What is the approximate density of a mineral with a mass of 262.2 grams that displaces 46 cubic centimeters of water?

1 Answer
May 21, 2017

Answer:

#d=5.7 " g"/"cm"^3#

Explanation:

#d=#?
#m=262.2" g"#

The volume of the mineral is equal to the volume of water it displaced, because the density of water is #1" g"//"cm"^3#
#V=46 " cm"^3#

Formula for density is

#d=m/V=(262.2" g")/(46 " cm"^3)=5.7 " g"/"cm"^3#