolecular compounds decompose, making it easier to clean the gas.

OK, great, thank you. I tried to edit your answer because I had some difficulties with the English, but I'm not a native speaker either, and I don't speak Chemist-English. Maybe someone else will jump in.


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Why does hydrogen form on such a long time scale?
If we were trying to figure out the time scale for a gas-phase reaction between two hydrogen atoms in a molecular cloud (which has density $~\pu{10^4cm^-3}$), apparently the reaction would happen on a time scale proportional to the inverse of the density multiplied by $10^{15}$ years.
Aside from the cloud not being dense and the probability that a collision will surpass the activation energy is small, is the time elongated because you need to induce a dipole moment between two hydrogen atoms to actually bind them together?
I recently read somewhere that even if the reactants would go into the unbound state of $\ce{H2}$ they can't emit a photon that would let it go into the bound state of $\ce{H2}$. I guess they are talking about quantum mechanics, but it is not obvious to me what the bound and unbound states of molecular hydrogen are. (If someone could clarify that too)

Is your timescale a rate, a half-life, the time it takes to reach a certain completion (i.e. 99%), the time it takes to reach equilibrium, etc.? If you can tell me what that timescale is, I will update my answer to be better.

Why is the time-scale so long?

The time-scale is so long because the density is not just low, it's virtually nonexistent.
The particle density of any gas can be determined at any temperature and pressure using the ideal gas law ($PV=nRT$). Let's use air as an example. At $\pu{25 o^\circ C}$ ($\pu{298 K}$), and 1 atmosphere, the particle density of air ($n/V$) is:
$$ \frac{n}V =\frac{RT}P $$
$$ \frac{n}V =\frac{(\pu{0.08026 L atm (K mol)^{-1})(\pu{298  K})}}{{\pu{1 atm}}} = \pu{23.9 mol L^-1} $$
Converting to molecules per $\pu{cm^3}$:
$$23.9 \frac{\mathrm{mol}}{\mathrm{L}} \times 6.022\times10^{23}\frac{\text{molecules}}{\text{mol}}\times\frac{1 \text{ L}}{1000 \text{ cm}^3}=1.44\times10^{22}\frac{\text{molecules}}{\text{cm}^3}$$
The density of gasses we are used to dealing with is 18 orders of magnitude higher than the density you quoted. If we have a gas of hydrogen atoms at air density, then the timescale for formation of H2 is:
$$ 10^{15} \frac{\text{years}\times\text{cm}^3}{\text{particles}} \times  \frac{1 \text{ cm}^3}{1.44\times10^{22}\text{ molecules}}=6.94\times 10^{-8}\text{ years} $$
$$6.94\times 10^{-8}\text{ years}= 2.54\times 10^{-5}\text{ days}=6.08\times 10^{-4} \text{ hours}=2.19 \text{ seconds}$$
At "normal" densities, the timescale is much shorter.
The second part of your question deals with the quantum mechanics of the activation barrier to this reaction, which I cannot do justice to at this time.


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Why does gas particle velocity affect rate of effusion?
I understand why smaller particles have more velocity, but I don't understand what velocity has to do with the rate of effusion:
My reasoning is thus:

Pressure is the number of impacts of particles in a given period of time.

If He and Ar are both in balloons at, say, 1.5 atm, both gasses have the same average Kinetic energy but He is moving faster (because it's smaller). Simple enough.

If both gasses are subject to the same pressure, they have the same number of impacts over a given area over a given period of time.

This should mean that each gas has an equal number of particles approaching a hole over a given period of time.

If a hole is big enough for both particles to fit through, the rate of effusion should be the same, since the same number of particles are approaching the exit hole.

For an analogy to explain my thinking ... if you have two lanes of cars going through a checkpoint, one has 30 cars per minute at 50 miles per hour and one has 30 cars per minute at 30 miles per hour, the resulting number of cars through the checkpoint should still be 30 cars per minute in each lane, regardless of their velocity.
My book (and professor, and Wikipedia...) say that it is the higher velocity of He particles that cause the faster rate of effusion; however, from my reasoning, that doesn't make sense except with small enough holes where the size of the particles would affect, i.e. the holes are very small in size and He could fit through but Ar could not, or there are simply more holes that He could fit through.
But Graham's law is referring to pinholes, which should be large enough for either atom to fit through.
So why does velocity affect the rate of effusion?

Since Helium has higher velocity don't you think that it will have a higher collision frequency than Argon which will shorten it's mean free path. As a result He will take more time to travel a given distance along a straight line. So velocity does affect effusion rate.

Your point 1 is mistaken (or incomplete)
$$P=\frac{F}{A}=\frac1A\frac{\Delta p}{\Delta t}=\frac1A\times\frac{2nmv}{\Delta t}$$
This is assuming all particles hit perpendicularly (you can always modify this for all colissions by taking the average component of velocity in the perpendicular direction).
So, pressure is proportional to number of impacts, mass, and velocity. Pressure is not simply "number of impacts", rather it is a combination of number of impacts with other stuff. Since the energies are equal, we can say that $v\propto m^{-\frac12}$, and we get $P\propto n m^{\frac12}$. P is also the same, so $n\propto\frac1{\sqrt m}\propto v$.
Thus, when avg KE and pressure are the same, rate of diffusion is inversely proportional to square root of mass.

But if I know this is pretty late ... but isn't it still the same number of impacts? i.e. a 5L balloon of helium has the same n as a 5L balloon of Xe. The difference is that the Xe is slower with more mass, but it should still be the same number of impacts shouldn't it?

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


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Sodium Perborate as a detergent ingredient
Sodium Perborate is a well known bleaching agent in detergents. What makes Sodium Perborate an effective bleaching agent and how is this related to its oxidation chemistry?

Connecting with water, sodium perborate releases hydrogen peroxide.
$$\ce{NaBO2*H2O2*3H2O->Na+ + BO2- + H2O2 + 3H2O}$$
$$\ce{H2O2->H+ + HOO-}$$
$$\ce{HOO- ->HO- + O}$$
Dissolution of sodium percarbonate in water at hand washing and the washing machine washing, leads to the formation of the perhydroxy anion. This anion is an effective bleach, but it is formed in sufficient quantities only at temperatures above 60 ° C.
From the perhydroxy anion obtain atomic oxygen, which has a strong oxidizing capacity.

I have still confusion because I cannot figure out the relation between oxidation chemistry and dirt. How dirt is detached from clothes.

Active oxygen accept electrons from the stain. The result: 1) oxygen cleavage of chemical bonds in the stain and breakdown it, after that fragments can be suspended in solution by surfactant action; 2) oxygen change in oxidation state of the stain, rendering it colorless.


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Clathrate Compounds with Quinone
I prepared a compound "$\ce{SO2}$-Quinolclathrate" by dissolving hydroquinone and sodium sulfite with hydrochloric acid in warm water.
After this, I tested a little sample of the product on a flame and it vaporised, following which I smelled the gas. I am supposed to define this gas and my prediction is that this could be $\ce{SO2}$(with orange flame).
Clathrate compounds include a trapping system with guest molecules (held "guest" in cages) and the host (hydroquinone here).
$\ce{SO2}$ is guest in the cages of hydroquinone and the gas is held due to crystal lattice forces. Does this mean there are no bonds?
What can be the reason for $\ce{SO2}$ vaporizing?
Additionally, the product was added to starch/iodine solution to find out if it is soluble in that solution and it is so. What can be the reason of this solubility.

Palin, D. E.; Powell, H. M. "Hydrogen Bond Linking of Quinol Molecules," Nature  156, 334 (1946), and DOI: 10.1039/DF9694800181
White aqueous starch/iod$ide$ tests for oxidizing agents by turning brown, liberated iodine and ionic oligomers intercalated within amylose starch (linear).  Amylopetcin (branched) gives red.  The mix is purple-brown.  Purple-brown aqueous starch/iod$ine$ tests for reducing agents by the reverse.


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Extraction of hydrogen and oxygen from water
The apparatus that I am using is a glass container, with two electrodes,two glass cylinders, water, and a battery. In the glass container will be filled with water and two electrodes kept inside the glass cylinder and then dipped in water.
The two electrodes are then connected with the positive and negative terminals of the battery and the container is sealed.  As the current is started the water will be separated into hydrogen and oxygen. After extraction of the gas, the two gases are passed separately.
In this process, will the hydrogen gas start burning immediately as soon as produced? If yes, then what to do to make it pass without burning?  The separated gas is passed through glass pipes, will the extraction work properly?

Please take any necessary precautions for handling the hydrogen.  If you have any hesitation about what the correct techniques are, please consult someone with the proper experience.

No, it won’t burn immediately. As your electrodes are generally separated by some distance, you should be able to isolate the produced oxygen and hydrogen gas by catching the respective bubble streams under a submerged test tube or the like. Alone, neither oxygen nor hydrogen will burn, however combining the oxygen and hydrogen gas will produce an explosive mixture. Mixing the hydrogen with air will also produce an explosive mixture as the atmosphere is about 22 % oxygen.
To get the hydrogen to burn, you need to supply the necessary activation energy, which can be supplied by a spark or flame. After this, the reaction will be self-sustaining due to its exothermicity. There are in fact entirely self-contained apparatus which can be purchased from scientific teaching shops that can electrolytically split water and then combust the hydrogen and oxygen using a piezoelectric ignition system, the ensuing tiny explosion causing water to spurt out of a small hole. Wikipedia mentions that the autoignition temperature of hydrogen in air is 500 degrees Celsius.
On a small scale, this reaction is quite safe, however you should always take safety precautions such as wearing protective eyewear and having fire-suppression equipment at hand.
Here’s an illustration of a reasonable water-splitting setup. Remember that your water should contain an electrolyte such as dissolved salt. Keep in mind that if your potential difference is high enough you will start to produce small quantities of chlorine gas and aqueous sodium hydroxide, both of which are hazardous in high concentrations.

I should draw attention to the fact that the $\ce{H2}$ headspace is about twice the volume of the $\ce{O2}$ headspace. This comes down to the stoichiometry of the reaction, and also means that the $\ce{H2}$ volume will increase at twice the rate of $\ce{O2}$. The wires I’ve illustrated connect to the terminals of a battery or lab PSU.
Addendum: Here’s a nice guide on how to construct and safely operate a water-splitting/combusting apparatus using household equipment, aimed at secondary school students.

Thanks for addressing the safety concerns further, I have never done this in a lab.

@jonsca - I've got the reaction going in a glass of salty water right now - 12 volt regulated power supply, two bits of wire and two mechanical pencil leads (graphite + clay) for electrodes. The electrodes rapidly disintegrate.


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Computation of pH when an acid and base are mixed in solution
I'm doing a basic chemistry course, and we are currently learning how to compute $\text{pH}$ from the acid dissociation constant (using $\left[\text{H}^{+}_{(\text{aq})}\right]=\sqrt{K_{a}\left[\text{HA}_{(\text{aq})}\right]}$) along with computing the $\text{pH}$ of strong bases by assuming full dissociation into $\text{OH}^{-}$ ions, and then using the ionic product of water to calculate the concentration  of protons.
The question I have been given is:

Calculate the pH of the solution obtained when $14.9\text{ cm}^{3}$ of $0.100\text{ mol dm}^{-3}$ sodium hydroxide solution has been added to $25.0\text{ cm}^{3}$ solution of methanoic acid of concentration $0.100\text{ mol dm}^{-1}$; $K_{a}=1.60\times 10^{-4}\text{ mol dm}^{-3}$

I'm not sure how I should go about solving this problem, I can calculate concentrations of hydrogen and hydroxide ions for each of the solutions but I'm unsure how to combine them (as presumably some of the $\text{OH}^{-}$ ions will react with the $\text{H}^{+}$ ions to form $\text{H}_{2}\text{O}$?)

The trick is: after the addition, a buffer solution is formed. Find the concentrations of the salt and acid and use Henderson's equation to arrive at the answer.

You are headed in the right direction.  Pretend that the $\ce{H+}$ ions and the $\ce{OH-}$ ions neutralize each other 1 for 1.  That will leave you with only one of those ions left.
The water ionization won't affect your problem as that ionization is repressed by the excess ion present in the solution.
Now you should be able to solve the problem.

Count the protons. Figure out the amount of pure methanoic acid. Consider the addition of the dilute acid to your mix as that amount of methanoic but include the water in the total volume. Consider it as pure water. Do the same with the lye i.e. find out how much pure lye and how much pure water are being added to the mix.
Guess an answer pH. Calculate the concentration of OH- and H+ at that pH and determine the number of protons that you would have to add (or subtract) to your volume of pure water to reach that pH. If your guessed pH is < 7 the H+ concentration will be higher and you will have to add protons to establish that level and neutralize the 10^-7 concentration at pH 7. Call this need for protons the "proton deficit". If the guess pH is > 7 the OH- level will go up and the H+ concentration goes down and the H+ concentration of 10^-7 at pH 7 represents a surfeit.
Assume the NaOH is completely dissociated. Thus you will need as many protons as there are OH- ions in the added NaOH to convert the OH- to water and the eqivalents of OH- become part of the proton deficit.
Now the tricky part. Use Henderson - Hallelbalch to compute the charge on the formic acid ion at your guess pH. To do this compute the dissociation ratio
r = 10^(pH - pK). This is the ratio of dissociated molecules to undissociated ones. If there was 1 mole of formic acid to start with and x remain undissociated the rx must be dissociated and x/(x + rx) = 1/(1 + r) is the fraction dissociated. The original acid is undissociated so in being added to a solution which has a certain pH it dissociates supplying C/(1 + r) protons where C is the number of moles of formic acid. This is a proton surfeit (with respect to the guessed pH).
So, assuming the guessed pH is < 7
1. Protons are required to neutralize the OH- in the pure water (deficit)
2. Protons are required to establish an H+ concentration in the water of 10^-pH (deficit)
3. Protons are required to neutralize the OH- ions from the NaOH (deficit)
4. Protons are released by the methanoic acid (surfeit)
Since no protons are generated or destroyed in this reaction the total of the deficits must equal the total of the surfeits when the reaction is complete. All you need to do is keep guessing pH's until you find the one that zeroes the sum. The easiest way to do this is to put your calculations into an Excel spreadsheet and use the Solver to find the pH. If you don't know about Solver, learn about it! If you don't want to it's pretty easy to zero in on the correct answer pretty quickly manually.


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What is the IUPAC nomenclature of alkane with large numbers of C atoms (more than 200)?
What is the IUPAC nomenclature for alkanes with large numbers of $\ce{C}$ atoms (more than 200)? For example, what is the IUPAC accepted name for $\ce{C_{205}H_{412}}$?

The definitive answer is, of course, in the rules established by the IUPAC’s “Commission on nomenclature in organic chemistry”. The reference you are looking for is:

“Extension of Rules A-1.1 and A-2.5 Concerning Numerical Terms Used in Organic Chemical Nomenclature”, Pure Appl. Chem., 1986, 58, 1693-1696

which can be found here as a PDF and here as an HTML version. In particular:

NT-1.1 - The fundamental numerical terms for use in hydrocarbon names or as multiplying prefixes for simple features are given in the following list:
\begin{array}{|c:c|c:c|c:c|c:c|}\hline
    1&\rm \text{mono- or hen-} &10 &\rm \text{deca-}&100&\rm \text{hecta-}& 1000&\rm \text{kilia-} \\\hline
    2&\rm \text{di- or do-}&20&\rm \text{icosa-}&200&\rm \text{dicta-}&2000&\rm \text{dilia-}\\\hline
    3&\rm \text{tri-}& 30&\rm \text{triaconta-}&300&\rm \text{tricta-}&3000&\rm \text{trilia-} \\\hline
    4&\rm \text{tetra-}&40&\rm \text{tetraconta-}&400&\rm \text{tetracta-}&4000&\rm \text{tetralia-} \\\hline
    5&\rm \text{penta-}&50&\rm \text{pentaconta-}&500&\rm \text{pentacta-}&5000&\rm \text{pentalia-}\\\hline
    6&\rm \text{hexa-}&60&\rm \text{hexaconta-}&600&\rm \text{hexacta-}&6000&\rm \text{hexalia-}\\\hline
    7&\rm \text{hepta-}&70&\rm \text{heptaconta-}&700&\rm \text{heptacta-}&7000&\rm \text{heptalia-}\\\hline
    8&\rm \text{octa-}&80&\rm \text{octaconta-}&800&\rm \text{ octacta-}&8000&\rm \text{octalia-}\\\hline
    9&\rm \text{nona-}&90&\rm \text{nonaconta-}&900&\rm \text{nonacta-}&9000&\rm \text{nonalia-}\\\hline
\end{array}

So, $\ce{C_205H_412}$ is the molecular formula of pentadictane. $\ce{C_7547H_15096}$ is heptatetracontapentactaheptaliane.

Presuming it is linear, then it should follow the trend of the higher alkanes.
For $\ce{C10}:  \ce{CH3(CH2)8CH3}$ - the name is decane (deca means ten).
For $\ce{C20}:  \ce{CH3(CH2)18CH3}$ - the name is icosane (icos means twenty)
For $\ce{C25}:  \ce{CH3(CH2)23CH3}$ - the name is pentacosane
After icosane, the higher alkanes follow the following:
For $\ce{C30}:  \ce{CH3(CH2)28CH3}$ - the name is triacontane
For $\ce{C40}:  \ce{CH3(CH2)38CH3}$ - the name is tetracontane
For $\ce{C50}:  \ce{CH3(CH2)48CH3}$ - the name is pentacontane
So...
For $\ce{C100}:  \ce{CH3(CH2)98CH3}$ - the name would be decacontane
For $\ce{C200}:  \ce{CH3(CH2)198CH3}$ - the name would probably be icosacontane
And for $\ce{C205}:  \ce{CH3(CH2)203CH3}$ - the name might be pentacosacontane

From wikipedia, the name for $C_{100}$ is hectane.


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Can the electrolytes be in gaseous state?

Possible Duplicate:
Will gaseous ionic compounds be free moving ions?

I know it sounds extremely expensive, but what if we could heat the ionic compound to evaporate it and achieve the lattice energy and break them into free ions, wouldn't the electrolysis be faster, as the ions can move faster freely than in liquid state?


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Is it not safe for health to finish a bottle of wine during a week
People said that wine is good for health. So I take a cup (less than 100 mL) of wine every evening. (Yes, wine is cheap where I live.) So to finish a bottle of wine (750 mL), it takes me a week.
(Of course I put it in the refrigerator.) I have to say that it doesn't taste good after 3 days (but still OK).
But I don't know if it is not safe to do so. Maybe the wine begins to produce something not good for health after a few days when it is opened. So, my question is: what chemical species are formed after opening a wine bottle, making it go bad? Are they only bad for the taste, or also bad unhealthy?

As a side note: box wine is very efficient for preserving wine from oxidation and spoilage. While in the past, only bad wine could be found in such packaging, nowadays you can find very decent wine in boxes…

The bad taste left after drinking wine can be attributed a number of agents found in wine, particularly alcoholic wines.
Alcoholic wines typically contain 8-15% ethanol. When ethanol mixes with oxygen in air it gets oxidised, gradually producing acetic acid (vinegar).
Furthermore, the reaction of ethanol with urea (naturally present in wine) to form ethyl carbomate (urethane) can produce a saline, bitter taste, although probably not toxic.
The ethanol in wine is produced through the fermentation of carbohydrates by yeast. This process also results in the production of acetaldehydes which contribute to its complex set of flavours.
Ethanol itself however is tasteless. It is however toxic and a class 1 carinogen according to the International Agency for Research on Cancer (IARC).
When consumed, ethanol is oxidised by alcohol dehydrogenase (ADH) enzymes in the body to produce acetaldehyde, which is 10-30 times more toxic than alcohol and possibly carcinogenic in humans.
Recent scientific research on cancer seems to indicate even low levels of alcohol consumption (1 cup per day) 'increases the risk of cancer of oral cavity and pharynx, esophagus and female breast' Source: http://annonc.oxfordjournals.org/content/early/2012/08/21/annonc.mds337.full
So a bad taste may not be the worst thing alcohol leaves in one's mouth.
With regard to the health effects of alcohol, please refer to the World Health Organisation report which is based on data collected scientifically from around the world over many years.

"When ethanol mixes with oxygen in air it gets oxidised, gradually producing acetic acid (vinegar)." - why doesn't this happen with hard liquor? Or is your description a simplification?

"Ethanol itself however is tasteless." How do you define "tasteless"? I have a bottle with 96% ethanol and it is definitely not neutral in my mouth (I have tried a drip on a teaspoon a few times).


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In sp² hybrid orbital, why does the p orbitals only have 3 sides instead of 4?
Usually two separate p orbitals would have '4 lobes' while in a $\ce{sp^2}$ hybrid those 2 p orbitals would only have '3 lobes'. why?
I'm referring to diagrams such as the one shown below:

First, an $sp^2$ orbital does have two lobes. One is large, and one is tiny. Secondly, $sp^2$ means "a 'mixture' of one $s$ and two $p$ orbitals". Like the following:

So, you plug in three orbitals, and you get out three more orbitals. When drawn in place, you get this:

There are a total of six lobes here, but they are distributed amongst three $sp^2$ orbitals.
The number of lobes need not stay constant--it's the number of orbitals that does, since each orbital corresponds to two electrons, and there's no way for new electrons to be added/removed (yet). What's happening is that the wavefunctions(orbitals) of the electrons are "mixing up" to distribute the electrons in a way that stimulates bonding (one per orbital). The lobes have a significance, but it's much deeper.


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Time required for a thin layer of ethanol to evaporate?
If someone were to pour about 5 ml of 94% pure ethanol on a flat surface and at room temperature, approximately how much time would it take for it to evaporate?
EDIT: In others words, it is to know approximately how much time it would take for a small amount (about 5 ml) of pure alcohol to evaporate if poured over a surface such as a table. Would it take about 15 seconds/30 seconds/2 minutes?

Is there any background for the experiment (for example, you're trying to make an ideal eyeglass cleaner or something)?  It might help someone to give you a better answer.  Otherwise, you could certainly give this a try at home with a lower purity ethanol and see if you can extrapolate.

What is the surface? Is the surface wettable by ethanol? Evaporation time of a thin layer is strongly dependent on the layer thickness/specific surface area. Suffice to say, ethanol evaporates quite rapidly with respect to water due to its relatively low specific heat capacity and high vapor pressure.

Basically, it is to know approximately how much time it would take for a small amount (about 5 ml) of pure alcohol to evaporate if poured over a surface such as a table. Would it take about 15 seconds/30 seconds/2 minutes?

Google for "evaporation number" and its definition

http://bayblab.blogspot.jp/2009/01/evaporation-rates-of-ethanol-solutions.html

In others words, it is to know approximately how much time it would take for a small amount (about 5 ml) of pure alcohol to evaporate if poured over a surface such as a table. Would it take about 15 seconds/30 seconds/2 minutes?

It be simpler and faster to do the experiment than to try to predict the time needed.
The kinetic molecular theory explains why liquids evaporate at temperatures lower than their boiling points. At any temperature, the molecules in a liquid have a range of kinetic energies described by a Boltzmann distribution. Some percentage of the molecules have enough kinetic energy to escape into the gaseous phase. These molecules contribute to the vapor pressure of the liquid. As the temperature increases, more molecules are above the gas threshold and the vapor pressure increases. As the vapor pressure increases.
In a closed system, equilibrium would be established with an unchanging ratio of liquid to vapor. Molecules would be exchanged between the liquid and the vapor, but their relative amounts would remain constant. The surface you describe is not closed - the vapor molecules can wander away by diffusion or convection. Equilibrium is disrupted, and Le Châtelier's Principle tells us that the equilibrium shifts to compensate. As more ethanol molecules escape, more vaporize to replace them until there are no molecules remaining in the liquid. In the following equation, $K$ is the equilibrium constant, $p$ is the partial pressure of ethanol vapor (the vapor pressure) and [$\ce{C2H6O}$] is the concentration of ethanol in the liquid.
$$\ce{C2H6O (l) <=> C2H6O (g)} $$
$$K=\frac{p_{\ce{C6H6O}}}{[\ce{C2H6O}]}$$
After establishment of equilibrium (fast), the rate-determining step of the evaporation is probably the diffusion of the gaseous ethanol molecules away. The average kinetic energy of a gas particle can be expressed as a function of mass ($m$, in kg) and root mean square velocity ($v^2_\text{rms}$) and separately as a function of temperature ($T$, in kelvin) times the Boltzmann constant ($k_\text{B} = 1.38 \times 10^{-23} \frac{\text{J}}{\text{K}}$). We can derive a formula for rms velocity.
$$\overline{E_\text{k}}=\frac{1}{2}mv^2_\text{rms}$$
$$\overline{E_\text{k}}=\frac{3}{2}k_\text{B} T$$
$$v^2_\text{rms}=\frac{3k_\text{B} T}{m}$$
$$v_\text{rms} = \sqrt{\frac{3k_\text{B} T}{m}}$$
You could calculate the velocity of ethanol particles escaping. If you set an arbitrary distance (one meter is probably fine), then you can calculate a time it takes one particle to travel that distance (on average). If we know the equilibrium constant, we can determine how much vapor is above the liquid and then calculate the amount of time it takes to move.
But how do we know the equilibrium constant? It varies with temperature! The value $\Delta G^\circ_\text{vap}$ in the equation below is the free energy change of vaporization of ethanol in the standard thermodynamic state. $R$ is the ideal gas constant.
$$K=\mathrm{e}^{-\frac{\Delta G^\circ_\text{vap}}{RT}}$$
The model above ignores complicating factors like mean free path, the fact that evaporation is endothermic (meaning the liquid cools as it evaporates and the vapor pressure decreases with time), the temperature and heat capacity of the surface govern how much kinetic energy is available to the liquid to begin with, and that any amount of air current in the vicinity will move the vapor away significantly faster than diffusion.
A complete model includes consideration of the following:

Constants

the Boltzmann constant
The ideal gas constant
the free energy change of vaporization of ethanol (not actually constant, but it varies only slightly over the temperature range)
the vapor pressure of ethanol as a function of temperature
the mass of an ethanol molecule
the heat capacity of the surface

variables

temperature of the air
volume of ethanol
temperature of the surface
atmospheric pressure (needed for mean free path corrections)
velocity of air currents

So, in principle, you could do it. Even so, the best answer cannot be had without some serious calculus. In practice, it would be faster to do the experiment (if often is). We often forget that science is empirical.

-1 for mixing equilibrium thermodynamics into a question for a kinetic process. Horrible

The point was to demonstrate that the theoretical prediction was hard while the experiment was trivial.

Please, PLEASE, give us the equations. Plugging in 5 variables looks so much easier than doing dozens of complicated experiments. To put into perspective: doing CAD costs ~1 week per model. Resin 3D printing costs 3-4 hours per print plus money. Are there books on this topic?

Essentially random.
The air movement anywhere is essentially random and the greater the air currents above the sample the quicker it will evaporate. Do the experiment: spill two samples of $5~\mathrm{ml}$ simultaneously, blow onto one of the two and check how much quicker it evaporates.
For a much more detailed answer, upvote Ben’s!


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Why can an s orbital only hold two electrons?
According to Pauli’s exclusion principle, an $s$ orbital contains at most two electrons with the opposite spin (up and down). Why can't an $s$ orbital contain a third electron whose state is the linear combination of spin up and down?

Electrons are magnets, they have magnetic fields. Those fields have only two possible orientations, and a single orbital can only be occupied by two electrons if those orientations are mutually opposed.
OK, but why? Why are electrons magnets? Why are there only two possible orientations for the fields? Why can the two electrons in the orbital only exist with opposite directions for the magnetic moments?
Well, there are answers to those questions, but they don't make sense a lot of sense. The magnetic moment of electrons is due to a property called "spin"; the spin value of an electron is 1/2, so it can adopt one of two spin quantum states, +1/2 and -1/2, which correspond to the "up" and "down" orientations of the magnetic moment. Since it has a half-integer spin, an electron is a member of a class of sub-atomic particles called "fermions" which obey rules called the Pauli Exclusion Principle and Fermi-Dirac statistics - one key result of these rules is that no two identical fermions can simultaneously occupy the same quantum state: you cannot have two electrons in the same orbital if they have the same spin orientation, so if one is +1/2, the other must be -1/2, and no more can be added because a third would have to adopt the same quantum state as the one of the first two.
OK, but why? Why does an electron have a spin value of 1/2? Why can't two fermions adopt the same quantum state (when integer spin particles, called bosons, can do so)? Why does the property of "spin" even exist? You need quantum mechanics to even begin to answer those questions -- again, that's more math than anybody is prepared to show you for at least five years -- and even then, the "why" of the Pauli Exclusion Principle is an empirical observation (it has no theoretical basis, it's just an after-the-fact thing that happens) without relativity, but there is a firm theoretical rationale and a reason "why" if one revisits the postulates of quantum mechanics and takes into general relativity into account. But relativistic quantum field theory is more math than any mortal should be forced to endure.
OK, but why? Why is quantum mechanics how the universe behaves? Why is general relativity the way the universe behaves?
You can play the "why" game for a very long time. None of your high school teachers will be able to give you more than one or two repetitions. University professors will be able to go a lot farther. Eventually, you get to questions to which the only possible answer is "nobody knows.... yet". And those are the questions that scientists -- including those university professors -- spend their day trying to answer.
Which is terrific fun. So keep asking "why", even to the answers your teachers give you. And if they can't answer, go ask someone else, or read about it, or go to university. And if they don't know, be a scientist, and figure it out yourself, and tell the rest of us.

Nice answer! By the way, if you are planning to contribute more answers - you can read more about formatting mathematical and chemical expressions here, here, and here. If you are already familiar with $\LaTeX$, then none of it will be new :)

I really loved this answer, in particular the last paragraph. Very nice motivation :) !

Because we can have only one electron per quantum state.  Spin up and spin down are two different states.  A linear combination of the two is not a new independent state.  It is obviously formed from the spin up and spin down states.

-1. First, it does not answer the question, it just paraphrases it in a different way. Secondly, it is just plain wrong. A linear combination of two states is a new state, this new state is obviously linearly dependent with the two states, but it is a different state.

The complete wave function of a two electron system is written using slater determinants. So, for helium it would look like this:
$$\begin{pmatrix}1s(1)\alpha(1) & 1s(1)\beta(1)\\\ 1s(2)\alpha(2) & 1s(2)\beta(2)\end{pmatrix}$$
For lithium, in what you are saying, it would look like this:
$$\begin{pmatrix}1s(1)\alpha(1) & 1s(1)\beta(1) & 1s(1)(\alpha(1)+\beta(1))\\\ 1s(2)\alpha(2) & 1s(2)\beta(2) & 1s(2)(\alpha(2)+\beta(2)) \\\ 1s(3)\alpha(3) & 1s(3)\beta(3) & 1s(3)(\alpha(3)+\beta(3))\end{pmatrix}$$
This obviously is zero, as can be seen from basic principles of matrices and determinants, which is not possible, as the wavefunction obviously cannot be zero.
(Note: 1s(i) refers to the spatial part of the wave function and $ \alpha,\beta $ represents the spin part.)


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What does the "D" group represent on a chiral center?
I am trying to figure out the configuration for this chiral carbon to answer a homework question, however they arbitrarily include a "D" group as shown in the image. My guess is that it is some higher priority group than the carbon, and lower priority than the halide, making the configuration (S). Is this correct? What does the D group represent?

I assume it's deuterium, the isotope of hydrogen that has a single neutron as well as a proton in the nucleus. As such, it has the same atomic number as hydrogen, which means you break the tie on atomic mass --  deuterium is higher priority than hydrogen but lower than everything else.


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Is it safe to hold silicon nitride ceramic without protection?
If I have a complete silicon nitride ball bearing unit, is it safe to hold without any protection? If it cracks, or disintegrates is it also safe?

I am not certain that it is safe, but where are you getting the idea that it might be unsafe to touch? Certainly if a powder or mist is appearing if it disintegrates it would probably make sense to avoid inhalation, right? Or if it appears to produce sharp edges that would be another reason to not handle it without protection.

@Leonardo to be honest, I don't understand anything about chemistry safety (except vague ideas about acids and radioactivity (so nothing really))

I have taken a pottery class and it is not 'unsafe' to touch the wet clay, some people are even known to consume small amounts of clay for nutrition. But, if this bearing is ceramic I am guessing it is already heated in a kiln, and therefore its surface is silicon-nitride glass. So I don't think it will melt your skin. It is however, not safe to inhale powders dealing with anything ceramic, and not always good to eat your lunch off of depending on the material (raku ware for example).

@Leonardo thanks for your answer, but are the ceramic materials similar enough to compare to the ones in pottery class? I saw a safety sheet for Silicon Nitiride powder and it had simple safety suggestions just like dealing with clay powder... but I'm still not sure.

I have done a minimal amount of research to try and help you understand this material. The credibility of this research is very debatable, as it basically involves me typing in silicon nitride into YouTube. That being said, understand that silicon nitride is first of all extremely strong, inert, according to that link, and being considered as a contemporary 'ideal' material for use in the human body. Inert just means it will not react which is generally safe and certainly not caustic to life.
Know however, that if this material is present in powder form I would highly recommend a breathing mask at the very least as it is usually inadvisable to inhale solid particles. In the scenario that it shatters it could potentially be sharp. However, if it is merely a bearing we are talking about, then, you probably have very little to worry about other than contaminating your materials if you are going to use the part in high performance devices, and you therefore would want to wear clean gloves so as not to get dust on the bearing.
I advise you to search you-tube and other sources for people handling the material, but more than likely it would be less beneficial to the bearing or your project to handle the material without gloves than it would be to your health.

I have a few silicon nitride balls in my drawer. They look pretty cool, that's why I keep them and touch them all of the time.

It is a very hard, glasslike material. It won't be easy to crack. There is nothing coming off the surface, ball bearings have extremely smooth surfaces (nm roughness), if anything would rub off, they would not be usable as ball bearings. Besides that, I don't think it is toxic. It is an extremely inert material, much more inert than steel, does not dissolve in much. It is even used in prosthetic parts that go into the body. I assume it needs a lot of certification for that.
If you break it (which won't be easy), you'll have some sharp pieces that might cut you, that's all.
The only risk would be from a ball bearing that was ground to a very fine dust.


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Non-Commercial Thermal Insulation
I am trying to create a small 10 to 20cm cube that can conserve the heat in 100 to 200ml of 60-90c water in a small Pyrex glass for about an hour. The problem is that I can not use commercial insulators like foam, bubble wrap, and the like. I have searched Google for ideas, but all the results are commercial. I also can not heat or cool the device beforehand, and I have to have a 2cm hole in the top.
I have tried cotton balls and a packed cotton balls with felt mixture, and cotton balls worked the best.

Shredded up newspaper might work

This sounds like an assignment (and a cool one) in a class one of my colleagues teaches.

http://en.wikipedia.org/wiki/Heat_capacity

Often the key with stopping the conduction of heat is to provide as many phase transitions as possible. Most thermal insulators achieve this by having voids of Air which is achieved in a variety of ways, bubble wrap, fibre glass etc.
For a "non comercial" I assume they mean stuff you have lying around the lab?
To attempt this I'd layer blue roll / cotton wool / glass wool with tin foil. The more layers the better.

Take a carton box in which the glass fits easily. Wrap the outside of the glass with a thin layer of plastic (sandwich bag). Now, buy some gap filler (PU spray foam, aerosol) to fill the rest of the space. The PU foam has good filling and isolation properties. Once it is cured (24h), you can take out the glass and use it when necessary.
Please read the safely instructions on the label !! Good luck.
PS: the justification for buying the PU foam, lies in the that you already spent some time trying to reach your goal.
EDIT: background info:
It concerns polyurethane chemistry. Here in Europe this type of canistered foam is widely used on the building site. Its purpose is predominantly to fill cavities and act as an isolation material. It's made out of an isocyante (R-NCO) and a polyol (R-(OH)n) forming a urethane bond (R-C=O-O-N-H-R). Together with a propellant (hydrocarbons), it is pressed out of a canister. The left-over isocyantes will react with ambient water, to form poly-urea bonds. This embodies the curing mechanism.
The reason for me to suggest this particular product is because it fills easily and shapes any form. It will cure within a couple of hour. Quit often this material is chosen to prepare surprises (packages with surprises in them....)


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Why are antibonding molecular orbitals formed?
When two atomic orbitals make one molecular orbital there must be an antibonding orbital also. Why should they make also an antibonding molecular orbital, such as the $2\mathrm b_2$ orbital in the scheme below?

also maybe there should be an anti-bonding orbital because suppose if there are a total of four electrons in the two orbitals of two atoms, it can only be distributed between BMO and AMO because one MO can only hold two electrons.

Why should they make also an antibonding molecular orbital?

Orbital symmetry must be conserved. Thus, when you mix orbitals, the number of orbitals must also be conserved. As a consequence, when two orbitals are operated on to produce new orbitals, two new orbitals must be created.
Think about it conceptually this way: When two orbitals that have the same symmetry and appropriately matched energy come together, a bonding orbital is formed. This bonding orbital contains all of the "good orbital stuff" from that pair of orbitals. But, because there were two orbitals originally, there has to be "orbital stuff" left over. This leftover "orbital stuff" forms the antibonding orbital.
Now, the screwy part is that orbitals are not real. They are mathematical constructs to describe the wave-like probabilistic properties of electrons. If there are no electrons in the orbital, then the orbital is not really "there". So, while an antibonding orbital is formed, it is not real unless it has electrons in it. In the case of water above, the antibonding orbitals are not populated, so they don't exist.

Good answer, except I'd remove your last paragraph entirely. It's not necessary for the OP's understanding, as it really is a philosophical question. What do you think?

@CHM: I think it's better to leave it there -- the fact that orbitals aren't "floating balloons waiting to be filled" can get rid of quite a few misconceptions. It's fine if an answer deviates a bit :)

What I mean is that the "reality" of orbitals has nothing to do with the depiction most high schoolers have of them. To me, they're just as "real" as anything else. I also happen not to agree that they spontaneously come into existence when the atom needs them. It's fine to point out that they're a solution to a mathematical equation, but I wouldn't go beyond that as the matter is a point of debate.

If antibonding orbitals do not exist, what do electrons get excited into when the molecule absorbs uv light?

@OscarLanzi a higher lying, multi electron excited state. At the end of the day, we know from Quantum mechanics that a multielectron state is not going to be describable exactly by a product, or linear combination of products, of one electron states unless you have an infinite number of them. But in that case, the choice of orbitals is basically arbitrary, so we can't say that one set of orbitals is more real than another.

@OscarLanzi maybe are not real because they represent a probabilities...and we can't say probalities exist, I suppose...


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Electronegativity of zinc vs copper in galvanic cell
I am reading up on how galvanic cell works and I realised that the flow of electrons is from Zn to Cu. But Zn is more electronegative compared to Cu according to periodic table trends
I read it somewhere that when comparing transition metals, it may be different, but it was not  further elaborated. Can someone explain this to me how Cu is more electronegative than Zn?

There are two common confusions in your question. I hope I can clear them up.

Electrons are not flowing from zinc to copper. Not exactly.

In the Zinc-Copper galvanic cell, the net reaction is:
$$\ce{Zn(s) + Cu}^{2+} (aq) \ce{-> Zn}^{2+} (aq)\ce{ + Cu(s)}$$
Zinc is being oxidized:
$$\ce{Zn(s)->Zn}^{2+}(aq) \ce{+2e-} $$
Copper is being reduced:
$$\ce{Cu}^{2+}(aq)\ce{+2e- ->Cu(s)}$$
Electrons, therefore are being transferred from zinc not to copper, but copper (II) ions, which are different than the copper atoms in Cu(s). We need to be comparing something about zinc atoms and copper (II) ions.

Electronegativity has nothing to do with electrochemistry.

Electronegativity is the attraction an atom has for a pair of electrons in a covalent bond.
Electrochemistry is about controlling the electron transfer that accompanies a redox reaction. Usually this involves ionic species, and so electronegativity is not the relevant quantity to compare. In this case, we care about the energy change that occurs when copper and zinc lose electrons (ionization energy) and the stability of the resulting ions in solution. Ionization data mined from this NIST publication.
For zinc:
$$\ce{Zn->Zn+ + e-} \ \ \ \ \ \Delta H= 9.585\times 10^2 \text{  kJ/mol}$$
$$\ce{Zn+->Zn}^{2+}\ce{ + e-} \ \ \ \ \ \Delta H= 1.733\times 10^3 \text{  kJ/mol}$$
$$\ce{Zn->Zn}^{2+} \ce{+ 2e-} \ \ \ \ \ \Delta H= 2.692\times 10^3 \text{  kJ/mol}$$
For copper:
$$\ce{Cu->Cu+ + e-} \ \ \ \ \ \Delta H= 7.454\times 10^2 \text{  kJ/mol}$$
$$\ce{Cu+->Cu}^{2+}\ce{ + e-} \ \ \ \ \ \Delta H= 1.958\times 10^3 \text{  kJ/mol}$$
$$\ce{Cu->Cu}^{2+} \ce{+ 2e-} \ \ \ \ \ \Delta H= 2.703\times 10^3 \text{  kJ/mol}$$

Why is zinc oxidized and copper reduced?

Because doing so leads to a lower energy state. It takes more energy to remove two electrons from copper than to remove two electrons from zinc. Thus, to get a negative energy change (spontaneous), when need to add the zinc reaction as written and the copper reaction in reverse:
$$\ce{Zn->Zn}^{2+} \ce{+ 2e-} \ \ \ \ \ \Delta H= 2.692\times 10^3 \text{  kJ/mol}$$
$$\ce{Cu}^{2+} \ce{+ 2e- -> Cu(s)} \ \ \ \ \ \Delta H= -2.703\times 10^3 \text{  kJ/mol}$$
$$\ce{Zn(s) + Cu}^{2+} (aq) \ce{-> Zn}^{2+} (aq)\ce{ + Cu(s)}\ \ \ \Delta H= -11 \text{  kJ/mol}$$
The estimate above assumes that the solvation energies of both ions are similar and their standard entropies are similar. I could not find a good source of those data.
In electrochemistry, we care more about the potential of the cell, $E_{cell}$, which is the potential for the cell to do electrical work. Potential, $E$, is related to free energy change $\Delta G$ by the following equation, where $n$ is the number of moles of electrons transferred and $F$ is Faraday's constant $F=96,485 \frac{\text{J}}{\text{V mol}}$.
$$\Delta G=-nFE$$
For a process to be spontaneous, $\Delta G$ must be negative. $\Delta G$ is negative when $E$ is positive. Thus, in a galvanic cell, electrons will flow in the direction that produces a positive cell potential.
How can we determine which direction that is? First, we could set up the zinc-copper cell, and connect it into a circuit and let it run. In this cell, a piece of copper metal serves as one electrode and a piece of zinc metal serves as the other. Whichever electrode increased in mass is the electrode at which reduction occurred: metals cations are converted to metal atoms, coming out of solution and increasing the mass of the electrode. The other electrode must be the one responsible for oxidation (metal atoms become cations and dissolve away); the mass of this electrode must decrease.
Or, we could do a simpler experiment. Dip a piece of copper metal into a solution of Zn2+. Nothing exciting happens (no reaction). Then, dip a piece of zinc metal into a solution of Cu2+. Zinc reduces the Cu2+ ions and copper metal plates onto the zinc.
Or, we could rely on tables of standard reduction potentials for half cells, measured against the standard hydrogen electrode SHE (for which E is defined to be ZERO). Standard reduction potentials come from Wikipedia, which mines them from a number of legitimate sources.
SHE:
$$\ce{H2(1 atm) + 2e- -> 2H+(1 M)} \ \ \ E^o =0 \text{ V}$$
For zinc:
$$\ce{Zn->Zn}^{2+} \ce{+ 2e-} \ \ \ E^o=-0.762 \text{ V}$$
For copper:
$$\ce{Cu->Cu}^{2+} \ce{+ 2e-}  \ \ \ E^o =+0.34 \text{ V}$$
To generate a spontaneous reaction, these half cells are combined in such a way so that the overall cell potential is positive:
$$\ce{Zn->Zn}^{2+} \ce{+ 2e-} \ \ \ E^o=+0.762 \text{ V}+$$
$$\ce{Cu}^{2+} \ce{+ 2e- -> Cu(s)} \ \ \ E^o=+0.34 \text{ V}$$
$$\ce{Zn(s) + Cu}^{2+} (aq) \ce{-> Zn}^{2+} (aq)\ce{ + Cu(s)}\ \ E^o=+1.10 \text{ V}$$
This analysis only applies to standard conditions. Under nonstandard conditions (for example [Zn2+]>>[Cu2+], then the reverse reaction could be spontaneous.

The answer by Ben Norris is nice but it seems to have an additional confusion.  The difference between reduction potentials of copper and zinc can be calculated based on the empirical ionization data that Ben describes, but this doesn't mean that the difference has nothing to do the electronegativity of the two different elements.  A nice discussion of this issue is provided at the following link.

Could you try to elaborate on this; maybe paraphrasing/summarizing the link here?

Electronegativity is tendency of an atom or a functional group to attract electrons (or electron density) towards itself. Adding an electron to copper is very easy because of its valence orbital which has one free electron, which wants to pair with another electron, whereas in zinc, the valence shell is completely filled, i.e., d-orbital.
Config of Cu: {Ar} 4s2, 3d9
Config of Zn: {Ar} 4s2 3d10


text:
Reduction and re-oxidation of TiO₂
I'm having a hard time finding data about an experiment that I know has been done a bunch of times:
I have a sheet of single crystal rutile $\ce{TiO_{2}}$ (110) that I want to reduce to various $\ce{TiO_{2-x}}$ forms where the x is what I want to vary. I want to try to avoid conversion to the other Magnéli phases if possible $(\ce{Ti_{n}O_{2n-1}})$. To do this reduction, I have a high temp oven that I can flow with 95% $\ce{N_2}$ and 5% $\ce{H_2}$ or simply evacuate. I can set the flow rate to cubic centimeters per second. I need a chart that will tell me what temperature, flow rate, and duration I need to convert my sheets surface to a desired reduced value (i.e., I want a specific x).
I then want to reoxidize the surface by using lower temperatures in a furnace that I can fill with oxygen at different pressures. I need a chart that will tell me what temp, partial pressure of $\ce{O_2}$, and duration I need to re-Oxidize the surface.
The goal here is not to simply get the surface, but to test other properties. I was hoping this step (finding the proper charts) would have been fairly straightforward, but my literature searches haven't turned up quite what I need. I was hoping someone here can point me to a resource that has this data.

I can't post a comment, so this is not a true answer but I think can help. If I've understand what you mean you want to perform a Temperature-programmed reduction (TPR) followed by a temperature-programmed oxidation (TPO). In this case I think this old book can help: Temperature-programmed Reduction for Solid Materials Characterization. I don't know much about single crystal but I think that many things from TiO2 studies of nanomaterials can be consider valid for your case, this online book can be helpful to understand all the variables.


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How dangerous is cobalt(II) chloride used in silica gel balls?
I was looking up silica gel beads on Wikipedia and noticed that some (and from experience most) are doped with a colouring agent to indicate that they have absorbed all that they can.
The Wikipedia page says that usually cobalt(II) chloride or methyl violet is used, both of which are toxic. But it mentions nothing about what quantity these become toxic? I assumed the quantity in a normal silica gel bag (that come in shoes for example) isn't enough to be harmful, otherwise why would they be using it?
Also, it mentions that

Crystalline silica dust can cause silicosis but synthetic amorphous silica gel is indurated, and so does not cause silicosis.

Are the silica gel beads synthetic amorphous, i.e. do not cause silicosis?

Same wiki page contains link to CoCl2 wiki page, that contains ld-50. It is more or less what should answer your first question. As for second question: no, silicosis is caused by silica DUST, and silica gel is amorphous, but compact.

@permeakra I'm answering as have not got the rep to comment yet. What if you crushed silica gel to dust, would this cause silicosis?

@Regie If you will breath in said dust on regular basis in significant amounts, then it possibly can. One-time exposure should not be a problem.

@permeakra Again, answering as I cannot comment :(. If Silica gel is amorphous, doesn't this mean it is not crystalline, and according to the wiki page only crystalline silica dust can cause sillicosis. Or am I mistaken?

Cobalt poisoning is apparently very rare. The only situation in which cobalt poisoning is a serious problem is when it is exposed directly to your bloodstream (Recently it has been an issue with people who have undergone certain hip implants -- see this page).
The amount of Cobalt chloride in silica gel balls would be absolutely harmless. According to most sources, you have to be exposed to significant amounts of cobalt (at least more than 20g for an average adult) for a long period of time for symptoms to appear.
Note: Cobalt is actually necessary for our body (in small amounts) in the form of Vitamin B12.


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If copper has 2 valance electrons and sulfur 6, why don't they bind in pairs?
Simple question, the average neutral copper atom has 2 valance electrons, and sulfur 6. Wouldn't the two electrons jump from copper and bind to sulfide?
The empirical formula is: $\ce{Cu + S -> Cu2S}$.
So apparently two copper atoms bind with every one sulfur atom. Does that mean that the copper atoms each gives one electron away and leaves them with 1 valance electron each?

the average neutral copper atom has 2 valance electrons

Nope. Only one:

In the transition metals, the previous shell is being filled. Generally, the valence shell can't have more than 8 electrons, but lower shells can. So, you have the two outer electrons filled in in Group 1 and 2, and the rest are filled in in the inner shell. With copper, we have 17 electrons in the inner shell, and 2 on the outer shell by this rule. But--it is sometimes more favorable for the inner shell to be "complete" with 18 electrons, leaving the outer shell with only one.
This is much better explained if you understand what atomic orbitals are. Think of these as as "subshells". In copper, we have two possible shell configurations: 10 electrons in $3d$ and one in $4s$, or 9 in $3d$, and 2 in $4s$. The numbers signify the shell, and the letters signify the orbital type. There are $5$ d orbitals per shell (not all shells have $d$ orbitals, though), and 1 $s$ orbital. Each orbital holds two electrons. Here, we have the trade-off between having a fully filled $3d$ orbital set and having a fully filled $4s$ orbital. Fully filled orbital sets lead to stability. Anyway, since both choices lead to a fully filled orbital set, a lot of other external factors decide which electronic configuration copper uses. So you have both $\ce{Cu^+}$ and $\ce{Cu^{2+}}$.
As a side note, $\ce{CuS}$ exists as well.

Interesting about the CuS, I didn't know that.

Still I don't understand how Cu's valence can equal 2. Its D-orbitals are all filled up, so how could Cu(OH)2 form..

I remember I once had an oral exam where my Prof asked me that. Why does Cu(2+) exist if it takes a lot of ionization energy to get from Cu(I) to Cu(II)? His answer back then was that the energy of hydration when Cu(II) is dissolved in water is much much higher and hence in solvated state Cu(II) get's stabilized more than Cu(I), while Cu(I) as solid state compound is quite stable.

The simple valency concept works well for many nonmetallic compounds and those containing alkaline metals, alkaline earth metals and some earth metals (groups 1, 2 and part of group 13). It is pretty much destined to fail for most other elements, especially any transition metals.
For one, no longer is the number of valence electrons (or eight minus that number) indicative of the element’s valency. Copper can form copper(I) compounds (e.g. $\ce{CuCl}$ or $\ce{Cu2S}$), copper(II) compounds (e.g. $\ce{CuCl2}$ or $\ce{CuS}$) and other more rare ones. Manganese, as an extreme example, is known in every oxidation state from $\pm0$ to $\mathrm{+VII}$ (and beyond).
For two, the concept of valence electrons is no longer clearly defined. Do you count only the s electrons of the outermost shell? Or do you count the former shell’s d-subshell, too? Do you count the actual ground state or a theoretical one according to the aufbau principle? Depending on your choice of definition, copper will have one, two or eleven valence electrons.
The bonding and electronic situations are much more complex than the simple planetary orbit model ManishEarth showed would lead us to believe. Both copper(I) and copper(II) are stable states but they are stable for different reasons. Going into these reasons would be too much for this page’s margin.
So the simple picture should stay in introductory courses where it does good work for nonmetallic molecules and then be wrapped away safely for the next class of beginners.