How many uranium atoms are there in 8.7 g of pure uranium? The mass of one uranium atom is 4x10^-26 kg;
2007-09-08 16:37:36 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Physics
The average atomic mass of Uranium found in nature (a mixture of isotopes, but mostly U238), is 238.02891. This means that an Avogadro's number (6.02 x 10^23) of Uranium atoms ("1 mole") weighs 238.02891 grammes. Therefore, 8.7g of Uranium is 8.7/238.02891 moles and contains 6.02 x 10^23 * 8.7/238.02891 = 2.200 x 10^22 atoms
The answer based on your input data for the weight of a Uranium atom would instead be 0.0087kg/(4x10^-26)kg = 2.175 x 10^23 atoms, which is too big by a factor of nearly 10 (your mass for a Uranium atom is wrong).
2007-09-08 16:55:03 · answer #1 · answered by David W 1 · 0⤊ 0⤋
Get the units together. 8.7g = .0087 kg. Then
.0087/4*10^-26 = 2.175*10^23
HTH
Doug
2007-09-08 16:43:34 · answer #2 · answered by doug_donaghue 7 · 0⤊ 0⤋
This is pretty simple division. I think you can figure this one out on your own.
2016-04-03 22:07:15 · answer #3 · answered by Anonymous · 0⤊ 1⤋
Uh-huh. And just what are you building?? Osama?
2007-09-08 16:41:10 · answer #4 · answered by Ham8888888888 3 · 0⤊ 0⤋