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gregori [183]
3 years ago
9

1. A sample of gold (Au) has a mass of 35.12 g.

Chemistry
1 answer:
Alecsey [184]3 years ago
4 0

1a) Answer is: the number of moles of gold (Au) in the sample is 0.178 mol.

m(Au) = 35.12 g; mass of gold.

M(Au) = 196.97 g/mol; molar mass of gold.

n(Au) = m(Au) ÷ M(Au).

n(Au) = 35.12 g ÷ 196.97 g/mol.

n(Au) = 0.178 mol; amount of gold.

1b) Answer is: the number of atoms of gold (Au) is 1.073·10²³.

N(Au) = n(Au) · Na.

N(Au) = 0.178 mol · 6.022·10²³ 1/mol.

N(Au) = 1.073·10²³.

2a) Answer is: 0.0035 moles of sucrose.

m(C₁₂H₂₂O₁₁) = 1.202 g; mass of sucrose.

M(C₁₂H₂₂O₁₁) = 12 · Ar(C) + 22 · Ar(H) + 11 · Ar(O) · g/mol.

M(C₁₂H₂₂O₁₁) = 12 · 12.01 + 22 · 1.01 + 11 · 16 · g/mol.

M(C₁₂H₂₂O₁₁) = 342.3 g/mol; molar mass of sucrose.

n(C₁₂H₂₂O₁₁) = m(C₁₂H₂₂O₁₁) ÷ M(C₁₂H₂₂O₁₁).

n(C₁₂H₂₂O₁₁) = 1.202 g ÷ 342.3 g/mol.

n(C₁₂H₂₂O₁₁) = 0.0035 mol; amount of sucrose.

2b) n(C) = 12·n(C₁₂H₂₂O₁₁).

n(C) = 12 · 0.0035 mol.

n(C) = 0.042 mol; amount of carbon in sucrose.

n(H) = 22·n(C₁₂H₂₂O₁₁).

n(H) = 22 · 0.0035 mol.

n(H) = 0.077 mol; amount of hydrogen in sucrose.

n(O) = 11·n(C₁₂H₂₂O₁₁).

n(O) = 11 · 0.0035 mol.

n(O) = 0.0385 mol; amount of oxygen atoms in sucrose.

2c) N(C) = n(C) · Na.

N(C) = 0.042 mol · 6.022·10²³ 1/mol.

N(C) = 2.53·10²²; number of carbon atoms in sucrose.

N(H) = n(H) · Na.

N(H) = 0.077 mol · 6.022·10²³ 1/mol.

N(H) = 4.63·10²²; number of hydrogen atoms in sucrose.

N(O) = n(O) · Na.

N(O) = 0.0385 mol · 6.022·10²³ 1/mol.

N(O) = 2.31·10²²; number of oxygen atoms in sucrose.

Na is Avogadro constant.

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