Answer: Kc= 4.653
Explanation:
From The equation of reaction
CO(g)+2H2(g)⇌CH3OH(g)
From the question
[CH3OH] =0.0679 moles
[CO]= 0.160moles
[H2]= 0.302
Kc = [CH3OH]/[CO] [H2]^2
Kc=[0.069]/[0.16]×[0.302]^2
Kc= 4.653
Best Answer: <span>We would write the molecular equation as ....
BaCl2(s) + H2SO4(aq) --> BaSO4(s) + 2HCl(aq)
You would write the ionic equation as....
BaCl2(s) + H+ + HSO4^- --> BaSO4(s) + 2H+ + 2Cl-
remove the H+ from both sides and write the net ionic equation as.....
BaCl2(s) + HSO4^- --> BaSO4(s) + H+ + 2Cl-
If you placed BaCl2 solid in sulfuric acid it would quickly acquire a coating of insoluble BaSO4 which will prevent the further reaction of the barium chloride.
The same is true if you took a block of BaCl2 and placed a drop of H2SO4 solution on it. The solid BaSO4 that forms will block further reaction. You would get HCl gas only if the small amount of solution became saturated in HCl.</span><span>
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This problem is being solved using Ideal Gas Equation.
PV = nRT
Data Given:
Initial Temperature = T₁ = 65.5 °C = 338.5 K
Initial Pressure = P₁ = 524 torr
Initial Volume = V₁ = 15.31 L
Final Temperature = T₂ = -15.8 °C = 257.2 K
Final Pressure = P₂ = 524 torr
Final Volume = V₂ = ?
As,
Gas constant R and Pressures are constant, so, Ideal gas equation can be written as,
V₁ / T₁ = V₂ / T₂
Solving for V₂,
V₂ = (V₁ × T₂) ÷ T₁
Putting Values,
V₂ = (15.31 L × 257.2 K) ÷ 338.5 K
V₂ = 11.63 L
Answer: n-Butanol are converted using SN2 and tert-butanol is converted using SN1
Explanation: For the conversion of n-butanol into butyl chloride using Hydrogen Chloride the reaction would follow SN2 mechanism.
SN2 reaction mechanism occurs only in the case of primary substrates as it is a one step mechanism that happens in a concerted manner. It involves backside attack of nucleophile on the substrate such that the nucleophile attacks from the back side and leaving group leaves from the front side.
In this reaction since hydroxy group (OH) is not a good leaving group hence firstly we need to convert it into a good leaving group. When we treat n-butanol with HCl hydroxy group is protonated and now it turns into a good leaving group as it can leave as H₂O.
Cl⁻ here acts as nucleophile and now attacks the primary carbon center from the back side which contains the protonated hydroxy group as a leaving group.
In the case of tertiary butanol the reaction follows SN1 mechanism and it is 2 step mechanism.
In the first step hydroxy group is protonated and as it becomes a good leaving group it leaves and leads to the formation of a stable tertiary carbocation as an intermediate.
In the second step this intermediate carbocation is attacked by the Cl⁻ nucleophile which leads to the formation of tertiary butyl chloride.
Kindly find in attachment the reaction mechanism for both the reactions.