Question

Answer 1

a.

\(M_{CuSO_4}=160\left(đvc\right)\)

\(\%Cu=\dfrac{64}{160}\cdot100\%=40\%\)

\(\%S=\dfrac{32}{160}\cdot100\%=20\%\)

\(\%O=\dfrac{64}{160}\cdot100\%=40\%\)

b.

\(M_{CuO}=80\left(đvc\right)\)

\(\%Cu=\dfrac{64}{80}\cdot100\%=80\%\)

\(\%O=\dfrac{16}{80}\cdot100\%=20\%\)

Answer 2

\(a.PTK_{CuSO_4}=64+32+4.16=160\left(đvC\right)\)

\(\%m_{Cu}=\dfrac{64}{160}.100=40\%\)

\(\%m_S=\dfrac{32}{160}.100=20\%\)

\(\%m_O=\dfrac{4.16}{160}.100=40\%\)

\(b.PTK_{CuO}=64+16=80\left(đvC\right)\)

\(\%m_{Cu}=\dfrac{64}{80}.100=80\%\)

\(\%m_O=\dfrac{16}{80}.100=20\%\)