The rate constant is mathematically given as
K2=2.67sec^{-1}
<h3>What is the Arrhenius equation?</h3>
The rate constant for a particular reaction may be calculated with the use of the Arrhenius equation. This constant can be stated in terms of two distinct temperatures, T1 and T2, as follows:
Therefore
KT1= 0.0110^{-1}
T1= 21+273.15
T1= 294.15K
T2= 200
T2=200+273.15
T2= 473.15K
Ea= 35.5 Kj/Mol
Hence, in j/mol R Ea is
Ea=35.5*1000 j/mol R
K2/0.0110 =e^(5.492)
K2/0.0110 =242.74
K2= 242.74*0.0110
K2=2.67sec^{-1}
In conclusion, rate constant
K2=2.67sec^{-1}
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The reaction of benzaldehyde with acetone and sodium hydroxide produces DibenzalacetoneThis is an example of an aldol condensation reaction.
Chemical reactions often involve color changes, temperature changes, gas evolution, or precipitate formation. Simple examples of everyday reactions are digestion, combustion, and cooking. As the name suggests, simple reactants produce or synthesize more complex products. The basic form of a synthetic reaction is A + B → AB. A simple example of a synthetic reaction is the formation of water from its elements hydrogen and oxygen: 2 H2(g) + O2(g) → 2 H2O(g).
A physical reaction is a reaction in which a change in the physical properties of matter or substances occurs. Physical properties include density, mass, and volume. The definition of a physical reaction is a reaction in which molecules undergo molecular rearrangements but do not change chemically.
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H2O2(I)
C6H6(O)
CO2(I)
C2H6(O)
HNO3(I)
Answer:
625 mL
Explanation:
From the question given above, the following data were obtained:
Volume of stock solution (V₁) = 250 mL
Molarity of stock solution (M₁) = 5 M
Molarity of diluted solution (M₂) = 2 M
Volume of diluted solution (V₂) =?
The volume of the diluted solution can be obtained by using the dilution formula as illustrated below:
M₁V₁ = M₂V₂
5 × 250 = 2 × V₂
1250 = 2 × V₂
Divide both side by 2
V₂ = 1250 / 2
V₂ = 625 mL
Therefore, the volume of the diluted solution is 625 mL.
You must know the concentration of the acetic acid. Suppose the concentration is 0.1 M. The solution is as follows:
CH₃COOH → CH₃COO⁻ + H⁺
I 0.1 0 0
C -x +x +x
E 0.1 - x x x
Ka = (x)(x)/(0.1 - x)
1.8×10⁻⁵ = x²/(0.1 - x)
Solving for x,
x = 1.333×10⁻³ = H⁺
pH = -log[H⁺] = -log(1.333×10⁻³)
pH = 2.88
