When The balanced equation is:
2Al + 3CuCl2 ⇒3 Cu + 2AlCl3
So, we want to find the limiting reactant:
1- no. of moles of 2Al = MV/n = (Wt * V )/ (M.Wt*n*V) = Wt / (M.Wt *n)
where M= molarity, V= volume per liter and n = number of moles in the balanced equation.
by substitute:
∴ no. of moles of 2Al = 0.2 / (26.98 * 2)= 0.003706 moles.
2- no.of moles of 3CuCl2= M*v / n = (0.5*(15/1000)) / 3= 0.0025 moles.
So, CuCl2 is determining the no.of moles of the products.
∴The no. of moles of 3Cu = 0.0025 moles.
∴The no.of moles of Cu= 3*0.0025= 0.0075 moles.
and ∵ amount of weight (g)= no.of moles * M.Wt = 0.0075 * M.wt of Cu
= 0.0075 * 63.546 =0.477 g
In earths surface or the bottom of the Ocean
The temperature decreases because there is no ozone in the mesosphere and the amount of air is decreasing.
Answer:
for more details are in the pic
Answer:
2.5×10⁶ s
Explanation:
From the question given above, the following data were obtained:
Rate constant (K) = 2.8×10¯⁷ s¯¹
Half-life (t½) =?
The half-life of a first order reaction is given by:
Half-life (t½) = 0.693 / Rate constant (K)
t½ = 0.693 / K
With the above formula, we can obtain the half-life of the reaction as follow:
Rate constant (K) = 2.8×10¯⁷ s¯¹
Half-life (t½) =?
t½ = 0.693 / K
t½ = 0.693 / 2.8×10¯⁷
t½ = 2.5×10⁶ s
Therefore, the half-life of the reaction is 2.5×10⁶ s
