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.
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When atoms are excited they emit light of certain wavelengths which correspond to different colors. Each element produces a unique set of spectral lines. Since no two elements emit the same spectral lines, elements can be identified by their line spectrum.
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Which of the following statements about electrical conductors is true?
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a:They have low resistance.
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Answer:
%age Yield = 20.12 %
Solution:
The Balance Chemical Reaction is as follow,
Cl₂ + 3 F₂ → 2 ClF₃
According to Equation ,
114 g (3 mole) F₂ produces = 184.88 g (2 moles) of ClF₃
So,
19.2 g of F₂ will produce = X g of ClF₃
Solving for X,
X = (19.2 g × 184.88 g) ÷ 114 g
X = 31.13 g of ClF₃ (Theoretical Yield)
As we know,
%age Yield = (Actual Yield ÷ Theoretical Yield) × 100
%age Yield = (6.3 g ÷ 31.3 g) × 100
%age Yield = 20.12 %