Answer:
Eºcell = -1.78 V
Explanation:
The Eº cell is an intensive property, i.e they do not depend on the quantity of material present and the desired reaction in our problem is exactly half the reverse of the one given, the Eºcell will then be the negative of the first then Eºcell is -1.78 V and the redox reaction will be non-spontaneous as opposed to the first.
To move around more. bounce off walls ect. they are more energised
Answer:
A Bunsen bumper,named after Robert Bunsen, is a kind of gas burner used as laboratory equipment;it produces a single open gas flame ,and used for heating ,sterilization,and combustion. The gas can be natural gas (which is mainly methane) or a liquefied petroleum gas,such as propane,butane,or a mixture
Explanation:
A compound inequality<span> is an equation with two or more </span>inequalities<span> joined together with either</span>
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>
