calculate the number of al ions present in 0.051 of Al2o3
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Answer:
MARK AS BRAINLIEST
Explanation:
Answer :
1 mole of Aluminium Oxide (Al2O3) = 2 × 27 + 3 × 16 = 102 g. ...
We know, The number of atoms present in given mass = (Given mass ÷ molar mass) × Avogadro number.
Then, 0.051 g of Aluminium Oxide (Al2O3) contains.
= (0.051 ÷ 102 ) × 6.022 × 1023 molecules of Aluminium Oxide (Al2O3)
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