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 distan ce.
(b) Qualitative. This is a value judgment. There is no numerical scale of measurement fo r artistic
excellence.
(c) Qualitative. If the numerical values for the densities of ice and water were given, it would be a
q
uantitative statemen t.
(d) Qualitative. Another value judgmen t.
(e) Qualitative. Even though numbers are involve d, they are not the result of measurement.
1.4 (a) h ypoth esis (b) law (c) th eory
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 ma tter
(d
ifferent 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 th eir
com
position.
(e) Chemical property. When uranium undergoes nucl ear decay, the products are chemically differen t
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 ba ttery.
(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 matte r.
(e) Physical change. The salt can be recovered unchanged by evaporatio n.Chemistry 10e Raymond Chang (Solutions Manual All Chapters, 100% Original Verified, A+ Grade) CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE 2 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) ele ment (b) com pound (c) ele ment (d) com pound
1.16 (a) hom ogeneous mixture (b) elem ent (c) com pound
(d) hom
ogeneous mixture (e) heterogeneous mixture (f) hom
ogeneous mixture
(g) het
erogeneous mixture
1.21 mass 586 g
volume 188 mL=== density 3.12 g/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. massdensityvolume=
Solution: mass = density × volume 0.798 g17.4 mL1m L=×= mass of ethanol 13.9 g
1.23 5C?C = ( F 3 2 F )9F°°° − ° ×° (a)5C?C = ( 9 5 3 2 ) F9F°°− ° × =°35 C° (b)5C? C=( 1 2 3 2 )F9F°°− ° × =°11 C−°
(c)5C? C = (102 32) F9F°°− ° × =°39 C°
(d)5C? C = (1852 32) F9F°°− ° × =°1011 C°
(e) 9F?F C 3 2 F5C°⎛⎞°=°× +°⎜⎟°⎝⎠
9F? F 273.15 C 32 F5C° ⎛⎞°=− °× + °=⎜⎟° ⎝⎠459.67 F−°
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 te xt. Substitute the temperature values given in the problem into the appropriate equation. (a) Co nversion from Fahrenheit to Celsiu s.
5C?C = ( F 3 2 F )9F°°° − ° ×° CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE 3 5C=( 1 0 5 3 2 )F9F°−° × =°?C 4 1 C°° (b) Conversion from Celsius to Fahrenheit. 9F?F C 3 2 F5C°⎛⎞°=°× +°⎜⎟°⎝⎠ 9F11.5 C 32 F5C° ⎛⎞=− °× + °=⎜⎟° ⎝⎠? F 11.3 F°° (c) Conversion from Celsius to Fahrenheit. 9F?F C 3 2 F5C°⎛⎞°=°× +°⎜⎟°⎝⎠ 3 9F6.3 10 C 32 F5C° ⎛⎞=× ° × + ° =⎜⎟° ⎝⎠4?F 1 . 1 1 0 F°× ° (d) Conversion from Fahrenheit to Celsius. 5C?C = ( F 3 2 F )9F°°° − ° ×° 5C= (451 32) F9F°−° × =°? C 233 C°° 1.25 1KK( C 2 7 3 C )1C=°+ °° (a) K = 113°C + 273°C = 386 K (b) K = 37°C + 273°C = 3.10 × 102 K (c) K = 357°C + 273°C = 6.30 × 102 K 1.26 (a) 1KK( C 2 7 3 C )1C=°+ °° °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 × 102 (c) 4.7764 × 104 (d) 9.6 × 10−2 1.30 (a) 10−2 indicates that the decimal point mu st be moved two places to the left. 1.52 × 10−2 = 0.0152 (b) 10−8 indicates that the decimal point must be moved 8 places to the left. 7.78 × 10−8 = 0.0000000778 CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE 4 1.31 (a) 145.75 + (2.3 × 10−1) = 145.75 + 0.23 = 1.4598 × 102 (b) 4
2279500 7.95 10=
2.5 10 2.5 10×=
××23.2 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 × 104) × (9.9 × 106) = 9.9 × 1010 1.32 (a) Addition using scientific notation. Strategy: Let's express scientific notation as N × 10n. When adding numbers using scientific notation, we must write each quantity w ith 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 10n by 103, so we must increase N by 103. 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 10n 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 × 10n. 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 × 102 Divide the N parts of the numbers in the usual way. 6.53 ÷ 5.75 = 1.14 Subtract the exponents, n. 1.14 × 10+2 − (−8) = 1.14 × 10+2 + 8 = 1.14 × 1010 (c) Subtraction using scientific notation. Strategy: Let's express scientific notation as N × 10n. 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.
