Consider the reaction in the lead-acid cell pb(s) pbo2(s) 2 h2so4(aq) 2 pbso4(aq) 2 h2o(l) for which e°cell = 2. 04 v at 298 k.
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Answer: The partial pressure of oxygen in the mixture if the total pressure is 525 mmHg is 310 mm Hg
Explanation:
mass of nitrogen = 37.8 g
mass of oxygen = (100-37.8) g = 62.2 g
Using the equation given by Raoult's law, we get:
= partial pressure of
= ?
= total pressure of mixture = 525 mmHg
Total moles = 1.94 + 1.35 = 3.29 moles
Thus the partial pressure of oxygen in the mixture if the total pressure is 525 mmHg is 310 mm Hg
Answer:
Explanation:
Given that:
From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:
Multiplying (2) with equation (4) ; we have:
From equation (1) ; multiplying (-1) with equation (1); we have:
From equation (2); multiplying (3) with equation (2); we have:
Now; Adding up equation (5), (6) & (7) ; we get:
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(According to Hess Law)