What do the Grade 11 Chemistry cards cover?

34 cards across 6 chapters of the national textbook: Atomic Structure and Periodic Properties of the Elements, Chemical Bonding, Physical states of matter, Chemical Kinetics, Chemical Equilibrium and Some Important Oxygen-Containing Organic Compounds. You can take any chapter one card at a time on the page itself.

Is there a national exam in Grade 11?

No. Ethiopia sets national exams in Grade 6, Grade 8 and Grade 12 only. These cards are for your school's own exams, and for the national exam that comes a few years later.

Where do these cards come from?

They are drafted from Grade 11 Chemistry, the Ministry of Education textbook for this grade. A second pass that cannot see the chapter then re-derives every formula, constant and table row, and anything it cannot confirm is held back instead of published.

Yes. Every card here is free to read and the printable sheet is free to download. Neither needs an account.

When was this last checked?

6 September 2026. Cards arrive chapter by chapter, and the line under each one says when that card was last read through.

Free short notes and revision cards

Grade 11 Chemistry

Grade 11 Chemistry on Temari has 34 revision cards, arranged by the chapters of the Ethiopian national curriculum. Every card says when the rule applies, what each symbol in it stands for, and the mistake students most often make with it. They are free to read and need no account.

34
Cards: 6
Chapters: 16
Formulas: 8
Reference tables: Grade 11 Chemistry
Textbook: Free
Price: 6 September 2026
Last checked

Everything34Formula13Reference table8Common mistake7Rule6

Atomic Structure and Periodic Properties of the Elements

de Broglie WavelengthFormula

\lambda = \frac{h}{m v}

\lambda = \frac{h}{m v}

$\lambda$: de Broglie wavelength $\text{m}$
$h$: Planck's constant $\text{J}\,\text{s}$
$m$: Mass of the particle $\text{kg}$
$v$: Velocity of the particle $\text{m}\,\text{s}^{-1}$

When you use it

Use when finding the wavelength associated with a moving particle such as an electron or atom.

Watch out

Mass must always be in kilograms, not grams or atomic mass units, to cancel units with Planck's constant properly.

Drafted from Grade 11 Chemistry, pages 2-66, then checked twice before it went up

Hydrogen Electron Transition EnergyFormula

\Delta E = -2.18 \times 10^{-18} \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)

\Delta E = -2.18 \times 10^{-18} \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)

$\Delta E$: Energy change of transition $\text{J}$
$n_i$: Initial principal quantum number $\text{dimensionless}$
$n_f$: Final principal quantum number $\text{dimensionless}$

When you use it

Use to calculate the energy absorbed or emitted when an electron in a hydrogen atom jumps between two orbits.

Watch out

A negative change in energy indicates emission of a photon, while a positive change indicates absorption.

Drafted from Grade 11 Chemistry, pages 2-66, then checked twice before it went up

Hydrogen Emission SeriesReference tableLyman · Balmer · Paschen · +1

Series	Final level ($n_f$)	Initial level ($n_i$)	Spectral region
Lyman	$1$	$2, 3, 4, \dots$	Ultraviolet
Balmer	$2$	$3, 4, 5, \dots$	Visible and ultraviolet
Paschen	$3$	$4, 5, 6, \dots$	Infrared
Brackett	$4$	$5, 6, 7, \dots$	Infrared

When you use it

Use to identify the spectral series and wavelength region for electron transitions in atomic hydrogen.

Watch out

Only transitions ending at n = 2 produce visible lines. Transitions ending at n = 1 are entirely ultraviolet.

Drafted from Grade 11 Chemistry, pages 2-66, then checked twice before it went up

Photoelectric Effect EquationFormula

\text{KE}_e = h\nu - h\nu_0

\text{KE}_e = h\nu - h\nu_0

$\text{KE}_e$: Kinetic energy of the ejected electron $\text{J}$
$h$: Planck's constant $\text{J}\,\text{s}$
$\nu$: Frequency of incident radiation $\text{s}^{-1}$
$\nu_0$: Threshold frequency of the metal $\text{s}^{-1}$

When you use it

Use when finding the kinetic energy or speed of electrons ejected from a metal illuminated by light above the threshold frequency.

Watch out

No electrons are emitted if the incident frequency is less than the threshold frequency, regardless of light intensity.

Drafted from Grade 11 Chemistry, pages 2-66, then checked twice before it went up

Photon Energy and WavelengthFormula

E = \frac{h c}{\lambda}

E = \frac{h c}{\lambda}

$E$: Energy of a photon $\text{J}$
$h$: Planck's constant $\text{J}\,\text{s}$
$c$: Speed of light $\text{m}\,\text{s}^{-1}$
$\lambda$: Wavelength $\text{m}$

When you use it

Use when converting between the wavelength of electromagnetic radiation and the energy carried by a single photon.

Watch out

Always convert wavelength from nanometers or micrometers to meters before calculating energy in joules.

Drafted from Grade 11 Chemistry, pages 2-66, then checked twice before it went up

The order the subshells fillRule

\mathrm{1s\;2s\;2p\;3s\;3p\;4s\;3d\;4p\;5s\;4d\;5p\;6s\;4f\;5d\;6p\;7s\;5f\;6d\;7p}

\mathrm{1s\;2s\;2p\;3s\;3p\;4s\;3d\;4p\;5s\;4d\;5p\;6s\;4f\;5d\;6p\;7s\;5f\;6d\;7p}

$\mathrm{s}$: one orbital, so the subshell holds 2 electrons
$\mathrm{p}$: three orbitals, so the subshell holds 6 electrons
$\mathrm{d}$: five orbitals, so the subshell holds 10 electrons
$\mathrm{f}$: seven orbitals, so the subshell holds 14 electrons
$2(2l+1)$: how many electrons a subshell holds, counted from its own quantum number

When you use it

Use whenever you write a ground state electron configuration. Electrons enter the lowest energy subshell first (Aufbau), an orbital takes at most two electrons and those two spin opposite ways (Pauli), and among orbitals of equal energy each takes one electron before any takes a second (Hund).

Watch out

After $\mathrm{3p}$ the next electron goes into $\mathrm{4s}$ and not $\mathrm{3d}$ , so potassium is $1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{1}$ . The trap is that the same textbook then prints scandium as $[\mathrm{Ar}]\,3d^{1}\,4s^{2}$ , because a finished configuration is written by shell number rather than by the order it filled. Fill in the order above, write the answer in shell order, and when the atom is ionised take the electrons out of $\mathrm{4s}$ first: $\mathrm{Fe^{2+}}$ is $[\mathrm{Ar}]\,3d^{6}$ and never $[\mathrm{Ar}]\,3d^{4}\,4s^{2}$ .

Chromium, copper and the half filled d subshellCommon mistakeChromium, Z = 24 · Copper, Z = 29 · Molybdenum, Z = 42 · +2

Element	Ground state	What to notice
Chromium, $Z = 24$	$[\mathrm{Ar}]\,3d^{5}\,4s^{1}$	the filling order predicts $[\mathrm{Ar}]\,3d^{4}\,4s^{2}$
Copper, $Z = 29$	$[\mathrm{Ar}]\,3d^{10}\,4s^{1}$	the filling order predicts $[\mathrm{Ar}]\,3d^{9}\,4s^{2}$
Molybdenum, $Z = 42$	$[\mathrm{Kr}]\,4d^{5}\,5s^{1}$	the same pattern as chromium, one row above it
Silver, $Z = 47$	$[\mathrm{Kr}]\,4d^{10}\,5s^{1}$	the same pattern as copper, one row above it
Niobium, $Z = 41$	$[\mathrm{Kr}]\,4d^{4}\,5s^{1}$	neither subshell ends up half filled or full

When you use it

Use when a d-block element's ground state refuses to match the plain filling order: one s electron shifts into the d subshell, because a half filled set ( $d^{5}$ ) or a completely filled set ( $d^{10}$ ) is the more stable arrangement. Check any configuration you write by adding its electrons up, since they have to total the atomic number.

Watch out

Students turn "half filled is stable" into a general licence and promote an s electron wherever the d subshell looks close to five or ten. Manganese is $[\mathrm{Ar}]\,3d^{5}\,4s^{2}$ and iron is $[\mathrm{Ar}]\,3d^{6}\,4s^{2}$ : count the electrons and the invented version comes out one short of the atomic number. The second half of the mistake shows up with ions. Once the atom ionises the $4s$ electron leaves first, so $\mathrm{Cu^{2+}}$ is $[\mathrm{Ar}]\,3d^{9}$ and $\mathrm{Cr^{3+}}$ is $[\mathrm{Ar}]\,3d^{3}$ , never $[\mathrm{Ar}]\,3d^{8}\,4s^{1}$ or $[\mathrm{Ar}]\,3d^{2}\,4s^{1}$ .

The s, p, d and f blocksReference tables · p · d · +1

Block	Outer configuration	Where it sits, and what it is like
s	$ns^{1-2}$	Groups 1 and 2, plus hydrogen and helium. All metals apart from those two, strongly electropositive with low ionization energy, and mostly ionic compounds; beryllium is the exception, since $\mathrm{BeCl_2}$ is covalent and $\mathrm{BeO}$ is amphoteric.
p	$ns^{2}\,np^{1-6}$	Groups 13 to 18. The only block that holds metals, non-metals and metalloids together, with 3 to 8 valence electrons and variable oxidation states; non-metallic character rises across a period and falls down a group.
d	$(n-1)d^{1-10}\,ns^{0-2}$	Groups 3 to 12. All metals, mostly hard and high melting, with variable oxidation states, coloured compounds and many catalysts; zinc, cadmium and mercury sit in the block but are not transition elements, because neither the atom nor its usual ion has a partly filled d subshell.
f	$(n-2)f^{1-14}\,(n-1)d^{0-1}\,ns^{2}$	The two rows printed below the main table: lanthanoids from La to Lu and actinoids from Ac to Lr, fifteen each. All metals, dense and high melting; every actinoid is radioactive, while among the lanthanoids only promethium is.

When you use it

Use when you are asked which block an element belongs to, or to write its outer configuration from its place in the periodic table. A block is named after the subshell that receives the last electron added, so read it off the configuration rather than off where the printed table puts the element.

Watch out

Students read the n in $(n-1)d$ and $(n-2)f$ as the number of the d or f subshell itself. It is the outermost shell, which is the period number. Iron sits in Period 4, so $n = 4$ and its configuration is $(n-1)d^{6}\,ns^{2} = 3d^{6}\,4s^{2}$ , never $4d^{6}\,4s^{2}$ ; cerium sits in Period 6, so $(n-2)f$ is $4f$ . Ask which period the element is in first, then subtract. The second trap is helium: it is printed above neon in Group 18 and gets written down as p-block, but $1s^{2}$ has no p electron at all, so helium is an s-block element.

Chemical Bonding

Bond Polarity versus Molecular PolarityCommon mistake

When you use it

Use this card when determining whether a molecule with polar bonds is overall polar or non-polar.

Watch out

Polar bonds do not always produce a polar molecule. In symmetric shapes like carbon dioxide or boron trichloride, individual bond dipoles cancel out to give zero net dipole moment.

Drafted from Grade 11 Chemistry, pages 67-144, then checked twice before it went up

Dipole MomentFormula

\mu = \delta \times d

\mu = \delta \times d

$\mu$: dipole moment $\text{C}\,\text{m}$
$\delta$: magnitude of partial charge $\text{C}$
$d$: distance separating the charges $\text{m}$

When you use it

Use this formula to calculate the dipole moment of a polar covalent bond or diatomic molecule.

Watch out

In non-SI units, 1 Debye equals 3.33564 multiplied by 10^-30 coulomb metres.

Drafted from Grade 11 Chemistry, pages 67-144, then checked twice before it went up

Exceptions to the Octet RuleReference tableLess than octet · Expanded octet · Odd-electron molecules

Category	Electrons on Central Atom	Examples
Less than octet	Fewer than 8 electrons	$\text{BeCl}_2$ , $\text{BF}_3$ , $\text{AlCl}_3$
Expanded octet	More than 8 electrons	$\text{PF}_5$ , $\text{SF}_6$ , $\text{XeF}_4$
Odd-electron molecules	Odd number of electrons	$\text{NO}$ , $\text{NO}_2$ , $\text{ClO}_2$

When you use it

Use this reference when identifying molecules that do not obey the standard eight-electron octet rule.

Watch out

Third period and lower non-metals expand their octet using available empty d orbitals, but second period elements cannot.

Drafted from Grade 11 Chemistry, pages 67-144, then checked twice before it went up

Resonance Hybrid MisconceptionCommon mistake

When you use it

Use this card when interpreting resonance structures and Lewis representations of molecules like ozone.

Watch out

Resonance does not mean a molecule flips back and forth between structures. The real molecule is a single, permanent hybrid structure with intermediate bonds.

Drafted from Grade 11 Chemistry, pages 67-144, then checked twice before it went up

Sigma and pi bonds from orbital overlapRule

When you use it

Use when a question asks how a covalent bond actually forms, or why the two bonds of a C=C are not alike. A bond appears where two half filled valence orbitals overlap and their electrons pair with opposite spins: head on overlap along the line joining the nuclei gives a sigma bond, while sideways overlap of two parallel p orbitals gives a pi bond whose density sits above and below that line.

Watch out

Students count a double bond as two of the same thing and draw the second bond of C=C as another head on overlap squeezed between the nuclei. Only the first bond between two atoms is a sigma bond, and every extra one is a pi bond, so C=C is one sigma plus one pi and a pi bond never stands alone. Its pair sits above and below the axis rather than between the nuclei, which is why it is the weaker bond and why the two ends of a double bond cannot rotate. The + and - marked on the p lobes are wave signs, not electric charges.

Counting sigma and pi bonds in a moleculeRule

n \text{ bonds between two atoms} \;=\; 1\,\sigma + (n-1)\,\pi

n \text{ bonds between two atoms} \;=\; 1\,\sigma + (n-1)\,\pi

$n$: the bond order: the number of electron pairs the two atoms share, 1 for a single bond, 2 for a double, 3 for a triple
$\sigma$: a sigma bond, from head on overlap along the line joining the nuclei
$\pi$: a pi bond, from sideways overlap of two parallel p orbitals

When you use it

Use when a question asks how many sigma and how many pi bonds a molecule holds. Count every line in the structure as one sigma bond, then add one pi for each extra line inside a double or a triple bond.

Watch out

Students count only the bonds inside the multiple bond and forget that every other line in the structure is a sigma bond too. Asked about ethene, C2H4, they answer 1 sigma and 1 pi. The real count is four C-H sigma bonds plus the C=C, which is itself one sigma and one pi, so 5 sigma and 1 pi. Ethyne, C2H2, is 3 sigma and 2 pi. The other habit to break is calling a double bond 2 sigma or a triple bond 3 sigma: there is never more than one sigma between the same two atoms.

Hybridization and the geometry it fixesReference tablesp · sp^2 · sp^3 · +2

Hybrid set	Orbitals mixed	Electron geometry	Bond angle	Examples
$sp$	$1s + 1p$	linear	180°	$\text{CO}_2$ , $\text{C}_2\text{H}_2$
$sp^{2}$	$1s + 2p$	trigonal planar	120°	$\text{BF}_3$ , $\text{C}_2\text{H}_4$
$sp^{3}$	$1s + 3p$	tetrahedral	109.5°	$\text{CH}_4$ , $\text{NH}_3$ , $\text{H}_2\text{O}$
$sp^{3}d$	$1s + 3p + 1d$	trigonal bipyramidal	120°, 90° and 180°	$\text{PCl}_5$
$sp^{3}d^{2}$	$1s + 3p + 2d$	octahedral	90° and 180°	$\text{SF}_6$

When you use it

Use when you have to name the shape around a central atom, or read its hybridization off a Lewis structure. Count the electron domains on that atom with the lone pairs included, and the count picks the row: the number of hybrid orbitals always equals the number of atomic orbitals mixed.

Watch out

Students read the geometry column as the shape of the molecule and write NH3 as tetrahedral at 109.5 degrees. The two answers come from different counts. Hybridization counts every electron domain, lone pairs included, so NH3 and H2O are both sp3 with tetrahedral electron geometry. The shape is named from the atoms alone, so NH3 is trigonal pyramidal at about 107 degrees and H2O is bent at about 104.5 degrees. The mirror image of the mistake is dropping the lone pairs and calling H2O sp2 because it has two bonds.

Bond energy against bond lengthReference tableH-H · Cl-Cl · H-Cl · +4

Bond	Bond order	Bond energy (kJ/mol)	Bond length (pm)
H-H	1	436	74
Cl-Cl	1	242	199
H-Cl	1	431	127
C-H	1	413	109
C-C	1	348	154
C=C	2	612	134
C≡C	3	837	120

When you use it

Use when a question compares two bonds for strength or for length, or asks how much energy breaking one mole of a bond costs. The first three rows are exact values for those diatomic molecules; the carbon rows are averages taken over many compounds, so another book may print a few units different.

Watch out

Students memorise that a shorter bond is a stronger bond and then use it across different pairs of atoms. Asked which is stronger, C-H at 109 pm or C=C at 134 pm, most answer C-H because it is shorter, and it is the weaker of the two at 413 against 612. The rule is only safe between the same two atoms, where C-C, C=C and C≡C get shorter and stronger together. H-H at 74 pm and C-C at 154 pm are different atoms, so the length tells you nothing about which is harder to break.

Molecular orbital bond orderFormula

\text{Bond order} = \tfrac{1}{2}\,(N_{b} - N_{ab})

\text{Bond order} = \tfrac{1}{2}\,(N_{b} - N_{ab})

$N_{b}$: the number of electrons sitting in bonding molecular orbitals
$N_{ab}$: the number of electrons in antibonding molecular orbitals, the ones written with a star

When you use it

Use when a question asks whether a diatomic molecule or ion can exist, how strong its bond is, or whether it is paramagnetic. Fill the molecular orbitals in energy order first, count the bonding and the antibonding electrons, then halve the difference: a bond order of 0 means the molecule does not form.

Watch out

Students count only the electrons they can see in the printed configuration and then subtract with the wrong totals. Decide once whether the 1s core is in or out and stay with it: N2 is half of 10 minus 4 with the core and half of 8 minus 2 without it, giving 3 either way, while half of 10 minus 2 mixes the two and gives 4. The second habit is pairing both pi 2p electrons into one orbital. B2 puts one in pi 2px and one in pi 2py with parallel spins, which is the whole reason B2 is paramagnetic. And do not carry the pi 2p below sigma 2p diagram past N2: O2 and F2 use the flipped order.

Physical states of matter

Combined Gas LawFormula

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

$P_1, P_2$: Initial and final Pressure $\text{Pa}$
$V_1, V_2$: Initial and final Volume $\text{m}^3$
$T_1, T_2$: Initial and final Temperature $\text{K}$

When you use it

Use when a fixed mass of gas undergoes simultaneous changes in pressure, volume, and temperature.

Watch out

Temperature must always be converted to Kelvin by adding 273 before substituting into the equation.

Drafted from Grade 11 Chemistry, pages 145-192, then checked twice before it went up

Graham's Law of DiffusionFormula

\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}

\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}

$r_1, r_2$: Rate of diffusion of gas 1 and gas 2 $\text{mol}\cdot\text{s}^{-1}$
$M_1, M_2$: Molar mass of gas 1 and gas 2 $\text{kg}\cdot\text{mol}^{-1}$

When you use it

Use to compare the rates of diffusion or effusion of two different gases at constant temperature and pressure.

Watch out

The rate ratio is inversely proportional to the square root of molar masses, so gas 1 rate is over gas 2, but molar mass 2 is over molar mass 1.

Drafted from Grade 11 Chemistry, pages 145-192, then checked twice before it went up

Ideal Gas EquationFormula

PV = nRT

PV = nRT

$P$: Pressure $\text{Pa}$
$V$: Volume $\text{m}^3$
$n$: Amount of substance $\text{mol}$
$R$: Universal Gas constant $\text{J}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}$
$T$: Temperature $\text{K}$

When you use it

Use to relate pressure, volume, temperature, and moles of an ideal gas under a single state condition.

Watch out

Ensure units match the constant R. When using R = 0.082 atm L per mol K, pressure must be in atm and volume in L.

Drafted from Grade 11 Chemistry, pages 145-192, then checked twice before it went up

Temperature Scale in Gas CalculationsCommon mistake

When you use it

Use when setting up temperature values for any gas law problem.

Watch out

Never insert temperatures in degrees Celsius into gas law equations. Always convert to Kelvin using T = degrees Celsius + 273.

Drafted from Grade 11 Chemistry, pages 145-192, then checked twice before it went up

Types of Crystalline SolidsReference tableIonic · Molecular · Covalent network · +1

Solid Type	Particles	Attractive Forces	Key Properties	Examples
Ionic	Cations and anions	Ionic bonds	High melting point, brittle, conducts in molten state	$\text{NaCl}, \text{CsCl}, \text{KCl}$
Molecular	Molecules or atoms	Intermolecular forces	Low melting point, soft, non-conductors	$\text{H}_2\text{O}, \text{CO}_2, \text{C}_{12}\text{H}_{22}\text{O}_{11}$
Covalent network	Atoms	Covalent bonds	Extremely high melting point, very hard, non-conductors	$\text{C (diamond)}, \text{SiO}_2, \text{Graphite}$
Metallic	Metal cations in electron sea	Metallic bonds	Variable melting point, malleable, conducts electricity	$\text{Cu}, \text{Ag}, \text{Au}, \text{Fe}, \text{Al}$

When you use it

Use to classify crystalline solids based on particle types, bonding forces, and physical properties.

Watch out

Ionic solids do not conduct electricity in the solid state, they only conduct when molten or dissolved in water.

Drafted from Grade 11 Chemistry, pages 145-192, then checked twice before it went up

Boyle's law and its two graphsFormula

P_1V_1 = P_2V_2

P_1V_1 = P_2V_2

$P_1$: the pressure the gas starts at $\mathrm{Pa},\ \mathrm{kPa},\ \mathrm{atm},\ \mathrm{mmHg}$
$V_1$: the volume it occupies at that pressure $\mathrm{m^{3}},\ \mathrm{L},\ \mathrm{cm^{3}}$
$P_2$: the pressure after the change $\mathrm{Pa},\ \mathrm{kPa},\ \mathrm{atm},\ \mathrm{mmHg}$
$V_2$: the volume it occupies afterwards $\mathrm{m^{3}},\ \mathrm{L},\ \mathrm{cm^{3}}$

What is given	What is asked	Working
200 kPa in 3.0 L	the pressure at 1.5 L	$P_2 = \dfrac{200 \times 3.0}{1.5} = 400$ kPa
1.5 atm in 2.0 L	the volume at 0.75 atm	$V_2 = \dfrac{1.5 \times 2.0}{0.75} = 4.0$ L
760 mmHg in 500 $\mathrm{cm^{3}}$	the pressure at 250 $\mathrm{cm^{3}}$	$P_2 = \dfrac{760 \times 500}{250} = 1520$ mmHg
100 kPa in 1.2 L	the volume at 60 kPa	$V_2 = \dfrac{100 \times 1.2}{60} = 2.0$ L

When you use it

Use for a closed container holding a fixed amount of gas whose temperature does not change, when squeezing it or letting it expand changes the pressure: a syringe pushed in, a bubble rising through water. Multiply the pressure by the volume you are given, then divide by whichever of the two the question changes.

Watch out

Students hear "inversely proportional" and draw the graph of $P$ against $V$ as a straight line sloping down. It is a curve, the rectangular hyperbola $PV = k$ , which bends towards both axes and never meets either one. Only $P$ against $1/V$ is a straight line through the origin, and its gradient is $k$ , so if a graph is straight and passes through the origin, read the horizontal axis, because it will say $1/V$ . The second trap is arithmetic. $P_1V_1 = P_2V_2$ balances only when both volumes carry one unit and both pressures carry one unit, so a question that gives 500 $\mathrm{cm^{3}}$ and asks about 1.5 L has to be converted first: 1 atm = 760 mmHg = 101.325 kPa, and 1 L = 1000 $\mathrm{cm^{3}}$ . Nothing has to be in SI, but the two sides must match.

Kinetic molecular theory of an ideal gasRule

E_{k} = \tfrac{3}{2}kT

E_{k} = \tfrac{3}{2}kT

$E_{k}$: the average translational kinetic energy of one gas particle $\mathrm{J}$
$k$: the Boltzmann constant $\mathrm{J\,K^{-1}}$
$T$: the absolute temperature, always in kelvin $\mathrm{K}$

When you use it

Use when a question asks why a gas fills any container it is put in, why it compresses so easily, or what heating does to its particles. The postulates describe an ideal gas: countless tiny particles in constant random motion, taking up no room of their own, pulling on nothing, colliding elastically, and carrying an average kinetic energy set by temperature alone. Real gases follow them closely at ordinary room pressure and temperature.

Watch out

Students read that kinetic energy is proportional to temperature and then put degrees Celsius into it. Heating a gas from 20 degrees Celsius to 40 degrees Celsius does not double the average kinetic energy: in kelvin that is 293 K to 313 K, a rise of about 7 percent, and T in this relation is always in kelvin. The second mistake is hearing that all gases at one temperature have equal average kinetic energy and answering that their molecules therefore move at the same speed. Equal energy with unequal mass means unequal speed, so at 25 degrees Celsius a hydrogen molecule averages roughly four times the speed of an oxygen molecule.

Chemical Kinetics

Effective Collision ConditionsRule

When you use it

Use to explain why collisions between reacting species succeed or fail to form products.

Watch out

A reaction requires both conditions simultaneously: reactant molecules must collide with proper orientation and with kinetic energy equal to or greater than the activation energy.

Drafted from Grade 11 Chemistry, pages 193-221, then checked twice before it went up

General Rate ExpressionFormula

\text{Rate} = -\frac{1}{a}\frac{\Delta [\text{A}]}{\Delta t} = -\frac{1}{b}\frac{\Delta [\text{B}]}{\Delta t} = \frac{1}{c}\frac{\Delta [\text{C}]}{\Delta t} = \frac{1}{d}\frac{\Delta [\text{D}]}{\Delta t}

\text{Rate} = -\frac{1}{a}\frac{\Delta [\text{A}]}{\Delta t} = -\frac{1}{b}\frac{\Delta [\text{B}]}{\Delta t} = \frac{1}{c}\frac{\Delta [\text{C}]}{\Delta t} = \frac{1}{d}\frac{\Delta [\text{D}]}{\Delta t}

$\text{Rate}$: Rate of reaction $\text{mol}\,\text{L}^{-1}\text{s}^{-1}$
$\Delta [\text{A}]$: Change in concentration of reactant A $\text{mol}\,\text{L}^{-1}$
$\Delta t$: Time interval $\text{s}$
$a$: Stoichiometric coefficient of reactant A $1$

When you use it

Use this to relate the rate of reaction to the consumption of reactants or the formation of products for any balanced chemical equation.

Watch out

Always divide the rate of concentration change of each species by its stoichiometric coefficient in the balanced equation.

Drafted from Grade 11 Chemistry, pages 193-221, then checked twice before it went up

Missing Negative Sign for ReactantsCommon mistake

When you use it

Use when writing rate expressions in terms of disappearing reactants.

Watch out

Reactant concentration decreases over time, making change in concentration negative. You must include a minus sign in front of reactant terms because rate of reaction is always positive.

Drafted from Grade 11 Chemistry, pages 193-221, then checked twice before it went up

Chemical Equilibrium

Le Chatelier Principle and Pressure ChangesRule

When you use it

Use to predict how a change in pressure or volume shifts the position of a gaseous equilibrium system.

Watch out

Increasing pressure shifts the reaction toward the side with fewer gas moles. Adding an inert gas at constant volume produces no shift.

Drafted from Grade 11 Chemistry, pages 222-263, then checked twice before it went up

Pure Solids and Liquids in Equilibrium ExpressionsCommon mistake

When you use it

Use when formulating equilibrium constant expressions for heterogeneous systems involving multiple physical states.

Watch out

Never include pure solids (s) or pure liquids (l) in KC or KP expressions because their concentrations are constant.

Drafted from Grade 11 Chemistry, pages 222-263, then checked twice before it went up

Relationship Between KP and KCFormula

K_P = K_C(RT)^{\Delta n}

K_P = K_C(RT)^{\Delta n}

$K_P$: equilibrium constant in terms of partial pressure $\text{atm}^{\Delta n}$
$K_C$: equilibrium constant in terms of molar concentration $(\text{mol}\,\text{L}^{-1})^{\Delta n}$
$R$: ideal gas constant $\text{L}\,\text{atm}\,\text{K}^{-1}\,\text{mol}^{-1}$
$T$: absolute temperature $\text{K}$
$\Delta n$: moles of gaseous products minus moles of gaseous reactants $1$

When you use it

Use when converting between the concentration equilibrium constant and the pressure equilibrium constant for gas phase reactions.

Watch out

Calculate Delta n using gaseous species only. When the number of moles of gaseous products equals reactants, Delta n is zero and KP equals KC.

Drafted from Grade 11 Chemistry, pages 222-263, then checked twice before it went up

Some Important Oxygen-Containing Organic Compounds

Temperature in Ethanol Acid DehydrationCommon mistake

When you use it

Use when determining whether heating ethanol with concentrated sulfuric acid produces an alkene or an ether.

Watch out

Do not confuse the reaction temperatures: elimination to ethene requires 170 °C, while condensation to diethyl ether occurs at 140 °C.

Drafted from Grade 11 Chemistry, page 264 onwards, then checked twice before it went up

Williamson Ether SynthesisFormula

R\text{-}X + R'\text{-}O^- \rightarrow R\text{-}O\text{-}R' + X^-

R\text{-}X + R'\text{-}O^- \rightarrow R\text{-}O\text{-}R' + X^-

$R\text{-}X$: Alkyl halide $none$
$R'\text{-}O^-$: Alkoxide ion $none$
$R\text{-}O\text{-}R'$: Ether $none$
$X^-$: Halide ion $none$

When you use it

Use when preparing symmetrical or unsymmetrical ethers from an alkyl halide and an alkoxide ion.

Watch out

This nucleophilic substitution attaches the alkoxide oxygen directly to the alkyl carbon that held the halogen.

Drafted from Grade 11 Chemistry, page 264 onwards, then checked twice before it went up

Telling the oxygen and nitrogen families apartReference tableAlcohol · Ether · Aldehyde · +5

Family	The group	How you recognise it
Alcohol	$-\mathrm{OH}$	The oxygen carries a hydrogen and is bonded to one carbon. Ethanol, $\mathrm{CH_3CH_2OH}$ .
Ether	$\mathrm{C}-\mathrm{O}-\mathrm{C}$	The oxygen sits between two carbons and carries no hydrogen. Methoxymethane, $\mathrm{CH_3OCH_3}$ .
Aldehyde	$-\mathrm{CHO}$	The carbonyl carbon is at the end of the chain and still holds a hydrogen. Ethanal, $\mathrm{CH_3CHO}$ .
Ketone	$\mathrm{R}-\mathrm{CO}-\mathrm{R}'$	The carbonyl carbon sits inside the chain, with a carbon on each side. Propanone, $\mathrm{CH_3COCH_3}$ .
Carboxylic acid	$-\mathrm{COOH}$	A carbonyl and a hydroxyl on the same carbon. Ethanoic acid, $\mathrm{CH_3COOH}$ .
Ester	$\mathrm{R}-\mathrm{COO}-\mathrm{R}'$	The carbonyl carbon is joined to an oxygen that carries a second carbon. Methyl ethanoate, $\mathrm{CH_3COOCH_3}$ .
Primary amine	$-\mathrm{NH_2}$	A nitrogen with three single bonds and no carbonyl beside it. Ethanamine, $\mathrm{CH_3CH_2NH_2}$ .
Amide	$-\mathrm{CONH_2}$	The nitrogen is bonded straight to a carbonyl carbon. Ethanamide, $\mathrm{CH_3CONH_2}$ .

When you use it

Use when a structure or a condensed formula is in front of you and you have to name its family before you can name the compound or say how it reacts. Read what the oxygen or the nitrogen is attached to, then find that description in this table.

Watch out

Students name the family from the presence of an oxygen instead of from where the oxygen sits, so all of these end up called alcohols. If the oxygen carries a hydrogen it is an alcohol, $\mathrm{CH_3CH_2OH}$ ; if it sits between two carbons it is an ether, $\mathrm{CH_3OCH_3}$ . Both are $\mathrm{C_2H_6O}$ : the same atoms in two different families. The same test runs down the rest of the table. A carbonyl at the end of the chain with a hydrogen on it is an aldehyde, the same carbonyl with a carbon on both sides is a ketone, and in $-\mathrm{COOH}$ the second oxygen holds a hydrogen while in $-\mathrm{COO}-$ it holds a carbon.

General formulas of the functional groupsReference tableAlcohol · Ether · Aldehyde · +4

Family	Functional group	General formula	Example
Alcohol	$-\mathrm{OH}$	$\mathrm{R}-\mathrm{OH}$	ethanol, $\mathrm{CH_3CH_2OH}$
Ether	$-\mathrm{O}-$	$\mathrm{R}-\mathrm{O}-\mathrm{R}'$	methoxymethane, $\mathrm{CH_3OCH_3}$
Aldehyde	$-\mathrm{CHO}$	$\mathrm{R}-\mathrm{CHO}$	ethanal, $\mathrm{CH_3CHO}$
Ketone	$-\mathrm{CO}-$	$\mathrm{R}-\mathrm{CO}-\mathrm{R}'$	propanone, $\mathrm{CH_3COCH_3}$
Carboxylic acid	$-\mathrm{COOH}$	$\mathrm{R}-\mathrm{COOH}$	ethanoic acid, $\mathrm{CH_3COOH}$
Primary amine	$-\mathrm{NH_2}$	$\mathrm{R}-\mathrm{NH_2}$	ethanamine, $\mathrm{CH_3CH_2NH_2}$
Haloalkane	$-\mathrm{X},\ \mathrm{X} = \mathrm{F},\ \mathrm{Cl},\ \mathrm{Br},\ \mathrm{I}$	$\mathrm{R}-\mathrm{X}$	chloroethane, $\mathrm{CH_3CH_2Cl}$

When you use it

Use when a question gives you a general formula such as $\mathrm{R}-\mathrm{CHO}$ or $\mathrm{R}-\mathrm{O}-\mathrm{R}'$ and asks which family it names, or when you have written a group down and want to check it against the family you meant. Each row is one homologous series: its members share this group and this general formula, and neighbouring members differ by one $-\mathrm{CH_2}-$ unit, which is 14 g/mol.

Watch out

Students write the aldehyde as $\mathrm{R}-\mathrm{COH}$ instead of $\mathrm{R}-\mathrm{CHO}$ . Read it letter by letter: $\mathrm{C}-\mathrm{O}-\mathrm{H}$ is a carbon holding a hydroxyl, which is an alcohol, so one swapped letter changes the whole family. The hydrogen comes before the oxygen. The same care separates the aldehyde from the ketone. Both hold a $\mathrm{C}=\mathrm{O}$ , but in an aldehyde that carbon still holds a hydrogen and therefore sits at the end of the chain, $\mathrm{CH_3CHO}$ , while in a ketone it is trapped between two carbons, $\mathrm{CH_3COCH_3}$ .

The same subject in other years

An exam paper keeps asking for what the year below taught. Those cards are here too.

Your other Grade 11 subjects

Everything else in this year that has cards so far.

Questions students ask

What do the Grade 11 Chemistry cards cover?: 34 cards across 6 chapters of the national textbook: Atomic Structure and Periodic Properties of the Elements, Chemical Bonding, Physical states of matter, Chemical Kinetics, Chemical Equilibrium and Some Important Oxygen-Containing Organic Compounds. You can take any chapter one card at a time on the page itself.
Is there a national exam in Grade 11?: No. Ethiopia sets national exams in Grade 6, Grade 8 and Grade 12 only. These cards are for your school's own exams, and for the national exam that comes a few years later.
Where do these cards come from?: They are drafted from Grade 11 Chemistry, the Ministry of Education textbook for this grade. A second pass that cannot see the chapter then re-derives every formula, constant and table row, and anything it cannot confirm is held back instead of published.
Is this free?: Yes. Every card here is free to read and the printable sheet is free to download. Neither needs an account.
When was this last checked?: 6 September 2026. Cards arrive chapter by chapter, and the line under each one says when that card was last read through.

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