Final answer:
To find the fraction of electrons removed from the copper ball, calculate the number of electrons in the ball and divide the net charge by the elementary charge. The fraction of electrons removed from the copper ball is approximately 1.29 x 10^-17.
Step-by-step explanation:
To find the fraction of electrons removed from the copper ball, we first need to calculate the number of electrons in the ball. Since each neutral copper atom has 29 electrons, we can calculate the number of atoms in the ball by dividing its mass by the atomic mass of copper. So, the number of copper atoms in the ball is 51 g / 63.5 g/mol = 0.803 moles. Since 1 mole contains Avogadro's number of particles (6.022 x 10^23), the number of copper atoms in the ball is 0.803 moles x 6.022 x 10^23 = 4.837 x 10^23 atoms.
The total number of electrons in the copper ball is the same as the number of copper atoms, so it is also 4.837 x 10^23 electrons. To find the fraction of electrons removed, we divide the net charge (1 μC) by the elementary charge (1.6 x 10^-19 C), which gives us 6.25 x 10^6. This is the number of excess protons in the ball, which is equal to the number of missing electrons.
Finally, to find the fraction of electrons removed, we divide the number of missing electrons by the total number of electrons: 6.25 x 10^6 / 4.837 x 10^23 = 1.29 x 10^-17. Therefore, the fraction of electrons removed from the copper ball is approximately 1.29 x 10^-17.
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