To determine the amount of bromine needed to convert 2 g of phenol into 2,4,6-tribromophenol, we follow this procedure:
Step 1: Calculate the molar mass of phenol (C6H5OH)
Molar mass of C = 12 g/mol
Molar mass of H = 1 g/mol
Molar mass of O = 16 g/mol
Molar mass of phenol = 6(12) + 6(1) + 16 = 94 g/mol
Step 2: Calculate the number of moles of phenol
Number of moles = mass / molar mass = 2 g / 94 g/mol ≈ 0.0213 mol
Step 3: Determine the reaction and molar mass of tribromophenol
Reaction: C6H5OH + 3 Br2 → C6H2Br3OH + 3 HBr
Molar mass of Br2 = 2(80) = 160 g/mol
Step 4: Calculate bromine required for the reaction
Bromine required per mol of phenol = 3 mols of Br2
Number of moles of Br2 = 3 × 0.0213 mol = 0.0639 mol
Mass of Br2 = moles × molar mass = 0.0639 mol × 160 g/mol = 10.224 g
Thus, the amount of bromine required is 10.22 g.
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).
Let \( T_r \) be the \( r^{\text{th}} \) term of an A.P. If for some \( m \), \( T_m = \dfrac{1}{25} \), \( T_{25} = \dfrac{1}{20} \), and \( \displaystyle\sum_{r=1}^{25} T_r = 13 \), then \( 5m \displaystyle\sum_{r=m}^{2m} T_r \) is equal to: