Chapter 2 - Solutions

Chapter 2 Scientific Notation 1. Write each of the following values in proper scientific notation. a) 0.2660 mg Answers: b) 19,370 m c) 0.00035900 g d) 87.420 s a) 2.660 x 10-1 mg b) 1.937 x 104 m c) 3.5900 x 10-4 g d) 8.7420 x 101 s Solutions First express the original quantity as a number between 1 and 10. Then multiply by the appropriate power of ten. This is determined by noting how many places, and in which direction, the decimal point was moved in the first step. For (a) and (c) the decimal point is moved to the right thereby making the number larger. This means that the power of ten to use is negative so that you are multiplying by a value smaller than 1. For (b) and (d) the decimal point is moved to the left thereby reducing the number. This means that the power of ten to use is positive so that you are multiplying by a value greater than 1. Note that the trailing zeros are significant for (a), (c) and (d) because of the presence of the decimal point 2. Write each of the following values in decimal form. a) 1.28 x 10-7 m Answers: b) 4.33 x 104 L a) 0.000000128 m b) 43300 L c) 5060 kg d) 0.00669 cm3 c) 0.0506 x 104 kg d) 669 x 10-5 cm3 Solutions a) Since the power of ten is a negative value, the number is less than one. Move the decimal point 7 places to the left using zeros to fill in six of the needed places. b) Since the power of ten is a positive value, the number is greater than one. Move the decimal point 4 places to the right. Because the number only has 3 significant figures do not indicate the decimal point. c) Since the power of ten is a positive value, move the decimal point 4 places to the right. Because the number only has 3 significant figures do not indicate the decimal point. d) Since the power of ten is a negative value, move the decimal point 5 places to the left using zeros to fill in two of the needed places. Significant Figures 1. How many significant figures are in each of the following measurements? a) 33.04 L Answers: b) 0.0099100 m c) 2500 s d) 6.1800 x 10-8 g a) 4 sig. fig. b) 5 sig. fig. c) 2 sig. fig. d) 5 sig. fig. Solutions Recall that all digits in a measurement are significant except trailing zeros in numbers without a decimal point and all leading zeros. a) All of the digits are significant. b) The leading zeros are not significant, but because of the decimal point the trailing zeros are significant, so there are 5 significant figures. c) Because there is no decimal point, the trailing zeros are not significant so there are just 2 significant figures. d) All of the digits shown in a quantity expressed in proper scientific notation are significant. 2. Round off each of the following to 2 significant figures. a) 2.65849 cm Answers: b) 472.8 s c) 0.028411 km d) 17069 ft a) 2.7 cm b) 470 s or 4.7 x 102 s c) 0.028 km d) 17000 ft or 1.7 x 104 ft Solutions a) The tenths place is rounded up because the next digits (5849) are not exactly 500000... b) The trailing zero is not significant because no decimal point is shown. c) The leading zeros are not significant and the thousandths place is left unchanged because the next digit (4) is less than 5. d) The trailing zeros are not significant because no decimal point is shown. 3. For each of the following, calculate the result and round it off to the proper number of significant figures. The units of these measurements have been omitted for clarity. Do not use a calculator. a) 6.127 x 105 + 4.53 x 102 = b) 9.45 x 10-2 - 2.66 x 10-3 = c) 6.018 x 10-3 ÷ 2.00 x 10-7 = Answers: a) 6.132 x 105 b) 9.18 x 10-2 c) 3.01 x 104 Solutions a) First covert 4.53 x 102 into a number having a power of ten exponent equal to 5. This requires increasing the exponent by three units (making a thousandfold increase in the power of ten) so the decimal number must be decreased a thousandfold. 4.53 x 102 ö 0.00453 x 105 Then add the decimal numbers and use the common power of ten. 6.127 x 105 5 + 0.00453 ________x 10 6.13153 x 105 ö 6.132 x 105 Note that the uncertain digit (in blue) in the first measurement is in the thousandths place while the uncertain digit in the second measurement is in the hundred thousandths place. Thus, the sum is rounded off to the (larger) thousandths place. Since the digits to be dropped (53) are greater than 50, the preceding digit is increased by 1. b) First covert 2.66 x 10-3 into a number having a power of ten exponent equal to -2. This requires increasing the exponent by one unit (making a tenfold increase in the power of ten) so the decimal number must be decreased tenfold. 2.66 x 10-3 ö 0.266 x 10-2 Then add the decimal numbers and use the common power of ten. 9.45 x 10-2 - 0.266 x 10-2 ________ 9.184 x 10-2 ö 9.18 x 10-2 Note that the uncertain digit (in blue) in the first measurement is in the hundredths place while the uncertain digit in the second measurement is in the thousandths place. Thus, the sum is rounded off to the (larger) hundredths place. Since the digit to be dropped (4) is less than 5, the preceding digit is unchanged. c) The calculated result is: 6.018 x10-3 = 3.009 x104 2.00 x10-7 Note the number of significant figures in each measurement in this calculation. 6.018 x 10-3 (4 sig. fig.) 2.00 x 10-7 (3 sig. fig.) Since 3 significant figures is the fewest number in any measurement, the result is rounded off to 3 significant figures. Since the first digit to be dropped (9) is greater than 5, the preceding digit is increased by 1. 3.009 x 104 ö 3.01 x 104 Unit Conversions 1. Write the solution map for the conversion of TL into nL. What is the conversion factor for each step? Answers: The solution map is TL ö L ö nL 1012 L The conversion factor for the first step is: . 1TL The conversion factor for the second step is: 1nL . 10-9 L Solutions Each step of the solution map requires a conversion factor. For TL ö L, we need a conversion factor with units of L into the numerator. Since 1 TL = 1012 L we use 1012 L . 1TL L which will cancel TL and introduce TL For L ö nL, we need a conversion factor with units of Since 1 nL = 10-9 L we use 2. nL to cancel L. L 1nL . 10-9 L How many meters does an automobile traveling 65 mph move in one second? Answer: 29 m Solution Using conversion factors that you readily know results in the following solution map. miles miles miles ft in cm m ö ö ö ö ö ö s s s s hr min s Putting all of the conversion factors together yields: 65 miles 1 hr 1 min 5280 ft 12 in 2.54 cm 10-2 m 29 m x x x x x x = 60s 1 cm s hr 60 min 1 mile 1 ft 1 in 3. Make each of the following conversion and express the answer in proper scientific notation. a) 411 nm to cm b) 1.19 g/mL to μg/kL Answers: a) 4.11 x 10-5 cm b) 1.19 x 1012 μg/kL Solutions a) The solution map for this calculation is: nm ö m ö cm 10-9 m The conversion factor needed in the first step is . 1nm The conversion factor needed in the second step is 1cm . 10-2 m Putting these together yields: 411 nm x b) 10-9 m 1cm x -2 = 4.11x10-5 cm 10 m 1 nm The solution map for this calculation is: g mL ö g L ö g kL ö μg kL The conversion factor needed in the first step is 1mL . 10-3 L The conversion factor needed in the second step is The conversion factor needed in the last step is 103 L . 1kL 1μg . 10-6 g Putting these together yields: 1.19 g 1 mL 103 L 1μg 1.19 x1012 μg x -3 x x -6 = 1kL 10 g kL mL 10 L 4. The price of gold recently reached $1780 per ounce. What is the value of 150 g of gold at this price? Note that gold is priced by the troy ounce (1 lb = 14.58 troy ounces). Answer: $8600 Solution What we know: g gold; $/troy oz gold; troy oz/lb; g/lb Desired answer: $ The solution map for this problem is: g gold ö lb gold ö troy oz gold ö $ The conversion factor needed in the first step is 1lb gold . 454 g gold The conversion factor needed in the second step is The conversion factor needed in the last step is 14.58 troy oz gold . 1lb gold $1780 . 1troy oz gold Putting these together yields: 150 g gold x 5. 1 lb gold 14.58 troy oz gold $1780 x x  $8600 1 troy oz gold 454 g gold 1 lb gold Express each of the following quantities as a value between 1 and 10 by converting them to the appropriate scaled SI unit. a) 0.0025 L Answers: b) 0.00000843 s c) 1600 m a) 2.5 mL b) 8.43 μs c) 1.6 km Solutions First write each value in scientific notation. Then replace the power of ten with the appropriate SI scaling prefix. a) 0.0025 L ö 2.5 x 10-3 L ö 2.5 mL b) 0.00000843 s ö 8.43 x 10-6 s ö 8.43 μs c) 1600 m ö 1.6 x 103 m ö 1.6 km 6. The average density of canola oil is 0.92 g/mL. What is the mass in grams of 1 gallon of canola oil. Answer: 3.5 x 103 g Solution What we know: gal canola oil; g canola oil/mL; L/gal; mL/L Desired answer: g canola oil The solution map for this problem is: gal canola oil ö L canola oil ö mL canola oil ö g canola oil The conversion factor needed in the first step is 3.78 L canola oil . 1gal canola oil The conversion factor needed in the second step is The conversion factor needed in the last step is 1mL canola oil . 10-3 L canola oil 0.92 g canola oil . 1mLcanola oil Putting these together yields: 1 gal canola oil x 7. 3.78 L canola oil 1 mL canola oil 0.92 g canola oil x -3 x = 3.5 x103 g canola oil 1 gal canola oil 10 L canola oil 1 mLcanola oil Suppose you have a 295-g sample of aluminum (density = 2.7 g/cm3) and a 438-g sample of iron (density = 9.0 g/cm3). Which sample will occupy the larger volume? Answer: 295 g aluminum Solution Use the density of each material to convert its mass into volume and compare the results. 1cm3 aluminum 295 g aluminum x = 110 cm3 aluminum 2.7 g aluminum 438 g iron x 1cm3 iron = 49 cm3 iron 9.0 g iron Thus the aluminum sample has the larger volume. 8. Pure silver has a density of 10.5 g/cm3 at room temperature. If a pure silver spoon is fully submersed in a graduated cylinder containing 26.50 mL of water and the water level rises to 28.11 mL, what is the mass of the spoon? Answer: 16.9 g Solution The volume of the spoon equals the difference between the final and initial water levels. Use the density of silver to convert the spoon volume into mass. Recall that 1 mL = 1 cm3. spoon volume = 28.11 mL - 26.50 mL = 1.61 mL = 1.61 cm3 1.61 cm3 silver x 9. 10.5g silver = 16.9 g silver 1 cm3 silver Imagine that you need special postage stamps that are sold in booklets at a cost of $43.20 per 5 booklets. If each booklet contains 3 sheets of 6 stamps each and a total of 210 stamps are needed, what will be the cost of the stamps? Answer: $100.80 Solution What we know: number of stamps; stamps/sheet; sheets/booklet; $/booklet Desired answer: $ The solution map for this problem is: numbers of stamps ö sheets of stamps ö booklets ö $ The conversion factor needed in the first step is 1sheet . 6stamps The conversion factor needed in the second step is The conversion factor needed in the last step is 1booklet . 3sheets $43.20 . 5 booklets Putting these together yields: 210 stamps x 1 sheet 1 booklet $43.20 x x = $100.80 5 booklets 6 stamps 3 sheets Problem-Solving Strategies 1. Suppose you wished to build a wooden box that will hold a ton of gravel. What additional information would you need in order to decide how much lumber to purchase for the construction of this box? Explain your reasoning. Solution The key to knowing how much lumber to purchase is an estimate of how big the box must be (the box volume). In order to make this estimate you will need an estimate of the volume of the ton of gravel. Since you know the sample mass and want the sample volume, the gravel density will serve as the necessary conversion factor. Since there are many different types of gravel, there is no single answer for the gravel density. A typical value is about 100 lb/ft3. This means a ton of gravel will occupy about 20 ft3 so a box 1 ft x 4 ft x 5 ft should work. This is probably smaller than you might have imagined. 3. The data in the table below show various Fahrenheit temperatures and their corresponding Kelvin temperatures. Use this information to determine the equivalent Fahrenheit temperature when the Kelvin temperature is zero. Assume that Fahrenheit temperature is the independent variable. EF K 212 373 68 293 32 273 -40 233 Answer: -459.4 EF Solution First, use software such as Excel to plot the Kelvin temperature versus the Fahrenheit temperature and fit the data with a straight line. The equation for this line allows you to solve for the Fahrenheit temperature (x) when the Kelvin temperature (y) is zero. x  y  255.22 K 0 K - 255.22 K    459.4 o F o 0.5556 K/ F 0.5556 K / o F