\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ n_{H_2}=\frac{3,36}{22,4}=0,15mol\\ n_{H_2}=n_{Fe}=0,15mol\\ \%m_{Fe}=\frac{0,15.56}{10}.100\%=84\%\\ \$m_{Cu}=100\%-84\%=16\%\)

Khi cho (\Fe\) vào (\H_2SO_4\) thì (\Fe\) tan dần vào dung dịch (\H_2SO_4\) và có khí không màu thoát ra

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

2p+n=18

3,5≤p≤18

5,12≤p≤6

⇒p=e=6;n=4

\(H_xN_yO_z\\ x:y:z=\frac{1,6}{1} : \frac{22,2}{14} : \frac{76,2}{16}\\ x:y:z=1,6:1,6:4,8\\ \Rightarrow x:y:z=1:1:3\\ \Rightarrow HNO_3\\ \Rightarrow A\)

\(57668778 - 7649 = 57661129\)

\(2p+n=18\\ \frac{18}{3,5} \leq p \leq\frac{18}{3}\\ 5,12\leq p\leq 6\\ \Rightarrow p=e=6; n=4 \)

\(2 + 4 + 6 + 8 + ... + 2x = 210\\ [\frac{2x-2}{2} + 1 ] . \frac{2x+2}{2} = 210\\ [ \frac{2x-2.1}{2} + 1 ] .\frac{2x+2.1}{2} = 210\\ [\frac{2(x-1)}{2} + 1 ] .\frac{2(x+1)}{2} = 210\\ [ ( x -1 ) + 1 ] . ( x + 1 ) = 210\\ x . ( x + 1 ) = 210\\ x . ( x + 1 ) = 14 . 15\\ \Rightarrow x = 14\\\)

\(x=+-3\)

\(4x^{2}-36=0\\ \rightarrow 4x^{2}=36\\ \rightarrow x^{2}=9\\ \rightarrow x=3\)

Nguyên tố: \(RO\)

Ta có:

\(R+16=80 \\ \Rightarrow R=64 (Cu) \)