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A piece of plastic has a net charge of +9.9 μC. How many more protons than electrons does this piece of plastic have? (e = 1.60

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Explanation:

Given that,

Charge $Q=+9.9\times10^{-6}\ C$

We know that,

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Where,

Q = total charge

n = number of protons

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Put the value into the formula

$n = \dfrac{Q}{e}$

$n=\dfrac{9.9\times10^{-6}}{1.6\times10^{-19}}$

$n=6.1875\times10^{13}\ C$

According to statement of question

Divide the answer by $10^{13}$

$n=\dfrac{6.1875\times10^{13}}{10^{13}}$

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