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### E1

Posted: Thu Oct 05, 2017 12:51 pm
The solutions manual wasn't very clear about this question...

Question: 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 144 pm, into one of these fibers by encapsulating them in carbon nanotubes . How long would the fiber extend?

### Re: E1

Posted: Thu Oct 05, 2017 1:28 pm
This problem is asking you for the length of the fiber---I imagine the fiber to be all the Ag atoms lined up in a row. The length each atom contributes to the total fiber length is its diameter. (eg: if you had 2 atoms each with a diameter of 2cm and you wanted to make a fiber, the fiber would be 4cm long)

There are three steps steps to this problem.
Firstly, 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

Secondly, calculate the diameter of each atom. Since the radius given is 144ppm, the diameter of each atoms is the radius times two.
144ppm*2= 288ppm is the diameter of each atom.

Lastly, multiply the diameter of each atom by the number of atoms to get the total length; then convert to meters (10^12ppm=1m) and then to kilometers (1000m=1km). Make sure the answer is the correct number of sig figs (this particular problem should have a 3 sig fig answer).
(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
The final answer shown in the book is: 1.73*10^11km

### Re: E1

Posted: Fri Oct 06, 2017 9:44 am
The solutions manual wasn't very clear about this question...

Question: 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 144 pm, into one of these fibers by encapsulating them in carbon nanotubes . How long would the fiber extend?

If you line up all the Ag atoms in 1.00 moles, how long would that line be if the radius of one atom was 144pm?
1. Calculate the number of atoms in 1.00 moles of Ag (Multiply by Avogadro's Number 6.022E23) = 6.022E23 atoms
2.The radius only gives half of the length of each atoms... multiply the radius by 2 to get the diameter = 288pm
3.Multiply the diameter of each atom by the total number of atoms to get the length of the atoms lined up... 6.022E23*288
4. Convert to scientific notation and check sig figs

### Re: E1

Posted: Wed Oct 02, 2019 11:02 pm
I don't fully understand why they did their unit conversions as such, or how we knew to convert it to atoms at all. Thanks!

### Re: E1

Posted: Wed Oct 02, 2019 11:55 pm
Responding to Brenna: The question is essentially telling us that we're lining up Ag atoms in a single-file line. You don't have to "convert" the problem to atoms because it is telling you that the unit we're working with is atoms right from the start.

### Re: E1

Posted: Wed Oct 09, 2019 10:45 am
How do we know that we are suppose to convert from pm to km to m?

### Re: E1

Posted: Thu Oct 10, 2019 1:13 pm
I don't believe it is necessary to convert from pm to km. Pm to m should be fine because a meter is the SI fundamental base unit. This can be done with 1 meter= 1*10^12 pm.

### Re: E1

Posted: Fri Oct 16, 2020 5:42 pm
Why should this answer be specifically three sig figs?

### Re: E1

Posted: Sat Oct 17, 2020 11:52 pm
Is there a reason why it was specifically converted to km in the answer key? I don't see the question asking for the answer to be in km

### Re: E1

Posted: Sun Oct 25, 2020 3:32 pm
This answer would be 3 sig figs because in the question, the smallest number of sig figs given was 3 (144 pm and 1.00 mol both have 3 sig figs). When you do calculations with multiplication or division, to determine how may sig figs your answer has, you use the same number of sig figs as the number with the lowest sig figs in the problem's question.
@lwong Dis1L