Answer:
Reaction type: Single displacement
Reactant: Magnesium
Product: Dihydrogen - H2
Answer:
The pKa of the conjugate acid is 17.7
Explanation:
If hydrogen is removed from water, the equilibrium concentration of the conjugate acid according to the information given in the question becomes,
Kₐ = [OH⁻]/[H₂O]
Now, we determine the equivalent pKa
pKa = -log[ka]
pKa = -log[100]
pKa = -2
Removal of hydrogen from water is reversible as shown below;
H₂O ⇄ OH⁻ + H⁺
15.7 -2
This reaction is reversible, and the difference in pKa = pKa[H₂O] - pKa[H⁺];
pKa of the conjugate acid = 15.7 - (-2) = 17.7
The pKa of the conjugate acid is 17.7
Answer is: the partial pressure of the helium gas is 0.158 atm.
p(mixture) = 0.48 atm; total pressure.
m(H₂) = 1.0 g; mass of hydrogen gas.
n(H₂) = m(H₂) ÷ M(H₂).
n(H₂) = 1.0 g ÷ 2 g/mol.
n(H₂) = 0.5 mol; amount of hydrogen.
m(He) = 1.0 g; mass of helium.
n(He) = 1 g ÷ 4 g/mol.
n(He) = 0.25 mol; amount of helium.
χ(H₂) = 0.5 mol ÷ 0.75 mol.
χ(H₂) = 0.67; mole fraction of hydrogen.
χ(He) = 0.25 mol ÷ 0.75 mol.
χ(He) = 0.33; mole fraction of helium.
p(He) = 0.33 · 0.48 atm.
p(He) = 0.158 atm; the partial pressure of the helium gas.
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