## Fundamentals E1

Jamie Lee 1H
Posts: 54
Joined: Fri Aug 09, 2019 12:15 am

### Fundamentals E1

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?
I know I start with converting the 1 mole of Ag atoms but where does the radius 144 pm come in?

Isabel Day 1D
Posts: 48
Joined: Fri Aug 09, 2019 12:15 am

### Re: Fundamentals E1

The radius is used to find the denominator of the atoms. Then if you multiply the denominator by the amount of atoms, you find the length of the fiber.

Louise Lin 2B
Posts: 97
Joined: Sat Aug 17, 2019 12:15 am

### Re: Fundamentals E1

Start by visualizing the molecules side by side, strung together! The 144 pm they give you for radius should be calculated into the diameter of the atom (so 288 pm).

First, like you said, convert 1 mole of Ag into number of atoms with Avogadro's number.

Next, multiply your answer by 288 pm to find out how long 1 mole of Ag strung together would be! :))

Posts: 100
Joined: Sat Jul 20, 2019 12:15 am

### Re: Fundamentals E1

You were correct in converting the moles into atoms. However, the next step is to convert the atoms into pm. Since it is stated in the question that the atoms are 144pm in radius, you know that they are 288pm in diameter (and it's easier to work with when it's in diameters). Since 1 atom of Ag is 288pm long, you multiply the moles by (288pm Ag)/(1 Ag atom), where then the atoms cancel out and you are left with pm. In the solutions manual they took it a step further and converted pm into m, but that isn't really necessary.

Shrayes Raman
Posts: 129
Joined: Sat Jul 20, 2019 12:15 am

### Re: Fundamentals E1

The 144pm is an atomic radius. The length of the atom is the diameter. Therefore, convert from radius to diameter (r*2 = d) to get 288pm. Then multiply by Avogadro Number to get total length.