Chemistry · Units 1–9 · Quick reference

General Chemistry

Grades 9–12+8 pages

Measurement, atoms and electrons, trends, bonding and shape, forces, naming, the mole and stoichiometry, reactions, solutions, gases, energy, rates, equilibrium, acids and bases, redox and nuclear chemistry.

1Measurement2Atoms3Electrons4Trends5Bonding6Shape & polarity7IMFs & phases8Naming9Mole & stoichiometry10Reactions11Solutions12Gases13Thermochemistry14Rates & equilibrium15Acids & bases16Redox & nuclear

1Measurement, Sig Figs & Units

SI prefixes
Prefix	Symbol	Factor
giga	G	10⁹
mega	M	10⁶
kilo	k	10³
centi	c	10⁻²
milli	m	10⁻³
micro	μ	10⁻⁶
nano	n	10⁻⁹
pico	p	10⁻¹²

SI base: kg (mass), m (length), s (time), K (temperature), mol (amount)
Accuracy: how close to the true value
Precision: how close repeated results are to each other

Rounding answers

× and ÷: keep the fewest sig figs of any factor.
+ and −: keep the fewest decimal places.
Round once, at the end.

EX4.56 × 1.4 = 6.384 → 6.4; 12.11 + 3.2 = 15.31 → 15.3

Counting significant figures

Nonzero digits always count.
Zeros between nonzero digits count: 304 has 3.
Leading zeros never count: 0.0045 has 2.
Trailing zeros count only with a decimal point: 45.00 has 4; 4500 is unclear (write 4.5 × 10³ or 4.500 × 10³).
Exact numbers (counted items, defined units) have unlimited sig figs.

Middle yes, front no, end needs a dot

Zeros in the middle always count, zeros in front never do, zeros at the end count only when the number shows a decimal dot.

Count them Worked

0.004 50 → 3; 30.040 → 5; 1.200 × 10³ → 4

Percent error Worked

measured 2.62 g/cm³, accepted 2.70 g/cm³

|2.62 − 2.70|2.70 × 100 = 3.0 %

Dimensional analysis

Multiply by conversion factors (each equal to 1) until only the wanted unit is left.

given × wanted unitgiven unit

Convert 65 mi/h to m/s Worked

65 mih × 1609 m1 mi × 1 h3600 s = 29 m/s1 mi = 1609.344 m

Density Worked

12.0 g stone raises water from 20.0 mL to 24.4 mL

d = mV = 12.0 g4.4 mL = 2.7 g/mL4.4 has 2 sig figs

Temperature

K = °C + 273.15

°F = 1.8(°C) + 32

EX37.0 °C = 310.2 K = 98.6 °F

2Atoms, Isotopes & Ions

Subatomic particles
Particle	Charge	Mass (u)	Where
proton	+1	1.007	nucleus
neutron	0	1.009	nucleus
electron	−1	0.000 549	outside

P picks, N nudges, E earns

Protons pick the element, neutrons nudge the mass, electrons earn the charge.

Nuclide symbol

Z = atomic number (protons); A = mass number (protons + neutrons); N = neutrons.

^A_ZX: A = Z + N

Count the particles Worked

⁵⁶Fe³⁺: Z = 26

26 p, 30 n, 23 e⁻lost 3 e⁻

Isotopes: same Z, different N (¹²C, ¹³C, ¹⁴C)
Ion: unequal p and e⁻; cation (+) lost e⁻, anion (−) gained e⁻
Atomic mass unit: 1 u = 112 the mass of a ¹²C atom

Average atomic mass

The weighted average printed on the periodic table.

Σ (fraction × isotope mass)

Chlorine Worked

75.78 % ³⁵Cl (34.969 u), 24.22 % ³⁷Cl (36.966 u)

0.7578(34.969) + 0.2422(36.966) = 35.45 u

Find the abundance Worked

Cu = 63.546 u from ⁶³Cu (62.930 u) and ⁶⁵Cu (64.928 u)

62.930x + 64.928(1 − x) = 63.546 → x = 0.691769.17 % ⁶³Cu

Models of the atom
Year	Scientist	Model
1808	Dalton	solid indivisible spheres
1897	Thomson	electrons in positive matter
1911	Rutherford	tiny dense nucleus
1913	Bohr	electrons in energy levels
1926	Schrödinger	orbitals (probability)

General ChemistryChemistry · Units 1–9 · continued

3Electrons, Light & Quantum Numbers

Light

c = 2.998 × 10⁸ m/s; h = 6.626 × 10⁻³⁴ J·s.

c = λν

E = hν = hcλ

Hydrogen’s red line, 656 nm Worked

ν = 2.998 × 10⁸656 × 10⁻⁹ = 4.57 × 10¹⁴ Hz

E = hν = 3.03 × 10⁻¹⁹ Jper photon

Bohr energy levels (hydrogen)

An electron dropping from a high level to a low one gives off a photon of exactly ΔE.

E_n = −2.18 × 10⁻¹⁸ Jn²

EXn = 3 → 2: ΔE = 2.18 × 10⁻¹⁸(14 − 19) = 3.03 × 10⁻¹⁹ J, the same 656 nm red line

Electromagnetic spectrum (low → high energy)

radio < microwave < infrared < visible (red → violet) < ultraviolet < X-ray < gamma

Longer wavelength means lower frequency and lower energy.

Quantum numbers
Number	Values	Tells
n	1, 2, 3, …	energy level, size
l	0 to n − 1 (s, p, d, f)	sublevel shape
m_l	−l to +l	orbital orientation
m_s	+½ or −½	spin

Which set is impossible? Worked

(2, 1, 0, +½) ✓; (3, 2, −2, −½) ✓

(2, 2, 0, +½) ✗l must be less than n

Sublevels: s 1 orbital (2 e⁻), p 3 (6), d 5 (10), f 7 (14)
Level n: holds 2n² electrons: 2, 8, 18, 32

2, 6, 10, 14: add four

Each new sublevel holds four more electrons: s 2, p 6, d 10, f 14.

Filling rules

Aufbau: fill the lowest energy first (diagonal chart).
Pauli: an orbital holds 2 e⁻ of opposite spin; no two e⁻ share all four numbers.
Hund: in equal orbitals, spread out one per orbital (same spin) before pairing.

Follow each arrow from its top end: 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p.

Sulfur (16 e⁻) Worked

1s² 2s² 2p⁶ 3s² 3p⁴ = [Ne] 3s² 3p⁴

3p: ↑↓ ↑ ↑ → 2 unpairedHund

Bromine (35 e⁻) Worked

[Ar] 4s² 3d¹⁰ 4p⁵

outer level n = 4: 2 + 5 = 7 valence e⁻group 17

Watch for

Cr = [Ar] 4s¹ 3d⁵ and Cu = [Ar] 4s¹ 3d¹⁰ (half- and full-d stability).
Transition-metal ions lose 4s before 3d: Fe²⁺ = [Ar] 3d⁶.
Full table with every configuration block: Periodic Table sheet.

4Periodic Trends

Pull on electrons rises to the upper right; size grows to the lower left.

Trend table
Property	Across →	Down ↓
atomic radius	smaller	larger
ionization energy	higher	lower
electronegativity	higher	lower
electron affinity	more negative (mostly)	less negative
metallic character	less	more

Up-right holds tight

Atoms toward the upper right hold electrons tightest: smallest radius, highest ionization energy and electronegativity (noble gases aside).

Families
Group	Family	Valence e⁻	Ion
1	alkali metals	1	1+
2	alkaline earth metals	2	2+
3–12	transition metals	—	varies
15	nitrogen family	5	3−
16	chalcogens	6	2−
17	halogens	7	1−
18	noble gases	8 (He 2)	none

Effective nuclear charge

Z_eff ≈ Z − core electrons

EXNa: 11 − 10 = 1; Cl: 17 − 10 = 7: Cl holds its electrons far tighter.

Rank the radius Worked

K, Na, Mg, Cl

K > Na > Mg > Cldown is bigger; right is smaller

Rank electronegativity Worked

C, N, O, F

F (3.98) > O (3.44) > N (3.04) > C (2.55)

Ionization energy: energy to remove an e⁻ from a gas atom
Electronegativity: pull on shared electrons in a bond (Pauling scale; F = 3.98)
Electron affinity: energy change when a gas atom gains an e⁻

Ion size

Cations are smaller than their atoms (lost a shell or e⁻).
Anions are larger (extra e⁻ repel).
Same electron count: more protons → smaller: O²⁻ > F⁻ > Na⁺ > Mg²⁺.

Read successive ionization energies Worked

Mg: 738, 1451, 7733 kJ/mol

huge jump after the 2nd → 2 valence e⁻group 2

Exceptions

Ionization energy dips at B (below Be) and O (below N).
Noble gases have no Pauling electronegativity (most form no bonds).

General ChemistryChemistry · Units 1–9 · continued

5Bonding & Lewis Structures

Bond type by ΔEN
ΔEN	Bond	Example
< 0.4	nonpolar covalent	C–H 0.35
0.4 – 1.7	polar covalent	H–Cl 0.96
> 1.7	mostly ionic	Na–Cl 2.23

Three kinds
Bond	Made of	Properties
ionic	metal + nonmetal ions	high melting, brittle; conducts melted or dissolved
covalent	nonmetals sharing e⁻	molecules; lower melting
metallic	metal cations in an e⁻ sea	conducts, malleable, shiny

Drawing a Lewis structure

Count valence e⁻ (add for − charge, subtract for +).
Connect atoms with single bonds; least electronegative in the center, H on the outside.
Complete octets on outer atoms.
Put leftovers on the center atom.
Check: center short of 8? Turn lone pairs into double or triple bonds.

Five Cs

Count, Connect, Complete, Center, Check.

CO₂ Worked

4 + 2(6) = 16 e⁻

O=C=O, two lone pairs on each Oall octets

NH₃ Worked

5 + 3(1) = 8 e⁻

3 N–H bonds + 1 lone pair on N

Formal charge

V valence e⁻, L lone-pair e⁻, B bonding e⁻. Best structure: charges near zero, any − on the more electronegative atom.

FC = V − L − B2

Carbonate, CO₃²⁻ Worked

4 + 3(6) + 2 = 24 e⁻; one C=O, two C–O

single-bonded O: 6 − 6 − 1 = −1; C and double-bonded O: 03 resonance forms

Octet exceptions

H holds 2; Be often 4; B often 6 (BF₃).
Period 3 and beyond can exceed 8: PCl₅ (10), SF₆ (12).
Odd electron counts: NO (11 e⁻).

Bond energies estimate ΔH Worked

H₂ + Cl₂ → 2 HCl; H–H 436, Cl–Cl 243, H–Cl 431 kJ/mol

ΔH ≈ broken − formed = 436 + 243 − 2(431) = −183 kJexothermic

Strength of ionic bonds (lattice energy)

Bigger ion charges → stronger attraction.
Smaller ions → closer together → stronger.

So MgO (2+, 2−) melts far higher than NaCl (1+, 1−).

Bond order: single 1, double 2, triple 3
Higher order: shorter and stronger: C≡C < C=C < C–C in length

Resonance: two or more valid Lewis structures; the real molecule is their average (O₃, CO₃²⁻)
Coordinate bond: both shared electrons come from one atom (NH₄⁺)

6Shape, Hybrids & Polarity

VSEPR shapes (domains = bonded atoms + lone pairs on the center atom; a double or triple bond counts once)
Domains	Lone	Shape	Angle	Hybrid	Example
2	0	linear	180°	sp	CO₂
3	0	trigonal planar	120°	sp²	BF₃
3	1	bent	< 120°	sp²	SO₂
4	0	tetrahedral	109.5°	sp³	CH₄
4	1	trigonal pyramidal	≈ 107°	sp³	NH₃
4	2	bent	≈ 104.5°	sp³	H₂O
5	0	trigonal bipyramidal	90°, 120°	sp³d	PCl₅
5	1	seesaw	< 90°, < 120°	sp³d	SF₄
5	2	T-shaped	< 90°	sp³d	ClF₃
5	3	linear	180°	sp³d	XeF₂
6	0	octahedral	90°	sp³d²	SF₆
6	1	square pyramidal	< 90°	sp³d²	BrF₅
6	2	square planar	90°	sp³d²	XeF₄

Is the molecule polar?

No polar bonds → nonpolar.
Polar bonds in a symmetric shape cancel → nonpolar (CO₂, BF₃, CCl₄).
Polar bonds in a lopsided shape (lone pairs on the center or mixed outer atoms) → polar (H₂O, NH₃, CHCl₃).

Lone pairs push harder

A lone pair spreads wider than a bond, squeezing the angles: 109.5° (CH₄) → 107° (NH₃) → 104.5° (H₂O).

Polar or not? Worked

CCl₄: tetrahedral, four equal pulls → nonpolar
H₂O: bent → polar
SO₂: bent (1 lone pair) → polar

σ bond: head-on overlap; every single bond
π bond: side-by-side overlap; a double bond is 1σ + 1π, a triple 1σ + 2π

Count σ and π in C₂H₄ Worked

4 C–H + 1 C–C σ = 5 σ

C=C adds 1 πeach C is sp²

Predict XeF₄ Worked

Xe: 8 valence e⁻; 4 bonds use 4 → 2 lone pairs

6 domains, 2 lone → square planar, nonpolarlone pairs opposite

General ChemistryChemistry · Units 1–9 · continued

7Intermolecular Forces & Phases

Intermolecular forces
Force	Between	Strength
London dispersion	all molecules; grows with size (more e⁻)	weakest per contact
dipole–dipole	polar molecules	moderate
hydrogen bond	H on N, O or F → lone pair on N, O or F	strongest IMF
ion–dipole	ions and polar molecules (salt in water)	stronger still

H bonds need F, O, N

Hydrogen bonds form only when H sits on F, O or N. (Traditional “H-FON”.)

Stronger IMFs mean

higher boiling and melting points
lower vapor pressure (less evaporation)
higher surface tension and viscosity

Rank boiling points Worked

CH₄ (−162 °C) < HCl (−85 °C) < H₂O (100 °C)

dispersion < dipole < hydrogen bonding

Halogens at room temperature Worked

F₂, Cl₂ gases; Br₂ liquid; I₂ solid

more electrons → stronger dispersion

Heating curve for water: temperature holds steady while the phase changes.

Phase changes
Change	Name	Heat
solid → liquid	melting (fusion)	absorbed
liquid → gas	vaporization	absorbed
solid → gas	sublimation	absorbed
gas → liquid	condensation	released
liquid → solid	freezing	released
gas → solid	deposition	released

Melt and warm 10.0 g of ice at 0 °C to 25.0 °C Worked

q = mΔH_fus + mcΔTΔH_fus = 334 J/g

= 10.0(334) + 10.0(4.184)(25.0) = 4.39 kJ3340 + 1046 J

Boiling point: vapor pressure equals the outside pressure
Triple point: all three phases together
Critical point: above it, liquid and gas merge

Phase diagram (water-like, not to scale): lines are where two phases coexist; T triple point, C critical point.

Boiling and pressure

Lower outside pressure (high altitude) → water boils below 100 °C.
A pressure cooker raises the pressure, so water boils hotter and food cooks faster.

8Naming Compounds

Ionic compounds

Cation name, then anion name (-ide for one-element anions).
Metals with more than one charge get a Roman numeral (Stock system).
Polyatomic ions keep their names; list: Periodic Table sheet.

Name it Worked

CaCl₂ → calcium chloride
Fe₂O₃ → iron(III) oxide2 Fe = +6 balances 3 O = −6
(NH₄)₂SO₄ → ammonium sulfate
Cu(NO₃)₂ → copper(II) nitrate

Writing the formula

Total charge must be zero.
Cross the charge numbers to subscripts, then reduce.
Use parentheses for 2 or more polyatomic ions.

EXAl³⁺ + O²⁻ → Al₂O₃; Ca²⁺ + PO₄³⁻ → Ca₃(PO₄)₂

Zero is the goal

Every ionic formula adds to zero charge: check it by multiplying each charge by its subscript.

Covalent prefixes (two nonmetals)

1 mono, 2 di, 3 tri, 4 tetra, 5 penta, 6 hexa, 7 hepta, 8 octa, 9 nona, 10 deca

Most-used polyatomic ions (full list: Periodic Table sheet)
Ion	Name	Ion	Name
NH₄⁺	ammonium	OH⁻	hydroxide
NO₃⁻	nitrate	SO₄²⁻	sulfate
CO₃²⁻	carbonate	PO₄³⁻	phosphate
HCO₃⁻	hydrogen carbonate	C₂H₃O₂⁻	acetate

Write the formula Worked

iron(III) sulfate → Fe₂(SO₄)₃2(+3) + 3(−2) = 0
sodium hydrogen carbonate → NaHCO₃baking soda
dinitrogen pentoxide → N₂O₅

Covalent names Worked

N₂O₄ → dinitrogen tetroxidedrop “a” before o
CO → carbon monoxideno mono- on the first element
PCl₅ → phosphorus pentachloride

Acids (in water)
Anion ends in	Acid name	Example
-ide	hydro___ic acid	HCl hydrochloric acid
-ate	___ic acid	HNO₃ nitric, H₂SO₄ sulfuric
-ite	___ous acid	HNO₂ nitrous, H₂SO₃ sulfurous

I ate something icky

-ate → -ic; -ite → -ous (“the sprITE was delicOUS”). (Traditional.)

Hydrate: water in the crystal: CuSO₄·5H₂O, copper(II) sulfate pentahydrate
Diatomic: H₂, N₂, O₂, F₂, Cl₂, Br₂, I₂ as elements

Common names: H₂O water, NH₃ ammonia, CH₄ methane, H₂O₂ hydrogen peroxide

General ChemistryChemistry · Units 1–9 · continued

9The Mole & Stoichiometry

Mole: 6.022 × 10²³ particles (Avogadro’s number)
Molar mass: grams per mole; add the atomic weights
Molar volume: 22.4 L per mole of gas at STP (0 °C, 1 atm)

The mole map

grams ÷ molar mass → moles × molar mass → grams
particles ÷ 6.022 × 10²³ ↔ moles × 6.022 × 10²³
liters of gas (STP) ÷ 22.4 ↔ moles
molarity × liters of solution → moles

Every road runs through moles

Never jump straight from grams of one thing to grams of another: go through moles.

Molar mass of Ca(NO₃)₂ Worked

40.078 + 2(14.007) + 6(15.999)

= 164.09 g/mol

25.0 g of water Worked

25.0 ÷ 18.015 = 1.39 mol

1.39 mol × 6.022 × 10²³ = 8.36 × 10²³ molecules

Percent composition

Use the element’s total mass in one mole of the compound.

% = element’s massmolar mass × 100

EXO in H₂O: 15.99918.015 × 100 = 88.81 %

Empirical formula from percents

Treat percent as grams (100 g sample).
Grams → moles of each element.
Divide by the smallest; round or multiply to whole numbers.
Molecular formula: multiply by molar massempirical mass.

40.0 % C, 6.71 % H, 53.3 % O; 180.16 g/mol Worked

C 3.33, H 6.66, O 3.33 mol → 1 : 2 : 1 → CH₂O

180.16 ÷ 30.026 = 6 → C₆H₁₂O₆

Steps

Balance the equation.
Given → moles.
Multiply by the mole ratio from coefficients.
Moles → asked-for unit.

Water from 10.0 g H₂ Worked

2 H₂ + O₂ → 2 H₂O
10.0 ÷ 2.016 = 4.96 mol H₂ × 2 H₂O2 H₂ = 4.96 mol H₂O
× 18.015 = 89.4 g H₂O

Limiting reactant

Work out the product from each reactant.
The one giving less product is limiting; that amount is the theoretical yield.
The other is in excess: subtract what reacted to find what is left.

The pantry rule

The ingredient that runs out first sets the size of the batch, no matter how much of the rest you have.

Ammonia: 28.0 g N₂ + 9.0 g H₂ Worked

N₂ + 3 H₂ → 2 NH₃
N₂: 0.9995 mol → 1.999 mol NH₃;  H₂: 4.46 mol → 2.98 mol NH₃
N₂ limits: 34.0 g NH₃× 17.031 g/mol
H₂ left: 4.46 − 3.00 = 1.47 mol = 3.0 g

Percent yield

% yield = actualtheoretical × 100

EXgot 25.5 g NH₃ of 34.0 g possible: 75.0 %

10Reactions & Net Ionic Equations

Balancing

Change coefficients only, never subscripts.
Balance metals, then nonmetals, then H, then O last.
Treat an unchanged polyatomic ion as one unit.

EXC₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O; 4 Al + 3 O₂ → 2 Al₂O₃

Five reaction types
Type	Pattern	Example
synthesis	A + B → AB	2 Na + Cl₂ → 2 NaCl
decomposition	AB → A + B	2 H₂O₂ → 2 H₂O + O₂
single replacement	A + BC → AC + B	Zn + 2 HCl → ZnCl₂ + H₂
double replacement	AB + CD → AD + CB	AgNO₃ + NaCl → AgCl + NaNO₃
combustion	fuel + O₂ → CO₂ + H₂O	CH₄ + 2 O₂ → CO₂ + 2 H₂O

Solubility in water
Ions	Soluble?	Exceptions
Group 1, NH₄⁺, NO₃⁻, C₂H₃O₂⁻	always	none
Cl⁻, Br⁻, I⁻	yes	Ag⁺, Pb²⁺, Hg₂²⁺
SO₄²⁻	yes	Ba²⁺, Pb²⁺, Ca²⁺, Sr²⁺
CO₃²⁻, PO₄³⁻, S²⁻	no	with Group 1 or NH₄⁺
OH⁻	no	Group 1, NH₄⁺; Ba²⁺, Sr²⁺, Ca²⁺ partly

Net ionic equation

Write the balanced molecular equation with states.
Split strong electrolytes (aq) into ions; keep solids, liquids, gases and weak acids whole.
Cross out spectator ions (same on both sides).

Spectators just watch

Ions that look the same on both sides watch the game and leave the final equation.

Yellow precipitate Worked

Pb(NO₃)₂(aq) + 2 KI(aq) → PbI₂(s) + 2 KNO₃(aq)

net: Pb²⁺(aq) + 2 I⁻(aq) → PbI₂(s)K⁺, NO₃⁻ watch

Any strong acid + strong base Worked

H⁺(aq) + OH⁻(aq) → H₂O(l)

Activity series (most → least active)

Li > K > Ba > Ca > Na > Mg > Al > Zn > Fe > Ni > Sn > Pb > (H) > Cu > Ag > Au

A metal replaces any ion below it: Cu + 2 Ag⁺ → Cu²⁺ + 2 Ag works; Cu + Zn²⁺ does not.

General ChemistryChemistry · Units 1–9 · continued

11Solutions & Concentration

Solute / solvent: what dissolves / what does the dissolving
Saturated: holds all it can at that temperature
Like dissolves like: polar with polar (salt, sugar in water), nonpolar with nonpolar (oil in hexane)
Electrolyte: makes ions in water: strong (NaCl, HCl), weak (acetic acid), non (sugar)

Molarity

M = mol soluteL solution

Dilution

M₁V₁ = M₂V₂

Mass percent Worked

15.0 g sugar in 135.0 g water

15.0150.0 × 100 = 10.0 %solution mass = 150.0 g

5.85 g NaCl in 250.0 mL Worked

5.85 ÷ 58.44 = 0.100 mol

0.100 mol ÷ 0.2500 L = 0.400 M

MV makes moles

Molarity × volume (L) = moles: the bridge for every solution problem.

500. mL of 1.50 M from 12.0 M HCl Worked

V₁ = 1.50 × 500.12.0 = 62.5 mLadd acid to water

Titration Worked

18.40 mL of 0.1000 M NaOH neutralizes 25.00 mL HCl
mol NaOH = 1.840 × 10⁻³ = mol HCl1 : 1
[HCl] = 1.840 × 10⁻³0.02500 L = 0.07360 M

Molality: m = mol solute ÷ kg solvent
Mass percent: (g solute ÷ g solution) × 100
ppm: (g solute ÷ g solution) × 10⁶

Colligative properties

Water: K_f = 1.86, K_b = 0.512 °C·kg/mol; i = particles per formula unit.

ΔT_f = i K_f m

ΔT_b = i K_b m

EX1.00 m NaCl (i = 2) freezes at −3.72 °C

Molality and freezing Worked

18.0 g glucose (180.16 g/mol) in 500. g water

0.0999 mol ÷ 0.500 kg = 0.200 m → ΔT_f = 1.86(0.200) = 0.372 °Ci = 1

Ions in solution Worked

0.10 M CaCl₂ → [Ca²⁺] = 0.10 M, [Cl⁻] = 0.20 M2 Cl⁻ per formula

Solubility and temperature

Most solids dissolve better when hot.
Gases dissolve better when cold and under higher pressure (an opened soda goes flat).

12Gas Laws

Before you start

Temperature in kelvin, always.
1 atm = 760 mmHg = 760 torr = 101.325 kPa.
STP: 0 °C (273.15 K) and 1 atm.

The gas laws
Law	Equation	Held constant
Boyle	P₁V₁ = P₂V₂	n, T
Charles	V₁T₁ = V₂T₂	n, P
Gay-Lussac	P₁T₁ = P₂T₂	n, V
Avogadro	V₁n₁ = V₂n₂	P, T
Combined	P₁V₁T₁ = P₂V₂T₂	n

Boyle is the one that flips

Squeeze a gas and pressure rises as volume falls: the only inverse pair. Every other pair moves together.

Boyle Worked

2.0 L at 1.0 atm → 0.50 atm

V₂ = 1.0 × 2.00.50 = 4.0 L

Charles Worked

3.00 L at 27 °C → 127 °C

V₂ = 3.00 × 400.15300.15 = 4.00 Lkelvin!

Combined: to STP Worked

2.50 L at 25 °C, 1.00 atm → 0 °C, 1.00 atm

V₂ = 2.50 × 273.15298.15 = 2.29 L

Ideal gas law

R = 0.08206 L·atm/(mol·K) = 8.314 J/(mol·K).

PV = nRT

Find P Worked

0.500 mol in 10.0 L at 25 °C

P = 0.500 × 0.08206 × 298.1510.0 = 1.22 atm

Molar mass from density Worked

M = dRTP: 1.25 g/L at STP

1.25 × 0.08206 × 273.15 ÷ 1 = 28.0 g/molN₂ or CO

Dalton: partial pressures

P_total = P₁ + P₂ + …

EXgas over water: 750. − 23.8 (water vapor at 25 °C) = 726 mmHg dry gas

Graham: effusion

rate₁rate₂ = √(M₂M₁)

EXHe vs O₂: √(32.00 ÷ 4.003) = 2.83 times faster

Gas stoichiometry Worked

Zn + 2 HCl → ZnCl₂ + H₂;  5.00 g Zn at 25 °C, 1.00 atm
5.00 ÷ 65.38 = 0.0765 mol H₂
V = nRTP = 0.0765(0.08206)(298.15) = 1.87 L

Kinetic-molecular theory

Particles are tiny compared with the space between them.
They move randomly and collide elastically.
Average kinetic energy ∝ kelvin temperature.
Real gases stray from ideal at high P (molecules take up room) and low T (attractions matter).

General ChemistryChemistry · Units 1–9 · continued

13Thermochemistry

Exo exits

In an exothermic change heat exits the system, so ΔH is negative and the surroundings warm up.

Heat

Exothermic: releases heat, ΔH < 0. Endothermic: absorbs heat, ΔH > 0. Water: c = 4.184 J/(g·°C).

q = mcΔT

EX150. g water, 22.0 → 35.5 °C: q = 150.(4.184)(13.5) = 8.47 kJ

Coffee-cup calorimeter Worked

50.0 g metal at 100.0 °C into 100.0 g water at 20.0 °C; final 23.7 °C
heat lost by metal = heat gained by water
c = 100.0(4.184)(3.7)50.0(76.3) = 0.41 J/(g·°C)

Hess’s law: C + ½O₂ → CO Worked

C + O₂ → CO₂   ΔH = −393.5 kJ
CO₂ → CO + ½O₂   ΔH = +283.0 kJreversed
add the steps and their ΔH: ΔH = −110.5 kJreverse a step → flip the sign

From enthalpies of formation

ΔH°_f of an element in its standard state = 0.

ΔH°_rxn = ΣnΔH°_f(products) − ΣnΔH°_f(reactants)

Burning methane Worked

CH₄ + 2 O₂ → CO₂ + 2 H₂O(l)ΔH°f: −74.8, 0, −393.5, −285.8
−393.5 + 2(−285.8) − (−74.8) = −890.3 kJ
16.0 g CH₄ (0.997 mol) releases 888 kJ

Gibbs free energy

ΔG < 0: spontaneous (product-favored). ΔS > 0: more disorder (gas made, solid dissolves).

ΔG = ΔH − TΔS

Sign chart
ΔH	ΔS	Spontaneous?
−	+	always
+	−	never
−	−	at low T
+	+	at high T

Ammonia at 298 K Worked

ΔH = −92.2 kJ, ΔS = −198.7 J/K
ΔG = −92.2 − 298(−0.1987) = −33.0 kJspontaneous
switches at T = ΔHΔS ≈ 464 K

14Reaction Rates & Equilibrium

Collision theory

Particles must collide,
with the right orientation,
and with energy ≥ the activation energy E_a.

SCAT speeds it up

Surface area, Concentration, Activation helpers (catalysts), Temperature: raise any of them and the rate rises.

Exothermic reaction: products sit lower (ΔH < 0). A catalyst (dashed) lowers Eₐ but not ΔH.

Rate law

Orders m and n come from experiments, not from coefficients. Overall order = m + n.

rate = k[A]^m[B]ⁿ

Orders from data Worked

double [A], rate × 4 → 2nd order in A
double [B], rate unchanged → 0 order in B
rate = k[A]²: 0.010 M/s at [A] = 0.10 M → k = 1.0 M⁻¹s⁻¹

By order
Order	Rate law	Units of k	Half-life
0	k	M/s	[A]₀ ÷ 2k
1	k[A]	s⁻¹	0.693 ÷ k
2	k[A]²	M⁻¹s⁻¹	1 ÷ (k[A]₀)

First-order half-life Worked

k = 0.0231 min⁻¹ → t_½ = 0.6930.0231 = 30.0 min

after 90.0 min (3 half-lives): ⅛ remains

Mechanism: the series of elementary steps
Rate-determining step: the slowest step sets the rate
Catalyst: used, then regenerated; offers a lower-E_a path

Equilibrium constant

For aA + bB ⇌ cC + dD. Leave out pure solids and liquids. K ≫ 1 favors products; K ≪ 1 favors reactants.

K = [C]^c[D]^d[A]^a[B]^b

Q versus K

Q uses the current amounts in the K expression.
Q < K: shifts right (makes products).
Q > K: shifts left.
Q = K: at equilibrium.

ICE table: H₂ + I₂ ⇌ 2 HI, K = 49.0 Worked

start 1.00 M H₂, 1.00 M I₂; change −x, −x, +2x
(2x)²(1.00 − x)² = 49.0 → 2x1.00 − x = 7.00 → x = 0.778
[HI] = 1.56 M; [H₂] = [I₂] = 0.222 M

Le Chatelier’s principle
Stress	Shift
add reactant or remove product	→ right
add product or remove reactant	← left
raise pressure (shrink volume)	toward fewer gas moles
raise temperature	toward the endothermic side (K changes)
add a catalyst	no shift; just faster

Push it and it pushes back

A system at equilibrium shifts to undo whatever stress you add.

Solubility product

For AgCl(s) ⇌ Ag⁺ + Cl⁻; the solid is left out.

K_sp = [Ag⁺][Cl⁻]

EXK_sp = 1.8 × 10⁻¹⁰ → s = √(1.8 × 10⁻¹⁰) = 1.3 × 10⁻⁵ M

General ChemistryChemistry · Units 1–9 · continued

15Acids, Bases, pH & Buffers

Three definitions
Theory	Acid	Base
Arrhenius	makes H⁺ in water	makes OH⁻ in water
Brønsted–Lowry	H⁺ donor	H⁺ acceptor
Lewis	e⁻-pair acceptor	e⁻-pair donor

Conjugate pairs Worked

NH₃ + H₂O ⇌ NH₄⁺ + OH⁻

base NH₃ / conjugate acid NH₄⁺; acid H₂O / conjugate base OH⁻

pH

[H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25 °C. pH < 7 acidic, 7 neutral, > 7 basic.

pH = −log[H⁺]

pH + pOH = 14.00

Strong acids: HCl, HBr, HI, HNO₃, H₂SO₄, HClO₄ (ionize 100 %)
Strong bases: Group 1 hydroxides; Ca(OH)₂, Sr(OH)₂, Ba(OH)₂

Three halogens, three oxyacids

The six strong acids: H + Cl, Br, I and the oxyacids nitric, sulfuric, perchloric. Everything else is weak.

25.0 mL of 0.100 M HCl titrated with 0.100 M NaOH: steep jump at 25.0 mL, pH 7 at equivalence.

pH practice Worked

0.0025 M HCl: pH = −log(0.0025) = 2.60
0.010 M NaOH: pOH = 2.00 → pH 12.00
pH 4.50 → [H⁺] = 10⁻⁴·⁵⁰ = 3.2 × 10⁻⁵ M

Weak acid: 0.10 M acetic acid Worked

Ka = 1.8 × 10⁻⁵ = x²0.10 − x
x ≪ 0.10: x²0.10 ≈ 1.8 × 10⁻⁵ → x = 1.3 × 10⁻³ M
pH = 2.87

Buffers

A weak acid with its conjugate base resists pH change: it soaks up added H⁺ or OH⁻.

pH = pK_a + log [A⁻][HA]

EX0.20 M acetate + 0.10 M acetic acid: 4.74 + log 2 = pH 5.04

16Redox, Cells & Nuclear Change

Assigning oxidation numbers
Rule	Oxidation number
element alone (Na, O₂)	0
monatomic ion	its charge
O in compounds	−2 (peroxides −1)
H	+1 (with metals −1)
F	always −1
sum	0, or the ion’s charge

Find the oxidation number Worked

Cr₂O₇²⁻: 2x + 7(−2) = −2 → Cr = +6

MnO₄⁻: x + 4(−2) = −1 → Mn = +7

OIL RIG

Oxidation Is Loss of electrons; Reduction Is Gain. (Traditional.)

Balancing in acid (half-reactions)

Split into oxidation and reduction halves.
Balance atoms other than O and H.
O with H₂O; H with H⁺; charge with e⁻.
Match electrons, add, cancel.

Permanganate and iron(II) Worked

MnO₄⁻ + 8 H⁺ + 5 e⁻ → Mn²⁺ + 4 H₂O
5 Fe²⁺ → 5 Fe³⁺ + 5 e⁻
MnO₄⁻ + 8 H⁺ + 5 Fe²⁺ → Mn²⁺ + 5 Fe³⁺ + 4 H₂Oatoms and charge balance

Standard reduction potentials
Half-reaction	E° (V)
F₂ + 2e⁻ → 2F⁻	+2.87
Ag⁺ + e⁻ → Ag	+0.80
Cu²⁺ + 2e⁻ → Cu	+0.34
2H⁺ + 2e⁻ → H₂	0.00
Fe²⁺ + 2e⁻ → Fe	−0.44
Zn²⁺ + 2e⁻ → Zn	−0.76
Li⁺ + e⁻ → Li	−3.04

Galvanic (voltaic) cell

Anode: oxidation; reduction at the cathode (traditional “An Ox, Red Cat”).
Electrons flow anode → cathode through the wire; a salt bridge closes the circuit.
E°_cell = E°_cathode − E°_anode > 0 means spontaneous.

Cell notation

anode | anode ion ‖ cathode ion | cathode (‖ = salt bridge)

Zn | Zn²⁺ ‖ Cu²⁺ | Cu Worked

E° = 0.34 − (−0.76) = 1.10 V

ΔG° = −nFE° = −2(96 485)(1.10) = −212 kJF = 96 485 C/mol

Electroplating copper Worked

2.00 A for 30.0 min: q = It = 3600 C

3600 ÷ 96 485 = 0.0373 mol e⁻ → 0.0187 mol Cu = 1.19 gCu²⁺ + 2e⁻

Radioactive decay
Decay	Particle	Change	Stopped by
alpha	⁴₂He	Z − 2, A − 4	paper, skin
beta	⁰₋₁e	Z + 1	aluminum
positron	⁰₊₁e	Z − 1	aluminum
gamma	⁰₀γ	none (energy)	lead, concrete

Alpha drops 4 and 2; beta bumps Z

An alpha takes 4 off the mass and 2 off Z. A beta turns a neutron into a proton, so Z goes up one and A stays.

Balance the nuclear equation Worked

²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂He238 = 234 + 4; 92 = 90 + 2

¹⁴₆C → ¹⁴₇N + ⁰₋₁e

Half-life

N = N₀(½)^{t / t½}

Carbon-14 dating, t_½ = 5730 yr Worked

25.0 % of the original ¹⁴C left = 2 half-lives

age ≈ 11 460 yr

Fission: a heavy nucleus splits (U-235 in power plants)
Fusion: light nuclei join (the sun’s hydrogen → helium)
Mass–energy: E = mc²: 1.00 g → 8.99 × 10¹³ J
Uses: medical imaging and treatment, smoke detectors, dating, power