Uranium–232 has a half-life of 68.9 years. A sample from 206.7 years ago contains 1.40 g of uranium–232. How much uranium was originally present?

Original amount of uranium is 11.3 g.

Explanation:

Given data:

Half life of uranium = 68.9 years

Time taken in decaying= 206.7 years

Amount left after decaying = 1.40 g

Original amount of uranium = ?

Solution;

Formula:

K = 2.303/t log[A₀]/[A] (1)

A₀ = initial amount

A = amount left after decaying

Value of K from first order kinetics.

Half life of uranium = 0.693 /K

68.9 years = 0.693 /k

k = 0.693/68.9 years

k = 0.0101 year⁻¹

Now we will put the values in formula (1)

K = 2.303/t log[A₀]/[A]

0.0101 year⁻¹ = 2.303/206.7 year log [A₀]/[1.40 g]

[A₀] = 11.3g

Original amount of uranium is 11.3 g.

11.2 is the answer, the guy above just copied it from a guy on another question I think. And they are both wrong. 11.3 is not the answer.

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.