A B and C are all chemical changes. paper tearing is not.
A) They have different masses
Answer:
pt1
The half-life of the element is 1,600 years.
pt2
k = ln(2)/(t1/2) = 0.693/(t1/2).
Where, k is the rate constant of the reaction.
t1/2 is the half-life of the reaction.
∴ k =0.693/(t1/2) = 0.693/(1,600 years) = 4.33 x 10⁻⁴ year⁻¹.
kt = ln([A₀]/[A]),
where, k is the rate constant of the reaction (k = 4.33 x 10⁻⁴ year⁻¹).
t is the time of the reaction (t = ??? year).
[A₀] is the initial concentration of the sample ([A₀] = 312.0 g).
[A] is the remaining concentration of the sample ([A] = 9.75 g).
∴ t = (1/k) ln([A₀]/[A]) = (1/4.33 x 10⁻⁴ year⁻¹) ln(312.0 g/9.75 g) = [8,000 year].
-Hops
