I just got this question; the answer is C: supersaturated. Hope this helps.
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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Answer:
3.49 g
Explanation:
The mass is the product of volume and density:
(8.96 g/cm³)(0.39 cm³) ≈ 3.49 g
The mass of a pure-copper penny would be 3.49 g.
A substance that cant be broken down by cheamical or physical means is an Element.
Hope this helps
Answer:
449730.879 cal/g
Explanation:
Given data:
Mass of sample = 4.9 g
Change in temperature = 2.08 °C (275.23 k)
Heat capacity of calorimeter = 33.50 KJ . K⁻¹
Solution:
C(candy) = Q/m
Q = C (calorimeter) × ΔT
C(candy) = C (calorimeter) × ΔT / m
C(candy) = 33.50 KJ . K⁻¹ × 275.23 K / 4.90 g
C(candy) = 9220.205 KJ / 4.90 g
C(candy) = 1881.674 KJ / g
It is known that,
1 KJ /g = 239.006 cal/g
1881.674 × 239.006 = 449730.879 cal/g
