Determine the original amount of ba-142 after 1.25 hours if 9.25 mg remains. the half-life of ba-142 is 10.6 minutes.

1246 mg = 1.246 g.

Explanation:

Where, k is the rate constant of the reaction.

t1/2 is the half-life time of the reaction (t1/2 = 10.6 minutes).

∴ k = ln2/(t1/2) = 0.693/10.6 = 6.54 x 10⁻² min⁻¹.

where, k is the rate constant of the reaction (k = 6.54 x 10⁻² min⁻¹.).

t is the time of the reaction (t = 1.25 hr x 60 = 75.0 min).

a is the initial concentration (original amount) (a = ??? mg).

(a-x) is the remaining concentration (a-x = 9.25 mg).

∴ kt = lna/(a-x)

(6.54 x 10⁻² min⁻¹)(75.0 min) = ln a/(9.25 mg)

4.903 =  ln a/(9.25 mg)

Taking e for the both sides:

134.7 = a/(9.25 mg)

∴ a = (134.7)(9.25 mg) = 1246 mg = 1.246 g.

i think   the answer is c hope i'm right

idk what the question is about