Answer:
antimony-121 has the highest percent natural abundance
Explanation:
percent natural abundance;
121.76 = 120.90 x + 122.90 (1 - x)
121.76 = 120.90 x + 122.90 - 122.90x
121.76 = -2x + 122.90
121.76 - 122.90 = -2x
x= 121.76 - 122.90/ -2
x= 0.57
Where x and 1 - x refers to the relative abundance of each of the isotopes
Percent natural abundance of antimony-121 = 57 %
Percent natural abundance of antimony-123 = (1 - 0.57) = 43%
Let us remember that isotopy refers to a phenomenon in which atoms of the same element have the same atomic number but different mass numbers. This results from differences in the number of neutrons in atoms of the same element.
We can clearly see that antimony-121 has the highest percent natural abundance.
Answer:
HCOOH(aq) + OH-(aq) —> HCOO-(aq) + H2O(l)
Explanation:
HCOOH is a weak acid and so will not ionised completely in solution.
KOH is a strong base and will ionised completely as shown below
KOH(aq) –> K+(aq) + OH-(aq)
The overall reaction can be written as follow:
HCOOH(aq) + K+(aq) + OH-(aq) —> HCOO-(aq) + K+(aq) + H2O(l)
Cancel out the K+ to obtain the net ionic equation as shown below
HCOOH(aq) + OH-(aq) —> HCOO-(aq) + H2O(l)
