Cram Unit 1: Atomic Structures and Properties

0.0(0)
studied byStudied by 1 person
0.0(0)
full-widthCall Kai
learnLearn
examPractice Test
spaced repetitionSpaced Repetition
heart puzzleMatch
flashcardsFlashcards
GameKnowt Play
Card Sorting

1/49

encourage image

There's no tags or description

Looks like no tags are added yet.

Study Analytics
Name
Mastery
Learn
Test
Matching
Spaced

No study sessions yet.

50 Terms

1
New cards

Atom

the smallest unit of matter; they are always neutral p+=e-

2
New cards

Nucleus

protons (+) and neutrons (neutral)

3
New cards

Electrons

Negatively charged; surround the nucleus.

4
New cards

Atomic number

The number of protons

5
New cards

Atomic mass

Protons + neutrons

6
New cards

Isotope

Different forms of atoms for the same element with same number of protons and different number of neutrons.

7
New cards

Mass spectroscopy

  • (% of abundance * mass of the isotope) +(% of abundance * mass of the isotope)…= average atomic mass of an element.

<ul><li><p>(% of abundance * mass of the isotope) +(% of abundance * mass of the isotope)…= average atomic mass of an element.</p></li></ul>
8
New cards

Estimating average atomic mass

You can add the numbers on top to find the percentage of abundance by multiplying (70 × 9/9+6+1)+(72 × 96/9+6+1)… OR find the in-between of the largest columns, here it would be 70 and 71. This means the amu should be around 71.

<p>You can add the numbers on top to find the percentage of abundance by multiplying (70 × 9/9+6+1)+(72 × 96/9+6+1)… <strong>OR </strong>find the in-between of the largest columns, here it would be 70 and 71. This means the amu should be around 71.</p>
9
New cards

Classifying Matter

knowt flashcard image
10
New cards

What are the pure substances?

They have invariant chemical composition and have distinct properties. elements and compounds

11
New cards

What are the mixtures?

Two or more pure substances that retain their original identities and can be separated by physical means. Homogeneous and heterogeneous

12
New cards

Element

fundamental substances; cannot be separated into simpler substances. (Cu, Zn, Ag or O2, H2, N2, O3)

<p>fundamental substances; cannot be separated into simpler substances. (Cu, Zn, Ag or O2, H2, N2, O3)</p>
13
New cards

Compound

two or more elements in fixed proportions and can be separated into simpler substances/elements by chemical means. (CuO, H2O, CH3CH2OH)

<p>two or more elements in fixed proportions and can be separated into simpler substances/elements by chemical means. (CuO, H2O, CH3CH2OH)</p>
14
New cards

Homogeneous

They have a uniform composition and properties throughout (also called a solution) (Saline water)

<p>They have a uniform composition and properties throughout (also called a solution) (<strong><em>Saline water)</em></strong></p>
15
New cards

Heterogeneous

They are not uniform in composition and properties all throughout. (Salad)

<p>They are not uniform in composition and properties all throughout. (<strong><em>Salad) </em></strong></p>
16
New cards

How do we separate mixtures that are completely soluble?

Through evaporation or distillation.

17
New cards

Molar mass (g/mol)

mass of any substance for 1 mole (Atomic mass)

18
New cards

Molar mass for a compound

Just add up the sum of the molar masses respective to the ratio.

19
New cards

Avogrado’s Number

6.022 X 10(23)

20
New cards

Mass

Number of moles * Molar mass

21
New cards

Number of particles

Number of moles * Avogrado’s Number

22
New cards

Number of moles

Mass of the sample / Molar mass

23
New cards

Empirical and Molecular Formula

Molecular C6H12O6 (6:12:6) → Empirical CH2O (1:2:1)

24
New cards

How to calculate molecular formula

Empirical formula + molar mass

  • Find the molar mass of the empirical formula

  • Divide the empirical formula molar mass by the molar mass given

  • Multiply the result to each of the atoms in the empirical formula

<p>Empirical formula + molar mass </p><ul><li><p>Find the molar mass of the empirical formula</p></li><li><p>Divide the empirical formula molar mass by the molar mass given </p></li><li><p>Multiply the result to each of the atoms in the empirical formula</p></li></ul>
25
New cards

Atomic structure

  • Atom= electrons, protons, and nucleus

  • Electrons are in specific shells (energy levels) and subshells (sublevels)

26
New cards

Energy shells

n= 1, n=2, n=3, n=4….

27
New cards

Valence electrons

Reside in the outermost shell

28
New cards

Subshells

s, p, d,f

  • Energy increases from s → f

29
New cards

Shell 1

s

30
New cards

Shell 2

s, p

31
New cards

Shell 3

s, p, d

32
New cards

Shell 4

s, p, d, f

33
New cards

s

2 electrons

34
New cards

p

6 electrons

35
New cards

d

10 electrons

36
New cards

f

14 electrons

37
New cards

s

1 orbital

38
New cards

p

3 orbitals

  • Hold two electrons each

39
New cards

Electron configuration

Aufbau principle: electrons first occupy those orbitals whose energy is the lowest.

<p>Aufbau principle: <span style="font-family: arial, sans-serif">electrons first occupy those orbitals whose energy is the lowest.</span></p>
40
New cards

Hund’s Rule

The lowest energy configuration for an atom is the one having the maximum number of unpaired electrons within the same energy sublevel.

<p>The lowest energy configuration for an atom is the one having the maximum number of unpaired electrons within the same energy sublevel.</p>
41
New cards

Photoelectron spectroscopy

Represents electron distribution and energy

  • X-axis: Binding energy (energy necessary to remove an electron)

  • To the left → Closer to the nucleus

  • To the right → Closer to the valence shell

  • Y-axis: # of electrons

<p>Represents electron distribution and energy</p><ul><li><p>X-axis: Binding energy (energy necessary to remove an electron)</p></li><li><p>To the left → Closer to the nucleus </p></li><li><p>To the right → Closer to the valence shell</p></li><li><p>Y-axis: # of electrons</p></li></ul>
42
New cards

Periodic Trends

Atomic radius, ionization energy, electron affinity, electronegativity

43
New cards

Coulomb’s Law

It indicates that forces of attraction depend on the distance and magnitude of the charges.

  • Greater distance → Less force

  • Greater magnitude → More force

44
New cards

Atomic radius

  • As we go down a group, atomic radius increases because there will be more energy levels, maximizing the distance between the nucleus and the valence electrons. Also explained by forces of repulsion as electrons increase, so does the electron shield.

  • As we go right a period, the atomic radius decreases because the number of protons also increases, creating greater magnitude, thus creating greater forces of attraction.

45
New cards

Ionic radius

  • If they lose electrons, they become smaller.

  • If they gain electrons, they become larger.

46
New cards

First ionization energy

The amount of energy required to remove an electron.

47
New cards

Ionization energy trend

  • As electrons are closer to the nucleus, energy increases because of the high nuclear charge.

  • As electrons are farther away from the nucleus, energy decreases because of the weaker nuclear charge.

48
New cards

It is likely that ionization energy will be lower if the subshell is not…

fully-filled.

49
New cards

Electron affinity

The energy change that occurs when an atom gains an electron.

50
New cards

Electronegativity

The ability of an atom to attract electrons.

  • Increases from left to right because effective nuclear charge increases

  • Decreases down a group because there are more shells, creating less attractions between the nucleus and electrons.