What is the concentration of each ion present when 25.0 mL of 0.73 M LiPO4 is mixed with 30.0 mL of 0.62 M NaPO4.
Added by Miguel B.
Step 1
- For LiPO4: Moles = Molarity × Volume = 0.73 M × 25.0 mL = 0.73 mol/L × 0.025 L = 0.01825 mol - For NaPO4: Moles = Molarity × Volume = 0.62 M × 30.0 mL = 0.62 mol/L × 0.030 L = 0.0186 mol Show more…
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