Question

Answer 1

\(M_{CaCO_3}=40+12+16.3=100\left(DvC\right)\)

\(\%Ca=\dfrac{40.1}{100}.100\%=40\%\\ \%C=\dfrac{12.1}{100}.100\%=12\%\\ \%O=100\%-40\%-12\%=48\%\)

Answer 2

\(M_{CuSO_4}=64+32+16.4=160\left(DvC\right)\\ \%Cu=\dfrac{64.1}{160}.100\%=40\%\\ \%S=\dfrac{32.1}{160}.100\%=20\%\\ \%O=\dfrac{16.4}{160}.100\%=40\%\)

Answer 3

còn lại giống như vậy nhá