Answer:
The heat of combustion is -25 kJ/g = -2700 kJ/mol.
Explanation:
According to the Law of conservation of energy, the sum of the heat released by the combustion reaction and the heat absorbed by the bomb calorimeter is equal to zero.
Qcomb + Qcal = 0
Qcomb = - Qcal
The heat absorbed by the calorimeter can be calculated with the following expression.
Qcal = C × ΔT
where,
C is the heat capacity of the calorimeter
ΔT is the change in temperature
Then,
Qcomb = - Qcal
Qcomb = - C × ΔT
Qcomb = - 1.56 kJ/°C × 3.2°C = -5.0 kJ
Since this is the heat released when 0.1964 g o quinone burns, the energy of combustion per gram is:

The molar mass of quinone (C₆H₄O₂) is 108 g/mol. Then, the energy of combustion per mole is:

<h3>
Answer:</h3>
20.62 Kilo-joules
<h3>
Explanation:</h3>
- The Enthalpy of combustion of ethyl alcohol is -950 kJ/mol.
- This means that 1 mole of ethyl alcohol evolves a quantity of heat of 950 Joules when burned.
Molar mass of ethyl ethanol = 46.08 g/mol
Therefore;
46.08 g of C₂H₅OH evolves heat equivalent to 950 kilojoules
We can calculate the amount of heat evolved by 1 g of C₂H₅OH
Heat evolved by 1 g of C₂H₅OH = Molar enthalpy of combustion ÷ Molar mass
= 950 kJ/mol ÷ 46.08 g/mol
= 20.62 Kj/g
Therefore, a gram of C₂H₅OH will evolve 20.62 kilo-joules of heat
The answer is B) The dependent variable relies upon the independent variable