Answer:
The disadvantages of each of the given model of electron configuration have been mentioned below:
1). Dot Structures - They take up excess space as they do not display the electron distribution in orbitals.
2). Arrow and line diagrams make the counting of electrons and take up too much space.
3). Written Configurations do not display the electron distribution in orbitals and help in lose counting of electrons easily.
Answer:
A --- (E)-oct-2-en-1-o1
B ----(E)-oct-2-enal
Explanation:
See the attached file for the structure.
Neon<span> is a rare inert </span>noble gas<span> and is</span><span> colorless and odorless </span><span>with about two-thirds the density of air. It </span><span>is chemically </span>inert <span>and does not form uncharged chemical compounds.</span>
Answer : The orbitals that are used to form each indicated bond in citric acid is given below as per the attachment.
Answer 1) σ Bond a: C 
O .
Explanation : The sigma bonds in the 'a' position involves carbon atom which undergoes hybridisation and has orbitals of carbon and as well as oxygen atoms involved in the bonding.
Answer 2) π Bond a: C has π orbitals and O also has π orbitals.
Explanation : The pi-bond at the 'a'position has carbon and oxygen atoms which undergoes pi-bond formation. And has pi orbitals of oxygen and carbon involved in the bonding process.
Answer 3) Bond b: O 
H has only S orbital involved in bonding.
Explanation : The bonding at 'b' position involves oxygen and hydrogen atoms in it. It has hybridized orbitals and S orbital of hydrogen involved in the bonding.
Answer 4) Bond c: C O
Explanation : The bonding involved at 'c' position has carbon and oxygen atoms involved in it. Both the atoms involves the orbitals of hybridized bonds.
Answer 5) Bond d: C C
Explanation : The bonding involved at the position of 'd' involves two carbon atoms. Therefore, they undergo hybridization. The orbitals involved in this hybridization are also .
Answer 6) Bond e: C1 containing O C2
Explanation : The bonding at the 'e' position involves two carbon atoms on is containing oxygen with double bonds and the C2 carbon atom. The carbon containing oxygen has hybridized orbitals involved in the bonding; whereas C2 carbon ha orbitals.
Answer:
2.1x10⁹ years
Explanation:
U-238 is a radioactive substance, which decays in radioactive particles. It means that this substance will lose mass, and will form another compound, the Pb-206.
The time need for a compound loses half of its mass is called half-life, and knowing the initial mass (mi) and the final mass (m) the number of half-lives passed (n) can be found by:
m = mi/2ⁿ
The mass of Pb-206 will be the mass that was lost by U-238, so it will be mi - m. Thus, the mass ration can be expressed as:
(mi-m)/m = 0.337/1
mi - m = 0.337m
mi = 1.337m
Substituing mi in the expression of half-life:
m = 1.337m/2ⁿ
2ⁿ = 1.337m/m
2ⁿ = 1.337
ln(2ⁿ) = ln(1.337)
n*ln(2) = ln(1.337)
n = ln(1.337)/ln2
n = 0.4190
The time passed (t), or the age of the sample, is the half-life time multiplied by n:
t = 4.5x10⁹ * 0.4190
t = 1.88x10⁹ ≅ 2.1x10⁹ years
