Answer:
The correct answer is option 3.
Explanation:
2A → B + C
First trial ;
[A] = x , Rate of the reaction = R
Rate law of the reaction can be written as:
![R=k[x]^a](https://tex.z-dn.net/?f=R%3Dk%5Bx%5D%5Ea)
..[1]
Second trial ;
[A] = 3x , Rate of the reaction = R' = 9R
Rate law of the reaction can be written as:
![R'=k[3x]^a](https://tex.z-dn.net/?f=R%27%3Dk%5B3x%5D%5Ea)
![9R=k[3x]^a](https://tex.z-dn.net/?f=9R%3Dk%5B3x%5D%5Ea)
..[2]
[1] ÷ [2]
![\frac{R}{9R}=\frac{k[x]^a}{k[3x]^a}](https://tex.z-dn.net/?f=%5Cfrac%7BR%7D%7B9R%7D%3D%5Cfrac%7Bk%5Bx%5D%5Ea%7D%7Bk%5B3x%5D%5Ea%7D)
a = 2
Rate of the reaction : ![R=k[A]^3](https://tex.z-dn.net/?f=R%3Dk%5BA%5D%5E3)
The order of the reaction is 2. This is because in the rate law expression the the power of the concentration of reactant A id 2.
The statement which describes the potential and the kinetic energy of the water is this: THE POTENTIAL ENERGY INCREASES AND THE AVERAGE KINETIC ENERGY REMAINS THE SAME.
At boiling point of water, the heat that is been added to the water is used to break the chemical bonds that holds the molecules of the water together so that they can turn into vapour. At this point, the average kinetic energy of the water remains the same while the potential energy increases.
Answer:
56.24g
Explanation:
To find the mass of N2O3 in 4.45 x 10^23 molecules, it must first be converted to moles by dividing the number of molecules in N2O3 by Avagadro's number (6.02 × 10^23).
number of moles in N2O3 = 4.45 x 10^23 ÷ 6.02 × 10^23
n = 4.45/6.02 × 10^(23 - 23)
n = 0.74 × 10^0
n = 0.74moles.
Using the formula below to find the mass of N2O3;
mole = mass ÷ molar mass
Molar mass of N2O3 = 14(2) + 16(3)
= 28 + 48
= 76g/mol
mass = mole × molar mass
Mass = 0.74 × 76
Mass = 56.24g
Answer:
75 mg
Explanation:
We can write the extraction formula as
x = m/[1 + (1/K)(Vaq/Vo)], where
x = mass extracted
m = total mass of solute
K = distribution coefficient
Vo = volume of organic layer
Vaq = volume of aqueous layer
Data:
m = 75 mg
K = 1.8
Vo = 0.90 mL
Vaq = 1.00 mL
Calculations:
For each extraction,
1 + (1/K)(Vaq/Vo) = 1 + (1/1.8)(1.00/0.90) = 1 + 0.62 = 1.62
x = m/1.62 = 0.618m
So, 61.8 % of the solute is extracted in each step.
In other words, 38.2 % of the solute remains.
Let r = the amount remaining after n extractions. Then
r = m(0.382)^n.
If n = 7,
r = 75(0.382)^7 = 75 × 0.001 18 = 0.088 mg
m = 75 - 0.088 = 75 mg
After seven extractions, 75 mg (99.999 %) of the solute will be extracted.
The answer to this question is a
