Agraduated cylinder if filled with water to a level of 40.0 ml. when a piece of paper of copper is lowered into the cylinder , the water level rises to 63.4ml. find the volume of the copper sample. if the density of the copper is 8.9g/cm³, what is the mass

The level of the water without the copper is 40.0 mL.
After placing the copper, the water reached to 63.4 mL. 

So the displacement  is
Displacement = 63.4 - 40.0 
Displacement = 23.4 mL

The displacement is equal to the volume of the copper.
The volume of the copper is 23.4 cm^3.

The formula of the density is
Density = Mass / Volume
8.9 g/cm^3 = Mass / 23.4 cm^3
Mass = 8.9 / 23.4
Mass = 0.38034 g
Mass = 3.8 x 10^-1 g

So the mass is 3.8 x 10^-1 g and the volume is 23.4 cm^3

Given:

Water level is 40mL

Water level rises to 63.4mL after copper is placed

Density is 8.9g/cm3

Required:

Mass of the object

Solution:

The displaced water level is the volume of the copper

63.4mL – 40mL = 23.4mL

 

Since 1mL = 1cm3, therefore 23.4mL is 23.4cm3

The density formula is D = M/V where D is density, M is mass of the object and V is the volume of the object. Rearranging we get M = DV.

M = DV

M = (8.9g/cm3)(23.4cm3)

M = 208 g

It's have to be 12 or idk

Aliquid is an answer i hope this