20071010 Day Lecture 14-2

20071010 Day Lecture 14-2 - Todays Plan Welcome to CHM...

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1 Welcome to CHM 1220/1225 Please turn off cell phones pages iPods Thank you! Today’s Plan Announcements Complete Chapter 5 • Pre-lecture Assignment 15 will be available at 2:00 p.m. today and is due by noon, Friday, October 12, 2007. • Chapter 5 Quiz is due by noon, Friday, October 12, 2007. • Chapter 5 Homework is due by 12:50 p.m. Friday, October 12, 2007 • Chapter 11 Quiz is due by noon, Monday, October 22, 2007. • Chapter 5 Homework is due by 12:50 p.m. Monday, October 22, 2007 Chapter 5 The Gaseous State The Gaseous State Gas Laws 1. Gas Pressure and Its Measurement 2. Empirical Gas Laws 3. The Ideal Gas Law 4. Stoichiometry Problems Involving Gas Volumes 5. Gas Mixtures; Law of Partial Pressures
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2 Kinetic-Molecular Theory 6. Kinetic Theory of Gases 7. Molecular Speeds; Diffusion and Effusion 8. Real Gases ? You prepare nitrogen gas by heating ammonium nitrite: NH 4 NO 2 ( s ) Æ N 2 ( g ) + 2 H 2 O( l ) If you collected the nitrogen over water at 23°C and 727 mmHg, how many liters of gas would you obtain from 5.68 g NH 4 NO 2 ? Molar mass NH 4 NO 2 = 64.04 g/mol P = 727 mmHg P vapor = 21.1 mmHg P gas = 706 mmHg T = 23°C = 296 K P = 727 mmHg P vapor = 21.1 mmHg P gas = 706 mmHg T = 23°C = 296 K P nRT V = () = mmHg atm mmHg ) K ( K mol atm L . mol . V 760 1 706 296 08206 0 0887 0 Molar mass NH 4 NO 2 = 64.04 g/mol 3 2 3 3 3 1 1 04 64 1 68 5 CaCO mol CO mol CaCO g . CaCo mol CaCO g . = 0.887 mol CO 2 gas P = 727 mmHg P vapor = 21.1 mmHg P gas = 706 mmHg T = 23°C = 296 K n = 0.8869 mol P nRT V = = mmHg atm mmHg ) K ( K mol atm L . mol . V 760 1 706 296 08206 0 0887 0 V = 2.32 L of CO 2 (3 significant figures) Kinetic-Molecular Theory (Kinetic Theory) A theory, developed by physicists, is based on the
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20071010 Day Lecture 14-2 - Todays Plan Welcome to CHM...

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