The partial pressure of oxygen given the total barometric pressure is : 108.15 mmHg
<u>Given data : </u>
Total barometric pressure = 515 mmHg
Assuming oxygen percentage = 21%
Barometric pressure dry at 37°C
<h3 /><h3>Determine the partial pressure of oxygen </h3>
Applying the relation below
Partial pressure = oxygen percentage * Barometric pressure
= 21% * 515 mmHg
= 108.15 mmHg
Hence we can conclude that the partial pressure of oxygen is 108.15 mmHg.
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Partial pressure of gas A is 1.31 atm and that of gas B is 0.44 atm.
The partial pressure of a gas in a mixture can be calculated as
Pi = Xi x P
Where Pi is the partial pressure; Xi is mole fraction and P is the total pressure of the mixture.
Therefore we have Pa = Xa x P and Pb = Xb x P
Let us find Xa and Xb
Χa = mol a/ total moles = 2.50/(2.50+0.85) = 2.50/3.35 = 0.746
Xb = mol b/total moles = 0.85/(2.50+0.85) = 0.85/3.35 = 0.254
Total pressure P is given as 1.75 atm
Pa = Xa x P = 0.746 x 1.75 = 1.31atm
Partial pressure of gas A is 1.31 atm
Pb = Xb x P = 0.254 x 1.75 = 0.44atm
Partial pressure of gas B is 0.44 atm.
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