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DENSITY is an important DENSITY is an important and useful physical propertyand useful physical propertyDENSITY is an important DENSITY is an important
and useful physical propertyand useful physical property
Gold
Mercury
13.6 g/cm13.6 g/cm33
19.3 g/cm19.3 g/cm33
Density Presentationby
Michael A. Russellfor the
Teaching and LearningCooperative
atMt. Hood
Community College
DENSITY is an important DENSITY is an important and useful physical propertyand useful physical propertyDENSITY is an important DENSITY is an important
and useful physical propertyand useful physical property
Density = mass(g) volume(cm3)
Gold
Mercury
13.6 g/cm13.6 g/cm33
19.3 g/cm19.3 g/cm33
DENSITY is an important DENSITY is an important and useful physical propertyand useful physical propertyDENSITY is an important DENSITY is an important
and useful physical propertyand useful physical property
Density = mass(g) volume(cm3)
Gold
Mercury
13.6 g/cm13.6 g/cm33
19.3 g/cm19.3 g/cm33
Many substances can be identified based solely
upon their density; very important to chemists
Density = mass(g) volume(cm3)
Density ProblemDensity ProblemProblemProblem: A piece of copper has a mass of : A piece of copper has a mass of
57.54 g. It is 9.36 cm long, 7.23 cm wide, 57.54 g. It is 9.36 cm long, 7.23 cm wide, and 0.95 mm thick. Calculate density and 0.95 mm thick. Calculate density (g/cm(g/cm33).).
Density Problem
SOLUTIONSOLUTION1. Get dimensions in common units.1. Get dimensions in common units.
0.95 mm • 1 cm10 mm
= 0.095 cm
Density Problem
SOLUTIONSOLUTION1. Get dimensions in common units.1. Get dimensions in common units.
0.95 mm • 1 cm10 mm
= 0.095 cm
2. Calculate volume in cubic centimeters.2. Calculate volume in cubic centimeters.
(9.36 cm)(7.23 cm)(0.095 cm) = 6.4 cm(9.36 cm)(7.23 cm)(0.095 cm) = 6.4 cm33
Density Problem
SOLUTIONSOLUTION1. Get dimensions in common units.1. Get dimensions in common units.
0.95 mm • 1 cm10 mm
= 0.095 cm
57.54 g6.4 cm3
= 9.0 g/cm3
3. Calculate the density.3. Calculate the density.
2. Calculate volume in cubic centimeters.2. Calculate volume in cubic centimeters.
(9.36 cm)(7.23 cm)(0.095 cm) = 6.4 cm(9.36 cm)(7.23 cm)(0.095 cm) = 6.4 cm33
Relative Densities of the ElementsRelative Densities of the Elements
DENSITYDENSITYStyrofoamStyrofoam BrickBrick
Be sure to notice that Be sure to notice that DENSITY is an DENSITY is an INTENSIVE INTENSIVE PROPERTYPROPERTY of of matter.matter.
INTENSIVEINTENSIVE - does not - does not depend on quantity of depend on quantity of matter.matter.
Contrast with Contrast with EXTENSIVEEXTENSIVE - - depends on quantity of depends on quantity of matter. Examples are matter. Examples are mass and volume.mass and volume.
PROBLEM: Mercury (Hg) has a density of 13.6 PROBLEM: Mercury (Hg) has a density of 13.6 g/cmg/cm33. What is the mass of 95 mL of Hg? In . What is the mass of 95 mL of Hg? In grams? In pounds?grams? In pounds?
Solve the problem using Solve the problem using DIMENSIONAL DIMENSIONAL ANALYSIS ANALYSIS - see the - see the Dimensional Analysis and Factor Label handouts
PROBLEM: Mercury (Hg) has a density of 13.6 PROBLEM: Mercury (Hg) has a density of 13.6 g/cmg/cm33. What is the mass of 95 mL of Hg?. What is the mass of 95 mL of Hg?
First, note that First, note that 1 cm1 cm33 = 1 mL = 1 mL
Then, use dimensional analysis to Then, use dimensional analysis to calculate mass.calculate mass.
PROBLEM: Mercury (Hg) has a density of 13.6 PROBLEM: Mercury (Hg) has a density of 13.6 g/cmg/cm33. What is the mass of 95 mL of Hg?. What is the mass of 95 mL of Hg?
95 cm3 • 13.6 g1 cm3
= 1.3 x103 g
First, note that First, note that 1 cm1 cm33 = 1 mL = 1 mL
Then, use dimensional analysis to Then, use dimensional analysis to calculate mass.calculate mass.
PROBLEM: Mercury (Hg) has a density of 13.6 PROBLEM: Mercury (Hg) has a density of 13.6 g/cmg/cm33. What is the mass of 95 mL of Hg?. What is the mass of 95 mL of Hg?
95 cm3 • 13.6 g1 cm3
= 1.3 x103 g
First, note that First, note that 1 cm1 cm33 = 1 mL = 1 mL
What is the mass in pounds? (1 lb = What is the mass in pounds? (1 lb = 454 g)454 g)
