# Question 30f8a

Jun 16, 2017

1.93 xx 10^4"kg"/("m"^3)

#### Explanation:

I'll assume you mean to ask what the density of gold is, in the units of "kg"/("m"^3).

We can use dimensional analysis to convert these units. The process is straightforward, multiplying by conversion factors until you have converted to the desired units. You're eliminating units along the way by converting them to other units. You might see what I mean by doing the work:

((19.3cancel("g"))/(1cancel("cm"^3)))((1color(red)("kg"))/(10^3cancel("g")))((100^3cancel("cm"^3))/(1color(blue)("m"^3))) = 1.93 xx 10^4(color(red)("kg"))/(color(blue)("m"^3))#

We changed from grams to kilograms by multiplying the conversion factor $1 \text{kg" = 10^3"g}$ by the given number of grams so that the unit "$\text{g}$" is eliminated.

The same process went for changing cubic centimeters to cubic meters. We know there are $100$ $\text{cm}$ in $1$ $\text{m}$, so, in three dimensions (to the power of three) there are ${100}^{\textcolor{b l u e}{3}}$ ${\text{cm}}^{3}$ in $1$ ${\text{m}}^{3}$.