Calculating Electrons in Silver Pin: 10g Mass, 47 e-/atom

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SUMMARY

This discussion focuses on calculating the number of electrons in a 10g silver pin, which contains approximately 2.62 x 1024 electrons based on its atomic structure (47 electrons per atom and a molar mass of 107.87 g/mol). For part (b), the challenge lies in determining how many additional electrons correspond to a charge of 1.00 mC, which equates to 6.25 x 1024 electrons. The conclusion drawn is that for every 109 electrons already present, approximately 2.38 electrons must be added to achieve the specified charge.

PREREQUISITES
  • Understanding of dimensional analysis in chemistry
  • Knowledge of atomic structure and electron configuration
  • Familiarity with the concept of electric charge and Coulombs
  • Basic proficiency in scientific notation and calculations
NEXT STEPS
  • Study the relationship between charge and the number of electrons using the formula Q = n * e, where Q is charge, n is the number of electrons, and e is the elementary charge.
  • Learn about the concept of milliCoulombs and how to convert between different units of electric charge.
  • Explore the principles of electrostatics and how charge distribution affects electron behavior in materials.
  • Investigate the implications of adding electrons to neutral objects and the resulting changes in electrical properties.
USEFUL FOR

This discussion is beneficial for physics students, electrical engineering students, and anyone interested in understanding the relationship between mass, charge, and electron count in conductive materials.

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Homework Statement



(a). Calculate the number of electrons in a small, electrically neutral silver pin that has a mass of 10g. Silver has 47 electrons per atom, and its molar mass is 107.87 g/mol.
(b). Imagine adding electrons to the pin until the negative charge has the very large value of 1.00 mC. How many electrons are added for every 10^9 electrons already present?

The Attempt at a Solution


I got the answers for both (a) and (b) (answers in back of book), but I still don't understand how to get (b). Heres my steps:

For (a), all I did was dimensional analysis to find the amount of electrons in 10 g of Ag, whic is about 2.62 x 10^24 e-. For part (b), I am confused, but what I found out how many electrons in 1 mC, which was 6.25 x 10^24 e-, and divided it by 2.62 x 10^24 e-, and I got 2.38 e- for every 10^9 electrons already present, but still this does not make sense to me. Can someone explain it to me or give me some helpful hints?
 
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How many elementary electrical charges are there in one milliCoulomb?
 
Carid said:
How many elementary electrical charges are there in one milliCoulomb?

6.25 x 10^24 e-... so what do I do with that?
 
How many groups of 10^9 electrons are there in the silver pin?
 
2.62 x 10^24 e-/10^9 e-=2.62 x 10^15
 
So if I divide the number of electrical charges in one milliCoulomb by the number of groups of 10^9 electrons, won't that supply the answer to the question?
 
why?
 
It seems to me that's exactly what they are asking for...

How many electrons are added for every 10^9 electrons already present?
 

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