Answer:
Ka3 for the triprotic acid is 7.69*10^-11
Explanation:
Step 1: Data given
Ka1 = 0.0053
Ka2 = 1.5 * 10^-7
pH at the second equivalence point = 8.469
Step 2: Calculate Ka3
pKa = -log (Ka2) = 6.824
The pH at the second equivalence point (8.469) will be the average of pKa2 and pKa3. So,
8.469 = (6.824 + pKa3) / 2
pKa3 = 10.114
Ka3 = 10^-10.114 = 7.69*10^-11
Ka3 for the triprotic acid is 7.69*10^-11
It results in a combustion reaction
Answer:
Average atomic mass of the vanadium = 50.9415 amu
Isotope (I) of vanadium' s abundance = 99.75 %= 0.9975
Atomic mass of Isotope (I) of vanadium ,m= 50.9440 amu
Isotope (II) of vanadium' s abundance =(100%- 99.75 %) = 0.25 % = 0.0025
Atomic mass of Isotope (II) of vanadium ,m' = ?
Average atomic mass of vanadium =
m × abundance of isotope(I) + m' × abundance of isotope (II)
50.9415 amu =50.9440 amu× 0.9975 + m' × 0.0025
m'= 49.944 amu
Explanation:
The answer is b, because if it gets colder then means more heat is exiting than it is entering.