## Chemistry and Chemical Reactivity (9th Edition)

a) $1.303\times10^{-2}\ mol$ b) $NiF_2$ c) Nickel (||) fluoride
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