Answer:
[tex]molar \: mass \: of \: copper(ii)nitrate = 64 + (14 + 48) \times 3 \\ = 250 \: g \\ 64 \: g \: of \: copper \: produces \: 250 \: g \: of \: copper \: nitrate \\ 10.36 \: g \: of \: copper \: will \: produce \: ( \frac{10.36 \times 250}{64} ) \: g \\ = 40.7 \: g[/tex]
