A) \[C{{H}_{3}}COOH\]
B) \[{{H}_{2}}S{{O}_{4}}\]
C) \[HCl\]
D) \[HN{{O}_{3}}\]
Correct Answer: D
Solution :
\[3Pb+\underset{(dil.)}{\mathop{8HN{{O}_{3}}}}\,\xrightarrow{\Delta }3Pb{{(N{{O}_{3}})}_{2}}+2NO\] \[+4{{H}_{2}}O\] \[\therefore \] Lead dissolves in dil. \[HN{{O}_{3}}\] but it becomes passive when it reacts with cone. \[HN{{O}_{3}}\].
You need to login to perform this action.
You will be redirected in
3 sec
