Question

The number of aluminum ions present in 0.051 g of aluminum oxide are ​

Answer 1

Answer:

To calculate the number of aluminium ions in 0.051g of aluminium oxide:

1 mole of aluminium oxide = molecules of aluminium oxide

1 mole of aluminium oxide=

$6.022 \times 10^{23} (\mathrm{Al}_{2} \mathrm{O}_{3})$

= 2 x Mass of aluminium + 3 x Mass of Oxygen

= (2x 27) + (3 x 16) = 54 +48 = 102g

1 mole of aluminium oxide =

$102g \: = \: 6.022 \times 10^{23}$

molecules of aluminium oxide

Therefore, 0.051g of aluminium oxide has =

$0.051 \times 6.022 \times 10^{23} / 102$

=

$3.011 \times 10^{20}$

molecules of aluminium oxide.

One molecule of aluminium oxide has 2 aluminium ions, hence the number of aluminium ions present in 0.051g of aluminium oxide =

$2 \times 3.011x 10^{20}$

molecules of aluminium oxide

=

$6.022 \times 10^{20}$

Hope this helps you

Answer 2

Answer:

Calculate the number of aluminium ions present in 0.051g of aluminium oxide. = 3.011 x 10^{20} molecules of aluminium oxide.

Explanation:

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