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What is the concentration of cl– ions that results from mixing 0.140 l of 0.25 m nh4cl with 0.380 l of 0.75 m cacl2?

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W0lf93
Answer: mole of Cl- in 0.140 L of 0.25 M NH4Cl = 0.140 * 0.25 mole of Cl- in 0.380 L of 0.75 M CaCl2 = 0.380 * 0.75 Total = 0.140 * 0.25 + 0.380 * 0.75 = 0.32

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