Answer:
The major product is 2-methyl-2-pentene [ CH₃-CH₂-CH=C(CH₃)₂ ] and a minor product 2-methyl-1-pentene [ CH₃-CH₂-CH₂-C(CH₃)=CH₂ ].
Explanation:
Dehydration reaction is a reaction in which a molecule loses a water molecule in the presence of a dehydrating agent like sulfuric acid (H₂SO₄).
<u>Dehydration reaction of 2-methyl-2-pentanol</u> gives a major product 2-methyl-2-pentene and a minor product 2-methyl-1-pentene.
CH₃-CH₂-CH₂-C(CH₃)₂-OH (2-methyl-2-pentanol)→ CH₃-CH₂-CH=C(CH₃)₂ (2-methyl-2-pentene, major) + CH₃-CH₂-CH₂-C(CH₃)=CH₂ (2-methyl-1-pentene, minor)
<u>Since more substituted alkene is more stable than the less substituted alkene. So, the trisubstituted alkene, 2-methyl-2-pentene is more stable than the disubstituted alkene, 2-methyl-1-pentene.</u>
<u>Therefore, the trisubstituted alkene, 2-methyl-2-pentene is the major product and the disubstituted alkene, 2-methyl-1-pentene is the minor product.</u>
Answer: I2 is the Oxidant; while the 2S2O3(-2) is the reductant.
Explanation:
An Oxidant is any substance that oxidizes, or receives electrons from, another; in so doing, it becomes reduced in oxidation number.
A Reductant thus exactly the opposite.
Note that the equation provided shows that Iodine (I2) received an electron to become NEGATIVELY CHARGED:
I2 --> 2I-.
The oxidation number reduced from 0 to -1.
In contrast, the oxidation number of 2S2O3(-2) increases from -4 to -2.
Thus, I2 is the Oxidant; while the 2S2O3(-2) is the reductant.
Answer:
Ecell = +0.25V
Explanation:
the half-cell reactions for a voltanic cell
cathode(reduction): 2H⁺(aq) + 2e⁻ ------- H₂(g)
anode(oxidation): 2AgCl(s) ------- 2Ag⁺(aq) + 2Cl⁻ + 2e⁻
we have the standard cell potential E⁺cell = 0.18V at 80C respectively
Q = [H⁺]/[Cl⁻]
sub for [H+] = 0.10M and [Cl-] = 1.5M
Q= 0.1M/1.5M
Q = 0.067
Ecell = E⁺cell - 
logQ
= 0.18 - 
log 0.067
0.18- 0.059(-1.174)
Ecell = +0.25V
