Treating (CH3)3C-Cl with a mixture of H2O and CH3OH at room temperature would yield: A) CH2=C(CH3)2 B) (CH3)3COH C) (CH3)3COCH3
D) All of these choices. E) None of these choices.
1 answer:
Answer:
All of the choices.
Explanation:
As the reaction is involving with a mixture of H₂O and CH₃OH, these two reagents can work as nucleophyle of the reaction. Both of them, are polar and promoves a Sn1/E1 reaction. When it reacts with water it will produce product B); it will form product C) when it reacts with methanol, and product A) will be formed when the reaction undergoes an E1 reaction.
In this case, the only way to show you this, it's doing the mechanism of reaction for each product. Picture attached show the mechanism for the formation of all these products.
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Answer:
The temperature difference of the body after 3 hours = 5.16 K
Explanation:
we know that the number of moles of O₂ inhaled are 0.02 mole/min⁻¹
or, 1.2 mole.h⁻¹
The average heat evolved by the oxidation of foodstuffs is then:
⇒ Q avg = = 7.2 kj.h⁻¹.Kg⁻¹
the heat produced after 3 h would be:
= 7.2 kj. h⁻¹.Kg⁻¹ x 3 h
= 21.6 kj. kg⁻¹
= 21.6 x 10³ j kg⁻¹
We know Qp = Cp x ΔT
Assume the heat capacity of the body is 4.18 J g⁻¹K⁻¹
⇒ ΔT =
⇒ ΔT =
⇒ ΔT = 5.16 K
Answer : The molar concentration of ethanol in the undiluted cognac is 8.44 M
Explanation :
Using neutralization law,
where,
= molar concentration of undiluted cognac = ?
= molar concentration of diluted cognac = 0.0844 M
= volume of undiluted cognac = 5.00 mL = 0.005 L
= volume of diluted cognac = 0.500 L
Now put all the given values in the above law, we get molar concentration of ethanol in the undiluted cognac.
Therefore, the molar concentration of ethanol in the undiluted cognac is 8.44 M
Answer:
The equilibrium concentration of CH₃OH is 0.28 M
Explanation:
For the reaction: CO (g) + 2H₂(g) ↔ CH₃OH(g)
The equilibrium constant (Keq) is given for the following expresion:
Keq= =14.5
Where (CH3OH), (CO) and (H2) are the molar concentrations of each product or reactant.
We have:
(CH3OH)= ?
(CO)= 0.15 M
(H2)= 0.36 M
So, we only have to replace the concentrations in the equilibrium constant expression to obtain the missing concentration we need:
14.5=
14.5 x (0.15 M) x = (CH₃OH)
0.2818 M = (CH₃OH)