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34kurt
2 years ago
15

A weather balloon was initially at a pressure of 0.900 atm, and its volume was 35.0 L. The pressure decreased to 0.750 atm, with

out loss of gas or change in temperature. What was the change in the volume?
Chemistry
1 answer:
marusya05 [52]2 years ago
3 0

Boyle's Law

P₁V₁=P₂V₂

0.9 x 35 = 0.75 x V₂

V₂ = 42 L

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A sample of O2(g) is placed in an otherwise empty, rigid container at 4224 K at an initial pressure of 4.97 atm, where it decomp
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Answer:

The value of K_p at 4224 K is 314.23.

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O_2(g)\rightleftharpoons 2O(g)

Initially

4.97 atm            0

At equilibrium

4.97 - p               2p

At initial stage, the partial pressure of oxygen gas = =4.97 atm

At equilibrium, the partial pressure of oxygen gas = p_{O_2}=0.28 atm

So, 4.97 - p = 0.28 atm

p = 4.69 atm

At equilibrium, the partial pressure of O gas = p_{O}=2p=2\times 4.69 atm=9.38 atm

The expression of K_p is given as :

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The value of K_p at 4224 K is 314.23.

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