Question

cứu t ạa :(((( 

Answer 1

\(M_{Cu\left(OH\right)_2}=64+\left(1+16\right).2=98amu\\ \%_{Cu}=\dfrac{64}{98}\cdot100\%\approx65,3\%\\ M_{CuSO_4}=64+32+16.4=160amu\\ \%_{Cu}=\dfrac{64}{160}\cdot100\%=40\%\\ M_{Fe\left(NO_3\right)_2}=56+\left(14+16.3\right).2=180amu\\ \%_{Fe}=\dfrac{56}{180}\cdot100\%\approx31,11\%\\ M_{FeCl_3}=56+35,5.3=162,5amu\\ \%_{Fe}=\dfrac{56}{162,5}\cdot100\%\approx34,46\%\\ M_{FeCl_2}=56+35,5.2=127amu\\ \%_{Fe}=\dfrac{56}{127}\cdot100\%\approx44,09\%\)