Given the reaction: cu(s) + 4hno3(aq) → cu(no3)2(aq) + 2no3(g) + 2h2o(l )as the reaction occurs, what happens to copper?
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Answer:
Explanation:
The half-life of Na-24 (15 h) is the time it takes for half of it to decay.
After one half-life, half (50 %) of the original amount will remain.
After a second half-life, half of that amount (25 %) will remain, and so on.
We can construct a table as follows:
No. of Fraction Mass
<u>half-lives t/da Remaining Remaining/g </u>
0 0 1 18
1 15 ½ 9.0
2 30 ¼ 4.5
3 45 ⅛ 2.2
4 60 ⅟₁₆ 1.1
5 75 ⅟₃₂ 0.56
6 90 ⅟₆₄ 0.28
We see that remain after five half-lives (75 h).