일반화학 7판 솔루션 (Chemistry 7th edition Steven S. Zumdahl (줌달)) 자료

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일반화학 7판 솔루션 (Chemistry 7th edition Steven S. Zumdahl (줌달))

줌달 일반화학 7판 솔루션입니다. 해답이아니라 풀이집입니다^^

Chapter 1
1. Various geometric formulas are given in Appendix E. (a) Expressing the radius of the Earth as

R = ( 6.37 × 106 m )(10?3 km m ) = 6.37 × 103 km, its circumference is s = 2π R = 2π (6.37 × 103 km) = 4.00 × 104 km. (b) The surface area of Earth is A = 4π R 2 = 4π ( 6.37 × 103 km ) = 5.10 × 108 km 2 .
2

(c) The volume of Earth is V =

4 π 3 4π R = 6.37 × 103 km 3 3

(

)

3

= 1.08 × 1012 km3 .

2. The conversion factors are: 1 gry = 1/10 line , 1 line = 1/12 inch and 1 point = 1/72 inch. The factors imply that 1 gry = (1/10)(1/12)(72 points) = 0.60 point. Thus, 1 gry2 = (0.60 point)2 = 0.36 point2, which means that 0.50 gry 2 = 0.18 point 2 . 3. The metric prefixes (micro, pico, nano, …) are given for ready reference on the inside front cover of the textbook (see also Table 1?2). (a) Since 1 km = 1 × 103 m and 1 m = 1 × 106 μm,

1km = 103 m = 103 m 106 μ m m = 109 μ m. The given measurement is 1.0 km (two significant figures), which i

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