Given:
\[\text{Power (P)} = V \cdot I\]
\[110 = 220 \times I\]
\[I = 0.5 \, \text{A}\]
Now, we know:
\[I = \frac{n \cdot e}{t}\]
Substitute the values:
\[0.5 = \frac{n \times (1.6 \times 10^{-19})}{t}\]
Rearrange to solve for \( n \):
\[\frac{n}{t} = \frac{0.5}{1.6 \times 10^{-19}}\]
\[\frac{n}{t} = 31.25 \times 10^{17}\]
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