Answer to Question #85713 in Classical Mechanics for Basheer Ahmad
2019-03-02T08:51:31-05:00
A solid cube of aluminum (density 2.7 g/cm3) has a volume of 0.2 cm3. It is known that 27 g of aluminum contains 6.02 x 1023 atoms. How many aluminum atoms are contained in the cube?
1
2019-03-05T11:07:33-0500
N / M = ( 6.02 ∙ 1 0 23 ) / 27 N/M=(6.02 ∙10^{23})/27 N / M = ( 6.02∙1 0 23 ) /27
n / m = N / M n/m=N/M n / m = N / M
Thus,
n = m N / M = ρ V N / M = ( 2.7 ) ( 0.2 ) ( 6.02 ∙ 1 0 23 ) / 27 = 1.2 ∙ 1 0 22 n=m N/M=ρV N/M=(2.7)(0.2)(6.02 ∙10^{23})/27=1.2∙10^{22} n = m N / M = ρ V N / M = ( 2.7 ) ( 0.2 ) ( 6.02∙1 0 23 ) /27 = 1.2∙1 0 22
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