Answer:
Structures are given below.
Explanation:
- Treatment of 2-bromo-2-methylbutane with KOH in ethanol will give elimination of HBr through E2 mechanism.
- H atoms adjacent to Br will be eliminated.
- 2-bromo-2-methylbutane has two possible adjacent H atoms that can be eliminated giving mixture of products.
- Product of this elimination reaction is alkene. Here saytzeff fule is followed during elimination. So most substituted alkene will be major product.
- Structure of alkenes are given below.
Answer:
4.48 - 6.48
Explanation:
A pH indicator works in a better way in a range of pH = pKa ± 1. That means we need to determine the pKa of the indicator propyl red to find the range over which it change its color. That is:
pKa = -log Ka
pKa = -log 3.3x10⁻⁶
pKa = 5.48
That means the range over propyl red will change from yellow to red or vice versa is:
4.48 - 6.48
Answer:
D.) Products are weakly favored
Explanation:
For the reaction:
2SO₃ ⇄ O₂ + 2SO₂ + 198kJ/mol
The kc is defined as:
kc = [O₂] [SO₂]² / [SO₃]²
As the kc is 8,1:
8,1 [SO₃]² = [O₂] [SO₂]²
The products are favored 8,1 times. This is a weakly favored because the usual kc are in the order of 1x10⁴. Thus, right answer is:
D.) Products are weakly favored
I hope it helps!
According to Avogadro constant 1 mole = 6.02 x 10^23 what about 9.25 x10 ^21
that is 1 mole x ( 9.25 x10 ^21) / (6.02 x10^23) = 0.0154 moles
