If you are studying cells and heredity bwhat classes are u in
1 answer:
You might be interested in
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).
Answer: (a) The solubility of CuCl in pure water is .
(b) The solubility of CuCl in 0.1 M NaCl is .
Explanation:
(a) Chemical equation for the given reaction in pure water is as follows.
Initial: 0 0
Change: +x +x
Equilibm: x x
And, equilibrium expression is as follows.
x =
Hence, the solubility of CuCl in pure water is .
(b) When NaCl is 0.1 M,
,
,
Net equation:
= 0.1044
So for,
Initial: 0.1 0
Change: -x +x
Equilibm: 0.1 - x x
Now, the equilibrium expression is as follows.
K' =
0.1044 =
x =
Therefore, the solubility of CuCl in 0.1 M NaCl is .