Agraduated cylinder contains 63.0 ml of water. a piece of gold, which has a density of 19.3 g/ cm3, is added to the water and the volume goes up to 64.5 ml. calculate the mass in grams of the gold that was added to the water. explain how you got your answer.

29.0 g

Explanation:

The mass of the piece of gold is given by:

m = dV

where

m is the mass

d is the density

V is the volume of the piece of gold

The density of gold is

d = 19.3 g/cm^3

while the volume of the sample is equal to the volume of displaced water, so

V = 64.5 mL - 63.0 mL = 1.5 mL

And since

1 mL = 1 cm^3

the volume is

V = 1.5 cm^3

So the mass of the piece of gold is:

m = (19.3 g/cm^3)(1.5 cm^3)=29.0 g

gas and liquid