2 Corsls provide treatments
A solution has a pOH of 7. 1 at 10∘c. Then the pH of the solution given that kw=2. 93×10−15 at this temperature is 7.4 .
It is given that,
pOH of solution = 7.1
Kw =2.93×10^(-15)
Firstly, we will calculate the value of pKw
The expression which we used to calculate the pKw is,
pKw=-log [Kw]
Now by putting the value of Kw in this expression,
pKw =−log{2.93×10^(-15)}
pKw =15log(2.93)
pKw=14.5
Now we have to calculate the pH of the solution.
As we know that,
pH+pOH=pKw
Now put all the given values in this formula,
pH+7.1=14.5
pH=7.4
Therefore, we find the value of pH of the solution is, 7.4.
learn more about pH value:
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Sorry if I'm wrong but I think that it is B.
Answer:
ΔS° = -268.13 J/K
Explanation:
Let's consider the following balanced equation.
3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)
We can calculate the standard entropy change of a reaction (ΔS°) using the following expression:
ΔS° = ∑np.Sp° - ∑nr.Sr°
where,
ni are the moles of reactants and products
Si are the standard molar entropies of reactants and products
ΔS° = [2 mol × S°(HNO₃(l)) + 1 mol × S°(NO(g))] - [3 mol × S°(NO₂(g)) + 1 mol × S°(H₂O(l))]
ΔS° = [2 mol × 155.6 J/K.mol + 1 mol × 210.76 J/K.mol] - [3 mol × 240.06 J/K.mol + 1 mol × 69.91 J/k.mol]
ΔS° = -268.13 J/K
