Answer: All solutions have two parts: the solute and the solvent. The solute is the substance that dissolves, and the solvent is the substance that dissolves the solute.
Hope that helps (:
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:
0.0253 M/s
Explanation:
From the reaction
N₂ + 3H₂ → 2NH₃
The rate of reaction can be written as
Rate = - ![\frac{d[N_2]}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BN_2%5D%7D%7Bdt%7D)
= - ![\frac{1}{3} \frac{d[H_2]}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B3%7D%20%5Cfrac%7Bd%5BH_2%5D%7D%7Bdt%7D)
= + ![\frac{1}{2} \frac{d[NH_3]}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%5Cfrac%7Bd%5BNH_3%5D%7D%7Bdt%7D)
From the above rate equation we can conclude that the rate of reaction of N₂ is equal to one third of the rate of reaction of H₂,
So,
Rate of reaction of molecular nitrogen = 
Upon calculation, we get rate of reaction of molecular nitrogen = 0.0253 M/s
