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A 9.0-mm-diameter copper ball is charged to 90 nC. Part A What fraction of its electrons have been removed? The density of copper is

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A 9.0-mm-diameter copper ball is charged to 90 nC. Part A What fraction of its electrons have been removed? The density of copper is 8900 kg/m. Nremoved No Submit E| Request Answer ?

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To solve this problem we need to take several steps 1 Calculate the volume of the copper ball 2 Calculate the mass of the copper ball using its volume and density 3 Calculate the number of copper atoms in the ball using its mass and the atomic mass of copper 4 Calculate the total number of electrons originally present in the copper ball 5 Calculate the number of electrons corresponding to the given charge 6 Calculate the fraction of electrons removed Step 1 Calculate the volume of the copper ball The volume V of a sphere is given by the formula V frac43pi r3 where r is the radius of the sphere The diameter of the copper ball is given as 90 mm so the radius r is 90 mm 2 45 mm Convert the radius to meters for consistency in SI units r 45 times 103 m V frac43pi 45 times 1033 Lets calculate the volume V frac43pi 45 times 1033 approx frac43pi times 91125 times 109 V approx 3817 times 109 m3 Step 2 Calculate the mass of the copper ball Mass m is density ho times volume V so m ho V The density of copper is given as 8900 kgm3 m 8900 times 3817 times 109 m approx 3397 times 103 kg Step 3 Calculate the number of copper atoms The number of atoms N in a given mass m can be calculated using the atomic mass MtextCu 63546 gmol for copper and Avogadros number NA 6022 times 1023 atomsmol N fracmMtextCu times NA First convert the mass of copper from kg to g since atomic mass is in grams per mole m 3397 times 103 kg 3397 times 103 g Now calculate N N frac3397 times 103 textg63546 text gmol times 6022 times 1023 text atomsmol N approx frac3397 times 10363546 times 6022 times 1023 N approx 3212 times 1025 atoms Step 4 Calculate the total number of electrons originally present Each copper atom ... blur-text-image

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