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.
,2 CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE
1.13 Li, lithium; F, fluorine; P, phosphorus; Cu, copper; As, arsenic; Zn, zinc; Cl, chlorine; Pt, platinum;
Mg, magnesium; U, uranium; Al, aluminum; Si, silicon; Ne, neon.
1.14 (a) K (b) Sn (c) Cr (d) B (e) Ba
(f) Pu (g) S (h) Ar (i) Hg
1.15 (a) element (b) compound (c) element (d) compound
1.16 (a) homogeneous mixture (b) element (c) compound
(d) homogeneous mixture (e) heterogeneous mixture (f) homogeneous mixture
(g) heterogeneous mixture
mass 586 g
1.21 density = = = 3.12 g/mL
volume 188 mL
1.22 Strategy: We are given the density and volume of a liquid and asked to calculate the mass of the liquid.
Rearrange the density equation, Equation (1.1) of the text, to solve for mass.
mass
density =
volume
Solution:
mass = density × volume
0.798 g
mass of ethanol = × 17.4 mL = 13.9 g
1 mL
5°C
1.23 ? °C = (°F − 32°F) ×
9°F
5°C
(a) ? °C = (95 − 32)°F × = 35°C
9°F
5°C
(b) ? °C = (12 − 32)°F × = − 11°C
9° F
5°C
(c) ? °C = (102 − 32)°F × = 39°C
9°F
5°C
(d) ? °C = (1852 − 32)°F × = 1011°C
9°F
⎛ 9° F ⎞
(e) ? °F = ⎜ °C × 5°C ⎟ + 32°F
⎝ ⎠
⎛ 9° F ⎞
? °F = ⎜ −273.15 °C × + 32°F = − 459.67°F
⎝ 5 °C ⎟⎠
1.24 Strategy: Find the appropriate equations for converting between Fahrenheit and Celsius and between
Celsius and Fahrenheit given in Section 1.7 of the text. Substitute the temperature values given in the
problem into the appropriate equation.
(a) Conversion from Fahrenheit to Celsius.
5°C
? °C = (°F − 32°F) ×
9°F
, CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE 3
5°C
? °C = (105 − 32)°F × = 41°C
9°F
(b) Conversion from Celsius to Fahrenheit.
⎛ 9° F ⎞
? °F = ⎜ °C × + 32°F
⎝ 5 °C ⎟⎠
⎛ 9° F ⎞
? °F = ⎜ −11.5 °C × + 32°F = 11.3 °F
⎝ 5 °C ⎟⎠
(c) Conversion from Celsius to Fahrenheit.
⎛ 9° F ⎞
? °F = ⎜ °C × + 32°F
⎝ 5°C ⎟⎠
⎛ 9 °F ⎞ 4
? °F = ⎜ 6.3 × 103 °C × ⎟ + 32°F = 1.1 × 10 °F
⎝ 5 °C ⎠
(d) Conversion from Fahrenheit to Celsius.
5°C
? °C = (°F − 32°F) ×
9°F
5°C
? °C = (451 − 32)°F × = 233°C
9°F
1K
1.25 K = (°C + 273°C)
1°C
(a) K = 113°C + 273°C = 386 K
2
(b) K = 37°C + 273°C = 3.10 × 10 K
2
(c) K = 357°C + 273°C = 6.30 × 10 K
1K
1.26 (a) 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
−8 2 4 −2
1.29 (a) 2.7 × 10 (b) 3.56 × 10 (c) 4.7764 × 10 (d) 9.6 × 10
−2
1.30 (a) 10 indicates that the decimal point must be moved two places to the left.
−2
1.52 × 10 = 0.0152
−8
(b) 10 indicates that the decimal point must be moved 8 places to the left.
−8
7.78 × 10 = 0.0000000778
, 4 CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE
−1 2
1.31 (a) 145.75 + (2.3 × 10 ) = 145.75 + 0.23 = 1.4598 × 10
79500 7.95 × 104
(b) = = 3.2 × 102
2.5 × 10 2
2.5 × 10 2
−3 −4 −3 −3 −3
(c) (7.0 × 10 ) − (8.0 × 10 ) = (7.0 × 10 ) − (0.80 × 10 ) = 6.2 × 10
4 6 10
(d) (1.0 × 10 ) × (9.9 × 10 ) = 9.9 × 10
1.32 (a) Addition using scientific notation.
n
Strategy: Let's express scientific notation as N × 10 . 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.
n 3 3
Let’s write 0.0095 in such a way that n = −3. We have decreased 10 by 10 , so we must increase N by 10 .
Move the decimal point 3 places to the right.
−3
0.0095 = 9.5 × 10
Add the N parts of the numbers, keeping the exponent, n, the same.
−3
9.5 × 10
−3
+ 8.5 × 10
−3
18.0 × 10
The usual practice is to express N as a number between 1 and 10. Since we must decrease N by a factor of 10
n
to express N between 1 and 10 (1.8), we must increase 10 by a factor of 10. The exponent, n, is increased by
1 from −3 to −2.
−3 −2
18.0 × 10 = 1.8 × 10
(b) Division using scientific notation.
n
Strategy: Let's express scientific notation as N × 10 . 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.
2
653 = 6.53 × 10
Divide the N parts of the numbers in the usual way.
6.53 ÷ 5.75 = 1.14
Subtract the exponents, n.
+2 − (−8) +2 + 8 10
1.14 × 10 = 1.14 × 10 = 1.14 × 10
(c) Subtraction using scientific notation.
n
Strategy: Let's express scientific notation as N × 10 . 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.
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.
,2 CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE
1.13 Li, lithium; F, fluorine; P, phosphorus; Cu, copper; As, arsenic; Zn, zinc; Cl, chlorine; Pt, platinum;
Mg, magnesium; U, uranium; Al, aluminum; Si, silicon; Ne, neon.
1.14 (a) K (b) Sn (c) Cr (d) B (e) Ba
(f) Pu (g) S (h) Ar (i) Hg
1.15 (a) element (b) compound (c) element (d) compound
1.16 (a) homogeneous mixture (b) element (c) compound
(d) homogeneous mixture (e) heterogeneous mixture (f) homogeneous mixture
(g) heterogeneous mixture
mass 586 g
1.21 density = = = 3.12 g/mL
volume 188 mL
1.22 Strategy: We are given the density and volume of a liquid and asked to calculate the mass of the liquid.
Rearrange the density equation, Equation (1.1) of the text, to solve for mass.
mass
density =
volume
Solution:
mass = density × volume
0.798 g
mass of ethanol = × 17.4 mL = 13.9 g
1 mL
5°C
1.23 ? °C = (°F − 32°F) ×
9°F
5°C
(a) ? °C = (95 − 32)°F × = 35°C
9°F
5°C
(b) ? °C = (12 − 32)°F × = − 11°C
9° F
5°C
(c) ? °C = (102 − 32)°F × = 39°C
9°F
5°C
(d) ? °C = (1852 − 32)°F × = 1011°C
9°F
⎛ 9° F ⎞
(e) ? °F = ⎜ °C × 5°C ⎟ + 32°F
⎝ ⎠
⎛ 9° F ⎞
? °F = ⎜ −273.15 °C × + 32°F = − 459.67°F
⎝ 5 °C ⎟⎠
1.24 Strategy: Find the appropriate equations for converting between Fahrenheit and Celsius and between
Celsius and Fahrenheit given in Section 1.7 of the text. Substitute the temperature values given in the
problem into the appropriate equation.
(a) Conversion from Fahrenheit to Celsius.
5°C
? °C = (°F − 32°F) ×
9°F
, CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE 3
5°C
? °C = (105 − 32)°F × = 41°C
9°F
(b) Conversion from Celsius to Fahrenheit.
⎛ 9° F ⎞
? °F = ⎜ °C × + 32°F
⎝ 5 °C ⎟⎠
⎛ 9° F ⎞
? °F = ⎜ −11.5 °C × + 32°F = 11.3 °F
⎝ 5 °C ⎟⎠
(c) Conversion from Celsius to Fahrenheit.
⎛ 9° F ⎞
? °F = ⎜ °C × + 32°F
⎝ 5°C ⎟⎠
⎛ 9 °F ⎞ 4
? °F = ⎜ 6.3 × 103 °C × ⎟ + 32°F = 1.1 × 10 °F
⎝ 5 °C ⎠
(d) Conversion from Fahrenheit to Celsius.
5°C
? °C = (°F − 32°F) ×
9°F
5°C
? °C = (451 − 32)°F × = 233°C
9°F
1K
1.25 K = (°C + 273°C)
1°C
(a) K = 113°C + 273°C = 386 K
2
(b) K = 37°C + 273°C = 3.10 × 10 K
2
(c) K = 357°C + 273°C = 6.30 × 10 K
1K
1.26 (a) 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
−8 2 4 −2
1.29 (a) 2.7 × 10 (b) 3.56 × 10 (c) 4.7764 × 10 (d) 9.6 × 10
−2
1.30 (a) 10 indicates that the decimal point must be moved two places to the left.
−2
1.52 × 10 = 0.0152
−8
(b) 10 indicates that the decimal point must be moved 8 places to the left.
−8
7.78 × 10 = 0.0000000778
, 4 CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE
−1 2
1.31 (a) 145.75 + (2.3 × 10 ) = 145.75 + 0.23 = 1.4598 × 10
79500 7.95 × 104
(b) = = 3.2 × 102
2.5 × 10 2
2.5 × 10 2
−3 −4 −3 −3 −3
(c) (7.0 × 10 ) − (8.0 × 10 ) = (7.0 × 10 ) − (0.80 × 10 ) = 6.2 × 10
4 6 10
(d) (1.0 × 10 ) × (9.9 × 10 ) = 9.9 × 10
1.32 (a) Addition using scientific notation.
n
Strategy: Let's express scientific notation as N × 10 . 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.
n 3 3
Let’s write 0.0095 in such a way that n = −3. We have decreased 10 by 10 , so we must increase N by 10 .
Move the decimal point 3 places to the right.
−3
0.0095 = 9.5 × 10
Add the N parts of the numbers, keeping the exponent, n, the same.
−3
9.5 × 10
−3
+ 8.5 × 10
−3
18.0 × 10
The usual practice is to express N as a number between 1 and 10. Since we must decrease N by a factor of 10
n
to express N between 1 and 10 (1.8), we must increase 10 by a factor of 10. The exponent, n, is increased by
1 from −3 to −2.
−3 −2
18.0 × 10 = 1.8 × 10
(b) Division using scientific notation.
n
Strategy: Let's express scientific notation as N × 10 . 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.
2
653 = 6.53 × 10
Divide the N parts of the numbers in the usual way.
6.53 ÷ 5.75 = 1.14
Subtract the exponents, n.
+2 − (−8) +2 + 8 10
1.14 × 10 = 1.14 × 10 = 1.14 × 10
(c) Subtraction using scientific notation.
n
Strategy: Let's express scientific notation as N × 10 . 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.
