Posted by **Hannah** on Sunday, December 4, 2011 at 1:26pm.

1) How many miles are covered in a 15km race? (1 mile=5280, 12 in=1 ft, 1in=2.54cm)

I know that I have to convert but I do not know where to start.

2) Silver (107.87amu) has two naturally occuring isotopes. In a typical sample, 51.84%of silver exists as Ag-107 (106.9051 amu). Calculate the atmoic mass of the other isotope.

I was going to set it up like this

(.5184 * 106.9051) + (x * 107.87) but I dont think this will work.

- Chemistry -
**DrBob222**, Sunday, December 4, 2011 at 1:55pm
Start with what you must convert. That is 15 km. Than multiply by the factors to convert to another unit. For example, you don't have the conversion in your post from km to meters or from meters to cm.

There are 1000 m in a km and 100 cm in a meter. Here is how you do them.

15 km x (1000 m/km) x (100 cm/m) x .....

Note that the conversion factor is placed so that the km cancels with km (which converts to meters), then the next factor converts m (m cancels) to cm. Now add the factor for cm to inches, then inches to feet, then feet to miles.

You're on the right tract for the Ag isotope problem.

If 51.84% is Ag 107, then 100-51.84 = 48.16% must be the other isotope. So

(0.5184*106.9051) + (0.4816*x) = 107.87

- Chemistry -
**Hannah**, Sunday, December 4, 2011 at 2:51pm
For #2 I got 55.41 from (.5184 * 106.9051) so

55.41 + 0.4816x = 107.87

-55.41 -55.41

then divide by 0.4816

The answer is 108.91amu but I did not get this.

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