CHEMISTRY COLLEGE

You are given a crushed sample that is a mixture of limestone (caco3), lime (cao), and sand. the calcium carbonate, limestone, is the only material present in the material that will decompose when heated. you subject a 6.4734 g sample of the mixture to strong heating and after the sample reaches a constant mass (no more mass is lost with additional heating), the sample has a final weight of 4.3385 g. what is the percentage of calcium carbonate present in the original mixture? (f.wt. caco3 = 100.1)

Answers

Answer 1
Answer: Answer is: percentage of calcium carbonate is 75%.
Chemical reaction: CaCO₃ → CaO + CO₂.
m₁(sample) = 6,4734 g.
m₂(sample) = 4,3385 g.
m(CO₂) = m₁(sample) - m₂(sample).
m(CO₂) = 6,4734 g - 4,3385 g.
m(CO₂) = 2,1349 g.
n(CO₂) = m(CO₂) ÷ M(CO₂).
n(CO₂) = 2,1349 g ÷ 44 g/mol.
n(CO₂) = 0,0485 mol.
From chemical reaction: n(CO₂) : n(CaCO₃) = 1 : 1.
n(CaCO₃) = 0,0485 mol.
m(CaCO₃) = 0,0485 mol · 100,1 g/mol.
m(CaCO₃) = 4,854 g.
percentage of calcium carbonate = 4,854 g ÷ 6,4734 g · 100%.
percentage of calcium carbonate = 75%.

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d. a substance in a solution that changes color in a narrow range of ph?

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             Option-C (A substance in a solution that releases and captures hydrogen ions, keeping the pH the same) is the correct answer.

Explanation:
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Answer: A substance in a solution that releases and captures hydrogen ions, keeping the pH the same

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Express your answer in terms of k and the necessary concentrations (e.g., k*[A]^3*[D]).

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Step1:Step2:X2Y2+Z2→X2Y2Z+Z (slow)X2Y2+Z→X2Y2Z (fast)
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Answers

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A)

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put eqn 2 in eqn 1

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= k'Kc*[A][B]^2

= K *[A][B]^2 {writing k'Kc = K}

Answer: K *[A][B]^2

B)

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