text:
Which is more basic, hydrazine or ammonia?
Which of the following will be more basic, hydrazine $\ce{H2N-NH2}$ or ammonia $\ce{NH3}$?
My chemistry teacher said that $\ce{NH3}$ is more basic since after giving $\ce{H+}$ to hydrazine results in $\ce{H3N+-NH2}$, and accommodation of the lone pair of $\ce{-NH2}$ is not possible by $\ce{-N+H3}$, since it doesn't have the space. So instead, $\ce{-NH2}$ will pull electrons from it making it unstable. Ammonia has no such problem so it must be more basic.
I am not so pleased with this argument. In $\ce{H3N+-NH2}$, although the lone pair cannot be accommodated, but the positive charge present on its sides , to an extent, should neutralize the intensity of the lone pair, making it somewhat stable. And also, not to forget, hydrazine has two spots where we can get the electrons, therefore, its ambident nature should also support it's basicity.
I need a bit of clarity on this.

Obligatory link to Evans' pKa table.
Ammonia is more basic than hydrazine, by about one order of magnitude.  This is expected, because the -NH2 group is more electronegative than -H or -CH3.
The second lone pair is not involved in the acid-base reaction, it does not point towards the -NH4+ group.  You can, however, force two lone pairs into close proximity.  This destabilizes the unprotonated form.  The keyword is "proton sponge".
1,8-Bis(dimethylamino)naphthalene has a pKa of 12.3, it's one of the strongest known amine bases.  Compare that to the pKa of aniline, which is something like 4.5.

Ok, I get yours and my teachers point, but please elaborate on why I am wrong.

How exactly were enol pKas determined?

What about the alpha effect? Will that not enhance the basicity of hydrazine?

Please dont give wrong pka values. The pka of the conjugate base of acid is 4.5, and not that of aniline.

What about nucleophilicity? I guess hydrazine is better.

When protonated, ammonia and hydrazine give their conjugated acids:
$$\ce{NH3 + H3O+ <=> H4N+ + H2O} \tag1$$
$$\ce{H2N-NH2 + H3O+ <=> H3N^+-NH2 + H2O} \tag2$$
Let's rewrite these conjugate acids: $\ce{H3N^+-H}$ and $\ce{H3N^+-NH2}$ . We all know that electran withdrawing ability ($-I$ effect) of $\ce{-NH2}$ group is higher than  that of $\ce{-H}$ group. Therefore, $\ce{-NH2}$ group in $\ce{H3N^+-NH2}$ destabilizes the positive charge more than $\ce{-H}$ group in $\ce{H3N^+-H}$. In other words, conjugate acid of $\ce{NH3}$ is more stable than that of $\ce{H2N-NH2}$. As a consequence, forward reaction of equation $(1)$ is favor than that in equation $(2)$. This means basicity of ammonia is greater compared to that of hydrazine.

We didnt consider the M effect of NH2 here because its not possible to donate right @MathewMahindaratne ?

I think these could be the reason-
For the first part,

...positive charge present on its sides , to an extent, should neutralize the intensity of the lone pair, making it somewhat stable

This is not possible because $\ce{NH_3^+}$(no vacant orbital) doesn't have any space at all to get involved with the lone pair.
...and the second part

...hydrazine has two spots where we can get the electrons, therefore, its ambident nature should also support it's basicity.

Abel already answered that at one time only one $\ce{-NH_2}$ takes part when we determine basicity and the second $\ce{-NH_2}$ plays no role.

To clarify the first part, I am not saying that the electrons will jump to the protonated nitrogen. I'm saying that the presence of a positive charge near the electrons will try to reduce its intensity and make it somewhat stable. For the second point you made, more number of nucleophilic sites would mean more chances of attack of an $H^+$, which adds to the basicity of Hydrazine.

You shouldn't compare the basicity of Hydrazine as a molecule. Just because it has two basic sites, it will not be more basic. It is akin to saying that just because Sulphuric acid has two acidic hydrogens, it is a stronger acid than Perchloro-acid, which is untrue. Remember, in any case, there will be only ONE protonation at a time. You should compare either Ka1, or Ka2, with the corresponding values for Ammonia.

I'm just saying that the probability of attack, and did not mean that it decreases it's $pK_b$ value.

Ammonia is more basic than hydrazine if you look at the neighbors you will see $NH_3$, and $NH_2-NH_2$ where Ammonia has hydrogen as third neighbor where hydrazine have N  as neighbors which gives more strong - I effect, after protonation.


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What are the conditions for making amines from alkyl halides?
From what I understand, to make an amine from an alkyl halide you need to heat the alkyl halide with a mixture of ammonia and ethanol. I have two questions:

Do both ammonia and ethanol need to be concentrated? Because I think I read somewhere that you use a 'concentrated mixture of ethanol and ammonia', although I'm not sure whether the 'concentrated' part refers to the ethanol or the ammonia. I'm guessing it refers to the ammonia.
What is the role of the ethanol? Does it act a a catalyst, or just a solvent, or both?

Ethanol is just a solvent often used for this reaction because 1) the reactants and products are soluble in it and 2) the reaction proceeds at a reasonable rate in boiling ethanol.  If your trying to prepare the corresponding primary amine, then a large excess of ammonia should be used.  Otherwise, if the concentration of the initial product (the primary amine) becomes comparable to the concentration of ammonia, then the primary amine could react with the starting alkyl halide and a complicated mixture of products could ensue (see this article for a nice description of all the possible products that could be formed if the ammonia concentration is not high enough).
The reaction is typically run at reflux (boiling ethanol).  Since ammonia is a gas at room temperature it would quickly escape from the reaction and we wouldn't have a high concentration of ammonia for long.  Therefore, the ammonia is condensed into a solution of the ethanol and alkyl halide (usually an alkyl bromide) in a glass tube at low temperature.  When the desired concentration of ammonia is obtained, the tube is sealed and then heated to allow the reaction to take place.

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How many unique molecules exist? And how many exist only by synthesis?
I'm interested in

How many  unique molecules (of all types: organic and inorganic, etc), exist  in the world accessible by man (let's exclude black holes and the centers of stars etc)?
What subset of those distinct types are only in existence because of synthesis by man?

To make it simpler, ignore variants such as molecules with different atom isotopes and different bond configurations (e.g. cis, trans). E.g. standard water and "heavy water" are both $\ce{H2O}$ and count as one.
I'm looking for a theoretical estimate, so orders of magnitude only are required. I'm not looking for a count of records in some database catalogue of chemicals.
Please explain methods for determining your estimates. This isn't really helpful.

I can't even begin to imagine how an appropriate answer can be offered to this question.  Given the vastness of the Universe and how much of it we don't actually know, and given the multitude of chemical environments that are possible, it would be difficult to answer this question without make a variety of gross approximations.  Also, what do you mean by distinct types?

Define "types of molecules".

Do you mean like "organic", "inorganic", "Biological"... ?

@LordStryker I don't know how to explain "distinct molecules" better than by expanding on the example in my question: H2O is one. O2 is two. CO2 is three. Ethanol C2H6O is four. Methyl parahydroxybenzoate CH3(C6H4(OH)COO) is five... Make sense?  Re "vastness", I'm only looking for orders of magnitude.

@Wildcat does my example for LordStryker help? What is not clear?

@AngusTheMan, no, not classes or general groups of molecules. Please see my comment to LordStryker. Does that make sense?

@poshest, is $\ce{SiO2}$ of the same type as $\ce{CO2}$?

@Wildcat no. They count as one each in the count. I think the word I should have used was "unique" molecules, not "distinct" (sorry, coming from a database background).

@poshest, then I have a feeling that your question is about the number of molecules with different molecular formulas. I doubt that one can estimate that.

Could we ask what is the motivation for this question? It seems like you want an estimate of synthetic molecules vs. all molecules. Why? Curiosity or something else?

@trb456 Yeah, I'm writing a paper, and wanted to start with some numbers to give perspective in the intro. It disturbed me that there was nothing on the web about this. But jerepierre gave me the secret code (search term) "chemical space"

I'm inclined to estimate over 9000.

Depending on the precise definition of the terms "molecule" and "unique", the answer can be either a very large finite number, or countable infinity, or even uncountable infinity.

The answer is likely to be uncountably big (bigger, for example, the the number of particles in the observable universe). For simple unsaturated hydrocarbons alone the number of structural isomers is thought to exceed 10^22 for just 50 carbons. And that is just two types of atom with restrictions on connectivity. The possibilities grow faster than exponentially for more atom types. And we can do longer chains.

@SherlockHolmes - my more refined estimate is over 9001...

Polymers make this question more or less unanswerable.
Consider human chromosome 1, which contains about 249,000,000 base pairs.  There is nothing that says we couldn't order those pairs in any way we like, so for DNA alone, in a quantity that exists in every person on Earth, there exists the possibility for 4249000000 different molecules.  Include things like alternative nucleotides (there are almost surely at least thousands of these that are possible) and longer chains and the number gets even more enormous, and that's only including DNA (and close relatives.)
There's no real way to include all of them, since there have to be an enormous number of ways to put these molecules together that we haven't found yet.
How many exist?  An awful lot.
How many only exist in our minds?  Almost all of them (the universe only contains about 1080 particles, after all.

That's a great explanation for a theoretical maximum. I was thinking about DNA combinations in the same way. As for actual numbers that exist, can you get any closer precision than "an awful lot"?  Any idea of the proportion of those that only exist because they're man-made?

I think amount of particles is very unstable in the universe. You know, black holes, black mass and stuff.

Awful lot? Actually, you mean infinite. There is no upper bound for chainlength of polymers, therefore their number itself is infinite. The only real upper bond is the size of universe and the number of atoms in it.

Well, that gives us a reasonable upper bound. Each atom is or is not bound to each other atom, so we're looking at ways to partition N=10^80 elements, i.e. the upperbound is Bell(10^80) = e^(e^(10^80)). Writing out that number is impossible in this universe, though, as it contains more than 10^80 digits (!)

@Greg If we're going to be picky, it wouldn't really be infinite, but it's going to be absurdly large.  Above a certain limit the molecule's behavior will be dominated by gravitational affects.  Make it large enough and it will collapse into a star and begin fusion, messing up the chemical bonds.  The best you could do for something that would qualify as a single molecule would be an upper mass limit for a white dwarf star.  That's going to limit your chain length to roughly 10^57 atoms total.  There are a ton of ways to arrange those, but it's not infinite.  I love this stuff...

@JasonPatterson You are correct. It is also an interesting non-chemical type of upper-bond.

Without some constraint on the number of atoms/molecular weight, the number of possible theoretical compounds is infinite, for the reason given by Jason Patterson: you can always extend a polymer by one more unit.
It has been estimated that there are 10^60 possible organic compounds with molecular weight less than 500 The art and practice of structure-based drug design: a molecular modelling perspective. CAS may have duplicates, but it is still the best database of all known compounds in the literature, and yet is not comprehensive. Almost all of the CAS database is wholly synthetic. A search for entries in CAS with molecular weight <500 gives 61,026,438 hits, about half of the entire database. The reference above suggests that life can exist with a few thousand small molecules (i.e. compounds with molecular weight <500). Nature produces far more compounds than the absolute minimum, but I expect that we have only scratched the surface of what is out there.
That should give you an idea of the orders of magnitude that we are talking about. Keep in mind that the numbers will grow exponentially as the constraint of MW < 500 is relaxed. In short, chemists have synthesized literally millions of different compounds, but essentially none relative to the possibilities.
I suggest you research the topic of "chemical space" if you want to know more about this area.

I think stratification into molecular weights is a good way to break the problem down. Who estimated the 10^60? And how? I don't have the book, sorry. +1 for "chemical space" too! The magic search formula! :)

This has some references to your numbers and a lot of other good stuff http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3447393/

Reymond (in the link above) suggested the 10^60, but others have agreed the number is probably in that range. These questions usually fall into the range of cheminformatics.

According to the Chemical Abstracts Service, CAS REGISTRY (SM) contains more than 90 million unique organic and inorganic chemical substances, such as alloys, coordination compounds, minerals, mixtures, polymers and salts, and more than 65 million sequences—more than any other database of its kind. It includes substances reported in the literature back to the early 1800s, and is updated daily with about 15,000 substances.
The CAS registry does not include substances yet to be thought of or discovered. It does indeed include your five examples.
This is about as close to an answer as I can think of.

well, I specifically asked not for a "count of records in some database catalogue of chemicals", but I do appreciate your answer. Is there any way of knowing the proportion of CAS that comprises wholly unnatural (synthetic) molecules?

I think the proportion of naturally occurring substances is a relatively small proportion of the total. My reasoning: Note that while many substances exist in nature, not very many exist in nature in pure, bulk form. Note also that "natural products chemistry" is a specialty within organic chemistry dealing with substances that are produced within living organisms. There are a lot of natural products out there that have not yet been discovered--but this is still a small subset of possible substances.

In a database of currently available synthetic compounds that i built, there are 9.8 milion synthetic and natural compounds available to purchase, and an additional 27 milion compounds that the vendors think they can produce if needed.
This explicitly includes only pure compounds that can be described by drawing a structure, so no mixtures, and no stuff like DNA where you can only guess at the exact structure.
I included only around 30 trustworthy vendors, and this set is specifically not the same as the CAS register, mind you, as that one is very heavily polluted with gazillions of duplicates and variations in notation. Also, as a nice to know: nearly all vendors lie about the number of compounds they have.

You built that? Nice work! The Alibaba of chemicals. :) This is a great angle on the commerical numbers. Any idea what proportion only exist synthetically?

Thanks for the compliment :-)
The natural compounds can be either synthetized to be exactly as they are in nature, or actually filtered out of a natural source. Still, those are only a very small fraction of the total available compounds.
I do not have a count at hand, but i would say an upper limit of 5% would be very royal.
Of all the others, most compounds could, by chance, exist 'in the wild' in very very small quantities. So there is our next definition hurdle: when do you declare that a compound 'exists' in nature? If you have one molecule?

It is not available online, sorry.
This data represents a lot of work and tools that are not free either.

You don't need to go to polymers. You can already create a huge amount of molecules in the realm of small molecules (MW < 900 Da). It is an interesting field since most of today's drugs are small molecules.
Here is the link to a research group in Bern, Switzerland that enumerates virtual compounds and tries to classify to many different categories. One is synthetic availability.
This is where they explain what they do
Here are the tools they offer
They have a lot of tools with fancy visualization for some. Happy playing.

Choose an upper limit on number of atoms in a molecule. (e.g. max_a=500)
Choose which of the atomic elements you wish to allow (e.g. els=1..92)
Select atom from els list
What possible number of bonds can this atom type make (e.g. b_el1 = {2,3,4})
Select how many bonds in this instance (e.g. current_b = 3)
Select next bond number  (bond_num = 1 in first iteration)
Select next add-on atom from els list, to be attached at current bond num
Check add-on atom for compatibility: bond type, electrons avail, space fit
If pass all tests, then add-on. Index atom counter +1.
If not pass, return to step 7.
If pass, return to step 6.
If all bonds full, return to step 7
If atom counter exceeds max_atoms, STOP
Eliminate redundancies (which will equal quant of termini-1)

For anyone with access to reaction compatibility data, this algorithm generates the total number of possible molecules up to a chosen limit.
Not all details are provided for an 'exhaustive search' procedure, such as tracking all branches, and how to handle double bonds. but these can be added in a straight forward manner. The 'set limit' feature is necessary to avoid putting your computer to millennia of number crunching.

This doesn't really answer the question, which is very difficult to answer anyway. Also, actually implementing this algorithm is a lot more difficult than you seem to imagine. For example Check add-on atom for compatibility: bond type, electrons avail, space fit is a highly non trivial task and there is also a question as to how compatible is 'compatible'.

Also, this algorithm can not create rings. As it happens, rings and fused rings make up the majority of carbon based natural compounds.

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Determining perovskite crystal structure

I've been reading that perovskite presents a sort of pseudocubic crystalline system, however it can be stretched and suffer modifications to become orthorhombic etc.
The thing is, if I consider it as cubic, it's sort of a combination between FCC and BCC, because it's an FCC with an atom in the center. I can't find a name for this sort of thing.
Can anyone help me out?

As to the FCC/BCC thing, I'd probably call it a pseudo-FCC lattice with an interstitial octahedral atom.

"Perovskite" IS what the structure is called.

At Geoff's request I'll make comments into an answer:
Since it is a multi-component system you need to think of a crystal structure with a basis. Not surprisingly, the basis here is $\ce{ABO3}$. In your picture from Wikipedia, start with just the blue atoms - they form the simple cubic. Then each blue atom shares 6 red atoms with other blues, so pick three for a given blue. Then place the green atom at the cube center to go with the blue atom at the lower right front blue atom. This is now your basis, and placing that basis unit at each corner of the cube (repeating to infinity) gives you the crystal structure
To explain further: each red has two blue neighbors, so each red is 'half-owned' by a blue. For a basis, you want only whole atoms to keep your mind a bit saner, so pick three to be 'wholly-owned' by a blue, and leave the others to be owned by another blue. Taking the left lower center blue atom, I'd pick the red atoms in the +x, +y, and +z directions, then the green atom in the (x,y,z) direction to make a nice basis.
I'll also add some on deviations from cubic symmetry.  If the green atom is at (1/2,1/2,1/2) than it all remains simple cubic.  If the greens are big fat atoms compared to the blues, they may want a little more room, which leads to distortions.  Simply stretching along one dimension is pretty easy, since it leaves 2/3's of the blue-red pairs at the same interatomic distance, stretching the other 1/3.  This creates an orthorhombic crystal, still with the same $\ce{ABO3}$ unit as a basis, but now slightly elongated along one direction (commonly taken as z).

To determine the crystal system we need to choose a type of atom. In your picture let's choose the blue atoms in the corners of the crystal cell. If those atoms repeat in the faces it is called FCC. If they repeat in the body the cell is called BCC.
For perowskites the atoms in the body or in the faces are different from the atoms in the corners. Thus, it is a primitive cubic. Perowskites are not BCC nor FCC. The space group of cubic perowskites is Pm3m, P indicates that is a primitive system.

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Spectroscopy and photometric error

What is meant by the lowest photometric error?
why it is important to work in the $0.2\ \mathrm{to}\ 0.7$ absorbance range for a single beam instrument?
Why $0.4\ \mathrm{to}\  1.4$ for double-beam instrument?

Start with a reading about the types of errors that arise in spectroscopy. Clearly photometric error $\delta I_{pe}$ is a device uncertainty. The suggestion is, its magnitude depends on the number of "beams". What changes as the number of beams increases? Path length? Number of gratings or optical components? What else? How do each of those factors contribute to an analytical expression for photometric uncertainty?
At the end, a reasonable result might suggest that, as you double the number of beams, you double the relative uncertainty in signal $\delta I_{pe} / I$, where $I$ is the signal intensity. To obtain the same absolute confidence, you therefore have to double the measurement intensity $I$.


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How to get back my beautiful crystals made from iron and sulphuric acid
I have put some old nails and screws in the 30% sulfuric acid to prepare an electrolyte for plating. After some bubbling and horribly irritating smell (probably $\ce{SO2}$), the reaction slowed down. Overnight, beautiful green/blue crystals formed:

I just love them. After another day, the reaction almost stopped. I suppose the acid was all bound to iron. To continue the process, I poured some concentrated sulfuric acid into the solution. That was a big mistake - the crystals reacted with concentrated acid and formed ugly white dust:

What did I create? What are the crystals and what's that dust? How to reverse the reaction? Does the dust dissolve to be electrolysed?

These are some lovely pictures.

I had to support my camera with phone flash-light to take them :)

how much time did it take after adding concentrated H2SO4 again to form that dust

Less than a minute.

I realize this question is over two years old, but I have to add the correct answer here: the iron (II) sulfate you had at the beginning was in the heptahydrate form.  SEVEN molecules of water for every iron (II) sulfate ion pair.  Now, what do you think would happen to that crystal structure when you add one of the most powerful dehydrating agents in existence?  That's right, it sucks all the water out of your crystals, and you're left with a powdery, noncrystalline form.  You could have just added less concentrated acid, that was diluted enough to not be such a strong dehydrating agent.
Incidentally, iron is a surprisingly strong reducer of sulfates.  I did this reaction recently and noticed gold flecks floating around in the solution.  I was shocked to see gold, since I obviously never added any, until I realized that the iron had reduced the sulfate all the way down to sulfide, and what I was seeing was iron pyrite -- fool's gold.  Adding more acid eliminated that almost immediately.

I seriously doubt, that concentrated sulfuric acid has any dehydrating effect after diluting it with an aqueous solution, as the OP stated.

OP claimed to have "poured" "concentrated sulfuric acid" into the solution.  Remember, crystals need a 7:1 ratio of H2O to Fe (II) SO4 to form.  There's certainly going to be some Fe (II) SO4 in solution still; possibly even an excess of what's necessary to form the heptahydrate salt.  Adding concentrated sulfuric acid would have pulled even more water out of the solution, and H2SO4 being heavier than water, would have sunk to the bottom and dehydrated the heptahydrate crystals preferentially.

A quick look at the Wiki suggests that you first formed iron(II) sulfate:
$\ce{Fe + H2SO4 → FeSO4 + H2}$
which is blue-green. It is then possible that you obtained iron(III) sulfate, which is pale. But you need to oxidize the iron(II) to iron(III), which usually requires oxygen, and sulfuric acid is not an oxidizing acid. Mere exposure to oxygen works:
$\ce{12 FeSO4 + 3 O2 → 4 Fe2(SO4)3 + 2 Fe2O3}$
but I would assume not that quickly. Still, this looks like the best explanation, and perhaps some impurities managed to produce the oxidation. Just a guess.
As to how to get back the iron(II), you would need to reduce the iron(III), and given that the oxidation direction is basically spontaneous in air, this might take some work. So why not just start again with more iron and dilute acid?

Yeah, I'll probably diluthe the solution, flush it and start over. But it took a day to form, so I'm a little sad.

Concentrated sulfuric acid is an oxidising acid but if he was just using 30% then that's not concentrated- 98%+ is oxidising.

I think you merely grew some Iron Sulfate crystals and then dissolved them by adding more solvent.
Depending on how much sulfuric acid you added the second time you could add a proportional amount of the same mixture of nails and screws to reform the crystals  except larger. Alternately you could evaporate the acid you added by putting it on a hot plate or something. or as trb said you could just start over.


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How does different diffraction grating effect resolution and sharpness of lines in a line spectrum?
In my lab notebook it asks us to think about the spectroscope further than the reading goes on.
It asks if we use a diffraction grating that contains 600 lines/mm in the spectroscope what do we expect would happen to the lines of the line spectrum in terms of sharpness and resolution.
Is the answer (assuming 600 is more than normal) that the lines would be both more sharper and have a higher resolution? My first thought was that the lines would be skinnier but more clear.

I think it would be useful if you familiarize yourself with the language of spectroscopy: "skinnier", "more clear","sharper" does not really have a distinct meaning in the field, and resolution may not always mean what you think it means.

Sorry i just don't understand this and there are not many resources. Does more lines mean the line on the spectrum is clearer?

As I said, "clear", "skinny", "sharp" is not spectroscopic terminology, though the last one is used informally.

Sharpness is colloquial for how narrow a peak is. Resolution is how well we can see two features as they move closer together. You might do well to re-read your book to learn these definitions better. As already noted, skinnier and more clear have no meaning in spectroscopy.
A search on terms such as diffraction grating equation will give you further reading to support or invalidate the guess (hypothesis) that you propose.


text:
Interpret Radiodecay Symbolism
$\binom{16}{8}$O (n,$\alpha$)$\binom{13}{6}$C  and $\binom{9}{4}$Be ($\alpha$,n)$\binom{12}{6}$C
I had initially thought the products & reactants may be O-16 $\to$ C-13 + He-2 + p$^+$ from alpha decay, but I'm leaning towards this form O-16 + C-13 $\to$ ... now. Any help clarifying & determining products would be much appreciated.

The given expression

$\ce{^{16}_{8}O}\,(\mathrm{n},\alpha)\,\ce{^{13}_{6}C}$

translates as
$\ce{^{16}_{8}O ->[+\ \mathrm{n}][-\ \alpha] ^{13}_{6}C}$
or
$\ce{^{16}O + n -> ^{13}C + ^4He}$
and the expression

$\ce{^{9}_{4}Be}\,(\alpha,\mathrm{n})\,\ce{^{12}_{6}C}$

translates as
$\ce{^{9}_{4}Be ->[+\ \alpha][-\ \mathrm{n}] ^{12}_{6}C}$
or
$\ce{^9Be + ^4He -> ^{12}C + n}$


text:
In this simple redox reaction, why does the Hydrogen oxidize the calcium metal?

I am understanding that the Hydrogen ions oxidize the calcium metal, neutralizing itself and creating a calcium cation. My question is why does the Hydrogen oxidize the calcium? Doesn't it prefer to have an empty outer shell? Does it want more to be neutral or to have an empty outer shell?

In one simple sentence, a more reactive metal is more likely to donate its electrons than a less reactive metal.
What makes it more reaction, for example francium, it has more electrons shells thus the nucleus Is less electronegative to the valence electrons.

I'm not concerned with why the calcium is oxidized, that makes sense to me. What I am wondering is why the Hydrogen takes the electron. Isn't the hydrogen more stable as an H+ ion?

No it has nothing to do with stability, it is just which is more willing to give up electron. And since calcium has a bigger atom. It tends to lose electrons due to weaker electronegativity. It gives the electrons to hydrogen.

This is a simple example of the reactivity series. More reactive metals will displace less reactive metal ions, and they switch places, with the more reactive metal becoming the ion and and the less reactive ion becoming a metal. This can be very cool in many situations: for example, copper metal can displace a silver salt to plate silver metal and produce a blue copper salt. See this video! In any case, hydrogen is consider a reference "metal" in the reactivity series, and it is less reactive than calcium, so the displacement reaction you cite occurs.


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What exactly is net ionic equation?
I once saw a problem, iron ions(II) titrated with potassium dichromate to produce Fe(II) and Cr(III). If I am to write the net ionic equation of this reaction, I got to make potassium dichromate on the right into dichromate ion since potassium is a spectator.
I got very confused over here. Why should you see potassium as a spectator if K(II) will be balanced as 2K on the right.

I'm not entirely sure what you're asking. But firstly, K(II) will not form, as potassium ionizes to a +1 charge. Potassium is generally deemed a spectator ion because it remains soluble in aqueous solution and will therefore not react to form a precipitate.

The molecular equation for this titration is $$\ce{K2Cr2O7 + 6Fe(NH4)2(SO4)2 + 7H2SO4 ->
3Fe2(SO4)3 + Cr2(SO4)3 + K2SO4 + 6(NH4)2SO4 + 7H2O}$$
The complete ionic equation is $$\ce{2K+ + Cr2O7^2- + 6Fe^2+ + 12NH4+ + 19SO4^2- + 14H+ ->
6Fe3+ + 14SO4^2- + 2Cr^3+  + 2K+  + 12NH4^+  + 7H2O}$$
When you cross out everybody who shows up on both sides in the same form, you get the net ionic equation for this titration: $$\ce{Cr2O7^2- + 6Fe^2+ + 14H+ -> 6Fe^3+ + 2Cr^3+ + 7H2O}$$
Just as the hydrogens in $\ce{H2SO4}$ become $\ce{2H+}$ so the potassiums in  $\ce{K2Cr2O7}$ become $\ce{2K+}$ and they don't change on the other side of the equation where they form $\ce{K2SO4}$.

Well first of all instead of  $\ce{K(II)}$ it is always $\ce{K(I)}$
now the use of net ionic equation are:
a) It saves your time in writing
b) It highlights the species that are undergoing change and bringing more clarity
c) spectator ions have little significance in reaction but are necessary for that ion to ion to be stable , also spectator ion are to show that to make that reaction happen you do not need only those spectator ions which are given in your reaction but any other ion of similar properties could do. So
d) you will see better pattern in remembering reaction.
I think instead of remembering these two reaction you would prefer to remember this one reaction only
$\ce{CuSO4 + Zn -> ZnSo4 + Cu}$
$\ce{Cu(NO3)2 + Zn -> Cu + Zn(NO3)2 }$
You would like to remember
$\ce{Cu^{2+} + Zn -> Zn^{2+} + Cu}$
also you could even see from here that the two reactions are clearly the same only spectator ion is changed


text:
What is the hybridization of the central carbon in buta‐1,2‐diene?

State the hybridisation of asterisked carbon in $\ce{CH3-CH=\overset{\ast}{C}=CH2}.$

I am not sure between $\mathrm{sp^2}$ and $\mathrm{sp}$ hybridization. Because of the presence of double bonds I think it should be $\mathrm{sp^2},$ but because of the two π-bonds I think $\mathrm{sp}$.
Which one is correct?

The molecule you've drawn belongs to a class of molecules called "allenes."  Here are some diagrams of the molecule's structure and bonding that may be helpful.

You must be clear that the hybridization is decided by the number of σ-electrons, so for $\ce{CH4}$ the number of σ-electrons is four and so its hybridization is $\mathrm{sp^3}.$
In case of $\ce{CH3-CH=\overset{\ast}{C}=CH2}$ the $\ce{\overset{\ast}{C}}$ is having two σ- and two π-bonds, so its hybridization is $\mathrm{sp}.$
While deciding hybridization we need to only care about bonds, and that too only the σ-bonds. Remember that possible hybridization for $\ce{C}$ are $\mathrm{sp^3},$ $\mathrm{sp^2}$ and $\mathrm{sp}.$

As the asterisked carbon has two neighbors, the hybridization of this asterisked carbon is $\mathrm{sp},$ i.e. the angle $\ce{−CH=\overset{\ast}{C}=CH2}$ is 180°.

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


text:
Finding the equilibrium temperature between two substances in a constant pressure calorimeter
I need to find the temperature two things equilibrate to in a constant pressure calorimeter given only their specific heats, masses, and starting temps.
The only thing I know for specific heat is that. $c = \frac{J}{G\times \Delta T}$, and how to work that around to find if one is missing, and well as $\frac{q}{\Delta T}$ for doing things with calorimeters, though a different kind. That's the only equation I know for this stuff. In this question I have neither the energy nor the change in temp. I know how to solver it if the two substances have equal specific heats, then you just find the mid point between temperatures. But two different substances, I have no idea. I tried to do it the hard way of just slowly taking of/adding one *C of each and matching up the energies required but that got me nowhere.

I  really wish people would comment on down voting and explain... If I knew what the problem was already I wouldn't have done it....

Welcome to Chemistry Stack Exchange! Please add what you have attempted towards solving the problem into the body of your question.  For more information, see the site's homework policy for how to ask homework questions. That's what someone was likely reacting to with their downvote.  I know you may be frustrated with the question, but if you put something forth (what equations do you know for specific heat, for example?) someone can steer you in the right direction.

Do you have the specific heat of the calorimeter given to you?  If not, is it made of something that doesn't absorb a lot of heat?

No. I says constant pressure calorimeter, and according to the book I don't have to worry about the amount of heat that type absorbs because it's really small.

Assuming the heat absorption of the calorimeter is negligible (so if it's made of an insulating material, the only changes in heat will be of the contents themselves).
Substance 1 (the one at the higher initial temperature) has a heat = $q_1$
Substance 2 has a heat of $q_2$
Since substance 1 will lose heat until it comes to equilibrium with substance 2, the heat lost by 1 will be absorbed by 2, so $q_1 = -q_2$
If $q_1 = m_1c_{p1}(T_{{1}_{initial}} - T_{{1}_{final}})$ and $q_2 = m_2c_{p2}(T_{{2}_{initial}} - T_{{2}_{final}})$ and $T_{1_{final}} = T_{2_{final}} = T_f$ you can solve for $T_f$.
If the heat absorption of the calorimeter was not negligible, then $q_1 + q_2 + q_{cal} = 0$

I get it if I had q1 or q2, but I don't. How in the word do you that with just m1xcp1x? m2xcp2x?

@Cr157 You have the masses, the specific heats at constant pressure ($c_p$), and the initial temperature of each, so you can fill in $m_1, m_2$ and the specific heats, make the two equal and solve for $T_f$.

@Cr157 See if that makes it a bit more clear.

Not there yet... but I'm trying to get it...

@Cr157 $q_1 = -q_2$, so it's just a matter of algebra at this point with $T_f$ as the unknown.

Yea, that's always my problem... Up until recently I had no algebra education, so I'm just starting and have a lot of holes to fill.


text:
Why doesn't calcium carbonate dissolve in water even though it is an ionic compound?
I studied the solubility of compounds in water. I have found that calcium carbonate doesn't dissolve in water. The teacher stated that the ionic compounds dissolve in water except some carbonates. What would be a clear elaboration of this phenomenon?

Is it possible that the solution the Calcium Carbonate is in is saturated? This would mean the maximum amount of solute has been dissolved and no more can dissolve.

Are you answering or asking?

I was asking. Is the solution saturated or unsaturated?

It depends if equilibrium is achieved. Btw it's a place for posting answers so this can be deleted.

Can you show me the sources behind this? I was taught that if the solution's already saturated, no more solute can dissolve. Furthermore, I was asking for clarification on your question because there can be numerous reasons why calcium carbonate wouldn't dissolve in water. Oh and btw, questions that have holes in them or aren't clear are even more frowned upon on this site than then asking a question for clarity.

Sure, the ionic compounds dissolve in water, except some oxides, some hydroxides, some fluorides, some chlorides, some bromides, some iodides, some sulfates, many carbonates, many sulfides, most silicates... did I forget anything?

As someone said here, this:

The teacher stated that the ionic compounds dissolve in water except
  some carbonates.

Is indeed an oversimplification. First of all, the distinction between an "ionic compound" to other compounds isn't too defined. What your teacher probably said, or didn't say but wanted to, is that some ionic compounds easily dissolve in water. Salt (halite - NaCl) is the best example.
Calcium carbonate, in nature, also commonly dissolves. It's just not as immediate as dissolution of the more soluble ionic compounds. You are probably familiar with this phenomenon:

This forms because calcium carbonate dissolves. The rock is limestone, which is usually composed of pure calcium carbonate. Acidic water greatly enhances the solubility of calcium carbonate, and it doesn't even need to be highly acidic. Rain or river water that come into contact with the atmosphere absorb the $\ce{CO2}$ as $$\ce{H2O + CO2 <=> H2CO3},$$ which then facilitates calcium carbonate dissolution with $$\ce{CaCO3 + H2CO3 <=> Ca^2+ + 2HCO3-}.$$

For an ionic substance to dissolve in water, there are two competing factors that determine the enthalpy of solution $\Delta{H}_\mathrm{sol}$ which is the enthalpy (energy) change  when a solute is dissolved in a solvent:

The lattice energy (LE), the energy of formation of the crystal between infinitely separated ions. As LE is proportional to the charges of its ions, for calcium carbonate, its LE would be roughly 4× the LE of sodium chloride. $$\ce{CaCO3 (s) -> Ca^2+ (g) + CO3^2- (g)}$$ This value is always positive, as energy is required to separate the ions.
The hydration energy of gaseous ions, which  is the enthalpy change when gaseous ions dissolve in sufficient water to give an infinitely dilute solution. $$\ce{Ca^2+ (g) + CO3^2- (g) -> Ca^2+ (aq) + CO3^2- (aq)}$$ These values are always negative, as energy is released upon hydration of ions.

As the LE of calcium carbonate is so large, a great amount of free energy would be required to break the strongly attracted ions apart, and this energy has to come from the enthalpy of hydration. In the case of calcium carbonate,  the enthalpy of hydration is not large enough to overcome the large lattice energy, hence it exists as a solid.
Do note that solubility of ionic compounds are not absolute. In fact, solubility in water is often given as a value in g/L of water.
However, generally, we define solubility as:

Soluble substances can form a 0.1 M solution at 25 °C.
Insoluble substances cannot form a 0.1 M solution at 25 °C.

You forgot entropy of solvation.

This answer goes into lots of detail, but it seems to focus on why calcium carbonate won't dissolve in the first place. (Which, to be fair, was the focus of the original question.) What about: Why does calcium carbonate precipitate out when you dissolve calcium chloride and sodium carbonate in water? Is it because all the ions are constantly combining and splitting, and this particular combination (due to the factors you list) is much less likely to split again once it has formed?

Teacher stated that the ionic compounds dissolve in water except some carbonates.

That is, at best, an oversimplification. Other ionic compounds such as silver sulfide are sparingly soluble in water. Note that this isn't a carbonate.
And sodium hydrogen carbonate - $\ce{NaHCO3}$ - is soluble in water.
Sodium carbonate ($\ce{Na2CO3}$) is also soluble in water.

I want a clear elaboration of this phenomenon can somebody help me??

Solubility depends on:
1) The strength of the intermolecular attractions within the solute and the strength of the solvent-solvent interactions. How strongly is the calcium ion and the carbonate ion bound together? You might think that the force of attraction between a +2 and a -2 ion is strong, but remember, covalent bonds are generally stronger than ionic interactions. How strong are the solvent-solvent molecule interactions - which must be broken before something can be solvated?
2) Solvation. How well can the solvent lower the energy of the dissociated solute ions? How well can calcium ion and carbonate ion be solvated (by water in this case?) How strong are the solvent-solute ion interactions? What about the entropy of solvation? (Solvation shells are highly ordered).
If you wish to predict the solubility of some compound, try using hard soft acid/base theory.

Solubility of salts is not a prediction that I like to make.  It all comes down to $\Delta{G}_{sol} < 0$.  That condition is fulfilled when $\Delta{H}_{sol} < 0$ and $T \Delta{S}_{sol} > 0$ or one factor outweighs the other.
There's two problems.  Lattice enthalpies are large, as are solvation enthalpies, and the common observation is that $\Delta{H}_{sol}$ in most cases is small.  We are calculating differences of large numbers, and the one fact we can confidently assert is that the difference is impossible to predict and generally meaningless.
Same thing for $\Delta{S}_{sol}$.  Lattice entropies are small, but entropies of solution are lower than one would think thanks to the short-range order the solvated cation imposes on the  solvent.  You are gaining translational entropy from dissolving the lattice and losing translational entropy from the solvent through the inner coordination shell.  You are calculating differences of numbers of equal magnitude again, and just by staring at reagents it's impossible to tell which way the pendulum is going to swing and if the result is positive or negative.

To be very honest with you, it's almost impossible to predict accurately, the solubility of compounds. At the end of the day, it's best to memorize them. Yes, there are some cases wherein a certain logic or reasoning can be used, but mostly its better to just memorize the thing. As Abel said, it comes down to a difficult interplay of lattice enthalpy, and hydration energy.


text:
Fick's Second Law of Diffusion
In the book "Foundations of Materials Science and Engineering" (Smith and Hashemi) the following is written:
For cases of non-steady state diffusion in which diffusivity is independent of time, Fick's second law of diffusion applies, which is
$$\frac{\partial c_{x}}{\partial t}=\frac\partial{\partial x}\left(D\frac{\partial c_{x}}{\partial x}\right)$$
In what cases are materials' diffusivity dependent on time, and in what cases are they not?

Diffusivity is not dependent on time. It is dependent on temperature (and material). Fick's second law holds because the diffusivity, $D$, is constant with respect to depth, $x$, and time, $t$, so it is treated as a constant in the equation.
$$\frac{\partial c_{x}}{\partial t}=\frac\partial{\partial x}\left(D\frac{\partial c_{x}}{\partial x}\right) = D\left(\frac{\partial^2c_x}{\partial x^2}\right)$$
Fick's second law of diffusion applies to non steady state diffusion, which concentration at a depth $x$, $c_x$ is a function of time.

In general D depends on both position x and time t, since it usually depends of the concentration c(x,t) as well as mobility factors.
This is why treating D as a constant (and thus putting it outside the last partial with respect to x) is a simplification, that is mostly used for getting an approximate solution for c(x,t). In many cases, however, this will be sufficiently close to the "correct" solution.


text:
Carburizing Calculation
The problem is stated as follows:
The surface of a steel gear made of 1022 steel (0.22 wt% C) is to be gas-carburized at $927$ °C. Calculate the time necessary to increase the carbon content to 0.30 wt% at 0.030 in below the surface of the gear. Assume the carbon content of the surface to be 1.20 wt% and $D_{927^{\circ}C}=1.28 \times 10^{-11}\frac{m^{2}}{s}$
The formula is $\frac{C_{s}-C_{x}}{C_{s}-C_{0}}=ERF(\frac{x}{2\sqrt{Dt}})$
The solution manual is setting $C_{x}=0.30$, but should it not be $C_{x}=0.30+0.22=0.52$, since the initial carbon content in the steel is $C_{0}=0.22$?

No.
In the equation $C_x$ simply means the concentration at depth x. It does not mean the change of concentration. There is no need to add its initial concentration.
Besides, concentration cannot be linearly added. (Does adding a 1M solution to a 2M solution give 3M?)

Thank you for your answer and for the analogy, but does this not mean that we are assuming that the initial carbon content is 0, and not 0.22? I am confused because the problem use the word "increase".

Nope. Its initial carbon content throughout the entire steel is 0.22wt%. You are not trying to increase it by 0.30wt% but increase it to 0.30wt%.


text:
Regioselective enolate formation
Our professor only told us that A is favored, and B is not getting formed. I thought about the resonance structures and saw a contradiction to that what the prof told us. To be honest I think my argument with the resonance structure is not strong enough, because it does not explain the prefered product. Why is structure B disfavored?

@jerepierre: I checked again. You're right! I'm sorry for the wrong re-drawing. Although the above discussed reaction with deprotonation was also dictactical important.
The correct starting material is in fact a alpha,beta-unsaturated ketone which is reduced by Li/NH3!
This is the correct reaction:

Given that the reaction is starting from an α,β-unsaturated ketone, the dissolving metal reduction is a very useful way of generating a particular enolate regioselectively. The mechanism dictates that only one enolate can be produced. Dissolving lithium metal in ammonia releases an electron that can add to the unsaturated system giving a radical anion. The radical anion is protonated by ammonia at the beta-position, leaving a resonance stabilized radical (resonance structure not shown). Filling in another electron gives the enolate. The enolate is typically trapped by alkylation, such as by methyl iodide.

This method avoids the issues discussed by Ron when there is a real selectivity issue when generating an enolate by deprotonation of an asymmetrical ketone. Of course, it means that the starting material must be an enone, which may or may not be practical for a given synthesis.

I do not understand, the question does not use an alpha beta unsaturated ketone.

@Martin He realized that the starting material isn't what was shown. The conditions didn't make sense for the reaction as drawn, which led me to speculate in the comments on some of the answers. Then laminin put an answer with the correct starting material. Any suggestions for how to clean this up?

I noticed that now. Sorry for the inconvenience. I flagged the answer and left a comment on the original post. I assume, that one of the moderators (or the OP) will take care of it within the day.

Unsymmetrical ketones can produce two different enolates, as shown in your reaction scheme.  The enolate pictured in the top row of your drawing is referred to as the thermodynamic enolate, while the enolate shown in the bottom row of your drawing is the kinetic enolate.
More highly substituted double bonds are thermodynamically more stable than less substituted double bonds. Therefore, the thermodynamically favored enolate is the one that gives rise to the most highly substituted double bond in the enolate intermediate. The thermodynamic enolate will be formed when reaction conditions are chosen (see below) to allow equilibration of all enolate intermediates; product formation is then thermodynamically controlled.
The kinetically favored enolate is the fastest formed enolate, the one formed by abstracting the most acidic proton (proton acidity follows primary > secondary > tertiary).  In your reaction, removal of the proton on the secondary carbon would be kinetically favored (more acidic) over removal of the proton on the tertiary carbon atom (less acidic).
Reaction conditions can be selected to favor kinetic (aprotic solvents, strong bases, low temperature) or thermodynamic (protic solvents, weaker bases, higher temperature) enolate formation, giving the chemist control over selecting which product will be formed.
If your reaction were run under conditions favoring kinetic enolate formation, then we would expect to observe product formed from following the bottom reaction line.  If the reaction were run under conditions favoring thermodynamic enolate formation, then we would expect to obtain products from following the top line of your reaction scheme.
So if your professor said that pathway "A", the top line in your drawing, is favored, then the reaction is being run under conditions favoring formation of the thermodynamic enolate.

An excellent description by @ron. Kinetic enolate formation is also favored by bulky bases (e.g., LDA) with at least one full equivalent of base. Small bases and substoichiometric amounts of base favor the thermodynamic enolate. I suspect that the conditions might have been LiNH2 (which would be a small base) instead of Li in NH3. Either that, or the starting material is an alpha,beta-unsaturated ketone that is reduced by Li/NH3.


text:
Is graphene chemically reactive?
Different internet sources say different things.
If possible, can someone explain why it is (or isn't) chemically reactive?

Graphene in its pristine form, a single sheet of pure sp2 hybridized benzenes, is (relatively speaking) one of the least chemically reactive materials known because of several reasons: one, its enormous size (we keep making larger and larger single "molecules", now thanks to IBM up to 4 inches!). Secondly, its physical hindrance: you can only do bottom/top down chemistry, granted this isn't that big of an issue but it does greatly reduce using large compounds. Last is it's stableness: this thing is crazy stable because of its gigantic p-butts and other effects.
The long story short is that you don't have any functional groups or easy points of entry. The way to think about it is "what does graphene look like and what makes those other molecules reactive?". The answer is...benzene and really any other hydrocarbon with no functional groups. The difference there is their size. But the same rules apply! You can still oxidize graphene, at the edges for example, much like you would benzene. And in the past years, we've seen even more radical methods, the most common being Hummer's Oxidizing, and some other clever ones like a Claisen Rearrangement directly onto the surface of graphene.
Hope that helps!
EDIT: after Greg's comment
Don't apologize! This'll get the discussion going, I was hoping to see more answers other than mine. I think you're right in terms of absolute stability but my statement is a phenomenon harder to describe than mere size. From experience, we noted that on very large graphene flakes, our functionalization of them would distort the sheets, but not break them, creating lots of hindrances and thereby decreasing its reactivity; it basically went from sheet shape to a jumbled up ball. However, on smaller (much smaller) sized flakes, the individual molecules (though hindered and flexed in a similar way) would, at the same concentrations of reactants, get functionalized more, ratio of functional groups to carbons on the sheet.
This is an indirect effect of the massive size of the graphene sheets. So I guess you're right in that size isn't exactly a stability factor but rather, again, it's all about hinderance. However, I wanted to point it out because the massive size of such a flat molecule is quite spectacular and does yield some interesting behavior we don't normally think of.

Sorry, but a molecule doesn't become more stable just because it is big. It is especially true for aromatic systems: in most cases the bigger the aromatic system, the LESSstable it is!


text:
Understanding Millimeter of Mercury
A Millimeter of Mercury(mmHg) is by definition the pressure exerted on a column of mercury 1 mm  in height.
I found this definition confusing because it doesn't say how the diameter of the container will affect the pressure.

Pressure is defined as force exerted per unit area. Therefore, the diameter of the mercury column is taken into account when we calculate the area of the column.  A mercury column with a large diameter will exert a greater force than a column of the same height, but with a smaller diameter - but the force divided by the area will be the same.

The different in height(u shape manometer) doesn't depend on the diameter of the tube. But with fix amounts of force, increasing diameter will reduce the difference in high

So you mean the pressure is the constant?

My textbook just told me diamt has nothing to do with height difference, but  how can the height not decrease with diameter decreasing if the volume of mercury stays the same?

Well, diameter does have something to do with it, it is part of the equation for the area.  But because diameter is included in the equation for pressure, we get the same pressure no matter what the diameter is.  A larger diameter will have more force on it (bigger area), but the force per unit area will be the same even if we use a smaller diameter.


text:
How to dissolve candle wax (paraffin)?
Wikipedia states that waxes can be dissolved by non-polar organic solvents.
I failed to dissolve candle wax in both acetone and ethanol, which are the only organic solvents I have at home.
I have some clothes with wax stuck on them. How can I dissolve the wax and drive it away without damaging my clothing?

Ethanol and acetone are not non-polar organic solvents. Each one has  a slight dipole moment; due to the difference of electronegativity between $\ce{H}$ and  $\ce{O}$ in ethanol and between $\ce{C}$ and  $\ce{O}$ in acetone.  Wax is composed of heavy, long-chain alkanes. And as "Like dissolves like" try to dissolve your wax in toluene or in xylene.

A proven way to get rid of candle wax in clothes is putting several sheets of newspaper on top and then apply a warm iron.

You can also put the clothes in the freezer and a substantial portion of he wax will crack off.

Being a chemist, I know that paraffin would require a lipophilic solvent as pointed out by others. It occurred to me that ordinary vegetable oil is a cheap and of course innocuous lipophilic substance. I spilled about 2 oz molten candle wax on a marble surface and after it solidified removed a good portion by scraping with a plastic spoon. This was followed by adding about a tablespoon of vegetable oil and mixing with the spoon. The wax dissolved in the oil with no problem and could be mopped up with a paper towel. Treatment of the oily residue with sink soap removed what was left. Hope this helps.

Gonna try that right now!

Just wanted to update - candle wax did not dissolve in sunflower vegetable oil.

Wow, this works great! This needs to be common household knowledge...

It’s a great tip! Now it doesn’t dissolve readily and requires some elbow grease but a health dose of canola oil, a cloth and some rubbing and it came off in a few minutes. Perfect for hardwood floors as I don’t have to worry about harsh chemicals on the finish.

Great answer!! Canola oil and rubbing with my thumbnail took the candle wax out of my tablecloth. Then I hit it with spray’n wash and put it in the washing machine to take the oil out.

A readily available mixture of various non-polar alkanes  would be petrol (gasoline).

http://en.wikipedia.org/wiki/White_spirit

However, I don't think this will result in clean clothes. You will clean the wax, but it will then stink of petrol.

Ordinary candle wax is readily removed by Methylated Spirits, especially if is on a candle stick. Scrape the bulk off carefully and then use a soft cloth with meths.

Dry clean your clothes. The solvents used in dry cleaning work well.
I had the airline industry pump my suitcase and suits full of jet fuel. Even with professional cleaning they still stunk. All had to be replaced.


text:
A thermos bottle containing milk is shaken vigorously
I have a conceptual physical chemistry question here.

A thermos bottle containing milk is shaken vigorously. Consider the milk as the system. Will the temperature rise as a result of the shaking?

My answer to this is yes, but I'm having trouble finding an equation that explains this result.
The best i have is the molecular kinetic energy formula that $$\mathrm{ke_{avg}= \frac{1}{2}mv^2=\frac{3}{2}kT}$$
So a change in kinetic energy is $$ \mathrm{\Delta ke = \frac{3}{2} k \Delta T}$$
Is this the right approach?

A conceptual approach might work better, since the shaking system is not going to convert mechanical energy into thermal energy as neatly as something like Joule's Paddle Wheel.  As the bottle is shaken, the liquid in it is swirled around into more or less randomly oriented vortices.  These cancel one another out and slow to rest via friction with their container, and the energy they contained has to go somewhere.  As you wrote, it winds up increasing the average velocity, and thereby temperature, of the particles in the milk.
This basic mechanical to thermal energy conversion is the source of the conversion factor from joules to calories.  Joules are used to measure mechanical energy and calories were used to measure thermal energy.  It was the paddle wheel experiment I linked above that first related the two to one another.
The challenge presented by trying to relate the work done on the bottle of milk to the thermal energy that winds up going into the milk is less mathematical than practical.  On any given shake we would first put energy into the bottle (move it upward, let's say) then take energy back out (slow it to a stop) before putting energy back into the bottle (moving back downward) and taking energy out again (slowing to a stop a second time.)  This repeats until we get tired of shaking, and if we measured the difference in all of the energy inputs and outputs that ought to equal the thermal energy added to the milk, but actually doing that for real would be a pain.

I don't think this picture is quite right--I think you might be conflating the kinetic energy of the bottle as a whole with that of its contents. Yes, when you accelerate it upward you are adding KE to the bottle as a whole, and when you decelerate it you are taking KE away from the bottle as a whole. But when it comes to warming, what matters is the effect of shaking on the KE of the contents. And for this, when you suddenly acclerate the bottle, you add KE to its contents (by swirling, etc., as you describe). But when you suddenly reverse the direction of shaking by applying acceleration...

...in the opposite direction, you are further adding KE to the contents.  I.e., while acceleration and deceleration increase and decrease the KE of the bottle as a whole (by which I mean 1/2 mv^2, where m is the mass of the bottle, and v is the velocity relative to your reference frame), respectively, acceleration and deceleration both increase the KE of the contents.   The analogy would be gas contained in a cylinder that has pistons at both ends.  Compressing either piston in the same way equally adds to the KE of the gas,  irrespective of the direction in which the work is being done.

The temperature of the milk increases because of work done on the milk molecules which cause , increase the velocities of the moecules and since velocity is directly proportional to the temperature.
Velocity    α  Temperature
So the temperature of milk increases due to the shaking process. Although this effect is very small but heat up the molecules.


text:
How do I calculate the molarity of the Na2CO3 solution?
My teacher isn't very good at helping me solve this problem. I keep getting the answer, but I don't know why I am getting it. Could you please help me understand this problem and why you should do this specific step?

A $\pu{1.00 g}$ sample of $\ce{Na2CO3.10H2O}$ was dissolved in $\pu{20.0 mL}$ of distilled water. Additional water was added so as to give $\pu{250 mL}$ of solution. What is the molar concentration of $\ce{Na2CO3}$?

Here is what I did and got the answer and don't understand why I need to do it:
\begin{align}
  \frac{\pu{1 g}~\ce{Na2CO3.10H2O}}{\pu{286 g}~\ce{Na2CO3.10H2O}} &= \pu{0.003447 mol}\\
\frac{\pu{0.003447 mol}}{\pu{0.020 L}} &= \pu{0.174825 M}\\
(\pu{0.174825 M})(\pu{0.020 L}) &= (\pu{0.250 L})(x)\\
x &= \pu{0.013986 M}
\end{align}
$\pu{0.013986 M}$ is the correct answer, but I don't know why and don't understand the question.
If you were to do the problem, could you explain why you did what you did to get to the answer.

@AwesomeFlame123 You clearly figured out the math aspect of the problem.  Where is your issue, specifically?  Which step is it that you are doing that you don't understand?  You also don't need the dilution step at all.  Once you've got moles, that can go directly into the molarity equation with 0.250L as your volume.  None of the solute ever leaves the solution, after all.  As written, you're dividing by 0.020 then immediately multiplying by 0.020.

@JasonPatterson I was trying to just use the formula M1V1 = M2V2

Technically you don't know the volume of the solution of 20ml of water and sodium carbonate. You don't know a "true" volume until you dilute to 250 ml.

What the question is asking is the molar concentration of the given substance after dilution with the help of values given. The first step of the answer is converting the given weight of $\ce{Na2CO3.10H2O}$ into amount of substance. For that we have a formula as
\begin{align}
\text{amount of substance} &= \frac{\text{mass of substance}}{\text{molecular mass of the substance}}\\
\text{amount of substance} &= \frac{
\pu{1 g}~\ce{Na2CO3.10H2O}}{\pu{286 g}~\ce{Na2CO3.10H2O}}
= \pu{0.003447 mol}.\\
\end{align}
We used this first step to determine the amount of substance of $\ce{Na2CO3.10H2O}$ present in the solution. It is necessary because even after dilution of the solution the amount of substance will remain constant.
The second step is converting the amount of substance into molarity. As,
\begin{align}
  \text{molarity} &= \frac{\text{amount of substance}}{\text{volume in L}}\\
  \text{molarity} &= \frac{\pu{0.003447 mol}}{\pu{0.020 L}} = \pu{0.174825 M}\\
\end{align}
The second step is done to find out the concentration of the substance in terms of molarity since the question has asked us to give the answer in molarity after dilution.
After that the given solution is diluted up to $\pu{250 mL}$, so we have to use dilution formula
$$\begin{multline}
  \text{initial concentration}\times \text{initial volume of solution} =\\
  \text{final concentration}\times \text{final volume of solution}.
\end{multline}$$
So we suppose that the final concentration of the solution is $x$ and we have values for others entities.
\begin{align}
(\pu{0.174825 M})(\pu{0.020 L}) &= (\pu{0.250 L})(x)\\
x &= \pu{0.013986 M}\\
\end{align}
This last step is carried out because after dilution the molarity of the solution changes but not the amount of substance as I stated earlier. So we equate the equation as $$\text{initial amount of substance} = \text{final amount of substance}.$$
And the final answer is $\pu{0.013986 M}$.

Thank You! I didn't understand that the Molarity of the final Na2CO3 is the same as the molarity of Na2CO3⋅10H2O.

Welcome. Well molarity is always calculated for the solution .So when crystals of $\ce{Na2CO3 * 10H2O}$ is dissolved in water then water of crystallization plays no role in determination of molarity since only no of moles of$\ce{Na2CO3}$ is taken in account.

This is needlessly complicated.

Please note that descriptive terms or names of quantities shall not be arranged in the form of an equation and that the quantity “amount of substance” shall not be called “number of moles”, just as the quantity “mass” shall not be called “number of kilograms”.

The first part makes more sense writing moles explicitly. The first step is determining how many moles of sodium carbonate are in the original $1.00\ \mathrm  g$ sample:
$$\frac{1.00\ \mathrm{g}}{286\ \mathrm{g/mol}} = 0.003447\ \mathrm{mol}$$
The second step is determining the concentration of the first $20.0\ \mathrm{mL}$ solution, i.e. what concentration results from dissolving the $0.00345\ \mathrm{mol}$ of sodium carbonate calculated above.
$$\frac{0.003447\ \mathrm{mol}}{0.0200\ \mathrm{L}} = 0.174825\ \mathrm{M}$$
The third part is simply a dilution – what concentration results when we dilute the first $20.0\ \mathrm{mL}$ solution to $250\ \mathrm{mL}$:
$$c_2 = \frac{c_1V_1}{V_2}$$
$$c_2 = \frac{(0.174825\ \mathrm{M})(0.020\ \mathrm{L})}{(0.250\ \mathrm{L})} = 0.0140\ \mathrm{M}$$

No arguments at all about what you've written here, but to be clear finding the 20mL concentration and using the dilution equation is just extra work.  Once we know the number of moles of sodium carbonate decahydrate, we can go straight to its molarity as the solute in 250mL of solution.

You're quite right, but he asked for an explanation of the steps he took.

Comments are not for extended discussion; this conversation has been moved to chat.


text:
How do I determine the mass percent of Barium nitrate and Sodium acetate in an original unknown solution
I am given $20 \mathrm {mL}$ of the unknown mixture of Barium Nitrate and Sodium Acetate dissolved in water. I am given the total mass of the two solids before they were put into the mixture.
I am given $0.200 \mathrm{M}$ of $\ce{NaOH}$, $0.200 \mathrm{M}$ of $\ce{Na2SO4}$, $0.200 \mathrm{M}$ of $\ce{NaCl}$ and ethanol to separate a precipitate.
I am learning about selective precipitation.

Please include what you have tried.

I plan to add Sodium Sulfate to the unknown mixture to obtain the precipitate of Barium Sulfate. But I am unsure of what to do after.

I have not yet conducted the experiment but I must come up with a laboratory procedure to follow.

This is not electrochemistry either but I could not come up with a more suitable tag.

Any help would be greatly appreciated

It sounds like you have a good idea of what to do already.  Once you've separated out the barium sulfate, dried it and know its mass, you've got the mass of the sodium acetate as well, right?  (Difference between total and original barium salt should be sodium salt, no?)  There aren't many options for precipitating sodium salts.

It occurred to me that you might have been asking how to complete this separation experimentally rather than conceptually, which is what the comments and previous answer all were pointing toward.  If that is the case, then once you've precipitated out the barium sulfate, you have to isolate it from all other substances in order to find its true mass.
The precipitate is a solid, so you could separate it from most of its contaminants by pouring the solution through a pre-weighed filter paper, making sure to wash the walls and bottom of the reaction vessel with distilled water and pouring this wash through the filter as well in order to collect all of the precipitate.
The filtrate would contain the bulk of the sodium acetate and sodium sulfate in the reaction mixture.  What remains in the filter paper would be barium sulfate and a small amount of reaction mixture, including the sodium salts.  To eliminate the sodium salts, add additional distilled water to the filter to wash the precipitate (and filter paper) clean.
Once you have a filter paper with clean barium sulfate, dry it thoroughly and weigh.  I don't believe that barium sulfate forms a hydrate, so the mass of the precipitate should all be barium sulfate.
Note: Barium salts are quite toxic, particularly highly soluble ones like barium nitrate.  Be careful using them and dispose of the product appropriately.

Start with your chemical equations: What's going to react and how; what will precipitate out of what sort of solution.
When get one component fully precipitated (as Jason Patterson points out) you can use the stoichiometry of the exchange reaction to determine the initial mass of one of the components.


text:
Central Atom in EXAFS
I read somewhere that EXAFS (Extended X-ray absorption fine structure) gives us the distances between central atom and its neighboring atoms. Is that true? Why can't we get the distance between two atoms, which none of them are central? How is EXAFS method distinguishing between central atom and other ones? So for example for CF4, I think we get one EXAFS peak for C...F distance and not F...F distance, why?

(Copied and edited from comments)
The central atom here refers to the center of attention (i.e. where the given peak assigned), the center of scattering which is often a center of an active site/coordination complex. I.e anything can be a central atom, if you tune your X-ray frequency to their peaks - you choose the central atom. In the given example theoretically you can choose to study C or F, and their closes neighbors. Theoretically you can see their neighbors; C-F distance is much shorter than F..F distance, though.
You should understand the use/limitations of the method: if you have chemically nonequivalent atoms of the same type in your sample, you will have their EXAFS peaks overlap. Both will give you a signal at the same place, which are hard to separate. In practice it means that you can use this analysis for e.g. central metal atom of complexes, but not for carbon atoms of a ligand as you generally have a lot of different carbon atoms in the sample.


text:
What is the strongest acid?
According to Wikipedia, $\ce{HeH+}$ and fluoroantimonic acid are the strongest.
According to a News article in Nature, Carborane acid is the strongest, but Wikipedia says fluoroantimonic acid is stronger.
Links:

$\ce{HeH+}$ / Wikipedia - stating HeH+ the second most acidic

$\ce{HeH+}$ / Wikipedia 2nd source - stating $\ce{HeH+}$ the most acidic (see in 'Notes' section)

Fluoroantimonic Acid / Wikipedia - stating it is the strongest acid (super_acid)

Carborane Acid / nature.com - According to Nature and even Google itself, it is the strongest acid created but according to wikipedia Flouroantimonic acid is stronger.

So, which is the strongest ?

HeH+ is a very acidic ion, but in order to exist in a condensed form there would have to be a counterion X−. If the counterion gets protonated by the HeH+, we end up with a molecular acid HX. This then becomes only as strong as the pKa of that acid. You cannot have a bottle of HeH+ so calling it an acid is in a way kind of nonsense. I might as well talk about having a bottle of protons, which would be at least as acidic as HeH+. Note that the free proton does not exist in aqueous solution, what we call H+(aq) is really H3O+ .

What do you mean by "According to...even Google itself"? Remember that Google - a search engine - cannot itself be the source for any information.

The problem with this question is that the exact answer is "it depends..."
First off, it depends on your definition of acidity and how you measure it. Everyone seems to be using Brønstead acids (i.e. $\ce{H+}$ donors). I see two different measures in other answers:

Proton affinity: This is a gas-phase measurement of $\ce{A^{−} + H^{+} -> HA}$
Hammett acidity ($H_0$): This is a solution measurement, given by $\mathrm pK_{\ce{BH^+}} - \log\left(\frac{[\ce{BH^+}]}{[\ce{B}]}\right)$

Secondly, it depends on medium as mentioned by LDC3.

Leveling effect: The solvent leveling effect reflects the lowest possible $\mathrm pK_\mathrm a$ in a particular solvent, based on the basicity of the conjugate base. So you need $\ce{HF}$ or fluorosulfuric acid to reach low $\mathrm pK_\mathrm a$.

So the problem in my mind is that carboranes and fluoroantimonic acid are solution measurements, but $\ce{HeH+}$ is a gas-phase measurement. It does have the highest gas-phase proton affinity. But I'd put my money on things I can use in lab.
Incidentally, the Reed group prepared the fluorinated carborane acid this year: Angew. Chem. Int. Ed. 2014, 53 (4), 1131–1134. So that compound, $\ce{H(CHB11F11)}$ wins the crown for strongest solution-phase Brønstead acid (for the moment, at least).
There's also a nice review article "Myths about the Proton. The Nature of H+ in Condensed Media" by the same group in Acc. Chem. Res. 2013, 46 (11), 2567–2575.

Well i reckon and also according to Wikipedia it has been discovered.  See helium compounds in Wikipedia.

So, is it more acidic than carborane. Also Wikipedia suggested that NeH+ has also been seen. So since Neon is even more inert so will it be the strongest acid ever known? If you read the HeH+ ion article in wikipedia, it is written it is the 2nd strongest acid known. So i was wondering!

I think it depends on your definition of "acidity" as I mention above. Are you asking about proton affinity? The linked Wikipedia article gives the "strongest known acid" comment. The list of proton affinities gives He above Ne.

So lets put it all down...which is the strongest acid regardless of medium and JUST for curiosity...?

Then I guess you'd have to go to the proton affinity definition, which points to $\ce{HeH+}$. But as I indicate in my answer, it's not like you can use it for actual chemistry.

@NeilRoy: Arguably, the strongest possible Brønsted acid is simply $\ce{H+(g)}$, i.e. a lone proton in gas phase, since anything stronger would be unstable and simply eject the proton (or break up in some other way). If this seems like a silly or trivial claim, note that species like $\ce{HeH+}$ are really just approximations of $\ce{H+}$; the helium and hydrogen in $\ce{HeH+}$ are barely bound together, and the helium, being almost completely inert, is little more than a spectator in any reaction between the proton and any Brønsted base it may react with.

@IlmariKaronen While you can argue that $\ce{H+}$ is the strongest possible Brønstead acid (its $H_0$ would be estimated at -68) it points to the "how do you define an acid" question, since you can't define a proton affinity for a proton.

You can, however, (sort of) define a proton affinity for the null molecule $\varnothing$ consisting of no nuclei or electrons; its proton affinity is trivially just 0 kJ/mol, since the "reaction" $\ce{\varnothing+H+->H+}$ does not involve any actual reaction at all, and thus no change in energy.

From Wikipedia
It depends on the medium.
For plasma: helium hydride ion
For liquids: fluoroantimonic acid
From Wikipedia
Fluoroantimonic acid - $H_0$ Value = −31.3
Carborane superacid - $H_0$ Value = −18.0
100% Sulfuric Acid - $H_0$ Value = -12.0
I didn't find any information about $\ce{HeH^+}$, I guess it's because it only exist in a plasma and difficult to study.

Carborane Acid  is the strongest acid.
The reason this is technically the world’s strongest acid is because it’s the strongest acid as a single molecule. Much stronger acids- the acids that melt through tables- can be created by mixing certain antimony compounds with hydrofluoric acid ($\ce{HF}$). For example when you mix $\ce{SbF5}$ with $\ce{HF}$, you get an acid with a pH of -31.

Reference: Chemistrytwig

What about that HeH+ ? I think it is stronger...

@neil_roy So we can say that HeH+ is strongest acid in ion form and Carborane Acid in molecule form. But it I can't figure which of them is the strongest. It seems HeH+ because it is in ion form!

yeah i agree that HeH+ is stronger also because it involves a noble gas... I was researching a li'l bit and found that NeH+ has also been found (Wikipedia : Neon Compounds) which makes it the strongest but it is so rare that...!

@NeilRoy The Ne proton affinity is lower than He. So no, NeH+ is not stronger than HeH+.

I think single molecule is an important distinction since the strongest Lewis acid is probably a very positively charged cathode ;)

@abinitio an anode maybe?


text:
Electrophilic addition of HX to 2,3-dimethylbut-2-ene
Generally, for the addition of dry HX to an alkene, it follows Markovnikov's rule. The hydrogen atom adds to the carbon with the most hydrogen atoms.
However, in the case of an alkene which does not have any hydrogen atom attached to its carbon, such as 2,3-dimethyl2-butene, would the electrophilic addition still take place?

Electrophilic addition would likely still occur. Remember that Markovnikov's rule is an observational heuristic for "in electrophilic addition reactions of alkenes in which carbocations form, the more/most stable carbocation forms preferentially, and thus the product originating from that cation is the major product".
In this case, a carbocation can form. Nothing about the tetrasubstituted pattern prevents addition. The molecule is still planar and can be aproached by the electrophile. The methyl groups are electron donating by induction. The methyl groups 1) increase the electron density of the alkene and make it more nucleophilic, and 2) stabilize the carbocation intermediate.
Thus, 2,3-dimethyl-2-butene is likely more reactive than other alkenes toward electrophiles. For example:
$$\ce{(CH3)2C=C(CH3)2 + HBr -> (CH3)2BrC-CH(CH3)2}$$


text:
Why did roasting of copper(II) sulfate pentahydrate hurt my nose?
I want to use copper(II) sulfate as rocket propellant additive in hope that it will alter the flame colour. Because water is not welcome in rocket fuel, I decided to dehydrate the crystals.
First I tried stainless steel mess tin, but that produced some smoke warning me that it's reacting. So I switched to aluminium mess tin, that one didn't smoke.
But still some fumes, presumably steam was produced. I continued roasting. Almost at the end, I noticed that my nose hurts in the inside a little.
I suppose copper sulfate has got in there. Do you think I risked badly, or is it just weak irritation? How can I avoid it next time, apart of using fumehood?

You probably just want to do it a lot slower, like in an oven at about 120 °C overnight or something like it.

I'm guessing that by roasting it, some of the sulfate converted to some sulfur gases?

I find this question somewhat disturbing with regards to the complete lack of safety awareness during the planning stage of the experiment.

@tschoppi You wouldn't believe how right you are. I had nasty accident with the same compound later. However now I'm afraid to ask why it happened, because I think I'll just get more comments suggesting how stupid am I.

@TomášZato Well, ask your questions anyway. You will surely get a snarky comment or two, but just to be clear: Behind that unpleasant-looking comment is a real human being that is genuinely concerned for another one. Please just learn from the mistakes you make, so you don't do the same error twice :)

Tomas, you can check for the presence of copper oxide by looking for a brown or black solid.  If present you probably formed sulfur trioxide as stated below.

Roasting of copper(II) sulfate at elevated temperatures generates sulfur trioxide, which would hurt your nose.
$$\begin{align}
\ce{$\underset{\text{(blue)}}{\ce{CuSO4.5H2O}}$ &->[\hphantom{\gt\ 200\ \mathrm{^\circ C}}][100\ \mathrm{^\circ C}] CuSO4.H2O + 4H2O}\\[6pt]
\ce{CuSO4.H2O &->[\hphantom{\gt\ 200\ \mathrm{^\circ C}}][\gt\ 200\ \mathrm{^\circ C}] $\underset{\text{(white)}}{\ce{CuSO4}}$ + H2O}\\[6pt]
\ce{CuSO4 &->[\hphantom{\gt\ 200\ \mathrm{^\circ C}}][] CuO + SO3}
\end{align}$$

For me, tschoppi really nails the right response to this question.
Before you handle any chemical you are unfamiliar with, you should first read AND UNDERSTAND all facets of the MSDS (SDS) for that chemical. You need to understand the hazard potential, risk mitigation steps, emergency response strategies and disposal management of all chemicals you are handling.
If you have the capacity to legally obtain chemicals such as copper sulfate pentahydrate, you should also be able to obtain the MSDS (SDS). Google searching will also provide you with this information.
Now, with regards to $\ce{CuSO4.5H2O}$, it has a GHS signal word DANGER, with the following key safety statements:
Hazard Statements

Toxic if swallowed.
Causes skin irritation.
Causes serious eye irritation.
Very toxic to aquatic life.

Precautionary Statements

Wash skin thoroughly after handling.
Do not eat, drink or smoke when using this product.
Avoid release to the environment.
Wear protective gloves/ eye protection/ face protection.

Response Statements

IF SWALLOWED: Immediately call a POISON CENTER or doctor/physician.
IF ON SKIN: Wash with plenty of soap and water.
IF IN EYES: Rinse cautiously with water for several minutes. Remove contact lenses, if present and easy to do. Continue rinsing.
If skin irritation occurs: Get medical advice/ attention.
If eye irritation persists: Get medical advice/ attention.
Take off contaminated clothing and wash before reuse.

Additionally, it advises to wear correct respiratory protection (P2 or N99) and avoid breathing vapours mist or gas.
Given this information, it is unsurprising that you experienced an irritation-type reaction to breathing the vapours. So, how to avoid this next time? Read and understand the MSDS prior to use. Employ the correct protective equipment. Carry out the procedure in a suitably ventilated area, with active ventilation. This could be outside with a strong fan pushing vapours away.
If you are unsure about how to handle this chemical correctly, forgo the pretty colours on your rocket.

Thank you for broad safety info. I was however aware of copper sulfate irritation danger in general. I was wearing safety glasses to avoid risk of eye irritation. I just didn't expect that any gas will form during the process. Your post literally does not contain any information I didn't know. Instead of acting negatively while mentoring me, you could've explained what exactly happened.

I want to use copper(II) sulfate as rocket propellant additive in hope that it will alter the flame colour.

It likely won't. AFAIK, common colored flame mixture using copper is copper benzoate/potassium perchlorate mix (pale blue trace), but using it involves known risks of explosive combustion in case of improper handling. In case you are interested in colored light compositions, you need specifically designed compositions, and most of them are not health-friendly.

I suppose copper sulfate has got in there. Do you think I risked badly, or is it just weak irritation? How can I avoid it next time, apart of using fumehood?**

There is no much risks here, though next time stick to ceramics. Copper salts are prone to hydrolysis, so I believe you smelled some sulfuric acid vapors. Not pleasant, but not deadly. In comparison, copper chloride would hydrolyze completely, producing much hydrochloric acid vapors, and that would be much worse.


text:
How many electrons with l = 1 does Si in its ground state have?
I was solving practice problems for electron configuration and periodic table, and I got stuck through a question:

An atom of silicon in its ground state has how many electrons with quantum number $l = 1$?
a) 14
b) 2
c) 8
d) 6
c) 28

According to the solution the answer is c: 8 electrons, but that wasn't my answer.
My analysis:

Silicon is a chemical element with symbol $\ce{Si}$ and atomic number 14, this is its electron configuration:

$$\mathrm{1s^22s^22p^63s^23p^2}$$

The quantum number $l = 1$ corresponds to $\mathrm{p}$ levels only. So we have:

$$\mathrm{2p^63p^2}$$

The total number of orbitals is: $6 + 2 = 8$ orbitals.
Since each orbital has 2 electrons, the total number of electrons is $8 \times 2 = 16$ electrons.

What's wrong with my answer? Maybe the question has a wrong answer?

You're on the right track, but those superscripts in the electronic configuration don't refer to orbitals. They refer to the number of electrons in that type of orbitals.
For more information, you might want to look at this question: What are the maximum number of electrons in each shell?

I was also stuck with questions like these in our exams and I learned through the internet that the superscripts are actually the number of electrons and that $l=1$ refers to p-orbital. In the electronic configuration of Silicon, consider only the p-orbital, since it is aking for electrons that have p-orbitals (l=1)
$2\mathrm{p}^6$ and $3\mathrm{p}^2$ are sub-orbitals of p-orbital. Since 2p have 6 electrons in it and 3p have 2 electrons in it, there are total of 8 electrons that have $l=1$ as one of their quantum numbers.


text:
Is my analysis of quantum numbers describing the ground state of At correct?
Which series of quantum numbers describes the highest (energy) occupied orbital in a ground state of At atom?
a) n = 6, l = 0
b) n = 6, l = 2
c) n = 5, l = 2
d) n = 4, l = 3
e) n = 6, l = 1
According to my teacher, the answer is e.
My analysis:

The electron configuration of At atom:

$$\begin{aligned}1s^22s^22p^63s^23p^6\ce{3d^10}4s^24p^64\ce{d^10}5s^25p^64\ce{f^14}\ce{5d^10}6s^26p^5
\end{aligned}$$
Knowing for levels 0,1,2,3 correspond to s,p,d,f respectively we can say that:
a) $$\begin{aligned}6s^2\end{aligned}$$
b) $$\begin{aligned}Nothing \end{aligned}$$
c) $$\begin{aligned}\ce{5d^10}\end{aligned}$$
d) $$\begin{aligned}\ce{4f^14}\end{aligned}$$
e) $$\begin{aligned}\ce{6p^5}\end{aligned}$$
What I thought here first is that d is the answer since 4f14 has the largest number of electrons, and thus will occupy the highest energy around.
Then I saw on another question that the orbitals are filled so that the ones of lowest energy are filled first.
$$1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s ...$$
In this case, e is the answer since 6p5 has the highest energy regardless of the number of electrons it is holding?

Your second way of thinking about it is correct.  The number of electrons in the orbital is not the determining factor here.  Well written question.

Yes , there is another name for it (n+l) rule or Aufbau principle
http://en.wikipedia.org/wiki/Aufbau_principle
Even though it is not correct to check this rule at such higher values of Z where the trend sometimes changes due to very less differnce between higher values of shell and orbitals .But if you are a beginner you should not heed to this statement.


text:
Can the electron configuration of Te be written that way?
Normally, the electron configuration of Te is known as:
$$\begin{aligned}  {[Kr]} 5s^2 \ce{4d^10} 5p^4 \end{aligned}$$
Then, one day I was asked in a exam if this can be written also as:
$$\begin{aligned}  {[Kr]} 5s^2 \ce{4d^10} 5p^3 6s^1\end{aligned}$$
I answered that it couldn't. But my answer turned out to be wrong.
If this is correct, why is it allowed to be written like this? And what general case can I learn when being asked such tricky questions?

The second configuration allows Te to take advantage of the relative stability of half-full shells. Having the 5p and 6s orbitals both half-populated is more stable than having 4 electrons in the 5p orbital.
Similar behavior is what allows carbon to form 4 bonds. The $\ce{s^1p^3}$ configuration is more stable than the $\ce{s^2p^2}$.
Tellurium shows the behavior while oxygen does not as a result of the overlap between the 5p and 6s orbitals in the larger atoms. The same degree of overlap does not occur with the 2p and 3s orbitals.

To add to this, consider the energy level diagram found here.  You can see at the lower energy levels the s and p orbitals are well separated and the distinction between the 1, 2, and 3 energy levels are also large.  Beyond that the various orbitals become quite close in energy, so lots of unexpected things occur.


text:
Why is LDL more easily oxidized/damaged than HDL?
Why are low-density lipoproteins (LDL) more easily oxidized than high-density lipoproteins (HDL)? By LDL and HDL, I mean the lipoproteins that carry cholesterol.
Oxidation of LDL ultimately contributes to heart disease.

Just a brief background about lipoproteins:
Lipids and proteins associate noncovalently to form lipoproteins, which function in the blood plasma as transport vehicles for triacylglycerols and cholesterol. Simple, unesterified fatty acids are merely bound to serum albumin and other proteins in blood plasma, but phospholipids, triacylglycerols, cholesterol, and cholesterol esters are all transported in the form of lipoproteins.
In general, lipoproteins are classified according to their densities. The densities are related to the relative amounts of lipid and protein in the complexes.  Thus, there are:

high-density lipoproteins (HDLs),
low-density lipoproteins (LDLs),
Intermediate density lipoproteins (IDLs),
very-low-density lipoproteins (VLDLs), and also
chylomicrons.

Chylomicrons have the lowest protein-to-lipid ratio and thus are the lowest-density lipoproteins. They are also the largest.

There are a couple of things to take into consideration:

How they function

LDL appears to be the major circulatory complex for cholesterol and cholesterol esters. At various target sites, particularly in the capillaries of muscle and adipose cells, these particles are degraded by lipoprotein lipase, which hydrolyses triacylglycerols. Lipase action causes progressive loss of triacylglycerol (and apoprotein) and makes the lipoproteins smaller. This process gradually converts VLDL particles to IDL and then LDL particles, which are either returned to the liver for reprocessing or redirected to adipose tissues.  This results in even more increase in LDL than other lipoproteins.

Environment

Another explanation is based on environment conditions -the LDL binds to specific LDL receptors, which cluster in domains of the plasma membrane which subsequently form coated vesicles which pinch off from the plasma membrane and form endosomes. In the low pH environment of the endosome, the LDL particles dissociate from their receptors. The endosomes then fuse with lysosomes, and the LDLs are degraded by lysosomal acid lipases into amino acids, fatty acids, and cholesterol.
Newly formed HDL contains virtually no cholesterol ester (however, over time, cholesterol esters are accumulated through the action of lecithin:cholesterol acyltransferase).
Another associated protein, cholesterol ester transfer protein, transfers some of these esters to VLDL and LDL. Alternatively, HDLs function to return cholesterol and cholesterol esters to the liver. This latter process apparently explains the correlation between high HDL levels and reduced risk of cardiovascular disease. High LDL levels, on the other hand, are correlated with an increased risk of coronary artery and cardiovascular disease.
Reference
Biochemistry (Grisham)
Biochemistry (Campbell and Farrel)

HDL might not be more easily oxidized than LDL.
From Lipids Health Dis. 2005[1]:

Our findings are in line with the studies by Bowry[4],
Suzukawa [10], Schnitzer [11], Ohmura
[12] and Raveh [9], who have come to the
conclusion that HDL is more easily oxidized than LDL. In other
studies, HDL has appeared to be less prone to oxidation and even to
protect LDL against copper-induced oxidation[3,13-15].

Reference:
[1]: Solakivi, T., Jaakkola, O., Salomäki, A. et al. HDL enhances oxidation of LDL in vitro in both men and women. Lipids Health Dis 4, 25 (2005). https://doi.org/10.1186/1476-511X-4-25


text:
Why is cytosine far more vulnerable to hydrolytic deamination than both adenosine and guanosine?
Why is cytosine far more vulnerable to hydrolytic deamination than both adenosine and guanosine? Here's the source for the claim (the half-life is 340 years at 298 K, which is much smaller than that of adenosine and guanosine).

Cytosine deaminates to form uracil at a finite rate in vivo
Pyrimidine bases e.g cytosine in DNA are more susceptible to spontaneous deamination than are the purine bases because there are more thermolabile.

This was demonstrated by the previous observation of significant
  cytosine deamination but no guanine deamination upon heating of the
  synthetic copolymer poly(dG-dC):poly(dG-dC) and the homopolymer
  poly(dG):poly(dC)

Reference: Journal of Biological Chemistry -  Nitric Oxide-induced Deamination of Cytosine and Guanine in Deoxynucleosides and Oligonucleotides
Another research article: Cytosine Deamination Plays a Primary Role in the Evolution of Mammalian Isochores agrees with the theory of temperature as the major cause of sponteneous demination in cytosine as it states:

Because a 10°C increase in temperature in vitro increases the rate of
  cytosine deamination 5.7-fold, cytosine deamination must be highly
  dependent on body temperature.

Cytosine residues in DNA (2) can undergo hydrolytic deamination to yield uracil residues (3)
The reaction may proceed either via attack of hydroxide on the neutral nucleobase or attack of water on the N3-protonated base:

In single-stranded DNA, and at base mismatches in duplex DNA,
  deamination proceeds much faster (t1/2 ~ 200 years) presumably due to
  increased solvent accessibility of the base.   Deamination of guanine
  and adenine residues in DNA is much slower, occurring at only 2–3% the
  rate of cytosine deamination.

Alkylation of the N3-position of cytosine and reactions that lead to saturation of the 5,6-double bond in cytosine and 5-methylcytosine accelerate deamination.
Activation-induced cytidine deaminases catalyse the conversion of cytosine residues to uracil residues in single-stranded regions of DNA.
Reference:  An Overview of Chemical Processes That Damage Cellular DNA: Spontaneous Hydrolysis, Alkylation, and Reactions with Radicals
Another factor which apparently "contributes" to susceptibilty of cytosine to hydroysis is its presence in both DNA and RNA.
DNA is not susceptible to alkaline hydrolysis while RNA is alkali labile and is readily hydrolysed by hydroxide ions. On the other hand RNA is relatively resistant to the effects of dilute acid, but gentle treatment of DNA
with acid leads to hydrolysis of purine glycosidic bonds and the loss of
purine bases from the DNA.
These different conditions  (i.e acidic and basic conditions) to which the bases are exposed to appears to have a "double effect" on cytosine. (and additionally , as a pyrimidine it is more thermolabile)
Other reference

Biochemisty (Grisham)


text:
Rate law and rate constant
How do we obtain the expressions for integrated rate laws ?!? What are the limits we integrate it under and with respect to what do we integrate it?

Remember first that the integrated rate laws that we are familiar with for zeroth, first, and second order reactions are based on systems involving a single reactant in their rate determining step (or those that have been manipulated to give very large excesses of one reactant such that its concentration remains essentially fixed during the reaction process.)
With that in mind, there are only two variables involved, the concentration of our reactant of interest, $[A]$, and time, $t$.  We can't directly integrate the differential rate law; it's a differential equation and must be solved as such.
For an example, consider the first order differential rate law:
$rate = k[A]$
The meaning of  $rate$ here is the rate of change of the concentration of reactant A with respect to time.  Since we want the rate to be positive, we use the opposite of that quantity.
$-\frac{d[A]}{dt} = k[A]$
This is a differential equation, specifically one that is separable (and thus fairly easy to solve.)  Multiply both sides by $-dt$ and we get:
$\frac{d[A]}{[A]} = -k\ dt$
$\int{\frac{1}{[A]}d[A]} = \int{-k\ dt}$
Integrating both sides and applying initial condition to find the constant gives the integrated first order rate law:
$ln[A]=-kt+ln[A_0]$
The other integrated rate laws are found in similar ways.