katelynn73
14.09.2019 •
Physics
A2.0-mm-diameter copper ball is charged to 40 nc . what fraction of its electrons have been removed? the density of copper is 8900 kg/m^3.
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Ответ:
0.02442 × 10⁻⁹
Explanation:
Given:
Diameter of copper ball = 2.00 mm = 0.002 m
Charge on ball = 40 nC = 40 × 10⁻⁹ C
Density of copper = 8900 Kg/m³
Now,
The number of electrons removed, n =
also, charge on electron = 1.6 × 10⁻¹⁹ C
Thus,
n =
or
n = 25 × 10¹⁰ Electrons
Now,
Mass of copper ball = volume × density
Or
Mass of copper ball = × 8900
or
Mass of copper ball = × 8900
or
Mass of copper ball = 0.03726 grams
Also,
molar mass of copper = 63.546 g/mol
Therefore,
Number of mol of copper in 0.03726 grams =
or
Number of mol of copper in 0.03726 grams = 5.86 × 10⁻⁴ mol
and,
1 mol of a substance contains = 6.022 × 10²³ atoms
Therefore,
5.86 × 10⁻⁴ mol of copper contains = 5.86 × 10⁻⁴ × 6.022 × 10²³ atoms.
or
5.86 × 10⁻⁴ mol of copper contains = 35.88 × 10¹⁹ atoms
Now,
A neutral copper atom has 29 electrons.
Therefore,
Number of electrons in ball = 29 × 35.88 × 10¹⁹ = 1023.37 × 10¹⁹ electrons.
Hence,
The fraction of electrons removed =
or
The fraction of electrons removed = 0.02442 × 10⁻⁹
Ответ:
31,500
Step-by-step explanation:
there you go hope it helps