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7 October, 12:15

How many moles of ions are in 285 ml of 0.0150 m mgcl2?

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  1. 7 October, 12:41
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    MgCl₂) = Mg²⁺ + 2Cl⁻

    V (MgCl₂) = 285cm³=0,285dm³

    c (MgCl₂) = 0,015 mol/dm³

    n (MgCl₂) = c·V = 0,015 mol/dm³ · 0,285dm³ = 0,0042 mol

    n (Mg²⁺) = n (MgCl₂) = 0,0042 mol

    n (Cl⁻) = 2n (MgCl₂) = 0,0084 mol
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