ISM_Chapter_01

ISM_Chapter_01 - CHAPTER 1 CHEMISTRY THE CENTRAL SCIENCE...

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CHAPTER 1: CHEMISTRY: THE CENTRAL SCIENCE CHAPTER 1 CHEMISTRY: THE CENTRAL SCIENCE 1.4 (a) Hypothesis - This statement is an opinion. (b) Law - Newton's Law of Gravitation. (c) Theory - Atomic Theory. 1.5 (a) Law - Newton's 2nd Law of Motion. (b) Theory - Big Bang Theory. (c) Hypothesis - It may be possible but we have no data to support this statement. 1.6 (a) O and H (b) C and H (c) H and Cl (d) N 1.7 (a) C and O (b) F and H (c) N and H (d) O 1.12 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 1.13 (a) K (potassium) (b) Sn (tin) (c) Cr (chromium) (d) B (boron) (e) Ba (barium) (f) Pu (plutonium) (g) S (sulfur) (h) A r (argon) (i) Hg (mercury) 1.14 (a) hydrogen: element (b) water: compound (c) gold: element (d) sugar: compound 1
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CHAPTER 1: CHEMISTRY: THE CENTRAL SCIENCE 1.15 (a) The sea is a heterogeneous mixture of seawater and various solids, but seawater, with the solids filtered out, is a homogeneous mixture (b) element (c) compound (d) homogeneous mixture (e) heterogeneous mixture (f) homogeneous mixture (g) heterogeneous mixture 1.16 (a) liquid (b) gas (c) mixture (d) solid 1.17 (a) element (b) compound (c) compound (d) element 1.23 Using Equation 1.4: = = = mL 188 g 586 V m d 3.12 g/mL 1.24. Rearrange Equation 1.4 to solve for mass: dV m = mass of ethanol = mL 4 . 17 mL 1 g 798 . 0 × = 13.9 g 1.25 Strategy: Find the appropriate equations for converting between Fahrenheit and Celsius and between Celsius and Fahrenheit given in Section 1.3 of the text. Substitute the temperature values given in the problem into the appropriate equation. Conversion from Fahrenheit to Celsius: 5 C ? C = ( F 32 F) 9 F H =P - H Conversion from Celsius to Fahrenheit: 9 F ? F C 32 F 5 C H H = + l (a) 5 C ? C = (95 F 32 F) 9 F H =P - = H 35 C . (b) 5 C ? C = (12 F 32 F) 9 F H DP - = H 11 C - (c) 5 C ? C = (102 F 32 F) 9 F H =P - = H 39 C . (d) 5 C ? C = (1852 F 32 F) 9 F H DP - = H 1011 C . (e) 9 F ? F 273.15 C 32 F 5 C H 8 = - + = 459.67 F - 1.26 (a) 5 C = (105 F 32 F) 9 F H D - = H ? C 41 C . (b) 9 F 11.5 C 32 F 5 C H = - + l = ? F 11.3 F . 2
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CHAPTER 1: CHEMISTRY: THE CENTRAL SCIENCE (c) 3 9 F 6.3 10 C 32 F 5 C H = + l = 4 ? F 1.1 10 F . (d) 5 C = (451 F 32 F) 9 F H - = H ? C 233 C . 1.27 Rearrange Equation 1.4 to solve for volume: d m V = volume of water = = g/mL 992 . 0 g 50 . 2 2.52 mL . 1.28 Rearrange Equation 1.4 to solve for volume: d m V = volume of platinum = = 3 g/cm 5 . 21 g 6 . 87 4.07 cm 3 . 1.29 Using Equation 1.1: K = ° C + 273.15 (a) K = 113 ° C + 273 = 386 K (b) K = 37 ° C + 273 = 3.10 × 10 2 K (c) K = 357 ° C + 273 = 6.30 × 10 2 K Note that because none of the temperatures given in ° C has any digits after the decimal point, we can add 273 rather than 273.15. 1.30
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ISM_Chapter_01 - CHAPTER 1 CHEMISTRY THE CENTRAL SCIENCE...

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