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 man
One approach to this can use the density of aluminum to find the mass of the1 cm? block. The density of aluminum is 2.70-g/cm? 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 10?? atoms Al / 1 mol Al) = 6.023 x 10?? atoms Al. Each Al, as you pointed out, contains 13 electrons so we have 6.023 x 10?? atoms Al x 13 electron/atom = 7.83 x 10?? electrons. 1 x 10?? pC = 1 C and 1 C = 6.24 x 10?? electrons=== 1.5 pC x (1C / 1 x 10?? pC) x ( 6.24 x 10?? electrons / 1 C) = 9.36 x 10? electrons a.) % removed = 9.36 x 10? / 7.83 x 10?? x 100% = 1.2 x 10??? % b.) Each electron has a mass of 9.11 x 10???-g , so the total mass removed =9.11 x 10???-g/elec x 9.36 x 10? electrons = 8.53 x 10???-g. Thus the % decrease is (8.53 x 10???-g/ 2.7-g) x 100% = 3.16 x 10??? %
