A J-shaped tube, closed at one end, is depicted in the figure at the right. It contains N2 gas that is trapped by an unknown liq
1 answer:
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The pressure of the gas in the flask (in atm) when Δh = 5.89 cm is 1.04 atm
<h3>Data obtained from the question</h3>The following data were obtained from the question:
- Atmospheric pressure (Pa) = 730.1 torr = 730.1 mmHg
- Change in height (Δh) = 5.89 cm
- Pressure due to Δh (PΔh) = 5.89 cmHg = 5.89 × 10 = 58.9 mmHg
- Pressure of gas (P) =?
The pressure of the gas can be obtained as illustrated below:
P = Pa + PΔh
P = 730.1 + 58.9
P = 789 mmHg
Divide by 760 to express in atm
P = 789 / 760
P = 1.04 atm
Thus, the pressure of the gas when Δh = 5.89 cm is 1.04 atm
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Missing part of question:
See attached photo
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