57gfe no3 3 0375mno3 00850 l molfe no3 l

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Unformatted text preview: n 0.125 M Fe(NO3)3 2.57 g 0.375 M 76.5 mL Work: ⎛ 0.125 mol Fe (NO3 )3 ⎞ ⎜ ⎝ L − ⎛ 241.88 g Fe (NO3 )3 ⎞ ⎛ 0.125 mol Fe (NO3 )3 ⎞ ⎛ 3 mol NO3 ⎞ − ⎟ = 2.57 g Fe (NO3 )3 ⎜ ⎟ = 0.375 M NO3 ⎟ ( 0.0850 L ) ⎜ ⎟⎜ ⎜ mol Fe (NO3 ) ⎟ ⎜ ⎟ L ⎠ ⎝ ⎠ ⎝ 1 m...
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This note was uploaded on 11/02/2013 for the course CHEM 102 taught by Professor Murphy during the Fall '09 term at Wisconsin Milwaukee.

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