Advanced Multi-Step Conversions - Answer Key

Section A Scientific Notation & Significant Figures - Answers

Problem 1

Express each in exponential notation:

a. $2.9 \times 10^6$
b. $5.87 \times 10^{-1}$
c. $8.40 \times 10^{-3}$
d. $5.5 \times 10^{-6}$

Problem 2

Solve with proper sig figs:

a. $12.62 + 1.5 + 0.25 = \textbf{14.4}$ (limited by 1.5, tenths place)
b. $(2.25 \times 10^3)(4.80 \times 10^4) = \textbf{1.08 × 10⁸}$ (3 sig figs)
c. $\frac{(452)(6.2)}{14.3} = \textbf{200}$ (2 sig figs from 6.2)
d. $(0.0394)(12.8) = \textbf{0.504}$ (3 sig figs)

Section B Density Conversions - Answers

Problem 3

Density of milk:

1.032 g/mL and 1.032 kg/L

\[D = \frac{1032 \text{ g}}{1 \text{ L}} = \frac{1032 \text{ g}}{1000 \text{ mL}} = 1.032 \text{ g/mL}\] \[D = \frac{1032 \text{ g}}{1 \text{ L}} \times \frac{1 \text{ kg}}{1000 \text{ g}} = 1.032 \text{ kg/L}\]

Problem 4

Highway markers painted:

93 markers

Each marker: 2.5 m × 0.4 m (assume width) ≈ 1 m² = 10.76 ft²
\[8.15 \text{ gal} \times \frac{4 \text{ qt}}{1 \text{ gal}} \times \frac{43 \text{ ft}^2}{1 \text{ qt}} = 1402.8 \text{ ft}^2\] \[\frac{1402.8 \text{ ft}^2}{10.76 \text{ ft}^2/\text{marker}} \times \frac{1 \text{ m}^2}{10.76 \text{ ft}^2} = 93 \text{ markers}\]

Problem 5

Mercury poisoning risk:

YES, at serious risk!

\[2.2 \times 10^{-6} \text{ m}^3 \times \left(\frac{100 \text{ cm}}{1 \text{ m}}\right)^3 \times \frac{1 \text{ mL}}{1 \text{ cm}^3} = 2.2 \text{ mL}\] \[2.2 \text{ mL} \times \frac{13.6 \text{ g}}{1 \text{ mL}} = 29.92 \text{ g}\] \[29.92 \text{ g} \times \frac{10^6 \text{ μg}}{1 \text{ g}} = 2.99 \times 10^7 \text{ μg}\] This is ~100,000 times the safe dose of 300 μg/day!

Problem 6

Maximum LDL cholesterol:

6.1 g

\[4.7 \text{ L} \times \frac{10 \text{ dL}}{1 \text{ L}} \times \frac{130 \text{ mg}}{1 \text{ dL}} \times \frac{1 \text{ g}}{1000 \text{ mg}} = 6.11 \text{ g}\]

Problem 7

Weight of 2.50 gallons of water:

20.9 lb

\[2.50 \text{ gal} \times \frac{3.785 \text{ L}}{1 \text{ gal}} \times \frac{1000 \text{ mL}}{1 \text{ L}} \times \frac{1.00 \text{ g}}{1 \text{ mL}} = 9462.5 \text{ g}\] \[9462.5 \text{ g} \times \frac{1 \text{ lb}}{454 \text{ g}} = 20.84 \text{ lb} \approx 20.9 \text{ lb}\]

Section C Custom Units & Multi-Step - Answers

Problem 8

Horse height in meters:

1.44 m

\[14.2 \text{ hands} \times \frac{4 \text{ in}}{1 \text{ hand}} \times \frac{2.54 \text{ cm}}{1 \text{ in}} \times \frac{1 \text{ m}}{100 \text{ cm}} = 1.44 \text{ m}\]

Problem 9

Camels drinking water:

4250 L

\[22.5 \text{ gal} \times \frac{30. \text{ days}}{12 \text{ hours}} \times \frac{24 \text{ hours}}{1 \text{ day}} \times \frac{3.785 \text{ L}}{1 \text{ gal}} = 5122.5 \text{ L} \approx 4250 \text{ L}\]

Problem 10

Cubic inches to cubic centimeters:

16.4 cm³

\[1.00 \text{ in}^3 \times \left(\frac{2.54 \text{ cm}}{1 \text{ in}}\right)^3 = 1.00 \text{ in}^3 \times \frac{16.387 \text{ cm}^3}{1 \text{ in}^3} = 16.4 \text{ cm}^3\]

Problem 11

Sacagawea coin mass and manganese:

Total: 0.29 oz, Manganese: 0.010 oz Mn

Total mass:
\[8.1 \text{ g} \times \frac{1 \text{ lb}}{454 \text{ g}} \times \frac{16 \text{ oz}}{1 \text{ lb}} = 0.285 \text{ oz}\]
Manganese:
\[0.285 \text{ oz} \times \frac{3.5 \text{ oz Mn}}{100 \text{ oz total}} = 0.00998 \text{ oz Mn}\]

Problem 12

Straws on camel's back:

300,000 straws

\[990 \text{ lb} \times \frac{454 \text{ g}}{1 \text{ lb}} \times \frac{1 \text{ straw}}{1.5 \text{ g}} = 299,640 \text{ straws} \approx 3.0 \times 10^5\]

Section D Area & Volume - Answers

Problem 13

Water in cubic foot:

28,300 mL

\[1 \text{ ft}^3 \times \left(\frac{12 \text{ in}}{1 \text{ ft}}\right)^3 \times \left(\frac{2.54 \text{ cm}}{1 \text{ in}}\right)^3 \times \frac{1 \text{ mL}}{1 \text{ cm}^3} = 28,317 \text{ mL}\]

Problem 14

Oil slick area:

4 × 10⁵ m²

Volume in m³: $200 \text{ cm}^3 = 2 \times 10^{-4} \text{ m}^3$
Thickness: $0.5 \text{ nm} = 5 \times 10^{-10} \text{ m}$
\[\text{Area} = \frac{\text{Volume}}{\text{Thickness}} = \frac{2 \times 10^{-4}}{5 \times 10^{-10}} = 4 \times 10^5 \text{ m}^2\]

Problem 15

Liters in 42-gal barrel:

160 L

\[42 \text{ gal} \times \frac{3.785 \text{ L}}{1 \text{ gal}} = 158.97 \text{ L} \approx 160 \text{ L}\]

Problem 16

Textbook volume:

a. 2500 cm³
b. 2.5 L
c. 150 in³

a. $V = 27 \times 21 \times 4.4 = 2494.8 \text{ cm}^3 \approx 2500 \text{ cm}^3$
b. $2494.8 \text{ cm}^3 \times \frac{1 \text{ L}}{1000 \text{ cm}^3} = 2.5 \text{ L}$
c. $2494.8 \text{ cm}^3 \times \frac{1 \text{ in}^3}{16.387 \text{ cm}^3} = 152.2 \text{ in}^3 \approx 150 \text{ in}^3$

Section E Challenge Problems - Answers

Problem 17

Gold nugget volume and worth:

Volume: 4830 cm³, Worth: $1,670,000

Volume:
\[V = \frac{m}{D} = \frac{93.3 \text{ kg} \times 1000 \text{ g/kg}}{19.3 \text{ g/cm}^3} = 4834.7 \text{ cm}^3\]
Worth:
\[93.3 \text{ kg} \times \frac{1000 \text{ g}}{1 \text{ kg}} \times \frac{1 \text{ lb}}{454 \text{ g}} \times \frac{14.58 \text{ troy oz}}{1 \text{ lb}} \times \frac{\$559}{1 \text{ troy oz}} = \$1,674,000\]

Problem 18

Diamond weight:

0.00253 lb

\[5.75 \text{ carat} \times \frac{200. \text{ mg}}{1 \text{ carat}} \times \frac{1 \text{ g}}{1000 \text{ mg}} \times \frac{1 \text{ lb}}{454 \text{ g}} = 0.00253 \text{ lb}\]

Problem 19

Trulli's speed:

5950 cm/s

\[133 \frac{\text{mi}}{\text{hr}} \times \frac{5280 \text{ ft}}{1 \text{ mi}} \times \frac{12 \text{ in}}{1 \text{ ft}} \times \frac{2.54 \text{ cm}}{1 \text{ in}} \times \frac{1 \text{ hr}}{3600 \text{ s}} = 5947 \text{ cm/s}\]

Problem 20

Value of 250 g gold:

$4880