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TO Master to PERFECTION before A’levels (Part 1)

Standard Definitions (Don’t Memorize. But appreciate and understand why key terms are important)

– Relative atomic, isotopic, molecular and formula mass, based on the 12C scale (just give mathematical expression)

– Mole in terms of the Avogadro constant
– VSEPR (2 assumptions)
– Basic assumptions of the kinetic theory as applied to an ideal gas
– Standard enthalpies (11 of them)
– Hess’ Law
– Entropy
– Standard electrode potential and standard cell potential
– Dynamic Equilibrium, LCP
– Strong and weak acids and bases
– Kc, KP, Ka, Kb, Kw, KSP,pH etc. (m. expression)
– Rate of reaction; rate equation; order of reaction; rate constant; (m. expression)
– Half life of a reaction
– Rate-determining step
– Activation energy
– Catalysts
– Transition metal, ligands, complex, coordination number
– Proteins 1o,2o,3o structure, Denaturation

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FAQ – Group II

For Gp II metals (e.g. magnesium, calcium, barium)

1. Reactivity (with water or other substances) increases down the group

Explanation:
• Increase in shielding effect outweighs increase in nuclear charge down the group due to increase
in number of electron shells.
• Hence ionization energy decreases
• Metals are able to form ions more easily.

For Gp II ionic compounds (e.g. carbonates, nitrates and hydroxides),

2. Melting point decreases down the group

Explanation:
• Increase in cation size
• leads to decrease in magnitude of lattice energy.
• Hence, ionic bonds are weaker and more easily broken.

3. Thermal decomposition temperature increases down the group

Explanation:
• Increase in cation size,
• leads to decrease in charge density of cation.
• Hence, electron cloud of anion is less distorted and hence less easily decomposed.

Note: Quote ionic radius values from Data Booklet to explain this. Calculate charge density if necessary.

4. Solubility of sulfates decreases down the group (not in syllabus but
may still be tested)

Explanation:
• Increase in cation size
• leads to significant decrease in magnitude of hydration energy (as compared to slight decrease in magnitude of lattice energy)
• Hence, enthalpy of solution (= LE – ΔHhyd) is less exothermic.

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FAQ – Group VII

For Group VII halogens (e.g. Cl2, Br2, I2)

1. Volatility decreases down the group (i.e. boiling point increases)

Explanation:
• Increase in electron cloud size hence more easily polarised
• leads to stronger dispersion forces between molecules

2. Oxidising power decreases down the group

X2 + 2e ⇔ 2X−

Explanation:
• Decrease in effective nuclear charge, hence electron affinity decreases
• Hence less easily accepts electrons (i.e. less easily reduced)
• Eθ value decreases (Quote from Data Booklet)

Two important proofs of this are:
• Any halide (e.g. I−) can be displaced by the halogen (e.g. Br2) above it.
• Reaction with sodium thiosulfate (explain change in oxidation state of S)

3. Reactivity with H2 decreases down the group
X2 + H2 → 2HX

Explanation:
• Atomic size of X increases
• Hence H – X bond becomes longer and weaker
• Product formed is less and less stable, hence reactivity decreases.

Note: Quote bond energy values to explain this, NOT Eθ values

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FAQ – Gases

Gases

Difference between assumptions and conditions

• 2 assumptions of ideal gas
o Negligible intermolecular forces
o Negligible particle volume compared to volume of container

• 2 conditions at which a gas acts most ideally
o High temperature
o Low pressure

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Colour Summary

Flame Test Colors
Li Deep red
Na Yellow
K Violet
Mg Bright white
Ca Orange-red
Sr Red
Ba Green
Cu Blue-green
P Pale blue-green
S Blue
Fe Gold
Pb Blue-white
Zn Blue-green

Aqueous Ion Colors
Cu+ Green
Cu2+ Blue
[CuCl4]2- Yellow
Cu(NH3)4 2+ Dark Blue; produced when ammonia is added to Cu2+ solutions

Fe2+ yellow-green (depending on the anion)
Fe3+ orange-red (depending on the anion)
FeSCN]2+ Red-brown, Wine-red to dark orange

Co2+ Pink
CoCl42- Blue (Co2+ with HCl will form a CoCl4 2- complex that is blue)

Cr3+ Violet (Cr(NO3)3 to Green (CrCl3)
CrO4 2- Yellow
Cr2O7 2- Orange

Ni2+ Green

Mn2+ Pink
MnO4 – Purple (Mn w/ +7 oxidation state is purple)
MnO4 2- Green

Pb3+ blue-green (Pb2+ and Pb4+ are colorless)

V2+ violet
V3+ blue-green

Ti(H2O)6 3+ Purple

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TO Master to PERFECTION before A’levels – Part 1

Standard Definitions (Don’t Memorize. But appreciate and understand why key terms are important)

– Relative atomic, isotopic, molecular and formula mass, based on the 12C scale (just give mathematical expression)
– Mole in terms of the Avogadro constant
– VSEPR (2 assumptions)
– Basic assumptions of the kinetic theory as applied to an ideal gas
– Standard enthalpies (11 of them)
– Hess’ Law
– Entropy
– Standard electrode potential and standard cell potential
– Dynamic Equilibrium, LCP
– Strong and weak acids and bases
– Kc, KP, Ka, Kb, Kw, KSP,pH etc. (m. expression)
– Rate of reaction; rate equation; order of reaction; rate constant; (m. expression)
– Half life of a reaction
– Rate-determining step
– Activation energy
– Catalysts
– Transition metal, ligands, complex, coordination number
– Proteins 1o,2o,3o structure, Denaturation

Standard Explanations (must be concise (save time), accurate and complete) – You must know this so well you have are absolutely confident of reproducing them under stressful exam conditions.

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TO Master to PERFECTION before A’levels – Part 2

1. Atomic Structure • Ionisation Energy (Trend across the period + 2 anomalies, Down the gp, successive IE,*TM)

Remark
• Predicting position from successive IE
Refer to AMS, Gases and Atomic Structure Revision Notes

2. Bonding
• Boiling point/melting point
• Volatility
• Electrical conductivity
• Solubility

3. Energetics (entropy)
• Discuss the effects on the entropy of a chemical system by the
following:
(i) change in temperature
(ii) change in phase
(iii) change in the number of particles (especially for gaseous
systems)
(iv) mixing of particles

Remark
• Predict the effect of temperature change on the spontaneity of a
reaction, given standard enthalpy and entropy changes disorderliness”/”ways to arrange particles” are key words.
Refer to entropy lect notes.

Using Gibbs equation ΔG=ΔHTΔS.
Make sure you are comfortable with putting your thoughts into words.

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TO Master to PERFECTION before A’levels – Part 3

4. Electrochemistry
• Predict qualitatively how the value of an electrode potential varies with the concentration of the aqueous ion
*Could be a disguised change e.g.adding NaOH (aq) to ppt. Mn+.

Refer to our consolidated list last term

5. Chem Eqm
• Apply LCP to deduce qualitatively (from appropriate information)
the effects of changes in concentration, pressure or temperature, on a system at equilibrium.
• Deduce whether changes in concentration, pressure or temperature or the presence of a catalyst affect the value of the equilibrium constant for a reaction

E.g. Given Kc ↑ when temp ↓, predict if reaction is exo/endo.
6. Ionic Eqm
• Explain the choice of suitable indicators for acid-base titrations,given appropriate data
• Explain how buffer solutions (i)control pH (ii) describe and explain
their uses, including the role of H2CO3/HCO3– in controlling pH in
blood

3 main points.
With equations with SINGLEHEADED arrows

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Glossary of Terms

1. State
A concise answer with little or no supporting argument, e.g. a numerical answer that can be obtained ‘by inspection’ is required.

2. List
A number of points, generally each of one word, with no elaboration is required. Where a given number of points is specified, this should not be exceeded

3. Explain
Reasoning or some reference to theory is required (depending on the context)

4. Describe
State in words (using diagrams where appropriate) the main points of the topic. It is often used with reference either to particular phenomena (where answers should include reference to observations associated) or to particular experiments.

5. Discuss
A critical account of the points involved in the topic should be provided.

6. Outline
Be concise i.e. restrict the answer to giving the essentials only.

7. Predict
Make a logical connection between other pieces of information. Such information may be wholly given in the
question or may depend on answers extracted in an early part of the question.

8. Deduce
Used in a similar way as predict except that some supporting statement is required, e.g. reference to a law/principle, or the necessary reasoning is to be included in the answer

9. Comment
It is an open-ended instruction, inviting one to recall or infer points of interest relevant to the context of the question, taking account of the number of marks available.

10. Suggest
It is used in two contexts, i.e either to imply that there is no unique answer (e.g. in chemistry, two or more substances may satisfy the given conditions describing an ‘unknown’), or to imply that candidates are
expected to apply their general knowledge to a ‘novel’ situation, one that may be formally ‘not in syllabus’.

11. Find
Can be interpreted as calculate, measure, determine etc

12. Calculate
A numerical answer is required. In general working should be shown.
Note:The misuse of units and/or significant figures is liable to penalty.

13. Determine
It implies that the quantity cannot be measured directly but is obtained by calculation, substituting measured and known values of other quantities into a standard formula.

14. Sketch
When applied to graph work, the shapes and/or position o the curve need only be qualitatively correct but some quantitative aspects (e.g. passing through the origin, having an intercept at a particular value) may be looked for.

In diagrams, a simple and freehand drawing is acceptable but care should be taken over proportions and the clear exposition of important details.

15. Construct
Often used in relation to chemical equations where one is expected to write a balanced chemical equation,not by factual recall but by analogy or by using information in the question.

16. Compare
Both similarities and differences between things or concepts should be provided.

17. Classify
Group things based on common characteristics.

18. Recognise
Often used to identify facts, characteristics or concepts that are critical (relevant/appropriate) o the understanding of the situation, event, process or phenomenon.

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Power Revision

A Level Chemistry – 2 hrs Each Lesson

1. Atoms Molecules and Stoichiometry – 2 lessons
2. Chemical Bonding – 2 lessons
3. Chemical Energetics- 2 lessons
4. Reaction Kinetics – 2 lessons
5. Chemical Equilibrium – 2 lessons
6. Ionic Equilibrium – 2 lessons
7. Introduction Organic/Alkanes/Alkenes – 2 lesssons
8. Arenes – 1 lessons
9. Halogen Derivatives – 2 lesons
10. Hydroxy Compounds – 2 lessons
11. Carbonyl Compounds – 2 lessons
12. Carboxylic Acids and Derivatives – 2 lessons

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Atoms, Molecules and Stoichiometry Part 1

1 Define the following terms:

(a) Isotopes are atoms of the same element with the same number of
protons but different number of neutrons.

(b) Relative isotopic mass is the number of times the isotope is heavier than 1/12 the mass of an atom of carbon-12.

(c) Relative atomic mass of an element is the average mass of its atoms in the isotopic mixture, compared to 1/12 the mass of an atom of carbon-
12.Symbol: Ar

(d) Relative molecular mass of a molecule is the average mass of its
molecule compared to 1/12 the mass of an atom of carbon-12.
Symbol: Mr

(e) Avogadro’s Law states that equal volumes of all gases under the same
conditions of temperature and pressure contain the same number of
molecules/atoms (valid for ideal gases or gases under ideal-like
conditions).

(f) Molecular formula gives the actual number of the number of atoms of
each element present in the compound.

2 Concept : Limiting Reagent and Dilution

The process of obtaining iodine from oil field brines involves the following three step process. In the first reaction, 250 g of sodium iodide (NaI) is reacted with 340 g of silver nitrate (AgNO3).

NaI + AgNO3 —> AgI + NaNO3
2AgI + Fe —> FeI2 + 2Ag
2FeI2 + 3Cl2 —> 2FeCl3 + 2I2

(a) Determine the limiting reagent in the first reaction.
No of mol NaI available = +25023 127 = 1.6667
No of mol AgNO3 available = ++340108 14 48 = 2.00
NaI ≡ AgNO3 ⇒ NaI is the limiting reagent

(b) Calculate the mass of iodine crystals that can be obtained from the whole process.
NaI ≡ AgI ≡ ½ FeI2 ≡ ½ I2
No mol of I2 obtained = ½ ( 1.6667 ) = 0.83335
Mass of I2 obtained = 0.83335 x 2(127) = 212 g

(c) The iodine crystals obtained was then dissolved in 50.0 m3 of organic solvent, trichloromethane. Calculate the concentration of iodine solution in g dm-3.

Concentration of I2 = 50000212 = 0.00423 g dm-3

(d) In a further experiment, a certain volume of organic solvent, trichloromethane was added to lower the concentration of the iodine solution calculated in (c) to 1.40 X 10-5 mol dm-3. Calculate the volume of trichloromethane added to achieve this effect.

CoVo = CdVd ⇒ (0.00423254)(50000) = (1.40 x 10-5) (Vd)
Vd = 59480 dm3
∴ Volume of trichloromethane = 59480 – 50000 ≈ 9.48 x 103 dm3

(e) State the assumption made in determining your answers in parts (c) and (d).

Iodine is soluble in organic solvent / trichloromethane

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Atoms, Molecules and Stoichiometry Part 2

3. (a) Define the term mole.

The mole is defined as the amount of substance that contains the same
number of particles as there are atoms in 12 g of pure carbon-12.

(b) Why is the phrase “the mass of one mole of oxygen” ambiguous?

It is because the statement can either refer to one mole of oxygen atoms or one mole of oxygen molecules (O2). The statement can also refer to 16O or 18O isotopes.

(c) A meteorological balloon of 2 m diameter has a volume of 4.19 m3. It floats since it is given an upthrust equal to the mass of air it displaces.

Calculate:
(i) the mass of hydrogen in the balloon,

No of moles of hydrogen gas = 4190 / 23 = 182.2
Mass of hydrogen = 18.2 x 2 = 364 g

(ii) the mass of air it displaces,

Volume of air displaced = 4190 dm^3
There should be 182.2 mol of air.
Mass of air displaced = 182.2 x 29 = 5280 g

(iii) the load the balloon can carry for it just to lift off from the ground.

Upthrust = mass of air displaced = mass of hydrogen + load
Load = 5283.8 – 364. = 4919.4 = 4.92 kg

4. The reaction of silicon tetrachloride with moist ethoxyethane produces two oxochlorides with the formulae Si2OCl6 and Si3O2Cl8. When 0.10 g of one of these oxochlorides completely reacted with water, all of its chlorine was converted into chloride ions, and produced 0.303 g of silver chloride precipitate when an excess of aqueous silver nitrate was added.
Deduce the identity of the oxochloride.

AgCl Ξ Cl
No of mol of Cl in 0.303 g of AgCl
= 0.303/(108+35.5)
= 2.11 x 10-3

Relative molecular mass of Si2OCl6
= 2×28.1+16+6×35.5
= 285.2

Relative molecular mass of Si3O2Cl8
= 3×28.1+2×16+8X35.5
= 400.3

No of mole of Cl in 0.1g of Si2OCl6
= (0.1/285.2) x 6
= 2.10 x 10-3

No of mole of Cl in 0.1g of Si3O2Cl8
= (0.1/400.3) x 6
= 2.00 x 10-3

The oxochloride is Si2OCl6.

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Atoms, Molecules and Stoichiometry Part 3

Concept: Determining Formula of compound

A metal hydroxide, M(OH)n, is one of the products formed in an electrochemical cell used to power golf trolleys. 50.0 cm3 of solution containing 0.028 mol of M(OH)n in 1 dm3 requires 21.00 cm3 of sulfuric acid for complete neutralisation. The sulfuric acid contains 0.2 g of hydrogen ions in 1 dm3.

(a) Calculate the number of moles of sulfuric acid that will react with 1 mole of M(OH)n.

H2SO4 ≡ 2H+

No. of moles of H+ in 1 dm3 = 0.2/1.0 = 0.200

[H2SO4] = 0.2002= 0.100 mol dm-3

No. of moles of H2SO4 in 21.10 cm3 = (21.00×0.100)/1000 = 2.10 x 10-3

No. of moles of M(OH)n in 50.0 cm3 = (50.00×0.028)/1000 = 1.40 x 10-3

No. of moles of H2SO4/No. of moles of M(OH)n
= 2.10 x 10-3/1.40 x 10-3
= 1.5

1 mol of M(OH)n reacts with 1.50 mol of H2SO4

(b) Hence, determine the value of n.

2M(OH)n (aq) + nH2SO4 (aq) → M2(SO4)n (aq) + 2nH2O (l)
Mole ratio: 2M(OH)n ≡ nH2SO4
M(OH)n ≡ (n/2)H2SO4
n/2 = 3/2
n = 3

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Atoms, Molecules and Stoichiometry Part 4

Concept: Back titration and calculation of Percentage Purity

The chemical used for detecting proteins, biuret reagent, H2NCONHCONH2, can be formed by heating urea, (NH2)2CO.

2(NH2)2CO → H2NCONHCONH2 + NH3 — (1)

3.88 g of impure sample of urea (NH2)2CO was heated strongly above its melting point. The ammonia liberated was absorbed in 32.0 cm3 of 2.00 mol dm-3 sulfuric acid. The resulting solution was made up to 500 cm3 with distilled water. 25.0 cm3 of the solution required 25.50 cm3 of 0.20 mol dm-3 sodium hydroxide for neutralization, using methyl orange as an indicator.
(a) Calculate the percentage purity of urea in the sample. [4]

2NH3 + H2SO4 (excess) → (NH4)2SO4 — (2)

H2SO4 (remaining) + 2NaOH → Na2SO4 + 2H2O — (3)

No. of moles of NaOH in 25.50 cm3 = 25.50 x (0.20/)1000 = 5.10 x 10-3

From (3) 2NaOH Ξ H2SO4 (remaining)

No. of moles of H2SO4 in 25.0 cm3 = (5.10 x 10-3)/2 = 2.55 x 10-3

No. of moles of H2SO4 in 500 cm3 = 2.55 x 10-3 x (500/25) = 0.0510

Initial no. of moles of H2SO4 in 32.0 cm3 = 2.00 x (32/1000) = 0.0640

No. of moles of H2SO4 reacted with NH3 = 0.0640 – 0.0510 = 0.0130

From (2) and (1) 2NH3 Ξ H2SO4 (reacted) Ξ 4(NH2)2CO

Mass of urea, (NH2)2CO produced = 0.0130 x 4 x 60 = 3.12 g

Percentage purity of urea, (NH2)2CO = (3.12/3.88) × 100 = 80.4 %

(b) Determine the maximum number of hydrogen atoms present in the biuret reagent after the impure sample of urea was heated. [2]

From (2) and (1) 2NH3 Ξ H2SO4 (reacted) Ξ 2H2NCONHCONH2

No. of moles of H2NCONHCONH2 = 0.0130 x 2 = 0.0260

H2NCONHCONH2 Ξ 5H

No. of H atoms in biuret reagent = 0.0260 x 5 x 6.02 x 10^23
= 7.83 x 10^22

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Atoms, Molecules and Stoichiometry

Practice Question (15 min)

The alums are a series of double salts formed when a cation, with a charge of +1, having a large radius and a cation, with a charge of +3, having a small radius combined with sulfate ions.

Ammonium chromium alum has the formula of (NH4)aCr(SO4)b.xH2O. 1.28 g of the salt was dissolved in 100 cm3 of 0.0500 mol dm-3 ammonium chloride solution and the solution was divided into two equal portions.

To one portion was added an excess of sodium hydroxide and the mixture was boiled. The ammonia that was evolved neutralized 25.60 cm3 of 0.150 mol dm-3 nitric acid.

(a) Write a balanced ionic equation for the reaction between sodium hydroxide and the ammonium ions in solution. [1]

(b) Calculate the number of moles of NH3 from the alum. [2]

To the other portion, an excess of zinc was added which reduced the Cr3+(aq) to Cr2+(aq). The mixture was then filtered and the filtrate was titrated with acidified potassium dichromate(VI). It was found that 22.35 cm3 of 0.0100 mol dm-3 acidified potassium dichromate(VI) was required for the titration.

(c) Write a balanced ionic equation for the reaction between Cr2+ ions and acidified potassium dichromate(VI). [1]

(d) Calculate the number of moles of chromium ions from the alum. [1]

(e) Use your answer to (b) and (d), calculate the values of a and b. [1]

(f) Hence, find the relative formula mass of ammonium chromium alum and the value of x. [2]

(g) Calculate the percentage of chromium in the sample. [2]

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