Answer:
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
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.
mass density volume =
Solution:
mass = density × volume
0.798 g 17.4 mL
1 mL mass of ethanol 13.9 g =× =
1.23 5 C ? C = ( F 32 F) 9 F
° ° °− ° × °
(a) 5 C ? C = (95 32) F 9 F
° ° − °× = ° 35 C°
(b) 5 C ? C = (12 32) F 9 F
° ° − °× = ° − ° 11 C
(c) 5 C ? C = (102 32) F 9 F
° ° − °× = ° 39 C°
(d) 5 C ? C = (1852 32) F 9 F
° ° − °× = ° 1011 C°
(e) 9 F ? F C 32 F
5 C
⎛ ⎞ ° ° =°× + ° ⎜ ⎟ ⎝ ⎠ °
9 F ? F 273.15 C 32 F
5 C
⎛ ⎞ ° °=− °× + ° = ⎜ ⎟ ⎝ ⎠ ° − ° 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 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 = (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
1.25 1 K K ( C 273 C) 1 C
=° + ° °
(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) 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 × 102 (c) 4.7764 × 104 (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
= 0.0152
(b) 10−8
indicates that the decimal point must be moved 8 places to the left.
7.78 × 10−8
= 0.0000000778
Explanation:
