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Katarina [22]
3 years ago
9

If the compound contains 35.06 % cl by mass, what is the identity of the metal?

Chemistry
1 answer:
Setler [38]3 years ago
4 0

Answer is: identity of the metal is gold (Au).

ω(Cl) = 35.06% ÷ 100%.

ω(Cl) = 0.3506; mass percentage of chlorine.

If we take 100 grams of the compound:

m(Cl) = ω(Cl) · m(compound).

ω(Cl) = 0.3506 · 100 g.

ω(Cl) = 35.06 g.

n(Cl) = m(Cl) ÷ M(Cl).

n(Cl) = 35.06 g ÷ 35.45 g/mol.

n(Cl) = 0.99 mol; amount of substance.

In molecule MCl₃: n(M) : n(Cl) = 1 : 3.

n(M) = 0.33 mol; amount of unknown metal.

M(M) = m(M) ÷ n(M).

M(M) = (100 g - 35.06 g) ÷ 0.33 mol.

M(M) = 196.8 g/mol; molar mass of the gold.

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where P_{1} and P_{2} are initial and final pressure respectively.

           V_{1}  and V_{2} are initial and final volume respectively.

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Here P_{1}=150.0kPa , V_{1}=1.75L , T_{1}=(273-23)K=250K, P_{2}=210.0kPa and V_{2}=1.30L

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