Chemistry Reference
In-Depth Information
13. [A. the pressure will increase and the mass will stay the same]—Boyle's Law
tells us that the pressure will increase, but why would the mass change?
Have you forgotten about conservation of mass?
14. [B. the volume will decrease and the mass will stay the same]—Charles's
Law tells us that the volume will decrease, and the Law of Conservation of
Mass tells us that the mass will remain the same.
15. [D. the sum of all of the partial pressures of the individual gases]—That is
what Dalton's Law teaches us.
16. [A. H]—Graham's Law shows us that the least massive gas will have the
greatest velocity, under identical conditions.
17. [A. 3.17 kPa]—We simply look up the water vapor pressure at 25.0
o
C in the
Vapor Pressure of Water table in Lesson 8-4.
18. [D. 82.3 kPa]—P
dry
H
2
= P
total
- PH
2
O = 85.5 kPa - 3.17 kPa = 82.33 kPa,
which we must round according to the rule for addition and subtraction of
significant digits.
V
1
P
1
T
2
P
2
T
1
19. [C.
]—This is the Combined Gas Law.
V
2
=
M
V
39.9 g
22.4 dm
3
20. [D. 1.78 g/dm
3
]—
D =
=
= 1.78125 g/dm
3
= 1.78 g/dm
3
21. [0.224 moles]—We use the ideal gas equation. Work shown following:
Convert: 23.0
o
C + 273 = 296 K
101.3 kPa
1 atm
1.45 atm =
= 147 kPa
Given: P = 147 kPa; V = 3.75 dm
3
; R = 8.31 dm
3
× kPa/mole × K;
T = 296 K
Find:
n
PV
RT
(1
4
7 kPa)(3.75 dm
3
)
(8.31 dm
3
× kPa/mol × K)(296 K)
= 0.224107 moles = .224 moles
Formula:
n
=
=
22. [234 cm
3
]—We use Boyle's Law. Work shown following:
V
2
=
P
1
V
1
(750 mm of Hg)(350 cm
3
)
(1120 mm of Hg)
=
= 234.375 cm
3
= 234 cm
3
P
2
23. [1:1.41]—We use Graham's Law. Work shown following:
V
M
32
.
9
u
Ne
Ar
=
=
=
1
.
975247
=
1
.
41
V
M
20
.
2
u
Ar
Ne
