The equation is:
3 O₂ + 4 Co → 2 Co₂O₃
Oxidation half reaction:
Co → Co³⁺ + 3 e
Reduction half reaction:
O₂ + 4 e → 2 O²⁻
To balance the equation number of electrons lost must be equal to number or electrons gained so we must multiply oxidation half time 4 and reduction half times 3
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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Answer:
MgO.
Explanation:
charges of both satisfy one another (balanced) -- producing a compound MgO.
Answer:
PubChem CID 16663
Structure Find Similar Structures
Chemical Safety Laboratory Chemical Safety Summary (LCSS) Datasheet
Molecular Formula C9H20
Synonyms 4-ETHYLHEPTANE 2216-32-2 Heptane, 4-ethyl- 4-ethyl-heptane 4-ethyl heptane
Explanation:
