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Is there a definition of rotatable bond?
Some intramolecular bonds are rotatable in the sense that the torsion angle around this bond is flexible (for example, $\psi$ and $\phi$ angles in a protein backbone). Others are essentially fixed (for example, the $\omega$ angle in the peptide bond in a protein backbone).
Is there an operational definition, or a published study where a general definition is given, which would allow me to recognize in a generic molecule (or at least an organic molecule) whether a given bond is rotatable or not?
If there is no general theory for this, what are the most common indicators of whether a bond can or cannot be rotated? Are there some rules of thumb?

If there is no fixing geometry like, e.g., in cyclohexane rings, then every single bond should be rotatable.

@PH13 The $\omega$ bond in a protein backbone has no fixing backbone (no rings), yet it is fixed at $\pm 180º$.

it has a fixing $\pi$-system

there is much resonance in the peptide bond between the CO's $\pi$-system and the N's lone pair, which fixes geometry http://webhost.bridgew.edu/fgorga/proteins/resonance.htm

@PH13 Oh, I don't understand your definition of fixing geometry. Maybe you can expand into an answer?

Is there an operational definition, or a published study where a
general definition is given, which would allow me to recognize in a
generic molecule (or at least an organic molecule) whether a given
bond is rotatable or not?

It depends on your purpose. Why do you want to identify some bonds and call the rotatable? For instance, PubChem, one of the biggest databases of molecules, defines rotatable bond as follows:

Rotatable bond is defined as any single non-ring bond, bounded to
nonterminal heavy (i.e., non-hydrogen) atom.

To some extent all this boils down to the following IUPAC definition of free rotation:

free rotation (hindered rotation, restricted rotation)
In a stereochemical context the rotation about a bond is called 'free'
when the rotational barrier is so low that different conformations are
not perceptible as different chemical species on the time scale of the
experiment. The inhibition of rotation of groups about a bond due to
the presence of a sufficiently large rotational barrier to make the
phenomenon observable on the time scale of the experiment is termed
hindered rotation or restricted rotation.

At usual lab conditions a rotation around a single non-ring bond, bounded to nonterminal heavy atom will be indeed almost free, thus, PubChem definition can be used as a very simple rule-of-thumb to identify such bonds. But in general it depends on conditions. And purposes.

With respect to protein backbones being rotatable or not: I'm not a speacialist in this area, but Wikipedia article on peptide bond claims that (emphasis mine)

Significant delocalisation of the lone pair of electrons on the
nitrogen atom gives the group a partial double bond character. The
partial double bond renders the amide group planar, occurring in
either the cis or trans isomers.

So, this partial double bond character might contribute in protein backbones being not quite rotatable. Besides, the Wikipedia article also claims that (emphasis again mine):

In the unfolded state of proteins, the peptide groups are free to
isomerize and adopt both isomers; however, in the folded state, only a
single isomer is adopted at each position (with rare exceptions).

So, at least in the unfolded state protein backbones are rotatable. The fact that only two configurations (cis and trans) are possible is irrelevant. The notion of rotatable bond does not imply having few specific possible configurations.

Under this definition, the $\omega$ angles of protein backbones are rotatable.

@becko, I expanded the answer (had troubles with an Internet connection). It it all about what do you want to achieve with that definition.

The IUPAC Gold Book states at least somthing to free rotation:

free rotation (hindered rotation, restricted rotation)
In a stereochemical context the rotation about a bond is called 'free' when the rotational barrier is so low that different conformations are not perceptible as different chemical species on the time scale of the experiment. The inhibition of rotation of groups about a bond due to the presence of a sufficiently large rotational barrier to make the phenomenon observable on the time scale of the experiment is termed hindered rotation or restricted rotation.

What they don't tell is what causes a sufficiently large rotational barrier. Those are, e.g., ring strains or resonance stabilized bonds.

For every bond, there is an energy barrier required to rotate around the bond. Would the thermal energy (at room temperature) be enough to allow rotation, it can be considered rotatable. For the reasonable rate constant I would consider $s ^{-1}$ or faster.
This means that single $\ce{C^{sp3}-C^{sp3}}$ bonds are rotatable, double bonds rarely, in other cases use common sense or quantum chemistry.
Medicinal chemists use more straightforward definition, http://pubs.acs.org/doi/abs/10.1021/jm020017n.

What references on quantum chemistry (maybe book chapters or papers) are relevant to this? Can you point me to some examples?

Have a look here: http://pac.iupac.org/publications/pac/36/1/0109/pdf/

@ssavec Could you please at least give a short summary or quote from the paper? Not everyone has access to ACS journal articles (e.g., from home, etc.).


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How to measure resistance of a gel electrolyte using a multimeter
How can I measure the resistance of a gel electrolyte using a multimeter? I know the formula to measure the conductivity of a gel electrolyte is k = (1/R) * (d/a)
where k is the conductivity (S/cm), R is the resistance (of the electrolyte), d is the distance between the two electrodes, and a is the effective area of the electrodes.
I wanted to measure the conductivity of the gel electrolyte with just a multimeter. Since this is a gel electrolyte, I was thinking to substitute the thickness of the gel electrolyte as d and substitute the area of the gel electrolyte as a. (I know the area (a) has to be that of the electrodes, but I want to measure the conductivity of the gel electrolyte without the electrodes). So now all that remains is the resistance of the gel electrolyte. To do this can I just measure its resistance using a multimeter?
(I understand that you have to use two similar electrodes sandwiched between the electrolyte in order to measure the conductivity of it, but is there a way around?)
Thank you.

I'm voting to close this question as off-topic because it doesn't really seem to be about chemistry. It seems to be more suited to physics.SE as it's just about measuring electrical properties.

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Do any substances work like consumers of hair moisturizer believe it to work?
Yesterday, my girlfriend remarked on how pleased she was with the effects of the hair moisturizer she has been using: "I started using it just a month ago, and look, it's healed all the damage." I complimented her, then said "...but hair is dead, so it couldn't have helped heal it." She replied, "Well it fills in the cracks". I accepted her answer; lots of hard-surface sealers work that way.
However, hair moisturizer seems odd.

Unlike hair moisturizer, the effects of hard-surface sealers typically don't need dozens of reapplications to obtain.
The exceptional sealers that do need several coats, typically do not need to cure before reapplication (although I suspect that some do). However, if a person were to apply coat after coat of hair moisturizer, they'd just get greasy hair.

I suspect that the porousness of hair necessitates the use of a "wetter" solvent than the one used in most hard-surface sealers, which protracts the curing time.
I also wonder whether the porousness of hair protracts the time it takes for the moisturizer to penetrate its fractures, and thus protracts the time the user ought to wait before reapplying: assuming the presence of the solvent along the path of penetration slows the penetration of the moisturizer then, the coats applied immediately after the first coat would penetrate the hair more slowly, and so they would cure closer to its surface, and would thus produce results that differed from the ones my girlfriend was so pleased with.
Are there substances that exhibit the properties that hair moisturizer seems to exhibit. At the molecular level, what explains why these substances exhibit those properties?

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text:
Can you predict the resistance of conjugated polymers as a function of degree of polymerization (DP)?
Electrical properties of very simple systems can be calculated using NEGF. Is there a way to calculate the resistance (and other electrical properties) of a conjugated polymer, as a function of degree of polymerization (DP)?

Short answer: No.
First off, the degree of polymerization does not directly relate to the charge transport properties of conducting polymers (more below). You can find highly conductive oligomers.
Secondly, in my opinion, NEGF is not a suitable method to understand the charge transport in polymers. It works for nanoscale devices and molecular wires, but most polymer devices are on the ~100 nm to micron scale.
That's not to say that different oligomer and polymer lengths don't have different transport properties. Absolutely, different molecular weight polymers have different charge mobility, etc. The effects are usually due to a number of hard to characterize factors:

Concentration of defects and traps
Morphology of the film (e.g., higher MW might yield smoother or more ordered films)
Polydispersity (e.g., broader MW distribution is often problematic)
Effective conjugation length
Packing

etc.
The electronic structure of oligomers and conjugated polymers can be simulated fairly well using first-principals methods, including DFT. This can often predict components like the energy levels, band gaps, band dispersion, etc. for crystals and parameters for amorphous films such as the Marcus reorganization energy which is often related to the charge mobility in many of these materials.
What you usually find is that as a function of chain length, charges become more delocalized and the reorganization energy drops (increasing mobility). In reality, this trend should saturate, although most DFT methods incorrectly delocalize charges completely.
Also, most devices with conjugated polymers are amorphous or polycrystalline, making device performance predictions difficult. Transport is not usually band-like, but some form of variable-range hopping between chains or sites within a chain.

Longer answer is "let's talk, this is what my group does for research."

Is DMRG suitable for taking "imperfections" into account?

I'm not sure you're following my answer. You can pick whatever electronic structure method you want, but that doesn't directly give you transport properties. You can use electronic structure methods to compute, say, orbital coupling if you have the solid state structure and the Marcus reorganization energy. Those might give you some indication of the ideal charge mobility, but those properties won't depend heavily on the polymer length and don't consider real problems like charge traps, grains, etc.


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How long would it take to dissolve someone using fluoroantimonic acid?
After writing an answer on worldbuilding.SE, I came here to ask a question. How long would it take to dissolve a human body using fluoroantimonic acid?
I understand that since the human body isn't homogenous, it's a bit tough to get the exact calculation, but lets say we create a mannequin using the material in the human body that fluoroantimonic acid takes the longest time to dissolve - how long would it take?

As for the question, my gut answer is no. Acids are not magic dissolving fluids. You'll certainly hydrolyze the living daylights out of esters and related bonds, of course.

@etherealflux we already know it takes 2 days (roughly) to dissolve a body in a vat of sulfuric acid. That tells me that it's at least possible to do the task at hand with an acid that basically dissolves anything not teflon.

@etherealflux Funnily enough, an acid weaker than fluoroantimonic acid was named "magic acid", because of its surprising ability to protonate even alkanes, meaning it could dissolve a candle.

Welp, that'll do the trick. I remember reading about that once - stabilizing carbocations is insane.

Related: http://chemistry.stackexchange.com/questions/34745/is-there-any-chemical-mixture-which-could-liquidize-a-person-in-a-couple-minutes

Unfortunately for the asker of the worldbuilding.se question, the result will not be all liquid -- there will be a lot of off-gassing gaseous.  This acid will dissolve organic compounds to produce lots of hydrogen, methane, steam, and/or HF.

@CurtF. and that is just fine, the only requirement for that other question was that the resulting mess be liquidy - it makes no mention of the amount of liquid required. The bigger question we're trying to get answered is how long it will take for the entire thing to happen.

Closed in accordance with the meta discussion.


text:
Does the delocalization of p-orbitals ensure conductivity?
For example, polyacetylene is a conjugated system, i.e. with de-localized p-orbitals, but accroding to Chiang et. al. (1977) polyacetylene is semiconducting, and becomes conductive only when doped with Cl/Br/I/AsF5.
Isn't the conjugation alone enough to allow conductivity?
(source)
UPDATE
One reason they mention in the paper is:

In the idealized situation of a uniform chain, the resulting
  conduction band would give rise to metallic behavior. However, such a
  system is unstable with respect to bond alternation, which causes the
  formation of an energy gap in the electronic spectrum.

But if this is true then (1) how the halogens improve the situation, and (2) why isn't there a similar phenomenon in all conjugated systems?

longer chain gives smaller bandgap but still non-zero bandgap

I bet Geoff could give an amazing answer to this if he has the time!

Delocalized $\pi$ orbitals do allow for mobile charges. The catch there is "allow." Just because these systems can have mobile charges does not make them conductive.
Conductivity can be defined as:
$\sigma = n e \mu$
where $e$ is the charge on an electron, $\mu$ is the charge mobility, and $n$ is the number of charges.
In the case of $doped$ conjugated polymers like you indicate above, the halogen vapors or other dopants usually oxidize the polymer and create more charges (i.e., $n$) and the conductivity increases. This does happen with pretty much every conjugated system I've seen: oxidize or reduce the neutral system and you'll have more charge carriers = greater conductivity. (NB: most conjugated organics are easier to oxidize than reduce, so you get + charges (holes) as the charge carrier, rather than electrons.)
That's the quick explanation as to why adding halogens increases the conductivity -- you get more charge carriers.
Incidentally, there are other ways to increase the charge carriers. Chemical or electrochemical doping usually makes a conductive system, but you can also use electrical bias (e.g., a field-effect transistor) or light (i.e., a photovoltaic) which will also increase carrier concentrations.
As to why trans-polyacetylene isn't a metal in the first place?
If we use a simple particle-in-a-box model, the difference in energy levels goes to zero as the box becomes very large. This would suggest trans-polyacetylene should be a metal (i.e., no band gap).
Consider also the Hückel model for polyacetylene. We can build up the nearly-infinite system by solving for ethylene, butadiene, etc.

Notice that the energy gap between levels becomes smaller and smaller. We can plot the HOMO-LUMO gap as a function of chain length:

The gap becomes smaller and smaller!
Wikipedia even gives the general formulas:
$$E_k = \alpha + 2\beta \cos \frac{k\pi}{(n+1)}$$
$$\Delta E = -4\beta \sin \frac{\pi}{2(n+1)}$$
So both particle in a box and Hückel predict no band gap when $n$ gets very large. Theory predicts a metal!
Why doesn't the real polyacetylene behave as a metal. For any one-dimensional system, there will be a Peierls distortion which lowers the total energy by distorting the geometry (long-short bond alternation) and increasing the band gap.1
In short, a truly one-dimensional metal is impossible, but doped polymers are certainly very good electrical conductors. Record films of trans-polyacetlyene have conductivity close to conventional metals.2

1 The long-short bond alternation is the Peierls distortion. It means some electrons are localized (shorter double bonds) and thus lower in energy. This creates a band gap (i.e., lower E for valence band).
2 The film isn't a 1D material, but the trans-polyacetylene chains are each 1D.

Thanks for the amazing explanation! 2 things are still unclear to me: (1) why long-short bond alteration causes a band gap, and (2) why polyacetylene film is a 1D material?

The long-short bond alternation is the Peierls distortion. It means some electrons are localized (shorter double bonds) and thus lower in energy. This creates a band gap (i.e., lower E for valence band). The film isn't a 1D material, but the trans-polyacetylene chains are each 1D.

Great work! I knew this would be smack-dab in the middle of your area of expertise, and polyacetylene is a classic and important system (Nobel worthy, even!). Towards organic metals!

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Beryllium hydride violates the octet rule
Beryllium hydride does not obey the octet rule. Isn't this impossible, unless there are more elements comprising the molecule?

It's really more like a guideline. ;)

Technically, it does not.
$\ce{BeH2}$ forms a network of $\ce{BeH4}$ tetrahedrons sharing corners. Since each hydrogen belongs to two such tetrahedron, the net formula is $\ce{BeH2}$. In this structure, each electron pair belongs to a triple of atoms $\ce{Be-H-Be}$. Such bonding is known as three-centre two-electron bonding, and is fairly common in electron-deficient compounds that would otherwise violate the octet rule. For example, this kind of bonding is essential for boron-hydrogen chemistry.
With that said, unambiguous violations of the octet rule are not unknown. The most obvious and reasonable stable example would be nitric oxide, $\ce{NO}$. However, this example again features strong 'extra' electron delocalization with 'extra' electron shared between the oxygen and nitrogen atoms. The molecule still shows tendency to dimerize at low temperatures. This kind of molecule is more of an exception and not a rule. The exact way such particles are stabilised is highly dependent on the particle.
Of course, if we are talking about transition elements or lanthanides, the situation is more complex, because of the involvement of fairly strongly bonded electrons in d- or f-orbitals. Still, the extra stability of complexes following the equivalent 18-electron rule is clearly observed.


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Why does Mg react vigorously in NaCl solution and less so in water?
When you put $\ce{Mg}$ into water a few $\ce{H2}$ bubbles appear. But when you put $\ce{Mg}$ into a $\ce{NaCl}$ solution there is a vigorous release of $\ce{H2}$, why is this and what reactions are taking place?

My first thought is that the salt is providing nucleation sites for $\ce{H_{2}}$, but I wasn't able to find definitive support with modest searching. I could be totally wrong. Do you have a reference or two that you can share, or is it your observation you are sharing?

It might be wrong but my hypothesis is different from the actual answer. When magnesium hydroxide reacts it forms a protective layer above pure magnesium, as in passived aluminum. When NaCl is added it increases hydroxide solubility (because of activities and so on); as a consequence, makes the reaction faster...

It would still be the same reaction.
\begin{align}
\ce{Mg (s) &-> Mg^2+ + 2 e-} & E^\circ&=\pu{2.36 V}\\
\ce{2 e- + 2H2O (l) &-> H2 (g) + 2OH-} & E^\circ&=\pu{-0.828 V} \times 2
\end{align}
The combined reaction would be spontaneous ($E > 0$), but $\ce{Mg(OH)2 (s)}$ is highly insoluble,
\begin{align}
\ce{Mg(OH)2 (s) &<=> Mg^2+ (aq)  + 2OH- (aq)}, &
K_\mathrm{sp}=7.1 \times 10^{-12},
\end{align}
leaving the solution insufficient ions to conduct the electrons necessary to fast complete the reaction (despite of the escaping hydrogen gas).
$$\ce{Mg (s) + 2H2O (l) -> H2 (g) + Mg(OH)2 (s)}$$
This is like an electrochemical cell with no salt bridge (or bridge with only de-ionized water). Hence a little salt ions would accelerate the reaction forward.
In contrast, calcium solid, having comparable standard reduction potential, will undergo a similar redox reaction in water and needs no help for additional ions because $\ce{Ca(OH)2 (s)}$ is about a million times more soluble ($K_\mathrm{sp}=6.5 \times 10^{-6}$) than magnesium hydroxide, providing sufficient free ions in the solution to drive the vigorous release of hydrogen gas. (There are a couple of YouTube video demos.)

Calcium hydroxide is not a million times more soluble than magnesium hydroxide.  The solubility product is a million times larger, yes, but with three ions forming this is proportional to the cube of the molarity.  The real molar ratio is thus about 100 to 1, but that is still enough to make a big difference in reactivity.

Where in the original question is there mention of calcium? You have not answered the question.

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Does relative amount of lithium and cobalt oxide change in recharged battery?
There is a question on my assignment about stoichiometry. I spent 2 whole days searching and review this topic but still clueless.
Reaction in lithium ion battery
$$
\ce{Li(s) + CoO2(s) → LiCoO2(s)} \, .
$$
When the battery is recharged, electrical energy will reverse the reaction. How do the relatives amounts of lithium and cobalt oxide change as the battery is recharged and how?

About maximum $20~\%$ (as I remember, perhaps more now) of the $\ce{Li}$ goes (as $\ce{Li+}$) out of the structure of $\ce{LiCoO2}$, through the elctrolyte, and into the electrode on the other side of the battery. That electrode consists mainly of graphite, which intercalates the $\ce{Li}$ in it's layers in a quasi-metallic form.
So it is $\ce{Li_{1-x}CoO2 + x~Li}$ after charging. $\ce{LiCoO2}$ has a special structure that does not break down when some $\ce{Li}$ is missing, except if you keep a charged battery at over $\mathrm{50~^\circ C}$ or so.
By the way, the related wikipedia article tells you all this. Two days of researching? ;-)


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Diazo transfer to 1,3-dicarbonyl with tosyl azide

What is the mechanism for the above reaction?

3+2 cycloaddition and retro-cyclo maybe but I'm not sure.

(I removed the fluff so that it's easier for future readers.)

This starting material is not actually an enone; it is a 1,3-dicarbonyl, confusingly drawn as the enol tautomer. 1,3-Dicarbonyls are mildly acidic ($\mathrm pK_\mathrm a \sim 10$) and can be deprotonated by weak bases, such as triethylamine ($\mathrm pK_\mathrm a \sim 9$), which was mentioned in the question. So, now, we have an enolate ion which is nucleophilic.
Tosyl azide, $\ce{TsN3}$, is an odd reagent. It's easy to think that it is acting as a nucleophile of some sort, because the azide ion is nucleophilic, but actually it is an electrophile.

The next part is the removal of the formyl group, which is a bit tricky. Here I am not 100% sure, so take it with a pinch of salt, but I think Mith was on the right track with the retro-(3+2) cycloaddition. This is my guess:

In fact, the cyclisation of the terminal nitrogen onto a carbonyl group has been proposed before, in mechanistic studies of electrophilic azidation by D. A. Evans.1 That makes me feel a bit more confident in my guess!

Anyway, the product above is simply a resonance form of the diazo product you want.

Reference

Evans, D. A.; Britton, T. C.; Ellman, J. A.; Dorow, R. L. The asymmetric synthesis of α-amino acids. Electrophilic azidation of chiral imide enolates, a practical approach to the synthesis of (R)- and (S)-α-azido carboxylic acids. J. Am. Chem. Soc. 1990, 112 (10), 4011–4030. DOI: 10.1021/ja00166a045. Non-paywall version available from the Evans group website.


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Does Temperature Affect Rate of Reaction of Electrolytic Cell
I completed an electrolysis experiment. I used copper electrodes and a copper sulphate solution, and my independant variable was the temperature of the electrolyte. I was just wondering if increasing the temperature has a positive correlation?

What were the results of your experiment?

The general trend was as the temperature increased, so did the rate of reaction. But only slightly. The experiment was controlled quite badly so this may be due to systematic error.


text:
Why does RCN attack the benzene ring in Hoesch Reaction?
In the Hoesch reaction, an alkyl cyanide reacts with a polyhydric benzene ring, and forms an imine salt by attacking the activated benzene ring. Most acid derivatives (and even alkyl cyanide itself with phenol) results in attack at the alcohol's oxygen, which is more nucleophilic than the benzene ring. In this case however, the major product doesn't form by attack on the alcohol.
Where lies the difference? If it were in the fact that benzene is more activated because of the many alcohol groups in the polyhydric compound, wouldn't it mean that even another acid derivative, like acid chloride, would react at the benzene instead and not at the O (thus carrying out a Friedel-Crafts acylation)?


text:
Assigning locants in haloalkanes and haloalkenes
In the first example below, if you add up the number of carbon which has a substituent, you will get 2 + 4 + 4 = 10. However, if you start from left to right, then you'd get 2 + 2 + 4 = 8. Why don't we follow the latter one when it gives us a lower number? (Lowest set of locant rule.)
The same applies for the second problem: we could've gotten 2 + 3 + 4, but we chose 2 + 4 + 5, why so?

The substituents for haloalkanes are named alphabetically. Chloro- before methyl-. Then use the lowest number first? Searching IUPAC nomenclature for haloalkanes will return lots of web results

The lowest sum of locants rule does not exist in the IUPAC recommendations.

Indeed, the first compound in your sketch is numbered incorrectly according to IUPAC. IUPAC rules for branched hydrocarbons say: "When series of locants containing the same number of terms are compared term by term, that series is "lowest" which contains the lowest number on the occasion of the first difference. This principle is applied irrespective of the nature of the substituents. " (emphasis mine). In your top case, this means that you have two numbering choices: either 2,2,4 or 2,4,4. The 2,2,4 choice has a lower locant at the first point of difference, so it is preferred. When you construct the name, you indicate the substituents in alphabetical order, so the full name of the top compound is 4-chloro-2,2-dimethylpentane.

If unsaturations are present, then these have a higher priority than alkyl and halogen substituents. From the rules for unsaturated hydrocarbons: "The chain is so numbered as to give the lowest possible numbers to the double bonds." This rule is sufficient to impose the numbering you report.

As Loong correctly pointed out, the links provided above refer to the text of the now obsolete 1979 recommendations. Both the 1979 and 1993 recommendations are available on the same site, maintained by ACD Labs with IUPAC's permission, and are in accordance with what shown above.
Unfortunately, the text of the most recent (2013) recommendations is not freely accessible (that I know of). This is a link to the electronic version of the book, published by The Royal Society of Chemistry. The relevant section is P-14.3.5 in Chapter P-1 (General Principles, Rules, and Conventions): "The lowest set of locants is defined as the set that, when compared term by term with other locant sets, each cited in order of increasing value, has the lowest term at the first point of difference; for example, the locant set '2,3,5,8' is lower than '3,4,6,8' and '2,4,5,7'."

But the numbering given above is correct (or at least it's given in the book).  That is why I am confused. It should be 2,2,4, but in the book, the name 2-chloro-4,4-dimethylpentane is given correct, for some reason.

I'm afraid that the book you are referring to might contain a mistake then. Here is the CAS number for that structure: 33429-72-0. You can search the web for the name it refers to, in order to check that it is, in fact, 4-chloro-2,2-dimethylpentane, e.g. from NIH TOXNET, and from a CRC reference book.

@MarcoB Please note that your links http://www.acdlabs.com/iupac/nomenclature/79/r79_36.htm and http://www.acdlabs.com/iupac/nomenclature/79/r79_53.htm lead to obsolete IUPAC recommendations of 1979.

@Loong Yes, and the same site contains the text of the 1993 recommendations, which are in agreement on this point. I'll add a link to the newest 2013 recommendations, but unfortunately those are behind a paywall. Fortunately, however, the interpretation of the point under contention here has not changed throughout the years, although the 1979 recommendations were the most direct about it.

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text:
Calculated 13C NMR Shifts of brominated Carbons
Since I calculate NMR spectra for a while by simply using one of recommendations by CHEmical SHIft REpository:
mPW1PW91/6-311+G(2d,p)-SCRF//B3LYP/6-31+G(d,p)

without thinking to much how it actually works*, there is a common misprediction that I'd like to understand.
It seems to be that the shift/shielding of the brominated carbons is always way off the measured shift. Compared to all other "normal" substituted carbons (C, H, O, N...) it is also far from being predicted good.
Luckily for me, as I don't want to show my own molecules :D, the Repository gives an example (Aplydactone) with a brominated carbon that shows exactly what I mean:

The overall $^1$H Mean Absolute Deviation (MAD) is about 0.08 ppm and for the $^{13}$C shifts the MAD is about 1.5 ppm but without the brominated carbon (right side of picture). The predicted shift has a failure of 12.25 ppm (measured 65.5 ppm vs. calculated 77.75 ppm) which is about ten times the MAD of all other carbon shifts.
Is there a common explanation for that? Is it simply that there were no brominated samples within the test set or is there a much deeper reason? Like Bromine must be treated with a different/super special basis set or sth similar.
This comes right now into my mind ... As there might already be some influences of the relativistic effect, could that be a reason?
Please tell me of your deeper insight into the calculation of NMR shieldings that could yield to this misprediction that I either understand how to correct those errors or that I at least have an explanation.
I did no further research throughout the internet yet, as I have not much time right now. But as this question keeps in my mind, maybe someone of you has done the research and would share it with me.

* I know that it uses a regression function which was calculated through comparing lots of experimental vs. predicted shifts for a bunch of small organic substances. What I mean is more the quantum theoretical aspect.

My suggestion would be to e-mail one of the authors your question directly. The review article suggests Tantillo as the corresponding author. I suspect they may have ideas.

The article suggests "In fact, error in computed chemical shifts for carbon atoms can approach several dozen ppm when, for example, three chlorine atoms, or fewer bromine or iodine atoms, are attached to the carbon atom in question."

The authors note in their review article* that such effects are expected as "heavy atom effects" due to electron-correlation effects and spin-orbit coupling: (references stripped for clarity)

This being said, one way to improve the computed chemical shifts of heavy-atom-bound carbon atoms is to utilize a method that better captures electron correlation effects, such as MP2 or higher-level methods along with correlation-consistent basis sets (see section 4.3). This will likely result in significant improvement for carbon atoms attached to sulfur or phosphorus as well as to halogens, although the improvement is expected to be greater for sulfur or phosphorus. Further improvement requires the use of methods that determine relativistic spin–orbit corrections.
  [...]
Another option for improving chemical shift calculations with heavy-atom errors is to utilize a linear regression method, as discussed in section 3.6. Due to the systematic nature of heavy-atom effects, they are amenable to improvement via this approach, provided that the linear regression data consists of nuclei similar to the nucleus of interest. That is, a linear correlation correction to carbon nuclei attached to a given number of bromine atoms, for example, should be derived from empirical data for other carbon nuclei attached to the same number and type of halogen atoms. In fact, this approach has been demonstrated specifically for such cases.

The latter section cites two relevant articles:

"Theoretical investigation on $\ce{^1H}$ and $\ce{^13C}$ NMR chemical shifts of small alkanes and chloroalkanes" J. Chem. Phys. (2006) 125, 144309.
"A hybrid quantum mechanical and empirical model for the prediction of isotropic $\ce{^13C}$ shielding constants of organic molecules" Phys. Chem. Chem. Phys., 2002, 4, 5498-5507.

Both articles suggest ways to use DFT methods with empirical corrections that work fairly well.

* Michael W. Lodewyk, Matthew R. Siebert, and Dean J. Tantillo, Chem. Rev., 2012, 112 (3), 1839–1862.


text:
Is 44.5 mg/L a measurement of weight or volume?
I have a test result that I am trying to fully grasp. I need to determine how much of a contaminant I need to reach that amount in 4,800 gallons. Making the conversion to liters is no problem, nor is the math necessary to figure how much I need by weight to get this much contaminant.
The one thing I am unclear on is my assumption that $44.5~\mathrm{mg~L^{-1}}$ is by weight.
In other words, if I have a result that measures $50~\mathrm{mg~L^{-1}}$ of a metal (nickel, for instance), that means if I were to be able to extract the metal and weigh it, it would weigh $50~\mathrm{mg}$.
Any help you can offer would be greatly appreciated.

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For such a low concentration one typically assumes that the the solute doesn't increase the volume of the solvent. In other words,  that the density of the solution is independent of the concentration of the solute.

$50~\mathrm{mg~L^{-1}}$ is an expression of concentration by mass. It means that if you took one litre of the solution and extracted all the metal, the resulting metal would weigh 50 milligrams.
The other common option is to express concentration in terms of moles per litre ($\mathrm{mol~L^{-1}}$) which means that if you extracted all the solute from one litre of solution, you would have 50 moles of solute.

Weight technically refers to a force exerted by a given mass, while the use of this term for masses is quite common, scientifically it is not correct. So the resulting metal has a mass of 50mg.

Very true. Edited the answer to reflect that.

The test result is mass per volume, or concentration. It just says there are 44.5 mg of (whatever) in each liter of solution.
If you're asking how much total contaminant there is in 4800 US gallons given a test result of 44.5 mg/L:
$$\left(44.5\,\mathrm{mg\over L}\right)\cdot\left({1\,\mathrm{L}\over 0.264172052 \,\mathrm{US\,gal}}\right)\cdot(4800\,\mathrm{US\,gal})\cdot\left({1\,\mathrm{g}\over 1000\,\mathrm{mg}}\right) = 808.6\,\mathrm{g}$$

Weight technically refers to a force exerted by a given mass, while the use of this term for masses is quite common, scientifically it is not correct.

Edited per your comments @Martin-マーチン - thanks!

In the given context, the quantity value $44.5\ \mathrm{mg\ l^{-1}}$ corresponds to the quantity mass concentration.
The mass concentration of substance $\ce{B}$ (recommended symbol: $\rho_{\ce{B}}$, alternative symbol: $\gamma_{\ce{B}}$, mass concentration of water: $w$) is defined as
$$\rho_{\ce{B}} = m_{\ce{B}}/V$$
where $m_{\ce{B}}$ is the mass of substance $\ce{B}$ and $V$ is the volume of the mixture.
The dimension of the mass concentration is
$$\dim \rho_{\ce{B}} = \mathsf{L}^{-3}\;\mathsf{M}$$
The coherent SI unit for mass concentration is ‘kilogram per cubic
metre’ (unit symbol: $\mathrm{kg/m^3}$).
In a different context, however, the quantity value $44.5\ \mathrm{mg\ l^{-1}}$ might as well correspond to the quantity mass density (or density).
The density $\rho$ is defined as
$$\rho = \mathrm dm/\mathrm dV$$
where $m$ is mass and $V$ is volume.
For example, the density of hydrogen at a temperature of $T = 20.0\ \mathrm{^\circ C}$ and a pressure of $p = 0.538\ \mathrm{bar}$ is $\rho = 44.5\ \mathrm{mg/l}$.
In general, quantities of the same dimension are not necessarily of the same kind.
Therefore, the unit symbol should not be used to provide specific information about the quantity, and should never be the sole source of information on the quantity. In plain language: it is important not to use the unit alone to specify the quantity.

I don't quite understand this formula: $\dim \rho_{\ce{B}} = \mathsf{L}^{-3};\mathsf{M}$ Is it $\mathsf{L}$ for a length and $\mathsf{M}$ for a mass?

@Martin-マーチン yes, the dimensions (represented by a single sans-serif capital) of the base quantities in the SI and ISQ are
$\mathsf{L}$ (length), $\mathsf{M}$ (mass), $\mathsf{T}$ (time), $\mathsf{I}$ (electric current), $\mathsf{\Theta}$ (thermodynamic temperature), $\mathsf{N}$ (amount of substance), and $\mathsf{J}$ (luminous intensity). Thus the dimension of any quantity $Q$ can be written in the form of a dimensional product $\dim Q = \mathsf{L}^\alpha;\mathsf{M}^\beta;\mathsf{T}^\gamma;\mathsf{I}^\delta; \mathsf{ \Theta}^\epsilon;\mathsf{N}^\zeta;\mathsf{J}^\eta$


text:
How to convert benzene to 1-bromo-3-iodobenzene?
Today, my chemistry teacher gave the following organic conversion as homework: benzene to 1-bromo-3-iodobenzene.
I tried something like:

I am not sure about it, can someone review my conversion (as I have got an exam tomorrow)?

It looks pretty good overall. The order of substitution is correct to achieve the required pattern but there are two points I would pick up on.

The final substitution of the diazonium salt is best done with $\ce{CuI}$ rather than $\ce{KI}$ as the copper(I) ions catalyse the reaction. Also heating is not required; the reaction can be run at room temperature (or possible slightly above but not much).
The diazonium salt is drawn incorrectly since it is ionic in nature. There is a triple bond between the nitrogens and a formal positive charge on the middle nitrogen, although in reality the charge is delocalised over both atoms:

Do you have an example using CuI for that reaction? I've actually only ever seen KI for the transformation of diazonium to iodide.

Interesting, I assumed that iodide would be similar to chloride and bromide but perhaps I am wrong. No I have never actually seen an example.


text:
Container to supercool water
The internet is full of videos like this showing water bottles for drinking used to supercool water. These bottles, not unexpectedly, appear a bit finicky, since they have lots of edges to serve as nucleation sites for ice crystals. What container would work best for this sort of demonstration?
Is the most important quality just to eliminate edges or does the material have an important effect? It's important in lab to add boiling stones as nucleation sites when boiling water in glassware. Could I use this trait for freezing too? Is the important quality of the glass the high surface energy or the smoothness of the surface? Alternatively, it seems like something like PTFE coating with super low surface energy might be effective?

Boiling stones has nothing to do with edges: they generate bubbles from the gas absorbed/stored in their pores. Also, most PET bottles has no edges, so it is not a particularly difficult condition to overcome.

Boiling stones most definitely do rely on having abrupt discontinuities... it would be in violation of conservation of matter if the "trapped gas" continually bubbled up as long as liquid were added.

Frankly I be inclined to use a bottle of soda under pressure.  I'd explain the cheat, but explain that you wanted the demo to work reliably. It's the rapid freezing that is the really neat part of the demo.

One way to approach this problem is think about what kind of material would best promote crystallization, or ice-nucleation, and then essentially use a material that does not have those characteristics.
Consider cloud seeding for example. The whole premise of this is that there is a dearth of ice-freezing-nuclei in the atmosphere, so in the presence of supercooled water droplets, you introduce a compound with a similar crystal structure to ice, like silver iodide for example. Even crystalline compounds with different structures like sodium chloride and dry ice are also effective. So, since we're trying to do the opposite of ice-nucleation, we want opposite characteristics of the surfaces in contact with the supercooled water; nothing crystal-like in nature, something as smooth and clean as possible on a very small scale.
Teflon is a highly hydrophobic synthetic polymer, meaning it's similarity to both liquid and solid water is minimal. A smooth, clean, unscratched Teflon surface is probably the ideal choice. A cheaper option using the same concept would be other organic polymers like polyethylene (probably what the bottles in your video link were made of), polypropylene, HDPE, etc.
Another option that is not crystal-like by definition is any type of consumer glass container. Again, any nano-scale defects or contamination will degrade it's ability to inhibit ice formation from supercooled water.

Well I think the point is to see that the liquid hasn't frozen. So the choice has to be transparent which leaves out Teflon.

@MaxW, yea, I lost track of that a bit with Teflon as my top recommendation, though there are glass containers with a thin layer of Teflon (still expensive and would look like freezing milk at best ;)  Something like a brand new 20 mL scintilation vial would probably be ideal, though I partly say that because that's something I always had access to for my demos. I also had the advantage of having tert-butyl alcohol available (MP 25C, supercools nicely). But it worked well with water too, pretty much only if brand new though...just the micro scratches from washing killed them for that purpose.


text:
Density of non-ideal vapor mixtures
I'm looking for a reasonably accurate density of a mixture of two gases that act quite unlike ideal gases (generally steam and water-miscible solvents). We can get a first order estimate by taking a molar average but that can be an order of magnitude off, as in the case of water/ammonia.
Literature is the obvious place to start but I wasn't able to find any general sources and only a very few individual pairs in a narrow temperature/pressure range.
I would guess you should be able to derive an excess volume from fugacity but I haven't been able to find a formula that doesn't take even more difficult-to-measure inputs.
Standard modeling references like Refprop seem to use an ideal mixing model which isn't right.
I can also imagine measuring them myself or doing molecular dynamics modeling but these are both quite a bit more work than I was hoping for.
Are there any literature sources for this data that I couldn't find? Anyone have an idea of another way to calculate it?

Could you please give an example of a mixture and the expected results that would satisfy your needs? Furthermore, could you clarify why you cannot accept Refprop results even though Refprop is supposed to use a departure function to account for the departure from ideal mixing?

@Loong According to Conde (which I think is a reputable source), the density of 75% molar concentration ammonia in water is 12.8 kg/m^3. If I do a simple molar average of the saturated vapor density of ammonia and water, I get 0.8. If I just use the ideal gas law, I get 0.6.  Refprop reports 0.6. Based on that Reprop seems to be using a simpler method.


text:
How to recognise the significant figures in given input values for the calculated output value?
If the transfer of $1~\mathrm{kJ}$ of energy to a quantity of water causes its temperature to rise $10~^\circ\mathrm{C}$, what is the mass of the water?  I know how to do the math and everything but I'm so confused with significant figures. If $1~\mathrm{kJ}$ or $100~\mathrm{J}$ only has 1 significant figure, then would I write $\pu{23.90 g}$ as $\pu{20 g}$?

You have a reasonable question: if the data supplied specified 1.00 $\mathrm{kJ}$, and the temperature change was given as $10.00~^{\circ}\mathrm{C}$, then "23.90 g" would be a justifiable answer.
However, since the data precision was given to only one digit for energy, then the mass would be "20 g".
Personally, given enough space, I'd write an explanation, of the sort you've give, e.g. "Though I calculated 23.90 g, it could be rounded to 20 g due to lack of significant digits in the data given."
BTW, you've shown good effort in thinking through the homework assignment, in my opinion, which is as important as a numerically correct answer.


text:
Preservation and bleaching of sliced mushrooms with hydrogen peroxide
According to the article Extending the Shelf Life of Fresh Sliced Mushrooms, it states that 5% (v/v) hydrogen peroxide is an effective means for the preservation of sliced mushrooms.
In regards to the preservation:

In the USA hydrogen peroxide
  is an approved bactericide for some dairy products and is used for disinfecting fruit
  and vegetables. Its activity is due to its oxidising effects on
  bacteria and it also bleaches mushrooms during the soaking period.

How does oxidising effects cause the bleaching, hence preservation of sliced mushrooms?

Hydrogen peroxide is quite a strong oxidizer. First, let's address the bleaching effect: It bleaches by oxidizing colored compounds, making them colorless. From Wikipedia:

An oxidizing bleach works by breaking the chemical bonds that make up the chromophore. This changes the molecule into a different substance that either does not contain a chromophore, or contains a chromophore that does not absorb visible light. This is the mechanism of bleaches based on chlorine.

This is also the mechanism of hydrogen peroxide as a bleach, since both chlorine-based bleach and hydrogen peroxide are oxidizers.
As for the antibacterial effect: hydrogen peroxide in this particular concentration (5%) may or may not be effective against bacteria (and viruses, fungi, etc.) due to the presence of the enzyme catalase in some bacteria and fungi. In those organisms which it is effective against, hydrogen peroxide would probably work by causing oxidative stress in the organism.


text:
% Fe in a sample
Seems so simple but is confusing me as when I take into account final weight I don't get one of the answers.
An iron ore was analysed by dissolving a $1.1324~\mathrm{g}$ sample in concentrated $\ce{HCl}$ and converting it to give $0.5394~\mathrm{g}$ of pure $\ce{Fe2O3}$. The $\mathrm{m\%}_\ce{Fe}$ in the sample is either $18.9~\%$, $25.2~\%$, $37.7~\%$ or $50.4~\%$.
My approach:
If I take the molar mass of iron as $55.85~\mathrm{g/mol}$ and oxygen as $15.99~\mathrm{g/mol}$. I have a $159.67~\mathrm{g/mol}$ weight for $\ce{Fe2O3}$. From this I calculated that amount of $\ce{Fe2O3}$ is$$\frac{0.5394~\mathrm{g}}{159.67~\mathrm{g/mol}}=3.378 \cdot 10^{-3}~\mathrm{mol}$$
The answer is then multiplied by ($2 \times 55.85~\mathrm{g/mol}$) giving $0.3773~\mathrm{g}$ iron but this is what is confusing.
Isn't the answer I got just for $0.5394~\mathrm{g}$ so shouldn't I take into account that I started off with $1.1324~\mathrm{g}$? If I do I get $33.32~\%$ so doesn't match an answer.

Welcome to Chemistry.SE. Take the [tour] to get familiar with this site. This appears to be a homework question, please share your thoughts and attempts towards the solution. It'll make us certain that we aren't doing your homework for you.

It's confusing us too. Could you provide the question and your approach in full? Your wording implies that we already know what question you're talking about.

Question from a past paper. If I take the Mass of Fe as 55.85 and O as 15.99.

Please edit in concise details of your attempts.

Question from a past paper. If I take the Mass of Fe as 55.85 and O as 15.99. I have a 159.67 weight for Fe2O3. From this I calculated that Mol Fe2O3 are 0.5394g1/159.67=3.378x10-3. The answer is then multiplied by (55.852) giving 0.3773gFe but this is what is confusing. Isn't the answer I got just for 0.5394g so shouldn't I take into account that I started off with 1.1324g? If I do I get 33.32% so doesn't match an answer.

"Question from a last paper." Please edit in the question in full!

Could there be an error in the question possibilities. I got Fe at 0.377g (coincidental number?) of the 0.5324g of Fe2O3. Which turns out as 33.4% of the overall sample. I get same result as @IbrahimGeorge

Looks like part (a) from this example from Fundamentals of Analytical Chemistry. The answer given in that case is 33.32%.

I have to admit I did come up with an answer of 33.32% but it didn't match any of the possible answerss

it is even easier than they do the calculation.$$x \cdot 1.1324~g = 70~% \cdot  0.5394~g \Rightarrow x = 33.32~%$$

Ok thanks, at least other people can't match the answers


text:
Alternative to water as solvent for lithium metaborate?
I want to prepare a sample of lithium metaborate ($\ce{LiBO2}$) and need to dissolve it for this purpose.
The only solvents I found so far are water and ethanol. In ethanol, one cannot dissolve enough of the solute. Water is quite nice from the solubility point of view, but its high surface tension and tendency of wetting forbids (more or less) its use.
Does anyone have an idea for another solvent as an alternative for water?

Please visit [this page](http://chemistry.stackexchange.com/help/notation " latex="" at="" chemistry.se""="), this page and this one on how to make your future posts better.


text:
Does sodium produce water when it reacts with water?
My 10th grade chemistry book states that:

"Metals react with water and produce a metal oxide and hydrogen gas. Metal oxides that are soluble in water dissolve in it to further form metal hydroxide. Metals like potassium and sodium react violently with cold water. In case of sodium and potassium, the reaction is so violent and exothermic, that evolved hydrogen immediately catches fire."

So $\ce{Na}$ should react like this:
(i) $\ce{2Na + H2O -> Na2O + H2}$
(ii) $\ce{Na2O + H2O -> 2NaOH}$

When hydrogen catches fire -- that is, when hydrogen reacts with $\ce{O2}$, right? -- does it form $\ce{H2O}$?
If water is not formed, then does the water level decrease?
Is (ii) an exothermic reaction?

Welcome to Chem.SE--thanks for posting your question here! The general questions at the end about combination and decomposition reactions should really be posted separately, though -- I'd recommend editing them out of this question and starting a new question with them.  Please check out the tour and help pages for more information about the site. Enjoy!

Also, the accepted answer at this question links to a fascinating video about the physics behind the explosive nature of the reaction of alkali metals with water: What exactly is happening when sodium, potassium, or other alkali metal explodes in water?

Please do not use capital letters for emphasis as they look like shouting. Emphasis ≠ SHOUTING

The reaction of sodium metal and water could (and usually does) indirectly produce more water, if enough heat is produced to spontaneously ignite the hydrogen gas which is produced by the initial reaction. This cyclic type of reaction is often found in explosive or extremely exothermic reactions, e.g. $\ce{K+ + I- + H2O2}$ will also react violently in a cyclic way.
Combustion of hydrogen in air is this:
$$\ce{2H2 + O2 -> 2H2O}$$
So, in a (well) closed system with excess oxygen, water could re-form and collect. It's worth noting that dissolution of $\ce{NaOH}$ in water ($\ce{NaOH -> Na+ + OH-}$) is also exothermic and adds even more heat to the system.

I have improved the formatting of your post using $\LaTeX$. For more information on how to do this yourself please see here and here.

This does produce water, but the water level decreases due to hydrogen escaping before sodium heats up. The reaction should also speed up and create more and more hydrogen. The force of the pop from the hydrogen exploding means more water escaping. The disassociation of $\ce{NaOH}$ is exothermic. Of course, if this was a closed system nothing would be lost.


text:
Molecular orbital (MO) diagram for N2 and N2^-
I have been taught that the MO diagram is different for molecules with 14 or less electrons than the one used for molecules with 15 or more electrons.
For $\ce{N2}$ the orbitals in increasing energy are:
$$\sigma_\mathrm{1s} < \sigma^*_\mathrm{1s} < \sigma_\mathrm{2s} < \sigma^*_\mathrm{2s} < \pi_{\mathrm{2p}_x}, \pi_{\mathrm{2p}_y} < \sigma_{\mathrm{2p}_z} < \pi^*_{\mathrm{2p}_x}, \pi^*_{\mathrm{2p}_y} < \sigma^*_{\mathrm{2p}_z}$$
because it has 14 electrons.
For $\ce{N2-}$ there are 15 electrons. Will the MO diagram be the same as that of $\ce{N2}$ or not? Why?
And because this is a good chance to ask, why is there this difference in orbital energies anyway?

There should be a difference in energies when you add another electron.
The new [say, spin up] electron (a) creates new repulsions with all the others and (b) this repulsion will be less intense with electrons of opposite spin due to Pauli exclusion principle.
So all other electrons will have their states changed (but not equally!) and so their orbital energies.

I have been taught that the MO diagram is different for molecules with
  14 or less electrons than the one used for molecules with 15 or more
  electrons.

This is (partly) wrong because the change in the order of $\mathrm{\sigma_{2p_{z}}}$ and $\mathrm{\pi_{2p_{xy}}}$ MOs to the left of $\ce{N2}$ is not directly related to the number of electrons. Rather, it is rationalized by a successive decrease of the $\mathrm{s}$-$\mathrm{p}$ interaction moving from $\ce{Li2}$ to $\ce{F2}$. The $\mathrm{s}$-$\mathrm{p}$ interaction is the bonding interaction between the $\mathrm{2s}$ orbital of one atom and the $\mathrm{2p_{z}}$ orbital of another atom which (among other things) increases the energy of the $\mathrm{\sigma_{2p_{z}}}$ MO relative to the hypothetical case without $\mathrm{s}$-$\mathrm{p}$ interaction.
Now the difference in energy between the $\mathrm{2s}$ and $\mathrm{2p_{z}}$ AOs increases from $\ce{Li2}$ to $\ce{F2}$ due to increasing nuclear charge and poor screening of the $\mathrm{2s}$ electrons by electrons in the $\mathrm{2p}$ subshell. As a result, the rising effect of $\mathrm{s}$-$\mathrm{p}$ interaction on $\mathrm{\sigma_{2p_{z}}}$ MO is getting less and less prominent, so that eventually to the right of $\ce{N2}$ $\mathrm{\sigma_{2p_{z}}}$ MO becomes lower in energy than $\mathrm{\pi_{2p_{x,y}}}$ MO.

Image courtesy of the UC Davis ChemWiki.
Now, back to the question.

For $\ce{N2-}$ there are 15 electrons. Will the MO diagram be the same as
  that of $\ce{N2}$ (because it is actually an ionized molecule of $\ce{N2}$) or not?
  Why?

The short answer is: we could not tell it using the primitive molecular orbital theory introduced in the general chemistry courses. In exact same way we could not tell why $\mathrm{\sigma_{2p_{z}}}$ MO becomes lower in energy than $\mathrm{\sigma_{2p_{z}}}$ MO to the left of $\ce{N2}$ and not to the left of, say, $\ce{C2}$. All this is simply because the primitive molecular orbital theory does not explain the things, rather it rationalizes them (for differences between explanation and rationalization, please, read here).
But then how do we know in the first place that $\mathrm{\sigma_{2p_{z}}}$ MO becomes lower in energy than $\mathrm{\sigma_{2p_{z}}}$ MO to the left of $\ce{N2}$ so that we can start to rationalize it? Well, we know that from a more advanced molecular orbital theory which is based on the laws of quantum mechanics. And the same advanced theory should be applied for the case of $\ce{N2-}$ to get the MO picture of this chemical species.
Now note that even in this advanced molecular orbital theory a bunch of approximations is introduced, and the answer in general depends on at which level of theory calculations are done. The simplest picture would be the so-called frozen-orbital approximation (in the Hartree-Fock picture) in which MOs of an anion (as well as that of a cation) are assumed to be the same as for the corresponding neutral molecule. But this approximate picture might be far away from the reality.
Update
The question in fact does not make a lot of sense since $\ce{N2-}$ is known to be electronically unstable, i.e. its energetically less favourable then the system of $\ce{N2}$ plus a free electron. There exist few metastable resonance states of $\ce{N2-}$ that correspond to attachment of the scattered electron to excited valence states1, but that's a different story which is beyond the scope of the question.

1) Krauss, M., Neumann, D., Valence Resonance States of N2, JResNatlStandSecA, 77A, 4, 1973. PDF.

The reason for the special order for molecules with 14 or less electrons is due to the s-p mixing that arises due to the s and p atomic orbitals being close in energy. It is not actually possible to definitely answer your question without looking at the actual orbital energies.
Despite $\ce{N2-}$ being isoelectronic to $\ce{O2+}$, the lower effective nuclear charge on nitrogen should make its s orbitals a little closer to the energies of the p orbitals than they would be in oxygen. Hence s-p mixing should occur. However there are other considerations that we need to take in account, such as the extra electron repulsion that occur due to the extra electron. Yet, as I have stated above we can't definitively say which effect is more important without looking at the actual orbital energies.

Which of the species in question ($\ce{N2}$ or $\ce{N2-}$) is "isoelectronic to  $\ce{O2-}$"? You probably meant that $\ce{N2-}$ is isoelectronic to  $\ce{O2+}$, not to $\ce{O2-}$.

Oh, yes. Sorry about that. I meant that $\ce{N2-}$ is isoelectronic to $\ce{O2+}$.


text:
Is HF the least acidic hydrogen halide?
Fluorine is the most electronegative halogen and therefore, there is larger difference in electronegativity between the atoms of $\ce{HF}$ than any other hydrogen halide, which means the positive charge on hydrogen atom is the greatest in this compound and hence a comparatively small negative charge is needed to attract it. If so, why is it classified as least acidic hydrogen halide if the $\ce{H+}$ is easy to remove?

related http://chemistry.stackexchange.com/questions/668/do-hydrogen-halides-hx-form-strong-or-weak-acids

First of all, as chipbuster says, $\ce{HF}$ in diluted solutions in water is nearly completely dissociated and therefore shouldn't be called weak. Wikipedia describes this nicely and cites several sources for this claim.
It was rather difficult to prove (spectroscopic methods were used), because hydronium ions created in dissociation are mostly bound to fluorine anions with hydrogen bonds, in what is called tight ion pairs. It prevents from detecting the true strength of $\ce{HF}$ with methods like acid–base titration - they show dissociation of ionic pair. Similar effect is present in ion exchange resins.
Strength of this acid is revealed in more concentrated solutions, where $\ce{HF}$ molecules replace hydronium ions, creating bifluoride anions, freeing them - it's homoassotiation mentioned by Wildcat.
More to the point - does it mean that $\ce{HF}$ isn't weaker acid than other hydrogen halides? The answer isn't simple, and acidity itself depends on many factors.
In terms of $K_a$ in diluted aqueous solutions $\ce{HF}$ is probably still weaker than $\ce{HCl}$ and the rest but getting exact value is greater problem than normally (generally values of $K_a$ for strong acids aren't precise). Decrease in acidity between hydrogen halides is described in answer to this question
For concentrated solutions one needs Hammett $H_0$ function and get estimates using stuff like nitrated aromatic bases, and it's still problematic to get real value. In an old paper, I found an estimate barely lower than -10 for pure $\ce{HF}$, while pure sulfuric acid has -12. A newer paper, on the other hand, gives value 15.1 - almost five orders of magnitude more acidic ($H_0$ is logarithmic like pH). The new value puts pure $\ce{HF}$ firmly in the superacidic region.
References

Herbert H. Hyman, Martin Kilpatrick, Joseph J. Katz; J. Am. Chem. Soc. 1957, 79, 14, 3668–3671. DOI:10.1021/ja01571a016.

Ronald J. Gillespie, Jack Liang; J. Am. Chem. Soc. 1988, 110, 18, 6053–6057. DOI:10.1021/ja00226a020.

One might argue that the dilute acid therefore is not HF.  The hydrogen fluoride has reacted with water to form (H3O)F.  That is the actual weak acid.

@OscarLanzi Yes, I was thinking more of equilibrium of breaking the h-bond and pair with it, but it's more reasonable to think of it as Brønsted acid. As far as "being HF", it's more of a HF then dilute HCl is HCl, I wager.

neither hydrogen halide is really the hydrogen halide in water.  The acid that's in aqueous HCl solution is (in simplified form) H3O+, which is weaker than HCl.

@OscarLanzi That was semi-jokingly, as H and F are still bound, just by h-bond :) Even "pure" H2SO4 isn't just H2SO4, at least in condensed phase.

Hydrofluoric acid is the least acidic hydrogen halide because of fluorine's electronegativity. Because of the fluoride ion's small size, it cannot disperse the negative charge over a larger space and will have an extremely high affinity for an electrophile (like $\ce{H+}$), and because of this it will remain mostly as $\ce{HF}$.
At high concentrations, through hydrogen bonding with its conjugate base, hydrofluoric acid is actually a significantly stronger acid due a process called homoassociation. (credit to Wildcat)

Fascinatingly, this paper claims that the HF actually dissociates nearly completely in water, but the resulting hydronium ion remains bound in an $\ce{H3O+ + F-}$ complex. I'm not certain exactly how that differs from $\ce{H2O + HF}$, but an interesting idea.

@chipbuster Of course it does I thought about asking this question just to say that ;)

HF is a strange molecule. I remember that Chemguide used to have an article trying to figure out why it was a "weak acid." He tried an electronegativity argument, an enthalpy of solvation argument, and a free energy argument, all of which showed that HF shouldn't really be weaker than any other halide acid. That section seems to have been removed and replaced with the above link though.


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Electron's Energy In The Atom
I know that the exact (OK most of the times not the exact) energy of an electron can be calculated by solving Schrodinger's differential equation, but can we explain an electron's energy of a specific orbital using the Penetration and Shielding effect?

Can we explain an electron's energy of a specific orbital using the
  Penetration and Shielding effect?

Yes, to some extent it is possible. Look no further than Slater's rules which specifies the values of the effective nuclear charge $Z^*$, and the effective principal quantum number $n^*$ that can be used to replace the nuclear charge $Z$ and the principal quantum number $n$ in the hydrogen-like atom orbital energy expression. The energy of each and every orbital can be approximated then by
$$
    E = - \left( \frac{Z^*}{n^{*}} \right)^{2}.
$$
But I would be careful calling this an explanation, since it is rather a rationalization. I wrote about the difference between those two here.


text:
Periodic Trends of Zeff and Electronegativity
The effective nuclear charge $Z_\mathrm{eff}$ increases from left to right and from top to bottom. Can you explain why it increases from top to bottom?
Also can we explain the periodic trend of electronegativity using the shielding effect?
EDIT:
The calculation of $Z_\mathrm{eff}$ for Li would be 1 (3 protons − 2 core electrons). For Na is 1 (11 protons − 10 core electrons). So far it doesn't seem to me that $Z_\mathrm{eff}$ should increase from top to bottom. Considering the fact though that for Na there are more electrons shielding the only valence electron than Li as well as Na's electron is relatively more distant than that of Li, the first should be experiencing less $Z_\mathrm{eff}$. This is not true however. As far electronegativity is concerned I can't come up with a neat explanation.

Looks like homework to me. Can you share what work you've done while working towards an answer?

Your misconception: Shielding by core electrons is perfect. Truth: shielding for Li is less than 2, shielding for Na is less than 10.

The Slater shielding rules give more quantitative information about the amount of shielding.
Put simply, valence electrons (as you go across a row) do not shield very well, roughly 35% of a full electron. So when you go from B to C to N, you keep increasing the nuclear charge by one proton, but the electrons don't fully shield the nucleus.
Core electrons, as suggested in comments and other answers do shield fully, since they're closer to the nucleus.
So yes, the shielding for Li and Na are similar. As you mention, the valence electron in Na is further from the nucleus, and that does effect the ionization energy and electronegativity as you suggest.

The electronegativity is the greatest at the top of the periodic table because fewer electrons are shielding the outermost electrons from the attraction of the nucleus.  As more electrons are added the electrons closer to the nucleus repel some of the outermost electrons and block the nucleus's attraction.  Within one left to right row the electronegativity increases as you move right because more protons exert an attractive pull on the electrons without adding more electrons that shield the charge of the nucleus.


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What happens to iron(III) sulfate (ferric sulfate) in aqueous solution?
Iron(II) sulfate (ferrous sulfate) is a salt that, in an aqueous solution, dissolves and yields $\ce{Fe^{2+}(aq)}$ and $\ce{SO4^{2-}(aq)}$. It then decomposes to ferric sulfate shortly thereafter.
But what happens to the iron(III) sulfate (ferric sulfate) in solution? Is the $\ce{Fe_2(SO_4)_3}$ compound preserved?

Fe2+ is oxidated by oxygen, Fe3+ can't be ox. this way.

Are you asking whether the ferric sulphate dissociates in solution into separate ions?

@bon yes. Or maybe some kind of reaction takes place?

Well hydrolysis may take place

In aqueous solution $\ce{Fe2(SO4)3}$ dissociates into $\ce{Fe^{3+} (aq)}$ and $\ce{SO4^{2-} (aq)}$ ions. The sulphate ions will be solvated by hydrogen bonding with the water and the iron ions will form the hexaaquairon(III) complex, $\ce{[Fe(H2O)6]^{3+}}$.
As @Mithoron mentioned, the $\ce{Fe^{3+}}$ ions cannot easily be further oxidised so no additional reaction takes place.


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Why does density decrease across the period 4 and 5 transition metals?
I know that density decreases across period 4 because the last two elements are liquids and gases which means that their densities are less but why are the densities increasing and then decreasing within the transition metals themselves?

Density is generally proportional to how far from a full octet (or in your case, an 18 electron set of outer-most orbitals) a material is, hence this effect, and its inverse - that earlier elements in a period are less dense. This does not apply only to transition metals, but all elements, as illustrated here. (Note: I am not sure what pressure is in that chart, so I suggest ignoring gas-phase elements, including O, F, and Cl.)
The reason for this is fairly simple. The further from a complete octet an element is, the more energetically favorable orbital overlap becomes. The more favorable orbital overlap becomes, the closer elements will pack at a given temperature. There is no energetic benefit to an element with a full octet (ideal gases) overlapping, so they typically do not, and thus form a gas. There is an intense energy benefit in elements with both half-full octets overlapping.


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What is the density of carbohydrates/protein/fat?
As in weight per volume? I mean.. you always see them specified by weight but they have to have a volume, right?
I suppose different fat(for example) molecules would have different densities but then how do they universally have the same energy density? Are they converted to like a standardized type of fat? Thats the fat I'm interested in.

What is "standardized fat"?

I get the sense you are interested in these compounds as they relate to dietary/nutritional science. Can you confirm/deny? That will help you get your question answered sooner.

I assume the "energy density" you are referring to are the widely known statistics that:

fat contains 9 "calories" per gram (actually kilocalories)
protein and carbohydrates both contain 4 "calories" per gram

How can all fats contain "9" kcal per gram but have different mass densities?  Here are a couple of reasons:

The 9 kcal/g figure is an approximation that assumes a typical dietary blend of fats.  The figure would if considering chemically pure fat molecules.
The mass densities of most common fats doesn't vary that much.  The densest "fat" molecules are probably the extremely short chain triglycerides like triformin and triacetin.  They have densities of approx. $1.3~ \frac{\mathrm g}{\mathrm{cm}^3}$.  These triglycerides have chain lengths of 1 or 2 carbons per fatty acid chain, a value far lower than common in natural fats.  Triolein is more representative of dietary fats and has a density of 0.95$~\frac{\mathrm g}{\mathrm{cm}^3}$.  So the mass density variation among different fat molecules is pretty small: a 30% increase is all that is really possible.


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What is the reason for the acidic character in phenols?
I came across this concerning phenols:

The hydroxyl group, in phenol is directly attached to
  the $\ce{sp^2}$
   hybridised carbon of benzene ring which acts as an
  electron withdrawing group. Due to the higher electronegativity of $\ce{sp^2}$
   hybridised carbon
  of phenol to which $\ce{–OH}$ is attached, electron density decreases
  on oxygen. This increases the polarity of $\ce{O–H}$ bond and results
  in an increase in ionisation of phenols than that of alcohols.

My concepts seem to be flawed, and hence I have few questions on the same.
When the $\ce{sp^2}$ carbon atom withdraws electrons from the oxygen atom, the oxygen's electron density decreases. How does this increase the polarity in the $\ce{O-H}$ bond? Shouldn't it be the opposite as oxygen becomes more electropositive due to the carbon atom? I'm missing something here.
Also, does the carbon atom in this case act as electron withdrawing because it is less electronegative than the oxygen atom?
Note that this question isn't about the comparison of acidic character between alcohols and phenols, and the relative stability of alkoxides and phenoxides.

Related: Inductive vs resonance effects and the acidity of phenol

possible duplicate of reason for the stronger acidic property of phenol than alcohol

@Mithoron I've mentioned in the end note that this question is not asking for " the comparison of acidic character between alcohols and phenols". I want to know why the sp2 carbon increases the polarity of the OH bond.

Explanation you're citing is rather bad one in these links you have better ones

I get the resonance part, which is what the link explains. I need the inductive explanation, which @Loong pointed to. But the way its explained in that link isnt so clear

I think your confusion is just over terminology. The sp2 carbon is more electronegative than the sp3 carbon, and as you said, oxygen is more electronegative than both. That means that oxygen will withdraw less electron density from the sp2 carbon than it does from the sp3 carbon. And that means that the phenol oxygen has less electron density than the alcohol oxygen. It's a little misleading to say that the carbon withdraws electron density from the oxygen, although loosely speaking it's okay.

When oxygen has less electron density, it will become "more electronegative", and the O-H bond will be polarised more towards O. Also, even if it is not explicit in your phrasing, your question is about comparison of phenols vs alcohols. Even the last sentence in your quote says this.

You are confusing the two terms electronegativity/electropositivity and higher/lower electron density.
The source you read states that the $\ce{sp^2}$ hybridised carbon in phenol is more electronegative than the $\ce{sp^3}$ hybridised one in (say) ethanol. This means, that this carbon has a greater desire to draw electrons towards it.
Therefore, even though any carbon is still more electropositive than oxygen, the phenolic oxygen will be able to draw the electrons towards it less effectively and therefore has a lower electron density than in ethanol.
But the oxygen atom in itself hasn’t changed; specifically, it is still hybridised the same way (assuming that one can speak of hybridisation for oxygens). Thus, oxygen’s electronegativity is still the same.
The combination of still having the same electronegativity but having less electron density means that any other electrons will be drawn towards it more strongly — specifically, this concerns the electrons in the $\ce{O-H}$ bond.
In a nutshell: The oxygen’s electronegativity didn’t change (it sounds weird to talk of electropositivity for oxygen if there is no fluorine around) it only has less electron density on one side due to more pulling. It combats that by pulling more from the other side.
This can be called an inductive effect.

One thing I don't understand is why does decreased electron density on oxygen increases  lit's acidity....

Or more properly affect it's acidity.Is there a logic behind this thought

@harambe Acidity or loss of a proton leaving a negative charge corresponds rather well with how stabilised the resulting anion is. And anions are typically most stable if there is a low electron density surplus. So if there are other groups drawing electrons away from the oxygen, this reduces the effective negative charge of an anion and thus makes the compound more acidic.


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Is the density of the solution significantly different from the density of the solvent?
I am working on a mathematical model for the dissolution of coffee in water. I need to know some precise details about the effects of dissolution on the density of the mixture of the substances (that is, the density of the solution). The system is isothermal and the pressure is constant. I just want to focus on mass, volume, density, and concentration as the relevant variables.
In the drawing (below), there is a graphical representation of the system in question (in the bottom image, I should have mentioned concentration of water and coffee instead of density). We have a container of water with a single coffee solid and I've drawn in 20 molecules for both the water and the coffee. The volumes are obviously not to scale (it looks like approximately 50% of the coffee has dissolved in the second pic, not 20%!).
Start
So at the beginning, $t = 0$, no coffee has dissolved so far and the solution liquid is simply the water. We have the following information for the density and the volume of the coffee, water, and the solution, respectively
$\rho_c = 1500$ kg m$^{-3}$
$\rho_w = 1000$ kg m$^{-3}$
$\rho_s = \rho_w = 1000$ kg m$^{-3}$
$V_c = 0.1$ m$^{-3}$
$V_w = 0.5$ m$^{-3}$
$V_s = V_w = 0.5$ m$^{-3}$
The solution is just the water at the start.
Finish
At, say, $t = 100$, 20% of the coffee has dissolved and has diffused out evenly into the solution and the system is in equilibrium.
Now the solution contains both water and coffee. I realise that the chemical reaction dissolution and interactions at a molecular level between the coffee and the water have an impact on the volume and density of the solution. I need information on the density of the solution to decide whether the compressible Navier-Stokes or the incompressible Navier-Stokes govern the flow in the solution.
What is the density  and volume of the solution? If experimental measurement is required for this can you instead give me an approximate answer.
I need to know how much of a change in solution density occurs from $t = 0$ to $t = 100$.

Your mathematical model looks awesome. Are you planning to do simulation in matlab or other software?.Can you upload the code here(If possible). It helps others to learn simulation easily

This depends a great deal both on the solute and the concentration.
At the concentrations that we normally drink coffee at, you can almost certainly ignore the difference in density of the solution. But ccording to your question you are trying to dissolve 0.1m3 coffee (I assume the minus sign is an error) in 0.5m3 water. That's 100L of coffee in 500L of water, a 100L/600L=16.667% volume/volume solution! A lot of very strong coffee!
As a general rule, the density rho is given as:
rho = (mass component 1 + mass component 2) / (volume component 1 + volume component 2 )
This assumes (correctly) that masses can be added. Strictly speaking, the volume of a solution is not equal to the sum of the volumes of its components, as there is a slight volume change on dissolution, but this is normally insignificant.
The hard part is to know the solid density of coffee, because it cannot be readily measured, as coffee comes in powder form. If you are not using instant coffee there is the problem that the husks of the grains will not dissolve. But in your case, you are told the density of the coffee is 1500kg/m3
So you can easily work out the masses of water and coffee, and then the density of your enormous vat of extra strong coffee.
Hint: it's going to be more than water, but not different enough to make any difference regarding the correct navier stokes model to follow.
EDIT: on re-reading, it seems you are assuming that only 20% of your 0.1m3 of coffee will dissolve: that is 0.02m3 in 0.5m3 of water, an approximate concentration of 4%. That's not quite as absurdly strong as before. I was a little confused because 0.1m3 is 20% of 0.5m3. However I'm not sure how you decided that you would reach equilibrium with 20% of the available coffee dissolved.


text:
In an O-H bond, will polarity increase/decrease if some of the electron density on the oxygen atom is reduced?
Like when an e- withdrawing group (sp2 carbon) is attached to oxygen of the OH bond, it attracts the electrons. So will the polarity of the OH bond increase or decrease?

The polarity on the oxygen atom will decrease. This is because of the sp2 hybridised carbon atom which is highly electronegative. This will result in a decreased polarity in the OH bond and hence it will disassociate more easily in solvents such as water. One example of this effect is the increased acidity of phenol compared to most alcohols. This is because the OH is bonded to sp2 hybridised carbon in phenol and hence will disassociate in water easily when compared to normal alcohols.

Yes, polarity can be changed. Example, between alcohol and phenol the O-H polarity is more in phenol nd less in alcohol. It is due to the extent of attraction of electron by oxygen from hydrogen which is more in phenol.

The electron density on oxygen decreasing will result in tye polarity of the OH bond increasing. This will increase the acidic strength of phenols when comapred to to alcohols, where the OH group is attached to the sp3 carbon.


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Properties of plasmas
In chemistry one can recognize that the four states of matter are solid, liquid, gas and plasma. The first is rigid, and has a definite shape and volume. The second doesn't have a shape, and assumes the shape of its container, but it has a fixed volume. The third doesn't have either a shape or a fixed volume and assumes the volume and shape of its container. What about the fourth one (plasma)?

"Like a gas, plasma does not have definite shape or volume. Unlike gases, plasmas are electrically conductive, produce magnetic fields and electric currents, and respond strongly to electromagnetic forces." Quoted from Wikipedia.

Plasma is made up of ionized gas, so molecules have a positive electric charge and valency electrons are totally or partially separated by their nuclei.
Plasma is different from a gas since it has a high temperature and is radiation emitting; think of the Sun and other stars, which are made up of plasma and show both these properties.
Plasma hasn't got a proper volume, like gases; e.g. stars can expand or contract under the opposite effects of gravity and nuclear fusion. For example, this property is important to comprehend the formation of white dwarfs and neutron stars, process caused by high pression due to gravity.
Little trivia: there is also a fifth state of matter, whose name is Bose-Einstein Condensate (BEC).


text:
Pepper escaping soap on water
Following a book of science experiments for children with my daughter, we did the following:

put some water in a shallow dish;
put some ground pepper floating on the water surface;
touch the water surface with the tip of a small stick previously dipped in liquid soap;
then the floating pepper escapes radially from the soap.

But I was at a loss to explain why. I mumbled something about “surface tension”, but didn't convince even myself.
What was going on?
I am pretty sure that the pepper in itself plays no role but showing the surface motion of water. Am I correct?

The "soap molecules" are so called amphiphiles: They've got one end that likes water ("polar" or "lipophobic") and the other ("lipophilic" or "fat-loving") doesn't.
These amphiphile molecules like to form a single-molecule thick film on water, all standing up in an orderly fashion (foot in the water), so if you put a drop of it on a water surface, it spreads very wide, until the film is monomolecular.
Or, until something pushes back. In your case the pepper, when its particles are pushed together so they also form a tightly packed surface. Waiting longer, the soap will likely act as a detergent, wrap the pepper particles in a bilayer, and drag them under the surface.
Search for, e.g., surfactants, detergents for some pictures.
As to your last question: Some water gets dragged along with the surfactant, but no, there is no large lateral movement of water on the surface. The spreading front you see is just the soap.