Answer:
a) [A⁻]/[HA] = 0.227
b) [A⁻]/[HA] = 0.991
c) [A⁻]/[HA] = 2.667
Explanation:
In the Henderson-Hasselbalch equation, HA stands from an acid an A⁻ stands from its conjugate base, as follows:
pH = pka + Log [A⁻]/[HA]
pH = 4.874 + Log[CH₃CH₂CO₂⁻]/[CH₃CH₂CO₂H]
4.23 = 4.874 + Log [A⁻]/[HA]
-0.644 = Log [A⁻]/[HA]
= [A⁻]/[HA]
0.227 = [A⁻]/[HA]
4.87 = 4.874 + Log [A⁻]/[HA]
-0.004 = Log [A⁻]/[HA]
= [A⁻]/[HA]
0.991 = [A⁻]/[HA]
5.30 = 4.874 + Log [A⁻]/[HA]
0.426 = Log [A⁻]/[HA]
= [A⁻]/[HA]
2.667 = [A⁻]/[HA]
Answer:
C15 H31 O4 S
Explanation:
molecular formula is also the same because the value of "n" is 1
Explanation:
nnbbnmkmknn bnnnbbtbbbbn' nn' t
Answer:
edfgkvisiaixiicciciviicsiaiaqwododc
<u>Answer:</u>
<em>The situation given here is imaginary such that the life of Rock has to be found using the half-life of the element lokium that has been found inside the rock. </em>
<u>Explanation:</u>
Half-life of any material is the amount of time taken by that particular material to decay. Now the amount of lokium found in rock can show after how many half-lives this amount has been left out.
The time elapsed will be log (L) atoms X half-life.
