The question is incomplete, here is the complete question:
Calculate the pH at of a 0.10 M solution of anilinium chloride 
. Note that aniline 
is a weak base with a 
of 4.87. Round your answer to 1 decimal place.
<u>Answer:</u> The pH of the solution is 5.1
<u>Explanation:</u>
Anilinium chloride is the salt formed by the combination of a weak base (aniline) and a strong acid (HCl).
To calculate the pH of the solution, we use the equation:
![pH=7-\frac{1}{2}[pK_b+\log C]](https://tex.z-dn.net/?f=pH%3D7-%5Cfrac%7B1%7D%7B2%7D%5BpK_b%2B%5Clog%20C%5D)
where,
= negative logarithm of weak base which is aniline = 4.87
C = concentration of the salt = 0.10 M
Putting values in above equation, we get:
![pH=7-\frac{1}{2}[4.87+\log (0.10)]\\\\pH=5.06=5.1](https://tex.z-dn.net/?f=pH%3D7-%5Cfrac%7B1%7D%7B2%7D%5B4.87%2B%5Clog%20%280.10%29%5D%5C%5C%5C%5CpH%3D5.06%3D5.1)
Hence, the pH of the solution is 5.1
Answer:
The reaction of perchloric acid and barium hydroxide yields to barium perchlorate (Ba(ClO4)2) and water (H2O). The resulting balanced equation is 2 HClO4 + Ba(OH)2 = Ba(ClO4)2 +2 H2O. This means, for every mole of barium perchlorate produced, 2 moles of perchloric acid and barium hydroxide are required.
Explanation:
Answer is: volume of carbon dioxide is 1,84·10⁸ l.
Chemical reaction: C + O₂ → CO₂.
m(C) = 100 t · 1000 kg/t = 100000 kg
m(C) = 100000 kg · 1000 g/kg = 10⁸ g.
n(C) = m(C) ÷ M(C).
n(C) = 10⁸ g ÷ 12 g/mol.
n(C) = 8,33·10⁶ mol.
From chemical reaction: n(C) . n(CO₂) = 1 : 1.
n(CO₂) = 8,33·10⁶ mol.
m(CO₂) = 8,33·10⁶ mol · 44 g/mol.
m(CO₂) = 3,66·10⁸ = 3,66·10⁵ kg.
V(CO₂) = 3,66·10⁵ kg ÷ 1,98 kg/m³ = 1,84·10⁵ m³.
V(CO₂) = 1,84·10⁵ m³ · 1000 l/m³ = 1,84·10⁸ l.
If you divide 8.1 by 1.8 that leads to 4.5
