Answer:
C
Explanation:
the n value must always be greater than the l value
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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Erm, I think when they are little. When they are just born.
Answer:HNO₃ and NO³⁻ would not function as buffer
Explanation:
The buffer solution are usually prepared by using any weak acid (which would partially dissociate) and mixing this weak acid with its own conjugate base or any weak base (which would partially dissociate) and mixing with with its conjugate acid.
A buffer solution is a solution which resists change in pH of the solution.
Since nitric acid is a very strong acid and hence neither nitric acid HNO₃ or its conjugate base NO³⁻ anionb is suitable for the preparation of buffer solution.
HCO³⁻ is a weak acid and hence it can form a buffer solution with its conjugate base CO₃²-. so they can be used to form buffer.
C₂H₅COOH is a weak acid and hence it can also form buffer solution with its conjugate base.
So only HNO₃and NO³⁻ would not be able to form buffer
So option a is the answer.