55= No (1/2)^55/57
55= No (1/2)^3.9
55= No (1/2)^4
55= No (1/16)
No= 880 g
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.
The volume that will occupy at STP is calculated as follows
by use of ideal gas equation
that is PV=nRT where n is number of moles calculate number of moles
n= PV/RT
p=0.75 atm
V=6.0 L
R = 0.0821 L.atm/k.mol
T= 35 +273= 308k
n=?
n= (o.75 atm x 6.0 L)/( 0.0821 L.atm/k.mol x 308 k)= 0.178 moles
Agt STP 1 mole= 22.4 L what obout 0.178 moles
= 22.4 x0.178moles/ 1moles =3.98 L( answer C)
Answer:
1. Theoretical yield = 2.03g
2. Actual yield 1.89g
Explanation:
Let us write a balanced equation. This is illustrated below:
Zn + 2HCI —> ZnCl2 + H2
Molar Mass of HCl = 1 +35.5 = 36.5g/mol
Mass of HCl from the balanced equation = 2 x 36.5 = 73g
Molar Mass of H2 = 2x1 = 2g/mol
1. From the equation,
73g of HCl produced 2g of H2.
Therefore, 74g of HCl will produce = (74 x 2)/73 = 2.03g
Therefore, theoretical yield = 2.03g
2. %yield = 93%
Theoretical yield = 2.03g
Actual yield =?
%yield = Actual yield /Theoretical yield x100
Actual yield = %yield x theoretical yield
Actual yield = 93% x 2.03 = (93/100)x2.03 = 1.89g
Actual yield =1.89g
