Kelly Nelson

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?

sirpsta3u

Expert

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 $M=107.87\frac{g}{m}ol$. It is possible to determine the number of atoms inside the pin
${N}_{Ag}=\frac{{m}_{\pi n}}{M}\cdot 6\cdot {10}^{23}$ atoms
${N}_{Ag}=5.56225\cdot {10}^{22}$ atoms
Inside every atom there a 47 electrons, it's necesarry to calculate the total amount of charge.
${N}_{e}=47{N}_{Ag}$
${N}_{e}=2.61\cdot {10}^{24}$ elektrons
${q}_{e}=47\cdot {N}_{e}$
${q}_{e}=435087C$
Step 2
It's necessary to calculate the number of electrons in 1 mC.
$N=\frac{1mC}{1.67\cdot {10}^{-19}C}=5.99\cdot {10}^{15}$ electrons
Now, we can determine the number of electrons for even ${10}^{9}$.
$n=\frac{N}{\frac{{N}_{e}}{{10}^{9}}}=2.29$ for ever ${10}^{9}$ electons

Ethan Sanders

Expert

Part A
Moles of silver $=10.0\frac{g}{107.8}\frac{g}{m}ol=0.093moles$
calculate number silver atom using the Avogadro law
$=0.093×6.023×{10}^{23}=5.58×{10}^{22}atoms$
number of electron is therefore number of atom x atomic number that is $5.58×{10}^{22}×47=2.62×{10}^{24}electrons$
Part B
we well know electron charge is $1.60×{10}^{-19}c$
convert 1.00mc to coulombs $=\frac{1}{1000}=1.0×{10}^{-3}c$
Number of electron in $1.0×{10}^{-3}$ is therefore
$\left\{\left(1.0×\frac{{10}^{-3}}{1.60}×{10}^{-19}\right)\right\}=6.25×{10}^{15}$
number of $\left({10}^{9}\right)$ electrons $\left\{\left(2.62×\frac{{10}^{24}}{{10}^{9}}\right)\right\}=2.62×{10}^{15}$
number of electron added per ${10}^{9}$ is therefore $=\left\{\left(6.25×\frac{{10}^{25}}{2.62}×{10}^{15}\right)\right\}=2.3per\left({10}^{9}\right)$

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