# Question #8ad0c

##### 1 Answer

Here's how you can do that.

#### Explanation:

As you know, **density** represents mass per unit of volume.

In your case, an unknown substance is said to have a density of **every**

#"density" = "0.384 g"/"1 cm"^3#

Now, your goal here is to convert the mass from *grams* to *pounds* and the unit of volume from *cubic centimeters* to *cubic feet*.

A well-known conversion factor that you can use to convert the mass of

#"1 kg = 2.20462 lbs"#

Now, you should already know that

#"1 kg" = 10^3# #"g"#

so you can say that you have

#0.384 color(red)(cancel(color(black)("g"))) * (1 color(red)(cancel(color(black)("kg"))))/(10^3color(red)(cancel(color(black)("g")))) * "2.20462 lbs"/(1color(red)(cancel(color(black)("kg")))) = 8.467 * 10^(-4)# #"lbs"#

At this point, you can rewrite the density as

#"density" = (8.467 * 10^(-4)color(white)(.)"lbs")/"1 cm"^3 color(white)(color(blue)( larr " equal to 0.384 g")/a#

Finally, to convert the volume, use the conversion factor

#"1 ft = 0.3048 m"#

A more useful form will be

#"1 ft"^3 = "0.3048 m" * "0.3048 m" * "0.3048 m"#

# = "0.028317 m"^3#

As you know, you have

#"1 m" = 10^2# #"cm"#

This means that

#"1 cm"^3 = "1 cm" * "1 cm" * "1 cm"#

# = 1 color(red)(cancel(color(black)("cm"))) * "1 m"/(10^2color(red)(cancel(color(black)("cm")))) * 1 color(red)(cancel(color(black)("cm"))) * "1 m"/(10^2color(red)(cancel(color(black)("cm")))) * 1 color(red)(cancel(color(black)("cm"))) * "1 m"/(10^2color(red)(cancel(color(black)("cm"))))#

# = "1 m" * "1 m" * "1 m" * 10^(-6)#

# = 10^(-6)# #"m"^3#

You can thus say that you have

#10^(-6) color(red)(cancel(color(black)("m"^3))) * "1 ft"^3/(0.028317color(red)(cancel(color(black)("m"^3)))) = 3.5314 * 10^(-5)# #"ft"^3#

At this point, the density of the substance is equal to

#"density" = (8.467 * 10^(-4)color(white)(.)"lbs")/(3.5314 * 10^(-5)color(white)(.)"ft"^3) color(white)(color(blue)( larr " equal to 0.384 g")/color(blue)(larr " equal to 1 cm"^3)#

To find the mass of *one unit of volume*, i.e. of

#"density" = color(darkgreen)(ul(color(black)(24.0 color(white)(.)"lbs ft"^(-3))))#

The answer must be rounded to three **sig figs**, the number of sig figs you have for the density of the substance.