Question 3 $1 \mathrm{pts}$ Iron has a density of $7.874 \mathrm{~g} / \mathrm{cm}^{3}$. To 4 significant figures, what is the mass of a rectangular block of iron with dimensions of $3.000 \mathrm{~cm}$ by $4.000 \mathrm{~cm}$ by $5.000 \mathrm{~cm}$ ? Note: Give your answer in units of grams, but do not explicitly include units in your answer. Only include a number in the box below. If you include a number and $a$ " $g$ " to represent the unit, your answer would be counted wrong.

Step 1 ：Given that the density of iron is \(7.874 \, \text{g/cm}^{3}\), and the dimensions of the rectangular block of iron are \(3.000 \, \text{cm} \times 4.000 \, \text{cm} \times 5.000 \, \text{cm}\).

Step 2 ：The volume of a rectangular block is calculated by multiplying its length, width, and height. So, \(\text{Volume} = 3.000 \, \text{cm} \times 4.000 \, \text{cm} \times 5.000 \, \text{cm} = 60.0 \, \text{cm}^{3}\).

Step 3 ：The mass of a rectangular block can be calculated by multiplying its volume by its density. So, \(\text{Mass} = \text{Volume} \times \text{Density} = 60.0 \, \text{cm}^{3} \times 7.874 \, \text{g/cm}^{3} = 472.44 \, \text{g}\).

Step 4 ：Final Answer: The mass of the rectangular block of iron is \(\boxed{472.44}\) grams.