### Density Worksheet Answers - Problems 16-20

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16) convert kg to grams: 57.0 kg x (1000 g / 1 kg) = 5.70 x 10^{4} g

determine volume of the copper wire: (5.70 x 10^{4} g) ÷ (8.94 g/cm^{3}) = 6375.8389 cm^{3}

convert mm to cm: 9.50 mm x (1 cm/10 mm) = 0.950 cm

determine length of wire: 6375.8389 cm^{3} = (3.14159) (0.475 cm)^{2} h; h = 8995 cm = 8.99 x 10^{3} cm

Note: for the step above, you need to know the formula for the volume of a cylinder. Note also the use of 0.475 cm for the radius.

convert cm to meters: 8.99 x 10^{3} cm x (1 m / 100 cm) = 89.9 m

17) assume the penny occupies 1.00 cm^{3}. This means:

copper occupies 0.06025 cm^{3} and zinc occupies 0.93975 cm^{3}.

calculate mass of copper: (0.06025 cm^{3}) (8.94 g/cm^{3}) = 0.538635 g

calculate mass of zinc: (0.93975 cm^{3}) (7.14 g/cm^{3}) = 6.709815 g

determine apparent density: 0.538635 g + 6.709815 g = 7.24845 g

since this mass is in 1.00 cm^{3}, the answer is 7.25 g/cm^{3}

18) calculate volume of ice: (1.42 x 10^{18} cm^{2}) (1.61 x 10^{5} cm) = 2.2862 x 10^{23} cm^{3}

calculate mass of ice: (2.2862 x 10^{23} cm^{3}) (0.92 g/cm^{3}) = 2.1 x 10^{23} g

19) Object A has a larger volume than Object B.

20) The solution to this problem involves the concept of buoyancy.

determine the mass of the cube: (45.0 mL) (0.900 g/cm^{3}) = 40.5 g

The cube will float when 40.5 g of liquid is displaced. We need to know what volume of the liquid weighs 40.5 g.

volume of liquid: (40.5 g) ÷ (1.36 g/mL) = 29.8 mL

This means that 29.8 mL of the cube is submerged (this is the answer to the question), displacing 40.5 g of the liquid. The rest of the cube (45.0 - 29.8) is above the surface of the liquid.

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