A) \[6.72\times {{10}^{-4}}A\]
B) \[6.72\times {{10}^{-5}}A\]
C) \[6.72\times {{10}^{-6}}A\]
D) \[6.72\times {{10}^{-7}}A\]
Correct Answer: D
Solution :
Given, \[C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{C}}\,H-C{{H}_{2}}-OH\] \[C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{C}}}\,-OH\] \[C{{H}_{3}}-C{{H}_{2}}-O-C{{H}_{3}}\] \[C{{H}_{3}}-C{{H}_{2}}-O-C{{H}_{2}}-C{{H}_{3}}\] and \[\underset{n=5/3}{\mathop{3{{I}_{2}}}}\,\xrightarrow{\,}\underset{n=5}{\mathop{IO_{3}^{-}}}\,+\underset{n=1}{\mathop{5{{I}^{-}}}}\,\] volts. The electric field in the block is V = 2 volts. The electric field in the block is \[\therefore \] The drift speed of electrons is \[Eq.\,\,wt\,\,of\,{{I}_{2}}=\frac{254}{5/3}=152.4\] \[-I\] Electron - current \[-I\] \[-N{{O}_{2}}>-Cl>-OC{{H}_{3}}\] \[{{r}_{n}}={{r}_{0}}\times \frac{{{n}^{2}}}{z}\]
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