Answer is: the percent purity of the sodium bicarbonate is 56.83 %.
1. Chemical reaction: 2NaHCO₃ + H₂SO₄ → 2CO₂ + 2H₂O + Na₂SO₄.
2. m(NaHCO₃) = 3.50 g
n(NaHCO₃) = m(NaHCO₃) ÷ M(NaHCO₃).
n(NaHCO₃) = 3.50 g ÷ 84 g/mol.
n(NaHCO₃) = 0.042 mol.
3. From chemical reaction: n(NaHCO₃) : n(CO₂) = 1 : 1.
n(CO₂) = 0.042 mol.
m(CO₂) = 0.042 mol · 44 g/mol.
m(CO₂) = 1.83 g.
4. the percent purity = 1.04 g/1.83 g ·100%.
the percent purity = 56.8 %.
Explanation:
The bond between C and O in CO₂ and O and H in H₂O
Therefore,
Option C is correct✔
The concentration of the solution reduces and the number of moles of solute isn't affected.
Data;
- V1 = 50mL
- C1 = 12.0M
- V2 = 200mL
- C2 = ?
<h3>Facts about the diluted solution</h3>
1. When the solution is diluted, the concentration changes and this time, the concentration reduces.
Using dilution formula

The concentration of the solution reduces.
2. The number of moles remains the same.
When a solution is diluted, the number of moles remains the same because there's no change in the mass of the solute.
Learn more on concentration of a solution here;
brainly.com/question/2201903
Answer:
It will be denatured.
Explanation:
Salivary amylase works best at pH 6.8.
At pH 2.5, its activity will decrease enormously as it becomes denatured by the high acidity of the stomach contents.