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.

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%

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