Chemistry Reference
In-Depth Information
3.
[764 mm of Hg]—Work shown following:
Given: V
1
= 5.56 dm
3
; T
1
= 288 K; P
1
= 755 mm of Hg; T
2
= 262 K;
V
2
= 5.00 dm
3
Find: P
2
Formula:
V
1
P
1
T
2
V
2
T
1
(5.56 dm
3
)(755 mm of Hg)(262 K)
(5.00 dm
3
)(288 K)
P
2
=
=
=763.766 = 764 mm of Hg
4.
[471 cm
3
]—Work shown following:
Given: V
1
= 5.00 × 10
2
cm
3
; T
1
= 295.5 K; P
1
= 103.3 kPa; T
2
= 273 K;
P
2
= 101.3 kPa
Find: V
2
Formula:
V
1
P
1
T
2
P
2
T
1
(5.00 × 10
2
cm
3
)(103.3 kPa)(273 K)
(101.3 kPa)(295.5 K)
V
2
=
=
= 471.0489 = 471 cm
3
5.
[2.29 L]—Work shown following:
Given: V
1
= 2.75 L; T
1
= 316.6 K; P
1
= 733 mm of Hg; T
2
= 273 K;
P
2
= 760 mm of Hg
Find: V
2
Formula:
V
1
P
1
T
2
P
2
T
1
(2.75 L)(733 mm of Hg)(273 K)
(760 mm of Hg)(316.6 K)
V
2
=
=
=2.287045 = 2.29 L
Lesson 8-6 Review
1.
[1:4.67]—Work shown following:
V
M
44
.
0
u
H
CO
2
=
2
=
=
21
.
78218
=
4
.
67
V
M
2
.
02
u
CO
H
2
2
2.
[1:4.44]—Work shown following:
V
M
39
.
u
H
Ar
2
19
.
75248
4
44
=
=
=
=
V
M
2
02
u
Ar
H
2
3.
[1:2]—Work shown following:
M
V
16
.
0
u
CH
He
4
00
2
.
00
=
4
=
=
=
V
M
4
00
u
CH
He
4
4.
[1:1.44]—Work shown following:
M
V
58
.
1
C
H
CO
4
10
2
.
075
1
.
44
=
=
=
=
V
M
28
.
0
u
C
H
CO
4
10
5.
[1:1.41]—Work shown following:
V
M
32
.
0
u
CH
O
2
.
00
1
.
41
4
=
2
=
=
=
V
M
16
.
0
u
O
CH
2
4
