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1 answer:
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Answer:
0.286 moles
Explanation:
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Answer:
59.8%
Explanation:
First find the Mr of manganese (III) nitrate.
Mr of Mn(NO₃)₃ = 54.9 + (14 × 3) + (16 × 3 × 3) = <u>240.9</u>
Since we have to find the percentage composition of oxygen, we need to find the Mr of oxygen in the compound, which is:
Mr of (O₃)₃ = (16 × 3) × 3 = <u>144</u>
Now we can find percentage composition / percentage by mass of oxygen.
% composition = × 100
% composition = × 100 = <u>59.776%</u>
∴ % compostion of oxygen in maganese(III)nitrate is 59.8% (to 3 significant figures).