Mootiek
- 2
- 0
AMU Too many variables...?
the atomic mass of lithium6 is 6.0151 and the amu of lithium-7 is 7.0160. calculate the natural abuncance of hese two isotopes. use the value of 6.941 amu as the value fr the average atomic mass of lithium
ave atomic weight = (% of isotope 1)(amu of isotope 1)+(% of isotope 2)*(amu of isotope 2)
Okay.. So the equation I set up ends up being
6.941 amu =X(7.0160)+Y(6.0151)
I have 2 variables, so here I get stuck... I have no idea how to solve when I have 2 variables...
Homework Statement
the atomic mass of lithium6 is 6.0151 and the amu of lithium-7 is 7.0160. calculate the natural abuncance of hese two isotopes. use the value of 6.941 amu as the value fr the average atomic mass of lithium
Homework Equations
ave atomic weight = (% of isotope 1)(amu of isotope 1)+(% of isotope 2)*(amu of isotope 2)
The Attempt at a Solution
Okay.. So the equation I set up ends up being
6.941 amu =X(7.0160)+Y(6.0151)
I have 2 variables, so here I get stuck... I have no idea how to solve when I have 2 variables...