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AysviL [449]
2 years ago
13

A pure gold ring weighs 23.5 grams. How many atoms of gold are in the ring?

Chemistry
1 answer:
Digiron [165]2 years ago
3 0

The number of atoms of gold in the pure ring are 7.18 × 10²² atoms.

<h3>HOW TO CALCULATE NUMBER OF ATOMS?</h3>

The number of atoms in a substance can be calculated by multiplying the number of moles of the substance by Avogadro's number.

The number of moles in the gold (Au) can be calculated by dividing the mass of gold by its molar mass (196.97g/mol).

no. of moles = 23.5g ÷ 196.97g/mol

no. of moles = 0.119mol

Number of atoms in Au = 0.119 × 6.02 × 10²³

no. of atoms = 7.18 × 10²² atoms.

Therefore, the number of atoms of gold in the pure ring are 7.18 × 10²² atoms.

Learn more about number of atoms at: brainly.com/question/15959704

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If 22.5L of nitrogen at 0.98 atm is compressed to 0.95 atm,what is the new volume?
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At constant temperature, if the pressure is compressed to the given value, the volume of the nitrogen gas increases to 23.2L.

<h3>What is Boyle's law?</h3>

Boyle's law simply states that "the volume of any given quantity of gas is inversely proportional to its pressure as long as temperature remains constant.

Boyle's law is expressed as;

P₁V₁ = P₂V₂

Where P₁ is Initial Pressure, V₁ is Initial volume, P₂ is Final Pressure and V₂ is Final volume.

Given that;

  • Initial volume of the gas V₁ = 22.5L
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  • Final pressure of the gas P₂ = 0.95atm
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P₁V₁ = P₂V₂

V₂ = P₁V₁ / P₂

V₂ = (0.98atm × 22.5L) / 0.95atm

V₂ = 22.05Latm / 0.95atm

V₂ = 23.2L

Therefore, at constant temperature, if the pressure is compressed to the given value, the volume of the nitrogen gas increases to 23.2L.

Learn more about Boyle's law here: brainly.com/question/1437490

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