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Answered

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what mass of iron(III) nitrate [Fe(NO3)3] is needed to make 400mL of a 0.3M solution?

What Mass Of IronIII Nitrate FeNO33 Is Needed To Make 400mL Of A 03M Solution class=

Sagot :

Answer:

29

Explanation:

.55.85+(14.01+48.00)3

=241.88g/mol

4g*.3m=0.12

241.857*0.12=29g

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