Respuesta :
Answer:
(3) 0.70 g
Explanation:
Half life of iron-53 = 8.5167 minutes
[tex]t_{1/2}=\frac {ln\ 2}{k}[/tex]
Where, k is rate constant
So,
[tex]k=\frac {ln\ 2}{t_{1/2}}[/tex]
[tex]k=\frac{ln\ 2}{8.5167}\ min^{-1}[/tex]
The rate constant, k = 0.08138 min⁻¹
Time = 25.53 minutes
Using integrated rate law for first order kinetics as:
[tex][A_t]=[A_0]e^{-kt}[/tex]
Where,
[tex][A_t][/tex] is the concentration at time t
[tex][A_0][/tex] is the initial concentration = 5.60 g
So,
[tex]\frac{[A_t]}{5.60}=e^{-0.08138\times 25.53}[/tex]
[tex][A_t]=\frac{5.6}{e^{2.0776314}}[/tex]
[tex][A_t]=0.70\ g[/tex]
Answer- (3) 0.70 g
Answer:
After 25.53 minutes there will remain 0.70 grams of the iron-53 sample. Option 3 is correct.
Explanation:
Step 1: Data given
Original mass of the original iron sample = 5.60 grams
Half life of iron-53 = 8.5167 minutes
Step 2: Calculate the rate constant
k = (ln 2)/(t1/2)
⇒ with k = the rate constant
⇒ with t1/2 = the half-life time = 8.5167 minutes
k = 0.08139 / min
Step 3: Calculate the mass after 25.53 minutes
At = A0*e^(-kt)
⇒ with At = the amount of iron sample after 25.53 minutes
⇒ A0 = the initial amount of iron sample =5.50 grams
⇒ with k = 0.08139 / min
⇒ with t = the time = 25.53 minutes
At = 5.50*e^(-0.08139*25.53)
At = 0.70 grams
After 25.53 minutes there will remain 0.70 grams of the iron-53 sample
