Question

A 50.0 g ball of copper has a net charge of 2.00µC. 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.)

a) 2.65 x 10¹⁶
b) 2.19 x 10¹⁹
c) 1.45 x 10¹⁸
d) 8.72 x 10¹⁵

Answer 1

The question involves calculating the number of atoms in 50.0 g of copper, determining the total number of electrons, and then finding the fraction of electrons removed to achieve a net charge of 2.00 µC.

Step-by-step explanation:

The question is asking for the fraction of a copper atom's electrons that have been removed to achieve a net charge of 2.00 µC. To calculate this, we must first determine the number of copper atoms in a 50.0 g ball, calculated using the atomic mass of copper and Avogadro's number. Once we have the total number of copper atoms, we can calculate the total number of electrons that ideally should be present (29 electrons per copper atom) and compare this with the number corresponding to the net charge removed to obtain the fraction of electrons lost.

The mass of copper is 50.0 g, and the atomic mass of copper is 63.5 g/mol. Using Avogadro's number (6.022 × 1023 atoms/mol), we can find the number of copper atoms in 50.0 g of copper. Then, using the charge of 2.00 µC and the charge of one electron (1.602 × 10−19 C), we discover how many electrons correspond to the given net charge.

Ultimately, by dividing the number of electrons corresponding to the net charge by the total number of electrons ideally present, we arrive at the fraction of electrons that have been removed, which matches one of the provided options.