Answer:
oxidation state of sulphur=x
Explanation:
Na2S4O6=2[+1]+4x+6[-2]=0
+2+4x-12=0
4x-10=0
4x=10
x=10/4=2.5
1) Data:
<span>initial decay rate, No =16,800 disintegrations/min
final decay rate, Nf = 10,860 disintegrations/min
t = 28.0 days
t half-life =?
2) Formulas
Radioactive disintegration =>
Nt = No * e ^ (-kt)
t half-life = ln (2) / k
3) Solution
From Nt = No * e^ (-kt) =>
Nt / No = e ^ (-kt)
=> -kt = ln (Nt / No)
=> kt = ln (No/Nt)
=> k = ln (No / Nt) / t
=> k = ln (16,800 / 10,860) / 28 days = 0.01558 days^ -1
From t half-life = ln(2) / k
t half-life = ln(2) / (0.01558 days^-1) = 44.5 days.
Answer: 44.5 days.
</span>
Answer:
valence shell, and the electrons found in it are called valence electrons.
Explanation: i think hope this helps u
there needs to be more info