The vapor pressure of water in kilopascals is here. Remember to convert to mmHg or atm. as needed.
65. 9.00 atm.
66. P_{He} = 0.300 x 4.00 atm = 1.20 atm. P_{Ar} = 4.00 - 1.20
67. 760.0 mmHg minus 55.3 mmHg
68. 96.00 kPa minus 12.33 kPa
69.
a) 480.0 g O_{2} / 32.0 g/mol
b) 80.00 g He / 4.00 g/mol
c) 35.0 moles
d) 15.0 mol O_{2} / 35.0 mol
e) 20.0 mol He / 35.0 mol
f) 7.00 atm x 0.4286
g) 7.00 atm x 0.5714
Keep in mind that once one partial pressure is calculated, the other can be arrived at by subtraction, if so desired.
70. Complete the following table
O_{2} | Ne | H_{2}S | Ar | Total | |
Moles | 5.00 | 3.00 | 6.00 | 9.00 | 18.00 |
Mole fraction | 5/18 = 0.278 | 3/18 = 0.167 | 6/18 = 0.333 | 4/18 = 0.222 | 1 |
Pressure fraction | 0.278 | 0.167 | 0.333 | 0.222 | 1 |
Partial Pressure | 1620 x 0.278 = 450.36 | 1620 x 0.167 = 270.54 | 1620 x 0.0.333 = 539.46 | 1620 x 0.222 = 359.64 | 1620.0 |
71. (14.0 g / 2.00 g/mol) + (84.0 g /28.0 g/mol) + (2.0 moles) = 12.0 moles total
2.0 / 12.0 = 0.167 of the total pressure. 78.00 is to 0.167 as the total pressure is to one, so 468 mmHg is the answer.
72. The solution is left to the reader.