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Unformatted text preview: bleiweiss (emb853) – H01: Chapt 16 17 – Mccord – (90540) 1 This printout should have 25 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. Chapter 16 sections 10 and 11, plus the first part of Chapter 17, dissolving solutes into solvents: Likes dissolve likes. 001 10.0 points What is the final concentration of NaOH when 335 mL of 0.75 M NaOH are mixed with 165 mL of 0.15 M NaOH? 1. 0.55 M correct 2. 1.67 M 3. 0.45 M 4. 0.18 M 5. 0.82 M Explanation: V 1 = 335 mL M 1 = 0.75 M V 2 = 165 mL M 2 = 0.15 M Molarity is moles solute per liter of solution. Two solutions are being mixed together in this problem. We find the moles of NaOH in each of the individual solutions: ? mol NaOH = 335 mL soln × 1 L soln 1000 mL soln × . 75 mol NaOH 1 L soln = 0 . 2512 mol NaOH ? mol NaOH = 165 mL soln × 1 L soln 1000 mL soln × . 15 mol NaOH 1 L soln = 0 . 0248 mol NaOH The total moles of NaOH in the new solu tion will be the sum of the moles in the two individual solutions: ? mol NaOH = 0 . 2512 mol + 0 . 0248 mol = 0 . 276 mol NaOH The total volume of the new solution will be the combined volume of the individual solu tions: ? mL new soln = 335 mL + 165 mL = 500 mL soln To calculate the molarity of NaOH in the new solution, we divide the total moles of NaOH in the new solution by the total volume of the new solution: ? M NaOH = . 276 mol NaOH 500 mL soln × 1000 mL soln 1 L soln = 0 . 552 M NaOH 002 10.0 points What is the mass of oxygen gas in a 16.8 L container at 23.0 ◦ C and 2.50 atm? Correct answer: 55 . 305 g. Explanation: T = 23 . ◦ C + 273 = 296 K P = 2 . 5 atm V = 16 . 8 L m = ? n = P V RT = (2 . 5 atm)(16 . 8 L) ( . 0821 L · atm mol · K ) (296 K) = 1 . 72828 mol O 2 m = (1 . 72828 mol) 32 g mol = 55 . 305 g O 2 003 10.0 points For the vaporization of water at some temper ature, which of the following is FALSE? 1. The process is endothermic. 2. Δ S is positive. 3. Δ G will increase with increasing temper ature. correct 4. Δ G can be negative or positive depending on the pressure. bleiweiss (emb853) – H01: Chapt 16 17 – Mccord – (90540) 2 Explanation: At standard phase change points, the pro cess is in equilibrium. At equilibrium Δ G = 0, but if we change the pressure or temper ature Δ G can be positive or negative. Va porization requires energy to be input into the system, so vaporization is endothermic. Gases are more disordered than liquids, so Δ S for vaporization is positive. If we increase the temperature, the liquid will spontaneously go to gas phase. Δ G for spontaneous changes is negative. 004 10.0 points A process CANNOT be spontaneous if 1. it is endothermic and there is a decrease in disorder. correct 2. it is exothermic and there is an increase in disorder....
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This note was uploaded on 09/01/2010 for the course CH 302 taught by Professor Holcombe during the Summer '07 term at University of Texas.
 Summer '07
 Holcombe

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