Answer:
is the value of the rate constant.
Explanation:

Let the order of the reaction be x.
The rate law of the reaction can be written as:
![R=k[H_2O_2]^x](https://tex.z-dn.net/?f=R%3Dk%5BH_2O_2%5D%5Ex)
1. Rate of the reaction when concentration changes from 0.882 M to 0.697 M in 0 seconds to 60 seconds.

..[1]
2. Rate of the reaction when concentration changes from 0.697 M to 0.566 M in 240 seconds to 360 seconds.

..[2]
[1] ÷ [2]
![\frac{0.00308 M/s}{0.00227 M/s}=\frac{k[0.697 M]^x}{k[0.236M]^x}](https://tex.z-dn.net/?f=%5Cfrac%7B0.00308%20M%2Fs%7D%7B0.00227%20M%2Fs%7D%3D%5Cfrac%7Bk%5B0.697%20M%5D%5Ex%7D%7Bk%5B0.236M%5D%5Ex%7D)
Solving fro x:
x = 0.92 ≈ 1
![R=k[H_2O_2]^1](https://tex.z-dn.net/?f=R%3Dk%5BH_2O_2%5D%5E1)
![0.00308 M/s=k[0.697 M]^1](https://tex.z-dn.net/?f=0.00308%20M%2Fs%3Dk%5B0.697%20M%5D%5E1)
![k=\frac{0.00308 M/s}{[0.697 M]^1}=0.00442 s^{-1}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B0.00308%20M%2Fs%7D%7B%5B0.697%20M%5D%5E1%7D%3D0.00442%20s%5E%7B-1%7D)
is the value of the rate constant.
Answer:
2
Step-by-step explanation:
A. Moles before mixing
<em>Beaker I:
</em>
Moles of H⁺ = 0.100 L × 0.03 mol/1 L
= 3 × 10⁻³ mol
<em>Beaker II:
</em>
Beaker II is basic, because [H⁺] < 10⁻⁷ mol·L⁻¹.
H⁺][OH⁻] = 1 × 10⁻¹⁴ Divide each side by [H⁺]
[OH⁻] = (1 × 10⁻¹⁴)/[H⁺]
[OH⁻] = (1 × 10⁻¹⁴)/(1 × 10⁻¹²)
[OH⁻] = 0.01 mol·L⁻¹
Moles of OH⁻ = 0.100 L × 0.01 mol/1 L
= 1 × 10⁻³ mol
B. Moles after mixing
H⁺ + OH⁻ ⟶ H₂O
I/mol: 3 × 10⁻³ 1 × 10⁻³
C/mol: -1 × 10⁻³ -1 × 10⁻³
E/mol: 2 × 10⁻³ 0
You have more moles of acid than base, so the base will be completely neutralized when you mix the solutions.
You will end up with 2 × 10⁻³ mol of H⁺ in 200 mL of solution.
C. pH
[H⁺] = (2 × 10⁻³ mol)/(0.200 L)
= 1 × 10⁻² mol·L⁻¹
pH = -log[H⁺
]
= -log(1 × 10⁻²)
= 2
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