Butan-1-ol contains an -OH group, allowing it to form hydrogen bonds, which increase intermolecular forces and raise the boiling point.
Ethoxyethane (ether) cannot form hydrogen bonds due to the lack of an -OH group, resulting in weaker intermolecular forces and a lower boiling point.
Hydrogen bonding is the primary reason for the stronger association of molecules in butan-1-ol compared to ethoxyethane.
Thus, both A and R are true, and R is the correct explanation of A.
(b.)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).
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