text:
How can the dipole moment of carbon monoxide be rationalised by molecular orbital theory?
Despite the fact that oxygen is much more electronegative than carbon, the bond in $\ce{CO}$ presents a weak dipole moment.
This observation can easily be explained using the concept of "dative bond", that is, one bond is formed with two electrons from oxygen, producing a polarization $\ce{O\bond{->}C}$ which equilibrates the expected polarization $\ce{C->O}$.
I would like to know if the molecular orbital model could be used to explain this phenomenon.
Here's the diagram for $\ce{CO}$:

Related http://chemistry.stackexchange.com/q/9510/4945

related: https://chemistry.stackexchange.com/a/66987/102629

Unfortunately, nothing in the bonding situation in carbon monoxide is easily explained, especially not the dipole moment. According to the electronegativities of the elements, you would expect the partial positive charge to be at the carbon and a partial negative charge at oxygen. However, this is not the case, which can only be explained by molecular orbital theory. A complete analysis of this can be found in Gernot Frenking, Christoph Loschen, Andreas Krapp, Stefan Fau,  and Steven H. Strauss, J. Comp. Chem., 2007, 28 (1), 117-126. (I believe it is available free of charge.)
Responsible for the dipole moment is the highest occupied molecular orbital, a $\pmb{\sigma}$ orbital, which has its largest coefficient at the carbon atom. In first order approximation, this orbital can be considered the lone pair of carbon. All other valence orbitals are more strongly polarised towards the oxygen. The orbital that can in first order approximation be considered as the oxygen lone pair has almost only s character and therefore contributes only little to the dipole moment.
\begin{align}
\ce{{}^{\ominus}\!:C#O:^{\oplus}} &&
\text{Dipole:}~|\mathbf{q}|=0.11~\mathrm{D} &&
\text{Direction:}~\longleftarrow
\end{align}
I have reproduced the MO scheme of carbon monoxide for you below. Please note, that the blue/orange coloured orbitals are virtual (unoccupied) orbitals, which should be taken with a grain of salt.

There are two possible decomposition schemes to explain the bonding, both of them involve donor-acceptor interactions. The term "dative bonding" should be avoided here, it is better to use it only for bonds, that consist purely of donor-acceptor interactions, as for example in $\ce{H3N\bond{->}BH3}$.
Below, the two decomposition schemes are reproduced from figure 6 (b & c) in the linked paper. Please note, that this decomposition does not include hybridised orbitals.

The left decomposition is a better description, since it retains the $C_{\infty{}v}$ symmetry of the molecule. We can see a donor-acceptor $\sigma$ bond and two electrons sharing $\pi$ bonds.
In the right configuration we assume an electron sharing $\sigma$ bond, an electron sharing $\pi$ bond and a donor-acceptor $\pi$ bond.
It is very important to understand, that the concept of a dative bond, that you are trying to employ here is only right by coincidence. The reason that the dipole moment is oriented towards the carbon is only to find in the weakly bonding HOMO, the lone pair of carbon.

I found some neat experimental evidence for this observation about the weakly bonding HOMO: The is an intense  peak in CO's UV photoelectron spectrum at 14.01 eV corresponding to (5σ)-1, the HOMO, and unlike the peaks at higher energies, the 0->0 transition predominates (very little vibrational fine structure), indicating that the Morse potential of CO+ after ionization of an electron from this orbital closely resembles that of CO, hence the orbital is primarily nonbonding. And we know from CO's qualitative interaction diagram that the density is on C because 5σ is closer in E to C2s than O2s.

The MO diagram shows that the $\ce{CO}$ molecule forms three filled MO's with σ symmetry and two MO's with π symmetry.  Of the five filled MO's (10 electrons) formed for $\ce{CO}$, only four of them can be half-filled from carbon electrons (4 valence electrons).  So one of the filled MO's must have two electrons that originally came from the oxygen atom.


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Why does the anode solution contain Sn2+ in a Sn-Cu voltaic cell?
Suppose you have the following voltaic cell: $\ce{Sn_{(s)}|Sn^{2+}_{(aq, 1.0 M)}||Cu^{2+}_{(aq, 1.0 M)}|Cu_{(s)}}$ and the salt bridge is $\ce{KNO_3}$. What I don't understand is why you need to have $\ce{Sn^{2+}}$ ions initially in the $\ce{Sn}$ half cell. As $\ce{Sn}$ is oxidized, electrons are supposed to travel through the wire connecting the two half cells and reduce the $\ce{Cu^{2+}}$ ions in solution, but if we have $\ce{Sn^{2+}}$ ions initially in the $\ce{Sn}$ half cell, the electrons might just as well reduce the $\ce{Sn^{2+}}$ ions. Why would the electrons travel all the way through the wire when there is $\ce{Sn^{2+}}$ ions right next to them that could be reduced?
I also don't understand why the $\ce{Cu}$ cathode has to exist. All we need is $\ce{Cu^{2+}}$ ions in solution that will be reduced, so why can't we just dip the wire directly into $\ce{Cu^{2+}}$ solutions. As the reaction proceeds, a $\ce{Cu}$ cathode will be created anyways as $\ce{Cu}$ metal precipitates on the wire.
I am obviously misunderstanding something, because according to my understanding the following voltaic cell should work: In one half cell, we have a $\ce{Sn}$ anode in water connected to a wire. The wire is submersed in the other half cell which is simply $\ce{Cu^{2+}}$ solution. The $\ce{Sn}$ gets oxidized and produces $\ce{Sn^{2+}}$ ions in the water. The electrons travel through the wire and reduce the $\ce{Cu^{2+}}$ ions in the cathode solution. The anode solution will become more positive and the cathode solution will become more negative, so we still need a salt bridge.

Please avoid using Latex in titles due to searching issues

Similar, but not a dupe to Is there always a need for a salt bridge to let a galvanic cell work continuously?. The link in my answer may be of help.

Also related http://chemistry.stackexchange.com/questions/16785/positive-or-negative-anode-cathode-in-electrolytic-galvanic-cell

A voltaic circuit needs to be a complete circuit first an foremost. So, to answer your 2 main concerns, think of a voltaic cell as a electric circuit:

In one half cell, we have a $\ce{Sn}$ anode in water connected to a wire.

Water is not a good conductor of electricity, dissolved ions are needed, such as the $\ce{Sn^{2+}}$ ions are needed to facilitate the movement of the electric current in the first place. Also, the presence of the ions allows the overall redox reaction be spontaneous to provide the electric current (ChemWiki).

The wire is submersed in the other half cell which is simply $\ce{Cu^{2+}}$ solution

According to the ChemWiki Voltaic Cells, the metals plates provide a surface for the reactions to occur on, a wire just submersed into the solution provides a very small surface area for the reaction to occur.
A diagram from the UC Davis ChemWiki page Electrochemical Cells helps explain this (the example uses zinc instead of tin, but the principle is the same):

Ahhh you are totally right about the second point, the cathode provides a larger surface area. As for the first point, can you please elaborate why you need $Sn^{+2}$ ions dissolved for the reaction to take place? You say they are needed to keep the current going, but I don't see why this is.

Water alone is a poor conductor of electricity (hence would 'break the circuit'), the charged ions provide a means for the circuit to be complete  - the questions and answers of Why can't pure water conduct electricity since it can be reduced at cathode and oxidised at anode? provide some insight - in short, to 'get around' water's high electrical resistance, charged ions are needed to complete the circuit.

But you get ions from the salt bridge which keep the electrode solutions electrically neutral, so why do you need the $Sn^{+2}$ ions? And is it not true that the $Sn^{+2}$ ions might accept the electrons from the $Sn$ electrode?

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Why does free silicon react with water?
Free silicon is not found in nature (in water, it might be present in the form $\ce{Si(OH)_{4}}$. Some compounds of silicon, such as silane ($\ce{SiH_{4}}$) will react spontaneously in an aqueous environment, while the analogous methane won't.
Is there an explanation for the tendency of silicon compounds to react with water?

Is there an explanation for the tendency of silicon compounds to react
  with water?

Reaction with water, oxygen, etc., the answer is the same and has to do with the strength of the $\ce{Si-O}$ bond. Let's compare the following two reactions:
$$\ce{CH4 + \frac{1}{2}O2 -> CH3OH}$$
$$\ce{SiH4 + \frac{1}{2}O2 -> SiH3OH}$$
In both reactions we break an $\ce{X-H}$ bond and an $\ce{O-O}$ bond, while we make an $\ce{X-O}$ bond and an $\ce{O-H}$ bond.  The $\ce{O-O}$ bond breaking and the $\ce{O-H}$ bond making are common to both reactions and therefore cancel out.  Let's compare the only differences, the $\ce{X-H}$ bond breaking and the $\ce{X-O}$ bond making.
The following Table contains the bond energies for these bonds.
\begin{array}{|c|c|c|c|} \hline
\ce{X} & \text{X-H Bond Strength} & \text{X-O Bond Strength} & \text{Overall Change}  \\ & \text{(kcal/mol)} & \text{(kcal/mol)} & \text{(kcal/mol)}\\ \hline
\ce{C} & 99 & 86 & -13\\ \hline
\ce{Si} & 75 & 110 & 35\\ \hline
\end{array}
Looking at the numbers it becomes clear that the $\ce{Si-H}$ bond is a lot weaker than the $\ce{C-H}$ bond, and the $\ce{Si-O}$ bond is a lot stronger than the $\ce{C-O}$ bond.  Consequently, the reaction involving silicon is more exothermic than the analogous reaction involving carbon.  This greater stability in the $\ce{Si-O}$ product is reflected in the transition state leading to it.  It lowers the energy of the transition state and consequently the reaction involving silicon is much more facile than the same reaction involving carbon.

This is interesting. As carbon and silicon are in the same group (14), do you know why the Si-O bond is stronger? I would expect it to be weaker since Si has a greater radius (and less superposition).

Most textbooks invoke $\ce{d \pi - p\pi}$ bonding to explain the strength of the $\ce{Si-O}$ bond. I'm not a fan of involving d-orbitals in non-transition metal bonding, but there must be some type of back bonding (electron transfer from the electronegative oxygen back to the electropositive silicon) that produces the high bond strength.

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text:
How important are those fields of chemistry in biotechnology and medicinal chemistry
At my university there is some fields of chemistry that i don't know if i should focus on them...
How important are in biotechnology and medicinal chemistry those fields of chemistry? :

Quantum chemistry
Physical chemistry
Computional chemistry

Computional chemistry - for computer aided drug design

I would say the three are linked and you can't study only one and ignore the other two. I mean, how are you going to do anything in quantum chemistry without computer calculations, for example?

Physical chemistry is a massive field with no clearly delineated boundaries. Arguably protein mass spec and protein NMR are disciplines of physical chemistry (and thus huge amounts of biotechnology). You could also argue that biological soft matter is a physical chemistry discipline.

Your question is broad and these chemistry which you have pointed out in your question is interlinked (Quantum ,Physical and Computational chemistry). The best way which I can explain this question is by giving links to main research carried out using these chemistry in biological science.
The study of quantum mechanics in photosynthesis is one such area and this nature paper(Nature 446, 782-786 (12 April 2007) is about that research
the highlights of the paper is the evidence for wavelike energy transfer through quantum coherence in photosynthetic systems

two-dimensional electronic spectroscopy investigations of the FMO
  bacteriochlorophyll complex, and obtain direct evidence for remarkably
  long-lived electronic quantum coherence playing an important part in
  energy transfer processes within this system. The quantum coherence
  manifests itself in characteristic, directly observable quantum
  beating signals among the excitons within the Chlorobium tepidum FMO
  complex at 77 K. This wavelike characteristic of the energy transfer
  within the photosynthetic complex can explain its extreme efficiency,
  in that it allows the complexes to sample vast areas of phase space to
  find the most efficient path.

Other useful links
Nature 431, 256-257 (16 September 2004)
Nature 446, 740-741 (12 April 2007)


text:
Velocity of electrons ejected from Cl atom
When neutral $\ce{Cl}$ atoms are bombarded by high energy photons, causing ejection of electrons, which subshells' ejected electrons will have highest velocity?
I chose $1s$ because I thought more energy is needed to ejected those electrons so they might have higher velocity. My second choice $3s$ is right, but I can't explain why.

It would be the $3s$ shell because less energy is required to bump it off course. Due to electron repulsion and the shielding effect, the $e-$ in the $3s$ is more prone to get dislodged. And of course because of this, it will have a higher velocity when it gets bumped off. Another concept to note: $e-$ will have more energy at higher energy levels.
A reason why it wouldn't be the $1s$ shell is because due to the strong pull on the $1s$ electron. Since it is on the first energy level, it is strongly held by the nucleus. Even it were to get dislodged it would move very slow because of the strong pull by the nucleus.

so it's because 3s electrons after bumped off will have more kinetic energy leftover right?

Exactly. You figured it out.

Why not $3p$?  Aren't they further out and higher energy than $3s$ electrons?


text:
About undocumented conventions(?) in PDB files
ATOM 22 N  GLN L 3 23.945 -2.849 106.974 1.00 15.00 N
ATOM 23 CA GLN L 3 23.125 -2.052 106.036 1.00 15.00 C
ATOM 24 C  GLN L 3 22.898 -0.648 106.616 1.00 15.00 C
ATOM 25 O  GLN L 3 22.613 -0.491 107.825 1.00 15.00 O

In every amino acid there are two O atoms; which one is described by the (in this example) fourth entry? (I guess it's the one that's next to the next amino acid's N, but then where are the coordinates of the other one or should this be obvious?)
ATOM 26 CB   GLN L 3 21.754 -2.682 105.749 1.00 15.00 C
ATOM 27 CG   GLN L 3 21.577 -3.299 104.379 1.00 15.00 C
ATOM 28 CD   GLN L 3 20.956 -4.676 104.431 1.00 15.00 C
ATOM 29 OE1  GLN L 3 20.940 -5.348 105.476 1.00 15.00 O
ATOM 30 NE2  GLN L 3 20.496 -5.137 103.279 1.00 15.00 N
ATOM 31 H    GLN L 3 24.339 -2.372 107.727 1.00 0.00  H
ATOM 32 HE21 GLN L 3 20.637 -4.559 102.496 1.00 0.00  H
ATOM 33 HE22 GLN L 3 20.049 -6.007 103.275 1.00 0.00  H

What's this entry's nr.31 supposed to describe? I thought that an entry without remoteness indicator (and branch designator) might be bound to the previous atom, but in this case the N's adjacent H atoms are already dealt with with entries 32 and 33.
Also, this doesn't quite follow the convention(?)* of increasing the branch designator by one each time a new branch emerges—is this usual when H atoms' positions are given as well?
* like here:
ATOM 1661 CG  TRP A 214 24.705  7.905 6.526 1.00 50.03 C
ATOM 1662 CD1 TRP A 214 23.435  7.531 6.857 1.00 49.65 C
ATOM 1663 CD2 TRP A 214 25.087  8.914 7.475 1.00 51.52 C
ATOM 1664 NE1 TRP A 214 22.999  8.246 7.947 1.00 51.68 N
ATOM 1665 CE2 TRP A 214 23.993  9.098 8.350 1.00 52.47 C
ATOM 1666 CE3 TRP A 214 26.251  9.675 7.670 1.00 52.68 C
ATOM 1667 CZ2 TRP A 214 24.019 10.027 9.407 1.00 54.73 C
ATOM 1668 CZ3 TRP A 214 26.279 10.600 8.724 1.00 53.86 C
ATOM 1669 CH2 TRP A 214 25.167 10.761 9.581 1.00 54.84 C

First, your PDB data seems to have some formatting issues, in that the columns are in the wrong places.  I'm not sure if it's actually messed up, or if it simply got messed up when you posted it.
"In every amino acid there are two O atoms":  True for a single amino acid; not true for an amino acid residue.  I presume you're referring to the two oxygens on the carboxyl group.  However, one of these is lost when the amino acid forms an amide linkage with the next amino acid.
"What's this entry's nr.31 supposed to describe? ":  I opened the PDB in PyMol (after I corrected the format), and it seems to be the amide hydrogen
Regarding the branch designator -- I've never heard of such a thing in a PDB file, and I don't see any mention of it in the format description.  Could you clarify?
ETA:  I've tried to post the correctly formatted PDB data, but it seems to be messing up the formatting.  I'll try to figure out how to do that.
Here's an image of the residue

Regarding the branch designator: That's a term I found here  http://haldane.bu.edu/needle-doc/new/atom-format.html; it's the number in the "atom name" field.
Thanks for the extensive answer! (and yes, the formatting got pretty messed up when inserting it here)


text:
Full Valence Shell (Octet Rule) vs Formal Charge
When does formal charge have more precedence over the octet rule / an atom having a full valence shell when drawing structures?
When you draw $\ce{BCl3}$ for example, there are no double bonds - just three single bonds. Because of this $\ce{B}$ does not end up with 8 valence electrons. The explanation I have seen for this is that a double bond makes it so that the formal charge is undesirable. When drawing structures which has a higher precedence: formal charge or ensuring the atoms have full valence shells?

First, keep in mind that Lewis-structures and formal charges are huge simplifications.  In reality, at the quantum level, charges and bonds are spread out and it's somewhat unrealistic to say "Atom X has a charge of Q" or "Atoms X and Y have an Nth order bond between them".  There's merely a cloud of electron density, and it's somewhat arbitrary to say which parts of that cloud belong to a given atom or bond.
It's perfectly valid to draw a structure with a double bond.  And in reality, I'm sure there is hyperconjugation that leads to a small amount of double bond behavior.  However, such a resonance structure puts a negative charge on an electropositive element, and a positive charge on a very electronegative element, and thus such a structure has a smaller contribution to the overall structure.

Boron is also probably the least appropriate element to try to learn the basics of any bond structure model - it misbehaves as a general rule of thumb... ;)


text:
Really how rare are hydrogen atoms?
Hasok Chang, a chemist, but perhaps more a philosopher recently published the book Is Water H2O? . Chang used the question and an extensive web of arguments ultimately as a platform to support a pluralistic approach to scientific endeavor. I don't know that I buy it all, but in a latter chapter he made some statements regarding 'fundamental' components of chemical reactions, and one in particular that interested me and brings me to the point of my question.
Hydrogen gas as I understand it naturally occurs as a hydrogen molecule - two hydrogen atoms stuck together as a diatomic molecule. If ionized, the electron is stripped from the atoms and you have hydrogen ions or rather protons. But what about single hydrogen atoms? Are these non-existent or just a rare thing? Is there some way I can arrange to have a container of single hydrogen atoms and prevent their grouping into hydrogen molecules without ionizing them?
For what answers that follow - do these apply in general to the stable nature of all diatomic gases?

https://en.wikipedia.org/wiki/Hydrogen_atom#Production_and_reactivity

While usually short-lived, isolated hydrogen atoms (also called hydrogen radicals) are well studied and chemically important species. Bond formation in simple species is always an exothermic reaction (usually highly so) with zero activation energy, which means bond formation is thermodynamically very favourable and has no kinetic barrier stopping it from happening. Simply put, the isolated atoms really want to react with each other, and they'll attract each other at a distance, so if at some moment you create a container with pure monoatomic hydrogen, the hydrogen atoms will react as soon as possible to regenerate dihydrogen molecules (hydrogen gas).
The best you can do is expose the dihydrogen molecules to energetic conditions, so that a fair amount of bonds are being constantly broken at a rate comparable to which they're formed. A simple way to do this is to strongly heat the hydrogen gas; enough thermal energy will cause the bonds to break, though the dihydrogen bond is unusually strong ($\rm{436\ kJ\ mol^{-1}}$) so even at several thousand degrees kelvin there will be only a small portion of free hydrogen atoms in the gas (see Mithoron's comment). A more interesting option is to shine a strong ultraviolet light into the gas, as the dihydrogen molecules can separate into their constituent atoms upon absorption of photons with energy around $\rm{4.52\ eV}$, which corresponds to a photon wavelength of $\rm{274\ nm}$.
Any substance can have its bonds broken through the above two methods (among others), so the same applies to all other diatomic molecules. The stronger the bond between the atoms, the more uncommon is it to find the isolated atoms. For substances with very weak bonds, such as molecular dibromine or diiodine, mere exposure to visible light is enough to trigger formation of a fair amount of isolated atoms, something which is exploited in several reactions.
Keeping in mind that isolated atoms are very reactive, they are rather abundant in space. The average density of matter in the Universe is very, very low (about 1 atom per 5 cubic metres of space!), so isolated atoms can go a long time before finding anything to react with. This makes them extremely important for understanding astrochemistry.

"isolated atoms [...] will attract each other at a distance" By what mechanism? They're electrically neutral, and gravitational interactions are negligible, aren't they?

The major interaction between neutral atoms is the van der Waals force, see http://en.wikipedia.org/wiki/London_dispersion_force

... which is very short distance. Plus two single atoms usually can't combine, they need a third partner to take the bonding energy with him.

Hydrogen atoms are the dominant species in the atmosphere beyond 2000 km.
Monoatomic oxygen concentration exceeds diatomic about 100km.
See Atmospheric Structure
See also Atomic Hydrogen Welding for a technique to produce atomic hydrogen as well as THE DISSOCIATION OF HYDROGEN INTO ATOMS.
At 3200K, hydrogen is about 50 percent dissociated in to atoms, at atmospheric pressure.   Lower pressure greatly favors dissociation, so in the extreme upper atmosphere, monoatomic hydrogen is favorable.


text:
Is a transition metal with 7 d electons ever diamagnetic?
A transition metal with 7 d electrons forms a complex ion with cyanide. I know that cyanide is a high field ligand, which means the electrons will pair before moving to the next energy level, but wouldn't there still be one unpaired electron? And wouldn't that make it paramagnetic? I was told that my prediction was wrong and that it was in fact diamagnetic, but I don't understand why.

With an octahedral or tetrahedral field, you cannot have a ground-state electron configuration containing 7 electrons and all of them paired; therefore, it cannot be diamagnetic.  Is it possible that the person to whom you were speaking meant your calculation of the valence electrons when s/he said your "prediction was wrong"?  Is there a specific compound in question?

It's a fake transition metal with 7 d electrons that forms a 3+ charge. The comment my evaluator left just said that my prediction was paramagnetic and that a correct prediction is not present. Was I incorrect in saying that there would be one unpaired electron? I'm not really sure what i did wrong

I think the question is whether there are 7d electrons in the atom or in the ion. For example, if you take 3 electrons from 7, you get 4d electrons in the $\ce{M^{3+}}$ ion which obviously could (and should) be diamagnetic.

It sounds like @GeoffHutchison has the right interpretation: a d7 metal, after forming a +3 charge, would have 4 electrons remaining and could be  diamagnetic.  If it is a hypothetical free ion, however, then the aufbau principle would still predict a paramagnetic species with four unpaired electrons (and one unfilled d-orbital).  Something is misleading about your evaluator's question.

As I noted in my comment above, I think the question is whether there are 7d electrons in the atom or in the ion.
For example, if you take 3 electrons from 7, you get 4d electrons in the $\ce{M^{3+}}$ ion which obviously could (and should) be diamagnetic.
This is all, of course, a guess given what little you posted.
Let's say I had a similar question in my class. Students might then assume this $\ce{M^{3+}}$ is octahedral. (Personally, I'd make it a bit more obvious in the question.)
Given that the ligand is high-field, you get an initial MO diagram like this:

But... we're not done. This electron configuration is subject to a weak Jahn-Teller distortion. So you'd end up with an MO diagram like this:

That, of course, is diamagnetic.

authors: Hąýįłę Møđęřņ Mąņ Bŕø


text:
Unit of time used in rate law equations
If doing a problem where you use the integrated rate law for a second order reaction
($\frac{1}{[\mathrm{A}]}-\frac{1}{[\mathrm{A_0}]}=kt$) to first find $k$, then use that to find the half-life. Does it matter seconds are used, or minutes, or whatever?

The unit of $k$ is $\frac{\mathrm{concentration}}{\mathrm{unit~time}}$. It is not necessary that you use any specific unit for concentration or time, but most often $t$ is measured in seconds, and the unit of $k$ is $\mathrm{mol~L^{-1}~s^{-1}}$.


text:
How to predict if hydrogen atoms could be placed in the interstials of a lattice
Palladium hydride is an example of a hydride which contains H atoms in palladium's crystal lattice (which is fcc).
I would like to know if it's possible to predict if another hydride, $\ce{NbH_{x}}$ could exhibit the same behavior.
I know that in an ionic solid, the fraction $$\frac{r_{+}}{r_{-}}$$
Can be used to predict the geometry around a cation.
Can I use this expression for niobium hydride? In the case of niobium, the crystal structure is bcc, so it contains tetrahedral and octahedral interstial sites.

Your intuition is correct; $\ce{NbH_{x}}$ retains the bcc Nb lattice, with the hydrides in the tetrahedral interstital sites, up to x = 2, when the compound switches to a fcc Nb lattice (the Fluorite structure). This is pretty typical behaviour for a transition metal hydride.
I don't think radius ratio rules are going to be useful in predicting the behaviour of these systems; there is a lot of covalency going on (hydride-hydride interactions and Nb-Nb interactions), and the metal lattices will end up being distorted by the correlated interstital occupancies of the hydrides. Hydride has a radius of around 130 pm while Niobium has a radius of around 150 pm so you wouldn't predict the observed behaviour based on radius ratios.
More information on this particular system can be found in Hauck (1976), Ordering of Hydrogen in Niobium Hydride phases.


text:
Exchange-correlation hole
I did a google and chemistry.exchange search and found several technical definitions of the exchange correlation hole.
The questions that I want to proposes are:

What the exchange-correlation hole means in common words?
How do you explain it to a non-literate person?

Could you add a little more context please: How "common" shall the words be that are used to explain it? No math at all or are some formulas allowed? How "non-literate" is the person supposed to be that it shall be explained to? Can any prior knowledge about quantum mechanics and density functional theory be assumed (especially: is it clear what exchange and correlation actually mean?) or has everything to be explained from scratch and in very simple words?

What the exchange-correlation hole means in common words?

Assuming that the probability is the common word, one could say that the exchange-correlation hole, is a region of space around an electron in which the probability of finding another electron is close to zero due to electron correlation.

How do you explain it to a non-literate person?

Long story short is that mathematically electron correlation arises as a consequence of the instantaneous Coulomb repulsive forces between electrons and also as a consequence of the fundamental principle of antisymmetry of an electronic (fermionic in general) wave function. Note that the first kind of electron correlation which is due to Coulomb repulsions is known as the Coulomb correlation while the second one which is due to antisymmetry of a wave function is knows as the exchange correlation. However, since the root of all ab-initio wave function methods, the HF method, already accounts for exchange correlation, the Coulomb correlation is usually also referred to as just the correlation.
With respect to these two different kinds of electron correlation note that exchange-correlation hole does not refer specifically to just the exchange correlation, rather it is the hole due to both types of the electron correlation mentioned. I think that one would better call it, say, exchange-Coulomb correlation hole to avoid an ambiguity, but I'm afraid we're stuck with the exchange-correlation hole term.

Long story
Consider the simplest many-electron system, a two-electron one, and let us examine the following two important events

$\vec{r}_{1}$ - the event of finding electron-one at a point $\vec{r}_{1}$;
$\vec{r}_{2}$ - the event of finding electron-two at a point $\vec{r}_{2}$.

Probability theory reminds us that in the general case the so-called joint probability of two events, say, $\vec{r}_{1}$ and $\vec{r}_{2}$, i.e. the probability of finding electron-one at point $\vec{r}_{1}$ and at the same time electron-two at point $\vec{r}_{2}$, is given by
\begin{equation*}
    \Pr(\vec{r}_{1} \cap \vec{r}_{2})
    =
    \Pr(\vec{r}_{1}\,|\,\vec{r}_{2}) \Pr(\vec{r}_{2})
    =
    \Pr(\vec{r}_{1}) \Pr(\vec{r}_{2}\,|\,\vec{r}_{1}) \, ,
\end{equation*}
where

$\Pr(\vec{r}_{1})$ is the probability of finding electron-one at point $\vec{r}_{1}$ irrespective of the position of electron-two;
$\Pr(\vec{r}_{2})$ is the probability of finding electron-two at point $\vec{r}_{2}$ irrespective of the position of electron-one;
$\Pr(\vec{r}_{1}\,|\,\vec{r}_{2})$ is the probability of finding electron-one at point $\vec{r}_{1}$, given that electron-two is at $\vec{r}_{2}$;
$\Pr(\vec{r}_{2}\,|\,\vec{r}_{1})$ is the probability of finding electron-two at point $\vec{r}_{2}$, given that electron-one is at $\vec{r}_{1}$.

The first two of the above probabilities are referred to as unconditional probabilities, while the last two are referred to as conditional probabilities, and in general $\Pr(A\,|\,B) \neq \Pr(A)$, unless events $A$ and $B$ are independent of each other.
If the above mentioned events $\vec{r}_{1}$ and $\vec{r}_{2}$ were independent, then the conditional probabilities would be equal to their unconditional counterparts
\begin{equation*}
    \Pr(\vec{r}_{1}\,|\,\vec{r}_{2}) = \Pr(\vec{r}_{1}) \, , \quad
    \Pr(\vec{r}_{2}\,|\,\vec{r}_{1}) = \Pr(\vec{r}_{2}) \, ,
\end{equation*}
and the joint probability of $\vec{r}_{1}$ and $\vec{r}_{2}$ would simply be equal to the product of unconditional probabilities,
\begin{equation*}
    \Pr(\vec{r}_{1} \cap \vec{r}_{2}) = \Pr(\vec{r}_{1}) \Pr(\vec{r}_{2}) \, .
\end{equation*}
Note that this is the picture in which electrons are sort of billiard balls, i.e. neutral macroscopic particles.
In reality, electrons are, of course, not like billiard balls; they are charged microscopic particles, and consequently,
\begin{equation*}
    \Pr(\vec{r}_{1}\,|\,\vec{r}_{2}) \neq \Pr(\vec{r}_{1}) \, ,
    \quad
    \Pr(\vec{r}_{2}\,|\,\vec{r}_{1}) \neq \Pr(\vec{r}_{2}) \, ,
\end{equation*}
and
\begin{equation}
    \Pr(\vec{r}_{1} \cap \vec{r}_{2}) \neq \Pr(\vec{r}_{1}) \Pr(\vec{r}_{2}) \, .
\end{equation}
And since the spatial probabilities of finding electrons are inevitably related to their states, the inequality above means that the state of electron-one is not independent of the state of electron-two and vice versa.
This interdependency of the states of electrons is referred to as the electron correlation.
The inequalities above, which are the mathematical essence of electron correlation, hold true for two reasons. First, electrons repel each other by Coulomb forces, and, as a consequence, at small distances between the electrons the conditional probability is less than half of the corresponding unconditional one
$$
    \Pr(\vec{r}_{1}\,|\,\vec{r}_{2}) < \Pr(\vec{r}_{1}) \, ,
    \quad
    \Pr(\vec{r}_{2}\,|\,\vec{r}_{1}) < \Pr(\vec{r}_{2}) \, ,
$$
while at large distances between the electrons the conditional probability is greater than half of the corresponding unconditional one
$$
    \Pr(\vec{r}_{1}\,|\,\vec{r}_{2}) > \Pr(\vec{r}_{1}) \, ,
    \quad
    \Pr(\vec{r}_{2}\,|\,\vec{r}_{1}) > \Pr(\vec{r}_{2}) \, ,
$$
In the extreme case when $\vec{r}_{1} = \vec{r}_{2}$, the Coulomb repulsion between the electrons becomes infinite, and thus, $\Pr(\vec{r}_{1}\,|\,\vec{r}_{1}) = 0$ and consequently $\Pr(\vec{r}_{1} \cap \vec{r}_{1}) = 0$, i.e. the probability of finding two electrons at the same point in space is zero.
The correlation due to Coulomb repulsion is termed as Coulomb correlation and one sometimes speaks about the presence of the so-called Coulomb hole around each electron - a region of space around it in which the probability of finding another electron is close to zero due to Coulomb repulsion.
Secondly, as a consequence of the Pauli exclusion principle, electrons in the same spin state can not be found at the same location in space, so that for electrons in the same spin state there is an additional contribution into the inequalities between conditional and unconditional probabilities above.
This effect is relatively localized as compared to one due to Coulomb repulsion, but bearing in mind the relationship between probabilities and wave functions and taking into account the continuity of the later, it is still noticeable when electrons are close to each other and not just at the same location in space.
This type of electron correlation, which is a consequence of the Pauli exclusion principle, is referred to as Fermi correlation, or exchange correlation, and one sometimes speaks about the presence of the so-called Fermi hole around each electron - a region of space around it in which the probability of finding another electron in the same spin state is close to zero.

is ECH is only for electron or it can be extended to proton also?

Well, protons are also fermions, so there will be Coulomb as well as exchange correlation, and consequently, I think the answer is yes, each proton is also surrounded by ECH, though, of a different kind than that of an electron.

What about protons inside nucleus? will they have ECH

@Eka, it does not matter where the protons are. Whenever you have a many particle system composed of charged fermions (protons, electrons, etc.) each and every particle will have the corresponding ECH around it.

In nucleus protons are "tightly packed" so its ECH will be lower comparing to ECH of free moving electron right? ECH is Zero probability of finding fermions around similar fermions.

@Eka, sorry, I'm a chemist, so I have no idea about nuclear structure. :D But I think, that ECH is always the same for each and every proton no matter how tight protons are "packed". ECH is the region around a charged fermion where the probability of finding another charged fermion of the same kind is close to zero. For a charged fermion the probability of finding another charged fermion is exactly zero only at one point, at the location of the particle, but it is also negligibly small in the close proximity to it.

The exchange-correlation hole refers to the fact the probability of finding at electron at a certain position (r1) given that there is an electron at another position (r2) decreases when the distance between the two electrons decrease.
Put another way, for a two electron system, the least likely position of one electron on top of the other

Just want to leave a note that the probability of finding two electrons at the same spatial spot is not zero. The argument of infinite repulsion does not fit. Details see here:
Can two electron occupy the same spatial spot in a statistical way?


text:
Dissociation Energy for CO2 per molecule
I'm trying to calculate the energy required to dissociate a single $\ce{CO2}$ molecule into its respective elements, and I'm given that it takes $ 373.6 \times 10^{3} \, \, \mathrm{\frac{joules}{gram}} $ where  $44.01$ grams is the mass of $6.022 \times 10^{23}$ molecules of $\ce{CO2}$.
By dimensional analysis:
$$\frac{44.01 \ \mathrm{grams}}{6.022 \times 10^{23}~\mathrm{molecules}} = 7.308 \times 10^{-23}\mathrm{\frac{grams}{molecule}}$$
$$\mathrm{\frac{joules}{molecule}} = \mathrm{\frac{joules}{gram} \cdot  \frac{grams}{molecule}} = (373.6 \times 10^{3}) \cdot (7.308 \times 10^{-23}) = 2.7303 \times 10^{-17} \mathrm{\frac{joules}{molecule}}$$
or,
$170.43\,\, \mathrm{eV/molecule}$
Is this derivation and the logical assumptions made herein correct?

Your calculations are right, but the starting numbers are not. This just can't be real. Suppose you make such-and-such calculations and found that the density of certain compound is 170 kg/L, or that the height of certain human is 170 feet. What would you think then?

$$\frac{44.01 \ \mathrm{grams}}{6.022 \times 10^{23}~\mathrm{molecules}} = 7.308 \times 10^{-23}\mathrm{\frac{grams}{molecule}}$$
$$\mathrm{\frac{joules}{molecule}} = \mathrm{\frac{joules}{gram} \cdot  \frac{grams}{molecule}} = (373.6 \times 10^{3}) \cdot (7.308 \times 10^{-23}) = 2.7303 \times 10^{-17} \mathrm{\frac{joules}{molecule}}$$
or,
$170.43\,\, \mathrm{eV/molecule}$


text:
How do I print property matrices in Avogadro?
I'd like to print the content of property matrices (i.e., the matrices shown after selecting View→Properties→X where X is one of the six options shown, including Angle properties..., Atom properties... and Bond properties...) in Avogadro, but I can't seem to find this option at Avogadro Wiki or other forms of Avogadro documentation.
e.g. this is the bond properties matrix.

Right, it doesn't exist at the moment. Probably the easiest thing would be to allow export as text or a CSV, but I'm not sure when the next release would be.

I'd also like to recommend future releases include a Windows version with python terminal as the only way I can access the python terminal is via installing Avogadro 1.0.1 python win32 (i.e., the file available here http://internode.dl.sourceforge.net/project/avogadro/avogadro/1.0.1/avogadro-1.0.1-python-win32.exe).

Yes, but without a Windows developer, those releases will be something of second-class citizens to Mac and Linux releases, sorry.

I am not aware of a menu button to print these tables.
There is however a different way to get your hands on the numbers.
Avogadro features objects: molecules, bonds and atoms are instances of classes, whose properties can be read and processed by programming.
The easiest way is the build-in Python-Terminal.
(Note that I'm using Avogadro 1.1.1 on Linux in German, your menus will have different labels.)

In Avogadro, open a Python Terminal (In German, this is done via
Einstellungen -> Werkzeugleisten -> Python-Terminal, which should be something like Settings -> Tool bars -> Python terminal

When run in the Python terminal of Avogradro, the following snippet writes

atom index
atomic number
x, y, z coordinates

for all atoms of the currently opened molecule to a csv file into an existing tmp directory in the user's home.

import os.path
import csv

home = os.path.expanduser('~')
data_dir = 'tmp'
atom_table_name = 'atoms.csv'
atom_table_path = os.path.join(home, data_dir, atom_table_name)
# generate full path for file with atom data

with open(atom_table_path, 'wt') as fout:
    writer = csv.writer(fout)
    writer.writerow(('Index', 'Atomic number', 'X', 'Y', 'Z'))

    mol = Avogadro.molecule

    for atom in mol.atoms:
        index = atom.index + 1
        x, y, z = atom.pos
        writer.writerow((index, atom.atomicNumber, x, y, z))

EDIT
I mentioned the Python Terminal first because I didn't want to introduce too much different things at once.
Once you have played a bit with the build-in Python Terminal, you will want to go one step beyond and try a more comfortable IDE with logging, tab-completion, build-in help, etc. My suggestion is to try IPython!
With IPython (or any other IDE of your choice), you can make use of Avogadro through its API without actually running it.
Given that your data file propen.cml looks like this:
<molecule>
 <atomArray>
  <atom id="a1" elementType="C" x3="-6.736896" y3="1.198243" z3="-0.001854"/>
  <atom id="a2" elementType="C" x3="-5.625957" y3="2.233390" z3="-0.108371"/>
  <atom id="a3" elementType="H" x3="-6.748313" y3="0.730420" z3="0.988429"/>
  <atom id="a4" elementType="H" x3="-6.615420" y3="0.408802" z3="-0.751312"/>
  <atom id="a5" elementType="H" x3="-7.710837" y3="1.671184" z3="-0.163704"/>
  <atom id="a6" elementType="C" x3="-4.268024" y3="1.629800" z3="0.106393"/>
  <atom id="a7" elementType="H" x3="-5.801148" y3="3.017025" z3="0.637857"/>
  <atom id="a8" elementType="H" x3="-5.668681" y3="2.695849" z3="-1.101309"/>
  <atom id="a9" elementType="C" x3="-3.137190" y3="2.345874" z3="0.062502"/>
  <atom id="a10" elementType="H" x3="-4.203709" y3="0.562202" z3="0.306350"/>
  <atom id="a11" elementType="H" x3="-3.147760" y3="3.413685" z3="-0.133506"/>
  <atom id="a12" elementType="H" x3="-2.172936" y3="1.873521" z3="0.222912"/>
 </atomArray>
 <bondArray>
  <bond atomRefs2="a1 a2" order="1"/>
  <bond atomRefs2="a1 a3" order="1"/>
  <bond atomRefs2="a1 a4" order="1"/>
  <bond atomRefs2="a1 a5" order="1"/>
  <bond atomRefs2="a2 a6" order="1"/>
  <bond atomRefs2="a2 a7" order="1"/>
  <bond atomRefs2="a2 a8" order="1"/>
  <bond atomRefs2="a6 a9" order="2"/>
  <bond atomRefs2="a6 a10" order="1"/>
  <bond atomRefs2="a9 a11" order="1"/>
  <bond atomRefs2="a9 a12" order="1"/>
 </bondArray>
</molecule>
you can load the CML file and iterate over the bonds to display the index numbers of the atoms in the bond and the bond length using:
import Avogadro as av

mol = av.MoleculeFile.readMolecule('propen.cml')
# don't forget to adjust the path for the data file!

for bond in mol.bonds:
    first = bond.beginAtom
    last = bond.endAtom
    print first.index, last.index, bond.length

That gave the output: Traceback (most recent call last):   File "<string>", line 1, in <module> IOError: [Errno 2] No such file or directory: 'C:\\Users\\Brenton\\tmp\\atoms.csv'

@BrentonHorne Does a tmp directory exist in your home?

Doesn't I'm afraid.

@BrentonHorne I'm sorry for the confusion. I just haven't made myself clear enough :) I wanted to avoid that your home folder gets littered up with stuff. Therefore, the code snippet was programmed to write the csv to a separate directory, tmp for all the temporary stuff that can be deleted later. If you create the tmp directory manually, the script - run from within Avogadro -  should work as intended :)

Thanks, but it appears that Avogadro still wants to be difficult, and is giving me this output: Traceback (most recent call last):   File "<string>", line 5, in <module> NameError: name 'Avogadro' is not defined

@BrentonHorne I just tried the first example again with a freshly started Avogadro, pasted the code, only changed the name of the output file and executed it. On my Linux machine with Avogadro 1.1.1, and Open Babel 2.3.2 installed from the repositories, it did work again and wrote another file.  From your first traceback, I reckon that you're on Windows. Maybe some other users have an idea on what's different there. I never used Avogadro under Windows, I don't even have a Windows machine available here. Maybe Geoff Hutchison has a good idea on how to solve this independent of the platform.

It'd be pretty easy to do this with Open Babel and Python if you want to go that route.

Hey one more thing, what about bond angles? How do I print them from Avogadro? Keeping in mind that most of the files I wish to calculate this for aren't CMLs -- they're CIFs and PDBs.

@GeoffHutchison your route sounds even better seeing how I'm guessing I could do that on my Windows system too. If you'd care to explain it (I realize here probably isn't appropriate as this question is about Avogadro, not OpenBabel so perhaps by email, I've emailed ya before, so you should have my address).


text:
Does a weak acid undergo it's weak ionization upon the addition of any kind of base?
For an example let's assume we have a weak acid  HA . And we add either a strong or a weak base. Then what I want to know is whether  the weak ionization of the acid HA taken place?.
Weak ionization of acid HA. ==>

Yes it takes place, the reaction you wrote is correct, what exactly do you want to know ?

So for a moment let's assume  we add  a base to this weak acid in order to reduce the H+ concentration, then what would really happen to  the the concentration of this weak acid HA and  the concentration of  conjugate base?.

A weak acid does not (contrary to what you show in your equation) predominantly ionize before the base is added. But keep in mind that proton transfer reactions (at least those between oxygen and/or nitrogen atoms) occur almost instantaneously in both directions, so the equilibrium you illustrated is established immediately (for a weak acid, with the big arrow pointing left). So when (strong) base is added, H3O+ is consumed immediately, then more acid is ionized to re establish the equilibrium--and this goes on as you add more base until you have consumed almost all of the weak acid. These reactions are controlled by the relative energies of the reactants and products, and not by reaction rates.


text:
Why is is this molecule called ketene?

The marking scheme says that the molecule has both alkene and ketone.
So there is carbon to carbon double bond.ok.check there is alkene.But what about ketone.Ketone has this structure:

So can we technically say that there is ketone in it?

This molecule is a ketene because it contains the ketene functional group:

Whilst the ketene appears to contain an alkene and a ketone functional group, which as you have suggested is where its name comes from, its chemistry is notably different from both of its 'constituent groups'.


text:
What is the mechanism of the uncatalysed epoxide opening with alcohols?
The Wikipedia page says this:

Ethoxylation is an industrial process in which ethylene oxide is added to alcohols and phenols to turn it into a surfactant.

How does the reaction proceed?
The answer to the question "Do epoxide rings react with bases?" regarding epoxide rings reacting with bases does not include alcohols, and does not answer my question.

Does this answer your question?

@KlausWarzecha Maybe that reaction might be one of the reaction steps. I don't know.

@Martin So intead of CH3O- of that answer, lp of CH3OH will attack? And Eka gave me a link - http://www.chemstations.com/content/documents/technical_articles/ethox.pdf and it is different.

The link you provided has does not explain anything about the mechanism and it is about a base catalysed reaction.

@ringo Thank you for fixing the layout. It is completely unnecessary, to include comments on what was edited (that can be seen in the history). The question should be standalone in any case.

@Martin-マーチン So am I correct in saying that the only difference is that over here, l.p of alcohol attacks?

@Larry The alcoholate, because it is deprotonated, has three lone pairs, one of them, the HOMO, will attack in a nucleophilic manner. An alcohol still has two lone pairs, and one of the is the HOMO, which will make the nucleophilic attack. The whole process might be slower, as it is less nucleophilic, but you can assume the mechanism to be almost identical.

Because of its special chemical struture Ethyelene oxide (EO) is highly reactive material thus easily participate in chemical reactions which includes polymerization. Mechanism of opening of EO's ring in acidic and alkaline medium is given below

If you see in both cases a molecule with alcoholic group ($\ce{-OH}$) is formed.If the epoxyethane is in excess, the reaction can continue and could lead to polymerization.
A good example is the reaction between EO and water

As explained in wikipedia ethoxylation is an industrial process to make surfactant from EO. Surfactants are chemical molecule which contain both hydrophobic groups (their tails) and hydrophilic groups (their heads).
The end result of reaction with EO and alchol is a long chain molecule with these hydrophobic ($\ce{-R}$) and hydrophilic ($\ce{-OH}$) group

In the first image, from where did the H+ and OH- come from? Where did they alcohol go.

It just shows reaction environment H+ from acid (HCl, H2SO4..etc) and OH- from base (NaOH etc) see this link

Where did the alcohol go?

@larry See the alcohol reaction here R-OH is the alcohol see this example

But that has an acid and alcohol. Epoxide and alcohol?

@Larry I am extremely sorry for that mistake , didnt properly seen the reaction see this reaction http://patentimages.storage.googleapis.com/US20040142177A1/US20040142177A1-20040722-C00007.png

But how do they react? That was my initial question. I understand that H+ and OH- are required along with alcohol to react with the epoxide.

@Larry the mechanism for a acid catalyzed epoxide ring opening also requires a nucleophile which will attack the epoxide and opens the ring. the opened epoxide ring then act as a nucleophile and attack other epoxides andor alcohol and form secondary products. see the first page in the pdf

What happened in the 2nd reaction which goes to the left side of that pdf? Did H+ of alcohol shift? Is this mechanism correct or the one provided here- http://chemistry.stackexchange.com/questions/27288/do-epoxide-rings-react-with-bases

@Larry You can find acidic and basic catalyzed ring opening in this link. Acidic and Bascic catlalyzed ring opening have different ring opening mechanisms. As of second reaction i have still doubts regarding it and i have to consult with my collegues


text:
Order of reactions in kinetics
In kinetics, zero order reactions are those reactions who are independent from the concentration of the reactant or the product. So their rate is equal to the constant k. But still we build the diagram where we find the correlation between concentration and time, where the slope is negative.
Why do we even write a diagram to find the correlation between concentration and time, since these reactions are completely independent from concentration ?

How would you know the correlation between concentration and  time is of zero order until you draw the diagram?

So their rate is equal to the constant k.

With other words, the change of concentration over time is linear:
\[ \frac{d[\ce{A}]}{dt} = k\]

So is that the answer ?


text:
Why might the ratio between two Q-band peaks in ZnTPP decrease with increasing temperature?
In ZnTPP (Zinc tetraphenylporphyrin), there are two Q-band emission peaks. When I run steady state emission measurements at different temperatures, the ratio of their respective peak intensities decreases with increasing temperature. This seems to me like an unusual behaviour compared to many other porphyrins, where the ratio remains constant.
Why is this?
The molecule has been around for a while so I would expect to find the answer in the literature, I am currently searching it and might post the answer if I find it before someone else answers.


text:
Reduce Decomposition Of Glycerin
Background: Glycerin, when heated with a metal coil, decomposes below its boiling point to acrolein, which in turn appears to decompose to formaldehyde and acetaldehyde. (see previous post)
Question: Is there a chemical that could be added, or a process that could be applied, that would lessen the decomposition of glycerine when vaporized?

Please note that there's a policy here that doesn't allow us to answer questions that look for medical advice. Ah well, I hope the answerer to this will be able to point to your issues with Wikipedia links and maybe some things like MSDS's.

"Noxious levels" is a subjective term and by no means a clear result from all ecigs. Besides those noxious compounds are also formed in cooking and it has so far proved impossible to show they cause harm in food (though many have tried). I think you've got carried away by anti-tobacco rhetoric which tends to lead to poor science getting a hearing.

At concentrations above 0.1 ppm in air formaldehyde can irritate the eyes and mucous membranes (http://en.wikipedia.org/wiki/Formaldehyde).That is not subjective, nor is it anti-tobacco rhetoric - it is an accepted scientific fact. Furthermore, e-liquids are usually vaporized and inhaled into the lungs - not eaten.

You could try the mechanism behind mist fountains or other "cold fog" machines:

These use an ultrasonic oscillating diaphragm to produce a fine mist of room temp water, no heating performed. The resulting vaping machine would be larger and require more power, but wouldn't have the harmful aldehydes inherent in current vaping liquids. Nicotine itself is hygroscopic but not really water soluble at room temp, but it can form salts that are water-soluble so it could be introduced to the body this way. Various flavors currently used might be more or less water-soluble. Finally, the fog, being simple water vapor, wouldn't be visible when exhaled which is one reason the glycerol compounds are used; they produce a "smoke" that dissipates more quickly than tobacco smoke.

Could you build a machine like this roughly the size of a home telephone ?

Should be feasible; a vaping hookah of sorts.

That looks very promising. Perhaps by mixing the glycerine with water, applying a little heat, and using a more heavy-duty mister, a reasonable "smoke" could be produced.

@KeithS Your answer implies that an ultrasonic oscillator couldn't be used for the production of vapor from a standard propylene glycol/vegetable glycerin-based 'e-juice.' If that was intentional on your part, what is the reason why it wouldn't be possible?

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


text:
Why does thallium form TlF and TlF3 but aluminium only forms AlF3?
Does this have something to do with the inert pair effect?

Yes, that's exactly right.

Ahh, Thallium is a very nice element. Because it's weird.

The +3 oxidation states are favorable except for the heavier elements, such as Tl, which prefer the +1 oxidation state due to its stability; this is known as the inert pair effect. Chemwiki.UCDavis
As an example in group 13 the +1 oxidation state of Tl is the most stable and Tl(III) compounds are comparatively rare. Inert pair effect, Wikipedia

So, why?
Inert pair effect was introduced to describe why the facts above are correct. However, the original explanation, provided by Sidgwick, relied on the fact that the $s$ subshell has an "inert" pair of electrons. This can't be correct, and Wikipedia uses the trend in ionization potentials of "$s$" electrons to prove it wrong:
$$\ce{In < Al < Tl < Ga}$$
A better explanation lies within the lower bond enthalpies of $\ce{MX}$ where M is one of the p-block heavy elements.
To reach a high oxidation number, elements will need to have a strong bond. They wanna bond in order to lose energy (i.e. become more stable) after all! If the bond is weak, this energy isn't provided. Thus, a heavy element is better off with being less oxidised.
See this and this. (Sorry, they might be behind a paywall)

-1, inert pair effect is definitely a real thing; it's due to the lanthanide contraction for the 6p elements + relativistic 6s contraction (the "inert pair"). Also, the thallium halides aren't molecular so invoking bond enthalpies doesn't give the full picture either.

Hmm...@J.LS I never said it's not real. Just that the description isn't very authentic. (Or rather, the first description) How do you explain the trend in ionization potentials?

@J.LS Figures out, the pattern in ionization potentials is due to the relativistic effects you mention.

No, the pattern in ionization potentials is due to the transition metal and lanthanide contractions and the associated decrease in shielding for Ga/Tl respectively; relativistic effects are only important for Tl and probably are of secondary importance.


text:
Why is picric acid more acidic than carbonic acid?
I just had a question in a test asking to compare the relative acidity of picric acid (TNP) and carbonic acid, $\ce{H2CO3}$ (not phenol). It turns out that TNP is more acidic. I have no idea why? Is there some rule I'm missing or is this one of those exceptions that I just have to remember?
I have been told that alcohols are generally less acidic than carboxyl groups. But in this case, doesn't the hydroxyl group form a chelate ring already? If the stability of the reactant is much greater than the the stability of the product, why would the reaction even take place? Sure, there is cross conjugation in carbonic acid, but shouldn't it be extremely hard to break the chelation in picric acid?

strongly related http://chemistry.stackexchange.com/questions/9074/reason-for-the-stronger-acidic-property-of-phenol-than-alcohol

To me, the asker just wants to know how and why a hydroxyl group can be more acidic than a carboxyl group.

Perhaps you misread what I wrote; carbonic acid ($\ce{H2CO3}$) contains a carboxyl! The acidic group in carbonic acid is the carboxyl group, R-COOH (for carbonic acid, R = OH). You've probably been told that hydroxyl groups (C-OH, present in alcohols and phenols) are weaker acids than carboxyl groups/carboxylic acids, yet picric acid is a substituted phenol and is quite a bit more acidic than carbonic acid and simple carboxylic acids. This is the source of your confusion?

Short answer could be: picryl is stronger stabilizer of negative charge on oxygen then carboxyl.

I think that by "chelation" you mean conjugation or aromaticity? You are right to think that a process which breaks conjugation would be less favourable, but the deprotonated phenol maintains the conjugation present in phenol. Have a look here.

Chelate is the ring form due to intramolecular hydrogen bonding. Because the hydrogen in picric acid is already hydrogen bonded, wont it be difficult to convert it into a non hydrogen bonded structure? As hydrogen bonding obviously contributes to stability.

Oh, you mean the intramolecular hydrogen bond between the hydroxyl and the nitro group? Indeed, that slightly lowers the energy of the reactant, which would make ionization a bit more difficult. However, the three nitro groups strongly lower the energy of the ionization product by an electron-withdrawing mesomeric effect, making deprotonation far more favourable. Have you studied this sort of stuff? Take a look at this section from the UC Davis ChemWiki, especially the part just above example 7.10. Does that help?

So you want to compare the acidity of different Brønsted–Lowry acids. This means ultimately you want to compare the equilibrium constants for reactions of the type $\ce{HX <=> H+ + X-}$. To do so, you'll have to analyse what stabilisation/destabilisation effects are present in each acid $\ce{HX}$ and each conjugate base $\ce{X-}$. This is easier said than done, especially quantitatively. There are many sorts of possible effects, often acting in a compounded fashion. Fortunately, there are some basic principles to help guide us.
Your question specifically asks to contrast carbonic acid, $\ce{H2CO3}$ and picric acid, more descriptively called 2,4,6-trinitrophenol. This already starts off as being a bit tricky, because they are not very similar substances; of course, the more similar the two substances are when being compared, the easier it will be to pick out the differences in stabilisation/destabilisation effects and the safer it will be to extrapolate from qualitative reasoning. With this in mind, it is best if we build up to the answer by making comparisons between several molecules, making small changes at each step.
First off, let us start with alcohols. These are substances containing hydroxyl groups ($-\ce{OH}$) linked to carbon atoms which are saturated (carbon atoms with four single bonds, i.e. $\rm{sp^3}$-hybridised carbon) and whose three other bonds are only to carbon or hydrogen atoms (in other words, linked to no heteroatoms other than the hydroxyl oxygen). This class represents neither of your substances, but is simple and is related to phenols, so it will be a useful benchmark. A simple alcohol is methanol, $\ce{CH3OH}$, which can act as an acid by the equilibrium:
$$\ce{H3C-OH_{(aq)} <=> H+_{(aq)} + H3C-O^{-}_{(aq)}}\ \ \ \ \ K_\mathrm a=3\times10^{-16}$$
The small equilibrium constant establishes that methanol is a very weak acid in water, and exists almost completely as the unionised reactant. Values of $\mathrm pK_\mathrm a$ around 15 are typical for simple alcohols.
Now let us take a step towards the structures we want, by examining at a hydroxyl linked to an unsaturated carbon atom, specifically a $\rm{sp^2}$-hybridised carbon atom linked to no heteroatoms other than the hydroxyl oxygen. What we have here is an enol. Simple enols are not very stable molecules, but for simplicity we shall disregard that. For the simple molecule propen-2-ol, the equilibrium is:
$$\ce{H2C=C(OH)-CH3_{(aq)} <=> H+_{(aq)} + H2C=C(O^{-})-CH3_{(aq)}}\ \ \ \ \ K_\mathrm a=1.3\times 10^{-11}$$
Evidently, enols are quite a bit more acidic than alcohols. There are two new effects at play here. First, $\rm{sp^2}$-hybridised carbon atoms show an electron-withdrawing effect compared to $\rm{sp^3}$-hybridised carbon. This ability to "suck away" electron density helps stabilise molecules when close to electron-rich regions such as the oxygen atom. This effect stabilises the reactant, but it stabilises the enolate product more (as it has extra electron density to be pulled), which drives the equilibrium to the right. A second effect is resonance. Some of the electrons from the oxygen atom are said to conjugate with the carbon-carbon double bond, creating resonance or canonical structures, which contribute to the true description of bonding in the molecule. Once again, spreading the electrons out contributes to decrease the energy of a molecule. The resonance structures for the reactant and the product are similar, but one of the resonance structures in the reactant contains a formal positive charge on oxygen (a very electronegative atom) and a formal negative charge on carbon (a not very electronegative atom), so this form of conjugation is not quite as favourable. In the product, though there is still a formal negative charge on carbon in one of the canonical structures, there is no formal positive charge on oxygen, making conjugation more favourable. Thus the resonance argument is also a part of the explanation for the acidity.
Next up is phenol, the parent compound of 2,4,6-trinitrophenol. The effects present here are much the same as in the enol above, except that conjugation happens in a larger extent in both the reactant and the product. Experimentally:
$$\ce{C6H5-OH_{(aq)} <=> H+_{(aq)} + C6H5-O^{-}_{(aq)}}\ \ \ \ \ K_\mathrm a=1.1\times10^{-10}$$
This represents only a small increase in acidity compared to propen-2-ol, which suggests conjugation is not the main factor in phenol acidity, but rather the inductive effect of the $\rm{sp^2}$ carbon.
Finally, we've built up to our first structure, 2,4,6-trinitrophenol. The new bit here is the addition of nitro groups ($\ce{-NO2}$), which are strongly electron-withdrawing both by induction and mesomerism (conjugation). Again, these effects decrease the energy of both the reactant and the product, but stabilize the product more because there is extra density to be spread around. As a quick interesting aside, the acidities of 2-nitrophenol and 4-nitrophenol are similar, and both are somewhat larger than 3-nitrophenol, which suggests most of the increase in acidity comes from electron-withdrawing by mesomerism rather than induction, represented by canonical structures similar to these. You also make a good point in mentioning the intramolecular hydrogen bond present only in 2,4,6-trinitrophenol and not its ionised form. Indeed, this helps stabilize the reactant a slight amount, and decreases its acidity. Let's see how all these effects add up:
$$\ce{(O2N)3C6H2-OH_{(aq)} <=> H+_{(aq)} + (O2N)3C6H2-O^{-}_{(aq)}}\ \ \ \ \ K_\mathrm a=0.4$$
The acidity increase from the electron-withdrawing effects of the three nitro groups far overwhelms the acidity decrease due to the formation of the intramolecular hydrogen bond. How could you have known this beforehand? Strictly speaking, you couldn't have deduced this from thin air (not without a computer, at any rate!), but generally you are expected to know the situation in similar cases and work out the answer based on previous knowledge. Intramolecular hydrogen bonding usually doesn't play a huge role in increasing or decreasing acidity, at least in simple molecules, as they are relatively weak interactions and so don't contribute massively to increasing/decreasing the energy of a structure. Furthermore, nitro groups are among the most electron-withdrawing there are, so if pulling electrons is going to have any role, it'll show some of its strongest effects with nitro groups.
To prove this answer comes to an end, we reach our final structure, carbonic acid. First, we notice it is related to carboxylic acids, and its acidic group is the $\ce{-COOH}$ in the structure. If you think about it, the carboxyl group looks like an enol where a $\ce{=CH2}$ group is replaced by an oxygen atom. Thus the analysis is similar to the enol. However, there is an increase electron-withdrawing effect due to induction (oxygen being more electronegative than carbon) and mesomerism (in the canonical structures, the negative formal charges are now on an oxygen atom, rather than a carbon atom as was the case for the enol). The instability of carbonic acid interferes, but leaving that aside, its dissociation would be something like:
$$\ce{H2CO3_{(aq)} <=> H+_{(aq)} + HCO3^{-}_{(aq)}}\ \ \ \ \ K_\mathrm a=2.5\times 10^{-4}$$
Comparing everything, the acidity increases in the order:
Simple alcohols < Simple enols and phenols < Carbonic acid and simple carboxylic acids < Picric acid
How would you know carbonic acid is a weaker acid than picric acid? Once again, it's not something you can deduce by pure logic; it requires some benchmarking and comparison with previous data and knowledge. The analysis above would have been helpful, but has its limits. In Chemistry, it is often far easier to explain than predict. Whether you should have been expected to know the answer to this question, I'm not sure. It would depend on how your chemistry course is conducted.

So, in general can I make an assumption that a strong withdrawing group, like SO3H or NO2, present at ortho or para positions of phenol is more acidic than carbonic acid? I ask this because I'm studying for a competitive exam and questions like this are sometimes asked.

If there is a $\ce{-SO3H}$ group, that itself will make the molecule a strong acid. A single $\ce{-NO2}$ group is not enough to make phenol more acidic than carbonic acid, and 2,4-dinitrophenol is also slightly weaker than carbonic acid. It takes three nitro groups to beat carbonic acid. Unfortunately when comparing sufficiently different molecules, there's no easy way to find the answer. Perhaps taking some time to acquaint yourself with the pKa table I linked here will help.


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Why is no bromine liberated in the reaction of potassium bromide and concentrated phosphoric acid?
$\ce{KBr}$ reacts with concentrated $\ce{H3PO4}$ to give $\ce{HBr}$ and ($\ce{KH2PO4}$ or $\ce{K3PO4}$) (not sure which one, if someone knows it, please tell).
Why isn't bromine gas liberated?

Think about the oxidation state of bromine.

@J.LS So it can't change from -1 to 0?

That would require a pretty strong oxidizing agent. Phosphoric acid is an oxidizing agent, but not strong enough. You could probably produce significant amounts of bromine if you used sulfuric acid.

Also, bromine exists as a liquid at room temperature.

@J.LS That explains. Thanks. How did you judge whether it will or won't?

If you are just after generating bromine, you can pass chlorine gas through a bed of KBr. This instantly precipitates pure bromine leaving KCl. The heat of the reaction vaporizes the bromine, so the reaction is best performed in a flask fitted with a condenser to liquify the bromine vapor.

Firstly, $\ce{H3PO4}$ is not a strong enough oxidizing agent to remove the electrons from the $\ce{Br-}$ ions in order for them to then form $\ce{Br2}$, which exists as a liquid at room temperature. Assuming we are working at room temperature, using a strong enough oxidizing agent such as $\ce{H2SO4}$ or a more reactive halogen like $\ce{Cl2_{(g)}}$ or $\ce{F2_{(g)}}$ would be enough to oxidize the $\ce{Br-}$ to $\ce{Br2_{(l)}}$.
If we desired to evolve $\ce{Br2_{(g)}}$, we would have to heat $\ce{Br2_{(l)}}$ to its boiling point of $58.8\mathrm{^oC}$ and then continue to supply it with enough heat to vaporize it. As docscience stated, however, some oxidizing reactions might be exothermic enough to vaporize $\ce{Br2_{(l)}}$ itself. In solution this would depend on $[\ce{HBr}]$ as to whether or not the heat evolved is enough to vaporize the $\ce{Br2_{(l)}}$, but in the solid form $\ce{Br2_{(l)}}$ is vaporized instantly in the extremely exothermic reaction of $\ce{KBr}$ with either $\ce{F2_{(g)}}$ or $\ce{Cl2_{(g)}}$.


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Protonation of ketone and ester
If there is a ketone and ester functional group in the same compound and I want to protonate with $\ce{HCl}$, which one will protonate first? Is it the carbonyl group in the ketone or the carbonyl group in the ester and why? Is it related to the $\mathrm{p}K_\mathrm{a}$ of the functional group?

The $\mathrm{p}K_\mathrm{a}$ values of protonated esters, protonated ketones and hydrochloric acid are close to each other (-7 to -6 according to  http://www2.onu.edu/~b-myers/organic/2511_Files/pKa%20table.pdf), so $\ce{HCl}$ will protonate both groups, perhaps with a slight preference for the ketone.

As was mentioned in the comments, with reference to the Myers $\mathrm{p}K_{\mathrm{a}}$ table, the $\mathrm{p}K_{\mathrm{a}}$ values of a protonated ester and a protonated ketone are largely similar (in the region of $-7$ to $-6$).
Therefore, it will likely depend on external factors, which group is protonated first. Are there electron withdrawing or electron donating groups in the vicinity of either? Can the proton on either maybe intramolecularly hydrogen bond to something?
All these effects will also slightly modify the $\mathrm{p}K_{\mathrm{a}}$ of the functional group you’re observing. The one with the higher $\mathrm{p}K_{\mathrm{a}}$ will be protonated first. Thus, it is impossible to give a general answer here, it depends on which molecule you have and which groups are where.
Side fact: $\ce{HCl}$ has almost the same $\mathrm{p}K_{\mathrm{a}}$, so the protonation will be very close to equilibrated between all three components.

Has nothing to do with pKA tables or external factors. pKa values are determined for a singular compound acting as a Bronsted acid.  When two similar species compete you can have behavior divergent from simple pKa analysis.
If you protonate the sp2-lone pair of a ketone, you are protonating a lone pair close to the nucleus which is less basic. No further stabilization occurs.
If you protonate the sp2-lone pair of an ester carbonyl oxygen, the conjugate acid is stabilized by resonance.  The resulting positive charge can be drawn on both the carbonyl oxygen and the alkoxy oxygen, which is a stabilizing factor.

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


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Manuals for a Cahn TG-171 Thermogravimetric Analyzer
Recently a Cahn TG-171 Thermogravimetric Analyzer was donated to my school. We did not get a service or user's manual with it, and some of it still needs to be put back together.
I have tried to contact Thermo Fisher who bought Cahn Instruments, but they no longer support the TG-171. Any sort of information on the setup of this instrument would be appreciated.
Update:
In particular, one of the main questions I now have is "what must I do to use the balance?" Currently there are eccentric screws (as shown in the photo) preventing the balance beam from moving so it is unable to register mass change. Do I loosen both sets and use a counterweight of a calibrated mass? Is there a common procedure for setting up balances of this type?

I asked about this on meta, but I didn't think I got a conclusive answer about whether it was on topic. http://meta.chemistry.stackexchange.com/q/1682/4813

although not really an answer, have you tried contacting a local university or possibly even the donor? It's possible that someone who works with these instruments (maybe not this model) maybe able to work out the details of your particular model.

Ask the material sciences department in Leuven, who have this instrument.


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Ionic bond and ionic radii in FeS

Which of the following statements is true regarding an ionic bond in $\ce{FeS}$?

Iron atoms gain electrons and sulfur atoms lose electrons.
Iron and sulfur ions bond by electrostatic attraction.
The ionic radius of an iron ion is equal to its atomic radius.
Breaking an ionic bond between $\ce{Fe^{2+}}$ and $\ce{S^2-}$ releases energy.

My attempt:
I know metals lose electrons. I thought the answer was iron and sulfur ions bond by electrostatic attraction, but a classmate and I have a disagreement, so now I am not sure. I don't think the ionic radius of an iron (II) ion is equal to its atomic radius.
I think it is either 2 or 4.

What is the disagreement about?

Iron and sulfur ions bond by electrostatic attraction.

This is the right answer. I see a hard time maneuvering around it.

Breaking an ionic bond between $\ce{Fe^2+}$ and $\ce{S^2-}$ releases energy.

Around the basic definition of forces and bonds, there are usually two big beliefs. One is a big misconception, and the other one is right.
The misconception: Some students fall in the trap of believing that bonds are like boxes of energy. So, breaking these boxes of energy should release energy, right? No! In reality, ionic bonds are nothing but forces. This is like imagining that when you pull a box, you're applying or somehow putting some kinda a box.
The reality: Ionic bonds are just forces that are a result of attraction between two species (such as two atoms). Why do these happen in the first place? Because for instance in \ce{NaCl} the sodium atom really likes to lose an electron and chlorine wants to earn one. This love of giving and/or taking an electron is because of reaching the stability. Stability in this case means lesser energy. So, ionic bonds happen because atoms want to have lesser energy.
Finally, what would breaking this bond result in? If the atoms bonded together to lower their energies, then what would removing the bond result in?

I don't think the ionic radius of an iron ion is equal to its atomic radius.

You're right! Think of the sodium chloride example again. The new sodium cation used to have 11 protons and 11 electrons. Now, after the lattice builds up, sodium crudely has only 10 electrons, with the same 11 protons. The electrons are outnumbered! They can't overpower protons anymore. So, what will happen to the distance electrons were able to travel to? (The distance they could "travel" further from the nucleus was called the atomic radius, and it's the ionic radius now)
So, is the atomic radius of iron equal to its ionic radius?

Rather pedantic and irrelevant, but there's no clear distinction between an ionic bond and a covalent bond. There are no "pure" ionic bonds.


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Does desalination work?
Does salt vaporise? If I boil salt-water and leave it there until half of the water vaporises, will it be twice as salty as in the beginning? Or exactly the same? Or somewhere in between?

Loss of salt will be very little - thats the point of distillation.

Are you saying that in the end the water will be almost twice as salty but not quite?

Think about salt lakes - a lot of water has evaporated, but the salt remains, so becomes more saline

Yes, you get me right.

In chemistry there is "salt" in particular and there are salts in general.  If by salt you mean "sodium chloride", then what everybody is saying is right.  But if you mean in general, there are salts for which vaporization can be significant.  E.g. ammonium bicarbonate or ammonium formate.

Pure salts do melt and then boil, if you heat them enough (e.g. NaCl has a boiling point of 1413°C) although that probably isn't what you were thinking of.

I think the main effect is that the bursting bubbles of the boiling salt water produce an aerosol of salt water droplets. Some salt will thus inevitably end up in the air, but this is not vaporization in the chemical sense. Nevertheless the remaining solution's salt concentration will be a small fraction lower than expected.

This is a classic science experiment. Where saltwater (using $\ce{NaCl}$) is heated, as it does, the water evaporates and most, if not all, of the salt remains - increasing the concentration of salt.
Evaporating half the water would approximately double the concentration of $\ce{NaCl}$.
Eventually, the salt concentration becomes supersaturated and crystallises. The reason is due to:

As the water evaporates less and less water molecules are present to keep the salt particles apart. The salt therefore recrystalises and can be collected.

Source
This process is also used in salt evaporation ponds, where

seawater or brine is fed into large ponds and water is drawn out through natural evaporation which allows the salt to be subsequently harvested.

As Curt F mentioned in the comments, different salts and mixtures of salts can have different outcomes in terms of vaporisation.

Certainly the vast majority of salt won't vaporize.  On the other hand, saturated salt water solutions tend to have an odor; is that caused by additives in salt, or do some H+ ions combine with Cl- and leave solution as HCl gas without being decomposed back into H+ and Cl- ?

@supercat that would be a good question unto itself

Note that pure NaCl does melt and evaporate, though at much higher temperatures (801ºC MP / 1413ºC BP)

Strictly speaking, $\ce{NaCl}$ does evaporate when brine is boiled. It even evaporates at room temperature on its own, just like any other substance at a temperature above absolute zero. That said, of course it evaporates very little when so far below its boiling point. However, I believe it is quite likely for some ions in very small clusters to be carried into the gas phase, solvated by a few water molecules, so I would expect some salt loss from evaporation of a brine solution.

Note Kevin's comment on another answer. The key observation here is that it is not a question of whether $\ce{NaCl}$ vaporises but whether there is a temperature range in which water vaporises and $\ce{NaCl}$ doesn't, the answer to which question is yes. The wider the gap between the solvent and the solute, the easier and more completely you can separate them.
$\ce{NaCl}$ will certainly vaporise but it also has a liquid phase, and in this phase is used to cool nuclear reactors. Going by the 600 degrees between Kevin's figures for melting point and boiling point it doesn't tend to sublimate.
Everything vaporises at any temperature above 4K. Atmospheric agitation is knocking bits off my keyboard even as I type, but for the most part they're big heavy lumps so they fall back down and either re-bond or sit there serving as ablative armour. Molecules small and light enough to float away mostly did so fairly promptly. Think new car smell. These are called volatiles, which simply means they are small and light enough to float away with modest stimulation.
(Why 4K? Helium liquefies. Drop the temperature even further and you get a Bose condensate which is super interesting stuff with directly observable quantum behaviour but doesn't have much to do with chemistry as we know it.)
So where was I? Oh yes, melting point and boiling point are inflection points in a continuum. They are temperatures which bound dominant behaviors.

Salt does not vaporise (but does melts and vaporise at high temperature) at room temperature.
Sea water which is (Salt+Water) is separated by evaporation process. Salt remains and water is vaporised. You can see that in sea side salt pans.

If you boil the salty water hard enough, droplets will fly out of it as bubbles rise and burst. These droplets will carry away some salt. Boil a pan of salt water on your stove, and you'll see a bit of white dust or crust appear around the pan.
If you just leave salt water out to evaporate without that kind of agitation, essentially all the salt will remain in the concentrated solution. (Yes, salt has a finite vapor pressure even at room temperature, so some will evaporate. No, you won't be able to measure it.)


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What is the molecularity of a reversible reaction?
I know that the definition of molecularity of a reaction is number of species reacting in an elementary step. But considering the theory of microscopic reversibility for elementary reactions, each reaction can be assumed to be reversible. In that case, if $$\ce{A <=> B + C}$$ then what is the molecularity of the reaction? Is it 1 or 2?

For a reversible reaction, the appropriate arrows are double arrows $\ce{<-->}$ (feature not implemented in MathJax's mhchem) or equilibrium arrows $\ce{<=>}$. The use of resonance indicating arrows $\ce{<->}$ is wrong. However, an elementary reaction can be reversed, but can practically not be in equilibrium - the system can be in one state or the other only.

For the reaction $$\ce {A <=> B + C}$$
its molecularity is unimolecular. If we were given $$\ce{B + C <=> A}$$
Its molecularity would be bimolecular. Thus we see molecularity is dependent on the number of reactant particles, even if it's in equilibrium.

Just to add onto the other answer.
Since the equilibrium constant also changes when we reverse a reaction, so does the molecularity of the reaction.
So many properties of reversible reactions are defined in the way we write down the reaction.


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Are glycans, such as glucose, considered alicyclic compounds?
That's my question. If true, it would make my job a lot easier. I will appreciate any feedback on the matter.

No - oxygen in the ring, see https://en.wikipedia.org/wiki/Alicyclic_compound

Mithoron you should make your comment into an answer...

Can you please tell us which job will be easier? We might be able to help you better.

Seems I underestimated this problem.
First passage from Wikipedia is crucial:

An alicyclic compound is an organic compound that is both aliphatic and cyclic.

And second is wrong:

They contain one or more all-carbon rings(...)

(Source)
Glucose is carbohydrate, in its circular form which is found in glycans (also glucose isn't a glycan, but glucose residue is found in many of them) it has oxygen atom in the ring - according to second passage from Wikipedia it therefore shouldn't be alicyclic, but one can easily find mention about alicyclic ethers for ex. here, with oxygen in ring. So I'm changing my answer to:
Yes, they are.

Glucose has only one cyclic form, what you called the aldehydic form. It cannot have a hemiacetalic form since it doesn't have a ketone in the open chain. Fructose does have a ketone (but no aldehyde) and makes a hemiacetalic form. For an aldohexose such as glucose, the C-1 aldehyde in the open-chain form of glucose reacts with the C-5 hydroxyl group to form an intramolecular hemiacetal. The resulting cyclic hemiacetal, a six-membered ring, is called pyranose because of its similarity to pyran. Similarly, a ketone can react with an alcohol to form a hemiketal.

The key phrase from wikipedia is "[Alicyclic compounds] contain one or more all-carbon rings".  Neither glucose nor fructose qualifies because their ring forms contain non-carbon atoms.  Would you mind editing the answer to reflect this?  I already upvoted you.

Thank you all for your answers. I knew from the beginning that this wasn't an easy question, I mean I knew there would be some debate. I am going to discuss this with my colleagues, and I will mention your comments as well. I still haven't made my mind on the matter. It's that oxygen atom on the ring that doesn't convince me about carbohydrates being alicylic.


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Activated complex theory vs. consecutive reactions
Activated complex theory, tells us that due to the collision between the molecules of the reactants, they form a transition specie before the product is formed, which is called active complex. On the other hand we have consecutive reactions on which is also formed a intermediate product before forming the actual product we're interested on. My question is :
Where is the difference between the activated complex and the intermediate product, since they are both formed before  the actual product ?

The key difference is that transition states occur at a maximum of the potential energy curve for the reaction whereas intermediates occur at a local minimum. Take this example of the reaction profile for an $\ce{S_{N}1}$ reaction:

You will see that the products and reactants occur at minima on the curve and the intermediate also occurs at a minimum, albeit a higher energy one. In between the minima are located the maxima where you find the transition states.  Unless the activation energy is zero (see this question for rare examples) there will always be a transition state located between any two minima.
Transition states are usually represented using dashed bonds to show bonds in the process of being broken or formed as opposed to being fully formed as in intermediates or products. Additionally the Hammond postulate says that the transition state will most resemble the stable species closest to it in energy, in this case the carbocation, and this can be used to help predict the structures of transition states.
Transition states are very short lived because they immediately 'roll downhill' on the potenetial energy curve to reach an intermediate or product. By contrast some intermediates are actually quite stable and can be isolated.

Thanks, so I assume there isn't any way we can know if some molecule is a transition state or a intermediate without seeing the diagram or having any energy value of them ?

@Ndrina Limani Edited my answer to give more information.

authors: Нияз Худайбергенов


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Theoretical Calculation of Lattice Parameter from Space Group
If I am given a compound which belongs to, say the cubic $\text{Ia}\bar 3\text{d}$ space group, is there a known way to calculate the lattice parameter of the compound? In general, can lattice parameters be calculate from knowing what space group the compound belongs to?

On the other hand, lattice parameters and crystal system can be known independent of space group and atoms by indexing a diffraction pattern.

No - the spacegroup only gives information about symmetry, not the lattice parameters.
The lattice parameters depend on what is in your unit cell, ie your compound - but even if you know your compound, this is not straightforward, as you don't know its arrangement. And if you do, you should already know your lattice parameters ;)
That being said: if you have additional information - e.g. the radius of a protein you crystallise - you might be able to get a rough estimate of what you are expecting as your minimal unit cell parameters. But this does not take into account any stoichiometry in the crystal, which is impossible to know in advance, so multiples thereof are just as likely.

Indeed, in full support of @Gerhard's answer, I will point out that there are a number of elemental metals that crystallize in the fcc structure, but they all have different lattice parameters. The symmetry of the interatomic potentials is distinct from the length scale of the potential.

So, the Wyckhoff positions don't determine the coordinates of where the atoms are in a compound that fits in a space group with those Wyckhoff positions?

No - the Wyckoff positions are still only part of the symmetry dictated by the spacegroup.

@Sam:I suspect that you are referring to the asymmetric unit: if you know what is in it, you could calculate the lattice parameters. But as I mentioned above, you do not know this in advance - otherwise you would already have determined the structure.


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Modified DNA bases in Pymol and Coot
When using Pymol or Coot molecular visualization software to solve x-ray crystallography structures, how can I mutate a cytosine to a 5-methylcytosine?

Pymol is a tool for visualization. You cannot modify the molecules with it (how should it know where exactly to place the methyl group? ). You need to modify your molecules source file by hand or with another tool.

I got the answer on a Pymol listserve:
Yes, it can be done in Pymol.
Choose the residue you want to mutate from cytosine to 5-methylcytosine.
Use Action -  Hydrogens - Add function to add hydrogens.
Change mouse selection to choose atoms.
Click on the hydrogen you want to replace.
Click builder (there is more work to do if you're on a Mac, see below) to open builder window.
Click C
If you're on a Mac, you won't have Builder functionality unless you change your MacPyMOL.app file name to PyMOLX11Hybrid.app


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In the Haworth projections of D and L glucose, is the stereochemistry at every carbon except the anomeric carbon just inverted?
I'm having a bit of trouble grasping sugar stereochemistry.
Using glucose as an example, if you cyclize D and L glucose, is the stereochemistry at every carbon just inverted from the other form?
Like for example, on carbon 2 for D glucose, the hydroxyl is down. On carbon 2 for L glucose, the hydroxyl would be up.
Is this trend true for every hydroxyl or $\ce{CH2OH}$ group except the one on the anomeric carbon?

Glucose is the hexose that projects its hydroxyl groups at positions C2, C3 and C4 as well as its CH2OH off C6 in the equatorial direction (in its low energy conformer that is). In the β and α configurations, its anomeric carbon C1 projects the hydroxyl in the equatorial and axial direction, respectively.
The D and L forms are mirror images in which all anomeric carbons have the opposite configuration between the two forms as:

Here are the equivalent structures using the Haworth projection.


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How did Millikan measure the mass of these oil droplets?
I was reading the R.A. Millikan's oil drop experiment and the following part confused me:

By measuring the rate of fall, Millikan was able to measure the fall of the oil droplets (which were dropped through a tiny hole in the upper electrical condenser).

In Newtonian physics, the rate of fall (or velocity) of a freely falling body does not depend on the mass of the object. So, how did Millikan measure the mass of these oil droplets? Does anyone know the equations he used?

Have you checked Wikipedia?
http://en.wikipedia.org/wiki/Oil_drop_experiment

You can find out here: https://www.youtube.com/watch?v=ijHKu6iXiRk

Have a look at http://www.aip.org/history/gap/PDF/millikan.pdf

By measuring the rate of fall, Millikan was able to measure the fall of the oil droplets (which were dropped through a tiny hole in the upper electrical condenser)

The rate of fall here means terminal velocity of oil, $v_\mathrm{t}.$ Oil was made to drop at constant rate so no acceleration. Drag force on oil drop is:
$$F_\mathrm{d} = 6\pi r\eta v_\mathrm{t},\tag{1}$$
where $r$ is the radius of oil drop and $\eta$ is the viscosity of air.
Assume oil droplet is spherical. Apparent weight of oil droplet in air is the true weight minus the upthrust (which equals the weight of air displaced by the oil drop). So, it can be written as:
$$w = \frac{4\pi r^3(\rho - \rho_\mathrm{air})g}{3}.\tag{2}$$
No acceleration is taking place so the total force acting on it must be zero and thus the two forces must cancel each other (i.e $F_\mathrm{d} = w).$ Solving for $r$ gives
$$r = \sqrt{\frac{9\eta v_\mathrm{t}}{2g(\rho - \rho_\mathrm{air})}}.\tag{3}$$
The density of oil is known so by known so by putting $r,$ you can calculate the mass of oil droplet as
$$m = \rho V = \rho\frac{4\pi r^3}{3}.\tag{4}$$


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Do all the precipitates undergo an equilibrium with its respective ions, why can't they dissociate fully?
I’m really confused about this. If we consider $\ce{Ca3(PO4)2}$, it can be seen it undergoes an equilibrium. But why it’s an equilibrium?
This is the equilibrium it undergoes $$\ce{Ca3(PO4)2_{(s)}<=> 3Ca^2+_{(aq)} + 2 PO4_{(aq)}^3-}$$
By the way I’m having a doubt of why some are using backward arrow for a dissociation of a particular precipitate while some are using a solid forward arrow for the same reaction. For an example let’s take $\ce{AgCl(s)}$
Now this is a place where a backward arrow is used,
$$\ce{AgCl_{(s)}<=> Ag+_{(aq)} + Cl_{(aq)}-}$$
But I have seen in some books where there is a solid forward arrow instead of back-ward arrow for the dissociation of $\ce{AgCl(s)}$.

In the two examples you mentioned, we have an equilibrium between the solid and the small quantity of the solid that is soluble in water and completely dissociated:
$$\ce{Ca3(PO4)2_{(s)}<=> 3Ca^2+_{(aq)}  + 2 PO4^{3-}_{(aq)} }$$
$$\ce{AgCl_{(s)}<=> Ag^+(aq)_{(aq)}  + Cl^{-}_{(aq)} }$$
I think it's wrong to write a solid forward reaction.
It's an heterogeneous equilibrium between the solid state (the precipitate) and the aqueous state (ions constituents of the precipitate). The constant of this equilibrium is the solubility product $K_{\mathrm{sp}}$ of the precipitate.

@ Youmen Then if some one wrote  Ag+  +  Cl^-  --->(solid forward arrow)AgCl ,  is it correct? Could you explain it also?

@On the way to success Please see my answer (I have added  some explanations).

@ Youmen  How can you  say constant of Ag+(aq)+Cl−(aq)<---> (AgCl(s)  is the inverse of solubility product of  AgCl(s)<-->.Ag+(aq)+Cl−(aq)

@ On the way to success It's the definition of Ksp =[Ag+][Cl-]. Ok?

@ Youmen Yes I do know Ksp =[Ag+][Cl-] but I'm confused with about the thing you said " constant of Ag+(aq)+Cl−(aq)<---> (AgCl(s) is the inverse of solubility product of AgCl(s)<-->.Ag+(aq)+Cl−(aq)"

Because it's the reverse reaction.


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When is Serpeck's process preferred over Bayer's process?
Both these processes are used in concentration of bauxite.
They use different reagents.
Which is more commonly used (wikipedia says Bayer's) and why?
When is one preferred over the other?
Which is more efficient and why?

From this source

Baeyer's process: This method is mainly applied when ferric oxide is present as chief impurity.
Serpeck's process: This process is used when silica is present in considerable amount of bauxide ore.

I couldn't find any information on the efficiency , but I suspect both are efficient in their own way. We cannot compare these two processes as they are used for two different impurities.

During extraction of Al from Bauxite containing more than 10% of silica as impurity, Baeyer's process will be uneconomical due to generation of insoluble sodium aluminium silicate. In such cases Serpeck's process can be preferred in which Silicon can be separated as volatile substance at higher temperature.

Baeyer's process is used to leach Red Bauxite, which has Fe2O3 as the chief impurity. Serpeck's process, on the other hand, is used to leach White Bauxite, which has Silica as main impurity. Both are efficient for their respective ores.
But Baeyer's process is more frequently used, as Bauxite, more often than not, is also a source of Gallium, which can be extracted from Baeyer's process.
I guess this is why Baeyer's process is more frequently used.
Source of Information : Wikipedia