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Transcript of 1 Chapter 2 Measurements 2.6 Writing Conversion Factors Basic Chemistry Copyright © 2011 Pearson...
1
Chapter 2 Measurements
2.6Writing Conversion Factors
Basic Chemistry Copyright © 2011 Pearson Education, Inc.
The mass of a packaged product is listed in both U.S. and metric units.
2
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
• Examples:
1 m = 1000 mm1 lb = 16 oz2.205 lb = 1 kg
Equalities
Basic Chemistry Copyright © 2011 Pearson Education, Inc.
3
Exact and Measured Numbers in Equalities
Equalities between units of
• the same system are definitions and are exact numbers
• different systems (metric and U.S.) use measured numbers and count as significant figures
Basic Chemistry Copyright © 2011 Pearson Education, Inc.
4
Some Common Equalities
Basic Chemistry Copyright © 2011 Pearson Education, Inc.
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Equalities on Food Labels
The contents of packaged foods
• are listed as both metric and U.S. units
• indicate the same amount of a substance in two different units
Basic Chemistry Copyright © 2011 Pearson Education, Inc.
The mass of a packaged product is listed in both U.S. and metric units.
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A conversion factor is• a fraction obtained from an equality
Equality: 1 in. = 2.54 cm• written as a ratio with a numerator and
denominator
• is inverted to give two conversion factors for every equality
1 in. and 2.54 cm 2.54 cm 1 in.
Conversion Factors
Basic Chemistry Copyright © 2011 Pearson Education, Inc.
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Write conversion factors for each pair of units:
A. liters and mL
B. hours and minutes
C. meters and kilometers
Learning Check
Basic Chemistry Copyright © 2011 Pearson Education, Inc.
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Write conversion factors for each pair of units:A. liters and mL (1 L = 1000 mL)
1 L and 1000 mL 1000 mL 1 L
B. hours and minutes (1 h = 60 min) 1 h and 60 min
60 min 1 h
C. meters and kilometers (1 km = 1000 m)1 km and 1000 m 1000 m 1 km
Solution
Basic Chemistry Copyright © 2011 Pearson Education, Inc.
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Factors with Powers
A conversion factor can be squared or cubed on both sides of the equality.
Equality 1 in. = 2.54 cm 1 in. and 2.54 cm
2.54 cm 1 in.Equality squared (1 in.)2 = (2.54 cm)2
(1 in.)2 and (2.54 cm)2
(2.54 cm)2 (1 in.)2
Equality cubed (1 in.)3 = (2.54 cm)3
(1 in.)3 and (2.54 cm)3
(2.54 cm)3 (1 in.)3
Basic Chemistry Copyright © 2011 Pearson Education, Inc.
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An equality and conversion factors • may be obtained from information in a word
problem• are for that problem only
Example: The price of one pound (1 lb) of red peppers is $2.39.
Equality: 1 lb peppers = $2.39Conversion factors:
1 lb red peppers and $2.39$2.39 1 lb red peppers
Conversion Factors in a Problem
Basic Chemistry Copyright © 2011 Pearson Education, Inc.
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A percent factor• gives the ratio of the part to the whole.
Percent(%) = part x 100 whole
• uses matching units to express the percent• uses the value 100 and a unit for the whole• Example: A food contains 30% (by mass) fat.
Equality: 100 g food = 30 g fatConversion factors:
30 g fat and 100 g food100 g food 30 g fat
Percent as a Conversion Factor
Basic Chemistry Copyright © 2011 Pearson Education, Inc.
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Percent Factor in a Problem
The thickness of the skin foldat the waist indicates 11% bodyfat. What equality and percentconversion factors can bewritten for body fat in kg? Equality (kg): 100 kg of body mass = 11 kg of fat
Percent conversion factors (kg) 11 kg fat and 100 kg mass 100 kg mass 11 kg fat
Basic Chemistry Copyright © 2011 Pearson Education, Inc.
The thickness of the skin fold at the waist is measured in millimeters (mm) to determine the percent of body fat.
ppm and ppb
Relationships of small percent values are• ppm (parts per million or mg/kg)
• ppb (parts per billion or g/kg)
Example:
A soil sample contains 2 ppm of lead.
Equality: 2 mg of lead = 1 kg of soil
Conversion factors: 2 mg lead and 1 kg soil 1 kg soil 2 mg lead
13 Basic Chemistry Copyright © 2011 Pearson Education, Inc.
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Learning Check
Write the equality and conversion factors for each
of the following:A. square meters and square centimeters
B. jewelry that contains 18% gold
C. one gallon of gas costing $3.29 D. a water sample with 55 ppb of chromium (Cr)
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Solution
A. 1 m2 = (100 cm)2
(1m)2 and (100 cm)2
(100 cm)2 (1m)2
B. 100 g of jewelry = 18 g of gold 18 g gold and 100 g jewelry
100 g jewelry 18 g goldC. 1 gal of gas = $3.29
1 gal and $3.29 $3.29 1 galD. 1 kg of water = 55 g of chromium
1 kg water and 55 g chromium
55 g chromium 1 kg water
Basic Chemistry Copyright © 2011 Pearson Education, Inc.