# chemistry

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Calculate the grams of K2CO3 required to prepare 546.0 mL of a 0.789 g/L K+ solution.

• chemistry -

moles K^+ needed = M x L = ?
Convert moles K ions to mol K2CO3.
moles K ion x (1 mol K2CO3/2 mol K ion) = ?
Then moles K2CO3 = grams/molar mass.

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