<em>K</em>_eq = 0.14
The chemical equation is
A ⇌ B
The equilibrium constant expression is
<em>K</em>_eq = [B]/[A]
If [A] = 7[B]
<em>K</em>_eq = [B]/{7[B]}= 1/7 = 0.14
Answer:
It is A. hydrogen -2 has one more proton
Answer:
1/32 of the original sample
Explanation:
We have to use the formula
N/No = (1/2)^t/t1/2
N= amount of radioactive sample left after n number of half lives
No= original amount of radioactive sample present
t= time taken for the amount of radioactive samples to reduce to N
t1/2= half-life of the radioactive sample
We have been told that t= five half lives. This implies that t= 5(t1/2)
N/No = (1/2)^5(t1/2)/t1/2
Note that the ratio of radioactive samples left after time (t) is given by N/No. Hence;
N/No= (1/2)^5
N/No = 1/32
Hence the fraction left is 1/32 of the original sample.
It is limited by the depth divers and equipment can go.
