A 5.91 g of octane, C8H18, reacts with excess oxygen in a bomb calorimeter. The heat capacity of the calorimeter is 6.97 kJ/°C and the temperature increases by 32.0°C. How much heat, in units of kJ/mol, was absorbed by the bomb calorimeter?

The heat absorbed by the bomb Calorimeter  Q = 5.07

Explanation:

Given data

Heat capacity of calorimeter = 6.97

Temperature Rise ΔT = 32 °c

Therefore the heat absorbed by the bomb calorimeter

 ΔT

Q = 6.97 × 32

Q = 223.04 KJ

Since  1 mole of propane weighs = 44 gm , So

Q = 5.07

Therefore the heat absorbed by the bomb calorimeter  Q = 5.07

the answer is the first option

explanation:

just googled it m8. not that hard

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answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :