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\(a,\%H=\dfrac{1}{63}.100\%=1,6\%\\\%N=\dfrac{14}{63}.100\%=22,2\%\\ \%O=100\%-1,6\%-22,2\%=76,2\%\\b,\%Al=\dfrac{54}{342}.100\%=15,8\%\\ \%S=\dfrac{96}{342}.100\%=28,1\%\\ \%O=100\%-15,8\%-28,1\%=56,1\% \%b,b,15,8\%\\ \)
\(M_{Fe_2\left(SO_4\right)_3}=400\left(g\text{/}mol\right)\)
\(\%Fe=\dfrac{56\cdot2}{400}\cdot100\%=28\%\)
\(\%S=\dfrac{32\cdot3}{400}\cdot100\%=24\%\)
\(\%O=100-28-24=48\%\)
\(PTK_{KNO_3}=101\left(đvC\right)\\ \Leftrightarrow\left\{{}\begin{matrix}\%_K=\dfrac{39}{101}\cdot100\%=38,61\%\\\%_N=\dfrac{14}{101}\cdot100\%=13,86\%\\\%_O=100\%-38,61\%-13,86\%=47,53\%\end{matrix}\right.\)
Trong hợp chất:
\(\left\{{}\begin{matrix}m_{Cu}=80\cdot80\%=64\left(g\right)\\m_O=80\cdot20\%=16\left(g\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}n_{Cu}=\dfrac{64}{64}=1\left(mol\right)\\n_O=\dfrac{16}{16}=1\left(mol\right)\end{matrix}\right.\)
Vậy CTHH A là \(CuO\)
\(M_{Fe(OH)_3}=56+17.3=107(đvC)\\ \%_{Fe}=\dfrac{56}{107}.100\%=52,34\%\\ \%_O=\dfrac{48}{107}.100\%=44,86\%\\ \%_H=100\%-52,34\%-44,86\%=2,8\%\)
\(\left\{{}\begin{matrix}\%Fe=\dfrac{56.1}{107}.100\%=52,336\%\\\%O=\dfrac{16.3}{107}.100\%=44,86\%\\\%H=\dfrac{1.3}{107}.100\%=2,804\end{matrix}\right.\)
\(\%m_{Cu}=\dfrac{M_{Cu}.1}{1.\left(M_{Cu}+M_O\right)}.100\%=\dfrac{64}{64+16}.100\%=80\%\\ \%m_O=100\%-80\%=20\%\)
%Zn=\(\frac{65}{65+32+16.4}.100\%=40,37\%\)
%S=\(\frac{32}{65+32+16.4}.100\%=19,87\%\)
%O=100-19,87-40,37=39,76%
Các bài khác tương tự
Gọi CTHH là: \(Cu_xO_y\left(x,y\in N\right)\)
Theo đề bài ta có: \(\dfrac{64x}{16y}=\dfrac{80\%}{20\%}=4\)
\(\Rightarrow4x=4y\Rightarrow x=y\)
Chọn x=y=1
Vậy CTHH là: \(CuO\)
Tham khảo
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