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8 January, 01:45

If 10. g of AgNo3 is available, what volume of 0.25 M AgNo3 can be prepared

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  1. 8 January, 02:02
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    The equation is L = m/M

    First, covert 10. grams of AgNO3 to moles which is 0.059 moles.

    Divide 0.059 moles by 0.25M which is 0.24 liters.
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