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Complexes formed by lanthanides and actinides
A very simple test for the presence of alcohols in the lab involves adding ceric(IV) ammonium nitrate solution which gives a pinkish red colour in the presence of alcohol.
When I looked up the structure of this compound it turns out to be quite interesting. So a couple of questions about lanthanides came up in my mind (a Google search only comes up with very complicated research papers):

Wouldn't this compound be highly unstable since there 12 donor atoms (forming an icosahedral structure) surrounding it (but it is quite stable)?
Is the formation of complexes in lanthanides and actinides similar to that of transition metals? Do $\ce{f}$ orbitals undergo crystal field splitting just like $\ce{d}$ orbitals?
Are higher coordination numbers like 12 more common (than octahedral and tetrahedral structures) among lanthanides and actinides because they have more orbitals for bonding, or is this compound an exception?

After a lot of digging around, I finally found something about this in J.D. Lee's Concise Inorganic Chemistry. I thought I'd post whatever I found over here anyway since it's very interesting:
Answer to question 2
Complex formation by lanthanides is different from that of actinides. In lanthanides, the $\mathrm{4f}$ orbitals are well shielded by the larger $\mathrm{5d}$ and $\mathrm{6s}$ orbitals and are deep inside the atom. So $\mathrm{f}$ orbitals do not participate in any bonding and complex formation is similar to that of transition metals. However in the actinides, the $\mathrm{5f}$ orbitals extend outwards and participate in bonding much more easily making the interactions much more complex than in the transition metals.
Answer to question 3
Higher coordination numbers are apparently extremely common among f-block elements (strange right?).
Octahedral (6) and tetrahedral (4) structures are very rare except when bulky ligands are present. Most common coordination numbers among lanthanides are

7 (Capped trigonal prismatic) yttrium acetylacetonate hydrate (Yttrium seems to be grouped with the f-block elements due to similar properties)
8 (Square antiprismatic and Dodecahedral) cerium acetylacetonate and holmium tropolonate
9 (Tri-capped trigonal prismatic) nonaaquaneodymium(III) complex $\ce{[Nd(H2O)9]^3+}$

10 (very complex) and 12 (icosahedral) are seen only in the larger elements like cerium and thorium with small ligands like $\ce{NO3-}$ and $\ce{SO4^2-}$
The actinides also commonly form exotic structures like chained tricapped trigonal prismatic in $\ce{[ThF8]^4-}$ and $\ce{[PaF7]^3-}$ distorted cubic in $\ce{[PaF8]^3-}$
The text mentions that the nature of bonding in $\ce{[Ce(NO3)6]^2-}$ is still not understood because it would imply bond orders of less than 1 or participation of f orbitals.
Could anyone explain anything about the first question since there is nothing mentioned on the stability of these compounds.

Maybe cerium's $f$ orbitals are more exposed early in the lanthanide series?

1) No, because it is not donor-acceptor relationship, but electrostatic one
2) No, f-orbitals of lanthanide and actinides are unavailable for valence interactions
3) No, they are common because amount of ligands in the case is determined by geometry, and not electronic structure. Lanthanide cations are big.

I don't understand what you meant by answers 1 and 3.. could you explain why having an electrostatic interaction makes it more stable.

The d-element complexes often have heavy dose of covalent bounding. This results in octahedral, tetrahedral or planar square coordination, as most closely resembling symmetry of d-orbitals. Complexes of lanthanides are purely electrostatic, and coordination number is determined by amount of ligands that can be in direct contact with central atom simulationely, the same way complexes of late alkali and alcali earths are built. Note: actinides are much trickier matter, as 5f- and 6d- orbitals in them have close energy.

If lanthanide complexes don't have crystal field splitting of the f orbitals can you explain why most lanthanide complexes are coloured? Crystal field splitting is ultimately just the lowering of symmetry leading to non-degenrate orbitals in the coulomb field, and this as the bonding in lanthanide complexes is generally considered to be ionic this will be the case here.

@IanBush Quick googling shows that crystal field splitting is minimal for f-orbitals in Ln complexes and optical $Ln^{3+}$ spectra have minnmal dependence on ligands. Apparently, their optical spectra are dominated by f-f transitions with energy difference arising from total angular momentum (including spin participation). I welcome any correction.

They form complexes compound but their tendency to form complexes is less their transition metal ions .the lanthanide ions spite a high charge have low charge density due to their large size.they form stable complexes with chelating ligands such as EDAT ,oxime beta-ketones.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. For more information in general have a look at the [help]. At the moment, while undoubtedly correct, this reads more like a comment than an actual answer, and only addresses a subset of the question. Please elaborate a little more. With a bit more rep, you will be able to post comments on any question/answer.


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Covalent Bond/Hydrogen Bond
During the formation of a molecule of water ($\ce{H2O}$), what kind of bond occurs between the elements? I'm studying this subject on my own and from some sources I saw that there is a covalent bond and others showed an hydrogen bond, responsible for the non-evaporation of water at $-90\ ^{\circ} \rm{C}$. Can you guys explain to me exactly how this bonds behave? Both of them occurs to the molecules of water?

Welcome to Chemistry.SE! This looks like an interesting question. We have some tools available to help make chemistry and maths look pretty in your posts. I edited your question to put them in. Check out our[faq#what-notation-and-symbols-are-commonly-used-here] for more info.

Covalent bonds are the bonds between the atoms within the same water molecule. Hydrogen bonds are the bonds between two water molecules. All molecules have covalent bonds, but only some molecules have hydrogen bonds. As an example, water has hydrogen bonds, but carbon dioxide does not.
One of the requirements for hydrogen bonding is that the molecule must be polar. Water molecules are polar because of two effects. The first is that oxygen is more electronegative than hydrogen, so each of the OH bonds will be polar. The electrons shared in each of those bonds will spend more time near the oxygen than near the hydrogen. The other effect is the geometry of the molecule: the oxygen atom has two free electron pairs in addition to the two bonds, so it assumes a tetrahedral arrangement. The $\ce{O-H}$ bonds are two corners of that tetrahedron, and the free electron pairs are the other two corners. This arrangement means that, if you only look at the atoms, the molecule has a bent geometry. Because of that, the polarity of the two bonds partially adds together. By contrast, $\ce{CO2}$ has a linear geometry, and the two polar bonds cancel each other out, leaving the molecule non-polar.
So the water molecule is polar, with a lot of excess positive charge around the hydrogen atoms, and a lot of negative charge on the side of the oxygen atom away from the hydrogen atoms. Hydrogen bond is the name given to the electrostatic interaction between the positive charge on a hydrogen atom and the negative charge on the oxygen atom of a neighboring molecule. The covalent bond is the electrostatic interaction between two atoms in the same molecule. Covalent bonds are much stronger than hydrogen bonds: the $\ce{O-H}$ has a strength of 467 kJ/mol, while the hydrogen bond is usually between 4 to 40 kJ/mol.

Just want to point out that there is a subtlety to the word "water":
Water can mean the molecule, $\ce{H2O}$, and those bonds (between central O and H's) may be covalent.
Water also means the bulk water, many molecules of water.  The interactions between two or more molecules is what dictates the bulk properties of a liquid, such as evaporation and freezing.  Here we would talk about hydrogen bonding.
Below is a picture of bonding of one configuration of "water" in "water."  This is computed from electronic structure theory, and the Quantum Theory of Atoms in Molecules:

Does this answer the question?

Water is a polar covalent molecule. Take a look at their electronegativity values, Hydrogen is 2.20 and Oxygen is 3.44. This is a difference of 1.24 meaning that it is Polar Covalent. A Covalent bond is present when there is a difference 1.7 +/- 0.5. Also there is Hydrogen Bonding in water also, this is because the hyrdogens will have a slightly psotove charge and the oxygen will have a slightly negative charge, which causes hydrogens in other water molecules to become attracted to the oxygens. I hope this helps

In H2O molecule, two water molecules are bonded by a Hydrogen bond but the bond between two H - O bonds within a water molecule are covalent.
Refer this diagram : http://goo.gl/84w9T
 The dotted lines represent a hydrogen bond and the solid lines represent a covalent bond.

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Database for molecular dipole transition moments?
I'm looking for the dipole transition matrix elements for electronic transitions between the ground and excited states of the carbon dioxide cation, $\text{CO}_2^+$, that is, objects like
$$\langle\text{CO}_2^+ \,\tilde{\text{A}}\,\Pi_\text{u}|\mathbf{r}|\text{CO}_2^+ \,\tilde{\text{X}}\,\Pi_\text{g}\rangle.$$
I feel this is a standard enough quantity that it ought to be easy to find in some appropriate database of chemical data. However, I've searched for quite some time and I can't even find something similar.
Where should I be looking?

For a triatomic molecule, I guess that would be difficult to come by in a database. Molecular Spectra and Molecular Structure III - Electronic Spectra and Electronic Structure of Polyatomic Molecules by G. Herzberg is a good place to start.

You can find some experimental information on dipole transition moments in the NIST Chemistry Webbook in the section on constants of diatomic molecules.
http://webbook.nist.gov/cgi/cbook.cgi?ID=B3000043&Units=SI&Mask=1000

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Are there computer searches for new theoretical minerals in a similar vein to Folding@Home or SETI@Home?
As a surprise gift to a friend I would like to discover a new theoretical mineral and have it named after them. I have some understanding that not all minerals have been discovered, and some minerals exist in theory but have not been found on earth.
Is there a computer search that looks for new minerals? Would it be difficult to write one, if one does not exist?

Done. I've added a bit more information about computing crystalline structures.

By Ross Snider's request, I have moved my comment into an answer - whilst it does not address computational searches for minerals, it touches on the complexities of actually getting a mineral officially named:
The International Mineralogical Association Commission on New Minerals, Nomenclature and Classification seems to require a physical specimen for consideration for official naming of a new mineral. Under current rules, anthropogenic minerals are also excluded, so if you came up with a hypothetical mineral it seems you would have to also find a naturally occurring specimen to get it named.
To add to this, I am not aware of any software for systematically searching for plausible minerals, however computational packages for the simulation of crystalline structures certainly exist. Determining the electronic structure of a crystal - which is what is really required to make conclusions about geometry, stability and so on - generally involves the use of plane wave QM codes such as VASP, CASTEP or ABINIT, the latter of which is free. The feasibility of mineralogical calculations using these methods is strongly dependent on how crystalline the mineral is and how ideal it is (from a perspective of not being heavily defected or substituted) as they work within the constraints of a periodic unit cell or supercell.


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Stir bar size and volume to stir
I'm looking for a table showing which size of bar to use for a given volume to stir.
Or a formula if there is one.

No chart I'm aware of.  However, an informal and incomplete survey indicates most chemists to this by trial and error.
For pipe-type stir bars in flat bottom flasks/beakers, and you want a 'fast' stir, generally you want one roughly two thirds the size of your magnet arm to minimum the motor moving ahead of the magnet spin and 'flinging' it around.  Obviously this is not possible if your flask is much smaller than the motor.  If your stir bar is too close to the size of your vessel, you have 'catching' issues. with the sizes.
For egg-type stir bars in round bottom flasks, the concavity of the surface will help, but not eliminate 'flinging'.If you need a fast stir, you need a bigger stir bar.  Personally, I've never needed larger than 3/4" because of my use preference, (also do not typically use roundbottoms larger than 1L).
Ultimately, however, remember stir bars are for mild to moderate stirring.  If you need vigorous, you are better served with a mechanical stirrer.  Additionally, if you are using very large (5L+) flasks, mechanical stirring is best.

And how would I know the size of the magnet "arm" without disassembling the device ?

The size of the 'plate' portion of the stir plate will give you a rough estimate.  The stirring arm will be a little smaller than this.


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Unique calling code for a chemical reaction
I'm writing a script and would like to use a "Gödel numbering" to search for specific reactions.
Is there any used procedure to assign an symbol (integer, possibly) to a chemical reaction, e.g.
$$\ce{CH4 + 2 O2 -> CO2 + 2 H2O} := 121231412?$$

To clarify, do you require reactions that are similar in some sense to generate similar numbers?

@RichardTerrett: Would be a feature, but isn't necessary. I don't expect the numbers/symbols to be accessible to a human.

My suggestion is to dabble in combining SMILES representations of the reactants and products and a stoichiometric matrix in a consistent fashion and then using a hashing function (Now that I think of it, I guess Godel numbering is an example of hashing) to assign the reaction a unique identifier. You need to use a representation of the reactants and products that unambiguously describes the molecular graph (including stereochemistry) so that stereo- and/or constitutional isomers cannot give rise to the same identifiers, and so that different orderings of atoms in a string describing identical molecules do not give rise to differing identifiers.
There exist a number of reaction ontologies that can be used to categorise and describe chemical reactions. KEGG RCLASS and RPAIR is one such example which gives rise to numerical identifiers, however I'm not totally clear on how it works and it appears to be tailored to metabolomics. It is however a good example of the kind of thing people are doing.
The CMLReact format is an XML format for representing chemical reactions that may be suitable for generating short identifiers from through hashing, however being XML it has enough cruft in it that chemically insignificant changes to a CMLReact file may give rise to differing derived identifiers. Open documentation of the format is sadly not terribly forthcoming on the internet however there is an library for reading and writing CMLReact in OpenBabel.

I don't know about Godel numbering specifically (though this is interesting, my being a math guy at heart), but there are existing textual systems for encoding molecules (not reactions that I know of).  From these, you may be able to encode reactions.
Two examples are InChI and SMILES.  These are Wikipedia links.

The existing answers are a little outdated. More recently there has been a good amount of work done towards extending the InChI identifiers (which encode chemicals) to the Reaction InChI, or RInChI (which encode reactions, as the name suggests). See e.g. Grethe, G.; Blanke, G.; Kraut, H.; Goodman, J. M. J. Cheminform. 2018, 10 (1), No. 22, for a very thorough description of how RInChI can be constructed.
Generally these tend to be quite long; so it was also considered desirable to perform some kind of hashing so create shorter keys called 'RInChIKeys', which is also described in the aforementioned paper. This will likely come closer to your idea of a 'symbol' representing a reaction, although in this case it is a string, not a number. Collisions are highly improbable, but in principle possible.
Software for generation of (R)InChI can be obtained from https://www.inchi-trust.org/downloads/; also, a small-ish database of reactions and RInChIs can be downloaded from http://www-rinchi.ch.cam.ac.uk/downloads/.

SMARTS are a canonical way of representing chemical reactions (and other pattern-matching problems). This is a formal language so you could assign a Goedel numbering to the SMARTS. If you need a library to handle SMARTS/SMILES etc., take a look at OpenBabel.

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Chemical reaction calculator
Is there any kind of program that is able to calculate the end result of a specific chemical reaction? Say I wanted to know what happens when you mix $\ce{HClO + NaOH}$, is there anything that could give me the answer to any question like this? Wolfram alpha works pretty good, but it doesn't give me an answer if it isn't a balanced equation, and I don't know what that means. $\ce{HClO}$ is an acid and $\ce{NaOH}$ is a base, so there has to be some reaction, balanced or not.

You could certainly try this: http://www.webqc.org/balance.php

This is a fundamentally very difficult question. Your example of of $\ce{HClO + NaOH}$ hides several assumptions, e.g., time frame and aqueous solution, which really means some number of $\ce{HClO}$, some number of $\ce{NaOH}$, and presumably a huge number of $\ce{H2O}$. When mixing all three components a wide variety of reactions take place, e.g., disproportionation of water ($\ce{H2O + H2O -> H3O^{+} + OH^{-}}$), solvation of $\ce{NaOH}$ etc.
In general, a full exploration of all possible reactions and their barriers have to be assessed in order to make a prediction. For instance, $\ce{HClO + H2O -> HCl + O2H2}$ is a possible reaction, or $\ce{NaOH + 2 HClO -> NaCl + H2O2}$ even in the absence of extra water. The number of possible recombinations of the input materials generally increases exponentially with the number of atoms (e.g., the number of possible isomers of an n-alkane).

I'd like to point out that water does not disproportionate in the reaction you have mentioned. There is no change in the oxidation state of any of the atoms involved in the reaction. A better term would be autoionization : http://en.wikipedia.org/wiki/Self-ionization_of_water


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Varying potential diffrence between two graphite strips in NaCl solution
I have two identical graphite strips of dimensions 3.5 cm by 1 cm. Both are 3/4 submerged in aqueous $\ce{NaCl}$ solution (0.34 mol/dm$^3$, 2% wt) separated by a distance of 7 cm. I have attached crocodile clips to each strip and a voltmeter is used to record the potential difference between the two graphite strips.
I have found that:
A. There is a potential difference that ranges between 1 mV and 100 mV; and
B. This potential difference can change quite dramatically within that range over time periods of an hour.
Does anyone know of any chemical or physical reason for these potential differences?

What type of voltmeter are you using? to verify the answer to my question, you need to determine 1) the precision and 2) the detection limit of the instrument .
The precision is a measure of the error in a typical measurement. The smaller the standard deviation in repeat measurements, the higher the precision. To get a guess at the precision, prepare a system that should generate a known potential difference in the same range as your current measurements (like 100 mV). If the variation in that value is similar to what you are seeing, then you are recording the noise.
The detection limit is the lowest value your instrument can reliably measure. Generally, you cannot reliably measure values smaller than the twice the precision of your instrument. If you determine a standard deviation of 50 mV when determining precision, then 100 mV is your the detection limit. If this is true, you are recording the noise.
The precision and detection limit of an instrument depend on quality (price), but also on purpose. A voltmeter designed to be used by an electrician will have high precision in the range 1-200 V, but low precision below 1 V (because that range is not as important). A sensitive voltmeter used for electrochemical measurements may have high precision between 1 uV and 1 V, and behave miserably outside that range.

I am using a Keithley 2000 with detection limit of 0.1μV. I have also examined the precision and it seems to be consistent to around 0.1 mV, so I dont think it is likely to be noise.

It is also possible that the previous history of the graphite strips is important.  The potential difference observed might, in part, be due to impurities in the graphite.

@PaulJ.Gans - that is a very good point. Graphite is not an inert material.

It might be asymmetry potential.
Asymmetry potentials are due to differences between the electrodes themselves, and are not due
to differences in potential within the NaCl solution.

Also don't rule out thermal gradients at the points where the probes/wiring are attached to the graphite strips. You can be measuring thermal electric effects as well.

What is the input impedance of your voltmeter? How much DC current does it pass to measure a potential?
In theory, the input impedance is infinite and the DC current is zero, but in practice, this isn't the case.
In the NaCl solution with graphite electrodes, any DC current has to pass through electrolysis or impurities. This will lead to the behavior you are seeing with the impurities making the potential fluctuate.
What are you really after?

Keithley 2000 has 10Gohm input resistance in relevant ranges.


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Anomalous electronic configurations
Cu has an anomalous electron configuration. $\ce{Cu ~=~ 1s^2~2s^2~2p^6~3s^2~3p^6~4s^1~3d^{10}}$, it does not follow the usual pattern. In this case, the 3d subshell is filled before the 4s, which usually happens in the reverse order ($\ce{1s^2~2s^2~2p^6~3s^2~3p^6~4s^2~3d^{9}})$.
My question is, can you tell by an element's position on the periodic table (Group #, Row #, Block, etc.) that it will have an anomalous electron configuration? Do I have to memorize which elements have this property?
I am noting that Cu is in the 1B Group of Transition metals which implies that it has one valence electron, this is consistent with its configuration. will this be true for all cases?
Also: $\ce{Fe ~=~ 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6}$ and is a transition metal as well (Group 8). This implies it must have 8 valence electrons. But following the definition of valence electrons (The number of electrons in outermost shell), $n = 4$ is the outermost shell and I get a conundrum. Obviously if I add $\ce{4s + 3d}$ electrons everything is good. Why is this?

Many related questions have been asked on this topic before. I suggest you read these previous posts:
Why do elements in columns 6 and 11 assume 'abnormal' electron configurations?
How can one explain Niobium’s weird electronic configuration?
The Aufbau Principle is a general rule of thumb to find the electronic configuration of an element but usually many exceptions arise in the d and f block elements due to electron-electron interactions.
The concept of valence electrons does get a little fuzzy when you consider $d$ block elements since the energies of the $\ce{3d}$ and $\ce{4s}$ subshells are quite similar so I think its better to avoid it. $\ce{d}$ block elements can show a staggeringly large number of oxidation states (for example manganese which shows almost all oxidation states from 0 to +7). This can't be explained using the concept of valence electrons and the octet rule.
$\ce{Fe}$ itself can lose 2 electrons to form $\ce{Fe^{2+}}$ or 3 electrons to form $\ce{Fe^{3+}}$ so now again you can see that you can't explain this by saying that $\ce{Fe}$ has 8 valence electrons.
Other exceptional configurations: (there are quite a few):
niobium, platinum, gold, cerium, gadolinium, etc.

Fe can have even more oxidation states than magnesium. All the way from -2 to +6. Example of iron in +6 oxidation state: Potassium ferrate, Example of Iron in -2 oxidation state: Disodium tetracarbonylferrate

First of all the configuration of $\ce{Cu}$ you see has anomalous electron configuration because completely filled or half filled sub shells are more stable than any other configuration and you can see the configuration at last it is $\ce{4s^13d^{10}}$ and the basic configuration is $\ce{4s^23d^{10}}$ so you can see that $\ce{4s}$ is half filled and $\ce{3d}$ is completely filled. It is said this configuration is more stable due to symmetry. And it has no connection to their position in periodic table, and you only have to remember the concept and be careful while writing the configuration to notice this type of conditions.
As much I know they both are to be taken as outer most shell because $\ce{4s}$ is a lower energy cell as compared to $\ce{3d}$ and it filled first. I don't know the details about this, I hope some one will give a better answer to the second part as I also wish to know about that.


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Why does oxygen not like to be a double anion?
The electron affinity of a neutral oxygen atom is −142 kJ (it releases this energy). The electron affinity for the now double anion $\ce{O^2-}$ is 710 kJ (work must be done on the atom).
My question is, why the large jump in electron affinity? We are taught that the electrons do not interact much with each other in the way of shielding. I think this to be incorrect, because otherwise, the effective nuclear charge on all of the valence electrons should be about the same, including the new ones added to ionize the atom. This would make the electron affinity about the same for each electron. Plus, doesn’t the oxygen atom want to complete its shell? So why is it putting up a fight?

The addition of an electron to any negatively charged ion will need to overcome the repulsive force of the negative ion. This is generally less favorable than adding an electron to a neutral atom, and the electron affinity will be positive. Even though the additional electron will complete the octet, as in the case of an  ion where the addition of electron will complete the  subshell, the electron affinity is a positive value.

Nothing really likes to be a double anion.  But there are mitigating factors.
First -- getting more charges onto the electronegative atom gives the opportunity for greater electrostatic energy in an (or a predominantly) ionic lattice.
Second -- with the specific electron configuration of oxygen or sulfur, the work required to get the second electron is less than what would be required, for instance, to get a chloride di-anion.
Add this up and the result is that the lattice energy is enough to make the di-anion in the condensed, generally solid compounds where we usually see it.


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Solubility of Calcium Fluoride in water
$\ce{CaF2}$ is insoluble in water, but $\ce{CaCl2}$, $\ce{CaBr2}$, $\ce{CaI2}$ are soluble. Why is this so?

Many factors contribute to the solubility of a compound in water. In this case, the main difference between the compounds is the bonds’ strength.
Fluoride is very electronegative, and thus the bonding forces between $\ce{Ca^2+}$ and each fluoride anions are very strong, they are tightly held together in the crystal form. A few water molecules do not have enough strength to get in between and dissolve it.
Chloride is less electronegative, thus $\ce{CaCl2}$ has weaker bonds, then comes $\ce{CaBr2}$ and at last $\ce{CaI2}$, the weakest of them all. You may also expect increase in bond length depending on the crystal structure, which may also contribute to the solubility properties of the given compound. So weaker the bond strength, greater the solubility – at least that’s the expected.
In fact, $\ce{CaCl2}$ is more soluble than $\ce{CaF2}$, and $\ce{CaBr2}$ is even more – as expected. However, I did some research and the Wikipedia article on $\ce{CaI2}$ claims it’s less soluble than $\ce{CaBr2}$, which is not the expected, but it does not provide a source for that information. It might be true, though, since there are a variety of factors that may or may not contribute to it’s solubility in water, such as it’s crystal structure and the particular characteristics of iodide.
You can do the experiment yourself to find out which one is more soluble, since all solubility facts and numbers we know come from experiments. I couldn’t find anything on the solubility of $\ce{CaI2}$ in my CRC Handbook of Chemistry and Physics either, so I can’t provide you with a precise answer on why it’s less soluble than $\ce{CaBr2}$ (if in fact it is), but I hope you can now understand some of the factors that make $\ce{CaF2}$ the least soluble compound on that list.

I have a thought of the crystal structures. When we look at $\ce{BeF2}$, $\ce{Be^2+}$ being a tiny cation couldn't simply accommodate 2 strong $\ce{F-}$ in its fluorite fitting. It would lead to negative repulsions and even the hydration enthalpy of $\ce{Be^2+}$ being high would consequently increase the chances of solubility. On the other hand, when the cation size increases, let it be $\ce{Mg^2+}$ or $\ce{Ca^2+}$, they could be large enough to accommodate $\ce{F-}$ ions in their tetrahedral voids making the lattice stable. This would reduce the chances of ionic dissociation. Now when we replace $\ce{F-}$ with larger $\ce{Cl-}$ and $\ce{Br-}$there would be again a change in the limiting radius ratio of the crystal which would enhance dissociation and hence solubility.

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How to rationalise the coordination number of CsCl versus NaCl?
Question:

Suggest an explanation for the coordination number in the $\ce{CsCl}$ lattice being different from those in $\ce{NaCl}$.

Standard answer:

Caesium ion is larger than sodium ion, hence it can be surrounded by greater number of chloride ions.

My questions:

The question didn’t specify whether it is asking for the coordination number of $\ce{Cs+}$ or $\ce{Cl-}$... Is it because the coordination numbers of both ions are the same?
The given answer sounds like each $\ce{Cl-}$ ion will be surrounded by less $\ce{Cs+}$ ions, how is it possible that the ratio of $\ce{Cs:Cl}$ is 1:1 (as from its empirical formula)?

You may also consider the radius ratio of the cation to anion. This dictates the type of coordination.

Since both compounds are of the form $\ce{AB}$ (and not $\ce{AB2}$ or $\ce{A2B3}$/etc), the coordination numbers will be equal for the ions. Similarly, for $\ce{AB2}$, the coordination number for $\ce{A}$ will be twice that of $\ce{B}$.
What happens is that each ion is connected to more ions of the opposite type in $\ce{CsCl}$. Each $\ce{Cs+}$ ion is connected directly to 8 $\ce{Cl-}$ ions, and vice versa. The net effect is that their ratio stays the same, they just have more connections.

Look at it this way: There are four men and four women in a room. First, each man holds hands with only one woman and vice versa. Now, we tell each man to hold the hands of two women, and we tell each women to do the opposite. Now, we shift it up to 3 hands (with the reasonable approximation that humans have an infinite supply of hands), and finally 4. Try this on a piece of paper -- you'll see that the man:woman ratio doesn't need to change as long as the "handholding number" (coordination number) for each person is the same.

Taking a look at the actual lattices:
For $\ce{CsCl}$ (body-centered cubic):

Note that each $\ce{Cs+}$ ion touches 8 $\ce{Cl-}$ ions and vice versa. Also note that the $\ce{Cs:Cl}$ ratio is indeed $1:1$. For each $\ce{Cs+}$ ion, there is a corresponding $\ce{Cl-}$ ion on its NE diagonal tilting into the plane of paper, and for each $\ce{Cl-}$ ion there is a corresponding $\ce{Cs+}$ ion on the SW diagonal tilting out of the plane of the paper.
Here's $\ce{NaCl}$ (face centered cubic, close packing):

Here each $\ce{Na+}$ ion touches 6 $\ce{Cl-}$ ions and vice versa. We can define a similar pairing as before -- each $\ce{Na+}$ ion has a corresponding $\ce{Cl-}$ ion to its east, and each $\ce{Cl-}$ ion has a corresponding $\ce{Na+}$ ion to its west.

‘with the reasonable approximation that humans have an infinite supply of hands’ — brilliant =D


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Is it possible to separate salt from food, or at least convert it into something that doesn't taste salty?
A couple months ago, one of my room mates accidentally used a cup of salt to make apple crumble. She asked if there was any way to separate it, probably because I had recently reduced the acidity of a tomato sauce by adding baking soda. I said "probably not", but ever since then I have wondered if there is a way to get the salt out.
At this point, I would settle for a difficult method that requires a lab and ruins the dish, as long as it is still harmless to eat and retains most of it's nutritional value.
Here are some of the ideas I have thought of:
First, scrape off the oat layer, because it is the only hard solid and the salt wasn't mixed with it.
Seperate it by mass:

The apple crumble could be puréed and spun in a centrifuge to separate the salty water, which could be poured off and replaced with sugar water.
It could be atomized and sprayed sideways into a cold vacuum chamber, and then if the salt fell in a consistent radius around the nozzle, we could scrape up the salt, mix the rest back together and eat it.

Alter the the properties of the salt:

Is there a way to encourage the salt to form large crystals, even when it is a relatively minor component of a complicated mixture? If so, those crystals could be filtered out of puréed apple crumble with a coarse filter.

Crazy biochemistry:

Are there proteins or enzymes that could break the salt apart and bind the sodium to something oily that would float up by itself, leaving the chlorine to evaporate? I saw a documentary about bull sharks which mentioned that they have a gland that can absorb salt against the salinity gradient, so that the shark can expel it's salt and move up into fresh water.
Are there bacteria that require lots of salt to survive, and by using it, convert it into something that doesn't taste salty, and wouldn't release tons of sodium in my stomach?

How about just temporarily change your sense of saltiness? There is a berry that changes your sense of bitterness so it tastes sweet. Soy sauce makes salty foods taste less  salty.

Even if there is a method to extract salt from food, specially from apple crumble, it’ll probably ruin it.
The simplest way I can think of doing this is dissolving the apple crumble in water and blending it. Filter it and then boil the filtrate.
Given that every solid ingredient but salt is organic, when all the water boils off you should be left with salt and burnt carbohydrates, since NaCl boiling point is way above 1000 °C and a stove will reach 400 °C tops. Now, if you ever burnt sugar you know the caramel sticks to the pan, so if you’re lucky just add water again to separate the salt crystals from the burnt carbs and transfer it to a different recipient. Boil the water to get the salt.
You can add the burnt carbs back to the apple crumble but I don’t think it’ll taste like apple crumble again.

Nice creative thinking, but I think that'll probably fail the "retains most of its nutritional value" test.

That is awesome, lol! I wish I could upvote. I was hoping to preserve some of the nutritional value that wasn't already lost in the baking process, e.g. the amino acids, but I think that even in an inert atmosphere or a vacuum, they would be destroyed at that temperature. I can't find out online though, all my searching leads to paywalled articles.


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What is the difference between a chemical state and just a combination?
I don't think a combination also can be called a "state" i.e. that the chemical state of vapour or solid would be just a combination. When I read Wikipedia about "chemical state" it refers to just a combination. But AFAIK, an entity that has just combination does not have a state. So how do we know what is a state and what is just a combination?

You seem to be confused with "state of matter"/"phase" and "chemical state".
The chemical state encompasses all the macroscopic parameters necessary to fully define a given system. For example, "pure $\ce{O2}$ at $1\:\mathrm{atm}$ and $100^{\mathrm o}\:\mathrm{C}$" is an example of chemical state. A more complicated example would be something like: "10 moles of liquid $\ce{O2}$ at $-100^{\mathrm o}\:\mathrm{C}$, mixed with 1 mole of gaseous $\ce{O2}$ at $1\:\mathrm{atm}$ and $100^{\mathrm o}\:\mathrm{C}$ and 2 moles of gaseous $\ce{N2}$ at $1\:\mathrm{atm}$ and $200^{\mathrm o}\:\mathrm{C}$" (Yes, this isn't at equilibrium)
On the other hand, the phase or sate of matter only applies to a homogeneous system. A gas is in gas phase, but a system consisting of a gas and a solid doesn't have a phase/state of matter. Note that if a gas/solid is dissolved in liquid, then the phase is liquid since solutions are homogeneous.

Thank you. It looks like a chemical state refers to a combination and also properties around it.


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How long can poly(ethylene oxide) molecules be?
In this video, Steve Spangler says the substance he is using is a poly(ethylene oxide): http://www.youtube.com/watch?v=m-88M75_PCI
He mentions that one molecule would go to the moon and back twice! I have a hard time believing this, but I guess it must be true.. Is it true that poly(ethylene oxide) molecules can be that long?

Short answer? Possibly, but these ones aren't.
So, the other thing mentioned in the video is that the specific material he's using has a molecular weight of approximately four million g/mol. If we assume it's ordinary poly(ethylene oxide), we can get the rough number of $\ce{(CH2)2O}$ subunits by just dividing by the molecular weight of the subunit:

$(12 \times 2) + (2 \times 2) + 16 = 44$
$\frac{4000000}{44} \approx 90900$

So, if we assume that each $\ce{C-O-C}$ back-bone unit is stretched out like a di-methyl ether molecule, a quick check on a fairly standard Ghemical force field gives me a subunit length of approximately 3.7 Å.

$3.7 \times 10^{-10}\ \mathrm{m}\times 90900 = 0.00003363\ \mathrm{m}$

So, these molecules should be about 0.03 millimetres long. Not exactly astronomical distances, but impressively long for a single molecule.
In theory, I don't think there's anything stopping you making a molecule as long as they claim, but these ones certainly aren't. I think also that the longest (synthetic) polymers in common use are of similar molecular weights and thus similar lengths: ultra-high molecular weight polyethene seems to top out around 4.5 million g/mol for commercial applications, from a quick search.
Biological molecules frequently get longer than this, though, especially DNA: the average human chromosome apparently contains a single DNA molecule approximately 5 centimetres long.
The man explaining in the video may have erroneously given the figure for the total length of all the molecules in that beaker.

Note that a while a millimetre (or kilometre) long molecule is surely allowed, you cannot straighten it out to that lenght. It would long before rupture under its own desire to coil up into a ball, unless it was cooled very close to absolute zero. Entropy rules.


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Lithography: Is isopropanol a developer and/or a stopper?
In the ebeam lithography with PMMA resists isopropanol is often used in a mixture 3:1 (IPA:MIBK) as a developer solution. To my knowledge the developing process is usually stopped by putting the sample in distilled water and drying it with a nitrogen airgun. But I also see collegeaus who use the 3:1 mixture for developing and then "stop" the development in pure IPA. I actually think they develop further, as I know some people also use pure IPA instead of 3:1 mixture for developing.
What am I missing, I know there many different recipes in lithography, but this IPA thing doesn't make sense to me, using IPA as a stopper means to me the developing process is strengthened further and the resist isotropically resolved, which you normally want to avoid with kind of written sharp edged rectangular structures written by ebeam

Could you expand abbreviations, please?

Methyl isobutyl ketone is the solvent for PMMA - the alcohol is just the carrier. MIBK has low solubility in water, so if you stopped with water, you would probably leave more MIBK residue in your PMMA.


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Delta G for Standard Hydrogen Electrode reaction
I’m dealing with electrochemistry problems.
For the following reaction
$$\ce{2H+ +2e- <=>H2}\quad E^\circ=0\ \mathrm{V}$$
Does it make sense to say
$$\Delta G^\circ=-nFE^\circ=0$$
Which would mean the reaction is in equilibrium?

You cannot apply the $\Delta{G}$ equation to a single electrode potential. It can be applied to a cell though so if the hydrogen electrode is connected to another electrode (say copper dipped in copper sulfate solution) then you can find the free energy.
It's better to remember the formula as:
$\Delta{G} = -nFE_{cell}^o$
Where
$E_{cell}^o = E_{cathode}^o - E_{anode}^o$ (using all standard reduction potentials)
That way you won't accidentally apply it anywhere else.
In this case, for $\ce{\Delta G}$ to be zero you would have to take Hydrogen electrodes for cathode as well as anode.

A More Complicated Explanation
Just in case you're curious about the reasons behind this (Since I was also very confused about this when I first learned Electrochemistry. It's ok if you don't understand this),
It is very difficult to measure the absolute potential of a single electrode and so the standard reduction potentials you are using are actually based on a reference system.
This is actually very similar to the system of atomic masses. The atomic masses you use in calculations are not the actual masses of the elements in grams but they are measured in unified atomic mass units. One unified atomic mass unit is defined as 1/12th the mass of a carbon-12 atom.
[Note: Now of course, after the mole was made a fundamental SI unit, the unified atomic mass unit has become equivalent to $1 g/mol$. This is just for the sake of explanation.]
So if you come across a situation where you have to use the mass of the atom (maybe some nuclear chemistry calculations) you can't use those normal atomic masses but you would have to convert it into kg (One unified atomic mass unit is about $1.6\times10^{-27}$ kg).
Similarly, the standard reduction potentials are actually measured with reference to a standard electrode (Whose reduction potential is arbitrarily taken to be zero) which is the standard hydrogen electrode.
This standard reduction potential of an electrode can be measured easily by connecting the electrode to a standard hydrogen electrode (SHE) and then measuring the potential difference using a voltmeter and applying the $E_{cell}^o$ equation. (If $H^+$ is being reduced to $H_2$, then $E_{cathode}^o$ is zero otherwise $E_{anode}^o$ is zero).
So this just means that when you say $E^o$ of the SHE is zero, you are actually saying that "the potential of the SHE with respect to itself is zero" which is quite obvious. That doesn't mean that the potential of the electrode itself is zero.
That is why you cannot apply the first equation in this case. But this does not apply to cells. If we assume the absolute potential of the SHE to be some $x$, then,
$E_{abs}^o = E^o + x$
Then,
$E^o_{cell} = E_{abs}(cathode) - E_{abs}(anode)$
$= E^o_{cathode} + x - [E^o_{anode} + x]$
$E^o_{cell} = E_{cathode}^o - E_{anode}^o$
The absolute potential of the SHE cancels out in the difference.


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Why does permanent marker writing come off a white board when coated with white board marker?
Why is it that dry erase markers allow permanent markings to come off?
Here's a demonstration.

It's apparently the non-polar solvent that the dry erase marker contains: http://en.wikipedia.org/wiki/Permanent_marker#Removal

The solvent of the marker dissolves the permanent ink and mixes it with the whiteboard ink, leaving a dark dust that doesn't stick to the board.
The link in Wikipedia doesn't name common ethyl alcohol (denatured or not) as being a solvent for old marks. That is what I have used.


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Energies of atomic orbitals on molecular orbital diagrams
When we plot molecular orbital diagrams we use a linear combination of atomic orbitals.
Where can I find the energies of particular atomic orbitals?
In the picture below, the energies of 2s and 2p orbitals of Oxygen are lower than Carbon's 2s and 2p orbitals of Carbon, what is the logic behind this?

It has to do with the fact that oxygen has a higher effective nuclear charge. Thus, the oxygen orbitals are "closer" to the nucleus and therefor more stable. If I can get something to back it up, I will make an answer.

Ben Norris's comment is correct.
An estimate of the atomic energies for orbitals may be gleaned from ionization energies, for example here:
http://periodictable.com/Properties/A/IonizationEnergies.html
Otherwise, the energies need to be calculated quantum mechanically.

This is just misleading. For elements greater in atomic number than lithium, the 2s orbital still has an ionization energy which increases with atomic number. The table that Lighthart linked is for outer shell ionization energies which is highly influenced by the shell structure of the atoms.

It is fairly obvious the OP's question refers to valence electrons, rendering this comment moot at best.


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Is it possible to produce graphene out of air?
In theory you should be able to turn $\ce{CO2}$ to $\ce{CO}$ and then to $\ce{C}$ and connect all the C-atoms together to form graphene.
That should be possible, because plants do something similar, they take $\ce{H2O}$ and $\ce{CO2}$ and link the C-atoms together as $\ce{C6H12O6}$.
Is it possible to produce graphene out of air? Here is a related article from the RCS: "Conjuring graphene oxide from thin air"

Seems like it should be as possible as for me to throw my clothes on my bed such that they fold themselves mid flight. Unfortunately it has not yet happened for me! Read more about entropy.

If this is a serious question, of which I am dubious, please gather some background information by reading an introductory bio or chem text first and revise your question with the facts you find.

A graphene sheet is easily obtained from a block of graphite using adhesive tape. Since graphite itself is essentially a pure grade of coal, it is the result of biological processes in vegetation some millions of years ago. As is pointed out in answers to this Biology.StackExchange question, the majority of a plant's carbon comes from the air via photosynthesis.
So in a sense, we already do that, but we just use the huge stores of graphite from millions of years of high pressure processing beneath our feet rather than start from the beginning.

Yes, any compound of carbon that can be synthesised, can be synthesised from carbon that comes from captured atmospheric $\ce{CO2}$.
We discussed related subjects recently on the new Sustainability Stack Exchange, on the options for plastics without oil.
There are at least three research outfits working on the atmospheric extraction of $\ce{CO2}$. Now, inevitably, capturing it and turning it back into carbon is going to take more energy than gets released from burning coal or gas in the first place to release the $\ce{CO2}$, but that doesn't make the chemistry impossible (and it isn't a deal-breaker economically either, because energy costs vary hugely by time and space.)
And as the article from the UK's Royal Chemistry Society from Jan 2012 that you linked to, highlights, this is already being done:

Graphene oxide can be easily reduced to graphene but is also useful in its own right ... Lee and colleagues produced graphene oxide through a two-step process: first, they fixed the $\ce{CO2}$ to ammonia borane ($\ce{NH3BH3}$) and then they heated the resulting solid to create a honeycomb structure of graphene oxide. In the first reaction, they placed $\ce{NH3BH3}$ into a stainless steel cell and then added $\ce{CO2}$ until the cell reached a target pressure. They then gradually heated the cell to 100°C over a few hours. At a pressure of about 30atm, the mass of the $\ce{NH3BH3}$ doubled thank to the addition of carbon dioxide, producing $\ce{OCH3}$, $\ce{HCOO}$ and aliphatic groups.

Here's the original paper on the research: Formation of Graphene Oxide Nanocomposites from Carbon Dioxide Using Ammonia Borane, Junshe Zhang et al, 2012

There are a few standard ways of making graphene (e.g. exfoliating graphite or chemical vapor deposition), but making it out of air isn't one of these ways (though I suppose the ability to do so would fix a few problems...)
No one has ever been able to do anything like this.  I suppose it wouldn't be a thermodynamically favorable process, and the concentration of CO2 in air isn't particularly high.

It's literally true that people don't make graphene out of air in the sense that there's no machine that can take in air and spit out graphene.  If you're talking in the grander sense (e.g. all of graphene is made of stardust from supernovae), then sure, graphene is made of carbon which can be traced back to air.  In the RSC article, they are taking high purity CO2 (probably produced by fermentation) and ammonia borane and putting it in some sort of oven.

yeah what, of course ChickenGod is right. Yeah, you can theoretically make graphene out of elemental carbon, but the application requires things that are just not feasible. It's not that it hasn't been done before, it's just that it requires a similar process to that of trying to make coal from air. Not only are the concentrations too low, but...why would you do it anyways? It would be like wanting to make drinkable water by harnessing oxygen and hydrogen from the air (good luck with the hydrogen) and then combining it, instead of going down to the river and filling a jug.

Graphene can be made from $\ce{CO2}$ by a process of special electrolysis known as SMD. This is a patent pending process which produces graphene in bulk from $\ce{CO2}$ with very little energy. The inventor Michael Nunnerley came across this by accident when he was working on a new system of methane synthesis from $\ce{CO2}$ and $\ce{H2O}$.


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Detection of potassium ions
I've bought a white powder. The label on the package states that the powder contains sodium chloride, potassium chloride and potassium citrate. How can I make sure that the powder does indeed contain potassium chloride? I'm mainly interested in the presence of potassium ions.

What kind of equipment do you have available, I assume you can't perform atomic absorption spectroscopy?

I don't know.  The standard college lab test was for years a flame test.  One took a platinum wire inserted into an insulated handle and stuck  the wire into a solution containing the potassium ion.  One then stuck the wire into a Bunsen burner flame.  The last step was to look at the flame using a deep red piece of glass.  If potassium was present a rather pretty lilac color was detected.  The red glass was needed because the bright yellow flame from sodium otherwise masked the color of the potassium.

@PaulJ.Gans That could be a great answer, really.

@PaulJ.Gans I agree. The flame test would be the easiest method to detect K. You should turn that into an answer. But I've heard that blue glass is usually used to block out the colour of sodium and not red. I've also read somewhere that Potassium Hexachloroplatinate (IV) $K_2[PtCl_6]$ is insoluble in water. Maybe if it is specific enough to Potassium, that could be used as a rough (though quite expensive) test.

$KClO4$ is very weakly soluble and $HClO_4$ can be used to detect $K$. It will form weakly soluble salts with big low-charge cations too, of course.

The glass is commonly known as "cobalt glass" which is a deep blue. You can probably get a small piece of dark blue glass from some shop that sells or makes stained glass windows. Going cheap wood splints can be used instead of a platinum wire. A bag of popsickle sticks is cheap. So buy some reduce Na salt at the grocery store which is a mixture of Na and K chlorides and try it out.

What do you want?

If you're just looking for a source of potassium ions you could just buy $\ce{KCl}$ or some other compound, they're relatively cheap if not pro analysis. If you actually want to test for $\ce{K+}$:

What else do you know about the product?

Tests are completely different for different amounts of $\ce{K+}$. Does the vendor specifies the amount of each compound on the product's label?

How sure do you need to be?

Do you need to do a QUALITATIVE analysis or a QUANTITATIVE analysis? If you need to be absolutely sure about the concentration then you need to guess it's ridiculously low - you'll probably need at least AAS (Atomic Absorption Spectroscopy) equipment, which can't be accessed by anyone nor find anywhere, but you can order an analysis. Alternatively, you can use an electrode - keep reading .
Some ways of testing that I can think of:
(Of course you'll need to dissolve the powder in at least distilled water to do all the following tests, and do a blank with the same water - the more pure the water the more accurate your results.)

For big amounts of $\ce{K+}$

The flame test suggested by Paul J. Gans on the comments above could work, although I never actually did it.
Another way is doing the following titration:
$$\ce{KCl(aq) + AgNO3(aq) -> AgCl(s) + KNO3(aq)}$$
$\ce{AgNO3}$ can be found at any store that sells chemicals. We used 0.1 M for small amounts of $\ce{Cl-}$.
What you are actually testing for is $\ce{Cl-}$ ions, but if the amount of K = the amount of $\ce{Cl}$ the job is done. Now, since the powder contains potassium citrate this is just a qualitative test, although it can be very accurate for small concentrations if properly executed (see Mohr method). Silver chloride is a white solid that is barely visible in small amounts, and a similar  titration's end point is just a few drops (Mohr method), and there are a lot of possible interference so you should not expect precision unless you do some pro work.

For small amounts of $\ce{K+}$

I think the best way of doing this is using a potassium-selective electrode. I never used one of those, but other ion-selective electrodes work great on water analysis, detection levels were a few mg/L on our old ones!
http://en.wikipedia.org/wiki/Potassium_selective_electrode
And of course, you can use AAS.


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What metabolite intermediate will accumulate in a mammalian cell-free extracts capable of glycolysis if lactate dehydrogenase is inhibited?
I know the reaction proceeds as follows:
Pyruvate accepts the electrons from NADH to form lactate. What would be the intermediate in this reaction? Would it be NADH or NAD+?

Please don't cross-post your questions.  I have recommended that the one on Biology.SE be closed and this one remain.  If, after a time you're not getting a response on one site, you can then ask for the question to be moved to the other.

As you probably know, lactate dehydrogenase catalyzes the interconversion of pyruvate and lactate, with electron transfer via NADH/NAD⁺. If lactate dehydrogenase is inhibited, the fate of pyruvate in the cell extract will depend on what cell structures as wells as enzymes and other compounds are available. In other words, the answer to your question will likely be strongly dependent on the organism, as well as the particular circumstances of the experiment you are performing.
To give you a speculative answer: Assuming that the metabolic pathways of the mammalian cell still function and that oxygen is available to accept electrons in respiration, pyruvate could be metabolized through the tricarboxylic acid cycle, forming energy-rich compounds and carbon dioxide. If other fermentative pathways are available, pyruvate could be funnelled into those.
In any case, predicting what compunds would accumulate would require more information than that which is given in your question.


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What chemicals are going to be used to make the smoke black or white?
For the papal election, white and black smoke will be used to signal success or failure in the voting process.
From what I understand, the ballots are burnt and chemicals are added to create the color.
But what chemicals are used to get black or white smoke?

I think Titanium Tetrachloride http://en.wikipedia.org/wiki/Titanium_tetrachloride would be a good option for white smoke (I've heard it's used quite often in smoke bombs).

This link could also be useful: http://thehappyscientist.com/science-experiment/black-and-white-smoke

@kaliaden, I read that link before posting, and it was an interesting read. But wouldn't Titanium tetrachloride be ruled out as poisonous?

It is quite dangerous because of the corrosive fumes released. But the link mentions that they have experimented with lots of (not especially safe) stuff before, so I wouldn't rule it out completely :)

I think Titanium Tetrachloride would be a good option for white smoke (I've heard it's used quite often in smoke bombs). But it releases quite a lot of $HCl$ gas and so it might not be very safe.
A quick Google search comes up with a lot of answers to your question though. This site even answers your question in the same context (The elections going on at the Vatican):
http://thehappyscientist.com/science-experiment/black-and-white-smoke
Apparently you can make white smoke using Magnesium powder as well. (The white colour is due to fine particles of Magnesium Oxide $MgO$ )
It seems the Vatican has experimented using many sources before (Military flares, Smoke bombs and so on) though apparently they didn't turn out too well...
EDIT
The Wikipedia article on Coloured smoke gives another possibility. You can add Titanium Dioxide ($TiO_2$) to a mixture of Potassium Chlorate ($KClO_3$) and Lactose for white smoke. $KClO_3$ is a powerful oxidizing agent which reacts with Lactose (which is a reducing sugar) to create an explosion. $TiO_2$ is safe and is used as a pigment in most white paints and as a food colouring as well.
EDIT2
The official recipes have been revealed by the Vatican press office. See this story on the New York Times website.
The previous edit was partially correct. They use a mixture of $KClO_3$, lactose and pine rosin for white smoke. For black smoke, they use Potassium perchlorate ($KClO_4$), anthracene (an aromatic compound made of three fused benzene rings) and sulfur.
The process (quoted from the site):
"The chemicals are electrically ignited in a special stove first used for the conclave of 2005, the statement said. The stove sits in the Sistine Chapel next to an older stove in which the ballots are burned; the colored smoke and the smoke from the ballots mix and travel up a long copper flue to the chapel roof, where the smoke is visible from St. Peter’s Square. A resistance wire is used to preheat the flue so it draws properly, and the flue has a fan as a backup to ensure that no smoke enters the chapel."

I just saw that NYT article, and was about to post that information as an answer. I'm glad to see someone beat me to it.

I suppose I was just in time.. I never thought we'd see the white smoke recipe in action so soon. :) We now have a new Pope (who already has a Wikipedia page within half an hour after being announced!) http://en.wikipedia.org/wiki/Pope_Francis


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What happens to the oxygen when carbohydrates turn into hydrocarbons?
According to the biogenic theory of hydrocarbon formation, hydrocarbons come from pressurized carbohydrates.
I am looking for an explanation of what happens to the oxygen in the process. How could the oxygen break away from the hydroxyl groups of molecules like glucose to form hydrocarbons?

The process likely involves pyrolysis, a generic term for the thermal degradation pathways of organic compounds.
In the case of carbohydrates, like glucose (shown below), the most common function groups are alcohols (R-OH). Alcohols undergo dehydration in the presence of strong acids, but the reaction can also be promoted by extreme heat in the absence of oxygen. The presence of oxygen leads to combustion (bad!).

In a dehydration elimination reaction, the OH on one carbon and the H on a neighboring carbon leave to form water. This usually follows an E1-like mechanism, involving a carbocation intermediate. In the geochemical version, the necessary acid could come from nearby slightly acidic mineral.
Protonation of the alcohol:
$$\ce{(CH3)2CH-OH + H+ -> (CH3)2CH-OH2+}$$
Loss of leaving group:
$$\ce{(CH3)2-CH-OH2+ -> (CH3)2CH+ + H2O}$$
Loss of proton (regeneration of acid):
$$\ce{CH3-CH+-CH2-{\bf{H}} -> CH3-CH=CH2 + {\bf{H+}}}$$
Overall:
$$\ce{(CH3)2CH-OH ->[\ce{H+}][\Delta] CH3-CH=CH2 + H2O }$$
In the case of carbohydrates, they tend to dehydrate to carbonyl compounds, which can then decompose further into hydrocarbons by reductive and radical pathways.
For example, fructose is converted into furan derivatives, like 5-hydroxymethylfurfural, by a series of dehydrations and rearrangements:
$$\ce{C6H12O6 ->[\ce{H+}][\Delta] C6H6O3 + 3H2O}$$

The image above was created by Wikipedia user V8rik and released under the GNU Free Documentation License and the Creative Commons Attribution-Share Alike 3.0 Unported License.

This will largely depend upon the reagents that perform the reduction.
When using $\ce{HI}$  (or a strong acid and heat or $\ce{H2}$ and $\ce{Pd/C}$) the reaction goes as follows.
$\ce{C6H12O6 ->[H^+] C6H14 + H2O}$ (organic products only)

As others answers have pointed out, the main pathways for reducing carbohidrates to hydrocarbons involves the dehydration elimination of the alcohol groups. This reaction is catalized by acidic conditions and can be carried out with linear alcohols (i.e. like ethanol, resulting in an olefin) in laboratory conditions, at temperatures higher than 160 °C and atmospheric pressure. The products are unsaturated compounds and water.
Even if the geochemical conditions don't allow a strongly acidic medium, the pressures and temperatures attained should be enough to pyrolitically dehydrate  sugars and cellulosic material which makes up most of the cell walls of vegetal biomass. Nevertheless, pyrolitic conversion of mainly-cellulosic biomass in sedimentary basins ussually doesn't produce aliphatic hydrocarbons. Instead, the carbon skeletons are further dehydrogenated and compressed forming coal deposits containing minor aromatic derivatives like phenols and aniline.
The main source of hydrocarbon deposits is the pyrolitic decarboxylation of fatty acids, in turn produced by the hydrolisis of lipids. Those lipids mostly come from microphytes (microalgae) which comprise most of the phytoplancton in the seas, and which have been accumulated at the bottom of ancient oceans. For example, the rich sedimentary basin of modern Middle East was once the Tethys Ocean seabed.
So, in essence, both in the dehydration of carbohidrates and in the decarboxylation of fatty acids, the oxygen goes away in the form of water or carbon dioxide.


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Stripeless cleaning of windows
Last week I was discussing with a friend how we thought the stripeless cleaning of windows is achieved when using a cleaner like Windex Glass Cleaner as opposed to just using hot water, in which case you do get stripes left behind.
At first we thought that the stripeless nature was caused by surfactants in the cleaner that allow the (polar) dirt on the windows to be collected in the water through micellation. However, when we use normal cleaner not specifically designed to clean glass/windows, we also get the stripes even though we are sure that there are also surfactants in the normal cleaner.
The question is: what is the difference between the glass cleaner and the regular cleaner that makes sure that windows become stripeless clean?
One thought we considered but couldn't really work out was that the stripe formation would be something caused by a combination of the marangoni effect and evaporation, similar to the coffee-stain effect which can be overcome by specific tuning of the surfactants as shown here.
Could someone explain how the stripeless cleaning works? Can fluid dynamics explain what happens or is it really something chemical going on?

When searching I also ran into a solution that (at least the manufacturers claim) will also give stripeless clean windows. The interesting thing is that they use pure $\ce{H2O}$ so without any surfactants but also without any dissolved salts like regular tap water has.

Cross-post: https://physics.stackexchange.com/q/56800/7433

After quick digging I found, that glass cleaner solutions usually contain alcohols and surfactants. This additions reduce surface tension and allows the solution to interact with glass more easily, distributing the solution over the entire surface.
The white stripes visible after water dried are usually solid salts. If the water does not contain any salts, but only compounds, that can quickly evaporate, it will not leave stripes. But it is not addition, but absence of additions that do the trick. For example, most bleacher/toilet cleaners contain hypoclorites, salts of unstalble $\ce{HOCl}$ acid and leave $\ce{NaCl}$ after evaporation
When tap water is allowed to dry, it usually leaves white stripes, but they are not visible on most surfaces
The only ideas I can get why the salts don't deposit if using a glass cleaner is that salts are removed (and if you have a cleaner not meant to be diluted, the water in it may be deionized) or to trap salt ions into organic ligands so resulted salts were not crystalline (possible, but I do not like the idea).

But this doesn't explain why regular cleaner does leave stripes. In regular cleaner we also have alcohol and surfactants.

The white stripes visible after water dried are usually solid salts. If the water does not contain any salts, but only compounds, that can quickly evaporate, it will not leave stripes. But it is not addition, but absence of additions that do the trick. For example, most bleacher/toilet cleaners contain hypoclorites, salts of unstalble $HOCl$ acid and leave $NaCl$ after evaporation.

Ok, that makes sense, thanks! But that brings to mind another question: normal tap water does contain salts (quite a bit in some locations) so how come this does not deposit?

@michielm The only ideas I can get is either to remove salts (and if you have a cleaner not meant to be diluted, the water in it may be deionized) or to trap salt ions into organic ligands so resulted salts were not crystalline (possible, but I do not like the idea).


text:
What is the procedure of designing a salt bridge for a galvanic cell?
I'm doing an experiment where I'm measuring the effect of changing the $\ce{Cu^{2+}}$ ion concentration on the voltage in a copper-zinc galvanic cell.
I'll be doing at least five trials for each of the five concentrations I'm testing for (from $0.2-1.0\:\mathrm{M}$), so in total that is 25 trials. However, do I need a new piece of filter paper as a salt bridge for each trial, or can I reuse the same piece of filter paper and dip it in a $\ce{KCl}$ solution each time? Do I really need to make 25 separate pieces of filter paper?

Can you tell us why you want to avoid using 25 pieces of filter paper?  It doesn't seem too excessive for the benefit of avoiding cross contamination.

I down voted your question because the title of the question in my opinion doesn't reflect the content. The question you are asking is in fact "Can I reuse a $KCl$ filter paper salt bridge more then once?" there is nothing about how to designing a salt bridge in what you are asking...

If you wish to do 25 trials with 5 trials per concentration, I would suggest the following:

You can use the same filter paper as long as the $\ce{Cu^{2+}}$ concentration remains constant. However, since you are changing the concentration 5 times, you will need 5 pieces of filter paper.
After every trial, thoroughly wash the filter paper with distilled water.

So I would say you need at least 5 pieces, but you would not need to go all the way to 25 pieces. Just make sure to follow step 2 after each trial.

You might also want to consider using string as a disposable salt bridge.  The general procedure would be to cut 6-8 inches of string, twist it and soak in the KCl solution.

do I need a new piece of filter paper as a salt bridge for each trial, or can I reuse the same piece of filter paper and dip it in a KCl solution each time? Do I really need to make 25 separate pieces of filter paper?

You need empirical evidence to resolve this question.

Perform entire experiment using 1 piece of filter paper dipped 25 times.
Perform entire experiment using 25 pieces of filter paper dipped 1 time.

If experiment #1 results differ greatly from experiment #2 then you cannot replace 25 pieces of filter paper dipped 1 time with 1 filter paper dipped 25 times.
Do not try to avoid scientific rigorousness. You're procedure should produce repeatable results.


text:
Selective etchant for Al2O3 (Does not etch Al, Cu, Zn, Sn)
I'm looking for a wet-etchant to remove aluminum oxide created by anodizing, but not etch the unoxidized metal (with greater than 10:1 selectivity)
I know about chromium(IV) oxide and fluorine-based etchants, but those are problematic for safety and disposal. Are there other etchants? It can be an acid or a base.

Did you actually manage to pull this off? Because etching isn't as hard as preventing a new layer of oxide forming in seconds in air... You'd basically have to work under an inert gas atmosphere, whatever is that you're trying to do thereafter. A more typical approach is to nickel coat the Al https://www.sae.org/publications/technical-papers/content/880446/

Fizz - no, I have not. Closest thing I found was zincating, but of course that does not satisfy the other constraints (no etch of other metals.) The problem is the raw material comes with an oxide.

In Fabrication of Microstructures by Wet Etching of Anodic
Aluminum Oxide Substrates Chem. Mater. 2005, 17, 4049-4052,
5% phosphoric acid is used and is selective relative to aluminum, but you need to check how it would do with respect to the other metals you mention

Finally, the AAO substrate was wet-etched in a 5 wt % H3PO4 solution
  at 20 °C for 8 h

Alumina can be dissolved in sodium carbonate. But it would expose the bare metal to aerobic corrosion.

While I do not agree that it does not attempt to answer the question, at the moment this reads more like a comment than an actual answer - could you elaborate a little more, perhaps include a link to a general procedure or outline it.


text:
Calculation of partial charges and protonation/deprotonation states depending on pH
What is the best method for calculating partial charges and protonation/deprotonation states of a small molecule such as phenol or caffeine, depending on the pH? For instance, it is well known that above pH=10 or so phenol is deprotonated, but there is data about its protonation around pH=2. For more complex molecules such as caffeine, similar things happen. So I am more interested in calculating partial charges of protonated and deprotonated forms than on pKas (sometimes, experimental data is available).
I head about Gaussian and the like, but I also wonder about software or methods that are easy to use.

What are partial charges, really? The charge of your molecule is 0 when protonated or -1 when deprotonated (that's for phenol; +1 and 0, correspondingly, for caffeine). What portion of the molecules will be at either state under given pH, is determined by pKa. That's just about it. If you want to know the atomic charges within any molecule or ion, well, these may be calculated, but it really takes some heavy quantum chemistry to find the figures that will be anywhere near reliable.


text:
How to rationalise the resonance structures and hybridisation of the nitrogen in a conjugated amine?
I was given the first structure, and then drew the other 5 resonance structures:

First of all, are they correct? ChemBioDraw had some complaints, but as far as I can see there's the same number of electrons, and no valence orbitals exceeding capacity.
My reasoning is that the first structure contributes more. There's no electron deficiency on any of them, so we can disregard that. There is charge separation on the last 5 though, and in combination with this, the negative charge is on the least electronegative atoms, thus making the last 5 structures noticably higher in energy, and thus they contribute less to the true electronic image of the molecule.
Hope I'm correct so far; here's my question:
In the first structure, nitrogen is sp³ hybridised, but on all others it's clearly sp² hybridised. So what does that mean? It's somewhere in between, but closer to sp³-hybridised?

First of all, are they correct? ChemBioDraw had some complaints, but as far as I can see there's the same amount of electrons, and no valence orbitals exceeding capacity.

Yes, these are the six most important resonance structures for this compound. The reason ChemDraw complains is that it is trying to act smarter than you, and it most certainly is not. It interprets that negative formal charge on the carbon atom as implying a lone pair, since carbon can only have the negative charge if it also has a lone pair. When you add the lone pair and the charge, ChemDraw is suddenly stupid and thinks you have exceeded the octet on carbon (3 bonds + 1 explicit LP + 1 implied LP from the charge).

In the first structure, nitrogen is $\mathrm{sp^3}$ hybridised, but on all others it's clearly $\mathrm{sp^2}$ hybridised. So what does that mean? It's somewhere in between, but closer to $\mathrm{sp^3}$-hybridised?

If the lone pair on a nitrogen (or on any atom)) participates in resonance, then that nitrogen (or whatever) atom must be $\mathrm{sp^2}$-hybridized so that the the lone pair is in a $\mathrm{p}$-like orbital to ensure appropriate symmetry for $\pi$-overlap. Put another way, if an atom is $\mathrm{sp^2}$-hybridized in one resonance structure, then it is $\mathrm{sp^2}$-hybridized in all of them. Atoms that are $\mathrm{sp^2}$-hybridized and $\mathrm{sp^3}$-hybridized have differing geometries, which is not permitted in the resonance phenomenon.
In truth, hybridization is an approximation we make to make quantum mechanics jive with molecular geometry. Geometry is real (empirically determinable) and hybridization is fictitious (not empirically determinable), but hybridization makes QM behave better conceptually and mathematically. Hybridization is also a useful predictor of chemical reactivity of various bonds and functional groups in organic chemistry. Experimentally, I would guess that the nitrogen atom is trigonal planar (or very close to it). Trigonal planar implies $\mathrm{sp^2}$-hybridized (not the other way around).
In molecular orbital theory, we do away with the need for both resonance and hybridization. This molecule would have 7 $\pi$ orbitals formed from linear combinations of the p orbitals on the 6 participating carbon atoms and the nitrogen atom. The probability density function plots of these 7 orbitals will be more complex than you might be used to, but they will suggest the same electron density and charge density as a resonance hybrid assembled from your six resonance structures. See the Wikipedia article on conjugated systems, which is not very good but will give you the idea.

My reasoning is that the first structure contributes more.

The five charge-separated resonance structures are more important than you think, since they imply that the five-membered ring is aromatic like the surprisingly stable cyclopentadienyl carbanion. However, you might not yet be this far along in your studies of organic chemistry. At the introductory level of understanding of resonance in organic molecules, the first structure is most important for the reasons you list.

Thanks for the comprehensive and lucid answer. Is M.O. more descriptive? Why use resonance and hybridisation at all?

Hybridization and resonance represent a "quick and dirty" or "back of the envelope" method that you could do anywhere. Generating good MO diagrams and representations for molecules even as complex as yours requires calculus, linear algebra, and a computer with appropriate software. The answer generated generated by hybridization/resonance is almost always the same as the MO theory answer, so choose the quick and easy option first. Use the more difficult theory only if the first approximation is not good enough.

@BenNorris If an atom participates in resonance why its hybridization must be $\mathrm{sp^2}$? I mean shouldn't be the weighted sum of the hybridizations in each resonance structure?

"Trigonal planar implies sp2-hybridized (not the other way around)"- did you mean that we see this trigonal planar atom, and we rationalize it by calling it sp2? And not that "sp2" (or any hybridisation for that matter) is an intrinsic property of atoms?

@HarryHolmes - Orbitals are mathematics constructs defining regions of space with high electron probability density. Geometry is an empirically measured property.

So ... That's a yes, right?

@HarryHolmes - That's a "yes."

I wanted to add a perspective to this since this can be a contentious issue, and I would still say knowing whether a nitrogen is $sp^2$ or $sp^3$ hybridized is not straightforward. I was originally hoping to add this under this question specifically about aniline, but that question is now closed and marked as duplicate, with a link to this question.
In a simple molecule like aniline, one might think that the hydrogens of the amine would be planar with the ring. However, we know from crystal structures of aniline that it is indeed pyramidal. If you have access to the Cambridge Structural Database (CSD), you can see this in structure BAZGOY. The improper dihedral angle of the hydrogens is ~26 degrees in that structure.
I was also curious to see what QM had to say about this. I performed some optimizations of analine with the improper dihedral angle of the hydrogens constrained at various angles. Here is a drawing of the improper dihedral I constrained.

I used the MP2 method and cc-pVTZ(-F) basis set, in implicit solvent (PBF) using the Jaguar software from Schrodinger. You can see in this plot and gif that the lowest energy angle was at ~35 degrees. That's pretty much completely pyramidal (in ethane, an equivalent angle is 33.4).

However, that being said, I also wouldn't say this is dogmatic. This is an implicit solvent calculation, and in real water with real hydrogens and oxygens to hydrogen bond to, things may be still different. NMR might be able to help figure this out, or it could further confuse if the timescale of flipping smears the signal to give a flat angle when it is not.
I should also point out that there are other crystal structures that contradict this. If you look at the crystal structure VOMFOS, which is of an adenosine analog, the amine hydrogens are planar with the ring. Is that just an artifact of the crystallization packing network, or a true representation of the hybridization of that nitrogen? Hard to say.
I think my main point here is to be careful what you assume. If you asked me on a test in undergrad whether the amine on aniline is $sp^2$ or $sp^3$, I would have said $sp^2$ with confidence. Experimental and computational evidence suggests that might not be the case.


text:
Changes in physical properties in homologous series : Solubility
In the text that I'm reading it is stated that :

As the molecular mass increases in any homologous series, a gradation in physical properties is seen. This is because the melting points and boiling points increase with increasing molecular mass. Other physical properties such as solubility in a particular solvent also show a similar gradation.

My question is regarding the last line; what kind of gradation in solubility is observed? Increase or decrease?

My question is regarding the last line, what kind of gradation in solubiltiy is observed? Increase or decrease?

Both
Solubility is not one single property, but a collection of related properties. Solubility in water is different from solubility in ethanol, which is different from solubility in benzene.
The cause of the gradation of (all of these) properties is a systematic change in intermolecular forces. The number and strength of these forces increase as molecular mass increases, which increases the boiling points and melting points. For solubility, we care about the interplay between the solute and the solvent, so we also need to know the type of intermolecular force (polar vs. nonpolar) in addition to the total strength.
For the homologous series of alcohols below, I have listed their water solubilities (mined from UC Davis's ChemWiki and from Wikipedia articles). Notice that as the number of $\ce{CH2}$ units increases, the water solubility decreases. This decrease occurs because, while the strength of nonpolar intermolecular forces (London-Dispersion) increases with each added $\ce{CH2}$ until the nonpolar interactions overpower the polar interactions (dipole-dipole and hydrogen-bonding) of the alcohol functional group. Water is a polar solvent, and with each added $\ce{CH2}$, the alcohol molecules begin to have stronger attraction to themselves than to water, and the solubility decreases.
Solubility of the homologous series of linear n-alkanols in water

Methanol $\ce{CH3OH}$ - miscible (infinite)
Ethanol $\ce{CH3CH2OH}$ - miscible (infinite)
n-propanol $\ce{CH3CH2CH2OH}$ - miscible (infinite)
n-butanol $\ce{CH3CH2CH2CH2OH}$ - 80 g / L
n-pentanol $\ce{CH3CH2CH2CH2CH2OH}$ - 22 g / L
n-hexanol$\ce{CH3CH2CH2CH2CH2CH2OH}$ - 5.9 g / L

At the same time, the solubility of this homologous series of alcohols increases in nonpolar solvents like hexane and benzene. Those data are not readily available, but a quick experiment will verify it. The increased amount of nonpolar interactions in the alcohols increases their affinity with nonpolar solvents.


text:
Why is this atom sp² hybridised?
I'm doing exercises on hybridisation, and I was given this molecule:

I'm wondering about this (electron deficient) oxygen atom. My intuition says it should be ${sp^2}$ like the answer says, but honestly I only see that it's $sp$ hybridised. Is it because you ask what it would look like if it was not electron deficient?

I might add, that since oxygen has a lone-pair I somehow missed, that means it is of course not electron deficient afterall.

This would become much clearer if you show the lone pair on the oxygen atom.
One $p$ orbital of the Oxygen atom forms the $\pi$ bond with Carbon, while the other three (remaining one $s$ and two $p$) orbitals are $sp^2$ hybridized.
Two $sp^2$ orbitals are used to form the $\sigma$ bonds while the last $sp^2$ orbital is occupied by the lone pair.

Note: I always like to stress that the s and p orbitals don't actually combine in any way to form the $n$ $sp^n$ hybrid orbitals (This is a pretty common mistake). Hybridization is just an approximation used in quantum mechanics to explain the geometry of the compound better.

Ah of course. Somehow missed the lone pair.

You can find the hybridization of the atom by finding its steric number:
steric number = no of atoms bonded (to the atom you are finding the hyb. of) + lone pairs with that atom.
if Steric no comes to be 4, there is $\ce{sp^3}$ hybridization in the atom.
if steric no comes to be 3, there is $\ce{sp^2}$ hybridization in the atom.
if steric no comes to be 2, there is $\ce{sp}$ hybridization in the atom.

and this certainly is not true for any period different than 2nd. Try to use it on PH3 and you will end with sp3, however the free electron pair is s-type orbital.

Well, it helps in basic studies.

+10 to kaliaden's comment.  The "reason" is because linear molecules are $\ce{sp}$, trigonal-like molecules are $\ce{sp^2}$, and tetrahedral are $\ce{sp^3}$.  Hybridization rules are completely bogus, and basically you know the answer by first observing the molecule experimentally, and post-hoc hammering it into these resonance structures.
This is nothing but a bunch of memorization. It is interesting to note that if you made $\ce{O}$ into $\ce{S}$, then it _may_not_ have the same shape.  (Actually, it probably does, but where it is different, we sound off the d-orbital participation mumbo-jumbo)


text:
Chemical formula of ethanol and other carbon compounds
Should chemical formula of ethanol be written as $\ce{CH3CH2OH}$ or $\ce{C2H5OH}$? And whats the logic behind it? I have seen the former being used more often, so why is ethane represented by $\ce{C2H6}$ than $\ce{CH3CH3}$?

$\ce{CH3CH2OH}$, and $\ce{C2H5OH}$ or $\ce{EtOH}$ (Ethyl OH) tells you specifically what are the bonds in the molecule. $\ce{CH3H6O}$, which would be the equivalent of writing ethane like $\ce{C2H6}$, on the other hand, tells you nothing, since it can be both ethanol or dimethyl ether. Bigger formulas can represent thousends of different compounds if written like that.
The reason why $\ce{C2H6}$ is used instead of $\ce{CH3CH3}$ is because it can only represent ethane - it's a very simple compound with no isomers, so you can write it both ways.
You can write ethanol both ways you suggested, although I've never seen $\ce{C2H5}$ instead of $\ce{Et}$ anywhere but in examples. What you shouldn't do is write it like $\ce{C2H6O}$.
If you were asking what is the logic of $\ce{CH3CH2OH}$:
What it's telling you is that the compound it is representing [ethanol] consists of a carbon atom bonded to 3 hydrogen atoms and to another carbon atom, which is bonded to 2 hydrogen atoms and 1 oxygen atom, which is finally bonded to 1 hydrogen atom, as follows:

Good. But didn't you mean ‘C₂H₆O' instead of ‘CH₃H₆O'(??) in the 1st paragraph, too?

I think that organic chemists tend do prefer $\ce{CH3CH2OH}$ (condensed) formula over $\ce{C2H5OH}$ (not mentioning favorite abbr. EtOH) because we/they like to see each carbon atom separated in the formula.*
$\ce{C2H5OH}$ looks like some ugly hybrid between molecular formula ($\ce{C2H6O}$) and nice condensed formula.
I speculate that most organic chemists are, for some reason, more comfortable with $\ce{C6H5\bond{-}}$ for phenyl (not speaking of $\ce{Ph\bond{-}}$) than $\ce{C2H5\bond{-}}$ for ethyl, even though it can mean multiple different structural isomers, e.g. $\ce{HC#C-CH2-C#C-CH2\bond{-}}$ (hexa-2,5-diyn-1-yl).
$\ce{C2H5\bond{-}}$ is unambiguous and represents only ethyl group. I think we should learn to be more comfortable with it.

*) In formulae like $\ce{(CH3)3C\bond{-}}$ (tert-butyl) or $\ce{CH3[CH2]7\bond{-}}$ (octyl), many carbons are not explicitly written (repeating units), but they are still written separately. No subscript index number $\ce{C_x}$ used.


text:
How do I figure out the hybridization of a particular atom in a molecule?
I'm learning how to apply the VSEPR theory to Lewis structures and in my homework, I'm being asked to provide the hybridization of the central atom in each Lewis structure I've drawn.
I've drawn out the Lewis structure for all the required compounds and figured out the arrangements of the electron regions, and figured out the shape of each molecule. I'm being asked to figure out the hybridization of the central atom of various molecules.
I found a sample question with all the answers filled out:
$\ce{NH3}$
It is $\mathrm{sp^3}$ hybridized.
Where does this come from? I understand how to figure out the standard orbitals for an atom, but I'm lost with hybridization.
My textbook uses $\ce{CH4}$ as an example.
Carbon has $\mathrm{2s^2 \,2p^2}$, but in this molecule, it has four $\mathrm{sp^3}$. I understand the purpose of four (there are four hydrogens), but where did the "3" in $\mathrm{sp^3}$ come from?
How would I figure out something more complicated like $\ce{H2CO}$?

In multiatom molecules, you have hybridized orbitals of the whole molecule. For example, I learned in Physical Inorganic Chemistry that in a nitrogen molecule, each p orbital of one atom hybridizes with a p orbital of the other atom to form a bonding orbital and an antibonding orbital, neither of which belongs distinctly to one atom. I think that in theory, even if you don't know much about a topic, it's possible for you to figure out given what's the truth about that topic what type of answer you would want and make it clear in your question. This is just an idea. I'm not telling you to do it

Multi is not Di Your explanation may explain N2 but fails when extended to isoelectronic CO. s character is important because it allows closer penetration to the nuclei for the bonding electrons. This brings nuclei closer and sp orbitals allow p orbital overlap in pi bonds.

If you can assign the total electron geometry (geometry of all electron domains, not just bonding domains) on the central atom using VSEPR, then you can always automatically assign hybridization. Hybridization was invented to make quantum mechanical bonding theories work better with known empirical geometries. If you know one, then you always know the other.

Linear - $\ce{sp}$ - the hybridization of one $\ce{s}$ and one $\ce{p}$ orbital produce two hybrid orbitals oriented $180^\circ$ apart.
Trigonal planar - $\ce{sp^2}$ - the hybridization of one $\ce{s}$ and two $\ce{p}$ orbitals produce three hybrid orbitals oriented $120^\circ$ from each other all in the same plane.
Tetrahedral - $\ce{sp^3}$ - the hybridization of one $\ce{s}$ and three $\ce{p}$ orbitals produce four hybrid orbitals oriented toward the points of a regular tetrahedron, $109.5^\circ$ apart.
Trigonal bipyramidal - $\ce{dsp^3}$ or $\ce{sp^3d}$ - the hybridization of one $\ce{s}$, three $\ce{p}$, and one $\ce{d}$ orbitals produce five hybrid orbitals oriented in this weird shape: three equatorial hybrid orbitals oriented $120^\circ$ from each other all in the same plane and two axial orbitals oriented $180^\circ$ apart, orthogonal to the equatorial orbitals.
Octahedral - $\ce{d^2sp^3}$ or $\ce{sp^3d^2}$ - the hybridization of one $\ce{s}$, three $\ce{p}$, and two $\ce{d}$ orbitals produce six hybrid orbitals oriented toward the points of a regular octahedron $90^\circ$ apart.

I assume you haven't learned any of the geometries above steric number 6 (since they are rare), but they each correspond to a specific hybridization also.

$\ce{NH3}$

For $\ce{NH3}$, which category does it fit in above? Remember to count the lone pair as an electron domain for determining total electron geometry. Since the sample question says $\ce{NH3}$ is $\ce{sp^3}$, then $\ce{NH3}$ must be tetrahedral. Make sure you can figure out how $\ce{NH3}$ has tetrahedral electron geometry.

For $\ce{H2CO}$

Start by drawing the Lewis structure. The least electronegative atom that is not a hydrogen goes in the center (unless you have been given structural arrangement).
Determine the number of electron domains on the central atom.
Determine the electron geometry using VSEPR. Correlate the geometry with the hybridization.
Practice until you can do this quickly.

If you know the geometry by VSEPR, then why bother to assign hybridization? What purpose does it fulfill? And as far as I know, quantum mechanics is the final word in chemistry so there isn't any need to "make it work". It already works.

Hybridization is given by the following formula:
$$H= \frac{1}{2} (V + X - C + A)$$
Where:

$V$ = number of valence electrons in central atom
$X$ = number of monovalent atoms around the central atom
$C$ = positive charge on cation
$A$ = negative charge on anion

$$H=4 \to \ce{sp^3},\;2\to \ce{sp,\;3}\to \ce{sp^2}...$$
e.g.: in $\ce{NH3}$, the hybridization of $\ce{N}$ atom is:
$$H= \frac{1}{2}(5+3-0+0)=4 \to \ce{sp^3}$$

@user12757  The problem with this method is that is does not always predict the correct hybridization.  For example, $\ce{PH3}$ is not $\ce{sp^3}$ hybridized; it is basically unhybridized with an H-P-H angle around 90°

@ron Wouldn't the currently accepted answer also suffer from the same fault? Unless anyone actually knows that the phosphorus in phosphine is unhybridized, they will always end up counting one lone pair and three sigma bonds for $sp3$ hybridisation...

@GaurangTandon  Yes, it would.

You can find the hybridization of an atom by finding its steric number:
The steric number = the number of atoms bonded to the atom + the number of lone pairs the atom has.
If the steric number is 4, the atom is $\mathrm{sp^3}$ hybridized.
If the steric number is 3, the atom is $\mathrm{sp^2}$  hybridized.
If the steric number is 2, the atom is $\mathrm{sp}$  hybridized.

You can find the hybridization by the AXE method; Here A represents the atom that we're interested in (usually the central atom), X is the number of atoms bonded to the central molecule or the one we're interested in, and E represents the non-bonding electrons (lone pairs). The total number of X and E is known as the steric number.

Formula
Shape (Name)
Bond Angles
Hybridization

AX2
Linear
180°
sp

AX3
Trigonal Planar
120°
sp²

AX4
Tetrahedral
109.5°
sp³

AX5
Triangular Bipyrimidal
120,90°
sp³d

AX6
Octahedral
90°
sp³d²

AX2N
Bent
120°
sp²

AX2N2
Bent
109.5°
sp³

AX3N
Trigonal Pyramidal
109.5°
sp³

You can also find hybridisation by the empirical formula
$$\frac{1}{2}(V + H - C + A)$$
Where,
V = Number of valance electrons in central atom
H = Number of surrounding monovalent atoms
C = Cationic charge
A = Anionic charge
Now try to figure out the hybridization of ethane yourself. Refer to this otherwise.


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Boiling and Melting point of Lithium (or any other element that does not occure in free state)
The melting and boiling point of Lithium as compared to other members of the Alkali group metals is high. But, Lithium does not occur in the free state.
So, what does it mean that it has high melting and boiling point? How are its BP and MP measured if it isn't in the free state?

Litium does not occur in the free state in nature. But it can be made in the free state if it is protected from reaction with air and water.

Your premise here is wrong, pure metal Lithium exists and can be handled without much difficulty.
However, materials which are extremely reactive can always be kept in an inert gas when you want to measure their physical properties. Of course, there are compounds which can don't need an excuse to explode (and  will explode in a vacuum/inert gas as well), in which case it is harder to find out BP/MPs (one can put them at low pressure and then extrapolate, but that's not as accurate).

It is written in my course book that it does not occur in free state.

@MuhammadRafique: What does it mean by "free state"? It may not occur in the free state naturally (as in, we need to synthesize it), but that is true for most elements.

Agreed. Very few elements exist naturally in their free states. All but a select few (francium comes to mind) can be prepared by decomposing their compounds. Lithium is prepared by electrolysis of lithium chloride in the presence of potassium chloride.

In fact I can't think of a single one outside of the noble gases that naturally occurs in pure elemental form. Most gaseous elements prefer to exist as diatoms in nature (oxygen gas is 100% oxygen, none of it "elemental" single atoms), and every other pure element I can think of is typically prepared from a naturally-occuring oxide or halide, with a few other chalcogen complexes in the mix such as sulfites/sulfates to make it interesting. There's just too much of everything else for any one thing to have the chance to remain in pure form.

Platinum metals, gold, silver, copper occur in elemental form.


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Reconciling multiple inorganic properties for the purposes of learning these elements
How can I reconcile all of the chemical properties, physical properties and peculiar behaviour of metals in inorganic chemistry for the purposes of studying these systematically?
It is very hard to learn all of these for each group.

Well, trying to learn the physical properties of the elements is tough since there are >110 of them, and while the properties are periodic, the trends are not obvious. I hope you haven't been asked to memorize them. Given the easy access to online and text references to these properties, memorization seems like a poor use of your time.

Yes, i have been asked to memorize them. I have to write them in my exam.

People who gravitate towards chemistry as a career seem to have a larger ability than most to keep encyclopediac information in their heads.  Good luck if you're not one of us.

Working with formulas and properties of molecules helps definitely getting some of the most relevant info into your head, but keeping it there is a whole new blank page...

You can't 'remember', there is too many of them. Each one has its own quirks, so there is no simple way here. Lucky enough, it is not required: most common chemistry happens to deal with elements up to bromine, second and third row of d-element are very similar, and f-elements almost do not occur outside special cases. Lucky enough, even that is not usually required, as each individual chemist usually focus on specific narrow class of compounds of several or even one element, so only a very shallow general knowledge is required.

first of all take one group a day.
right whatever you have to learn and then try to identify some relationship. Then reread the material and write down whatever you remember . Now look what you miss.
everyday take a new group and do the same.
Also write whatever you remember of the previous groups you have learned.