How many oxygen atoms are present in 4.28 g of LiBroz sample?

0.43×10²³ atoms

Explanation:

Given data:

Mass of LiBrO₂ = 4.28 g

Number of atoms of oxygen = ?

Solution:

Number of moles = mass/molar mass

Number of moles = 4.28 g/ 118.84 g/mol

Number of moles = 0.036 mol

We can see 1 mole of LiBrO₂ contain 2 mole of oxygen atm.

0.036 mol × 2 = 0.072 mol

1 mole contain 6.022×10²³ atoms

0.072 mol × 6.022×10²³ atoms / 1mol

0.43×10²³ atoms

a 20-cm-long, cylindrical graphite (pure carbon) rod is inserted into an oxidizing atmosphere at 1145 k and 1.013 x 105 pa pressure. the oxidizing process is limited by the diffu- sion of oxygen counterflow to the carbon monoxide that is formed on the cylindrical surface.