Answer is: pH of aniline is 9.13.<span>
Chemical reaction: C</span>₆H₅NH₂(aq)+
H₂O(l) ⇌ C₆H₅NH₃⁺(aq) + OH⁻(aq).
pKb(C₆H₅NH₂) = 9.40.
Kb(C₆H₅NH₂) = 10∧(-9.4) = 4·10⁻¹⁰.
c₀(C₆H₅NH₂) = 0.45 M.
c(C₆H₅NH₃⁺) = c(OH⁻) = x.
c(C₆H₅NH₂) = 0.45 M - x.
Kb = c(C₆H₅NH₃⁺) · c(OH⁻) / c(C₆H₅NH₂).
4·10⁻¹⁰ = x² / (0.45 M - x).
Solve quadratic equation: x = c(OH⁻) = 0.0000134 M.
pOH = -log(0.0000134 M.) = 4.87.
pH = 14 - 4.87 = 9.13.
Answer:
Both are B. Hope this helps.
The question is incomplete. The complete question is :
C. Balance these fossil-fuel combustion reactions. (1 point)
C8H18(g) + 12.5O2(g) → ____CO2(g) + 9H2O(g) + heat
CH4(g) + ____O2(g) → ____CO2(g) + ____H2O(g) + heat
C3H8(g) + ____O2(g) → ____CO2(g) + ____H2O(g) + heat
C6H6(g) + ____O2(g) → ____CO2(g) + ____H2O(g) + heat
Solution :
C8H18(g) + 12.5O2(g) → __8__CO2(g) + 9H2O(g) + heat
When 1 part of octane reacts with 12.5 parts of oxygen, it gives 8 parts of carbon dioxide and 9 parts of water along with liberation of energy.
CH4(g) + __2__O2(g) → __1__CO2(g) + __2__H2O(g) + heat
When 1 part of methane reacts with 2 parts of oxygen, it gives 1 part of carbon dioxide and 2 parts of water along with liberation of energy.
C3H8(g) + __5__O2(g) → __3__CO2(g) + __4__H2O(g) + heat
When 1 part of propane reacts with 5 parts of oxygen, it gives 3 part of carbon dioxide and 4 parts of water along with liberation of energy.
C6H6(g) + __1/2__O2(g) → __6__CO2(g) + __3__H2O(g) + heat
When 1 part of propane reacts with 1/2 parts of oxygen, it gives 6 part of carbon dioxide and 3 parts of water along with liberation of energy.
