A) 10mA
B) 15mA
C) 20 mA
D) 5 mA
Correct Answer: D
Solution :
For \[\frac{0.1}{0.8}={{\left[ \frac{50}{400} \right]}^{n-1}}\] \[\log \frac{0.1}{0.8}=\left( n-1 \right)\log \frac{50}{400}\] For \[\log \frac{1}{8}=\left( n-1 \right)\log \frac{1}{8}\] \[0.90=\left( n-1 \right)0.90\] \[\Rightarrow \] \[2Na+2{{H}_{2}}O\to 2NaOH+{{H}_{2}}\uparrow \] = 5 mAYou need to login to perform this action.
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