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An unknown radioactive substance has a half-life of 3.20 hours. If 39.6 g of the substance is currently present, what mass Ao

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Answer #1

First of all we know that all the nuclear reactiond are of first order reactios.

For a first order reaction rate constant ,

k = ( 2.303 /t )x log ( Ao / A)
Where
Ao = initial mass= ?
A= Mass left after time t =39.6 g
t = time = 8.0 hours
k= rate constant=0693/half life

= 0.693/(3.20hours)

= 0.217 hour^-1

Plug the values we get log(Ao/A)=(kt)/2.303

=0.752

Ao/A=10^0.752=5.65

Ao= 5.65*A

= 5.65*39.6 g

= 223.8 g

Part C:-

For a first order reaction rate constant ,

k = ( 2.303 /t )x log ( Ao / A)
Where
Ao = initial mass
A= Mass left after time t = Ao- 34.0%Ao= 66%Ao=0.66Ao
t = time = ?
k= rate constant=0.693/half life

= 0.693/432 years

= 1.604*10^-3year^-1

Substitute the values we get

t=(2.303/k)*log(Ao/A)

t= 259 years

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