The simple trick which one can consider in such problem where it is asked for positron emission is :
<span>When the atomic number goes DOWN by one and mass number remains unchanged, then a positron is emitted.
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<span>a. </span>

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Here the atomic number decreases by one.
Similarly, options b and d are eliminated.
Option c is also not the answer.
For c, Count the atomic number on left side and compare it with right side. You will see it is 9 on left and 8 on right. Atomic no. did go down by 1. But the atomic mass is changed as well.
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A.3 moles H2
1 mole N2
I think it is correct
Answer:
0.0468 g.
Explanation:
- The decay of radioactive elements obeys first-order kinetics.
- For a first-order reaction: k = ln2/(t1/2) = 0.693/(t1/2).
Where, k is the rate constant of the reaction.
t1/2 is the half-life time of the reaction (t1/2 = 1620 years).
∴ k = ln2/(t1/2) = 0.693/(1620 years) = 4.28 x 10⁻⁴ year⁻¹.
- For first-order reaction: <em>kt = lna/(a-x).</em>
where, k is the rate constant of the reaction (k = 4.28 x 10⁻⁴ year⁻¹).
t is the time of the reaction (t = t1/2 x 8 = 1620 years x 8 = 12960 year).
a is the initial concentration (a = 12.0 g).
(a-x) is the remaining concentration.
∴ kt = lna/(a-x)
(4.28 x 10⁻⁴ year⁻¹)(12960 year) = ln(12)/(a-x).
5.54688 = ln(12)/(a-x).
Taking e for the both sides:
256.34 = (12)/(a-x).
<em>∴ (a-x) = 12/256.34 = 0.0468 g.</em>