Chemistry: Molecular Approach (4th Edition)

Published by Pearson
ISBN 10: 0134112830
ISBN 13: 978-0-13411-283-1

Chapter 2 - Exercises - Page 83: 131

Answer

25.06 g/mol

Work Step by Step

Use weighted masses. The abundance of the first isotope is given, so subtracted 100 with that number. This number is then solved for using the ration they give you.. 7899*23.98504+.1101*25.98259+.1x=24.312 x=25.06 g/mol
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