Answer : The enthalpy for the reaction is 49.1 kJ/mol
Explanation :
According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.
The formation reaction of
will be,

The intermediate balanced chemical reaction will be,
(1)

(2)

(3)

Now we will reverse the reaction 1 and divide by 2, multiply reaction 2 by 6 and reaction 3 by 3/2 then adding all the equations, we get :
(1)

(2)

(3)

The expression for enthalpy of formation of
will be,



Therefore, the enthalpy for the reaction is 49.1 kJ/mol
Answer:it will be c on edg 2021
Explanation:
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:

Answer: 0.156 mol
Explanation:
To find the moles of 17.2 g K₂S, we need to know the molar mass to convert.
