# Question ad05a

Aug 31, 2016

Here's what I got.

#### Explanation:

I'm assuming that you're starting with a piece of iron that has a known volume and are interested in finding its mass in pounds.

The thing to keep in mind density is that it can be used as a conversion factor to help you go from mass to volume or vice versa.

Let's say that in your case, you have a piece of iron of volume $V$ ${\text{cm}}^{3}$. The density of iron is said to be equal to ${\text{7.9 g cm}}^{- 3}$. This tells you that ${\text{1 cm}}^{3}$ of iron has a mass of $\text{7.9 g}$.

As a result, the piece of iron of volume $V$ will have a mass of

V color(red)(cancel(color(black)("cm"^3))) * overbrace("7.9 g"/(1color(red)(cancel(color(black)("cm"^3)))))^(color(blue)("the given density")) = (7.9 * V)" g"

Now all you have to do is use the given conversion factor

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{1 lb " = " 453.6 g}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

to convert the mass from grams to pounds

(7.9 * V) color(red)(cancel(color(black)("g"))) * "1 lb"/(453.6color(red)(cancel(color(black)("g")))) = (0.0174 * V)" lb"#

As a numerical example, let's say that $V = {\text{122 cm}}^{3}$. The mass in pounds of this volume of iron will be

$\text{mass in lb" = 0.0174 * 122 = "2.12 lb}$