summer 2006 exam2 practice

Summer 2006 exam2 - Form B 1 NAME General Chemistry 115 Exam 2 h = 6.63 x 10-34 J s RH = 2.18 x 10-18 R = 0.0821 L atm K mol 1 L atm = 101.3 J

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Form B NAME: ______________________ SECTION:_____ General Chemistry 115 Exam 2 April 5, 2004 h = 6.63 x 10 -34 J . s 1 nm = 1 x 10 -9 m c = 3.00 x 10 8 m/s R H = 2.18 x 10 -18 s of H 2 O = 4.184 J/g C R = 0.0821 L . atm/ K . mol = 8.314 J/ K . mol 1 L . atm = 101.3 J 1 atm = 760 torr 1. A scientist studying the properties of hydrogen at low temperatures takes a 1.00 liter volume of hydrogen at 1.00 atm and a temperature of 25.00 degrees C and cools the gas at constant pressure to -200.00 degrees C. Predict the volume of the hydrogen after this process. a. 4.08 L b. 56.9 L c. - 0.800 L d. 0.844 L e. 1.33 x 10 7 L f. 9.33 x 10 -2 L g. 1.66 L h. 0.245 L 2. Concentrated nitric acid acts on copper to give nitrogen dioxide and dissolved copper ions, according to the following balanced equation. Cu (s) + 4HNO 3 (aq) 2 NO 2 (g) + Cu(NO 3 ) 2 (aq) + 2 H 2 O (l) If 13.3 grams copper is reacted with excess nitric acid at a constant the pressure of 1.10 atm and temperature of 25 degrees C, what was the work done on the surroundings by this process? Assume the copper metal and reacting nitric acid have negligible volume. a. - 18.3 J b. 26.6 J c. - 1.04 x 10 3 J d. 895 J e. - 1.88 x 10 6 J f. 13.8 J g. - 1.22 J h. - 2.99 x 10 -6 J 1 1
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Form B 3. Consider the following equation: NH 4 NO 3 (s) N 2 O (g) + 2H 2 O (g) Assume that NH 4 NO 3 (s) has negligible volume. Assume the reaction is carried out at constant temperature and an external pressure of one atmosphere. What is the sign of work? (negative or positive?)__________ Is the work done on the surroundings or on the system? _____________ If the reaction of one mole of NH 4 NO 3 (s) under these condition releases 36 kJ of heat, what is the sign of E? a.
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This note was uploaded on 05/04/2008 for the course CHEM 115 taught by Professor Blank during the Spring '08 term at Rutgers.

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Summer 2006 exam2 - Form B 1 NAME General Chemistry 115 Exam 2 h = 6.63 x 10-34 J s RH = 2.18 x 10-18 R = 0.0821 L atm K mol 1 L atm = 101.3 J

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