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anonymous

  • 5 years ago

If a sample of Al2(SO4)3 is found to contain 5.57 · 10-3 mol of Al3+ ions how many moles of Al2(SO4)3 formula units are present? Express your answer in scientific notation.

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  1. anonymous
    • 5 years ago
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    You know that one molecule of aluminium sulfate can produce two cations of Al3+, so\[2 Al^{3+} \iff Al_2(SO_4)_3\]which means one Al3+ is equivalent to 1/2 of one aluminium sulfate:\[Al^{3+} \iff \frac{1}{2}Al_2(SO_4)_3\]You have 5.55 x 10^-3 mol of Al3+, so you have\[5.57 \times 10^{-3}mol Al^{3+} \iff \frac{1}{2}5.57 \times 10^{-3}molAl_2(SO_4)_3\]that is, you have\[2.785 \times 10^{-3}molAl_2(SO_4)_3\]

  2. anonymous
    • 5 years ago
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    To keep to the number of significant figures, the number of moles of aluminium sulfate is \[2.79 \times 10^{-3}mol\]

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