Exam_2_Review_LE_v2_KEY

Exam_2_Review_LE_v2_KEY - Chem 102A-01 Exam 2 Review Exam...

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Chem Chem 102A 102A - - 01 01 Exam 2 Review Exam 2 Review
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Exam Time and Location for Exam Time and Location for Chem Chem 102A 102A - - 01 01 Regular schedule: Thursday, October 23, 2008 Furman 114 from 7-9 PM Students approved by ODC for time and half : Stevenson Center 5502 from 6-9 PM
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Exam Format Exam Format Outline : ( 1) True/False (5 questions; 15 pts) (2) Multiple choice (7 questions; 21 pts) (3) Matching (6 questions; 12 pts) (4) Greater than, less than or equal to (6 questions; 12 pts) (5) Problems (6 questions ; 40 pts) What will be provided with the exam? (1) Periodic table (2) Constants ( R , Avogadro, Planck’s, Rydberg’s, C s of water and speed of light) (3) Δ H f What do you need to bring? (1) A fresh mind (2) Pencils and erasers (3) Calculator without any stored information
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Chapter 6 Chapter 6 Heat Capacity Heat Capacity ± State functions: ± E and H are state functions; q and w are NOT ± Perform calculations with specific heat capacity (J/g·K) or (J/g· ° C) ± Molar heat capacity (J/mol·K) = specific heat capacity × molar mass Example: What is the molar heat capacity of water? ( C s = 4.18 J/g·K) ± Heat ( q ) = mass × specific heat capacity × temperature changes = m × C s ×Δ T ± Specific heat capacity is inversely proportional to temperature changes ( Δ T ) Example: Which of the following substance would show the greatest temperature change upon absorbing 100.0 J of heat? 10 g of Ag, C s = 0.235 J/g·K 10 g of Au, C s = 0.128 J/g·K 10 g of Fe, C s = 0.449 J/g·K s q C mT = K mol J 3 . 75 mol 1 O H g .02 18 K g J 18 . 4 2 = ×
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where sys Eq w w P V Δ= + = Δ Example: The air within a piston equipped with a cylinder absorbs 565 J of heat and expands. Assume all of the heat is transferred to air inside the cylinder. Suppose the initial volume the air in the cylinder is 0.10 L against an external pressure of 1 atm. Determine the final volume of the air within the piston. Solution 1 L.atm 565 J 5.58 L.atm 101.3 J PV × = () 1 atm ( ) 5.58 L atm fi VV ×− = piston air Chapter 6 Chapter 6 Internal Energy, Heat and Work Internal Energy, Heat and Work 5.58 L atm 0.10 L 5.58 L 1 atm f V −= = 5.58 0.10 5.68 L f V =+= Since there is no energy exchange between the system and the surroundings, therefore q P Δ V = 0 or P Δ V = q = 565 J.
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± Predicting Endothermic and Exothermic Reactions ± Exothermic reaction: Heat is released into the surroundings, therefore the temperature of the surroundings increases (hot) ± Endothermic reaction: Heat is absorbed from the surroundings, therefore the temperature of the surroundings decreases (cold) ± Stoichiometry Involving Δ H rxn Problem 6.57: How much heat would be released by a complete combustion of 177 mL of acetone (C 3 H 6 O) according to the reaction below? The density of acetone is 0.788 g/mL. Solution: Chapter 6 Chapter 6 Enthalpy and Enthalpy and Stoichiometry Stoichiometry C 3 H 6 O( l ) + 4 O 2 ( g ) 3 CO 2 ( g ) + 3 H 2 O( g ) Δ H = 1790 kJ 3 0.788 g 1 mol 1790 kJ 177 mL 4.30 10 kJ released mL 58.08 mL 1 mol ×× × = ×
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Example: Dissolving lithium chloride in 150.0 g of water at room temperature in a styrofoam cup calorimeter with a heat capacity of 52.5 J/K, raises the temperature
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This note was uploaded on 02/27/2009 for the course CHEM 102a taught by Professor Hanusa during the Fall '06 term at Vanderbilt.

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Exam_2_Review_LE_v2_KEY - Chem 102A-01 Exam 2 Review Exam...

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