Chapter 1, Problem 1.53P

Carry out each calculation, paying special attention to significant figures, rounding, and units:

1. $\frac{(4.32\times {10}^7\text{g}}{\frac{4}{3}\left(3.1416\right){\left(1.95\times {10}^2\text{cm}\right)}^3}\left(\text{The term}\frac{4}{3}\text{is exact}\text{.}\right)$ $\frac{\left(1.84\times {10}^2\text{ g}\right){\left(44.7\text{m/s}\right)}^{\text{2}}}{2}\left(\text{The term 2 is exact}\text{.}\right)$ $\frac{{\left(1.07\times {10}^{-4}\text{ mol/L}\right)}^2\left(3.8\times {10}^{-3}\text{mol/L}\right)}{\left(8.35\times {10}^{-5}\text{mol/L}\right){\left(1.48\times {10}^{-2}\text{mol/L}\right)}^3}$