Moles / Light

Question 1

1 mole

Answer 1

• 6.022 x 10²³ atoms
• As many particles as there are in 12.0 g of ¹²C
• Avagadro’s number

Question 2

Molar conversions

Answer 2

Question 3

A mole of a compound (covalent, ionic)

Answer 3

• Covalent: Avogadro’s number of molecules
  
  • 1 mole CO2 = 6.022 x 10²³ CO₂ molecules
• Ionic: Avogadro’s number of formula units

Question 4

How many moles of C, H, O in C₆H₁₂O₂?

Answer 4

• C) 6 moles
• H) 12 moles
• O) 6 moles

Question 5

How to find molar mass

Answer 5

• Add for each element:
  
  • Number of moles in the element * atomic mass (from periodic table, g/mole)

Question 6

Electromagnetic radiation

Answer 6

• Generated by moving electrons
• Can transport energy without a medium (unlike sound which requires matter to move through)
• Displays wave properties
  
  • Diffraction
  • Interference

Question 7

Electromagnetic spectrum

Answer 7

Question 8

Features of waves

Answer 8

• Wavelength: the distance between two identical points on two asjacent waves
  
  • Symbol: lambda λ
  • Unit: m, nm
    
    • 10⁹ nm in m

Question 9

Frequency

Answer 9

• The amount of crests that pass a point in one second
• Symbol: f, nu, ν
• Unit: s^-1, Hz

Question 10

Relationship between wavelength and frequency

Answer 10

• All EM waves travel at speed of light
• c = 3.0 x 10⁸ m/s
• c = λv
• Long wavelength = low frequency
• Short wavelength = high frequency

Question 11

Max Planck

Answer 11

• Solids emit radiation as heated
• Study relationships between wavelength and intensity of radiation emitted and temperature
• Energy is quantizes
• Lower frequency = intense light

Question 12

Quantization of energy

Answer 12

• Energy is released or absorbed in quanta
• Energy radiation is directly proportional to frequency (specific allowable energies)
• Quantized - values are restricted to certain quantities

Question 13

Energy equation (Planck’s constant)

Answer 13

• E = hv
• h = Planck’s constant = 6.626 x 10^-34 J S
• v = frequency

Question 14

Wavelength vs energy vs frequency (red vs violet)

Answer 14

• RED: Large wavelength = low frequency = low energy
• VIOLET: Small wavelength = high frequency = high energy

Question 15

Photoelectric effect

Answer 15

• When light is shone on a metal, electrons are ejected from the metal
• The ejected electrons have a ceratin amount of kinetic energy

Question 16

Classical physics prediction

Answer 16

• Higher amplitude (brighter) = more energy
• Bright red = eject electrons, dim red = no electrons
• Bright blue = eject electrons, dim blue = no electrons
• Brighter light = higher energy of electrons popped off

Question 17

Blue vs red light actually observed

Answer 17

• Bright red = no electrons, dim red = no electrons
• Bright blue = eject electrons, dim blue = eject electrons (but fewer)
• Suggests that the ability of light to eject electrons depends on its frequency, not brightness
• Planck - energy and frequency directly related
• Einstein uses this to explain the photoelectric effect

Question 18

Einstein explains photoelectric effect

Answer 18

• Light as particles, each with its own quantum (specific amount) of energy, depending on its frequency
• Particles called photons
• Only those photons with the minimum amount of energy needed to eject electrons from metal will do so

Question 19

Photon

Answer 19

• A particle of electromagnetic radiation having zero mass and carrying a quantum of energy
• Light has a dual nature

Question 20

Classical view of atom

Answer 20

• An electron traveled about the nucleus in an orbit
• Any orbit should be possible for any electron
• Did not know why the electrons did not fall into the nucleus

Question 21

Neils Bhor

Answer 21

• Built a simple model of an atom
• Based on an understanding of the sharp line emission spectrum of excited atoms

Question 22

Line emission spectra

Answer 22

• When atoms or molecules absorb energy, that energy is often released as light energy (fireworks, etc)
• When that light is passed through a prism. a pattern is seen that is unique to that type of atom or molecule — pattern called an emission spectrum
  
  • ​Non-continuous
  • Can be used to identify the material
    
    • Flame tests

Question 23

Line emission spectra of excited atoms

Answer 23

• Excited atoms emit light of only certain wavelengths
• The wavelengths of emitted light depend on the element

Question 24

Ground state

Answer 24

Electron in lowest energy level possible

Question 25

Atomic spectra and Bhor

Answer 25

• e- can only exist in certain discrete orbits of specific energy (quantized energy states)
  
  • e- have a specific amount of energy that is equal to the PE of their location
  • PE of e- is determined by its distance from the nucleus
  • Closer to nucleus = lower PE
• Each type of photon has an energy equal to the energy lost by specific jumps in energy levels
• Nobel prize 1922
• Problem: only works for hydrogen –> quantum/wave mechanics

Question 26

Balmer series

Answer 26

• The series of visible lines in the hydrogen atom spectrum

Question 27

L de Broglie (+ equations)

Answer 27

• Proposed (1924) that all moving objects have wave properties
• For light: E = mc²
  
  • E = hν = hc / λ
  • mc = h / λ
• For particles: c = velocity
  
  • (mass)(velocity) = h / λ
  • mv = h / λ
  • λ = h / mv

Question 28

Quantum / wave mechanics (+ equation)

Answer 28

• Using deBroglie’s equation λ = h/mv can calulcate the wavelength for moving objects
  
  • λ = wavlength (meters)
  • v = velocity (m/sec)
  • m = mass (kg)
  • h = Planck’s constant

Question 29

Uncertainty principle

Answer 29

• Problem of defining nature of electrons in atoms
  
  • Solved by W. Heisenberg
• Cannot simultaneously define the position and momentum (m*v) of an electron
• We define e- energy exactly but accept limitation that we do not know exact position
• m = kg
• v = m/s

Question 30

Double slit experiment

Answer 30

• Demonstration that light and matter can display characteristics of both classically defined waves and particles
• Electrons act like waves and particles
• Interference pattern

Question 31

E Schrodinger

Answer 31

• Applied idea of e- behaving as a wave to the problem of electrons in atoms
• Developed the wave equation
• Solution gives set of math expressions called wave functions
• Each wave function describes an allowed energy state of an e- which corresponds to an orbital - the region of space within which an electron is found
  
  • Area of probability, not exact location

Question 32

Orbital

Answer 32

• For an electron with a given energy, can describe a region in the atom of high probability of finding it
• Many of the properties of atoms are related to the energies of the electrons

Question 33

s orbital

Answer 33

• 1 direction spherical

Question 34

p orbital

Answer 34

• 3 directions
• Three p orbitals lie 90 degrees apart in space

Question 35

d orbital

Answer 35

• 5 directions

Question 36

f orbital

Answer 36

• 7 directions

Question 37

g orbital

Answer 37

• 9 directions
• Hypothetical

Question 38

Principal quantum number

Answer 38

• Symbol: n
• The main energy level occupied by the electron
• Values of n are positive integers and the value of n=1 has the lowest energy

Question 39

Sublevels and relative energy

Answer 39

• 4 sublevels for the elements discovered to date
• Sublevel indicates the shape of the orbital
  
  • First energy level, n = 1 has 1 sublevel
  • Fourth energy level, n = 4 has 4 sublevels
• Levels split because negative e- next to each other
• Sublevels broken up into orbitals

Question 40

The Aufbau Rule

Answer 40

• The lowest energy levels are filled first

Question 41

The Pauli Exclusion Principle

Answer 41

• Orbitals can have max of 2 electrons
• e- in same orbital must have opposite spins

Question 42

Hund’s Rule

Answer 42

• When orbitals of identical energy (same shape) are available, e- enter those orbitals singly before any spin pairing takes place

Question 43

Energy emitted by e- energy level change

Answer 43

E in J

Question 44

Wavelength emitted by e- energy level change

Answer 44