For this reaction:
\[K = \frac{2.303}{t} \log \frac{[A]_0}{[A]}\]
Given: \( k = 4.6 \times 10^{-2} \, \text{s}^{-1}, \, [A]_0 = 1 \, \text{M}, \, [A] = 0.1 \, \text{M} \)
\[4.6 \times 10^{-2} = \frac{2.303}{t} \log \frac{1}{0.1}\]
\[4.6 \times 10^{-2} = \frac{2.303}{t} \times 1\]
\[t = \frac{2.303}{4.6 \times 10^{-2}} \approx 50 \, \text{sec.}\]
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32