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CHEM 120 Quiz 4 (Version 1), CHEM 120: Introduction to General, Organic & Biological Chemistry with Lab, Verified and Correct Answers, Chamberlain College of Nursing.$7.99
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CHEM 120 Quiz 4 (Version 1), CHEM 120: Introduction to General, Organic & Biological Chemistry with Lab, Verified and Correct Answers, Chamberlain College of Nursing.
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Course
CHEM 120, Chamberlain College of Nursing
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CHEM 120, Chamberlain College Of Nursing
CHEM 120 Quiz 4 (Version 1), CHEM 120: Introduction to General, Organic & Biological Chemistry with Lab, Verified and Correct Answers, Chamberlain College of Nursing.
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 Chapter 6; Using Boyle’s law, P1V1 = P2V2. We have V1 (520 mL), P1
Explanation: (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 Chapter 6; Using Charles’ Law, (V1/T1) = (V2/T2). First, convert
Explanation: 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 Chapter 6; 1 mol of a gas at STP has a volume of 22.4L. x molar mass =
Explanation: 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 Chapter 6; We given that n = 1.5 mol; temperature is 50 degrees , which
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