Quiz 4: chem 120

Chem. 120: week 4 Question 1: (TCO 6) A sample of helium gas occupies 520 mL at 79 mmHg. For a gas sample at constant temperature, determine the volume of helium at 45 mmHg. Student Answer: Boyles Law P1V1=P2V2 Temp is constant 79 mmHg x 520 mL= 45 mmHg x V2 41080= 45 mmHg x V2 912.88 mL=V2 Instructor Explanation: Chapter 6; Using Boyle’s law, P1V1 = P2V2. We have V1 (520 mL), P1 (79 mmHg) and P2 (45 mmHg). We want to determine V2, so V2 = (P1 x V1)/P2 = (520 mL * 79 mmHg)/45 mmHg) = 912.8 mL. Points Received: 6 of 6 Comments: Question 2.Question : (TCO 6) A gas at a temperature of 65 degrees C occupies a volume of 225 mL. Assuming constant pressure, determine the volume at 150 degrees C. Student Answer: Charles Law (V1/T1)=(V2/T2) K=273+C K= 338 K= 423 (225 mL/ 338K) = (V2/423K) 95175=338 V2 281.58 mL= V2 Instructor Explanation: Chapter 6; Using Charles’ Law, (V1/T1) = (V2/T2). First, convert temperature to KELVIN (T1 = t1 +273) Thus, T1 = 65 + 273 = 338 K. We have V1 (225 mL) & T2 = (150 + 273) = 423 K. V2 = (V1*T2)/T1 = (225 mL*423 K)/338 K = 281.6 mL. Points Received: 6 of 6 Comments: Question 3.Question : (TCO 6) For a gas at standard temperature and pressure with a density of 3.5 g/L, determine its molar mass. Student Answer: Molar Mass= 3.5 g/L x (22.4 L/ 1 mol) = 78.4 g/mol Instructor Explanation: Chapter 6; 1 mol of a gas at STP has a volume of 22.4L. x molar mass = 3.5 g/L x (22.4 L/1 mol) = 78.4 g/mol. Points Received: 5 of 5 Comments: Question 4.Question : (TCO 6) Calculate the volume, in liters, of a 1.5 mol H2S(g) at 50 degrees C and 2 atm. Student Answer: Ideal Gas Law PV=nRT K=273+C K= 323 2 atm(V)= 1.5 mol x 0.0821 x 323K V= 19.88 L Instructor Explanation: Chapter 6; We given that n = 1.5 mol; temperature is 50 degrees , which is 323 K; P = 2 atm; First, convert your temperatures to KELVIN: T = t + 273. Place the known values in this equation: PV= nRT. So, V = nRT/P = (1.5 mol * 0.0821 L-atm/mol-K *323K)/2 atm = 19.9 L Points Received: 5 of 5