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 is: the energy of exactly one photon of this light is 4.75·10⁻¹⁹ J.
Photon energy equation: E = h·ν.
E - energy of one photon.
ν- frequency.
h - Planck's constant.
ν = 7.17·10¹⁴ Hz.
h = 6.63·10⁻³⁴ J·s.
E = 6.63·10⁻³⁴ J·s · 7.17·10¹⁴ Hz.
E = 4.75·10⁻¹⁹ J.
Explanation:
The given data is as follows.
Mass of apple sauce mixture = 454 kg
Heat added (Q) = 121300 kJ
Heat capacity (
) of apple sauce at 
= 4.0177 
So, Heat given by heat exchanger = heat taken by apple sauce
Q = 
or, Q = 
Putting the given values into the above formula as follows.
Q =
121300 kJ = 
= 
Thus, we can conclude that outlet temperature of the apple sauce is .
The solution is as follows:
K = [Partial pressure of isoborneol]/[Partial pressure of borneol] = 0.106
The molar mass of isoborneol/borneol is 154.25 g/mol
Mol isoborneol = 15 g/154.25 = 0.0972 mol
Mol borneol = 7.5 g/154.25 = 0.0486 mol
Use the ICE approach
borneol → isoborneol
I 0.0972 0.0486
C -x +x
E 0.0972 - x 0.0486 + x
Total moles = 0.1458
Using Raoult's Law,
Partial Pressure = Mole fraction*Total Pressure
[Partial pressure of isoborneol] = [(0.0972-x)/0.1458]*P
[Partial pressure of borneol] = [(0.0486+x/0.1458)]*P
0.106 = [(0.0972-x)/0.1458]*P/ [(0.0486+x/0.1458)]*P
Solving for x,
x = 0.0832
Thus,
<em>Mol fraction of borneol = (0.0486+0.0832)/0.1458 = 0.904</em>
<em>Mol fraction of isoborneol = (0.0972-0.0832)/0.1458 = 0.096</em>
