Grade 11 Science: Chemistry

In this unit, 'Some basic concepts of chemistry and atomic structure', you learn to use the mole and quantum numbers. You work with exam-style questions and written reasoning. You see it, say it, then do it alone.

• Observe: see, touch, measure or watch a safe demonstration linked to 'Some basic concepts of chemistry and atomic structure' and write what you notice.
• Explain: use the key words of the unit to say what is happening and why, in a few lines.
• Use it: use the mole and quantum numbers. Use a diagram, table or short calculation where it fits.
• Test and conclude: say what evidence supports your answer and what you would change.

Step by step.

1. Step 1, Calculate moles. 5.85 g of NaCl has molar mass 58.5 g/mol. n = 5.85 / 58.5 = 0.1 mol. Particles = 0.1 x 6.022 x 10^23 = 6.022 x 10^22 formula units.
2. Step 2, Find a percentage composition. Water H2O is 18 g/mol, with H = 2/18 = 11.1 per cent and O = 16/18 = 88.9 per cent. Empirical formula from 40 per cent C, 6.7 per cent H, 53.3 per cent O: moles 3.33, 6.7, 3.33, ratio 1:2:1, so CH2O.
3. Step 3, Do stoichiometry. 2H2 + O2 gives 2H2O. 4 g of H2 is 2 mol and needs 1 mol of O2 (32 g), giving 2 mol of water (36 g). Limiting reagent: with 2 g H2 and 32 g O2, H2 limits and 18 g of water forms.
4. Step 4, Calculate molarity. 4 g of NaOH (40 g/mol) is 0.1 mol. In 250 mL, M = 0.1 / 0.250 = 0.4 mol/L.
5. Step 5, Assign quantum numbers for a 3p electron. n = 3, l = 1, m can be -1, 0, +1, and spin is +1/2 or -1/2. Maximum electrons in shell n is 2n^2, so n = 3 holds 18.
6. Step 6, Your turn: find moles in 11 g of CO2 (44 g/mol) and the number of molecules. Answer: 0.25 mol; 0.25 x 6.022e23 = 1.5 x 10^23 molecules.

Activity. About 40 to 50 minutes of class work, then one short practice task for home. Keep evidence: a timed paper and an error log.

Standard to meet. Use the mole and quantum numbers: at least 70 percent on an exam-style task, with the method shown.

In this unit, 'Periodic table and chemical bonding', you learn to predict trends and bonding. You work with exam-style questions and written reasoning. You see it, say it, then do it alone.

• Observe: see, touch, measure or watch a safe demonstration linked to 'Periodic table and chemical bonding' and write what you notice.
• Explain: use the key words of the unit to say what is happening and why, in a few lines.
• Use it: predict trends and bonding. Use a diagram, table or short calculation where it fits.
• Test and conclude: say what evidence supports your answer and what you would change.

Step by step.

1. Step 1, Predict trends across period 3: Na, Mg, Al, Si, P, S, Cl. Atomic radius decreases because nuclear charge grows with the same shell. Ionisation enthalpy generally rises, with dips at Al and S. Electronegativity rises to Cl.
2. Step 2, Compare down group 1: Li, Na, K. Atomic radius increases, ionisation enthalpy falls, so reactivity with water rises. K reacts more violently than Na.
3. Step 3, Draw Lewis structures. For NH3, N has 5 valence electrons, forming 3 bonds with H and keeping 1 lone pair. VSEPR gives a trigonal pyramidal shape with HNH angle about 107 degrees. CH4 is tetrahedral, 109.5 degrees.
4. Step 4, Calculate bond order from molecular orbital theory. O2 has 12 valence electrons: 8 bonding, 4 antibonding, bond order (8 - 4)/2 = 2, and it is paramagnetic with 2 unpaired electrons. N2 has bond order 3.
5. Step 5, Compare electronegativity. Using rounded textbook values, HCl has a difference of 3.0 - 2.1 = 0.9, so it is polar covalent. NaCl has 3.0 - 0.9 = 2.1, so it is ionic. Hybridisation of carbon in ethene is sp2 and in ethyne sp.
6. Step 6, Your turn: give the shape and bond angle of H2O and the bond order of N2. Answer: bent with about 104.5 degrees; bond order 3.

Activity. About 40 to 50 minutes of class work, then one short practice task for home. Keep evidence: a timed paper and an error log.

Standard to meet. Predict trends and bonding: at least 70 percent on an exam-style task, with the method shown.

In this unit, 'States of matter and thermodynamics', you learn to apply gas laws and enthalpy. You work with exam-style questions and written reasoning. You see it, say it, then do it alone.

• Observe: see, touch, measure or watch a safe demonstration linked to 'States of matter and thermodynamics' and write what you notice.
• Explain: use the key words of the unit to say what is happening and why, in a few lines.
• Use it: apply gas laws and enthalpy. Use a diagram, table or short calculation where it fits.
• Test and conclude: say what evidence supports your answer and what you would change.

Step by step.

1. Step 1, Use the ideal gas law PV = nRT. For 2 mol of gas at 300 K in 10 L: P = nRT/V = 2 x 0.0821 x 300 / 10 = 4.93 atm. Use R = 0.0821 L atm per mol per K.
2. Step 2, Use Boyle's law. A gas at 1 atm and 6 L is compressed at constant temperature to 2 L. P2 = P1 V1 / V2 = 1 x 6 / 2 = 3 atm. Charles's law: 300 K and 2 L heated to 450 K gives V = 2 x 450 / 300 = 3 L.
3. Step 3, Apply the first law, delta U = q + w. A gas absorbs 500 J of heat and does 200 J of work on the surroundings, so w = -200 J and delta U = 500 - 200 = +300 J.
4. Step 4, Calculate enthalpy. For CH4 + 2O2 gives CO2 + 2H2O(l), use rounded textbook formation enthalpies: CH4 -74.8, CO2 -393.5, H2O(l) -285.8 kJ/mol. delta H = [-393.5 + 2(-285.8)] - [-74.8] = -890.3 kJ/mol. Exothermic.
5. Step 5, Predict spontaneity with delta G = delta H - T delta S. For delta H = +40 kJ and delta S = +100 J/K at 500 K: delta G = 40000 - 500 x 100 = -10000 J, so it is spontaneous at 500 K.
6. Step 6, Your turn: find the volume of 1 mol of ideal gas at 273 K and 1 atm. Answer: V = 0.0821 x 273 / 1 = 22.4 L.

Activity. About 40 to 50 minutes of class work, then one short practice task for home. Keep evidence: a timed paper and an error log.

Standard to meet. Apply gas laws and enthalpy: at least 70 percent on an exam-style task, with the method shown.

In this unit, 'Equilibrium and redox reactions', you learn to apply Le Chatelier's principle and balance redox equations. You work with exam-style questions and written reasoning. You see it, say it, then do it alone.

• Observe: see, touch, measure or watch a safe demonstration linked to 'Equilibrium and redox reactions' and write what you notice.
• Explain: use the key words of the unit to say what is happening and why, in a few lines.
• Use it: apply Le Chatelier's principle and balance redox equations. Use a diagram, table or short calculation where it fits.
• Test and conclude: say what evidence supports your answer and what you would change.

Step by step.

1. Step 1, Meet a reversible reaction: N2 + 3H2 <=> 2NH3, with energy released (delta H = -92 kJ per mole of reaction). At equilibrium the forward and backward rates are equal, so concentrations stay steady while reactions continue. Kc = [NH3]^2 / ([N2][H2]^3). A large Kc means products are favoured.
2. Step 2, Apply Le Chatelier's principle: a system opposes a change. Raise the pressure: there are 4 moles of gas on the left and 2 on the right, so the equilibrium shifts right to make fewer moles. Raise the temperature: the forward reaction gives out heat, so it shifts left. Remove NH3 and it shifts right. A catalyst only speeds both directions.
3. Step 3, Calculate Kc for H2 + I2 <=> 2HI. Suppose that at equilibrium [H2] = 0.10, [I2] = 0.10 and [HI] = 0.40 mol per dm3, so Kc = 0.40^2 / (0.10 x 0.10) = 0.16 / 0.01 = 16. If a mixture has [HI] = 0.20, then Q = 0.04 / 0.01 = 4, which is less than 16, so it shifts right.
4. Step 4, Track oxidation numbers. In Zn + Cu2+ -> Zn2+ + Cu, zinc goes from 0 to +2 (loses electrons, oxidised) and copper goes from +2 to 0 (gains electrons, reduced). Remember OIL RIG: oxidation is loss, reduction is gain. Cu2+ is the oxidising agent, and Zn is the reducing agent.
5. Step 5, Balance in acid by half-equations. MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O. Fe2+ -> Fe3+ + e-, times 5. Add: MnO4- + 8H+ + 5Fe2+ -> Mn2+ + 4H2O + 5Fe3+. Charge check: -1 + 8 + 10 = 17 on the left and 2 + 15 = 17 on the right. Safety: do this titration only with a teacher, wearing goggles and gloves.
6. Step 6, Your turn: balance Cr2O7 2- + Fe2+ -> Cr3+ + Fe3+ in acid, and say how removing NH3 changes the ammonia equilibrium. Answer: Cr2O7 2- + 14H+ + 6Fe2+ -> 2Cr3+ + 7H2O + 6Fe3+ (charge 24 both sides); removing NH3 shifts the equilibrium right.

Activity. About 40 to 50 minutes of class work, then one short practice task for home. Keep evidence: a timed paper and an error log.

Standard to meet. Apply Le Chatelier's principle and balance redox equations: at least 70 percent on an exam-style task, with the method shown.

In this unit, 'Hydrogen, s-block and p-block elements', you learn to compare properties and uses. You work with exam-style questions and written reasoning. You see it, say it, then do it alone.

• Observe: see, touch, measure or watch a safe demonstration linked to 'Hydrogen, s-block and p-block elements' and write what you notice.
• Explain: use the key words of the unit to say what is happening and why, in a few lines.
• Use it: compare properties and uses. Use a diagram, table or short calculation where it fits.
• Test and conclude: say what evidence supports your answer and what you would change.

Step by step.

1. Step 1, Compare hydrogen isotopes: protium, deuterium and tritium with 0, 1 and 2 neutrons. Preparation in the lab: Zn + 2HCl gives ZnCl2 + H2. Industrially, CH4 + H2O at high temperature with nickel catalyst gives CO + 3H2.
2. Step 2, Compare s-block metals. Sodium reacts: 2Na + 2H2O gives 2NaOH + H2. Flame tests: sodium golden yellow, potassium lilac, calcium brick red, barium apple green. Reactivity rises down the group.
3. Step 3, Compare uses: sodium carbonate in glass and washing, sodium hydroxide in soap, calcium oxide (quicklime) in cement, calcium hydroxide in whitewash, and magnesium in alloys.
4. Step 4, Describe p-block. Boron forms borax; aluminium oxide Al2O3 is amphoteric; carbon exists as diamond (sp3) and graphite (sp2). Silica SiO2 is a giant covalent solid. Group 17 halogens oxidising power falls Cl > Br > I.
5. Step 5, Predict a reaction: Cl2 + 2KBr gives 2KCl + Br2, because chlorine is a stronger oxidiser than bromine. Ammonia is made by the Haber process N2 + 3H2 gives 2NH3 at about 450 C with iron catalyst.
6. Step 6, Your turn: write the reaction of calcium with water and the flame colour of calcium. Answer: Ca + 2H2O gives Ca(OH)2 + H2; brick red.

Activity. About 40 to 50 minutes of class work, then one short practice task for home. Keep evidence: a timed paper and an error log.

Standard to meet. Compare properties and uses: at least 70 percent on an exam-style task, with the method shown.

In this unit, 'Organic chemistry: basic principles and hydrocarbons', you learn to name compounds and describe reactions. You work with exam-style questions and written reasoning. You see it, say it, then do it alone.

• Observe: see, touch, measure or watch a safe demonstration linked to 'Organic chemistry: basic principles and hydrocarbons' and write what you notice.
• Explain: use the key words of the unit to say what is happening and why, in a few lines.
• Use it: name compounds and describe reactions. Use a diagram, table or short calculation where it fits.
• Test and conclude: say what evidence supports your answer and what you would change.

Step by step.

1. Step 1, Name by IUPAC. Choose the longest chain with the principal group. CH3CH(CH3)CH2CH3: the longest chain has 4 carbons, with a methyl on carbon 2, so 2-methylbutane. CH3CH=CHCH3 is but-2-ene.
2. Step 2, Draw isomers of C4H10: butane (straight chain) and 2-methylpropane (branched). They are structural isomers with the same molecular formula and different structures. C5H12 has three isomers.
3. Step 3, Write substitution on alkanes. CH4 + Cl2 under UV gives CH3Cl + HCl by free radical chain steps: initiation, propagation, termination. Chlorine radicals form from Cl2 by UV.
4. Step 4, Write addition on alkenes. CH2=CH2 + Br2 gives CH2Br-CH2Br, decolourising bromine water (a test for unsaturation). Markovnikov: propene + HBr gives mainly 2-bromopropane because the more stable secondary carbocation forms.
5. Step 5, Write combustion and calculate: C2H4 + 3O2 gives 2CO2 + 2H2O. Ethene of 28 g (1 mol) needs 3 mol O2 (96 g) and gives 88 g CO2. Benzene undergoes electrophilic substitution, not addition.
6. Step 6, Your turn: name the alkane CH3CH2CH(CH3)CH3 by the longest chain rule and write the main product of propene with HBr. Answer: 2-methylbutane; 2-bromopropane by Markovnikov addition.

Activity. About 40 to 50 minutes of class work, then one short practice task for home. Keep evidence: a timed paper and an error log.

Standard to meet. Name compounds and describe reactions: at least 70 percent on an exam-style task, with the method shown.

In this unit, 'Environmental chemistry', you learn to explain pollution and green chemistry. You work with exam-style questions and written reasoning. You see it, say it, then do it alone.

• Observe: see, touch, measure or watch a safe demonstration linked to 'Environmental chemistry' and write what you notice.
• Explain: use the key words of the unit to say what is happening and why, in a few lines.
• Use it: explain pollution and green chemistry. Use a diagram, table or short calculation where it fits.
• Test and conclude: say what evidence supports your answer and what you would change.

Step by step.

1. Step 1, Identify pollutants. Air: CO from incomplete combustion, SO2 and NOx (acid rain), particulates. Water: nutrients causing eutrophication, heavy metals. Soil: pesticides. Name each with a source.
2. Step 2, Explain acid rain. SO2 + 1/2 O2 gives SO3 and SO3 + H2O gives H2SO4. Rain is acidic when its pH is below 5.6. Normal rain is about 5.6 because dissolved CO2 forms carbonic acid.
3. Step 3, Explain the greenhouse effect. CO2, CH4 and N2O absorb infrared radiation re-emitted by Earth. A rise from roughly 280 ppm before industry to well above 400 ppm raises mean temperature; check the current figure.
4. Step 4, Explain ozone depletion. In the stratosphere, CF2Cl2 gives Cl atoms under UV. Cl + O3 gives ClO + O2 and ClO + O gives Cl + O2. A single Cl can destroy many O3 molecules in a catalytic cycle.
5. Step 5, Use water quality numbers. Clean water has a low BOD and heavily polluted water a much higher one (textbook tables often use about 5 ppm and 17 ppm as cut-offs, so check the figures in your syllabus), because microbes use dissolved oxygen to break down waste. Green chemistry principles: prevent waste, use safer solvents, high atom economy, and catalysis.
6. Step 6, Your turn: write the reaction that forms sulphuric acid from SO3 and say why rain at pH 4.5 is acidic. Answer: SO3 + H2O gives H2SO4; pH below 5.6 means extra strong acids dissolved.

Activity. About 40 to 50 minutes of class work, then one short practice task for home. Keep evidence: a timed paper and an error log.

Standard to meet. Explain pollution and green chemistry: at least 70 percent on an exam-style task, with the method shown.

On the slide: L1 Basic · Solids, liquids and gases

Notes: Name the three states of matter and say what happens when a substance is heated. Notes: warm ice in a beaker of warm tap water and record the temperature each minute. Use no flame, and handle warm water with care. Learners say which change is melting.

On the slide: L2 Awareness · Atoms, elements and compounds

Notes: Describe atoms, elements and compounds and read simple chemical formulae. Notes: for water, H2O, learners count 2 hydrogen and 1 oxygen, which is 3 atoms in one molecule. They sort symbols into elements and compounds on cards. Wear goggles whenever any chemical is out, even for show-and-tell.

On the slide: L3 Foundation · Balanced equations and the mole

Notes: Calculate relative formula mass and balance simple equations. Notes: balance 2H2 + O2 gives 2H2O by counting atoms on both sides. Water has Mr 18, so 36 g is 2 mol. Learners repeat this for carbon dioxide, Mr 44, and check that 88 g is 2 mol.

On the slide: L4 Practitioner · Acid-base titration

Notes: Apply stoichiometry to find an unknown concentration from a titration. Notes: 25.0 cm3 of sodium hydroxide needs a mean 20.0 cm3 of 0.100 mol/dm3 hydrochloric acid. Moles of acid are 0.00200, equal to moles of alkali, so the alkali is 0.0800 mol/dm3. Wear goggles; use dilute solutions and rinse spills at once.

On the slide: L5 Specialist · Energy changes and rates

Notes: Analyse a calorimetry result and link energy change to bond breaking and making. Notes: 50 g of water warms by 6.5 K, so q equals 50 times 4.18 times 6.5, which is 1358.5 J. For 0.025 mol that gives about 54.3 kJ/mol released. Learners list heat loss as the main error and suggest a lid.

This course: Grade 11 Science Chemistry is at this level. Its units include Some basic concepts of chemistry and atomic structure; Periodic table and chemical bonding; States of matter and thermodynamics; Equilibrium and redox reactions and more.

On the slide: L6 Advanced · Thermodynamics and kinetics

Notes: Apply Gibbs energy and rate laws to predict whether and how fast a reaction goes. Notes: with change in H of +40 kJ/mol and change in S of +110 J/K/mol at 298 K, G is +7.22 kJ/mol, not spontaneous. It becomes spontaneous above 40000/110, about 364 K. Learners explain why a positive G says nothing about speed.

On the slide: L7 Super expert · Method validation in analysis

Notes: Evaluate an analytical method for accuracy, precision and detection limit, and design a validation plan. Notes: with blank noise of 0.002 absorbance and a calibration slope of 0.250 per mg/L, the limit of detection is 3.3 times 0.002 over 0.250, which is 0.0264 mg/L. Write a hazard assessment and waste plan before any lab work.