Question

Excess electrons are placed on a small lead sphere with a mass of 8.30 g so that its net charge is −3.70×10−9 C .

(a) Find the number of excess electrons on the sphere.(b) How many excess electrons are there per lead atom? The atomic number of lead is 82, and its atomic mass is 207 g/mol?

Answer 1

To calculate the number of excess electrons on the lead sphere with a net charge of -3.70×10⁻¹ C, divide the net charge by the charge of one electron. Then, to determine the number of excess electrons per lead atom, divide the total excess electrons by the number of lead atoms calculated from the mass of the sphere, the atomic mass of lead, and Avogadro's number.

Step-by-step explanation:

To find the number of excess electrons on a lead sphere with a net charge of −3.70×10⁻¹ C, we can use the elementary charge of an electron, which is approximately −1.60×10⁻ C.

1. For
2. electrons per lead atom
3. , we consider the atomic mass of lead (207 g/mol) and Avogadro's number (6.022×10¹³ mol⁻¹):

After calculating, we will know how many excess electrons there are per lead atom.

Answer 2

(a)
`2.3125* 10^(10)\ electrons`

(b)
`9.577* 10^(-13)\ electrons/atom`

Step-by-step explanation:

(a) Let the number of the excess electrons on sphere is x.

The charge of the electron (e) =
`-1.60* 10^(-19)\ C`.

Given, the net charge (q) =
`-3.70* 10^(-9)\ C`

Number of excess electrons is:

`n=\frac {q}{e}=\frac {-3.70* 10^(-9)\ C}{-1.60* 10^(-19)\ C}=2.3125* 10^(10)\ electrons`

(b) Given that the mass of the sphere = 8.30 g

Molar mass = 207 g/mol

Since, 1 mole of Pb contains 6.022×10²³ atoms of Pb

Also, 1 mole mass = 207 g

207 g of Pb contains 6.022×10²³ atoms of Pb

1 g of Pb contains 6.022×10²³ / 207 atoms of Pb

So,

8.30 g of Pb contains (6.022×10²³ / 207)*8.30 atoms of Pb

No of atoms of Pb in 8.30g = 2.4146×10²² atoms

`Excess\ electrons\ per\ lead\ atom=\frac {2.3125* 10^(10)\ electrons}{2.4146* 10^(22) atoms}=9.577* 10^(-13)\ electrons/atom`