Chemistry: Molecular Approach (4th Edition)

by Tro, Nivaldo J.

Published by Pearson

ISBN 10: 0134112830

ISBN 13: 978-0-13411-283-1

Chapter 1 - Exercises - Page 38: 52

Answer

a. 0°C b. -321.1°F c. -78.333333333°C d. 310.15 K

Work Step by Step

Formulas: K = 273.15+ °C °C = (°F − 32) × 5/9 K = (°F − 32) × 5/9 + 273.15 a. (32°F − 32) × 5/9 = 0°C b. (77 K − 273.15) × 9/5 + 32 = -321.1°F c. (-109°F-32)×5/9 = -78.333333333°C d. (98.6°F − 32) × 5/9 + 273.15 = 310.15 K

Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.

GradeSaver will pay $15 for your literature essays

GradeSaver will pay $25 for your college application essays

GradeSaver will pay $50 for your graduate school essays – Law, Business, or Medical

GradeSaver will pay $10 for your Community Note contributions

GradeSaver will pay $500 for your Textbook Answer contributions