Answer:
an increase in 1-butene was observed when t-butoxide was used
Explanation:
When a base reacts with an alkyl halide, an elimination product is formed. This reaction is an E2 reaction.
Here we are to compare the reaction of two different bases with one substrate; 2-bromobutane. Both reactions occur by the E2 mechanism but follow different transition states due to the size of the base.
The Saytzeff product, 2-butene, is obtained when the methoxide is used while the non Saytzeff product, 1-butene, is obtained when t-butoxide is used.
The Saytzeff rule is reliable in predicting the major products of simple elimination reactions of alkyl halides given the fact that a small/strong bases is used for the elimination reaction. Therefore hydroxide, methoxide and ethoxide bases give similar results for the same alkyl halide substrate. Bulky bases such as tert-butoxide tend to yield a higher percentage of the non Saytzeff product and this is usually attributed to steric hindrance.
18)
a. Ra
d. Ag
19)
a. C
b. Br
c. Mg
20)
a. S
b. Br
c. O
Sorry wasn’t sure about some of them in #18
Hope the notes are helpful.
Answer:
0.185M sulfuric acid
Explanation:
Based on the reaction:
H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O
<em>1 mole of sulfuric acid reacts with 2 moles of KOH</em>
Initial moles of H₂SO₄ and KOH are:
H₂SO₄: 0.750L ₓ (0.470mol / L) = <em>0.3525 moles of H₂SO₄</em>
KOH: 0.700L ₓ (0.240mol / L) = <em>0.168 moles of KOH</em>
The moles of sulfuric acis that react with KOH are:
0.168mol KOH ₓ (1 mole H₂SO₄ / 2 moles KOH) = 0.0840 moles of sulfuric acid.
Thus, moles that remain are:
0.3525moles - 0.0840 moles = <em>0.2685 moles of sulfuric acid remains</em>
As total volume is 0.700L + 0.750L = 1.450L, concentration is:
0.2685mol / 1.450L = <em>0.185M sulfuric acid</em>
Answer:
compound, but I could be wrong
Answer:
Gas:
- No fixed shape or volume
- Molecules are very loosely packed
- Flows in all the directions
Liquid:
- No fixed shape but has volume
- Molecules are closely packed
- Always flows from higher to lower level
