Amixture of h2 and water vapor is present in a closed vessel at 20°c. the total pressure of the system is 755.0 mmhg. partial pressure of water vapor at 20°c equals 17.5 mmhg. what is the partial pressure of h2? mmhg

The partial pressure of H2 is 737.47 mmHg

Explanation:

Let's apply the Ideal Gas Law to discover the total mols.

We assume that the closed vessel has 1 L of volume

P. V  = n . R . T

We should convert mmHg to atm.

760 mmHg ___ 1 atm

755 mmHg ___ (755/760) = 0.993 atm

0.993 atm . 1L = n . 0.082 L.atm/mol.K . 293 K

(0.993 atm . 1L) / ( 0.082mol.K/L.atm . 293K) = n

0.0413 mols = n

These are the total moles.

Now we can kwow the moles of water vapor, to find out the molar fraction of it.

P. V  = n . R . T

760 mmHg ___ 1 atm

17.5 mmHg ___ (17.5 mmHg / 760 mmHg) = 0.0230 atm

0.0230 atm . 1L = n . 0.082 L.atm/mol.K . 293 K

(0.0230 atm . 1L) / (0.082mol.K/L.atm . 293K)= n

9.58x10⁻⁴ mols = n

Molar fraction = mols of gas/total mols

Molar fraction water vapor = 9.58x10⁻⁴ mols / 0.0413 mols

Sum of molar fraction = 1

1 - 9.58x10⁻⁴ mols / 0.0413 mols = molar fraction H2

0.9767 = molar fraction H2

H2 pressure / Total pressure = molar fraction H2

H2 pressure / 755 mmHg = 0.9767

h2 pressure = 755 mmHg . 0.9767 = 737,47 mmHg

answer: a tank of height h is filled with water and sits on the ground. water squirts from a hole at a height y above the ground and has a range r. for two y values, 0 and h, r is zero. determine the value of y for which the range will be a maximum.

explanation:

a 20-cm-long, cylindrical graphite (pure carbon) rod is inserted into an oxidizing atmosphere at 1145 k and 1.013 x 105 pa pressure. the oxidizing process is limited by the diffu- sion of oxygen counterflow to the carbon monoxide that is formed on the cylindrical surface.