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\(a,\%Na=\dfrac{23}{85}.100\%=27,06\%\\ \%N=\dfrac{14}{85}.100\%=16,47\%\\ \%O=100\%-27,06\%-16,47\%=56,47\%\\ b,\%Al=\dfrac{54}{234}.100\%=27,1\%\\ \%C=\dfrac{36}{234}.100\%=15,4\%\\ \%O=100\%-27,1\%-15,4\%=57,5\%\)
\(c,\%N=\dfrac{28}{79}.100\%=35,4\%\\ \%H=\dfrac{4}{79}.100\%=5,1\%\\ \%O=100\%-35,4\%-5,1\%=59,5\%\)
\(M_{NaNO_3}=23+14+16.3=85\left(\dfrac{g}{mol}\right)\\ \Rightarrow\%m_{Na}=\dfrac{23.100\%}{85}=27\%\\ \%m_N=\dfrac{14.100\%}{85}=16,47\%\\ \Rightarrow\%m_O=100\%-\left(16,47\%+27\%\right)=56,53\)
\(M=242+132=374(đvc)\\ \%_{Fe}=\frac{56}{374}.100=14,97\%\\ \%_{N}=\frac{14.3+14.2}{374}.100=18,72\%\\ \%_H=\frac{1.4.2}{374}.100=2,14\%\\ \%_S=\frac{32}{374}.100=8,57\%\\ \%_O=55,6\%\)
\(Fe(NO_3)_3\\ \%_{Fe}=\frac{56}{242}.100\%=23,14\%\\ \%_N=\frac{14.3}{242}.100\%=17,36\%\\ \%_O=595,5\%\\ (NH_4)_2SO_4\\ \%N=\frac{14.2}{132}=21,21\% \\ \%H=\frac{(1.4).2.}{132}=6,06\% \\ \%S=\frac{32}{132}=24,24\% \\ \%O=\frac{16.4}{132}=48,49\%\\ \)
\(M=242+132=374(đvc)\\ \%_{Fe}=\frac{56}{374}.100=14,97\%\\ \%_N=\frac{14.3+14.2}{374}.100=18,72\%\\ \%_H=\frac{1.4.2}{374}=2,14\%\\ \%_S=\frac{32}{374}.100=8,57\%\\ \%_O=55,6\%\)
a.nFe2O3=\(\frac{32}{160}\)=0,2 nCO\(\frac{6}{7}\) nCuSO4=0,1
\(\rightarrow\)\(\text{nFe=0,4 nCu=0,1}\)
\(\rightarrow\)\(\text{mFe=22,4 mCu=6,4}\)
b. %mFe=\(\frac{22,4}{32}\)=70%\(\rightarrow\)%mO=30%
\(\text{mCuSO4=0,1.160=16}\)
\(\rightarrow\)%mCu=\(\frac{6,4}{16}\)=40%
nS=nnCuSO4=0,1\(\rightarrow\)%mS=\(\frac{0,1.32}{16}\)=20%
\(\rightarrow\)%mO=40%
nC=nCO=6/7
\(\rightarrow\)%mC=12.6/7/24=42,86%
\(\rightarrow\)%mO=57,14%
c. Theo kết quả câu b thì hàm lượng O trong CO cao nhất
\(M_{NaNO_3}=23+14+3.16=85\)
% m Na = \(\dfrac{23}{85}.100\%=27,1\%\)
% m N = \(\dfrac{14}{85}.100\%=16,5\%\)
% m O = 100% - 27,1% - 16,5% = 56,4%
\(M_{K_2CO_3}=2.39+12+3.16=138\)
% m K = \(\dfrac{2.39}{138}.100\%=56,5\%\)
% m C = \(\dfrac{12}{138}.100\%=8,7\%\)
% m O = 100% - 56,5% - 8,7% = 34,8%