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CHAPTER 1

CHEMISTRY: THE STUDY OF CHANGE

Problem Categories

Biological : 1.24, 1.48, 1.69, 1.70, 1.78, 1.84, 1.93, 1.95, 1.96, 1.97, 1.105.

Conceptual : 1.3, 1.4, 1.11, 1.12, 1.15, 1.16, 1.54, 1.62, 1.89, 1.101, 1.103.

Environmental : 1.70, 1.87, 1.89, 1.92, 1.98.

Industrial : 1.51, 1.55, 1.72, 1.81, 1.91.

Difficulty Level

Easy : 1.3, 1.11, 1.13, 1.14, 1.15, 1.21, 1.22, 1.23, 1.24, 1.25, 1.26, 1.29, 1.30, 1.31, 1.32, 1.33, 1.34, 1.54, 1.55, 1.63,

1.64, 1.77, 1.80, 1.84, 1.89, 1.91.

Medium : 1.4, 1.12, 1.16, 1.35, 1.36, 1.37, 1.38, 1.39, 1.40, 1.41, 1.42, 1.43, 1.44, 1.45, 1.46, 1.47, 1.48, 1.49, 1.50,

1.51, 1.52, 1.53, 1.56, 1.57, 1.59, 1.60, 1.61, 1.62, 1.70, 1.71, 1.72, 1.73, 1.74, 1.75, 1.76, 1.78, 1.79, 1.81, 1.82, 1.83,

1.85, 1.94, 1.95, 1.96, 1.97, 1.98.

Difficult : 1.58, 1.65, 1.66, 1.67, 1.68, 1.69, 1.86, 1.87, 1.88, 1.90, 1.92, 1.93, 1.99, 1.100, 1.101, 1.102, 1.103, 1.104,

1.105, 1.106.

1.3 (a) Quantitative. This statement clearly involves a measurable distance.

(b) Qualitative. This is a value judgment. There is no numerical scale of measurement for artistic

excellence.

(c) Qualitative. If the numerical values for the densities of ice and water were given, it would be a

quantitative statement.

(d) Qualitative. Another value judgment.

(e) Qualitative. Even though numbers are involved, they are not the result of measurement.

1.4 (a) hypothesis (b) law (c) theory

1.11 (a) Chemical property. Oxygen gas is consumed in a combustion reaction; its composition and identity are

changed.

(b) Chemical property. The fertilizer is consumed by the growing plants; it is turned into vegetable matter

(different composition).

(c) Physical property. The measurement of the boiling point of water does not change its identity or

composition.

(d) Physical property. The measurement of the densities of lead and aluminum does not change their

composition.

(e) Chemical property. When uranium undergoes nuclear decay, the products are chemically different

substances.

1.12 (a) Physical change. The helium isn't changed in any way by leaking out of the balloon.

(b) Chemical change in the battery.

(c) Physical change. The orange juice concentrate can be regenerated by evaporation of the water.

(d) Chemical change. Photosynthesis changes water, carbon dioxide, etc., into complex organic matter.

(e) Physical change. The salt can be recovered unchanged by evaporation.

5 C

= (105 32) F

9 F

− ° × =

? ° C 41 C °

(b) Conversion from Celsius to Fahrenheit.

9 F

? F C 32 F

5 C

° = ° × + °

9 F

11.5 C 32 F

5 C

= − ° × + ° =

? ° F 11.3 F °

(c) Conversion from Celsius to Fahrenheit.

9 F

? F C 32 F

5 C

° = ° × + °

3 9 F

6.3 10 C 32 F

5 C

= × ° × + ° =

4 ? ° F 1.1 × 10 ° F

(d) Conversion from Fahrenheit to Celsius.

5 C

? C = ( F 32 F)

9 F

° ° − ° ×

5 C

= (451 32) F

9 F

− ° × =

? ° C 233 C °

1 K

K ( C 273 C)

1 C

(a) K = 113 °C + 273 °C = 386 K

(b) K = 37 °C + 273 °C = 3.10 × 10

2 K

(c) K = 357 °C + 273 °C = 6.30 × 10

2 K

1.26 (a)

1 K

K ( C 273 C)

1 C

° C = K − 273 = 77 K − 273 = − 196 ° C

(b) ° C = 4.2 K − 273 = − 269 ° C

(c) ° C = 601 K − 273 = 328 ° C

1.29 (a) 2.7 × 10

− 8 (b) 3.56 × 10

2 (c) 4.7764 × 10

4 (d) 9.6 × 10

− 2

1.30 (a) 10

− 2 indicates that the decimal point must be moved two places to the left.

1.52 × 10

− 2 = 0.

(b) 10

− 8 indicates that the decimal point must be moved 8 places to the left.

7.78 × 10

− 8 = 0.

1.31 (a) 145.75 + (2.3 × 10

− 1 ) = 145.75 + 0.23 = 1.4598 × 10

2

(b)

4

2 2

×

× ×

2 3.2 × 10

(c) (7.0 × 10

− 3 ) − (8.0 × 10

− 4 ) = (7.0 × 10

− 3 ) − (0.80 × 10

− 3 ) = 6.2 × 10

− 3

(d) (1.0 × 10

4 ) × (9.9 × 10

6 ) = 9.9 × 10

10

1.32 (a) Addition using scientific notation.

Strategy: Let's express scientific notation as N × 10

n

. When adding numbers using scientific notation, we

must write each quantity with the same exponent, n. We can then add the N parts of the numbers, keeping the

exponent, n , the same.

Solution: Write each quantity with the same exponent, n.

Let’s write 0.0095 in such a way that n = −3. We have decreased 10

n by 10

3 , so we must increase N by 10

3 .

Move the decimal point 3 places to the right.

0.0095 = 9.5 × 10

− 3

Add the N parts of the numbers, keeping the exponent, n , the same.

9.5 × 10

− 3

+ 8.5 × 10

− 3

18.0 × 10

− 3

The usual practice is to express N as a number between 1 and 10. Since we must decrease N by a factor of 10

to express N between 1 and 10 (1.8), we must increase 10

n by a factor of 10. The exponent, n , is increased by

1 from −3 to −2.

18.0 × 10

− 3 = 1.8 × 10

− 2

(b) Division using scientific notation.

Strategy: Let's express scientific notation as N × 10

n

. When dividing numbers using scientific notation,

divide the N parts of the numbers in the usual way. To come up with the correct exponent, n , we subtract the

exponents.

Solution: Make sure that all numbers are expressed in scientific notation.

653 = 6.53 × 10

2

Divide the N parts of the numbers in the usual way.

6.53 ÷ 5.75 = 1.

Subtract the exponents, n.

1.14 × 10

• 2 − (−8) = 1.14 × 10

• 2 + 8 = 1.14 × 10

10

(c) Subtraction using scientific notation.

Strategy: Let's express scientific notation as N × 10

n

. When subtracting numbers using scientific notation,

we must write each quantity with the same exponent, n. We can then subtract the N parts of the numbers,

keeping the exponent, n , the same.

(b) Subtraction

Strategy: The number of significant figures to the right of the decimal point in the answer is determined by

the lowest number of digits to the right of the decimal point in any of the original numbers.

Solution: Writing both numbers in decimal notation, we have

0.00326 mg

− 0.0000788 mg

0.00318 12 mg

The bolded numbers are nonsignificant digits because the number 0.00326 has five digits to the right of the

decimal point. Therefore, we carry five digits to the right of the decimal point in our answer.

The correct answer rounded off to the correct number of significant figures is:

0.00318 mg = 3.18 × 10

− 3 mg

(c) Addition

Strategy: The number of significant figures to the right of the decimal point in the answer is determined by

the lowest number of digits to the right of the decimal point in any of the original numbers.

Solution: Writing both numbers with exponents = +7, we have

(0.402 × 10

7 dm) + (7.74 × 10

7 dm) = 8.14 × 10

7 dm

Since 7.74 × 10

7 has only two digits to the right of the decimal point, two digits are carried to the right of the

decimal point in the final answer.

(d) Subtraction, addition, and division

Strategy: For subtraction and addition, the number of significant figures to the right of the decimal point in

that part of the calculation is determined by the lowest number of digits to the right of the decimal point in

any of the original numbers. For the division part of the calculation, the number of significant figures in the

answer is determined by the number having the smallest number of significant figures. First, perform the

subtraction and addition parts to the correct number of significant figures, and then perform the division.

Solution:

(7.8 m 0.34 m) 7.5 m / (1.15 s 0.82 s) 1.97 s

3.8 m s

1.37 Calculating the mean for each set of date, we find:

Student A: 87.6 mL

Student B: 87.1 mL

Student C: 87.8 mL

From these calculations, we can conclude that the volume measurements made by Student B were the most

accurate of the three students. The precision in the measurements made by both students B and C are fairly

high, while the measurements made by student A are less precise. In summary:

Student A: neither accurate nor precise

Student B: both accurate and precise

Student C: precise, but not accurate

1.38 Calculating the mean for each set of date, we find:

Tailor X: 31.5 in

Tailor Y: 32.6 in

Tailor Z: 32.1 in

From these calculations, we can conclude that the seam measurements made by Tailor Z were the most

accurate of the three tailors. The precision in the measurements made by both tailors X and Z are fairly high,

while the measurements made by tailor Y are less precise. In summary:

Tailor X: most precise

Tailor Y: least accurate and least precise

Tailor Z: most accurate

1.39 (a)

1 dm 22.6 m 0.1 m

? dm = × = 226 dm

(b)

0.001 g 1 kg 25.4 mg 1 mg 1000 g

= × × =

5 ? kg 2.54 10 kg

− ×

(c)

3 1 10 L 556 mL 1 mL

− × ? L = × = 0.556 L

(d)

3 2

3

10.6 kg 1000 g 1 10 m

1 m 1 kg^ 1 cm

×

= × × ⎜ ⎟ =

3 3

g ? 0.0106 g/cm

cm

1.40 (a)

Strategy: The problem may be stated as

? mg = 242 lb

A relationship between pounds and grams is given on the end sheet of your text (1 lb = 453.6 g). This

relationship will allow conversion from pounds to grams. A metric conversion is then needed to convert

grams to milligrams (1 mg = 1 × 10

− 3 g). Arrange the appropriate conversion factors so that pounds and

grams cancel, and the unit milligrams is obtained in your answer.

Solution: The sequence of conversions is

lb → grams → mg

Using the following conversion factors,

453.6 g

1 lb 3

1 mg

1 10 g

− ×

we obtain the answer in one step:

3

453.6 g 1 mg 242 lb 1 lb (^1 10) g −

= × × =

×

8 ? mg 1.10 × 10 mg

Check: Does your answer seem reasonable? Should 242 lb be equivalent to 110 million mg? How many

mg are in 1 lb? There are 453,600 mg in 1 lb.

(d)

Strategy: The problem may be stated as

? lb = 28.3 μg

A relationship between pounds and grams is given on the end sheet of your text (1 lb = 453.6 g). This

relationship will allow conversion from grams to pounds. If we can convert from μg to grams, we can then

convert from grams to pounds. Recall that 1 μg = 1 × 10

− 6 g. Arrange the appropriate conversion factors so

that μg and grams cancel, and the unit pounds is obtained in your answer.

Solution: The sequence of conversions is

μg → g → lb

Using the following conversion factors,

6 1 10 g

1 g

− ×

μ

1 lb

453.6 g

we can write

6 1 10 g 1 lb 28.3 g 1 g 453.6 g

− × = μ × × = μ

8 ? lb 6.24 10 lb

− ×

Check: Does the answer seem reasonable? What number does the prefix μ represent? Should 28.3 μg be a

very small mass?

1255 m 1 mi 3600 s

1 s 1609 m 1 h

× × = 2808 mi/h

1.42 Strategy: The problem may be stated as

? s = 365.24 days

You should know conversion factors that will allow you to convert between days and hours, between hours

and minutes, and between minutes and seconds. Make sure to arrange the conversion factors so that days,

hours, and minutes cancel, leaving units of seconds for the answer.

Solution: The sequence of conversions is

days → hours → minutes → seconds

Using the following conversion factors,

24 h

1 day

60 min

1 h

60 s

1 min

we can write

24 h 60 min 60 s = 365.24 day 1 day 1 h 1 min

× × × =

7 ? s 3.1557 × 10 s

Check: Does your answer seem reasonable? Should there be a very large number of seconds in 1 year?

6 8

1.609 km 1000 m 1 s 1 min (93 10 mi) 1 mi 1 km (^) 3.00 10 m 60 s

× × × × × =

×

8.3 min

1.44 (a)

1 mi 5280 ft 12 in 1 min

13 min 1 mi 1 ft 60 s

? in/s = × × × = 81 in/s

(b)

1 mi 1609 m

13 min 1 mi

= × =

2 ? m/min 1.2 × 10 m/min

(c)

1 mi 1609 m 1 km 60 min

13 min 1 mi 1000 m 1 h

? km/h = × × × = 7.4 km/h

1 m 6.0 ft 3.28 ft

× = 1.8 m

453.6 g 1 kg 168 lb 1 lb 1000 g

× × = 76.2 kg

55 mi 1.609 km

1 h 1 mi

? km/h = × = 88 km/h

62 m 1 mi 3600 s

1 s 1609 m 1 h

× × =

2 1.4 × 10 mph

6

0.62 g Pb 0.62 ppm Pb

1 10 g blood

×

3 6

0.62 g Pb 6.0 10 g of blood

1 10 g blood

× × =

×

3 3.7 10 g Pb

− ×

1.49 (a)

8 365 day 24 h 3600 s 3.00 10 m 1 mi 1.42 yr 1 yr 1 day 1 h 1 s 1609 m

×

× × × × × =

12 8.35 × 10 mi

(b)

36 in 2.54 cm 32.4 yd 1 yd 1 in

× × =

3 2.96 × 10 cm

(c)

10 3.0 10 cm 1 in 1 ft

1 s 2.54 cm 12 in

×

× × =

8 9.8 × 10 ft/s

1.50 (a)

9 1 10 m 185 nm 1 nm

− × = × =

7 ? m 1.85 10 m

− ×

(b)

9 365 day^ 24 h^ 3600 s (4.5 10 yr) 1 yr 1 day 1 h

= × × × × =

17 ? s 1.4 × 10 s

(c)

3 3 0.01 m 71.2 cm 1 cm

= × ⎜ ⎟ =

3 5 3 ? m 7.12 10 m

− ×

(d)

3

3

2 3

1 cm 1 L 88.6 m 1 10 m 1000 cm

−

= × ⎜ ⎟× =

×

4 ? L 8.86 × 10 L

1.58 You are asked to solve for the inner diameter of the tube. If you can calculate the volume that the mercury

occupies, you can calculate the radius of the cylinder, V cylinder = π r

2 h ( r is the inner radius of the cylinder,

and h^ is the height of the cylinder). The cylinder diameter is 2^ r.

mass of Hg volume of Hg filling cylinder density of Hg

3 3

105.5 g volume of Hg filling cylinder 7.757 cm

13.6 g/cm

Next, solve for the radius of the cylinder.

Volume of cylinder = π r

2 h

volume

π ×

r h

3 7.757 cm 0.4409 cm 12.7 cm

π ×

r

The cylinder diameter equals 2 r.

Cylinder diameter = 2 r = 2(0.4409 cm) = 0.882 cm

1.59 From the mass of the water and its density, we can calculate the volume that the water occupies. The volume

that the water occupies is equal to the volume of the flask.

mass volume density

Mass of water = 87.39 g − 56.12 g = 31.27 g

3

mass 31.27 g

density (^) 0.9976 g/cm

3 Volume of the flask 31.35 cm

343 m 1 mi 3600 s

1 s 1609 m 1 h

× × = 767 mph

1.61 The volume of silver is equal to the volume of water it displaces.

Volume of silver = 260.5 mL − 242.0 mL = 18.5 mL = 18.5 cm

3

3

194.3 g

18.5 cm

3 density 10.5 g/cm

1.62 In order to work this problem, you need to understand the physical principles involved in the experiment in

Problem 1.61. The volume of the water displaced must equal the volume of the piece of silver. If the silver

did not sink, would you have been able to determine the volume of the piece of silver?

The liquid must be less dense than the ice in order for the ice to sink. The temperature of the experiment must

be maintained at or below 0°C to prevent the ice from melting.

4

3 3

mass 1.20 10 g

volume (^) 1.05 10 cm

×

×

3 density 11.4 g/cm

mass Volume density

3

3

1.20 10 g

0.53 g / cm

×

3 3 Volume occupied by Li 2.3 × 10 cm

1.65 For the Fahrenheit thermometer, we must convert the possible error of 0.1°F to °C.

5 C

? C = 0.1 F 0.056 C

9 F

° ° × = °

The percent error is the amount of uncertainty in a measurement divided by the value of the measurement,

converted to percent by multiplication by 100.

known error in a measurement Percent error 100% value of the measurement

= ×

For the Fahrenheit thermometer,

0.056 C

38.9 C

= × =

percent error 0.1%

For the Celsius thermometer,

0.1 C

38.9 C

= × =

percent error 0.3%

Which thermometer is more accurate?

1.66 To work this problem, we need to convert from cubic feet to L. Some tables will have a conversion factor of

28.3 L = 1 ft

3 , but we can also calculate it using the dimensional analysis method described in Section 1.9 of

the text.

First, converting from cubic feet to liters:

(^3 3 ) 7 3 9 3

12 in 2.54 cm 1 mL 1 10 L (5.0 10 ft ) 1.42 10 L 1 ft 1 in (^) 1 cm 1 mL

− ⎛ ⎞ ⎛ ⎞ (^) × × × (^) ⎜ ⎟ × (^) ⎜ ⎟ × × = ×

⎝ ⎠ ⎝ ⎠

The mass of vanillin (in g) is:

11 2.0 10 g vanillin (^9 ) (1.42 10 L) 2.84 10 g vanillin 1 L

− × (^) − × × = ×

The cost is:

(2.84 10 g vanillin) 50 g vanillin

− × × = $0.064 = 6.4¢

9 F

? F = C + 32 F

5 C

° ° × °

Let temperature = t

= + 32 F

t t °

19 2.205 lb^ 1 ton (4.8 10 kg) 1 kg 2000 lb

= × × × =

16 mass NaCl (tons) 5.3 × 10 tons NaCl

1.72 First, calculate the volume of 1 kg of seawater from the density and the mass. We chose 1 kg of seawater,

because the problem gives the amount of Mg in every kg of seawater. The density of seawater is given in

Problem 1.71.

mass volume density

1000 g volume of 1 kg of seawater 970.9 mL 0.9709 L 1.03 g/mL

In other words, there are 1.3 g of Mg in every 0.9709 L of seawater.

Next, let’s convert tons of Mg to grams of Mg.

4 2000 lb^ 453.6 g 10 (8.0 10 tons Mg) 7.26 10 g Mg 1 ton 1 lb

× × × = ×

Volume of seawater needed to extract 8.0 × 10

4 ton Mg =

10 0.9709 L seawater (7.26 10 g Mg) 1.3 g Mg

× × =

10 5.4 × 10 L of seawater

1.73 Assume that the crucible is platinum. Let’s calculate the volume of the crucible and then compare that to the

volume of water that the crucible displaces.

mass volume density

3

860.2 g Volume of crucible 21.45 g/cm

3 40.10 cm

3

(860.2 820.2)g Volume of water displaced

0.9986 g/cm

3 40.1 cm

The volumes are the same (within experimental error), so the crucible is made of platinum.

1.74 Volume = surface area × depth

Recall that 1 L = 1 dm

3

. Let’s convert the surface area to units of dm

2 and the depth to units of dm.

2 2 8 2 1000 m^ 1 dm 16 2 surface area (1.8 10 km ) 1.8 10 dm 1 km 0.1 m

= × × ⎜ ⎟ × ⎜ ⎟ = ×

3 1 dm 4 depth (3.9 10 m) 3.9 10 dm 0.1 m

= × × = ×

Volume = surface area × depth = (1.8 × 10

16 dm

2 )(3.9 × 10

4 dm) = 7.0 × 10

20 dm

3 = 7.0 × 10

20 L

1.75 (a)

31.103 g Au 2.41 troy oz Au 1 troy oz Au

× = 75.0 g Au

(b) 1 troy oz = 31.103 g

1 lb 453.6 g ? g in 1 oz 1 oz 16 oz 1 lb

= × × = 28.35 g

A troy ounce is heavier than an ounce.

Volume of sphere 3

= π r

3 4 15 cm 3 3 Volume 1.77 10 cm 3 2

= π = × ⎜ ⎟ ⎝ ⎠

3 3

3

22.57 g Os 1 kg mass volume density (1.77 10 cm ) 1 cm 1000 g

= × = × × × =

1 4.0 × 10 kg Os

1 2.205 lb 4.0 10 kg Os 1 kg

× × = 88 lb Os

1.77 (a)

|0.798 g/mL 0.802 g/mL| 100% 0.798 g/mL

× = 0.5%

(b)

|0.864 g 0.837 g| 100% 0.864 g

× = 3.1%

1.78 62 kg = 6.2 × 10

4 g

O : (6.2 × 10

4 g)(0.65) = 4.0 × 10

4 g O N : (6.2 × 10

4 g)(0.03) = 2 × 10

3 g N

C : (6.2 × 10

4 g)(0.18) = 1.1 × 10

4 g C Ca : (6.2 × 10

4 g)(0.016) = 9.9 × 10

2 g Ca

H : (6.2 × 10

4 g)(0.10) = 6.2 × 10

3 g H P : (6.2 × 10

4 g)(0.012) = 7.4 × 10

2 g P

1.79 3 minutes 43.13 seconds = 223.13 seconds

Time to run 1500 meters is:

1 mi 223.13 s 1500 m 1609 m 1 mi

× × = 208.01 s = 3 min 28.01 s

1.80? ° C = (7.3 × 10

2 − 273) K = 4.6 × 10

2 ° C

2 9 F

(4.6 10 C) 32 F

5 C

= × ° × + ° =

2 ? ° F 8.6 × 10 ° F

3 34.63% Cu^ 1000 g (5.11 10 kg ore) 100% ore 1 kg

= × × × =

6 ? g Cu 1.77 × 10 g Cu

4 2000 lb Au^ 16 oz Au^ $ (8.0 10 tons Au) or 1 ton Au 1 lb Au 1 oz Au

× × × × =

12 $2.4 × 10 2.4 trillion dollars

1.89 (a) homogeneous

(b) heterogeneous. The air will contain particulate matter, clouds, etc. This mixture is not homogeneous.

1.90 First, let’s calculate the mass (in g) of water in the pool. We perform this conversion because we know there

is 1 g of chlorine needed per million grams of water.

4 7 2 2

3.79 L 1 mL 1 g (2.0 10 gallons H O) 7.58 10 g H O 1 gallon 0.001 L 1 mL

× × × × = ×

Next, let’s calculate the mass of chlorine that needs to be added to the pool.

7 2 6 2

1 g chlorine (7.58 10 g H O) 75.8 g chlorine 1 10 g H O

× × =

×

The chlorine solution is only 6 percent chlorine by mass. We can now calculate the volume of chlorine

solution that must be added to the pool.

100% soln 1 mL soln 75.8 g chlorine 6% chlorine 1 g soln

× × =

3 1.3 × 10 mL of chlorine solution

22 20

1 yr (2.0 10 J)

1.8 10 J

× × =

×

2 1.1 × 10 yr

1.92 We assume that the thickness of the oil layer is equivalent to the length of one oil molecule. We can calculate

the thickness of the oil layer from the volume and surface area.

2 2 1 cm 5 2 40 m 4.0 10 cm 0.01 m

× ⎜ ⎟ = ×

0.10 mL = 0.10 cm

3

Volume = surface area × thickness

3 7 5 2

volume 0.10 cm thickness 2.5 10 cm surface area (^) 4.0 10 cm

− = = = ×

×

Converting to nm:

7 9

0.01 m 1 nm (2.5 10 cm) 1 cm (^1 10) m

− −

× × × =

×

2.5 nm

1.93 The mass of water used by 50,000 people in 1 year is:

2 13 2 2

150 gal water 3.79 L 1000 mL 1.0 g H O 365 days 50, 000 people 1.04 10 g H O/yr 1 person each day 1 gal 1 L 1 mL H O 1 yr

× × × × × = ×

A concentration of 1 ppm of fluorine is needed. In other words, 1 g of fluorine is needed per million grams of

water. NaF is 45.0% fluorine by mass. The amount of NaF needed per year in kg is:

13 2 6 2

1 g F 100% NaF 1 kg (1.04 10 g H O) 10 g H O 45% F^ 1000 g

× × × × =

4 2.3 × 10 kg NaF

An average person uses 150 gallons of water per day. This is equal to 569 L of water. If only 6 L of water is

used for drinking and cooking, 563 L of water is used for purposes in which NaF is not necessary. Therefore

the amount of NaF wasted is:

563 L

569 L

× = 99%

1.94 (a)

(^3 3 )

3

$1.30 1 ft 1 in 1 cm 1 mL

15.0 ft 12 in^ 2.54 cm^ 1 mL^ 0.001 L

× ⎜ ⎟ × ⎜ ⎟× × =

3 $3.06 10 /L

− ×

(b)

3

3

0.304 ft gas $1. 2.1 L water 1 L water (^) 15.0 ft

× × = $0.055 = 5.5¢

1.95 To calculate the density of the pheromone, you need the mass of the pheromone, and the volume that it

occupies. The mass is given in the problem. First, let’s calculate the volume of the cylinder. Converting the

radius and height to cm gives:

1609 m 1 cm 4 0.50 mi 8.05 10 cm 1 mi 0.01 m

× × = ×

12 in 2.54 cm 3 40 ft 1.22 10 cm 1 ft 1 in

× × = ×

volume of a cylinder = area × height = π r

2 × h

volume = π(8.05 × 10

4 cm)

2 × (1.22 × 10

3 cm) = 2.48 × 10

13 cm

3

Density of gases is usually expressed in g/L. Let’s convert the volume to liters.

13 3 10

3

1 mL 1 L (2.48 10 cm ) 2.48 10 L

1 cm 1000 mL

× × × = ×

8

10

mass 1.0 10 g

volume (^) 2.48 10 L

− × = = =

×

19 density 4.0 10 g/L

− ×

1.96 First, convert 10 μm to units of cm.

4 1 10 cm 3 10 m 1.0 10 cm 1 m

− × − μ × = × μ

Now, substitute into the given equation to solve for time.

2 3 2

7 2

(1.0 10 cm)

(^2) 2(5.7 10 cm /s)

−

−

×

×

0.88 s

x

D

t

It takes 0.88 seconds for a glucose molecule to diffuse 10 μm.

1.97 The mass of a human brain is about 1 kg (1000 g) and contains about 10

11 cells. The mass of a brain cell is:

8

11

1000 g 1 10 g/cell 1 10 cells

− = × ×