Suppose a piece of dirt in an electrostatic precipitator has 1.05 times 10^17 pr

Question

Suppose a piece of dirt in an electrostatic precipitator has 1.05 times 10^17 protons in it and carries a net charge of -4.9 nC. How many electrons does it have? Numeric: A numeric value is expected and not an expression. N_0 = _______ A 51 g cube of copper has a net charge of 2.1 mu c. What fraction of the copper's electrons has been removed? Each copper atom has 29 protons, and copper has an atomic mass of 63.5. Numeric: A numeric value is expected and not an expression. N_removed/N_initial = _______

Explanation / Answer

1. Net charge = (np - ne) (e)

-4.9 x 10^-9 = (np - ne) / (1.6 x 10^-19)

np - ne = - 3.0625 x 10^10

ne = 1.049999694 x 10^17

2. mole = 51 / 63.5

number of atoms = mole x NA

= (51 / 63.5) (6.022 x 10^23)

number of electrons = (51 / 63.5) (6.022 x 10^23) (29)

electrons removed = (2.1 x 10^-6) / (1.602 x 10^-19)

fraction = electron removed / number of elecron

= 9.346 x 10^-13

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