Answer:
V = 22.42 L/mol
N₂ and H₂ Same molar Volume at STP
Explanation:
Data Given:
molar volume of N₂ at STP = 22.42 L/mol
Calculation of molar volume of N₂ at STP = ?
Comparison of molar volume of H₂ and N₂ = ?
Solution:
Molar Volume of Gas:
The volume occupied by 1 mole of any gas at standard temperature and pressure and it is always equal to 22.42 L/ mol
Molar volume can be calculated by using ideal gas formula
PV = nRT
Rearrange the equation for Volume
V = nRT / P . . . . . . . . . (1)
where
P = pressure
V = Volume
T= Temperature
n = Number of moles
R = ideal gas constant
Standard values
P = 1 atm
T = 273 K
n = 1 mole
R = 0.08206 L.atm / mol. K
Now put the value in formula (1) to calculate volume for 1 mole of N₂
V = 1 x 273 K x 0.08206 L.atm / mol. K / 1 atm
V = 22.42 L/mol
Now if we look for the above calculation it will be the same for H₂ or any gas. so if we compare the molar volume of 1 mole N₂ and H₂ it will be the same at STP.
Potential energy can be calculated by the formula Pe=mgh. Plug in your values:
Pe=mgh
Pe=(6 kg)(9.8m/s^2)(100 m)
Pe=5880 kg x m^2/s^2, or 5880 Joules
Distilled water has the a neutral ph. Pure water always has a ph of 7.
The property of potential energy that distinguishes it from kinetic energy are Shape and position
Answer:
a. 1.23 V
b. No maximum
Explanation:
Required:
a. Is there a minimum standard reduction potential that the half-reaction used at the cathode of this cell can have?
b. Is there a maximum standard reduction potential that the half-reaction used at the cathode of this cell can have?
The standard cell potential (E°cell) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E°cell = E°red, cat - E°red, an
If E°cell must be at least 1.10 V (E°cell > 1.10 V),
E°red, cat - E°red, an > 1.10 V
E°red, cat - 0.13V > 1.10 V
E°red, cat > 1.23 V
The minimum standard reduction potential is 1.23 V while there is no maximum standard reduction potential.
