## Chemistry: Molecular Approach (4th Edition)

Published by Pearson

# Chapter 1 - Exercises - Page 41: 127

#### Answer

$\underline{1.28\times {{10}^{11}}\text{ km}}$ and $\underline{2.4\times {{10}^{19}}\text{ km}}$

#### Work Step by Step

$1\text{ pm}={{10}^{-12}}\ m$ Diameter of a hydrogen atom in meters is $212\times {{10}^{-12}}\ m$. Now, $1\text{ km}={{10}^{3}}m$. The length in kilometers of a row of $6.02\times {{10}^{23}}$ atoms is calculated as follows: $\left( 6.02\times {{10}^{23}}\text{ atoms} \right)\left( \frac{212\times {{10}^{-12}}\ \text{m}}{1\text{ atom}} \right)\left( \frac{1\text{ km}}{{{10}^{3}}\text{ m}} \right)=1.28\times {{10}^{11\text{ }}}\text{km}$ Since $1\text{ cm}={{10}^{-2}}\ m$, diameter of a ping pong ball in meters is $4\times {{10}^{-2}}\ m$. So, the length in kilometers of a row of $6.02\times {{10}^{23}}$ ping pong balls is calculated as follows: $\left( 6.02\times {{10}^{23}}\text{ balls} \right)\left( \frac{4.0\times {{10}^{-2}}\text{ m}}{1\text{ ball}} \right)\left( \frac{1\text{ km}}{{{10}^{3}}\text{ m}} \right)=2.4\times {{10}^{19\text{ }}}\text{km}$ The length of a row of hydrogen atoms is $\underline{1.28\times {{10}^{11}}\text{ km}}$ and that of ping pong balls is $\underline{2.4\times {{10}^{19}}\text{ km}}$.

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