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HIGHER CHEMISTRY REVISION .

HIGHER CHEMISTRY REVISION. Unit 1:- The Mole. 1. The balanced equation for the decomposition of hydrogen peroxide into water and oxygen is: 2H 2 O 2 (l)  2H 2 O(l) + O 2 (g) . 24000 cm 3 of gas = 1 mol of O 2 So 40 cm 3 = 0.0167 mol

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HIGHER CHEMISTRY REVISION .

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  1. HIGHER CHEMISTRY REVISION. Unit 1:- The Mole 1. The balanced equation for the decomposition of hydrogen peroxide into water and oxygen is: 2H2O2 (l)  2H2O(l) + O2 (g) 24000 cm3 of gas = 1 mol of O2 So 40 cm3 = 0.0167 mol 1 mole of O2 weighs 32g So 0.0167 mol weighs 0.534g Using information from the above graph, calculate the mass of hydrogen peroxide used in the reaction, assuming all the hydrogen peroxide decomposed. (Take the molar volume of oxygen to be 24 litres mol-1)

  2. 2. . A student electrolysed dilute sulphuric acid using the apparatus shown in order to estimate the volume of one mole of hydrogen gas. (a) Q= I x t = 0.5 x 14 x 60 = 420C 2H+ + 2e - H2 2 x 96500C  1 mol 193000C  1 mol 420C gives 52 cm3 So 193000C gives 23 895 cm3 or 23.895 litres (b) Introduce a variable resistor so that current can be kept constant. Current = 0.5 A Time = 14 minutes. Volume of hydrogen collected = 52 cm3. (a) Calculate the molar volume of hydrogen gas. (b) What change could be made to the apparatus to reduce a possible significant source of error?

  3. 3. The symbol for the Avogadro Constant is L. Identify the two true statements. A 64.2 g of sulphur contains approximately L atoms. B 16.0 g of oxygen contains approximately L molecules. C 6.0 g of water contains approximately L atoms. D 1.0g of hydrogen contains approximately L protons. E 2.0 litres of 0.50 mol l-1 sulphuric acid contains approximately L hydrogen atoms. F 1.0 litres of 1.0 mol l -1 barium hydroxide solution contains approximately L hydroxide atoms. C and D 4. A student heated a compound which gave off carbon dioxide gas and water vapour. The volume of carbon dioxide collected was 240 cm3. Calculate the number of molecules in this volume. (Take the molar volume of carbon dioxide to be 24 litres mol-1.) 24 litres = 24000 cm3. 24000 cm3 contains 6.02 x 1023 CO2 molecules. So 240 cm3 contains 240/24000 x 6.02 x 1023 = 6.02 x 1021 CO2 molecules.

  4. 5. A student added 0.20 g of silver nitrate, AgNO3, to 25 cm3 of water. This solution was then added to 20 cm3 of 0.0010 mol l-1 hydrochloric acid as shown in the diagram The equation for the reaction which occurs is: AgNO3 (aq) + HCl(aq)  AgCl(s) + HNO3 (aq) (a) Name the type of reaction which takes place. (b) Show by calculation which reactant is in excess. • Precipitation. • No of moles of HCl = C x V(litres) = 20/1000 x 0.001 • = 2 x 10-5 mol • No of moles of AgNO3 = mass/gfm = 0.2/169.9 • = 1.2 x 10-3 mol • So the silver nitrate is in excess.

  5. 6. When sodium hydrogencarbonate is heated to 112oC it decomposes and the gas carbon dioxide is given off: 2NaHCO3(s)  Na2CO3(s) + CO2(g) + H2O(g) The following apparatus can be used to measure the volume of carbon dioxide produced in the reaction. (a) Why is an oil bath used and not a water bath? (b) (i) Calculate the theoretical volume of carbon dioxide produced by the complete decomposition of 1.68 g of sodium hydrogencarbonate. Take the molar volume of carbon dioxide to be 23 litres mol-1 (ii) Assuming that all of the sodium hydrogen carbonate is decomposed, suggest why the volume of carbon dioxide collected in the measuring cylinder would be less than the theoretical value. (a) Water boils at 100oC and so could not raise the temperature to 112oC. (b) (i) From equation 2mol of NaHCO3 gives 1 mol of CO2. i.e. 168 g NaHCO3  44g CO2. So 1.68 g  0.44g b) (ii) Some of the carbon dioxide dissolves in water.

  6. 7. The concentration of a solution of sodium thiosulphate can be found by reaction with iodine. The iodine is produced by electrolysis of an iodide solution using the apparatus shown. The current is noted and the time when the indicator detects the end-point of the reaction is recorded. Iodine is produced from the iodide solution according to the following equation: 2I-(aq)  I2(aq) + 2e- (a) Calculate the number of moles of iodine generated during the electrolysis given the following results. Current = 0.010 A Time = 1 min 37 s Q = I x t = 0.01 x 97 = 0.97C 2I-(aq)  I2(aq) + 2e- 1 mol 2F 1 mol 2 x 96500 C 1 mol 193000C So 0.97 C  0.97/193000 x 1 = 5 x 10-6 mol

  7. (b) (i) Starch solution. (ii) I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62-(aq) 1 mol 2 mol So 1.2 x 10-5 mol of I2(aq) reacts with 2.4 x 10-5 mol of S2O32-(aq Concentration of thiosulphate = no of moles/volume (litres) = = 0.008 mol l-1. 2.4 x 10-5 0.003 7. (b) The iodine produced reacts with the thiosulphate ions according to the equation: I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62-(aq) iodine thiosulphate ions At the end-point of the reaction, excess iodine is detected by the indicator. (i) Name the indicator which could be used to detect the excess iodine present at the end-point. (ii) In a second experiment it was found that 1.2 x 10-5 mol of iodine reacted with 3.0 cm3 of the sodium thiosulphate solution. Use this information to calculate the concentration of the sodium thiosulphate solution in mol l-1.

  8. 8. Aluminium is manufactured in cells by the electrolysis of aluminium oxide dissolved in molten cryolite. What mass of aluminium is produced each hour, if the current passing through the liquid is 180 000 A? Al3+(l) + 3e- Al(l) 3 F  1 mol 3 x 96500C  27 g 289500C  27 g Q = I x t = 180 000 x 60 x 60 = 648 000 000 C 289500C  27 g So 648 000 000 C  648 000 000/289500 x 27 = 60 435 g = 60.435 kg of Al

  9. 9. Calcite is a very pure form of calcium carbonate which reacts with nitric acid as follows: CaCO3(s) + 2HNO3 (aq)  Ca(NO3) 2(aq) + H2O(l) + CO2(g) A 2.14 g piece of calcite was added to 50.0 cm3 of 0.200 mol l-1 nitric acid in a beaker. (a) Calculate the mass of calcite, in grams, left unreacted. (b) Describe what could be done to check the result obtained in (a) • No. of moles of calcite = mass/gfm = 2.14/100 = 0.0214 mol. • No. of moles of acid = C x V(litres) = 0.2 x 50/1000 = 0.01 • From equation 0.0214 mol of calcite reacts with 0.0418 mol of acid. • So all the acid is used up • 0.01 mol of acid reacts with 0.005 mol of calcite = 0.005 x 100 = 0.5 g • Mass of calcite left over = 2.14 – 0.5g = 1.64 g. • (b) Filter off any unreacted calcite, dry and then weigh it.

  10. 10. Diphosphine, P2H4, is a hydride of phosphorus. All of the covalent bonds • in diphosphine molecules are non-polar because the elements present have • the same electronegativity. • What is meant by the term”electronegativity”? • (b) The balanced equation for the complete combustion of diphosphine is: • 2P2H4(g) + 7O2 (g)  P4O10(s) + 4H2O(l) • What volume of oxygen would be required for the complete combustion o • 10 cm3 of diphosphine? • (c) Calculate the volume occupied by 0.330 g of diphosphine. • (Take the molar volume to be 24.0 litres mol-1.) • (a) Electronegativity is a measure of the attraction of an atom • for electrons it shares with other atoms. • (b) 2P2H4(g) + 7O2 (g)  P4O10(s) + 4H2O(l) • 2 mol 7 mol • 2 vol 7 vol • 10cm3 35cm3 • 1 mole of P2H4 weighs 66g • 66g occupies 24.0 litres • So 0.330g occupies 0.33/66 x 24 = 0.12 litres

  11. 11. In 1996, the scientists Robert Curl, Harold Kroto and Richard Smalley won the Nobel Prize in Chemistry for their contribution to the discovery of new forms of carbon called fullerines. (a) In what way does the structure of fullerines differ from the other forms of carbon, diamond and graphite? (b) One form of fullerine, C60, forms a superconducting crystalline compound with potassium. Its formula can be represented as K3C60. A sample of this compound was found to contain 2.88 g of carbon. (i) Calculate the number of moles of fullerine used to make this compound. (ii) Calculate the mass of potassium, in grams, in the sample. • (a) Fullerines are covalent molecular solids – diamond and graphite are covalen • network solids. • 1 mole of K3C60 contains 60 x 12 = 720g of carbon. • So 2.88g of carbon = 2.88/720 = 0.004 mol of fullerine. • 1 mole of K3C60 contains 3 x 39 g of potassium. • So 0.004 moles contains 3 x 39 x 0.004 g • = 0.468 g of potassium

  12. 12. Ionisation energies provide information about the structure of atoms. (a) Write the equation, showing state symbols, for the first ionisation energy of sodium. (b) Calculate the number of electrons lost when one mole of boron atoms is converted into one mole of boron atoms with a charge of 3+. • Na(g)  Na+(g) + e- • B(g)  B3+(g) + 3e- • 1 mol3 mol • 3 x 6.02 x 1023 • 1.806 x 1024 electrons 13. The ion-electron equation for the production of zinc in an electrolysis cell is Zn2+ + 2e- Zn If a current of 2000 A is used in the cell, calculate the mass of zinc, in kg, produced in 24 hours. Zn2+ + 2e- Zn Q= I x t 2 F  1 mol = 2000 x x 24 x60 x 60 193000C  65.4 g = 172.8 x 106C So 172.8 x 106 C  58555g or 58.55 kg

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