MCQ
In the above compound $Cl$ will liberated easily in the form of


- A$Cl^{\oplus}$
- ✓$Cl^-$
- C$Cl^{\bullet}$
- D$Cl^{2+}$


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Considering the above chemical reaction, identify the product $"X"$
$2C_6H_6 (l) + 15O_2 (g) \longrightarrow 12CO_2 (g) + 6H_2O(l)$
at $300\, K$ is ....$J\,mol^{-1}$ ($R = 8.314\, J\, mol^{-1}\, K^{-1}$)
$MnO_4^ - + 5F{e^{2 + }} + 8{H^ + } \to M{n^{2 + }} + 5F{e^{3 + }} + 4{H_2}O$
Here $10\,ml$ of $0.1\,M$ $KMn{O_4}$ is equivalent to