Answer:
The total pressure of the mixture in the tank of volume 6.25 litres at 51°C is 1291.85 kPa.
Explanation:
For N2,
Pressure(P₁)=125 kPa
Volume(V₁)=15·1 L
Temperature (T₁)=25°C=25+273 K=298 K
Similarly, for Oxygen,
Pressure(P₂)= 125 kPa
Volume(V₂)= 44.3 L
Temperature(T₂)=25°C= 298 K
Then, for the mixture,
Volumeof the mixture( V)= 6.25 L
Pressure(P)=?
Temperature (T)= 51°C = 51+273 K=324 K
Then, By Combined gas laws,
or, 
or, 
or, 
∴P=1291.85 kPa
So the total pressure of the mixture in the tank of volume 6.25 litres at 51°C is 1291.85 kPa.
Answer:
0.37atm
Explanation:
Given parameters:
Initial pressure = 0.25atm
Initial temperature = 0°C = 273K
Final temperature = 125°C = 125 + 273 = 398K
Unknown:
Final pressure = ?
Solution:
To solve this problem, we use a derivative of the combined gas law;
= 
P and T are pressure and temperature
1 and 2 are initial and final values
= 
P2 = 0.37atm
Explanation:
Moles of metal,
=
4.86
⋅
g
24.305
⋅
g
⋅
m
o
l
−
1
=
0.200
m
o
l
.
Moles of
H
C
l
=
100
⋅
c
m
−
3
×
2.00
⋅
m
o
l
⋅
d
m
−
3
=
0.200
m
o
l
Clearly, the acid is in deficiency ; i.e. it is the limiting reagent, because the equation above specifies that that 2 equiv of HCl are required for each equiv of metal.
So if
0.200
m
o
l
acid react, then (by the stoichiometry), 1/2 this quantity, i.e.
0.100
m
o
l
of dihydrogen will evolve.
So,
0.100
m
o
l
dihydrogen are evolved; this has a mass of
0.100
⋅
m
o
l
×
2.00
⋅
g
⋅
m
o
l
−
1
=
?
?
g
.
If 1 mol dihydrogen gas occupies
24.5
d
m
3
at room temperature and pressure, what will be the VOLUME of gas evolved?
Answer:
I don't understand what you are asking
