Find the mass of the empirical formula.
You must be given a sample of some kind to calculate the weight or know how many moles are present. Then you figure out what one mol would be. The key step is multiplying the empirical formula numbers by what it takes to make 1 mol.
It would be clearer if we were working from some choices.
Answer:
The true statements are given below.
Explanation:
1 D glucose is a reducing sugar
2 The oxidation of reducing sugar forms a carboxylic acid sugar.
D glucose is a reducing sugar because glucose contain a free hydroxyl group (-OH)in its anomeric carbon.
The oxidation of reducing sugar result in the conversion of -CHO group in case of aldose sugar and -CH2OH group in case of ketose sugar into carboxylic acid(-COOH).
Answer:
0.0905 M
Explanation:
Let's consider the neutralization reaction between H2SO4 and KOH.
H₂SO₄ + 2 KOH → K₂SO₄ + 2 H₂O
22.87 mL of 0.158 M KOH react. The reacting moles of KOH are:
0.02287 L × 0.158 mol/L = 3.61 × 10⁻³ mol
The molar ratio of H₂SO₄ to KOH is 1:2. The reacting moles of H₂SO₄ are 1/2 × 3.61 × 10⁻³ mol = 1.81 × 10⁻³ mol
1.81 × 10⁻³ moles of H₂SO₄ are in 20.0 mL. The molarity of H₂SO₄ is:
M = 1.81 × 10⁻³ mol / 0.0200 L = 0.0905 M