Chemistry: The Molecular Nature of Matter and Change, Ninth Edition PDF

Chemistry The Molecular Nature of Matter and Change Ninth Edition Martin S. Silberberg and Patricia G. Amateis © 2021 McGraw Hill. All rights reserved. Authorized only for instructor use in the classroom. No reproduction or further distribution permitted without the prior written consent of McGraw...

Chemistry The Molecular Nature of Matter and Change Ninth Edition Martin S. Silberberg and Patricia G. Amateis © 2021 McGraw Hill. All rights reserved. Authorized only for instructor use in the classroom. No reproduction or further distribution permitted without the prior written consent of McGraw Hill.  Chapter 1: Keys Studying Chemistry 1.1 Some Fundamental Definitions. 1.2 The Scientific Approach: Developing a Model. 1.3 Measurement and Chemical Problem Solving. 1.4 Uncertainty in Measurement: Significant Figures. © McGraw Hill  Chemistry Chemistry is the study of matter, its properties, the changes that matter undergoes, and the energy associated with these changes. © McGraw Hill  Definitions Matter: anything that has both mass and volume – the “stuff” of the universe: books, planets, trees, professors, students. Composition: the types and amounts of simpler substances that make up a sample of matter. Properties: the characteristics that give each substance a unique identity. © McGraw Hill  The States of Matter A solid has a fixed shape and volume. Solids may be hard or soft, rigid or flexible. A liquid has a varying shape that conforms to the shape of the container, but a fixed volume. A liquid has an upper surface. A gas has no fixed shape or volume and therefore does not have a surface. © McGraw Hill  The Physical States of Matter Figure 1.1 © McGraw Hill  Physical and Chemical Properties Physical Properties. Properties a substance shows by itself without interacting with another substance. Color, melting point, boiling point, density. Chemical Properties. Properties a substance shows as it interacts with, or transforms into, other substances. Flammability, corrosiveness. © McGraw Hill  The Distinction Between Physical and Chemical Change Figure 1.2 © McGraw Hill (A) © Paul Morrell/Stone/Getty Images; (B) © McGraw-Hill Education/Stephen Frisch, photographer  Sample Problem 1.1 – Problem and Plan Visualizing Change on the Atomic Scale PROBLEM: The scenes below represent an atomic-scale view of substance A undergoing two different changes. Decide whether each scene shows a physical or a chemical change. PLAN: We need to determine what change is taking place. The numbers and colors of the little spheres that represent each particle tell its “composition.” If the composition does not change, the change is physical, whereas a chemical change results in a change of composition. © McGraw Hill  Sample Problem 1.1 – Solution 1 SOLUTION: Each particle of substance A is composed of one blue and two red spheres. Sample B is composed of two different types of particles – some have two red spheres while some have one red and one blue. As A changes to B, the chemical composition has changed. A → B is a chemical change. © McGraw Hill  Sample Problem 1.1 – Solution 2 SOLUTION: Each particle of C is still composed of one blue and two red spheres, but the particles are closer together and are arranged in a regular pattern. The composition remains unchanged, but the physical form is different. A → C is a physical change. © McGraw Hill  Temperature and Change of State A change of state is a physical change. Physical form changes, composition does not. Changes in physical state are reversible. By changing the temperature. A chemical change cannot simply be reversed by a change in temperature. © McGraw Hill  Some Characteristic Properties of Copper Table 1.1 Some Characteristic Properties of Copper Source: (copper) ©Dgstudiodg/iStock/Getty Images, ©McGraw-Hill Education/Stephen Frisch, photographer; (candlestick) ©Willard/iStock/Getty Images; (copper carbonate, copper reacting with acid, copper and ammonia) ©McGraw-Hill Education/Stephen Frisch, photographer © McGraw Hill  Sample Problem 1.2 – Problem and Plan Distinguishing Between Physical and Chemical Change PROBLEM: Decide whether each of the following processes is primarily a physical or a chemical change, and explain briefly: (a) Frost forms as the temperature drops on a humid winter night. (b) A cornstalk grows from a seed that is watered and fertilized. (c) A match ignites to form ash and a mixture of gases. (d) Perspiration evaporates when you relax after jogging. (e) A silver fork tarnishes slowly in air. PLAN: “Does the substance change composition or just change form?” © McGraw Hill  Sample Problem 1.2 – Solution (a) Frost forms as the temperature drops on a humid winter night – physical change: water vapor changes to ice crystals. (b) A cornstalk grows from a seed that is watered and fertilized – chemical change: the seed undergoes changes in composition. (c) A match ignites to form ash and a mixture of gases – chemical change: substances in the match head are converted into other substances. (d) Perspiration evaporates when you relax after jogging – physical change: liquid water changes to gaseous water. (e) A silver fork tarnishes slowly in air – chemical change: silver changes to silver sulfide by reacting with sulfur-containing substances in the air. © McGraw Hill  Energy in Chemistry Energy is the ability to do work. Potential Energy is energy due to the position of an object. Kinetic Energy is energy due to the movement of an object. Total Energy = Potential Energy + Kinetic Energy © McGraw Hill  Energy Changes Lower energy states are more stable and are favored over higher energy states. Energy is neither created nor destroyed. It is conserved. It can be converted from one form to another. © McGraw Hill  Potential Energy is Converted to Kinetic Energy 1 A gravitational system. The potential energy gained when a weight is lifted is converted to kinetic energy as the weight falls. A lower energy state is more stable. Figure 1.3 © McGraw Hill  Potential Energy is Converted to Kinetic Energy 2 A system of two balls attached by a spring. The potential energy gained by a stretched spring is converted to kinetic energy when the moving balls are released. Energy is conserved when it is transformed. Figure 1.3 © McGraw Hill  Potential Energy is Converted to Kinetic Energy 3 A system of oppositely charged particles. The potential energy gained when the charges are separated is converted to kinetic energy as the attraction pulls these charges together. Figure 1.3 © McGraw Hill  Potential Energy is Converted to Kinetic Energy 4 A system of fuel and exhaust. A fuel is higher in chemical potential energy than the exhaust. As the fuel burns, some of its potential energy is converted to the kinetic energy of the moving car. Figure 1.3 © McGraw Hill  The Scientific Approach: Developing a Model Figure 1.4 © McGraw Hill  SI Base Units Table 1.2 SI Base Units Physical Quantity (Dimension) Unit Name Unit Abbreviation Mass kilogram kg Length meter m Time second s Temperature kelvin K Amount of substance mole mol Electric current ampere A Luminous intensity candela cd Table 1.2 © McGraw Hill  Common Decimal Prefixes Used With SI Units Table 1.3 Common Decimal Prefixes Used with SI Units* Exponential Example [using gram ( g )† or meter ( m ) ] †† Prefix∗ Symbol Conventional Notation Notation tera (T) 1,000,000,000,000 1×1012 1 teragram (Tg) = 1×1012 g giga (G) 1,000,000,000 1×109 1 gigagram (Gg) = 1×109 g mega (M) 1,000,000 1×106 1 megagram (Mg) = 1×106 g kilo (k) 1000 1×103 1 kilogram (kg) = 1×103 g hecto (h) 100 1×102 1 hectogram (hg) = 1×102 g deka (da) 10 1×101 1 dekagram (dag) = 1×101 g − − 1 1×100 deci (d) 0.1 1×10−1 1 decimeter (dm) = 1×10−1 m centi (c) 0.01 1×10−2 1 centimeter (cm) = 1×10−2 m milli (m) 0.001 1×10−3 1 millimeter (mm) = 1× 10−3 m micro (µ) 0.000001 1×10−6 1 micrometer (µm) = 1× 10−6 m nano (n) 0.000000001 1×10−9 1 nanometer (nm) = 1×10−9 m pico (p) 0.000000000001 1×10−12 1 picometer (pm) = 1×10−12 m femto (f) 0.000000000000001 1×10−15 1 femtometer (fm) = 1×10−15 m *The prefixes most frequently used by chemists appear in bold type. †The gram is a unit of mass. ††The meter is a unit of length. © McGraw Hill Table 1.3  Common SI-English Equivalent Quantities Table 1.4 Common SI–English Equivalent Quantities Table 1.4 Access the text alternative for slide images. © McGraw Hill  Some Volume Relationships in SI 1 Figure 1.5 © McGraw Hill  Common Laboratory Volumetric Glassware Figure 1.6 © McGraw Hill  Quantities of Length (A), Volume (B), and Mass (C) Figure 1.7 © McGraw Hill  Chemical Problem Solving All measured quantities consist of, a number and a unit. Units are manipulated like numbers: 12 ft 2 3 ft × 4 ft = 350 mi 50 mi or 50 mi ⋅ h −1 7h 1h © McGraw Hill  Conversion Factors A conversion factor is a ratio of equivalent quantities used to express a quantity in different units. The relationship 1 mi = 5280 ft gives us the conversion factor: 1 mi 5280 ft = =1 5280 ft 5280 ft © McGraw Hill  Conversion Factor Problem A conversion factor is chosen and set up so that all units cancel except those required for the answer. PROBLEM: The height of Angel Falls is 3212 ft. Express this quantity in miles (mi) if 1 mi = 5280 ft. PLAN: Set up the conversion factor so that ft will cancel and the answer will be in mi. 1 mi SOLUTION: 3212 ft × = 0.6083 mi 5280 ft © McGraw Hill  Systematic Approach to Solving Chemistry Problems State Problem. Plan. Clarify the known and unknown. Suggest steps from known to unknown. Prepare a visual summary of steps that includes conversion factors, equations, known variables. Solution. Check. Comment. Follow-up Problem. © McGraw Hill  Sample Problem 1.3 – Problem and Plan Converting Units of Length PROBLEM: To wire your stereo equipment, you need 325 centimeters (cm) of speaker wire that sells for $0.15/ft. What is the price of the wire? PLAN: We know the length (in cm) of wire and cost per length ($/ft). We have to convert cm to inches and inches to feet. Then we can find the cost for the length in feet. © McGraw Hill  Sample Problem 1.3 – Solution SOLUTION: Length (in) = length (cm) × conversion factor 1 in = 325 cm × = 128 in 2.54 cm Length (ft) = length (in) × conversion factor 1 ft = 128 in × = 10.7 ft 12 in Price ($) = length (ft) × conversion factor $0.15 = 10.7 ft × = $1.60 1 ft © McGraw Hill  Sample Problem 1.4 – Problem and Plan Converting Units of Volume PROBLEM: A graduated cylinder contains 19.9 mL of water. When a small piece of galena, an ore of lead, is added, it sinks and the volume increases to 24.5 mL. What is the volume of the piece of galena in cm3 and in L? PLAN: The volume of the galena is equal to the difference in the volume of the water before and after the addition. © McGraw Hill  Sample Problem 1.4 – Solution SOLUTION: (24.5 − 19.9)mL = volume of galena = 4.6 mL 1 cm3 4.6 mL × = 4.6 cm 3 1 mL 10−3 L 4.6 mL × = 4.6 × 10−3 L 1 mL © McGraw Hill  Sample Problem 1.5 – Problem Converting Units of Mass PROBLEM: Many international computer communications are carried out by optical fibers in cables laid along the ocean floor. If one strand of optical fiber weighs 1.19 × 10−3 lb/m, what is the mass (in kg) of a cable made of six strands of optical fiber, each long enough to link New York and Paris ( a distance of 8.94 × 103 km ) ? © McGraw Hill  Sample Problem 1.5 – Plan and Solution PLAN: We convert the length of one fiber from km to m and then find its mass (lb) by using the conversion factor 1 m. Then we multiply the fiber mass by 6 to get the 1.19 × 10−3 lb = cable mass and convert lb to kg. SOLUTION: Converting length of fiber from km to m: 103 m 3 8.84 × 10 km × = 8.84 × 106 m 1 km Converting length of fiber (m) to mass (lb): −3 1.19 × 10 lb 8.84 × 106 m × = 1.05 × 104 lb 1 m © McGraw Hill  Sample Problem 1.5 – Solution Finding the mass of the cable: 1.05 × 104 lb 6 fibers 4 lb × = 6.30 × 10 1 fiber 1 cable cable Converting mass of cable from lb to kg: 6.30 × 104 lb 1 kg × = 2.86 × 104 kg / cable 1 cable 2.205 lb © McGraw Hill  Sample Problem 1.6 – Problem and Plan Converting Units Raised to a Power PROBLEM: A furniture factory needs 31.5 ft2 of fabric to upholster one chair. Its Dutch supplier sends the fabric in bolts that hold exactly 200 m2. How many chairs can be upholstered with three bolts of fabric? PLAN: We know the amount of fabric in one bolt in m2; multiplying the m2 of fabric by the number of bolts gives the total amount of fabric available in m2. We convert the amount of fabric from m2 to ft2 and use the conversion factor 31.5 ft2 of fabric = 1 chair to find the number of chairs (see the road map). © McGraw Hill  Sample Problem 1.6 – Solution SOLUTION: Converting from number of bolts to amount of fabric in m 2 : 200 m 2 Amount ( m 2 ) of fabric = 3 bolts × = 600 m 2 1 bolt Converting the amount of fabric from m 2 to ft 2 : ft , we have ( 0.3048 ) m (1) 2 2 2 Since 0.3048m 1= ft 2 ,so 1 ft 2 Amount ( ft 2 ) 2 of fabric = 600 m × = 6460ft 2 ( 0.3048) 2 m2 Finding the number of chairs: 1 chair Number of chairs = 6460 ft × 2 2 = 205 chairs 31.5 ft © McGraw Hill  Density Density is the mass of a sample divided by its volume: mass density = volume At a given temperature and pressure, the density of a substance is a characteristic physical property and has a specific value. © McGraw Hill  Densities of Some Common Substances Table 1.5 Densities of Some Common Substances* Density ( g / cm ) 3 Substance Physical State Hydrogen gas 0.0000899 In the following table, Oxygen gas ‘g/cm3' read as gram0.00133 per Grain cubic alcoholcentimeter. liquid 0.789 Water liquid 0.998 Table salt solid 2.16 Aluminium solid 2.70 Lead solid 11.3 Gold solid 19.3 *At room temperature (20°C) and normal atmospheric pressure (1 atm). © McGraw Hill  Sample Problem 1.7 – Problem and Plan Calculating Density from Mass and Volume PROBLEM: (a)Lithium, a soft, gray solid with the lowest density of any metal, is a key component of advanced batteries. A slab of lithium weighs 1.49 × 103 mg and has sides that are 20.9 mm by 11.1 mm by 11.9 mm. Find the density of lithium in g/cm3. (b) What is the volume (in mL) of a piece of lithium with a mass of 895 mg? PLAN: (a) Density is expressed in g/cm3 so we need to convert the mass from mg to g. We convert the lengths of the three sides from mm to cm and multiply them to obtain the volume in cm3. (b) We convert the mass of sample from mg to g and use its density as a conversion factor between mass and volume in cm3. Finally, we convert cm3 to mL. © McGraw Hill  Sample Problem 1.7 –Plan © McGraw Hill  Sample Problem 1.7 – Solution 1 SOLUTION: −3 10 g (a) 1.49 × 103 mg × = 1.49 g 1 mg 1 cm 20.9 mm × = 2.09 cm 10 mm Similarly the other sides will be 1.11 cm and 1.19 cm, respectively. Volume = 2.09 cm × 1.11 cm × 1.19 cm = 2.76 cm3 1.49 g 3 Density = of Li = 0.540 g / cm 2.76 cm3 © McGraw Hill  Sample Problem 1.7 – Solution 2 SOLUTION: −3 (b) 895 mg × 10 g = 0.895 g 1 mg 1 cm 3 1 mL 0.895 g × × 3 = 1.66 mL 0.540 g 1 cm © McGraw Hill  Temperature Temperature is a measure of how hot or cold one object is relative to another. Heat is the energy that flows from an object with a higher temperature to an object with a lower temperature. © McGraw Hill  Temperature Scales Celsius (oC) – The Celsius scale is based on the freezing and boiling points of water. This is the temperature scale used most commonly around the world. Kelvin (K) – The “absolute temperature scale” begins at absolute zero and has only positive values. The Celsius and Kelvin scales use the same size degree although their starting points differ. Note that the kelvin is not used with the degree sign (o). Fahrenheit (oF) – The Fahrenheit scale is commonly used in the U.S. The Fahrenheit scale has a different degree size and different zero points than both the Celsius and Kelvin scales. © McGraw Hill  Some Interesting Temperatures Figure 1.8 © McGraw Hill  Freezing and Boiling Points of Water Figure 1.9 © McGraw Hill  Temperature Conversions T = ( in K ) T ( in °C ) + 273.15 ( in °C ) T ( in K ) − 273.15 T= 9 T ( in °F ) = T ( in °C ) + 32 5 5 T = ( in °C ) T ( in °F ) − 32 9 © McGraw Hill  The Three Temperature Scales 1 Table 1.6 The Three Temperature Scales Scale Unit Size of Degree Freezing Boiling T at Conversion (Relative to K) Point Point Absolute of H2O of H2O Zero Kelvin kelvin — 273.15 K 373.15 K 0K to °C (absolute) (K) (Equation 1.2) Celsius Celsius 1 0°C 100°C −273.15° to K (Equation degree C 1.3) (°C) to °F (Equation 1.4) Fahrenhei Fahren 5 32°F 212°F −459.67° to °C t heit F (Equation 1.5) 9 degree (°F) © McGraw Hill  The Three Temperature Scales 2 Table 1.6 The Three Temperature Scales Scale Unit Size of Degree Freezing Boiling T at Conversion (Relative to K) Point Point Absolute of H2O of H2O Zero Kelvin kelvin — 273.15 K 373.15 K 0K to °C (absolute) (K) (Equation 1.2) Celsius In Celsius the following 1 table, 0°C ‘5/9'100°C read as−273.15° 5 over 9.to K (Equation degree C 1.3) (°C) to °F (Equation 1.4) Fahrenheit Fahren 5 32°F 212°F −459.67° to °C heit F (Equation 1.5) 9 degree (°F) © McGraw Hill  Sample Problem 1.8 – Problem, Plan and Solution Converting Units of Temperature PROBLEM: A child has a body temperature of 38.7°C, and normal body temperature is 98.6°F. Does the child have a fever? What is the child’s temperature in kelvins? PLAN: We have to convert °C to °F to find out if the child has a fever. We can then use the °C to Kelvin relationship to find the temperature in Kelvin. SOLUTION: 9 Converting from °C to °F: ( 38.7 °C ) + 32= 101.7°F 5 Yes, the child has a fever. Converting from °C to K 38.7℃ + 273.15= 311.8 K © McGraw Hill  Extensive and Intensive Properties Extensive properties are dependent on the amount of substance present; mass and volume, for example, are extensive properties. Intensive properties are independent of the amount of substance; density is an intensive property. © McGraw Hill  Significant Figures Every measurement includes some uncertainty. The rightmost digit of any quantity is always estimated. The recorded digits, both certain and uncertain, are called significant figures. The greater the number of significant figures in a quantity, the greater its certainty. © McGraw Hill  The Number of Significant Figures in a Measurement Figure 1.11 © McGraw Hill  Determining Which Digits Are Significant All digits in a measurement are significant. Except zeros used only to position the decimal point. 0.0055 has 2 significant figures. Zeros that end a number are significant. Whether they occur before or after the decimal point. As long as a decimal point is present. 1.030 mL has 4 significant figures. 5300. L has 4 significant figures. If no decimal point is present. Zeros at the end of the number are not significant. 5300 L has only 2 significant figures. © McGraw Hill  Sample Problem 1.9 – Problem and Plan Determining the Number of Significant Figures PROBLEM: For each of the following quantities, underline the zeros that are significant figures (sf), and determine the number of significant figures in each quantity. For (e) to (g), express each in exponential notation first. (a) 0.0030 L (b) 0.1044 g (c) 53.069 L (d) 3040 kg (e) 0.00004715 m (f) 57,600. s (g) 0.0000007160 cm3 PLAN: We determine the number of significant figures by counting digits, paying particular attention to the position of zeros in relation to the decimal point, and underline zeros that are significant. © McGraw Hill  Sample Problem 1.9 - Solution SOLUTION: (a) 0.0030 L has 2 sf (b) 0.1044 g has 4 sf (c) 53,069 mL has 5 sf (d) 3040 kg has 3 sf (e) 0.00004715 m = 4.715 × 10−5 m has 4 sf (f) 57,600. s = 5.7600 × 104 s has 5 sf (g) 0.0000007160 cm3 = 7.160 × 10−7 cm 3 has 4 sf © McGraw Hill  Rules for Significant Figures in Calculations 1 1. For multiplication and division. The answer contains the same number of significant figures as there are in the measurement with the fewest significant figures. Multiply the following numbers: 9.2 cm × 6.8 cm × 0.3744= cm3 23 cm 3 cm 23.4225 = Two significant figures are allowed in the answer because two of the values in the calculation have only two significant figures. © McGraw Hill  Rules for Significant Figures in Calculations 2 2. For addition and subtraction. The answer has the same number of decimal places as there are in the measurement with the fewest decimal places. Example: adding two volumes 83.5 mL + 23.28 mL = 106.78 mL = 106.8 mL The answer has only one digit to the right of the decimal as does 83.5. Example: subtracting two volumes 865.9 mL − 2.8121 mL = 863.0879 mL = 863.1 mL The answer has only one digit to the right of the decimal as does 865.9. © McGraw Hill  Rules for Rounding Off Numbers 1 1. If the digit removed is more than 5, the preceding number increases by 1. 5.379 rounds to 5.38 if 3 significant figures are retained. 2. If the digit removed is less than 5, the preceding number is unchanged. 0.2413 rounds to 0.241 if 3 significant figures are retained. 3. If the digit removed is 5 followed by zeros or with no following digits, the preceding number increases by 1 if it is odd and remains unchanged if it is even. 17.75 rounds to 17.8, but 17.65 rounds to 17.6. If the 5 is followed by other nonzero digits, rule 1 is followed: 17.6500 rounds to 17.6, but 17.6513 rounds to 17.7. © McGraw Hill  Rules for Rounding Off Numbers 2 4. Be sure to carry two or more additional significant figures through a multistep calculation and round off the final answer only. © McGraw Hill  Significant Figures in the Lab The measuring device used determines the number of significant digits possible. Figure 1.12 © McGraw Hill  Exact Numbers Exact numbers have no uncertainty associated with them. Numbers may be exact by definition: 1000 mg= 1 g. 60 min = 1 hr. 2.54 cm = 1 in. Numbers may be exact by count: Exactly 26 letters in the alphabet. Exact numbers do not limit the number of significant digits in a calculation. © McGraw Hill  Sample Problem 1.10 – Problem and Plan Significant Figures and Rounding PROBLEM: Perform the following calculations and round each answer to the correct number of significant figures: 16.205 cm 2 − 1.5 cm 2 (a) 7.081 cm 4  1g  4.80 × 10 mg   1000 mg  (b) 11.55 cm3 PLAN: We use the rules for rounding presented in the text: (a) We subtract before we divide. (b) We note that the unit conversion involves an exact number. © McGraw Hill  Sample Problem 1.10 - Solution SOLUTION: 16.205 cm 2 − 1.5 cm 2 14.7 cm 2 (a) = = 2.08 cm 7.081 cm 7.081 cm The answer to the subtraction (14.705) is rounded to 14.7 with one decimal place because 1.5 has only one decimal place. 4  1g  4.80 × 10 mg   1000 mg  48.0 g (b) = = 4.16 g / cm 3 11.55 cm3 11.55 cm3 © McGraw Hill  Precision, Accuracy, and Error Precision refers to how close the measurements in a series are to each other. Accuracy refers to how close each measurement is to the actual value. Systematic error produces values that are either all higher or all lower than the actual value. This error is part of the experimental system. Random error produces values that are both higher and lower than the actual value. Random error always occurs. © McGraw Hill  Precision and Accuracy in a Laboratory Calibration © McGraw Hill