The balanced reaction is:
\[\text{C}_6\text{H}_5\text{NH}_2 + \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- \rightarrow \text{C}_6\text{H}_5\text{N} = \text{N} -\text{C}_6\text{H}_5 \, (\text{Aniline yellow}).\]
Given:
\[\text{Molar mass of aniline} = 93 \, \text{g/mol}, \quad \text{Given mass} = 279 \, \text{g}.\]
Moles of aniline:
\[n = \frac{279}{93} = 3 \, \text{mol}.\]
Mass of aniline yellow:
\[\text{Molar mass of product} = 197 \, \text{g/mol}.\]
\[\text{Mass} = 3 \cdot 197 = 591 \, \text{g}.\]
Final Answer:
\[591 \, \text{g}.\]
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 \( 4.606 \times 10^{-2} \text{ s}^{-1} \).
A hydrocarbon which does not belong to the same homologous series of carbon compounds is
Let $ f(x) = \begin{cases} (1+ax)^{1/x} & , x<0 \\1+b & , x = 0 \\\frac{(x+4)^{1/2} - 2}{(x+c)^{1/3} - 2} & , x>0 \end{cases} $ be continuous at x = 0. Then $ e^a bc $ is equal to
Total number of nucleophiles from the following is: \(\text{NH}_3, PhSH, (H_3C_2S)_2, H_2C = CH_2, OH−, H_3O+, (CH_3)_2CO, NCH_3\)