Answer:
Correct answer is: "Cold Water"
Explanation:
EDGE 2020
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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Answer:
ΔU = −55.45 kJ
Explanation:
From first law of thermodynamics in chemistry, we have;
ΔU = Q + W
where;
ΔU is change in internal energy
Q is the net heat transfer
W is the net work done
We are given;
Q = 74.6 kJ
But Q will be negative since heat is released
Thus;
ΔU = -74.6 kJ + W
We are given;
Constant pressure; P = 35 atm = 35 × 101325 = 3546375 N/m²
Volume before reaction; Vi = 8.2 L = 0.0082 m³
Volume after reaction; V_f = 2.8 L = 0.0028 m³
Now,
W = -P(V_f - V_i)
W = - 3546375(0.0028 - 0.0082)
W = 19.15 KJ
Thus;
ΔU = Q + W
ΔU = -74.6 kJ + 19.15 KJ =
ΔU = −55.45 kJ
