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$A.$ Surface tension is the outcome of equal attractive and repulsion forces acting on the liquid molecule in bulk.
$B.$ Surface tension is due to uneven forces acting on the molecules present on the surface.
$C.$ The molecule in the bulk can never come to the liquid surface.
$D.$ The molecules on the surface are responsible for vapour pressure if the system is a closed system.
$Hex - 3 - ynal\xrightarrow[\begin{subarray}{l}
(ii)\,PB{r_3} \\
(i)\,Mg/ether \\
(i)\,C{O_2}/{H_3}{O^ + }
\end{subarray} ]{{(i)\,NaB{H_4}}}\,?$
Given : $K_b(H_2O) = 0.51\, Kkg \,mol^{-1}$
$N_A = 6 × 10^{23}$
$[Li = 7, \,\,Na = 23$, $K = 39,\,\, Rb = 85.5$ , $Cs = 133, \,\,Cl = 35.5]$