Gpcinee Second, we find the density of the unknown rectangular solid. volume of the rectangular block. 5.00 cm x 2.55 cm x 1.25 cm = 15.9 cm³ bu 139.443 g 15.9 cm3 = 8.77 g/cm3 The thickness of a sheet of metal foil is too thin to measure with a ruler. However, we Can find the thickness indirectly by calculation. Given the mass, length, and width of a metar Toll, we can use the density of the metal to calculate the thickness of the foil. Example Exercise 4 • Thickness of an Aluminum Foil A piece of aluminum foil has a mass of 0.212 g and measures 5.10 cm by 10.25 cm. ont ol Given the density of aluminum, 2.70 g/cm3, calculate the thickness of the foil. Solution: To calculate the thickness of the foil, we must first find the volume. The volume can be calculated using a density factor as follows. 1 cm3 0.212 g X 2.70 g 0.0785 cm3 %3D n-0.01o The thickness is found after dividing the volume by its length and width. 0.0785 cm3 diod 0.00150 cm (1.50 x 10-3 cm) (5.10cm) (10.25 cm) Density of Liquids and Solids 35
Gpcinee Second, we find the density of the unknown rectangular solid. volume of the rectangular block. 5.00 cm x 2.55 cm x 1.25 cm = 15.9 cm³ bu 139.443 g 15.9 cm3 = 8.77 g/cm3 The thickness of a sheet of metal foil is too thin to measure with a ruler. However, we Can find the thickness indirectly by calculation. Given the mass, length, and width of a metar Toll, we can use the density of the metal to calculate the thickness of the foil. Example Exercise 4 • Thickness of an Aluminum Foil A piece of aluminum foil has a mass of 0.212 g and measures 5.10 cm by 10.25 cm. ont ol Given the density of aluminum, 2.70 g/cm3, calculate the thickness of the foil. Solution: To calculate the thickness of the foil, we must first find the volume. The volume can be calculated using a density factor as follows. 1 cm3 0.212 g X 2.70 g 0.0785 cm3 %3D n-0.01o The thickness is found after dividing the volume by its length and width. 0.0785 cm3 diod 0.00150 cm (1.50 x 10-3 cm) (5.10cm) (10.25 cm) Density of Liquids and Solids 35
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 44E: Although the preferred SI unit of area is the square meter, land is often measured in the metric...
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Question
An aluminum foil weighs 0.500 g and measures 10.00 cm by 10.00 cm. Given the density of aluminum is 2.70 g/cm3, refer to Example Excercise 4 and find the thickness of the foil.
Transcribed Image Text:Gpcinee
Second, we find the density of the unknown rectangular solid.
volume of the rectangular block.
5.00 cm x 2.55 cm x 1.25 cm = 15.9 cm³
bu 139.443 g
15.9 cm3
= 8.77 g/cm3
The thickness of a sheet of metal foil is too thin to measure with a ruler. However, we
Can find the thickness indirectly by calculation. Given the mass, length, and width of a metar
Toll, we can use the density of the metal to calculate the thickness of the foil.
Example Exercise 4 • Thickness of an Aluminum Foil
A piece of aluminum foil has a mass of 0.212 g and measures 5.10 cm by 10.25 cm.
ont ol Given the density of aluminum, 2.70 g/cm3, calculate the thickness of the foil.
Solution: To calculate the thickness of the foil, we must first find the volume. The
volume can be calculated using a density factor as follows.
1 cm3
0.212 g X 2.70 g
0.0785 cm3
%3D
n-0.01o
The thickness is found after dividing the volume by its length and width.
0.0785 cm3
diod
0.00150 cm (1.50 x 10-3 cm)
(5.10cm) (10.25 cm)
Density of Liquids and Solids
35
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