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 %.
The stoichiometric ratio of CuCl2 to NaCl is 1 is to 2. The stoichiometric ratio of 31.0 g CuCl2 is 26.95 grams of NaCl by converting the amount of CuCl2 to mole and multiplying by 0.5 and molar mass of NaCl.This amount is equal to 78.65% yield.
I think <span>500 m but i just think i dont know </span>
<u>Answer:</u> The molality of magnesium chloride is 1.58 m
<u>Explanation:</u>
To calculate the molality of solution, we use the equation:

Where,
= Given mass of solute (magnesium chloride) = 75.0
= Molar mass of solute (magnesium chloride) = 95.21 g/mol
= Mass of solvent = 500.0 g
Putting values in above equation, we get:

Hence, the molality of magnesium chloride is 1.58 m
Answer: Fluorine
Explanation: It belongs in the same group as Bromine