The rate constant of the reaction K we can get it from this formula:
K=㏑2/ t1/2 and when we have this given (missing in question):
that we have one jar is labeled t = 0 S and has 16 yellow spheres inside and the jar beside it labeled t= 10 and has 8 yellow spheres and 8 blue spheres and the yellow spheres represent the reactants A and the blue represent the products B
So when after 10 s and we were having 16 yellow spheres as reactants and becomes 8 yellow and 8 blue spheres as products so it decays to the half amount so we can consider T1/2 = 10 s
a) by substitution in K formula:
∴ K = ㏑2 / 10 = 0.069
The amount of A (the reactants) after N half lives = Ao / 2^n
b) so no.of yellow spheres after 20 s (2 half-lives) = 16/2^2 = 4
and the blue spheres = Ao - no.of yellow spheres left = 16 - 4 = 12
c) The no.of yellow spheres after 30 s (3 half-lives) = 16/2^3 = 2
and the blue spheres = 16 - 2 = 14
Answer: 9.09 %
Explanation:
To calculate the percentage concentration by volume, we use the formula:
Volume of ethanol (solute) = 30 ml
Volume of water (solvent) = 300 ml
Volume of solution= volume of solute + volume of solution = 30+ 300 = 330 ml
Putting values in above equation, we get:
Hence, the volume percent of solution will be 9.09 %.
Answer:
C) Modernization of humans
Answer:
C). The Bohr-Rutherford model
Explanation:
The 'Bohr-Rutherford model' of the atom failed to elaborate on the attraction between some substances. It essentially targeted hydrogen atoms and failed to explain its stability across multi-electrons. The nature and processes of the chemical reactions remained unillustrated and thus, this is the key drawback of this model. Thus, <u>option C</u> is the correct answer.
Answer :
(A) The rate expression will be:
![Rate=-\frac{1}{2}\frac{d[HBr]}{dt}=+\frac{d[H_2]}{dt}=+\frac{d[Br_2]}{dt}](https://tex.z-dn.net/?f=Rate%3D-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BHBr%5D%7D%7Bdt%7D%3D%2B%5Cfrac%7Bd%5BH_2%5D%7D%7Bdt%7D%3D%2B%5Cfrac%7Bd%5BBr_2%5D%7D%7Bdt%7D)
(B) The average rate of the reaction during this time interval is, 0.00176 M/s
(C) The amount of Br₂ (in moles) formed is, 0.0396 mol
Explanation :
Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.
The given rate of reaction is,

The expression for rate of reaction :
![\text{Rate of disappearance of }HBr=-\frac{1}{2}\frac{d[HBr]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20disappearance%20of%20%7DHBr%3D-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BHBr%5D%7D%7Bdt%7D)
![\text{Rate of disappearance of }H_2=+\frac{d[H_2]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20disappearance%20of%20%7DH_2%3D%2B%5Cfrac%7Bd%5BH_2%5D%7D%7Bdt%7D)
![\text{Rate of formation of }Br_2=+\frac{d[Br_2]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20formation%20of%20%7DBr_2%3D%2B%5Cfrac%7Bd%5BBr_2%5D%7D%7Bdt%7D)
<u>Part A:</u>
The rate expression will be:
![Rate=-\frac{1}{2}\frac{d[HBr]}{dt}=+\frac{d[H_2]}{dt}=+\frac{d[Br_2]}{dt}](https://tex.z-dn.net/?f=Rate%3D-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BHBr%5D%7D%7Bdt%7D%3D%2B%5Cfrac%7Bd%5BH_2%5D%7D%7Bdt%7D%3D%2B%5Cfrac%7Bd%5BBr_2%5D%7D%7Bdt%7D)
<u>Part B:</u>
![\text{Average rate}=-\frac{1}{2}\frac{d[HBr]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BAverage%20rate%7D%3D-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BHBr%5D%7D%7Bdt%7D)


The average rate of the reaction during this time interval is, 0.00176 M/s
<u>Part C:</u>
As we are given that the volume of the reaction vessel is 1.50 L.
![\frac{d[Br_2]}{dt}=0.00176M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BBr_2%5D%7D%7Bdt%7D%3D0.00176M%2Fs)
![\frac{d[Br_2]}{15.0s}=0.00176M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BBr_2%5D%7D%7B15.0s%7D%3D0.00176M%2Fs)
![[Br_2]=0.00176M/s\times 15.0s](https://tex.z-dn.net/?f=%5BBr_2%5D%3D0.00176M%2Fs%5Ctimes%2015.0s)
![[Br_2]=0.0264M](https://tex.z-dn.net/?f=%5BBr_2%5D%3D0.0264M)
Now we have to determine the amount of Br₂ (in moles).



The amount of Br₂ (in moles) formed is, 0.0396 mol