chemistry conversions and denisty

Question

(a) Atoms are very small compared to objects on the macroscopic scale. The radius of a gold atom is 144 pm. What is this value in meters and in centimeters?  m   cm

(b) The mass of a single gold atom is 3.27×10^-22 g. Suppose enough Au atoms were lined up like beads on a string to span a distance of 33.5 cm ( 13 inches). How many atoms would be required?  atoms

What mass in grams of Au would be used?  g

Could you weigh out this amount of gold using a typical laboratory balance? _____yesno

(c) Taking the density of gold metal to be 19.3 g/cm³, calculate the mass of metal needed to form a piece of Au wire with the same length as the distance in b, but with a diameter of 1.50 mm. Hint: The volume of a cylinder is π times its radius squared times its height. (V = π r² h)  g

How many gold atoms does this represent?  atoms

Answer 1

a) 144 pm x 1 m / 10x10¹² pm = 144x10^-12 m = 1.44x10^-10 m

1.44x10^-10 m x 10² cm / m = 1.44x10^-8 cm

b) 33.5 cm x 1 Au atom / 1.44x10^-8 cm = 2.33x10⁹ Au atoms

2.33x10⁹ atoms x 3.27x10^-22 g/atom = 7.13x10^-14 g

This could not be weighed on a typical laboratory balance. It is too small.

You can do the rest, or maybe someone else will answer. I've got to run.

Answer 2

Hello, Hannah,

I "thought" I had answered this question already, but it hasn't appeared. I believe because the table I entered exceeded the space requirement. I tried again, but the table won't fit and I am out of time for a written explanation. I've copied and pasted a worksheet into Google Sheets: https://docs.google.com/spreadsheets/d/1HTGpvmWmYplE2LvmC7bI_Gw-tGZi_Seu4LFfZKyj-78/edit?usp=sharing

I hope you'll find the entries make enough sense to see what I did. Units are important in this interpretation - this problem is one primarily of unit conversions.

The line numbers with the answers are: a: 4,5. b) 8, 10, No c) 18, 21

Sorry I can't add more detail. Ask if you have questions, and I can repond late today.

Bob