show the calculations that demonstrate that dissolving 0.2810 g Fe(NH4)2(SO4)2•6

Question

show the calculations that demonstrate that dissolving 0.2810 g Fe(NH4)2(SO4)2•6H2O in 1.000 L of solution results in a solution that is 40 ppm Fe and calculate the molar concentration of Fe show the calculations that demonstrate that dissolving 0.2810 g Fe(NH4)2(SO4)2•6H2O in 1.000 L of solution results in a solution that is 40 ppm Fe and calculate the molar concentration of Fe show the calculations that demonstrate that dissolving 0.2810 g Fe(NH4)2(SO4)2•6H2O in 1.000 L of solution results in a solution that is 40 ppm Fe and calculate the molar concentration of Fe

Explanation / Answer

40 ppm means 40mg of Fe+2 present in 1L solution

0.040 g Fe+2 present in 1 L solution

0.04 /55.845 = 0.00072 moles present in 1 L solution

molarity = 0.00072

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