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18 May, 19:16

A graduated 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

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  1. 18 May, 21:04
    0
    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
  2. 18 May, 23:14
    0
    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
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