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Solution Manual For Chemistry: The Molecular Nature Of Matter And Change, 7th Edition

Solution Manual For Chemistry: The Molecular Nature Of Matter And Change, 7th Edition is here to help you with textbook problems, offering clear solutions and helpful explanations.

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    1-1 CHAPTER 1 KEYS TO THE STUDY OF CHEMISTRY FOLLOW–UP PROBLEMS 1.1A Plan: The real question is “Does the substance change composition or just change form?” A change in composition is a chemical change while a change in form is a physical change. Solution: The figure on the left shows red atoms and molecules composed of one red atom and one blue atom. The figure on the right shows a change to blue atoms and molecules containing two red atoms. The change is chemical since the substances themselves have changed in composition. 1.1B Plan: The real question is “Does the substance change composition or just change form?” A change in composition is a chemical change while a change in form is a physical change. Solution: The figure on the left shows red atoms that are close together, in the solid state. The figure on the right shows red atoms that are far apart from each other, in the gaseous state. The change is physical since the substances themselves have not changed in composition. 1.2A Plan: The real question is “Does the substance change composition or just change form?” A change in composition is a chemical change while a change in form is a physical change. Solution: a) Both the solid and the vapor are iodine, so this must be a physical change. b) The burning of the gasoline fumes produces energy and products that are different gases. This is a chemical change. c) The scab forms due to a chemical change. 1.2B Plan: The real question is “Does the substance change composition or just change form?” A change in composition is a chemical change while a change in form is a physical change. Solution: a) Clouds form when gaseous water (water vapor) changes to droplets of liquid water. This is a physical change. b) When old milk sours, the compounds in milk undergo a reaction to become different compounds (as indicated by a change in the smell, the taste, the texture, and the consistency of the milk). This is a chemical change. c) Both the solid and the liquid are butter, so this must be a physical change. 1.3A Plan: We need to find the amount of time it takes for the professor to walk 10,500 m. We know how many miles she can walk in 15 min (her speed), so we can convert the distance the professor walks to miles and use her speed to calculate the amount of time it will take to walk 10,500 m. Solution: Time (min) = 10,500 m � 1 km 1000 m � � 1 mi 1.609 km � � 15 min 1 mi � = 97.8869 = 98 min Road map: 1000 m = 1 km 1.609 km = 1 mi Distance (m) Distance (km) Distance (mi)

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    1-2 1 mi = 15 min 1.3B Plan : We need to find the number of virus particles that can line up side by side in a 1 inch distance. We know the diameter of a virus in nm units. If we convert the 1 inch distance to nm, we can use the diameter of the virus to calculate the number of virus particles we can line up over a 1 inch distance. Solution: No. of virus particles = 1.0 in � 2.54 cm 1 in � � 1 x 10 7 nm 1 cm � � 1 virus particle 30 nm � = 8.4667 x 10 5 = 8.5 x 10 5 virus particles Road map: 1 in = 2.54 cm 1 cm = 1 x 10 7 nm 30 nm = 1 particle 1.4A Plan: The diameter in nm is used to obtain the radius in nm, which is converted to the radius in dm. The volume of the ribosome in dm 3 is then determined using the equation for the volume of a sphere given in the problem. This volume may then be converted to v olume in μL. Solution: Radius (dm) = diameter 2 = 9 21.4 nm 1 m 1 dm 2 0.1 m 1 x 10 nm             = 1.07 x 10 –7 dm Volume (dm 3 ) = ( ) ( ) 3 3 7 4 4 3.14159 1.07 x 10 dm 3 3 − π = r = 5.13145 x 10 –21 = 5.13 x 10 –21 dm 3 Volume ( μ L) = ( ) 21 3 3 6 1 L 1 μL 5.13145 x 10 dm (1 dm) 10 L − −             = 5.13145 x 10 –15 = 5.13 x 10 –15 μ L Road map: diameter = 2 r V = 4/3π r 3 Diameter (dm) Radius (dm) Volume (dm 3 ) Time (min) Length (in) Length (cm) Length (nm) No. of particles

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    1-3 1 dm 3 = 1 L 1 L = 10 6 μL 1.4B Plan: We need to convert gallon units to liter units. If we first convert gallons to dm 3 , we can then convert to L. Solution: Volume (L) = 8400 gal � 3.785 dm 3 1 gal � � 1 L 1 dm 3 � = 31,794 = 32,000 L Road map: 1 gal = 3.785 dm 3 1 dm 3 = 1 L 1.5A Plan: The time is given in hours and the rate of delivery is in drops per second. Conversions relating hours to seconds are needed. This will give the total number of drops, which may be combined with their mass to get the total mass. The mg of drops will then be changed to kilograms. Solution: Mass (kg) = 3 3 60 min 60s 1.5 drops 65 mg 10 g 1 kg 8.0 h 1 h 1 min 1 s 1 drop 1 mg 10 g −                              = 2.808 = 2.8 kg Road map: 1 hr = 60 min 1 min = 60 s 1 s = 1.5 drops 1 drop = 65 mg 1 mg = 10 3 g Volume (μL) Time (hr) Time (min) Time (s) No. of drops Mass (mg) of solution Mass (g) of solution Volume (gal) Volume (dm 3 ) Volume (L)

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    1-4 10 3 g = 1 kg 1.5B Plan: We have the mass of apples in kg and need to find the mass of potassium in those apples in g. The number of apples per pound and the mass of potassium per apple are given. Convert the mass of apples in kg to pounds. Then use the number of apples per pound to calculate the number of apples. Use the mass of potassium in one apple to calculate the mass (mg) of potassium in the group of apples. Finally, convert the mass in mg to g. Solution: Mass (g) = 3.25 kg � 1 lb 0.4536 kg � � 3 apples 1 lb � � 159 mg potassium 1 apple � � 1 g 10 3 mg � = 3.4177 = 3.42 g Road map: 0.4536 kg = 1 lb 1 lb = 3 apples 1 apple = 159 mg potassium 10 3 mg = 1 g 1.6A Plan : We know the area of a field in m 2 . We need to know how many bottles of herbicide will be needed to treat that field. The volume of each bottle (in fl oz) and the volume of herbicide needed to treat 300 ft 2 of field are given. Convert the area of the field from m 2 to ft 2 (don’t forget to square the conversion factor when converting from squared units to squared units!). Then use the given conversion factors to calculate the number of bottles of herbicide needed. Convert first from ft 2 of field to fl oz of herbicide (because this conversion is from a squared unit to a non-squared unit, we do not need to square the conversion factor). Then use the number of fl oz per bottle to calculate the number of bottles needed. Solution: No. of bottles = 2050 m 2 � 1 ft 2 (0.3048) 2 m 2 � � 1.5 fl oz 300 ft 2 � � 1 bottle 16 fl oz � = 6.8956 = 7 bottles Road map: (0.3048) 2 m 2 = 1 ft 2 Mass (kg) of solution Mass (kg) of apples Mass (lb) of apples No. of apples Mass (mg) potassium Mass (g) potassium Area (m 2 ) Area (ft 2 )

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    1-5 300 ft 2 = 1.5 fl oz 16 oz = 1 bottle 1.6B Plan: Calculate the mass of mercury in g. Convert the surface area of the lake form mi 2 to ft 2 . Find the volume of the lake in ft 3 by multiplying the surface area (in ft 2 ) by the depth (in ft). Then convert the volume of the lake to mL by converting first from ft 3 to m 3 , then from m 3 to cm 3 , and from cm 3 to mL. Finally, divide the mass in g by the volume in mL to find the mass of mercury in each mL of the lake. Solution: Mass (g) = 75,000 kg � 1000 g 1 kg � = 7.5 x 10 7 g Volume (mL) = 4.5 mi 2 � (5280) 2 ft 2 1 mi 2 � � 35 ft � � 0.02832 m 3 1 ft 3 � � 1 x 10 6 cm 3 1 m 3 � � 1 mL 1 cm 3 � = 1.2 x 10 14 mL Mass (g) of mercury per mL = 7.5 x 10 7 g 1.2 x 10 14 mL = 6.2 x 10 –7 g/mL Road map: 1 kg = 10 3 g 1 mi 2 = (5280) 2 ft 2 V = area (ft 2 ) x depth (ft) 1 ft 3 = 0.02832 m 3 1 m 3 = 10 6 cm 3 1 cm 3 = 1 mL Volume (fl oz) No. of bottles Mass (kg) Mass (g) Area (mi 2 ) Area (ft 2 ) Volume (ft 3 ) Volume (m 3 ) Volume (cm 3 ) Volume (mL) Mass (g) of mercury in 1 mL of water divide mass by volume

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    1-2 1.7A Plan: Find the mass of Venus in g. Calculate the radius of Venus by dividing its diameter by 2. Convert the radius from km to cm. Use the radius to calculate the volume of Venus. Finally, find the density of Venus by dividing the mass of Venus (in g) by the volume of Venus (in cm 3 ). Solution: Mass (g) = 4.9 x 10 24 kg � 10 3 g 1 kg � = 4.9 x 10 27 g Radius (cm) = � 12,100 km 2 � � 10 3 m 1 km � � 10 2 cm 1 m � = 6.05 x 10 8 cm Volume (cm 3 ) = 4 3 𝜋𝑟 3 = 4 3 (3.14)(6.05 x 10 8 cm) 3 = 9.27 x 10 26 cm 3 Density (g/cm 3 ) = 4.9 x 10 27 g 9.27 x 10 26 cm 3 = 5.3 g/cm 3 Road map: 1 kg = 10 3 g d = 2r 1 km = 10 3 m 1 m = 10 2 cm V = 4 3 𝜋𝑟 3 1.7B Plan: The volume unit may be factored away by multiplying by the density. Then it is simply a matter of changing grams to kilograms. Solution: Mass (kg) = ( ) 3 3 7.5 g 1 kg 4.6 cm 1000 g cm            = 0.0345 = 0.034 kg Road map: multiply by density (1 cm 3 = 7.5 g) Volume (cm 3 ) Mass (g) Mass (kg) Mass (g) Diameter (km) Radius (km) Radius (m) Radius (cm) Volume (cm 3 ) divide mass by volume Density (g/cm 3 ) divide mass by volume

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    1-2 10 3 g = 1 kg 1.8A Plan: Using the relationship between the Kelvin and Celsius scales, change the Kelvin temperature to the Celsius temperature. Then convert the Celsius temperature to the Fahrenheit value using the relationship between these two scales. Solution: T (in °C) = T (in K) – 273.15 = 234 K – 273.15 = –39.15 = –39°C T (in °F) = 5 9 T (in °C) + 32 = 5 9 (–39.15°C) + 32 = –38.47 = –38°F Check: Since the Kelvin temperature is below 273, the Celsius temperature must be negative. The low Celsius value gives a negative Fahrenheit value. 1.8B Plan: Convert the Fahrenheit temperature to the Celsius value using the relationship between these two scales. Then use the relationship between the Kelvin and Celsius scales to change the Celsius temperature to the Kelvin temperature. Solution: T (in °C) = 5 9 (T (in °F) – 32) = 5 9 (2325 ° F – 32) = 1273.8889 = 1274 ° C T (in K) = T (in °C ) + 273.15 = 1274 °C + 273.15 = 1547.15 = 1547 K Check: Since the Fahrenheit temperature is large and positive, both the Celsius and Kelvin temperatures should also be positive. Because the Celsius temperature is greater than 273, the Kelvin temperature should be greater than 273, which it is. 1.9A Plan: Determine the significant figures by counting the digits present and accounting for the zeros. Zeros between non-zero digits are significant, as are trailing zeros to the right of a decimal point. Trailing zeros to the left of a decimal point are only significant if the decimal point is present. Solution: a) 31.070 mg; five significant figures b) 0.06060 g; four significant figures c) 850.°C; three significant figures — note the decimal point that makes the zero significant. Check: All significant zeros must come after a significant digit. 1.9B Plan: Determine the significant figures by counting the digits present and accounting for the zeros. Zeros between non-zero digits are significant, as are trailing zeros to the right of a decimal point. Trailing zeros to the left of a decimal point are only significant if the decimal point is present. Solution: a) 2. 000 x 10 2 mL; four significant figures b) 3.9 x 10 –6 m; two significant figures — note that none of the zeros are significant. c) 4.01 x 10 –4 L; three significant figures Check: All significant zeros must come after a significant digit. 1.10A Plan: Use the rules presented in the text. Add the two values in the numerator before dividing. The time conversion is an exact conversion and, therefore, does not affect the significant figures in the answer. Solution: The addition of 25.65 mL and 37.4 mL gives an answer where the last significant figure is the one after the decimal point (giving three significant figures total): 25.65 mL + 37.4 mL = 63.05 = 63.0 mL When a four significant figure number divides a three significant figure number, the answer must round to three significant figures. An exact number (1 min / 60 s) will have no bearing on the number of significant figures. 63.0 mL 1 min 73.55 s 60 s       = 51.394 = 51.4 mL/min Mass (kg)

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    1-3 1.10B Plan: Use the rules presented in the text. Subtract the two values in the numerator and multiply the numbers in the denominator before dividing. Solution: The subtraction of 35.26 from 154.64 gives an answer in which the last significant figure is two places after the decimal point (giving five significant figures total): 154.64 g – 35.26 g = 119.38 g The multiplication of 4.20 cm (three significant figures) by 5.12 cm (three significant figures) by 6.752 cm (four significant figures) gives a number with three significant figures. 4.20 cm x 5.12 cm x 6.752 cm = 145.1950 = 145 cm 3 When a three significant figure number divides a five significant figure number, the answer must round to three significant figures. 119.38 g 145 cm 3 = 0.8233 = 0.823 g/cm 3 END–OF–CHAPTER PROBLEMS 1.1 Plan: If only the form of the particles has changed and not the composition of the particles, a physical change has taken place; if particles of a different composition result, a chemical change has taken place. Solution: a) The result in C represents a chemical change as the substances in A (red spheres) and B (blue spheres) have reacted to become a different substance (particles consisting of one red and one blue sphere) represented in C. There are molecules in C composed of the atoms from A and B. b) The result in D represents a chemical change as again the atoms in A and B have reacted to form molecules of a new substance. c) The change from C to D is a physical change . The substance is the same in both C and D (molecules consisting of one red sphere and one blue sphere) but is in the gas phase in C and in the liquid phase in D. d) The sample has the same chemical properties in both C and D since it is the same substance but has different physical properties . 1.2 Plan: Apply the definitions of the states of matter to a container. Next, apply these definitions to the examples. Gas molecules fill the entire container; the volume of a gas is the volume of the container. Solids and liquids have a definite volume. The volume of the container does not affect the volume of a solid or liquid. Solution: a) The helium fills the volume of the entire balloon. The addition or removal of helium will change the volume of a balloon. Helium is a gas . b) At room temperature, the mercury does not completely fill the thermometer. The surface of the liquid mercury indicates the temperature. c) The soup completely fills the bottom of the bowl, and it has a definite surface. The soup is a liquid , though it is possible that solid particles of food will be present. 1.3 Plan: Apply the definitions of the states of matter to a container. Next, apply these definitions to the examples. Gas molecules fill the entire container; the volume of a gas is the volume of the container. Solids and liquids have a definite volume. The volume of the container does not affect the volume of a solid or liquid. Solution: a) The air fills the volume of the room. Air is a gas . b) The vitamin tablets do not necessarily fill the entire bottle. The volume of the tablets is determined by the number of tablets in the bottle, not by the volume of the bottle. The tablets are solid . c) The sugar has a definite volume determined by the amount of sugar, not by the volume of the container. The sugar is a solid . 1.4 Plan: Define the terms and apply these definitions to the examples. Solution: Physical property – A characteristic shown by a substance itself, without interacting with or changing into other substances.

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    1-4 Chemical property – A characteristic of a substance that appears as it interacts with, or transforms into, other substances. a) The change in color (yellow–green and silvery to white), and the change in physical state (gas and metal to crystals) are examples of physical properties . The change in the physical properties indicates that a chemical change occurred. Thus, the interaction between chlorine gas and sodium metal producing sodium chloride is an example of a chemical property . b) The sand and the iron are still present. Neither sand nor iron became something else. Colors along with magnetism are physical properties . No chemical changes took place, so there are no chemical properties to observe. 1.5 Plan: Define the terms and apply these definitions to the examples. Solution: Physical change – A change in which the physical form (or state) of a substance, but not its composition, is altered. Chemical change – A change in which a substance is converted into a different substance with different composition and properties. a) The changes in the physical form are physical changes . The physical changes indicate that there is also a chemical change . Magnesium chloride has been converted to magnesium and chlorine. b) The changes in color and form are physical changes . The physical changes indicate that there is also a chemical change . Iron has been converted to a different substance, rust. 1.6 Plan: Apply the definitions of chemical and physical changes to the examples. Solution: a) Not a chemical change, but a physical change — simply cooling returns the soup to its original form. b) There is a chemical change — cooling the toast will not “un–toast” the bread. c) Even though the wood is now in smaller pieces, it is still wood. There has been no change in composition, thus this is a physical change , and not a chemical change. d) This is a chemical change converting the wood (and air) into different substances with different compositions. The wood cannot be “unburned.” 1.7 Plan: If there is a physical change, in which the composition of the substance has not been altered, the process can be reversed by a change in temperature. If there is a chemical change, in which the composition of the substance has been altered, the process cannot be reversed by changing the temperature. Solution: a) and c) can be reversed with temperature; the dew can evaporate and the ice cream can be refrozen. b) and d) involve chemical changes and cannot be reversed by changing the temperature since a chemical change has taken place. 1.8 Plan: A system has a higher potential energy before the energy is released (used). Solution: a) The exhaust is lower in energy than the fuel by an amount of energy equal to that released as the fuel burns. The fuel has a higher potential energy. b) Wood , like the fuel, is higher in energy by the amount released as the wood burns. 1.9 Plan: Kinetic energy is energy due to the motion of an object. Solution: a) The sled sliding down the hill has higher kinetic energy than the unmoving sled. b) The water falling over the dam (moving) has more kinetic energy than the water held by the dam. 1.10 Alchemical: chemical methods – distillation, extraction; chemical apparatus Medical: mineral drugs Technological: metallurgy, pottery, glass 1.11 Combustion released the otherwise undetectable phlogiston. The more phlogiston a substance contained; the more easily it burned. Once all the phlogiston was gone, the substance was no longer combustible.

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    1-5 1.12 The mass of the reactants and products are easily observable quantities. The explanation of combustion must include an explanation of all observable quantities. Their explanation of the mass gain required phlogiston to have a negative mass. 1.13 Lavoisier measured the total mass of the reactants and products, not just the mass of the solids. The total mass of the reactants and products remained constant. His measurements showed that a gas was involved in the reaction. He called this gas oxygen (one of his key discoveries). 1.14 Observations are the first step in the scientific approach. The first observation is that the toast has not popped out of the toaster. The next step is a hypothesis (tentative explanation) to explain the observation. The hypothesis is that the spring mechanism is stuck. Next, there will be a test of the hypothesis. In this case, the test is an additional observation — the bread is unchanged. This observation leads to a new hypothesis — the toaster is unplugged. This hypothesis leads to additional tests — seeing if the toaster is plugged in, and if it works when plugged into a different outlet. The final test on the toaster leads to a new hypothesis — there is a problem with the power in the kitchen. This hypothesis leads to the final test concerning the light in the kitchen. 1.15 A quantitative observation is easier to characterize and reproduce. A qualitative observation may be subjective and open to interpretation. a) This is qualitative . When has the sun completely risen? b) The astronaut’s mass may be measured; thus, this is quantitative . c) This is qualitative . Measuring the fraction of the ice above or below the surface would make this a quantitative measurement. d) The depth is known (measured) so this is quantitative . 1.16 A well-designed experiment must have the following essential features: 1) There must be two variables that are expected to be related. 2) There must be a way to control all the variables, so that only one at a time may be changed. 3) The results must be reproducible. 1.17 A model begins as a simplified version of the observed phenomena, designed to account for the observed effects, explain how they take place, and to make predictions of experiments yet to be done. The model is improved by further experiments. It should be flexible enough to allow for modifications as additional experimental results are gathered. 1.18 Plan: Review the definitions of mass and weight. Solution: Mass is the quantity of material present, while weight is the interaction of gravity on mass. An object has a definite mass regardless of its location; its weight will vary with location. The lower gravitational attraction on the Moon will make an object appear to have approximately one-sixth its Earth weight. The object has the same mass on the Moon and on Earth. 1.19 The unit you begin with (feet) must be in the denominator to cancel. The unit desired (inches) must be in the numerator. The feet will cancel leaving inches. If the conversion is inverted the answer would be in units of feet squared per inch. 1.20 Plan: mass Density = volume . An increase in mass or a decrease in volume will increase the density. A decrease in density will result if the mass is decreased or the volume increased. Solution: a) Density increases . The mass of the chlorine gas is not changed, but its volume is smaller. b) Density remains the same . Neither the mass nor the volume of the solid has changed. c) Density decreases . Water is one of the few substances that expands on freezing. The mass is constant, but the volume increases. d) Density increases . Iron, like most materials, contracts on cooling; thus the volume decreases while the mass does not change.

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    1-6 e) Density remains the same . The water does not alter either the mass or the volume of the diamond. 1.21 Plan: Review the definitions of heat and temperature. The two temperature values must be compared using one temperature scale, either Celsius or Fahrenheit. Solution: Heat is the energy that flows between objects at different temperatures while temperature is the measure of how hot or cold a substance is relative to another substance. Heat is an extensive property while temperature is an intensive property . It takes more heat to boil a gallon of water than to boil a teaspoon of water. However, both water samples boil at the same temperature. Convert 65°C to °F: T (in °F) = 9 5 T (in °C) + 32 = 9 5 (65°C) + 32 = 149°F A temperature of 65°C is 149°F. Heat will flow from the hot water (65°C or 149°F) to the cooler water (65°F). The 65°C water contains more heat than the cooler water. 1.22 There are two differences in the Celsius and Fahrenheit scales (size of a degree and the zero point), so a simple one-step conversion will not work. The size of a degree is the same for the Celsius and Kelvin scales; only the zero point is different so a one-step conversion is sufficient. 1.23 Plan: Review the definitions of extensive and intensive properties. Solution: An extensive property depends on the amount of material present. An intensive property is the same regardless of how much material is present. a) Mass is an extensive property . Changing the amount of material will change the mass. b) Density is an intensive property . Changing the amount of material changes both the mass and the volume, but the ratio (density) remains fixed. c) Volume is an extensive property . Changing the amount of material will change the size (volume). d) The melting point is an intensive property . The melting point depends on the substance, not on the amount of substance. 1.24 Plan: Review the table of conversions in the chapter or inside the back cover of the book. Write the conversion factor so that the unit initially given will cancel, leaving the desired unit. Solution: a) To convert from in 2 to cm 2 , use ( ) ( ) 2 2 2.54 cm 1 in ; to convert from cm 2 to m 2 , use ( ) ( ) 2 2 1 m 100 cm b) To convert from km 2 to m 2 , use ( ) ( ) 2 2 1000 m 1 km ; to convert from m 2 to cm 2 , use ( ) ( ) 2 2 100 cm 1 m c) This problem requires two conversion factors: one for distance and one for time. It does not matter which conversion is done first. Alternate methods may be used. To convert distance, mi to m, use:             km 1 m 1000 mi 1 km 609 . 1 = 1.609x10 3 m/mi To convert time, h to s, use:             s 60 min 1 min 60 h 1 = 1 h/3600 s Therefore, the complete conversion factor is 3 1.609 x 10 m 1 h 1 mi 3600 s     =           0.4469 m•h mi•s . Do the units cancel when you start with a measurement of mi/h? d) To convert from pounds (lb) to grams (g), use lb 2.205 g 1000 .

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    1-7 To convert volume from ft 3 to cm 3 use, ( ) ( ) ( ) ( ) 3 3 3 3 1 ft 1 in 12 in 2.54 cm             = 3.531x10 –5 ft 3 /cm 3 . 1.25 Plan: Review the table of conversions in the chapter or inside the back cover of the book. Write the conversion factor so that the unit initially given will cancel, leaving the desired unit. Solution : a) This problem requires two conversion factors: one for distance and one for time. It does not matter which conversion is done first. Alternate methods may be used. To convert distance, cm to in, use:       1 in 2.54 cm To convert time, min to s, use:       1 min 60 s b) To convert from m 3 to cm 3 , use ( ) ( ) 3 3 100 cm 1 m ; to convert from cm 3 to in 3 , use ( ) ( ) 3 3 1 in 2.54 cm c) This problem requires two conversion factors: one for distance and one for time. It does not matter which conversion is done first. Alternate methods may be used. To convert distance, m to km, use:       1 km 1000 m To convert time, s 2 to h 2 , use: ( ) ( ) ( ) ( ) 2 2 2 2 60 s 60 min 1 min 1 h             = 2 2 3600 s h d) This problem requires two conversion factors: one for volume and one for time. It does not matter which conversion is done first. Alternate methods may be used. To convert volume, gal to qt, use:       4 qt 1 gal ; to convert qt to L, use:       1 L 1.057 qt To convert time, h to min, use:       1 h 60 min 1.26 Plan: Use conversion factors from the inside back cover: 1 pm = 10 –12 m; 10 –9 m = 1 nm. Solution: Radius (nm) = ( ) 12 9 10 m 1 nm 1430 pm 1 pm 10 m − −             = 1.43 nm 1.27 Plan: Use conversion factors from the inside back cover: 10 –12 m = 1 pm; 1 pm = 0.01 Å. Solution: Radius (Å) = ( ) 10 12 1 pm 0.01 Å 2.22x10 m 1 pm 10 m − −             = 2.22 Å 1.28 Plan: Use conversion factors: 0.01 m = 1 cm; 2.54 cm = 1 in. Solution: Length (in) = ( ) 1 cm 1 in 100. m 0.01 m 2.54 cm             = 3.9370x10 3 = 3.94x10 3 in 1.29 Plan: Use the conversion factor 12 in = 1 ft to convert 6 ft 10 in to height in inches. Then use the conversion factors 1 in = 2.54 cm; 1 cm = 10 mm. Solution:

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    1-8 Height (in) = ( ) 12 in 6 ft 10 in 1 ft   +     = 82 in Height (mm) = ( ) 2.54 cm 10 mm 82 in 1 in 1 cm           = 2.0828x10 3 = 2.1x10 3 mm 1.30 Plan: Use conversion factors (1 cm) 2 = (0.01 m) 2 ; (1000 m) 2 = (1 km) 2 to express the area in km 2 . To calculate the cost of the patch, use the conversion factor: (2.54 cm) 2 = (1 in) 2 . Solution: a) Area (km 2 ) = ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 0.01 m 1 km 20.7 cm 1 cm 1000 m             = 2.07x10 –9 km 2 b) Cost = ( ) ( ) ( ) 2 2 2 2 1 in $3.25 20.7 cm 1 in 2.54 cm               = 10.4276 = $10.43 1.31 Plan: Use conversion factors (1 mm) 2 = (10 –3 m) 2 ; (0.01 m) 2 = (1 cm) 2 ; (2.54 cm) 2 = (1 in) 2 ; (12 in) 2 = (1 ft) 2 to express the area in ft 2 . Solution: a) Area (ft 2 ) = ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 3 2 2 2 2 2 2 2 2 10 m 1 cm 1 in 1 ft 7903 mm 1 mm 0.01 m 2.54 cm 12 in −                          = 8.5067x10 –2 = 8.507x10 –2 ft 2 b) Time (s) = ( ) 2 2 45 s 7903 mm 135 mm         = 2.634333x10 3 = 2.6x10 3 s 1.32 Plan: Use conversion factor 1 kg = 2.205 lb. The following assumes a body weight of 155 lbs. Use your own body weight in place of the 155 lbs. Solution: Body weight (kg) = ( ) 1 kg 155 lb 2.205 lb       = 70.3 kg Answers will vary, depending on the person’s mass. 1.33 Plan: Use conversion factor 1 short ton = 2000 lb; 2.205 lb = 1 kg; 1000 kg = 1 metric ton. Solution: Mass (T) = ( ) 15 3 2000 lb 1 kg 1 T 2.60 x10 ton 1 ton 2.205 lb 10 kg                 = 2.35828x10 15 = 2.36x10 15 T 1.34 Plan: Mass in g is converted to kg in part a) with the conversion factor 1000 g = 1 kg; mass in g is converted to lb in part b) with the conversion factors 1000 g = 1 kg; 1 kg = 2.205 lb. Volume in cm 3 is converted to m 3 with the conversion factor (1 cm) 3 = (0.01 m) 3 and to ft 3 with the conversion factors (2.54 cm) 3 = (1 in) 3 ; (12 in) 3 = (1 ft) 3 . The conversions may be performed in any order. Solution: a) Density (kg/m 3 ) = ( ) ( ) 3 3 3 1 cm 5.52 g 1 kg 1000 g cm 0.01 m                     = 5.52x10 3 kg/m 3 b) Density (lb/ft 3 ) = ( ) ( ) ( ) ( ) 3 3 3 3 3 2.54 cm 12 in 5.52 g 1 kg 2.205 lb 1000 g 1 kg cm 1 in 1 ft                            = 344.661 = 345 lb/ft 3

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    1-9 1.35 Plan: Length in m is converted to km in part a) with the conversion factor 1000 m = 1 km; length in m is converted to mi in part b) with the conversion factors 1000 m = 1 km; 1 km = 0.62 mi. Time is converted using the conversion factors 60 s = 1 min; 60 min = 1 h. The conversions may be performed in any order. Solution: a) Velocity (km/h) = 8 3 2.998 x10 m 60 s 60 min 1 km 1 s 1 min 1 h 10 m                           = 1.07928x10 9 = 1.079x10 9 km/h b) Velocity (mi/min) = 8 3 2.998 x10 m 60 s 1 km 0.62 mi 1 s 1 min 1 km 10 m                           = 1.11526x10 7 = 1.1x10 7 mi/min 1.36 Plan: Use the conversion factors (1 μm) 3 = (1x10 –6 m) 3 ; (1x10 –3 m) 3 = (1 mm) 3 to convert to mm 3 . To convert to L, use the conversion factors (1 μm) 3 = (1x10 –6 m) 3 ; (1x10 –2 m) 3 = (1 cm) 3 ; 1 cm 3 = 1 mL; 1 mL = 1x10 –3 L. Solution: a) Volume (mm 3 ) = ( ) ( ) ( ) ( ) 3 6 3 3 3 3 3 1x10 m 1 mm 2.56 μm cell 1 μm 1x10 m − −                             = 2.56x10 –9 mm 3 /cell b) Volume (L) = ( ) ( ) ( ) ( ) ( ) 3 6 3 3 3 5 3 3 3 2 1x10 m 1 cm 2.56 μm 1 mL 1x10 L 10 cells cell 1 mL 1 cm 1 μm 1x10 m − − −                                           = 2.56x10 –10 = 10 –10 L 1.37 Plan: For part a), convert from qt to mL (1 qt = 946.4 mL) to L (1 mL = 1x10 –3 L) to m 3 (1 L = 10 –3 m 3 ). For part b), convert from gal to qt (1 gal = 4 qt) to mL (1 qt = 946.4 mL) to L (1 mL = 10 –3 L). Solution: a) Volume (m 3 ) = ( ) 3 3 3 946.4 mL 10 L 10 m 1 qt 1 qt 1 mL 1 L − −                  = 9.464x10 –4 m 3 b) Volume (L) = ( ) 3 4 qt 946.4 mL 10 L 835 gal 1 gal 1 qt 1 mL −                 = 3.160976x10 3 = 3.16x10 3 L 1.38 Plan: The mass of the mercury in the vial is the mass of the vial filled with mercury minus the mass of the empty vial. Use the density of mercury and the mass of the mercury in the vial to find the volume of mercury and thus the volume of the vial. Once the volume of the vial is known, that volume is used in part b. The density of water is used to find the mass of the given volume of water. Add the mass of water to the mass of the empty vial. Solution: a) Mass (g) of mercury = mass of vial and mercury – mass of vial = 185.56 g – 55.32 g = 130.24 g Volume (cm 3 ) of mercury = volume of vial = ( ) 3 1 cm 130.24 g 13.53 g         = 9.626016 = 9.626 cm 3 b) Volume (cm 3 ) of water = volume of vial = 9.626016 cm 3 Mass (g) of water = ( ) 3 3 0.997 g 9.626016 cm 1 cm       = 9.59714 g water Mass (g) of vial filled with water = mass of vial + mass of water = 55.32 g + 9.59714 g = 64.91714 = 64.92 g 1.39 Plan: The mass of the water in the flask is the mass of the flask and water minus the mass of the empty flask. Use the density of water and the mass of the water in the flask to find the volume of water and thus the volume of the flask. Once the volume of the flask is known, that volume is used in part b. The density of chloroform is used

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    1-10 to find the mass of the given volume of chloroform. Add the mass of the chloroform to the mass of the empty flask. Solution: a) Mass (g) of water = mass of flask and water – mass of flask = 489.1 g – 241.3 g = 247.8 g Volume (cm 3 ) of water = volume of flask = ( ) 3 1 cm 247.8 g 1.00 g         = 247.8 = 248 cm 3 b) Volume (cm 3 ) of chloroform = volume of flask = 247.8 cm 3 Mass (g) of chloroform = ( ) 3 3 1.48 g 247.8 cm cm       = 366.744 g chloroform Mass (g) of flask and chloroform = mass of flask + mass of chloroform = 241.3 g + 366.744 g = 608.044 g = 608 g 1.40 Plan: Calculate the volume of the cube using the relationship Volume = (length of side) 3 . The length of side in mm must be converted to cm so that volume will have units of cm 3 . Divide the mass of the cube by the volume to find density. Solution: Side length (cm) = ( ) 3 2 10 m 1 cm 15.6 mm 1 mm 10 m − −             = 1.56 cm (convert to cm to match density unit) Al cube volume (cm 3 ) = (length of side) 3 = (1.56 cm) 3 = 3.7964 cm 3 3 3 mass 10.25 g Density (g/cm ) volume 3.7964 cm = = = 2.69993 = 2.70 g/cm 3 1.41 Plan: Use the relationship c = 2 πr to find the radius of the sphere and the relationship V = 4/3 πr 3 to find the volume of the sphere. The volume in mm 3 must be converted to cm 3 . Divide the mass of the sphere by the volume to find density. Solution: c = 2 πr Radius (mm) = 32.5 mm = 2 2 π π c = 5.17254 mm Volume (mm 3 ) = 3 4 3 π r = 3 4 (5.17254 mm) 3 π       = 579.6958 mm 3 Volume (cm 3 ) = ( ) 3 3 3 3 2 10 m 1 cm 579.6958 mm 1 mm 10 m − −               = 0.5796958 cm 3 3 3 mass 4.20 g Density (g/cm ) volume 0.5796958 cm = = = 7.24518 = 7.25 g/cm 3 1.42 Plan: Use the equations given in the text for converting between the three temperature scales. Solution: a) T (in °C) = [ T (in °F) – 32] 5 9 = [68°F – 32] 5 9 = 20.°C T (in K) = T (in °C) + 273.15 = 20.°C + 273.15 = 293.15 = 293 K b) T (in K) = T (in °C) + 273.15 = –164°C + 273.15 = 109.15 = 109 K T (in °F) = 9 5 T (in °C) + 32 = 9 5 (–164°C) + 32 = –263.2 = –263°F c) T (in °C) = T (in K) – 273.15 = 0 K – 273.15 = –273.15 = –273°C T (in °F) = 9 5 T (in °C) + 32 = 9 5 (–273.15°C) + 32 = –459.67 = –460.°F

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