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
Answer:
1. earth metal 2. halogen 3.The elements are arranged in seven horizontal rows, called periods or series, and 18 vertical columns, called groups. ... Elements in the periodic table are organized according to their properties.
Inventor: Dmitri Mendeleev please name most brainliest
Answer:
F2 is the limiting reactant
27.6 grams of NaF is produced.
Explanation:
Balance the equation first.
2Na+ F2 ---> 2NaF
To find the limiting reactant, solve for how much NaF can be produced with Na and F2
12.5g F2 x (1 mole F2/ 38.00 grams F2)x (2 mole NaF/ 1 mole F2)
=0.658 moles NaF
16.2g Na x (1 mole Na/ 22.99 grams Na)x (2 mole NaF/ 2 mole Na)
=0.705 moles NaF
Since F2 produced the least NaF, F2 is the limiting reactant.
Now, to find how much NaF there is, use the moles solved above with F2 as the limiting reactant.
0.658 moles NaF x (41.99 grams NaF/ 1 mole NaF)= 27.6 moles NaF
27.6 moles of NaF would be theoretically produced.
