Câu trả lời:
\(2M+2nH_2SO_{4đ,n}\rightarrow M_2\left(SO_4\right)_n+nSO_2+2nH_2O\)
\(n_{SO_2}=\dfrac{4.2}{22.4}=0.1875\left(mol\right)\)
\(\dfrac{7}{M}=\dfrac{0.1875\cdot2}{n}\)
n | 1 | 2 | 3 |
M | 18.6 | 37.3 | 56(nhận) |
Vậy M là Fe
\(2M+2nH_2SO_{4đ,n}\rightarrow M_2\left(SO_4\right)_n+nSO_2+2nH_2O\)
\(n_{SO_2}=\dfrac{4.2}{22.4}=0.1875\left(mol\right)\)
\(\dfrac{7}{M}=\dfrac{0.1875\cdot2}{n}\)
n | 1 | 2 | 3 |
M | 18.6 | 37.3 | 56(nhận) |
Vậy M là Fe
B
số giao tử=\(8\cdot24\cdot4=768\)
6. B
7. A
8. A
9. D
10. B
11. B
A
6. B
8. C
18. D
22. C
23. A
218
\(2Y+2xH_2SO_4\rightarrow xSO_2+2xH_2O+Y_2\left(SO_4\right)_x\)
\(n_{SO_2}=\dfrac{0.896}{22.4}=0.04\)
\(\Rightarrow n_Y=\dfrac{0.08}{x}=\dfrac{2.6}{M_Y}\)
x | 1 | 2 |
M | 32.5 | 65 (nhận) |
\(\Rightarrow C\left(kẽm\right)\\ \)