Answer:
Therefore it will take 7.66 hours for 80% of the lead decay.
Explanation:
The differential equation for decay is
Integrating both sides
ln A= kt+c₁
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The initial condition is A(0)= A₀,
.........(1)
Given that the has half life of 3.3 hours.
For half life putting this in equation (1)
[taking ln both sides, ]
⇒k= - 0.21
Now A₀= 1 gram, 80%=0.8
and A= (1-0.8)A₀ = (0.2×1) gram = 0.2 gram
Now putting the value of k,A and A₀in the equation (1)
⇒ t≈7.66
Therefore it will take 7.66 hours for 80% of the lead decay.
The final temperature, t₂ = 30.9 °C
<h3>Further explanation</h3>
Given
24.0 kJ of heat = 24,000 J
Mass of calorimeter = 1.3 kg = 1300 g
Cs = 3.41 J/g°C
t₁= 25.5 °C
Required
The final temperature, t₂
Solution
Q = m.Cs.Δt
Q out (combustion of compound) = Q in (calorimeter)
24,000 = 1300 x 3.41 x (t₂-25.5)
t₂ = 30.9 °C
Answer:
Explanation:
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