1.
-Water levels are dangerously high for wildlife and humans.
-Animals seem to be lost, like the cow and the sheep especially.
2.
-There are not many trees near the water, meaning less areas for wildlife to live.
-There is not much wildlife in general.
Inferences
1. The wildlife shown will move relocate and adapt to another area.
2. Industry — emissions are visible in top left— will continue to hurt the environment. CO2 emissions will increase.
Good luck!
Answer:
(a) oxygen
(b) 154g (to 3sf)
(c) 79.9% (to 3sf)
Explanation:
mass (g) = moles × Mr/Ar
note: eqn means chemical equation
(a)
moles of P = 84.1 ÷ 30.973 = 2.7152 moles
moles of O2 = 85÷2(16) = 2.65625 moles
Assuming all the moles of P is used up,
moles of O2 / moles of phosphorus = 5/4 (according to balanced chemical eqn)
moles of O2 required = 5/4 × 2.7152moles = 3.394 moles (more than supplied which is 2.65625moles)
therefore there is insufficient moles of O2 and the limiting reactant is oxygen.
(b)
moles of P2O5 produced
= 2/5 (according to eqn) × 2.7152
= 1.08608moles
mass of P2O5 produced
= 1.08608 × [ 2(30.973) + 5(16) ]
= 154.164g
= approx. 154g to 3 sig. fig.
(c)
% yield = actual/theoretical yield × 100%
= 123/154 × 100%
= 79.870%
= approx. 79.9% (to 3sf)
Answer:
189.71 secs
Explanation:
We know that decomposition is a first order reaction;
So;
ln[A] = ln[A]o - kt
But;
[A]o = 1.00 M
[A] = 0.250 M
t =135 s
Hence;
ln[A] - ln[A]o = kt
k = ln[A] - ln[A]o/t
k = ln(1) - ln(0.250)/135
k =0 - (-1.386)/135
k = 1.386/135
k= 0.01
So time taken now will be;
ln[A] - ln[A]o = kt
t = ln[A] - ln[A]o/k
t = ln (3) - ln(0.450)/0.01
t = 1.0986 - (-0.7985)/0.01
t = 1.0986 + 0.7985/0.01
t = 189.71 secs
