These three chapters of the Study Guide and Student Solutions Manual for the
Seventh Edition of Seager/Slabaugh's Chemistry for Today: General, Organic and
Biochemistry are provided as a preview only. Students can purchase the Student
Solutions Manual using ISBN 0-538-73458-2.
Chapter 1: Matter, Measurements, and Calculations CHAPTER OUTLINE 1.1 What Is Matter? 1.2 Properties and Changes 1.3 A Model of Matter 1.4 Classifying Matter 1.5 Measurement Units 1.6 The Metric System 1.7 Large and Small Numbers ­
1.8 Significant Figures 1.9 Using Units in Calculations 1.10 Calculating Percentages 1.11 Density LEARNING OBJECTIVES/ASSESSMENT re
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When you have completed your study of this chapter, you should be able to: 1. Explain what matter is. (Section 1.1; Exercise 1.2) 2. Explain differences between the terms physical and chemical as applied to: a. Properties of matter (Section 1.2; Exercises 1.10 b & c) b. Changes in matter (Section 1.2; Exercises 1.8 a & b) 3. Describe matter in terms of the accepted scientific model. (Section 1.3; Exercise 1.12) 4. On the basis of observation or information given to you, classify matter into the correct category of each of the following pairs: a. Heterogeneous or homogeneous (Section 1.4; Exercise 1.22) b. Solution or pure substance (Section 1.4; Exercise 1.24) c. Element or compound (Section 1.4; Exercise 1.18) 5. Recognize the use of measurement units in everyday activities. (Section 1.5; Exercise 1.28) 6. Recognize units of the metric system, and convert measurements done using the metric system into related units. (Section 1.6; Exercises 1.30 and 1.40) 7. Express numbers using scientific notation, and do calculations with numbers expressed in scientific notation. (Section 1.7; Exercises 1.48 and 1.60) 8. Express the results of measurements and calculations using the correct number of significant figures. (Section 1.8; Exercises 1.64 and 1.66) 9. Use the factor‐unit method to solve numerical problems. (Section 1.9; Exercise 1.82) 10. Do calculations involving percentages. (Section 1.10; Exercise 1.92) 11. Do calculations involving densities. (Section 1.11; Exercise 1.98) rP
SOLUTIONS FOR THE END OF CHAPTER EXERCISES Fo
WHAT IS MATTER? (SECTION 1.1) ;1.2 All matter occupies space and has mass. Mass is a measurement of the amount of matter in an object. The mass of an object is constant regardless of where the mass is measured. Weight is a measurement of the gravitational force acting on an object. The weight of an object will change with gravity; therefore, the weight of an object will be different at different altitudes and on different planets. 1.4 The distance you can throw a bowling ball through the air will change more than the distance you can roll a bowling ball on a flat, smooth surface. When throwing a ball through the air, gravity pulls the ball toward the ground and air resistance slows its decent. The gravitational force on the moon is approximately 1/6th the gravitational force that is present on the earth; therefore, when throwing a ball on the moon, you should be able to throw it further than you can on earth. The moon does not have air resistance. When rolling a ball, friction helps to slow down the ball. If the flat, smooth surface is the same on the earth and the moon, the amount of friction should remain constant. 1 2 Chapter 1 1.6 The attractive force of gravity for objects near the earth’s surface increases as you get closer to the center of the earth (Exercise 1.5). If the earth bulges at the equator, the people at the equator are further from the center of the earth than people at the North Pole. If two people with the same mass were weighed at the equator and at the North Pole, the person at the equator would weigh less than the person at the North Pole because the gravitational force at the North Pole is stronger than the gravitational force at the equator. Fo
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PROPERTIES AND CHANGES (SECTION 1.2) 1.8 ;a. The two pieces of the stick still have the same chemical composition as the original stick. This was a change that did not involve composition; therefore, it is a physical change. ;b. As the candle burns, it produces carbon dioxide, water, soot, and other products. This is a change that involves composition; therefore, it is a chemical change. c. The pieces of rock salt have the same chemical composition as the original larger piece of rock salt. This was a change that did not involve composition; therefore, it is a physical change. d. Many tree leaves are green in the spring and summer because of the green chlorophyll that is used in photosynthesis to produce energy for the tree. During these seasons, the tree stores the extra energy so that in autumn when the days grow shorter, the chlorophyll is no longer needed. As the leaves in the cell stop producing chlorophyll, the other colors present in the leaves become more visible. This change involves composition; therefore, it is a chemical change. 1.10 a. The phase of matter at room temperature is a physical property because the composition does not change while making this observation. ;b. The reaction between two substances is a chemical property because the composition of the products differs from the reactants. The products for the reaction between sodium metal and water are sodium hydroxide and hydrogen gas. (Note: Predicting the products for this type of chemical reaction is covered in Section 9.6.) ;c. Freezing point is a physical property because the composition does not change while making this observation. d. The inability of a material to form new products by rusting is a chemical property because rust would have a different chemical composition than gold. Attempting to change the chemical composition of a material is a test of a chemical property regardless of whether the attempt is successful. e. The color of a substance is a physical property because the composition does not change while making this observation. A MODEL OF MATTER (SECTION 1.3) ;1.12 a. Yes, the succinic acid molecules have been changed by the process. The molecules of succinic acid released at least one atom each in the form of a gas. Without those atoms, the molecules cannot be succinic acid molecules. Also, if they were still succinic acid molecules, the melting point of the remaining solid would still be 182°C. b. No, the white solid that remains after heating is not succinic acid. The melting point of succinic acid is 182°C, but the melting point of this new solid is not 182°C. c. The succinic acid molecules contain more atoms than the molecules of the white solid produced by this process. Some of the atoms that were originally part of the succinic acid molecules are given off as a gas. That leaves fewer atoms to be a part of the molecules of new white solid. Matter, Measurements, and Calculations 3 d. Succinic acid is heteroatomic. Some of the atoms were able to leave the succinic acid molecule in the form of a gas. Other atoms remained as part of a new molecule. If all of the atoms were the same type, they would have all turned into a gas or they would have all remained as a solid. 1.14 Carbon dioxide is heteroatomic. If oxygen and carbon atoms react to form one product, then carbon dioxide must contain these two types of atoms. 1.16 Water is heteroatomic. If breaking water apart into its components produces both hydrogen gas and oxygen gas, then water must contain two types of atoms. y
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CLASSIFYING MATTER (SECTION 1.4) ;1.18 a. Substance A is a compound because it is composed of molecules that contain more than one type of atom. b. Substance D is an element because it is composed of molecules that contain only one type of atom. c. Substance E is a compound because it is a pure substance that can break down into at least two different materials. Substances G and J cannot be classified because no tests were performed on them. 1.20 a. Substance R might appear to be an element based on the tests performed. It has not decomposed into any simpler substances based on these tests; however, this is not an exhaustive list of tests that could be performed on substance R. Substance R cannot be classified as an element or a compound based on the information given. b. Substance T is a compound. It is composed of at least two different elements because it produced two different substances on heating. c. The solid left in part b cannot be classified as an element or a compound. No tests have been performed on it. It is homogeneous because it has the same composition ;1.22 a. A gold chain throughout. It is homogeneous because it has the same composition b. Liquid eye drops throughout. It is heterogeneous because it does not have the same composition c. Chunky peanut butter throughout (peanut chunk vs. smooth regions). It is heterogeneous because it does not have the same composition d. A slice of watermelon throughout (rind, meat, and seeds). It is homogeneous because it has the same composition e. Cooking oil throughout. It is heterogeneous because it does not have the same composition f. Italian salad dressing throughout (oil, vinegar, and seasonings). It is homogeneous because it has the same composition g. Window glass throughout. If the chain is only made of gold, then it is a pure substance. Most ;1.24 a. A gold chain jewelry is made of gold alloyed with another metal for strength. If this chain is made of a gold alloy, then it is a solution. 4 Chapter 1 b. Liquid eye drops e. Cooking oil g. Window glass This is a solution because it contains saline (water and sodium chloride) as well as other substances. It can be a solution or a pure substance depending on the oil. Some oils only contain a single compound (pure substances); however, most oils are a mixture of several different compounds (solution). It is a solution because it contains a mixture of silicon dioxide and other components like iron. Fo
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MEASUREMENT UNITS (SECTION 1.5) 1.26 Modern society is complex and interdependent. Accomplishing projects like building a bridge, constructing a house, or machining an engine may require many different people to participate. Some people design the project, others supply the necessary materials, and yet another group does the construction. In order for the project to be successful, all of these people need a common language of measurement. Measurement is also important for giving directions, keeping track of the time people work, and keeping indoor environments at a comfortable temperature and pressure. ;1.28 The amount of weight that a horse could carry or drag might have been measured in stones. It could also be used to measure people or other items in the 50‐500 pound range. It is likely that a large stone was picked as the standard weight for the “stone” unit. Stones may have also been used as counterweights on an old‐fashioned set of balances. THE METRIC SYSTEM (SECTION 1.6) ;1.30 The metric units are (a) degrees Celsius, (b) liters, (d) milligrams, and (f) seconds. The English units are (c) feet and (e) quarts. 1.32 Meters are a metric unit that could replace the English unit feet in the measurement of the ceiling height. Liters are a metric unit that could replace the English unit quarts in the measurement of the volume of a cooking pot. ⎛ 1 μL ⎞
6
1.34 a. 1.00 L ⎜⎜ −6 ⎟⎟ = 1.00 × 10 μL 10
L
⎝
⎠
⎛ 1000 watts ⎞
4
b. 75 kilowatts ⎜⎜
⎟⎟ = 7.5 × 10 watts 1
kilowatt
⎝
⎠
6
⎛ 10 hertz ⎞
⎟ = 1.5 × 107 hertz c. 15 megahertz ⎜
⎜ 1 megahertz ⎟
⎝
⎠
⎛ 10 −12 meters ⎞
⎟ = 2.00 × 10 −10 meters 200 picometers ⎜
⎜ 1 picometer ⎟
⎝
⎠
d. 1.36 ⎛ 240 mL
1 cup ⎜
⎜ 1 cup
⎝
⎞⎛
1L
⎟ ⎜⎜
⎟ ⎝ 1000 mL
⎠
⎞
⎟⎟ = 0.240 L
⎠
⎛ 240 mL
or 1 cup ⎜
⎜ 1 cup
⎝
⎞ ⎛ 1cm 3
⎟ ⎜⎜
⎟ ⎝ 1 mL
⎠
⎞
3
⎟⎟ = 240 cm ⎠
Matter, Measurements, and Calculations ;1.40 ⎛ 2.20 lbs. ⎞
⎟ = 8.8 lbs. 4.0 kg ⎜
⎜ 1 kg ⎟
⎝
⎠
⎛ 3 ft ⎞ ⎛ 12 in ⎞
⎟
= 3.4 in a. 1m = 1.094 yd, so : 1m − 1yd = 1.094 yd − 1yd = 0.094 yd ⎜
⎜ 1 yd ⎟ ⎜⎜⎝ 1 ft ⎟⎟⎠
⎝
⎠
b. The size of 1°C is the same as 1 K; therefore, a change of 65°C is also a change of 65 K. ⎛ 1kg ⎞
c. 5 lbs. ⎜⎜
⎟⎟ = 2.27 kg = 2 kg with significant figures ⎝ 2.20 lbs. ⎠
y
1.38 ⎞
⎟ = 1.0 × 10 −3 kg or ⎟
⎠
⎛ 1 dm 3
1.0 cm 3 ⎜
⎜ 1000 cm 3
⎝
⎞ ⎛ 1 kg
⎟⎜
⎟ ⎜ 1.0 dm 3
⎠⎝
⎞ ⎛ 1000 g ⎞
⎟⎜
⎟ = 1.0 g ⎟ ⎜ 1 kg ⎟
⎠⎝
⎠
⎛ 1.057 qt
2.0 L ⎜
⎜ 1L
⎝
c. ⎛
1mg
5 grain ⎜
⎜ 0.015 grain
⎝
1.44 ⎞ ⎛ 32 fl oz ⎞
⎟ = 68 fl oz ⎟⎜
⎟ ⎜ 1 qt ⎟
⎠⎝
⎠
b. °F =
9
5
( °C ) + 32
O
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a. ⎞ ⎛ 1kg
⎟⎜
⎟ ⎜ 1.0 dm 3
⎠⎝
ew
1.42 ⎛ 1 dm 3
1.0 cm 3 ⎜
⎜ 1000 cm 3
⎝
⎞
⎟ = 333 mg = 3 × 10 2 mg with significant figures ⎟
⎠
°F =
9
5
( 36.1°C ) + 32 = 97.0°F
°F =
9
5
( 37.2°C ) + 32 = 99.0°F Fo
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LARGE AND SMALL NUMBERS (SECTION 1.7) 1.46 a. 02.7 x 10‐3 Improper form because no leading zero is necessary. (2.7 x 10‐3) 2
b. 4.1 x 10 Correct. c. 71.9 x 10‐6 Improper form because only one digit should be to the left of the decimal point. (7.19 x 10‐5) 3 d. 10
Improper form because a nonexponential term should be written before the exponential term. (1 x 103) ‐2
e. 0.0405 x 10 Improper form because one nonzero digit should be to the left of the decimal point. (4.05 x 10‐4) f. 0.119 Improper form because one nonzero digit should be to the left of the decimal point and an exponential term should be to the right of the nonexponential term. (1.19 x 10‐1) ;1.48 a. 14 thousand = 14,000 = 1.4 x 104 b. 365 3.65 x 102 c. 0.00204 2.04 x 10‐3 d. 461.8 4.618 x 102 e. 0.00100 1.00 x 10‐3 f. 9.11 hundred = 9.11 x 100 = 9.11 x 102 1.50 a. 186 thousand mi/s 186 x 1000 = 1.86 x 105 mi/s b. 1100 million km/h 1100 x 1,000,000 = 1.1 x 109 km/h 5 6 Chapter 1 1.52 0.000 000 000 000 000 000 000 105 g The decimal point has been moved 22 places to the left. This places 21 zeros to the right of the decimal point and before the numbers 105 g. 1.54 a. b. c. d. e. (8.2 x 10‐3)(1.1 x 10‐2) (2.7 x 102)(5.1 x 104) (3.3 x 10‐4)(2.3 x 102) (9.2 x 10‐4)(2.1 x 104) (4.3 x 106)(6.1 x 105) = 9.02 x 10‐5 = 1.377 x 107 = 7.59 x 10‐2 = 1.932 x 101 = 2.623 x 1012 1.56 a. b. c. d. (144)(0.0876) (751)(106) (0.0422)(0.00119) (128,000)(0.0000316) = (1.44 x 102)(8.76 x 10‐2) = (7.51 x 102)(1.06 x 102) = (4.22 x 10‐2)(1.19 x 10‐3) = (1.28 x 105)(3.16 x 10‐5) 1.58 a. b. c. d. e. 3.1 × 10 −3
1.2 × 10 2
7.9 × 10 4
3.6 × 10 2
4.7 × 10 −1
7.4 × 10 2
0.00229
3.16
119
3.8 × 10 3
= 3.131578947 x 10‐2 = 3.1 x 10‐2 with SF ;1.60 a. ( 5.3 )( 0.22 )
( 6.1)( 1.1)
= 1.7377 x 10‐1 = 1.7 x 10‐1 with SF b. = 1.025 x 10‐9 = 1.0 x 10‐9 with SF = 2.59459 x 106 = 2.6 x 106 with SF = 2.335279 x 100 = 2.3 x 100 with SF = 1.520989 x 107 = 1.5 x 107 with SF = 2.2 x 102 with SF = 6.35135 x 10‐4 = 6.4 x 10‐4 with SF = 7.2468354 x 10‐4 = 7.25 x 10‐4 with SF ew
6.3 × 10 3
4.8 × 106
c. ( 7.4 × 10 )( 2.5 × 10 )
d. 5.6
( 0.022 )( 109 )
Fo
= 2.194 x 102 −2
−4
3
= 2.6 x 10‐5 with SF re
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( 3.8 × 10 )(1.7 × 10 ) −4
= 2.583 x 10‐5 rP
e. = 1.26 x 101 with SF = 7.96 x 104 with SF = 5.02 x 10‐5 with SF = 4.04 x 100 with SF O
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= 1.26144 x 101 = 7.9606 x 104 = 5.0218 x 10‐5 = 4.0448 x 100 y
= 9.0 x 10‐5 with significant figures = 1.4 x 107 with significant figures = 7.6 x 10‐2 with significant figures = 1.9 x 101 with significant figures = 2.6 x 1012 with significant figures ( 4.6 × 10 )( 2.3 × 10 ) ( 7.4 × 10 )( 9.4 × 10 )
−3
2
−4
−5
SIGNIFICANT FIGURES (SECTION 1.8) 1.62 a. A ruler with smallest scale marking of 0.1 cm b. A measuring telescope with smallest scale marking of 0.1 mm c. A protractor with smallest scale marking of 1° d. A tire pressure with smallest scale marking of 1 lb/in2 ;1.64 a. Exactly 6 mL of water measured with a graduated cylinder 0.01 cm 0.01 mm 0.1° 0.1 lb/in2 6.00 mL Matter, Measurements, and Calculations that has smallest markings of 0.1 mL. b. A temperature that appears to be exactly 37 degrees using a thermometer with smallest markings of 1°C. c. A time of exactly nine seconds measured with a stopwatch that has smallest markings of 0.1 second. d. Fifteen and one‐half degrees measured with a protractor that has 1 degree scale markings. 37.0°C 9.00 seconds 15.5° b. Measured = percentages Exact = 5 players 1.68 a. 0.0400 b. 309 c. 4.006 1.70 ( 3.71)(1.4 ) b. ( 0.0851)( 1.2262 ) c. ( 0.1432 )( 2.81)
( 0.7762 )
5.06 lb.
lb.
lb.
= 0.31625
= 0.316
with SF 16 potatoes
potato
potato
71.2% + 66.9% + 74.1% + 80.9% + 63.6%
= 71.34% with SF 5 players
y
a. Measured = 5.06 lbs. Exact = 16 potatoes 5.194 a. 2 SF 4 SF 5 SF = 5.2 with significant figures 0.10434962 = 0.104 with significant figures 0.518412780211 = 0.518 with significant figures d. ( 3.3 × 10 )( 3.09 × 10 ) 101.97 = 1.0 x 102 with significant figures e. ( 760 )( 2.00 )
2.52491694352 x 10‐18 = 2.52 x 10‐18 with significant figures (assuming 0 in 760 is significant) re
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a. 0.208 + 4.9 + 1.11 b. 228 + 0.999 + 1.02 c. 8.543 − 7.954 d. ( 3.2 × 10 −2 ) + ( 5.5 × 10 −1 ) rP
1.72 −3
4
6.02 × 10
(Hint: Write in decimal form first, then add.) e. 336.86 − 309.11 f. 21.66 − 0.02387 1.74 a. ( 0.0267 + 0.00119 )( 4.626 )
Fo
d. 4.4 x 10‐3 e. 1.002 f. 255.02 ew
3 SF (0.0400) 3 SF 4 SF O
nl
;1.66 28.7794
b. 212.6 − 21.88
86.37
c. 27.99 − 18.07
4.63 − 0.88
d. 18.87 18.07
−
2.46
0.88
(HINT: Do divisions first, then subtract.) = 6.218 = 230.019 = 0.589 = 6.2 with significant figures = 2.30 x 102 with significant figures = 0.589 with significant figures = 0.582 = 0.58 with significant figures = 27.75 = 21.63613 = 27.75 with significant figures = 21.64 with significant figures = 0.004483037867 = 0.00448 with significant figures = 2.20817413454 = 2.208 with significant figures = 2.6453333 = 2.65 with significant figures = ‐12.8633592018 = ‐13 with significant figures 7 8 Chapter 1 e. ( 8.46 − 2.09 )( 0.51 + 0.22 )
( 3.74 + 0.07 )( 0.16 + 0.2 )
= 3.3902741324 = 3 with significant figures f. 12.06 − 11.84
0.271
= 0.811808118081 = 0.81 with significant figures 2
A = 3.18 cm × 2.55 cm = 8.109 cm = 8.11 cm 2
Green 2
A = 13.22 cm × 0.68 cm = 8.9896 cm 2 = 9.0 cm 2 Orange P = 2 ( 3.18 cm ) + 2 ( 2.55 cm ) = 11.46 cm P = 2 ( 13.22 cm ) + 2 ( 0.68 cm ) = 27.80 cm b. Length Black ⎛ 1m
12.00 cm ⎜⎜
⎝ 100 cm
Red ⎛ 1m
20.20 cm ⎜⎜
⎝ 100 cm
Width ⎞
⎟⎟ = 0.1200 m ⎠
⎞
⎟⎟ = 0.2020 m ⎠
⎛ 1m ⎞
3.18 cm ⎜⎜
⎟⎟ = 0.0318 m ⎝ 100 cm ⎠
Orange ⎛ 1m ⎞
13.22 cm ⎜⎜
⎟⎟ = 0.1322 m ⎝ 100 cm ⎠
Green ⎛ 1m ⎞
10.40 cm ⎜⎜
⎟⎟ = 0.1040 m ⎝ 100 cm ⎠
⎛ 1m ⎞
2.42 cm ⎜⎜
⎟⎟ = 0.0242 m ⎝ 100 cm ⎠
⎛ 1m ⎞
2.55 cm ⎜⎜
⎟⎟ = 0.0255 m ⎝ 100 cm ⎠
⎛ 1m ⎞
0.68 cm ⎜⎜
⎟⎟ = 0.0068 m ⎝ 100 cm ⎠
Perimeter ( P = 2 ( l ) + 2 ( w ) ) re
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Area ( A = l × w ) Black A = 0.1200 m × 0.1040 m = 0.01248 m Red A = 0.2020 m × 0.0242 m = 0.0048884 m
P = 2 ( 0.1200 m ) + 2 ( 0.1040 m ) = 0.4480 m 2
= 0.00489 m
rP
= 8.11 × 10
−4
2
m
P = 2 ( 0.2020 m ) + 2 ( 0.0242 m ) = 0.4524 m 2
A = 0.0318 m × 0.0255 m = 8.109 × 10 −4 m 2
Green 2
= 9.0 × 10
−4
m
2
P = 2 ( 0.0318 m ) + 2 ( 0.0255 m ) = 0.1146 m A = 0.1322 m × 0.0068 m = 8.9896 × 10 −4 m 2
Orange P = 2 ( 0.1322 m ) + 2 ( 0.0068 m ) = 0.2780 m c. No, the number of significant figures in the answers remains constant. The numbers of places past the decimal are different; however, that could be fixed by rewriting all of the answers in scientific notation. Fo
P = 2 ( 20.20 cm ) + 2 ( 2.42 cm ) = 45.24 cm y
A = 20.20 cm × 2.42 cm = 48.884 cm = 48.9 cm 2
Red P = 2 ( 12.00 cm ) + 2 ( 10.40 cm ) = 44.80 cm A = 12.00 cm × 10.40 cm = 124.8 cm 2 Black Perimeter ( P = 2 ( l ) + 2 ( w ) ) a. Area ( A = l × w ) O
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1.76 ew
USING UNITS IN CALCULATIONS (SECTION 1.9) 0.015 grains
1.78 a. 20 mg to grains 1 mg
b. 350 mL to fl oz 1.80 ⎛
1 km
26 miles ⎜⎜
⎝ 0.621 miles
0.0338 fl oz
1 mL
c. 4 qt to liters d. 5 yd to meters ⎞
⎟⎟ = 41.8679549114 km = 42 km ⎠
1L
1.057 qt
1m
1.094 yd
Matter, Measurements, and Calculations ;1.82 ⎛ 0.0338 fl oz ⎞ ⎛ 1 cup ⎞
⎟⎜
⎟ = 1.05625 cups = 1.1 cups 250 mL ⎜
⎜
⎟ ⎜ 8 fl oz ⎟
1 mL
⎝
⎠⎝
⎠
(Note: Cups are not measured in 0.1 increments.) 1.84 ⎛ 2.20 lbs. ⎞
⎟ = 39.6 lbs The bag is not overweight. 18.0 kg ⎜
⎜ 1 kg ⎟
⎝
⎠
mg ⎛
1g
⎜
dL ⎜⎝ 1000 mg
⎞ ⎛ 10 dL
⎟ ⎜⎜
⎟⎝ 1 L
⎠
⎞
g
⎟⎟ = 1.31 L
⎠
;1.92 Total = 987.1 mg + 213.3 mg + 99.7 mg + 14.4 mg + 0.1 mg = 1314.6 mg IgG =
987.1 mg
1314.6 mg
× 100 = 75.09%; IgA =
IgD =
14.4 mg
1314.6 mg
× 100 = 1.10%; IgE =
213.3 mg
1314.6 mg
0.1 mg
1314.6 mg
× 100 = 16.23%; IgM =
Volume = ( 3.98 cm ) = 63.0 cm 3
rP
1.96 3
Density =
⎛ 1 mL
280 g ⎜
⎜ 0.736 g
⎝
Fo
;1.98 718.3 g
mass
=
= 11.4
volume ( 3.98 cm )3
99.7 mg
1314.6 mg
× 100 = 7.58%;
× 100 = 0.008%
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DENSITY (SECTION 1.11) 39.6 g
1.94 a. g
= 0.792 mL 50.0 mL
243 g
b. g
= 1.03 mL 236 mL
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CALCULATING PERCENTAGES (SECTION 1.10) $32.75
1.88 × 100 = 7.000% $467.80
mg
1.90 1.0 day
× 100 = 71% mg
1.4 day
y
131
O
nl
1.86 g
cm 3
⎞
⎟ = 380 mL ⎟
⎠
ADDITIONAL EXERCISES ⎛ 1000 m ⎞ ⎛ 1000 mm ⎞
6
1.100 a. 4.5 km ⎜⎜
⎟⎟ ⎜⎜
⎟⎟ = 4.5 × 10 mm 1
km
1
m
⎝
⎠⎝
⎠
⎛ 1g ⎞
6
3
b. 6.0 × 10 mg ⎜⎜
⎟⎟ = 6.0 × 10 g 1000
mg
⎝
⎠
c. d. 39.54 g
g
= 1.98 L 20.0 L
222.5 g
g
= 8.90 cm 3 25.0 cm 3
9 10 Chapter 1 1.102 1.104 ⎛ 1 kg
175 lbs. ⎜
⎜ 2.2 lbs.
⎝
⎛
14 lbs. fat
170 lbs. body weight ⎜
⎜ 100 lbs. body weight
⎝
= 54 days with significant figures
⎞ ⎛ 4500 kcal
⎟⎜
⎟ ⎜ 1 lb. fat
⎠⎝
⎞ ⎛ 1 day
⎟⎜
⎟ ⎜⎝ 2000 kcal
⎠
⎞
⎟⎟ = 53.55 days
⎠
O
nl
y
⎛ 1 km ⎞
15
12
c. 9.86 × 10 m ⎜⎜
⎟⎟ = 9.86 × 10 km 1000
m
⎝
⎠
⎛ 1000 g ⎞ ⎛ 1000 mg ⎞
−4
2
d. 1.91 × 10 kg ⎜⎜
⎟⎟ ⎜⎜
⎟⎟ = 1.91 × 10 mg 1
kg
1
g
⎝
⎠⎝
⎠
⎛ 1 g ⎞ ⎛ 1000 mg ⎞
−6
5.0 ng ⎜⎜ 9
⎟⎟ ⎜⎜
⎟⎟ = 5.0 × 10 mg 10
ng
1
g
e. ⎝
⎠⎝
⎠
⎞ ⎛ 12 mg ⎞
⎟⎜
⎟ = 954.54 mg
⎟ ⎜ 1 kg ⎟
⎠⎝
⎠
2
= 9.5 × 10 mg with significant figures
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ALLIED HEALTH EXAM CONNECTION 1.106 The physical properties include (b) boiling point and (d) osmolarity because these properties can be observed without changing (or attempting to change) the chemical composition of the materials. re
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1.108 Only (b) rice and beans is a mixture. 1.110 °F =
9
5
( °C ) + 32
°F =
9
5
( 25°C ) + 32 = 77°F (d) 25°C is approximately 77°F. 1.112 1.114 A calorie is a form of (b) heat. ⎛ 10 −9 m ⎞ ⎛ 1 pm ⎞
5
4.50 × 10 2 nm ⎜⎜
⎟⎟ ⎜ −12
⎟ = 4.50 × 10 pm 1
nm
10
m
⎝
⎠
⎝
⎠
4.50 x 102 nm is (d) 4.50 x 105 pm. Fo
1.116 The freezing point of water is (d) 273 K. rP
1.118 ⎛ 10 −9 m ⎞
−7
4.50 × 10 2 nm ⎜⎜
⎟⎟ = 4.50 × 10 m 1
nm
⎝
⎠
2
4.50 x 10 nm is (c) 4.50 x 10‐7 m. 1.120 (b) 1,000,000 = 106 1.122 (d) 0.05620 = 5.620 × 10 −2 Matter, Measurements, and Calculations 11 1.124 ⎛ 1 g CHO ⎞
(c) 36 calories ⎜⎜
⎟⎟ = 9 g CHO ⎝ 4 calories ⎠
1.126 10% solution =
10 g glu cos e
× 100 100 mL solution
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A 10% glucose solution contains (d) 10 grams of glucose per 100 milliliters of solution. CHEMISTRY FOR THOUGHT 1.128 a. To separate wood sawdust and sand, I would add water. The sawdust will float, while the sand will sink. The top layer of water and sawdust can be poured off into a filter. The water will run through the filter leaving the sawdust on the filter. The sawdust can then be allowed to dry. The remainder of the water and sand can be poured off into a filter and the sand can be allowed to dry. b. To separate sugar and sand, I would add water to dissolve the sugar. I would then filter the mixture to isolate the sand. I would evaporate the water to isolate the sugar. c. To separate iron filings and sand, I would use a magnet. The iron filings will be attracted to the magnet, while the sand will not be attracted to the magnet. d. To separate sand soaked with oil, I would pour the mixture through a filter. The oil will go through the filter and leave the sand behind on the filter. 1.130 ⎛ 2.2 lbs. ⎞
⎟ = 97.9 lbs.
44.5 kg ⎜
⎜ 1 kg ⎟
⎝
⎠
44.5 ÷ 2.2 = 20.2
This student should have used the relationship 2.2 lbs. = 1 kg to multiply 44.5 kg by 2.2 lbs./kg to find a weight of 97.9 lbs. The mistake she made appears to be that she divided 44.5 kg by 2.2 rather than multiplying by it. Consequently, she found a weight of only 20.2 lbs. Since she knows 2.2 lbs. = 1 kg, she was expecting the pound value to be larger than the kilogram value and she determined she had made a calculation error. 1.132 Hang gliding confirms that air is an example of matter because air occupies space and has mass. If air did not occupy space or have mass, the hang glider would fall to the ground rather than gliding through the air. 240.8 g
240.8 g
1.134 g
=
= 8.76 mL density =
60.1 mL − 32.6 mL 27.5 mL
The density of the object is only 8.76 g/mL; therefore, it does not have the same density as silver and is not silver. 1.136 When two teaspoons of sugar are dissolved in a small glass of water, the volume of the resulting solution is not significantly larger than the original volume of the water because as they dissolve, the sugar molecules are separated from one another and surrounded by water molecules. The sugar molecules fit in between the water molecules and do not significantly increase the volume of the solution. 12 Chapter 1 ADDITIONAL ACTIVITIES Section 1.1 Review: Is air matter? Describe how you might test your answer. Which of these devices will give the same readings for an object at the top of a mountain, at the bottom of a valley, and on another planet? Explain. O
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Section 1.2 Review: Identify the following as chemical or physical AND as properties or changes: Chemical or Physical Property or Change a. the boiling point of water b. mixing baking soda and vinegar to produce carbon dioxide gas c. freezing vinegar d. milk reacts with lemon juice rP
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Section 1.3 Review: Prefix review. Scientists make use of many prefixes. Complete the table below. Prefix Meaning Meaning when combined with “–atomic molecule” mono‐ monoatomic molecule* di‐ diatomic molecule tri‐ triatomic molecule poly‐ polyatomic molecule homo‐ homoatomic molecule hetero‐ heteroatomic molecule * This term is not actually used. Can you think of a reason why that might be? Fo
Draw examples of a: a. heteroatomic diatomic molecule b. homoatomic triatomic molecule c. polyatomic molecule Matter, Measurements, and Calculations 13 Section 1.4 Review: Complete the following organizational chart to visualize the relationships between the definitions related to the classification of matter. Matter
_______________
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Mixture
Compound
____________________
(e x. oil a nd wa te r)
____________________
atoms
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____________________
(e x. sa lt wa te r)
________________
(e x. H2 )
(e x. C)
________________
(e x. H2 O)
formula units
(e x. Na Cl)
introduced in Section 4.5
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Section 1.5 Review: Identify the measurement units in the following paragraphs. Sally Smith drove 4.3 miles to the grocery store to pick up a few items. The trip was fairly uneventful, but she was stopped by 4 red lights, and while driving through a school zone, she was very careful not to exceed 25 mph. re
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Once at the store Ms. Smith picked up a dozen eggs, a ½ gallon of milk, a 2‐L bottle of sparkling water, a case of soda, a box of animal crackers, a pint of ice cream, 2 tomatoes, and a bunch of bananas. At the checkout, the grocery clerk placed the tomatoes and the bananas on the scale. They weighed 1.28 lbs. and 2.41 lbs., respectively. The tomatoes cost a $1.50/lb. and the bananas were on sale for 4 lbs./$1. Her total bill was $25.38. Fo
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On the way home, Sally stopped at the gas station because she had less than ¼ tank of gas. The mid‐
grade gas was $2.89/gallon and she filled her tank for $28.73. She reset her trip meter, which had read 218.4 miles before it returned to zero. Section 1.6 Review: Two nutrition labels are shown below. Answer the following questions based on those labels. Caffeine-Free Lemon-Lime Beverage
Fat Free Milk
Nutrition Facts
Nutrition Facts
(1) How many nutritional calories are in 355 mL of fat free milk? (2) How many kilocalories are in 355 mL of both beverages? Serving Size
1 can (355 mL)
Amount per serving
Serving Size
Calories 160
Calories 90
% Daily Value*
Total Fat 0 g
Saturated Fat 0 g
1 cup (240 mL)
Amount per serving
0%
0%
Polyunsaturated Fat 0 g
% Daily Value*
Total Fat 0 g
Saturated Fat 0 g
0%
0%
(3) How many Joules are in 355 mL of both beverages? Polyunsaturated Fat 0 g
Monounsaturated Fat 0 g
Monounsaturated Fat 0.1 g
Cholesterol 0 mg
0%
Cholesterol less than 5 mg
1%
Sodium 75 mg
3%
Sodium 130 mg
5%
Total Carbohydrates 13 g
4%
Total Carbohydrates 40 g
Dietary Fiber 0 g
Sugars 40 g
Protein 0 g
13%
0%
Dietary Fiber 0 g
Sugars 12 g
Protein 9 g
0%
(4) Which beverage provides more energy per 355 mL serving? 14 Chapter 1 (5) Is energy the only consideration when selecting foods? (Preview of Chapter 22) Section 1.7 Review: How many pennies are needed to have $58.39? Write this number in both expanded form and scientific notation. If you have 6.4 x 10‐2 pennies, is this more, less, or the same as having $58.39? O
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Section 1.8 Review: Record your measurements based on each of the following instruments. Pay careful attention to significant figures and units. When appropriate, read from the bottom of the meniscus. Underline the digit in your measurement that contains uncertainty. Close‐up: Close‐up: ew
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10‐mL graduated cylinder 50‐mL buret ‐10°C – 110°C thermometer Fo
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60‐minute stopwatch (when reset, both hands point to 60) top‐loading balance Matter, Measurements, and Calculations 15 80 60 100
40 75
50 20 25
125‐mL Erlenmeyer flask 5‐cm ruler* y
100‐mL beaker O
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*Many rulers have the centimeter scale marked with mm. On a ruler that has both inches and centimeters, you can double check whether the markings are millimeters or centimeters by remembering: 2.54 cm = 1 inch. Section 1.9 Review: Convert the units in the Section 1.8 review as shown below: • volumes to cubic centimeters and liters • temperature to degrees Fahrenheit and Kelvin • length measurement to inches, feet, millimeters, meters, and kilometers • mass to kilograms and milligrams • time to hours, minutes, and microseconds. ew
Report your answers in both expanded form and scientific notation. Section 1.10 Review: A solution of ammonia in water has a concentration of 27%(w/w). mass of ammonia
× 100 mass of solution
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%(w / w) =
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Section 1.11 Review: The density of 27%(w/w) ammonia in water is 0.90 g/mL. What is the mass in grams of 75 mL of this solution? What is the volume in milliliters of 8.4 g of this solution? Tying It All Together with a Laboratory Application: Fo
A chemistry student performs an experiment to separate a 2.503 g mixture of sand, salt, and calcium carbonate. The sample has the same consistency throughout; it is a (1) __________________ mixture. The student adds 20.8 mL of water to the mixture and stirs. The salt dissolves in the water, but the sand and calcium carbonate do not. Salt dissolving in water is a (2) _________________ process. The student allows the mixture to stand and the sand and calcium carbonate fall to the bottom of the container. This is a (3) ___________________ mixture. The student filters the mixture. The salt water goes through the filter paper, but the sand and calcium carbonate do not. The salt water is a (4) ___________________ mixture. The student pours the salt water into a beaker with a mass of 25.842 g, then places the beaker on a hotplate. The water boils and leaves salt crystals in the beaker. The student used a (5) _______________ change to isolate the salt from the water. The mass of the salt crystals and the beaker are 26.048 g. The 16 Chapter 1 ew
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mass of salt in the original mixture is (6) ___________________, which is (7) ____________ % of the original mixture. In order to isolate the sand from the sand and calcium carbonate, the student adds 20.0 mL of 3.00 M hydrochloric acid (HCl). The student notices that the addition of HCl produces a slight fizzing. Once the fizzing ends, the student observes that the white calcium carbonate has disappeared, but the sand remains in the container. The reaction of hydrochloric acid with calcium carbonate to produce carbon dioxide, water, and calcium chloride is a (8) ________________ property. The student decants (pours off) the liquid from the (9) __________________ mixture of sand, water, and calcium chloride. The student washes the sand with distilled water and decants it into the same container as the water and calcium chloride. The student dries the sand in a container on the hotplate. The mass of the sand is 1.942 x 10‐3 kg, which is (10) _____________ g, or (11) _____________% of the original mixture. The student adds 20.00 mL of 1.25 M potassium carbonate to the clear calcium chloride and water mixture. A white precipitate of calcium carbonate forms immediately. This was a (12) _______________________ change. The student filters the mixture and the calcium carbonate remains on the 425 mg piece of filter paper. After allowing the calcium carbonate and filter paper to dry, the student finds their combined mass is 765 mg. The mass of the calcium carbonate is (13) ______________ mg or (14) _______________ g, which is (15) ________________% of the original mixture. The total mass of the recovered substances, sand, salt, and calcium carbonate, is (16) __________________________ which is (17) _________________ (greater than, less than, or the same as) the mass of original mixture. Which of the following might have occurred (18)? _________________ A. All of the components were completely recovered. B. Some of the material was lost during the many transferring steps of this procedure. C. One of the components was not dried thoroughly. re
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SOLUTIONS FOR THE ADDITIONAL ACTIVITIES rP
Section 1.1 Review: Air is matter because it has mass and occupies space. The mass of air can be verified by finding the mass of an empty balloon, blowing up the balloon, and taking the mass of the filled balloon. The difference in the mass of the full balloon and the empty balloon is the mass of the air in the balloon. The volume of the air is visible because the balloon is filled. The lab balance will give the same readings for an object regardless of where the measurement is taken because it measures mass. The bathroom scale will not be consistent because it measures weight. Weight is the measurement of the gravitational force on an object, not the mass of the object. Fo
Section 1.2 Review: a. the boiling point of water b. mixing baking soda and vinegar to produce carbon dioxide gas c. freezing vinegar d. milk reacts with lemon juice Chemical or Physical physical chemical Property or Change property change physical chemical change property Section 1.3 Review: Prefix Meaning Meaning when combined with “–atomic molecule” mono‐ 1 monoatomic molecule* a molecule containing 1 atom di‐ 2 diatomic molecule a molecule containing 2 atoms tri‐ 3 triatomic molecule a molecule containing 3 atoms Matter, Measurements, and Calculations 17 poly‐ many polyatomic molecule a molecule containing many atoms homo‐ same homoatomic molecule a molecule containing only one type of atom hetero‐ different heteroatomic molecule a molecule containing at least 2 types of atoms * A molecule must contain at least two atoms, otherwise it is called an atom. Draw examples of a: a. heteroatomic diatomic molecule homoatomic triatomic molecule c. polyatomic molecule This could be any molecule containing 3 of the same type of atoms. y
b. This could be any molecule containing 2 different types of atoms. O
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This could be any molecule containing 4 or more atoms. Section 1.4 Review: ew
Matter
Homogeneous
Heterogeneous
Element
Compound
(e x. oil a nd wa te r)
atoms
molecules
molecules
(e x. C)
(e x. H2 )
(e x. H2 O)
formula units
(e x. Na Cl)
introduced in Section 4.5
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(e x. sa lt wa te r)
Pure Substance
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Mixture
Section 1.5 Review: The measurement units were miles (twice), mph, gallon, L, pint, and lbs. (twice). The non‐measurement units were red lights, dozen, case, box, tomatoes, bunch, $/lb., lbs./$, $ (twice), tank, and $/gallon. Fo
Section 1.6 Review: (1) 355 mL milk ×
90 Cal
= 133.125 Cal = 130 Calories 240 mL
355 mL soda ×
160 Calories 1 kilocalorie
= 160 kcal
×
355 mL
1 Calorie
(2)
90 Cal 1 kilocalorie
= 130 kcal
×
355 mL milk ×
240 mL
1 Calorie
18 Chapter 1 355 mL soda ×
(3)
160 Cal 1 kilocalorie 1000 cal 4.184 J
×
×
×
= 669440 J = 6.7 × 10 5 J
1 cal
355 mL
1 Calorie
1 kcal
90 Cal 1 kilocalorie 1000 cal 4.184 J
×
×
×
355 mL milk ×
= 556995 J = 5.6 × 10 5 J
1 cal
240 mL
1 Calorie
1 kcal
(4) The soda provides more energy per 355 mL serving. y
(5) Energy is not the only consideration when selecting food. Nutritional value is also important. The milk has more nutritional value than the soda. O
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Section 1.7 Review: 100 pennies
$58.39 ×
= 5839 pennies = 5.839 × 10 3 pennies $1.00
6.4 x 10‐2 = 0.064 pennies; This value is less than one penny and much less than $58.39. 29.2 mL thermometer 73.3°C ⎛ 1 cm 3 ⎞
3
1
3
29.2 mL ⎜⎜
⎟⎟ = 29.2 cm = 2.92 × 10 cm 1
mL
⎝
⎠
⎛
⎞
1L
−2
29.2 mL ⎜⎜
⎟⎟ = 0.0292 L = 2.92 × 10 L 1000
mL
⎝
⎠
°F = 95 ( 73.3°C ) + 32 = 164°F re
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buret Section 1.9 ⎛ 1 cm 3 ⎞
3
0
3
7.74 mL ⎜⎜
⎟⎟ = 7.74 cm = 7.74 × 10 cm 1
mL
⎝
⎠
⎛
⎞
1L
−3
7.74 mL ⎜⎜
⎟⎟ = 0.00774 L = 7.74 × 10 L 1000
mL
⎝
⎠
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Section 1.8 and 1.9 Review: Instrument Section 1.8 graduated 7.74 mL cylinder 73.3°C + 273 = 346.3 K 14.2 sec ⎛ 1 min . ⎞ ⎛ 1 hour ⎞
−3
14.2 sec . ⎜⎜
⎟⎟ ⎜⎜ 60 min . ⎟⎟ = 0.00394 hr. = 3.94 × 10 hr. 60
sec
.
⎠
⎝
⎠⎝
⎛ 1 min . ⎞
−1
14.2 sec . ⎜⎜
⎟⎟ = 0.237 min . = 2.37 × 10 min . 60
sec
.
⎝
⎠
6
⎛ 10 μ s ⎞
7
14.2 sec . ⎜⎜
⎟⎟ = 14200000 μ s = 1.42 × 10 μ s 1
sec
.
⎝
⎠
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stopwatch balance beaker 42.381 g 47 mL ⎛ 1 kg
42.381 g ⎜
⎜ 1000 g
⎝
⎞
⎟ = 0.042381 kg = 4.2381 × 10 −2 kg ⎟
⎠
⎛ 1000 mg ⎞
⎟ = 42381 mg = 4.2381 × 10 4 mg 42.381 g ⎜
⎜ 1g ⎟
⎝
⎠
⎛ 1 cm 3 ⎞
3
1
3
47 mL ⎜⎜
⎟⎟ = 47 cm = 4.7 × 10 cm 1
mL
⎝
⎠
⎛
⎞
1L
−2
47 mL ⎜⎜
⎟⎟ = 0.047 L = 4.7 × 10 L 1000
mL
⎝
⎠
Matter, Measurements, and Calculations 19 40 mL ruler 3.66 cm ⎛ 1 cm 3 ⎞
3
1
3
40 mL ⎜⎜
⎟⎟ = 40 cm = 4.0 × 10 cm 1
mL
⎝
⎠
⎛
⎞
1L
−2
40 mL ⎜⎜
⎟⎟ = 0.040 L = 4.0 × 10 L ⎝ 1000 mL ⎠
⎛ 0.394 in. ⎞
0
3.66 cm ⎜⎜
⎟⎟ = 1.44 in. = 1.44 × 10 in. 1
cm
⎝
⎠
⎛ 0.394 in. ⎞ ⎛ 1 ft. ⎞
−1
3.66 cm ⎜⎜
⎟⎟ ⎜⎜ 12 in. ⎟⎟ = 0.120 ft. = 1.20 × 10 ft. 1
cm
⎠
⎝
⎠⎝
⎛ 10 mm ⎞
1
3.66 cm ⎜⎜
⎟⎟ = 36.6 mm = 3.66 × 10 mm 1
cm
⎝
⎠
⎛ 1m ⎞
−2
3.66 cm ⎜⎜
⎟⎟ = 0.0366 m = 3.66 × 10 m 100
cm
⎝
⎠
⎛ 1 m ⎞ ⎛ 1 km ⎞
−5
3.66 cm ⎜⎜
⎟⎟ = 0.0000366 km = 3.66 × 10 km ⎟⎟ ⎜⎜
1000
m
100
cm
⎠
⎝
⎠⎝
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Erlenmeyer flask Section 1.11 Review: 0.90 g
= 67.5 g = 68 g
1 mL
8.4 g ×
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Section 1.10 Review: 575 g × 27% = 155.25 g = 160 g ammonia; 575 g − 155.25 g = 419.75 g = 420 g water
419.75 g
× 100 = 73%(w / w)
or
100% − 27% = 73%(w / w)
575 g
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Tying It All Together with a Laboratory Application: (1) homogeneous (6) 0.206 g (11)
(2) physical (7) 8.23% (12)
(3) heterogeneous (8) chemical (13)
(4) homogeneous (9) heterogeneous (14)
(5) physical (10) 1.942 g (15)
77.59% chemical 340 mg 0.340 g 13.6% 1 mL
= 9.3 mL = 9.3 mL 0.90 g
(16) 2.488 g (17) less than (18) B SELF‐TEST QUESTIONS Fo
Multiple Choice 1. Which of the following involves a chemical change? a. stretching a rubber band c. lighting a candle b. breaking a stick d. melting an ice cube 2. A solid substance is subjected to a number of test and observations. Which of the following would be classified as a chemical property of the substance? a. It is gray in color. c. It dissolves in acid and a gas is liberated. b. It has a density of 2.04 g/mL. d. It is not attracted to either pole of a magnet. 3. Which of the following terms could not be properly used in the description of a compound? a. solution b. polyatomic c. pure substance d. heteroatomic 20 Chapter 1 4. Which of the following is an example of heterogeneous matter? a. water containing sand c. a pure sample of iron b. a sample of salt water d. a sample of pure table salt 5. When a substance undergoes a physical change, which of the following is always true? a. It melts. c. Heat is given off. b. A new substance is produced. d. The molecular composition is unchanged. 6. Which of the following is not a chemical change? a. burning magnesium c. exploding nitroglycerine b. pulverizing sulfur d. rusting iron 7. Which of the following is the basic unit of length in the metric system? a. centimeter b. meter c. millimeter d. kilometer 8. Which of the following is a derived unit? a. calorie b. cubic decimeter d. kilogram O
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11. Twenty‐one (21) students in a class of 116 got a B grade on an exam. What percent of the students in the class got B’s? a. 21.0 b. 22.1 c. 15.3 d. 18.1 rP
12. What single factor derived from Table 1.3 would allow you to calculate the number of quarts in a 2.0 L bottle of soft drink? 1L
1.057 quarts
0.0338 fl oz
1 mL
1L
1 mL
d. 0.0338 fl oz
a. b. c. 1.057 quarts
Fo
13. On a hot day, a Fahrenheit thermometer reads 97.3°F. What would this reading be on a Celsius thermometer? a. 118°C b. 22.1°C c. 36.3°C d. 143°C 14. The density of a 1 mL sample of a patient’s blood is 1.08 g/mL. The density of a pint of blood taken at the same time from the same patient would be _________________ 1.08 g/mL. a. greater than c. equal to b. less than d. more than one possible answer 15. A 125 mL urine specimen weighs 136.0 g. The density of the specimen is: g
g
g
g
a. 1.09 mL b. 0.919 mL c. 261 mL d. 11.0 mL Matter, Measurements, and Calculations 21 Fo
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Matching Match the type of measurement on the right to the measurement units given on the left. 16. Kelvin a. mass 17. milliliter b. volume 18. gram c. length 19. centimeter d. temperature 20. cubic decimeter e. density 21. kilometer 22. pounds per cubic foot True‐False 23. The mass of an object is the same as its weight. 24. A physical property can be observed without attempting any composition changes. 25. The cooking of food involves chemical changes. 26. The smallest piece of water that has the properties of water is called an atom. 27. Carbon monoxide molecules are diatomic and heteroatomic. 28. The prefix milli‐ means one thousand times. 29. One meter is shorter than one yard. 30. A pure substance containing sulfur and oxygen atoms must be classified as a compound. 31. The calorie and Joule are both units of energy. 32. In scientific notation, the exponent on the 10 cannot be larger than 15. 33. The correctly rounded sum resulting from adding 13.0, 1.094, and 0.132 will contain five significant figures. 34. If an object floats in water, it must have a higher density than water. 35. Most gases are less dense than liquids. 22 Chapter 1 ANSWERS TO THE SELF‐TEST QUESTIONS 1. 2. 3. 4. 5. 6. 7. C C A A D B B 8. 9. 10. 11. 12. 13. 14. B B C D A C C 15. 16. 17. 18. 19. 20. 21. A D B A C B C 22. 23. 24. 25. 26. 27. 28. E F T T F T F 29. 30. 31. 32. 33. 34. 35. F T T F F F T Fo
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