Answer:
23.6 moles
Explanation:
From the question given above, the following data were obtained:
Mass of air = 3.6 Kg
Mass percentage of O₂ = 21%
Number of mole of O₂ =?
Next, we shall convert 3.6 Kg of air to grams (g). This can be obtained as follow:
1 kg = 1000 g
Therefore,
3.6 Kg = 3.6 Kg × 1000 / 1 kg
3.6 Kg = 3600 g
Next, we shall determine the mass of O₂ in the air. This can be obtained as follow:
Mass of air = 3600 g
Mass percentage of O₂ = 21%
Mass of O₂ =?
Mass of O₂ = 21% × 3600
Mass of O₂ = 21/100 × 3600
Mass of O₂ = 756 g
Finally, we shall determine the number of mole of O₂ in the sample of air. This can be obtained as follow:
Mass of O₂ = 756 g
Molar mass of O₂ = 2 × 16 = 32 g/mol
Number of mole of O₂ =?
Mole = mass /Molar mass
Number of mole of O₂ = 756 / 32
Number of mole of O₂ = 23.6 moles
Thus, the number of mole of O₂ in the
sample of air is 23.6 moles
It becomes an anion because it has a negative charge.
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Answer:
The balanced reaction is given by,
⇒ 
Explanation:
The reaction is as given.
Lets count the number of each elements in the reaction.
<em>In reactant side, number of sodium atoms are 1 , lead are 1, nitrogen are 1 and oxygen are 4.</em>
<em>in product side, number of sodium atoms are 2 , lead are 1 , nitrogen are 2 and oxygen are 7.</em>
<em>So we need to balance sodium and oxygen atoms in the reaction.</em>
<em>There is deficient of sodium and oxygen atoms on reactant side</em>.
Thus, multiply (NaNO3) by 2.
<em>Thus, sodium atoms become 2 , nitrogen 2 and oxygen 6. Total 7 oxygen atoms.</em>
Thus, the balanced reaction is,
⇒
