MCQ
During detection of phosphorus in an organic compound, yellow precipitate is formed due to the formation of
- A$(NH_4)_3PO_4$
- B$Mg_2P_2O_7$
- ✓$(NH_4)_3PO_4.12MoO_3$
- D$(NH_4)_2MoO_4$
The aqueous solution so obtained is boiled with concentrated nitric acid, and ammonium molybdate solution is added to it.
A yellow solution or precipitate indicates the presence of phosphorus in the organic compound.
The yellow precipitate is of ammonium phosphomolybdate
$2 \mathrm{P}+3 \mathrm{Na}_{2} \mathrm{O}_{2}+\mathrm{O}_{2} \stackrel{\text { fusion }}{\mathrm{air}} \rightarrow 2 \mathrm{Na}_{3} \mathrm{PO}_{4}$
$\mathrm{Na}_{3} \mathrm{PO}_{4}+12\left(\mathrm{NH}_{4}\right)_{3} \mathrm{MoO}_{4}+21 \mathrm{HNO}_{3}$
$\begin{aligned}\left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4} \cdot 12 \mathrm{MoO}_{3}+& 21 \mathrm{NH}_{4} \mathrm{NO}_{3} \\ \text { canary yellow } &+12 \mathrm{H}_{2} \mathrm{O} \end{aligned}$
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$K_1 = 1.6 \times 10^3$ and
$[Ag(NH_3)]^+ + NH_3 \rightleftharpoons [Ag(NH_3)_2]^+$ ;
$K_2 = 6.8 \times 10^3$ .
Then formation constant of $[Ag(NH_3)_2]^+$ is :
$H _{2( g )}+ Br _{2( g )} \rightarrow 2 HBr _{( g )}$
Given that bond energy of $H _{2}$ and $Br _{2}$ is $435 \;kJ mol ^{-1}$ and $192 \;kJ mol^{-1}$, respectively, what is the bond energy (in $kJ mol ^{-1}$ ) of $HBr?$