4A +18HNO3 --> 4A(NO3)3 +3NO +3NO2 +9H2O (1)

vì sau phản ứng khối lượng trong bình giảm => mhh=1,42(g)

nhh=0,045(mol)

=>Mhh=31,56(g/mol)

giả sử trong 1 mol hh có x mol NO

y mol NO2

=>\(\left\{{}\begin{matrix}x+y=1\\30x+46y=31,56\end{matrix}\right.=>\left\{{}\begin{matrix}x=0,9025\left(mol\right)\\y=0,0975\left(mol\right)\end{matrix}\right.\)

lập tỉ lệ :

\(\dfrac{0,9025}{3}>\dfrac{0,0975}{3}\)

=> NO2 hết ,NO dư => tính theo NO2

theo (1) : nA=4/3nNO2=0,13(mol)

=> 5,2/M_A=0,13=> M_A=40(g/mol)

=>A:Ca

mik sửa lại đề : Để hòa tan 3,9g kim loại X cần dùng thể tích(ml) dd HCl và có 1,344 l H2 bay ra (đktc) . mặt khác để hòa tan 3,2g oxit kim loại Y cần dùng thể tích dd HCl ở trên . hỏi X,Y là kim loại gì ?

2X +2nHCl --> 2XCln +nH2 (1)

YxOy + 2yHCl --> xYCl2y/x + yH2 (2)

nX=3,9/M_X (MOL)

nH2=0,06(mol)

theo (1): nX=2/n .nH2=0,12/n(mol)

=> 3,9/M_X=0,12/n => M_X=32,5n(g/mol)

Xét => X :Zn

nZn=0,06(mol)

nYxOy=3,2/xM_Y+16y(mol)

theo (2) : nYxOy=1/2y.nHCl=0,06/y(mol)

=> \(\dfrac{3,2}{xMY+16y}=\dfrac{0,06}{y}\)=> MY=\(\dfrac{56}{3}.\dfrac{2y}{x}\) (g/mol)

Xét => Y :Fe

sai đề r

P=\(\dfrac{\sqrt{x+2-4\sqrt{x-2}+\sqrt{x+2+4\sqrt{x-2}}}}{2\left(x-2\right)}\)

=\(\dfrac{\sqrt{\left(\sqrt{x-2}-2\right)^2+\sqrt{\left(\sqrt{x-2}-2\right)^2}}}{2\left(x-2\right)}\)

* Nếu \(\sqrt{x-2}-2\ge0\Leftrightarrow x\ge6\) thì :

\(\dfrac{\sqrt{x+2-4\sqrt{x-2}+\sqrt{x-2}-2}}{2\left(x-2\right)}\)

=\(\dfrac{\sqrt{x-3\sqrt{x-2}}}{2\left(x-2\right)}\)

* nếu \(\sqrt{x-2}-2< 0\Leftrightarrow x< 6\) thì :

\(\dfrac{\sqrt{x+2-4\sqrt{x-2}-\sqrt{x-2}+2}}{2\left(x-2\right)}\)

=\(\dfrac{\sqrt{x-5\sqrt{x-2}+4}}{2\left(x-2\right)}\)

\(\dfrac{5+\sqrt{5}}{5-\sqrt{5}}+\dfrac{5-\sqrt{5}}{5+\sqrt{5}}-\sqrt{10}\)

=\(\dfrac{\left(5+\sqrt{5}\right)^2}{20}+\dfrac{\left(5-\sqrt{5}\right)^2}{20}-\sqrt{10}\)

=\(\dfrac{30+10\sqrt{5}+30-10\sqrt{5}}{20}-\sqrt{10}\)

=\(3-\sqrt{10}\)

\(\dfrac{1}{\sqrt{2}+1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+\dfrac{1}{\sqrt{4}+\sqrt{3}}\)

=\(\dfrac{\sqrt{2}-1}{1}+\dfrac{\sqrt{3}-\sqrt{2}}{1}+\dfrac{2-\sqrt{3}}{1}\)=\(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+2-\sqrt{3}=1\)

\(\dfrac{1-x\sqrt{x}}{1-\sqrt{x}}=\dfrac{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}+x\right)}{1-\sqrt{x}}=1+\sqrt{x}+x\)

\(\sqrt{\left(x-5\right)^2}=8\)

ĐKXĐ: \(x-5\ge0\Leftrightarrow x\ge5\)

\(\Leftrightarrow\left|x-5\right|=8\)\(\Leftrightarrow\left\{{}\begin{matrix}x-5=8\\x-5=-8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=13\\x=-3\end{matrix}\right.\)

\(2\sqrt{27}-3\sqrt{12}\sqrt{\left(2-\sqrt{3}\right)^2}\)

=\(6\sqrt{3}-6\sqrt{3}\left(2-\sqrt{3}\right)=6\sqrt{3}-12\sqrt{3}+18\)

=\(-6\sqrt{3}+18\)

a) Ba(OH)2 +Na2SO4 --> BaSO4 +2NaOH (1)

b) nBaSO4 =0,1(mol)

theo (1) : nNa2SO4=nBaSO4=0,1(Mol)

=> mNa2SO4=14,2(g)

=> mNaCl=25,9-14,2=11,7(g)

c) theo (1) : nBa(OH)2=nBaSO4=0,1(mol)

=>mBa(OH)2=17,1(g)

=>mddBa(OH)2=\(\dfrac{17,1.100}{20}=85,5\left(g\right)\)

=>mdd_{sau phản ứng}=25,9+200+85,5-23,3=288,1(g)

C%dd NaCl=11,7/288,1 .100=4,061(%)

C%dd Na2SO4=14,2/288,1 .100=4,93(%)

C%dd Ba(OH)2=17,1/288,1.100=5,935(%)

theo (1) : nNaOH=2nBaSO4=0,2(mol)

=>C%dd NaOH=\(\dfrac{0,2.40}{288,1}.100=2,777\left(\%\right)\)