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22 answers across CBSE, ICSE, and State boards - Class 1 to 12, every subject.

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GeneralClass 12CBSE

P°(hexane) = 408 Torr, P°(heptane) = 141 Torr. x(hexane) = 0.300. Find Y₆ and Y₇.

Step-1: Partial pressures:P(hexane) = 0.300 × 408 = 122.4 TorrP(heptane) = 0.700 × 141 = 98.7 TorrStep-2: Total pressure:P_total = 122.4 + 98.7 = 221.1 TorrStep-3: Vapour mole fractions:Y₆ (hexane) = 122.4 / 221.1 ≈ 0.554Y₇ (heptane) = 98.7 / 221.1 ≈ 0.446Answer: Y₆ ≈ 0.554, Y₇ ≈ 0.446

GeneralClass 12CBSE

2N₂O₅(g) → 4NO₂(g) + O₂(g). Initial P = 50 mmHg; P at 30 min = 87.5 mmHg. Find P at 60 min.

Step-1: Let N₂O₅ decrease by 2p. Total pressure = 50 + 3p:50 + 3p = 87.5 ⇒ p = 12.5 mmHgP(N₂O₅) at 30 min = 50 − 2(12.5) = 25 mmHgStep-2: Rate constant (first order):k = (1/30) × ln(50/25) = ln(2)/30 = 0.02310 min⁻¹Step-3: P(N₂O₅) at 60 min (= two half-lives = 25% of initial):P(N₂O₅) at 60 min = 50/4 = 12.5 mmHgStep-4: Total pressure at 60 min:Decrease = 50 − 12.5 = 37.5

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