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insens350 [35]
2 years ago
14

Determine the number of Au atoms found in 5.18 grams of gold.

Chemistry
1 answer:
Agata [3.3K]2 years ago
3 0

Answer:

1.58× 10²² atoms of Au

Explanation:

Given data:

Mass of gold = 5.18 g

Number of atoms of gold = ?

Solution:

First of all we will calculate the number of moles of gold.

Number of moles = mass/molar mass

Number of moles = 5.18 g/196.96 g/mol

Number of moles = 0.0263 mol

Number of atoms:

Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

1 mole = 6.022 × 10²³ atoms of Au

0.0263 mol × 6.022 × 10²³ atoms of Au / 1mol

0.158× 10²³ atoms of Au

1.58× 10²² atoms of Au

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finlep [7]

Answer:

The total vapor pressure is 84.29 mmHg

Explanation:

Step 1:  Data given

Solution = 40.00 (v/v) % benzene in CCl4

Temperature = 20.00 °C

The vapor pressure of pure benzene at 20.00 °C = 74.61 mmHg

Density of benzene is 0.87865 g/cm3

The vapor pressure of pure carbon tetrachloride is 91.32 mmHg

We suppose the total volume = 100 mL

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40 % benzene = 40 mL

60 % mL CCl4 = 60 mL

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Mass = density * volume

Mass of benzene = 40.00 mL *  0.87865 g/mL = 35.146 g

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Moles = mass / molar mass

Number of moles of benzene  = 35.146 grams / 78 g/mol  = 0.45059 mol

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Mass of CCl4 = 60 mL * 1.5940 g/mL = 95.64 g

Step 6: Calculate moles CCl4

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Total number of moles = moles benzene + moles CCl4

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Mole fraction of benzene = 0.45059 / 1.07163 = 0.4205

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Partial pressure of benzene = 0.4205 * 74.61 = 31.37 mmHg

Partial pressure of CCl4      = 0.5795 * 91.32 = 52.92 mmHg

Total vapor pressure = 31.37 + 52.92 = 84.29 mmHg

The total vapor pressure is 84.29 mmHg

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2 years ago
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heat (Q) = moles x heat of vaporization

= 29.1 kj/mole  x 0.757 moles = 22.03 kj
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