# Answers_to_Resource_Sheet-3[1] - 3 P(10 6 cm 3 air =(x cm 3...

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C H E M / E N V S 3 8 0 Answers to Resource Sheet-3 Example-3: The concentration of CO 2 in the atmosphere is currently about 380 ppm. (a) If you have an air sample in a container having total pressure or 1 atm, what is the partial pressure exerted by CO 2 in your sample? Concentration of CO 2 in the air = 380 ppm = air of atm CO of atm 1 380 2 µ Therefore, the answer is 380 μatm (or 380 x 10 -6 atm). (b) If your sample of air contained a total of 2 million gas molecules, how many of them will be CO 2 ? 380 ppm means that there are 380 molecules of CO 2 in every million molecules of air. Therefore, in 2 million molecules of air, there will be 760 molecules of CO 2 . (c) The molecular weight of CO 2 is 44. What is the mass of CO 2 in your sample? g mol g mol molecules CO molecules CO of weight molecular Number s Avogadro CO molecules 20 1 1 23 2 2 2 2 10 55 . 5 44 10 02 . 6 760 ' # CO of Mass × = × × = ×
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Unformatted text preview: 3 P)/(10 6 cm 3 air) = (x cm 3 P)/(25 cm 3 air) Solving for x gives 3.0 x 10--4 cm 3 Example-5: If P has a molecular weight of 36 g mol-1 , how many grams of P is present in this sample? Assume 1 atm and 25 degC. We know the volume of P from Example-3. If we can determine how many molecules of P occupy this volume, then using the molecular weight, we can determine the mass of P. To convert the volume of P to moles of P, use the ideal gas law: PV = nRT, where P = 1 atm, T = 25 + 273 = 298 deg. K, and R = 0.0821 L atm mol-1 K-1 . n = PV/RT = (1 atm) (3.0 x 10--7 L) /{(0.082 L atm mol-1 K-1 )(298 K)} = 1.2 x 10-8 mol P When using the Ideal Gas Law, don’t forget to express the volume in L. Recall that 1L = 1000 cm 3 , so 3.0 x 10--4 cm 3 is equivalent to 3.0 x 10--7 L. From molecular weight of P, compute final answer: (1.2 x 10-8 mol P)(36 g mol P-1 ) = 4.3 x 10-7 g T. Komada 1/26/2010...
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