0.737 KL would be the answer, pls give brainliest.
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
Answer: generic material and protein coat. Have a great day
Explanation:
The answer should be neutralization
