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Ann [662]
2 years ago
15

A 36.41-g sample of calcium carbonate (CaC O 3 )

Chemistry
1 answer:
Lunna [17]2 years ago
7 0
M(O) = m(CaCO₃) - m(Ca) - m(C).
m(O) = 36.41 g - 14.58 g - 4.36 g.
m(O) = 17.47 g.
ω(Ca) = m(Ca) ÷ m(CaCO₃) · 100%.
ω(Ca) = 14.58 g ÷ 36.41 g · 100%.
ω(Ca) = 40 %; mass percent of calcium.
ω(C) = m(C) ÷ m(CaCO₃) · 100%.
ω(C) = 4.36 g ÷ 36.41 g · 100%
ω(C) = 12%; mass percent of carbon.
ω(O) = 100% - ω(Ca) - ω(C).
ω(O) = 100% - 40% - 12%.
ω(O) = 48%; mass percent of oxygen.
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Answer:

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Explanation:

Tartaric acid is a dyprotic acid. It reacts to water like this:

H₂Tart  +  H₂O  ⇄  H₃O⁺   +  HTart⁻         Ka1

HTart⁻  +  H₂O  ⇄  H₃O⁺   +  Tart⁻²           Ka2

When we anaylse the base, we have

Tart⁻²   +  H₂O  ⇄  OH⁻  +  HTart⁻       Kb1

HTart⁻  +  H₂O  ⇄  OH⁻  +    H₂Tart          Kb2

Remember that Ka1 . Kb2  = Kw, plus pKa1 + pKb2 = 14

Kb2 = Kw / Ka1  →   1×10⁻¹⁴ / 9.20×10⁻⁴  = 1.08×10⁻¹¹

so pKb = - log Kb2   → - log 1.08×10⁻¹¹ = 10.96  

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