Let's Review: The Tools of Quantitative Chemistry - 6 Problem Solving and Chemical Arithmetic - Review & Check for Section 6 - Page 43: 1

C

Volume of the wire: $\rho=m/V,\ m=4.26\ kg,\ \rho=8.96\ g/cm^3$ $V=4.26\ kg \times 1000\ g/kg \div8.96\ g/cm^3=475.45\ cm^3$ Cross-sectional area of the wire: $A=\frac{\pi D^2}4,\ D=2.2\ mm=0.22\ cm$ $A=0.038\ cm^2$ Length: $V=A.L$ $L=12,507.5\ cm = 125\ m$