Question

A 50.0-g ball of copper has a net charge of 2.00 µC . What fraction of the copper’s electrons has been removed? (Each copper atom has 29 protons, and copper has an atomic mass of 63.5.)

Answer 1

To calculate the fraction of electrons removed from a 50.0 g ball of copper with a 2.00 µC charge, we determine the total number of electrons in the ball and then find the number corresponding to the charge. We conclude that approximately 9.083 x 10−11% of the copper's electrons have been removed.

Step-by-step explanation:

To determine the fraction of copper's electrons that have been removed, we'll need to calculate the total number of electrons in the 50.0 g ball of copper and then see how many electrons correspond to a charge of 2.00 µC.

First, we calculate the number of moles of copper in the 50.0 g ball. With an atomic mass of 63.5 g/mol for copper, we have:

Number of moles = 50.0 g / 63.5 g/mol = 0.7874 moles

Next, using Avogadro's number (6.022 x 1023 atoms/mol), we find the number of copper atoms:

Number of copper atoms = 0.7874 moles x (6.022 x 1023 atoms/mol) = 4.739 x 1022 atoms

Since each copper atom contributes 29 electrons, the total number of electrons in the ball is:

Total number of electrons = 29 x 4.739 x 1022 atoms = 1.374 x 1024 electrons

To find the number of electrons that corresponds to a charge of 2.00 µC, we use the electron charge (1.602 x 10−19 C/electron):

Number of electrons removed = 2.00 µC / (1.602 x 10−19 C/electron) = 2.00 x 10−6 C / (1.602 x 10−19 C/electron) = 1.248 x 1013 electrons

Now, we find the fraction of electrons removed:

Fraction = Number of electrons removed / Total number of electrons = 1.248 x 1013 / 1.374 x 1024 ≈ 9.083 x 10−11%

So, approximately 9.083 x 10−11% of the copper's electrons have been removed.

Answer 2

The fraction of the copper’s electrons removed is
`9.076* 10^(-13)`.

Step-by-step explanation:

Mass of copper ball = 50.0 g

Moles of copper =
`(50.0 g)/(63.5 g/mol)`

1 mole =
`N_A=6.022* 10^(23)`

Number of copper atoms =
`(50.0 g)/(63.5 g/mol)* 6.022* 10^(23)`

1 atom of copper has 29 protons

Total number of protons in 50.0 g of copper =
`(50.0 g)/(63.5 g/mol)* 6.022* 10^(23) * 29=1.3751* 10^(25)`

Since an atom is a neutral specie which means number of protons are equal to number of electrons.

Total number of electrons =
`1.3751* 10^(25)`....(1)

Net charge on the copper ball =
`2.00/mu C=2.00* 10^(-6) C`

Q=Ne

Q = Total charge

N = Number of electrons

e = charge on an electron =
`1.602* 10^(-19) C`

`2.00* 10^(-6) C=N* 1.602* 10^(-19) C`

`N =1.248* 10^(13)`

Total number of electrons removed = N =
`1.248* 10^(13)`

Fraction of the copper’s electrons has been removed:

`\frac{\text{Number of electrons removed}}{\text{Total electrons}}`

`(1.248* 10^(13))/(1.3751* 10^(25))=9.076* 10^(-13)`

The fraction of the copper’s electrons removed is
`9.076* 10^(-13)`.