Answer:
2.82 g
Explanation:
Step 1: Write the balanced precipitation reaction
3 Ba(NO₃)₂ (aq) + Al₂(SO₄)₃ (aq) ⇒ 3 BaSO₄(s) + 2 Al(NO₃)₃(aq)
Step 2: Calculate the reacting moles of Ba(NO₃)₂
45.0 mL (0.0450 L) of 0.548 M Ba(NO₃)₂ react.
0.0450 L × 0.548 mol/L = 0.0247 mol
Step 3: Calculate the moles of Al₂(SO₄)₃ that react with 0.0247 moles of Ba(NO₃)₂
The molar ratio of Ba(NO₃)₂ to Al₂(SO₄)₃ is 3:1. The reacting moles of Al₂(SO₄)₃ are 1/3 × 0.0247 mol = 8.23 × 10⁻³ mol
Step 4: Calculate the mass corresponding to 8.23 × 10⁻³ moles of Al₂(SO₄)₃
The molar mass of Al₂(SO₄)₃ is 342.2 g/mol.
8.23 × 10⁻³ mol × 342.2 g/mol = 2.82 g
Answer:
The answer is "5.18
".
Explanation:
It's a question of chemistry. Therefore, the following solution is provided:
We will first determine the solution's pOH. The following can possible:
The concentration of Hydroxide ion ![[OH^{-}] = 1.5\times 10^{-9}\ M](https://tex.z-dn.net/?f=%5BOH%5E%7B-%7D%5D%20%3D%201.5%5Ctimes%2010%5E%7B-9%7D%5C%20M)
![pOH =?\\\\pOH = -\log [OH^{-}]\\\\pOH = -\log 1.5\times 10^{-9}\\\\pOH = 8.82\\\\](https://tex.z-dn.net/?f=pOH%20%3D%3F%5C%5C%5C%5CpOH%20%3D%20-%5Clog%20%5BOH%5E%7B-%7D%5D%5C%5C%5C%5CpOH%20%3D%20-%5Clog%201.5%5Ctimes%2010%5E%7B-9%7D%5C%5C%5C%5CpOH%20%3D%208.82%5C%5C%5C%5C)
Furthermore, the pH of the solution shall be established. It was provided as follows:
When collecting all the like terms:
Therefore, the solution of pH is 5.18.
From the group I A of the periodic table of the element can the chemist infer the element.
If it's to balance the equation then it's already balanced.
