The state of energy that is forbidden is 2p orbital. The correct option is b).
<h3>
What is the energy state of electrons?</h3>
The energy state of an electron depends upon the presence of the electron on the orbitals. Lower the energy they will be in the lower orbital. When they get higher energy they move to the higher orbital.
By using the Selection Rules for Electron Transitions
1.) ?l = +/- 1 and
2.) ?m = 0, +/- 1
The conservation of angular momentum is required by these laws. A photon's inherent angular momentum is 1. As 4p is higher than 2p and the electron is lowering its energy. So, it will go down to 2p orbital.
Thus, the correct option is b). 2p orbital.
The question is incomplete. Your full question is given below:
a) 3d
b) 2p
c) 1s
d) 2s
To learn more about the energy state of electrons, refer to the link:
brainly.com/question/4138621
#SPJ4
Answer:
0.06022 × 10²³ eggs
Explanation:
Given data:
Moles of eggs = 0.01 mol
Number of eggs = ?
Solution:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
For example,
1 mole of egg = 6.022 × 10²³ eggs
0.01 mol × 6.022 × 10²³ eggs / 1 mol
0.06022 × 10²³ eggs
Thus in 0.01 moles of eggs have 0.06022 × 10²³ eggs number of eggs.
Answer: 9.68 x 10^10 grams.
Explanation:
Given that:
Mass of CO2 = ?
Number of molecules of CO2 = 2.2x10^9 molecules
Molar mass of CO2 = ? (let unknown value be Z)
For the molar mass of CO2: Atomic mass of Carbon = 12; Oxygen = 16
= 12 + (16 x 2)
= 12 + 32 = 44g/mol
Apply the formula:
Number of molecules = (Mass of CO2 in grams/Molar mass)
2.2x10^9 molecules = Z/44g/mol
Z = 2.2x10^9 molecules x 44g/mol
Z = 9.68 x 10^10g
Thus, the mass of 2.2x10^9 molecules of CO2 is 9.68 x 10^10 grams.
Answer:
THE PRESSURE OF THE TIRE ON THE TRIP HOME AT THE ROAD SURFACE TEMPERATURE OF 32°C IS 160 kPa.
Explanation:
Initial Pressure = 75 kPa
Initial temperature = 15 °C
Final temperature = 32 °C
Final pressure = unknown
Using the combined equation of gases;
P1V1/T1 = P2V2/ T2
Since the tire will have the same volume of air in it showing that volume of constant both at the repair shop and on the road surface.
The relationship between pressure and temperature is used with constant volume.
P1/T1 = P2/ T2
75 kPa / 15 °C = P2 / 32 °C
P2 = 75 kPa * 32 °C / 15 °C
P2 = 2400 kPa °C / 15 °C
P2 = 160 kPa.
So therefore, the pressure of the tire on the trip home when the temperature of the road surface is 32°C is 160 kPa.
Answer:
The resonance forms of O3 are attached as an image.
Explanation:
A compound with different contributing structures comes together and forms a resonating or intermediate structure that best describes the properties of that compound.
The given compound is ozone, having the chemical formula
O3 = 6 electrons * 3 = 18 electrons
O → prefers to have a complete their octet
The bonding electrons and lone pair electrons are radical electrons that are present on the oxygen atoms tend to delocalize and results in various resonating forms of O3.
