Answer
B
Work Step by Step
Heat removal required: $100\ g\times 4.20\ J/g.K \times 20\ K=8400\ J=8.4\ kJ$
Amount of dry ice required: $8.4\ kJ \div 25.2\ kJ/mol\times 44.01\ g/mol=14.67\ g$
Chemistry and Chemical Reactivity (9th Edition)
by Kotz, John C.; Treichel, Paul M.; Townsend, John R.; Treichel, David A.
Published by
Cengage Learning
ISBN 10:
1133949649
ISBN 13:
978-1-13394-964-0
Chapter 12 The Solid State - 12-6 Phase Changes Involving Solids - Review & Check for Section 12-6 - Page 463: 2
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