[admin]

A-Level Chemistry Tuition Singapore/H2 Chemistry Tuition/JC Chemistry Tutor

Redox – Exam Practice Question
3 (Time allocation: 12 min)
(a) Mixtures of CaCl2 and NaCl are used for salting roads to prevent ice formation. 1.95 g sample of such mixture was dissolved in water and excess aqueous Na2C2O4 was added to completely precipitate the Ca2+ ions as CaC2O4.

The CaC2O4 formed was separated from the solution and dissolved in sulphuric acid. The resulting H2C2O4 solution was titrated with 37.8 cm3 of 0.102 mol dm-3 KMnO4 solution.

(i) Given that C2O42- is oxidized to form carbon dioxide, write a balanced equation for the reaction of C2O42- and MnO4-. [1]

(ii) Calculate the number of moles of H2C2O4 reacted with KMnO4. [1]

(iii) Calculate the mass of CaCl2 in the original sample. [2]

(iv) Hence, calculate the percentage mass of NaCl in the original sample.[2]

(b) 20.0 cm3 of 0.100 mol dm-3 vanadium (II) solution, V2+(aq), reacted with 40.0 cm3 of a 0.100 mol dm-3 iron (III) solution. Iron (III) ions was reduced to form iron (II) ions in the reaction. Calculate the final oxidation state of vanadium in the above reaction. [2]

For worked solution, please contact @9863 9633

[admin]

O Level Chemistry Tuition Singapore/Chemistry O Level Tuition/Tutor

Alkenes

1. Alkenes are unsaturated hydrocarbons that contain one or more carbon-carbon double bond

2. The general formula of alkenes with one double bond is CnH2n

3. Naming of alkenes
Ethene C2H4
Propene C3H6
Butene C4H8

If you need help in the Organic Chemistry, please contact Angie @96790479 or Mr Ong 98639633

[admin]

O Level Chemistry Tuition Singapore/Chemistry O Level Tuition/Tutor

Alkanes

1. Saturated hydrocarbon are hydrocarbons with only single bonds between carbon atoms.

2. Unsaturated hydrocarbons are hydrocarbons with one or more double bond or triple bond between carbon atoms.

3. The alkanes are saturated hydrocarbons with the general molecular formula CnH2n+1 where n = 1, 2, 3, ……

4. Alkanes contain only single bonds between carbon atoms. Each carbon atom in an alkane molecule uses all its valence electrons in forming single bonds with four other atoms.

5. First four members of the alkane
Methane CH4
Ethane C2H6
Propane C3H8
Butane C4H10

If you need help in the Organic Chemistry, please contact Angie @96790479 or Mr Ong 98639633

[admin]

O Level Chemistry Tuition Singapore/Chemistry O Level Tuition/Tutor

Important Definitions – Alkanes & Alkenes

6. Isomerism is the existence of two or more organic compounds with the same molecular formula but different structural formulae. These different compound called isomers.

7. A substitution reaction is a reaction in which an atom (or a group of atoms) in an organic molecule is replaced by another atom ( or group of atoms)

8. An addition reaction is a reaction in which an unsaturated organic compound combines with another substance to form a single product.

9. Cracking is a process in which a large hydrocarbon molecules are broken down into smaller molecules.

10. Polyunsaturated compounds are organic molecules that contain more than c=c bond.

If you need help in the Organic Chemistry, please contact Angie @96790479 or Mr Ong 98639633

[admin]

A-Level Chemistry Tuition Singapore/H2 Chemistry Tuition/JC Chemistry Tutor

(Time allocation: 9 min)
When sodium is burned in air, a mixture of sodium oxide, Na2O, and sodium peroxide, Na2O2, is formed. The mixture reacts with water according to the following equations.
Na2O + H2O —> 2NaOH
Na2O2 + 2H2O —> 2NaOH + H2O2
The following information will allow you to calculate the relative amounts of the two oxides produced when sodium is burned.
• The mixture obtained by burning a sample of sodium was dissolved in distilled water and made to 100 cm3 to give solution H.
• A 25.0 cm3 of solution H was titrated with 0.100 mol dm-3 HCl. 22.50 cm3 of acid was required to reach the end-point. (A)
• The H2O2 content of solution H was found by titration of another 25.0 cm3 portion with 0.0200 mol dm-3 KMnO4. 10.0 cm3 of KMnO4 solution was required to reach the end-point. (B)

(a) Using the results of HCl titration, calculate the total number in moles of NaOH in 100 cm3 of solution H.

(b) (i) Write a balanced equation for the oxidation of H2O2 with KMnO4 in an acidic medium.

(ii) Using the results of the KMnO4 titration, calculate the amount in moles of H2O2 in 100 cm3 of solution H.

(c) Hence, calculate the amount in moles of Na2O and Na2O2 formed during the burning of the sodium sample.
Na2O + H2O —> 2NaOH …..(1)
Na2O2 + 2H2O —> 2NaOH + H2O2 ….(2)

For worked solution, please contact @9863 9633

 

[admin]

A-Level Chemistry Tuition Singapore/H2 Chemistry Tuition/JC Chemistry Tutor

1. Concept: Redox Half Equations

(a) Define the term disproportionation. [1]

2. (Time allocation: 12 min)
(a) Mixtures of CaCl2 and NaCl are used for salting roads to prevent ice formation. 1.95 g sample of such mixture was dissolved in water and excess aqueous Na2C2O4 was added to completely precipitate the Ca2+ ions as CaC2O4.

The CaC2O4 formed was separated from the solution and dissolved in sulphuric acid. The resulting H2C2O4 solution was titrated with 37.8 cm3 of 0.102 mol dm-3 KMnO4 solution.

(i) Given that C2O42- is oxidized to form carbon dioxide, write a balanced equation for the reaction of C2O42- and MnO4-. [1]

(ii) Calculate the number of moles of H2C2O4 reacted with KMnO4. [1]

(iii) Calculate the mass of CaCl2 in the original sample. [2]

(iv) Hence, calculate the percentage mass of NaCl in the original sample.
[2]

(b) 20.0 cm3 of 0.100 mol dm-3 vanadium (II) solution, V2+(aq), reacted with 40.0 cm3 of a 0.100 mol dm-3 iron (III) solution. Iron (III) ions was reduced to form iron (II) ions in the reaction. Calculate the final oxidation state of vanadium in the above reaction. [2]

For worked solution, please contact @9863 9633

[admin]

A-Level Chemistry Tuition Singapore/H2 Chemistry Tuition/JC Chemistry Tutor

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]

For worked solution, please contact @9863 9633

[admin]

O Level Chemistry Tuition Singapore/Chemistry O Level Tuition/Tutor

Important Definitions – Alkanes & Alkenes

1. Saturated hydrocarbon are compounds containing only carbon and hydrogen and having single bonds between carbon atoms.

2. Unsaturated hydrocarbon are compounds containing only carbon and hydrogen and having double bonds between carbon atoms.

3. Alkane series has a general formula CnH2n+2

4. Alkenes series has a general formula CnH2n

5. Structural formula is the chemical formula that shows how the atoms are joined together in a molecue.

If you need help in the Organic Chemistry, please contact Angie @96790479 or Mr Ong 98639633

[admin]

O Level Chemistry Tuition Singapore/Chemistry O Level Tuition/Tutor

Petroleum

1. Petroleum is a non-renewable resource and there is only a limited amount of petroleum in the Earth.

2. Moet petroleum is used as fuel. Only a small amount is used as feedstock to make useful petrochemicals.

3. Gasohol is an alternative energy sources used as a fuel in motor vehicles.

4. Biogas is also used as an alternative energy source. It contain about 50% ethane.

5. Hydrogen can also be used as a fuel for cars.

If you need help in the Organic Chemistry, please contact Angie @96790479 or Mr Ong 98639633

[admin]

O Level Chemistry Tuition Singapore/Chemistry O Level Tuition/Tutor

The Atmosphere and Environment – Important Definitions

1. Air pollution is the condition in which air contains substances that are harmful to living things or the environment. These substances are known as air pollutants.

2. A photochemical reaction is a chemical reaction caused by light or ultraviolet radiation.

3. Acid rain is rainwater with pH 4 or less than 4

4. Flue gases are waste gases produced when fossil fuels undergo commutation.

5. Chlorofluorocarbons CFCs are compounds containing the elements carbon, fluorine and chlorine.

6. The carbon cycle is the movement of carbon from carbon dioxide in the in the air through living things and back to the air again.

7. Greenhouse effect is the absorption of infrared radiation by some gases in the air which leads to atmospheric warming.

8. Global warming is the gradual warming of the Earth’s atmosphere due to the increased concentration of greenhouse gases in the atmosphere.

If you need help in the above topic, please contact Angie @96790479 or Mr Ong 98639633

[admin]

A-Level Chemistry Tuition Singapore/H2 Chemistry Tuition/JC Chemistry Tutor

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

For exam based questions with solutions please contact @9863 9633

[admin]

A-Level Chemistry Tuition Singapore/H2 Chemistry Tuition/JC Chemistry Tutor

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.

For exam based questions with solutions please contact @9863 9633

[admin]

O Level Chemistry Tuition Singapore/Chemistry O Level Tuition/Tutor

Ammonia – Important Definitions

1. A reversible reaction is a chemical reaction that can proceed in both the forward and reverse directions.

2. Haber process is an industrial process for making ammonia from nitrogen and hydrogen.

Operating conditions in the Haber process
Temperature – 450 deg C
Pressure – 200 – 300 atm
catalyst – Finely divided iron

If you need help in the above topic, please contact Angie @96790479 or Mr Ong 98639633

 

[admin]

O Level Chemistry Tuition Singapore/Chemistry O Level Tuition/Tutor

Speed of Reaction – Important definitions

1. the speed of reaction is the amount of reactant used up or the amount of product obtained per unit time.

2. Activation energy is the minimum energy that the reacting particles must possess for a reaction to occur.

3. An effective collision is a collision that successful in producing a chemical reaction.

4. A catalyst is a substance that increases the speed of a chemical reaction.

5. An enzyme is a substance that catalyses biochemical reactions.

If you need help in the above topic, please contact Angie @96790479 or Mr Ong 98639633

[admin]

A-Level Chemistry Tuition Singapore/H2 Chemistry Tuition/JC Chemistry Tutor

Atoms, Molecules and Stoichiometry

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

For exam based questions with solutions please contact @9863 9633

[Author]

Posts