Em chụp anh thấy cứ bị xén đi 1 góc nội dung á em!

\(\dfrac{2A}{2A+16.5}=\dfrac{43,66}{100}\)

=> \(200A=43,66.\left(2A+16.5\right)\)

=> \(200A-87,32A=3492,8\)

=> \(112,68A=3492,8\)

=> A= 31 

Cái đó là tìm ra A là bn ạ

Câu 4:

4.1/ Ta có: \(n_{NaCl}=2,5.0,4=1\left(mol\right)\)

\(\Rightarrow m_{NaCl}=1.58,5=58,5\left(g\right)\)

4.2/ Ta có: \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

_____0,2___________0,2____0,2 (mol)

b, \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)

c, \(m_{ZnCl_2}=0,2.136=27,2\left(g\right)\)

Bạn tham khảo nhé!

MgO

K2O

Al(OH)3

SO2

a)

ta có \(n_{N_2}=\dfrac{4,2}{28}=0,15\left(mol\right)\)

\(\Rightarrow V_{N_2}=0,15.22,4=3,36l\)

ta có \(n_{O_2}=\dfrac{3,2}{32}=0,1\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,1.22,4=2,24l\)

b) 

ta có \(n_{SO_2}=\dfrac{32}{64}=0,5\left(mol\right)\)

\(\Rightarrow V_{SO_2}=0,5.22,4=11,2l\)

ta có \(n_{O_2}=\dfrac{32}{32}=1\left(mol\right)\)

\(\Rightarrow V_{O2}=1.22,4=22,4l\)

c) 

ta có \(n_{H_2}=\dfrac{2}{2}=1\left(mol\right)\)

\(\Rightarrow V_{H_2}=1.22,4=22,4l\)

ta có \(n_{O_2}=\dfrac{8}{32}=0,25\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,25.22,4=5,6l\)

d) 

ta có \(n_{N_2}=\dfrac{8,4}{28}=0,3\left(mol\right)\)

\(\Rightarrow V_{N_2}=0,3.22,4=6,72l\)

ta có \(n_{O_2}=\dfrac{6,4}{32}=0,2\left(mol\right)\)

\(\Rightarrow V_{N_2}=0,2.22,4=4,48l\)

Cảm mơn bạn nha !

Bài 5
CuO<Cu(OH)₂<CuCl₂<CuSO₄
Bài 6
\(\%R=\dfrac{2R}{2R+48}\cdot100\%=52,94\%\\ \Rightarrow R=27,Al\)
Bài 7
\(CTPT\left(Z\right):H_xS_y\\ M_Z=17.2=34g/mol=x+32y\\ \%H=\dfrac{x}{x+32y}\cdot100\%=5,88\%\\ \Leftrightarrow\dfrac{x}{34}\cdot100\%=5,88\%\\ \Rightarrow x\approx2\\ y=\dfrac{34-2}{32}=1\\ \Rightarrow CTPT\left(X\right):H_2S\)