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:
Yes it would
Explanation: Well it kinda depend on the voltage and how the battery has been in use or based on the condition
I honestly have no clue on this one
Answer:
KI
Explanation:
From the question, we can see that a qualitative analysis of the compound shows that it has a lilac flame colour. The lilac flame colour corresponds to the potassium ion (K^+).
Again, the test of addition of HNO3(aq) and AgNO3(aq) to a solution is a test for halogens. If the result is a green precipitate, then the ion present is the iodide ion (I^-).
Hence, the compound must be KI.