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What is the ratio of Al3+ ions to S2- ions in a neutral compound? A. 1:1 B 2:1 C.2:3 D. 3:2
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Let the number of \(Al^{3+}\) be x and let the number of \(S^{2-}\) ions be y. Then their charges times their number of atoms must be 0 \((+3)x+(-2)y=0\) To avoid math, you can just "cross" their charges to be their coefficients, let me illustrate: |dw:1409942355934:dw| so you end up with \(Al_2S_3\) This makes sense because 2 \(Al^{3+}\) is 6+ and 3 \(S^{2-}\) are -6 This adds up to 0
@aaronq so it would be A. Correct?
no, the ratio is the number of atoms - i.e. their coefficients
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