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18 March, 13:34

Analysis of a compound indicates that it contains 1.04 grams K 0.70 g Cr and 0.86 g O. Find its empirical formula

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  1. 18 March, 13:43
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    1.04gK*1molK/39.01g K = 0.0267 mol K

    0.70gCr*1mol/52.0g Cr = 0.0135 mol Cr

    0.86 gO * 1 mol/16.0 g O = 0.0538 mol O

    0.0267 mol K/0.0135 = 2 mol K

    0.0135 mol Cr / 0.0135 = 1 mol Cr

    0.0538 mol O/0.035 = 4 mol Cr

    K2CrO4
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