text:
What is the carbon dioxide content of a soda can or bottle?
I sort of know how carbonated beverages are carbonated: a lot of $\ce{CO2}$ gets pushed into the liquid, and the container is sealed. There are at least two things I don't know. First, how much carbon dioxide is actually dissolved in the liquid? Second, what is the resulitng partial pressure of $\ce{CO2}$ in the headspace and the total pressure in the headspace? I'm interested in cans, plastic bottles, and glass bottles. I know from experience that there is some variation among manufacturers even for the same beverage, so I will be happy with general numbers or a good estimate.

General estimates have placed a can of Coca-Cola to have 2.2 grams of $\ce{CO2} $ in a single can. As a can is around 12 fluid ounces, or 355 ml, the amount of $\ce{CO2}$ in a can is:
$$\text{2.2 g} \ \ce{CO2}* \frac{\text{1 mol} \ \ce{CO2}}{\text{44 g} \ \ce{CO2} } = 0.05 \ \text{mol}$$
$$ \text{355 mL} * \frac{\text{1 L}}{\text{1000 mL}} = 0.355 \ \text{L} $$
So here we can see we have about 0.05 mol/0.355 L or about 0.14 mol of carbon dioxide per liter of soda. Of course this value varies by manufacturer, type of drink, container, etc.
Looking at Wikipedia, inside Coca-Cola is:

Carbonated water, Sugar (sucrose or high-fructose corn syrup depending on country of origin), Caffeine, Phosphoric acid, Caramel color (E150d), Natural flavorings
A can of Coke (12 fl ounces/355 ml) has 39 grams of carbohydrates (all from sugar, approximately 10 teaspoons), 50 mg of sodium, 0 grams fat, 0 grams potassium, and 140 calories.

Thus, we can calculate the pressure of $CO_2$ gas using the Ideal Gas equation if we store our coke at, say, 20 Celsius:
$$ P = \frac{nRT}{V} $$
$$ P = \frac{\text{0.05 mol} * \text{0.08206} \frac{\text{L} \cdot \text{atm}}{\text{mol} \cdot \text{K} } * \text{293.15 K}}{\text{0.355 L}}$$
$$ P = 3.39 \ \text{atm} $$
According to this website, http://hypertextbook.com/facts/2000/SeemaMeraj.shtml:

On average, the 12 ounce soda cans sold in the US tend to have a pressure of roughly 120 kPa when canned at 4 °C, and 250 kPa when stored at 20 °C.

$$\text{3.39 atm} * \frac{\text{760 torr}}{\text{1 atm}} * \frac{\text{133 Pa}}{\text{1 torr}} * \frac{\text{1 kPa}}{\text{1000 Pa}} = 342.66 \ \text{kPa}$$
Water vapor exerts it's own partial pressure. Looking at standard tabulated values for water vapor pressure, water exerts a pressure of 17.5 torr at 20 Celsius.
$$\text{17.5 torr}* \frac{\text{133 Pa}}{\text{1 torr}} * \frac{\text{1 kPa}}{\text{1000 Pa}} = 2.3275 \ \text{kPa} $$
Knowing that our total pressure is the sum of all our pressures:
$$P_{total} = \text{342.66 kPa} + \text{2.3275 kPa} = 344.99 \ \text{kPa}$$
Here, we are roughly about 100 kPa off from the data provided by the website. This is just an approximation. A more accurate way would be to calculate the moles of each product inside the soda, and knowing the total pressure or partial pressure of one of the parts, we can calculate the pressures more accurately. However, that information is proprietary. It's their secret recipe!
Headspace:
http://extension.psu.edu/food/preservation/news/2012/why-allow-headspace
We have to determine the volume of the headspace - again, I am not sure of exact data - which is between 1/2 inch to 1 and 1/2 inches depending on the container and what it holds. I will assume that the headpsace occupies 6% of the total volume of the can which is 21.3 mL.
At manufacturing and at storage, the can is at different temperatures. Taking the above data, we'll say it is manufactured at $4^\circ C$ and say, stored at $20^\circ C$
Furthermore, when carbon dioxide is solubilized in water, it forms carbonic acid. I will neglect that as the ionization constant is small.
Assuming our 2.2 grams of carbon dioxide is the maximum amount of carbon dioxide that can be placed inside, some of the carbon dioxide is soluble in water while the rest exerts pressure inside the headspace to force the carbon dioxide inside the liquid. The pressure is necessary inside this closed volume as, once you open the cap, the carbon dioxide tries to achieve equilibrium.
$$ \ce{CO_2 (solution) <=> CO2 (g)} $$
In general, the solubility of gases increases at lower temperatures and decreases at higher temperatures. A notable exception is the noble gases. In regards to pressure, Henry's law states that the solubility of a gas in a liquid is directly proportional to the pressure of that gas above the surface of the solution.
The solubility of carbon dioxide in water shifts according to Le Chatlier's principle.
For the solution to this problem, we would need to know several things about the manufacturing of soda's, or rather, more quantitative data. At manufacturing, the pressure is high and temperature low, so as much carbon dioxide as possible is solubilized in the water.
Once shipped out and stored, the pressure inside the headspace increases as the pressure is decreased from manufacturing, the carbon dioxide then leaves solution and enters the headspace. Increasing temperature also decreases solubility.
If you can provide me with more information, I will be more than happy to help. :)

Why are you adding water vapor partial pressure to the $\ce{CO2}$ pressure inside the soda ?

@mannaia A mixture of gases that do not react with one another behaves like a single pure gas. The total pressure results from the sum of the pressure of carbon dioxide and water vapor. If I wanted to find just the pressure exerted by carbon dioxide, I would find the difference of the total pressure and water vapor.

That's fine, but water vapor is above (headspace) while the $\ce{CO2}$ you take (2.2 g) is mainly below, dissolved in the soda. Shouldn't you calculate the $\ce{CO2}$ vapor pressure rising from $\ce{CO2}$ dissolved in the soda and add it to the water vapor pressure ?

@mannaia Ah, I didn't know what headspace was. Apparently, the pressure inside is maximized in the headspace in order to force more $\ce{CO_2}$ in the drink. I'll need to look up some numbers for that however.

I appreciate the detailed exploration, but you should really cite a source for amount of CO2 in a can as that answer seems critical to this question.

Through an interesting approach, authors estimate $\ce{CO2}$ pressure inside carbonated beverages, by measuring the freezing point ($fp$) depression caused by $\ce{CO2}$. In other words, when $\ce{CO2}$ is dissolved in water, a
solution is formed and the freezing point is lowered.
The molality of $m_{\ce{CO2}}$ in water can be obtained, according to:
$$
m_{\ce{CO2}}=\Delta{T_{fp}}/k_{fp}
$$
with $k_{fp}$ for water $=-1.86\,^{o}C \, kg \, mol^{-1}$. For a particular brand of sparkling water, a value of $\Delta{T_{fp}}$ = $-0.42^oC$ was measured, thus yielding a value for $m_{\ce{CO2}}$ equal to $0.23$ $mol \, kg^{-1}$. This first result allows to estimate the mass of ${\ce{CO2}}$ dissolved in one liter of that kind of beverage.
Said mass of ${\ce{CO2}}$ is approximately equal to $10\,g$ per liter (or $3.6\,g$ per can content volume).
Applying Henry's law,
$$
P_{\ce{CO2}}=m_{\ce{CO2}}/k_{H}
$$
with $k_{H}$ being Henry's law constant ($~0.077\,mol\,kg^{-1}\,atm^{-1}$ at $0^{o}C$), one finally gets the pressure $P_{\ce{CO2}}$ exerted by the gas over the liquid, at $0^{o}C$:
$$
P_{\ce{CO2}} = 3.0\,\text{atm}
$$
that corresponds to a mass of about $0.1\,g$ of ${\ce{CO2}}$ inside the headspace volume of a can (about $15\,ml$), mass calculated via $PV=nRT$.
EDIT: for most carbonated beverages, $\Delta{T_{fp}}$ should fall at around $-0.2^oC$ (as reported by Brooker): that would give a $P_{\ce{CO2}} = 1.5\,\text{atm}$ at $0^{o}C$, that rises at about $3\,\text{atm}$ at $25^{o}C$.

It depends from drink and time, according to S. Teerasong, Analytica Chimica Acta 668 (2010) 47–53 for a normal Cola is about $3.1 \frac{g_{CO_2}}{L}$ (take a look even to Glevitzky, Chem. Bull. "POLITEHNICA" Univ. (Timişoara) Volume 50 (64), 1-2,2005 but his esteem is very high).
Regarding the pressure in the headspace this is very difficult to esteem theoretically because you have to take in account not only the temperature but the whole system. :you can shake the bottle and increase the pressure dramatically.

@G M Shaking the bottle does not increase the pressure. http://www.acs.org/content/dam/acsorg/education/resources/highschool/chemmatters/chemmatters-december-2007.pdf

@Jun-GooKwak interesting do you have the link for the next month? Sincerely I'm still skeptic I think the author doesn't know the time the the equilibrium needs! Thanks a lot!

@G M Hi, here is the link http://www.acs.org/content/dam/acsorg/education/resources/highschool/chemmatters/chemmatters-feb-2008.pdf

@Jun-GooKwak Thanks a lot for correcting me! I will edit my answer I can't vote more then one time for your answer but If I could I will do it gladly!

@G M Haha! Thanks, no problem! I initially thought the same until I saw that article. Right now my answer is still incomplete as I have to refer to the volume of the headspace and the pressure therin. In doing so, I have to make quite a few assumptions!


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Oxyhydrogen applied to a catalytic converter
Recently people have been reporting that if you blow 'HHO' gas (Oxyhydrogen) into a used catalytic converter it gives off excessive heat.
Look here for any number of examples.
There is a suggestion that it is cold fusion.
What could this reaction actually be?

I am not aware of any current claim on "cold fusion" that could hold up against a scientific re-examination. Previous reports by Pons and Fleischmann now seem to be (mis)led by wishful thinking and dirt effects,  while those by Taleyarkhan were debunked as fraud.
As far as the inventions of Andrea Rossi are concerned, I'd be very careful with his claims. So far, he is apparently more "famous" for leaving behind stockpiled waste and huge tax debts.
The more recent findings of Rossi and Focardi and the related so-called "independent study" were heavily criticized by other researchers.
In summary, all this nicely fits in the category of pathological science.

...but what chemical reaction is happening here?

@DJA - one reaction I can imagine that will give of heat is simply the recombination of $\ce{O_2}$ and $\ce{H_2}$ into water. Combustion of hydrogen will provide quite some heat and a catalyst can easily make this occur at low temperatures

@Michiel Exactly! And if you don't want to "burn" it, run it through a fuel cell. But $\eta$ is not >> 1 here ;)

If you expose a stoichiometric mixture (("Brown's gas," or any mixture between LEL and UEL) of hydrogen and oxygen to a catalyst, $BOOM!!!$.  A static electricity spark or even a cosmic ray can do it. The Hindenberg was not a pre-mixed flame.
Conservation of mass-energy (First Law of Thermodynamics) is enforced by Noether's theorems and the homogeneity of time.  If you want to diddle that conservation, you must break the symmetry or falsify mathematics.  Good luck on locally doing either for any fuel cycle.
http://physwiki.ucdavis.edu/Classical_Mechanics/Conservation_of_Mass_and_Energy/Noether's_Theorem_for_Energy
(copy and paste the entire link)
http://www.sjsu.edu/faculty/watkins/noetherth.htm
  The mathematics
http://www.lhup.edu/~dsimanek/museum/unwork.htm
 Perpetual motion machines

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Reason for the stronger acidic property of phenol than alcohol

In phenol, pulling the $\mathrm{p}_z$ electrons from the oxygen atom into the ring causes the hydrogen atom to be more partially positive than it is in aliphatic alcohols. This means it is much more easily lost from phenol than it is from aliphatic alcohols, so phenol has a stronger acidic property than ethanol.

Could someone please explain the link between the $\mathrm{p}_z$ electrons of oxygen overlapping with the cloud of delocalised electrons of the phenol ring and the increase in positivity of the bonded hydrogen … which in turns allows phenol to lose a proton more easily?

If there's one thing to be learned from the nice question and the answers, it is: Draw, draw, draw :) Keep a pen ready and draw resonance structures on paper, beer mats, etc.

In fact I find a more simple reasoning with resonance structure. When phenol loses the $\ce{H+}$ the phenolate ion is stabilized due to the resonance effect, as shown below:

The energy of the dissociated form is lower and so phenol has more chance to be in the solution dissociated with the phenolate ion. Aliphatic alcohols are not stabilized by resonance so they are not very prone to dissociation.

While this is by far the most common explanation for the acidity of phenol, it may not be the largest factor. Comparing a simple saturated alcohol with phenol, there is also an inductive effect operating due to the $sp^2$ carbon in the latter case, as opposed to a $sp^3$ carbon in an alcohol, and it must be taken into account. Of course, the importance of the mesomeric effect can be increased by substitution, and picric acid is a clear example of this.

You are assuming that the phenol and alcohol are in water. In a less polar solvent I suspect no or very little dissociation of phenol occurs, thus the polar nature of the solvent has a real role to play in stabilising the ions and driving the reaction what ever the nature of the 'resonance'. BTW is there any experimental evidence for these resonance structures even though they are commonly drawn?

The effect is indeed amazing, if you compare the $\mathrm{p}K_\mathrm{a}$ of tert-butanol (17.0) with that of phenol (9.95).
Deprotonation is facilitated when the reaction goes downhill (energywise). In order to obtain stabilization of the anion, the negative charge needs to be distributed over a larger number of centres. This distribution (only) is possible if the participating orbitals overlap in space.
So, which orbitals may play a role here? On the one hand, there is the $\pi$ system of the benzene, formed from the $\mathrm{p}_z$ orbitals on the carbon atoms. On the other hand, there's  the $\mathrm{p}_z$ on the oxygen with the same spatial orientation.
Now, can we get comparable molecules, more acidic than phenol itself?
The answer is yes, if we allow further substituents that (a) pull the negative off the oxygen by inductive effects or (b) even allow for a further distribution of the negative charge by additional resonance.

An impressive example for the latter is 2,4,6-trinitrophenol, also known as picric acid with a $\mathrm{p}K_\mathrm{a}$ of 0.29.

Here's a simple explanation
The acidity of any molecule is decided by the stability of it's ANION.
We know that one of the lone pairs of oxygen in phenol gets conjugated with pie e- of the ring, distributing the charge, making phenoxide ANION very stable when H+ leaves in aq.
In alcohols, alkyl groups rather pushes the e- on already negative oxygen, concentrating the charge on ANION rather making less stable hence  less acidic.

This doesn't add anything that hasn't already been covered by the previous answers.

Phenol is more acidic than ethanol because in phenol lone pair of electrons are utilized in resonance stabilization so bond length between O and H atom increases and $\ce{H+}$ ion is easily released. While in case of alcohol no resonance stabilization takes place so no release of $\ce{H+}$ ion.

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Decomposition of potassium permanganate under prolonged heating
When we heat potassium permanganate, the equation of the decomposition of it is:
$$\ce{2KMnO4 ->[\Delta] K2MnO4 + MnO2 + O2}$$
So  what will happen to potassium manganate ($\ce{K2MnO4}$) when we continue heating?

There are numerous mixed oxide of Mn, so I believe structures like $\ce{KMnO2}$ will be formed.

Potassium manganate is produced by reacting manganese dioxide with potassium hydroxide and oxygen gas :
$\ce{2 MnO2 + 4 KOH + O2 → 2 K2MnO4 + 2 H2O}$
So I would suspect that the reverse process will occur when further heating. (Of course when there is no water available you might get other products, maybe potassium oxide instead of potassium hydroxide)

I expected the same, but my results are different.

The reaction is given here.
$$\ce{3K2MnO4 ->[\Delta] 2K3MnO4 + MnO2 + O2}$$

The thermal decomposition of potassium manganate to produce potassium
  hypomanganate, manganese(IV) oxide and oxygen. This reaction takes
  place at a temperature of 190-500°C.

I cannot confirm this result. I tested this in reduction atmosphere by mixing the manganate with carbon (charcoal, cleaned) powder and heating to about 700C. The result is reddish brown solid, does not dissolve in or react with water. Neither manganese oxide, nor the hypomanganate look like that.


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What is the canonical reference for studying stereochemistry?
I am well versed with all the named reactions and all of organic chemistry.  I am quite weak in stereochemistry.
What canonical text can I use to study this topic and connect with other parts of the chemistry?

The classical book on stereochemistry is Eliel: Stereochemistry of Organic Compounds, first edition appeared 1962.
You will find overwhelming amount of information there. I would recommend it, regardless your age and level. It is so well written, that it is worth at least of fast reading over most important parts.
You should find it in any reasonable library, no need to get the latest edition, even the older ones should cover what you need.

i like something less elaborte for quick reading like i 3-4 days.

Than try Khan academy topic, https://www.khanacademy.org/science/organic-chemistry/stereochemistry-topic
Not that all-covering, but shorter and concise.

http://osf1.gmu.edu/~bbishop1/CHEM%20814-579%20Stereochemistry%20Lecture%20slides.pdf
   Start here.  Now, the fun stuff off-syllabus.
It is possible to create chiral centers - including planar chiral centers - whose handedness cannot be named, even in principle, e.g., $\ce{CR4}$ and $\ce{R2>C=C<R2}$. The pictured acyclic carbons are chiral centers.  Show their non-superposable enantiomers that illustrate the anomaly.  (Answer to be posted end of the week - though you already know it, but may not know you know it.)
Optical rotation is not a measure of chirality, or even an indication of chirality. Solution optical rotations ignore atomic mass distribution[1].  Silver thiogallate, $\ce{AgGaS2}$ in achiral space group ${I}$-42${d}$, rotates 522°/mm along [100] at 497.4 nm[2].  ${P3_121}$ quartz and ${P3_221}$ berlinite, with opposite sense three-fold screw axes ($3_1$ right-handed, $3_2$ left-handed), are both levorotatory[3].  CIP notation flips with composition at constant geometry (L-cysteine versus most other L-amino acids).
[1] http://www.mazepath.com/uncleal/norbors.gif
http://www.mazepath.com/uncleal/norone.png
[2] Appl. Cryst. 33, 126 (2000)
[3] J. Appl. Crystallogr. 19, 108 (1986)
J. Math. Phys. 40, 4587 (1999)
http://petitjeanmichel.free.fr/itoweb.petitjean.freeware.html#QCM
Quantitative geometric chirality can be calculated in any number of dimensions for any collection of a countable number of anonymous points with overall finite principle moments of inertia.  Mathematically perfectly chiral arrays are rotationally symmetric but lacking all ${S_n}$ improper rotation axes.

The enantiomer of CR4 is CS4 (Cahn-Ingold-Prelog notation).  A chiral substituent may not be cut by a mirror plane.  R and S reflect or invert as S and R respectively, so those are not allowed symmetries.  We thus see two planar chiral centers in the olefin.  It is nice as a test extra credit question.


text:
Does the outcome of aromatic nitrations depend on the concentrations of nitric acid and sulfuric acid?
I came across a few equation and observed that when concentration of the $\ce{HNO3}$ and $\ce{H2SO4}$ with benzene ($\ce{C6H6}$) or its derivative varied the product changed. I did not perform any experiment and all of this I ask here is from the stuff I read in various books.please tell me about this nitrating solution and give me all the cases if possible.

Could you please augment your question with the findings you have collected so far?

okay i can try but its difficult .

Nitration occurs as electrophilic aromatic substitution and activity of substrate varies depending on exact structure of ring system and substitutes in the ring. In general electron acceptors (nitro-, acyl-, nitrile- and sulfogroups) deactivates the reaction, while electron donors (hydroxy-, aryl-, alkenyl- and to lesser extent alkylgroups) activates the reaction. As nitration occurs step by step, it will advance to dinitroarene only if mononitroarene is not deactivated too much by introduced nitrogroup, and same for other nitroarenes. However, different nitration agents have different nitration activity with 100% nitric acid in oleum being one of the strongest, usual nitration mixture (70% nitric acid + 98 % sulfuric acid) being strong, 70% nitric acid being moderate, diluted nitric acid being gentle and sodium nitrite in acidic solution being one of the gentlest. The picture is further complicated because strong nitration agents are also strong oxidizers, capable of breaking apart delicate molecules like phenols. Given that, nitration is often performed in several steps, with first 'gentle' nitration to introduce the first nitrogroup without breaking the molecule and then further nitration with stronger agents (see, for example, synthesis of styphnic acid).
Exhaustive review of the topic is beyond what a simple post can cover. For further insight into the reaction I recommend to consult a decent university-grade organic chemistry textbook, like 'March's Advanced Organic Chemistry'.


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Is alcohol Lewis base or Brønsted base?
I know that Lewis' concept of acid base deals in terms of electrons and Brønsted in terms of $\ce{H+}$ and base which can provide electrons or $\ce{OH-}$. If I am wrong (partially or fully) please explain Lewis and Brønsted-Lowry concept along.
According to me it is a Lewis base because of the lone pair on oxygen of the alcohol and a Brønsted acid.

Obviously you have some ideas about Lewis and Bronsted theory. So, what do you think an alcohol is? Is it one or the other or is can it be both at the same time? Please include your own thoughts on the matter into the question, so that we can help you to clear any misunderstandings.

Alcohols can act as both.  Many reactions of alcohols are believed to start with their protonations or deprotonation. However, isolating salts where they act as bases is much harder, while alcoholates are quite easy both to make and isolate. Protonated alcohols are very active alkylating agents and if fitting $\beta$-hydrogen is present, are prone to beta-elimination.

Alcohols overall are weak Brønsted acids, (water and ethanol pKa 15.7 and 16).  They are middling Lewis (hard) bases (HCl gas ionization in anhydrous ethanol solution).
Relevant page
Consider Al(O-tert-Bu)${_3}$ and AlMe${_3}$ as simultaneous Brønsted bases and Lewis acids.  Cf: hard and soft acids and bases, solvo-cations and solvo-anions, and even as a redox reaction. It works solid state as intercalation, J. Chem. Ed. 81(6) 819 (2004).

Alcohols are weak acids and donate protons to the strong bases forming alkoxide.
Alkoxide ion is a strong conjugate base which shows that alcohols are weak acid but alcohols are strong lewis bases and donate electrons to hydrogen in reaction with HX.

Alcohols is a brønsted base because of the presence of unpaired electron over Oxygen atom which make them proton acceptor.


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Single Displacement Reactions: A + AX =?
This seems like a stupid question, but how would I find the product of a single displacement reaction like the following?
$$\ce{Pb + Pb(NO3)2 -> \ ?}$$
Normally, a single displacement reaction follows the path of $\ce{A + BX -> AX + B}$. However, what will the equation look like when A and B are the same element?

Try this link. Maybe that helps some. (Don't know if it's valuable advice, though...)

If Pb(I) were stable, 2PbNO3.  Nitrate itself might be a sufficient  oxidant to prevent that.  Cu(0) + CuCl2 gives 2CuCl when refluxed in acetonitrile (soft Lewis base ligand).  A blue solution with red metal dust turns opaque brown-green, then goes clear and colorless.

Well, nitpicking involved, there could be the isotopic statistical equilibrium: $\ce{^{206}Pb(s) + ^{208}Pb(NO3)2(aq) <=> ^{208}Pb(s) + ^{206}Pb(NO3)2(aq)}$

Nothing. All you would have is just a mixture of white powder and lead metal.
In general, lead (II) nitrate is synthesized by:
$$\ce{PbO (s) + 2 HNO3 (aq) -> Pb(NO3)2 (aq) + H2O (l)}$$
Lead (II) nitrate is soluble in water.
Alternatively, you could do:
$$\ce{PbCO3(s) + 2HNO3(aq) -> Pb(NO3)2(aq) + H2O(l) + CO2(g)} $$
Single displacement reactions are a kind of redox reaction. For single displacement reactions to occur:
1.) A and B must either be different metals OR
$$ \ce{2AgNO3(aq) + Zn(s) -> 2Ag(s) + Zn(NO3)2(aq)}$$
2.) A and B must be a halogen
$$\ce{Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g)}$$
This is necessary as for the redox reaction to occur, you need two differing metals on the reactivity series in order for the single replacement to work, or alternatively, use compounds that contain halogens.
Even so, in combining two compounds with metals or halogens, if the element is less reactive than the element in the compound, the single replacement reaction will not proceed.
Some examples are:
$$\ce{Ag (s) + Cu(NO3)2 (aq) -> } \text{No reaction}$$
$$ \ce{I2 + 2KBr ->} \text{No reaction} $$

If you consider reaction of type A + AX ->, of course they can be achieved. However, they won't be considered displacement reaction but rather disproportionation reaction. A handful of interhalogen compound synthesis proceeds through this reaction format (if I consider X to be the more electronegative halogen than A).
$$\ce{Cl2 + ClF3 ->[300 °C] 3ClF}$$
$$\ce{Br2 + BrF3 -> 3BrF}$$
$$\ce{I2 + IF3 ->[-78 °C][CCl3F] 3IF}$$

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text:
Two chemistry inquiries
Here is my second question, I know 1000 grams of water is 1 liter. But does the density affect the result? What if the substance is mercury?
When you add seed crystal in to a supersaturated solution, the solute comes out. Does this make it saturated? Is what came out the extra solute that make it supersaturated? Or is it all the solute? This is my first question.

These are completely unrelated questions and should be asked separately.

$1000 \, \text{g}$ of water occupy a volume of (approximately) $1 \, \text{L}$ because the density of water is (approximately) $1000 \, \text{g/L}$.
If you had $1000 \, \text{g}$ of mercury, whose density is $13534 \, \text{g/L}$, it would occupy a volume of $0.0739 \, \text{L}$.
Usually you can only dissolve so much of a solute into a solvent until a certain threshold is reached.
When this threshold is reached you have a saturated solution.
After that point, it is thermodynamically more favourable (under the given conditions, e.g. temperature, pressure, etc.) for any extra solute not to get dissolved in the solvent but clump together and precipitate.
But if you use some special techniques you can prepare a supersaturated solution that contains more solute than it should, thermodynamically speaking.
A supersaturated solution is a meta-stable state, meaning that it is only a local energy minimum for the system, not the global minimum.
But it would like to get to the global minimum if it gets the chance.
The only thing preventing that, is a small energy barrier that has to be surmounted.
So, if you disrupt the supersaturated solution in any way, like shaking it or adding a crystal seed, the solution can surmount the energy barrier and attain its global minimum state, i.e. the saturated state.

Ok, after the crystal seed was added, the supersaturated solution becomes saturated?

@user132522 Yes.

And also, talk about recrystalization, or any other way to retrieve the solvated particles.

@user132522 I'm not sure, if I understand you right, but if you want to retrieve the solvated particles you have to get rid of the solvent or change the condition such that the solubility of your solute decreases and it precipitates.

Do you know the graph of solubility of salts.

At the same temp, how could you decrease the concentration of the solvent in a solution?

One of the ways I could think of, is lower the temperature of the solution, drop a crystal to turn it saturated, and reheat to the previous point of temperature.

@Apler I haven't said anything about not raising the temperature. You could distill the solvent and thus seperate solute and solvent. This is a completely different question by the way.

And then you'll end up with the a solution at the same temp as it was at the supersaturated state, but with a lower concentration.

I ll probably make another post in the future. Thanks

@Apler Yeah, that might be best. I'm not sure if I understand what you are getting at and in a new question you can explain better what you want to do and why.

@Apler And by the way, since you're new to this site, have a look at the about page to familiarize yourself with the workings of Chemistry.SE.


text:
How does the activity of Sn chemical potential vary as a function of composition at 900 C?
This is referring to the Cu-Sn phase diagram in which composition varies along a horizontal line at a certain temperature. I understand that as you move along that line different intermetallic or liquid phases form, but from a activity point of view does the activity go up because of different crystallographic arrangements that are present at different compositions?

Do you mean $\ce{Sn}$ (as you wrote in the title) or $\ce{Si}$ (as you wrote in the body of the text? Furthermore, it would be helpful if you could include a picture of and/or a link to the phase diagram you are talking about.

http://www.doitpoms.ac.uk/tlplib/microstructural_exam/cualloys.php

Just above the line, in the two-phase field where $\epsilon$ is in equilibrium with the liquid, the chemical potentials of both Cu and Sn are constant (very likely not equal to each other mind you).  This is a consequence of the tie-line on the Gibbs free energy diagram - check out a book on binary alloy thermodynamics such as Porter and Easterling, Phase Transformations in Metals and Alloys.
At the peritectic point itself, the tie line connects all three phases, $\epsilon$, $\eta$ and the liquid, so the chemical potentials for each element are equal. Dropping the temperature just a smidge leads to two separate two-phase regions, one of $\epsilon$ with $\eta$, and one with $\eta$ and liquid.  The chemical potential of Sn (or Cu) is different in each of those two-phase regions.
Again, Porter and Easterling would be a good reference.


text:
Why does the equilibrium constant depend on the temperature but not on pressure and concentration?
According to my textbook (and intuitively) certain changes when the aforementioned 3 variables are altered occur in accordance with Le Chatelier's Principle. However, what I don't understand is what makes temperature change alter the equilibrium constant while the other two do not. All three cause changes in the equilibrium position (the state of concentrations at which equilibrium is achieved with lowest possible stress in the system), so why don't they all also change $K_{eq}$?

Please have a look at this answer of mine. It contains the derivation of the defining formula of the equilibrium constant.
\begin{equation}
\log \underbrace{\prod_i [a_{i}]^{\nu_{i}}}_{= \, K} = -\frac{\Delta G^{0}}{RT} \qquad \Rightarrow \qquad \log K = -\frac{\Delta G^{0}}{RT} \ .
\end{equation}
This formula shows the temperature dependence explicitely, so it is clear that $K$ must depend on $T$.
Apart from the temperature this formula also relates $K$ to the standard Gibbs free energy of reaction $\Delta G^{0}$.
This quantity is defined for specific standard conditions involving standard pressure and equilibrium concentrations.
Since there aren't any other quantities contained in the formula (apart from the constant $R$) it follows that $K$ is defined for this specific pressure and these specific concentrations too and so it has no pressure or concentration dependence.


text:
Calculating concentration of an ionic compound
$n$ liters of $k$ M $\ce{NaCl}$ are poured into $m$ liters of $l$ M $\ce{NaCl}$. Determine $\ce{[NaCl]}$, $\ce{[Na^+]}$, and $\ce{[Cl^{-}]}$.

Hi Okarin, please show us your efforts, what have you done so far? Have a look to the homework policy in this site!

Keep it as simple as possible :)
If both initial solutions are not saturated and all the $\ce{NaCl}$ is dissolved, this will not change upon mixing.
If in doubt with the result of a calculation, have a look at the units!
Concentrations are given as the quotient of moles and volume, $c = \frac{n}{V}$, the unit apparently is $\mathrm{mol \cdot l^{-1}}$. The absolute amount of $\ce{NaCl}$ in the first solution is therefore given as $n = c\cdot V$. (Note that this is not the n in your question ;).
The new $\ce{NaCl}$ concentration is the sum of the moles, divided by the new total volume:
$c = \frac{(k\cdot n)+(l\cdot m)}{m+n}$
Unless given further information, it is safe to assume that $\ce{NaCl}$ still is fully dissociated after mixing.

Well in my opinion you do need the dissociatiion constant for NaCl
                         `NaCl          ->Na    +        Cl`

initial conc.             k*n+m*l          0              0
change at equilibrium     -a(k*n+m*l)    +a(k*n+m*l)     +a(k*n+m*l)
final conc.               k*n+m*l(1-a)   k*n+m*l(1+a)    k*n+m*l(1+a)

where a is the degree of dissociation
Kb=  2*(k*n+m*l)(1+a)/k*n+m*l(1-a)=   2*(1+a)/(1-a)
where kb is the dissociation constant for nacl
so when you have the "kb" the you can find "a" and then put it in the formula under Na or Cl at final conc. to get their concentration


text:
why do battery leaks still happen
Battery leaks was an issue in the 80s and 90s, and since quality and innovation is constantly improving, I thought battery leak might be slowly going away. But I found that even nowadays, name brand alkaline batteries still leak after a while, even if originally concealed in a tight package of the metal shell. Is it true that the chemical in an alkaline battery inherently expands in volume over time, and make the alkaline battery leak issue unavoidable?
also why does the alkaline expand in volume over time?


text:
lanthanide contraction
"The Lanthanide Contraction refers to the fact that the 5s and 5p orbitals penetrate the 4f sub-shell so the 4f orbital is not shielded from the increasing nuclear change."
First, 5s and 5p are filled before 4f, so if the atom wants to give an electron out, which energy shell does the electron come out of? How 5s and 5p penetrate 4f? Isn't 4f outer than 5s and 5p?

Have a look here. This should give you an idea why 5s is so contracted that it "penetrates" 4f (I'm not completely sure whether it is right to say that 5p penetrates 4f too, but if so the principle is the same).

According to Aufbau principle, for multi-electron atoms, in the ground state of an atom, an electron enters the orbital of lowest energy first and subsequent electrons are fed in the order of increasing energies.
For multi-electron atoms, the following sequence is observed for orbitals in the increasing order of energy: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s.
Here the numbers 1, 2, 3, 4.. indicates shell number. It may be remembered that, though filling of electrons takes place according energy of orbitals, when electrons are removed from the neutral atom, electrons with highest value of n are removed first. So, electrons from 5p or 5s is removed first than from 4f.
To understand lantanide contraction, lets consider an example. If you consider Pd (of 5th period) and Pt (of 6th period), you would expect Pt to have relatively large radii than Pd. In practice, they have roughly same radii. This is because, Pd has 4d electrons and Pt has 5d and 4f electrons. Due to diffused shaped of 4f electrons, they can't shield 5d electrons of Pt. Thus, contracting the size of Pt roughly same as Pd.
Considering your last question, no 4f subshell lies closer to the nuclei than  5s and 5p. For more info, I would suggest you to read this page: Lanthanide Contraction

So does it have anything to do with 5s and 5p penetration?


text:
Pressure and LeChatelier's Principle
We know that if a system's equilibrium is exposed to a stress, the system shifts to relieve that stress. According to my teacher, certain kinds of pressures are stresses and others are not.
Normally, if we just say that the total pressure of a container with a reaction occuring inside is increased, the reaction shifts toward the side with a small sum of the mole ratios (i.e. $\rm 2A +3B \longleftrightarrow C + D$, if increasing pressure reaction shifts to the right to decrease pressure). I understood this formerly, but another case erases this understanding.
My teacher also said that if we add a noble gas to the container in which this reaction is occurring, although the pressure increases because of a new gas, the equilibrium does not shift to counteract this pressure increase. She says it has to do with the partial pressures of the reactants and products of only the particles In the reaction, but not in the total balloon. But this makes no sense to me because by increasing the total pressure of the system as described before, how does this cause an equilibrial stress and this case doesn't? How does increasing total pressure even cause an equilibrium shift? By increasing total pressure we have a proportional increase in the collision rates of all particles involved, do we not? If this is true, then why wouldn't the reaction rates both be doubled, and hence cause the equilibrium to not shift?
If I'm missing some fundamental principle, please let me know.
Thanks!

Le Chatelier's principle is not about pressure, it is about concentrations and temperature (Please also see the good article on Wikipedia for Le Chatelier's principle).
Le Chatelier's principle is also called "The Equilibrium Law", that can be used to predict the effect of a change in conditions on a chemical equilibrium. Chemical equilibrium means that reaction is reversible: it can go as forward, so backwards.
$$\rm 2A+3B \longleftrightarrow C+D;\space \Delta H=-100\,kJ\cdot mol^{-1} (for\space example) $$
Lets imagine that $\rm A,\,B,\,C,\,D$ are gases. So you wrote correctly that a pressure increase will cause the reaction equilibrium to shift to the "right" - more products $\rm C,\,D$ will be formed. When we increase the pressure in the system, it means that we reduce volume. We squeeze our system, and there is less volume for molecules to be in. They hit each other more frequently, and the probability of product formation increases. When you reduce the volume by increasing the pressure, you basically increase the concentration of reagents. The same amount of atoms or molecules occupy lesser volume.

I'll make a silly comparison. Let's take this reaction:
$$\rm Boy+Girl\longleftrightarrow Pair$$
If there is too little space on the dance-floor to dance on your own it could be a good idea to find a partner to reduce the volume you both occupied to feel much more comfortable. But when the other people start to leave the club you can unpair with your partner...

Lets take 2 mol of $\rm A$, 3 mol of $\rm B$, 1 mol of $\rm C$, 1 mol of $\rm D$ at room temperature. We will put it in the jar of $\rm 156.8\,L$ ($\rm (2+3+1+1)\,mol \cdot 22.4\, L\cdot mol^{-1}=156.8\,L$). Pressure in the jar is $\rm 1\,atm$.

$P(A)=X(A)\cdot P(total)=2/7\cdot1=0.286\,atm$
$P(B)=X(B)\cdot P(total)=3/7\cdot 1=0.439\,atm$
$P(C)=X(C)\cdot P(total)=1/7\cdot 1=0.143\,atm$
$P(D)=X(D)\cdot P(total)=1/7\cdot 1=0.143\,atm$

Our system is in equilibrium, so we can calculate Equilibrium constant: $$K_P=\dfrac{P(C)^1P(D)^1}{P(A)^2P(B)^3}=\dfrac{0.143\cdot0.143}{0.286^2\cdot0.439^3}=3.17\,atm^{-3}$$
$$K_P=\dfrac{(X(C)\cdot P(total))^1(X(D)\cdot P(total))^1}{(X(A)\cdot P(total))^2(X(B)\cdot P(total))^3}=\dfrac{X(C)^1X(D)^1P^{\Delta n}}{X(A)^2X(B)^3};\,\Delta n=n(C)+n(D)-n(A)-n(B)=1+1-2-3=-3$$
Now lets double the pressure in the jar, by reducing the volume, the temperature did not change so $K_P$ stays the same. Lets calculate the molar concentrations from:
$$K_P=3.17\,atm^{-3}=\dfrac{X'(C)^1X'(D)^1\cdot (2\,atm)^{-3}}{X'(A)^2X'(B)^3},$$
 where

$\upsilon'(A)=2\,mol-2x\,mol$
$\upsilon'(B)=3\,mol-3x\,mol$
$\upsilon'(C)=1\,mol+x\,mol$
$\upsilon'(D)=1\,mol+x\,mol$

Total mols: $\sum=(1+x)+(1+x)+(2-2x)+(3-3x)=(7-3x)\,mol$

$X'(A)=(2-2x)/(7-3x)$
$X'(B)=(3-3x)/(7-3x)$
$X'(C)=(1+x)/(7-3x)$
$X'(D)=(1+x)/(7-3x)$

Solving the eqution gives $x=0.325\,mol$, so now:

$\upsilon'(A)=1.35\,mol$
$\upsilon'(B)=2.03\,mol$
$\upsilon'(C)=1.33\,mol$
$\upsilon'(D)=1.33\,mol$

But it the begining we has $A$=2 mol, $B$=3 mol, $C$=1 mol, $D$=1 mol. The reactions shifted to the right.

$P'(A)=0.45\,atm$
$P'(B)=0.67\,atm$
$P'(C)=0.44\,atm$
$P'(D)=0.44\,atm$

Ok, but when we add inert gas, what happens:
$$\rm 2A+3B+inert\,gas \longleftrightarrow C+D+inert\,gas $$
It does not do anything because even though the pressure increased, the probabilty for 2 atoms $\rm A$ to hit 3 atoms $\rm B$ does not change. The partial pressures stay the same. Why is it so? It can be easily shown. For example we have this reaction:
$$\rm A_g+B_g\longleftrightarrow C_g$$
Lets say we put this reaction in a jar of 134.4 L volume at room temperature, we have 3 mol of $\rm A$, 2 mol $\rm B$, and one 1 mol $\rm C$. 6 mol of gases being put in 134.4 L at RT will give us 1 atm pressure.
$$6\,{\rm mol} \cdot 22.4\,{\rm L\cdot mol^{-1}}=134.4\,{\rm L}$$
The partial pressures are:

$ P(A)=3/6\cdot P_{total}=0.5\cdot1\,atm$
$ P(B)=2/6\cdot P_{total}=0.33\cdot1\,atm$
$P(C)=1/6\cdot P_{total}=0.17\cdot1\,atm$

Now we will add 1 mol of inert gas:
$$\rm A_g+B_g+inert\,gas\longleftrightarrow C_g+inert\,gas$$
Now, the partial pressures are (pressure in the jar increased, because quantatiy of molecules increased, but volume stays the same):

$P(A)=3/7\cdot P_{total}=3/7\cdot 1.17\,atm=0.5\,atm$
$P(B)=2/7\cdot P_{total}=2/7\cdot 1.17\,atm=0.33\,atm$
$P(C)=1/7\cdot P_{total}=1/7\cdot 1.17\,atm=0.17\,atm$

As you can see, the partial pressures did not change (concentration did not change). That is why there is no effect.

Let's take this reaction:
$$\rm Boy+Girl+Cat\longleftrightarrow Pair+Cat$$
You cannot dance with a cat. Adding extra cats on the dance-floor will increase the pressure in the crowd, but will not increase the probability of formation of a dancing pair.

Let's imagine that $\rm A$ and $\rm B$ are liquids, and see what a pressure increase will cause. Usually it does not cause a lot, because it is very difficult to squeeze liquids. No matter how you try, the volume which molecules of liquid occupy stays the same. So increasing the pressure in reaction:
$$\rm A_\ell+B_\ell\longleftrightarrow C_\ell$$
will not do practically anything, because concentrations of reagents will not increase under the pressure.
When I wrote $\Delta H=-100\,kJ\cdot mol^{-1}$ (the enthalpy is negative), it means that in reaction the heat is released.
$$\rm 2A+3B \longleftrightarrow C+D;\space \Delta H=-100\,kJ\cdot mol^{-1} (for\space example) $$
So if we decrease temperature in the jar, reaction will move to the right. It can be understood using this primitive logic. Assume that heat is some substance, that could be treated as "product". If you remove this so called "product" (heat, by temperature reduction) you constantly force reactin to happen from left to right. In reality the equilibrium is just shifted from left to right.
If we increase temperature, it is like we added more of this so called "product" (heat) in the jar, so reaction is moved to the left.
If $\Delta H>0$, you need to think about heat not as "product", but as "reagent".
For better understanding try to "dive" into Equilibrium thermodynamics.

I like the astute observation about the Principle's not applying to pressure. Pressure simply influences concentrations, so this is why the shift occurs. Also, though, I need to correct you in the second example about partial pressures. The end idea is still right, but you had an arithmetic error - Ptotal=2/7⋅1.17atm=0.5atm should be = 0.33 atm.

@user11629 Thank you for correction. And one remark, chemistry is about concentrations always, if something can change concentration it will make effect on reaction. Chemistry is also about activation energies (about temperatures or light).

But one thing that you haven't explained is why decreasing the volume does not affect both reactants and products equally, so why does pressure even cause an equilibrial stress if the likelihood that both reactions will occur is doubled? This is the same effect as a catalyst, which makes both the forward and backward reactions occur faster - and this effect has no net change on the equilibrium position.

@user11629 I did not catch what you are asking?

Two rates are equal. More reactants than products exist at equilibrium. We double pressure. Why does this cause a change in the equilibrium position at all? You said it increases the partial pressures of the reactants, which is true, but it also increases the partial pressure of the products, so the ratio between pressures of reactants and products are the same.

@user11629 see the answer. Hope it is clear now.

@saldenisov When you say "lets double the pressure" you mean in instantaneousy?

I finally figured it out using simple logic for myself as follows. Thank you for the answer though!
Consider an equilibrium as described:
$K_{eq} = \dfrac{[A]^n}{[B]^m}$
If we double the pressure of the system at large, i.e., halving the volume, we double the concentrations of products and reactants. If $n=m$, Le Chatelier's Principle is proven correct because no change occurs, as the mole sum is the same for products and reactants. If $n\neq m$, Le Chatelier's Principle is also correct; doubling and squaring, for instance, will not yield the same resulting as doubling and cubing the concentrations.
In this way we have a short, comprehensive, and intuitive understanding of the reasoning behind Le Chatelier's Principle with regard to concentration changes caused by pressure.

Well, but one thing is to change the overall pressure by changing the volume (that has an effect on the equilibrium, since partial pressures/concentrations do vary), another is to change the overall pressure by adding an inert gaz without changing the volume (that has no effect on the equilibrium, since partial pressures/concentrations do not vary). So, I do not think your reasoning (which is correct) applies to the arguments you were discussing in your question.


text:
Which two substances react spontaneously?

Which two substances react spontaneously?
(A) $\ce{Ag}$ and $\ce{Cu}$
(B) $\ce{Ag+}$ and $\ce{Cu}$
(C) $\ce{Ag}$ and $\ce{Cu^2+}$
(D) $\ce{Ag+}$ and $\ce{Cu^2+}$
$\tiny{\text{Taken from 2012 Chem Olympiad}}$

The given answer is B, yet I am unclear as to why. I believe it has something to do with reduction potentials/redox or the gibbs free energy. Can someone please explain?

Do you have data regarding the reduction potentials of these species and elementary idea about what these values suggest? This can only be solved with known reduction potential data.

This was from the 2012 chem olympiad and they do not provide a table of reduction potentials (and you obviously can't bring your own :P). So in the absence of that info how would you solve it?

You then need an idea about their reduction potentials, or equivalently, their relative position in the electrochemical/activity series.

From the faint outline of electrochemical series I remember $Cu$ is above $Ag$. The substance which is above is able to displace the substance below from its aqueous solution(obviously lower substance is in ionic form).
Hence $(B)$ is true.
We were made to learn this outline in lower class :
K Na Ca Mg Al Zn Fe Pb H Cu Hg Ag Au Pt

By displace you mean act as a reducing agent?

Yep...btw kids are taught in the way of simple displacement reaction. In higher classes, they introduce the electrode potentials and other stuff. Like : $Cu + AgNO_3$ ...


text:
Steam distillation- how it actually works?
To obtain phenylamine after it is formed from the reduction of nitrobenzene with tin, we carried out steam distillation.
This is what I learned in class:

"During the process,steam is injected into the flask (containing the reaction mixture) to create artificial vapour pressure on the surface of the liquid. When vapour pressure is equal to atmospheric pressure, the liquid starts to boil at 95 degrees"

This didn't make much sense to me. So I searched it up on the net and I  got another explanation (source):

As the hot steam passes through the mixture it condenses, releasing heat. This will be enough to boil the mixture of water and phenylamine at 98°C provided the volume of the mixture isn't too great

Here, it is saying that the heat produced from the condensation of the steam will allow the mixture to boil. I am really confused, I mean I carried out steam distillation but how is it different from normal distillation and how is phenylamine collected at a lower temperature through this process?

There is similar question here: http://chemistry.stackexchange.com/questions/6985/internal-vs-external-steam-distillation
Have you looked there?

You boil the water and the product co-distills.  There is an anomalous selective interaction between steam and many hydrophobes.  If you heat the pot you deplete the water contained and may pyrolyze contents on the hot wall.  Blowing in steam limits temp to near 100 C, moves a lot of energy into the flask and out (540 cal/g enthalpy of vaporization), and maintains the water content.
You could filter off the solids, acidify the aqueous raffinate,  extract non-basics, then alkalinize and extract the aniline.  Since aniline is air sensitive to oxidation and extremely toxic by skin contact, you can appreciate advantages of steam distillation.

Water and any rather non-polar organic substance form a "positive" azeotrope, which lowers the boiling point, and makes a steam that contains both substances. You take a large excess of water, so you are quite far from the point where you have the actual azeotropic mixture. Thus the whole thing boils essentially like water, i.e. you don't go far from 100°C.
It doesn't matter if the two substances (water and diphenylamine in your case) mix in the condensed phase, or if one is even a solid. Except that a solid will tend to clog your condenser, unfortunately. ;-)
One well known example is extracting water from an esterification reaction by boiling it off with toluene: They separate again after condensation, and you can let the lighter toluene phase flow back into the reaction vessel. Same principle, only water is the minor compound in this case.

Hm, perhaps we are in fact at (or very close to) the azotropic mixture. After all, it has the lowest boiling point. Comments?


text:
Is there a way to get hot water without altering plumbing or electrical wiring too much?
Is there a way (other than collecting water and heating it on the stove ) to get hot/warm water from the tap without having to alter existing plumbing or electrical wiring ?
Perhaps use a chemical reaction to provide the heat to warm the water ?

Welcome to Chemistry.SE. This question does not appear to be about chemistry, as defined in our help center. It is likely to be put on hold for that reason. If your intent was to use a chemical reaction to provide the heat to warm the water, we need to know that in the question and it would be on topic. Otherwise, perhaps this question belongs at Home Improvement?

@BenNorris I edited the question to include the verbiage you used in your comment. Hope it is sufficiently on topic now.

Crystallization of a supersaturated solution of sodium acetate will release heat - once. But then it is just a solid. It works nicely for pocket warmers, but since you have to heat the flexible container in boiling water prior to re-use, this and other comparable processes are of little use for a water supply.
At the end of the day, you won't get around plumbing to tap an energy source.
Your options would be:

solarthermal
geothermal
heat released from the fermentation of organic waste

Yes I did think of hand warmers but the idea of pre-boiling it was not appealing. Have you heard of coffee joulies? (joulies.com?) These things lower/maintain temperature. Wonder if a similar thing exists for increasing the temperature of water/fluids.

You can always dissipate the heat and loose it to the environment, but harvesting energy against a temperature gradient does not work. Putting your feet in a cold creek to warm them up while the water cools down is not an option. Thermodynamics is a harsh mistress ;)


text:
What are some examples of a clear, liquid, UV photoactive polymer?
Something that is not reactive to lesser wavelengths of light, specifically IR.

Do you want a material that does not react with IR radiation (most materials do not react with IR)? Or do you want something that is transparent to IR?

"Lesser wavelength" or IR: You can only have one. Do you really want a photoactive polymer or a photoinitiated polymerizaton?

Methacrylates make nice starting materials for the lightinduced polymerization.
In methyl methacrylate (MMA), camphorquinone (a typical initiator) has an absorption maximum around $\lambda$ = 473 nm.

You mean "monomer" or "oligomer" not "polymer."  You are making the polymer by polymerization or crosslinking.  UV-curing optical adhesives.  You choose nice (mix of) free radical-cure monomers, then dissolve a photoinitiator like camphorquinone, a diarylketone, a benzoin ether, or a benzil ketal.  The molecule fragments to free radicals after absorbing a UV photon. That sets off the polymerization.  Exclude oxygen (triplet diradical quench).  Cage recombination of the intiator is your enemy.
A microwave-pumped electrodeless mercury bulb the size of your finger can output a kilowatt of UV.  Beware of pinholes in your irradiation chamber, including entry and exit if continuous.  And now, a second of silence for the Trommsdorff effect.
If you want a polymer as such to do be photoactive, then something like Prog. Polym. Sci. 28 729 (2003).  Pull it up in a university library to get the graphics.


text:
Is ion activity in a salt solution pH-dependent?
:) Without being a chemist myself, I need some basic understanding of the ion activities in my buffer solutions.
Now I am wondering whether the ion activity of the K$^+$ and Cl$^-$ ions is different for the buffers with different pH-values.
When I google for that, all I get is ion activity for the hydrogen ion, which -I think- does not interest me at all.
If the ion activity actually IS different, could you point me to some table to look up the correction values?
Thank you!
.
edit: tl&dr:
The buffer solutions contain:
Buffer A: 250mM KCl, 10mM MOPS
Buffer B: 020mM KCl, 10mM MOPS
pH is set to 5,6,7 resp., using unknown amounts of Tris.
Is the ion activity of K$^+$ and Cl$^-$ affected by the pH?

Considering the effect of $\text{pH}$ on ion activity, one should take into account that $\ce{K+}$ and $\ce{Cl-}$ are respectively conjugated acid of a strong base and conjugated base of a strong acid. They are therefore both weak, with no influence on the $\text{pH}$.
As a consequence, I would basically assume that also the other way round is true and that $\text{pH}$ has no effect on their activity.


text:
How to determine which gas would have the lowest pressure?
If the volume of each sample of gas was reduced to one-tenth of its original size and the temperature remained the same, which gas would have the lower pressure. I'm getting confused because my options are $\ce{C6H4}$, $\ce{C3H4}$, $\ce{N2}$, $\ce{O2}$, and $\ce{Cl2}$. Our teacher told us to pick the polar compound but the problem is all of them are non polar. Would the answer be $\ce{C3H4}$? I have a feeling I might be drawing the lewis structure wrong because I'm positive the other four are non polar.

Let's not forget allene as the alternative structure for $\ce{C3H4}$, and - at least in theory - $\ce{C6H4}$ could be bis-cyclopropenylidene ;)

Not all of them are polar. The gases  $\rm N_2$, $\rm O_2$,  $\rm Cl_2$ are not polar.
The $\rm HC\equiv C-CH_3$ is polar, but $\rm H_2C=C=CH_2$ should not be and $\rm C_6H_4$ is also polar, if it is right chemical formula you wrote. But I'm not sure which is more polar, but I will say that it is $\rm C_6H_4$
.
In more polar compound there will more interaction between molecules and that is why pressure will be smaller, simply because they "want" to be liquid form - not in gas.

sorry I meant to say that they were all non polar. my bad. I'll look into the polarity of both the molecules. Thanks for your help!

I'm quite sure benzyne is not the intended compound for $\ce{C6H4}$, as it only exists briefly as a reaction intermediate. I'm don't know what other structure is more plausible, though (but as Klaus mentions, there are other possibilities, all of them rather weird). I'm half-convinced that $\ce{C6H4}$ is actually supposed to be $\ce{C6H14}$, which could be good old hexane (among other isomers). Though really, the teacher should have realized that only giving the formulae is a bad idea.

@ Nicolau Saker Neto But you can buy it! ^) http://www.chemspider.com/Chemical-Structure.109690.html

When it comes to chemical suppliers, there's a world of a difference between offering a product and providing a product!

@saldenisov The supplier listed at ChemSpider does not really offer it. I am not surprised ;)

A homodiatomic gas cannot be polar, by symmetry arguments. At best it can be polarizable by dispersion forces overall.  The lowest pressure would be the most strongly attracting molecules if any (e.g., water and hydrogen bonding), the most inert molecules, the smallest molecules (van der Walls equation coefficients), and then "anomalous" behaviors.  Look up the coefficients (CRC Handbook)to get the real world answer.
One would guess nitrogen for its being very tightly bound and all electrons bonded.  Look it up, as above.
/_\PV = energy, 101.325 J/liter-atm. Adiabatically compressing a gas tenfold will heat it.  A diesel cylinder compresses 15 - 22 to one, minimum 550 C at the top of the stroke.

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


text:
Approximate time for pure water pH to equilibrate in open air?
I'm a complete non-chemist, so I don't know how to calculate this,  or even what to google for (my searches yielded esoteric research papers).
I'm doing some hobby experiments, and I'd  like to know what approximate timeframe to expect for relatively pure (distilled or double distilled) water left in open air at around 24 °C/75F to reach whatever the pH equilibrium point is (I read around 5.5 somewhere).
I'm not looking for a giant equation please,  just a quick approximate number.
Thanks

Assume you have poured your distilled water in a 1 liter container.
We write the flux of $n_{\ce{CO2}}$ between air towards your container, as
$$
J_{\ce{CO2}}=\frac{dn_{\ce{CO2}}}{dt}\times\frac{1}{A}=\frac{d[\ce{H2CO3}]^{*}}{dt}\times\frac{V}{A}
$$
where $n_{\ce{CO2}}$ is the number of ${\ce{CO2}}$ moles in water, $A$ the water surface and $V$ the water volume.
Assuming $\ce{CO2}$ atmospheric pressure equal to $10^{-3.5}\,atm$, we obtain the concentration $[\ce{CO2}]_{air}$ approximately equal to $10^{-5}\,M$ (the latter value obtained via $PV=nRT$).
We call $[\ce{H2CO3}]^{*}$ the sum of $[{\ce{CO2}}]_{water}$ and $[\ce{H2CO3}]_{water}$ concentrations in water: at the beginning, we set it equal to $0$. Note that $[\ce{H2CO3}]^{*}$ could be reasonably written as:
$$
[\ce{H2CO3}]^{*}\approx[{\ce{CO2}}]_{water}
$$
indicating that the majority of carbon dioxide in water is not converted into carbonic acid, being the kinetics for such a conversion very slow, as pointed out by @Nicolau Saker Neto, .
By expressing $J_{\ce{CO2}}$ as:
$$
J_{\ce{CO2}}=\frac{D_{{\ce{CO2}}}}{Z}\times ([\ce{CO2}]_{air}-[\ce{H2CO3}]^{*})
$$
with $D_{\ce{CO2}}$ the diffusion coefficient of ${\ce{CO2}}$ equal to $7.2\times10^{-4}\,dm^{2}h^{-1}$and $Z$ the thickness of superficial layer through which the exchange occurs. For the latter, we take $40\times10^{-5}\,dm$.
Putting previous equations together (with $A/V=1\,dm^{-1}$), the following is obtained:
$$
\frac{d[\ce{H2CO3}]^{*}}{dt}=1.8\times10^{-5}-1.8\times[\ce{H2CO3}]^{*}
$$
which yields:
$$
[\ce{H2CO3}]^{*}=1\times10^{-5}(1-e^{-1.8t})
$$
The graph below shows that about $2$ hours later, a plateau is established, for $[\ce{H2CO3}]^{*}= 1\times10^{-5} M$.
By considering the acid-base equilibrium ($pK_{a}=6.3$):
$$
\text{H}_{2}\text{C}\text{O}_{3}^{*} \ce{<=> HCO3- + H+}
$$
we obtain a $\text{pH}=5.7$, at the plateau.

UPDATE 1: the value for $Z$ has of course a big impact on time. A larger value will increase the time needed to reach the plateau. I have used $40\times10^{-5}\,dm$, considered a "typical value" by the authors of this book (sorry in French, page 181, 3rd edition). Any more accurately derived value for $Z$ is welcome !
UPDATE 2: I used the notation $\text{H}_{2}\text{C}\text{O}_{3}^{*}$ to represent the two species $\ce{CO2}$ and $\ce{H2CO3}$, in water.

I'm not too good with diffusion and kinetics, but this answer seems to assume that acidification of pure water is limited by the diffusion of $\ce{CO2}$ into the liquid. However, the actual reaction $\ce{CO2 + H2O → H2CO3}$ is quite slow, having a rate constant of $0.039\ s^{-1}$. Do the reaction kinetics have no impact on the acidification speed?

@NicolauSakerNeto Definitely you are right: I should have used (and I will do) the notation $[\ce{H2CO3}]^{*}$ as suggested by http://en.wikipedia.org/wiki/Carbonic_acid to indicate the overall concentration of the two species, $\ce{H2CO3}$ and  $\ce{CO2}$, the latter being definetely dominating, as you say. But this will not alter the final $\text{pH}$, being $\ce{CO2}$ pressure the only factor, determining the composition of the solution (http://en.wikipedia.org/wiki/Carbonic_acid#pH_and_composition_of_carbonic_acid_solutions)


text:
Ground state electron configuration of chromium
What is the ground state electron configuration of chromium?
Is it $\ce{[Ar]}4s^23d^4$ or
Is it $\ce{[Ar]}4s^13d^5$

At worst, your chances are 50:50. If you remember the rules concerning half-filled and filled d orbitals, they are much better ;)

@Philipp my textbook says the answer is
[Ar]4s2 3d4
But when i googled it, the answer was opposite.
So i am confused..

@Googleuser Hmm, sorry then. What textbook are you using? Usually it is a rather well known thing that chromium has $\ce{[Ar] 4s^1 3d^5}$. Maybe you should consider using a different textbook.

See http://chemistry.stackexchange.com/questions/151/why-do-elements-in-columns-6-and-11-assume-abnormal-electron-configurations

WebElements page about chromium (and a number of resources) agree with the comment by @Philipp:

The ground state electron configuration of ground state gaseous neutral chromium is  $\ce{[Ar]}3d^54s^1$

Which in some resources is written as $\ce{[Ar]}4s^13d^5$

Based on the Royal Society of Chemistry article The trouble with the aufbau principle:

it appears that the most stable configuration for atoms of chromium, copper, niobium, molybdenum, ruthenium, rhodium, silver, platinum and gold involves only moving one electron into an $s$ orbital.

Chromium is one of a handful of Transition elements that share this electron configuration.

Chromium and copper are examples of elements with "anomalous" electron configurations, meaning that they don't follow the normal rules we use for populating the configurations of other elements.
The commonly given reason for this is that the energy of a shell is minimized when the number of electrons with the same spin is maximized (Hund's rule). As a result, when the energy levels of two successively filled sub-shells are already close together (as they are with 4s and 3d sub-shells), the slightly favored half-filled configuration can "win" over the energy increase needed to move an electron to an even-more-slightly higher energy level. In the case of chromium, this means that one of the 4s electrons will go to the 3d orbital, resulting in two half-filled sub-shells where all electrons within each sub-shell have the same spin.
In the case of copper, a similar thing happens. The difference is that the 4s electron moves into an almost-filled 3d shell in order to completely fill it. You get a slight energy decrease when all electrons are paired within a sub-shell. This, in combination with the decrease obtained from achieving a half-filled s orbital, ends up being enough to overcome the increase in energy required to move that electron to the 3d orbital in the first place.
It would be nice if these empirical rules were consistent across the entire table, but unfortunately they are not. If you look up the actual electron configuration for other d- and f-block elements, you will see there are some patterns, and similar things happen for other elements, but because they are so dependent on the delicate balance between energy levels, it is not possible to reliably predict them with simple rules for all elements. In "real life" we use spectroscopy and quantum mechanical calculations to find the actual electron configurations.
However, since chromium and copper are common enough and reliably predictable with simple rules, we tend to use those as classroom examples to demonstrate that the reality of electron configurations is more complex than the simple rules we give you in school.


text:
How to chose wavelength in spectrophotometry? How to use Beer-Lambert Law?
How is the wavelength chosen to measure the concentration of an ion determined? and What does the wavelength correspond to in an atom/ion?
When light is shined on a solution what happens to the ions in solution? Include how and why light is absorbed or emitted from a substance.
Would a high or low concentration absorb more light?
How does the beer-lambert law relate to concentration and information from the spectrophotometer?

Welcome to Chemistry Stack Exchange. I asked for a modification of your title. Could you please read information page to be aware of Stack Exchange standards. In order to keep a good Q/A quality you must redact your question or answer in a fashion that is suitable and valuable to everybody.

Thanks @jlandercy, that was a wording goof on my part :)

Working wavelength is chosen by analysing the spectrogram $A(\lambda)$. Such diagram is obtained by measuring absorbance $A$ in function of wavelength $\lambda$. Practically, you test a moderate concentration of your analyte and sweep the complete wavelength bandwidth of your analyser. Then you select wavelength where a maximum of absorbance occurs. If multiple maximum are present, you may prefer the highest one (best sensitivity) or a smaller one with less interferences (best selectivity). Both can happens simultaneously. The main criterion is:
$$ \underset{\lambda}{\max}\left\{A(\lambda)\right\} $$
For information: the wavelength is linked to photon energy $E$ by the following relation:
$$ E = h \nu = h \frac{c}{\lambda} $$
Where $\nu$ is the frequency of the EM wave, $c$ the speed of light and $h$ the Plank constant.
Absorbance is a property of matter, when an atom/ion/molecule absorbs a photon (visible or nearly visible IR/UV), it can promote electron to a highest energy level (UV, highest energy) or increase vibrational energy (IR, lowest energy) of the compound and the photon is destroyed.
A spectrophotometer assess the transmittance $T$ (ratio of transmitted $\phi$ and incident flux $\phi_0$ expressed as powers) of light beam through a sample cell.
$$ T = \frac{\phi}{\phi_0} $$
And transmittance (dimensionless) is linked to absorbance (dimensionless) in the following way (functional scaling to achieve linearisation between signal and physical quantity):
$$ A = -\log_{10}(T)$$
Beer-Lambert Law is a limit law conditioned by several hypothesis and it is only valid for low concentrations. When Beer-Lambert applies you can linearly link concentration $x$ with absorbance.
$$ A(x) = \epsilon(\lambda)lx $$
Where $\epsilon$ is a proportional coefficient (with suitable units) called extinction coefficient which describes how matter absorbs light at wavelength $\lambda$. The quantity $l$ stands for the optical path length through the sample cell.
Numerous substances validate Beer-Lambert Law but some does not. Therefore you must assess linearity before quantifying unknown concentrations. This is usually done when performing calibration curve. Then you know your dynamic range and sensitivity of your analytical method and you can regress concentration from absorbance.
I suggest you to read more about those concepts in a reference book such as Skoog & West, Fundamentals of Analytical Chemistry or Vogels, Textbook of Quantitative Chemical Analysis. There is also a free electronic book available on internet called Analytical Chemistry 2.0.


text:
How to trace an energy versus time diagram?
I'm looking for data to create a simple graph of an exothermic or endothermic reaction but can't find any.
Could someone help me with some enough data to create a simple graph displaying energy as function of time.

Thermodynamics only tells us about starting and ending conditions and nothing about what happens in between, i.e. it does not measure rate constants or activation energies and  for this you need chemical kinetics.

Assuming you want to plot heat exchanged by reaction versus time, you need at least to know what reaction you want to study. Then you will have to assess two different quantities: enthalpy if the process is isobaric (thermodynamic) and reaction rate (kinetic). This kind of data can be found on NIST Database.
If you do not find reaction enthalpy $\Delta_{R} H$, you can assess it by using enthalpies of formation:
$$ \Delta_{R} H = \sum\limits_i \nu_i h_{\mathrm{f},i} $$
Where $i$ indexes reactants and products, $\nu_i$ are stoichiometric coefficients (note: they are negative for reactant) and $h_{\mathrm{f},i}$ are enthalpies of formation.
Then you need to find out which is the kinetic $v$ of your reaction and solve the related differential equation. If you do not find any information you might try the activated complex model:
$$ v = \frac{1}{V}\frac{\mathrm{d}\xi}{\mathrm{d}t} = k(T)\prod_j x_j^{\alpha_j} $$
Where $\xi$ is the molar coordinate of your reaction (De Donder variable), $k(T)$ the kinetic constant, $x_j$ are reactant concentrations and $\alpha_j$ are partial orders of reaction (they can be taken equals to $|\nu_j|$). Solving this differential equation - for a given set of initial conditions - will provide you a relation, maybe explicit, between $\xi$ and $t$.
Merging both step results will lead to enthalpy-time expression which can be plotted. Of course those calculation might be perform numerically.
If you were looking for an hand-made and pedagogic Energy Plot like this, you have not searched so far.
Method exposed above does not create this kind of plot which shows internal energy of sum of compounds versus coordinate of a molecular transformation. Gap between the initial and final state is the energy relaxed/absorbed by the reaction (often pure heat). Such plot are arbitrarily plotted because the activated state is often unknown and definitely non measurable. Furthermore everything between initial and final state is also not-well-known because we just have access on macroscopic data so we ignore microscopic details. Last hint: activation energy (maximum) is always higher than the higher state, you need extra energy to initiate the transformation.


text:
Which of these reduction reagents would me most appropriate to yield this product?
I am very confused, I think using reduction would be a good start, but LiAl would go straight to an OH.

1) As you say*, the ketone must be protected.  Protect it. Any propagating naughtiness is irrelevant to the next step.
2) Reduce the ester. Beat on it.
3) Deprotect the ketone.
*Note that you begin with an alpha-ketoester.  In the old Morrison and Boyd that was a French joke.  LAH would have the same problem.  How the mighty have fallen.


text:
Reaction of a vicinal diol with acetone
I have the first part correct; however, I can't seem to understand what the 2nd part's answer is. I realize the water leaves, but then what?

I changed the title of your question - completely. Don't get confused with the apparatus used to trap the water, look at the reactants instead. You have a diol and a ketone :)

That is an interesting product.  What is the reducing agent?  Absent that, how does it proceed to preserve the chiral center?

There was this guy who recently protected a ketone in the presence of an ester. You might have heard about him ;)


text:
What does the equilibrium equation look like when calculating the solubility from Ksp?
For example, given a reaction
$\ce{NaCl -> Na+ + Cl-}$
and its $Ksp$ value, I know that you can set up
$$Ksp\; =\; \left[ \ce{Na^{+}} \right]\left[ \ce{Cl^{-}} \right]$$ and solve from there.
My question is why do you not put $\ce{[NaCl]}$ in the denominator, as isn't that part of the equilibrium expression also?

The reason why $K_{sp}$ does not include $[\ce{NaCl}]$ is because it is solid:
$$\ce{NaCl(s) -> Na+ (aq) + Cl- (aq)}$$
The equilibrium constant $K$ would include $[\ce{NaCl}]$:
$$K=\dfrac{[\ce{Na+}][\ce{Cl-}]}{[\ce{NaCl}]}$$
However, the concentration of sodium chloride in solid sodium chloride is constant, thus we can write an equation for the observed rate constant that is based only on the variable concentrations. $K_{sp}$ is this observed rated constant:
$$K_{sp}=K_{obs}=K[\ce{NaCl}]=[\ce{Na+}][\ce{Cl-}]$$

But since $\ce{NaCl}$ is soluble shouldn't you write it as $Na\mbox{C}l_{\left( aq \right)}$?

But then we're not talking about sodium chloride's solubility product constant, we're talking about its dissociation constant. Also, there is a limit to sodium chloride's solubility in water (359 g/L according to Wikipedia)


text:
Determining Fluorescence Quantum Yield of C60 in Toluene
For my experiment, I am trying to determine the fluorescence quantum yield of C60-fullerene in toluene (at room temperature).  In order to do that, I need to have a standard sample with a valid and known fluorescence quantum yield. I have been looking for a standard sample with a known quantum yield in toluene, however, I am not sure of what to use.
Would it be better if I used a different solvent in determining the quantum yield and if so what solvent and standard should I use.

A fairly popular standard is Rhodamine 6G, in some sort of alcohol. Be aware that if you decide to use a different solvent, you'll have to account for differences in refractive index.

Toluene instead of benzene can be a poor choice when H-atom abstraction by the excited state is possible. Here, you're safe.
Try to get a copy of Fluorescence spectra and quantum yields of buckminsterfullerene (C60) in room-temperature solutions. No excitation wavelength dependence.
The authors apparently used 9-cyanoanthracene in toluene as a standard to determine $\Phi_f$ of C60.
You might also want to have a look for review articles by Dirk Guldi from the time around 2000; I remember him giving a number of talks on C60 around that time.

In addition you should use a standard with emission in the same region as you sample to avoid effects of detector etc effects with wavelength. Porphyrins are probably good here as there are standard corrected spectra published.  Automatic correction, as used in some fluorimeters, is not always as good as its made out to be.


text:
Enthalpy of Formation problems using Hess's Law
How do I solve problems where I'm asked for $\Delta H^\circ_f$ and given $\Delta H^\circ_f$ values for equations with the involved elements? For example:

Find the $\Delta H^\circ_f$ of the following reaction:
  $\ce{C + O2 -> CO2}$, given:
$\ce{SrO + CO2 -> SrCO3}$ ; $\Delta H^\circ_f = \pu{-234 kJ}$
$\ce{2SrO -> 2Sr + O2}$ ; $\Delta H^\circ_f =  \pu{1184 kJ}$
$\ce{2SrCO3 -> 2Sr + 2C + 3O2}$ ; $\Delta H^\circ_f = \pu{2440 kJ}$

I know that I'm somehow supposed to manipulate the equations to make them "add up" (adding straight down) to the target equation, but I'm not sure how, exactly, to achieve this.
Also, is there a different and/or better method for doing these?
I've made some progress on my own, using this About.com page. What I'm wondering now, is if elements are locked into compounds, can I split them up, as long as I keep them on the same side? In the example above, I've manipulated it to where I have $\ce{2CO2 + O2 -> 2C + 3O2}$; am I allowed to split the $\ce{2CO2}$ into $\ce{2C + 2O2}$?
I've made it as far as getting to $\ce{2C + 2O2 -> 2CO2}$. If you take these as molar ratios, it's the same as $\ce{C + O2 -> CO2}$, which is the target equation. What do I have to do to the enthalpies to go from the first to the second?

Hess's law isn't too difficult once you know what to look for. What may have caused you difficulty is 1.) either you are missing a reaction, or 2.) one of those listed reactions is incorrect. In order to properly use Hess's law, you need at least one reaction where $\ce{CO2}$ is present.

@Jun-GooKwak Am I approaching it the right way with the manipulating equations method?

Not quite. You can either multiply those reactions by integers, or use the opposite reaction. Essentially you are trying to derive the equation that you have above by canceling out common elements on both sides of the products and reactants. In addition, you are keeping track of the enthalpies and find the sum. However, I do not see any reaction with carbon dioxide as a product or reactant.

@Jun-GooKwak You're right; on the problems sheet I got the question from, it's actually a $CO _{2}$ in the first one, not an $O _{2}$.

Ah, excellent. It's solvable now. I'll do this example for you and I think it will be quite clear soon.

@Jun-GooKwak Am I allowed to split compounds as described in the edit?

@Jun-GooKwak Please double check that my $\Delta H^{circ_f}$ notation is correct.  I was only guessing based on the context.

@jonsca Looks good so far! I think you missed one of them however. It should have a \ in front of it. I fixed it accordingly.

Find the $\Delta H^\circ_f$ of the following reaction:
$$\ce{C + O2 -> CO2}$$

We have our list of reactions with known enthalpies:
$\ce{SrO + CO2 -> SrCO3} \quad \Delta H^\circ_f = \pu{−234 kJ}$
$\ce{2SrO -> 2Sr + O2} \quad \Delta H^\circ_f = \pu{1184 kJ} $
$\ce{2SrCO3 -> 2Sr + 2C + 3O2} \quad \Delta H^\circ_f = \pu{2440 kJ} $
Hess's Law takes advantage of the fact that enthalpy is a state function, looking at our reactions, we need to find a way to derive what we want:
Carbon dioxide is on the reactants side, however, we want it on the products, so we can run that reaction reverse; however we need to change the sign of enthalpy if we do that:
$\ce{SrCO3 -> SrO + CO2 } \quad \Delta H^\circ_f = \pu{234 kJ}$
We need to get rid of $\ce{SrO}$, thus, we use our second equation:
$\ce{2SrO -> 2Sr + O2 } \quad \Delta H^\circ_f = \pu{1184 kJ} $
Since, there's only one strontium oxide in the first reaction, we divide by the integer 2 across the chemical equation including the enthalpy of formation:
$\ce{SrO -> Sr + 1/2O2 } \quad \Delta H^\circ_f = \pu{592 kJ} $
Now, with equation 3, we need carbon and oxygen on the reactants side, and need to remove $\ce{SrCO3}$ and also divide again by 2, since we only have one strontium.
$\ce{Sr + C + 3/2O2 -> SrCO3 } \quad \Delta H^\circ_f = \pu{-1220 kJ} $
Now we add up all our reactions and enthalpies, and cancel out common elements on both sides:
$\ce{SrCO3 -> SrO + CO2 } \quad \Delta H^\circ_f = \pu{234 kJ}$
$\ce{SrO -> Sr + 1/2O2 } \quad \Delta H^\circ_f = \pu{592 kJ} $
$\ce{Sr + C + 3/2O2 -> SrCO3 } \quad \Delta H^\circ_f = \pu{-1220 kJ} $

$\ce{C + O2 -> CO2} \quad \Delta H^\circ_f= (\pu{234 kJ}) + (\pu{592 kJ}) + (\pu{-1220 kJ}) = \pu{-394 kJ}$


text:
How do you have pKa for a base?
The $\mathrm{p}K_\text{a}$ is the $-\log_{10}$ of the $K_\text{a}$ value for the disassociation of an acid. So how do you get a $\mathrm{p}K_\text{a}$ value for a base? The base produces no hydrogen ions so can’t you only have a $\mathrm{p}K_\text{b}$?

For bases, you can derive the $\mathrm{p}K_\text{a}$ from the $\mathrm{p}K_\text{b}$.
Knowing that
$$\mathrm{p}K_\text{a} + \mathrm{p}K_\text{b} = \mathrm{p}K_\text{w} = 14$$
Let’s examine ammonia.
Ammonia has a $K_\text{b}$ of $1.8 \times 10^{-5}$.
Taking the negative logarithm, we can derive $\mathrm{p}K_\text{b}$:
$$\mathrm{p}K_\text{b} = -\log( 1.8 \times 10^{-5}) = 4.74$$.
Knowing that $\mathrm{p}K_\text{a} + \mathrm{p}K_\text{b} = \mathrm{p}K_\text{w} = 14$:
We can find $\mathrm{p}K_\text{a}$ by manipulating that expression, thus:
$$\mathrm{p}K_\text{a} = 14 - \mathrm{p}K_\text{b} $$
$$\mathrm{p}K_\text{a} = 14 - 4.74 $$
$$\mathrm{p}K_\text{a} = 9.26 $$
Even in very basic solutions, there will be minute quantities of hydronium ions. By taking the anti-logarithm of the $\mathrm{p}K_\text{a}$, you can derive the $K_\text{a}$.

Yes, I know that you can manipulate the pKb to get a pKa. However, what meaning does the pKa have for bases?

$pK_a$ is the acid dissociation constant on a logarithmic scale. And the acid dissociation constant $K_a$ is just a measure of the strength of the acid in solution. No base will ever have a $pK_b$ of 14, thus we can find the $pK_a$ of the base. In the example above, it's a very, very small number.

Many bases do not deprotonate (e.g. $\ce{NaOH}$), making the notion of $\mathrm{p}K_\mathrm{a}$ based on deprotonation (Brønsted-Lowry theory) extremely difficult to measure/apply. However, there are some bases which are also acids: for instance $\ce{HSO4-}$ is a base because it can accept a proton:
$$\ce{HSO4- + H2O <=> H2SO4 + OH-}$$
and an acid because it can donate a proton:
$$\ce{HSO4- + H2O <=> SO4^2- + H+}$$
There are many bases which can act as acids and therefore those “bases” have a $\mathrm{p}K_\mathrm{a}$ for the deprotonation. In that case, the “base” would be acting as an “acid”. Measuring $\mathrm{p}K_\mathrm{a}$ for a specific substance assumes that substance undergoes acid dissociation.
Subtracting the $\mathrm{p}K_\mathrm{b}$ of $\ce{HSO4-}$ from 14 does not give the $\mathrm{p}K_\mathrm{a}$ of $\ce{HSO4-}$: it gives the $\mathrm{p}K_\mathrm{a}$ of the conjugate acid $\ce{H2SO4}$ in water.
If you take this conjugate acid $\ce{H2SO4}$ and ask what its $\mathrm{p}K_\mathrm{a}$ is, you are assuming it is behaving as an acid by losing a proton to form $\ce{HSO4-}$. So set up the following chemical equation
$$\ce{H2SO4 + H2O <=> HSO4- + H3O+}$$
If you are asking what is the $\mathrm{p}K_\mathrm{b}$ of $\ce{H2SO4}$, you are assuming it is behaving as a base by gaining a proton to form $\ce{H3SO4+}$, a likely unstable species. Set up the following chemical equation
$$\ce{H2SO4 + H2O <=> H3SO4+ + OH-}$$
Therefore, when calculating concentrations of acid/base/conjugate based on $K_\mathrm{a}$ or $K_\mathrm{b}$ values, you need to make sure you have the appropriate chemical equilibrium written out which matches either protonation or deprotonation of a specific chemical species in a specific solvent (likely water). Use this correct chemical equation to fill in the equilibrium constant expression and go from there.

The "$\mathrm{p}K_\mathrm{a}$" of a base is just the $\mathrm{p}K_\mathrm{a}$ of the conjugate acid of that base.
Consider the following reaction:
$$\ce{B + H2O <=> HB+ + OH- }$$
it has the equilibrium constant:
$$K_\mathrm{b} = \frac{(\ce{HB+})(\ce{OH-})}{(\ce{B})}.$$
Now consider the reaction with the conjugated acid $\ce{HB+}$:
$$\ce{HB+ + H2O <=>  B + H3O+}$$
the equilibrium constant is:
$$K_\mathrm{a} = \frac{(\ce{B})(\ce{H3O+})}{(\ce{HB+})}.$$
The product of the two gives:
$$K_\mathrm{b} \times K_\mathrm{a} = \ce{\frac{(HB+)(OH^{-})}{(B)}} \times \ce{\frac{(B)(H3O+)}{(HB+)}} = \ce{(OH^{-}).(H3O+)} = 10^{-14} = K_\mathrm{w}$$
and
$$\mathrm{p}K_\mathrm{b}+\mathrm{p}K_\mathrm{a}=\mathrm{p}K_\mathrm{w}$$
thus
$$\mathrm{p}K_\mathrm{b}=\mathrm{p}K_\mathrm{w}-\mathrm{p}K_\mathrm{a}=14-\mathrm{p}K_\mathrm{a}.$$


text:
How do you find the average rate and instantaneous rate given a kinetics plot?
What is average rate? What is instantaneous rate? How do you find these? My teacher described finding the tangent to a graph. I am not sure what is meant by this.

This is more of a question of calculus than chemistry, but in a first order reaction
$\ce{A -> B}$, the average rate is:
$$\frac{\Delta A}{\Delta t} = \frac{A_2 - A_1}{t_2 - t_1}$$

 Graciously borrowed from here
So, the average rate is the slope of a secant line between the two points,
$$ \frac {\Delta A}{\Delta t} = \frac{1 - 0.5}{400 - 200}$$
The instantaneous rate is the derivative of the plot:
$$\frac{dA(t)}{dt}$$ or $$\lim_{t\rightarrow 0} \frac{\Delta A}{\Delta t}$$
For the above first order rate law, the equation is in the form:
$$ A(t) = e^{-kt + ln(A_0)} $$ so the derivative is simply
$$ \frac{dA(t)}{dt} = -ke^{-kt + ln(A_0)} $$

Doubtless I've mangled something, so please correct my math!


text:
Diels Alder reaction with maleic anhydride. Reactivity of different dienes
I don't understand question 6, although I took notes and the professor mentioned an explanation in class.
Can someone please help using easier terms?

I took the liberty to change the title of your question. Please note that the reaction is named after Otto Diels and Kurt Alder. There's no 's in here.

a) The compound can rotate into a trans configuration that is not suitable for Diels Alder
b) The compound is locked into a cis configuration that is favorable. Furthermore, a carbon group is slightly electron donating
c) The compound has an electron withdrawing group.
d) What compound is that? It's compounds labelled b) and c)
Comment: Normal demand Diels Alder is enhanced by an electron withdrawing group in the dienophile and an electron donating group in the diene. Look at reaction mechanism tab (http://en.wikipedia.org/wiki/Diels%E2%80%93Alder_reaction)
I am not sure as to why that is. I have forgotten.

Does c) react slower than a) ?


text:
Can someone please explain buffers to me?
This is what I understand so far:
The concept of a buffer is to minimize any swings in pH. In order to do this you need to create a solution with both acidic and basic compounds.
To create one, you start with an acid:
$\ce{HA + H2O -> A- + H3O+}$
[My first question is why for buffers the acid needs to be weak.]
The buffer is then created by adding in a salt containing the anion of the base which allows for both the acid and its conjugate base to exist in the solution.
Here, why doesn't the conjugate base react with the acid? Second, why does this help minimize pH changes. I see that since that you have a weak acid the conjugate base will be strong and can react with any acid added. However, if you add base what is going to react with it? Isn't the acid too weak to do so? Third, what happens to the $\ce{H3O+}$ that exists in the solution?

Writing out the equilibrium expression for a weak acid in solution:
$$[\ce{H3O+}] = K_\mathrm a\frac{[\ce{HA}]}{[\ce{A-}]}$$
Buffers need to keep the concentration of the weak acid to its conjugate base similar in order for it be effective.
Let’s say you add a small amount of base. Some of the $\ce{HA}$ molecules are converted into $\ce{A-}$ ions. The ratio $\frac{[\ce{HA}]}{[\ce{A-}]}$ changes but not by a large amount.
If we add a small amount of acid, we generate more $\ce{HA}$. Again, the ratio $\frac{[\ce{HA}]}{[\ce{A-}]}$ changes but not by a large amount.

Why do buffers for the acid needs to be weak?

If you look at the expression, the concentration of hydronium ions is directly related to the ratio of the conjugate acid and its base. If you take a strong acid for instance, hydrochloric, it will almost completely deprotonate in water. Thus the ratio of $\frac{[\ce{HA}]}{[\ce{A-}]}$ is close to 0 and is negligible, and we use the $K_\mathrm a$ to directly find the concentration of hydronium ions. As strong acids have very large $K_\mathrm a$ values, we can directly take the concentration and assume that will be the concentration of our hydronium ions.

Why doesn’t the conjugate base react with the acid?

It depends on the acid. In buffer solutions, competition arises between acid and the conjugate base. For instance, if the $K_\mathrm a$ is large relative to $10^{-7}$, the acid ionization will proceed and you will have an equilibrium with the conjugate base reacting with its acid resulting in no buffer action.
If however, the tendency of the acid in donating hydrogen ions, say, to water, is outcompeted by the tendency of the base to accept those hydrogen ions from water, this net reaction produces $\ce{OH-}$ ions, the $K_\mathrm a$ of this acid-base pair is less than $10^{-7}$, and we use $K_\mathrm b$ to find equilibrium.
Derivation of the Henderson–Hasselbalch equation:
Let’s use acetic acid and water as an example:
$$\ce{AcOH + H2O <=> AcO- + H3O+} $$
$$ K_\mathrm a = \frac{[\ce{AcO-}][\ce{H3O+}]}{[\ce{AcOH}]}$$
Taking the log of both sides:
$$ \log(K_\mathrm a) = \log\left(\frac{[\ce{AcO-}][\ce{H3O+}]}{[\ce{AcOH}]}\right)$$
Using the law of logs where $\log(xy) = \log(x) + \log(y) $:
\begin{aligned}
 \log\left(K_\mathrm a\right) &= \log\left(\frac{[\ce{AcO-}]}{[\ce{AcOH}]}\right) + \log([\ce{H3O+}])\\
 -\log([\ce{H3O+}]) &= \log\left([\ce{AcO-}][\ce{AcOH}]\right) - \log(K_\mathrm a) \\
 \mathrm{pH} &= \log\left(\frac{[\ce{AcO-}]}{[\ce{AcOH}]}\right) + \mathrm{p}K_\mathrm a \\
 \mathrm{pH} &= \mathrm{p}K_\mathrm a + \log\left(\frac{[\ce{B}]}{[\ce{A}]}\right) \\
\end{aligned}
With buffer systems, you can calculate the $\mathrm{pH}$ of any buffer solution. The Henderson–Hasselbalch equation has several caveats however.

You need have both the acid and its conjugate base in solution.
The extent of ionization must be small enough so that the
concentrations of hydronium and hydroxide ions are small relative
$[\ce{HA}]_0$ and $[\ce{A-}]_0$.

When you have a buffered solution and add a base, why doesn't the base react with the H3O+ ion (which would change the pH)? Why does it prefer to react with the acid instead?

@user2612743: The base would most probably react with $H_3O^+$ as you suggest, but very soon (nanoseconds at most), the acid will "create a new" $H_3O^+$. It is just equilibrium in zillions of molecules.

For the part about acids not reacting with conjugate base, is it because the only way the react is $\ce{HA + A- -> A- + HA}$ resulting in no net change of concentration?

And why must only weak acids be used for buffers? Why can't you use a strong one?

@user2612743 I readdressed those issues and clarified them in my post.

ok as here this discussed extensively here what will make a more effective buffer a weak acid or strong acid


text:
Red Cabbage Indicator colour change with Sodium Metabisulphite
I made some pH indicator by boiling and straining cabbage juice and in an effort to keep it from spoiling for a while without having to store it at fridge temperatures I added some wine making stabiliser which consists primarily of Sodium Metabisulphite. I had previously tested some of it to check if adding it would affect the pH of the indicator since I wanted a pH neutral preservative. Well strangely after leaving it for a while I came back to find that the indicator had turned a weird shade of purple I had never seen the indicator take before. At the concentration it is at the original colour was a very dark purple - almost black. Now I know that the metabisulphite ion decomposes to produce SO2 gas which is acidic under aqueous conditions but I would expect that to change the indicator to its normal purple for a not very concentrated acid. But the colour it currently is is very pale and in fact when I pipette out some of the indicator it is almost totally clear. So I guess really my question is does Potassium Metabisulphite have some kind of bleaching effect I'm not aware of?

http://mylespower.co.uk/2012/04/06/homemade-ph-indicator/

Now we add the reducing agent (model compound).

That lessens conjugation and bleaches the anthocyanine.  "But the colour it currently is is very pale" If the new compounds' residual optical absorbances are in the green, the solution will be iodine purple by transmission.

You're answer is very thorough and the linked article very interesting. Thank you very much


text:
Can an organic compound have an ionic bond?
Is there any organic compound that has another type of bond other than the covalent bond?

You see. The question is difficult to answer since the line that distinguishes organic compounds from inorganic ones isn't quite definite.

There's a large number of salts with organic anions (e.g. potassium hydrogen tartrate) or organic cations (e.g. tetrabutylammonium hexafluorophosphate)
There are even salts, in which both the anion and the cation are organic compounds with covalent bonds. This is often used in pharmaceuticals. Here the active component appears in protonated form and is served as a salt with the conjugated bases of carboxyclic acids (e.g. citrates, succinates, etc.)

Of course, organic can make an ionic bond. For example: Acetylene
Acetylene is relatively acidic, even it's less acidic than water or alcohols. If acetylene reacts with strong base, it will form acetylide. For example: reaction between acetylene and sodium amide ($\ce {Na+ (NH2)-}$) will form sodium acetylide ($\ce {Na+ (H-C#C:)-}$). And this sodium acetylide is ionic and a organic compound.
Another compound is a organometallic compound like n-Butyllithium ($\ce{(CH3-CH2-CH2-CH2:)- Li+}$), Grignard reagents ($\ce{(R:)- (MgX)+}$), etc.

n-Butyllithium and Grignard reagents have polar covalent C-Mg and C-Li bonds

React maleic anhydride with a mole of $\ce{H_2N-CH_2CH_2-N(CH_3)_3^+}$ to afford the zwitterionic amide.
Protein - with every kind of bonding interaction from the variety of amino acid side groups.  Hydrogen bonding, ionic bonding, covalent bonding, pi-stacking, zinc-fingers, charge transfer, etc.


text:
Is there a (carbon) heterocyclic compounds named from number three to infinity?
For example, oxepane is a 7-atom ring. Can there be 8-atom ring or 9-atom ring … or million-atom ring to infinity?

Million atom sized rings do not make a lot of sense to me.

Yes, as @Klaus-DieterWarzecha has stated, there are certain rules that govern how organic compounds are formed, such that a million atom sized ring is not plausible.

To my knowledge, the Hantzsch-Widman nomenclature defines stems for ring sizes up to 10 (ecine for unsaturated, ecane for saturated heterocycles).
Oxepane is not the general stem for a seven-membered heterocyclic ring, but the suffix for one containing an oxygen atom.

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), heteromonocyclic compounds with up to and including ten ring members are named by the extended Hantzsch–Widman system; for example

oxirane ($\ce{C2H4O}$)
oxetane ($\ce{C3H6O}$)
oxolane ($\ce{C4H8O}$)
oxane ($\ce{C5H10O}$)
oxepane ($\ce{C6H12O}$)
oxocane ($\ce{C7H14O}$)
oxonane ($\ce{C8H16O}$)
oxecane ($\ce{C9H18O}$)

For monocyclic rings with eleven and more ring members, skeletal replacement (‘a’)
nomenclature is used; for example

oxacycloundecane ($\ce{C10H20O}$)
oxacyclododecane ($\ce{C11H22O}$)
oxacyclotridecane ($\ce{C12H24O}$)
oxacyclotetradecane ($\ce{C13H26O}$)
oxacyclopentadecane ($\ce{C14H28O}$)
etc.

Theoretically speaking, I don't see why not. In theory, it should be possible for a million-atom ring to exist. But, of course, in practice, and regarding reality as we currently know it, it isn't something that has been observed or produced, so far.
For example, here is a ring of 36 carbon atoms called cyclohexatriacontane:

And with IUPAC nomenclature, you can name such compounds or rings with a very high number of carbon atoms.
Regarding a ring of carbon infinity, that definitely does not exist in our universe. It sounds closer to religion, or at least multi-verse theory, than chemistry to me, for example.


text:
Comb and hair, is that an example for ionic boding?
As i understand, when brushing hair, some electron transfer from hair to the comb. Thus, make the comb have different charged from hair and they attract each other.
It's look likely what i have learn from school about ionic bond. One electron from Na is transfer to F and so we have NaF . Isn't it ???

No, not at all. It is an example for triboelectric charging. A mechanical example for this type of electrostatic generators would be a Van de Graaff generator.
But all this has nothing to do with ionic bonds.


text:
How can the number of molecules and atoms be the same in a gas?

A sample of certain mass of gas was collected and it was found that the number of atoms and the number of molecules in the collected sample are equal. You are required to state the conclusion drawn from this observation.

I am not sure of the answer. I think it is related to the quantity of a mole being constant.

The gas was one (or a mixture) of the noble gases, as every other gas is composed of more atoms than molecules.
Example

Helium: $1~\text{mol}~\ce{He}~\text{gas} = 1~\text{mol}~\ce{He}~\text{atoms}$
Nitrogen: $1~\text{mol}~\ce{N2}~\text{gas} = 2~\text{mol}~\ce{N}~\text{atoms}$

The conclusion is that the gas is monoatomic - helium, neon, argon, krypton, xenon; or something clever done in a physics lab with low temperature, high vacuum, and lasers. It is a lot of fun to do a little chair parade, learn new things, and present a second answer that is beyond your teacher's understanding.  Then, smile.  Knowledge wants to propagate.

If in a given volume of gas the number of atoms and molecules is same ,then the only conclusion that can be drawn is that the mixture was of the noble gases.E.g.-
He,Ne,Ar,Kr,Xe etc.