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Answers
Explanation:
To calculate the number of aluminium ions in 0.051g of aluminium oxide:
1 mole of aluminium oxide = 6.022 x 10²³ molecules of aluminium oxide
1 mole of aluminium oxide (Al2O3) = 2 x Mass of aluminium + 3 x Mass of Oxygen
= (2x 27) + (3 x16) = 54 +48 = 102g
1 mole of aluminium oxide = 102g = 6.022 x 10²³ molecules of aluminium oxide
Therefore, 0.051g of aluminium oxide has = 0.051 x 6.022 x 10²³ / 102
= 3.011 x 10²⁰ molecules of aluminium oxide
One molecule of aluminium oxide has 2 aluminium ions, hence number of aluminium ions present in 0.051g of aluminium oxide = 2 x 3.011x 10²⁰molecules of aluminium oxide
= 6.022 x 10²⁰
It is non terminating and repeating decimal. for a no. To end with zero it should be in the form of2^n and 5^m but in this digit 3 ...