Answer:
1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹
2. 0.58 mol
Explanation:
1.Given ΔO₂/Δt…
2H₂O₂ ⟶ 2H₂O + O₂
-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt
d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹
d[H₂O]/dt = 2d[O₂]/dt = 2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = 6.6 × 10⁻³mol·L⁻¹s⁻¹
2. Moles of O₂
(a) Initial moles of H₂O₂
(b) Final moles of H₂O₂
The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.
(c) Moles of H₂O₂ reacted
Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol
(d) Moles of O₂ formed
The answer is 100%
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Answer:
Oxygen in hydrogen peroxide oxidizes from -1 to 0.
Explanation:
Oxidation is the loss of electrons. The specie which is oxidized has has elevation in its oxidation state as compared in the reactant and the products.
The given reaction is shown below as:
Manganese in 
has oxidation state of +7
Manganese in 
has an oxidation state of +2
It reduces from +7 to +2
Oxygen in hydrogen peroxide has an oxidation state of -1.
Oxygen in molecular oxygen has an oxidation of 0.
Thus, oxygen in hydrogen peroxide oxidizes from -1 to 0.
Answer:
Explanation:
Hello,
For the given chemical reaction:
We first must identify the limiting reactant by computing the reacting moles of Al2S3:
Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:
Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:
Finally, we compute the percent yield with the obtained 2.10 g:
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Answer: Mass Of CFC that needs to evaporate for the freezing of water = 328.24 g
Explanation: Heat gained by the CFC = Heat lost by water
Heat lost by water = Heat required to take water's temperature to 0°c + Heat required to freeze water at 0°c
Heat required to take water's temperature from 33°c to 0°c = mCΔT
m = 201g, C = 4.18 J/(gK), ΔT = 33
mCΔT = 201 × 4.18 × 33 = 27725.94 J
Heat required to freeze water at 0°c = mL
m = 201g, L = 334 J/g
mL = 201 × 334 = 67134 J
Heat gained by CFC to vaporize = mH = 27725.94 + 67134 = 94859.94 J
H = 289 J/g, m = ?
m × 289 = 94859.9
m = 328.24 g
QED!!
