Answer:
Kc = 168.0749
Explanation:
initial mol: 0.822 0 0
equil. mol: 2(0.822 - x) x x
∴ [ HI ]eq = 0.055 mol/L = 2(0.822 - x) / (1.11 L )
⇒ 1.644 - 2x = 0.055 * 1.11
⇒ 1.644 = 2x + 0.06105
⇒ 2x = 1.583
⇒ x = 0.7915 mol equilibrium
⇒ [ H2 ] eq = 0.7915mol / 1.11L = 0.7130 M = [ I2 ] eq
⇒ Kc = ([ H2 ] * [ I2 ]) / [ HI ]²
⇒ Kc = ( 0.7130² ) / ( 0.055² )
⇒ Kc = 168.0749
Explanation:
It is given that,
The time period of artificial satellite in a circular orbit of radius R is T. The relation between the time period and the radius is given by :

The radius of the orbit in which time period is 8T is R'. So, the relation is given by :



So, the radius of the orbit in which time period is 8T is 4R. Hence, this is the required solution.
Explanation:
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