Answer : The current passing between the electrodes is, 
Explanation :
First we have to calculate the charge of sodium ion.
where,
q = charge of sodium ion
n = number of sodium ion = 
e = charge on electron = 
Now put all the given values in the above formula, we get:
Now we have to calculate the charge of chlorine ion.
where,
q' = charge of chlorine ion
n = number of chlorine ion = 
e = charge on electron =
Now put all the given values in the above formula, we get:
Now we have to calculate the current passing between the electrodes.
Thus, the current passing between the electrodes is,
Answer:
ΔG=ΔG0+RTlnQ where Q is the ratio of concentrations (or activities) of the products divided by the reactants. Under standard conditions Q=1 and ΔG=ΔG0 . Under equilibrium conditions, Q=K and ΔG=0 so ΔG0=−RTlnK . Then calculate the ΔH and ΔS for the reaction and the rest of the procedure is unchanged.
Explanation:
Answer:
2Cu2S + 3O2 + 2C -------> 4Cu + 2SO2 + 2CO
Explanation:
Equation 1 should correctly be written as;
2Cu2S + 3O2-----> 2Cu2O + 2SO2
Equation 2 should be correctly written as;
2Cu2O + 2C -----> 4Cu + 2CO
The overall reaction equation is;
2Cu2S + 3O2 + 2C -------> 4Cu + 2SO2 + 2CO
Note that species that are intermediates are cancelled out .
Answer:
V = 42.6 L
Explanation:
Given data:
Number of moles of Cl₂ = 1.9 mol
Temperature and pressure = standard
Volume occupy = ?
Solution:
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
By putting values,
1 atm × V = 1.9 mol ×0.0821 atm.L /mol.K × 273.15 k
V = 42.6 atm.L / 1 atm
V = 42.6 L
