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A soluble iodide was dissolved in water. Then an excess of silver(I) nitrate, AgNO3 was added to completely precipitate out all the iodide ion as silver(I) iodide, AgI. if 1.545 grams of soluble iodide gave 2.185g of silver(I) iodide, then

a) how many moles of I(I-) are in the original iodide sample

b) how many grams of I(I-) are in the original iodide sample

c) what is the percentage by mass of I(I-) in the original soulble iodide sample

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