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Fundamentals of General, Organic, and Biological Chemistry 5th Edition. Chapter Two. Measurements in Chemistry. James E. Mayhugh Oklahoma City University  2007 Prentice Hall, Inc. Outline. 2.1 Physical Quantities 2.2 Measuring Mass 2.3 Measuring Length and Volume - PowerPoint PPT Presentation
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Chapter Chapter Two Two Measurements in Chemistry Fundamentals of General, Organic, and Biological Chemistry 5th Edition James E. Mayhugh James E. Mayhugh Oklahoma City University Oklahoma City University 2007 Prentice Hall, Inc. 2007 Prentice Hall, Inc.
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Page 1: Chapter  Two

ChapterChapter Two TwoMeasurements in

Chemistry

Fundamentals of General, Organic, and Biological Chemistry

5th Edition

James E. MayhughJames E. MayhughOklahoma City UniversityOklahoma City University2007 Prentice Hall, Inc.2007 Prentice Hall, Inc.

Page 2: Chapter  Two

OutlineOutline► 2.1 Physical Quantities► 2.2 Measuring Mass► 2.3 Measuring Length and Volume► 2.4 Measurement and Significant Figures► 2.5 Scientific Notation► 2.6 Rounding Off Numbers► 2.7 Converting a Quantity from One Unit to Another► 2.8 Problem Solving: Estimating Answers► 2.9 Measuring Temperature► 2.10 Energy and Heat► 2.11 Density► 2.12 Specific Gravity

Chapter Two 2

Page 3: Chapter  Two

2.1 Physical Quantities2.1 Physical Quantities

Physical properties such as height, volume, and temperature that can be measured are called physical quantities. Both a number and a unit of defined size is required to describe physical quantity.

Chapter Two 3

Page 4: Chapter  Two

► A number without a unit is meaningless.► To avoid confusion, scientists have agreed on a standard set

of units.► Scientists use SI or the closely related metric units.

Prentice Hall © 2007 Chapter Two 4

Page 5: Chapter  Two

► Scientists work with both very large and very small numbers.

► Prefixes are applied to units to make saying and writing measurements much easier.

► The prefix pico (p) means “a trillionth of.” ► The radius of a lithium atom is 0.000000000152

meter (m). Try to say it.► The radius of a lithium atom is 152 picometers

(pm). Try to say it.

Chapter Two 5

Page 6: Chapter  Two

Frequently used prefixes are shown below.Frequently used prefixes are shown below.

Chapter Two 6

Page 7: Chapter  Two

2.2 Measuring Mass2.2 Measuring Mass

► Mass is a measure of the amount of matter in an object. Mass does not depend on location.

► Weight is a measure of the gravitational force acting on an object. Weight depends on location.

► A scale responds to weight.► At the same location, two objects with identical

masses have identical weights.► The mass of an object can be determined by

comparing the weight of the object to the weight of a reference standard of known mass.

Chapter Two 7

Page 8: Chapter  Two

a) The single-pan balance with sliding counterweights. (b) A modern electronic balance.

Chapter Two 8

Page 9: Chapter  Two

Relationships between metric units of mass and the mass units commonly used in the United States are shown below.

Chapter Two 9

Page 10: Chapter  Two

2.3 Measuring Length and Volume2.3 Measuring Length and Volume

► The meter (m) is the standard measure of length or distance in both the SI and the metric system.

► Volume is the amount of space occupied by an object. A volume can be described as a length3.

► The SI unit for volume is the cubic meter (m3).

Chapter Two 10

Page 11: Chapter  Two

Relationships between metric units of length and volume and the length and volume units commonly used in the United States are shown below and on the next slide.

Chapter Two 11

Page 12: Chapter  Two

Chapter Two12

A m3 is the volume of a cube 1 m or 10 dm on edge. Each m3 contains (10 dm)3 = 1000 dm3 or liters. Each liter or dm3 = (10cm)3 =1000 cm3 or milliliters. Thus, there are 1000 mL in a liter and 1000 L in a m3.

Page 13: Chapter  Two

The metric system is based on factors of 10 and is much easier to use than common U.S. units. Does anyone know how many teaspoons are in a gallon?

Chapter Two 13

Page 14: Chapter  Two

14

Exact NumbersExact Numbers

An exact number is obtained when you count objects or use a defined relationship.

Counting objects are always exact2 soccer balls4 pizzas

Exact relationships, predefined values, not measured1 foot = 12 inches1 meter = 100 cm

For instance is 1 foot = 12.000000000001 inches? No 1 ft is EXACTLY 12 inches.

Page 15: Chapter  Two

15

Learning Check Learning Check

A. Exact numbers are obtained by 1. using a measuring tool

2. counting3. definition

B. Measured numbers are obtained by 1. using a measuring tool

2. counting3. definition

Page 16: Chapter  Two

16

SolutionSolution

A. Exact numbers are obtained by 2. counting

3. definition

B. Measured numbers are obtained by 1. using a measuring tool

Page 17: Chapter  Two

17

Learning CheckLearning Check

Classify each of the following as an exact or ameasured number.

1 yard = 3 feet

The diameter of a red blood cell is 6 x 10-4 cm.

There are 6 hats on the shelf.

Gold melts at 1064°C.

Page 18: Chapter  Two

18

Classify each of the following as an exact (1) or ameasured(2) number. This is a defined relationship.A measuring tool is used to determine length.The number of hats is obtained by counting.A measuring tool is required.

SolutionSolution

Page 19: Chapter  Two

2.4 Measurement and Significant 2.4 Measurement and Significant FiguresFigures

► Every experimental measurement has a degree of uncertainty.

► The volume, V, at right is certain in the 10’s place, 10mL<V<20mL

► The 1’s digit is also certain, 17mL<V<18mL

► A best guess is needed for the tenths place.

Chapter Two 19

Page 20: Chapter  Two

20

What is the Length?

1 2 3 4 cm

►We can see the markings between 1.6-1.7cm►We can’t see the markings between the .6-.7►We must guess between .6 & .7►We record 1.67 cm as our measurement►The last digit an 7 was our guess...stop there

Page 21: Chapter  Two

Learning CheckLearning Check

What is the length of the wooden stick?1) 4.5 cm 2) 4.54 cm 3) 4.547 cm

Page 22: Chapter  Two

22

8.00 cm or 3 (2.2/8)?

Page 23: Chapter  Two

► To indicate the precision of a measurement, the value recorded should use all the digits known with certainty, plus one additional estimated digit that usually is considered uncertain by plus or minus 1.

► No further insignificant digits should be recorded.

► The total number of digits used to express such a measurement is called the number of significant figures.

► All but one of the significant figures are known with certainty. The last significant figure is only the best possible estimate.

Chapter Two 23

Page 24: Chapter  Two

Chapter Two 24

Below are two measurements of the mass of the same object. The same quantity is being described at two different levels of precision or certainty.

Page 25: Chapter  Two

► When reading a measured value, all nonzero digits should be counted as significant. There is a set of rules for determining if a zero in a measurement is significant or not.

► RULE 1. Zeros in the middle of a number are like any other digit; they are always significant. Thus, 94.072 g has five significant figures.

► RULE 2. Zeros at the beginning of a number are not significant; they act only to locate the decimal point. Thus, 0.0834 cm has three significant figures, and 0.029 07 mL has four.

Chapter Two 25

Page 26: Chapter  Two

► RULE 3. Zeros at the end of a number and after the decimal point are significant. It is assumed that these zeros would not be shown unless they were significant. 138.200 m has six significant figures. If the value were known to only four significant figures, we would write 138.2 m.

► RULE 4. Zeros at the end of a number and before an implied decimal point may or may not be significant. We cannot tell whether they are part of the measurement or whether they act only to locate the unwritten but implied decimal point.

Chapter Two 26

Page 27: Chapter  Two

Practice Rule #1 Zeros

45.8736

.000239

.00023900

48000.

48000

3.982106

1.00040

6

3

5

5

2

4

6

•All digits count

•Leading 0’s don’t

•Trailing 0’s do

•0’s count in decimal form

•0’s don’t count w/o decimal

•All digits count

•0’s between digits count as well as trailing in decimal form

Page 28: Chapter  Two

2.5 Scientific Notation2.5 Scientific Notation

► Scientific notation is a convenient way to write a very small or a very large number.

► Numbers are written as a product of a number between 1 and 10, times the number 10 raised to power.

► 215 is written in scientific notation as: 215 = 2.15 x 100 = 2.15 x (10 x 10) = 2.15 x 102

Chapter Two 28

Page 29: Chapter  Two

Chapter Two 29

Two examples of converting standard notation to scientific notation are shown below.

Page 30: Chapter  Two

Chapter Two 30

Two examples of converting scientific notation back to Two examples of converting scientific notation back to standard notation are shown below. standard notation are shown below.

Page 31: Chapter  Two

► Scientific notation is helpful for indicating how many significant figures are present in a number that has zeros at the end but to the left of a decimal point.

► The distance from the Earth to the Sun is 150,000,000 km. Written in standard notation this number could have anywhere from 2 to 9 significant figures.

► Scientific notation can indicate how many digits are significant. Writing 150,000,000 as 1.5 x 108 indicates 2 and writing it as 1.500 x 108 indicates 4.

► Scientific notation can make doing arithmetic easier. Rules for doing arithmetic with numbers written in scientific notation are reviewed in Appendix A.

Chapter Two 31

Page 32: Chapter  Two

2.6 Rounding Off Numbers2.6 Rounding Off Numbers

► Often when doing arithmetic on a pocket calculator, the answer is displayed with more significant figures than are really justified.

► How do you decide how many digits to keep?► Simple rules exist to tell you how.

Chapter Two 32

Page 33: Chapter  Two

► Once you decide how many digits to retain, the rules for rounding off numbers are straightforward:

► RULE 1. If the first digit you remove is 4 or less, drop it and all following digits. 2.4271 becomes 2.4 when rounded off to two significant figures because the first dropped digit (a 2) is 4 or less.

► RULE 2. If the first digit removed is 5 or greater, round up by adding 1 to the last digit kept. 4.5832 is 4.6 when rounded off to 2 significant figures since the first dropped digit (an 8) is 5 or greater.

► If a calculation has several steps, it is best to round off at the end.

Chapter Two 33

Page 34: Chapter  Two

Practice Rule #2 RoundingPractice Rule #2 Rounding

Make the following into a 3 Sig Fig numberMake the following into a 3 Sig Fig number

1.5587

.0037421

1367

128,522

1.6683 106

1.56

.00374

1370

129,000

1.67 106

Your Final number must be of the same value as the number you started with,129,000 and not 129

Page 35: Chapter  Two

Examples of RoundingExamples of RoundingFor example you want a 4 Sig Fig number

4965.03

780,582

1999.5

0 is dropped, it is <5

8 is dropped, it is >5; Note you must include the 0’s

5 is dropped it is = 5; note you need a 4 Sig Fig

4965

780,600

2000.

Page 36: Chapter  Two

RULE 1. RULE 1. In carrying out a multiplication or division, the answer cannot have more significant figures than either of the original numbers.

Chapter Two 36

Page 37: Chapter  Two

►RULE 2. In carrying out an addition or subtraction, the answer cannot have more digits after the decimal point than either of the original numbers.

Chapter Two 37

Page 38: Chapter  Two

Multiplication and divisionMultiplication and division

32.27 1.54 = 49.6958

3.68 .07925 = 46.4353312

1.750 .0342000 = 0.05985

3.2650106 4.858 = 1.586137 107

6.0221023 1.66110-24 = 1.000000

49.7

46.4

.05985

1.586 107

1.000

Page 39: Chapter  Two

Addition/SubtractionAddition/Subtraction

25.5 32.72 320 +34.270 0.0049‑ + 12.5 59.770 32.7151 332.5

59.8 32.72 330

Page 40: Chapter  Two

__ ___ __

Addition and SubtractionAddition and Subtraction

.56 + .153 = .713

82000 + 5.32 = 82005.32

10.0 - 9.8742 = .12580

10 – 9.8742 = .12580

.71

82000

.1

0

Look for the last important digit

Page 41: Chapter  Two

Mixed Order of OperationMixed Order of Operation

8.52 + 4.1586 18.73 + 153.2 =

(8.52 + 4.1586) (18.73 + 153.2) =

239.6

2180.

= 8.52 + 77.89 + 153.2 = 239.61 =

= 12.68 171.9 = 2179.692 =

Page 42: Chapter  Two

TryTry

i

i

n

xx

Find the standard deviation for the following numbers: 7.691 g, 7.23 g, 7.892 g

2

12

1

n

xx

s ii

Page 43: Chapter  Two

TryTry

i

i

n

xx

7.691 g 7.23 g 7.892 g22.813 g

22.81 g = 7.603 g 3

7.691 g, 7.23 g, 7.892 g

Page 44: Chapter  Two

TryTry

= 7.603 g

7.691 g – 7.603 g = .088 g 7.23 g – 7.603 g = -.37 g 7.892 g – 7.603 g = .289 g

.01 g

2

12

1

n

xx

s ii

Page 45: Chapter  Two

TryTry

2

12

1

n

xx

s ii

.25

.01 = .01 g

007.

2

01.2/12

s

Page 46: Chapter  Two

2.7 Problem Solving: Converting a 2.7 Problem Solving: Converting a Quantity from One Unit to AnotherQuantity from One Unit to Another

► Factor-Label Method: A quantity in one unit is converted to an equivalent quantity in a different unit by using a conversion factor that expresses the relationship between units.

Chapter Two 46

(Starting quantity) x (Conversion factor) = Equivalent quantity

Page 47: Chapter  Two

Chapter Two 47

Writing 1 km = 0.6214 mi as a fraction restates it in the form of a conversion factor. This and all other conversion factors are numerically equal to 1.

The numerator is equal to the denominator. Multiplying by a conversion factor is equivalent to multiplying by 1 and so causes no change in value.

Page 48: Chapter  Two
Page 49: Chapter  Two

When solving a problem, the idea is to set up anequation so that all unwanted units cancel, leaving only the desired units.

Chapter Two 49

Page 50: Chapter  Two
Page 51: Chapter  Two

2.8 Problem Solving: Estimating 2.8 Problem Solving: Estimating AnswersAnswers

► STEP 1: Identify the information given.► STEP 2: Identify the information needed to answer.► STEP 3: Find the relationship(s) between the known

information and unknown answer, and plan a series of steps, including conversion factors, for getting from one to the other.

► STEP 4: Solve the problem.► BALLPARK CHECK: Make a rough estimate to be

sure the value and the units of your calculated answer are reasonable.

Chapter Two 51

Page 52: Chapter  Two
Page 53: Chapter  Two
Page 54: Chapter  Two

54

How many minutes are in 1.4 days?Initial unit: 1.4 days

Unit plan: days hr min

Set up problem: 1.4 days x 24 hr x 60 min = 2.0 x 103 min

1 day 1 hr

Example: Problem SolvingExample: Problem Solving

Page 55: Chapter  Two

55

An adult human has 4650 mL of blood. How many gallons of blood is that?Unit plan: mL qt gallon

Equalities: 1 quart = 946 mL 1 gallon = 4 quarts

Learning Check Learning Check

Page 56: Chapter  Two

56

Unit plan: mL qt gallon

Setup: 4650 mL x 1 qt x 1 gal = 1.23 gal

946 mL 4 qt 3 SF 3 SF exact 3 SF

SolutionSolution

Page 57: Chapter  Two

►150 pounds (American) is how many stones (British)?

1 pound = 265 dram1 Gram = 1.71 dram 1.13 pennyweights = 1 Gram1.21 scruples = 1 pennyweight17.2 scruples = 1 stone

57

?

Page 58: Chapter  Two

►The following relationships are British liquid units.1 hogshead = 7 firkin18 pottle = 1 firkin140 pottle = 1 puncheon504 pottle = 1 tun

How many hogsheads in 12.5 tuns?

58

pottle

firkin

18

1hogsheads0.50

Page 59: Chapter  Two

►A 60 watt light bulb is the how many horsepower?3.41 Btu/hr = 1 watt4.20 calorie/minute = 1 Btu/hr.0514 Foot-pound-force/second = 1 calorie/minute1.825×10-3 horsepower = 1 Foot-pound-force/second

59

Page 60: Chapter  Two

Learning CheckLearning Check

►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?

►The dosage ordered is 485 mg of Erythromycin four times a day (q.i.d). If the oral suspension contains 200 mg Erythromycin/5 mL, how many mL will be given in a day?

►The dimensions of a box are 12 inch by 11 inch by 5.5 inch. Calculate the volume of the box in cm3. There is 2.54 cm per inch.

60

Page 61: Chapter  Two

2.9 Measuring Temperature2.9 Measuring Temperature

► Temperature is commonly reported either in degrees Fahrenheit (oF) or degrees Celsius (oC).

► The SI unit of temperature is the Kelvin (K).► 1 Kelvin, no degree, is the same size as 1 oC.► 0 K is the lowest possible temperature, 0 oC =

273.15 K is the normal freezing point of water. To convert, adjust for the zero offset.

► Temperature in K = Temperature in oC + 273.15► Temperature in oC = Temperature in K - 273.15

Chapter Two 61

Page 62: Chapter  Two

Freezing point of H2O Boiling point of H2O

32oF 212oF0oC 100oC

212oF - 32oF = 180oF covers the same range of temperature as 100oC - 0oC = 100oC covers. Therefore, a Celsius degree is exactly 180/100 = 1.8 times as large as a Fahrenheit degree. The zeros on the two scales are separated by 32oF.

Chapter Two 62

Page 63: Chapter  Two

Fahrenheit, Celsius, and Kelvin temperature scales.Chapter Two 63

Page 64: Chapter  Two

► Converting between Fahrenheit and Celsius scales is similar to converting between different units of length or volume, but is a little more complex. The different size of the degree and the zero offset must both be accounted for.

► ooF = (1.8 x F = (1.8 x ooC) + 32C) + 32► ooC = (C = (ooF – 32)/1.8F – 32)/1.8

Chapter Two 64

Page 65: Chapter  Two

2.10 Energy and Heat2.10 Energy and Heat

► Energy: The capacity to do work or supply heat.► Energy is measured in SI units by the Joule (J); the

calorie is another unit often used to measure energy.► One calorie (cal) is the amount of heat necessary to

raise the temperature of 1 g of water by 1°C.► A kilocalorie (kcal) = 1000 cal. A Calorie, with a capital

C, used by nutritionists, equals 1000 cal.► An important energy conversion factor is:

1 cal = 4.184 J1 cal = 4.184 J

Chapter Two 65

Page 66: Chapter  Two

► Not all substances have their temperatures raised to the same extent when equal amounts of heat energy are added.

► One calorie raises the temperature of 1 g of water by 1°C but raises the temperature of 1 g of iron by 10°C.

► The amount of heat needed to raise the temperature of 1 g of a substance by 1°C is called the specific heat of the substance.

► Specific heat is measured in units of cal/gC

Chapter Two 66

Page 67: Chapter  Two

Chapter Two 67

► Knowing the mass and specific heat of a substance makes it possible to calculate how much heat must be added or removed to accomplish a given temperature change.

► (Heat Change) = (Mass) x (Specific Heat) x (Temperature Change)

► Using the symbols Δ for change, H for heat, m for mass, C for specific heat, and T for temperature, a more compact form is:

ΔH = m×C×Δ T

Page 68: Chapter  Two

Learning CheckLearning Check

1. How much energy is required to change the temperature of 15.0 g Fe from 18.5 C to 56.8 C? The specific heat of iron is 0.451 J/g·K.

2. (2.67) copper has specific heat of .092 cal/(g °C). ∙When 52.7 cal of heat is added to a piece of copper, the temperature increases from 22.4 °C to 38.6 °C. What is the mass of the piece of copper?

3. If 34.8 J is required to change the temperature of 10.0 g of mercury by 25 K, what is the specific heat of mercury?

68

Page 69: Chapter  Two

Learning CheckLearning Check

1. How much energy is required to change the temperature of 15.0 g Fe from 18.5 C to 56.8 C? The specific heat of iron is 0.451 J/g·K.

69

ΔH = 15.0 g×0.451 J×(56.8-18.5)C g·K

ΔH = m×C×Δ T

ΔH = 259 J

Page 70: Chapter  Two

Learning CheckLearning Check

2. (2.67) copper has specific heat of .092 cal/(g °C). ∙When 52.7 cal of heat is added to a piece of copper, the temperature increases from 22.4 °C to 38.6 °C. What is the mass of the piece of copper?

70

ΔH = m×C×Δ T

52.7 cal = m×.092 cal× (38.6-22.4)C gC

Page 71: Chapter  Two

Learning CheckLearning Check

3. If 34.8 J is required to change the temperature of 10.0 g of mercury by 25 K, what is the specific heat of mercury?

71

ΔH = m×C×Δ T

34.8 J = 10.0 g×C×25 K

Page 72: Chapter  Two

2.11 2.11 DensityDensity

Density relates the mass of an object to its volume. Density is usually expressed in units of grams per cubic centimeter (g/cm3) for solids, and grams per milliliter (g/mL) for liquids.

Chapter Two 72

Density = Mass (g)

Volume (mL or cm3)

Page 73: Chapter  Two

►Which is heavier, a ton of feathers or a ton of bricks?

►Which is larger?

►If two objects have the same mass, the one with the higher density will be smaller.

Chapter Two 73

Page 74: Chapter  Two

Mercury has a density of 13.6 g/mL. How many milliliters of mercury weigh 475 grams?

1. 0.000155 mL2. 0.0286 mL3. 34.9 mL4. 6460 mL

Page 75: Chapter  Two

Learning CheckLearning Check

►(2.75) What is the density of lithium metal ( in g/cm3) if a cylindrical wire with a diameter of 2.40 mm and a length of 15.0 cm has a mass of .3624 g; vcyln=r2l.

►The density of acetic acid is 1.05 g/mL. What is the volume of 275 g of acetic acid?

►A cube of iron has a mass of 15.37 g. If each side of the cube has dimensions of 1.25 cm, what is the density of iron?

75

Page 76: Chapter  Two

Learning CheckLearning Check

►(2.75) What is the density of lithium metal ( in g/cm3) if a cylindrical wire with a diameter of 2.40 mm and a length of 15.0 cm has a mass of .3624 g; vcyln=r2l.

76

Density = Mass (g)

Volume (mL or cm3)

Page 77: Chapter  Two

Learning CheckLearning Check

►The density of acetic acid is 1.05 g/mL. What is the volume of 275 g of acetic acid?

77

Density = Mass (g)

Volume (mL or cm3)

Page 78: Chapter  Two

Learning CheckLearning Check

►A cube of iron has a mass of 15.37 g. If each side of the cube has dimensions of 1.25 cm, what is the density of iron?

78

Density = Mass (g)

Volume (mL or cm3)

Page 79: Chapter  Two

2. 12 Specific Gravity2. 12 Specific Gravity

Specific gravity (sp gr): density of a substance divided by the density of water at the same temperature. Specific gravity is unitless. The density of water is so close to 1 g/mL that the specific gravity of a substance at normal temperature is numerically equal to the density.

Chapter Two 79

Density of substance (g/ml)Density of substance (g/ml)

Density of water at the same temperature (g/ml)Density of water at the same temperature (g/ml)Specific gravity =Specific gravity =

Page 80: Chapter  Two

The specific gravity of a liquid can be measured using an instrument called a hydrometer, which consists of a weighted bulb on the end of a calibrated glass tube. The depth to which the hydrometer sinks when placed in a fluid indicates the fluid’s specific gravity.

Chapter Two 80

Page 81: Chapter  Two

►Galileo’s Galileo’s ThermometerThermometer

As temperature As temperature changes so do changes so do the density's of the density's of the solutions in the solutions in the floating the floating bulbs.bulbs.

81

Page 82: Chapter  Two

Chapter SummaryChapter Summary► Physical quantities require a number and a unit.► Preferred units are either SI units or metric units.► Mass, the amount of matter an object contains, is

measured in kilograms (kg) or grams (g). ► Length is measured in meters (m). Volume is

measured in cubic meters in the SI system and in liters (L) or milliliters (mL) in the metric system.

► Temperature is measured in Kelvin (K) in the SI system and in degrees Celsius (°C) in the metric system.

Chapter Two 82

Page 83: Chapter  Two

Chapter Summary Cont.Chapter Summary Cont.► The exactness of a measurement is indicated by

using the correct number of significant figures.► Significant figures in a number are all known with

certainty except for the final estimated digit.► Small and large quantities are usually written in

scientific notation as the product of a number between 1 and 10, times a power of 10.

► A measurement in one unit can be converted to another unit by multiplying by a conversion factor that expresses the exact relationship between the units.

Chapter Two 83

Page 84: Chapter  Two

Chapter Summary Cont.Chapter Summary Cont.

► Problems are solved by the factor-label method.► Units can be multiplied and divided like numbers.► Temperature measures how hot or cold an object is.► Specific heat is the amount of heat necessary to

raise the temperature of 1 g of a substance by 1°C. ► Density relates mass to volume in units of g/mL for

a liquid or g/cm3 for a solid.► Specific gravity is density of a substance divided by

the density of water at the same temperature.

Chapter Two 84

Page 85: Chapter  Two

End of Chapter 2End of Chapter 2

Chapter Two 85


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