Answer:
The correct option is;
D. (2)(56 g)
Explanation:
MgCl₂ + 2KOH → Mg(OH)₂ + 2KCl
From the balanced chemical reaction equation, we have;
One mole of MgCl₂ reacts with two moles of KOH to produce one mole of Mg(OH)₂ and 2 moles of KCl
Therefore, the number of moles of KOH that react with one mole of KCl = 2 moles
The mass, m, of the two moles of KOH = Number of moles of KOH × Molar mass of KOH
The molar mass of KOH = 56.1056 g/mol
∴ The mass, m, of the two moles of KOH = 2 moles × 56.1056 g/mol = 112.2112 grams
The amount in grams of KOH that react with one mole of MgCl₂ = 112.2112 grams ≈ 112 grams = (2)(56 g).
Explanation:
sorry I am class ninthonly
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:
brainly.com/question/24311846
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