2 HIGHER CHEMISTRY REVISION. Unit 1:- The MoleThe 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 O2So 40 cm3 = mol1 mole of O2 weighs 32gSo mol weighs 0.534gUsing information from the above graph, calculate the mass of hydrogen peroxide used in the reaction, assuming all the hydrogen peroxidedecomposed.(Take the molar volume of oxygen to be 24 litres mol-1)
3 order to estimate the volume of one mole of hydrogen gas. 2. . A student electrolysed dilute sulphuric acid using the apparatus shown inorder to estimate the volume of one mole of hydrogen gas.(a) Q= I x t= 0.5 x 14 x 60= 420C2H+ + 2e H22 x 96500C mol193000C 1 mol420C gives 52 cm3So C gives cm3or litres(b) Introduce a variable resistorso that current can be keptconstant.Current = 0.5 ATime = 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 possiblesignificant source of error?
4 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 approximatelyL hydrogen atoms.F 1.0 litres of 1.0 mol l -1 barium hydroxide solution containsapproximately L hydroxide atoms.C and D4. A student heated a compound which gave off carbon dioxidegas 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 = cm3.24000 cm3 contains 6.02 x 1023 CO2 molecules.So 240 cm3 contains 240/24000 x 6.02 x = 6.02 x 1021 CO2 molecules.
5 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 mol l-1 hydrochloric acid as shown in the diagramThe 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 molNo of moles of AgNO3 = mass/gfm = 0.2/169.9= 1.2 x 10-3 molSo the silver nitrate is in excess.
6 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 dioxideproduced 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 completedecomposition 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, suggestwhy the volume of carbon dioxide collected in the measuring cylinder wouldbe less than the theoretical value.(a) Water boils at 100oC andso could not raise thetemperature to 112oC.(b) (i) From equation2mol of NaHCO3gives 1 mol of CO2.i.e. 168 g NaHCO3 44g CO2.So 1.68 g 0.44gb) (ii) Some of the carbon dioxidedissolves in water.
7 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 thereaction is recorded.Iodine is produced from the iodide solution according to the following equation:2I-(aq) I2(aq) e-(a) Calculate the number of moles of iodine generated during the electrolysis giventhe following results.Current = A Time = 1 min 37 sQ = I x t = x 97 = 0.97C2I-(aq) I2(aq) e-1 mol F1 mol x C1 mol CSo 0.97 C / x 1 = 5 x 10-6 mol
8 7. (b) The iodine produced reacts with the thiosulphate ions according to the equation: I2(aq) S2O32-(aq) 2I-(aq) S4O62-(aq)iodine thiosulphate ionsAt 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 presentat the end-point.(ii) In a second experiment it was found that 1.2 x 10-5 mol of iodine reacted with3.0 cm3 of the sodium thiosulphate solution.Use this information to calculate the concentration of the sodium thiosulphatesolution in mol l-1.(b) (i) Starch solution.(ii) I2(aq) S2O32-(aq) 2I-(aq) S4O62-(aq)1 mol molSo 1.2 x 10-5 mol of I2(aq) reacts with 2.4 x 10-5 mol of S2O32-(aqConcentration of thiosulphate = no of moles/volume (litres)== mol l-1.2.4 x 10-50.003
9 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 passingthrough the liquid is A?Al3+(l) + 3e- Al(l)3 F 1 mol3 x 96500C 27 g289500C 27 gQ = I x t= x 60 x 60= C289500C 27 gSo C / x 27= g= kg of Al
10 CaCO3(s) + 2HNO3 (aq) Ca(NO3) 2(aq) + H2O(l) + CO2(g) 9. Calcite is a very pure form of calcium carbonate which reacts with nitricacid 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 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/ = mol.No. of moles of acid = C x V(litres) = 0.2 x 50/ = 0.01From equation mol of calcite reacts with mol of acid.So all the acid is used up0.01 mol of acid reacts with mol of calcite = x 100 = 0.5 gMass of calcite left over = 2.14 – 0.5g = 1.64 g.(b) Filter off any unreacted calcite, dry and then weigh it.
11 10. Diphosphine, P2H4, is a hydride of phosphorus 10. Diphosphine, P2H4, is a hydride of phosphorus. All of the covalent bondsin diphosphine molecules are non-polar because the elements present havethe same electronegativity.What is meant by the term”electronegativity”?(b) The balanced equation for the complete combustion of diphosphine is:2P2H4(g) O2 (g) P4O10(s) H2O(l)What volume of oxygen would be required for the complete combustion o10 cm3 of diphosphine?(c) Calculate the volume occupied by g of diphosphine.(Take the molar volume to be 24.0 litres mol-1.)(a) Electronegativity is a measure of the attraction of an atomfor electrons it shares with other atoms.(b) 2P2H4(g) O2 (g) P4O10(s) H2O(l)2 mol mol2 vol vol10cm cm31 mole of P2H4 weighs 66g66g occupies 24.0 litresSo g occupies 0.33/66 x = litres
12 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 ofcarbon called fullerines.(a) In what way does the structure of fullerines differ from the other forms ofcarbon, diamond and graphite?(b) One form of fullerine, C60, forms a superconducting crystalline compound withpotassium.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 covalennetwork solids.1 mole of K3C60 contains 60 x 12 = 720g of carbon.So 2.88g of carbon = 2.88/ = mol of fullerine.1 mole of K3C60 contains 3 x 39 g of potassium.So moles contains 3 x 39 x g= g of potassium
13 is converted into one mole of boron atoms with a charge of 3+. 12. Ionisation energies provide information about the structure of atoms.(a) Write the equation, showing state symbols, for the first ionisationenergy of sodium.(b) Calculate the number of electrons lost when one mole of boron atomsis converted into one mole of boron atoms with a charge of 3+.Na(g) Na+(g) + e-B(g) B3+(g) e-1 mol mol3 x 6.02 x 10231.806 x 1024 electrons13. The ion-electron equation for the production of zinc in anelectrolysis cell isZn e- ZnIf a current of 2000 A is used in the cell, calculate the mass ofzinc, in kg, produced in 24 hours.Zn e- Zn Q= I x t2 F mol = 2000 x x 24 x60 x 60193000C g = x 106CSo x 106 C g or kg