A) zero
B) 3 m
C) 6 m
D) 9 m
Correct Answer: A
Solution :
\[=2\times 2=\,In\,2\] \[=2\times 2\times 0.693\] \[[\because \,I{{n}^{2}}\,or\,{{\log }_{e}}2=0.693]\] Velocity at \[FeC{{l}_{3}}+3NaOH\xrightarrow{\,}\,Fe{{(OH)}_{3}}+3NaCl\] is \[NaOH\xrightarrow{\,}N{{a}^{\oplus }}+O{{H}^{\odot -}}\] Velocity at \[O{{H}^{\odot -}}\] is \[Fe{{(OH)}_{3}}\] According to work-energy theorem. Work done by all the forces = Change in KE \[Fe{{(OH)}_{3}}\]You need to login to perform this action.
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