Schrodinger developed a famous equation that allows the solutions for electron wave functions to be found given a specific potential. For the case of an atom, Schroginger's equation allows the determination of electron wave functions. These wave functions tell us how electrons are distributed in space around the atom.
The isotope that is more abundant, given the data is isotope Li7
<h3>Assumption</h3>
- Let Li6 be isotope A
- Let Li7 be isotope B
<h3>How to determine whiche isotope is more abundant</h3>
- Molar mass of isotope A (Li6) = 6.02 u
- Molar mass of isotope B (Li7) = 7.02 u
- Atomic mass of lithium = 6.94 u
- Abundance of A = A%
- Abundance of B = (100 - A)%
Atomic mass = [(mass of A × A%) / 100] + [(mass of B × B%) / 100]
6.94 = [(6.02 × A%) / 100] + [(7.02 × (100 - A)) / 100]
6.94 = [6.02A% / 100] + [702 - 7.02A% / 100]
6.94 = [6.02A% + 702 - 7.02A%] / 100
Cross multiply
6.02A% + 702 - 7.02A% = 6.94 × 100
6.02A% + 702 - 7.02A% = 694
Collect like terms
6.02A% - 7.02A% = 694 - 702
-A% = -8
A% = 8%
Thus,
Abundance of B = (100 - A)%
Abundance of B = (100 - 8)%
Abundance of B = 92%
SUMMARY
- Abundance of A (Li6) = 8%
- Abundance of B (Li7) = 92%
From the above, isotope Li7 is more abundant.
Learn more about isotope:
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The answer is true, particles in the gaseous state are the furthest apart
Answer: 4.08183 or you can round it to 4
Explanation:
I mol of anything is 6.02 * 10^23 (in this case molecules of water)
6 mols of water = x
1/6 = 6.02 * 10^23 / x Cross multiply
x = 6 * 6.02* 10^23
x = 3.612 * 10^24 molecules in 6 mols of water
