Tuesday
August 4, 2015

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 410^−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.

Answer this Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Members