Answer:
6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.
Explanation:
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:
Concentration is decreased to 1.56 % which means that 0.0156 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)
= 0.0156
Thus,
![\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)
kt = 4.1604
The expression for the half life is:-
Half life = 15.0 hours
Where, k is rate constant
So,
<u>6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.</u>
Answer:
A. The balloons will increase to twice their original volume.
Explanation:
Boyle's law states that the pressure exerted on a gas is inversely proportional to the volume occupied by the gas at constant temperature. That is:
P ∝ 1/V
P = k/V
PV = k (constant)
P = pressure, V = volume.
Let the initial pressure of the balloon be P, i.e. 
, initial volume be V, i.e. 
. The pressure is then halved, i.e. 
Therefore the balloon volume will increase to twice their original volume.
Yes ...................................
Answer:
Alumminum hydroxide reacts with sulfuric acid as follows: 2Al(OH)3+H2SO4-->Al2(SO4)+6H2O.
Explanation:
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Answer:
Well could you please show the answers?
Explanation:
I need to see the statements
