Step 1: Factorizing the numerator and denominator \( x^3 - 27 = (x-3)(x^2 + 3x + 9). \) \( x^2 - 9 = (x-3)(x+3). \)
Step 2: Cancel common terms \( \frac{(x-3)(x^2+3x+9)}{(x-3)(x+3)}. \) For \( x \neq 3 \), canceling \( (x-3) \), \( \lim_{x \to 3} \frac{x^2+3x+9}{x+3}. \)
Step 3: Substitute \( x = 3 \) \( \frac{3^2+3(3)+9}{3+3} = \frac{9+9+9}{6} = \frac{27}{6} = \frac{9}{2}. \)
Which of the following are ambident nucleophiles?
[A.] CN$^{\,-}$
[B.] CH$_{3}$COO$^{\,-}$
[C.] NO$_{2}^{\,-}$
[D.] CH$_{3}$O$^{\,-}$
[E.] NH$_{3}$
Identify the anomers from the following.

The standard Gibbs free energy change \( \Delta G^\circ \) of a cell reaction is \(-301 { kJ/mol}\). What is \( E^\circ \) in volts?
(Given: \( F = 96500 { C/mol}\), \( n = 2 \))