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7 July, 02:55

A 3.682 g sample of kclo3 is dissolved in enough water to give 375. ml of solution. what is the chlorate ion concentration in this solution?

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  1. 7 July, 03:21
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    MKClO₃: 39g+35,5g + (16g*3) = 122,5 g/mol

    1 mol - - 122,5g

    X mol - - 3,682g

    X = 3,682/122,5

    X = 0,03 mol

    375 ml = 0,375 l = 0,375 dm³

    C = n/V

    C = 0,03/0,375

    C = 0,08 mol/dm³

    [ClO₃-] = [KClO₃] = 0,08 mol/dm³
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