Answer : The standard enthalpy of formation of ethylene is, 51.8 kJ/mole
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 equations, we get :
(1) 
(2) 
(3) 
The expression for enthalpy of formation of will be,
Therefore, the standard enthalpy of formation of ethylene is, 51.8 kJ/mole
Volume<span> of matter </span>decreases<span> under </span>pressure<span> ... -under </span>pressure<span>, the </span>particles<span> in a </span>gas<span> are </span>forced closer together<span> ... </span>factors<span> affecting </span>gas pressure<span> ... -</span>if pressure<span> in a sealed container is </span>lower than<span> outside, </span>gas will<span> rush in ...</span>
Physicist Ernest Rutherford<span> established the nuclear theory of the atom with his </span>gold-foil experiment<span>. When he shot a beam of alpha particles at a sheet of </span>gold foil<span>, a few of the particles were deflected. He concluded that a tiny, dense nucleus was causing the deflections.</span>
1) you want to increase friction when it gets cold. If you're outside and it's really cold, you're going to rub your hands to warm them up, therefore friction is increasing
I'm not do sure about decreasing.
