Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 16 - Exercises and Problems - Page 310: 47

Answer

Please see the work below.

Work Step by Step

We know that $(\frac{125Kcal}{mil})(26.2mil)=3275Kcal$ $(\frac{1gram}{9Kcal})(3275Kcal)=364grams$
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