The unsaturated zone is the portion of the subsurface above the groundwater table. The soil and rock in this zone contains air as well as water in its pores. ... Unlike the aquifers of the saturated zone below, the unsaturated zone is not a source of readily available water for human consumption
N = 4 moles of Ar2, P = 1.90 atm, V = ?
T = 50C = 273 + 50K = 323K
PV = nRT --> V = nRT/P
V = (4)(.0821)(323)/1.90
V = 106.07/ 1.9
V = 55.8 L
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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Absorbance is related to the concentration of a substance using the Beer-Lambert's Law. According to this law, absorbance is linearly related to concentration. However, this is only true up to a certain concentration depending on the substance. For this case, we assume that the said law is applicable.
A = kC
Using the first conditions, ewe solve for k.
0.26 = k (0.10)
k = 2.6
A = kC
A = 2.6 (0.20) = 0.52
Therefore, the absorbance at a concentration of 0.20 M and wavelength of 500nm is 0.52.
