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# Exam4 - Version 430 Exam 4 Sutclie(53120 This print-out...

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Version 430 – Exam 4 – Sutcliffe – (53120) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Which statement would be the best interpre- tation of the First Law of Thermodynamics? 1. The total amount of entropy in the uni- verse is increasing. 2. The total amount of energy in the uni- verse is constant. correct 3. The total amount of energy in the uni- verse is increasing. 4. The total amount of matter in the uni- verse is constant. Explanation: The first law is the law of conservation of energy and mass (not matter). The second law deals with entropy. 002 10.0 points We find that 61.1 mL of an HCl solution reacts exactly with 74.2 mL of 1.371 M KOH solution. Calculate the molarity of the HCl solution. 1. 1.66 M correct 2. 0.832 M 3. 1.17 M 4. 3.33 M 5. 0.667 M Explanation: V 1 = 61 . 1 mL V 2 = 74 . 2 mL M 2 = 1 . 371 M The balanced equation for the reaction is HCl + KOH KCl + H 2 O Molarity is moles solute per liter of solution. We know the volume of the HCl solution. If we could find the moles of HCl in the solution we could calculate the molarity. We start by calculating the moles of KOH present: ? mol KOH = 74 . 2 mL soln × 1 L soln 1000 mL soln × 1 . 371 mol KOH 1 L soln = 0 . 1017 mol KOH Using the mole-to-mole ratio from the bal- anced chemical equation we calculate the moles of HCl needed to react with this amount of KOH: ? mol HCl = 0 . 1017 mol KOH × 1 mol HCl 1 mol KOH = 0 . 1017 mol HCl This is the number of moles of HCl needed to react with the KOH and therefore the num- ber of moles present in the 61.1 mL of HCl solution. We can calculate the molarity of the HCl solution by dividing the moles of HCl by the volume of the HCl solution: ? M HCl = 0 . 1017 mol HCl 0 . 0611 L soln = 1 . 66 M HCl 003 10.0 points The heat of combustion of butane (C 4 H 10 ) is 2878 kJ/mol. Assume a typical cigarette lighter holds about 1.75 g of butane. How much heat energy would be released if ALL of the butane in the lighter were combusted? 1. 86.63 kJ correct 2. 40.61 kJ 3. 95.59 kJ 4. 1.05 kJ Explanation: MW butane = 58.14 g/mol

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Version 430 – Exam 4 – Sutcliffe – (53120) 2 ? kJ = 2878 kJ mol · (1 . 75 g) · 1 mol 58 . 14 g = 86 . 6271 kJ 004 10.0 points The sublimation of solid carbon dioxide is a spontaneous process. Predict the sign (+, , or 0) of Δ G r , Δ H r , and Δ S r , respectively. 1. , +, + correct 2. , , 3. , +, 4. , 0, + 5. 0, +, + Explanation: Δ G is negative for a spontaneous reaction. Sublimation requires energy to facilitate the solid becoming a gas, so the process is en- dothermic (Δ H is positive). Finally, the en- tropy of a gas is more than that of a solid, so disorder increases (Δ S is positive). 005 10.0 points Consider the following substances: HCl(g) F 2 (g) HCl(aq) Na(s) Which response includes ALL of the sub- stances listed that have Δ H 0 f = 0? 1. Na(s) and F 2 (g) correct 2. HCl(g), Na(s) and F 2 (g) 3. HCl(g) and Na(s) 4. HCl(g), Na(s), HCl(aq) and F 2 (g) 5. Na(s) Explanation: Δ H 0 f = 0 for elements in their standard states. This would be true for both Na(s) and F 2 (g).
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