Answer: [N2]₀ = 10M and [H2]₀ = 11M
Explanation: To calculate the initial concentration, you would have to set up an ICE table, which is an organized way of tracking known quantities or the ones you want to find. ICE stands for:
I is initial amount;
C is change in concentration;
E is for equilibrium concentration;
For the mixture,
N2 3H2 2NH3
I [N2]₀ [H2]₀ 0
C - x -3x +2x
E [N2]₀ - x =8 [H2]₀ - 3x =5 2x =4
With the product, we can find "x":
2x=4
x=2M
With x=2, find the concentrations:
[N2]₀ - x = 8
[N2]₀ = 10M
[H2]₀ - 3x = 5
[H2]₀ = 11M
The initial concentrations of nitrogen gas [N2] is 10.0 M and of hydrogen gas [H2] is 11.0 M.
Answer:
volume is 7.0 liters
Explanation:
We are given;
- Molarity of the aqueous solution as 2.0 M
- Moles of the solute, K₂S as 14 moles
We are required to determine the volume of the solution;
We need to know that;
Molarity = Moles ÷ volume
Therefore;
Volume = Moles ÷ Molarity
Thus;
Volume of the solution = 14 moles ÷ 2.0 M
= 7.0 L
Hence, the volume of the molar solution is 7.0 L
Answer:
Explanation:
To calculate the cell potential we use the relation:
Eº cell = Eº oxidation + Eº reduction
Now in order to determine which of the species is going to be oxidized, we have to remember that the more the value of the reduction potential is negative, the greater its tendency to be oxidized is. In electrochemistry we use the values of the reductions potential in the tables for simplicity because the only thing we need to do is change the sign of the reduction potential for the oxized species .
So the species that is going to be oxidized is the Aluminium, and therefore:
Eº cell = -( -1.66 V ) + 0.340 V = 5.06 V
Equally valid is to write the equation as:
Eº cell = Eº reduction for the reduced species - Eº reduction for the oxidized species
These two expressions are equivalent, choose the one you fell more comfortable but be careful with the signs.
Answer: 4.7432 L
Explanation:
Use stoichiometry: .4235 mol CuCl2 (1 mol I2 / 2 mol CuCl2)(22.4 L / 1 mol I2) = 4.7432 L :)
Chadwick, Thompson, Rutherford, Bohr
