## Question E1(6th Edition)

Danny Zhang 4L
Posts: 62
Joined: Fri Sep 28, 2018 12:26 am

### Question E1(6th Edition)

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 (see Chapter 6). How long would the fiber extend?

The way I approached this problem was 1*6.022x10^(23)*144x10^(-12), which got me 8.67x10^(13) m.
However, the textbook suggests that the answer is 1.73x10^(11) km, and I do not know how to get that answer.
If anyone else is able to get 1.73x10^(11) km, I would like to know how you approached this problem.
Thank you!

Leela_Mohan3L
Posts: 44
Joined: Fri Sep 28, 2018 12:26 am

### Re: Question E1(6th Edition)

I began this problem by multiplying 1.00 mol of Ag by Avogadro's number to get 6.022x10^23 atoms of Ag. Then I multiplied that by 2(144 pm)--144 pm is the radius of the atom, so to get the total length of the atom you have to double it. The end result was 1.73x10^26 pm, which is equal to 1.73x10^11 km. Hope that helped!

Nicolette_Canlian_2L
Posts: 77
Joined: Fri Sep 28, 2018 12:25 am
Been upvoted: 1 time

### Re: Question E1(6th Edition)

Can we leave our answer in pm, or do we have to convert it to km?

Danny Zhang 4L
Posts: 62
Joined: Fri Sep 28, 2018 12:26 am

### Re: Question E1(6th Edition)

Considering what Dr.Lavelle said in class on Monday, I believe that it is fine for the answers' units to be in any form for problems in homework and tests when it is not specified. He mentioned that we may want our units to be simplified a certain way only if we are in a lab setting.

Dayna Pham 1I
Posts: 98
Joined: Fri Sep 28, 2018 12:16 am
Been upvoted: 3 times

### Re: Question E1(6th Edition)

Nicolette_Canlian_3G wrote:Can we leave our answer in pm, or do we have to convert it to km?

Hi, Nicolette!

Even though the textbook answer is in kilometers, I would write my answer in meters, since meters is the standard SI unit for length. In this case though, I assume they converted all the way to kilometers since it was a large answer. Either way, it shouldn't be an issue since the textbook didn't specify which units to convert to.

Hope this helped!

harperlacroix1a
Posts: 43
Joined: Fri Sep 28, 2018 12:19 am

### Re: Question E1(6th Edition)

Leela_Mohan3L wrote:I began this problem by multiplying 1.00 mol of Ag by Avogadro's number to get 6.022x10^23 atoms of Ag. Then I multiplied that by 2(144 pm)--144 pm is the radius of the atom, so to get the total length of the atom you have to double it. The end result was 1.73x10^26 pm, which is equal to 1.73x10^11 km. Hope that helped!

why do we multiply 2(144) by the atoms of Ag? I understood the problem up until that point.

Brandon Mo 4K
Posts: 70
Joined: Fri Sep 28, 2018 12:15 am

### Re: Question E1(6th Edition)

harperlacroix1a wrote:
Leela_Mohan3L wrote:I began this problem by multiplying 1.00 mol of Ag by Avogadro's number to get 6.022x10^23 atoms of Ag. Then I multiplied that by 2(144 pm)--144 pm is the radius of the atom, so to get the total length of the atom you have to double it. The end result was 1.73x10^26 pm, which is equal to 1.73x10^11 km. Hope that helped!

why do we multiply 2(144) by the atoms of Ag? I understood the problem up until that point.

Since the question is asking for the length made by the atoms of Ag. You want to multiply 1.00 mol of Ag by Avogadro's number to get the number of atoms of Ag. The reason you multiply this by 2(144 pm) is that the radius of 1 atom of Ag is 144 pm. You multiply by 2 to get the diameter (d = 2r) of the Ag atom so that you can find the length of the fiber. I think you can imagine all of the 6.022x10^23 atoms of Ag being lined up.

Stevin1H
Posts: 89
Joined: Fri Sep 28, 2018 12:17 am

### Re: Question E1(6th Edition)

For this problem, we need to know that Avogadro's # = 6.022x10^23. And given that the radius is 144pm, the radius is only half the distance, so we would multiply it by 2 in order to get the full length. 2(144) = 288pm. Given that we have 1.00 mol Ag, we convert the mol to atoms by multiplying it by Avogadro's #. (1mol Ag)(6.022x10^23 atoms) = 6.022x10^23 atoms Ag.

Now that we have the number of atoms of Ag, we multiply the length of each atom (288pm) to the total number of atoms that we previously found and convert it to meters. (6.022x10^23 atoms Ag)(288pm)(1x10^-12m) = 1.73x10^14m Ag.

However, the answer was in kilometers so I converted again. (1.73x10^14m Ag)(1x10^-3km)=1.73x10^11 km.