A) \[N{{H}_{2}}\]
B) \[{{N}_{3}}H\]
C) \[N{{H}_{3}}\]
D) \[{{N}_{2}}{{H}_{4}}\]
Correct Answer: D
Solution :
In an unknown compounds containing N and H given % of H = 12.5% \[\therefore \]% of N = 100 ? 12.5 = 87.5%Element | Percentage | Atomic ratio | Simple ratio |
H | 12.5% | \[\frac{12.5}{1}=12.5\] | \[\frac{12.5}{6.25}=2\] |
N | 87.5 | \[\frac{87.5}{14}=6.25\] | \[\frac{6.25}{6.25}=1\] |
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