Answer
a) $1.303\times10^{-2}\ mol$
b) $NiF_2$
c) Nickel (||) fluoride
Work Step by Step
a) Volume of the foil: $(1.25\ cm)^2\times0.055\ cm=8.59\times10^{-2}\ cm^3$
Mass of the foil: $8.902\ g/cm^3\times8.59\times10^{-2}\ cm^3=0.765\ g$
Atomic weight of Ni: $58.6934\ g/mol$
Number of moles used: $0.765\ g\div58.6934\ g/mol= 1.303\times10^{-2}\ mol$
b) Mass of fluorine used $1.261-0.765= 0.496\ g$
Number of moles used: $0.496\ g\div18.998\ g/mol= 2.611\times10^{-2}\ mol$
Ratio to the number of moles of Ni: $2.00$
Empirical formula: $NiF_2$
c) Nickel (||) fluoride
