Calcium-39 undergoes positron decay. Each positron carries 5.49 MeV of energy. How much energy will be emitted when 0.0025 mol of calcium-39 decays? A) 13.2 kJ B) 1.32 × 10⁴ kJ C) 1.32 × 10⁶ kJ D) 1.32 × 10⁹ kJ E) None of these choices are correct.

Question

Quizplus · Accepted Answer

First, we need to calculate the total amount of calcium-39 present in 0.0025 mol.

1 mol of calcium-39 has a mass of 39 g, so 0.0025 mol will have a mass of:

0.0025 mol x 39 g/mol = 0.0975 g

Next, we need to convert the mass of calcium-39 into the number of nuclei present:

0.0975 g x (1 mol/39 g) x (6.02 x 10^23 nuclei/mol) = 1.55 x 10^21 nuclei

Now, we can calculate the total energy released:

Total energy = number of nuclei x energy released per decay

Total energy = (1.55 x 10^21) x (5.49 MeV)

Total energy = 8.51 x 10^10 MeV

Finally, we need to convert MeV to kJ:

1 MeV = 1.602 x 10^-13 kJ

Total energy = (8.51 x 10^10 MeV) x (1.602 x 10^-13 kJ/MeV)

Total energy = 1.36 x 10^1 kJ = 1.32 x 10^16 kJ (to two significant figures)

Therefore, the answer is choice C.

Calcium-39 Undergoes Positron Decay