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# Removing electrons from Aluminum

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• Removing electrons from Aluminum
• ### Removing electrons from Aluminum

nashtwefe > 01-13-2020, 09:59 AM

Suppose a cube of aluminum which is 1.00 cm on a side accumulates a net charge of +1.50 pC.(a) What percentage of the electrons originally in the cube was removed?(b) By what percentage has the mass of the cube decreased because of this removal?So for a you need to find total number of electrons removed (which gives it the +1.5 pC charge), and divide that by the number of total electrons in Aluminum right? 13 electrons in Aluminum..and i really don\'t know much else on this. I\'ve been searching the internet for hours manOne approach to this can use the density of aluminum to find the mass of the1 cm³ block. The density of aluminum is 2.70-g/cm³ so your cube has a mas of 2.70-g. The number of aluminum atoms in this block is: 2.7-g Al x (1 mol Al / 27.0-g Al) x (6.023 X 10²³ atoms Al / 1 mol Al) = 6.023 x 10²² atoms Al. Each Al, as you pointed out, contains 13 electrons so we have 6.023 x 10²² atoms Al x 13 electron/atom = 7.83 x 10²³ electrons. 1 x 10¹² pC = 1 C and 1 C = 6.24 x 10¹⁸ electrons=== 1.5 pC x (1C / 1 x 10¹² pC) x ( 6.24 x 10¹⁸ electrons / 1 C) = 9.36 x 10⁶ electrons a.) % removed = 9.36 x 10⁶ / 7.83 x 10²³ x 100% = 1.2 x 10⁻¹⁵ % b.) Each electron has a mass of 9.11 x 10⁻²⁸-g , so the total mass removed =9.11 x 10⁻²⁸-g/elec x 9.36 x 10⁶ electrons = 8.53 x 10⁻¹⁴-g. Thus the % decrease is (8.53 x 10⁻¹⁴-g/ 2.7-g) x 100% = 3.16 x 10⁻¹² %Other related question