This print-out should have 20 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.The due time is Centraltime.C6H5COOH = 122.1 g/mol(benzoic acid)C3H8= 44.1 g/mol(propane)ozone is O3001(part 1 of 1) 10 pointsWhat mass of propane (C3H8(g)) must beburned to supply 2775 kJ of heat? The stan-dard enthalpy of combustion of propane at298 K is-2220 kJ·mol-22.214.171.124 g2.102 g3.35.3 g4.55.1 gcorrect5.75.9 gExplanation:2775 / 2220 = 1.25 mol of propane1.25 mol×44.1 g/mol = 55.1 g of propane002(part 1 of 1) 12 pointsThe temperature of 2.00 mol CO2(g) is in-creased from 10◦C to 150◦C at constant pres-sure. Calculate the change in the entropy ofcarbon dioxide. Assume ideal behavior.1.+17.71 J/K2.-11.70 J/K3.+11.70 J/K4.-23.30 J/KT1= 10◦C = 283 KT2= 150◦C = 423 Kn= 2 molCp,m= 8.314Jmol·KΔS=nCp,mlnparenleftbiggT2T1parenrightbigg= (2 mol)72(8.314Jmol·K)×lnparenleftbigg423 K283 Kparenrightbigg= 23.3912 J/K003(part 1 of 1) 10 pointsAir is cooled until it freezes and the solid air iscooled to nearly 0 K. What value does the en-tropy of the resulting solid mixture approach
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