A chemical reaction does happen between sodium and water to form Sodium Hydroxide (NaOH) and Hydrogen Gas (H-2)
C(HClO) = 0,3 M.
<span>V(HClO) = 200 mL = 0,2 L.
n(HClO) = </span>c(HClO) · V(HClO).
n(HClO) = 0,06 mol.<span>
c(KClO</span>) =
0,2 M.
<span>V(KClO) = 0,3 L.
n(KClO) = 0,06 mol.
V(buffer solution) = 0,2 L + 0,3 L = 0,5 L.
ck</span>(HClO) = 0,06 mol ÷ 0,5 L = 0,12 M.
cs(KClO) = 0,06 mol ÷ 0,5 L = 0,12 M.<span>
Ka(HClO</span>) =
2,9·10⁻⁸.<span>
This is buffer solution, so use Henderson–Hasselbalch
equation:
pH = pKa + log(cs</span> ÷ ck).<span>
pH = -log(</span>2,9·10⁻⁸) + log(0,12 M ÷ 0,12 M).<span>
pH = 7,54 + 0.
pH = 7,54</span>
Answer: Option (a) is the correct answer.
Explanation:
A miscible solution is defined as the one in which two or more than two components are soluble with each other.
So, when hexane is added to heptane then both of them will mix with each other and hence they are miscible. This mixing will lead to formation of more number of ions into the solution as a result, more will be the disorder present in the solution.
As entropy is the degree of randomness present.
Hence, we can conclude that a solution of hexane and heptane will lead to an increase in entropy.
Answer:
She saw you at the elderly N*de swimming event.
Explanation: You tell me...