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Chemistry: The Molecular Nature of Matter and Change Canadian Edition (2013) Solution Manual

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    Silberberg, Lavieri, & Venkateswaran, Chemistry: The Molecular Nature of Matter and Change , Canadian Edition Page 1-1 Solution Manual CHAPTER 1 KEYS TO THE STUDY OF CHEMISTRY 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. 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.

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    Silberberg, Lavieri, & Venkateswaran, Chemistry: The Molecular Nature of Matter and Change , Canadian Edition Page 1-2 Solution Manual 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. 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).

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    Silberberg, Lavieri, & Venkateswaran, Chemistry: The Molecular Nature of Matter and Change , Canadian Edition Page 1-3 Solution Manual 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 The unit you begin with (kilometres) must be in the denominator to cancel. The unit desired (centimetres) must be in the numerator. The kilometres will cancel leaving centimetres. If the conversion is inverted the answer would be in units of kilometres squared per centimetre. 1.19 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 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 (km to m) and one for time (h to s). It does not matter which conversion is done first and alternate methods may be used. To convert distance, km to m, use: 1000 m 1 km       = 10 3 m/km To convert time, h to s, use: 1 h 1 min 1 h 60 min 60 s 3600 s    =       Therefore, the complete conversion factor is: 1000 m 1 h 1 m h km 3600 s 3 6 km s . ⋅    =    ⋅    Do the units cancel when you start with units of km/h? d) To convert from kg/m 3 to g/cm 3 requires two conversion factors: To convert mass, kg to g:

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    Silberberg, Lavieri, & Venkateswaran, Chemistry: The Molecular Nature of Matter and Change , Canadian Edition Page 1-4 Solution Manual 3 1000 g g 10 1 kg kg   =     To convert volume from cm 3 to m 3 use, ( ) ( ) 3 3 6 3 3 1 m m =10 cm 100 cm −         . The complete conversion is: 3 3 3 6 3 3 3 g m g m 10 10 10 kg cm kg cm − −     ⋅ =     ⋅    Do the units cancel when you start with units of kg/m 3 ? 1.20 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 mm, use: 10 mm 1 cm       To convert time, s to min, use: 1 min 60 s       The complete conversion is: 10 mm 1 min 1 mm min 1 cm 60 s 6 cm s ⋅    =    ⋅    b) To convert from m 3 to cm 3 , use ( ) ( ) 3 3 100 cm 1 m c) This problem requires two conversion factors: one for distance and one for time squared. 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       = 10 -3 km/m To convert time, s 2 to h 2 , use: 2 2 60 min 60 s 1 h 1 min             = 1.296 x 10 7 s 2 /1 h 2 Therefore, the complete conversion factor is 3 10 km 1.296 10 s 1 m 1 h    × × ⋅ =       ⋅    7 2 10 2 2 2 1.296 10 km s m h . Do the units cancel when you start with a measurement of m/s 2 ? 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, mL to L, use: 1 L 1000 mL       To convert time, s to min, use: 60 s 1 min       The complete conversion factor is: 1 L 60 s 6 L s 1000 mL 1 min 100 mL min ⋅    =    ⋅   

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    Silberberg, Lavieri, & Venkateswaran, Chemistry: The Molecular Nature of Matter and Change , Canadian Edition Page 1-5 Solution Manual 1.21 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. 1.22 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. 1.23 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. e) Density remains the same . The water does not alter either the mass or the volume of the diamond. 1.24 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 K: T (in K) = T (in °C) + 273.15 = (65°C) + 273.15 = 338 K A temperature of 65°C is 338 K. Heat will flow from the hot water (65°C or 338 K) to the cooler water (65 K). The 65°C water contains more heat than the cooler water. 1.25 When we have a set of ratios, the units will cancel out as they are multiplicative. For example, m=d/V; if we express mass in kg and density in kg/m 3 , it would give us the identical volume (but in different units) than if we used mass in g and density in g/cm 3 . In the first case, the volume will have units of m 3 and in the second it will have units of cm 3 ; however, the actual volume will be the same (if you convert one to the other). When temperature is one of the variables, however, the conversion between Celsius and Kelvin is an additive conversion, not multiplicative. Hence, in the equation PV=nRT, we must have the units for T match the units for R (which contains the temperature unit to be cancelled). This temperature, with only one or two exceptions, will be a temperature in kelvin. 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

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    Silberberg, Lavieri, & Venkateswaran, Chemistry: The Molecular Nature of Matter and Change , Canadian Edition Page 1-6 Solution Manual 1.27 Plan: Use conversion factors from the inside back cover: 10 –12 m = 1 pm; 10 -9 m = 1 nm. Solution: Radius (Å) = ( ) ( ) 10 2 12 10 9 1 pm 2.22x10 m 2.22 10 pm 10 m 1 nm 2.22x10 m 0.222 nm 10 m − − − −   = ×       =     1.28 Plan: Use conversion factors: 1 m = 10 -9 nm Solution: Length (nm) = ( ) 9 11 10 nm 100. m 10 nm 1 m   =       1.29 Plan: Use the conversion factor 1 km = 10 6 mm to convert km to height in mm. Solution: Height (mm) = ( ) 6 10 mm 0.00196 km =1960 mm 1 km         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 10 mm $3.25 20 7 cm 1 mm 1 cm       ×         3 . = $6.73 10 1.31 Plan: Use conversion factors (1 mm) 2 = (10 –3 m) 2 ; (0.01 m) 2 = (1 cm) 2 ; Solution: a) Area (m 2 ) = ( ) ( ) ( ) 2 3 2 2 10 m 7903 mm . 1 mm 3 2 7 903 10 m − −     = ×       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 g = 1 paper clip. Solution: total mass (g) = ( )( ) mass per clip number of clips ( ) 3 3 3.56×10 g 3.56×10 clips mass per clip 1 g/cli Total mass g Number of Clips p = = = 1.33 Plan: Use conversion factor 1000 kg = 1 metric ton. Solution: Mass (T) = ( ) 21 3 1 kg 1 T 2.36 x10 g 1000 g 10 kg              = 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 mg

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    Silberberg, Lavieri, & Venkateswaran, Chemistry: The Molecular Nature of Matter and Change , Canadian Edition Page 1-7 Solution Manual in part b) with the conversion factors 1000 mg = 1 g. Volume in cm 3 is converted to m 3 with the conversion factor (1 cm) 3 = (0.01 m) 3 and to mm 3 with the conversion factors (10 mm) 3 = (1 cm) 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 (mg/mm 3 ) = ( ) ( ) 3 3 3 3 1 cm 5.52 g 1000 mg 5.52 mg / mm 1 g cm 10 mm         =             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 (cm/min) = 8 12 2.998 x10 m 60 s 100 cm 1.799x10 cm / min 1 s 1 min 1m     =            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 mL to L (1 mL = 1x10 –3 L) and then to m 3 (1 L = 10 –3 m 3 ). For part b), convert from mm 3 to L using the following conversion factors: (10 mm) 3 = (1 cm) 3 , 1 mL = 1 cm 3 and 1 mL = 10 –3 L. Solution: a) Volume (m 3 ) = 3 3 3 10 L 10 m 946.4 mL 1 mL 1 L − −             = 9.464x10 –4 m 3 b) Volume (L) = ( ) 3 3 3 3 3 (1 cm) 1mL 10 L 835 mm 1 mL (10 mm) 1 cm −                         . x L 4 8 35 10 − = 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

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    Silberberg, Lavieri, & Venkateswaran, Chemistry: The Molecular Nature of Matter and Change , Canadian Edition Page 1-8 Solution Manual 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 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 π r 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

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    Silberberg, Lavieri, & Venkateswaran, Chemistry: The Molecular Nature of Matter and Change , Canadian Edition Page 1-9 Solution Manual 1.42 Plan: Use the equations given in the text for converting between the three temperature scales. Solution: a) T (in K) = T (in °C) + 273.15 = 18°C + 273.15 = 291.15 = 291 K b) T (in K) = T (in °C) + 273.15 = –164°C + 273.15 = 109.15 = 109 K c) T (in °C) = T (in K) – 273.15 = 0 K – 273.15 = –273.15 = –273°C 1.43 Plan: Use the equations given in the text for converting between the three temperature scales. Solution: a) T (in K) = T (in °C) + 273.15 = 37 ° C + 273.15 = 310.15 = 310 K b) T (in K) = T (in °C) + 273.15 = 3410°C + 273 = 3683 K c) T (in ° C) = T (in K) –273.15 = 6.1x10 3 K – 273 = 5.827x10 3 = 5.8 x 10 3 °C 1.44 Plan: Find the volume occupied by each metal by taking the difference between the volume of water and metal and the initial volume of the water (25.0 mL). Divide the mass of the metal by the volume of the metal to calculate density. Use the density value of each metal to identify the metal. Solution: Cylinder A: volume of metal = [volume of water + metal] – [volume of water] volume of metal = 28.2 mL – 25.0 mL = 3.2 mL Density = mass 25.0 g = volume 3.2 mL = 7.81254 = 7.8 g/mL Cylinder A contains iron . Cylinder B: volume of metal = [volume of water + metal] – [volume of water] volume of metal = 27.8 mL – 25.0 mL = 2.8 mL Density = mass 25.0 g = volume 2.8 mL = 8.92857 = 8.9 g/mL Cylinder B contains nickel . Cylinder C: volume of metal = [volume of water + metal] – [volume of water] volume of metal = 28.5 mL – 25.0 mL = 3.5 mL Density = mass 25.0 g = volume 3.5 mL = 7.14286 = 7.1 g/mL Cylinder C contains zinc . 1.45 Plan: Use 1 nm = 10 –9 m to convert wavelength in nm to m. To convert wavelength in pm to nm, use 1000 pm = 1 nm. Solution: a) Wavelength (m) = ( ) 9 10 m 247 nm 1 nm −         = 2.47x10 –7 m b) Wavelength (Å ) = ( ) 1 nm 6760 pm 1000 pm       = 6.76 nm 1.46 Plan: The liquid with the larger density will occupy the bottom of the beaker, while the liquid with the smaller density volume will be on top of the more dense liquid. Solution: a) Liquid A is more dense than water; liquids B and C are less dense than water. b) Density of liquid B could be 0.94 g/mL . Liquid B is more dense than C so its density must be greater than 0.88 g/mL. Liquid B is less dense than water so its density must be less than 1.0 g/mL.

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    Silberberg, Lavieri, & Venkateswaran, Chemistry: The Molecular Nature of Matter and Change , Canadian Edition Page 1-10 Solution Manual 1.47 Plan: Calculate the volume of the cylinder in cm 3 by using the equation for the volume of a cylinder. The diameter of the cylinder must be halved to find the radius. Convert the volume in cm 3 to dm 3 by using the conversion factors (1 cm) 3 = (10 –2 m) 3 and (10 –1 m) 3 = (1 dm) 3 . Solution: Radius = diameter/2 = 0.85 cm/2 = 0.425 cm Volume (cm 3 ) = πr 2 h = π (0.425 cm) 2 (9.5 cm) = 5.3907766 cm 3 Volume (dm 3 ) = ( ) 3 3 2 3 1 10 m 1 dm 5.3907766 cm 1 cm 10 m − −                 = 5.39078x10 –3 = 5.4x10 –3 dm 3 1.48 Plan: Use the percent of copper in the ore to find the mass of copper in 5.01 kg of ore. Convert the mass in kg to mass in g. The density of copper is used to find the volume of that mass of copper. Use the volume equation for a cylinder to calculate the height of the cylinder (the length of wire); the diameter of the wire is used to find the radius which must be expressed in units of cm. Length of wire in cm must be converted to m. Solution: Mass (kg) of copper = ( ) 66% 5.01 kg Covellite 100%       = 3.3066 kg copper Mass (g) of copper = ( ) 1000 g 3.3066 kg 1 kg       = 3.3066 x10 3 g Volume (cm 3 ) of copper = ( ) 3 3 cm Cu 3.3066x10 g Cu 8.95 g Cu         = 369.453 cm 3 Cu V = π r 2 h Radius (cm) = 0.1601 mm 1 cm 2 10 mm          = 8.005x10 –3 cm Height (length) in cm = 2 π V r = ( ) ( ) 3 2 3 369.453 cm 8.005 x10 cm − π = 1.835 x 10 6 cm Length (m) = ( ) 2 6 10 m 1.835x10 cm 1 cm −         = 1.835x10 4 = 1.84x10 4 m 1.49 An exact number is defined to have a certain value (exactly). There is no uncertainty in an exact number. An exact number is considered to have an infinite number of significant figures and, therefore, does not limit the digits in the calculation. 1.50 Random error of a measurement is decreased by (1) taking the average of more measurements. More measurements allow a more precise estimate of the true value of the measurement. Calibrating the instrument will allow greater accuracy but not necessarily greater precision. 1.51 a) If the number is an exact count then there are an infinite number of significant figures. If it is not an exact count, there are only 5 significant figures. b) Other things, such as number of tickets sold, could have been counted instead. c) A value of 15,000 to two significant figures is 1.5x10 4 . Values would range from 14,501 to 15,499. Both of these values round to 1.5x10 4 . 1.52 Plan: Review the rules for significant zeros. Solution: a) No significant zeros (leading zeros are not significant) b) No significant zeros (leading zeros are not significant) c) 0.0410 (terminal zeros to the right of the decimal point are significant) d) 4.01 00 x10 4 (zeros between nonzero digits are significant; terminal zeros to the right of the decimal point are significant)

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    Silberberg, Lavieri, & Venkateswaran, Chemistry: The Molecular Nature of Matter and Change , Canadian Edition Page 1-11 Solution Manual 1.53 Plan: Review the rules for significant zeros. Solution: a) 5.08 (zeros between nonzero digits are significant) b) 508 (zeros between nonzero digits are significant) c) 5.080x10 3 (zeros between nonzero digits are significant; terminal zeros to the right of the decimal point are significant) d) 0.05080 (leading zeros are not significant; zeros between nonzero digits are significant; terminal zeros to the right of the decimal point are significant) 1.54 Plan: Review the rules for rounding. Solution: (significant figures are underlined) a) 0.000 35 54: the extra digits are 54 at the end of the number. When the digit to be removed is 5 and that 5 is followed by nonzero numbers, the last digit kept is increased by 1: 0.00036 b) 35.83 48: the extra digits are 48. Since the digit to be removed (4) is less than 5, the last digit kept is unchanged: 35.83 c) 22.4 555: the extra digits are 555. When the digit to be removed is 5 and that 5 is followed by nonzero numbers, the last digit kept is increased by 1: 22.5 1.55 Plan: Review the rules for rounding. Solution: (significant figures are underlined) a) 231.5 54: the extra digits are 54 at the end of the number. When the digit to be removed is 5 and that 5 is followed by nonzero numbers, the last digit kept is increased by 1: 231.6 b) 0.00 84 5: the extra digit is 5 at the end of the number. When the digit to be removed is 5 and that 5 is not followed by nonzero numbers, the last digit kept remains unchanged if it is even and increased by 1 if it is odd: 0.0084 c) 14 4,000: the extra digits are 4000 at the end of the number. When the digit to be removed (4) is less than 5, the last digit kept remains unchanged: 140,000 (or 1.4x10 5 ) 1.56 Plan: Review the rules for rounding. Solution: 19 rounds to 20: the digit to be removed (9) is greater than 5 so the digit kept is increased by 1. 155 rounds to 160: the digit to be removed is 5 and the digit to be kept is an odd number, so that digit kept is increased by 1. 8.3 rounds to 8: the digit to be removed (3) is less than 5 so the digit kept remains unchanged. 3.2 rounds to 3: the digit to be removed (2) is less than 5 so the digit kept remains unchanged. 2.9 rounds to 3: the digit to be removed (9) is greater than 5 so the digit kept is increased by 1. 4.7 rounds to 5: the digit to be removed (7) is greater than 5 so the digit kept is increased by 1. 20 x 160 x 8 3 x 3 x 5       = 568.89 = 6x10 2 Since there are numbers in the calculation with only one significant figure, the answer can be reported only to one significant figure. (Note that the answer is 560 using the original numbers.) 1.57 Plan: Review the rules for rounding. Solution: 10.8 rounds to 11: the digit to be removed (8) is greater than 5 so the digit kept is increased by 1. 6.18 rounds to 6.2: the digit to be removed (8) is greater than 5 so the digit kept is increased by 1. 2.381 rounds to 2.38: the digit to be removed (1) is less than 5 so the digit kept remains unchanged. 24.3 rounds to 24: the digit to be removed (3) is less than 5 so the digit kept remains unchanged. 1.8 rounds to 2: the digit to be removed (8) is greater than 5 so the digit kept is increased by 1. 19.5 rounds to 20: the digit to be removed is 5 and the digit to be kept is an odd number, so that digit kept is increased by 1. 11 x 6.2 x 2.38 24 x 2 x 20       = 0.1691 = 0.2 Since there is a number in the calculation with only one significant figure, the answer can be reported only to

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    Silberberg, Lavieri, & Venkateswaran, Chemistry: The Molecular Nature of Matter and Change , Canadian Edition Page 1-12 Solution Manual one significant figure. (Note that the answer is 0.19 with original number of significant figures.) 1.58 Plan: Use a calculator to obtain an initial value. Use the rules for significant figures and rounding to get the final answer. Solution: a) ( )( ) 2.795 m 310 m 6.48 m = 133.71 = 134 m (maximum of 3 significant figures allowed since two of the original numbers in the calculation have only 3 significant figures) b) V = ( ) 3 4 17.282 mm 3   π     = 21,620.74 = 21,621 mm 3 (maximum of 5 significant figures allowed) c) 1.110 cm + 17.3 cm + 108.2 cm + 316 cm = 442.61 = 443 cm (no digits allowed to the right of the decimal since 316 has no digits to the right of the decimal point) 1.59 Plan: Use a calculator to obtain an initial value. Use the rules for significant figures and rounding to get the final answer. Solution: a) 2.420 g 15.6 g 4.8 g + = 3.7542 = 3.8 (maximum of 2 significant figures allowed since one of the original numbers in the calculation has only 2 significant figures) b) 7.87 mL 16.1 mL 8.44 mL − = 1.0274 = 1.0 (After the subtraction, the denominator has 2 significant figures; only one digit is allowed to the right of the decimal in the value in the denominator since 16.1 has only one digit to the right of the decimal.) c) V = π (6.23 cm) 2 (4.630 cm) = 564.556 = 565 cm 3 (maximum of 3 significant figures allowed since one of the original numbers in the calculation has only 3 significant figures) 1.60 Plan: Review the procedure for changing a number to scientific notation. There can be only 1 nonzero digit to the left of the decimal point in correct scientific notation. Moving the decimal point to the left results in a positive exponent while moving the decimal point to the right results in a negative exponent. Solution: a) 1.310000x10 5 (Note that all zeros are significant.) b) 4.7x10 –4 (No zeros are significant.) c) 2.10006x10 5 d) 2.1605x10 3 1.61 Plan: Review the procedure for changing a number to scientific notation. There can be only 1 nonzero digit to the left of the decimal point in correct scientific notation. Moving the decimal point to the left results in a positive exponent while moving the decimal point to the right results in a negative exponent. Solution: a) 2.820x10 2 (Note that the zero is significant.) b) 3.80x10 –2 (Note the one significant zero.) c) 4.2708x10 3 d) 5.82009x10 4 1.62 Plan: Review the examples for changing a number from scientific notation to standard notation. If the exponent is positive, move the decimal back to the right; if the exponent is negative, move the decimal point back to the left. Solution: a) 5550 (Do not use terminal decimal point since the zero is not significant.) b) 10070. (Use terminal decimal point since final zero is significant.) c) 0.000000885 d) 0.003004

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    Silberberg, Lavieri, & Venkateswaran, Chemistry: The Molecular Nature of Matter and Change , Canadian Edition Page 1-13 Solution Manual 1.63 Plan: Review the examples for changing a number from scientific notation to standard notation. If the exponent is positive, move the decimal back to the right; if the exponent is negative, move the decimal point back to the left. Solution: a) 6500. (Use terminal decimal point since the final zero is significant.) b) 0.0000346 c) 750 (Do not use terminal decimal point since the zero is not significant.) d) 188.56 1.64 Plan: In most cases, this involves a simple addition or subtraction of values from the exponents. There can be only 1 nonzero digit to the left of the decimal point in correct scientific notation. Solution: a) 8.025x10 4 (The decimal point must be moved an additional 2 places to the left: 10 2 + 10 2 = 10 4 ) b) 1.0098x10 –3 (The decimal point must be moved an additional 3 places to the left: 10 3 + 10 – 6 = 10 – 3 ) c) 7.7x10 –11 (The decimal point must be moved an additional 2 places to the right: 10 – 2 + 10 – 9 = 10 – 11 ) 1.65 Plan: In most cases, this involves a simple addition or subtraction of values from the exponents. There can be only 1 nonzero digit to the left of the decimal point in correct scientific notation. Solution: a) 1.43x10 2 (The decimal point must be moved an additional 1 place to the left: 10 1 + 10 1 = 10 2 ) b) 8.51 (The decimal point must be moved an additional 2 places to the left: 10 2 + 10 – 2 = 10 0 ) c) 7.5 (The decimal point must be moved an additional 3 places to the left: 10 3 + 10 – 3 = 10 0 ) 1.66 Plan: Calculate a temporary answer by simply entering the numbers into a calculator. Then you will need to round the value to the appropriate number of significant figures. Cancel units as you would cancel numbers, and place the remaining units after your numerical answer. Solution: a) ( ) ( ) 34 8 9 6.626 x10 J s 2.9979 x10 m/s 489 x10 m − −  = 4.062185x10 –19 J 4.06x10 –19 J (489x10 –9 m limits the answer to 3 significant figures; units of m and s cancel) b) ( )( ) 23 2 6.022 x 10 molecules/mol 1.23 x 10 g 46.07 g/mol = 1.6078x10 24 molecules 1.61x10 24 molecules (1.23x10 2 g limits answer to 3 significant figures; units of mol and g cancel) c) ( )( ) 23 18 2 2 1 1 6.022 x 10 atoms/mol 2.18 x 10 J/atom 2 3 −   −     = 1.82333x10 5 J/mol 1.82x10 5 J/mol (2.18x10 –18 J/atom limits answer to 3 significant figures; unit of atoms cancels) 1.67 Plan: Calculate a temporary answer by simply entering the numbers into a calculator. Then you will need to round the value to the appropriate number of significant figures. Cancel units as you would cancel numbers, and place the remaining units after your numerical answer. Solution: a) ( ) ( ) 7 3 2 4.32 x10 g 4 3.1416 1.95x10 cm 3 = 1.3909 = 1.39 g/cm 3 (4.32x10 7 g limits the answer to 3 significant figures) b) ( ) ( ) 2 2 1.84x10 g 44.7 m/s 2 = 1.8382x10 5 = 1.84x10 5 g·m 2 /s 2 (1.84x10 2 g limits the answer to 3 significant figures)

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    Silberberg, Lavieri, & Venkateswaran, Chemistry: The Molecular Nature of Matter and Change , Canadian Edition Page 1-14 Solution Manual c) ( ) ( ) ( ) ( ) 2 4 3 3 5 2 2 1.07 x10 mol / L 3.8 x 10 mol / L 8.35 x10 mol / L 1.48 x 10 mol / L − − − − = 1.9248 x 10 3 = 1.9 x 10 3 L 2 /mol 2 (3.8x10 –3 mol/L limits the answer to 2 significant figures; mol 3 /L 3 in the numerator cancels mol 5 /L 5 in the denominator to leave mol 2 /L 2 in the denominator or units of L 2 /mol 2 ) 1.68 Plan: Exact numbers are those which have no uncertainty. Unit definitions and number counts of items in a group are examples of exact numbers. Solution: a) The height of Horseshoe Falls is a measured quantity. This is not an exact number. b) The number of planets in the solar system is a number count. This is an exact number. c) The number of grams in a pound is not a unit definition. This is not an exact number. d) The number of millimeters in a meter is a definition of the prefix “milli–.” This is an exact number. 1.69 Plan: Exact numbers are those which have no uncertainty. Unit definitions and number counts of items in a group are examples of exact numbers. Solution: a) The speed of light is a measured quantity. It is not an exact number. b) The density of mercury is a measured quantity. It is not an exact number. c) The number of seconds in an hour is based on the definitions of minutes and hours. This is an exact number. d) The number of provinces and territories is a counted value. These are exact numbers. 1.70 Plan: Observe the figure, and estimate a reading the best you can. Solution: The scale markings are 0.2 cm apart. The end of the metal strip falls between the mark for 7.4 cm and 7.6 cm. If we assume that one can divide the space between markings into fourths, the uncertainty is one-fourth the separation between the marks. Thus, since the end of the metal strip falls between 7.45 and 7.55 we can report its length as 7.50 ± 0.05 cm . (Note: If the assumption is that one can divide the space between markings into halves only, then the result is 7.5 ± 0.1 cm.) 1.71 Plan: You are given the density values for five solvents. Use the mass and volume given to calculate the density of the solvent in the cleaner and compare that value to the density values given to identify the solvent. Use the uncertainties in the mass and volume to recalculate the density. Solution: a) mass 11.775 g Density (g/mL) volume 15.00 mL = = = 0.7850 g/mL. The closest value is isopropanol . b) Ethanol is denser than isopropanol. Recalculating the density using the maximum mass = (11.775 + 0.003) g with the minimum volume = (15.00 – 0.02) mL, gives mass 11.778 g Density (g/mL) volume 14.98 mL = = = 0.7862 g/mL. This result is still clearly not ethanol. Yes , the equipment is precise enough. 1.72 Plan: Calculate the average of each data set. Remember that accuracy refers to how close a measurement is to the actual or true value while precision refers to how close multiple measurements are to each other. Solution: a) I avg = 8.72 g 8.74 g 8.70 g 3 + + = 8.7200 = 8.72 g II avg = 8.56 g 8.77 g 8.83 g 3 + + = 8.7200 = 8.72 g III avg = 8.50 g 8.48 g 8.51 g 3 + + = 8.4967 = 8.50 g

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    Silberberg, Lavieri, & Venkateswaran, Chemistry: The Molecular Nature of Matter and Change , Canadian Edition Page 1-15 Solution Manual IV avg = 8.41 g 8.72 g 8.55 g 3 + + = 8.5600 = 8.56 g Sets I and II are most accurate since their average value, 8.72 g, is closest to the true value, 8.72 g. b) To get an idea of precision, calculate the range of each set of values: largest value – smallest value. A small range is an indication of good precision since the values are close to each other. I range = 8.74 g – 8.70 g = 0.04 g II range = 8.83 g – 8.56 g = 0.27 g III range = 8.51 g – 8.48 g = 0.03 g IV range = 8.72 g – 8.41 g = 0.31 g Set III is the most precise (smallest range), but is the least accurate (the average is the farthest from the actual value). c) Set I has the best combination of high accuracy (average value = actual value) and high precision (relatively small range). d) Set IV has both low accuracy (average value differs from actual value) and low precision (has the largest range). 1.73 Plan: Remember that accuracy refers to how close a measurement is to the actual or true value; since the bull’s- eye represents the actual value, the darts that are closest to the bull’s-eye are the most accurate. Precision refers to how close multiple measurements are to each other; darts that are positioned close to each other on the target have high precision. Solution: a) Experiments II and IV — the averages appear to be near each other. b) Experiments III and IV — the darts are closely grouped. c) Experiment IV and perhaps Experiment II — the average is in or near the bull’s-eye. d) Experiment III — the darts are close together, but not near the bull’s-eye. 1.74 Plan: If it is necessary to force something to happen, the potential energy will be higher. Solution: a) b) a) The balls on the relaxed spring have a lower potential energy and are more stable. The balls on the compressed spring have a higher potential energy, because the balls will move once the spring is released. This configuration is less stable. b) The two + charges apart from each other have a lower potential energy and are more stable. The two + charges near each other have a higher potential energy, because they repel one another. This arrangement is less stable. P otential Energy P otential Energy + + + +

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