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What are radicals and how do they act in radical chain reactions?
What really is a radical? For example, when $\ce{Cl2}$ receives enough energy to dissociate the bond and turn into $\ce{ 2Cl.}$, the two chlorine atoms each carry one of the electrons that form the bond. What is the difference between the $\ce{Cl}$ radical and the original $\ce{Cl}$? They have exactly the same number of electrons. And how do they act as a chain carrier?

free radical is a molecule with one or more unpaired electrons. Despite this definition, d-element compounds with unpaired electrons on d-element, like $\ce{CrCl3.6H2O}$ are not usually considered as free radicals because their unpaired electrons are relatively unreactive. Free radicals are especially reactive, because their reactions usually go through activated complex with relatively small energy. The reasons behind it are not that easy do describe.
In general most reaction with free radicals in organic chemistry falls into two categories: one, where free radical reacts with exposed hydrogen/heteroatom and one where it reacts with exposed multiple bond. In both cases, orbitals of interests are three orbitals of the radical in consideration and two atoms of the organic molecules. MO theory says, that such interaction results in three MOs: bonding, nonbonding and antibonding. With three electrons occupying this orbitals, the energy of the system is mostly same as for the radical and the molecule separated, so the activation energy of the reaction is near zero.
In case of nucleophilic/electrophilic reactions in , the interaction in activated complex are similar, with two electrons on bonding and zero or two electrons on nonbonding orbitals. However, in case of aliphatic derivatives, the reaction usually goes throw sterically hindered state with 5-coordinated carbon  and/or requires generation of hi-energy particles that are not preserved after the molecule reacts. In cases when hi-energy intermediate may be preserved (for example, ionic polymerization), and/or reaction goes through sterically unhindered intermediates (for example, nucleophilic substitution in acyl halides), and/or one of the reagents easily produces active intermediates (for example, nitrating mixture in aromatic substitution) reactions without free radicals as intermediates may proceed swiftly.
Example of chain reaction for aliphatic radical substitution:
$\ce{Cl2 + CH4 -> CH3Cl + HCl}$
in case if $\ce{CH4}$ is presented in huge excess, resulting in reduced rate of multichlorination, the reaction proceeds as following:

Chain initiations: $\ce{Cl2 -> 2.Cl}$
Chain growth 1: $\ce{.Cl + CH4 -> HCl + .CH3}$
Chain growth g 2: $\ce{.CH3 + Cl2 -> CH3Cl + .Cl}$

In case of ionic polymerization

Chain initiation: $\ce{H2SO4 + CH2=CH-Ph -> HSO4- + H-CH_2-\stackrel{+}{C}H-Ph}$
Chain growth: $\ce{H-(CH2-CHPh-)_{n}-CH2-\stackrel{+}{C}H-Ph + CH2=CH-Ph ->} $  $\ce{H-(CH2-CHPh-)_{n + 1}-CH2-\overset{+}{C}H-Ph}$

Both reactions proceed smoothly at room temperature if source of active particles is available (UV-light for breaking $\ce{Cl-Cl}$ bond for the former and concentrated sulfuric acid for the latter)
Compare nucleophilic substitution $\ce{KOH + CH3I -> CH3OH + KI}$

$\ce{KOH -> K+ + OH-}$
$\ce{OH- + CH3-I -> (HO-CH3-I)- -> HO-CH3 + I-}$
$\ce{I- + K+ -> KI}$

As you see, the reaction proceeds through sterically hindered state with 5-coordinated carbon. Compare to acyl halide case $\ce{KOH + CH3COCl -> CH3COOH + KCl}$

$\ce{KOH -> K+ + OH-}$
$\ce{OH- + CH3-(C=O)-Cl -> CH3-C(O^{-})(OH)Cl -> CH3-COOH + Cl-}$
$\ce{Cl- + K+ -> KCl}$

in this case the reaction proceeds through sterically unhindered state with 4-coordinated carbon, so acetylchloride reacts with water at room temperature.
TL;DR : free radicals are particles with unpaired electron, readily available for covalent interactions, they reacts with molecules with easy because they are ones of big family of 'active particles' with low energy of activation of many reactions, and, unlike many other active species, they usually regenerate themselves in reaction, allowing chain reactions to proceed. Other active particles or plainly active substrates also exists, so free radicals are not special. The fact of presence of unpaired electron does not necessary make the particle an active free radical, many relatively inactive free radicals exists, like $\ce{NO}$.
Further reading: MO theory (case of 3-atomic interactions) and kinetics (case of chain reactions) parts of your favorite advance chemistry textbook, parts on reaction mechanisms and active particles of your favorite advanced organic chemistry textbook.
Despite above answer based exclusively on organic chemistry examples, everything stated above is true for inorganic chemistry. However the set of possible kinds of elementary acts, active particles and interactions is much wider than for basic organic chemistry, leading to many special and unique cases and consequently to much less predictable reactions in complicated cases.

A radical is a species with an unpaired electron. The chlorine molecule has a covalent bond joining the 2 chlorine atoms, there are no free or unpaired electrons. Because of this, the chlorine molecule is a stable species and energy must be put into the system to break the bond. If you break that bond you now have a chlorine atom with an unpaired electron. Such a radical with the unpaired electron is a very reactive species because of the unpaired electron. When such a radical reacts with a double bond a new free radical is generated at the other end of the double bond and chain propagation continues.

The difference between the Cl2 and the Cl (chlorine and chloride respectively) is that the in the chlorine MOLECULE the chlorine atoms are covalently bonded to each other thus filling all the occupied orbitals as one 3p electron from each Chlorine (2 in total) is shared between the two. As a result, one can take each chlorine atom's electron configuration to be 1s2, 2s2, 2p6, 3s2, 3p6. (identical to the unreactive noble gas Argon) When the covalent bond is broken, typically using UV light, it is broken by homolytic fission which means that one of the electrons from the two used in the covalent bond goes to each chlorine atom leaving them uncharged. However, if one takes a look at the electron configuration of the atom now - 1s2, 2s2, 2p6, 3s2, 3p5 - it's clear that the 3p orbital is incomplete (one p orbital is only occupied by one electron). This renders the Chlorine atom a Chlorine radical; there is no distinction between a chlorine atom and a chlorine radical. This also explains why chlorine is diatomic (is mostly present as a Cl2 molecule) with a single covalent bond. This is similar the same as all group 7 elements. Also similar to elements like oxygen which are di-radical as there are two unpaired electrons in the the 3p subshell. This explains why oxygen molecules have a double bond  and are also diatomic. I know "two un**paired** electrons" sounds counter intuitive but the p sub-shells are filled in such a way as to avoid pairing the electrons in an orbital until absolutely necessasry (i.e once three electrons are already in the p sub-shell). N2 (Nitrogen) follows the pattern, forming triple bonds in a diatomic molecule to pair electrons in orbitals.
Radicals are extremely reactive as a result. They're chain carriers in reactions called radical substitution reactions (search "radical substitution chemguide" on google). An example is the production of halogenoalkanes from alkanes which are useful for synthesis of alcohols and amines.


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Is greater relative AA battery capacity at high currents indicative of greater capacity at low currents?
I originally posted this on EE.SE but might be better here:
Battery showdown compared the mAh and mWh capacity of different AA batteries discharged at 200mA.
Would the relative capacities of each battery type (within the same chemistry type) be the same at much lower current draw, e.g. 1mA? If not, why not? Are there things battery manufacturers do to 'tune' batteries to a certain discharge range?
Furthermore, I've noted that as current goes down alkaline batteries approach lithium batteries in capacity. Is there a current below which alkaline batteries have more than lithium?
Image from battery showdown:

This is a great question and I'm sorry it wasn't answered sooner. I'll give the caveat that I am not a card-carrying electrochemist, and I'm definitely not a battery researcher, even though I have worked with them in the past. I'll give my best effort.
As you note, in principal, the capacity of a battery is tracked with units of mA$\cdot$hr. In theory, this should be constant no matter the current draw.
That doesn't mean the capacity is uniform across a particular size/shape designation. The "AA" type specifies a particular shape/volume dimension, but different chemistry will give different electrochemical capacities. For example, I have plastic shells that you put a AA battery to fit "C" or "D" types - the downside is less electrochemical capacity (~2500 mA$\cdot$hr for a typical AA vs. ~8000 mA$\cdot$hr for a C cell).
One part of the electrochemical capacity is how much of the volume is filled with electro-active materials, rather than electrolyte, fillers designed to dissipate heat, or for other safety margins.
In practice, as seen in the showdown, nothing ever matches the theoretical capacity, and different batteries and different chemistries yield different practical capacities depending on the current draw.

Results @ 200 mA
Results @ 1000 mA

We see that the lithium cells get ~3000 mA$\cdot$hr under either current. The alkaline cells get ~2500 mA$\cdot$hr at low current, and much less under high current ~1500 mA$\cdot$hr.
So what's the difference?
First, as we mentioned, not every battery has the same capacity, even of the same battery type (size/shape/volume). Assume for the sake of argument that we have two 3000 mA$\cdot$hr batteries: one lithium (of some sort) and one alkaline.
Above a certain current threshold, lower currents will let batteries get closer to their theoretical capacities.
The problem is actually simple. Besides electrons, in every battery ions move through an electrolyte.

Alkaline cells use mobile $\ce{OH-}$ ions.
Lithium cells use mobile $\ce{Li+}$ ions.

At the end of the day, discharge requires ionic conductivity to match the electrical current. So which is likely to have a higher ionic conductivity? Well, $\ce{Li+}$ is smaller and usually has higher ionic conductivity.
In principal, at low current, an alkaline cell with the same capacity rating as a lithium cell, say 3000 mA$\cdot$hr should have about the same capacity. I'm not sure I've ever seen an alkaline AA battery with such high capacity -- they usually come around 2500-2700 mA$\cdot$hr theoretical capacity.
Finally, you ask if manufacturers "tune" the batteries. Put simply, yes. Besides safety considerations, different cells add compounds to minimize self-discharge, corrosion, shelf-life, etc. These can affect the amount of electrochemical capacity.
UPDATE
I realized I should clarify the very low current comment. If the current is low enough, then the rate of ionic mobility will be high enough for the chemical part of a battery to keep up with the electrical draw. So decreasing current draw will improve the capacity, up to this point - and then the real capacity should remain the same no matter how low the current draw.

Many thanks for returning to this question and giving an easily understood but detailed answer.

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Enantiomers or Identical Compounds?
I don't know why my solutions manual says these are enantiomers:

The Fisher projections I drew suggest that these are identical compounds.

The solutions manual is correct, they are enantiomers.  The compound on the left is (R)-2-chloro-(S)-3-butanol and the compound on the right is (S)-2-chloro-(R)-3-butanol.  Try to use the "R\S" nomenclature system on the 2 chiral carbons, that should help figuring out whether they are identical [(R,R and R,R), (S,S and S,S), (R,S and R,S), (S,R and S,R)] enantiomers [(R,R and S,S), (R,S and S,R), (S,R and R,S)] or diastereomers [(R,R and R,S), (R,R and S,R), (S,S and S,R), (S,S and R,S)].

Did I draw the Fischer projections wrong?

Yes, I believe the chlorine should be on the "inside" and the hydroxyl on the "outside" in the drawing on the left.

Hm, that suggests that I was looking at the molecule incorrectly when drawing the Fischer projections. From which perspective should I look at the molecules?

For me, it is easier to draw and understand Newman projections rather than Fisher projections.  I know Newman projections have been discussed previously in your posts, but have you tried them?

I think my issue was not keeping track of the carbons in the Fischer projections. If I suppose the carbon with the hydroxyl group as c2 and the other carbon with the chlorine is c3 (I don't know the actual numbering; nomenclature of alcohols has not been introduced) the the Fischer projections start making sense.

In the 3-dimensional views of the question, the 4 carbons are arranged in a trans (anti) conformation.  To draw the Fischer projection you should mentally rotate C-C bonds to all be in the cis (syn) conformation.

@ron but aren't RS and SR optically inactive? So shouldn't they be identical rather than enantiomers?

In the following figure, everything on the left is identical and everything on the right is identical.

See this page for how to draw Fischer Projections.

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Some questions on stereochemistry
I have a couple of basic questions on stereochemistry that confused me. Please help me understand them.
1) Why does a molecule become achiral if it has a plane of symmetry even though it has steriocentes?
2) When working with Fischer Projections, they say you can rotate a Fischer projection 180 degrees, but not by 90 degrees. I don't understand why.
Moreover, can you please suggest me a good resource wherein I can learn stereochemistry (not to very advanced level)?

1) Why does a molecule become achiral if it has a plane of symmetry
  even though it has stereocenters?

If a plane of symmetry bisects a molecule such that one half of the bisected molecule is the mirror image of the other half of the bisected molecule, then the entire molecule must be superimposable on its mirror image, it must be achiral.  Said differently, if mirror symmetry exists within a molecule, then the mirror image of the entire molecule will be superimposable upon the original drawing of the entire molecule.
Here's an example of a molecule ($\ce{A}$) with 2 chiral groups ($\ce{G_{R}}$ and ($\ce{G_{S}}$).  There is a plane of symmetry that bisects $\ce{A}$, it is perpendicular to the plane of the screen and contains the central carbon atom and the 2 methyl groups.  Now, when we draw the mirror image of $\ce{A}$ (let's call it ($\ce{\bar{A}}$), we find that it is superimposable upon the original drawing of $\ce{A}$ by a simple 180-degree rotation.
Even though $\ce{A}$ contains 2 chiral substituents, the molecule as a whole is achiral.

If you rotate the substituents on a Fischer projection by 90 degrees you change the stereochemistry of each substituent.
The up/down positions on a Fischer projection indicate substituents projected away from you. The left and right positions indicate substituents projected toward you.
Here's a rotated tetrahedral molecule to help you see what I mean by the substituents are projected away/toward you.

We can represent toward using a solid wedge and away using a hatched wedge:

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Rationalizing Strength of Binary Halogen Acids with MO Theory

Help me walk through the logic here:
1) HF is a weaker acid than HCl.
2) This is as far as I can understand. How does sigma star affect the acidities of the binary halogen acids?

If a covalent bond was formed, you would expect a decreasing acidity
  down the periodic table of the halogens, because the energy of the σ∗
  would increase forming a less attractive LUMO (lowest unoccupied
  molecular orbital), which would result in less hydrogen atoms being
  pulled of the halide (i.e. lower acidity, higher pKa).

Here are some questions about the above excerpt:
1) Covalent bond formed between what?
2) Why would the sigma star orbital be populated if the hydrogen were removed and a bond formed in place of it? Or is a bond going to be formed with the hydrogen still present?

Update:
https://usflearn.instructure.com/courses/986898/files/31117039

Some initial comments on the above:
1) Bonding in the $\ce{HX}$ series is a varying mix of ionic and covalent bonding.  $\ce{HF}$ has the largest covalent bonding component and $\ce{HI}$ the smallest.  This trend goes against what electronegativities alone would suggest, but is consistent with what overlap integrals (e.g. overlap between a $\ce{1s}$ orbital and a $\ce{2sp^3}$ orbital is better than overlap between a $\ce{1s}$ orbital and a $\ce{5sp^3}$ orbital) would predict.  Therefor, the latter effect appears to predominate.
2) An MO diagram depicts only the covalent portion of the bonding situation.  Therefor it is unlikely to accurately predict acid strength over the entire $\ce{HX}$ series where the (covalent\ionic) contribution to bonding is varying.
3) The MO diagram should probably have hybridized the $\ce{Cl}$ AO's prior to mixing them with the $\ce{H}$ AO.  That is, I'd guess that the covalent portion of the $\ce{HX}$ bond is between a $\ce{H}$ $\ce{1s}$ orbital and a $\ce{Cl}$ $\ce{3sp^3}$ orbital, not a $\ce{Cl}$ $\ce{3p}$ orbital.  This could be assessed by looking at the photoelectron spectrum of $\ce{HCl}$ and analyzing it (e.g. how many chlorine lone pair signals?) in terms of hybridization.
4) $\ce{\sigma^{\ast}}$ orbitals don't really need to be brought in to the discussion since they are unoccupied throughout the $\ce{HX}$ dissociation process.
5) Increased $\ce{HX}$ acidity correlates inversely with $\ce{HX}$ bond strength. $\ce{HF}$ with the best overlap and strongest bond is the weakest acid; $\ce{HI}$ with the poorest overlap and weakest bond is the strongest acid.

Response to OP update
I'm not sure I understand just what % ionic means in the Table you posted.  When bonding is discussed, usually 3 types of bonding are presented (see this link for example) 1) covalent bonding where electrons are roughly equally shared, 2) polar bonding where we have a covalent bond with unequal electron sharing and 3) ionic bonding where electrons are not shared.  For $\ce{HX}$ you can draw 2 resonance structures, one covalent and one ionic that both contribute to the true picture of the covalent bond in $\ce{HX}$.  Is the % ionic in the Table referring to the polarization in the $\ce{HX}$ covalent bond, or is referring to a truly ionic bond?  How does the increasing covalency suggested by the Table as you move down the $\ce{HX}$ series square with decreased bond strength down the series?  Since bromobenzene and methyl bromide have dipole moments, the equation used to generate the % ionicity in the Table could also generate a percent ionicity for these compounds, but they are certainly not "ionic" compounds.  I suspect the Table is measuring the polarization of a covalent bond and not referring to the ionic character found in a salt.

Thanks! Why would overlap be better with F and not I? What do the overlap integrals mean, physically?

When we say "overlap" that means we're talking covalent, not ionic, bonding.  The electron density in iodine's $\ce{5 sp^3}$ orbital is much more diffuse (spreadout over a larger volume) than it would be in fluorine's much smaller $\ce{2 sp^3}$ orbital.  Therefor, overlap (covalent bonding) should be much stronger (e.g. the overlap integral should be larger) between hydrogen with its $\ce{1s}$  orbital and fluorine.  Think of overlaying a small circle with another circle of similar size and one of much larger size.  Say there is one electron uniformly distributed in each of the 3 circles,

which situation will lead to the greatest absolute electron density in common between the overlapping circles?

The former example. This is very interesting. My prof only rationalizes binary halide acid strengths based on the stabilities of their conjugate bases. I'll have to bring this up to him. He doesn't like quantum mechanics but this concept of overlap originated with Pauling if I remember correctly (whom he adores).

This also reminds me of another question I asked some time ago - why is HBr considered polar. Someone brought up bromobenzene and reaction rates. Your explanation could be another justification.

Wait are you sure that ionic character increases down group 7? OP updated.

Posted a response in my original answer.

" Think of overlaying a small circle with another circle of similar size and one of much larger size. Say there is one electron uniformly distributed in each of the 3 circles," but electron density isn't uniformly dispersed around a nucleus!

Doesn't matter, replace uniform density with s-orbital density pattern and you get the same result.  See top of the page at http://books.google.com/books?id=XxGfF3MgPlEC&pg=PA10&lpg=PA10&dq=angular+dependence+of+p+orbital+overlap&source=bl&ots=hdPu48fPiG&sig=X3NGkSm_ZjbPmHERqLq_0M4O8kg&hl=en&sa=X&ei=btWmU876D9O2yASC44DYBw&ved=0CGEQ6AEwCA#v=onepage&q=angular%20dependence%20of%20p%20orbital%20overlap&f=false


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Why is H2S more polar than HCl?
From EN alone we might expect $\ce{HCl}$ to be more polar but a book says that the bent structure of $\ce{H2S}$ gives rise to its polarity - which is also apparently more than $\ce{HCl}$. Why? Does this have to do with the lone pairs? Lone pairs supply electron density, right?

How are you quantifying polarity? Wikipedia says the dipole moment for $\ce{H2S}$ is smaller than for $\ce{HCl}$ (0.97 D vs 1.05 D, respectively).

That's a good question; I'm not sure why my book says that hydrogen sulfide is more polar than HCl.

I could be way off on this, but wouldn't the geometry H2S mean that it's net dipole moment is the vector sum of both H-S dipoles?

What about the lone pairs?

I have some additional comments. As mentioned previously, by the most straightforward measure of polarity for small molecules (net molecular dipole moment), $\ce{HCl}$ is undoubtedly more polar than $\ce{H2S}$. If some other measure of polarity is being considered, then it needs to be stated for clarity.
I suspect the authors of the book and some of its readers are taking the higher boiling point of $\ce{H2S}$ compared to $\ce{HCl}$ to mean the former is a more polar molecule. In fact, this behaviour holds for all of the group 15/16/17 molecular hydrides. However, this is only an indirect measure of polarity, and there are other confounding factors. In this particular case, I believe the reason $\ce{H2S}$ has a higher boiling point does not have to do with the strength of each intermolecular interaction, but the average number of intermolecular interactions which can be held at once, compared to its neighbouring hydrides ($\ce{PH3}$ and $\ce{HCl}$).
A hand-waving explanation is that the group 16 hydrides have a better "balance" of partially charged atoms compared to the group 15 and 17 hydrides in the same period. For example, looking at the hydrogen bonded molecules of the second period and considering an optimal hydrogen bonded network, $\ce{NH3}$ is capable of being the donor atom for three H-bonds and being the acceptor atom for only one H-bond. In $\ce{HF}$, the situation is inverted; now there are three H-bond acceptor sites, and only one H-bond donor. $\ce{H2O}$, however, is a convenient match of being a H-bond double donor and double acceptor. Thus, water performs more hydrogen bonds per molecule on average than either ammonia or hydrogen fluoride, and has a significantly higher boiling point, even though the actual hydrogen bonds are arguably strongest for $\ce{HF}$. The logic would be similar for the group 15/16/17 hydrides of the third period.
Lastly, I'd like to clarify the role of non-bonding pairs in the dipole moment of a molecule. Electrons in pure atomic orbitals are symmetric with respect to the nucleus and produce no net dipole. In a hybridization picture, one might say that non-hybridized $s$ and $p$ orbitals possess no dipole moment, while hybrid orbitals are not symmetric and do possess a dipole moment. In the case of the second period element hydrides, the lone pairs in hybridized $sp^3$ orbitals would provide a net dipole moment. However, for the hydrides in the third period and below, the molecular geometries are consistent with no hybridization (bond angles close to 90°), and therefore the lone pairs would be present in effectively $s$ and $p$ atomic orbitals, with no effect on a molecule's polarity.
In a more accurate picture without invoking hybridization, the molecular orbitals which contain the lone pairs are almost the same as the central atom's atomic orbitals in the heavier group 15/16/17 hydrides (and hence produce no net dipole), while for the second period hydrides the molecular orbitals are not as symmetric and hence actually produce a dipole.

Clearly, dipole of $\ce{HCl}$ (1.05D) is more than $\ce{H2S}$ (0.97D).
This might be an error in your book or your book's author might have a different hypothetical theory about Dipole Moment. It might also be possible that the author is quantifying the dipole in different perspective (Not quite sure about what it can be).
See this thread. The guys over there had quite a lot discussion on the very same topic. Everyone came up with same result, Dipole of $\ce{HCl}$ > $\ce{H2S}$ in every possible perspective.


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How do I make a simple home distillation device?
Like this:

I have two medium sized glass flasks. I understand I need a T-shaped adapter from the first flask that will connect the thermometer and the condenser. So, I also need the condenser and a L-type adapter that will connect the last flask with the condenser.
Is the condenser really necessary for a completely beginner set up?
Do I really need the adapters? I'm worried they will not fit. How high are the costs for these parts?
Thanks for any replies. I know some theoretical chemistry stuff, but with this glassware I've got no idea what I'm doing.
EDIT: Thanks guys, I accidentally found a very nice chemical equipment store and a nice lady there sold me all the parts. I only didn't get the thermometer because it was way too expensive (and I can make something similar myself). It cost 80 eur. Flask (1l), adapter, condenser, another adapter, mount for the flask and rubber pipes. The thermometer was extra 35.

This depends on what you need this setup to accomplish, as the amount of solutions that this setup can manipulate vary greatly.
The basic consensus is that if you would want an optimal apparatus you would need:

Round Bottom Flask
Thermometer
A Column
Condenser (with Water In/Water Out nozzles)
Collection Flask

It's good to know that condenser is pretty damn expensive, so it's good to find a way around as soon as you're just "home experimenting".


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What is the fire point of sucrose with an average particle diameter of 60 micrometres?
What is the fire point of sucrose with an average particle diameter of 60 micrometres? I can't seem to find any documentation on this, online.

"This question was flagged as low-quality because of its length and content." This is an automated process. It would be good if you could add any research you have already done, so that helpful volunteers do not need to do it again.

http://www.wiley.com/legacy/wileychi/rhodestechnology2e/explosion_hazards.html suggests that fine particle sugar can explode.

160 F judging by a MSDS- too explosive for feeble burning to matter.

I managed to find the answer in the MSDS of sugar given by the US Department of Labour, after some searching.
In the MSDS found here there is an ignition temperature (which I imagine to be the same as the fire point) of 350 C, with a minimum explosion concentration of 0.045 g/l in air. There also seems to be a possible ignition from a dramatic increase in pressure (mentioned in the MSDS to be 5,000 psi/sec), though no further dangers are expressed.

I'd say that's explosion point. It burns too feebly to have a fire point- that requires 5 seconds of burning. Its however to find disasters at sugar processing plants.


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Interpretation of the following IR?
I have the following IR Spectrum that I am trying to interpret:

So far I am thinking this compound have:

a primary amine due to two peaks at 3624 and 3533.
The sharp peak 3416 could indicate an alkyne(sp C-H).
And the other sharp peak at 1716 is an aldehyde or ketone.

Is this analysis correct ? And what have I missed ? Thanks for any help !

The sharp, intense band at 1716 $\mathrm{cm^{-1}}$ is indeed characteristic for a carbonyl group ($\ce{C=O}$ stretch, in this case aldehyde or ketone). The medium and broad bands between 2732-2957 $\mathrm{cm^{-1}}$ suggest that it is more likely an aldehyde. In this case, they would be assigned to the $\ce{=C-H}$ stretch, and the intensity of the broad band fits better to a saturated aldehyde, i.e. there is no double or triple bond in $\alpha, \beta$-position to the $\ce{CHO}$ group.
Your assignment of the peaks at 3624 and 3533 $\mathrm{cm^{-1}}$ to a primary amine is reasonable, since primary amines have two bands, while a secondary amine would have only one, weaker band at lower wavenumbers. One of the bands in the area of 700-940 $\mathrm{cm^{-1}}$ can then be assigned to the $\ce{N-H}$ wagging vibration mode.
The sharp band at 3416 $\mathrm{cm^{-1}}$ could indicate a terminal alkyne ($\ce{C \equiv C-H}$ stretch), the $\ce{C-H}$ stretch band of this group at approximately 2900 $\mathrm{cm^{-1}}$ would then be included in the broad band at 2957 $\mathrm{cm^{-1}}$. The weak band in the area between 2000-2250 $\mathrm{cm^{-1}}$ can then be assigned to the $\ce{C \equiv C}$ stretch.
Sources: 1, 2, 3, 4

The band at 3416 might be the aldehyde C-H.  http://goodmolecules.blogspot.com/2013/06/dear-students-this-is-tutorialon.html

@brinnb On the linked page, it is stated that the bands at ~2700 and ~2800 belong to the aldehyde CH stretch (which is in range of the bands between 2700-2900 in the spectrum of the OP), and that the band at around 3500 would probably be an overtone of the carbonyl band. Furthermore, this suspected overtone band is significantly less sharp than the 3416 band in the spectrum of the OP.


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Free Radical Reaction of Methane and Chlorine
The problem statement:

Lets say you have $1 L$ of $2 M$ methane and the same amount of chlorine. Lets also say that both are liquids since those are most likely to react. Now the only way they can both be liquids at STP is if the temperature is as cold as an antarctic winter so this is not aqueous. Gases more often bump the wrong way and solids don't react unless it is oxidization or dissolving.
Now the initiation step is forming the first molecule of $\ce{HCl}$ and Methyl.
Now the methyl and chlorine atom really want to react and form chloromethane

Relevant equations
$\ce{CH4 + Cl2 -> HCl + CH3Cl}$
  (this continues up to tetrachloromethane)
$\ce{2 CH3Cl -> Cl2 + C2H6}$
  (this can continue for much longer than the previous one can)

The attempt at a solution
$\ce{2 CH4 + 2 Cl2 -> 2 HCL + 2 CH3 + 2 Cl}$
$\ce{2 Cl + 2 CH3 -> 2 CH3Cl}$
$\ce{2 CH3Cl + 2 CH3Cl -> 1 C2H6 + 1 Cl2}$
$\ce{2 HCl -> 2 H2 + 2 Cl2}$
This obviously can't happen because then we have more chlorine than we started out with. Why? Well that $1 M\; \ce{Cl2}$ from ethane plus $2 M\; \ce{Cl2}$ from $\ce{HCl}$ is $ 3 M\; $$\ce{Cl2}$ and we started with $2 M \ce{Cl2}$. Just like the number of each element the molarity has to be balanced. What I find confusing is figuring out the molarity of each compound at each step of the process not the compounds per se.

OK, let's focus on the stoichiometry of your last 4 equations:
$\ce{2 CH4 + 2 Cl2 -> 2 HCL + 2 CH3 + 2 Cl~~~~}$  this looks OK
$\ce{2 Cl + 2 CH3 -> 2 CH3Cl~~~~~~~~~~~}$ this looks OK
$\ce{2 CH3Cl + 2 CH3Cl -> 1 C2H6 + 1 Cl2~~~~}$ this should be $\ce{1 CH3Cl + 1 CH3Cl -> 1 C2H6 + 1 Cl2}$
$\ce{2 HCl -> 2 H2 + 2 Cl2~~~~~~~}$ this should be $\ce{2 HCl -> 1 H2 + 1 Cl2}$
I'm not sure I agree with all of your chemical reactions, but at least now you don't wind up with more chlorine then you started with.


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Why is hydrogen sulfate put together as it is?
I recently looked up hydrogen sulfate ($\ce{HSO4-}$) to see how it's put together, and found this image of methyl hydrogen sulfate:

I'd assume that hydrogen sulfate looks the same, but without the methyl group attached. However I'm confused, how can it be constructed like that? This shows the sulfur having a total of 12 valence electrons, but shouldn't the maximum be 8?
I might be asking something that'll be obvious to me later on in chemistry, but this (and other groups that act as atoms) appear early on. I probably don't need to know it yet but I'd prefer to know how it works if I can as opposed to just accepting it.

I think it's time for you to know about Octet Expansion rule. The molecules that expand their octet are known as Hypervalent Molecules.

Normally, atoms in group 2 have 8 electrons and can thus form 4 bonds.  This is mainly because they have s and p orbitals. However, in elements of group 3, they have extra orbitals called d orbitals that can be used to form more bonds.  when forming extra bonds, the atom promotes electrons into the d orbitals and allows the atom to from more than 4 bonds.

Same thing happens in the case of sulfur. Sulfur belongs to 3rd period and therefore it has a vacant 3d orbital and thus can accommodate more than 8 electrons in its valence cell.

I see. I know about orbitals so I suppose that makes sense. I hadn't heard of hypervalent molecules before though, so thanks!

The involvement of d orbitals in sulfur compounds is marginal (below 1%). Hypervalency and hybridisation are the most famous misunderstood concepts in chemistry.

@Martin : So what are you proposing here? o.O

As $\ce{SO4^{2-}}$ is isoelectronic with $\ce{PO4^{3-}}$ I would simply suggest an analogous bonding picture. I.e. Only sigma single bonds to the oxygens, each carrying a negative charge and sulfur being positively charged. The bonding picture of $\ce{HSO4-}$ can then be derived accordingly.

@Martin : Well thanks for information. but I am still thinking that my answer is appropriate for OP or not. Since I wrote what I was taught about Hypervalency in 12th grade. I am not sure that OP is capable of taking any higher concepts. Should I take down my answer? :)

No definitely not. The concept you provide is/was, however outdated it is now, in use. Unfortunately it will almost certainly still be taught at some schools and it is hard to get rid of these myths - but then again some teach creation theory.

I'm starting out with my final year of high school chemistry (to be honest I'm redoing this from last year as I previously failed it), and I'm not sure whether hypervalency will be taught here or not. I think I understand Martin's explanation as well though. In any case my main goal was to find some explanation, and make sure I wasn't overlooking a simple one I should know. Whether I actually understand it now or later on isn't a huge concern.

Hypervalency is an obsolete concept - it was used to explain bonding situations of molecules, that seem to exceed the octet rule. It was accomplished by the help of $\ce{spd}$ hybrid orbitals. It is now known, that for almost all molecules that were described this way the contribution of $\ce{d}$ orbitals is negligible (<1%).
Focussing on the question at hand, the bonding situation of the core $\ce{SO4^{2-}}$ anion is isoelectronic to the phosphate anion $\ce{PO4^{3-}}$.
For the methylsulfate this situation may be extended. A qualitative analysis of the natural charges (BP86/cc-pVDZ) reveals no difference. All of the oxygen atoms carry a negative charge, while sulfur is highly positive charged. Further more, NBO analysis confirms that the terminal oxygens have three lone pairs, while the other oxygens have two lone pairs. A more accurate Lewis structure would represent that fact and only use single bonds.

While I can understand the hypervalency concept, if that's obsolete then this might be a bit too advanced for me. My current study material hints towards using the d orbital but still doesn't explain it, simply states that "some" atoms can hold up to 18 electrons, including sulphur and phosphorus, and that others like boron and beryllium are content with less than 8. My main issue is because they are exceptions to the octet rule when drawing Lewis structure, but at this point I don't think there's much I can do besides memorising them. All the same, thanks a lot!


text:
What the balanced chemical equation for the conversion of starch to dextrin by dry heating?
This question is related to the toasting of bread. When we toast a slice of bread, the starch in it breaks down into dextrin. I want to know the balanced chemical equation for the process.

Let's take a look at the structure of both molecules. First, starch:

Now, dextrin:

Do you see the difference? They are the same molecule (both are glucose polymers), the only difference is the molecular weight, or the length of the polymer chain.
Starch, by convention, is 300-600 monomers long. Dextrin is something less than that.
Since the number of dextrin molecules you get will depend on the polydispersity (size distribution), average chain length, and extent of the reaction, there is no way to write a balanced reaction equation for a real reaction.
You could assume that a starch molecule of a given length breaks into dextrin molecules of a given length, and write an equation. It would look like this:
$\ce{C6H11O5[C6H10O5]_{m}C6H11O6 + (\frac{$m-n$}{2})H2O->} \frac{(m+2)}{(n+2)}\ce{C6H11O5[C6H10O5]_{n}C6H11O6}$
This would only work for one length of starch and dextrin (in reality there could be 10 - 1000) and would also only be true if $m > n$, and $m$ is evenly divisible by $n$.
Also, in the question you specify dry heat, but there is no way to balance the equation without including water (unless you use hydrogen and oxygen gas). I suspect that even under dry heat conditions, there is moisture on the bread, and it is consumed in the reaction. This could be part of the reason that toasting drys bread.


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Number of orbitals in full configuration interaction
I am using one quantum chemistry package where I am supposed to assign a number of active orbitals I want to be included in the FCI (full configuration interaction) calculation. For example if I have  ${H_2}$ molecule (singlet) and I use $cc-pV6Z$ basis set of Dunning, what would be the maximum number of active orbitals in the FCI? Can somebody provide me insight into this problem?

In simple terms, Full CI uses all of the possible orbitals (a "full" active space). Thus the number of active orbitals should equal the size of your basis set. If you have 10 basis functions, and then perform a Hartree-Fock calculation, you get 10 orbitals out (some occupied, some virtual).
Complete Active Space (CAS) methods, on the other hand, only use some of the orbitals and then perform a Full CI on that subset of orbitals, leaving the other orbitals (inactive space) alone.
Full CI is CAS in the total orbital / total active space limit.


text:
What does the adenosine do in ATP?
This is an extremely basic question about biochemistry from someone who doesn't know much about organic chemistry. I'm aware of how organisms store energy ATP and release it by dephosphorylating it into ADP. My question is about the role that the adenosine part of the ATP molecule plays in this process.
Naively, it seems as if the triphosphate group could be attached to any old thing, and you'd get the same energy by dephosphorylating it. But as far as I'm aware, it's only ATP that's used for this purpose in biology, and not any other triphosphate. Why is this? Is the fact that the triphosphate is attached to an adenosine just a "frozen accident" like the genetic code, or does it play an important role in the phosphorylation/dephosphorylation reactions themselves?

In biology, good amount of questions has one simple answer: it first the first solution to appear and its existence prevented any other to appear because, once some solution become widespread, it also achieves perfected state, so other (inefficient before long development, but may be better after it) solutions will fail competition before long development.

The adenosine part is needed to make DNA and RNA.  It is logical for the triphospate to be attached to ADP, GDP, UDP or CDP since they are needed to make DNA and RNA.
In some instances, GTP is used instead of ATP.  In fact the Krebs cycle does involve phosphorylation of GDP.   GTP is used in the formation of microtubes by tubulin.
I've seen some give the rate of hydrolysis being greater for ATP than GTP as a reason for ATP being more utilized.
There is a related question here: https://biology.stackexchange.com/questions/11286/why-is-atp-the-preferred-choice-for-energy-carriers


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Stereochemistry of meso compounds
I got tripped up by this question; I thought these were all enantiomers when in fact they were all meso compounds.
How can one prevent this from happening?

How does one rotate the Fischer projection, for example, in the first problem; I understand that one has to hold one group in place, but then what?

To rotate on a single bond in a Fischer Projection, all 3 groups move in the same direction. $\frac {1} {2\ \ \ 3} \rightarrow \frac {2} {3\ \ \ 1}$. See this page for more information.

Exchanging two groups in fischer projection changes it to its enantiomer and further exchange reverts it back.After exchanging as follows gives meso compound with internal plane of symmetry:

Note that these are for the compunds on the left.
  Do similiarly for others on right

a) $\ce{CH3<->H}$ THEN $\ce{H<->OH}$
b)Rotate nearer $C$ by $\pi/3$ rad. anticlock wise.
c)Imagine rotation about $C_2-C_3$ by $\pi$ rad., the $OH$ in the right back of the plane comes in front projecting out towards us hanging down and thus we have a vertical plane of symmetry.

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


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Lewis base strength and the hydrogen proton
My prof told me that we can compare Lewis base strength based on electron donation to the $\ce{H+}$ proton - i.e sigma bond formation with the $\ce{H+}$ proton - and proceeded to conclude that $\ce{S^{2-}}$ is a weaker Lewis base than $\ce{O^{2-}}$, since the former is a weaker base than the latter. He then proceeded to generalize that the oxide ion is always a stronger Lewis base than the sulfide ion.
Is this a valid method of comparing the strength of Lewis bases? I have two issues:

Isn't Lewis acid/base strength quantified using kinetics, not thermodynamics? Don't stronger LAs react faster, and weaker LAs react slower?
Even if this comparison were valid, would the comparison still be valid outside of water solution?

Does this mean that this question is invalid? No medium is specified.

I think he knows what he's talking about. In explaining the much greater stability of silver sulfide ($K_{\mathrm{sp}} \approx 6 \cdot 10^{-51}$) vs. $\ce{Ag2O}$ ($K_{\mathrm{sp}} = 3.5 \cdot 10^{-16}$), he notes that charge density does not explain this disparity. In fact, charge density would imply that the oxide salt should be more stable.

For instance, although $\ce{O^2-}$ is a stronger B/L base than $\ce{S^2-}$, $\ce{S^2-}$ can be (and usually is) a stronger Lewis base than $\ce{O^2-}$ since S has a lower EN than O. Furthermore, recall that Lewis base strength depends on the Lewis acid with with the Lewis base interacts.

The text goes on to mention other factors such as having a metal cation with a surfeit of valence d electrons, and how that can populate the empty valence 3d orbitals of $\ce{S^2-}$. (S has 3d orbitals?)
This makes sense; a good nucleophile has less stabilized electrons - i.e. electrons that are further from the nucleus. It still seems that the questions posted above put B/L acidity ahead as the main deciding factor for sigma-donor strength.

1) There is no, actually, a definitive, wide acknowledged measure of Lewis base/acid strength. For Brønsted acids/bases there is $\mathrm pK_\mathrm a$/$\mathrm pK_\mathrm b$ , which is thermodynamical (and actually depends on solvent, so there is no absolute scale of acid strength). [1]
Keeping what said above in mind, a qualitative measure of Lewis base strength would be its ability to donate its lone pair. Obviously, it grows with negative charge of the particle and lowering its electronegativity, there is nothing new with it. It should grow with shrinking of negative ion, also understandable. The most common Lewis acid is proton, so comparing  strength of conjugated acid we can roughly estimate order of strength of Lewis base. [2]
2) Quantitatively - no. Qualitatively - yes, it would in most cases
[1] On a side not, in some cases there IS a detectable difference between thermodynamical and kinetic Brønsted acidity. For details see 'sterically hindered bases' in context of organic synthesis.
[2] This, however, comes with a little twist: a Lewis base working as a strong base for one element may work as a weaker base for another element. This depends on relative size of donating and receiving orbitals size/energy. Some food for thoughts may be found in solubility table, comparing solubility of, for example, Be, Ba, Hg and Pb salts.

Speaking of sterically hindered bases one might also be interested in "frustrated" Lewis acid/bases.

Speaking also of Lewis acid/base strength, can we kinetically quantify LA/LB strength?

@Dissenter highly doubt it.

One of the keys in the definition of Lewis acids and Lewis bases is that they react to form a Lewis adduct.  To compare the relative strengths of different Lewis bases, you would allow them to react with the same Lewis acid, form the required adduct and measure the enthalpy of formation of the adduct. The adduct with the largest enthalpy of formation would contain the strongest Lewis base.
1) So Lewis acid and base strength is about thermodynamics.
2) You could run the reaction in any solvent (or gas phase) as long as all of the Lewis acids or bases you want to compare are able to form an adduct in that solvent.  There is no reason why the order of a series of enthalpies of bond formation has to remain constant when you move from solvent to solvent, so your relative strengths could well change from one solvent to another.  The same applies to oxide \ sulphide relative Lewis base strengths if you substitute some other Lewis acid in place of the proton.  Again the relative ordering might well change.

So the comparison would not be necessarily valid outside water solution?

The relative ordering of the comparison may turn out different when water is replaced with another solvent.

That's odd because I've seen test questions regarding "strongest" electron donator and I'm fairly sure he expects us to rationalize electron donation strength through acid/base strengths. I've put an example question above in the OP.

What Lewis acid are they reacting with $\ce{BF_3 ,~ H^{+}}$, what solvent is this taking place in?  I'm sure the standard assumption is proton, water, but it is poor practice not to specify them.  Test questions are just another opportunity to teach.  Calling out the solvent and the Lewis acid might make a student think - "ah, so there could be other choices".

Wait, where do you see boron trifluoride? Also, it would be interesting if you had an example in which the strength of two Lewis acids/bases flipped depending on the solvent. I'll try to think of one myself too. I'd imagine that BF3 would be a good electrophile in water solution but a poorer electrophile in say a hydrocarbon solvent by nature of the solvent being less polar.

Also I'm not clear if my prof himself understands solvent effects; we were talking about metal-ligand coordination bonds one day and he brought up sigma bond donation with the hydrogen proton in water. I'll find the video and possibly link to it on the OP. I do remember him explicitly saying that it all goes back to acid/base strength and told us to look at our Ka/Kb table (which implies water as the solvent).

Oops I take back what I said about BF3 being a good electrophile in water; that doesn't make any sense; I'd imagine that BF3 would be a poorer electrophile in water because it'd be reacting with water and unavailable to react with other molecules.


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Do isotones share any similarities?
A trivial research in atoms and their basic theories led me to this term: "isotones"
Nuclides sharing the same number of neutrons but different atomic numbers.
A simple request or shall I say, quest:
please indicate me to any feasible similarities existing between isotones and please, DO NOT use jargon.

Chemistry is determined entirely by electron interactions. Since the number of electrons in an atom is determined by the number of protons in the nucleus, the atomic number (number of protons) is what defines the chemical behavior of a given element. This means that in terms of chemical properties, isotones would have no similarities unless they happened to be in the same group (column) of the periodic table, but that would be a coincidence and would have nothing to do with the number of neutrons.
As DavePhD and ron said, the nuclear stability could be similar, since nuclear stability has a lot to do with the total number of nucleons (neutrons + protons) as well as the relative proportion of protons to neutrons. The wikipedia article on isotones discusses stability in a little more detail.

I suspect that having the same number of neutrons may mean something at the nuclear chemistry and nuclear physics level, but in terms of general chemistry, including inorganic, organic and biochemistry, it's all about the number of electrons and protons.  So I don't think that isotones would show any differences in terms of (non-nuclear) reaction chemistry.

In chemistry, if an atom has a closed valence electron shell, as do the noble gases, it is particularly stable.
Similarly, there is a shell model of the nucleus.
Certain numbers of neutrons (and also protons) are referred to as magic numbers: 2,8,20,28,50,82,126, all even numbers, because nuclei with these numbers of neutrons (or protons) are particularly stable.
The similarity between isotones is that stability of a nucleus is somewhat correlated with the number of neutrons.


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is there any difference between glycerin and glycerol?
In laboratory setting, is there a difference between glycerin and glycerol?
There are some conflicting info on this topic.

From what I have heard, glycerine is almost the same thing as glycerol.

Nope, there is no chemical difference between glycerol, glycerin or glycerine.  All 3 names refer to the same compound, propane-1,2,3-triol.

glycerols are the triol compound used for many purposes in pure or mixed form , but glycerine is the commercial name of glycerol, which is not pure ,which contain mostly 95% of glycerol , it can't be used when pure glycerol is required .

Glycerin and glycerol are both names for the same molecule. However, depending on where you are getting the glycerol from, it could be more or less pure.

As far as I know, glycerin and glycerol both refer to the same compound: propantriol.


text:
If all chemical processes are equilibria at least to some extent, where does the activation energy come from?
If all chemical processes are equilibria at least to some extent, then where does the energy to overcome the activation energy come from?
Is a process with a very high activation energy (i.e. the ionization of sodium vapor into sodium ion) still considered an equilibrium process at least to some extent?
If so, where might the energy to furnish the forward reaction come from naturally - i.e. with no human input of energy?

I don't think the statement that all processes are in equilibrium to some extent is true. By definition, dynamic equilibrium means that the rate of change in one direction is equal to the rate of change in the opposite direction.
This is certainly not true for all processes, otherwise there would not be a process - nothing would appear to change.
On the other hand, for any given reaction equation that is known to occur, we can say:
for a reaction at equilibrium, neither the concentration of reactants nor the concentration of products will be zero
I think that may be what you meant in your question?
If that is the case, then let's look at your example:

Is a process with a very high activation energy (i.e. the ionization of sodium vapor into sodium ion) still considered an equilibrium process at least to some extent?

It's not an equilibrium process. However, the reaction:
$\ce{Na(g)->Na(g)+ + e-}$
could be at equilibrium. In other words, the ratio of reactants to products could be such that the Gibbs free energy is at a minimum for this temperature and pressure. A more mathematical way to say that is:
$Q = K_{eq}$
where $Q=\frac{[Na^+]}{[Na]}$ and $K_{eq}$ is the equilibrium constant under these conditions. (Brackets mean "partial pressure" - not concentration here.)
Under normal lab conditions (1 atm, 298 K), this is not thermodynamically favorable. The reaction quotient heavily favors the reactants, or another way to say it is $K_{eq}$ is really close to zero. So close you may not be able to measure it. The activation energy is extremely high (equal to the ionization energy), and so any electrons that are removed would have to get that energy from somewhere - which was your next question:

If so, where might the energy to furnish the forward reaction come from naturally - i.e. with no human input of energy?

In an idealized, isolated system, you only have one source of energy for things like that: kinetic. Thermal motion of the atoms could conceivably put enough energy into a single atom that an electron could be "kicked off" and result in an ion. However - the odds of that happening are extremely low, given that the distribution of speeds would have a vanishingly small, although still non-zero, probability of occurring at the magnitude needed to overcome the activation energy barrier. Since the reaction is thermodynamically unfavorable, the probability that the electron would jump right back on is very high - much higher than it would be for leaving in the first place.
Now when you have $6.022 \cdot 10^{23}$ opportunities each femtosecond, very low odds can still work out. But the total amount that would be ionized at any one time is still very, very low. Possibly so low that you would have to wait for an essentially infinite amount of time to see a measurable amount of it.
In a real system, a much larger fraction would be ionized (and this may or may not be reflected in $K_{eq}$, depending on how it is measured). The reason is that there is always ionizing radiation present on earth, and there will be more or less depending on where you are located. Statistical thermodynamics doesn't take this into account, so any prediction of the relative amounts of reactants and products would not be correct. This is kind of a unique example in that respect - most equilibrium reactions, even with very slow kinetics, deal with systems where the thermal energy of the system has a much bigger effect than radiation.

@ thomij  "I don't think it would be a Boltzmann distribution"  Why?

What's a Boltzmann distribution?

@ron - because it's a solid, Maxwell-Boltzmann distributions are derived for the gas phase and apply reasonably well to liquids. I could be wrong though.

@Dissenter - A Maxwell-Boltzmann distribution. Often times people will refer to one or the other when in a hurry.

Actually, I just read the question more carefully and saw that he was talking about gas phase sodium - I will edit my answer.

It's interesting how even though I didn't know anything about the Boltzmann distribution I still had an intuition that the process would still "go" if only sporadically - i.e. when the activation energy barrier is overcome through the kinetic energy of the particles and their collisions.

@Dissenter - that's good! You are thinking in terms of very large collections of randomly moving particles - a very important concept for statistical mechanics/thermodynamics!

Thanks, but it's unfortunate that even though I have an intuitive grasp, I'm not as smart as Boltzmann who actually quantified the relationship.

Also if it's true that the activation energy can be overcome, then why would there be "spontaneous" and non-spontaneous processes if all processes "go" to some extent?

It has to do with the relative free energy on either side of the activation energy. For a spontaneous reaction, the product side is lower. For a non-spontaneous reaction, the reactant side is lower. Some of the molecules that can overcome the activation energy in the spontaneous direction have enough energy to overcome it in the opposite direction, but all molecules that can overcome it in the non spontaneous direction have enough energy to go back.

Maybe this is another case where semantics enters in.  It seems that if a reaction coordinate connects two species, then over time an equilibrium will be achieved between them.  "Over time" could mean microseconds or millennia.  The kinetic energy distribution of a  collection of molecules is described by a Boltzmann distribution.  Therefor even at room temperature, there will be some molecules with enough energy to surmount a high barrier along the reaction coordinate.  If there is a high barrier connecting two molecules, then it will just take longer for equilibrium to be achieved.

I think there are a couple conceptual problems here you should consider:
1) It is not true that all reactions are equilibrium in the normal sense of equilibrium: just burn any complex organic compound, and now try to re-make the material from carbon-dioxide and water. In other words: entropy does mater. Technically any elementary reaction can go both direction, but this is a statement about microscopic processes, and you should careful about it when you try to conclusions about ensembles.
2) You are talking about reactions and equilibrium in a macroscopic sense, while you are asking about microscopic mechanism (transfer of activation energy). In microscopic scale there is a fluctuation of the energy what a molecule can get from its environment. The kinetic energy, vibration energies etc of molecules are not a single value, rather a distribution of values, so there is a non-zero probability that two molecules collide with enough energy or enough energy is stored in a vibration mode so the reaction can go.


text:
what's the mass of the original sample (g)?
here is a question that i don't know how it is solved :
A sample that is 75 % chloride by mass is dissolved in water and treated with an excess of AgNO3. If the mass of the AgCl precipitate that forms is 2.013 g, what was the mass of the original sample?
I hope you give me the the proper way of solving it

Hi and welcome to Chemistry StackExchange. To best answer your question, rather than just giving the answer, it would be useful to know what part of the question you are having problems with. Could you describe what attempts you have already made, and what bits of information you think might be useful in helping answer the problem.

thank you ! as far as i understood the problem, i was thinking that there will an equation like : xcl + agno3 --> agcl + xno3 // where x is an element // and the compound xcl is the limiting reactant because it is stated that " excess of agno3 " // therefore, i can get the moles number with the formula = mass/molarMass .. from there i will use the moles number to get xcl mass // I hope the information is clear and my "English is Ok :( " .. in short; i want the equation the describes the reaction.

I have two questions to ask, which may help you. In 2.013g of AgCl, how much Chloride is there? If Chloride makes up 75% of the mass of the original sample, how much original sample was there?

the first one: I dont know ! .. the second : the original sample will be 100% .. cl 75% and the other component 25% ??

It is first of all important to recognise that all the chloride in the original sample will precipitate upon treatment with excess AgNO3. This will give you a starting point, which is a measurable amount of AgCl.
The molecular mass of AgCl = (107.87+35.45)=143.32g
Therefore, the proportion of chloride in AgCl = (35.45/143.32) = 0.2473
Therefore in 2.013g of AgCl, there is (0.2473 x 2.013) = 0.4978g of chloride.
If 0.4978g of chloride constitutes 75% of the mass of the original sample, then the amount of original sample is (0.4978 / 0.75) = 0.6637g

I dont understand the way you used to solve this problem.. but first let me ask you: what does " treatment with or treat with " mean ?

It means to "mix with" or "add to". If I treat a solution with excess AgNO3, I add, or mix an excess of AgNO3 to it. See definition 6 at [http://www.thefreedictionary.com/treat]

ok got it .. if i ask you to write an equation to this reaction .. will you be able to write with the given information?

There is no equation to write, other than the generic equation you have given in your comments above. If you make some assumptions, it is possible to guess that a molecule that is 75% chloride by weight and that is water soluble is likely to be MgCl2, but that should not form part of this question.

Ohh yeeah !! I just got it !Thank you very muuch

Glad I could help!

there are some questions that are still unanswered ..you may  check my profile .. I will appreciate it !


text:
Use the given information to write the formula of this compound
the question says :
Two elements, A and B, combine to form two binary compounds. In the first
compound, 3.5 g of A combines with 8.00 g of B. In the second compound,
5.0 g of A combines with 17.1 g of B. If the formula of the first compound is
AB2, then the formula of the second compound would be ?
I can't even make the first step :(

As for your previous question, a question back to you to get you thinking: If the molecular formula AB2 gives a compound of 3.5g A, and 8.0g B, can you derive a simple A:B mass ratio ie for every 1g of A, how many grams of B?

by dividing ; 1g of A will give 2.29 g of B ?

No, this still represents a ratio of A:2B. You need to bring it back to give you a unity ratio A:B.

will it be 1g of A for each 4.57 g of B ?

Oops. Going the wrong way there. I hope you understand why we try to get you to do some working out before presenting answers to these types of questions.

For a molecule AB2, 3.5g of A represents one molar fraction, and 8.00g B represents 2 molar fractions (or 4.0+4.0). Therefore, a direct ratio can be given as 3.5:4.0, or 1:1.14.  This means a molecule AnBm will give a mass ratio for A:B of n:1.14xm
For a molecule AB, for every 1g of A, you will have 1.14g of B.
For a molecule AB2, for every 1g of A, you will have 2.28g of B.
For a molecule A2B3, for every 1g of A, you will have (1.14x3/2) 1.71g of B.
etc etc
So for 5.0g of A, in order to have 17.1g of B, you will have 1.14 x (m/n) = 17.1/5
=> m/n = 17.1 / 5 / 1.14 = 3
This means m/n=3, or AB3 (alternatively it could also be A2B6)


text:
Flipping a molecule vs. mirror behind compound
Why aren't these enantiomers?
If you place a mirror behind the compound and look into the mirror, you get the second drawing.

If you place a mirror behind the compound and look into the mirror,
  you get the second drawing

"If you place a mirror behind compound 1 and look into the mirror" - If by that you mean, generate the mirror image of compound 1, well OK, but that does not produce compound 2. It generates the molecule labeled "mirror image" in the following diagram, and it is a non-superimposable mirror image.

Thanks! How does one get "good" at this?

Models help a lot, especially in the beginning.

What about just assigning R/S configurations? Would that always be able to tell me whether I'm looking at enantiomers or diastereomers?

Yes, that would do it.


text:
what's the ratio of atoms of these two isotopes?
here is a question says
:

I know that atom's number is found by =  $6.022*10^{23} * mol$
but how this formula is used here is what i dont know
what's the $idea$ of this question?

You need more information.  They probably expect you to look in a book and see that the atomic weight (which is a weighted average over all isotopes) of $\ce{Tl}$ is 204.383.  Now if x is the amount of $\ce{^203Tl}$ with atomic weight 202.97 and (1-x) is....
Does that get you started?
$$\ce{(x*202_{.}97) + ((1-x)*204_{.}97)~=~?}$$
Can you get to the answer from here?

could you write one more line?

I think i got it. The final answer is 2.51, right?

Set the above equation equal to 204.383 and solve for x, the fractional amount of the TL 203 isotope.  You'll find that x=29%.  So about 29% of the 203 isotope and (1-.29) or 71% of the 205 isotope.


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React these iron oxide, ammonium nitrate, and sucrose
Two questions: what reaction forms between ammonium nitrate, black iron oxide, and sucrose; and a story related question.
So here's the whole story:
So I was playing around a few months ago and stumbled upon a mixture that generated something similar to Black Snakes (firework).
I recently decided to demo this mixture to someone, but I could not find my notes about the experiment, so I decided to try to replicate the mixture from what I remembered.
I am certain that the mixture contained ammonium nitrate, black iron oxide, and some sort of fuel (probably sucrose, impure in the form of powdered sugar).
Combining these three ingredients (one part by volume, whatever particle sizes I have), and igniting with a lighter, the mixture does form black snaking ash, and turns red and crumbles as it cools.
Q1: So, what kind of reaction has taken place between the black iron oxide, ammonium nitrate, and sucrose?
If I leave out any one of the ingredients, the ash snake does not form, so all three compounds are definitely involved. If we leave out either the ammonium nitrate or sucrose, little happens with the application of heat.
Here's a picture after the fact.
I haven't had a very good batch since I lost my notes, so maybe I am missing an ingredient, or maybe I just haven't got the right ratios.
Q2: Any Suggestions?^
Edit:
I have been having more success with this mixture (by volume):
1 part ammonium nitrate
1 part potassium nitrate
2 parts black iron oxide
2 parts sucrose
Mix contents and melt over medium indirect heat (skillet is fine); once it is all melted, immediately remove from heat and form into pellet as it cools. The mixture will reduce in volume by several times.
I suggest doing this in small batches since it can self ignite if you are not careful, or do not heat it at the right rate.
If the ammonium nitrate is omitted or substituted by potassium nitrate, the ash does not expand or cohere. So the ammonium nitrate is doing something more than just providing oxygen here.
The reduction of iron oxide seems to make the ash less monolithic, and moderate reductions in iron oxide forms snakes that fall apart quickly during production.

I'd say that's a quasi-dehydration reaction with Iron oxide as a catalyst forming nitric acid from the ammonium nitrate which removes the water equivalents from the sucrose.

Disclaimer: what I offer here is an educated guess.
$\ce{NH4NO3}$ is known to decompose quite easily, especially if proper catalyst is present. Thus, one of the reactions is
$\ce{NH_4NO_3 = N_2 + 2H_2O + [O]}$
generated oxygen may be used to oxidize sucrose or black iron oxide. I doubt that $\ce{N2O}$ is formed, since it is quite strong oxidizer and decomposes quite easily, especially if redox catalyst is present
$\ce{C_12H_22O_11 + 24[O] = 12 CO_2 + 11 H_2O}$
$\ce{2Fe3O4 +[O] = 3 Fe2O3}$
However, as you said, the snake is black, so the second reaction does not occur
the final process is dehydration of sugar, producing carbon (actually, the carbon may contain significant amount of hydrogen and oxygen)
$\ce{C_12H_22O_11 = 12 C + 11 H_2O}$
at the same time numerous other processes, producing gaseous reducing organics, happens, so I advise to use a fume hood.
When cooled, reducing atmosphere from sugar decomposition vanishes, and black iron oxide reacts with air
$\ce{4Fe3O4 +O2 = 6 Fe2O3}$
Since black iron oxide acts as a catalyst and does not melt on reaction conditions, it is good to use as fine one as you can get and maybe it may be used in smaller amounts. Since ammonium nitrate acts as main energy source here, it is likely that it can be used in larger amounts. Sucrose forms the body of the 'snake', so it may be used in smaller amount.
The 'snake' consists of (partially) carbonized/caramelized sucrose binding other ashes together and leftovers from iron oxide. The expansion of volume results from gases emitted in the processes of decomposition of ammonium nitrate and decomposition of sucrose, though decomposition of ammonium nitrate is probably the main source here.
BTW, thanks for recipe, I think it would come handy for me this year.

If you make it, I suggest melting it together over low indirect heat. It will shrink in volume considerably.


text:
Enantiomers or Diastereomers?
The key tells me these are enantiomers.

I however don't think they are; their chiral carbons' configurations aren't inverted - i.e. only one chiral carbon's R/S configuration changes! This suggests diastereomers to me! Plus the Sawhorse projections I draw don't seem to be mirror images.

They are diastereomers.  Both of the "front" chiral carbons (the one with the bromine attached) are "S".  Rotate the one on the right 120 degrees CCW and it is identical to the one on the left.  But the "back" carbon (the one with the hydroxyl attached) is "S" on the left figure and "R" on the right figure - diastereomers.

I get the opposite of what you get for the back carbon for both images.

Are you "holding" the hydrogen such that the "steering wheel" is between your eye and the hydrogen?

I'm not holding the hydrogen; I'm holding the group with priority 2 and rotating the molecule so the hydrogen is pointing away. ETA: And so I am holding the hydrogen so that the steering wheel is between my eye in the hydrogen.

Try holding the lowest priority group, in this case hydrogen, so that the 3 remaining groups form a circle (steering wheel) between your eye and hand.  When you travel around the circle going from the highest priority group (1) to the next to lowest priority group (3) by way of (2), if you travel CW then your chiral center has the "R" configuration, if you travel CCW, then the "S" configuration.

I'm still traveling clockwise for the left figure. Will insert picture of model in OP.

The OH group has priority 1, the carbon chain priority 2, the methyl group priority 3, the hydrogen priority 4 - right?

Yep, got that. Model inserted.

what color ball represents what in your model?

Sorry, it's on the paper. Y is H, B is methyl, G is -OH, O is side chain.

Your model does not represent the Newman projection pictured at the very top-left of your post.  You need to swap the chain and the methyl.

What does that do? The order is still green --> orange --> blue.

If you swap groups 2 and 3 then the direction of travel (CW \ CCW) has to change.

Thanks, I see it now! How did I manage to build the model wrong and draw the Fischer projection incorrectly?

Great!  Glad to help.

Yes! I think I also figured out my Newman to Fischer woes - I have to eclipse the Newman before converting to Fischer. Guess I learn something everyday.


text:
amu and g/mol relation
Do we have that $\pu{1 g/mol} = \pu{1 amu}$ ?
Because we have, for the mass of an atom of carbon 12, call it $m(\ce{^12C})$, that
$$m(\ce{^12C}) = \pu{12 amu}$$
and furthermore
$$\pu{1 mol} \cdot m(\ce{^12C}) = \pu{12 g}$$
therefore
$$m(\ce{^12C}) = \pu{12 amu} = \pu{12 g/mol}$$
So finally we get that $\pu{1 g/mol} = \pu{1 amu}$ .
However, my chemistry teacher is telling me that those are two completely different things and that I am confused between the mass per atom and the mass per $6.022\cdot10^{23}$ atoms. I can't understand how, and this is really bugging me, so help is very appreciated.
Note that this requires the mole to be a number (or a "constant"), which may be where I'm wrong.

You are correct, but to make it a little more clear you can include the assumed "atom" in the denominator of amu:
$$
\begin{align}
m_{\ce{C}^{12}} &= \pu{12amu atom^-1} \\ \\
m_{\ce{C}^{12}} &= \pu{12g mol^-1} \\ \\
\pu{12amu atom^-1} &= \pu{12g mol^-1} \\ \\
\pu{1amu atom^-1} &= \pu{1g mol^-1}
\end{align}
$$
In other words, the ratio of amu/atom is the same as the ratio of g/mol. The definitions of amu and moles were intentionally chosen to make that happen (I'm surprised your teacher didn't explain this, actually). This allows us to easily relate masses at the atomic scale to masses at the macroscopic scale.
To check this, look at the mass of an amu when converted to grams:
$\pu{1amu}= \pu{1.6605E-24 g}$
Now divide one gram by one mole:
$\pu{1g mol^-1}= \frac{\pu{1 g}}{\pu{6.022E23 atom}} = \pu{1.6605E-24 g atom^-1}$
It's the same number! Therefore:
$\pu{1g mol^-1}= \pu{ 1 amu atom^-1}$

Thanks for the clarification. Do you know know any reputable reference where this is stated? I searched but couldn't find any.

I think any general chemistry textbook should have it - I know that Tro's Chemistry, 3rd ed. has it on page 71.

It should be noted that amu as a unit is deprecated. Now the unified atomic mass unit u should be used instead.

@Martin that is a good point, but I think that for the most part 'amu' is understood to be the unified atomic mass unit based on carbon-12, not oxygen-16. I at least have always seen 'amu' and not 'unified amu' in publications within the last couple of decades.

@thomij I'm really glad you are still with us here, I missed you quality answers recently. The confusion is one of the worst, and it is all IUPAC/IUPAP's fault. They should have used something completely different. However, your argument does not lose any value through that change.

@thomij"$m_{\ce{C}^{12}}"$:  means mass of one carbon atom or atomic mass of carbon?so :
$$m_{\ce{C}^{12}} = \pu{12amu  } = 12\times\pu{1.6605E-24g atom^-1 } =\pu{1.9926E-23g    atom^-1}$$

This is an abuse of notation. Your second line of reasoning is actually more correct than your first. In the first part, strictly speaking you've only shown this for C-12. That is, each place you've written "atom", you are really meaning "atoms of C12" or "Per atom of C12". Your final line is really " grams per  mol of C12 = amu per atom of C12". Really it follows from the fact that the # of atoms in a mol of a substance is equal to the # of grams per amu.

You need to be more careful with your units. The erroneous result is that you are equating a value in amu (a measure of mass, like grams) with a value in grams per mole (an invariant property of an element or compound, regardless of the amount you have).

I am equating them since they are both the mass of an atom of carbon, and I do believe that the mass of an atom of carbon is equal to itself. What is wrong with that? It is nothing exceptional to have units expressed in terms of others.

I support this statement, technically the comparison is not accurate, I would upvote this, if there were a little more explanation.

It's like comparing a speedometer and an odometer - they measure 2 different things. Saying a car has gone 100 miles is not the same as saying it has gone 100 miles per hour.

There are two things that routinely mystify science students:

anything to do with amount of substance (now to be called "chemical amount"), the mole, and the Avogadro constant (or the Avogadro number), and
anything to do with the now-you-see-me-now-you-don't radian.  Let me address the first.

If we have a general number of entities of kind X (e.g. X is the chemical symbol) represented by N(X), the corresponding chemical amount of X is denoted by n(X), which is an aggregate of N(X) entities.  In symbols:
n(X) = N(X) ent, where ent represents an amount of one entity (atom, molecule, ion, sub-atomic particle, . . .), i.e. the entity itself.
The Avogadro number is the (dimensionless) ratio of one gram to one "atomic mass unit" (now called dalton, Da): g/Da.  One mole is an an Avogadro number of entities: mol = (g/Da) ent. Thus we have the important relationship: Da/ent = g/mol = kg/kmol, exactly.  In other words, at the atomic level, the appropriate unit for amount-specific mass ("molar" mass) is dalton per entity--and, because of the mole definition as an Avogadro number of entities, dalton per entity is exactly equal to the macroscopic units gram per mole or kilogram per kilomole.
The critical problem is that IUPAC does not have a recognised symbol for one entity.  It is sometimes (incorrectly) thought of as the (dimensionless) number one.  In which case the "mole" is simply another name for the Avogadro number: "mol = g/Da".  In this case we have the (incorrect) relationship: "Da = g/mol".  Tables of "atomic weights" list the numerical values of atomic-scale masses in daltons--e.g. Ar(O) = ma(O)/Da = 16.  The corresponding amount-specific mass is M(O) = 16 Da/ent; and this is (exactly) equal to 16 g/mol or 16 kg/kmol.


text:
Why are atom spherical in shape ?
I am learning about structure of atom, in which i saw right from J.J. Thomsom atomic model to modern nuclear atomic model all are spherical in shape.
I have seen how different discoveries help to develop the atomic model but i have't came through a single discovery which proves that atoms are spherical in shape.
So, Why atoms are spherical in shape?

Related: http://chemistry.stackexchange.com/q/9561

atoms are not perfectly spherical.

Only isolated atoms can be considered spherical, and even then as approximation. E.g. Hydrogen atoms in $\mathrm{H_2O}$ are very far from being spherical in shape.

This is a delightfully interesting question, and @DavePhD has put up an excellent answer. I will try and contribute my own perspective to it, although more or less in line with what has already been said; I have also included some illustrations to make concepts a bit more tangible, and perhaps, palatable
Preliminaries
How do we determine if something is spherical? Well, we do so by "looking at", measuring its dimensions etc. For large macroscopic objects, that usually means photons of light scatter off the object, our ruler etc. and are captured by our eyes. The signals generated are converted into visual information by our brains. Essentially, all processes involved are quantum mechanical, however, for macroscopic objects their unintuitive, quantum weirdness does not manifest itself.
Similar, without going into details, to determine the "spherical-ness" of an atom we need to probe some physical observable, with which we associate an operator $\hat{\mathcal{O}}$. For a given quantum state $|\psi\rangle$, the expectation value for this observable, i.e. something we expect to get would be given by $\langle \psi|\hat{\mathcal{O}} |\psi\rangle$ (depending on your level of education, you may need to look this up. You can read some details here). Anyway, we can also talk about the probability density of finding an electron in some region of space, and that would be given by
$\langle \psi|\psi\rangle \equiv |\psi|^2 $
So essentially, one never really sees the "wavefunction".
Also, one needs to point out is that space is isotropic. Basically, this means there is nothing special about any particular direction in three-dimensional space. Also, even from classical electrostatics we know that Coulombic interactions behave the same way: if you have a negative charged place at some point in space, it will attract positive charges all around it in the same fashion regardless of whether they are placed to its right or two its left.
Now, we know electrons in atoms occupy "orbitals" which are obtained by solving Schrödinger's equation. It is complicated to solve for many-electron systems, but we can do it for hydrogen. Writing down the equation, we see,
$\left( -\frac{\hbar^2}{2\mu} \nabla^2 - \frac{e^2}{4\pi \epsilon_o}\right)\psi (r,\theta, \phi) = E \psi (r,\theta, \phi) $
Here, $\mu$ is the reduced mass of the system, this allows us to treat the problem in centre of mass coordinates (remember Coulomb forces are central). $\nabla^2$ is the laplacian operator, which basically contains the second-derivatives along x,y,z in cartesian space. It has a more complicated form in spherical coordinates which is what we are using for our wavefunction $\psi$, but basically, the important fact is that this first term is kinetic energy, and then the second term is the coulombic potential due to the interaction of electron and nucleus.
This is a separable, partial differential equation which can be solved in terms of special functions. The normalised position wave functions, given in spherical coordinates are
$\psi_{nlm} (r,\theta, \phi) = \sqrt{\left(\frac{2}{na_0^*}\right)^3 \left(\frac{(n-l-1)!}{(2n+l)!}\right)} e^{-\rho/2}\rho^l L^{2l+1}_{n-l-1} (\rho) Y^m_l( \theta, \phi) $
Where, $\rho = \frac{2r}{na_{0}^{*}}$,
$a_{0}^{*} = \frac{4 \pi \epsilon _{0}\hbar ^{2}}{\mu e^{2}}$ (reduced Bohr radius)
$L$ and $Y$ denote the generalised Laguerre Polynomial and Spherical Harmonic respectively, and n,l,m are the quantum numbers you must be familiar with.
Anyway, we are interested only in the functional forms, so we can ignore all the constants out in front. We also note that the wave function is a product of two-functions, radial and angular.
$\psi_{nlm} (r,\theta, \phi) = R(r)Y^m_l(\theta, \phi)$
Now, essentially, spherical symmetry demands that there will be no angular dependance, and any properties measured or probability of finding electrons etc. only depends on distance from the centre r. This means whenever you measure $\langle \psi|\hat{\mathcal{O}} |\psi\rangle$, it will again only depend on $r$ and not on the angular variables.
We only need to focus our attention on the Spherical Harmonics, as the contain the angular information.
For the purpose of this discussion, we can crudely assume that even in many-electron atoms, the spatial wave functions are similar to hydrogenic wave functions (which is only a one electron system). This is because it is only possible to obtain analytic, functional solutions for the hydrogen atom.
Main Argument
For say an $s$-orbital like l = 0, m = 0. It is trivial to see that the wave function, and consequently $|psi|^2$ is spherically symmetric. i.e. depends only on $r$ because $Y^0_0 = \frac{1}{2\sqrt{ \pi}}$.
Now, let us look at the $p$-orbitals where $l = 1$ and $m$ takes values $-1,0,1$
$$ Y^0_1 = \frac{1}{2} \sqrt{\frac{3}{\pi}} cos \left( \theta \right)$$
$$ Y^{\pm 1}_1 = \mp \frac{1}{2} \sqrt{\frac{3}{\pi}} sin \left( \theta \right)e^{\pm i \phi}$$
Now, clearly $\sum\limits_{j} |Y^j_1|^2  = \frac{3}{4\pi}$ is independent of $\theta$ and $\phi$
The same analysis can be repeated for $d,f$ -orbitals as well. A table of spherical harmonics can be found here
So essentially atoms with electrons in valence $s$-orbitals and/or completely filled or half-filled $p,d,f$ shells would be spherically symmetric.
Now, what if we have something like $2p^1$ type situation. A single $p$-orbital is not spherical. But given that space is isotropic, we have no external fields etc., all the $p$-orbitals corresponding to different $m$ values are essentially the same (degenerate). Nature can't tell the difference, and neither can you when you perform an experiment to measure the "Spherical-ness" of an electron.
This does not mean we are naively taking a superposition of the $p$-orbitals. If you see proof by pictures section, this is clearly not true
Proof by Pictures
Note: The spherical harmonics are complex valued functions, so they are hard to plot, the individual $p$ orbitals conventionally plotted are generated by taking linear combinations of them which are still solutions to the Schrödinger's equation.
Given below is $p_z$ orbital, $p_x$ and $p_y$ look the same just are oriented along the $x$ and $y$ axis.

A naive sum of $p_x$, $p_y$ and $p_z$ orbitals.

Plot of $\sum\limits_{j} |Y^j_1|^2 $. Remember that observable quantities are related to $|\psi|^2$.

I think you should think this question a little through:

The easy answer is that atoms are not spherical in shape. Sphere is a solid object, atoms to not have well defined boundaries with "solid cutoffs". Atoms are more like clouds consisting distributions of electrons, neutrons and protons.
Atoms, per definition, are electrons + a single nucleus, ie. the huge positive charge of a point-like object (nucleus) pulls together the whole thing. That result is that electrons approximately move in a central field. Such central field generates spherical symmetry of the electronic states, and that is why you have e.g. spherical harmonics when you describe the atomic orbitals. Check the solution of the Schrodinger equation how it leads to states, and how wave-functions with only radial dependence (i.e. sphere-like) appears.

+++++ Incorrect part of the answer was deleted based on the comments of DavePhD.

well I don't agree as High resoultion electron microscopy shows surfaces of elements like Pt to be spherically bumpy, and with STED it is even more conforming

Why "only the atoms with field or half-field electron systems"? and what do you mean by "S states"? What is the connection between being half-field (filled?) and being an S state?

Thanks, corrected the typo. Electronic states have symmetry symbols, and some of them has higher, some of them has lower symmetries (http://en.wikipedia.org/wiki/Term_symbol). In case of isolated atoms, typically only filled and half-filed orbitals will lead to S states due to the Hund-rule (http://en.wikipedia.org/wiki/Hund%27s_rules).

@Greg, ok but you still haven't explained the only part of "only the atoms with filled or half-filled...", why the others are not spherical by the electron wavefunctions being in a superposition of all the degenerate orbitals of the partially filled subshell, why you disagree with the statement in J. Chem Ed. vol. 47, page 397 that "all isolated atoms are spherical". http://pubs.acs.org/doi/abs/10.1021/ed042p397

@DavePhD  OK, I see you point now about the multireference nature of L>0 states. Thank you for the correction. Sorry, I haven't read through your answer previously, as your answer was added months later. You are right.

Essentially because the Coulomb interaction is isotropic. Physics would have been pretty interesting if a spherically symmetric potential results in a less symmetric wavefunction -- what is there to skew the distribution of electrons?

Spherical symmetry applies when one ignores the fine details of real-world. Those symmetric wavefunctions are derived by assuming the nucleus is a point-like particle, and Coulomb interaction taken classically. Remember the physicists' joke about spherical cows. Real world is far from spherical symmetries.


text:
What are the other end products in sulfur oxoacid formation?
I got this reaction and I can't figure out the end products
$$\ce{SO2 + S + CaO + Na2SO4 -> Na2S2O3 +  ~?}$$
I am guessing some calcium sulfur salt should form, but which one?

I am not one hundred percent certain, but I will try to give you some guidelines. There is the well known reaction of sulfur dioxide with calcium oxide.
$$\ce{SO2 + CaO -> CaSO3}$$
Under aerobic conditions, this would further react with oxygen to form sulfate, since we have excess sulfur here I think it is fair to assume it will react to form thiosulfate.
$$\ce{1/8S8 + CaSO3 -> CaS2O3}$$
Adding sodium sulfate is probably just a way to control/ buffer the pH during the reaction, I fail to see the actual relevance. But judging from the side reaction
$$\ce{SO2 + 1/8S8 (s) + 3H2O <=> S2O3^{2−} + 2 H3+O}$$
it is actually necessary to have a buffer system based on sulfate.
$$\ce{Na2SO4 + H3+O <=> 2Na+ + HSO4- + H2O}$$
I am not aware of any reaction of sulfur with sulfate.
In total I would write your equation like this:
$$\begin{multline}\ce{SO2 (g) + S (s) + CaO  (s) + Na2SO4 (s) ->[\ce{H2O}]\\ 2Na+ (aq) + Ca^{2+} (aq) + S2O3^{2-} (aq) +  SO4^{2-} (aq)}\end{multline}$$

Why is the meaning of $\ce{H3+O}$? Did you mean $\ce{H3O+}$?

@GaurangTandon in an attempt to display that the positive charge is not at the oxygen, I found it more suitable to write it this way.

First thing that came to my mind was calcium sulfate.
The unbalanced equation would be$$\ce{SO2 +S +CaO +NaSO4 ->Na2S2O3 + CaSO4}$$


text:
A sample of copper(II) sulfate pentahydrate (CuSO4·5H2O) contains 0.360 g of water. What is the total number of atoms in the compound
A sample of copper(II) sulfate pentahydrate (CuSO4·5H2O) contains 0.360 g of water.
What is the total number of atoms in the compound
what's the idea of this question?
what i know is that atoms number is found by multiplying the Avogadro number by the mol's number
??

You have to calculate the amount of substance with the mass of water.
$n[H_2O]=\frac{m}{M}=\frac{0.360\,g}{18\,g} \cdot mol = 0.02\, mol$
$1~ n[CuSO_4\cdot 5 H_2O] = 5~n[H_2O]$
$1~ n[CuSO_4\cdot 5 H_2O] = 0.004\, mol$
In 0.004 mol copper sulfate pentahydrate you have 0.36 g water.
1 molecule of copper sulfate pentahydrate is composed of 21 atoms and the Avogadro constant is $N_A = 6.022 \cdot 10^{23}\, mol^{-1}$.
Now you can simply multiply all values.
$\text{#}Atoms = 0.004\, mol \cdot 6.022 \cdot 10^{23}\, mol^{-1} \cdot 21 = \underline{5.06 \cdot 10^{22}}$


text:
What is the formula of potassium ammonium sulfate?
I tried to get the formula of potassium ammonium sulfate. What I got is $\ce{K(NH4)2SO4}$ but I don't think this the right formula. I searched on the internet and found an answer ($\ce{KNH4SO4}$), but I don't think it is correct.
So what is the correct formula of potassium ammonium sulfate?

It would help if you explained your logic for why you think $\ce{K(NH4)SO4}$ is incorrect and why $\ce{K(NH4)2SO4}$ should be the answer.

your answer doesn't have zero net charge $+1(\ce K)+2(\ce{(NH4)2})-2(\ce{SO4})=1\ne0$ but $+1(\ce K)+1(\ce{NH4})-2(\ce{SO4})=0$

oh yes ! you are right ! I didn't pay attention for the net charge

$\ce{K(NH_4)SO_4}$ is the correct formula. You have two unipositive cations and a single dinegative anion. The total charge must be zero.