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.
3 L will be the final volume for the gas as per Charle's law.
Answer:
Explanation:
The kinetic theory of gases has two significant law which forms the backdrop of motion of gases. They are Charle's law and Boyle's law. As per Charle's law, the volume of any gas molecule at constant pressure is directly proportional to the temperature of the molecule.
V∝ T
Since, here two volumes are given and at two different temperatures with constant pressure. Then as per Charle's law, the relation between the volumes of air at different temperature will be
So in this case, V1 = 6 L and T1 = 80° C. Similarly, T2 = 40° C. So we have to determine the V2.
So, 3 L will be the final volume for the gas as per Charle's law.
Answer:
V = 12.5 L
Explanation:
Given data:
Volume of NO = 15.0 L
Temperature and pressure = standard
Volume of nitrogen gas produced = ?
Solution:
Chemical equation:
6NO + 4NH₃ → 5N₂ + 6 H₂O
Number of moles of NO:
PV = nRT
n = PV/RT
n = 1 atm × 15.0 L / 0.0821 atm.L /mol.K × 273.15 K
n = 15.0 atm.L / 22.43 atm.L /mol
n = 0.67 mol
now we will compare the moles of No and nitrogen gas.
NO : N₂
6 : 5
0.67 : 5/6×0.67 = 0.56
Volume of nitrogen gas:
PV = nRT
1 atm × V = 0.56 mol × 0.0821 atm.L /mol.K × 273.15 K
V = 12.5 atm.L / 1 atm
V = 12.5 L
Explanation:
23 are the number of atoms
