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

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

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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