We can use two equations for this problem.<span>
t1/2 = ln
2 / λ = 0.693 / λ
Where t1/2 is the half-life of the element and λ is
decay constant.
20 days = 0.693 / λ
λ = 0.693 / 20 days
(1)
Nt = Nο eΛ(-λt) (2)
Where Nt is atoms at t time, No is the initial amount of substance, λ is decay constant and t is the time
taken.
t = 40 days</span>
<span>No = 200 g
From (1) and (2),
Nt = 200 g eΛ(-(0.693 / 20 days) 40 days)
<span>Nt = 50.01 g</span></span><span>
</span>Hence, 50.01 grams of isotope will remain after 40 days.
<span>
</span>
2KClO₃ → 2KCl + 3O₂
mole ratio of KClO₃ to O₂ is 2 : 3
∴ if moles of O₂ = 5 mol
then moles of KClO₃ =
![\frac{5 mol * 2}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B5%20mol%20%20%20%2A%20%20%202%7D%7B3%7D)
= 3.33 mol
Mass of KClO₃ needed = mol of KClO₃ × molar mass of KClO₃
= 3.33 mol × ((39 × 1) + (35.5 × 1) + (16 × 3) g/mol
= 407.93 g