Chapter 1 Measurements 1.6 Writing Conversion Factors 1

Equalities use two different units to describe the same measured amount. are written for relationships between units of the metric system, U.S. units, or between metric and U.S. units. For example, 1 m = 1000 mm 1 lb = 16 oz 2.20 lb = 1 kg 2

Exact and Measured Numbers in Equalities Equalities between units in the same system of measurement are definitions that use exact numbers. different systems of measurement (metric and U.S.) use measured numbers that have significant figures. Exception: The equality 1 in. = 2.54 cm has been defined as an exact relationship. Thus, 2.54 is an exact number. 3

Some Common Equalities in qt 946 mL = 1 qt

Equalities on Food Labels The contents of packaged foods in the U.S. are listed in both metric and U.S. units. indicate the same amount of a substance in two different units. 5

Conversion Factors A conversion factor is obtained from an equality. Equality: 1 in. = 2.54 cm written as a fraction (ratio) with a numerator and denominator. inverted to give two conversion factors for every equality. 1 in. and 2.54 cm 2.54 cm 1 in. 6

Learning Check Write conversion factors from the equality for each of the following. A. liters and mL B. hours and minutes C. meters and kilometers 7

Conversion Factors in a Problem A conversion factor may be obtained from information in a word problem. is written for that problem only. Example 1: The price of one pound (1 lb) of red peppers is $ lb red peppers and$2.39 $2.391 lb red peppers Example 2: The cost of one gallon (1 gal) of gas is $ gallon of gasand $2.89 $2.891 gallon of gas 8

Percent as a Conversion Factor A percent factor gives the ratio of the parts to the whole. % = parts x 100 whole uses the same unit in the numerator and denominator. uses the value 100. can be written as two factors. Example: A food contains 30% (by mass) fat. 30 g fat and100 g food 100 g food 30 g fat 9

Percent Factor in a Problem The thickness of the skin fold at the waist indicates 11% body fat. What factors can be written for percent body fat (in kg)? Percent factors using kg: 11 kg fat and 100 kg mass 100 kg mass11 kg fat 10

Smaller Percents: ppm and ppb Small percents are shown as ppm and ppb. Parts per million (ppm) = mg part/kg whole Example: The EPA allows 15 ppm cadmium in food colors 15 mg cadmium = 1 kg food color Parts per billion ppb =  g part/kg whole Example: The EPA allows10 ppb arsenic in public water 10  g arsenic = 1 kg water 11

Arsenic in Water Write the conversion factors for 10 ppb arsenic in public water from the equality 10  g arsenic = 1 kg water. Conversion factors: 10  g arsenic and 1 kg water 1 kg water10  g arsenic 12

Study Tip: Conversion Factors An equality is written as a fraction (ratio). provides two conversion factors that are the inverse of each other. 13

Learning Check Write the equality and conversion factors for each of the following. A. meters and centimeters B. jewelry that contains 18% gold C. One gallon of gas is $

Risk-Benefit Assessment A measurement of toxicity is LD 50 or “lethal dose.” the concentration of the substance that causes death in 50% of the test animals. in milligrams per kilogram (mg/kg or ppm) of body mass. in micrograms per kilogram (  g/kg or ppb) of body mass. 15

Learning Check The LD 50 for aspirin is 1100 ppm. How many grams of aspirin would be lethal in 50% of persons with a body mass of 85 kg? A. 9.4 g B. 94 g C g 16

Chapter 1 Measurements 1.7 Problem Solving 17

Given and Needed Units To solve a problem, identify the given unit. identify the needed unit. Example: A person has a height of 2.0 meters. What is that height in inches? The given unit is the initial unit of height. given unit = meters (m) The needed unit is the unit for the answer. needed unit = inches (in.) 18

Learning Check An injured person loses 0.30 pints of blood. How many milliliters of blood would that be? Identify the given and needed units given in this problem. Given unit= _______ Needed unit = _______ 19

Study Tip: Problem Solving Using GPS The steps in the Guide to Problem Solving (GPS) are useful in setting up a problem with conversion factors. 20

Problem Setup Unit 1 x Unit 2 = Unit 2 Unit 1 Given x Conversion= Needed unit factor unit 21

Setting Up a Problem How many minutes are in 2.5 hours? Given unit= 2.5 h Needed unit=min Plan=h min Set Up Problem Given Conversion Needed unit unitfactor 22 Copyright © 2009 by Pearson Education, Inc.

Learning Check A rattlesnake is 2.44 m long. How many cm long is the snake? 1) 2440 cm 2)244 cm 3)24.4 cm 23

Using Two or More Factors Often, two or more conversion factors are required to obtain the unit needed for the answer. Unit 1 Unit 2Unit 3 Additional conversion factors are placed in the setup problem to cancel each preceding unit. Given unit x factor 1 x factor 2 = needed unit Unit 1 x Unit 2 x Unit 3 = Unit 3 Unit 1 Unit 2 24

Example: Problem Solving How many minutes are in 1.4 days? How many ounces are in 1.0 kg? 25

Study Tip: Check Unit Cancellation Be sure to check the unit cancellation in the setup. The units in the conversion factors must cancel to give the correct unit for the answer. What is wrong with the following setup? 1.4 day x 1 day x 1 h 24 h 60 min = day 2 /min is not the unit needed Units don’t cancel properly. 26

Learning Check A bucket contains 4.65 L of water. Write the setup for the problem and calculate the gallons of water in the bucket. plan: Equalities: Set Up Problem: 27

Learning Check If a ski pole is 3.0 feet in length, how long is the ski pole in mm? If your pace on a treadmill is 65 meters per minute, how many minutes will it take for you to walk a distance of 7500 feet? 28

Learning Check How many lb of sugar are in 120 g of candy if the candy is 25% (by mass) sugar? 29

Chapter 1Measurements 1.8 Density 30 Copyright © 2009 by Pearson Education, Inc.

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 31

Densities of Common Substances 32 (at 4 °C)

Learning Check 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 33

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 mL mL = 9.5 mL = 9.5 cm 3 34

Density Using Volume Displacement The density of the zinc object is then calculated from its mass and volume. Density = mass = g = 7.2 g/cm 3 volume 9.5 cm 3 35

Learning Check 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 36 object 33.0 mL 25.0 mL

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. 37

Learning Check 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) K K W W W V V V K

Learning Check The density of octane, a component of gasoline, is g/mL. What is the mass, in kg, of 875 mL of octane? 1) kg 2) 614 kg 3) 1.25 kg 39

Study Tip: Density as a Conversion Factor 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 40

Learning Check 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 41

Learning Check 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 42

Learning Check Which of the following samples of metals will displace the greatest volume of water? g of aluminum 2.70 g/mL 45 g of gold 19.3 g/mL 75 g of lead 11.3 g/mL