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Given that carbonates are insoluble (with the exce
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Answer #1

1. Reaction,

2AgNO3 + Na2CO3 ---> Ag2CO3 + 2NaNO3

2. balanced equation,

6HF + 3NaAlO2 = Na3AlF6 + 3H2O + Al2O3

moles of Na3AlF6 produced = 12 g/209.94 g/mol = 0.057 mols

3 moles of NaAlO2 gives 1 mole of Na3AlF6

So mass of NaAlO2 required = 3 x 0.057 mol x 81.97 g/mol = 14.017 g

3. Oxidation states in,

(a) CdS

Cd = +2

S = -2

(b) AlH3

Al = +3

H = -1

(c) S2O3^2-

S = +2

O = -2

4. moles of H2O = 1.61 g/18 g/mol = 0.09 mol

moles of H = 2 x 0.09 mol = 0.18 mol

mass of H = 0.18 x 1 = 0.18 g

moles of CO2 = 2.818 g/44 g/mol = 0.064 mol

moles of C = 0.064 mol

mass of C = 0.06 x 12 = 0.77 g

mass of O = 1 - 0.77 = 0.23 g

mols of O = 0.23 g/16 g = 0.014 mol

Divide by smallest factor,

C = 0.06/0.014 = 4

H = 0.18/0.014 = 12

O = 0.014/0.014 =1

Empirical formula = C4H13O

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