MCQ
Conjugate acid of $HPO_4^{-2}$ is
- A$H_3PO_4$
- ✓${H_2}PO_4^\Theta $
- C$PO_4^{ - 3}$
- D${H_2}PO_3^\Theta $
$H _2 PO _4^{-} \rightarrow H ^{+}+ HPO _4^{2-}$
The base and its conjugate acid differ by one proton only.
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(Atomic numbers : $Fe=26,Co=27,Mn=25,Cr=24$)
$C(s) + {O_2}(g) \to C{O_2}(g),\Delta H = - 394\,kJ$
$2{H_2}(g) + {O_2}(g) \to 2{H_2}O(l),\,\Delta H = 568\,kJ$
$C{H_4}(g) + 2{O_2}(g)\,\, \to \,\,C{O_2}(g) + 2{H_2}O(l)$ $\Delta H = - 892\,kJ$ Heat of formation of $C{H_4}$ is.....$kJ$

