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The radioactive decay obeys first order kinetics
the rate law expression for radioactive decay is
![ln\frac{[A_{0}]}{[A_{t}]}=kt](https://tex.z-dn.net/?f=ln%5Cfrac%7B%5BA_%7B0%7D%5D%7D%7B%5BA_%7Bt%7D%5D%7D%3Dkt)
Where
A0 = initial concentration
At = concentration after time "t"
t = time
k = rate constant
For first order reaction the relation between rate constant and half life is:

Let us calculate k
k = 0.693 / 72 = 0.009625 years⁻¹
Given
At = 0.25 A0

time = 144 years
So after 144 years the sample contains 25% parent isotope and 75% daughter isotopes**
Simply two half lives
Answer:
"0.60 g" is the appropriate solution.
Explanation:
The given values are:
Volume of base,
= 30 ml
Molarity of base,
= 0.05 m
Molar mass of acid,
= 400 g/mol
As we know,
⇒ 
On substituting the values, we get
⇒ 
⇒ 
⇒
hence,
⇒ 
On substituting the values, we get
⇒ 
⇒ 
⇒ 
We assume that we have Ka= 4.2x10^-13 (missing in the question)
and when we have this equation:
H2PO4 (-) → H+ + HPO4-
and form the Ka equation we can get [H+]:
Ka= [H+] [HPO4-] / [H2PO4] and we have Ka= 4.2x10^-13 & [H2PO4-] = 0.55m
by substitution:
4.2x10^-13 = (z)(z)/ 0.55
z^2 = 2.31x 10^-13
z= 4.81x10^-7
∴[H+] = 4.81x10^-7
when PH equation is:
PH= -㏒[H+]
= -㏒(4.81x10^-7) = 6.32
A
The number of electrons/protons is the one that determines the atomic number of an element.