Remember that in this case pressure is equal to 1.00 atm and temperature is equal to 273.15K. So,
P
V
=
n
R
T
→
n
=
P
V
R
T
=
1.00
a
t
m
⋅
7.0
L
0.082
a
t
m
⋅
L
m
o
l
⋅
K
⋅
273.15
K
=
0.31
Since we know hydrogen's molar mass (
2.0
g
m
o
l
), we can determine the mass
m
H
2
=
n
⋅
m
o
l
a
r
.
m
a
s
s
=
0.31
m
o
l
e
s
⋅
2.0
g
m
o
l
=
0.62
g
If indeed you are dealing with STP, remember that, under these conditions, 1 mole of any ideal gas occupies
22.4
L
. So,
n
=
V
V
m
o
l
a
r
=
7.0
L
22.4
L
=
0.31
moles
And, once again,
m
=
0.31
⋅
2.0
=
0.6
Answer:
677.7 mmHg
Explanation:
The first empirical study on the behaviour of a mixture of gases was carried out by John Dalton. He established the effects of mixing gases at different pressures in the same vessel.
Dalton's law states that,the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of the individual gases present in the mixture of gases. When a gas is collected over water, the gas also contains some water vapour. The partial pressure of the gas will now be given as; total pressure of gas mixture - saturated vapour pressure of water (SVP) at that temperature.
Given that;
Total pressure of gas mixture = 692.2 mmHg
SVP of water at 17°C = 14.5 mmHg
Therefore, partial pressure of oxygen = 692.2-14.5
Partial pressure of oxygen = 677.7 mmHg
Answer:
C
Explanation:
KE at B is max and PE is 0
KE at C is 0 and PE is max
so when student swings from C to B
its KE increases
and PE decreases
