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1 1.8 Density Chapter 1Measurements Copyright © 2009 by Pearson Education, Inc.
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2 Density compares the mass of an object to its volume. is the mass of a substance divided by its volume. Density Expression Density = mass = g or g = g/cm 3 volume mL cm 3 Note: 1 mL = 1 cm 3 Density
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3 Densities of Common Substances (at 4 °C)
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4 Osmium is a very dense metal. What is its density in g/cm 3 if 50.0 g of osmium has a volume of 2.22 cm 3 ? 1) 2.25 g/cm 3 2) 22.5 g/cm 3 3) 111 g/cm 3 Learning Check
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5 Given: mass = 50.0 g volume = 2.22 cm 3 Plan: Place the mass and volume of the osmium metal in the density expression. 2) D = mass = 50.0 g volume2.22 cm 3 calculator answer = 22.522522 g/cm 3 final answer = 22.5 g/cm 3 Solution
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6 Volume by Displacement A solid completely submerged in water displaces its own volume of water. The volume of the object is calculated from the difference in volume. 45.0 mL - 35.5 mL = 9.5 mL = 9.5 cm 3 Copyright © 2009 by Pearson Education, Inc.
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7 Density Using Volume Displacement The density of the zinc object is then calculated from its mass and volume. Density = mass = 68.60 g = 7.2 g/cm 3 volume 9.5 cm 3 Copyright © 2009 by Pearson Education, Inc.
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8 What is the density (g/cm 3 ) of 48.0 g of a metal if the level of water in a graduated cylinder rises from 25.0 mL to 33.0 mL after the metal is added? 1) 0.17 g/cm 3 2) 6.0 g/cm 3 3) 380 g/cm 3 Learning Check object 33.0 mL 25.0 mL
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9 Solution Given: 48.0 g Volume of water = 25.0 mL Volume of water + metal = 33.0 mL Need: Density (g/mL) Plan: Calculate the volume difference in cm 3 and place in density expression. 33.0 mL - 25.0 mL= 8.0 mL 8.0 mL x 1 cm 3 = 8.0 cm 3 1 mL Set Up Problem: Density = 48.0 g = 6.0 g = 6.0 g/cm 3 8.0 cm 3 1 cm 3
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10 Sink or Float Ice floats in water because the density of ice is less than the density of water. Aluminum sinks because its density is greater than the density of water. Copyright © 2009 by Pearson Education, Inc.
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11 Which diagram correctly represents the liquid layers in the cylinder? Karo (K) syrup (1.4 g/mL); vegetable (V) oil (0.91 g/mL); water (W) (1.0 g/mL) 1 2 3 K K W W W V V V K Learning Check
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12 1) vegetable oil 0.91 g/mL water 1.0 g/mL Karo syrup 1.4 g/mL K W V Solution
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13 The density of octane, a component of gasoline, is 0.702 g/mL. What is the mass, in kg, of 875 mL of octane? 1) 0.614 kg 2) 614 kg 3) 1.25 kg Learning Check
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14 Density can be written as an equality. For a substance with a density of 3.8 g/mL, the equality is 3.8 g = 1 mL From this equality, two conversion factors can be written for density. Conversion 3.8 g and 1 mL factors1 mL 3.8 g Study Tip: Density as a Conversion Factor
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15 Solution 1) 0.614 kg Given: D = 0.702 g/mL V= 875 mL Plan: mL g kg Equalities: density 0.702 g = 1 mL and 1 kg = 1000 g Setup: 875 mL x 0.702 g x 1 kg = 0.614 kg 1 mL 1000 g density metric factor factor
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16 If olive oil has a density of 0.92 g/mL, how many liters of olive oil are in 285 g of olive oil? 1) 0.26 L 2) 0.31 L 3) 310 L Learning Check
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17 Solution 2) 0.31 L Given: D = 0.92 g/mL mass = 285 g Need: volume in liters Plan: g mL L Equalities: 1 mL = 0.92 g and 1 L = 1000 mL Set Up Problem: 285 g x 1 mL x 1 L = 0.31 L 0.92 g 1000 mL density metric factorfactor inverted
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18 A group of students collected 125 empty aluminum cans to take to the recycling center. If 21 cans make 1.0 lb aluminum, how many liters of aluminum (D=2.70 g/cm 3 ) are obtained from the cans? 1) 1.0 L2) 2.0 L3) 4.0 L Learning Check
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19 Solution 1) 1.0 L 125 cans x 1.0 lb x 454 g x 1 cm 3 x 1 mL x 1 L 21 cans 1 lb 2.70 g 1 cm 3 1000 mL density factor inverted = 1.0 L
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20 Which of the following samples of metals will displace the greatest volume of water? 1 2 3 25 g of aluminum 2.70 g/mL 45 g of gold 19.3 g/mL 75 g of lead 11.3 g/mL Learning Check
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21 Solution 25 g of aluminum 2.70 g/mL 1) Plan: Calculate the volume for each metal and select the metal sample with the greatest volume. 1) 25 g x 1 mL = 9.3 mL aluminum 2.70 g 2) 45 g x 1 mL = 2.3 mL gold 19.3 g 3) 75 g x 1 mL = 6.6 mL lead 11.3 g density factors
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