Solve Physics Question: 3.0g Copper Penny & 38uC Positive Charge

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Homework Help Overview

The problem involves a 3.0g copper penny that has a positive charge of 38μC, and the original poster seeks to determine the fraction of electrons lost due to this charge. The subject area includes concepts of charge, atomic structure, and stoichiometry related to copper.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the fraction of lost electrons using a formula involving the mass of copper and Avogadro's number but encounters issues with the result. Some participants suggest checking the units and considering the charge of an electron to clarify the calculations.

Discussion Status

The discussion includes attempts to clarify the calculations and units involved in the problem. Some guidance has been offered regarding the necessary steps to approach the question systematically, although the original poster later indicates they have resolved the issue independently.

Contextual Notes

There may be constraints related to the assumptions about the charge distribution in the copper penny and the need for accurate unit conversion in the calculations.

Jodi
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Hi; Could somebody please help me with this question: A 3.0g copper penny has a positive charge of 38uC. What fraction of its electrons has it lost?
The method I tried was: (3g/63.55g) x 6.02E23 x A(what I'm solving for) = 38xE-6. Than I took my answer and divided it by 29. However this will not give me the correct answer, and the correct answer is 1/(3.5E9). Can somebody please help me. Thank you so much.
 
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Look at your units. You have a dimensionless number on the left side and coulombs on the right side. You need to factor in the charge on an electron.
 
Try a more systematic approach:

- How many electrons does it take to make up -38\mbox{\mu C} charge? (hint: use the electron charge)

- How many electrons are in 3.0g copper (when neutral)? (29 per copper atom)
 
Thanks...

Thanks guys, I figured this one out.
 

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