AgBr precipitate (FM 5 187.77)? (Answer: 0.425 5 g)
- How Many Grams Of Br2 Were In A Sample That Produced 1 000 G Of Agbr Precipitate Fm 5 187 77 Answer 0 425 5 G 1 (94.27 KiB) Viewed 54 times
How many grams of Br2 were in a sample that produced 1.000 g of AgBr precipitate (FM 5 187.77)? (Answer: 0.425 5 g)
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How many grams of Br2 were in a sample that produced 1.000 g of AgBr precipitate (FM 5 187.77)? (Answer: 0.425 5 g)
How many grams of Br2 were in a sample that produced 1.000 g of
AgBr precipitate (FM 5 187.77)? (Answer: 0.425 5 g)
EXAMPLE A Gravimetric Calculation A 10.00-ml solution containing CI was treated with excess AgNO3 to precipitate 0.436 8 g of AgCl. What was the molarity of Cl in the unknown? Solution The formula mass of AgCl is 143.321. Precipitate weighing 0.436 8 g contains 0.436 8 g Age 143.321 g AgCl/mol AgCl 3.048 x 10-3 mol AgCI Because 1 mol of AgCl contains 1 mol of CI", there must have been 3.048 x 10-3 mol of cl in the unknown. [CI] 3.048 x 10-3 mol 0.010 00 L 0.304 8 M TEST YOURSELF How many grams of Br" were in a sample that produced 1.000 g of AgBr precipitate (FM = 187.77)? (Answer: 0.425 5 g)
AgBr precipitate (FM 5 187.77)? (Answer: 0.425 5 g)
EXAMPLE A Gravimetric Calculation A 10.00-ml solution containing CI was treated with excess AgNO3 to precipitate 0.436 8 g of AgCl. What was the molarity of Cl in the unknown? Solution The formula mass of AgCl is 143.321. Precipitate weighing 0.436 8 g contains 0.436 8 g Age 143.321 g AgCl/mol AgCl 3.048 x 10-3 mol AgCI Because 1 mol of AgCl contains 1 mol of CI", there must have been 3.048 x 10-3 mol of cl in the unknown. [CI] 3.048 x 10-3 mol 0.010 00 L 0.304 8 M TEST YOURSELF How many grams of Br" were in a sample that produced 1.000 g of AgBr precipitate (FM = 187.77)? (Answer: 0.425 5 g)
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