To be honest, I can’t really see the question. So please next time just type it out lol
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 %.
Answer:
Carboxylic acid
A carboxylic acid is an organic acid that contains a carboxyl group attached to an R-group. The general formula of a carboxylic acid is R−COOH or R−CO₂H, with R referring to the alkyl, alkenyl, aryl, or other group. Carboxylic acids occur widely. Important examples include the amino acids and fatty acids.
There are approximately 4 resonance structures that can be drawn for N205 (no N-N bond) (minimal formal charge).
Answer:
0.286 moles
Explanation:
i hope this is helpful for you