A) 28 m/s2
B) 22 m/s2
C) 12 m/s2
D) 10 m/s2
Correct Answer: B
Solution :
Given that \[=\frac{0.30\times 15\times 60}{96485}\times \frac{197}{3}=0.184g\] \[E_{cell}^{0}=E_{cathode}^{0}-E_{anode}^{0}\]at\[=0.40-(-0.44)=0.84V\] \[\underset{(1-\alpha )}{\mathop{nA}}\,\underset{\frac{\alpha }{n}}{\mathop{An}}\,\]C = 0 When particle is 2m away from the origin, then \[i=1-\alpha +\frac{\alpha }{n}\] \[=\frac{0.75}{253}\]\[V(mL)\]\[STP=\frac{0.75\times 22400}{253}\]\[=66.40mL\] \[K\propto \frac{1}{t}\] \[\log \frac{{{k}_{2}}}{{{k}_{1}}}=\log \frac{{{t}_{1}}}{{{t}_{2}}}=\frac{{{E}_{a}}}{R}\left( \frac{{{T}_{2}}-{{T}_{1}}}{{{T}_{1}}.{{T}_{2}}} \right)\] For\[\log \frac{48}{4}=\frac{{{E}_{a}}}{8.314}\left( \frac{10}{300\times 290} \right)\] \[{{E}_{a}}=78.0\text{kJ}\,\text{mo}{{\text{l}}^{-1}}\]
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