Chemistry: The Central Science (13th Edition)

Published by Prentice Hall
ISBN 10: 0321910419
ISBN 13: 978-0-32191-041-7

Chapter 3 - Chemical Reactions and Reaction Stoichiometry - Exercises: 3.32

Answer

1 - $NaHCO_3$ 2 - $CO_2$ 3 - $Ne$

Work Step by Step

1. Find the number of moles in 42g of $NaHCO_3$ - Molar Mass: 23 + 1 + 12 + (16*3) = 84g/mol - n(moles) = $\frac{mass}{molarmass}$ - n(moles) = $\frac{42}{84}$ = 0.5 moles. But, since there is 6 atoms for each molecule, the number of moles of atoms is 0.5 * 6 = 3 2. The number of moles of $CO_2$ is 1.5, therefore, the number of atom moles is 1.5 * 3 = 4.5. 3. Convert $6 \times 10^{24}$ atoms of $Ne$ into number of moles. $n(atoms) = n(moles) * 6.0 \times 10^{23}$ $6 \times 10^{24} = n(moles) * 6.0 \times 10^{23}$ $n(moles) = \frac{6\times 10^{24}}{6\times 10^{23}} = 10$ Therefore, in order of increasing number of atoms, this samples are: $NaHCO_3$, $CO_2$ and $Ne$.
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.