Wednesday
March 29, 2017

Post a New Question

Posted by on .

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.

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question