Adding a catalyst as this would speed up the reaction and the rest would slow it down
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>
Answer:
The answer to your question is 432 g of CO₂
Explanation:
Data
CaCO₃ = 983 g
CaO = 551 g
CO₂ = ?
Balanced reaction
CaCO₃ (s) ⇒ CaO (s) + CO₂ (g)
This reaction is balanced, to solve this problem just remember the Lavoisier Law of conservation of mass that states that the mass of the reactants is equal to the mass of the products.
Mass of reactants = Mass of products
Mass of CaCO₃ = Mass of CaO + Mass of CO₂
Solve for CO₂
Mass of CO₂ = Mass of CaCO₃ - Mass of CaO
Mass of CO₂ = 983 g - 551 g
Simplification
Mass of CO₂ = 432 g
