An unknown radioactive substance has a half-life of 3.20hours . if 46.2g of the substance is currently present, what mass a0 was present 8.00 hours ago? express your answer with the appropriate units.

americium-241 is used in some smoke detectors. it is an alpha emitter with a half-life of 432 years. how long will it take in years for 34.0% of an am-241 sample to decay?

express your answer with the appropriate units.

a fossil was analyzed and determined to have a carbon-14 level that is 80% that of living organisms. the half-life of c-14 is 5730 years. how old is the fossil?

express your answer with the appropriate units.

a) a0 was 46.2 grams

b) It will take 259 years

c) The fossil is 1845 years old

Explanation:

An unknown radioactive substance has a half-life of 3.20hours . If 46.2g of the substance is currently present, what mass A0 was present 8.00 hours ago?

A = A0 * (1/2)^(t/h)

⇒ with A = the final amount = 46.2 grams

⇒ A0 = the original amount

⇒ t = time = 8 hours

⇒ h = half-life time = 3.2 hours

46.2 = Ao*(1/2)^(8/3.2)

Ao = 261.35 grams

Americium-241 is used in some smoke detectors. It is an alpha emitter with a half-life of 432 years. How long will it take in years for 34.0% of an Am-241 sample to decay?  

t = (ln(0.66))-0.693) * 432 = 259 years

It will take 259 years

A fossil was analyzed and determined to have a carbon-14 level that is 80% that of living organisms. The half-life of C-14 is 5730 years. How old is the fossil?

t = (ln(0.80))-0.693) * 5730 = 1845

The fossil is 1845 years old

every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. therefore, there must be the same number of atoms of each element on each side of a chemical equation.