When $XO_2$ is fused with an alkali metal hydroxide in the presence of an oxidizing agent such as $KNO_3$,a dark green product is formed which disproportionates in an acidic solution to afford a dark purple solution. $X$ is
- A$Mn$
- B$Cr$
- C$V$
- D$Ti$
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Similar Questions
The alloy that contains copper and $Zn$ is $x$ and the one that contains copper and $Ni$ is $y$. What are $x$ and $y$ respectively?
The chief constituents of steel made in India are
Easy
View SolutionMatch Column $A$ with Column $B$:
Column $A$ Column $B$ $(1)$ Brass $(a)$ $Ni$ $(60\%)$,$Cr$ $(40\%)$ $(2)$ Bronze $(b)$ $Cu$ $(80\%)$,$Sn$ $(20\%)$ $(3)$ Cupronickel $(c)$ $Cu$ $(90\%)$,$Sn$ $(10\%)$ $(4)$ Nichrome $(d)$ $Cu$ $(70\%)$,$Zn$ $(30\%)$ $(e)$ $Cu$ $(75-85\%)$,$Ni$ $(15-25\%)$
| Column $A$ | Column $B$ |
|---|---|
| $(1)$ Brass | $(a)$ $Ni$ $(60\%)$,$Cr$ $(40\%)$ |
| $(2)$ Bronze | $(b)$ $Cu$ $(80\%)$,$Sn$ $(20\%)$ |
| $(3)$ Cupronickel | $(c)$ $Cu$ $(90\%)$,$Sn$ $(10\%)$ |
| $(4)$ Nichrome | $(d)$ $Cu$ $(70\%)$,$Zn$ $(30\%)$ |
| $(e)$ $Cu$ $(75-85\%)$,$Ni$ $(15-25\%)$ |
Medium
View Solution$I$. $4FeCr_2O_4 + 16NaOH + 7O_2 \rightarrow 8Na_2CrO_4 + 2Fe_2O_3 + 8H_2O$ (where $A = Na_2CrO_4$)
$II$. $2Na_2CrO_4 + H_2SO_4 \rightarrow Na_2Cr_2O_7 + Na_2SO_4 + H_2O$ (where $B = H_2SO_4$)
Compounds $(A)$ and $(B)$ are :
$II$. $2Na_2CrO_4 + H_2SO_4 \rightarrow Na_2Cr_2O_7 + Na_2SO_4 + H_2O$ (where $B = H_2SO_4$)
Compounds $(A)$ and $(B)$ are :
Medium
View Solution$A$ compound of mercury used in cosmetics,in Ayurvedic and Yunani medicines and known as vermilion is
Medium
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