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6 January, 05:41

Q 3.6: calculate the number of po43 - ions present in 23.7 g of ca3 (po4) 2.

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  1. 6 January, 06:50
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    Moles of ca3 (po4) 2 = 23.7 / 310.17 = 0.076

    moles of (PO4) 3 - = 0.076 x 2 = 0.152

    now, no. of ions = 0.152 x 6.022 x 10^{23}

    = 9.2 x 10^{22}
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