# How Many Electrons Are Removed from a Charged Copper Ball?

• mysticbms
In summary, a 50.0 g ball of copper with a net charge of 2.00 µC has removed 6.25×10^18 protons, equal to the number of electrons in 63.5 g of copper. This means that all the electrons in the copper ball have been removed, leaving a positive charge.
mysticbms

## 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.

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.

Last edited:

I would approach this problem by first converting the given charge of 2.00µC into coulombs, the SI unit of charge. This can be done by using the conversion factor provided in the homework equations, where 1.00 C is equal to 6.25×10^18 protons. This gives us a charge of (2.00×10^-6 C) × (6.25×10^18 protons/1 C) = 1.25×10^13 protons.

Next, I would calculate the number of electrons present in the 50.0 g ball of copper by using the atomic mass of copper (63.5 g/mol) and Avogadro's number (6.02×10^23 particles/mol). This gives us (50.0 g)/(63.5 g/mol) × (6.02×10^23 particles/mol) = 4.76×10^23 electrons.

Finally, I would calculate the fraction of electrons removed by dividing the number of protons removed (1.25×10^13 protons) by the total number of electrons in the copper ball (4.76×10^23 electrons). This gives us a fraction of 2.63×10^-11, meaning that only a very small fraction (0.0000000000263%) of the copper's electrons have been removed.

Overall, this problem requires knowledge of basic chemistry concepts such as atomic mass and Avogadro's number, as well as the ability to convert between units and use conversion factors.

## 1. What is the Fraction of Electrons Removed?

The Fraction of Electrons Removed refers to the percentage or proportion of electrons that have been taken away or removed from a given substance or material.

## 2. How is the Fraction of Electrons Removed calculated?

The Fraction of Electrons Removed is calculated by dividing the number of electrons that have been removed by the total number of electrons in the substance or material, and then multiplying by 100 to convert it into a percentage.

## 3. What is the significance of knowing the Fraction of Electrons Removed?

Knowing the Fraction of Electrons Removed is important in understanding how a substance or material may behave in different environments. It can also provide insight into the electrical conductivity and chemical properties of the substance.

## 4. How does the Fraction of Electrons Removed affect chemical reactions?

The Fraction of Electrons Removed can impact the rate and outcome of chemical reactions, as it determines the number of electrons available for bonding and reactions to occur. It can also affect the stability and reactivity of molecules.

## 5. How can the Fraction of Electrons Removed be controlled or manipulated?

The Fraction of Electrons Removed can be controlled or manipulated through various methods, such as applying an external electrical current, introducing catalysts, or adjusting the temperature or pressure of the environment. These factors can alter the number of electrons available for bonding and reactions to occur, thus changing the Fraction of Electrons Removed.

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