(a) Calculate the number of electrons in a small, electrically neutral silver pin that has...

Kelly Nelson

Kelly Nelson

Answered

2022-01-08

(a) Calculate the number of electrons in a small, electrically neutral silver pin that has a mass of 10.0 g. 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 1.00 mC. How many electrons are added for every 109 electrons already present?

Answer & Explanation

sirpsta3u

sirpsta3u

Expert

2022-01-09Added 42 answers

Step 1
a) It's necessary to determine the number of electrons in a small, electrically neutral silver pin of mass 10g. Molar mass of silver is known M=107.87gmol. It is possible to determine the number of atoms inside the pin
NAg=mπnM61023 atoms
NAg=5.562251022 atoms
Inside every atom there a 47 electrons, it's necesarry to calculate the total amount of charge.
Ne=47NAg
Ne=2.611024 elektrons
qe=47Ne
qe=435087C
Step 2
It's necessary to calculate the number of electrons in 1 mC.
N=1mC1.671019C=5.991015 electrons
Now, we can determine the number of electrons for even 109.
n=NNe109=2.29 for ever 109 electons
Ethan Sanders

Ethan Sanders

Expert

2022-01-10Added 35 answers

Part A
Moles of silver =10.0g107.8gmol=0.093moles
calculate number silver atom using the Avogadro law
=0.093×6.023×1023=5.58×1022atoms
number of electron is therefore number of atom x atomic number that is 5.58×1022×47=2.62×1024electrons
Part B
we well know electron charge is 1.60×1019c
convert 1.00mc to coulombs =11000=1.0×103c
Number of electron in 1.0×103 is therefore
{(1.0×1031.60×1019)}=6.25×1015
number of (109) electrons {(2.62×1024109)}=2.62×1015
number of electron added per 109 is therefore ={(6.25×10252.62×1015)}=2.3per(109)

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