- Sep 5, 2000
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Originally posted by: Heen05
any idea how to go from 1.24 x 102 eV to kg? i hate this...damn these equations
Originally posted by: Heen05
Now if your lab partner determined that there were 1.672 x 1019 atoms of H in an unknown sample of C2H4Cl2, how many milligrams would the unknown sample weigh?
any ideas? heh
Originally posted by: Heen05
Now if your lab partner determined that there were 1.672 x 1019 atoms of H in an unknown sample of C2H4Cl2, how many milligrams would the unknown sample weigh?
Originally posted by: HamburgerBoy
Originally posted by: Heen05
Now if your lab partner determined that there were 1.672 x 1019 atoms of H in an unknown sample of C2H4Cl2, how many milligrams would the unknown sample weigh?
92kg.
Originally posted by: EarthwormJim
Originally posted by: Heen05
Now if your lab partner determined that there were 1.672 x 1019 atoms of H in an unknown sample of C2H4Cl2, how many milligrams would the unknown sample weigh?
any ideas? heh
That's just sad that you have to ask for help on a problem that easy...
Do your own homework. You obviously need the practice.
Originally posted by: Heen05
Now if your lab partner determined that there were 1.672 x 1019 atoms of H in an unknown sample of C2H4Cl2, how many milligrams would the unknown sample weigh?
any ideas? heh
