




Reaction mechanism: Lithium borohydride (LiBH$_4$) is a selective reducing agent that reduces esters (CO$_2$Et) to alcohols while leaving carboxylic acids (CO$_2$H) unchanged.
Step-by-step process:
Option (A) The ester group (CO$_2$Et) is reduced by LiBH$_4$ in ethanol (EtOH) to form a primary alcohol (-CH$_2$CH$_2$OH).
Option (B) The carboxylic acid group (CO$_2$H) remains unchanged during the reaction.
Option (C) Protonation with H$_3$O$^+$ ensures the stability of the final product.
The final product contains an unchanged carboxylic acid group and a newly formed primary alcohol group, corresponding to option (4).
Calculate the potential for half-cell containing 0.01 M K\(_2\)Cr\(_2\)O\(_7\)(aq), 0.01 M Cr\(^{3+}\)(aq), and 1.0 x 10\(^{-4}\) M H\(^+\)(aq).

A square loop of sides \( a = 1 \, {m} \) is held normally in front of a point charge \( q = 1 \, {C} \). The flux of the electric field through the shaded region is \( \frac{5}{p} \times \frac{1}{\varepsilon_0} \, {Nm}^2/{C} \), where the value of \( p \) is: