How Many Electrons Are Removed from a Charged Copper Ball?

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    Electrons Fraction
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To determine the fraction of electrons removed from a charged copper ball weighing 50.0 g with a net charge of 2.00 µC, one must first calculate the number of copper atoms in the ball using its atomic mass and Avogadro's Number. Each copper atom has 29 protons, and the total charge can be converted to the number of electrons removed. The discussion highlights the initial confusion about how to approach the problem, with participants suggesting starting with the number of atoms. Ultimately, the solution involves calculating the total number of electrons and the fraction removed based on the given charge. Understanding these calculations is essential for solving the problem accurately.
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Homework Statement


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.)

Homework Equations


1.00 C× (1proton)/(1.60×10^−19C)=6.25×10^18 protons
C = coloumb, the SI unit of charge

The Attempt at a Solution



I don't know where to begin. I haven't taken Chemistry in a long time so I'm pretty stumped.
 
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mysticbms said:

Homework Statement


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.)

Homework Equations


1.00 C× (1proton)/(1.60×10^−19C)=6.25×10^18 protons
C = coloumb, the SI unit of charge

The Attempt at a Solution



I don't know where to begin. I haven't taken Chemistry in a long time so I'm pretty stumped.
How many copper atoms are there in 63.5 grams of copper?
 
The atomic mass tells you the number of grams making up a mole of a substance. A mole is a specific number of particles (look up Avogadro's Number). So start by determining how many atoms comprise the 50.0 g ball of copper.EDIT: Beaten to the punch by SammyS!
 
Got it thanks.
 
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