The half-life of cesium-130 is 30 min. What percentage of the initial activity w

Question

The half-life of cesium-130 is 30 min. What percentage of the initial activity will be present after 2.0 hours, assuming radioactive decay occurs by a first order rate mechanism? 50 6.25 25 12.5 The decomposition of nitrogen dioxide to nitrogen and oxygen is second order with a rate constant, k = 12.5 M^-1 1s^-1. What is the half-life for the reaction if [NO_2]0 = 0.00260 M? 30.8 22.8 10.6 42.4 The t_1/2 of a second-order decomposition (A rightarrow products) is 26 hours when the initial concentration of A is 0.256 M. How many hours would it take for the concentration of A to drop from 0.256 M to 0 064 M? 52 hours 78 hours 9.2 hours 36 hours

Explanation / Answer

11.

Apply half life equation

A = A0*(1/2)^(t/HL)

HL = 30 min

t = 2 hours = 2*60 = 120 min

then

A/A0 = (1/2)^(t/HL)

A/A0 = (1/2)^(120/30) = 0.0625

this is fracitonal

so change to %

A/A0 = 0.0625*100% = 6.25 %

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