Question

Question #67e63

Answer 1

The block has a mass of `"33,000 g"`.

Explanation:

In order to be able to solve this problem, you need to know osmium's density, which is listed at `"22.59 g/cm"""^3`.

Now, your block can be thought of as a rectangular prism, which has a volume of

`color(blue)(V_"rectangular prism" = "length" xx "height" xx "width")`

Since osmium's density is given in grams per cubic centimeters, you'll need to convert your dimensions from inches to centimeters by using the fact that `"1 inch" = "2.54 cm"`.

This means that you can write

`5.0color(red)(cancel(color(black)("in"))) * "2.54 cm"/(1color(red)(cancel(color(black)("in")))) = 5.0 * "2.54 cm"`

You can write the volume of the block as

`V = 5.0 * "2.54 cm" * 4.0 * "2.54 cm" * 4.5 * "2.54 cm"`

`V = 5.0 * 4.0 * 4.5 * 2.54""^3 "cm"^3`

`V = "1475 cm"""^3`

Density is defined as mass per unit of volume

`color(blue)(rho = m/V)`

This means that you can determine the mass of the block by

`m = rho * V = 22.57"g"/color(red)(cancel(color(black)("cm"^3))) * 1475color(red)(cancel(color(black)("cm"^3))) = "33291 g"`

You need to round this off to two sig figs, the number of sig figs you have for the dimensions of the block

`m = color(green)("33,000 g")`