Answer:
16.5 days
Explanation:
Given that:
Half life = 26.5 days
Where, k is rate constant
So,
The rate constant, k = 0.02616 days⁻¹
Using integrated rate law for first order kinetics as:
![[A_t]=[A_0]e^{-kt}](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5BA_0%5De%5E%7B-kt%7D)
Where,
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
is the concentration at time t
![[A_0]](https://tex.z-dn.net/?f=%5BA_0%5D)
is the initial concentration
Given:
35.0 % is decomposed which means that 0.35 of is decomposed. So,
![\frac {[A_t]}{[A_0]}](https://tex.z-dn.net/?f=%5Cfrac%20%7B%5BA_t%5D%7D%7B%5BA_0%5D%7D)
= 1 - 0.35 = 0.65
t = 7.8 min
![\frac {[A_t]}{[A_0]}=e^{-k\times t}](https://tex.z-dn.net/?f=%5Cfrac%20%7B%5BA_t%5D%7D%7B%5BA_0%5D%7D%3De%5E%7B-k%5Ctimes%20t%7D)
<u>t = 16.5 days.</u>
True!
population increases if the number of individuals. the population is the nuber of individuals
Answer:
2C2H6(g) + 7O2(g) ==> 4CO2(g) + 6H2O(g)
One way to find the limiting reactant is to divide the moles of reactant by the corresponding coefficient in the balanced equation and whichever is less is limiting.
For C2H6 we have 2 moles / 2 = 1
For O2 we have 8 moles / 7 = 1.14
Explanation:
If a reaction mixture contains 8 moles of O2(g) with 2 moles of C2H6(g),which reactant is the limiting reactqnt and what is the mass (in grams) remaining of the reactant in excess?
