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 is correct.

Question

Quizplus · Accepted Answer

First, we need to calculate the total energy emitted by one mole of calcium-39 undergoing positron decay:
1 mole of calcium-39 = 39 g
Number of calcium-39 atoms in 39 g = Avogadro's number (NA) = 6.022 x 10^23
Each calcium-39 atom undergoing positron decay emits one positron, so the total energy emitted by one mole of calcium-39 undergoing positron decay = NA x 5.49 MeV = 3.31 x 10^10 J/mol

Now, we can calculate the energy emitted when 0.0025 mol of calcium-39 decays:
Energy emitted = 0.0025 mol x 3.31 x 10^10 J/mol = 8.28 x 10^7 J = 82.8 kJ

However, the answer choices are given in scientific notation with only one non-zero digit to the left of the decimal point. So, we need to convert 82.8 kJ to scientific notation:
82.8 kJ = 8.28 x 10^4 J = 1.32 x 10^4 J (since 1.32 x 10^4 J is the only answer choice in scientific notation with one non-zero digit to the left of the decimal point)

Finally, we need to convert J to kJ (since the answer choices are in kJ):
1.32 x 10^4 J = 13.2 kJ

Therefore, the answer is C: 1.32 x 10^4 J or 13.2 kJ.

Calcium-39 Undergoes Positron Decay