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Evgesh-ka [11]
3 years ago
14

A 1.45 g sample of phosphorus burns in air and forms 2.57 g of a phosphorus oxide. Calculate the empirical formula of the oxide.

Express your answer as a chemical formula.
Chemistry
1 answer:
bija089 [108]3 years ago
5 0

Answer:

P2O3

Explanation:

Firstly, we know that the phosphorus would absorb oxygen from air to form its oxide. We already know the mass of the phosphorus, we can get the mass of the oxygen by subtracting the mass of the phosphorus from the mass of the oxide.

The mass of the oxygen is 2.57 - 1.45 = 1.12g

From here, we covert these masses to moles by dividing by the atomic masses of phosphorus and oxygen respectively.

The atomic mass of phosphorus is 31 while that of oxygen is 16.

P = 1.45/31 = 0.046774193548

O = 1.12/16 = 0.07

We now divide by the smallest which is that of phosphorus.

P = 0.046774193548/ 0.046774193548= 1

O = 0.07/ 0.046774193548 = 1.5

We then multiply the answers by 2 for conversion to whole numbers. Making P = 2 and O = 3

The empirical formula is thus P2O3

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Answer:

We can not swim in 1.00 × 10²⁷ molecules of water

Explanation:

The given number of molecules of water = 1.00 × 10²⁷ molecules

The Avogadro's number, N_A, gives the number of molecules in one mole of a substance

N_A ≈ 6.0221409 × 10²³ molecules/mol

Therefore

Therefore, we have;

The number of moles of water present in 1.00 × 10²⁷ molecules, n = (The number of molecules of water) ÷ N_A

∴ n = (1.00 × 10²⁷ molecules)/(6.0221409 × 10²³ molecules/mol) = 1,660.53902857 moles

The mass of one mole of water = The molar mass of water = 18.01528 g/mol

The mass, 'm', of water in 1,660.53902857 moles of water is given as follows;

Mass = (The number of moles of the substance) × (The molar mass of the substance)

∴ The mass of the water in the given quantity of water, m = 1,660.53902857 moles × 18.01528 g/mol ≈ 29.9150756 kg.

The density pf water, ρ = 997 kg/m³

Volume = Mass/Density

∴ The volume of the water present in the given quantity of water, v = 29.9150756 kg/(997 kg/m³) ≈ 30.0050909 liters

The volume of the water present in 1.00 × 10²⁷ molecules of water ≈ 30.0 liters

The average volume of a human body = 62 liters

Therefore, we can not swim in the given quantity of 1.00 × 10²⁷ molecules = 30.0 liters water

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