MCQ
On charging the lead accumulator cell :
- A$PbO _2$ dissolves
- B$PbSO _4$ deposits on the lead electrode
- ✓$H _2 SO _4$ is formed again
- DThe amount of $H _2 SO _4$ is less
✓
Answer
Correct option: C.
$H _2 SO _4$ is formed again
c
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Choose the correct answer from the options given below:
In the reaction $A$ is
→Conductivity $k,$ is equal to $......$
$a. \frac{1}{\text{R}}\frac{1}{\text{A}}$
$b. \frac{\text{G}^*}{\text{R}}$
$c. \wedge_{\text{m}}$
$d. \frac{\text{l}}{\text{A}}$
→$a. \frac{1}{\text{R}}\frac{1}{\text{A}}$
$b. \frac{\text{G}^*}{\text{R}}$
$c. \wedge_{\text{m}}$
$d. \frac{\text{l}}{\text{A}}$