Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Chapter 1 Chemistry and Measurements 1 What Is Chemistry? Chemistry • is the study of composition, structure, properties, and reactions of matter • happens all around you everyday Antacid tablets undergo a chemical reaction when dropped in water. Matter is another word for all substances that make up our world. • Antacid tablets are matter. • Water is matter. • Glass is matter. • Air is matter. 2 Branches of Chemistry The field of chemistry is divided into branches, such as • organic chemistry, the study of substances that contain carbon • inorganic chemistry, the study of all substances except those that contain carbon • general chemistry, the study of the composition, properties and reactions of matter 3 Chemistry + Other Sciences Chemistry is often combined with other sciences: Geology + Chemistry = Geochemistry Biology + Chemistry = Biochemistry Physical Science + Chemistry = Physical Chemistry Biochemists analyze lab samples. 4 Chemicals Chemicals are • substances that have the same composition and properties wherever found • often substances made by chemists that you use everyday Toothpaste is a combination of chemicals. 5 Units of Measurement Scientists use the metric system of measurement and have adopted a modification of the metric system called the International System of Units as a worldwide standard. International System of Units (SI) is an official system of measurement used throughout the world for units in length, volume, mass, temperature, and time. 6 Length, Meter (m) and Centimeter (cm) 1 m = 100 cm 1 m = 39.4 in. 7 1 m = 1.09 yd 2.54 cm = 1 in. Volume, Liter (L) and Milliliter (mL) 1L = 1000 mL 1L = 1.06 qt 946 mL = 1 qt We use graduated cylinders to measure small volumes. 8 Mass, Gram (g) and Kilogram (kg) 1 kg = 1000 g 1 kg = 2.20 lb 454 g = 1 lb The mass of a nickel is 5.01 g on an electronic scale. 9 Temperature, Celsius (oC) and Kelvin (K) Water freezes: 32 oF 0 oC The Kelvin scale for temperature begins at the lowest possible temperature, 0 K. A thermometer is used to measure temperature. 10 Time, Second (s) The second is the correct metric and SI unit for time. The standard measure for 1 s is an atomic clock. A stopwatch is used to measure the time of a race. 11 Writing a Number in Scientific Notation Numbers written in scientific notation have three parts: coefficient power of 10 unit To write 2400 m in correct scientific notation: • the coefficient is 2.4 • the power of 10 is 3 • the unit is "m" 2400. m = 2.4 x 1000 = 2.4 x 103 m 3 places 0.00086 g = 4 places coefficient x power of 10 unit 8.6 10,000 = 8.6 x 10−4 g coefficient x power of 10 unit 12 Measurements in Scientific Notation Diameter chickenpox virus = 0.0000003 m = 3 x 10−7 m 13 Scientific Notation and Calculators Number to enter: Enter: Display: 4 x 106 4 EXP (EE) (x10x) 6 4 06 or 406 4 E06 Number to enter: Enter: Display: 2.5 x 10−4 2.5 EXP (EE) (x10x) +/− 4 2.5 −04 or 2.5−04 2.5 E−04 14 Some Powers of 10 15 Measured Numbers Measured numbers are the numbers obtained when you measure a quantity such as your height, weight, or temperature. To write a measured number, • observe the numerical values of marked lines • estimate value of number between marks • the estimated number is the final number in your measured number 16 Writing Measured Numbers for Length The lengths of the objects are measured as (a) 4.5 cm (b) 4.55 cm (c) 3.0 cm 17 Rules for Significant Figures 1) All non-zero digits are significant. 2) All zeros between non-zero digits are significant. 3) All zeros to the right of the decimal and to the right of the last non-zero digit are significant. 4) All zeros to the left of the first non-zero digit are NOT significant. 5) Zeros to the right of the first non-zero digit and to the left of the decimal may or may not be significant. They must be written in scientific notation. 6) Some numbers have infinite significant figures or are exact numbers. A number is a significant figure (SF) if it is Example a. not a zero 4.5 g 2 SF b. a zero between digits 205 m c. a zero at the end of a 50. L decimal number d. in the coefficient of a 4.8 x 105 m number written in scientific notation 3 SF 2 SF 2 SF A number is not significant if it is Example a. at the beginning of a decimal number b. used as a placeholder in a large number without a decimal point 0.0004 s 1 SF 850 000 m 2 SF 19 Learning Check Identify the significant and nonsignificant zeros in each of the following numbers: A. 0.002650 m B. 43.026 g C. 1,044,000 L 20 Solution Identify the significant and nonsignificant zeros in each of the following numbers: A. 0.002 650 m • The zeros preceding 2 are not significant. • The digits 2, 6, 5 are significant. • The zero in last decimal place is significant. 4 SF B. 43.026 g • The zeros between nonzero digits or at the end of decimal numbers are significant. 5 SF C. 1 044 000 L • The zeros between nonzero digits are significant. • The zeros at end of a number with no decimal are not significant. 4 SF 21 Exact Numbers Exact numbers are • numbers obtained by counting • in definitions that compare two units in the same measuring system 8 cookies 6 eggs 1 qt = 4 cups 1 kg = 1000 g Any enumerated value. We can’t have 27 ½ students. We can’t have 6 ½ cats (unless we’re in a biology lab!) 22 Rules for Rounding Off 1. If the first digit to be dropped is 4 or less, then it and all the following digits are dropped from the number. 2. If the first digit to be dropped is 5 or greater, then the last retained digit of the number is increased by 1. 23 Learning Check Select the correct value when 3.1457 g is rounded to: A. three significant figures B. two significant figures 24 Solution Select the correct value when 3.1457 g is rounded to: A. To round 3.1457 to three significant figures, • drop the final digits, 57 • increase the last remaining digit by 1. The answer is 3.15 g. B. To round 3.1457 g to two significant figures, • drop the final digits 457. • do not increase the last number by 1 since the first of these digits is 4. The answer is 3.1 g. 25 Rules for Multiplication and Division In multiplication or division, the final answer is written so it has the same number of significant figures as the measurement with the fewest significant figures (SFs). Example 1: Multiply the following measured numbers: 24.66 cm x 0.35 cm = 8.631 (calculator display) = 8.6 cm2 (2 significant figures) Multiplying 4 SFs by 2 SFs gives us an answer with 2 SFs. 26 Multiplication and Division with SFs Example 2: Multiply and divide the following measured numbers: 21.5 cm x 0.30 cm = 1.88 cm Put the following into your calculator: 21.5 x 0.30 ÷ 1.88 = 3.430851063 = 3.4 cm (2 significant figures) Multiplying 4 SFs by 2 SFs gives us an answer with 2 SFs. 27 Multiplication and Division with SFs Example 3: Multiply and divide the following measured numbers: 6.0 g = 2.00 g Put the following into your calculator: 6.0 ÷ 2.00 = 3 (calculator display) = 3.0 g (2 significant figures) Add one zero to give 2 significant figures. 28 Learning Check Perform the following calculation of measured numbers. Give the answer in the correct number of significant figures. 5.00 cm x 3.408 cm = 2.00 cm Solution: (3 SF x 4 SF ÷ 3 SF ) = 8.52 cm calculator display and correct significant figures. 29 Addition and Subtraction In addition or subtraction, the final answer is written so it has the same number of decimal places as the measurement with the fewest decimal places. Example 1: Add the following measured numbers: 2.012 g three decimal places 61.09 g two decimal places + 3.0 g one decimal place 66.102 g (calculator display) = 66.1 g answer rounded to one decimal place Example 2: Subtract the following measured numbers: 65.09 g two decimal places − 3.0 g one decimal place 62.09 g (calculator display) = 62.1 g answer rounded to one decimal place 30 Learning Check Add the following measured numbers: 82.409 mg + 22.0 mg Solution: Add the following measured numbers: 82.409 mg three decimal places + 22.0 mg one decimal place 104.409 mg (calculator display) = 104.4 mg answer rounded to one decimal place 31 Prefixes 32 A special feature of the SI as well as the metric system is that a prefix can be placed in front of any unit to increase or decrease its size by some factor of ten. For example, the prefixes milli and micro are used to make the smaller units: milligram (mg) microgram (μg) Prefixes and Equalities • A special feature of the SI as well as the metric system is that a prefix can be placed in front of any unit to increase or decrease its size by some factor of ten. • For example, the prefixes milli and micro are used to make the smaller units: milligram (mg) microgram (μg) • The relationship of a prefix to a unit can be expressed by replacing the prefix with its numerical value. • For example, when the prefix kilo in kilometer is replaced with its value of 1000, we find that a kilometer is equal to 1000 meters. kilometer = 1000 meters kilogram = 1000 grams 33 Prefixes That Increase/Decrease Unit Size 34 Measuring Length, Mass & Volume Examples of some Length equalities.. 1 m = 100 cm = 1 x 102 cm 1 m = 1000 mm = 1 x 103 mm 1 cm = 10 mm = 1 x 101 mm Examples of Some Mass Equalities 1 kg = 1000 g = 1 x 103 g 1 g = 1000 mg = 1 x 103 mg 1 g = 100 cg = 1 x 102 cg 1 mg = 1000 μg = 1 x 103 μg Examples of Some Volume Equalities 1 L = 10 dL = 1 x 101 dL 1 L = 1000 mL = 1 x 103 mL 1 dL = 100 mL = 1 x 102 mL 35 The Cubic Centimeter 1 cm3 = 1 cc = 1 mL 10 cm x 10 cm x 10 cm = 1000 cm3 = 1000 mL = 1 L 36 Equalities 37 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, 1m = 1 lb = 16 oz 2.20 lb = 1000 mm 1 kg Exact and Measured Numbers in Equalities 38 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 and therefore 2.54 is an exact number. Some Common Equalities 39 Equalities & Conversion Factors An equality • is written as a fraction (ratio) • provides two conversion factors that are the inverse of each other A conversion factor is • obtained from an equality and written in the form of a fraction with a numerator and denominator Equality: 1 in. = 2.54 cm • inverted to give two conversion factors for every equality 1 in. 2.54 cm and 2.54 cm 1 in. Conversion Factors in a Problem 41 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 $2.39. 1 lb red peppers and $2.39 $2.39 1 lb red peppers Example 2: The cost of one gallon (1 gal) of gas is $2.89. 1 gal gas and $2.89 $2.89 1 gal gas 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 of 100 • can be written as two factors Example: A food contains 30% (by mass) fat: 30 g fat and 100 g food 100 g food 30 g fat 42 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 mass 11 kg fat 43 Smaller Percents: ppm and ppb Small percents are given as ppm and ppb. • Parts per million (ppm) = mg part kg whole Example: The EPA allows 15 ppm cadmium in food colors. 15 mg of cadmium = 1 kg of food color • Parts per billion (ppb) = μg part kg whole Example: The EPA allows 10 ppb arsenic in public water. 10 μg of arsenic = 1 kg of water 44 Given and Needed Units 45 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.) 5-Step Problem Solving Method 1. What am I being asked for? - What answer does the problem want me to give? - Beware of multi-step problems! 2. What do I know? - What information does the problem give me or allow me to solve for? - This includes information from common sources/tables. 3. What are the relationships between what I’m being asked for and what I know? - What formulas relate “What I know” with “What I’m being asked for”? 4. Are there any unit conversions I have to make? - Make sure the units are compatible and that you are giving the answer in the proper units. 5. Always solve the problem for the variable of interest BEFORE plugging in any values!! Study Tip: Problem Solving Using GPS 47 The steps in the Guide to Problem Solving (GPS) are useful in setting up a problem with conversion factors. Setting Up a Problem 48 Question: How many minutes are in 2.5 hours? Solution: Step 1 State the given and needed quantities. Given unit: 2.5 hours Needed unit: min Step 2 Write a unit plan. Plan: hours min Step 3 State equalities and conversion factors to cancel units. 60 min = 1 h 60 min and 1h Step 4 Set up problem to cancel units. Given Conversion Needed unit unit factor 2.5 h x 60 min = 150 min (2 SF) 1h 1h 60 min Learning Check 49 A rattlesnake is 2.44 m long. How many centimeters long is the snake? Solution 50 A rattlesnake is 2.44 m long. How many centimeters long is the snake? Step 1 State the given and needed quantities. Given unit: 2.44 m Needed unit: cm Step 2 Write a unit plan. Plan: meters centimeters Step 3 State equalities and conversion factors to cancel units. 1 m = 102 cm 102 cm and 1m 1m 102 cm Step 4 Set up problem to cancel units. Given Conversion Needed unit unit factor 2.44 m x 102 cm = 244 cm (3 SF) 1m Learning Check 51 How many minutes are in 1.4 days? Solution 52 How many minutes are in 1.4 days? Step 1 State the given and needed quantities. Given unit: 1.4 days Needed unit: minutes Step 2 Write a unit plan. Factor 1 Plan: days Factor 2 h min Solution 53 How many minutes are in 1.4 days? Step 3 State equalities and conversion factors to cancel units. 1 day = 24 hours 24 hours and 1 day 1 day 24 hours 1 hour = 60 minutes 60 min and 1 h 1h 60 min Solution 54 How many minutes are in 1.4 days? Step 4 Set up problem to cancel units. Given Conversion Conversion unit factor factor Needed unit 1.4 days x 24 h x 60 min = 2.0 x 103 min 1 day 1h (rounded) 2 SF Exact Exact = 2 SF Learning Check 55 How many pounds of sugar are in 120 g of candy if the candy is 25% (by mass) sugar? Solution: Step 1 State the given and needed quantities. Given units: 120 g of candy 25% by mass sugar Needed unit: pounds sugar Step 2 Write a unit plan. Conversion factor Plan: grams Percent factor pounds candy pounds sugar Solution 56 How many pounds of sugar are in 120 g of candy if the candy is 25% (by mass) sugar? Step 3 State equalities and conversion factors to cancel units. 1 pound = 454 grams 454 g and 1 lb 1 lb 454 g 25 pounds of sugar = 100 pounds of candy 25 lb sugar and 100 lb candy 100 lb candy 25 lb sugar Solution 57 How many pounds of sugar are in 120 g of candy if the candy is 25% (by mass) sugar? Step 4 Set up problem to cancel units. Given unit Conversion factor Percent factor 120 g candy x 1 lb candy x 25 lb sugar 454 g candy 100 lb candy = 0.066 lb of sugar Density 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 volume mL cm3 or g/cm3 Note: 1 mL = 1 cm3 58 Calculating Density 59 Learning Check Osmium is a very dense metal. What is its density in g/cm3 if 50.0 g of osmium has a volume of 2.22 cm3? 1) 2.25 g/cm3 2) 22.5 g/cm3 3) 111 g/cm3 60 Solution Step 1 State the given and needed quantities. Given: 50.0 g; 22.2 cm3 Need: density, g/cm3 Step 2 Write the density expression. D = mass volume Step 3 Express mass in grams and volume in mL or cm3. Mass = 50.0 g Volume = 22.2 cm3 Step 4 Substitute mass and volume into the density expression and calculate. D = 50.0 g = 22.522522 g/cm3 2.22 cm3 = 22.5 g/cm3 (rounded to 3 SFs) 61 Volume by Displacement 62 A solid • completely submerged in water displaces its own volume of water • has a volume calculated from the volume difference 45.0 mL − 35.5 mL = 9.5 mL = 9.5 cm3 Density Using Volume Displacement 63 The density of the zinc object is calculated from its mass and volume. Density = mass = 68.60 g = 7.2 g/cm3 volume 9.5 cm3 Learning Check What is the density (g/cm3) of a 48.0-g sample 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/cm3 25.0 mL 2) 6.0 g/cm3 3) 380 g/cm3 33.0 mL object 64 Solution Step 1 State the given and needed quantities. Given: 48.0 g Volume of water = 25.0 mL Volume of water + metal = 33.0 mL Need: Density Step 2 Write the density expression. Density = mass of metal volume of metal 65 Solution Step 3 Express mass in grams and volume in mL or cm3. Mass = 48.0 g Volume of the metal is equal to the volume of water displaced. Volume of water + metal − Volume of water Volume of metal = 33.0 mL = 25.0 mL = 8.0 mL Step 4 Substitute mass and volume into the density expression and calculate the density. Density = 48.0 g = 6.0 g = 6.0 g/mL 8.0 mL 1 mL 66 Problem Solving Using Density 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 factors 1 mL and 1 mL 3.8 g 67 Problem Solving Using Density 68 Learning Check The density of octane, a component of gasoline, is 0.702 g/mL. What is the mass, in kg, of 875 mL of octane? A. 0.614 kg B. 614 kg C. 1.25 kg 69 Solution The density of octane, a component of gasoline, is 0.702 g/mL. What is the mass, in kg, of 875 mL of octane? Step 1 State the given and needed quantities. Given: Density of octane = 0.702 g/mL Volume = 875 mL Needed: Mass of octane Step 2 Write a plan to calculate the needed quantity. Density Plan: milliliters Conversion factor grams kilograms 70 Solution The density of octane, a component of gasoline, is 0.702 g/mL. What is the mass, in kg, of 875 mL of octane? Step 3 Write equalities and their conversion factors including density. density 0.702 g = 1 mL and 1 kg = 1000 g Step 4 Set up problem to calculate the needed quantity. 875 mL x 0.702 g x 1 kg = 0.614 kg 1 mL 1000 g Answer is A, 0.614 kg. 71