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Calculating solubility! The molar solubility of Ag2CO3 is 1.2 x 10^-4 M at 25 degrees Celsius (room temperature). Express this value in grams per 100.0 ml.
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the solubility is given in \(M\), or \(\frac{mol}{L}\), you need to convert that into a \(different\) unit, \(\frac{g}{100.mL}\)
At the end when I get grams, can I just divide it by 100 mL, would that work?
you actually divide it by 10, because in 1000mL, there are 10 100mL increments
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