It is believed that two c-12 nuclei can react in the core of a supergiant star to form na-23 and h-1. calculate the energy released from this reaction for each mole of hydrogen formed. the masses of c-12, na-23, and h-1 are 12. u, 22.989767 u, and 1.007825 u, respectively. 126c + 126c →2311na + 11h 2.16 × 1015 kj

a. 2.16 × 105 kj
b. 2.16 × 1011 kj
c. 2.16 × 108 kj
d. 2.16 × 1014 kj

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 correct answer is:

a. formation of precipitation