A) \[3.7904\text{ }kJ\,mo{{l}^{-1}}\]
B) \[37.904\text{ }kJ\text{ }mo{{l}^{-1}}\]
C) \[41.00\text{ }kJ\text{ }mo{{l}^{-1}}\]
D) \[AgI{{O}_{3}}\]
Correct Answer: C
Solution :
Let h be the height of a tower \[\because \] \[\angle AOB={{60}^{o}}\] \[\therefore \] \[\Delta OAB\]is equilateral. \[\therefore \]\[OA=OB=AB=a\] Now, in\[\Delta OAC\], \[\tan {{30}^{o}}=\frac{h}{a}\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{a}\Rightarrow h=\frac{a}{\sqrt{3}}\]You need to login to perform this action.
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