Describe the kinetic energy/ movement of the particles in plasma.
2 answers:
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The ideal gas law may be written as
where
p = pressure
ρ =density
T = temperature
M = molar mass
R = 8.314 J/(mol-K)
For the given problem,
ρ = 0.09 g/L = 0.09 kg/m³
T = 26°C = 26+273 K = 299 K
M = 1.008 g/mol = 1.008 x 10⁻³ kg/mol
Therefore
Note that 1 atm = 101325 Pa
Therefore
p = 2.2195 x 10⁵ Pa
= 221.95 kPa
= (2.295 x 10⁵)/101325 atm
= 2.19 atm
Answer:
2.2195 x 10⁵ Pa (or 221.95 kPa or 2.19 atm)
where
p = pressure
ρ =density
T = temperature
M = molar mass
R = 8.314 J/(mol-K)
For the given problem,
ρ = 0.09 g/L = 0.09 kg/m³
T = 26°C = 26+273 K = 299 K
M = 1.008 g/mol = 1.008 x 10⁻³ kg/mol
Therefore
Note that 1 atm = 101325 Pa
Therefore
p = 2.2195 x 10⁵ Pa
= 221.95 kPa
= (2.295 x 10⁵)/101325 atm
= 2.19 atm
Answer:
2.2195 x 10⁵ Pa (or 221.95 kPa or 2.19 atm)
Answer : The actual cell potential of the cell is 0.47 V
Explanation:
Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.
The given redox reaction is :
The balanced two-half reactions will be,
Oxidation half reaction :
Reduction half reaction :
The expression for reaction quotient will be :
In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.
Now put all the given values in this expression, we get
The value of the reaction quotient, Q, for the cell is, 13.6
Now we have to calculate the actual cell potential of the cell.
Using Nernst equation :
where,
F = Faraday constant = 96500 C
R = gas constant = 8.314 J/mol.K
T = room temperature = 316 K
n = number of electrons in oxidation-reduction reaction = 2 mole
= standard electrode potential of the cell = 0.51 V
= actual cell potential of the cell = ?
Q = reaction quotient = 13.6
Now put all the given values in the above equation, we get:
Therefore, the actual cell potential of the cell is 0.47 V