4. How many moles are in 54.0 grams of Silver?
1 answer:
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Answer:
0.0119
Explanation:
There was a part missing. I think this is the whole question:
<em>Before any reaction occurs, the concentration of A in the reaction below is 0.0510 M. What is the equilibrium constant if the concentration of A at equilibrium is 0.0153 M?</em>
A (aq) ⇌ 2B (aq) + C(aq)
<em>Remember to use correct significant figures in your answer. Do not include units in your response.</em>
First, we have to make an ICE Chart, which stands for initial, change and equilibrium. We will call "x" unknown concentrations.
A (aq) ⇌ 2B (aq) + C (aq)
I 0.0510 0 0
C -x +2x +x
E 0.0510-x 2x x
Since the concentration at equilibrium of A is 0.0153 M, we get
We can use the value of x to calculate the concentrations at equilibrium.
The equilibrium constant, Kc, is the ratio of the equilibrium concentrations of products over the equilibrium concentrations of reactants each raised to the power of their stoichiometric coefficients.
The equilibrium constant for this reaction at equilibrium is 0.0119.
You can learn more about equilibrium here: brainly.com/question/4289021
Answer:
32.7 grams of Zn will remained in the crucible after cooling.
Explanation:
..[1]
..[2]
Adding [1] + 2 × [2] we get:
..[3]
Moles of ZnS in crucible = 0.50 mol
According to reaction [3]. 2 moles of ZnS gives 2 moles of Zn.
Then 0.50 moles of ZnS will give:
of Zn.
Mass of 0.50 moles of Zn =
= 0.50 mol × 65.4 g/mol =32.7 g
32.7 grams of Zn will remained in the crucible after cooling.