What mass of aluminum has a volume of #6.3# #cm^3#?

1 Answer
Dec 13, 2015

Answer:

The mass of aluminum with a volume of #6.3 \ cm^3# is #17 \ grams#.

Explanation:

We can figure out this out by using aluminum's density. Density is the ratio of the amount of mass that can be fit into a certain volume.

The equation for density is #D = M / V#.

#M# stands for mass, #V# stands for volume, and #D# stands for density.

According to Elmhurst College, the density of aluminum is #(2.7 \ grams)/(mL)#.

In order to use this density with a volume in #cm^3#, we need to convert it into #(grams)/(cm^3)#. Fortunately, #cm^3# and #mL# are equal to each other, so the density of aluminum is #(2.7 \ grams)/(cm^3)#

The equation for density can be rearranged to solve for mass: #V * D = M#

To get the value of #M#, we can multiply #(2.7 \ grams)/(cm^3)# and #6.3 \ cm^3# together, and you are left with the mass of the aluminum.

#6.3 \ cm^3 * (2.7 \ grams)/(cm^3) = 17.01 \ grams#

The #cm^3# both cancel, and you are left with #grams#.

From there, we use significant figures to show the precise digits of our answer. Both #6.3 \ cm^3# and #(2.7 \ grams)/(cm^3)# have only 2 significant figures, so we round our answer to 2 significant figures as well.

#17 \ grams# of aluminum is the final answer.