Chapter 1

# Chapter 1 - CHAPTER 1 INTRODUCTION 1.7(a(b(c(d(e Chemical...

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CHAPTER 1 INTRODUCTION 1.7 (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. 1.8 (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. 1.9 (a) extensive (b) extensive (c) intensive (d) extensive 1.10 (a) extensive (b) intensive (c) intensive 1.11 (a) element (b) compound (c) element (d) compound 1.12 (a) compound (b) element (c) compound (d) element 1.17 The density of the sphere is given by: 4 33 1.20 10 g 1.05 10 cm × == = × 3 11.4 g/cm m V d 1.18 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 13.6 g 95.8 mL 1mL = 3 mass of Hg 1.30 10 g ×

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CHAPTER 1: INTRODUCTION 2 1.19 (a) 5C = (105 F 32 F) 9F ° °− ° × = ° ?C 4 1C °° (b) 11.5 C 32 F ° ⎛⎞ =− °× + ° = ⎜⎟ ° ⎝⎠ ?F 1 1 . 3F (c) 3 6.3 10 C 32 F ° ° × + ° = ° 4 1 . 1 1 0 F ° (d) = (451 F 32 F) ° = ° 2 3 3C 1.20 Strategy: Find the appropriate equations for converting between Fahrenheit and Celsius and between Celsius and Fahrenheit given in Section 1.5 of the text. Substitute the temperature values given in the problem into the appropriate equation. Solution: (a) 1K K( C 2 7 3 C ) =° + ° ° (i) K = 113 ° C + 273 ° C = 386 K (ii) K = 37 ° C + 273 ° C = 3.10 × 10 2 K (iii) K = 357 ° C + 273 ° C = 6.30 × 10 2 K (b) C 2 7 3 C ) ° ° (i) ° C = K 273 = 77 K 273 = 196 ° C (ii) ° C = 4.2 K 273 = 269 ° C (iii) ° C = 601 K 273 = 328 ° C 1.21 (a) 2.7 × 10 8 (b) 3.56 × 10 2 (c) 9.6 × 10 2 1.22 Strategy: Writing scientific notation as N × 10 n , we determine n by counting the number of places that the decimal point must be moved to give N , a number between 1 and 10. If the decimal point is moved to the left, n is a positive integer, the number you are working with is larger than 10. If the decimal point is moved to the right, n is a negative integer. The number you are working with is smaller than 1. (a) Express 0.749 in scientific notation. Solution: The decimal point must be moved one place to give N , a number between 1 and 10. In this case, N = 7.49 Since 0.749 is a number less than one, n is a negative integer. In this case, n = 1.
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## This note was uploaded on 10/11/2009 for the course CHE 102 taught by Professor Maggie during the Fall '09 term at Wisc Whitewater.

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Chapter 1 - CHAPTER 1 INTRODUCTION 1.7(a(b(c(d(e Chemical...

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