# If an object has a density of #8.65# #g##/cm^3#, what is its density in units of #kg##/m^3#?

##### 1 Answer

#### Answer:

#### Explanation:

Your goal here will be to use two **conversion factors**, one to take you from *grams* to *kilograms* and one to take you from *cubic centimeters* to *cubic meters*.

The first conversion factor is pretty straight forward

#color(purple)(|bar(ul(color(white)(a/a)color(black)("1 kg" = 10^3"g")color(white)(a/a)|)))#

For the second conversion factor, use the fact that

#"1 m"^3 = "1 m" xx "1 m" xx "1 m"#

Now, you will have to go from *centimeter* to *decimeter*, then from *decimeter* to *meter* by using the fact that you have

#{: ("1"color(white)(a)"m " = " 10 dm "), ("1 dm " = " 10 cm") :}} implies "1 m" = 10 xx "10 cm" = 10^2"cm"#

Therefore,

#"1 m"^3 = overbrace(10^2"cm")^(color(blue)("= 1 m")) " "xx" " overbrace(10^2"cm")^(color(blue)("= 1 m")) " "xx" " overbrace(10^2"cm")^(color(blue)("= 1 m")) #

and so

#color(purple)(|bar(ul(color(white)(a/a)color(black)("1 m"^3 = 10^6"cm"^3)color(white)(a/a)|)))#

The given density will thus be equivalent to

#8.56 color(red)(cancel(color(black)("g")))/(color(red)(cancel(color(black)("cm"^3)))) * "1 kg"/(10^3color(red)(cancel(color(black)("g")))) * (10^6color(red)(cancel(color(black)("cm"^3))))/"1 m"^3 = color(green)(|bar(ul(color(white)(a/a)color(black)(8.56 * 10^3color(white)(a)"kg m"^(-3))color(white)(a/a)|)))#