One of the best buffer choice for pH = 8.0 is Tris with Ka value of 6.3 x 10^-9.
To support this answer, we first calculate for the pKa value as the negative logarithm of the Ka value:
pKa = -log Ka
For Tris, which is an abbreviation for 2-Amino-2-hydroxymethyl-propane-1,3 -diol and has a Ka value of 6.3 x 10^-9, the pKa is
pKa = -log Ka
= -log (6.3x10^-9)
= 8.2
We know that buffers work best when pH is equal to pKa:
pKa = 8.2 = pH
Therefore Tris would be a best buffer at pH = 8.0.
Answer:
the atmosphere supports life by giving us simple things like wood.
Answer:
Glaciers moving rocks and soil through a valley
Explanation:
The definition of Erosion is "the process of eroding or being eroded by wind, water, or other natural agents". Which in turn means the movement of glaciers is causing erosion on the local valley.
Answer:
<em><u>To determine the number of significant figures in a number use the following 3 rules:</u></em>
<em><u>To determine the number of significant figures in a number use the following 3 rules:Non-zero digits are always significant.</u></em>
<em><u>To determine the number of significant figures in a number use the following 3 rules:Non-zero digits are always significant.Any zeros between two significant digits are significant.</u></em>
<em><u>To determine the number of significant figures in a number use the following 3 rules:Non-zero digits are always significant.Any zeros between two significant digits are significant.A final zero or trailing zeros in the decimal portion ONLY are significant.</u></em>
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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