q5pf09key - 1. Assuming the apparatus itself absorbs no...

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1. Assuming the apparatus itself absorbs no heat, what will be the final temperature of a bomb calorimeter's heat sink consisting of 100 mL of water at 15 °C if the reaction releases 6.276 kJ of heat? 1. 30 K 2. 303 °C 3. 30 °C 4. 0 °C 5. -303 °C 6. 273 K q = m∙c∙∆T 6,276 J = 100 g∙4.184 j∙g -1 ∙K -1 ∙∆T ∆T = 15 K T f = ∆T + T i = 15 K + 288 = 303 K = 30 °C 2. Calculate the change in enthalpy for the reaction below based on the provided data. N 2 H 4 (l) + H 2 (g) 2NH 3 (g) ∆H = -201.1 kJ∙mol -1 N 2 (g) + 3H 2 (g) 2NH 3 (g) ∆H = -91.8 kJ∙mol -1 CH 3 OH(l) CH 2 O(g) + H 2 (g) ∆H = 85.2 kJ∙mol -1 -------------------------------------------------------------------------- CH 2 O(g) + N 2 (g) + 3H 2 (g) N 2 H 4 (l) + CH 3 OH(l) ∆H = ? 1. ∆H = -28.4 kJ∙mol -1 2. ∆H = -207.7 kJ∙mol -1 3. ∆H = 194.5 kJ∙mol -1 4. ∆H = -378.1 kJ∙mol -1 5. ∆H = 24.1 KJ∙mol -1 In order for the three provided reactions to cancel to result in the unknown reaction (the
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This note was uploaded on 03/04/2011 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas.

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q5pf09key - 1. Assuming the apparatus itself absorbs no...

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