Calculate the number of aluminium ions (Al
3+
) in 0.056g of aluminium oxide
(Al2O3
). 3
Answers
Answer:
23 because the corrosion will be underestimation of the purpose
Explanation:
answer: 66.2\times 10^{19}66.2×10
19
ions of Al^{3+}Al
3+
Moles of Al_2O_3=\frac{\text{ given mass}}{\text{ molar mass}}Al
2
O
3
=
molar mass
given mass
Moles of Al_2O_3=\frac{0.056g}{102g/mol}=5.5\times 10^{-4}molesAl
2
O
3
=
102g/mol
0.056g
=5.5×10
−4
moles
Al_2O_3\rightarrow 2Al^{3+}+3O^{2-}Al
2
O
3
→2Al
3+
+3O
2−
thus 1 mole of Al_2O_3Al
2
O
3
contains =2\times 6.023\times 10^{23}=12.046\times 10^{23}2×6.023×10
23
=12.046×10
23
ions of Al^{3+}Al
3+
5.5\times 10^{-4}moles5.5×10
−4
moles of Al_2O_3Al
2
O
3
contains =\frac{12.046\times 10^{23}}{1}\times 5.5\times 10^{-4}=66.2\times 10^{19}
1
12.046×10
23
×5.5×10
−4
=66.2×10
19
ions of Al^{3+}Al
3+