Question

38) Determine the final temperature of a gold nugget (mass 376 g) that starts at 398 K and loses 4.85 kJ of heat to a snowbank when it is lost. The specific heat capacity of gold is 0.128 J/g°c A) 133 K B) 398 K C) 187 K D) 297 K E) 377 K 39) A 6.55 g sample of aniline (C6H5NH2, molar mass 93.13 g/mol) was combusted in a bomb calorimeter. If the temperature rose by 32.9°C, use the information below to determine the heat capacity of the calorimeter (C). 4 C6H5NH2()+35 02(g) 24 CO2(e) + 14 H20(g)+ 4 NO2(g) AHrxn-1.28 x 104 k A) 97.3 kJ/*C B) 38.9 kl/C C) 5.94 kJ/C D) 6.84 kJ/ C E) 12.8 kl/*C 40) How many molecules of butane are contained in 25.0 mL of butane? The density of butane is 0.6011 g/mL and the molar mass is 58.12 g/mol. A) 2.59x 1023 molecules butane B) 1.46x 1027 molecules butane C)6.87x 1023 molecules butane D) 1.56x 1023 molecules butane E) 7.14 x 1025 molecules butane X 41) A 6.55 g sample of aniline (C6H5NH2, molar mass 93.13 g/mol) was combusted in a bomb calorimeter with a heat capacity of 14.25 kJ/C. If the initial temperature was 32.9"C, use the information below to determine the value of the final temperature of the calorimeter. 4 C6H5NH2()+35 O2(g) 24 CO2 (g) + 14 H20(g) + 4 NO2(g) AH'rxn =-1.28x 104 kJ A) 257"C B) 46.6 C C) 48.7 C D) 41.9°C E) 931 C
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Answer 1

38) Q = mcat 4.85 x10*5 = 3769 x 0.128 Bike * (396 K – Tr) • Tp : 214 x 16 I molt BUTANE HAS 0.259 moli HAS = 6.02 X10²s molecules 23 0.258 motx 6.02x10 3 1 more 1.56 x 10 molecules = of moles OF ANILINE MASS OF ANILINE MW OF ANILINE 6.559 Of Butawe = 93.13g moet 6.555 93.13 g moel 41) No.of moles of ANILINE = = 0.07 mor HEAT ROLEASED: 0.07 mar x 1.28 x 10' 4 mol = 224 kJ HEAT CAPAUTY OF CNORIMETERS 224 AJI 32.9°C - - = 0.07 more HEAT RELEASED = 0.07moe 1.28x104 kt I 4moe = 224 kJ HEAT RELEASED IN HEAT ABSORBED REACTION BY CALORIMETER 224 kJ = CCAL AT 22485 = 14.25 kJ < (Tp - 32.8) 40) mass of BUTANE = 25.0m x 0.60119 = 15:02759 NO. OF molis OF BUTME 1502759 -0.259 moe T = 48.7°C 58.12 8 nol