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djyliett [7]
2 years ago
9

Find the percent composition of OXYGEN in Manganese (III) nitrate, Mn(NO3)3.

Chemistry
1 answer:
BaLLatris [955]2 years ago
8 0

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 = \frac{Mr\ of\ oxygen\ in\ compound}{Mr\ of\ compound} × 100

% composition = \frac{144}{240.9} × 100 = <u>59.776%</u>

∴ % compostion of oxygen in maganese(III)nitrate is 59.8% (to 3 significant figures).

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Final temperature of water and aluminium = 30.0 °C

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⇒ mass of aluminium = 73.6 grams

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⇒ ΔT(aluminium) = The change of temperature = T2 - T1 = 30 .0 °C - 95.0 °C = -65.0°C

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