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SVEN [57.7K]
3 years ago
10

PLEASE HELP!!! The gold foil Rutherford used in his scattering experiment had a thickness of approximately 4×10−3 mm . If a sing

le gold atom has a diameter of 2.9×10−8 cm, how many atoms thick was Rutherford's foil?
Chemistry
2 answers:
san4es73 [151]3 years ago
8 0

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

Ilya [14]3 years ago
3 0

Answer:

16 atoms thick

Explanation:

4*10-3= 37

2.9*10-8=21

37-21=16

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