Answer to Question #58011 in General Chemistry for Ashton
n(NiF2)=V*CM(NiF2) = 0.5 L*0.123 M= 0.0615 mol
The mass of NiF2 is:
m(NiF2) = M(NiF2)* n(NiF2) = 97 g/mol* 0.0615 mol = 5.9655 g
We need to add 5.9655 g of nickel(II) fluoride
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