Answer: 78.69 g of Fe
Explanation: Iron(III) oxide is Fe2O3. It has a molar mass of 159.7 g/mole. That consists of:
2 x Fe at 55.85 = 11.69 or 69.94%
3 x O at 16.0 = 48.0 or 30.06%
Therefore 112.5 grams will contain (112.5 g)*(69.94%) = 78.69 g of Fe
633.97 L
Explanation:
Well use the combined gas law;
P₁V₁T₁ = P₂V₂T₂
We need to change the temperatures into Kelvin;
18.9°C= 292.05 K
5.9°C = 279.05 K
756 * 512 * 292.05 = 639 * V₂ * 279.05
113,044,377.6 = 178,312.95 V₂
V₂ = 113,044,377.6 / 178,312.95
V₂ = 633.97 L
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