The combustion of 1.771 g of propanol ( C 3 H 7 OH ) increases the temperature of a bomb calorimeter from 298.00 K to 302.34 K . The heat capacity of the bomb calorimeter is 13.70 kJ/K . Determine Δ H for the combustion of propanol to carbon dioxide gas and liquid water.

ΔH for the combustion of propanol to carbon dioxide gas and liquid water is 1980 kJ

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

Q = Heat absorbed by calorimeter =?

C = heat capacity of calorimeter = 13.70 kJ/K

Initial temperature of the calorimeter  = = 298.00 K

Final temperature of the calorimeter  =  = 302.34 K

Change in temperature ,

Putting in the values, we get:

As heat absorbed by calorimeter is equal to heat released by combustion of propanol

Heat released by 0.030 moles of propanol = 59.4 kJ

Heat released by 1 mole of propanol =

ΔH for the combustion of propanol to carbon dioxide gas and liquid water is 1980 kJ/mol

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