February 25, 2017

Homework Help: Chemistry (urgent)

Posted by Bella on Tuesday, September 13, 2011 at 5:09am.

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 x10^-8cm , how many atoms thick was Rutherford's foil?
Express your answer using one significant figure.

convert 4x10^-3mm to cm, and you get .04cm.

gold foil thickness divided by atom diameter.
.04cm/2.9x10^-8cm= 1379310 atoms

Is that correct? Also how am I suppose to put that into 1 significant figure? I've already tried 1x10^6 and that was wrong.

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