When 0.0901 mol of an unknown hydrocarbon is burned in a bomb calorimeter, the calorimeter increases in temperature by 2.19°C. If the heat capacity of the bomb calorimeter is 1.229 kJ/°C, what is the heat of combustion for the unknown hydrocarbon?

The heat of combustion for the unknown hydrocarbon is -29.9 kJ/mol

Explanation:

Step 1: Data given

Number of moles of the unknown hydrocarbon = 0.0901 moles

The temperature in the calorimeter rises with 2.19 °C

The heat capacity of the calorimeter is 1.229 kJ/°C

Step 2: Calculate heat

Q = c(calorimeter) * ΔT

⇒with Q = the heat transfer = TO BE DETERMINED

⇒with c = the heat capacity of the calorimeter = 1.229 kJ/°C

⇒with ΔT = The change in temperature = 2.19 °C

Q = 1.229 kJ/°C * 2.19 °C

Q = 2.6915 kJ

Step 3: Calculate the heat of combustion for the unknown hydrocarbon

Heat of combustion = Q/moles

Heat of combustion = 2.6915 kJ / 0.0901 moles

Heat of combustion = 29.9 kJ/mol ( the sign should be negative )

Heat of combustion = -29.9 kJ/mol

The heat of combustion for the unknown hydrocarbon is -29.9 kJ/mol

you need to have a picture of the atom that way we can answer it