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Answer: The mass of lead deposited on the cathode of the battery is 1.523 g.
Explanation:
Given: Current = 62.0 A
Time = 23.0 sec
Formula used to calculate charge is as follows.

where,
Q = charge
I = current
t = time
Substitute the values into above formula as follows.

It is known that 1 mole of a substance tends to deposit a charge of 96500 C. Therefore, number of moles obtained by 1426 C of charge is as follows.

The oxidation state of Pb in
is 2. So, moles deposited by Pb is as follows.

It is known that molar mass of lead (Pb) is 207.2 g/mol. Now, mass of lead is calculated as follows.

Thus, we can conclude that the mass of lead deposited on the cathode of the battery is 1.523 g.
Answer : When a parallel circuit is built the voltage across each of the components remains the same, also the total current passed is the equal to sum of the currents passing through each components in the circuits.
When 2 or more components are tried to be connected in parallel they maintain the same potential difference (in voltage) across their ends of the circuit.
The potential differences across the components are the observed to be same in magnitude, and they have identical polarities between them.
Then, this same voltage is applicable to all circuit components connected in parallel.
So, if each bulb is wired to the battery in a separate loop, the bulbs will be in parallel series.