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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The question above can be solved by using this equation:
CAVA =CBVB
Where:
CA =Concentration of acid = 1.0 M
VA = Volume of acid = ?
CB = Concentration of base = 1.0 M
VB = Volume of base = 25 ml
VA = CBVB / CA
VA = [1 * 25] / 1 = 25 / 1 = 25
VA = 25 ml
Therefore, the volume of acid that is required to completely neutralize the base is 25 ml.<span />
80 protons,121 Neutrons, and 80 electrons the element is Mercury
Answer:
One electron is the answer if you need work i can show
