Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning
ISBN 10: 1133949649
ISBN 13: 978-1-13394-964-0

Chapter 2 Atoms, Molecules, and Ions - Study Questions - Page 95j: 152


a)$7.010\times10^{-4}\ mol$ $U_2O_5$ Uranium (V) oxide $3.58\times10^{-4}\ mol$ b)238 c) 6

Work Step by Step

a) Number of moles of uranium: $0.169\ g\div238.02891\ g/mol=7.010\times10^{-4}\ mol$ Number of moles of oxygen: $(0.199-0.169)\div15.9994=1.876\times10^{-3}\ mol$ Dividing the oxygen value by uranium's: $1.876\times10^{-3}/7.010\times10^{-4}=2.67\approx5/2$ The empirical formula is then : $U_2O_5$ which is called Uranium (V) oxide whose molecular weight is $556.06\ g/mol$ Number of moles obtained: $0.199\ g/mol\div556.06\ g/mol=3.58\times10^{-4}\ mol$ b) The isotope with mass number of 238 because the atomic weight is closest to this mass number. c) Molecular weight of: $UO_2(NO_3)_2$:$394.04\ g/mol$ $H_2O$: $18.015\ g/mol$ The number of moles must have been the same after the dehydration: $0.679\ g\div394.04\ g/mol= 1.72\times10^{-3}\ mol$ Before: $0.865\ g\div1.72\times10^{-3}\ mol=501.98\ g/mol$ $501.98=394.04+z\times18.015$ $z=6$
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