Docstoc

gases

Document Sample
gases Powered By Docstoc
					                                           CHEMISTRY 115

                                            THE GAS LAWS

I.    Introduction to this Handout:
      In this handout, I will attempt to demonstrate that the only equation necessary to solve gas law
      problems is the general equation:
           PV = nRT                                                                                      (1)

II.   Rearranging the Ideal Gas Law:
      Be careful when you manipulate the equation algebraically!! Remember that you can perform any
      operation you want to one side of an equation, as long as you perform the identical operation to the
      other side.
      A)   Solving for P:
           If we want to solve for the pressure, P, given that we have known values for n, V, and T,
           simply divide both sides of the equation by the volume, V:
                  PV nRT
                    
                  V   V
           This gives the following relationship for the pressure:
                      nRT
                 P       .                                                                              (2)
                       V
      B)   Solving for V:
           If we want to solve for the volume, V, given that we have known values for all the other
           parameters, we simply divide both sides by P:
                  PV nRT
                     
                   P   P
           This gives the following relationship for the volume:
                      nRT
                 V                                                                                      (3)
                       P
      C)   Solving for n:
           Remember my recommendations for solving problems: First, balance any chemical reaction
           equations, if any. Next, always convert any information you can into moles!! If the
           temperature, pressure and volume all have known values, then the only unknown in the Ideal
           Gas Law will be the pressure, (or the partial pressure, in some cases). To solve for n, the
           number of moles of gas present, simply divide both sides of the Ideal Gas Law by RT:
                  PV nRT
                     
                  RT   RT


                                                   –1–                                    Raynor: 12/4/2011
          This gives the following form for the number of moles, n:
                 PV
                    n.
                 RT
                      PV
          I.e., n       .                                                                                  (4)
                      RT

III. Using the Ideal Gas Law to Determine MWs:
     The book derives a special formula to use in order to determine the molecular weight of a gas from
     its density, the pressure it exerts and the temperature. However, we are often given the mass of the
     gas and the volume of the container it is in, instead of the density. Of course, we can calculate the
     density of the gas from the mass and volume, using d = m / V. However, rather than memorize
     specific formulas for specific circumstances, I will demonstrate that you can always use the Ideal
     Gas Law in its normal form, regardless of what information is given to us. For example, given the
     volume of the container and the mass of the gas contained within that volume, as well as the
     pressure and temperature, we can determine the molar mass of the gas by following the steps
     below:
          1)     As always, whenever possible, the first step in any problem should be to solve for the
                 number of moles present of a substance. Since P, V, and T all have known values, we
                 can simply use the Ideal Gas Law to solve for n, the number of moles of gas present in
                                                PV
                 the container, i.e., using n     .
                                                RT
          2)     Now we know how many moles are present. Since we were told the mass of the gas in
                 the container, m, we can now directly solve for the MW of the gas. [Recall always that
                 MW is a ratio of mass in grams to moles – i.e., it has units of g/mole]:
                               m   g
                        MW                                                                                (5)
                               n mole
                 Note how easy this was!

IV. Using the Ideal Gas Law to Determine Densities:
     First, recall that the density of a substance is the ratio of the mass of a sample of that substance
     divided by the volume it is contained in:
                m
           d     .                                                                                         (6)
                V
     The book derives a special formula to use in order to determine the density of a gas from known
     values of the following: P, T, and the MW of the gas. They derive a formula that involves all of
     these values. Instead, let’s proceed as usual by looking at which values in the Ideal Gas Law are
     known, and which are unknown. In the equation below, I have underlined all known parameters,
     (including the gas constant, R):
           PV  nRT .                                                                                       (7)


                                                    –2–                                      Raynor: 12/4/2011
Note that two parameters are unknown – n and V. There are two ways we can proceed from here:
1)   Whenever two parameters have unknown values, you can choose a convenient value for one
     of them and then use this value in order to solve for the other. [Be careful: you must use the
     same assumed value throughout the problem!]
     In this case, let’s assume a value for V and solve for n, as usual. The most convenient choice
     is to use V = 1 liter, (exactly). [By assuming an exact value, we are not affecting the number
     of sig figs in the problem. The sig figs will be determined by the other data that was given.]
     We can now rearrange Eq. (7) in order to solve for the number of moles present:
                P 1 liter 
           n                  .                                                                     (8)
                   RT
     Since we know the molecular weight of the substance, we can now use it and the number of
     moles to find the mass of gas in our 1 liter container:
           m  n  MW .                                                                              (9)
     [It is easiest to follow the units here, in order to avoid dividing when you should multiply,
     etc. I.e., moles  g/mole  g.]
     Now we have a known value for the mass of the gas, m, and a known value for the volume it
     is in, (we assumed a value of V = 1 liter), so we can now determine the density of the gas in
     g/L, using Eq. (6) – i.e., using d = m/V.
2)   We can also determine the density algebraically, by starting with the Ideal Gas Law, as
     expressed in Eq. (7). Let’s rearrange Eq. (7) until all our unknowns are on the left-hand side
     and all are known values are on the right-hand side, by dividing both sides by RT and also by
     V. This gives the relationships shown below:
            P   n                  n   P
                                      .                                                         (10)
           RT V                    V RT
     Note that we are now able to determine a density in the form of moles/L. Multiplying our
     result by the MW of the substance, (with units of g/mole), will give us the ratio we are
     seeking, with units of g/L. I.e.,
                    n
           d  MW                                                                               (11)
                    V 
     [Don’t memorize this formula – use unit analysis. I.e., g/mole  mole/L yields g/L – the
     density of the gas.]




                                             –3–                                     Raynor: 12/4/2011
V.   Determining MW from the Density of a Gas:
     The book goes one step further and derives the relationship between MW and known values for the
     density of a gas, the pressure it exerts and the temperature of the gas. Once again, it is not
     necessary to memorize another formula! Let’s again note which parameters are known and which
     unknown in the Ideal Gas Law, keeping in mind that our goal is to determine the MW of the gas:
           PV  nRT .                                                                                   (12)
     Rearranging this equation to place all unknowns on one side and known values on the other, we get
     Eq. (10), [which is repeated below as Eq. (13)]:
          n   P
               .                                                                                       (13)
          V RT
     As before, there are two different ways we can choose to solve this problem.
     1)   First, we can choose any value we want for one of the unknown parameters in Eq. (13) and
          then use the chosen value to determine the other one. The simplest choice is to choose V
          identical to 1 liter. Then we can use Eq. (13) to solve for the number of moles of this gas that
          would be present in a 1 liter sample of the gas.
          Next, we can apply the same value we chose to determine the mass present in our 1 liter
          sample, using the value we were given for the density of the gas:
                     m     m
                d                m.                                                                  (14)
                     V 1 liter 

          Now we have determined the mass of our sample and the number of moles in the same
          sample and can use those values to determine the MW of the gas:
                        m     g 
                MW                                                                                  (15)
                        n    moles 
     2)   We can also solve the problem algebraically. Eq. (13) gives us the ratio of moles to the
          volume, (in liters), using the known values of P and T. Next, we note that a numerical value
          for the density has also been given to us, (in g/L). We can now use the value we determined
          for the ratio of moles to liters with the density to solve for the MW, (with units of g/mole).
          This is most easily accomplished using unit analysis:
                                    d      g   L 
                MW (g/mole)                
                                  n / V   L mole 
                                                                                                        (16)
                                                    
     The following example from the textbook shows how you can solve such problems without having
     to use anything other than the Ideal Gas Law in its normal form.




                                                  –4–                                     Raynor: 12/4/2011
Example: [Example 5.9 in text]
A chemist synthesized a gaseous compound of chlorine and oxygen and finds its density is 7.71 g/L
at 36 C and 2.88 atm. Calculate its molar mass and determine its molecular formula.
     The book uses a special formula to solve this problem. I will show below how we can use the
     above described methods to solve it using PV = nRT to determine the molar mass:
     Method 1:
     Since density is uniform, we can choose any size sample we want for the volume and use it to
     solve for the mass of gas that must be present. Let’s choose a sample with a volume given by
     V = 1 L, exactly. [Since this is an exact value, it has an infinite number of sig figs. Thus the
     other numerical data in the problem will determine how many sig figs are in our answer.] We
     can now use the density to convert our known volume of the gas into a mass of this gas:
           V = 1 L  7.71 g /1 L = 7.71 g.
     Now consider PV = nRT. We have numerical values for all data except for the number of
     moles of gas in our 1 L sample. [As always, solve for the number of moles present as soon as
     possible!]
           n = PV / RT = (2.88 atm) (1 L) / (0.0821 L-atm/mol-K) (309K)
           n = 0.1135 moles
     We now know the mass in our 1 L sample and the number of moles present in that same
     sample. Since the molar mass is the ratio of mass divided by moles, (its units are g/mol), we
     can now calculate the molar mass:
           MW (units = g/mol)  MW = mass/moles = 7.71 g / 0.1135 moles = 67.9 g/mol
     Method 2:
     Let’s use the Ideal Gas Law to determine the ratio of moles to volume – i.e., the ratio of the
     two unknowns in the Ideal Gas Law. [We are given values for P and T, but not for n or V.]
           n   P          2.88 atm
                                          0.1135 mol/L
           V RT 0.0821 L-atm/mol-K  309 K
     Next, we can use unit analysis to convert the density into units of g/mol, using the result
     given above for mol/L. [It tells us that 0.1135 mol of the gas occupy 1 Liter.]
                                  1L
           MW = 7.71 g/L                 67.9 g/mol
                              0.1135 mol
     Final Step: Determining Molecular Formula from Molar Mass:
     Once we know the molar mass, we can use it to help us determine the molecular formula.
     Usually, we are given (or have determined) an empirical formula for the substance. However,
     in this case we are told only that the molecule contains only 2 elements – chlorine and
     oxygen. Thus, the molecular formula must have the form ClxOy, where x and y have integer
     values. Thus the molar mass must have the following form relative to the atomic masses of
     chlorine, (35.45), and oxygen, (16.00):

                                             –5–                                     Raynor: 12/4/2011
               MW = 67.9 = (35.45) x + (16.00) y
         First, x cannot be larger than 1. [If it were equal to 2, the contribution from chlorine to the
         molar mass would be 70.9, which is larger than the entire molar mass.] Thus, x = 1 is the only
         possible outcome. We now have only one integer to determine – the value of y:
               67.9 = (35.45) (1) + (16.00) y → 32.4 = 16.00 y → y = 2
         Finally, we can deduce that the molecular formula must be ClO2.


VI. Boyle’s Law Problems, Charles’ Law Problems, etc.:
    It is not necessary to memorize Boyle’s Law or Charles’ Law, etc., in order to solve problems that
    involve them. The easiest way to proceed in any of these types of problems is to first isolate all
    parameters that vary on one side of the equation and all those that remain constant on the other.
    The easiest method is to underline all parameters being held constant first, and then rearrange the
    equation, if necessary. I will illustrate a couple of types of problems each below:
    A)   Boyle’s Law Problems:
         These are problems where the pressure and volume of a gas is varying from one experiment
         to the next, but the temperature and sample size remain unchanged. [Note: since the sample
         of gas is the same in both experiments, the number of moles of gas is the same for both
         experiments.] In the Ideal Gas equation written below, I have underlined the termperature and
         numbers of moles of gas, (as well as the gas constant, R):
               PV  n RT                                                                              (17)
         The next step is to rearrange the equation in order to place all parameters that vary on one
         side and all the parameters that remain constant on the other. Note that we already have the
         desired form in this case.
         Since everything on the right-hand side is held constant, we can rewrite the equation as
         follows:
               PV  constant                                                                          (18)
         This means that the product of P times V will be the same for both experiments. I.e.,
               PV1  constant and PV2  constant
                1                  2                                                                  (19)
         This leads directly to Boyle’s Law:
               PV1  PV2 .
                1     2                                                                               (20)
    B)   Charles’ Law Problems or Any Other Similar Problem:
         Alternatively, you might be told that the pressure is held constant while the volume and
         temperature change or that the volume is held constant while the pressure and temperature
         change. In either of these cases, the easiest way to solve them is to follow the same approach
         we used for Boyle’s Law problems.
         For example, suppose you are told that the pressure is held constant as the volume and
         temperature change. The easiest way to proceed is to first write down the Ideal Gas Law,

                                                 –6–                                    Raynor: 12/4/2011
     underlining all parameters that remain constant – the pressure and number of moles of gas, in
     this example.
           PV  n RT                                                                                (21)
     Next, we need to rearrange the equation in order to place all constants on one side and all
     variables on the other. I.e., we will divide both sides by P, and will also divide both sides by
     T, as shown below:
           V nR
                 constant                                                                         (22)
           T   P
     Since everything on the right-hand side is held constant, the entire right-hand side will be
     equal to a constant value, for both experiments. Thus we can use Eq. (22) for each of the
     experiments, giving
           V1                V2
               constant and     constant                                                          (23)
           T1                T2
     This leads directly to the desired relationship between the two experiments:
           V1 V2
              .                                                                                    (24)
           T1 T2
     Another algebraic approach is to write down the form for the Ideal Gas Law for each
     experiment. In the example above, we would begin by writing the following:
           PV1  n RT1 and PV2  n RT2                                                              (25)
     The underlined data are those which remain unchanged during experiments 1 and 2.
     Next, divide the 2nd equation by the 1st one and then cancel out any data that appears
     identically in numerator and denominator:
           P V2 n R T2            V2 T2
                                   .                                                             (26)
           P V1 n R T1            V1 T1
C)   An Alternative Approach:
     Since two of the parameters are unknown and are held constant during these experiments, one
     approach that can be used is to choose a convenient value for one of the two parameters and
     then use its value and the values given in Experiment 1 for the other data to determine a value
     for the other unknown parameter. The values for these constant parameters are then used with
     the data given for Experiment 2 to determine the required answer.
     For example, suppose you were told, (either directly or indirectly), that the sample size, (i.e.,
     n), and the volume were held constant as the pressure and temperature changed from one set
     of values to another set. As before, I have underlined the parameters that are being held
     constant, (i.e., the values which do not change from one experiment to the next). For the
     pressure and temperature, I show them with subscripts to indicate that they relate to the initial
     data, (i.e., the data from Expt. 1):    P V  n RT1
                                              1                                                   (25)


                                              –7–                                    Raynor: 12/4/2011
The value for R is of course known, so the only unknowns in this equation are the volume
and the number of moles of gas, both of which remain constant. We can choose a value for
either of these parameters and then solve for the value the other must have. Let’s choose
V = 1 Liter, exactly. We can now rearrange Eq. (25), in order to solve for the only remaining
unknown value – that for n:
              P V P 1 L 
      nn     1
                   1                                                                        (26)
              RT1     RT1
Once we have determined values for both constant parameters, we can use these values with
the data given for Expt. 2 to solve for the answer. In this example, we would either be told a
value for the temperature in Expt. 2 and asked to find the pressure; or we would be told the
pressure in Expt. 2 and asked for the new temperature. Thus, for Expt. 2, we would have the
following form, (where I have underlined all data for which we have known values):
      P V  n RT2
       2              P 1 L  n RT2
                        2                                                                    (27)

[Note that we have a known value for n from the previous step.]
In the example above, either we would be given a value for T2 and asked for P2 or given a
value for P2 and asked for the value of T2. The example below illustrates this process:


Example: A sample of gas was placed in a sealed container at 300 K. The pressure of the
gas was found to be 1.00 atm. If this container was then heated to a temperature of 400 K,
what would be the pressure exerted by the gas in the container at this new temperature?
First, note that the same container is used at both temperatures and it contains the same
       sample of gas. Thus V2 = constant. We will assume 1 L.
Let’s make a list of the data we know for each set of conditions, (Expt. 1 and 2):
           T1 = 300 K                            T2 = 400 K
           P1 = 1.0 atm                          P2 = ?
           V1 = 1 L                              V2 = 1L
           n1 = ?                                n2 = ?
           P1V1 = n1RT1                          P2V2 = n2RT2
This means that only one unknown remains for Expt. 1 – the number of moles present in our
     1 L container. Solving for the number of moles gives:

           n
                        1 atm 1 L            0.0406 moles
                 0.0821 L-atm/mol-K 300K 
Note that the same sample was used in the same container under the conditions given for the
     2nd experiment. Thus, n2 = n1 = n = 0.0406 moles. The only unknown left for Expt. 2 is
     the pressure when the temperature is raised to 400 K:
           PV  nRT  P 1 L   0.0406 moles  0.0821 L-atm/mol-K  400K 
                       2


     Solving for the pressure, we get            P2  1.33 atm

                                         –8–                                   Raynor: 12/4/2011
          The example below shows a case where we cannot make any assumptions, because only one
          unknown exists for one of the sets of conditions. It is actually a straight-forward use of the
          Ideal Gas Law. The only “wrinkle” is that you must realize that the same sample of gas is
          used with both sets of conditions.
          Example (from the text): A small bubble rises from the bottom of a lake where T = 8 C and
          P = 6.4 atm, to the surface where T = 25 C and P = 1.0 atm. Calculate the final volume of
          the bubble if its initial volume was 2.1 mL.
                First, let’s list what we know about each set of conditions. I’ll use a subscript 1 for the
                initial conditions, (at the bottom of the lake), and subscript 2 for the final ones, (at the
                surface):
                       T1 = 8 C + 273 = 281 K                T2 = 25 C + 273 = 298 K
                       P1 = 6.4 atm                           P2 = 1.0 atm
                                                –3
                       V1 = 2.1 mL = 2.1 × 10 L               V2 = ?
                       n1 = ?                                 n2 = ?
                       P1V1 = n1RT1                           P2V2 = n2RT2
                Notice that there is only one unknown value for the initial conditions – the number of
                moles of gas in the bubble, n1. Let’s use that all important rule – always convert any
                data you can into moles before you do anything else! Rearranging the Ideal Gas Law
                for the initial conditions gives us the following solutions for n1:
                       n1 = P1V1/RT1 = (6.4 atm) (2.1 × 10–3 L) / (0.0821 L-atm/mol-K) (281 K)
                       n1 = 5.82 × 10–4 mol
                Now, note that the mass of the bubble remains unchanged and thus the number of moles
                of gas in the bubble remains unchanged. Thus, n2 = n1 = 5.82 × 10–4 mol. Our only
                unknown for the final conditions is now just the volume of the bubble, V2. We can now
                solve for the volume using the Ideal Gas Law with the known final condition values:
                       P2V2 = n2RT2
                       V2 = (n2RT2) / P2 = (5.82 × 10–4 mol) (0.0821 L-atm/mol-K) (298K) / (1.0 atm)
                       V2 = 1.42 × 10–2 L × 1 mL/(10–3 L) = 14.2 mL

VII. Partial Pressures:
     The partial pressure of a gas is defined as the pressure that gas would exert if it were alone in the
     container. If a mixture of gases is contained inside a vessel, then all of the gases will occupy the
     same volume, V, and all the gases will be at the same temperature, T. The only factors that will
     distinguish them are then the number of moles of each gas that is present in the container and the
     partial pressure each gas exerts within the container. Using the definition of partial pressures, we
     can use Eq. (2) to determine the relationship between the partial pressures of each gas and the
     number of moles of each gas that occupy the same container. I.e., the partial pressure of a gas A,
     PA is related to the number of moles of gas A that are present in the container:
                  nA RT
           PA          .                                                                                (28)
                    V
     Since all the gases that are mixed together in the same container will each occupy the same
     volume, V, and will all be at the same temperature, T, (the temperature of the container), then we

                                                   –9–                                     Raynor: 12/4/2011
can rewrite Eq. (28) in order to bundle together all the values that will be identical for the gases,
and separate those values from the values that can vary, (the partial pressures and numbers of
moles for each):
               RT             RT             RT 
      PA  nA      , PB  nB      , PC  nC     ,                                              (29)
               V              V              V 
Note that the factor given by (RT/V) will be the same for all the gases. We need only calculate it
once. Then, the partial pressures of each gas can simply be found by multiplying that factor by the
number of moles of that particular gas.


Example: A 0.13 g sample of an unknown gas is collected over water at 30 C. It is found that this
gas occupied 85 mL at an atmospheric pressure of 750 mmHg. Calculate the molar mass of this
gas.
     First, we need to look up the vapor pressure of water at 30 C. It is discovered that its vapor
     pressure is 31.82 mmHg at that temperature. We can now use this information to determine
     the partial pressure exerted by the unknown gas:
           PTOT = Punk + Pwater
           750 mmHg = Punk + 31.82 mmHg
           Punk = 718 mmHg × (1 atm / 760 mmHg) = 0.945 atm
     Now we can use the Ideal Gas Law to determine the number of moles of unknown gas that is
     present:
           n = PV / RT = (0.945 atm) (0.085 L) / (0.0821 L-atm/mol-K) (303 K)
           n = 3.2 × 10–3 mol
     Since we were given the mass of this sample, 0.13 g, and we now know how many moles of
     molecules are within that sample, we can determine its molecular weight:
           MW = 0.13 g / 3.2 × 10–3 mol = 41 g/mol


The next example demonstrates Dalton’s Law when combined with a chemical reaction.


Example: Initially a flask contains N2 and H2 with partial pressures of 0.300 atm and 0.850 atm,
respectively. What will be the total pressure and the partial pressures of N2, H2 and NH3 after the
following reaction is complete: N2 + 3 H2 → 2 NH3. [Assume the temperature remains constant.]
     Since all gases are in the same container, at the same temperature, we can note that the partial
     pressures of each gas will be proportional to the number of moles of that gas:
           PA = nA (RT / V) = nA × c [where c is the same constant for all gases]
     We can therefore use the stoichiometry in terms of the partial pressures – i.e., if 1.0 atm of N2
     reacts with 3.0 atm of H2, they will form 2.0 atm of NH3.

                                              – 10 –                                   Raynor: 12/4/2011
          First, we need to determine the limiting reagent by calculating the amount of NH3 that would
          form if all of each gas was consumed:
                If all the N2 reacts, then we get 0.300 atm N2 × (2 atm NH3 / 1 atm N2) = 0.600 atm
                NH3 formed
                If all the H2 reacts, then we get 0.850 atm H2 × (2 atm NH3 / 3 atm H2) = 0.567 atm
                NH3 formed
          Thus, H2 will be the limiting reagent – it will form the least amount of product. We can now
          immediately conclude the following final values for the partial pressures of H2, (which will
          be all used up), and of NH3:
                PH2 = 0 atm
                PNH3 = 0.567 atm
          Before we can determine the partial pressure due to excess N2 that is present, we first need to
          determine how much of it will react, (using the stoichiomety):
                PN2 reacted = 0.850 atm H2 × (1 atm N2 / 3 atm H2) = 0.283 atm N2 reacted
          The amount of N2 remaining will be the difference between the amount we started with and
          the amount that reated:
                PN2 remaining = 0.300 atm – 0.283 atm = 0.017 atm
                PTOT = PN2 + PH2 + PNH3 = 0.017 + 0.000 + 0.567 atm = 0.584 atm
          An alternative approach would be to choose any values you like for the temperature and
          volume and then convert all the partial pressures into moles, using
                nA = PA (V / RT)
          Then use moles to find the limiting reagent, determine the amount of NH3 formed and the
          moles N2 remaining. Finally, using the same values of T and V, convert the moles back into
          partial pressures:
                PA = nA (RT / V)
          This route requires far more calculations, but will ultimately lead to the same final results.
          [Try it on your own using V = 1 L and T = 100 K, or any values you like.]


VIII. Problems involving STP:
     STP stands for “standard.” temperature and pressure. The temperature is 0 C and the pressure is 1
     atm. [Note that the “standard” temperature is not a value close to room temperature, which would
     be 25 C, but is instead the freezing point of water, 0 C!]
     The book would have you memorize that 1 mol of gas occupies 22.4 L at STP to solve problems.
     However, this is quite unnecessary, as shown below. Simply learn that STP means 0 C and 1 atm
     and use the Ideal Gas Law.




                                                  – 11 –                                   Raynor: 12/4/2011
Example: [Example 5.4 in textbook]
    Calculate the volume in L that is occupied by 7.40 g of NH3 at STP.
    For either method, the 1st step, (as always!), is to convert the mass of NH3 into moles NH3,
    using its molar mass:
         7.40 g NH3 × 1 mol NH3 / 17.03 g NH3 = 0.434 mol NH3
    If you use the fact that 1mol occupies 22.4 L at STP as a conversion factor, you get
         0.434 mol NH3 × 22.4 L / 1 mol = 9.73 L
    Alternatively, we can use the Ideal Gas Law directly, using T = 273 K and P = 1 atm:
         (1 atm) V = 0.434 mol × 0.0821 L-atm/mol-K × 273 K
         V = 9.73 L
    Notice that we get the same result, without memorizing that 1 mol of gas occupies 22.4 L at
    STP. This relationship is true ONLY at STP – not at other temperatures or pressures. [If we
    ask the same question, but at room temperature, (298 K), then we get V = 10.6 L instead! Try
    it on your own!]




                                           – 12 –                                 Raynor: 12/4/2011

				
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
views:3
posted:12/4/2011
language:English
pages:12