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: 150

Answer

a) $3.3\times10^{22}\ atoms$ b) $1.82\times10^{-23}\ cm^3$ $0.193\ nm$

Work Step by Step

a) Volume of the cube: $1.000\ cm^3$ Mass of the cube: $11.35\ g$ Atomic weight of lead: $207.2\ g/mol$ Number of moles: $11.35\ g\div207.2\ g/mol=5.48\times10^{-2}\ mol$ $5.48\times10^{-2}\ mol\times6.022\times10^{23}\ atoms/mol=3.3\times10^{22}\ atoms$ b) 60% of the volume: $0.6\ cm^3$ Atomic volume: $0.6\ cm^3\div3.3\times10^{22}\ atoms= 1.82\times10^{-23}\ cm^3$ Radius: $\sqrt[3]{3\times1.82\times10^{-23}\ cm^3\div4\pi}=1.93\times10^{-8}\ cm=0.193\ nm$
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.