Carbon-11 is a radioactive isotope of carbon. Its half-life is 20 minutes. What amount of the initial mass of 70.0g of C-11 atoms in a sample will be left after 80 minutes? a)5.68 g
b)4.38 g
c)4.20 g
d)3.96 g
e)50.00 g

Option B. 4.38 g

Explanation:

The following data were obtained from the question:

Half life (t½) = 20 mins

Original amount (N₀) = 70 g

Time (t) = 80 mins

Amount remaining (N) =.?

Next, we shall determine the decay constant. This can be obtained as illustrated below:

Half life (t½) = 20 mins

Decay constant (K) =.?

Decay constant (K) = 0.693/ half life (t½)

K = 0.693/20

K = 0.03465 min¯¹

Finally, we shall determine the amount remaining after 80 mins as follow:

Original amount (N₀) = 70 g

Time (t) = 80 mins

Decay constant (K) = 0.03465 min¯¹

Amount remaining (N) =.?

Log (N₀/N) = kt/2.303

Log (70/N) = (0.03465 × 80)/2.303

Log (70/N) = 2.772/2.303

Log (70/N) = 1.2036

Take the anti log of 1.2036

70/N = antilog (1.2036)

70/N = 15.98

Cross multiply

70 = N × 15.98

Divide both side by 15.98

N = 70/15.98

N = 4.38 g

Therefore, the amount of the isotope remaining after 80 mins is 4.38 g

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120,630,890

step-by-step explanation:

110+10=120,620+10=630,880+10=890