Respuesta :
Answer:
c. 2.16 × 10^8 kJ
Explanation:
In the given question, 2 C-12 nuclei were used for the reaction and the mass of C-12 is 12.0000 amu. Therefore, for 2 C-12 nuclei, the mass is 2*12.0000 = 24.0000 amu.
In addition, a Na-23 and a H-1 were formed in the process. The combined mass of the products is 22.989767+1.007825 = 23.997592 amu
The mass of the reactant is different from the mass of the products. The difference = 24.0000 amu - 23.997592 amu = 0.002408 amu.
Theoretically, 1 amu = 1.66054*10^-27 kg
Thus, 0.002408 amu = 0.002408*1.66054*10^-27 kg = 3.99858*10^-30 kg
This mass difference is converted to energy and its value can be calculated using:
E = mc^2 = 3.99858*10^-30 *(299792458)^2 = 3.59374*10^-13 J
Furthermore, 1 mole of hydrogen nuclei contains 6.022*10^23 particles. Thus, we have:
E = 3.59374*10^-13 * 6.022*10^23 = 2.164*10^11 J = 2.164*10^8 kJ
The energy that is released by the nuclear reaction as shown is 2.164*10^8 kJ
What is nuclear reaction?
Mass of the C-12 nuclei = 2*12.0000 = 24.0000 amu.
Mass of Na-23 and H-1 = 23.997592 amu
The mass defect of the reaction = 24.0000 amu - 23.997592 amu = 0.002408 amu.
We know that;
1 amu = 1.66054*10^-27 kg
Hence;
0.002408 amu = 0.002408*1.66054*10^-27 kg = 3.99858*10^-30 kg
Using the Eienstein equation;
E = mc^2 = 3.99858*10^-30 *(299792458)^2 = 3.59374*10^-13 J
Since, 1 mole of hydrogen nuclei must have 6.022*10^23 particles :
E = 3.59374*10^-13 * 6.022*10^23 = 2.164*10^11 J
= 2.164*10^8 kJ
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