Chapter 3: The Metric System • The English system was used primarily in the British Empire and wasn’t very standardized. • The French organized a committee to devise a universal measuring system. • After about 10 years, the committee designed and agreed on the metric system. • The metric system offers simplicity with a single base unit for each measurement. © 2011 Pearson Education, Inc. Chapter 3 1 Metric System Basic Units Base units: -temperature: kelvin (K) -amount of substance: mole (mol) -heat: joules (J) - (1 cal = 4.184 J ) -power: watts (W) - (J/s) © 2011 Pearson Education, Inc. Chapter 3 2 1 Original Metric Unit Definitions • A meter was defined as 1/10,000,000 of the distance from the North Pole to the equator. • A kilogram (1000 grams) was equal to the mass of a cube of water measuring 0.1 m on each side. • A liter was set equal to the volume of one kilogram of water at 4 °C. © 2011 Pearson Education, Inc. 3 Chapter 3 Metric System Advantage • Another advantage of the metric system is that it is a decimal system. • It uses prefixes . • For example: base unit is meters. – A kilometer is 1000 meters. – A millimeter is 1/1000 of a meter. © 2011 Pearson Education, Inc. Chapter 3 4 2 Metric System Prefixes • The following table lists the common prefixes used in the metric system: © 2011 Pearson Education, Inc. Chapter 3 5 Metric Prefixes, Continued • For example, the prefix kilo- increases a base unit by 1000: – 1 kilogram is 1000 grams. • The prefix milli- decreases a base unit by a factor of 1000: – There are 1000 millimeters in 1 meter. © 2011 Pearson Education, Inc. Chapter 3 6 3 Metric Symbols • The names of metric units are abbreviated using symbols. Use the prefix symbol followed by the symbol for the base unit, so: – Nanometer is abbreviated nm. (nanotechnology) – Microgram is abbreviated μg. – Deciliter is abbreviated dL. – Gigasecond is abbreviated Gs. © 2011 Pearson Education, Inc. Chapter 3 7 Metric Equivalents • We can write unit equations for the conversion between different metric units. • The prefix kilo- means 1000 basic units, so 1 kilometer is 1000 meters. • The unit equation is 1 km = 1000 m. • Similarly, a millimeter is 1/1000 of a meter, so the unit equation is 1000 mm = 1 m. © 2011 Pearson Education, Inc. Chapter 3 8 4 Metric Unit Factors • Since 1000 m = 1 km, we can write the following unit factors for converting between meters and kilometers: 1 km 1000 m or 1000 m 1 km • Since 1000 mm = 1 m, we can write the following unit factors: 1000 mm 1m © 2011 Pearson Education, Inc. or 1m . 1000 mm 9 Chapter 3 Metric–Metric Conversions • Unit analysis method - learned in Chapter 2 for metric–metric conversion. • Remember, there are three steps: 1. . 2. . 3. . © 2011 Pearson Education, Inc. Chapter 3 10 5 Metric–Metric Conversion Problem What is the mass in grams of a 325 mg aspirin tablet? • Step 1: We want . • Step 2: We write down the given: . • Step 3: We apply a unit factor ( and round to three significant figures. ) = x © 2011 Pearson Education, Inc. 11 Chapter 3 Two Metric–Metric Conversions A hospital has 125 deciliters of blood plasma. What is the volume in milliliters? • Step 1: We want the answer in • Step 2: We have . • Step 3: We need to first convert convert : © 2011 Pearson Education, Inc. . Chapter 3 and then 12 6 Two Metric–Metric Conversions, Continued • Apply both unit factors, and round the answer to three significant digits. • Notice that both dL and L units cancel, leaving us with units of mL. © 2011 Pearson Education, Inc. Chapter 3 13 Another Example The mass of the Earth’s moon is 7.35 × 1022 kg. What is the mass expressed in megagrams, Mg? • We want Mg; we have 7.35 x 1022 kg. • Convert kilograms to grams, and then grams to megagrams. 7.35 x 1022 kg × © 2011 Pearson Education, Inc. 1000 g 1 Mg x = 7.35 x 1019 Mg 1000000 g 1 kg Chapter 3 14 7 Metric and English Units • The English system is very common in the US. • We often have to convert between English and metric units. © 2011 Pearson Education, Inc. Chapter 3 15 Metric–English Conversion The length of an American football field, including the end zones, is 120 yards. What is the length in meters? • We want meters • We have 120 yd • Given 1 yd = 0.914 m). © 2011 Pearson Education, Inc. Chapter 3 16 8 English–Metric Conversion A half-gallon carton contains 64.0 fl oz of milk. How many milliliters of milk are in a carton? • We want mL; • we have 64.0 fl oz. • Use 1 qt = 32 fl oz, and 1 qt = 946 mL. © 2011 Pearson Education, Inc. Chapter 3 17 Compound Units • Some measurements have a ratio of units. • For example, the speed limit on many highways is 55 miles per hour. How would you convert this to meters per second? • Convert one unit at a time using unit factors. 1. First, miles → meters 2. Next, hours → seconds © 2011 Pearson Education, Inc. Chapter 3 18 9 Compound Unit Problem A motorcycle is traveling at 75 km/hour. What is the speed in meters per second? • We want m/s • Have km/hr • Use 1 km = 1000 m and 1 h = 3600 s. © 2011 Pearson Education, Inc. Chapter 3 19 Volume by Calculation • The volume of an object is calculated by multiplying the length (l) by the width (w) by the thickness (t). volume = l x w x t • All three measurements must be in the same units. • If an object measures 3 cm by 2 cm by 1 cm, the volume is 6 cm3 (cm3 is cubic centimeters). © 2011 Pearson Education, Inc. Chapter 3 20 10 Cubic Volume and Liquid Volume • The liter (L) is the basic unit of volume in the metric system. • One liter is defined as the volume occupied by a cube that is 10 cm on each side. © 2011 Pearson Education, Inc. Chapter 3 21 Cubic and Liquid Volume Units • 1 liter is equal to 1000 cubic centimeters. – 10 cm x 10 cm x 10 cm = 1000 cm3 • 1000 cm3 = 1 L = 1000 mL. • Therefore, 1 cm3 = 1 mL. © 2011 Pearson Education, Inc. Chapter 3 22 11 Cubic–Liquid Volume Conversion An automobile engine displaces a volume of 498 cm3 in each cylinder. What is the displacement of a cylinder in cubic inches? • We want in3; • we have 498 cm3. • Use 1 in = 2.54 cm three times. © 2011 Pearson Education, Inc. Chapter 3 23 Volume by Displacement • If a solid has an irregular shape, its volume cannot be determined by measuring its dimensions. • You can determine its volume indirectly by measuring the amount of water it displaces. • This technique is called volume by displacement. • Volume by displacement can also be used to determine the volume of a gas. © 2011 Pearson Education, Inc. Chapter 3 24 12 Solid Volume by Displacement You want to measure the volume of an irregularly shaped piece of jade. • Partially fill a volumetric flask with water and measure the volume of the water. • Add the jade, and measure the difference in volume. • The volume of the jade is 10.5 mL. © 2011 Pearson Education, Inc. Chapter 3 25 Gas Volume by Displacement You want to measure the volume of gas given off in a chemical reaction. • The gas produced displaces the water in the flask into the beaker. The volume of water displaced is equal to the volume of gas. © 2011 Pearson Education, Inc. Chapter 3 26 13 The Density Concept • The density of an object is a measure of its concentration of mass. • Density is defined as the mass of an object divided by the volume of the object. mass = density volume © 2011 Pearson Education, Inc. Chapter 3 27 Density • Density is expressed in different units. It is usually grams per milliliter (g/mL) for liquids, grams per cubic centimeter (g/cm3) for solids, and grams per liter (g/L) for gases. © 2011 Pearson Education, Inc. Chapter 3 28 14 Calculating Density What is the density of a platinum nugget that has a mass of 224.50 g and a volume of 10.0 cm3 ? Recall, density is mass/volume. 224.50 g = 22.5 g/cm3 10.0 cm3 © 2011 Pearson Education, Inc. Chapter 3 29 Densities of Common Substances © 2011 Pearson Education, Inc. Chapter 3 30 15 Estimating Density • We can estimate the density of a substance by comparing it to another object. • A solid object will float on top of a liquid with a higher density. • Object S1 has a density less than that of water, but larger than that of L1. • Object S2 has a density less than that of L2, but larger than that of water. © 2011 Pearson Education, Inc. Chapter 3 31 Density as a Unit Factor • We can use density as a unit factor for conversions between mass and volume. • An automobile battery contains 1275 mL of acid. If the density of battery acid is 1.84 g/mL, how many grams of acid are in an automobile battery? – We want grams; have 1275 mL © 2011 Pearson Education, Inc. Chapter 3 32 16 Critical Thinking: Gasoline The density of gasoline is 730 g/L at 0 ºC (32 ºF) and 713 g/L at 25 ºC (77 ºF). What is the mass difference of 1.00 gallon of gasoline at these two temperatures (1 gal = 3.784L)? • The difference is about 60 grams (about 2 % how?). © 2011 Pearson Education, Inc. Chapter 3 33 Temperature • Temperature is a measure of the average kinetic energy of the individual particles in a sample. • There are three temperature scales: 1. Celsius 2. Fahrenheit 3. Kelvin • Kelvin is the absolute temperature scale. © 2011 Pearson Education, Inc. Chapter 3 34 17 Temperature Scales • On the Fahrenheit scale, water freezes at 32 °F and boils at 212 °F. • On the Celsius scale, water freezes at 0 °C and boils at 100 °C. These are the reference points for the Celsius scale. • Water freezes at 273 K and boils at 373 K on the Kelvin scale. © 2011 Pearson Education, Inc. 35 Chapter 3 Temperature Conversions • This is the equation for converting °C to °F. °C x ( 180°F 100°C ) = °F + 32°F = °F • This is the equation for converting °F to °C. ( ) 100°C = °C (°F - 32°F) x 180°F • To convert from °C to K, add 273. °C + 273 = K © 2011 Pearson Education, Inc. Chapter 3 36 18 Fahrenheit–Celsius Conversions • Body temperature is 98.6 °F. What is body temperature in degrees Celsius? In Kelvin? © 2011 Pearson Education, Inc. Chapter 3 37 Heat • Heat is the flow of energy from an object of higher temperature to an object of lower temperature. Heat measures the total energy of a system. • Temperature measures the average energy of particles in a system. • Heat in SI units of joules(J) or calories(cal) the more familiar everyday unit. 1 cal heat required to raise the temperature of 1 g water 1°C • 1 cal = 4.184 J © 2011 Pearson Education, Inc. Calorie in food is kcal(1000cal) Chapter 3 38 19 Heat Versus Temperature • Although both beakers below have the same temperature (100 ºC), the beaker on the right has twice the amount of heat because it has twice the amount of water. © 2011 Pearson Education, Inc. Chapter 3 39 Energy conversion • When 1.00 g of gasoline burns and yields about 10.3 kcal what is the energy in SI units? • Unit Analysis: [1000 cal/1 kcal & 4.184 J/1 cal ] • want SI energy joules (J). • Given 10.3 kcal per g gasoline • = © 2011 Pearson Education, Inc. Chapter 3 40 20
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