Chemistry: The Central Science (13th Edition)

by Brown, Theodore E.; LeMay, H. Eugene; Bursten, Bruce E.; Murphy, Catherine; Woodward, Patrick; Stoltzfus, Matthew E.

Published by Prentice Hall

ISBN 10: 0321910419

ISBN 13: 978-0-32191-041-7

Chapter 10 - Gases - Exercises - Page 435: 10.43b

Answer

9.40L

Work Step by Step

At STP condition, temp= 273K, P = 1 atm So by using the combined gas law PV = nRT and the n from question 10.43a, (n=0.42mol) (1 atm)(V) = (0.42 mol)(0.08201)(273K) Solving for V, V = (0.42)(0.08201)(273)/(1) = 9,4L

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