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babunello [35]
4 years ago
11

Mercury has a density of 13.56 g/ml. What is the volume in L of 52 kg of mercury?

Chemistry
1 answer:
sweet [91]4 years ago
5 0

Hello!

Given the density of mercury being 13.57 g/mL, and the mass of 52 kilograms, we need to find the volume.

To find the volume, we need to divide mass by density (V = m/d).

Notice that you are given 52 kilograms, but not grams. To convert kilograms to grams, you need to multiply it by 1000.

52 x 1000 = 52000 grams

With the correct measurements, we can find the volume.

V = 52000 grams / 13.56 grams/milliliter

V ≈ 3834.80826

Therefore, the volume of the mercury is about 3,834.81 mL.

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Mass of SO₄⁻² = 0.123 g.

Mass percentage of SO₄⁻² = 41.2%

Mass of Na₂SO₄ = 0.0773 g

Mass of K₂SO₄ = 0.1277 g

Explanation:

Here we have

We place Na₂SO₄ = X and

K₂SO₄ = Y

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Therefore since the BaSO₄ is formed from BaCl₂, Na₂SO₄ and K₂SO₄ we have

Amount of BaSO₄ from Na₂SO₄ is therefore;

X\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, Na_2SO_4}

Amount of BaSO₄ from K₂SO₄ is;

Y\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, K_2SO_4}

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Na₂SO₄ = 142.04 g/mol

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X\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, Na_2SO_4} = X\times\frac{233.38 }{142.04} = 1.643·X

Y\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, K_2SO_4} = Y\times\frac{233.38 }{174.259 } = 1.339·Y

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1.643·X + 1.339·Y = 0.298 g.....(2)

Solving equations (1) and (2) gives

The mass of SO₄⁻² in the sample is given by

Mass of sample = 0.298

Molar mass of BaSO₄ = 233.38 g/mol

Mass of Ba = 137.327 g/mol

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Mass fraction of SO₄⁻² in BaSO₄ = 96.05 g/233.38 g = 0.412

Mass of SO₄⁻² in the sample is 0.412×0.298 = 0.123 g.

Percentage mass of SO₄⁻² = 41.2%

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