<span> 1C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O, the ∆H is –2200 kJ
1mol </span>C3H8(g) ---- –2200 kJ
2mol C3H8(g) ----2(–2200 kJ)=-4400kJ
Answer : The standard enthalpy of formation of ethylene is, 52.4 kJ
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, multiply reaction 2 and 3 by 2 then adding all the equation, we get :
(1)

(2)

(3)

The expression for enthalpy of formation of
will be,



Therefore, the standard enthalpy of formation of ethylene is, 52.4 kJ
Answer:
Blackbody radiation
Explanation:
This phenomenon described by Susan is know as black body radiation. Different materials by color reacts to incident radiation on them in diverse manner.
- A white surface will reflect all the wavelength of light incident on it.
- A black surface will absorb all the wavelength of light on it.
This absorption causes the black surface to be hotter compared to other surfaces.