## Chemistry (7th Edition)

Published by Pearson

# Chapter 1 - Chemical Tools: Experimentation and Measurement - Worked Example - Page 15: 7

#### Answer

(a) $E_k$ = $7.4 \times 10^{-13}$ J (b) $0.74$ pJ.

#### Work Step by Step

(a) $E_k = \frac{1}{2}mv^2$ $E_k = \frac{1}{2}(6.6 \times 10^{-27}kg)(1.5 \times 10^7\frac{m}{s})^2$ $E_k = 7.4 \times 10^{-13}$ $\frac{kg*m^2}{s^2}$ = $7.4 \times 10^{-13}$ J (b) The appropriate prefix is pico (p), because it is the closest to $10^{-13}$ since it represents $10^{-12}$ $7.4 \times 10^{-13}$ J$= 7.4 \times 10^{-13}\frac{1}{10^{-12}}$ pJ = $0.74$ pJ.

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