Answer:
There are three significant figures
Explanation:
When counting sig figs you don't count the zeros unless it is between a number greater than zero. The two zeros aren't between the greater numbers so there are only 3.
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>
When 440.23 grams of iron(III) oxide are reacted with hydrogen gas, the amount of iron produced will be 307.66 grams
<h3>Stoichiometric calculation</h3>
From the equation of the reaction:
The mole ratio of iron(III) oxide to produced iron is 1:2.
Mole of 440.23 iron(III) oxide = 440.23/159.69 = 2.76 moles
Equivalent mole of produced iron = 2.76 x 2 = 5.52 moles
Mass of 5.52 moles of iron = 5.52 x 55.8 = 307.66 grams
More on stoichiometric calculations can be found here; brainly.com/question/27287858
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Answer:
MnO4⁻ (aq) + 8H⁺ (aq) + 5Fe³⁺ (aq) →Mn(aq)²⁺ + 4H2O (l) + 5Fe²⁺(aq)
Explanation:
a)
MnO4⁻ (aq) + 8H⁺ (aq) + 5e⁻ → Mn(aq)²⁺ + 4H2O (l)
b)
5Fe³⁺ (aq) +5e⁻ → 5Fe²⁺(aq)
c)
MnO4⁻ (aq) + 8H⁺ (aq) + 5Fe³⁺ (aq) →Mn(aq)²⁺ + 4H2O (l) + 5Fe²⁺(aq)
