text:
What mass of fragrances does a cubic meter of air contain on average?
What mass of fragrance are airborne in scented environments for example in a perfume deparment?
What is the minimum g/m3 of fragrance that a human can detect, and how much is present when it evaporates from an oven, in a cookie shop, from a cut lawn, near a perfumed lady and average levels from a day of walking.
What range of fragrance concentrations do humans typically have to cope with.

Unless I misunderstood you in the first,  you might want to talk about (mostly terpenoid) fragrances rather than about (the structure class) of aromatic compounds in order to avoid further misunderstandings.

Okay, perhaps the question is too difficult because some fragrances can be detected in volumes 1000 times greater than others. it's possible to evaporate scents and weigh them, to know the quantities that represent different levels. in general the weight of fragrance of an essential oil for a cubic meter is less than a microgram. i figure that's in the range of 1ppm.

This question will be incredibly difficult to answer. The first part requires a lot of of information about the given environment, and any analytical chemist will tell you that there is no general answer that can be given. it will vary from department to department, as well as on what perfumes they carry, how often they're sprayed, and a ton of other factors. The second question, the first part might have an answer if a study was done, but definitely varies person to person. The second part of question 2, again has many circumstantial variables. Third, is answerable if there's been a study.


text:
UV-Vis emission from charge-separated states?
Are there examples of UV-Vis emission from charge-separated states? The cases I have encountered have only absorption. Is the recombination process of a nonradiative kind by necessity?

There are at least cases in which charge recombination results in an excited species that deactivates via fluorescence.
Imagine the ionization of a hydrocarbon ($\ce{RH}$) to yield a radical cation and a solvated electron.
\[\ce{RH ->[\mathrm{e-beam}] RH^{+\cdot} + e-}\]
Trap the electron by reducing an aromatic hydrocarbon ($\ce{ArH}$) to the coresponding radical anion.
\[ \ce{ArH +e- -> ArH^{-\cdot}}\]
Charge recombination between the radical cation and the radical anion can yield $\ce{ArH}$ in an excited state.
\[\ce{RH^{+\cdot} + ArH^{-\cdot} -> RH + ArH^{\ast}} \]
Back in the days, Alexander D. Trifunac did a lot of work on that at ANL.

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text:
Can any atom donate more than 1 lone pair as a coordinate bond to the same metal?
This would mean the formation of a σ and one or two π bonds, all coordinate. I haven't heard of this happening, like a fluorine anion or water with two lone pairs on oxygen will donate only one lone pair, but I wanted to know why it's not possible.

What do you mean by "at the same time"? And I think you mean the oxygen atom gets $\ce{sp_2}$ hybridised, not $\ce{H2O}$ itself

It is possible for pi-donating ligands, like NH2- (amide anion), but not really possible for water (neutral, oxygen more electronegative).

Yes, I'm sorry, I did mean O getting sp2 hybridised.

It's probably more common than I'm aware, but I know that osmium makes several compounds with a coordinated nitrogen containing a triple bond to the metal centre. An example is the nitridoosmate(VIII) anion, $\ce{OsO3N^{-}}$. Look here for a few more examples. Is this what you wanted?

Are they all coordinate bonds in those structures?

Water has 2 lone pairs on O, so can it donate both (one as a pi bond) just like those compounds? If not, why?

@Nicolau I think this would make a good answer.

"Water has 2 lone pairs on O, so can it donate both (one as a pi bond) just like those compounds? If not, why?"

As I posted:
"It is possible for pi-donating ligands, like $\ce{NH2^{-}}$ (amide anion), but not really possible for water (neutral, oxygen more electronegative)."
Pi-donation from water would lead to oxygen having +2 formal charge, thus becoming very acidic and losing proton - turning into hydroxide ligand, but I think it's the other way around; it first loses proton and then hydroxide pi-donates (becomes doubly bonded). Therefore pi-donor ligands, are often anions. Also unsaturated ligands have this property.

All these cases above seem like special cases - why aren't pi donor ligand more often? I am in high school, and although I have studied about pi acceptor ligands (such as CO), I haven't studied any pi donor ligands' complexes.

No, it's nothing unusual, if ligand is an anion with two or more lone pairs like I- it can have even strong pi-donating properties, check this link http://www.chm.davidson.edu/vce/coordchem/Ligands.html

There is a class of compounds, called carbodiphosphoranes, which formally has a divalent carbon (0) central atom. Employed as a ligand it acts as a sigma and pi donor. This capability is also one of the major features to distinguish these compounds and derivatives from carbenes, which are sigma donors and pi acceptors. The following is reproduced from Thi Ai Nhung Nguyen and Gernot Frenking, Chem. Eur. J., 2012, 18, 12733-1748:

Similarly carbodicarbenes (substitute the phosphine with a carbene), also known as bent allenes, are also double donors.
This post shall only serve as an example, I am quite certain, there are more compound with this kind of bonding motive out there. Nicolau mentioned $\ce{OsO3N^{-}}$ and other metal nitrido complexes, which he will hopefully extend in an answer.

This is a very ubiquitous process. In most cases, the second bond will be considerable weaker than the first one, not only because of the general weakness of π bonds compared to σ bonds.
In many answers of mine, including this one, I have included the basic molecular orbital scheme of an octahedral complex surrounded only by σ type ligands. If we add π type orbitals on the ligands, they will transform as $\mathrm{t_{1g} + t_{1u} + t_{1g} + t_{2u}}$. Only the $\mathrm{t_{2g}}$ type can properly interact with the metal’s orbitals since it is the only one with fitting symmetry, and the only fitting orbitals it can interact with are the metal $\mathrm{t_{2g}}$ orbitals: $\mathrm{d}_{xy}, \mathrm{d}_{yz}$ and $\mathrm{d}_{xz}$. The improved scheme will look something like this:

Figure 1: Molecular orbital scheme of an octahedral complex with σ and π ligand-metal interactions.
Now imagine a halido ligand such as chloride in hexachloridoferrate(III) $\ce{[FeCl6]^3-}$. All of the ligand orbitals are of course fully populated, and each of iron’s 3d orbitals contains a single electron. The $\mathrm{t_{2g}}$ orbitals will contain two electrons each, the $\mathrm{t_{2g}^*}$ ones one each. This can be thought of one additional electron per orbital being donated from the ligand to the metal.
It gets more pronounced if we go to tetrahedral complexes. Consider tetrachloridotitanium(IV), often called titanium tetrachloride. Yes, it is technically closer to a molecule, but we can also consider it a very strong coordination complex. Its orbital scheme would look something like this:

Figure 2: General molecular orbital scheme of a tetrahedral complex with σ and π type ligand-metal interaction. Originally taken from this answer of mine.
Originally, there were no electrons in titanium(IV) — the metal side — but each ligand orbital — chloride — was again fully populated. The discussion is a lot more complex since both species have two sets of $\mathrm{t_2}$ orbitals resulting in four orbitals of that kind altogether, however, we can safely say that $\mathrm{a_1}$ and one of the $\mathrm{t_2}$’s represent the σ type interactions. Thus the other $\mathrm{t_2}$ and $\mathrm{e}$ represent π type interactions — from five of the eight ligand p-orbitals with π symmetry along the $\ce{L-M}$ bond (the other three are represented by $\mathrm{t_1}$ and are nonbonding). Strictly speaking, each orbital is contributing to bonding, and also each of these interactions is a ligand-donated electron pair, so you can consider five electron pairs donated in a π manner from the four chlorides; together with the four σ electron pairs that gives you an overall bond order of $2.25$ lone pairs donated per chloride. More than one.
Finally, let’s consider the pentaaquavanadyl(IV) ion $\ce{[VO(H2O)5]^2+}$. This is an octahedral complex similar to the one in figure 1, but it’s special. First, let’s neglect the π contributions of the five water ligands altogether (they’re weak anyway). We would arrive at the most basic of all octahedral orbital schemes as in the figure in the linked answer above. You see that there are three $\mathrm{t_{2g}}$ orbitals ready for accepting π interactions from any ligand available. If we assume the oxido ligand to be sitting in positive $z$ direction, then these two are $\mathrm{d}_{xz}$ and $\mathrm{d}_{yz}$. And the oxido ligand has two ready and filled p-type orbitals orthogonal to each other that have the same perfect symmetry: $\mathrm{p}_x$ and $\mathrm{p}_y$. Each of these two can now form an interaction with the central vanadium ion, resulting in two donated π coordinate bonds in addition to the $\mathrm{d}_{z^2}^\unicode[Times]{x3c3}\mathrm{p}_z^\unicode[Times]{x3c3}$ $\ce{V-O}$ bond. The best way to represent the ion would be $\ce{{^{+}}V#O+}$ — a triple bond and a formal positive charge on oxygen. All three electron pairs of this bond are oxygen-donated coordinate bonds.
This strong triple bond also explains the extremely short vanadium—oxygen distance of less than $160~\mathrm{pm}$ (for comparison: $\ce{C-C}$ bonds are typically $154~\mathrm{pm}$, the other aqualigands have bonds lengths of c. $210~\mathrm{pm}$) and also explains why the oxido ligand does not get protonated although we are in aquaeous medium where oxides usually do not survive.
Acknowledgement
Most of the knowledge presented herein was generated by Professor Klüfers in his coordination chemistry courses at the LMU Munich.

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Why does egg yolk turn white when washed with water?
I recently noticed that whenever I washed the egg yolk stuck on some utensils the water becomes milky white.
It has to be the egg yolk that turns white and give off the white color to water right?


text:
Why is fat nonpolar?
Why is fat nonpolar while water is polar?
In water there is the $\ce{H-O}$ bond and the $\ce{O}$ is more negative, we have
$\text{dipole = charge}* \text{distance}$
A fat molecule is like
$\ce{H-O-C-C-C-C-C-C...}$
but the carbon atoms don't matter, the polarity is still ${q*d}$, as in the water molecule, so they should have the same amount of polarity.
So why do people say fat is a nonpolar molecule?

Welcome to chemistry.SE! If you had any questions about the policies of our community, you can visit the help center or take a tour of the website.

The polarity of a single bond might indeed be the same for some different molecules. When speaking of polarity of molecules however, the whole molecule should be taken into account, since the macro properties are dictated by all the parts of a molecule.
Regarding the fatty acids: The 'head' of such a molecule is indeed polar, and in specific cases this means that the interactions and reactivity will be like polar molecules. This effect however is only contained to the head of the molecule. Fatty acids as a whole are still considered non-polar because this type of molecule will not be dissolved in water. Note that the main reason for this is not due to the interactions of water with the fatty acid. The fatty acid is just as 'happy' interacting with water as it is with other fatty acid chains. Water, on the other end, does not 'want' to interact with fatty acids, since it will have to rearrange around the fatty acid molecule to ensure the same number of hydrogen bonds are made. This is an entropic cost, that makes the total Gibbs Free Energy change unfavorable.
The result is that water molecules do not rearrange the hydrogen bonds with other water molecules and the fatty acids interact with each other. Due to this effect, water not being willing to rearrange hydrogen bonds, fatty acids are considered non-polar.
Hope this has helped, and if I used to much terminology or jargon, please let me know!

"Water, on the other end, does not 'want' to interact with fatty acids, since it cannot form nearly as many hydrogen bonds as it can with other water molecules." - this is probably not the case; most hydrophobic compounds have favourable enthalpy of solution, but unfavourable entropy of solution due presumably to highly ordered water structure around the solvated hydrocarbon chain.

You are right in the sense that for many small non-polar entities the enthalpy is still favorable. Sorry for taking that shortcut. In this case the Entropy is the reason for the total Gibbs Free Energy to be unfavorable. Would you like to see an edit on that, stating the reason for not 'wanting' to interact is an entropic effect?
Further more I would like to add that this holds for small molecules. In the literature I consulted, an example case of neopentane was discussed. I could imagine, and I do not have numbers to back this up, that for long fatty acids, this does not still hold.

The hydrophobic effect is really complex and most explanations are heavy with conjecture. For a long-chain fatty acid I would expect positive enthalpy of solution; for example from this paper:
"In other words, a large positive enthalpy of mixing of a
non-polar solute with water was compensated by a
large negative enthalpy of iceberg formation that
resulted in a higher solubility than that for a regular
solution. This explanation should be applicable
to the dissolution of a fatty acid in water at a temperature
below 150 degrees C."

"Iceberg formation" just refers to the reorganization of the hydrogen bonding network to a more ordered state.

Okay, so the paper states the numbers I did not have (or looked for). So under 150 Celsius. All clear I would say. Thanks for looking  that up.
Regarding the question, I can imagine that this is a bit to deep into the subject, though it is of course best to have a proper, full explanation is stead of an incomplete one. I'll make an edit to my answer.

While there is nothing wrong with @eljee's answer there is a more obvious reason why fats are not as polar as you assume.
Fats are not long chain alcohols: they are the triglyceride esters of long chain carboxylic acids. So, rather than being terminated by a relatively polar carboxylic acid or alcohol group, they are terminated by a relatively non polar glycerol ester group. Only when saponified (which breaks the free acids from the ester) do they become the more schizophrenic long chain acids which act as surfactants because one end of the molecule is polar enough to like water while the other is fatty enough to prefer oily, non-polar environments. Hence soap!
The wikipedia page summarises this well.


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'Emulating' calcium hydroxide
I posed this question at the Seasoned Advice SE and thought I might get some more information here.
Traditionally prepared corn is cooked in a calcium hydroxide solution to make it more digestible, but when my search for $\ce{Ca(OH)2}$ hit a dead end, I began thinking: could I make a substitute with stuff I already have (sodium hydroxide, which I use for pretzel/soap making; and calcium chloride, for brewing)?
My thought process was that, upon dissolving, sodium hydroxide plus calcium chloride would provide the same ions (though not necessarily in the same concentration) as calcium hydroxide and sodium chloride. Since I would be adding salt to this recipe at some point anyway, I didn't see any harm to this.
Is this feasible, or am I off-base in my chemical understanding about this? More importantly, is there some chemical consideration I've failed to take into account?

Your idea seems mostly sound, but one thing you should consider is the solubility of calcium hydroxide (a.k.a. slaked lime).  It is fairly low.
That low solubility provides you with an opportunity.  If you have calcium chloride and sodium hydroxide, you can mix saturated solutions of both and a white precipitate should form.  The precipitate is calcium hydroxide.  You can then remove the precipitated calcium hydroxide, wash it with ice cold water, and then you'll have $\ce{Ca(OH)2}$!  A YouTube demo of this idea provides some more details.
$$\ce{CaCl2(aq) +  2NaOH(aq) -> 2NaCl(aq) + Ca(OH)2(s)}$$
I think this method is preferred because the low solubility of calcium hydroxide compared to sodium hydroxide naturally limits its alkalinity and corrosiveness.  It would be too easy to use undesirable (and possibly even dangerous) levels of NaOH if you made a mistake in how much you used for nixtamalification.  Additionally, simply mixing NaOH and calcium chloride and using the resulting milky liquid as a substitute for calcium hydroxide will have far too much salt (NaCl) present to be a good substitute.  Or conversely, if you lower the amount of NaOH and calcium chloride to where the saltiness is tolerable, you will have too little alkalinity.

I see. Thanks for the advice. How complete would this reaction be? As in, if I measured precisely, would I end up with fairly close to 100% Ca(OH)2 and salt water at the end? Or would there be significant amounts of calcium, hydroxide, sodium and chloride ions left in the water?

You could calculate this if you wanted from the solubility product of calcium hydroxide.  By a very rough back of the envelope calculation, if you used 1 molar calcium chloride and 2 molar of sodium hydroxide, the concentration of calcium ions left in the solution would be about 10 millimolar.  That is, the reaction should be about 99% complete.

Am I right that calcium hydroxide's solubility decreases as temperature rises? If so could I push the reaction even farther by boiling the solution?

If the purpose is purely for experimentation, the above answer is the way to go. However if you just need calcium hydroxide, you can purchase it in a food grade form called Cal Mexicana, which is used to remove the husks from corn prior to making corn flour for tortillas with an added benefit of increasing the bioavailability of niacin in corn.
I know the question is old, but someone who just needs calcium hydroxide for say hydroponic/aquaponic pH adjustment, this is a safe and inexpensive solution.

Thanks, the issue was always availability. I knew what I was looking for was pickling lime or cal or whatever, it just was not available at any stores near me and so I figured rather than order it online, I would see if I could make it with what I had on hand (so it was a bit of an experiment).


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How quickly does sulfur hexafluoride mix with the atmosphere?
A common science demonstration is to float a "boat" on sulfur hexafluoride.
How long would the sulfur hexafluoride stay in the container (in the experiment linked above) if the container were to be left uncovered and undisturbed at room temperature?
Also, I'm wondering what would happen if you scaled the experiment up to the size of a lake, ignoring the time it would take to fill said lake and the impossibility of obtaining that much sulfur hexafluoride.  Let's also assume that there is no wind to blow it away. Could you have a large permanent pool of sulfur hexafluoride or would it eventually dissipate and mix with the atmosphere?
Would this change if a different dense (non-reactive) gas were used, like xenon?

Up front: this depends on what you mean by "stay in". It would diffuse out of the container and, given the surroundings are closed, obtain a small equilibrium concentration.
Not that I could provide an exact answer to your question (it is rather a physics question), but this paper (paywalled) deals with the diffusion of the $\ce{N2-SF6}$ pair. The binary diffusion coefficient is reported to be 0.0945 cm2/s. In their experiments they used a container of d = 0.5 m, l = 1 m, introduced a small amount (2-5%) of $\ce{SF6}$ at the bottom of the vessel and measured the time it took to obtain the equilibrium concentration of the mixture. For this setup, this was ~15 h, with a reported time constant of 4-5 hours (I believe this is the time it takes to reach a value of [local $\ce{SF6}$ concentration]/[equilibrium concentration] of 0.5).
They used a lengthy formula derived from a one-dimensional model for the comparison of calculated and measured concentrations and in my opinion the measured data agreed quite well with the calculated ones.


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Calculating pH for titration of weak base with strong acid

Calculate the pH at the equivalence point for the titration of $\pu{0.130 M}$ methylamine ($\ce{CH3NH2}$) with $\pu{0.130 M}$ $\ce{HCl}$. The $K_\mathrm{b}$ of methylamine is ${5.0 \cdot 10^{–4}}$.

So I started with the equation:
$$\ce{HCl + CH3NH2 <=> CH3NH3+ + Cl-}$$
and then I knew that
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \left(\frac{\ce{[base]}}{\ce{[acid]}} \right)$$
So, I put ${\log \left(\frac{0.130}{0.130}\right) = \log 1 = 0}$ and then added that to the $\mathrm{p}K_\mathrm{a}$, which I got from the equation
$$\mathrm{p}K_\mathrm{a} = \frac{K_\mathrm{w}}{K_\mathrm{b}} \quad \rightarrow
 \quad \mathrm{p}K_\mathrm{a} = -\log(K_\mathrm{a})$$
However, after I plugged those in to get a $\mathrm{pH}$, it turned out to be wrong and then comments said that when titrated a weak base with a strong acid, the volume is doubled at equivalence point and the concentrations are halved.
Why is this? I now know that my original equation was wrong and it should be
$$\ce{CH3NH3+ <=> H+ + CH3NH2}$$
and from there I should make an ICE table with the concentration of $\pu{0.0650 M}$.

First of all, when you titrate  a weak base (methylamine) with a strong acid, the equation of titration is:$$\ce{H+(aq) + CH3NH2 -> CH3NH3+ }$$The reaction of titration, as you can see is total and quantitative. The constant of the aforementioned equilibrium is:  $$ K=\frac{K_\mathrm b}{K_\mathrm w}= 5\times 10^{10} \gg 10^3$$
According to the stoichiometry of the neutralization equation, at the equivalence point: $n_{\mathrm{acid}}=n_{\mathrm{base}}$ $$ C_{\mathrm{acid}}.V_{\mathrm{acid}}=C_{\mathrm{base}}.V_{\mathrm{base}}$$
As the initial concentration of the base equals the initial concentration of the acid, the volume of the acid at the equivalence point equals the volume of the base:  $$ V_{\mathrm{acid}}^{equiv.}=V_{\mathrm{base}}$$
You can see that the total volume is doubled.
Now, let's calculate the $\mathrm{pH}$ at the equivalence point: We have a weak acid at the initial concentration $$C_{\ce{CH3NH3+}}=\frac{C_{\mathrm{base}}.V_{\mathrm{base}}}{V_{\mathrm{base}}+V_{\mathrm{acid}}}=\frac{C_{\mathrm{base}}.V_{\mathrm{base}}}{2V_{\mathrm{base}}}=\frac{C_{\mathrm{base}}}{2}= \frac{0.13}{2} = \pu{0.065 M}$$
It's in equilibrium with its conjugated base:
$$\ce{CH3NH3+<=>H+(aq) + CH3NH2}$$ The acid is partially dissociated:$$ [\ce{CH3NH3+}]= 0.065-x$$ $$ [\ce{CH3NH2}]=x$$ $$ [\ce{H+(aq)}]=x$$
Let's write the constant of this equilibrium:
$$K_\mathrm a= 0.2\times 10^{-10}= \frac{x^2}{0.065-x}$$
We solve this equation of second order to find $x$, the concentration of ion hydronium.
We find: $\mathrm{pH}= 5.94$


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Double Replacement Reactions form Precipitates
Why do most all double replacement reactions result in a precipitate and an aqueous molecule? Is it because the individual solute molecules always form stronger bonds such as how in single replacement the molecule with more energy bonds with the molecule of the opposite charge?

Precipitation of silver chloride is a common example of a double replacement reaction. e.g. $$\ce{AgNO3~_{(aq)} + NaCl_{(aq)} -> AgCl_{(s)} + NaNO3~_{(aq)}}$$
Initially all four ions are present in solution. When the solutions are mixed the silver and chloride ions bond together and precipitate, leaving the sodium and nitrate ions remaining in solution along with a small number of silver and chloride ions because the precipitation is an equilibrium so a small number of dissolved ions will still be present (thanks to @Martin for pointing this out). The reaction happens because the precipitation of silver chloride removes these ions from the solution equilibrium, driving the reaction to completion.
Silver chloride is insoluble because the $\ce{Ag-Cl}$ bond has a high degree of covalent character due to the high polarizing ability of the silver cation and the high polarizability of the chloride anion and the similar size of both ions which allows a shorter bond length and so a stronger bond requiring more energy to break apart in order to dissolve the solid. See this question for more detail on this point.

Most of the double displacement reactions form precipitates but not all of them.
precipitates are substances formed after a chemical reaction which are insoluble in water.the products formed by double displacement may or may not be soluble in water

You seem to be repeating what the question said. If this is an answer, you should add more.

Precipitation helps drive the reaction in one direction by removing some of the reactants from solution.
Another example of double displacement is $\ce{FeS + H2SO4 -> FeHSO4 + H2S}$ where the reaction is driven in the forward direction by the loss of gaseous $\ce{H2S}$.


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Why do dianions (such as malonate) bind cations more strongly than anions?
Why does a dianion (such as malonate) bind cations more strongly than its equivalent anion (acetate)?
Is it simply because of the proximal availability of another $\ce{O-}$ group that can bind to cations? Does the second $\ce{COO-}$ group on malonate distribute charge in a better way that can allow for stronger binding?

This is known as the chelate effect.
The main reason why you observe this is that cations in solution have an ordered solvent shell around them, especially in polar solvents where there will be defined solvent geometry around the shell; often octahedral for metal ions in water. The formation of a complex with a bidentate ligand such as malonate is more favourable than that with a monodentate ligand such as acetate because the same number of solvent ligands are lost with half the number of ligands complexed. This results in a greater increase in entropy - you can see this in terms of the degrees of freedom gained by the water molecules more than compensating for those lost by the new ligands.

Is it simply because of the proximal availability of another $\ce{O-}$ group that can bind to cations ?

This is likely also a factor, although in most cases less important. In the case of the monodentate ligands there is a small energetic cost associated with bringing the anionic oxygen centres close to complex the metal centres. In the case of the bidentate ligand, some of this cost is "already paid" in the enthalpy of formation of the ligand - it is preorganized. This contribution becomes dominant for macrocyclic ligands.


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Manganese(VII) oxidation state
A transition element is one which forms one or more stable ions which have partially filled d-orbitals. Manganese (Mn) could have an oxidation state of 7+, which means that it will lose all of the $\ce{3d^5}$ and $\ce{4s^2}$ electrons. So how could it make a 7+ ion in that case (which means that it will lose all of the $\ce{4s^2}$ and $\ce{3d^5}$ electrons) and still be defined as a transition element?

It has plenty of other transition states too. Look back at your definition.

It would be permanganate - it's impossible to have bare +7 cation, even +3 would be problematic - probably not really bare even if it would seem so.

yeah i know that manganese has various states, does it depend to whats covalently bonded to manganese atom (how much of its outer electrons are shared).?

Yes, you're right.

The fact that manganese has a stable oxidation state in which it has (formally) lost all its $d$ electrons does not mean that it cannot be considered a transition metal, because it still has a lot of other oxidation states (like +II in $\ce{Mn^2+}$ or +VI in $\ce{MnO4^2-}$) with partially filled $d$ orbitals. Moreover, a naked $\ce{Mn^7+}$ ion does not exist in solution, it is only stable in the form of the oxocomplex $\ce{MnO4-}$.


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Computer killed by insecticide; what compounds are likely?
A friend of mine sprayed "Black Flag Home Defense" insecticide under her
computer. It died immediately. I'm trying to figure out why, and how
fixable it might be.
I found an MSDS, but it only says what the active ingredient is,
deltamethrin, 0.02%. Apparently everything else is proprietary and/or
non-hazardous.
Deltamethrin is not water-soluble, so I'm thinking that this probably
doesn't contain any water; more like nonpolar hydrocarbon sorts of
compounds. I'm thinking there's some particular nonpolar solvent that
makes up the bulk of it.
So, my questions:

Information or informed speculation as to the likely solvent
involved.
For those with knowledge of electronics, information or informed
speculation about the effect of such solvents on PC boards and
electronic components.

Answers to 1 alone are also useful, as I can then ask on
electronics.se what those compounds would do.
UPDATE: I don't have access to the computer myself. I passed on some of
the suggestions to my friend, and she will bring it in to a shop and see
how repairable the damage seems to be. Thanks for all your thoughts.
UPDATE 2: I spoke with the tech who looked at it. Apparently the
computer was very dirty inside; years of dust, cigarette smoke residue,
and insects. She's not sure exactly what happened, but there seems to be
a short somewhere on the motherboard, and that may have fried the power
supply as well. Given the dirt, it was probably not in good shape even
before the incident; the heat sinks were thoroughly clogged.
Overall conclusion: keep your computer clean! Keeping unknown chemicals
away from it is a good idea too.

Two questions which might help: 1. How much did she spray? From one quick spray to a full can. 2. Do the label or MSDS mention the aerosol used?

When you say computer died, could you elaborate a little more? Does it power up at all? Any lights internally or externally? Do you see anything on the screen? I realize those are not chemistry type questions, but they would help with fixability.

Can you open up the computer case? Could it be that the insect driven by the insecticide enters the computer and causes a short circuit?

Ingredients
So we could probably guess what's in the insecticide based on what's used in some of their other products.
Black Flag Flying Insect Killer
Tetramethrin - Sounds like the "active ingredient" here. By "active ingredient", I mean poison.
D-phenothrin - Another poison. This product sounds like it kills more than one insect.
Petroleum Distilates - Likely a solvent. Unspecified
Liquefied Petroleum Gas - Propane or butane. An aerosol propellant. Was your product from a bucket or can? If a can, then it probably has this stuff.
So what happened?
My guess is that one of the products created a short circuit

A short circuit is an abnormal connection between two nodes of an electric circuit intended to be at different voltages. This results in an excessive electric current limited only by the Thévenin equivalent resistance of the rest of the network and potentially causes circuit damage, overheating, fire or explosion.

Typically, when something like a computer experiences a short circuit, it will stop functioning, "die", once the irreversible damage is done. As specified in the above text, circuit damage is likely done, and the circuits will need to be rewired. If there was no damage, then removing any remaining substances through a solvent, followed by thoroughly drying the computer, would remove the short circuit and allow it to function normally. It's possible that if the short circuit was created by a solvent, that the solvent has already been removed by evaporation, due to common organic solvents' melting points.

Nobody comes here to get guesses!

Given that the product is proprietary, and I'm not about to do a separation and NMR of the mixture, it's difficult to specify exact contents of the insecticide. Same with knowing exactly what happened given the amount of information OP has provided.

@JohnSnow: This seems like the best answer so far. However, I don't see how a non-polar solvent or alkane could cause a short circuit. Would such compounds conduct electricity well? It seems unlikely.

The only reason for a short could be that by spraying a lot
of that can some moisture from air condensed due to the
refrigerating action of the spray.
But: Does the computer start again after some time has passed now?
Usually all supply voltages in Computers are short-Proof
and shut down on a short.
After dryinfg the computer should start again.

An interesting point that had not occurred to me. But the computer still didn't start half a day later, so there's probably other damage as well.

Did You switch off mains (or pull the plug) if there is no real mains switch?

From the answer of John Snow I saw that it contained Petroleum Distilates
From my years as IT tech it was absolutely forbidden to use any Petrolium/oil based product for cleaning out the computer parts, because they cause corrosion/react with the compounds.
The acidity of crude oil based products is just too high. It's also a damn good rust remover(try it, it makes your bike all sparkly new again). Even when refined it's still highly reactive(cleaning petrol, lamp oil, etc...)
So spraying it in your computer where there are heaps of soft, easely corrodable metals is generally a bad idea.
For your reading enjoyment
How to fix it? Depends, if pieces have melted I have bad hopes for you being to fix it. You might try dousing it in IsoPropyl Alcohol to flush away any solvents, but i'm afraid the damage has been done and the chips and resistors are dead.
You might be able to replace resistors if they are damaged. You can recognise those by them bulging out instead of having a flat top.
Damaged chips are recognisable by being burnt. Might need a magnifying glass to see it.
But usually in these cases the damage is nigh invisible.

Because the problem was described as immediately failure, I'd put my money on a short over instant corrosion (although in any case, as you say, cleaning it is important to at minimum prevent further damage). Petroleum jelly, at least, is actually great at preventing corrosion on electronics, but can corrode certain insulators over time, like the insulation of some types of wiring. And oil in general makes a great insulator for example.

Petroleum based products are highly reactive. Just put some oil on a cloth and wipe it over some rusted metal. it comes off almost instantly. It will cause a reaction within a second, and if you have a nice hot computer with hot components to stimulate reaction, it's not entirely unfeasable. Just look at the chemical reactions in these gifs http://www.iflscience.com/10-chemical-reaction-gifs-will-amaze-you now these are pretty huge objects and they react instantly. Now imagine a small hot metallic wire reacting to a corrosive substance. its nigh instant reaction.

You're speaking of petroleum based solvents (like, e.g. gasoline) rather than petroleum based products in general (some of which are solvents, others of which are not only safe to use on electronics but widely used as insulators). But I admit to nit picking and losing sight of your point. Sorry. Distillates, like naptha or whatever, could definitely do some damage.

Crude oil "based" products are not necessarily acidic nor do they remove rust.  They are excellent removers of dirt, but not of iron oxides.

Such petroleum distillates are not acidic or corrosive at all!


text:
Why does the mixture's temperature change while phase transition?
Consider a mixture made up of alcohol and water at room temperature. If I freeze this mixture, why does it change its temperature during the phase change?
I don't understand the difference between the case when you have a pure substance (only water or only alcohol) and that one. How could it be different (I mean, in molecular thinking...)?

The concentration of water decreases as the mixture cools, so one is left with fairly pure ice, freezing at ~273 K, and pure(r) ethanol, freezing ~159 K (well, the eutectic mixture, about 93% ethanol). Since the ratio of the components are changing as it freezes, there is no distinct freezing point (unless you start with the eutectic mixture). See the phase diagram for ethanol-water mixture. Instead, the slush mix has a range of solidus and liquidus temperature. BTW, this is a method to make apple jack or brandy by freezing out water from cider or wine.
The same is noticed with solder, where it crystallizes gradually and dully, except for the sudden freezing of a eutectic mix (~63% Sn, ~37% Pb), making a shiny joint.


text:
What exactly is Indium's ability to stabilize non-ferrous metals?
I was reading the History section about Indium on Wikipedia when I came across the following sentence:

In 1924, indium was found to have a valued ability to stabilize non-ferrous metals, which was the first significant use for the element.

I'm wondering if the ability is to improve crystallinity, structural stability of alloys, resistance to corrosion, or something else.

All I did was click on the reference from the History wikipedia link above for the sentence you were talking about. The sentence comes from a paragraph of a manuscript which is found here: http://pubs.acs.org/doi/abs/10.1021/ed011p270
Note the bolded part, that was what is meant.
" Its[Indium's] identification as well as its name came from the beautiful indigo-blue line which it so prominently displays in its spectrum. Its commercial development came as the result of the knowledge that it should be a valuable stabilizer of non-ferrous metals much as chromium is a stabilizer of the ferrous metals. "

Above is quote from:
French, Sidney J. (1934). "A story of indium". Journal of Chemical Education 11 (5): 270. Bibcode:1934JChEd..11..270F. doi:10.1021/ed011p270.

This doesn't really explain how or why indium works.


text:
What is the explanation of the changes in stability going down a group for carbonates, bicarbonates, fluorides, and chlorides?
For carbonates and bicarbonates, I know that stability increases down the group, and for chlorides and fluorides, stability decreases down the group. Why does this happen?  Can someone explain this in detail?
(I am talking about S block alkali metals)

What do you mean by "stability"?

Well how should i explain :-P!I mean less reactive :-)!

Reactive with what?  I'm not trying to be difficult; the terms 'stable' and 'reactive' encompass a lot of different areas & answering your question well depends on exactly what you're referring to.  Can you provide more context to your question?

Ok this is a doubt that occured due to my textbook and here is the direct quotation :  "As the electropositive character
increases down the group, the stability of the
carbonates and hydorgencarbonates increases."

http://www.ncert.nic.in/NCERTS/l/kech203.pdf

Now i guess you'll get the context ! Search for "stabilty" in the pdf

Indeed, that definitely helps.  I may not be able to give a satisfactory answer; this link should assist others in answering, though.

If there is relevant information in the pdf, please include it in the question. Responders shouldn't have to search for it.

I already quoted necessary lines to explain the concept.

"...and for chlorides and fluorides, stability decreases down the group" : This is incorrect. Down the group, thermal stability decreases for fluorides, but increases for chlorides, fluorides and bromides.

Carbonates
The quote from your text:

Carbonates of alkaline earth
  metals are insoluble in water and can be
  precipitated by addition of a sodium or
  ammonium carbonate solution to a solution
  of a soluble salt of these metals. The solubility
  of carbonates in water decreases as the atomic
  number of the metal ion increases. All the
  carbonates decompose on heating to give
  carbon dioxide and the oxide. Beryllium
  carbonate is unstable and can be kept only in
  the atmosphere of CO2. The thermal stability
  increases with increasing cationic size.

So the stability that you are referring to is thermal stability. This is an important detail. So what is thermal stability? It's how resistant a molecule is to decomposition at higher temperatures.
What's happening to cause thermal instability?
So, lets look at the carbonate ion here:

This is just an illustration, and in reality the negative charge we see on the two $\ce{O}$ atoms is localized due to resonance.
Below the illustration shows where the negative charge is likely to be concentrated (colored in red).

So, when we create a carbonate complex like the example below, the negative charge will be attracted to the positive ion.

Because of this polarization, the carbon dioxide will become more stable and energetically favorable.
How does going down a group play into this?
Well as you go down the group, the charged ion becomes larger. The larger the ion, we see a lower charge density. Charge density is basically the amount of charge in a given volume. So, if a small ion has the same charge as a larger ion, the charge density will be greater for that small ion. Greater charge density, means a greater pull on that carbonate ion, and a greater pull causes the delocalized ions, and a more stable $\ce{CO2}$ molecule. So, the larger the ion, the lower the charge density, the less polarizing of an effect, and reduced stability of a $\ce{CO2}$ molecule, favoring the $\ce{CO3}$.
Chloride and fluoride stability
Stability of fluorides, chlorides, and other halogens, are likewise related to thier size. The halogens, specifically fluouride, is known for their electronegativity. Electronegativity, is the tendency to attract electrons to itself. As you move up the group, you see an increase in electronegtivity. This results in the creation of polar bonds.
Illustrated below, you see that as charge of the positive ions increase, polarizability increases (left), and as the halogen ion increases, polarizability and electronegativity decrease (right). When the ions electron cloud, is less polarized, the bond is less strong, leading to a less stable molecule.

Information and illustrations on carbonate ions were sourced from here.

As per this question and answer, it seems, the stability of alkali metal fluorides decreases down the group whereas it increases for alkali metal chlorides, bromides and iodides. In short the trend of enthalpy of formation and hence stability is reversed when comparing fluorides with rest of the halides.

"...as the halogen ion increases, polarizability and electronegativity decrease (right)" : This is incorrect. Polarizability of halides increases down the group due to increase in size and no. of electrons.

Also your answer treats fluorides and chlorides in the same way, in spite of the fact that down the group, thermal stability decreases for fluorides, but increases for chlorides, fluorides and bromides.

By Fajan's Rule you should be getting the answer and then more electropositive metal will have more ionic character and then that will increase stability.

Welcome to chemistry.SE! Could you please be a little more elaborate? Your answer might sound comment-like to some people, and I don't think it will solve the OP's problem, really.


text:
Will the sulfur-iodine cycle and Haber Method work at lower pressures and temperatures?
I have a year-long research project next year in my senior year of high school, in which we have to make some product or prototype to solve some issue.
My interest was in the Liquid Fluoride Thorium Reactor but considering that I do not have the access I need, I opted to invest in the heat it produces.
One use as proposed by Kirk Sorensen proposed using the heat from the reactor to fuel the sulfur-iodine cycle to split water into $\ce{H2}$ and $\ce{O2}$.  Then use those combined with carbon collected from the waste of coal and fossil fuel plants and nitrogen from the atmosphere to make methanol dimethyl-ether and ammonia.
My teacher and I have ruled out the first two, but decided that we might be able to make ammonia.  My question is whether it would be feasible and practical to use the sulfur-iodine cycle in combination with the Haber process to split water and make ammonia at a much lower temperature and pressure to ensure safety.

Yes this is possible but you must remember it is a reversible reaction so if more pressure creates ammonia, applying less will cause the reaction to reverse creating some of the reactants again.


text:
How to calculate the osmotic pressure of a sucrose solution?

The boiling point of an aqueous solution containing sucrose, $\ce{C12H22O11}$, is $101.45~^\circ\mathrm{C}$. Calculate the osmotic pressure of this solution at $35~^\circ\mathrm{C}$, at which the solution density is $\pu{1.036 g/mL}$. The boiling point elevation constant is $k_b = 0.512~^\circ\pu{C\cdot m^{-1}}$.
  The expected answer is
  $\Pi = 37.6~\mathrm{atm}$.

I tried it this way:

I used $\Delta T_b = k_b \cdot m$ and substituted $T_b = 101.45$ and $k_b = 0.512$
and got $m = 198.144~\mathrm{mol/kg}$.
Density equal to $\pu{1.036 g/ml}$ equals $\pu{1.036 kg/L}$, which means $\pu{1.036 kg}$ for $\pu{1 L}$ and $\pu{1 kg} = \pu{0.965 L}$.
$M = \pu{198.144 mol}/\pu{0.965 L} = \pu{205.33 mol/L}$
The osmotic pressure equation $\Pi = MRT = 205.33 \cdot 0.08206 \cdot (35+273)
  = \pu{640 atm}$.

Where have I gone wrong?

I did it in this step: First I use ΔTb=kbm And substitute following Tb=101.45 and kb=0.512 I get m=198.144 mol/kg

This should be the first clue that something went wrong.  198 mol of sucrose is about 67 kilograms of sucrose, so your implied molality means that for every kg of water, there are 67 kg of sucrose.  That can't be right.  (It would be about the same as adding a tablespoon or two of water to a 3-lb bag of sucrose and expecting the whole thing to dissovlve.)
I think you forgot to use the change in boiling point temperature in your $\Delta T_b = k_b m$ equation.  The change is only 1.45 °C, not 101.45 °C.
What happens if you plug in that value and redo your calculations?
Update: Now that you've done that, your answer is at least chemically reasonable but there is still one problem.

Density equal to 1.036 g/ml equals 1.036 kg/L, which means 1.036 kg for 1 L and 1 kg=0.965 L.

This is true, but keep in mind the definition of molality: it is moles of solute per kg of solvent.  However the density is obviously referring to the kg of total solution.  So, it is inappropriate to find the molarity by simply dividing the molality by the density.  As I described in a past answer on chem.se, if the molarity is $C$, the molality is $\frac{1000}{f}\frac{\frac{C f}{\rho}}{1-\frac{Cf}{\rho}}$, where $f$ is the formula weight of the solute and $\rho$ is the density.  So you will have to invert this relationship to find the molarity from the molality.

how can you know its boiling point is 100 degrees? because it's solution can be anything not only water

Your question says "aqueous solution".

Thank you! but after did what you suggest I still doesn't get 37 atm instead I got 74 atm. Is there any of my calculation went wrong again?

Thanks for the comment Champ!  I realize that there is an additional point of confusion so I will update my answer.

After more than five years, it seems safe to assume that an explicit answer will no longer serve to furnish the OP with a homework solution, so I will post numerical values for all the steps already given in Curt F.'s answer:
Molality of the solution
The change in boiling temperature is $\pu{1.45 K}$. Together with the boiling point elevation constant, we can calculate the molality of the solution as $\pu{2.83 mol/kg}$.
Concentration of the solution
To figure out the concentration, let's assume we have a solution containing $\pu{1 kg}$ of solvent. So we would have $\pu{2.83 mol}$ or $\pu{0.9694 kg}$ of sucrose. We can get the mass of the solution as the sum of the mass of solvent and solute, and the volume of the solution because we also know the density. The concentration comes out as $\pu{1.49 M}$. If you write down the calculation, you will see that the mass of solvent cancels out, giving you the following relationship between the molality $b$ and the concentration $c$:
$$c = \frac{b \cdot \rho}{1 + M_\mathrm{sucrose} \cdot b}$$
with $M_\mathrm{sucrose}$ the molar mass of sucrose and $\rho$ the density of the solution at the relevant temperature. Density and concentration are temperature-dependent, while molality is not.
Osmotic pressure
The osmotic pressure comes out as $\pu{37.6 atm}$ when you plug in the correct concentration, temperature and the universal gas constant into $\Pi = c R T$.
Step-by-step math
$ΔT = 1.45\  \mathrm{K}$
$k_{\mathrm{b}} = 0.512\  \frac{\mathrm{K}\ \mathrm{kg}}{\mathrm{mol}}$
$b = \dfrac{ΔT}{k_{\mathrm{b}}}$
$\ \ \ =\dfrac{1.45\  \mathrm{K}}{0.512\  \frac{\mathrm{K}\ \mathrm{kg}}{\mathrm{mol}}}$
$\ \ \ =2.83\  \frac{\mathrm{mol}}{\mathrm{kg}}$
$ρ = 1.036\  \frac{\mathrm{kg}}{\mathrm{L}}$
$m_{\mathrm{solvent}}$ $= 1\  \mathrm{kg}$
$c = \dfrac{ρ}{\dfrac{1}{b} + M_{\mathrm{\ce{C12H22O11}}}}$
$\ \ \ =\dfrac{1.036\  \frac{\mathrm{kg}}{\mathrm{L}}}{\dfrac{1}{2.83\  \frac{\mathrm{mol}}{\mathrm{kg}}} + 342.30\  \frac{\mathrm{g}}{\mathrm{mol}}}$
$\ \ \ =\dfrac{1.036\  \frac{\mathrm{kg}}{\mathrm{L}}}{0.3531\  \frac{\mathrm{kg}}{\mathrm{mol}} + 342.30\  \frac{\mathrm{g}}{\mathrm{mol}}}$
$\ \ \ =\dfrac{1.036\  \frac{\mathrm{kg}}{\mathrm{L}}}{0.6954\  \frac{\mathrm{kg}}{\mathrm{mol}}}$
$\ \ \ =1.490\  \frac{\mathrm{mol}}{\mathrm{L}}$
$R = 0.08205\  \frac{\mathrm{atm}\ \mathrm{L}}{\mathrm{K}\ \mathrm{mol}}$
$T = 35.\  \mathrm{°aC}$
$Π = c \cdot R \cdot T$
$\ \ \ =1.490\  \frac{\mathrm{mol}}{\mathrm{L}} \cdot 0.08205\  \frac{\mathrm{atm}\ \mathrm{L}}{\mathrm{K}\ \mathrm{mol}} \cdot 35.\  \mathrm{°aC}$
$\ \ \ =0.12224\  \frac{\mathrm{atm}}{\mathrm{K}} \cdot 308.\  \mathrm{K}$
$\ \ \ =37.7\  \mathrm{atm}$

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


text:
heat capacity and thermal equilibrium
A $52\ \mathrm{g}$ piece of iron at $250\ \mathrm{^\circ C}$ is added to a mixture of $100\ \mathrm{g}$ of ice and $100\ \mathrm{g}$ of liquid water at thermal equilibrium $(0\ \mathrm{^\circ C})$. Calculate the mass of ice that melts (not all of the ice will melt, so you know the final temp of everything mixed together).
I think that the temperature of the mixture should stay same, if is at thermal equilibrium. Because then it less willing to accept change in temperature. But I am not sure how to use heat of fusion and heat capacity of liquid water to solve this.
My attempt:
$Q$ lost from iron = $Q$ gained by water + ice mixture
$(0.93\ \mathrm{mol}\ \ce{Fe})(25.1\ \mathrm{J\ mol^{-1}\ K^{-1}})(T_\text{f}-250\ \mathrm{^\circ C}) = -(200\ \mathrm{g}\ \text{mixture})(\text{what is }C\text{?})(T_\text{f}-0\ \mathrm{^\circ C})$

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i attempted it, can you tell me if it's right please

Is the first paragraph part of the homework question? Is it given that all ice won't melt? And if all ice doesn't melt, what should be the final temperature of the mixture?

Seems like you need the heat capacities of iron and ice

You'll also need heat of fusion of water


text:
Do Faraday's gold colloidal solutions still exist?
The Royal Institution seems to claim that they still possess some of Faraday's original gold colloids: "These liquids are some of the first examples of metallic gold colloids, made by Michael Faraday over 150 years ago."
However, in a publication in Gold Bulletin (D. Thompson, Gold Bull. 2007, 40(4), 267-269.), the following is stated: "Faraday's original gold colloids were very stable and at least one example survived until the second World War, when it was accidentally destroyed by enemy action [...]."
Do Faraday's gold colloidal solutions still exist?

The Royal Institution of Great Britain has a Facebook page, and I shared this question there, hoping they would respond here. Instead they answered my inquiry on Facebook itself. Therefore I am quoting it here:

Hi Martin. I'm afraid David Thompson is a little mistaken. We have multiple examples of Faraday colloids with the evidence to prove their provenance. We have over 10 samples and a further 700+ slides of his colloidal solutions which have been fixed to glass.
For more information, you can look at the research of Ryan Tweney who has looked at the solutions and slides extensively: http://ma.ecsdl.org/content/MA2009-02/3/202.full.pdf
There's also more information on our website:
http://www.rigb.org/our-history/iconic-objects/iconic-objects-list/faraday-gold-colloids?hc_location=ufi

Please note that I haven't seen the sample myself!
However, the Royal Institution claims that a bottle with the collidal gold solution is on display at Michael Faraday's Magnetic Laboratory (see bottle at the bottom of the image) in the museum of the Royal Institution.
Update
This might be nitpicking and English isn't my native language, but David Thompson does not explicitly state in his article in Gold Bulletin that all samples are lost.
On the other hand, Ryan D. Tweney in his article Discovering Discovery: How Faraday Found the First Metallic Colloid, published in Perspectives on Science, 2006, 14, 97-121 (DOI) not only describes the finding of 700 surviving microscopic slides in Faraday's reconstructed magnetic with gold film samples and numbered in Faraday's hand and that the numbers correspond to the numbering in Faraday's 1856 Diary (lab journal) but continues:

It had long been known that some of Faraday's gold colloids survive from the 1856 research: Four bottles of these are on display at the Royal Institution in London […]. These few colloids are all that survive of many dozens referred to in the Diary.


text:
Why does the molar conductivity decrease with increasing charge density?
This is a question in a problem sheet I have been set. Is it do do with the following equation: $\Lambda=\Lambda_0-a\sqrt c$? Surely charge density is proportional to concentration so therefore molar conductivity would decrease linearly with the square root of concentration (and $c$ $\alpha$ charge density?). Also, what is the physical origin of this equation; my lecture notes simply say that the ions interact with each other, is it as simple as that?
The second part to this question asks "Does the actual conductivity decrease?". Presumably by "actual conductivity", he means $\kappa$ where $\kappa=\Lambda c$. I have no idea about this part so any help would be much appreciated.


text:
Determining which hydrate of calcium chloride I have?
I have calcium chloride in the form of small (1-2 mm) round white pellets, originally bought at a home-brew shop. The packaging gives no indication what hydrate it is. Since each water molecule adds ~16% of the weight of anhydrous CaCl2 this introduces a big error when using it to treat water to a specified amount (in ppm).
Naturally I figured I could throw it in the toaster-oven for a while at 350°F and weigh it every 15 minutes, to figure out how much water was driven off. Results:
0 min.  -  1.15 g.
15 min. -  0.92 g.
30 min. -  0.90 g.
45 min. -  0.90 g.

I originally assumed this would be anhydrous, but after reading that calcium chloride dihydrate decomposes at 347°F, and calcium chloride monohydrate decomposes at 500°F, I'm not so sure.
I couldn't find any other information on figuring out the hydrate online  or indicating what state calcium chloride is typically in (at least that I could understand). Am I correct in thinking I'm left with the monohydrate? Or is it actually anhydrous?

I don't think your answer is there, but have you seen this?

The mass of your sample decreased by 22%.
If you originally had $\ce{CaCl2.4H2O}$ and quantitatively dehydrated that to $\ce{CaCl2.H2O}$, then you should have seen a decrease in mass of 27%.
If you originally had $\ce{CaCl2.4H2O}$ and quantitatively dehydrated that to $\ce{CaCl2.2H2O}$, then you should have seen a decrease in mass of 18%.
Although it's tempting to say that you just have some mix of mono-and-dihydrates, I think that we just can't ignore the stability of your mass measurements and the temperature of your sample, which due to the radiative heating from the toaster heating coils was probably even greater than $\mathrm{350^oF}$. The most reasonable option remaining is that you originally had primarily $\ce{CaCl2.4H2O}$, but also some $\ce{CaCl2.6H2O}$ mixed in, and now you have just the monohydrate remaining.


text:
What is palindromic DNA?
What is palindromic DNA and why isn't every complementary strand palindromic?
I know AGCT is palindromic, but what is an example of a strand that isn't?

I'm voting to close this question as off-topic because it seems to be more about biology than pure chemistry. Better move to Biology.SE

It came from a biochem course. I was eavesdropping in some friends when they were studying.

This is a chemistry question, not a biology question.  DNA is a complex heteropolymer and whether or not comething is a palindrome can affect the polymer's reactivity and behavior.  Also, since when have questions needed to be about "pure" chemistry to be on topic around here? :-)

A palindromic stretch of DNA is a strand whose reverse complement is itself.  So 5'-AAAT-3' is not palindromic.  It's reverse complement is 5'-ATTT-3'.  Those two pieces of DNA are not identical.  However, 5'-GGATCC-3' is palindromic, because the reverse complement is identical.

It has nothing to do with the complementarity (of the other strand).
Try this to figure out whether some sequences are palindromic:
#!/usr/bin/env python3
def is_palindromic(seq):
    translation_table = str.maketrans('ACGT', 'TGCA')
    translation = seq.translate(translation_table)
    #print(seq, translation[::-1])
    return seq == translation[::-1]

is_palindromic('GATTACA')


text:
Allotropy in Nitrogen and Bismuth of Group 15
Which elements of Group 15 of the periodic table have allotropes?
P, As and Sb have well-known allotropes:
P: White, Red and Black;
As: Yellow, Black and Gray (Grey);
Sb: White, Yellow and Black.
What about Nitrogen and Bismuth?

The same source suggests that Nitrogen has an allotrope named Dinitrogen while Bismuth has none.
Chemwiki says:

$\ce{N2}$ does not have any allotropes.

and doesn't comment on Bismuth.
Royal Society of Chemistry says yes for Nitrogen and no for Bismuth.
Inorganic Reactions and Methods: The Formation of Bonds to N, P, As, Sb, Bi (Part 1), Volume 7 has an article which suggests that Bismuth allotrope exists.
This not so reliable source says no for Bismuth.
Wikipedia doesn't mention anything for Bismuth but for Nitrogen:

At atmospheric pressure, molecular nitrogen condenses (liquefies) at $\pu{77 K}$ ($\pu{−195.79^\circ C}$) and freezes at $\pu{63 K}$ ($\pu{−210.01^\circ C}$) into the beta hexagonal close-packed crystal allotropic form.
Unstable allotropes of nitrogen consisting of more than two nitrogen atoms have been produced in the laboratory, like $\ce{N3}$ and $\ce{N4}$. Other (as yet unsynthesized) allotropes include hexazine and octaazacubane.

According to me, nitrogen has allotropes and bismuth doesn't, but the above sources contradict. Any clarifications and reasons for them showing or not showing allotropy would be appreciated.

I feel that the question has become so dense that it might already have the answer.

Solid nitrogen exists in 2 states Alpha nitrogen having cubical structure. Beta nitrogen having hexagonal crystalline structure.

Bismuth certainly has other known crystal structures at elevated pressures and temperatures, at least 4 others in addition to the rhombohedral structure stable at room temperature and pressure.  One place to start would be an article from NIF on shock physics of bismuth. For a scholarly article, you might start with 'Phase Diagrams of Arsenic, Antimony, and Bismuth at Pressures up to 70 kbars', W. Klement et al., Phys Rev 131(2) 632-637 (1963).
As for nitrogen, it too would appear to exhibit a variety of solid diatomic phases under pressure.  A starting point to explore might be 'High-pressure amorphous nitrogen', E. Gergoryanz et al., Phys Rev. B64 052103 (2001), which mentions a variety of phases in addition to the titular amorphous phase they explore.
Both Physical Review articles should be available to the general public...


text:
How is charge density a useful measure of polarising power and polarisability?
How do you use charge density ($Z/r$) as a measurement of polarising power and polarisability, and hence hard/soft acid character?
So far I've been looking at different polarising power and polarisability and how that affects a substance in whether it is a hard or soft acid/base, but I don't understand how charge density comes into play, and how we can use it to understand the characteristics.

What does higher charge density of a cation means? It means that the positive charge on a cation is concentrated in a small volume. Small, concentrated charge can cause greater distortion in the anion and therefore they have greater polarising power, or greater power to cause attract the negative electron cloud of the anion and hence greater distortion. The greater the positive charge on a cation, the larger the charge density will be (it is directly proportional to $Z$). Greater positive charge will cause greater distortion in the electron cloud of anion.
How do small and concentrated charge cause greater distortion? What will affect you more? A force applied on a very very small area of your body, or the same amount of force applied on whole of your body? Of course, the former one will affect us more, i.e., a small and concentrated force.
What about the anion? When the electron cloud of the anion is small, it is closer to the nucleus and experiences a greater attraction from the nucleus, therefore it is not easy to distort it. But, if the electron cloud is large, the inward force experienced by it from the nucleus is less, hence it can be polarised much easily. The inference is that the greater the radius, the larger the polarisability of the anion will be. What does greater radius mean? The greater the radius, the smaller the charge density will be (it's inversely proportional to $r$, or proportional to $1/r$).


text:
What forces keep a cotton fiber together?
According to Wikipedia, a single cellulose molecule is:

a linear chain of several hundred to many thousands of β(1→4) linked
  D-glucose units.

which means a cellulose molecule is made of ~1μm of glucose units.
According to Trotman, 1984,

Each cotton fibre is a unicellular hair [...] Each fibre, however long
  or short it may be, consists of one single complete vegetable cell
  [...] Average fibre length 13-51 mm.
(6th ed., pages 27-33)

(source: Trotman, page 31)

So if cellulose molecules are only ~1μm in length, what forces keep the cellulose molecules together inside the S1,S2,S3 structures throughout the fiber? How ~1μm units end up giving cotton its mechanical properties?

A glucose molecule is roughly 1 nanometer long (ref).  There are roughly 9,000-15,000 glucose molecules in a cellulose fiber (ref, see Table 5).  If we use 10,000 as the number of glucose units, then the length of a cellulose fiber will be
$$10^{-9}~ \mathrm{m/molecule} \cdot 10^4 ~\mathrm{molecules/fiber}=10^{-5}~\mathrm{m/fiber} $$
or ~10 microns/fiber.
The mechanical strength of the fiber comes from two sources.

Structural: As your diagram illustrates there are many walls in the cellulose fiber.  All of these walls reinforce one another and build up the mechanical stability of the fiber. Each wall is composed of highly-organized cellulose chains.  "Highly-organized" is another way of saying "crystalline".  Cotton has an average crystallinity around 75%.  This allows the cellulose fibers to pack closely together further increasing mechanical stability.  The cellulose fibers in the different walls are orientated in different directions, again serving to increase the fiber strength.
Chemical: Here is a drawing of 2 glucose molecules from a cellulose chain.

Note that each glucose molecule has 3 equatorial hydroxyl groups oriented out and away from the polymer chain.  Because of the high crystallinity mentioned above, the cellulose chains are positioned very close to one another.  This means that thousands of hydrogen bonds can form between hydroxyl groups on different chains.  As a rule of thumb, a hydrogen bond might have a strength around 5 kcal/m; if we have thousands of these bonds holding the cellulose chains together, then we have the equivalent of hundreds of normal chemical bonds holding the chains together!  These hydrogen bonds account for much of the stability of the cotton fiber.

so the only thing preventing fibers to "slide off" (shear) of each other is the hydrogen bonding? (I imagine this as plenty of small needles layered on top of each other)

Hmmm, I though you were asking about what gives a single fiber its mechanical properties ("what forces keep the cellulose molecules together inside the S1,S2,S3 structures throughout the fiber?").

yeah, and you gave a good answer. this is just another part of the question...

I know that the fiber surface is smooth so cotton fibers can't hook onto one another.  While hydrogen bonding between fibers may help hold them together, I think the actual manufacturing process of making cotton is what is most important.  In the manufacturing process cotton fibers are aligned (combing) and then twisted with each other (spinning).  I think this is the main reason cotton fibers hold together so well.


text:
Is there a way to efficiently estimate the enthalpy of formation for an organic molecule in computational chemistry?
I'm exploring the platonic hydrocarbons and one of the methods I'd like to use is to compare their enthalpies of formation.
The issue is that carbon's standard form is a covalent lattice and I can't think of a way to efficiently obtain its electronic and thermal enthalpies without resorting to optimising a giant PAH.
FWIW I'm using GaussView/Gaussian.

I would say yes, you could estimate standard enthalpies of formation for molecules more or less accurately by application of quantum and statistical mechanics, although there is a 'but': the usual procedure works for gas-phase heats of formation only.
And thus, if the standard state of the material of interest corresponds to the condensed phase, you have to resort to some procedure which allows you to calculate a condensed phase heat of formation given the corresponding gas-phase one. Such procedures are based on a trivial application of the infamous Hess's law, which for the case of solids can be written as follows
$$
   \Delta H_{\mathrm{solid}} = \Delta H_{\mathrm{gas}} - \Delta H_{\mathrm{subl}} \, ,
$$
where $\Delta H_{\mathrm{subl}}$, the heat of sublimation, is estimated usually from experiementally available data on simillar compounds by a correlation scheme. However, as with any empirically fitted scheme the reliability and the accuracy of any such approach are doubtful.

Alternatively, in many cases, it is enough to compare the gas-phase heat of formations, even if the standard state of material is solid or liquid. To calculate the gas-phase heat of formation for a substance in Gaussian all you need is to run frequencies calculations on the previously optimised geometry of a single molecule.
If you would like to give the gas-phase heats of formations a try, I suggest running G4 calculations first. In the route section just write
# G4

and that would do it. Alternatively, in GaussView's Gaussian Calculation Setup window first choose "Energy" at the "Job Type" tab and then at the "Method" tab choose "Compound..." in the second combobox and "G4" in the third one as shown below.

I did do calculations using atomic carbon but the enthalpies made no sense; the value for tetrahedrane was -2920 kJ/mol.

@gsurfer04, what do you mean by "using atomic carbon"? Also, the level of theory is important. For thermodynamical quantities you could try G3/G4 composite methods first.

@gsurfer04, by G3/G4 I mean use G4 if you have Gaussian 09 and G3 only if you still use the previous Gaussian 03 in which G4 is not available.

G4 is much more expensive, maybe G3 is good enough

@user1420303, may be. I do not remember now, since once I switched to G4, I've never looked back. :D It works surprisingly well.

@Wildcat It will depends of the requirements/resources of the user. I think that for the number of atoms in those molecules, the time difference will be very significative. G4 includes HF hartree-fock calculations up to a modified aug-cc-pV5Z basis set, and specially coupled cluster (CCSD(T)) which scales very bad with system size. It can be interesting to know the computational resolurces available for the job.

I got a copy of Gaussian 09 from my uni. It's a bit late for me to run calculations now but I appreciate the help. I used B3LYP/6-31(d,p).


text:
What happens if don't use MgCl₂ in buffer solution in Ca²⁺/Mg²⁺ titration?
What happens to a complexometric titration of calcium carbonate (analyte) with $\ce{EDTA}$ (titrant) when no magnesium(II) chloride crystals are added to the buffer ($\ce{NH3/NH4^+ + EDTA}$) solution? In a water analysis (where water sample is the analyte and $\ce{EDTA}$ is the titrant)?

You need to give more details to get any answer. What else is used in the experiment? What is the pH of the solution? Unless you elevate the pH to 12 or above, EDTA will also chelate $\ce{Mg^2+}$.

Oh I'm sorry, I forgot to tell that there is an addition of NH3-NH4+ buffer with a pH 10 to EDTA. EBT is the indicator used for the experiment.

Summary The Magnesium is added because the color change is easier to spot with it, even though calcium also changes the color of the solution in presence of TBT.
Long explanation and experiment
Even though your question might looks silly at first, I know it makes sense because I work with Soil Analysis, and the titrimetric determination of $\ce{Ca^2+/Mg^2+}$ is performed using the same pH 10 $\ce{NH3/NH4^+}$ buffer, with the addition of a small amount $\ce{MgSO4}$ and $\ce{EDTA}$ to the buffer.
As far as I know, and according to a fast online search, the addition of magnesium is to ensure the visibility of the end point when using Eriochrome Black T (EBT) as indicator. Supposedly, the color change wouldn't be easily spotted without the magnesium addition. This is how it's supposed to happen:
Considering there will be no change of color in EBT only due to Calcium, a small amount of magnesium is necessary to interact with the indicator and dye the solution. The equivalent number of moles of EDTA is added to the buffer, otherwise the results would be overestimated. Upon completion of the titration, the remaming $\ce{Mg^2+}$ complexed with TBT would be chelated by EDTA and the endpoint detected.
This is important in basically two scenarios:

There is no magnesium in your sample.
The buffer used is not pH 10, but pH 12, exactly to remove magnesium and detect only calcium.

In both cases, you need some (known) amount of magnesium to see the color change.
The problem with this is that Calcium is supposed to color TBT. It's even the example given on TBT's Wikipedia page, and there's also a full article about the interactions of magnesium and TBT. So why would you add magnesium?
To answer your queston - which also became my own - I did a qualitative experiment, because I'm in the lab right now. And the result was interesting.
I added small amounts of Calcium Chloride and Magnesium Sulphate to two different beakers containing purified water, and then added a few drops of TBT. I also added TBT to the purified water as a blank. This is what I got:

As you can see, the $\ce{Ca^2+}$ solution did change the color of the indicator, but not nearly as much as $\ce{Mg^2+}$ did. If you think about real-sample analysis, where usually the water (or soil extract) contains other coloured materials, making sure the endpoint is easily visible is very important. As a colorblind, I already find this change in color a bit hard, and for $\ce{Ca^2+}$ alone I suppose it would be even harder without the magnesium.
So, finally: The Magnesium is added because the color change is easier to spot, even though calcium changes the color of the solution in presence of TBT.


text:
Could we possibly discover new extraterrestrial elements and minerals?
So, after dissolving a good portion of my brain by watching a ton of sci-fi, I started thinking about all of the strange new materials that we will supposedly have in the future. For example, the ever-present "di-lithium" in Star Trek. But, as I started thinking about it, I thought it was a little ridiculous (I know science-FICTION) that all these exotic elements could exist. No matter where a species comes from in the galaxy or universe, they would be working with the exact same Table of Elements as us. Furthermore, so long as they are similar to human-life, i.e. they breathe oxygen, need water, etc., and therefore live on similar planets, wouldn't the compounds be the same as well?
Let's say we are 250 million years into the future and have discovered superluminal space-flight at 100,000,000c. Say we cruise around to inhabitable planets throughout the galaxy and universe. Upon landing on these new worlds, the elements and compounds we find would be the same that occur on Earth, right? Would we still find diamonds, rubies, uranium, iron, etc.? Or is it possible that there are combinations of the elements yet unknown? Is it for certain that we have catalogued every element and isotope that occurs naturally? I understand that new compounds could be formed, but as far as the base elements and the naturally occurring minerals, have we found them all?

I think you'll find Stack Exchange World Building pretty fascinating.

Actually we flew 4 times faster then you say but was quite an achievement ;)

Could someone comment on different "elements" by means of metastable nuclear isomers? Obviously, with insanely long half-lives (I know nothing about nuclear physics, are those possible?). Could this exist somewhere out there? While the chemistry would probably not be that different, the spectroscopy could be.

@TAR86 Better late than never! Nuclear isomers are just nuclei excited above their ground state, and much like atoms/molecules in excited electronic states, they are almost always very unstable. The key word is almost, however. I think you'll be fascinated to learn about tantalum-180m, the only metastable nucleus with significant natural occurrence. The table of radionuclides still has many gaps (especially metastable states), but it is unlikely any gaps hide very surprising species.

One thing which can be said with absolute certainty is that no species  in the Universe (or even all their combined efforts) will even come close to exhausting the daunting enormity of chemical space. When you drag combinatorics into a problem, you can easily stumble on massive numbers, and this is the case here; it has been roughly calculated that there are on the order of $10^{60}$ "chemically reasonable" distinct molecules below a molar mass of $\mathrm{500\ g\ mol^{-1}}$! This means that almost all chemical substances which can exist, never will exist, be it from natural or artificial production.
No matter how far chemistry develops, we will always be interested in investigating the chemical composition of new samples, and will always find many new compounds doing so. Most of the time we'll find the same of what we do on Earth; there's no escaping the fact that $\ce{N_2}$ and $\ce{Al2O3}$ are very stable substances, for example, and can exist in a wide range of conditions. But there will always be something new for us waiting out there. Organic substances are tremendously varied thanks to the concatenability of carbon atoms. One might think inorganic substances are far less variable, but just the class of silicate minerals is enormous, and many polymorphs of the same substances can exist.
Having said that, at least one area of chemistry seems far closer to being exhausted; the elements themselves. There is no possibility of us having "missed" any element among the first 112, and we're still filling in the gaps at higher atomic numbers, but each heavier element discovered involves ever more complicated procedures and lower yields. Much of the experimentally determined properties of elements above $\mathrm{Z=104}$ come from the analysis of a tiny amount of atoms, as little as tens of them. Each new element is significantly more difficult to synthesize, and is less stable, which is hampering progress. We don't yet know very well how to get past $\mathrm{Z=122}$ or so even in principle, and are unsure how accessible the ultraheavy elements will prove to be. There is still great academic interest in several aspects of the elements and their isotopes, but current research deals with the production of very shortlived species, which have limited applicability.
In all, we will always find something new to explore, but whether we will always find something strange depends on what you mean by the word. There are still many, many substances to find which will display impressive biochemical properties, for example, and many lethal or crippling diseases today might still become curable in the future. But there is zero chance of finding a substance which makes magic real, which unlocks faster-than-light travel, which has endochronic properties or which can be used to make overunity engines and perpetual motion devices. Ultimately, Chemistry is a form of applied quantum electrodynamics, a subsection of Physics, and quantum electrodynamics is a very developed theory with incredible precision. Thus, there is little room for surprises, and they likely will not come from a chemical system unless it is very specially tailored, rather than randomly found in nature.

It is worth mentioning the hypothetical 'island of stability' explicitly https://en.m.wikipedia.org/wiki/Island_of_stability. The prediction of stable,  ultra heavy, yet-to-be-discovered elements warrants this great question.


text:
How can I measure the internal temperature of a solid metal?
I’m heating up several metal blocks as part of an experiment to determine their specific heat capacities. In a nutshell, I’m heating the blocks of known mass to 100 °C, then adding them to an insulated vessel containing water of known temperature and measuring the increase. Pretty basic stuff.
I’m trying to work out how I can measure the temperature of the blocks, so that I know they’re at 100 °C. Leaving them in the boiling water for a long period of time is not ideal – I want to be sure that they are at the desired temperature that I can record.
How can I do this effectively? I thought about using an infrared thermometer, but wouldn’t that just measure the temperature on the surface of the block?
I also thought about something to do with measuring the resistance of the metals, but is this actually accurate and easy to do? Any advice or suggestions are appreciated.

Do you really have to measure the internal temperature? I'd rather keep the experimental setup as simple as possible. You could vary the heating time in the 100 °C bath to see if that makes a difference. Imagine that a metal cube is in contact with a heating block at 100 °C. Could you roughly calculate how long it would take for the opposite face to reach the same temperature, depending on the dimension of the cube and the material, but neglecting heat loss through the rest of the surface? This rough estimate might be enough to justify a simple setup.

Thermal diffusivities vary between different metals - I'd note that in particular steels are pretty poor conductors of heat.  I say this so if you are using different metals they may equilibrate over different time scales (10x or so). However, you could drill a hole to the center and use a thermocouple to directly measure the temperature at the center.

What metal is that and what's the volume of one block?

Direct Answer
You are correct in saying an infrared thermometer would just measure the surface of the metal, and in addition infrared thermometers don't measure metal surfaces particularly well anyway (metals typically have a low emissivity). Measuring electrical resistance is better, but minor alloyed contamination would heavily skew your measurement.
Ultrasonic thermometry (as permiakra suggests) is one of the only ways to measure the internal temperature of anything without actually drilling into it. Technically it's mesuring the average temperature along the path of the sound, but that's still something.
Practical Answer
You absolutely do not need to measure the internal temperature of these metal blocks. Metals are ridiculously good conductors of heat. If you have a good measurement of the temperature of the heating bath, you can consider the metal to be the same temperature after about a minute at most, more likely a dozen seconds. On the size scales you're working with even the most non-heat-conductive metals (bismuth, a few aerospace superalloys, plutonium...) will have a temperature within the error tolerance of the other parts of your setup (e.g. convection in the heating bath) within a matter of maybe a minute.
If you really need to convince yourself that metals are that heat conductive, go get a about half a foot of smooth coper pipe*, a piece of notebook paper, and a candle.

Light the candle.

Place the pipe in the middle of the paper and pick up the pipe using the paper as a sling.

Hold this assembly over the candle so that the flame touches the section of paper that is in contact with the pipe

As long as any part of the copper pipe is cool enough to touch, the paper touching any other part will not burn**. This is because as fast as the fire is dumping heat into the pipe, the pipe is spreading it away. Only once the whole pipe has risen significantly in temperature - far past the point where you can touch it without getting burned - will the paper start to burn.
* If the pipe is very thin-walled this might not work, however I have never found a copper pipe that thin.
** There may be a buildup of black soot from the candle on the paper, but this will only be on the side in contact with the flame, the opposite side will show no such buildup.

You can use ultrasound thermometry. It uses change in speed of sound under heat. AFAIK, there are industrial-grade solutions on the market.


text:
While electrolyzing concentrated aqueous sodium chloride, why is it that chlorine is discharged but not sodium?
Hydrogen and hydroxide both exceed sodium and chlorine in terms of reduction and oxidation potential respectively.
While electrolyzing a concentrated solution of aqueous NaCl, it is known that chlorine is discharged at the anode (contaminated with traces of oxygen). However, at the cathode, hydrogen is charged in place of sodium. Why is this so? I have read many explanations, but they never provide the reasoning.
I suspect these particular properties of factoring in preferential discharge:

Atomic mass (and consequently weight).
Magnitude of oxidation & reduction potentials.

Let's take a look at a table of standard electrode potentials and find the relevant half-reactions:
Reductions (there is also a half-reaction for water reduction that doesn't involve a proton at -0.83 V, but for the sake of simplicity, we can ignore it):
$$\begin{align*}
&\ce{Na+ + e- <=> Na} & &E°=-2.71\ \mathrm{V}\\
&\ce{2H+ + 2e- <=> H2} & &E°=0\ \mathrm{V}
\end{align*}$$
Oxidations (reduction reactions reversed):
$$\begin{align*}
&\ce{2H2O <=> O2 + 4H+ + 4e-} & &E°=-1.229\ \mathrm{V}\\
&\ce{2Cl- <=> Cl2 + 2e-} & &E°=-1.36\ \mathrm{V}
\end{align*}$$
Now, if you know the concentrations of the species involved, you can use the Nernst equation to figure out the actual potentials, but we can reason around it without knowing the exact numbers.
The big thing to notice is how close the potentials of each of the reductions and oxidations are to one another—the reduction reactions are separated by almost 3 V, but the oxidation reactions are very similar. This means that for reduction, it is far less energetically costly to produce hydrogen gas than to reduce sodium, and in fact, if any sodium were produced, it would instantly react with the water to produce hydrogen gas, so if there is any water around, that half-reaction will be favoured. In the real world, the exact potential hydrogen is produced depends a great deal on the electrode used (see overpotential), but it's more or less impossible to electrodeposit metals with much lower reduction potentials than zinc in aqueous solutions. To actually produce metallic sodium, electrolysis of molten NaCl without any water present is the most common method.
Looking at the oxidations, the story's quite different. It doesn't take much more energy to oxidize chloride than it does water (and if the chloride concentration is high, the oxidation potential will go up). This means that (subject to the effects of the electrodes on overpotential), some amount of both half-reactions will occur and because chlorine is a gas and therefore leaves the solution and the oxidations potentials are so close (leaving little energy to overcome kinetic barriers), the chlorine produced will not significantly oxidize the water.

In your answer, you mentioned another possible reduction half reaction for water with a having a standard potential of -0.83V. But that can be ignored anyway as it is less than 0V. Similarly, there is an oxidation reaction directly from Hydroxide ions, i.e. 4OH->O2 + 2H2O + 4e. Why don't we consider THIS reaction at the anode? Comparing the standard electrode potentials, this one is much more likely to occur than the oxidation of chloride that you mentioned. So why does Chlorine still get liberated?

Should I explain my doubt better with a separate question or could you understand what I was saying? Thanks

@newton Remember that these are standard electrode potentials. OH- is going to be at much lower concentration than either H2O or Cl-. This is where you must use the Nernst equation with the actual activities of all the species to get the right potentials.

I get your point, but you used the H+ reaction at the cathode right? And not the one involving H2O. We just got lucky that using either results in the same conclusion at the cathode.

@Newton Yeah, I just picked that one because its potential is 0 V so it's easy to see the huge difference versus sodium reduction. It will really be a mix of several reactions, including the H+ one and in the real world, things like the electrode material you use (which determines the overpotential for each reaction) can make a big difference.

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


text:
Orbitals involved in Ni(CO)₄
Which orbitals are filled? Is it diamagnetic? According to CFT, $\ce{CO}$ is a strong field ligand, so will it pair the electrons and be diamagnetic?

I can't understand this answer(http://chemistry.stackexchange.com/questions/7883/how-do-i-determine-crystal-field-splitting-for-example-for-nico4). It says d10 while nickel has 8 d electrons.

It is because $4s$ electrons of $\ce{Ni}$ atom are "pushed" into $3d$ orbitals in the strong field of $\ce{CO}$ ligands. See, for example, here for details.

So, yes, you are perfectly right. The strong field of $\ce{CO}$ ligands will "push" and "pair up" the $\ce{Ni}$ electrons in $3d$ orbitals and $\ce{Ni(CO)4}$ is indeed diamagnetic.

But that push will be against aufbau principle @Wildcat

Yes, it does. And it is a well-known "breakage" of the Aufbau principle similar to what happens in the transition metals cations. See, for instance, here.


text:
Does hydrogen react with air at room temperature and pressure?
I have a container with 3% hydrogen and the rest is air. How fast will the hydrogen get oxidized by air?
How much will the reaction speed up with increased temperature?

The combustion of hydrogen is famous for being an extremely thermodynamically favourable process held back by a very large kinetic barrier. I've heard that mixtures of oxygen and hydrogen can last centuries without appreciable reaction, but I'd be pleased to see a calculation.

The 3% is below the lower explosive limit (LEL), so it won't go BOOM!...

@NicolauSakerNeto Thanks for the info. I wonder, what temperature is necessary for them to react reasonably fast?

@JonCuster Yes, LEL for hydrogen is 17% and LFL is 4% according to http://environmentalchemistry.com/yogi/periodic/H.html

UV can also speed up the reaction. In fact, a mix of hydrogen and chlorine can be exploded with a tiny UV light -- see https://www.youtube.com/watch?v=NN82GoBG98s.

According to the Encyclopedia Britannica, to somewhat paraphrase this non-technical source, the reaction rate of $\ce{O2(g)}$ and $\ce{H2(g)}$ at atmospheric pressure is not measurable at any mixing ratio at temperatures below $\ce{300^oC}$.  Unfortunately that source does not report what kind of experimental detection limits define what "not measurable" means.
These class notes from the University of Toronto Institute for Aerospace Studies, AER1304 course state this in more dramatic fashion:

The reaction between hydrogen and oxygen is a good example of a
multicomponent kinetic system. To describe the system properly, we
should consider eight major species and at least 16 reactions.
[The overall] reaction is exothermic, but mixtures of gaseous hydrogen and oxygen are quite stable at
atmospheric conditions. Any conceivable direct reaction between the
two gases is zero.

The reaction half-time at atmospheric [pressures] has been estimated to be much larger than the age of the universe.
If the reaction is initiated by some free-radical species, then the reaction proceeds very rapidly and violently.

So the manner in which the reaction rate of $\ce{O2(g)}$ with $\ce{H2(g)}$ increases with temperature is very complex, though basically negligible in the absence of a catalyst until you reach the auto-ignition temperature.
According to this Wikipedia article:

The hydrogen autoignition temperature, the temperature of spontaneous
ignition in air, is 500 °C.

Notice that this says air not oxygen. The $\ce{H2}$ concentrations for the basis of this statement range from 4% to 74%.  This seems like quite a large range to give a single result for autoignition temperature, but given the great complexities of the $\ce{H2}$ + $\ce{O2}$ reaction, some estimations are clearly necessary in order to discuss this question in a reasonably concise fashion.
One note about the question regarding the composition and contents of the container; I'd be sure that it never contained any high surface-area platinum powder or be coated with any sort of perchlorate residues or any such ridicules thing!  The whole issue with the difficulty in getting this highly exothermic reaction to proceed is kinetics.  As eluded to in some of the comments, there are chemical catalysts that can help the reaction get over this large energy barrier, it doesn't have to be a spark!


text:
Can carbon monoxide contaminate water?
Just wondering if you fitted a cold water tank overflow pipe into a car park at the height of a car exhaust , if the carbon monoxide fumes went back up the oveflow pipe into the tank would this contaminate the water?

My question is "Will Carbon Monoxide saturated water emit Carbon Monoxide from it's surface?

Contamination would occur if a sufficient quantity of carbon monoxide wouldn't just bubble through but

would react with water to form an unhealthy product or
is dissolved in water

Both isn't really true:

$\ce{CO}$ does not react under these conditions.
The solubility of $\ce{CO}$ in water at 20 °C is as low as $\mathrm{30\,mg\cdot L^{-1}}$ (GESTATIS database)

Therefore a small amount of CO in the air gap above the water would really be seen as sufficient to penetrate the water surface ?

Honestly, my laziness and a stupid cold keep me from doing any calculations on that. But I'd probably drink from the water and wouldn't worry about it.

For comparison, the allowable maximum for benzene is 0.005 mg/l, so soluability of 30 mg/l doesn't mean no effect. This guide says there is no data on CO in drinking water.

Basically I am a plumber I went to clean a tank for a customer and my co meter alarm activated when I opened the lid. I couldnt understand why there would be co in the tank as I have maintained the tank for a few years. On investigation I found that another contractor had run the overflow pipe to the car park and a car was left running. The co has come back up the overflow into the tank.

I decided to drain the tank and ventilate until no more co was reading on the meter. I think chlorinated the tank and reinstated, just hoping its ok.

@DanG Then this is a question about safety of an actual procedure. You should be consulting an expert with appropriate training and certification, not random strangers on the internet.

Thanks for your comment David , yes I did consult the local ukas approved laboratory but felt like getting a second opinion.

@DanG I'd be more concerned about the other components in exhaust fumes which may contaminate the water, heavy metals for a start.


text:
Reason behind ZnO color
Why is zinc oxide yellow when hot and white when cold?
Is it yellow because of the flame used to make it hot?

Welcome to Chemistry.SE! There is a search function on this site, please feel free to give it a try ;)

@KlausWarzecha Admittedly, even though I was sure I'd read a zinc oxide colour question before, I couldn't find it myself after a quick search! Unfortunately the SE search function is not stellar.

@NicolauSakerNeto I simply entered ZnO and could find the question. But I also remembered that I had answered it :D

@Nicolau Apparently the search function is the worst thing about SE for many people. They're really trying hard to improve it.

Sorry. It didn't show up in the search. And thank you for the link. @KlausWarzecha


text:
Partial hydrolysis of XeF6
What are the products of partial hydrolysis of $\ce{XeF_6}$?
Is $\ce{XeOF_4}$ formed or $\ce{XeO_2F_2}$ is formed?

Both can be formed. It depends on the amount of water added:
\begin{align}
&&\ce{XeF6 + H2O  &->  XeOF4 + 2HF}\\
&\text{or}&
\ce{XeF6 + 2H2O &-> XeO2F2 + 4HF}\\
&\text{or even}&
\ce{XeF6 + 3H2O &-> XeO3 + 6HF}
\end{align}

@Martin - マーチン Why Xenic acid is not produced ?


text:
Which species can't appear in the rate expression for this reaction scheme?

Question
  Which species can't appear in the rate expression for this reaction scheme?
  Here is the reaction scheme in steps:
  \begin{align}
\ce{A + B~(slow) &\rightarrow C} \\
\ce{C~(fast) &\rightarrow D + B}\\
\ce{D + E~(fast) &\rightarrow F}\\
\ce{F~(fast) &\rightarrow G}\\
\end{align}

The answers are apparently F and G, but I don't see why D and E couldn't also be valid (considering the first step is the slowest). A and B will inevitably be in the rate equation since they are reactants in the rate determining step. C produces B (alongside D) and so will also be in the rate equation as a result. However, why can't we rule out D and E of being in the rate equation, considering they don't seem to have any effect on the concentrations of A and  B.

B seems to me like a catalyst, and C is an intermediate.

Answering this question requires assuming that all steps are "elementary", e.g. that $\ce{A + B -> C}$ is an irreversible bimolecular reaction that is first-order in $[\ce{A}]$ and first-order in $[\ce{B}]$.  (Obviously we are also assuming that activities can approximated by concentrations.)
With that assumption, the fact that $\ce{C}$ conversion is "fast" means that we can replace the existence of $\ce{C}$ in the scheme entirely:
\begin{align}
\ce{A + B~(slow) &\rightarrow D + B} \\
\ce{D + E~(fast) &\rightarrow F}\\
\ce{F~(fast) &\rightarrow G}\\
\end{align}
Likewise if $\ce{F}$ conversion is "fast", then it can be replaced:
\begin{align}
\ce{A + B~(slow) &\rightarrow D + B} \\
\ce{D + E~(fast) &\rightarrow G}\\
\end{align}
Hopefully this makes it clear why $\ce{E}$ must appear in any rate law for formation of $\ce{G}$: there is no source of it in the two reactions we are left with.
For $\ce{D}$, I agree that the situation is a bit more complex.  It might appear in the rate equation, or it might not.  Necessary reactants at the beginning of the reaction if we are ever to see product $\ce{G}$ are $\ce{A}$, $\ce{B}$, and $\ce{E}$, so it would make sense to eliminate $\ce{D}$ from the rate law if possible.  However, that could be algebraically difficult, or perhaps $\ce{D}$ is easier to measure, etc.  There is no reason $\ce{D}$ can't appear in the rate law.
Huge caveat: "fast" and "slow" are relative terms and as such without further context they are ambiguous.  The problem as constructed sounds a bit sloppy.  For example, is the "fastness" of $\ce{F -> G}$ faster or less fast than the fastness of $\ce{D + E -> F}$?  The problem doesn't say but the simplifications I did above actually depend on these sorts of details.


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Why does this reaction take place in terms of electronic structure?
I need to know, in terms of electronic charge - the electrophiles and nucleophiles involved - how the electronegativity of the potassium and hydrogen atoms alongside the hydroxide and H+ ions result in this reaction taking place. You could also think of this as asking why this reaction is exothermic from an organic chemistry perspective - why are the products more stable in terms of their electronic arrangement. It might be helpful to have the mechanism explained perhaps.
$$\ce{2K + 2H2O -> 2KOH + H2}$$
Of course, I have learnt before that potassium is higher up the reactivity series than hydrogen, but what does this mean on a deeper level - what factors mean that potassium is more reactive than hydrogen in terms of electron arrangement.

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[[I edited the question to balance the reaction.  Only one mole of H2 is produced per 2 moles of K.]]