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A+B------>C
Na2O+H2O-->2NaOH
C+D--->E+G
2NaOH+2Zn-->2NaZnO2+H2
G+K---->B
H2+O2-->H2O
B+F--->Ca(OH)2
H2O+CaO-->Ca(OH)2
Chúc bạn học tốt
\(CO_2+CaO\underrightarrow{t^o}CaCO_3\)
\(CaCO_3+BaCl_2\rightarrow CaCl_2+BaCO_3\downarrow\)
\(Ca+\frac{1}{2}O_2\underrightarrow{t^o}CaO\)
\(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)
Fe + O2 = Fe3O4
Fe3O4 + HCl = FeCl2 + FeCl3 + H2O
FeCl2 + NaOH = Fe(OH)2 + NaCl
FeCl3 + NaOH = Fe(OH)3 + NaCl
Fe(OH)2 + O2 + H2O = Fe(OH)3
Fe(OH)3 = Fe2O3 + H2O
=> A=Fe3O4;B=FeCl2;C=FeCl3;D=Fe(OH)2;E;Fe(OH)3;G=NaCl;F=Fe2O3
\(2KClO_3\rightarrow3O_2+2KCl\)
\(5O_2+4P\rightarrow2P_2O_5\)
\(P_2O_5+3H_2O\rightarrow2H_3PO_4\)
\(3Zn+2H_3PO_4\rightarrow Zn_3\left(PO_4\right)_2+3H_2\)
\(2H_2+O_2\rightarrow2H_2O\)
\(CaCO_3\rightarrow CO_2+CaO\)
\(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)
lần sau đừng lấy C, K vì dễ lẫn vs cữ viết tắt của cacbon và kali
a.\(M_A=23.2=46\) ( g/mol )
b.\(M_B=2,7.16=43,2\) ( g/mol )
c.\(M_C=2.29=58\) ( g/mol )
d.\(M_D=2.17=34\) ( g/mol )
e.\(M_E=1,32.44=58,08\) ( g/mol )
f.\(M_F=2,71.34=92,14\) ( g/mol )
g.\(M_G=1,5.32=48\) ( g/mol )
h.\(M_H=0,41.71=29,11\) ( g/mol )
a) \(M_{Ca\left(OH\right)_2}=40+\left(16+1\right).2=74\left(DvC\right)\)
\(\%Ca=\dfrac{40.1}{74}.100\%=54\%\)
\(\%O=\dfrac{16.2}{74}.100\%=43\%\)
\(\%H=100\%-54\%-43\%=3\%\)
a) \(\left\{{}\begin{matrix}\%m_{Ca}=\dfrac{40.1}{74}.100\%=54,054\%\\\%m_O=\dfrac{16.2}{74}.100\%=43,243\%\\\%m_H=\dfrac{2.1}{74}.100\%=2,703\%\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\%m_{Ba}=\dfrac{137.1}{208}.100\%=65,865\%\\\%Cl=\dfrac{35,5.2}{208}.100\%=34,135\%\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}\%m_K=\dfrac{39.1}{56}.100\%=69,643\%\\\%m_O=\dfrac{16.1}{56}.100\%=28,571\%\\\%m_H=\dfrac{1.1}{56}.100\%=1,786\%\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{27.2}{102}.100\%=52,94\%\\\%m_O=\dfrac{16.3}{102}.100\%=47,06\%\end{matrix}\right.\)
e) \(\left\{{}\begin{matrix}\%m_{Na}=\dfrac{23.2}{106}.100\%=43,396\%\\\%m_C=\dfrac{12}{106}.100\%=11,321\%\\\%m_O=\dfrac{16.3}{106}.100\%=45,283\%\end{matrix}\right.\)
g) \(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{56.1}{72}.100\%=77,78\%\\\%m_O=\dfrac{16.1}{72}.100\%=22,22\%\end{matrix}\right.\)
h) \(\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{65.1}{161}.100\%=40,373\%\\\%m_S=\dfrac{32.1}{161}.100\%=19,876\%\\\%m_O=\dfrac{16.4}{161}.100\%=39,751\%\end{matrix}\right.\)
i) \(\left\{{}\begin{matrix}\%m_{Hg}=\dfrac{201.1}{217}.100\%=92,627\%\\\%m_O=\dfrac{16}{217}.100\%=7,373\%\end{matrix}\right.\)
k) \(\%m_{Na}=\dfrac{23.1}{85}.100\%=27,06\%;\%m_N=\dfrac{14.1}{85}.100\%=16,47\%\%;\%m_O=\dfrac{16.3}{85}.100\%=56,47\%\)
\(a,2H_2O\underrightarrow{đp}2H_2+O_2\\ Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(b,2H_2O\underrightarrow{đp}2H_2+O_2\\ 2KClO_3\xrightarrow[MnO_2]{t^o}2KCl+3O_2\)
\(c,S+O_2\underrightarrow{t^o}SO_2\\ SO_2+H_2O\rightarrow H_2SO_3\)
\(d,PbO+H_2\underrightarrow{t^o}Pb+H_2O\\ e,CaCO_3\underrightarrow{t^o}CaO+CO_2\\ CaO+H_2O\rightarrow Ca\left(OH\right)_2\)
e, làm ròi
A: O2
B: KCl
C: P
D: P2O5
E: H2O
F: H3PO4
G: H2
I: CO2
J: CaO
K: Ca(OH)2
Các PTHH:
\(2KClO_3\underrightarrow{^{to}}2KCl+3O_2\)
\(4P+5O_2\underrightarrow{^{to}}2P_2O_5\)
\(P_2O_5+3H_2O\rightarrow2H_3PO_4\)
\(3Zn+2H_3PO_4\rightarrow Zn_3\left(PO_4\right)_2+3H_2\)
\(2H_2+O_2\underrightarrow{^{to}}2H_2O\)
\(CaCO_3\underrightarrow{^{to}}CaO+CO_2\)
\(CaO+H_2O\underrightarrow{^{to}}Ca\left(OH\right)_2\)
Làm rồi nhé