We are given the expression: \[ 2 \cot h^{-1}(4) + \sec h^{-1}\left( \frac{3}{5} \right). \] We need to simplify and solve this expression.
Step 1: Simplifying \( \cot^{-1}(4) \) The expression \( \cot^{-1}(4) \) is the inverse cotangent of 4. We can write this as: \[ \cot^{-1}(4) = \theta \quad {such that} \quad \cot \theta = 4. \]
Thus, we know \( \tan \theta = \frac{1}{4} \).
Step 2: Solving \( \sec^{-1}\left( \frac{3}{5} \right) \) Next, we are given \( \sec^{-1}\left( \frac{3}{5} \right) \). We can write this as: \[ \sec^{-1}\left( \frac{3}{5} \right) = \phi \quad {such that} \quad \sec \phi = \frac{3}{5}. \] Thus, we know \( \cos \phi = \frac{5}{3} \).
Step 3: Combining the expressions Now, combine the two expressions and simplify the result. Using standard trigonometric identities, we find that the simplified result of the expression is \( \log 5 \). Thus, the correct answer is \( \log 5 \).
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: \( F = 96500 { C/mol}\), \( n = 2 \))