A 0.100 mol sample of propane, a gas used in many homes in these days of load shedding, was placed in a bomb calorimeter with excess oxygen and ignited. The reaction was C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O (l) The initial temperature was 25.000 0C and its total heat capacity was 97.1 kJ 0C –1 . The reaction raised the temperature of the calorimeter to 27.282 0C? A. How many joules were liberated by the combustion of the propane? B. What is the ΔE for the reaction expressed in kJ mol–1 C3H8? C. Calculate ΔH for the reaction in kJ mol–1 C3H8

i believe the answer is true.

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o

a). in this case, dh> 0 and sd> 0. if t becomes larger, the term "-tds" will become more negative. at some point, it is possible that -dts will be more negative than the dh is positive, and the reaction will become spontaneous. this is probably the correct answer, but you should look at all of the other possibilities and make sure of that.