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.

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.