(A) The extraction of aluminum is achieved through the Hall-Héroult process, where Al2O3 is obtained as a precipitate by bubbling CO2 through a solution of sodium aluminate, resulting in the reaction:
\(2\text{Na}[ \text{Al}(\text{OH})_4](\text{aq.}) + \text{CO}_2 \rightarrow \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} + 2\text{Al}(\text{OH})_3(\downarrow) \text{ or } \text{Al}_2\text{O}_3 \cdot 2\text{H}_2\text{O} (\text{ppt})\)
(B) The electrolytic reduction of pure alumina occurs in a steel box lined with carbon (cathode), along with the presence of cryolite (Na3AlF6) and fluorspar (CaF2), which decrease the melting point and enhance the electrolyte's conductivity.
(C) In the electrolysis process of the Hall-Héroult method, graphite rods serve as anodes. At the cathode, aluminum ions are reduced:
\(\text{Al}^{3+} + 3e^- \rightarrow \text{Al}\)
At the anode, the liberated oxygen reacts with the carbon of the anode to form CO and CO2:
\(\text{C} + \text{O}_2^- \rightarrow \text{CO} + 2e^-\)
\(\text{C} + 2\text{O}_2^- \rightarrow \text{CO}_2 + 4e^-\)
(D) In this context, the cathode consists of a steel vessel with a carbon lining.
Let $ a_0, a_1, ..., a_{23} $ be real numbers such that $$ \left(1 + \frac{2}{5}x \right)^{23} = \sum_{i=0}^{23} a_i x^i $$ for every real number $ x $. Let $ a_r $ be the largest among the numbers $ a_j $ for $ 0 \leq j \leq 23 $. Then the value of $ r $ is ________.
Let $ \mathbb{R} $ denote the set of all real numbers. Then the area of the region $$ \left\{ (x, y) \in \mathbb{R} \times \mathbb{R} : x > 0, y > \frac{1}{x},\ 5x - 4y - 1 > 0,\ 4x + 4y - 17 < 0 \right\} $$ is
The center of a disk of radius $ r $ and mass $ m $ is attached to a spring of spring constant $ k $, inside a ring of radius $ R>r $ as shown in the figure. The other end of the spring is attached on the periphery of the ring. Both the ring and the disk are in the same vertical plane. The disk can only roll along the inside periphery of the ring, without slipping. The spring can only be stretched or compressed along the periphery of the ring, following Hooke’s law. In equilibrium, the disk is at the bottom of the ring. Assuming small displacement of the disc, the time period of oscillation of center of mass of the disk is written as $ T = \frac{2\pi}{\omega} $. The correct expression for $ \omega $ is ( $ g $ is the acceleration due to gravity): 