When we wish to convert a gas to liquid we have to either
a) decrease temperature
b) increase pressure
In case of fire extinguisher the CO2 is found to be in liquid state, this is as the CO2 is pressurized at high pressure which keeps CO2 in liquid state
the ideal pressure and temperature conditions when CO2 gas can be converted to CO2 gas
Pressure = 5 - 73 atm
Temperature = -57 to 31 degree Celsius
Using ideal gas equation,
P\times V=n\times R\times T
Here,
P denotes pressure
V denotes volume
n denotes number of moles of gas
R denotes gas constant
T denotes temperature
The values at STP will be:
P=100 kPa
T=293 K
R=8.314472 L kPa K⁻¹ mol⁻¹
Number of moles of gas=3.43 mole
Putting all the values in the above equation,
V=83.55 L
So the volume will be 83.55 L.
83.55 L of radon gas would be in 3.43 moles at room temperature and pressure (293 K and 100 kPa).
Answer : The cell emf for this cell is 0.118 V
Solution :
The half-cell reaction is:
In this case, the cathode and anode both are same. So, 
is equal to zero.
Now we have to calculate the cell emf.
Using Nernest equation :
![E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cl^{-}{diluted}]}{[Cl^{-}{concentrated}]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B0.0592%7D%7Bn%7D%5Clog%20%5Cfrac%7B%5BCl%5E%7B-%7D%7Bdiluted%7D%5D%7D%7B%5BCl%5E%7B-%7D%7Bconcentrated%7D%5D%7D)
where,
n = number of electrons in oxidation-reduction reaction = 1
= ?
![[Cl^{-}{diluted}]](https://tex.z-dn.net/?f=%5BCl%5E%7B-%7D%7Bdiluted%7D%5D)
= 0.0222 M
![[Cl^{-}{concentrated}]](https://tex.z-dn.net/?f=%5BCl%5E%7B-%7D%7Bconcentrated%7D%5D)
= 2.22 M
Now put all the given values in the above equation, we get:
Therefore, the cell emf for this cell is 0.118 V
The number of grams of Ag2SO4 that could be formed is 31.8 grams
<u><em> calculation</em></u>
Balanced equation is as below
2 AgNO3 (aq) + H2SO4(aq) → Ag2SO4 (s) +2 HNO3 (aq)
- Find the moles of each reactant by use of mole= mass/molar mass formula
that is moles of AgNO3= 34.7 g / 169.87 g/mol= 0.204 moles
moles of H2SO4 = 28.6 g/98 g/mol =0.292 moles
- use the mole ratio to determine the moles of Ag2SO4
that is;
- the mole ratio of AgNo3 : Ag2SO4 is 2:1 therefore the moles of Ag2SO4= 0.204 x1/2=0.102 moles
- The moles ratio of H2SO4 : Ag2SO4 is 1:1 therefore the moles of Ag2SO4 = 0.292 moles
- AgNO3 is the limiting reagent therefore the moles of Ag2SO4 = 0.102 moles
<h3> finally find the mass of Ag2SO4 by use of mass=mole x molar mass formula</h3>
that is 0.102 moles x 311.8 g/mol= 31.8 grams
