7. Calculate how many grams of each solute would be required in order to make the given solution. a. 3.40 L of a 0.780 M solution of iron(III) chloride, FeCl3 b. 60.0 mL of a 4.10 M solution of calcium acetate, Ca(CH3COO)2

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Chapter7: Solutions And Colloids
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Problem 7.21E
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b. 61.8 g of NH3 in enough water to make 7.00 L of solution
c. 100. mL of ethanol (C2H5OH) in 500. mL of solution (The density of ethanol is 0.789 g/mL.)
6. How many moles of KF are contained in 180.0 mL of a 0.250 M solution?
7. Calculate how many grams of each solute would be required in order to make the given solution.
a. 3.40 L of a 0.780 M solution of iron(III) chloride, FeCl3
b. 60.0 mL of a 4.10 M solution of calcium acetate, Ca(CH3COO)2
Transcribed Image Text:b. 61.8 g of NH3 in enough water to make 7.00 L of solution c. 100. mL of ethanol (C2H5OH) in 500. mL of solution (The density of ethanol is 0.789 g/mL.) 6. How many moles of KF are contained in 180.0 mL of a 0.250 M solution? 7. Calculate how many grams of each solute would be required in order to make the given solution. a. 3.40 L of a 0.780 M solution of iron(III) chloride, FeCl3 b. 60.0 mL of a 4.10 M solution of calcium acetate, Ca(CH3COO)2
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