Suppose you have 100 grams of radioactive plutonium-239 with a half-life of 24,000 years. how many grams of plutonium-239 will remain after
a.12,000 years
b.24,000 years
c.96,000 years?

To solve this problem, let us first calculate for the rate constant k using the half life formula:

t1/2 = ln 2 / k

where t1/2 = half life period = 24,000 years, therefore k is:

k = ln 2 / 24,000

k = 2.89 x 10^-5 / yr

Now we use the rate equation:

A = Ao e^(-k t)

where,

A = mass of Plutonium-239 after number of years

Ao = initial mass of Plutonium-239

t = number of years

A. t = 12,000 years, find A

A = 100g e^(- 2.89 x 10^-5 * 12,000)

A = 70.7 g

B. t = 24,000 years, find A

A = 100g e^(- 2.89 x 10^-5 * 24,000)

A = 50 g

C. t = 96,000 years, find A

A = 100g e^(- 2.89 x 10^-5 * 96,000)

A = 6.24 g

answer is d because it is trust me

"hydrogen bonding is the attraction of h atoms in the water molecule (slight + charge) to o atoms in a neighboring water molecule (slight - charge); it is a weak interatomic force. but it is still strong enough to the water molecules "stick" together."