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Answer:
The equilibrium constant for the reversible reaction = 0.0164
Explanation:
At equilibrium the rate of forward reaction is equal to the rate of backwards reaction.
The reaction is given as
A ⇌ B
Rate of forward reaction is first order in [A] and the rate of backward reaction is also first order in [B]
The rate of forward reaction = |r₁| = k₁ [A]
The rate of backward reaction = |r₂| = k₂ [B]
(Taking only the magnitudes)
where k₁ and k₂ are the forward and backward rate constants respectively.
k₁ = 0.010 s⁻¹
k₂ = 0.0610 s⁻¹
|r₁| = 0.010 [A]
|r₂| = 0.016 [B]
At equilibrium, the rate of forward and backward reactions are equal
|r₁| = |r₂|
k₁ [A] = k₂ [B] (eqn 1)
Note that equilibrium constant, K, is given as
K = [B]/[A]
So, from eqn 1
k₁ [A] = k₂ [B]
[B]/[A] = (k₁/k₂) = (0.01/0.0610) = 0.0163934426 = 0.0164
K = [B]/[A] = (k₁/k₂) = 0.0164
Hope this Helps!!!
Answer:
Mean rate of reaction produced = 0.533 g/sec (approx.)
Explanation:
Given:
Reaction produced = 1.6 gram
Time taken = 30 sec
Find:
Mean rate of reaction produced
Computation:
Mean rate of reaction produced = Reaction produced / Time taken
Mean rate of reaction produced = 1.6 / 30
Mean rate of reaction produced = 0.533 g/sec (approx.)
Answer: alcohols
Explanation:
The carbonyl group refers to C=O. It is contained in aldehyde, Ketones, carboxylic acids , esters, amides and acyl chlorides. They are not found in alcohols. The alchols are generally ROH. They do not contain any carbon-oxygen unsaturated bond in their structure hence the answer.
Answer:
Explanation:
H ₂ S O ₄ + 2 N a O H ⟶ 2 H ₂ O + N a ₂ S O ₄
29.09 mL of 0.639 M N a O H is mixed with 213.8 mL of H ₂ S O ₄
Let the concentration of H ₂ S O ₄ be S₂ .
In terms of normal or equivalent solution is will be 2 N solution
From the formula S₁ V₁ = S₂ V₂
= 29.09 x .639 = 213.8 x S₂
S₂ = .087 N solution
In terms of molar solution it will be .087 / 2 M
= .0435 M
