nhóm Halogen

T

thupham22011998

Ta có: n $CO_2$=0,02 mol

--> $n M_2CO_3 + n MHCO_3=0,02 mol$

[TEX] \Rightarrow \frac{2M+60+M+61}{2} = \frac{1,9}{0,02}[/TEX]

[TEX]\Leftrightarrow M=23[/TEX]

Vậy M là $Na$
 
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