Answer:
44.91% of Oxygen in Iron (III) hydroxide
Explanation:
To solve this question we must find the molar mass of Fe(OH)3 and the molar mass of the oxygen in this molecule. Percent composition will be:
<em>Molar mass Oxygen / molar mass Fe(OH)3 * 100</em>
<em />
<em>Molar mass Fe(OH)3 and oxygen:</em>
1Fe = 55.845g/mol*1 = 55.845
3O = 16.00g/mol*3 = 48.00 - Molar mass of Oxygen
3H = 1.008g/mol*3 = 3.024
55.845 + 48.00 + 3.024 =
106.869g/mol is molar mass of Fe(OH)3
% Composition of oxygen is:
48.00g/mol / 106.869g/mol * 100 =
<h3>44.91% of Oxygen in Iron (III) hydroxide</h3>
The balanced chemical reaction for this would be written as:
2Mg + O2 = 2MgO
We use this reaction and the amount of the reactant given to calculate for the amount of magnesium oxide that is produced. We do as follows:
1.5 g Mg (1 mol / 24.31 g) ( 2 mol MgO / 2 mol Mg ) (40.30 g /1 mol ) = 2.49 g MgO produced
Answer:
The volume of blood plasma in adults is 3.1 L. it's density is 1.03 g/cm3.
Explanation:
