Answer
$Al_2O_3$: 101.96 g/mol
$Na_3AlF_6$: 209.95 g/mol
Work Step by Step
$ Al_2O_3 $ : ( 26.98 $\times$ 2 )+ ( 16.00 $\times$ 3 )= 101.96 g/mol
$ Na_3AlF_6 $ : ( 22.99 $\times$ 3 )+ ( 26.98 $\times$ 1 )+ ( 19.00 $\times$ 6 )= 209.95 g/mol
Chemistry: Atoms First (2nd Edition)
by Zumdahl, Steven S.; Zumdahl, Susan A.
Published by
Cengage Learning
ISBN 10:
1305079248
ISBN 13:
978-1-30507-924-3
Chapter 5 - Exercises - Page 240d: 49
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