The sample of smoke described above can be described as a heterogeneous mixture. This type of mixture do not have uniform properties and composition. So, getting a certain small sample would not represent the whole mixture since it does not have uniform composition.
The amount in grams of Al₂O₃ produced is approximately 6.80 g.
Aluminium reacts completely with oxygen(air) to produce Al₂O₃. The reaction can be represented with a chemical equation as follows:
AL + O₂ → Al₂O₃
Let's balance it
4AL + 3O₂ → 2Al₂O₃
4 moles of Aluminium reacts with 3 moles of Oxygen molecules to produce 2 moles of Aluminium oxide. Therefore,
Since, aluminium reacts completely, it is the limiting reagent in the reaction. Therefore,
Atomic mass of AL = 27 g
Molar mass of Al₂O₃ = 101.96 g/mol
4(27 g) of AL gives 2(101.96 g) of Al₂O₃
3.6 g of AL will give ?
cross multiply
mass of Al₂O₃ produced = 3.6 × 203.92 / 108 = 734.112 / 108 = 6.797
mass of Al₂O₃ produced = 6.80 g.
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Answer:
2.64 ×
J
Explanation:
I think you should mark it a physics question instead but anyway.
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The Planck equation should be applied:
E = hv , while E is energy of proton; h is Planck constant; and v is frequency.
E = 6.6 × 
× 4 × 
= 6.6 × 4 × 
= 2.64 × J
