Examples of polar covalent bonds
1 answer:
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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
Considering the Charles's law, the sample of carbon dioxide gas will occupy 308.72 mL.
<h3>Charles's law</h3>Charles's law establishes the relationship between the temperature and the volume of a gas when the pressure is constant. This law says that the volume is directly proportional to the temperature of the gas: for a given sum of gas at constant pressure, as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases.
Mathematically, Charles's law states that the ratio between volume and temperature will always have the same value:
Considering an initial state 1 and a final state 2, it is fulfilled:
In this case, you know:
- V1= 250 mL
- T1= 25 C= 298 K (being 0 C=273 K)
- V2= ?
- T2= 95 C= 368 K
Replacing in Charles's law:
Solving:
<u><em>V2= 308.72 mL</em></u>
Finally, the sample of carbon dioxide gas will occupy 308.72 mL.
Learn more about Charles's law:
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