Pure- table salt
Impure- vegetable oil
Answer:
aqueous acid is used as a reagent.
Explanation:
Addition of Grignard reagent in aldehyde and followed by the acidification give rise to the primary or secondary alcohol. when the formaldehyde is used than the primary alcohol is formed otherwise secondary alcohol is formed.
in this reaction we also use the aqueous acid for the acidification as a reagent. We add aqueous acid when ethanol is present. This is because ethanol is get converted in the presence of aqueous acid into the chloroethane.
Answer:
1,500 mm
Explanation:
if 1 meter = 1000 mm, 0.5 meters is 500 mm, so 1.50 meters is 1,500 mm
A watering the plant. Plants wilt or droop when there’s not enough water because it helps reduce the amount of a surface area
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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