The neutralization pH of an HCI solution with KOH solution is 7
Explanation:
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Answer:
Kc = 105062.5 at 298K
Explanation:
First of all, we state the equilibrium:
3O₂(g) ⇄ 2O₃(g)
We know data about equilibrium concentration of oxygen. We suppose 1 mol of oxygen at the begining. During the reaction, x moles have reacted.
As ratio is 2:3, we can determine how many moles of ozone have been produced.
(x . 2)/3
So we have the final concentration of oxygen, so, let's find out x
1 - x = 0.016 moles
x = 0.984 moles
Then, the [O₃] in equilibrium will be (0.984 . 2) /3 = 0.656.
We supose a volume of 1 L, so we have molar concentration to determine Kc. Let's state the expression for it:
Kc = [O₃]² / [O₂]³
Kc = 0.656² / 0.016³ → 105062.5
Answer:
136.36 mL
Explanation:
Here we have to use the dilution formula
From C1V1= C2V2
Where;
C1= initial concentration of the solution= 12.0 M
C2= final concentration of the solution= 2.20 M
V1 = initial volume of the solution= 25.0 ml
V2= final volume of the solution= ?????
Then recall;
C1V1=C2V2
V2 = C1V1/C2
Substituting values from the parameters given;
V2= 12.0 × 25.0 / 2.20
V2= 136.36 mL
Answer:
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Explanation: