Moles and Molar Mass

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Cecilia Jardon 1I
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Moles and Molar Mass

Postby Cecilia Jardon 1I » Thu Oct 04, 2018 10:47 pm

Hey guys, I was wondering if anyone could please explain how to do E1. I was having trouble understanding how to start off. Is there a certain equation that needs to be used?
E1: The field of the technology 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?

Chem_Mod
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Re: Moles and Molar Mass

Postby Chem_Mod » Thu Oct 04, 2018 10:58 pm

Think of what you can do to convert the moles into atoms (hint: look at Avogadro's number, and see if you can use it somehow). From there, each atom has a certain radius, and you can determine the length of the string of atoms, as they are side-by-side.

Helen Mejia 1I
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Re: Moles and Molar Mass

Postby Helen Mejia 1I » Thu Oct 04, 2018 11:00 pm

Hey, so consider using Avogadros number (6.022 x 10^23) and multiplying it by twice the radius (which would be 288 pm).


6.022 x 10^23) Ag x (288 pm x 10^-12)= 1.73 x 10^14

Lynsea_Southwick_2K
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Re: Moles and Molar Mass

Postby Lynsea_Southwick_2K » Thu Oct 04, 2018 11:03 pm

Helen Mejia 3F wrote:Hey, so consider using Avogadros number (6.022 x 10^23) and multiplying it by twice the radius (which would be 288 pm).


6.022 x 10^23) Ag x (288 pm x 10^-12)= 1.73 x 10^14


Why do you multiply the radius twice?

Mariam Baghdasaryan 4F
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Re: Moles and Molar Mass

Postby Mariam Baghdasaryan 4F » Thu Oct 04, 2018 11:04 pm

So you are given 1.00 mol of Ag, but what you are supposed to find is the number of Ag atoms from this given info. because only then can you use the radius given (144pm) to figure out how long the fiber is.
Step 1: you would convert the mol of Ag to number of Ag atoms
1.00 mol Ag x (6.0221x10^23)/mol = 6.02 x 10^23 Ag (the mol cancels out)

Step 2: multiply number of Ag atoms that you found in step 1 by the diameter of each atom. You need to multiply the radius by 2 to get the diameter so you can find how long the fiber will be. So 2(144)=288 pm
6.02 x 10^23 Ag x (288pm/1 Ag) = 1.73 X 10 ^26 pm (Ag cancels out)

Step 3: Now the question didn't specify the units but the answer in the back of the book was in km so the pm has to be converted to km.
1.73 X 10 ^26 pm x (1m/10^12 pm) x (1km/10^3 m) = 1.73 x 10^11 km (this is the answer)

Cecilia Jardon 1I
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Re: Moles and Molar Mass

Postby Cecilia Jardon 1I » Thu Oct 04, 2018 11:22 pm

Thank You. I did the problem the way that you guys told me but I seem to be getting a different answer. I got 1.73 x 10 ^38.
This is what I did.
1.0 mol Ag x (6.0221x10^23 / 1 mol Ag) x (288 pm /1 atom) x (1m/ 10 ^-12 pm) = 1.73 x10 ^38 m.

Helen Mejia 1I
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Re: Moles and Molar Mass

Postby Helen Mejia 1I » Thu Oct 04, 2018 11:26 pm

Mariam Baghdasaryan 4F wrote:So you are given 1.00 mol of Ag, but what you are supposed to find is the number of Ag atoms from this given info. because only then can you use the radius given (144pm) to figure out how long the fiber is.
Step 1: you would convert the mol of Ag to number of Ag atoms
1.00 mol Ag x (6.0221x10^23)/mol = 6.02 x 10^23 Ag (the mol cancels out)

Step 2: multiply number of Ag atoms that you found in step 1 by the diameter of each atom. You need to multiply the radius by 2 to get the diameter so you can find how long the fiber will be. So 2(144)=288 pm
6.02 x 10^23 Ag x (288pm/1 Ag) = 1.73 X 10 ^26 pm (Ag cancels out)

Step 3: Now the question didn't specify the units but the answer in the back of the book was in km so the pm has to be converted to km.
1.73 X 10 ^26 pm x (1m/10^12 pm) x (1km/10^3 m) = 1.73 x 10^11 km (this is the answer)




The solution manual says that the answer is 1.73 x 10^14 m = 1.73 x10^11 km.

Helen Mejia 1I
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Re: Moles and Molar Mass

Postby Helen Mejia 1I » Thu Oct 04, 2018 11:27 pm

Lynsea_Southwick_3F wrote:
Helen Mejia 3F wrote:Hey, so consider using Avogadros number (6.022 x 10^23) and multiplying it by twice the radius (which would be 288 pm).


6.022 x 10^23) Ag x (288 pm x 10^-12)= 1.73 x 10^14


Why do you multiply the radius twice?



Because you need to know how long the fiber will be.

Cecilia Jardon 1I
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Re: Moles and Molar Mass

Postby Cecilia Jardon 1I » Thu Oct 04, 2018 11:29 pm

Actually I saw what I did it was a calculator mistake.

Jovian Cheung 1K
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Re: Moles and Molar Mass

Postby Jovian Cheung 1K » Thu Oct 04, 2018 11:29 pm

@Cecilia
Glad that you fixed it but was planning to say 1m is 10^12pm, not 10^-12pm; I think that's the only problem there, otherwise nothing's wrong with it! :-)

Cecilia Jardon 1I
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Joined: Fri Sep 28, 2018 12:16 am

Re: Moles and Molar Mass

Postby Cecilia Jardon 1I » Thu Oct 04, 2018 11:43 pm

@Jovian thank you! :)


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