Homework problem E1
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Homework problem E1
Could someone please exercise E1. I looked at the solutions but I am still a bit confused.
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Re: Homework problem E1
Basically it's asking the for the length of a chain of 1 mol of atoms strung together. Since there's 6.02 x 10^23 atoms in a mol (avogadro's number), you would multiply the radius of one Ag atom by the number of atoms in 1 mol, which gives you the total length.
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Re: Homework problem E1
One tip for that problem is that they give you the radius of one atom, but to find the total length you must use the diameter, so multiply by two. You'll need to convert moles to atoms first, as the radius and diameter are for one atom, not one mole. Then convert pm to m using 1 m = 1 x 10^12 pm.
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Re: Homework problem E1
This problem is basically asking you to find the the length of all of the fiber from 1.00 moles of Ag atoms. Knowing this you can multiple 1.00 mol Ag by Avogadro's number to get the atoms of Ag. From there, since the radius of one Ag atom is 144 pm, the diameter is 288 pm which you need in order to find the length of the fiber when all the Ag atoms are lined up. Multiplying the atoms of Ag by 288 pm/atoms of Ag gives you the length of the fiber in pm. Convert pm to meters by dividing by 10^12 and you will get the answer 1.73x10^14 meters.
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Re: Homework problem E1
Since you want to find the length of the atoms in 1 mol of a element, you would mutiple the given radius by 2 to get the diameter and then mutiple it by the total number of atoms (6.02 x 10^23).
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Re: Homework problem E1
E1.
Step 1: 1 mol Ag * (6.022*10^23 Ag atoms / 1 mol Ag) = 6.022*10^23 atoms of Ag
Step 2: Multiply 144 by 2 to get 288 (convert radius to diameter), and then multiply by the number of atoms of Ag. --> 288 * 6.022*10^23 = 1.73*10^26pm
Step 3: Convert our answer from picometers (pm) to meters (m). --> (1.73*10^26 pm) * (1 km) / (1*10^15 pm) = 1.73 * 10^11 m
Step 1: 1 mol Ag * (6.022*10^23 Ag atoms / 1 mol Ag) = 6.022*10^23 atoms of Ag
Step 2: Multiply 144 by 2 to get 288 (convert radius to diameter), and then multiply by the number of atoms of Ag. --> 288 * 6.022*10^23 = 1.73*10^26pm
Step 3: Convert our answer from picometers (pm) to meters (m). --> (1.73*10^26 pm) * (1 km) / (1*10^15 pm) = 1.73 * 10^11 m
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Re: Homework problem E1
For me, the easiest way to approach this problem was simply as a dimensional analysis problem. As with any dimensional analysis, I started by establishing conversion factors.
It is important to note that the problem gives the radius of an Ag atom, but since we are thinking of a string of atoms lined up end-to-end, we actually need the diameter of each atom. If each atom has a radius of 144pm, the diameter is 288pm.
The conversion factors we need to solve the problem, therefore, are 6.022 * 10^23 atoms/mole; 288pm length/atom, and 1m/10^12pm. Starting with 1.0 mole of Ag atoms, multiply these conversion factors, in order, straight across so that all the units cancel and you are left with a 1.73 *10^14 meter fiber.
It is important to note that the problem gives the radius of an Ag atom, but since we are thinking of a string of atoms lined up end-to-end, we actually need the diameter of each atom. If each atom has a radius of 144pm, the diameter is 288pm.
The conversion factors we need to solve the problem, therefore, are 6.022 * 10^23 atoms/mole; 288pm length/atom, and 1m/10^12pm. Starting with 1.0 mole of Ag atoms, multiply these conversion factors, in order, straight across so that all the units cancel and you are left with a 1.73 *10^14 meter fiber.
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Re: Homework problem E1
Basically like everyone else has said, you take Avogadro's Number and multiple it by the radiusx2 so you are multiplying the diameter. The only confusing part for me was having to convert from pm to m in which you have to divide the number that you got by 10^12 because "p" stands for pico which is the base unit x 10^-12
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Re: Homework problem E1
I imagined the atoms just as units of lengths rather than actual atoms. Once you convert the mole(s) of Ag to atoms w/ Avogadro's number, you'll basically have half of the actual length of the chain. The atoms' diameter rather than radius has to be accounted for, so you double the amount you calculated.
Re: Homework problem E1
The problem is asking how long the chain is where the chain is 1 mole of argon atoms long. It gives you the length of one of the argon atoms and you would then just multiply that by how many atoms are in a mole (Avogadro's number) to find how long the chain is.
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