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:
It corresponds to Charles' Law
Explanation:
Charles's Law corresponds to one of the gas laws, where temperature and volume are related, to constant pressure. That is, according to said equation, the volume of a gas varies directly with the temperature, under conditions of constant pressure.
A.) temperature
By raising the temperature the gas molecules speed up causing collisions with the volume of a container if kept constant.
Answer:
0.550
Explanation:
The absorbance (A) of a substance depends on its concentration (c) according to Beer-Lambert law.
A = ε . <em>l</em> . c
where,
ε: absorptivity of the species
<em>l</em>: optical path length
A 45 mM phosphate solution (solution A) had an absorbance of 1.012.
A = ε . <em>l</em> . c
1.012 = ε . <em>l</em> . 45 mM
ε . <em>l</em> = 0.022 mM⁻¹
We can find the concentration of the second solution using the dilution rule.
C₁ . V₁ = C₂ . V₂
45mM . 11mL = C₂ . 20.0 mL
C₂ = 25 mM
The absorbance of the second solution is:
A = (ε . <em>l</em> ). c
A = (0.022 mM⁻¹) . 25 mM = 0.55 (rounding off to 3 significant figures = 0.550)
