odugerry odugerry

03-02-2019
Chemistry

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

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Аноним Аноним

03-02-2019

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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