MCQ
In $Xe{O_3}$ and $Xe{F_6}$ the oxidation state of $Xe$ is
- A$+4$
- ✓$+6$
- C$+1$
- D$+3$
$\mathop {Xe{O_3}}\limits^{ * \,\,\,\,\,\,\,\,\,\,} $ $\mathop {Xe{F_6}}\limits^{ * \,\,\,\,\,\,\,\,\,} $
$x - 2 \times 3 = 0$ $x - 6 = 0$
$x = + 6$ $x = + 6$
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[Atomic mass : Silver $=108$, Bromine $=80]$
$x\ MnO + y\ PbO_2 + z\ HNO_3 \to a\ HMnO_4 + b\ Pb(NO_3 )_2 + c\ H_2O$