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aleksandrvk [35]
3 years ago
10

how many grams of solid silver nitrate would you need to prepare 200.0 mL of a 0.150 M AgNO3 solution

Chemistry
1 answer:
aleksklad [387]3 years ago
4 0
0.150 M AgNO3 = x mol / 0.200 Liters
x mol = 0.03 mol AgNO3
0.03 mol AgNO3 (169.9g AgNO3 / 1 mol AgNO3) We are converting moles to grams here with stoichiometry.

Final answer = 5.097 grams, but if you want it in terms of sig figs then it is 5.09 grams.
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Answer:

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Explanation:

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This problem:

                          FeCl3(aq)          +    Na3PO4(aq)   => FePO4(s) + 3 NaCl(aq)

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moles:        27.8g/162.2g/mole      1.9g/163.94g/mole

                    = 0.1714 mole               = 0.0116 mole

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smaller value is LR => =>  =>  =>  => => LR is Na₃PO₄              

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3 0
3 years ago
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shtirl [24]

Answer:

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Explanation:

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Answer:

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