Step 1: Determining the amount of acid neutralized by ammonia gas.
Moles of H\(_2\)SO\(_4\) taken:
\[ \text{Moles} = M \times V = 0.2 \times \frac{100}{1000} = 0.02 \, \text{moles} \]
Moles of NaOH used to neutralize excess H\(_2\)SO\(_4\):
\[ \text{Moles} = 0.5 \times \frac{40}{1000} = 0.02 \, \text{moles} \]
Since 2 moles of NaOH neutralize 1 mole of H\(_2\)SO\(_4\), moles of unreacted H\(_2\)SO\(_4\) are:
\[ \frac{0.02}{2} = 0.01 \, \text{moles} \]
Moles of H\(_2\)SO\(_4\) reacted with NH\(_3\):
\[ 0.02 - 0.01 = 0.01 \, \text{moles} \]
Moles of nitrogen in NH\(_3\):
\[ 2 \times 0.01 = 0.02 \, \text{moles} \]
Step 2: Calculating the mass of nitrogen.
Mass of nitrogen:
\[ 0.02 \times 14 = 0.28 \, \text{g} \]
Given that nitrogen is 46% of the organic compound:
\[ x = \frac{0.28}{0.46} = 0.608 \, \text{g} \]
Thus, the value of \( x \) is 0.608 g.
Find the least horizontal force \( P \) to start motion of any part of the system of three blocks resting upon one another as shown in the figure. The weights of blocks are \( A = 300 \, {N}, B = 100 \, {N}, C = 200 \, {N} \). The coefficient of friction between \( A \) and \( C \) is 0.3, between \( B \) and \( C \) is 0.2 and between \( C \) and the ground is 0.1.
A man of mass 70 kg jumps to a height of 0.8 m from the ground, then the momentum transferred by the ground to the man is
(g = 10 m/s\(^{-2}\)):