Azue

Theo đề bài ta có : \(nH2=\dfrac{4,368}{22,4}=0,195\left(mol\right)\)

Gọi x ,y lần lượt là số mol của Mg và Al

Ta có PTHH :

\(\left(1\right)Mg+2HCl->MgCl2+H2\uparrow\)

x mol..... 2 x mol...... x mol........ x mol

\(\left(2\right)2Al+6HCl->2AlCl3+3H2\uparrow\)

y mol....... 6 y mol........ y mol......3/2y mol

Ta có HPT : \(\left\{{}\begin{matrix}24x+27y=3,78\\x+\dfrac{3}{2}y=0,195\end{matrix}\right.=>\left\{{}\begin{matrix}x=0,045\\y=0,1\end{matrix}\right.\)

a) Ta có : \(\left\{{}\begin{matrix}nMg=x=0,045\left(mol\right)\\nAl=y=0,1\left(mol\right)\end{matrix}\right.=>\left\{{}\begin{matrix}\%mMg=\dfrac{0,045.24}{3,78}.100\%\approx28,57\%\\\%mAl=100\%-28,57\%=71,43\%\end{matrix}\right.\)

b) Ta có : nHCl = 2nH2 = 2. 0,195 = 0,39(mol)

=> CMHCl = \(\dfrac{0,39}{0,1}=3,9\left(M\right)\)

\(\sqrt{4}=2\left(\text{đ}\text{úng}\right)\)

\(x^2=4< =>x=\sqrt{4}=>x=2\left(\text{đ}\text{úng}\right)\)

Vì x^2 > 0

a) \(\sqrt{x}=15=>x=15^2=225\)

Bài 1 :

a) \(\sqrt{4\left(a-3\right)^2}+2\sqrt{\left(a^2+4a+4\right)}\)

= \(2\left|a-3\right|+2\left|a+2\right|\)

\(=2.\left(-a+3\right)+2\left(-a-2\right)\)

b) có sai đề ko ?

c) \(4x-\sqrt{8}+\dfrac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}=4x-\sqrt{8}+\sqrt{\dfrac{x^2\left(x+2\right)}{x+2}}=4x-2\sqrt{4}+x=3x-2\sqrt{4}\)