Solved: An aluminum engine block has a volume of 4.77 L and a mass of 12.88 kg. What is

Step 1 of 2

Given:   Mass = 12,88 kg  and Volume = 4.77L

Conversion of unit:

We know,

1 kg =1000 g

1000 mL = 1L

1 cm3 =1 mL

The conversion factors are :\(\frac{1000 \mathrm{g}}{1 \mathrm{kg}}, \frac{1000 \mathrm{mL}}{1 L}, \frac{1 \mathrm{cm} 3}{1 \mathrm{ml}}\)

Volume of Al will be \(=4.77 \mathrm{L} \times \frac{\frac{1000 \mathrm{mL}}{1 L}}{1 \mathrm{L}}=4.77 \times 10^{3} \mathrm{mL}\)

Mass of the Al will be \(=12.88 \mathrm{kg} \times \frac{1000 \mathrm{g}}{1 \mathrm{kg}}=12.88 \times 10^{3} \mathrm{g}\)

Conversion of volume mL to cm3.

Volume of Al will be: 

\(\begin{array}{l} =4.77 \mathrm{L} \times \frac{1000 \mathrm{mL}}{1 \mathrm{L}} \\ =4.77 \times 10^{3} \mathrm{mL} \times \frac{1 \mathrm{cm} 3}{1 \mathrm{mL}} \\ =4.77 \times 10^{3} \mathrm{cm}^{3} \end{array}\)