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
Answer:
What do <u>YOU</u> think?
Explanation:
This question is asking for an opinion. The word should is included.
Answer:
Total percent of magnesium in sample = 25.5%
Explanation:
Given:
Mass of magnesium = 24 gram
Mass of chlorine = 70 gram
Find:
Total percent of magnesium in sample = ?
Computation:
Total mass of sample = Mass of magnesium + Mass of chlorine
Total mass of sample = 24 gram + 70 gram
Total mass of sample = 94 gram
Total percent of magnesium in sample = [Mass of magnesium / Total mass of sample]100
Total percent of magnesium in sample = [24/94]100
Total percent of magnesium in sample = [0.255]100
Total percent of magnesium in sample = 25.5%
The rates of the forward and reverse reaction depends on the temperature on which the reaction will proceed, either endothermic of exothermic. it also depends of the concentration of the reactants and products. if the reaction is exothermic, so if the reaction temperature is increased then it will favor the forward reaction, then if the reaction is lowered then it will favor the reverse reaction
