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containing 400-mg of Al(OH)3

*Al= 29.98 amu, O= 16.00 amu, and H= 1.008 amu

your help with step by step detail please, thank you.

2007-03-12 15:02:59 · 1 answers · asked by xxxladyxxx 1 in Science & Mathematics Chemistry

**Correction Al= 26.98 amu

2007-03-13 10:59:21 · update #1

1 answers

You've got the Ar of Al wrong. Check this.

Add up: 1 x Al, 3 x O and 3 x H.

Divide 0.400 by this sum for your answer.

2007-03-12 20:31:43 · answer #1 · answered by Gervald F 7 · 0 0

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