Answer : The standard enthalpy of formation of methane is, -74.8 kJ/mole
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, multiply reaction 3 by 2 then adding all the equations, we get :
(1)

(2)

(3)

The expression for enthalpy of formation of
will be,



Therefore, the standard enthalpy of formation of methane is, -74.8 kJ/mole
I believe the answer is thermal expansion.
Answer:
Mass of nitrogen dioxide produced = 4.6 g
Explanation:
Given data:
Volume of ammonia = 2.30 L
Mass of nitrogen dioxide produced = ?
Solution:
Chemical equation:
4NH₃ + 7O₂ → 4NO₂ + 6H₂O
Number of moles of ammonia at STP:
PV = nRT
n = PV/RT
n = 1 atm × 2.30 L / 0.0821 atm.L/K.mol × 273 K
n = 2.30 atm .L / 22.414 atm.L/mol
n = 0.1 mol
Now we will compare the moles of ammonia with nitrogen dioxide from balance chemical equation.
NH₃ : NO₂
4 : 4
0.1 : 0.1
Mass of NO₂:
Mass = number of moles × molar mass
Mass = 0.1 mol × 46 g/mol
Mass = 4.6 g