Why only electrons stick to oil drop in millikan's experiment however protons are also present there?
1 answer:
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Answer:
<em><u>≅ 50, 909 nm</u></em>
Explanation:
Wavelength, λ =
Given:
speed of light =
f = 5.89 ×
∴ λ = 3× / 5.89 ×
= 0.0000509 m <em><u>≅ 50, 909 nm</u></em>
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First a balanced reaction equation must be established:
→ 
Now if mass of aluminum = 145 g
the moles of aluminum = (MASS) ÷ (MOLAR MASS) = 145 g ÷ 30 g/mol
= 4.83 mols
Now the mole ratio of Al : O₂ based on the equation is 4 : 3
[4Al + 3 O₂ → 2 Al₂O₃]
∴ if moles of Al = 4.83 moles
then moles of O₂ = (4.83 mol ÷ 4) × 3
= 3.63 mol (to 2 sig. fig.)
Thus it can be concluded that 3.63 moles of oxygen is needed to react completely with 145 g of aluminum.
Now if mass of aluminum = 145 g
the moles of aluminum = (MASS) ÷ (MOLAR MASS) = 145 g ÷ 30 g/mol
= 4.83 mols
Now the mole ratio of Al : O₂ based on the equation is 4 : 3
[4Al + 3 O₂ → 2 Al₂O₃]
∴ if moles of Al = 4.83 moles
then moles of O₂ = (4.83 mol ÷ 4) × 3
= 3.63 mol (to 2 sig. fig.)
Thus it can be concluded that 3.63 moles of oxygen is needed to react completely with 145 g of aluminum.