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.
Explanation:
It is often desirable to determine the mass percent of elements in a given compound.
To determine the mass percent of elements:
- Evaluate the formula mass of the compound. This is done by summing the atomic masses of the atoms in the compound together.
- The mass percentage is determined by pacing the mass contribution of each element or group to the formula mass of the compound and multiply by 100.
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The number of chlorine atoms needed would simply be the
ratio of distance and diameter. But first convert 200 pm to mm:
<span>200 pm = 2 E-7 mm </span>
So the number of chlorine atoms needed is:
<span>1.0 mm / (2 E-7 mm) = 5,000,000 Chlorine atoms = 5 E6
atoms</span>
