Given the molar mass of Nitrogen is 14.01g/mol you can use that to solve for the moles of nitrogen.
0.235g(1mol/14.01g) = .0168 moles.
Answer:
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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:
-0.050 kJ/mol.K
Explanation:
- A certain reaction is thermodynamically favored at temperatures below 400. K, that is, ΔG° < 0 below 400. K
- The reaction is not favored at temperatures above 400. K, that is. ΔG° > 0 above 400. K
All in all, ΔG° = 0 at 400. K.
We can find ΔS° using the following expression.
ΔG° = ΔH° - T.ΔS°
0 = -20 kJ/mol - 400. K .ΔS°
ΔS° = -0.050 kJ/mol.K
