3. Determine the enthalpy of formation for propane. 3C(s, gr) + 4H2(g) ---> C3H2(g) CzH3(g) AH = -2219.9 kJ C(s, gr) AH = -393.5 kJ H,(g) AH = -285.8 kJ a. -205.7 /mol b. -103.8 kJ/mol Å C. + 205.7 /mol d. + 103.8 /mol
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Ответ:
b I hope this is right good luck
Ответ:
The pH is 10.6
Explanation:
pH = pKa
pKb = - log (5.0 ×
)
pKb = 3.30
But pKa + pKb = 14 at equivalence point
∴ pKa = 14 - 3.30 = 10.598 ≈ 10.6
⇒pH = 10.6
N.B : What we calculate is not the pH at the end point - but the theoretical pH at the equivalence point.