Ch 1 HW

72. The density of pure silver is 10.5 g/cm3 at 20 deg. C. If 5.25 g of pure silver pellets is added to a graduated cylinder containing 11.2 mL of water, to what volume level will the water in the cylinder rise?

-Convert the mass into cm then mL 5.25 g x (1 cm^3/10.5 g)= 0.500 cm^3 = 0.500 mL The volume in the cylinder will rise to 11.7 mL (11.2 mL + 0.500 mL = 11.7 mL)

48* A children's pain relief elixir contains 80. mg acetaminophen per 0.50 teaspoon. The dosage recommended for a child who weighs between 24 and 35 lb is 1.5 teaspoons. What is the range of acetaminophen dosages, expressed in mg acetaminophen/kg body weight, for children who weigh between 24 and 35 lb?

1.5 teaspoons × (80. mg acet / 0.50 teaspoon ) = 240 mg acetaminophen (2SFs) They want it expressed in mg/kg, so how many mg in ONE kg (240 mg acet/ 24 lb) x (1 lb/ 0.454 kg) = 22 mg acetaminophen/kg (240 mg acet/ 35 lb) x (1 lb/ 0.454 kg) = 15 mg acetaminophen/kg The range is from 15 to 22 mg acetaminophen per kg of body weight

65. A material will float on the surface of a liquid if the material has a density less than that of the liquid. Given that the density of water is approximately 1.0 g/mL, will a block of material having a volume of 1.2 x 10^4 in3 and weighing 350 lb float or sink when placed in a reservoir of water?

Density < 1.0 g/cm^3 to float Volume: 1.2 x 10^4 in^3 x (2.54 cm/1 in) x (1 ml/1 cm^3) =196644.268 mL Mass: 350 lb x (453.59/1 lb) = 158756.56 g d = m/V ---> 158756.56 g /196644.268 mL = 0.807 (2SFs) =0.81 g/cm^3 ---> will float

55 * Mercury poisoning is a debilitating disease that is often fatal. In the human body, mercury reacts with essential enzymes leading to irreversible inactivity of these enzymes. If the amount of mercury in a polluted lake is 0.4 μg Hg/mL, what is the total mass in kilograms of mercury in the lake? (The lake has a surface area of 100 mi2 and an average depth of 20 ft.)

Lake volume: Convert to ft so we can convert to cm and then mL 100 mi^2 x 20ft x (5280 ft / 1 mi)^2 =5.5756 x 10^10 (1 SFs) = 6 x 10^10 ft^3 6 x 10^10 ft^3 x ( 12 in/ 1 ft x 2.54 cm / 1 in)^3 x (1 mL / 1 cm^3) x (0.4 μg/1 mL) =7 x 10^14 μg 7 x 10^14 μg x (1 g/10^6 μg) x (1 kg/10^3 g) = 7 x 10^5 kg

98. * a 25.00-g sample of a solid is placed in a graduated cylinder, and then the cylinder is filled to the 50.0-mL mark with benzene. The mass of benzene and solid together is 58.80 g. Assuming that the solid is insoluble in benzene and that the density of benzene is 0.880 g/cm3 , calculate the density of the solid

Mbenzene= 58.80 g - 25.00 g = 33.80 g Vbenzene= m/d = 33.80 g / (0.880 g / cm^3) = 38.4cm^3 Vsolid = 50.0 cm^3- 38.4 cm^3 = 11.6 cm^3 Dsolid = m/v = ( 25.00 g ) / ( 11.6 g/cm^3)= 2.16 g/cm^3

51 * Would a car traveling at a constant speed of 65 km/h violate a 40 mi/h speed limit?

No, since 65 km/ hr is barely 40 mi/hr: 65 km x (1 mile/1.609 km) =40.397 (2 SFs) -40.

56.* Carbon monoxide (CO) detectors sound an alarm when peak levels of carbon monoxide reach 100 parts per million (ppm). This level roughly corresponds to a composition of air that contains 400,000 μg carbon monoxide per cubic meter of air (400,000 μg/m3 ). Assuming the dimensions of a room are 18 ft x 12 ft x 8 ft, estimate the mass of carbon monoxide in the room that would register 100 ppm on a carbon monoxide detector

Vroom= 12 ft x 18 ft x 8 ft = 1728 ft^3 -Convert ft^3 to m^3 1728 ft^3 x (12 in / 1 ft)^3 x (2.54 cm/1 m)^3 =50 m^3 -m^3 to μg 50m^3 x (400,000 ug/1 m^3)= 2 x 10^7 μg -μg to g 2 x 10^7 μg x (1 g/10^6 μg)= 20 g CO

34. 3 beakers with different precisions hold a volume of water. The estimated volume of the first beaker is 32.7 mL, the estimated volume of the middle beaker is 33 mL, and the estimated volume in the last beaker is 32.73 mL. Is it possible for each of the three beakers to contain the exact same amount of water? If no, why not?

Yes, all volumes could be identical to each other because the more precise volume readings can be rounded to the other volume readings. But because the volumes are in three different measuring devices, each with its own unique uncertainty, we cannot say with certainty that all three beakers contain the same amount of water

41. a. Congratulations! You and your spouse are the proud parents of a new baby, born while you are studying in a country that uses the metric system. The nurse has informed you that the baby weighs 3.91 kg and measures 51.4 cm. Convert your baby's weight to pounds and ounces and her length to inches (rounded to the nearest quarter inch). b. The circumference of the earth is 25,000 mi at the equator. What is the circumference in kilometers? in meters? c. A rectangular solid measures 1.0 m by 5.6 cm by 2.1 dm. Express its volume in cubic meters, liters, cubic inches, and cubic feet.

a. -3.91 kg x (10^3 g / 1 kg) x (1 lb / 453.59 g)= 8.62 lb -0.62 lb x (16 oz/1 lb)= 9.9 oz - 51.4 cm x (1 in /2.54 cm) =20.2 in b. 25,000 mi x (1.609 km / 1 mile) = 40225 km = 4.0 x 10^4 km - 4.0 x 10^4 km x (10^3 m/1 km) = 4.0225 x 10^7 = 4.0 x 10^7 m c. *Cubic meters* Convert to m, multiply all 1.0 m x 0.056 m x 0.21 m = =0.012 m^3 *Liters* Convert to cm -0.21 m x (10^2 cm/ 1 m ) = 21 cm -1.0 m x (10^2 cm/1m)=1.0 x 10^2 cm Multiply all - 5.36 cm x 1.0 x 10^2 cm x 21 cm = 11760 cm^3= 1.2 x 10^4 (track 2 SFs) Convert from cm to mL to L 11760 cm^3 x (1 ml/1 cm^3) x (1 L/10^3 mL)= 12 L *inches* Convert from cm to in 11760 cm^3 x (1 in/2.54 cm)^3 = 720 in^3 *Feet* 720 in^3 x (1 ft/12 in)^3 = 0.42 ft^3

29. How many significant figures are there in each of the following values? a. 6.07 x 10^-15 b. 0.003840 c. 17.00 d. 8 x 10^8

a. 3 ; the zero between the 6 & 7 counts as significant b. 4 ; if there is a decimal in the number, the zero at the end of a number is significant. Leading zeros are not significant c. 4 ; if there is a decimal in the number, the zero at the end of a number is significant d. 1 ; only 8 is significant

31. Round off each of the following numbers to the indicated number of significant digits, and write the answer in standard scientific notation. a. 0.00034159 to three digits b. 103.351 x 10^2 to four digits c. 17.9915 to five digits d. 3.365 x 10^5 to three digits

a. 3.42 x 10^-4 b. 1.034 x 10^4 ; In order for this to be in scientific notation, t needed to be between 1 & 10. If we placed the decimal back, we need to increase the degree of 10 from 10^2 to 10^4 c. 1.7992 × 10^1 ; remember, values need to be between 1 & 9.99 d.3.37 x 10^5

39. Perform each of the following conversions. a. 8.43 cm to millimeters b. 2.41 x 10^2 cm to meters c. 294.5 nm to centimeters d. 1.445 x 104 m to kilometers e. 235.3 m to millimeters f. 903.3 nm to micrometers

a. 8.43 cm x (10^1 mm/ 1 cm) =84.3 mm b. 2.41 x 10^2 cm x (1 m / 10^2 cm) =2.41 m c. 294.5 nm x (1 cm /10^7 nm) =2.945 x 10^-5 d. 1.445 x 10^4 m x (1 km/10^3 m) =14.45 km e. 235.3m x (10^3 mm/1 m) =2.353 x 10^5 mm f. 903.3 nm x (1 μm /10^3 nm) = 0.9033 μm

37. Perform the following mathematical operations, and express the result to the correct number of significant figures a. (2.526/ 3.1 ) + (0.470/ 0.623) + (80.705/0.4326) b. (6.404 x 2.91) / (18.7 - 17.1) c. 6.071 x 10^-5 - 8.2 x 10^-6 - 0.521 x 10^-4 d. (3.8 x 10^-12 + 4.0 x 10^-13)/(4 x 10^12 + 6.3 x 10^13) e. (9.5 + 4.1 + 2.8 + 3.175) / 4 (Assume that this operation is taking the average of four numbers. Thus 4 in the denominator is exact.) f. (8.925 - 8.905 / 8.925 ) x 100 (This type of calculation is done many times in calculating a percentage error. Assume that this example is such a calculation; thus 100 can be considered to be an exact number.)

a. Here, apply the multiplication/division rule first; then apply the addition/subtraction rule to arrive at the one-decimal-place answer. We will generally round off at intermediate steps in order to show the correct number of significant figures. However, you should round off at the end of all the mathematical operations in order to avoid round-off error. The best way to do calculations is to keep track of the correct number of significant figures during intermediate steps, but round off at the end (2.526 / 3.1) + (0.470 / 0.623) + (80.705 / 0.4326) = 0.8148 + 0.7544 + 186.558 =188.1 b. Here, the mathematical operation requires that we apply the addition/subtraction rule first, then apply the multiplication/division rule =(6.404 x 2.91) / (18.7- 17.1) =(6.404 x 2.91) / 1.6 =12 c. You cant add values with different bases & exponents so we need to give them the same base before we add them. THEN, we can divide without worrying if they have the same base or not 6.071 × 10^−5 − 8.2 × 10^−6 − 0.521 × 10^−4 = 60.71 × 10^−6 − 8.2 × 10^−6 − 52.1 × 10^-6 = 0.41 × 10^−6 = 4 × 10^−7 d. see above (3.8 x 10^-12 + 4.0 x 10^-13)/(4 x 10^12 + 6.3 x 10^13) =(38 x 10^-13 + 4.0 x 10^-13)/(4 x 10^12 + 63 x 10^12) =42 x 10^-13 / 67 x 10^12 =6.3 x 10^-26 e. (9.5 + 4.1 + 2.8 + 3.175) / 4 =19.575 / 4 =4.89 =4.9 f. (8.925 - 8.905 / 8.925 ) x 100 =(0.020 / 8.925) x 100 =0.2241 =0.22

99. For each of the following, decide which block is more dense: the orange block, the blue block, or it cannot be determined. Explain your answers

a. Volume × density = mass; the orange block is more dense. Because mass (orange) > mass (blue) and because volume (orange) < volume (blue), the density of the orange block must be greater to account for the larger mass of the orange block. --SInce the orange block is smaller but heavier than the blue block, it is more dense b. Which block is more dense cannot be determined. Because mass (orange) > mass (blue) and because volume (orange) > volume (blue), the density of the orange block may or may not be larger than the blue block. If the blue block is more dense, its density cannot be so large that its mass is larger than the orange block's mass. --Just because the orange block is bigger and heavier than the blue block, it doesn't mean that the blue block is less dense c. The blue block is more dense. Because mass (blue) = mass (orange) and because volume (blue) < volume (orange), the density of the blue block must be larger in order to equate the masses. --The blue block is more dense since its smaller but is as heavy as the larger orange block d. The blue block is more dense. Because mass (blue) > mass (orange) and because the volumes are equal, the density of the blue block must be larger in order to give the blue block the larger mass --The blue block is more dense since both blocks are roughly the same size but the blue one is bigger

81. What is the difference between homogeneous and heterogeneous matter? Classify each of the following as homogeneous or heterogeneous.

a. a door b. the air you breathe c. a cup of coffee (black) d. the water you drink f. your lab partner e. salsa

69. Diamonds are measured in carats, and 1 carat = 0.200 g. The density of diamond is 3.51 g/cm3 a. What is the volume of a 5.0-carat diamond? b. What is the mass in carats of a diamond measuring 2.8 mL?

a. mass: 5.0 carat x 0.200 g = 1.0 g V= m/d = 1.0 g / (3.51 g/cm^3) = 0.28 cm^3 b. 2.8 mL x (1 cm^3/1 mL) x (3.51 g /1 cm^3) x (1 carat/0.200 g) = 49 carats -Use density as conversion factor -Read question carefully, they're asking for mass in carats

83. Classify each of the following as a mixture or a pure substance.

a. water b. blood c. the oceans d. iron i. table salt e. brass f. uranium g. wine h. leather i. table salt

77. *The density of osmium (the densest metal) is 22.57 g/cm3 . If a 1.00-kg rectangular block of osmium has two dimensions of 4.00 cm x 4.00 cm, calculate the third dimension of the block

d = 22.57 g / cm^ m= 1.00 kg ----> 1.00 kg x (10^3 g /1 kg)= 1.00 x 10^3 g V = l x w x h ----> V= 4.00 cm x 4.00 cm x h V= m/d = 1.00 x 10^3 / 22.57 = 44.30 (3 SFs) 44.30 = l x w x h h= 44.30 / 4.00 cm x 4.00 cm = 2.77 cm

78.* a copper wire (density = 8.96 g/cm^3) has a diameter of 0.25 mm. If a sample of this copper wire has a mass of 22g, how long is the wire?

d = 8.96 g/cm^3 m = 22 g V=? V=m/d = 22 g / (8.96 g /cm^3)= 2.45 cm^3 d = 0.25 mm --> 0.25 mm x (cm/10 mm)= 0.025 cm r= 0.025 cm / 2 = 0.0125 cm r^2= 0.0015625 cm^2 V= pi r^2 h h= V / pi r^2 = 2.45 cm^2 / pi 0.0015625 cm^2 = 4991.09 cm = 5.0 x 10^3

29. How many significant figures are there in each of the following values? e. 463.8052 f. 300 g. 301 h. 300.

e. 7; all numbers are significant including the trapped zero f. 1 ; since there is no decimal, the zeros are not significant g. 3 ; all numbers are significant including the trapped zero h. 3 ; since there is a decimal after the 300, specifically after two zeros, they are significant

115. * Sterling silver is a solid solution of silver and copper. If a piece of a sterling silver necklace has a mass of 105.0 g and a volume of 10.12 mL, calculate the mass percent of copper in the piece of necklace. Assume that the volume of silver present plus the volume of copper present equals the total volume. Refer to Table 1.5

let x = mass of copper, let y = mass of silver 105.0 g = x +y *together x & y supply the total mass of the necklace 10.12 mL = (x / 8.96 g / mL) + (y / 10.5 g /mL ) *The sum of their masses over their densities give the total volume of the necklace y = 10.50 - x *Y in terms of x 10.12 = (x /8.96) + (105.0 - x / 10.5) *Replace y (8.96 x 10.5) 10.12 = [(x / 8.96) + (105.0 - x / 10.5)] (8.96 x 10.5) *Multiply 8.96 & 10.5 to lose denominator 952.1 = 10.5x + 940.18 - 8.96x 11.3 = 1.54x x= 7.3 g ---> mass of copper Mass % = mass of copper / total mass = 7.3 / 10.5 x 100 = 70% copper

49 * Science fiction often uses nautical analogies to describe space travel. If the starship U.S.S. Enterprise is traveling at warp factor 1.71, what is its speed in knots and in miles per hour? (Warp 1.71 = 5.00 times the speed of light; speed of light = 3.00 x 10^8 m/s; 1 knot = 2030 yd/h.)

warp speed = 5c =5 (3.00 x 10^8) (1.5 x 10^9 m/s ) x (10^2 cm/ 1 m) x (1 in x 2.54 cm) x (1 ft / 12 in) x (1 yd / 3 ft) x (3600 s/ 1 hr) x (1 knot / 2030 yd/hr) = 2.91 x 10^9 knot (1.5 x 10^9 m/s ) x (10^2 cm/ 1 m) x (1 in x 2.54 cm) x (1 ft / 12 in) x (3600 s/ 1 hr) x (1 mile/5280 ft) =3.36 x 10^9 mile/hr