Question

How many oxygen atoms are present in 2.2 mg of carbon dioxide

Answer 1

1 g = 1000 mg

=> 2.2 mg = 2.2 / 1000 = 0.0022 g or 2.2 * 10⁻³ g

I mole = 44 g of CO₂

=> X moles = 2.2 * 10⁻³ g of CO₂

=> x = 2.2 * 10⁻³ / 44

=> x = 0.05 * 10⁻³ moles

1 mole of CO₂ has 2 moles of O₂

=> In 0.05 * 10⁻³ moles we have = 0.1 moles of O₂

1 mole of O₂ = 6.022 * 10²³ atoms

=> 0.1 * 10⁻³ moles of O₂ = 0.1 * 6.022 * 10²³ * 10⁻³

=> Number of atoms = 6.022 * 10¹⁹ atoms.

Hope it helped !

Answer 2

Answer:Oxigen atoms are present in 3.2 mg of carbon dioxide are 0.10×6.022×10^23= 0.6022×10^23.

Explanation:1g=1000mg. 2.2mg=. 2.2/1000=0.0022gm or day 2.2×10^-3gm. We know that, 1 mole=44g of co2. So, X moles =2.2×10^-3g of co2. X=2.2 ×10^-3/44 X=0.05×10^-3 moles. From above we have seen that x=0.05×10^-3 moles then now 1 mole of co2 have 2 moles of a O2 in 0.05 ×10^-3 moles of O2 = 0.1 ×6.022×10^23

oxygen atoms are present in 2.2 mg of carbon dioxide is 0.1×6.022×10^23.