text:
Why is astatine so radioactive?
I have heard and read about astatine's radioactivity, and it makes no sense to me. Why is it so unstable?  Uranium has a half life ranging from 70 to 4.5 billion years (natural isotopes), and it has 7 extra protons to deal with. I know that atomic nuclei can be stabler if they are spherical, implying that if you make a nucleus that isn't spherical you could destabilize it, but astatine has a maximum half life of 8 hours. Can someone explain this to me?
Information came from a few sources, including SciShow and Wikipedia.

related http://chemistry.stackexchange.com/questions/28695/why-is-tin-112-stable-but-indium-112-radioactive

That is useful, but doesn't actually answer why astatine is so short lived compared to its neighbors, as like I have mentioned: Astatine has a longest half life of around 8 hours. Using the logic from the linked question, couldn't we just shove less neutrons into the nucleus to make it less unstable? This wouldn't work, because of the way the strong nuclear force works but this si chemistry stack exchange, not physics stack exchange.

I wouldn't say there's sth unusual. After Pb with "magical" configuration, nuclei lower they stability. There's simply local minimum for this number of protons.

@Crimsonvale That’s why Mith wrote related and not duplicate ;)

Promethium sticks out much more in the PSE than astatine. Actinium, francium, protactinium are also very unstable compared to their neighbours. An odd number of protons is often not very favourable. But really this is no chemistry question.

What about francium? It has a half-life of 22 minutes

I don't think there is an actual answer to that. What we know about the radioactivity is based mostly on empiric observation and quantum physics is still not able to give precise answers about the process of radioactive decay.
A similar question could be asked about technetium: why is it unstable, even though it stands in the periodic table between stable elements? Probably it's a matter of their proton and neutron configuration that for some reason are particularly unstable compared to their neighbourhood, but there is still no equation capable of giving the half-life of an isotope.
This has already been discussed and probably we will have to wait until the atom nucleus is better studied before being able to answer this question.


text:
Why is SnCl2 produced when tin reacts with hydrochloric acid?
$$\ce{Sn + 2HCl -> SnCl2 + H2}$$
Our teacher told us that it's because $\ce{Sn^2+}$ is more stable, but I can't find that in any textbook.
Why is $\ce{Sn^2+}$ in the chloride product over any other oxidation state?

$\ce{Sn^2+}$ is more stable than $\ce{Sn^4+}$ due to inert pair effect

First, let's consider possibility, that indeed some cation is more stable than others. Let's look at ionization energies of tin (in eV)
\begin{array} {|r|r|}
\hline
1 &7.34 \\
\hline
2 &14.63 \\
\hline
3 &30.50 \\
\hline
4 &40.73 \\
\hline
5 &72.28 \\
\hline
\end{array}
As you can see, the ionization energies increase monotonically, meaning that each subsequent step to higher charged cation is progressively harder. This means, that if only cation stability (in vacuum) was considered, first all available tin should produce tin (I) compound and only then move further. This is, obviously, not the case.
Still, if you are lazy, it is perfectly OK to go with notion of 'stable' valence states, that are quite easy to memorize. Standard electrode potentials are sometimes quite puzzling from uninvolved point of view and it is easier to use the tables without much thoughts about them.
Still, if we want a reasonably acceptable explanation. We can see, that difference between 2nd and 1st ionization energies (~7 eV) and between 3rd and 4th (~10) is lower than between 2nd and 3rd (~15). This suggests that (II) state should oxidize disproportionally harder. Since cations are stabilized by ion-dipole or ion-ion interactions that strengthens linearly with charge, this suggests extra stability of tin (II) ionic compounds comparing to (I) and (III).
The oxidation state (IV) of tin is a completely different matter. Cations of charge $3+$ and higher very rarely truly bear said charge. Typically, they are involved in Lewis acid-base interactions reducing the charge considerably. Tin (IV) compounds have Tin bound covalently. In this case stability of compound is guarded by 8 (18 for d-elements) electron rule.
Mind you, explanation above are not very convincing from quantum mechanical POV, but it is good enough for most cases
.

So tin(IV) never has 4+ charge but only +4 oxidation state?

@user5281300 Yes. The mulliken charge on tin atom in $\ce{SnCl4}$ is calculated around 1.7 and it definitely overestimate effective charge.

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license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

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text:
I have 6% weight by vol of hydrogen peroxide. How much water should i add to make it 3%
i read that to bleach hair we should use 3 percent hydrogen peroxide. the bottle i got is 100ml and says 6% w/v h2o2.so how much percentage of peroxide does it contain. how should i dilute it

Welcome to Chemistry! This is a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed. Please [edit] in your full reasoning or thoughts on this.

If you mix equal masses of 6% hydrogenperoxide and water you will get 3% hydrogenperoxide.
Since the density of both liquids is rougly the same you can also mix equal volumes.

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text:
How to identify a pseudo-first-order reaction?
We can easily identify a pseudo first-order reaction with the integrated rate equation, but what if only the reaction is given? For example, what if I were given a question like the following on the hydrolysis of acetic anhydride?

Identify the type of this reaction: $$\ce{(CH3CO)2O + H2O -> 2CH3COOH}$$

A. Pseudo-first-order reaction
B. Pseudo-second-order reaction
C. Zero-order reaction
D. Third-order reaction

I would like to know how I could identify the correct answer if such a question were asked.

Look at the mechanism of the reaction

There are four steps:

Identify the mechanism.
Find the rate-determining step.
Write the corresponding rate equation according to the stoichiometric coefficients of that step.
Remove abundant reactants.

1. Identify the mechanism

Assuming that there is no acid to catalyze the reaction, it proceeds via a nucleophilic addition/elimination reaction.
An acetic anhydride molecule $(1)$ is attacked by the lone pairs on the water to form the complex $(2)$. This is the addition step.
The complex tautomerizes to give $(3)$, which is converted to our products, namely 2 acetic acid molecules $(4)$. This is the elimination step.
2. Find the rate-determining step.
Presumably, it is the first step.
3. Write the rate equation.
In the first step, one molecule of acetic anhydride reacts with one molecule of water. Therefore, it is a first-order reaction with respect to both of them:

$\text{Rate} = k [\ce{Ac2O}] [\ce{H2O}]$

4. Remove abundant reactants.
Water is abundant, so it is removed from the equation:

$\text{Rate} = k [\ce{Ac2O}]$

Conclusion
It is a pseudo-first-order reaction [A].

The stoichiometric equation does not indicate what the reaction mechanism is unless by accident. Thus if you don't know all the individual equilibrium, unimolecular and bimolecular reactions you cant work out what is going on without doing experiments. For example by performing experiments your reaction you may turn out to be acid catalysed, but this does appear in the equation you give.


text:
How are the two formulas for enantiomeric excess/optical purity equal?
I’ve been given two formulas to use for enantiomeric excess or optical purity:
$$\frac{\text{observed rotation}}{\text{specific rotation of pure enantiomer}}\:\:\:\:\:\times\:100\%\tag{1}$$
$$\frac{|d-l|}{d+l}\:\times\:100\%\tag{2}$$
However, I cannot figure out how they are equal to each other. You could rewrite equation (1) as
$$\frac{\alpha_d-\alpha_l}{[\alpha]}\:\times\:100\%$$
and because $\alpha=[\alpha]cL$,
$$(c_dL-c_lL)\:\times\:100\%$$
However, that’s as far as I got. I can’t see how you would eliminate L, and even doing so would not bring it to resemble the second equation. I would appreciate it if someone could offer a mathematical proof or—if this isn’t even a matter of mathematics at all but rather a conceptual misunderstanding—clarify things for me.
EDIT
$d$ is the mass of the dextrorotary enantiomer and $l$ is the mass of the levorotary enantiomer. $c$ refers to concentration, $L$ the pathlength travelled by light, [$\alpha$] specific rotatio, and $\alpha$ (without the brackets) means observed rotation.

The better formulation of equation (1) is $$\dfrac{[\alpha]{\mathrm{sample}}}{[\alpha]\mathrm{pure\ enantiomer}}$$ Units need to cancel, and with this version they do.

@BenNorris Equation (1) as I’ve written it also has units that cancel. Both the numerator and denominator are measured in degrees.

Observed rotation $\alpha$ has units of $^\circ$, but specific rotation $[\alpha]$ must have units of $^\circ \mathrm{mL\cdot dm \cdot g^{-1}}$ in order for the equation $\alpha =[\alpha]cL$ to be valid. For convenience, we shorten specific rotation to degrees.

The equations are trivially equal since each is equal to the enantiomeric excess. To prove this, we need only show each equal to the enantiomeric excess.

Start with the second equation. Because both enantiomers have the same molar mass, we can divide both numerator and denominator by the molar mass of the compound and rewrite the equation as $$\text{EE} = \frac{|\text{difference in moles of enantiomers}|}{\text{total moles of enantiomers}} = |\text{difference in mole fractions of enantiomers}|,$$ which is the definition of enantiomeric excess. This equation is not very useful, since most of the time the masses of each of the enantiomers in solution are not known.
The first equation, as you have it, is incorrect. Following Ben Norris's suggestion, we write $$\text{EE} = \frac{|[\alpha]_\text{obs}|}{[\alpha]_\text{pure}} = \frac{|[\alpha]_\text{obs}|}{[\alpha]_\text{pure}} = \frac{|\chi[\alpha]_\text{pure}-(1-\chi)[\alpha]_\text{pure}|}{[\alpha]_\text{pure}} = |\chi-(1-\chi)|,$$
where $\chi$ represents the mole fraction of one enantiomer, and we once again obtain the same result.

Had your original equation been right, $L$ would have cancelled in your derivation and you would have obtained $(c_d-c_l)/(c_d+c_l)$, which is readily brought to equation $(2)$.


text:
What does this sentence mean?
I was reading the book: " Nanotechnology: A Gentle Introduction to the Next Big Idea" and I came across this sentence:

For these molecules to perform useful functions, they must be easy to
  assemble and easy to recognize and bind to by other molecules.

Who can explain this sentence? I have problem with "easy to assemble and easy to recognize" part?
It was in the paragraph below:-

Generally, the molecules found in nature are complex and the source of
  much dismay to beginning organic chemistry students. For these
  molecules to perform useful functions, they must be easy to assemble
  and easy to recognize and bind to by other molecules. They must also
  be made by biological processes and have variable properties.

https://en.wikipedia.org/wiki/Molecular_recognition

This might refer to enzymes, for example - proteins which catalyze one specific reaction. "Easy to assemble", then, could imply that the body can only produce molecules in physological condition, i.e. no temperature above 42 °C, no high pressures, no highly corrosive or toxic compounds, etc.

I think the sentence means that the molecule the nanotechnician is using has to be quite easy for other molecules to interact with, either by chemically bonding to it or acting like a catalyst (or otherwise).


text:
Why are ketones acidic?
I have found various sources saying that ketones are acidic (pKa=20), and then relate it to the formation of the enolate ion. However, isn't an enolate ion of a ketone basic due to the oxygen's negative charge and all the pairs of electrons surrounding it?

If an acid loses its proton, it becomes a base and thus basic.

I think you need to recall the definition of an acid (in the Bronsted-Lowry scheme). An acid is a proton donor. A base is a proton acceptor. When a acid dissociates, it forms a hydrogen ion and the conjugate base of the acid. Remember, acid and base are only relative terms. The species which loses a proton in the reaction is acting as an acid and that which gains a proton is a base.

In the example above, the ketone is acting as an acid because it donates a proton. The hydride anion is acting a base because it accepts a proton. The resulting enolate anion is stabilised by delocalisation of the negative charge onto the oxygen. In the reverse reaction, the enolate would act as a base, accepting a proton from hydrogen (this is unlikely to happen though because the hydrogen will be liberated as a gas).

It will shift the places of the double bond and one of the hydrogen atoms, which will end up in a compound with a double bond between two of the carbon atoms.
That makes an enol which isn't as stable as a Ketone. The enol and Ketone are in a equilibrium which causes the enol to lose an hydrogen ion which then - the enol will become an enolate.

Enol formation is not required before the ketone can be deprotonated. Both routes yield the same result - an enolate ion.


text:
Why do better π-acceptor ligands cause smaller Δ(T) d-orbital splitting?
Consider a formazanate ligand. Then imagine an electron-withdrawing group added to it. I expect a small σ-donating effect, and good π-accepting properties.
My supervisor tells me that an electron-withdrawing group causes a lowering of the π*-orbital energy, leading to a smaller $\Delta_\mathrm{T}$ in the d-orbital splitting of the resulting iron-based molecular orbitals. Why would this be the case?

@Jan The ligand is a formazanate ligand on which we change one of the substituents into a more electron withdrawing group.

I recommend you give slightly more details about the ligand in question.

Since not many people know the ligand formazanate (I hadn’t heard of it before this question), let me quickly supply an image of what it looks like:

Scheme 1: General formazane structure.
Coodination happens from the lower two nitrogens, one of which is deprotonated (hence -ate) and both of which chelate the metal with $\mathrm{sp^2}$-type orbitals. The five formazane atoms form a π system, which can be extended by appropriate $\ce{R}$-groups. And one of these $\ce{R}$-groups is now modified to include an electron-withdrawing group.
I already previously gave a qualitative orbital scheme for a tetrahedral complex including σ and π interactions as given below.

Figure 1: Qualitative MO-scheme of a tetrahedral $\mathrm{d^0}$ complex with σ and π interactions. Originally created for this answer.
The σ donor orbitals transform according to $\mathrm{a_1 + t_2}$, leaving $\mathrm{e + t_1 + t_2}$ for the π ligands. Unfortunately, the picture is a tad clouded, since both σ and π interactions have a $\mathrm{t_2}$ contribution.
Adding an electron-withdrawing group to your ligand will leave the σ interaction unchanged in first approximation, since mainly the ligand’s π system would be affected. The orbital energies of the π system would be reduced. Since we are talking about a metal with a low oxidation state, its orbitals are higher than drawn in the qualitative scheme (which was designed for permanganate). And thus also, the most important π interaction of the ligand will be, as you stated, accepting electrons into its π system.
The exact nature of any effect observed will depend on the original orbital energies. However, the $\mathrm{e}$ orbitals are only affected by the π interaction while the $\mathrm{t_2}$ orbitals are also affected by the σ interaction. The magnitude of π interaction is expected to be the same for both groups of orbitals, since the start from the same value on each side. Thus, the difference between $\mathrm{e}$ and $\mathrm{t_2}$ stems mainly from σ interactions which only affect $\mathrm{t_2}$ due to symmetry.
Because the σ interaction starts with the ligand orbitals being much lower in energy than the metal orbitals, reducing their energy even more will result in a weaker interaction, therefore a lowering of the $\mathrm{t_2}$ level and thus a smaller $\Delta_\mathrm{T}$ or $10~\mathrm{Dq}$ value.


text:
When does a reaction equation require the reversible sign and when does it not?
Sometimes my professor puts the $\to$ symbol and sometimes he uses $\leftrightharpoons$. How do I know which to use? I get that one means the reaction only heads one way while the other means that the reaction holds at an "equilibrium" where the reactants are converting into products and the products are converting into reactants. But if I'm only given a word problem that lists 2 reactants and their concentrations, how do I know which to use?

Probably you should ask some guys who know teacher's preferences, which may be quite artificial.

@Mithoron That would connote a teacher the needs to be educated. ;-)

You just need to learn and know. Luckily, there is a certain logic to the question which reactions can be reversible and which cannot. However, a full list is impossible and would exceed the space of this margin.
For most mechanisms, there is (at least) one point-of-no-return step. Learn to identify this step. All other arrows should probably be $\ce{<=>}$, if you’re strict. These non-reversible steps are often (but not always!) carbon-carbon bond forming reactions, as a very rough guideline on which is which.

Hmm ok. But there seems to be a particular brand of equations for which my professor always uses the single arrow. I think they were called dissolving/ionization equations? I may be wrong though.

I wonder who downvoted me and why?

Ask yourself if it is reasonable to give thought to the backward reaction. If not, because it is not relevant for the particular question, use the single arrow.
E.g. for partial equations, usually the single arrow is used, because statements about the equillibrium only make sense for the full reaction.


text:
Organic polyether with carboxyl moiety reacting with aldehyde, following reaction with sulfuric acid
I want to react the following molecule with conc. sulfuric acid in the presence of methanol at room temperature:

If I am right, this would make a methyl ester to be formed on the carboxyl.
The aim afterwards is to let it react with vanillin (in the presence of left over sulfuric acid) at 90°C, so it will give a coloured derivative.
Below is the structure of vanillin. As you can see, this has an aldehyde group and a hydroxy group.

Is it possible to let an aldehyde and ester react at higher temperatures?

What do you hope to get as a result of this reaction? (Also, I'm not sure the first reaction would give you the ester in good yield - there are a number of -OH that can easily be protonated, and then leave to form a carbocation, and do a number of things from there. For instance, that allylic hydroxyl looks particularly tasty for concentration sulfuric acid.)

I found these steps in a post-column derivatization reaction in HPLC. It gives a coloured molecule, but  I don't know how this would look like. I would like to know the principle, so I could use this for similar reactions in the future.

I just checked chemaxon, and it seems that the hydroxy groups have pka's that are in the range of -3 ; 12.7. So it seems that these groups will not be protonated with conc. sulfuric acid as this is only a semi-strong acid.

Regarding your first comment: do you have a link to a paper describing this? It might provide some more insight.

I don't really understand your second comment. What is the "semi-strong acid" you refer to? Sulfuric acid has a Hammett acidity of around -12 if I remember correctly - thus making it a very very strong acid. Depending on how concentrated the sulfuric acid is, it can also produce species such as pyrosulfuric acid, which is even MORE acidic. The question isn't whether or not the hydroxyls will get protonated - it's whether something else will happen faster, and lead to a more useful product.

I found a paper where they react the molecule immediately with vanillin in conc. sulfuric acid. "The derivatization reagent consisted of 20 mL concentrated sulfuric acid, 950 mL methanol and 30 g vaniline." Some papers: [link]https://www.researchgate.net/publication/259003919_The_Determination_of_Six_Ionophore_Coccidiostats_in_Feed_by_Liquid_Chromatography_with_Postcolumn_Derivatisation_and_SpectrofotometricFluorescence_Detection [link]https://www.researchgate.net/publication/10932237_Determination_of_monensin_residue_in_chicken_by_HPLC_with_post-column_derivatization

Here are the links of 2 reagents that do the trick: [link]https://www.pickeringlabs.com/products/analysis-products/reagents-chemicals/p-dimethylaminobenzaldehyde/  [link]https://www.pickeringlabs.com/products/analysis-products/reagents-chemicals/vanillin/

The papers you linked say nothing of concentrated sulfuric acid other than the following sentence: "The derivatization reagent consisted of 20 mL concentrated sulfuric acid, 950 mL methanol and 30 g vaniline." Note the volumes of the reagents!!! This is not at all what you say: "I want to react the following molecule with conc. sulfuric acid in the presence of methanol at room temperature." Rather, a more correct sentence would be to say: "I want to dissolve the following compound in acidified methanol." There should not be significant reactions going on in the first step, however, some will.

I tried to follow the references to find the first paper that mentions this vanillin/$\ce{H2SO4}$ staining, and eventually found a 1972 paper: "Automated photometric microbiological assay for monensin in poultry rations." published in "Journal - Association of Official Analytical Chemists" (which no longer exists, by the way) by Kavanagh FW and Willis M. I could not find the body of this paper - very likely it never existed in electronic form.

If I were to guess what reaction happens, it is that the carbocation at the allylic hydroxyl forms (by way of water leaving), then the ring in vanillin attacks the carbocation, forming a vanillin derivative. However, I'm not sure such a compound would be colored, as usually coloring suggests significant conjugation. The electron withdrawing effect of the carbonyl may be enough to make the compound colored, but again, I'm not 100% sure.


text:
Why oxygen leaks from tires faster than nitrogen?
In critical applications tires are inflated using pure nitrogen (link).

Nitrogen molecules have a more difficult time escaping through the
  microscopic spaces that exist between a tire's rubber molecules. [...]
  Nitrogen reduces the loss of tire pressure due to permeation through
  rubber over time by about 1/3. (source)

According to WebElements, nitrogen bond length (109.76 pm) is smaller than oxygen bond length (120.74 pm), so why nitrogen permeation through rubber is said to be lower than oxygen?

I don't believe oxygen leaks any faster. The problem with oxygen is that it oxidizes things, which nitrogen does not.

I agree with @Ivan Neretin. You show that $O_2$ and $N_2$ are similar in size and both very small. Hydrogen and helium would leak out faster though. Given time, oxidation possibly round the glue on the rim may be important. Carbon dioxide may be a better gas to use.

There'll be small differences, the one that can be more easily polarised (i guess oxygen?) will diffuse slightly faster, i think. But a factor three is superstition, perhaps by people who are afraid to think of their tyres igniting. Which might be a real concern, depending on your style of driving. ;-)

As mentioned by @Ivan Nertin, oxygen is more reactive... leading to rusting of steel wheels, particularly if a water-based tire sealer/inflater is used. In the form of ozone, it does damage rubber, but usually to the outside of the tire. Mostly, it's hype to sell nitrogen.

As oxygen reacts with rubber or iron it is removed from the air/tire... so it did not leak out, it was consumed. Oils and rubber are polymers of carbon. Oxygen attacks the carbons and become bound/removed from the air. Since air is 20% O2, the tire will deflate as O2 is consumed.

@porphyrin: Carbon dioxide passes ordinary rubber about 5 times faster than oxygen and about 15 times faster than nitrogen.

@aventurin That sounds in line with the "material compatibility" listed here. Could you please post a link to a detailed reference?

See e.g. http://nvlpubs.nist.gov/nistpubs/ScientificPapers/nbsscientificpaper387vol16p327_A2b.pdf or http://people.ds.cam.ac.uk/jae1001/cus/teaching/materials/M6_Lecture_6.pdf.

I'll point out that tires are not really made of "natural rubber" anymore. There are dozens(?) of various synthetic rubber compositions.

We don't know that the rate of leakage of anything through the material of tyres is significant at all. The original experimental rates were determined on thin sheets of balloon rubber. Tyres are not thin sheets and the actual rate of gas permeation in them could easily be insignificant.

The rate of oxygen, $\ce{O2}$, permeation through tire rubbers is about 3.4 times faster than nitrogen.  The reference for this is "Permeability Properties of Plastics and Elastomers" by L. W. McKeen.
The reason is that $\ce{O2}$ is a smaller molecule than $\ce{N2}$.  The size of a molecule is not determined by only the bond lengths between atoms but the size of the electron cloud around the atoms.  The kinetic diameter of $\ce{O2}$ is 0.346 nanometers and $\ce{N2}$ is 0.364 nanometers.  So, $\ce{N2}$ is larger than $\ce{O2}$ even though the $\ce{N-N}$ bond length is shorter than the $\ce{O-O}$ bond length.

Can't we apply Graham's law? That would indicate N2 diffuses faster

If it is a solid empirical finding, then we have to accept it. But those statistics on the size of the molecules do not convincingly explain the large scale of the empirical difference in rates. Is something other than pure diffusion happening here?

Actually, it is worth reading the original source for the permeation numbers. It speculates that the issue is solubility not diffusion. Moreover the experiments were conducted in the early 1900s on thin rubber balloons (or balloon materials) and might not apply to thick modern tyres. Most likely the permeation in modern tyres is so low (because of their thickness) it would be hard to measure. I suspect it is insignificant and some people just didn't read the original source to check for relevance.

Moreover, the small difference in size is seriously insufficient to explain the large difference in permeation. This is clearly something more complex than diffusion making the statement that "the reason is that O2 is a smaller molecule" just wrong.

I would hazard a guess that surface adsorption to the rubber surface is also quite different between nitrogen and oxygen. I would think that has a greater role to play than the atomic size.

While the comments are basically correct, this is a teaching opportunity. You can't just look at bond lengths for determining the size of the molecule as the atoms also have radii which have to be considered. The longest part of the molecule (or widest surface) then can be simply be "bond length + 2x(atomic radius)".
I was able to get some covalent radii data from this website, and if we use it to calculate we find:
Width of dioxygen = 1.21 + 0.73 x 2 = 1.21 + 1.46 = 2.67 Angstroms
Width of dinitrogen = 1.10 + 0.75 x 2 = 1.10 + 1.50 = 2.60 Angstroms
So the original difference of 11 angstroms is down to 7 angstroms. Not much of an improvement, and Oxygen is still larger than Nitrogen. However, if you incorrectly used the atomic radius...
Width of dioxygen = 1.21 + 0.48 x 2 = 1.21 + 0.96 = 2.17 Angstroms
Width of dinitrogen = 1.10 + 0.56 x 2 = 1.10 + 1.12 = 2.22 Angstroms
Suddenly Nitrogen is bigger.

Good point. It'd be nice to see some work on the prediction of diffusivity, based on molecule size, polarity, polarisabilty etc. That's been done for sure.

Yes to checking the molecular sizes properly. But this does precisely nothing to illuminate the issue because small differences like that would be barely detectable if the process were basically diffusion.


text:
Why does more heat transfer take place in a reversible process than in a irreversible process?
We have defined $$\mathrm dS=\frac{\delta Q_\text{reversible}}{T}$$
and the second law of thermodynamics (Clausius inequality) says
$$\mathrm dS\geq\frac{\delta Q}{T}$$
(Here $\delta Q$ is just the differential of heat regardless of it being reversible or irreversible) and the equality holds for it being a reversible process (as defined).
Now,
$$\mathrm dS\geq\frac{\delta Q}{T}$$
If I put $\mathrm dS$ on the left as $\frac{\delta Q_\text{rev}}{T}$,
I will get $\frac{\delta Q_\text{rev}}{T}\geq\frac{\delta Q}{T}$,
or $\delta Q_\text{rev}\geq\delta Q$.
This implies that in a reversible process, more heat transfer takes place than in an irreversible one. Why is this so?

because the internal irreversibilities of the system generate excess heat and that dissipative effect is always positive, so when a cycle is completed the internally generated heat is part of the total heat exchange.

expanding irreversibly (spontaneously) into a vacuum does no work, as the resisting pressure is zero, so no heat absorbed from surroundings. Expanding reversibly against pressure $ p$ results in work $\int pdV$ so heat is absorbed.

In the Clausius inequality, the temperature in the denominator is supposed to be the temperature at the boundary of the system where the heat transfer is taking place.  This important point is often omitted from many thermodynamics textbooks.  In a reversible process, the temperature at the boundary is essentially equal to the temperature of the system, because the system temperature is uniform throughout.  But, in an irreversible process, the temperature of the system is not uniform, and the temperature at the boundary can differ substantially from even the average temperature of the system.  So the Clausius inequality does not always require that $\delta Q_{rev}\geq \delta Q$.

So is the first answer in http://chemistry.stackexchange.com/questions/19735/what-does-the-clausius-inequality-mean-in-thermodynamics-entropy?rq=1 wrong?

the order got changed, i meant the one with 7 upvotes

The answer you referred to does not make the important distinction regarding the T in the Clausius inequality being the value at the boundary of the system where heat transfer is occurring.  Otherwise, it seems pretty OK.

Well, there he is saying "Therefore δqrev−δq=δw−δwrev≥0" which would mean δqrev≥δq. Morever, it say work in a reversible process is lesser than in a irreversible process. What about isothermal irreversible compression compared to a reversible one

Suppose you are operating your process in a cycle where $\Delta U$ and $\Delta S$ are both zero.  Both the reversible process and the irreversible process start and end at the same state.  The irreversible process uses a great big cycle with large W and large Q, while the reversible process uses a tiny little cycle with small W and small Q.  You certainly can't say that Q for the reversible cycle is greater than the Q for the irreversible cycle.  However, for the irreversible process cycle, the integral of $\delta Q/T$ at the heat transfer boundary(ies) is less than zero.

re temperature at boundary. What about irreversible expansion of a gas into a vacuum? No work is done, (as pressure resisting expansion is zero) so no heat is absorbed from the surroundings, so temperature change, boundary or otherwise.

There is still a temperature at the system boundary even though there is no heat transfer there.  So, in this case $\Delta S >0$.

I try to explain as follows. To illustrate using the Clausius inequality consider a cyclic process. In the first step 1 to 2 the system is isolated and undergoes a spontaneous irreversible change and in the second step 2 to back to 1 the system interacts with the environment and returns by any reversible path. Because S is a state function $\Delta S = 0$ for this whole process. Thus $$ \Delta S = 0 > \int _1^2 \frac{\delta q_{irr}}T +  \int _2^1 \frac{\delta q_{rev}}T$$
 and the inequality is because of the irreversible step (note change in limits). The first integral is zero because the system is isolated, and the second integral is $S_1 - S_2 $ as entropy is a state function, thus $0> S_1 - S_2 $,  which produces $$ \Delta S = S_2-S_1 > 0$$
thus we see that the entropy increases when the isolated system goes from state 1 to 2 by a general spontaneous (irreversible) process.
Thus you see that more heat is transferred in a reversible than irreversible process as you suggest. Hope this helps.
(From this if we consider the universe to be an isolated system and all naturally occurring processes are irreversible, one statement of the second law is that the entropy of the universe is constantly increasing. Clausius summarised the first two laws as ' The energy of the world is constant' and 'The entropy is tending to a maximum'). (See MCQuarrie &  Simon, Physical Chemistry)


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Complete transfer of Heat into Work
During an isothermal reversible expansion of a gas, all heat absorbed by the gas from the reserviour is being used up to push the piston up. But doesn't this contradict something like the second law?
When I push a brick on a rough table, the motion will eventually stop and some heat will be generated due to friction, but If I get a bunsen burner and supply the same amount of heat, it brick won't move back to where it started.

I'll have to think if exactly this it allowed at all, but generally such a thermodynamical process can only be reversible if it is done infinitely slow, in which case you never get any work out of it. (?)

And that's why there's no such thing as fully reversible process.

only in a cycle a complete conversion of absorbed heat to work is prohibited.

Yes, the above comment by hyportnex is correct. Levine's Physical Chemistry 6th ed. writes: http://i.stack.imgur.com/i3NzD.png

In an isothermal reversible process the total entropy change of piston and energy reservoirs (surroundings) must be zero, $\Delta S_T = 0$. Thus the entropy of the expanding gas may increase but only if it is compensated for by a decrease in entropy of the reservoirs. So no contradiction of second law. (The heat absorbed from the surroundings is $T\Delta S $ and if this is positive more work can be done by a chemical reaction than the heat of reaction.)
In the second question you will realise that you can turn all the work into heat but not all the heat into work. Its entropy again.


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Molecules and atoms
Which contains the fewest molecules?
Which contains greatest number of atoms?

0.5 g of hydrogen
4.0 g of oxygen
11.0 g of carbon dioxide

This seems like a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed.

4.0g of Oxygen has the fewest amount of atoms since H2 has more atoms than O2 (2 atoms) and Carbon (12/18 (if we include the electrons) atoms) also has more atoms than O2 (2 atoms).
0.5g of Hydrogen has the most amount of atoms - since Hydrogen is the least dense from them all.
Hydrogen density - 0.08988 g/L
Oxygen density - 1.429 g/L
Carbon density - 2.25 grams per cubic centimeter
I tried to explain with the best of my ability.

For homework questions we have a policy where you do not simply post the answer without explanation but rather that you give hints that lead the asker towards their own solution. We are not a homework completion service.

Oh, thanks for warning me. I'll keep that in mind for future answers of mine. Sorry.

This is not a homework just doing some revision and wanted to check that Im doing right. Thanks for the help anyway

No problem, I also added the explanations as Bon said.


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Alkali and alkali earth metal carbonates and bicarbonates decomposition

Which of the following can not decompose on heating to give $\ce{CO_2}$?

$\ce{Li_2CO_3}$
$\ce{Na_2CO_3}$
$\ce{KHCO_3}$
$\ce{BaCO_3}$

What I have read till now: Lithium carbonate is not so stable to heat and bicarbonates of alkali metals decompose at approximately $\pu{300 ^\circ C}$. The stability of alkaline earth metals increases down the group and they decompose on red heating.
This only leaves us with one option as correct (2nd), but what is the reaction of decomposition of $\ce{Na_2CO_3}$?

From your question, the heating temperature is unclear. See these links for sodium and lithium carbonates, the hydrocarbonate or barium.
They are all also mentioned on Wikipedia.
In other words, they all seem to decompose at some point, so what is meant by heating?
The reaction for sodium carbonate decompositon at melting point should be:
$$\ce{Na2CO3 -> Na2O + CO2}$$

Is there any temperature where $Na_2CO_3 $ doesn't give $CO_2$ on thermal decomposition?

Take a look at this paper: http://oasis.postech.ac.kr/bitstream/2014.oak/11644/1/OAIR002402.pdf
They discuss the decomposition of sodium and lithium carbonate in detail.

@KartikWatwani sorry for the late follow-up, but I wanted to add this: so you're asking as to whether if you heat up sodium carbonate, at what temperatures does is decompose WITHOUT producing carbon dioxide? Aside from some arcane extreme conditions (extreme pressures and temperatures combined, for example), I think that if it decomposes, it will probably produce carbon dioxide. That's just my opinion though, I am happy to be corrected.

Here the comparison is highly relative(or subjective). The stability of group 1 and 2 carbonates increases down the group therefore you can expect that the lower the metal in the periodic table the more stable is its corresponding carbonate.
Also, for the same period the group 1 carbonate is relatively more stable.
Not getting into the energetics one can safely state that the reason for these trend is the polarising power of the metal cation.
We know that in the carbonate ion, the charge is delocalized and is thus highly polarizable.
Group one and two cations have a very high charge to radius ratio(e/r) which implies that their polarizing powers are fairly high. This makes the C-O bond weaker allowing the -CO2  group to depart easily.
This e/r decreases down the group making the compound relatively stable.
Thus the most stable species here shall be BaCO3

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


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Empirical formula, simplifying problems
Need some help with second part.....can't get it to be balanced and therefore  can't find the formula and name! Thanks in advance!
What is the empirical formula of each of the following compounds?
a) A liquid containing 2.0g of hydrogen, 32.1g of sulfur and 64.0g of oxygen.
•   Hydrogen = 2.0g / 1.0g = 2 / 1 = 2
•   Sulfur = 32.1g / 32.1g = 1 / 1 = 1
•   Oxygen = 64.0g / 16.0g = 4 / 1 = 4

Empirical formula of this compound is: $\ce{SH_2O_4}$ and the name of the compound is Sulfuric Acid.
b) A white solid containing 4.0g of calcium, 3.2g of oxygen and 0.2g of hydrogen.
****•   Calcium = 4.0g / 40.1g = 0.09975062344 =
•   Oxygen = 3.2g / 16g = 0.2 =
•   Hydrogen = 0.2g / 1g = 0.2 =****

Empirical formula of this compound is:
c) A white solid containing 0.243g of magnesium and 0.710g of chlorine
•   Magnesium = 0.243g / 24.3g = 0.01 / 0.01 = 1
•   Chlorine = 0.710g / 35.5g = 0.02 / 0.01 = 2

Empirical formula of this compound is: $\ce{MgCl_2}$ and the name of the compound is Magnesium Chloride.

You know about rounding and significant figures, right?

For the first question, it should $\ce{H_2SO_4}$ because usually when we write acid we always start with hydrogen first

You find #2 difficult because #1 and #3 are ridiculously easy!
In these two questions the mass of every element present just happens to match exactly with the atomic weight of that element.  So when you divide each mass by the corresponding atomic weight, the resulting ratios are very close to integers, or integers divided by 100.  So the (almost) final step of dividing each ratio by the smallest ratio goes by almost unnoticed.
But consider for the moment what would happen if you used three or four decimal place atomic weights.  All those nice neat exact answers would disappear in a haze of "sort of" whole numbers, and you would have to use some judgement as to whether a number is close enough to a whole number.
In #2, the questioner made the "drastic" change of using 4.0 and 40.1 as the two numbers for an element.  If you had carried on mechanically, dividing all ratios by the smallest ratio, you would have gotten:

Ca ... $1.0$
O .....$2.005$
H......$2.005$

Is $2.005$ close enough to a whole number?  Or should you multiply everything by $1000$ to give $1000, 2005, 2005$ as the empirical formula?


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Melting points of alkali metal halides

The correct order of melting point of alkali metal halides is:

$\ce{MF}>\ce{MCl}>\ce{MBr}>\ce{MI}$
$\ce{MI}>\ce{MBr}>\ce{MCl}>\ce{MF}$
$\ce{MCl}>\ce{MF}>\ce{MBr}>\ce{MI}$
$\ce{MI}>\ce{MF}>\ce{MCl}>\ce{MBr}$

The correct answer to this is option (1), the reason I could think of is that as the size of anion in ionic compound increases the covalent character increases and hence the melting point decreases.

As with all the periodic trends concerning differences inside a single group, it will have something to do with the size of the atoms involved.
You already noted that fluorine is the smallest of the halogen atoms and iodine is the largest. This is due to there being only two core electrons in fluorine while iodine has 46. While adding an electron to any of these will increase their size, the size differences between them will remain the same due to said core electrons; hence, fluoride will be the smallest and iodide will be the largest.
This means that fluoride has a high charge-per-volume ratio while iodide’s is much lower (compare that to Pearson’s HSAB concept — it’s the same idea). A higher negative charge density means that it draws cations towards it better and thus stabilises the crystal structure. A secondary effect is that the smaller size of fluoride means that it must have a lower coordination number in the crystal structure, thus having less positively charged neighbours, thus attracting them more.
Together, these influences mean that it is just harder to break up the high ordering of the crystal structure to melt the salts the higher up you are in the group.
A higher degree of covalent bonding for iodine may play a role, but I would deem it neglectable.


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Acid Catalyzed Hydration of an Alkene Synthesis Problem

Devise syntheses of the following three compounds, starting in each case with any alkene that contains four carbons or fewer. You do not have to write mechanisms, although at this point it may be very helpful for you to do so.

$\ce{(CH3)3COH}$

$\ce{(CH3)2CHOH}$

$\ce{CH3CH2CHOHCH3}$

This is the question I'm trying to tackle. I'm having a bit of trouble with the third part of it, which is the synthesis of butan-2-ol. The text solutions says that I should use but-2-ene, which I can understand but I don't quite understand why my answer is also not plausible. Can someone help me out with this?

It looks right but using but-2-ene would mean you'd only get the desired product. Your method gives the product as the major product not the exclusive product. Assuming no other, more elaborate, side reactions.

@RobChem but for my method, with my product, only a secondary carbocation forms because the primary carbocation is too unstable to form, no? So that means that my method does give the desired product as the exclusive product, not just the major product, no?

Because from what i see, there are only two carbocations that could form and the primary carbocation itself is too unstable, in this reaction anyway, to form?

The main point here I think here is that using  but-2-ene eliminates any arguments about selectivity of carbocation formation because of the symmetry of the alkene. Yes the secondary carbocation is more stable in your proposed synthesis but I suspect there may be small amounts of the undesired terminal alcohol or perhaps rearrangement products. Using but-2-ene is just better because there is no possibility of these complications (no matter how small) so why would you use your method. Your reasoning is sound but the other way is just better

Oh okay so in general, for synthesis questions, the thing that i have to worry about is whether i get my desired product exclusively and which starting compound would help me achieve that?

Well chemists always look to maximise yield so look for the most efficient method yes. Things like atom economy, number of steps etc play a role too in which synthesis is preferable

You propose to use but-1-ene rather than but-2-ene as a starting material to synthesise butan-2-ol via addition of water to the double bond. There is no reason why it would not happen the way you drew.
However, using but-2-ene allows only one possible reaction pathway. With the relatively high symmetry of the molecule ($C_\mathrm{2h}$, assuming trans), attacking either side of the double bond will lead to an identical product: The molecule’s intrinsic $C_2$ axis will transform one into the other. Thus, it is only possible to generate butan-2-ol.
Starting from but-1-ene, you have two non-equivalent first steps, generating either the primary or the secondary carbocation. These obviously cannot be transformed into each other by symmetry (the alkene’s symmetry is now $C_\mathrm{s}$). Thus, you will get a side product: butan-1-ol. These products will be difficult to separate. If anything, careful distillation would do the job.
As RobChem already noted in the comments, (organic) chemists typically attempt to avoid side reactions and different products, hence why the book suggests using the more symmetric alkene.

Of course, i can understand what the book was thinking when it gave that compound as the starting material. Like, i could understand why the book gave that as the answer and that's because i wrote out the mechanism for but-2-ene.

I was just wondering if there was anything wrong with the starting material i used haha. I didn't realize that the product from the primary carbocation would form in the ever so slightest amounts, despite how unstable it is.


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What happens with enol and ketone
What process happens between an enol and a ketone, so that the ketone will enolate and become acidic?
Please, explain what happens with them in this process?

Because the enol and ketone aren't stable, equilibrium happens; therefore the enol has a chance of losing a hydrogen ion and because of that, the enol becomes an enolate.
Also keep in mind that an acid is a molecule that can donate a proton.
The enolate is a base and it's gathering/accepting a proton from the hydrogen.
I assume this is what you meant when you asked the question.

No problem, glad I could help. There was a similar question to this before, so next time before you ask something, remember to see if a similar question to it has already been answered.

"Enol and Ketone aren't stable, the process called Equilibrium happens"  - you should be more careful with reasoning. 1) What would it even mean they aren't stable? 2) Equilibrium isn't a process.

@Mithoron I am aware that it isn't a process, I used that word on accident. Was going to edit my post but it got accepted, so I thought it wasn't needed. Thanks though.

I really don't get what you are trying to say with your first paragraph. Equilibrium is not a type of chemical reaction... Furthermore the deprotonation can occur directly from the ketone and tautomerisation to the enol form isn't required for deprotonation


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Entropy change of a rubber band
When I stretch a rubber band, the fibers straighten out causing lesser randomness and more order. This should imply the entropy of the rubber band has decreased, but how does the entropy of the whole universe increase here? Does this mean a rubber band gets colder on stretching? I have seen an experiment where it gets hotter. Doesn't entropy of the whole universe increase only in spontaneous processes? Here I have to pull it hence its not spontaneous? In a nutshell, I'm really confused.

You are treating the rubber band as an isolated system when in fact you are acting upon it.  The entropy increase occurs in your body as glucose is oxidize into water and carbon dioxide to produce the energy needed for your muscles to contract to stretch the rubber band.

The force you are exerting on the rubber to stretch it is in the same direction as the displacement.  So you are doing positive work on the rubber, and it is doing negative work on you (its surroundings).  So, from the first law of thermodynamics, if the process is adiabatic, the internal energy of the rubber is increasing.  The internal energy of rubber is known from observations to be nearly a function only of temperature.  So, when you stretch the rubber adiabatically, its temperature should rise.  For an adiabatic reversible process, the change in entropy of the system has to be zero.  So the decrease in configurational entropy of the rubber must be offset by the increase in entropy resulting from the rise in temperature.

Let $L$ denote the length of the substance then you can always write
$$\left ( \frac{\partial T}{\partial L} \right )_{S} = - \left ( \frac{\partial T}{\partial S} \right )_{L} \left ( \frac{\partial S}{\partial L} \right )_{T}$$
The term $\left ( \frac{\partial T}{\partial S} \right )_{L} $ is always positive for being the reciprocal of the heat capacity per unit temperature. Thus $\left ( \frac{\partial T}{\partial L} \right )_{S}$ is positive or negative whether $\left ( \frac{\partial S}{\partial L} \right )_{T}$ is negative or positive. Therefore, if entropy decreases upon isothermal stretching then the temperature must increase during adiabatic stretching.


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Why do atoms need 8 electrons to stabilize?
As the title says. I have surfed all of the net but could never find the answer to this question. Why do atoms need 8 electrons to stabilize? I mean why not 7 or 5 or 10 electrons? Why specifically 8? And what does stabilization of atoms even mean? Are the atoms going to burst without 8? Why do they need 8 electrons?
This is a question which is not often discussed in chemistry and I need an answer which really makes any sense. Not just `it is because the atoms need to fill their octet.

Unfortunately while this is a very fundamental question, a good answer to this is not easy to write. There are far too many hidden sins in the phrasing that you quote ("8 electrons", "stabilise") and a great deal of elaboration on those points is needed. Perhaps someone will be up to the task.

@orthocresol so this is a very deep and fundamental question that only the professionals can understand?

Some good pointed questions, like this one, are tricky to answer because it gets just close enough to the philosophy of knowledge to force people to get very precise.  You are looking for what we, in philosophy, call a "justification" for an idea.  It is clear you have already refuted the justification of "we've done this experiment millions of times, and here's what we've seen," so you are asking for an even better justification than that.  Such a justification will invariably stray into the world of quantum physics.

The valence orbitals of atoms are composed of suborbitals (s and p)  there is 1 s suborbital which is spherical and can hold 2 electrons (one with up spin and one with down spin). There are 3 p suborbitals which are dumbbell shaped (look like two balloons tied together at the ends) and align along the x,y&z axies and hold a total of 6 electrons (2 per axis, 3 with up spin and 3 with down spin).  Since an atom is energetically satisfied when all of the electrons are paired and 2+6 = 8, an atom must have 8 electrons in its valence shell to pair all of the electrons.
The octet rule does not apply to all atoms though, for transition metals and later elements there is a d suborbital which can hold an addition 10 electrons, meaning that now an atom may have 18 electrons in its valence shell to be satisfied.  This allows for materials such as metal complexes and the sulfate poly-atomic-ion to exist despite having more than eight electrons in the valence shell.

what do you mean by energetically satisfied? How can an atom know if it is satisfied? And what causes this satisfaction? And why is there even a need for satisfaction in atoms?

@Adi I'm assuming means a low energy state.

@DraggyWolf And as I said in your question...what does low energy state has to do with stabilization?

@Adi see http://physics.stackexchange.com/questions/47253/why-does-the-nature-always-prefer-low-energy-and-maximum-entropy

@Adi High energy states (tend to) have many ways of becoming lower energy states, but very few ways to remain in a high energy state. Consider a tall tower with improper supports. There are many ways to fall, but only a few ways to stay upright.

@A.K.'s answer is correct. Have a look at an undergraduate level book on atomic spectroscopy or physical chemistry, most any one will do, and will give you details of atomic structure and you will see where these 'rules' come from.

I think the most important thing here is to point out some details which I don't believe get emphasized enough in chemistry classes, and yet are extremely important in approaching questions like this.

I mean why not 7 or 5 or 10 electrons? Why specifically 8?

I've been in this situation before. The one where I'm asking a question which I believe to be a question about the fundamental nature of matter, but the more I explore it, it just seems to be a meaningless question. The actual number of electrons required to stabilize a system is not a question we should really care about. The question we care about is whether or not our theories can predict that number. After all, in science, we are asking nature the questions, and nature gives us the answers. So in this case, nature has told us 8 electrons, so the answer is 8 electrons. I think that's one thing which is emphasized a lot more in physics than in chemistry. Nature is the arbiter. We simply try to rationalize the arbiter's decisions.

Why do atoms need 8 electrons to stabilize?

Now that's a question which can be answered by a theory! This can be answered in lots of detail or a little detail, but I get the feeling you are going to ask questions about why we don't just keep filling up the orbitals with more electrons? This is basically what you're asking about A.K.'s answer when saying,

what do you mean by energetically satisfied? How can an atom know if it is satisfied?

This leads us to perhaps the greatest disservice done to chemistry students in their classes: the habit of personifying chemical systems. Talking about the "desires" of the system makes it easier to understand when first learning, but this ought to quickly be abandoned (fluorine just loves electrons!). The answer to your question here is that the system doesn't know what it wants, and that is often a very good question to ask! These sorts of questions can be very productive when studying something like quantum entanglement. But, for the matter at hand, something simpler suffices. Chemical systems are dynamic. They are changing in time by colliding with other molecules, vibrating, rotating, absorbing photons and emitting photons, and doing all these things at once! So then, what it means for a system to be "energetically satisfied" is that all these things have found some kind of balance. It is perfectly possible that at any given moment some atom might lose an electron and another one picks it up, but again we must come back to what nature tells us. Nature says that when we have two atoms flying around, and one of them has seven electrons, that atom is going to take an electron from something else.
This need not be reserved for chemical systems though. In all of physics, we observe that systems tend towards their lowest energy state. This isn't dictated absolutely by some law, but rather it is a general rule. After all, we can find systems that aren't at the global minimum of their potential energy surface, but if the activation energy can be ignored, we will almost always find that system in its lowest energy state.
It's almost as if nature is probing these five, seven, and ten electrons states which you suggest, but simply does not stay there because that is not the way the world works. To be technical, systems tend towards equilibrium (you must convince yourself of this because the mathematics to show it can be involved), so when I find an atom with ten electrons, the forces in the system are out of balance. There might be eight positive charges in the nucleus, but ten negatively charged electrons. Thus, when something with seven electrons comes around, it is likely that system will take one of the extra electrons because this balances out the forces (as well as can be done in this example). Some kind of equilibrium has been established. This is obviously a simplistic example, but you could imagine then that if you had an ensemble of Avogadro's number of particles, that all these things would be exchanging their imbalances and staying around the equilibrium. This equilibrium, again, is observed to be the low energy state.
Concluding Remarks:
If and when you ever study quantum mechanics, you'll find that these things all arise quite naturally out of the theoretical framework laid there. When you find that angular momentum is quantized (and that the electron can have zero angular momentum), you naturally ask what this waving electron thing looks like (say 95% of the time). So, you plot this using the equations that you've found from your theory, and you get out the orbitals that we learn all about in chemistry classes. The next question leads into what you're asking here, both about the number of electrons each atom will naturally have around it, and about the relative stability of each of those cases. If you really want to be satisfied about those questions, study quantum mechanics. Otherwise, what we've said here ought to suffice. All in all, there comes a point when you either to need to look deeper, and be willing to grapple with some complicated ideas in quantum mechanics, or you need to give up on asking epistemological questions because the answer to those kinds of questions in science is almost always, "because mother nature says so."

thank you for an answer but I believe that everything has a reason behind it..even the most fundamental order of the universe has a reason. Nothing just happens. There is no Because mother nature says so. But for this matter I think that I have to put this question on hold until I learn quantum mechanics as right now I am only a 10th grade student.

I think that the more you study science you will understand what I mean. Physics and chemistry are simply the act of doing experiments and explaining those observations. An experiment is just asking nature what is true. So, in that sense, there is definitely a "because mother nature says so." And you're right that nothing just happens, but I never said that...

I agree with jheindel... We observe what happens in nature and we construct sensible explanations to rationalize what we observe. Technically, a pertinent atom doesn't consciously "desire" 8 electrons, because it can't actually think. But for early education purposes, it's easier to think that they have this "desire". In other words, if we wanted, we could consider the $n$, $l$, $m_l$, and $m_s$ quantum numbers to rationalize our own constructed octet rule, but atoms know nothing of our constructed octet rule.

@jheindel i understand.But one day I believe I will find the answer.

Atoms generally do not need 8 electrons to stabilize.
Using quantum physics and its models of atoms and bonds we can define four quantum numbers. $n$ being principal quantum number describes the electron shell, $l$ being azimuthal number describes the orbital shape, $m$ being magnetic number describe degeneration of the orbital and $s$ being spin number describes the spin. Pauli's principle says that 2 electrons (fermion) cannot exist in the same (energetic) state and quantum numbers describe such energetic states of electron in the atom.
It was also found, that $n=1,\ldots,~ l=0,\ldots ,n-1, ~m=-l,\ldots,l,~s=\pm1/2$.
Another rule claims that electron is preferably in the state with lowest possible energy and therefore the states with lower $n$ are preferred before the others with similar energy.
Al this together we obtain the sequence of occupying the energetic levels as follows:
$$\rm 1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^{10}, 4p^6, 5s^2, 4d^{10}, 5p^6, 6s^2, 4f^{14}, 5d^{10}, 6p^6,\ldots$$
where number is principal quantum number, $\rm s,p,d,f$ stands for $l=0, 1, 2, 3$, respectively, superscript shows the count of available $m$ and $s$ pairs.
The atom is considered stable when all orbitals prior $n\mathrm s$ are fully occupied (noble gasses) and those electrons do not interact in chemical bonds. The electrons with highest $n$ beyond this limit are valence electrons and they can form chemical bonds. Their energies in the atom are highest. For example Lithium electronic configuration can be written as:
\begin{align}_3\ce{Li}&:\mathrm{1s^2\ 2s^1}\\ _3\ce{Li}&: [_2\ce{He}]\ \mathrm{ 2s^1}\end{align}
Now for the question, the most common atoms have valence electrons in $\rm s$ and $\rm p$ orbitals and 8 electrons can fill them up, That's why there is the "magic 8 rule" because only 4 atoms do not stabilise in 8-valence configuration:

Hydrogen stabilises in $_0\ce{H^+}$ or $\ce{_2H^{-}[_2He]}$ configuration;
Helium is stable as $\ce{_2He}$;
Lithium stabilises in $\ce{_3Li^{+}[_2He]}$;
Beryllium stabilises in $\ce{_4Be^{2+}[_2He]}$;
Boron stabilises in $\ce{_5B^{3+}[_2He]}$;
Carbon stabilises in $\ce{_6C^{4+}[_2He]}$ or $\ce{_6C^{4-}[_{10}Ar]}$

According the question in DraggyWolf's comment, why is not 1 electron per atom considered as stable?
Let's consider that energy electron relative to the core is zero when it is not bound. If proton ($\ce{H^+}$) accepts electron they form hydrogen and release $13.6~\rm  eV$ of energy (photon). Energy of electron in hydrogen atom is $-13.6~\rm eV.$ In case of Helium the energies are $-54.4$ and $24.6~\rm eV.$ For Lithium we obtain energy $-30~\rm eV$ for $\rm 1s^2$ state and $\rm -5 eV$ for $\rm 2s^1$.
It is clearly seen how strong the core-electron bond near the core is and that it neglect the electron-electron repulsion - The core with all positive charge is "located" in space 23,000 - 145,000 times smaller that whole electron cloud.

My 2 cents: as noted by others, the stability is really only there for n = 1, 2 and 3, beyond that the situation becomes complicated (and He has only 2 electrons, so really we are only talking about neon and argon).
That means that the orbitals in question are: 1s, 2s, 2p, 3s and 3p. For each possible occupation of these orbitals by up to 18 electrons, energy levels can be obtained for the system as a whole. It's important to remember that only discrete energy levels exist.
Turns out that energy levels within the two periods are minimal, respectively, for electron configurations $1s^2 2s^2 2p^6$ and $1s^2 2s^2 2p^6 3s^2 3p^6$.
Incidently, that's 8 electrons in the outer shell.
As for the other questions:
And what does stabilization of atoms even mean? - In this context, it means that certain atoms are more likely to give up or take additional electrons, nothing else. It does not describe a switchlike situation - "is it stabilized or not", it only says that some are more stable that others.
Why do atoms need 8 electrons to stabilize? I mean why not 7 or 5 or 10 electrons? Why specifically 8? - This is already answered - different electron configurations have different stability.
Are the atoms going to burst without 8? - No.
Why do they need 8 electrons? - They try to obtain it because in that configuration, they have the maximum possible stability (in other words, minimum energy).

The octet rule is overcomplicated.
Real answer is - for many organic chemistry-relevant atoms, a shell with 8 electrons is by far the one with low energy state if one performs the quantum mechanical calculations.
However, the octet rule doesn't apply for other types of elements.

Why aren't atoms on the lowest energy with only 1 electron?And what does being on lowest energy have to do with the stabilization of atoms?

@Adi Sorry, I cant explain this better.

this does not answer the question in any meaningful way. You effectively say , its 8 because its 8.

and so down vote. What are ' organic-chemistry relevant atoms'. Not just C, what about H (1 electron) , S, P and many metals. Are not chlorophyll (with a Mg atom) and vitamin B12 (with Cobalt) not organic molecules?

Please excuse my rather layman reply, my specialism is not in chemistry or physics, but I felt this question is asking more why than how.
Why do atoms need 8 electrons to stabilize? I mean why not 7 or 5 or 10 electrons? Why specifically 8?
I suspect this more has to do with the fact that force is equalised. Even numbers. An odd number would suggest an unpaired electron which would suggest an unstable state. Think of it like having two cars pushing against each other evenly, or two equally attracted magnets. I don't think anyone knows why it's specifically 8, as opposed to 6 or 4 or 2.
And what does stabilization of atoms even mean?
It means they are in a low energy state. A high energy state is where an atom holds a potential to react. Take for example sodium - when introduced to water it reacts violently and explodes. That is because, in simplistic terms, it transfers electrons between sodium and water. When an atom achieves 8 electrons, there's no more electrons for it to gain or lose (IE if it was less than 8 it would have to gain additional electrons from some other material, and more than 8, it could lose some - essentially a reaction would occur).
So when we say stabilisation has occurred, we are essentially saying that conventionally the atom won't react. It has become stable.
Are the atoms going to burst without 8?
The atom itself won't burst, per se, but like in the example, the sodium will.
Edit: Not sure why the downvotes. No feedback, no correction. Electron transfer produces reactions:
https://en.wikipedia.org/wiki/Electron_transfer
https://en.wikipedia.org/wiki/Redox

The downvotes (at least mine) is because of the very unscientific answer that simplifies in the wrong direction, states wrong facts etc.

While I agree that this is not helpful, I do not agree, that it formally does not qualify as an answer. Such answers should be down votes, but not deleted.

@Jan I upvoted your honesty, but what facts are wrong or unscientific? Do not chemical reactions result due to electron transfer, which is often due to a difference in the number of electrons in given shells, and thus the energy differential transfers, that is, the reaction occurs as a result of the energy attempting to equalise? And if so, would not a stable set of 8 electrons be a result of an atom reaching it's energy resting point?

Your centering around eight electrons being stable. It is not as simple as that. In some cases, eight electrons are stable, in others they simply are not. Stable in itself is also a highly problematic term implying thermodynamic stability, which is only really true for the most stable state of a given element. On the other hand there are metastable states, which don’t react for quite some time, no matter whether they have eight, seven or a different number of electrons per atom. Also, quantum mechanics tell us very clearly why eight is stable and not six, four or ten.

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


text:
What are the bond angles of propane?
In an organic chemistry book, I find out that the observed C-C-C bond angle is approximately $111.66^\circ$. What about the C-C-H and H-C-H bonds shown in the picture below?

According to “Structure of Free Molecules in the Gas Phase”, in CRC Handbook of Chemistry and Physics, 97th Edition (2016), William M. Haynes, ed., CRC Press/Taylor and Francis, Boca Raton, FL., the bond angles in propane are
$\angle\ce{CCC}=112^\circ$
and
$\angle\ce{HCH}=107^\circ$
Therefore, the remaining bond angles should each be about
$\angle\ce{CCH}\approx109^\circ$

Presumably because there is a small amount of steric interaction between the two methyl groups and this opens up the angle from the ideal tetrahedral ones. The HCH angle is less because the other is larger and gets 'squashed' a bit!

There are two recent publications of propane crystal structures determined at various conditions. Determined geometry is also in a very good agreement with the listed angular parameters provided in Loong's answer.

From [1]: $\ce{Mo}K_\alpha$; $T = \pu{30 K}$ (Lindemann capillary in $\ce{He}$ stream); H-atom positions were taken from a difference Fourier map and refined without constraints and isotropically.
$\angle \ce{C-C-C} = 112.14^\circ$; $\angle \ce{H-C-H} = 107.26^\circ$; $\angle_\mathrm{avg.} \ce{C-C-H} = 109.33^\circ$.

From [2]: $\ce{Mo}K_\alpha$; $T = \pu{295 K}$; $p = \pu{3.2 GPa}$ (DAC); H-atom positions were located from molecular geometry.
$\angle \ce{C-C-C} = 113.24^\circ$; $\angle \ce{H-C-H} = 107.72^\circ$; $\angle_\mathrm{avg.} \ce{C-C-H} = 108.94^\circ$.

References

Boese, R.; Weiss, H.-C.; Bläser, D., Angewandte Chemie International Edition 1999, 38 (7), 988–992 DOI: 10.1002/(SICI)1521-3773(19990401)38:7<988::AID-ANIE988>3.0.CO;2-0.
Podsiadło, M.; Olejniczak, A.; Katrusiak, A., J. Phys. Chem. C 2013, 117 (9), 4759–4763 DOI: 10.1021/jp311747m.


text:
Orange dust floating ontop of Electrolysis solution
While I was doing an Electrolysis experiment with stain less Anode/Cathodes on aluminium plate, Orange dust started floating on top of the water solution.
What is this chemical's name and is it toxic or not?

You realize that your description of the experimental setup is a bit scanty, right?

@IvanNeretin I'm guessing he/she only wants to know what the Chemical inside the solution is.

@DraggyWolf I mean electrolize on aluminium plaet with stain less rods

Could be a chromate or dichromate, which is very likely to be rather toxic. These compounds tend to have yellow and orange colors. Stainless steel is an alloy of iron and chromium, the latter of which can could cause this.
I would recommend NOT using stainless steel electrodes, and find a way to safely dispose of the mixture.
Here is a link which might help explain why this is not a great idea.
However, it might also be possible that the chemical you are electrolysis is causing this. What is in the solution?
Assuming that it is producing chromates, you might be able to stop this by turning down the voltage.

This answer assumes that the "stain less" you refer to is actually stainless steel, and I assume that English is not your first language.


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Reaction between phosphorus pentachloride and sulfuric acid?
I am a novice in chemistry. While reading a magazine, I stumbled on some information about phosphorus pentachloride and sulfuric acid. Phosphorous pentachloride and sulfuric acid are both good dehydrating agents. What is the reaction between them (at standard temperature and pressure).

An invoice? Do you mean a novice?

Are you sure that you want to ask about phosphorous pentachloride $(\ce{PCl5})$ rather than phosphorus pentoxide $(\ce{P2O5})$ as a dehydrating agent?

Now that would be an epic battle. I guess $\ce{PCl5}$ would wrestle some water from sulfuric acid and drive it to $\ce{SO3}$, or maybe even all the way to $\ce{SO2Cl2}$.

@IvanNeretin Brace yourselves! We need meme pictures! ;)

First phosphorus pentachloride reacts with sulfuric acid to form the intermediate(thanks @bon) chlorosulfonic acid. See here. Also Wikipedia
$$\ce{H2SO4 + PCl5 → POCl3 + HSO3Cl + HCl~~~~~~~~~~~~~~~~~~(1)}$$
Chlorosulfonic acid is actually an intermediate between  sulfuryl chloride  and sulfuric acid. After the formation of the intermediate, the reaction will proceed forward to form stable sulfuryl chloride. (Here)
$$\ce{HSO3Cl + PCl5 → POCl3 + SO2Cl2 + HCl~~~~~~~~~~~~~~~~~~~(2)}$$
Adding (1) and (2), we get the final reaction:
$$\ce{H2SO4 + 2PCl5 ->2POCl3 + SO2Cl2 + 2HCl}$$
My reaction is same as that of Vishnu JK's reaction and bon.

This seems to be a much more complicated reaction than one might initially guess. A variety of products may be obtained depending on the temperature and the relative proportions of the reactants.
Chlorosulfuric acid, $\ce{HSO3Cl}$, was first discovered by Williamson in 1854[1] from the reaction of $\ce{PCl5}$ and $\ce{H2SO4}$. He states that (emphasis mine):

as a result of numerous experiments with the most varied proportions of pentachloride and acid ... the first action of the pentachloride consists in ... forming the compound $\ce{HSO3Cl}$, which is strictly intermediate between the hydrated acid and the final product $\ce{SO2Cl2}$.

A more recent source[2] suggests that chlorosulfuric acid decomposes at higher temperatures and at 170 °C an equilibrium exists (also mentioned here[4]): $$\ce{2HSO3Cl <=> H2SO4 + SO2Cl2}$$
It also says that:

chlorosulfonic acid, by boiling in the prescence of ... catalysts, decomposes quantitatively into sulfuryl chloride and sulfuric acid.

It would seem that formation of $\ce{SO2Cl2}$ by this route is favoured by using excess $\ce{PCl5}$.
Other recent works suggest that it is possible to prepare $\ce{HSO3Cl}$ by careful distillation of the product. One book[3] suggests that it may be distilled at 161 °C and gives the following two reactions (the first is also mentioned by another source[4]): $$\ce{H2SO4 + PCl5 -> HSO3Cl + POCl3 + HCl}$$ $$\ce{POCl3 + 2H2SO4 -> 2HSO3Cl + HCl + HPO3}$$
It mentions that the 'calculated quantity of phosphorous pentachloride' should be used but declines to say what this is but from the equations one would expect that this route is favoured by using excess sulphuric acid.
References:
[1] http://www.jstor.org/stable/111726?seq=4#page_scan_tab_contents
[2] https://books.google.co.uk/books?id=roI_CwMEcHMC&pg=PA1&lpg=PA1&dq=phosphorus+pentachloride+and+sulphuric+acid&source=bl&ots=FeM9kqKg8-&sig=BRMAXCcVormdf6nAvy8dI0yP4Do&hl=en&sa=X&ved=0ahUKEwiFqsCbu-TNAhUBuhoKHREeD244ChDoAQg2MAU#v=onepage&q=phosphorus%20pentachloride%20and%20sulphuric%20acid&f=false
[3] https://books.google.co.uk/books?id=IDsXBQAAQBAJ&pg=PA401&lpg=PA401&dq=phosphorus+pentachloride+and+sulphuric+acid&source=bl&ots=M4Gpbcn4tO&sig=bvYuWQ9i2hmGbanLt9x8sFt443s&hl=en&sa=X&ved=0ahUKEwiFqsCbu-TNAhUBuhoKHREeD244ChDoAQgzMAQ#v=onepage&q=phosphorus%20pentachloride%20and%20sulphuric%20acid&f=false
[4] http://sulphur.atomistry.com/chlorosulphonic_acid_cl.html

The reaction occur as follows

thus the products are sulfuryl cholride,phosphoryl chloride and hydrochloric acid.
This might clear all the doubt between Nilay Ghosh and my answer.

And Thank@bon for pointing out the mistake
sources
first link
2nd link
and also my text book
4th question pdf

That is not thionyl chloride, it is sulfuryl chloride. Also, I don't think either of those sources are reliable.


text:
Melting point of ocean salt: eutectic?
I'm interested in the melting point of ocean salt. While sodium chloride has a melting point of 801 °C, ocean salt contains 15 % other ions, mainly sulphate, magnesium, calcium and potassium (the exact composition seems to differ depending on the water source).
Does ocean salt form a eutectic mixture with a lower melting / boiling point? Are there perhaps publications actually testing ocean salt melting points? Thanks!

Without any hard data, my feeling is that not all the "salts" sea water are mutually soluble. In additoin, bicarbonates may decompose before the melting point of the sodium chloride, and anhydrous magnesium sulfate decomposes at 1,124°C rather than melting.

Excellent point. This would lead, then, to oxides instead of sulphates and carbonates, in the mixture, but they would be a minority, so the question still stands, as I see it.

Your should read up on what exactly "Eutecticum" means. Wikipedia?

I presume you refer to the German term... it is "eutectic mixture" or "system" in English. I agree that I am not using the term correctly in the sense of describing a phase equilibrium - it's to get the idea across; freezing-point depression seems to be less clear in this context.

Yes, the melting point of ocean salt is going to be lower than pure NaCl at 801 °C.
In sea water you don't have compounds but rather dissolved ions. Other than the most abundant Cl- and Na+ (in this order), sea water contains Mg2+, K+, Ca2+ and SO42- (not in this order). Thus, we can assume that the solid residue would be NaCl, MgCl2, CaCl2 and KCl.
Let's have a look at several phase diagrams, taken from http://www.crct.polymtl.ca/fact/pdweb.php:

So yes, adding additional cations other than Na+ will reduce the melting point. Note however, that you will get a lower melting point (solidus) and a higher melting point (liquidus) that in between you will have an equilibrium mixture of solid crystals and melt.
Other than Cl- there's also SO42- and CO32- (as both carbonate and bicarbonate). This will also lower the melting point:

The melting point of the sea salt mixture will be lower than the decomposition of the sulfates and carbonates, so you will dissolve them into the liquid salt rather than degas them as SO2 and CO2 (at least initially).
It's also interesting to see that as you heat the mixture, you will end up with intermediate compounds, such as (Na,K)Cl and (Na,Mg)Cl. As to the exact melting point of the mixture, I couldn't find any experimental data in my quick search. It is possible that thermodynamic modeling would be able to give a prediction. But remember, as the mixture is eventually dominated by NaCl it will not be too far off from the melting point of the pure compound. My uneducated guess is that you will only start seeing any significant melting at around 700 °C and above.
Note of caution - if you do decide to actually melt sea salt, be sure to take the necessary precautions. Even though SO2 will mostly dissolve in the melt, some of it will degas. Molten halides are extremely corrosive and will react with atmospheric moisture, sometimes explosively. They will react with your crucible, possibly causing leakage of really nasty stuff. Unless you know what and how you're doing it, don't do it.

Excellent answer - thank you very much for the time and work spent to collect and write down all of this!


text:
What guarantees that there is a thermodynamic equilibrium in a closed chemical reaction network independently of the mass-action kinetics?
I would like to know according to which we can state that there is a unique equilibrium of a closed chemical reaction network (at constant circumstances) without supposing mass-action kinetics if time is increasing beyond all limits. Can we derive it from a non-equilibrium thermodynamic law or is it a postulate? Thank you for your answer!

I think what you're describing isn't a law, but a convenient way of introducing equilibrium thermodynamics. There are examples of non-equilibrium thermodynamics that are not time-invariant, such as clock reactions https://en.m.wikipedia.org/wiki/Chemical_clock

The equilibrium is determined by the thermodynamics, not by the kinetics.  The rate constants for the reaction kinetics must be consistent with the equilibrium determined by the thermodynamics.  For example, for a single reversible reaction, the reverse reaction rate constant must be equal to the forward rate constant divided by the equilibrium constant.


text:
Why does rotating a double bond break the bond?
It is well known that double bonds are "stiff" and make particles less likely to rotate around the bond. But why? What makes it so that the bond has to be orientated a specific way?

Reduce overlap.

What do you know about how the second bond of a double-bond forms?

https://youtu.be/nxVO2cz4OCM

Besides the thermodynamic barrier with respect to double bond rotation, we could also look at it from a symmetry perspective.
Rotating a double bond, without breaking the $\pi$ bond first, requires that you rotate the left or right orbitals that participate in the $\sigma$ and $\pi$ bonds (but not both atoms' orbitals, because that would be rotating the molecule, not the bond).
Suppose that we are looking at a $2p_x-2p_x$ $\pi$ overlap for a $\text{C}=\text{C}$ bond. If we rotate the $2p_x$ orbital $90^o$ about the internuclear ($z$) axis, we can transform the $2p_x$ orbital into a $2p_y$ orbital. Ethene, for example, is a molecule of $D_{2h}$ symmetry, and contains a $\text{C}=\text{C}$ bond.
According to this D2h character table, in ethene, the $2p_x$ orbital transforms under the $B_{3u}$ IRREP, but the $2p_y$ orbital transforms under the $B_{2u}$ IRREP.
Two orbitals transforming under different IRREPs cannot overlap, so the $\pi$ bond can no longer be made if one of the orbitals is rotated.
Beyond that, if you rotate one of the $2p_x$ orbitals about the internuclear axis by some small angle (instead of specifically $90^o$), you would change the orientation of the x-axis for only that orbital, and they would technically not be the same orbital anymore because each would be following a different x- (and y-)axis convention.

What you say about symmetry is true, but in essence the lack of  rotation is about the stability of the double bond vs. thermal energy available. The double bond's strength is far greater than this and so is stable wrt rotation. The bond has to have two similarly oriented p orbitals as these otherwise do not overlap, i.e. we say that px, py , pz etc are orthogonal to one another.

It is quite easy to break a double bond with a photon of the appropriate energy. This promotes an electron into an 'antibonding' LUMO from which rotation is far easier as $\pi$ electron overlap is reduced.

@porphyrin therefore can we say that cis-trans forms are interconvertible at high temperatures

Yes, technically they are interconvertible any any temperature  but both the rate constants will become bigger as the temperature increases, as per Arrhenius for example.


text:
Which elements exhibit allotropy?
I was just reading this article about allotropes which relates allotropy with some specific elements and also this wiki article. Also some other articles some of which relate allotropy with catenation some with electronegativity and some with electronic configuration. I was understanding the relation of allotropy with all these factors but I am unable to get that how allotropy is dependent on them exactly. Can we predict the allotropy in elements (and also the extent to which they exhibit allotropy) just by concerning their chemical properties?

see this:-http://chemistry.stackexchange.com/questions/34223/difference-between-catenation-and-allotropy

Given proper conditions, any element could have multiple allotropes For example, under sufficient pressure, hydrogen has a metallic phase, which may have been observed at Harvard recently.
The extreme example is degenerate matter, formed under the pressure of a collapsed star, though it could be argued that the elements are no longer the same, since nucleons themselves are eventually pooled in a "sea".
If you mean by your question "having allotropes at standard temperature and pressure (STP)", then the list is shorter, of course. Since this depends on the electronic configuration of an element and on the minimal-energy configuration of atomic arrangement, calculation is difficult, but not impossible, and there is software available for crystal structure prediction.
Now if I could only find a more convenient way to change graphite to diamond, without Superman's help...

As usual, the acid test for predicting elemental crystal structures is Manganese - getting the cbcc and cubic structures is actually quite a chore without lots of tweaking of 'first principles' methods. (Or actinides, although I rate them as clearly odd based on f electrons.)


text:
Toothpaste and water
I noticed when brushing my teeth that toothpaste causes water to spread away from it. If you put a bit of toothpaste water mixture on your finger and touch it to a wet sink you will see the water on the sink travel away from your finger.
My question is what chemical in toothpaste is causing this to happen and in particular what property of the chemical is causing it.

It's all about surfactants. Soap would do the same, maybe stronger.

Drops or patches of water in your sink are basically water that is not evenly spread out. The surface tension of the outermost water molecules acts a bit like a "balloon" containing the inner molecules. To a small degree it is thus possible for the water molecules to be stacked on top of each other because the water's surface tension is stronger than the gravitational pull.
Toothpaste contains detergents like sodium dodecyl sulfate which act as a surfactant reducing the surface tension of water.
When you coat your finger in a mixture of water and a weak detergent like toothpaste or a stronger one like soap and touch the water drops in the sinks, the water mixes with the detergent. This lowers the surface tension and gravity spreads the drop out. The water is thus not really moving away from your finger, but the drops are being spread out under and around your finger.


text:
Net conductivity of mixtures
If we have a solution, of conductivity k1, and another solution of same volume having conductivity k2, then what will be the conductivity if I mix the 2?
I think that net conductance should be the sum of individual conductance of the 2 solutions in the final beaker (where volume is doubled). But then won't the individual conductance of the 2 solutions in double the volume be different from that in the initial volume?
Also, can we say that conductivity is the conductance of unit volume of solution?

This concept of electrolytic solution conductivity is complex and dependent on a variety of factors aside from mere concentration of electrolytes. There are separate laws used to describe the relationship between solution conductance and concentration dependent on the dissociative strength of electrolytes and the extent of overall solution saturation.
In your simple case, just recognize that the net conductance of the final solution may not work out to be the mere sum of conductances in the original solutions
From Wikipedia:

The specific conductance of a solution containing one electrolyte
  depends on the concentration of the electrolyte. Therefore, it is
  convenient to divide the specific conductance by concentration. This
  quotient, termed molar conductivity, is denoted by Λm
  $$\Lambda _{m}={\frac  {\kappa }{c}}$$

Hence, since conductivity is proportional to concentration, and concentrations aren't additive, you cannot expect conductivities to be additive.

First let me remark that at low concentrations the conductivity will be proportional to concentration as noted. However if you extrapolate this line to higher concentration then the linear relationship overestimates  the conductance. In general the greater the concentration the worse the relative error.

Formula for centration mixing - letting $c_m$ be the concentration of the mixture then: $$c_m = \dfrac{c_1v_1 + c_2v_2}{v_1 + v_2}$$


text:
Decomposition products of magnesium ethanoate
What does magnesium ethanoate decompose to? The Wikipedia page states that one of the products is magnesium oxide. What's the other product? I can come up with some guesses:

Ethanoic anhydride ($\ce{MgC4H6O4(s) -> C4H6O3(g) + MgO(s)}$)
Propanone and carbon dioxide ($\ce{MgC4H6O4(s) -> C3H6O(g) + CO2(g) + MgO(s)}$)

Since calcium acetate (ethanoate) decomposes into propanone, I'd say that it is the case here as well. Also, at least some metal oxides, particularly ThO2, catalyze carboxylic acids' decomposition into ketones. The process seems to be pretty generic, even if requiring quite elevated temperatures.

EDIT: Ok, here's your answer (Thermochimica Acta, 78 (1984) 17-27): It decomposes into the oxide, carbon dioxide and propanone:
$Mg(C_2H_3O_2)_{2(s)} \rightarrow MgO_{(s)} + CO_{2(g)} + CH_3COCH_{3(g)}$
PRE-EDIT:
The Russian Wikipedia favors the second reaction. This seems to support the idea.
This site mentions magnesium carbonate and propane-2-one as products, but this seems to be questionable, given the other sources.


text:
Preparation of a fluoride solution using CaF2
I need to prepare several solutions with different concentrations of $\ce{F-}$, in order to carry out some investigations on fluoridated water. Would this be achievable by dissolving calcium fluoride, $\ce{CaF2}$, in water? Wikipedia writes that $\ce{CaF2}$ is insoluble in water, so I am not sure.

Calcium fluoride has a $K_\mathrm{sp}=3.9\times 10^{-11}$. That means a saturated solution of calcium fluoride will have a fluoride concentration of $4.28\times 10^{-4}\mathrm{\ M}$, which, assuming the density of this dilute solution is approximately that of water, is $\mathrm{8.13\ mg/L}$ fluoride.
Fluoridated water has fluoride concentration ranging from $\mathrm{0.5\ mg/L}$ to $\mathrm{1.5\ mg/L}$, so your saturated calcium fluoride solution is right in that range. If you need to test lower concentrations, dilute your solution. If you need to test higher concentrations, you are out of luck. You cannot make more concentrated fluoride solutions with calcium fluoride.


text:
Variation in pressure for a gas phase reaction

Predict how the total pressure varies during the gas-phase reaction
  $\ce{N_2}(g)+ 3 \ce{H2}(g) \rightarrow 2 \ce{NH3} (g) $ in a constant volume container.

Reference: Atkin's Physical Chemistry, Tenth Edition, Chapter 20, Exercise 20A.1 (b) on page 870
My attempt at a solution: Assuming ideal gas behaviour we know that pressure is proportional to number of moles (volume and temperature held constant). Also, by observing the reaction stoichiometry the final pressure should be half the initial pressure as the no. of gas moles changes from 4 to 2
Initial no. of moles $n_i = n_{\ce{N2}} + n_{\ce{H2}}$ with no ammonia present.
at an arbitrary instant in the reaction $\alpha$ is the fraction of $\ce{N2}$ that has reacted.
So total no. of gas moles: $n = [n_{\ce{N2}}(1-\alpha)] + [n_{\ce{H2}}-3\alpha n_{\ce{N2}}]+ 2\alpha n_{\ce{N2}} $ ; the first two terms being the change in no. of moles of reactant and the last one being the no. of moles of ammonia produced.
This simplifies to, $n = n_{\ce{N2}} + n_{\ce{H2}}-2\alpha n_{\ce{N2}}$
What I think the question wants from me is a $P$ at an arbitrary instant given as a fraction of $P_i$ i.e $P_t= $$P_i$*[term depending on  $\alpha$]. Also substituting $\alpha$ = 0 or 1 should yield $P_i$ and $\frac{1}{2}P_i$ respectively.
For that, I am guessing the expression for $n$ should be expressed as a fraction of $n_i$.  However, I can't seem to think of the necessary algebra to obtain such an expression.
EDIT: Inspiration struck, what if I do the following:
$n = n_{\ce{N2}} + n_{\ce{H2}}-2\alpha n_{\ce{N2}}$
gives,
$n = (n_{\ce{N2}} + n_{\ce{H2}})(1-2\alpha \frac{n_{\ce{N2}}}{n_{\ce{N2}} + n_{\ce{H2}}})$
which is the same as, $(n_{\ce{N2}} + n_{\ce{H2}})(1-2\alpha .\chi_{\ce{N2}})$ where $\chi_{\ce{N2}}$ is the initial mole fraction of nitrogen in the reaction mixture
so, $n = n_i(1-2\alpha .\chi_{\ce{N2}})$
and consequently, $P_t = P_i(1-2\alpha .\chi_{\ce{N2}})$
Would this be a valid expression for describing the variation in pressure during the gas phase reaction described above?


text:
Software for visualization of 3D models of chemical structures with atomic orbitals
A number of programs are available that let you build, visualize, and analyze 3D models of chemical structures (I use Chem3D/ChemBio3D). Current versions usually include various possibilities for the calculation and graphical representation of molecular orbitals.

However, the resulting images can be overloaded or unclear. Introductory explanations using such images might be overdoing it. Therefore, many textbooks on organic chemistry use simple chemical structure diagrams based on atomic orbitals.

In particular for didactic purposes, I am looking for an intermediate solution. I would like to build, optimize, and visualize a 3D model of a chemical structure and show a graphical representation of the original atomic orbitals in the 3D model. Unfortunately, this option doesn’t seem to be available in Chem3D/ChemBio3D. Does any software exist that lets you build a 3D model of a chemical structure and visualize atomic orbitals?

I haven't come across something like that but I have been wanting something like it as well. I could imagine that it might be possible to achieve this with Avogadro if you are proficient enough with it and C++ to write some sort of plugin.

Does any software exist that lets you build a 3D model of a chemical structure and visualize atomic orbitals? || to my knowledge, no. However, QC packages often offer 'natural orbital analisys', that among other things prints participation of atomic orbitals in molecular orbitals. You can also visualize individual atomic orbitals in some cases and copy-paste them.

I'm not entirely sure what you want. For example, you say "can be overloaded or unclear" and "might be overdoing it." In my experience, most students can grasp molecular orbitals (or the natural orbital transformations) well. You just need to pick rendering options (e.g., partial transparency) that suit your needs.

To follow on the other comments, besides doing a natural orbital analysis (a good option - you can then use most software to view the results), I am not aware of any such software. However, there are probably ways to hack this even without coding a rendering option. If you want to pursue it, consider contacting me.

You could try JMOL as its free and has the ability to draw orbitals as well as bonds. It is fairly easy to construct structures given coordinates. See molecule-viewer.com for some examples of what you can do.

The "overloading" likely arises from default parameters for the isosurface used to display. Just increase that and the MOs will look less crazy.

As porphyrin mentions in the comments, have a look at Jmol.

It indeed does the rendering as in your original version, but with the many options (see this mo command documentation) and particularly interactivity provided by the scripting, I think you may pedagogically achieve what you are looking for. Proteopedia is a nice website where the efficiently use Jmol's scripting feature for educational purposes (try the green links).


text:
Alkali metal hydrides
I know that stability of alkali metal hydrides decreases down the group,Hence the lowest member in the group easily breaks into its cation but does this easiness of breaking into cation has any relation with its reducing nature.I also wanted to know what is order of reducing character in alkali metal group?

Alkali metal hydrides can typically be viewed as labile hydride anions. The extent of lability can correlate with the electronegativity of the metal. Hence, less electronegative metals in alkali metal hydrides yield more labile hydride anions and more "reducing" nature.