Ba2+ and Cu2+, and Sr2+ and Li+
Answer:
54 days
Explanation:
We have to use the formula;
0.693/t1/2 =2.303/t log Ao/A
Where;
t1/2= half-life of phosphorus-32= 14.3 days
t= time taken for the activity to fall to 7.34% of its original value
Ao=initial activity of phosphorus-32
A= activity of phosphorus-32 after a time t
Note that;
A=0.0734Ao (the activity of the sample decreased to 7.34% of the activity of the original sample)
Substituting values;
0.693/14.3 = 2.303/t log Ao/0.0734Ao
0.693/14.3 = 2.303/t log 1/0.0734
0.693/14.3 = 2.6/t
0.048=2.6/t
t= 2.6/0.048
t= 54 days
I disagree with the answer pick of D. If you have a neutral pH 7 solution and you proceed to add a base even with an relatively insignificantly low Kb your solution would still be more basic then acidic. The answer should be b which is true that the base only ionizes slightly in aqueous solution. This is also truer to the definition of what Kb represents.
