Answer:
HOMO of 1,3-butadiene and LUMO of ethylene
HOMO of ethylene LUMO of 1,3-butadiene
Explanation:
1,3 - butadiene underogoes cycloaddition reaction with ethylene to give cyclohexene.
According to Frontier molecular orbital theory HOMO of 1,3 butadiene and LUMO of ethylene and HOMO of ethylene and LUMO of ethylene underoges (4 + 2) in thermal or photochemical condition.
Answer:
C
Explanation:
Looking at the periodic table, we can see that sodium is in group 1, so a sodium ion would be Na⁺, with a charge of +1. Oxygen is in group 16, so an oxygen ion would be O²⁻, with a charge of -2.
A compound formed only by a single sodium ion and a single oxygen ion would thus have a charge of -1, and in order to have a stable ionic compound its charge must be zero.
A volumetric flask is used to contain a predetermined volume of substance and only measures that volume, for example 250 ml.
Conical flasks can be used to measure the volume of substances but the accuracy they provide is usually up to 10ml. Conical flasks are used in titrations, reactions where the liquid may boil, and reactions which involve stirring.
Pippettes are of two types, volumetric and graduated. Pippettes are used where high accuracy is required and volumetric pippettes come in as little as 1 ml. Pippettes are usually used in titrations.
Graduated cylinders come in a wide variety of sizes and their accuracy can be down to as much as 1 ml. They are used to contain liquids.
Explanation:
Formula for work done is as follows.
W =
where, k = proportionality constant = 


d = separation distance = 0.45 nm = 
Now, we will put the given values into the above formula and calculate work done as follows.
W =
=
= 
Thus, we can conclude that work required to increase the separation of the two ions to an infinite distance is
.
Answer:
AC₄ will precipitate out first.
Explanation:
A solid will precipitate out if the ionic product of the solution exceeds the solubility product.
Let us check the ionic product
a) A₂B₃
Ionic product = [A]²[B]³
[A] = say "s"
[B] = 0.05 , [B]³ = (0.05)³ = 0.000125
2.3 X 10⁻⁸ = [A]²(0.000125)
[A] = 0.0136
b) AC₄
Ionic product = [A] [C]⁴
[A] = "s"
[A][0.05]⁴ = 4.10 X 10⁻⁸
[A]=0.00656 M
So for ionic product to exceed solubility product, we need less concentration of A in case of AC₄.