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PLEASE HELP!!! The gold foil Rutherford used in his scattering experiment had a thickness of approximately 4×10−3 mm . If a single gold atom has a diameter of 2.9×10−8 cm, how many atoms thick was Rutherford's foil?

Respuesta :

timeptest

Answer:

16 atoms thick

Explanation:

4*10-3= 37

2.9*10-8=21

37-21=16

RajatDutt

Answer : the number of gold atoms is 1.38 ×10^5

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