E.1 7th edition

Moderators: Chem_Mod, Chem_Admin

Ziyan Wang 3J
Posts: 51
Joined: Wed Sep 18, 2019 12:22 am

E.1 7th edition

Postby Ziyan Wang 3J » Thu Oct 03, 2019 4:31 pm

E.1 The field of nanotechnology offers some intriguing possibilities, such as the creation of fibers one atom wide. Suppose you were able to string together 1.00 mol Ag atoms, each of radius 144pm, into one of these fibers by encapsulating them in carbon nanotubes. How long would the fiber extend?

Samuel Tzeng 1B
Posts: 103
Joined: Sat Aug 24, 2019 12:15 am

Re: E.1 7th edition

Postby Samuel Tzeng 1B » Thu Oct 03, 2019 4:38 pm

1.73*10^14m

005206171
Posts: 107
Joined: Wed Sep 18, 2019 12:20 am

Re: E.1 7th edition

Postby 005206171 » Thu Oct 03, 2019 4:43 pm

Calculate the number of atoms in 1 mol Ag using Avagadro's number
1 mol Ag * (6.022*10^23 atoms Ag/1 mol Ag)= 6.022*10^23 Ag atoms

calculate the diameter of each atom where 2r=d
144ppm*2= 288ppm

multiply the diameter of each atom by the number of atoms to get the total length then convert to meters
(288pm)*(6.022*10^23)=1.73*10^26ppm
1.73*10^26ppm*(1m/10^12ppm)= 1.73*10^14m
1.73*10^14m*(1km/1000m)=1.73*10^11km

Julie Park 1G
Posts: 100
Joined: Thu Jul 25, 2019 12:15 am

Re: E.1 7th edition

Postby Julie Park 1G » Thu Oct 03, 2019 4:47 pm



Step 1: Find the number of Ag atoms in 1.00mol Ag atoms.
Step 2: Multiply the # of Ag atoms by the length (diameter is 144 x 2 = 288pm) of each atom.
Step 3: Convert total length from pm to m, and if needed, convert to km
--> 1.73x10^11km

Noah Canio 3C
Posts: 55
Joined: Wed Sep 18, 2019 12:20 am

Re: E.1 7th edition

Postby Noah Canio 3C » Thu Oct 03, 2019 4:53 pm

This problem appears to be asking for the length of an entire line of Ag atoms stringed together.
We're given the amount of atoms in moles (which again is a unit of measuring many atoms). If we want the total number of atoms in actual atoms and not moles, then we'd multiply 1 mol of Ag by avogadro's number: 6.0221 x 10^23:

1 mol Ag X (6.0221x10^23/1 mol) = 6.0221x10^23 Ag atoms. (This is the total # of Ag atoms in 1 mole of Ag.)

Next, we're given that each atom has a radius of 144pm (Keep in mind that p=pico, which is 10^-12 meters).
However, the radius only gives us half of the distance through an atom (think of a circle). If we want to get that full distance, we'd find the diameter:

144pm X 2 = 288pm.

Now, we have the total number of atoms (6.0221x10^23) and how "long" each atom would be if you were to put one side by side (288pm).
Just multiply those two numbers together:

6.0221x10^23 Ag atoms X 288pm X (1m/10^12pm) Note: an additional step would be to convert picometers to meters.
= 1.73x10^14 meters.


Return to “SI Units, Unit Conversions”

Who is online

Users browsing this forum: No registered users and 1 guest