What percent of magnesium bromide, MgBr2, is magnesium?
1 answer:
You might be interested in
Answer:
0.011 mol/L
Explanation:
This can be solved with something called an ICE table.
I = initial
C = change
E = equilibrium
Initially, there is 0.04 M of N₂, 0.04 M of O₂, and 0 M of NO.
x amount of N₂ reacts. Since the stoichiometry is 1:1, x amount of O₂ also reacts. This produces 2x of NO.
After the reaction, there is 0.04-x of N₂, 0.04-x of O₂, and 2x of NO.
Here it is in table form:
Now we can use the equilibrium constant:
Kc = [NO]² / ( [N₂] [O₂] )
Substituting:
0.10 = (2x)² / ( (0.04 - x) (0.04 - x) )
Solving:
0.10 = (2x)² / (0.04 - x)²
√0.10 = 2x / (0.04 - x)
(√0.10) (0.04 - x) = 2x
(√0.10)(0.04) - (√0.10)x = 2x
(√0.10)(0.04) = 2x + (√0.10)x
(√0.10)(0.04) = (2 + √0.10)x
x = (√0.10)(0.04) / (2 + √0.10)
x = 0.0055
At equilibrium, the concentration of NO is 2x. So the answer is:
[NO] = 2x
[NO] = 0.011
The equilibrium concentration of NO is 0.011 mol/L.
Answer:
<h2>1.23 moles</h2>Explanation:
To find the number of moles in a substance given it's number of entities we use the formula
where n is the number of moles
N is the number of entities
L is the Avogadro's constant which is
6.02 × 10²³ entities
From the question we have
We have the final answer as
<h3>1.23 moles</h3>Hope this helps you