A) 2 and 3
B) 2 and 6
C) 1 and 3
D) 3 and 8
Correct Answer: B
Solution :
Balanced equation for producing NO and \[L=\frac{nh}{2\pi }=\frac{2h}{2\pi }=\frac{h}{\pi }\] respectively are \[\because \] ....(i) \[Cu+4HN{{O}_{3}}\xrightarrow[{}]{{}}Cu{{(N{{O}_{3}})}_{2}}+2N{{O}_{2}}+2{{H}_{2}}O\] ...(ii) Here NO and \[\therefore \] are evolved in equal volumes, therefore adding Eqs. (i) and (ii) We get \[\lambda \alpha \frac{1}{p}\]or, \[\Rightarrow \]Thus , coefficients x and y of Cu and \[\frac{\Delta p}{p}=-\frac{\Delta \lambda }{\lambda }\] respectively are 2 and 6.You need to login to perform this action.
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