MCQ
The modern atomic weight scale is based on
- ✓${C^{12}}$
- B${O^{16}}$
- C${H^1}$
- D${C^{13}}$
It is equal to $1 / 12$ of the mass of an atom of carbon-$12 .$
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${N_2}{H_4}(l)\,+\,2{H_2}{O_2}(l)\,\, \to \,{N_2}(g)\,\, + \,4{H_2}O(l);\,\,{\Delta _r}H_1^o\, = \, - 818\,\,kJ/mol$
${N_2}{H_4}(l)\,\, + {O_2}(g)\,\, \to \,{N_2}(g)\,\, + \,2{H_2}O(l);\,\,{\Delta _r}H_2^o\, = \, - 622\,\,kJ/mol$
${H_2}(g)\,\, + 1/2{O_2}(g)\,\, \to \,{H_2}O(l);\,\,{\Delta _r}H_3^o\, = \, - 285\,\,kJ/mol$
......$kJ/mol$