Two compounds containing six carbon atoms on treatment with <span>either with BH</span>₃<span> in THF followed by H</span>₂<span>O</span>₂<span> /OH</span>⁻<span> (OR) with H</span>₂O <span>/ Hg</span>²⁺<span> / H</span>₃O⁺ gives the same products.
The two alkynes, their two different types of reactons and products are given below,
Answer:The volume increases by 11%
Explanation:
When the pressure is decreased, the volume must definitely increase, to maintain the same pressure, the temperature must then also be increased as we see in the question. The volume was increased just as the pressure decreased and the temperature had to be increased to maintain a constant pressure. We have applied the general gas equation in solving the problem.
Given data:
Hydrogen (H) = 3.730 % by mass
Carbon (C) = 44.44%
Nitrogen (N) = 51.83 %
This means that if the sample weighs 100 g then:
Mass of H = 3.730 g
Mass of C = 44.44 g
Mass of N = 51.83 g
Now, calculate the # moles of each element:
# moles of H = 3.730 g/ 1 g.mole-1 = 3.730 moles
# moles of C = 44.44/12 = 3.703 moles
# moles of N = 51.83/14 = 3.702 moles
Divide by the lowest # moles:
H = 3.730/3.702 = 1
C = 3.703/3.702 = 1
N = 3.702/3.702 = 1
Empirical Formula = HCN