# Question #67e63

##### 1 Answer

#### Answer:

The block has a mass of

#### Explanation:

In order to be able to solve this problem, you need to know osmium's density, which is listed at

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

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")#