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Kryger [21]
3 years ago
6

Calculate the number of gold atoms in a sample of gold(III) chloride . Be sure your answer has a unit symbol if necessary, and r

ound it to significant digits.
Chemistry
1 answer:
Fudgin [204]3 years ago
7 0

Answer: N = 2.78 × 10^23 atoms

There are N = 2.78 × 10^23 atoms in 70g of Au2cl6

Completed Question:

Calculate the number of gold atoms in a 70g sample of gold(III) chloride . Be sure your answer has a unit symbol if necessary, and round it to significant digits

Explanation:

Given:

Molar mass of Au2cl6 = 303.33g/mol

Mass of Au2cl6 = 70g

Number of moles of Au2cl6 = 70g/303.33g/mol = 0.231mol

According to the chemical formula of Au2cl6,

1 mole of Au2cl6 contains 2 moles of Au

Number of moles of Au = 2 × 0.231mol = 0.462mole

There are 6.022 × 10^23 atoms in 1 mole of an element.

Number of Atom of gold in 0.462 mole of gold is:

N = 0.462 mol × 6.022 × 10^23 atoms/mol

N = 2.78 × 10^23 atoms

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pav-90 [236]

Answer:

concentration of bromide (Br⁻) = 4234 mg/L = 4234 ppm

Explanation:

ppm (parts per million) concentration is defined as the mass (in milligrams) of a substance dissolved in one liter of solution.

In our case we have:

mass of MgBr₂ = 12.41 g

volume of water (which is equal to the final solution volume) = 2.55 L

Now we devise the following reasoning:

if         12.41 g of MgBr₂ are dissolved in 2.55 L of water

then         X g of MgBr₂ are dissolved in 1 L of water

X = (1 × 12.41) / 2.55 = 4.867 g of MgBr₂

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Y = (4.867 × 160) / 184 = 4.232 g of bromide (Br⁻)

4.232 g of bromide (Br⁻) = 4234 mg of bromide (Br⁻)

concentration of bromide (Br⁻) = 4234 mg/L = 4234 ppm

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