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Revision for salt, electrochemistry, carbon compounds , thermochemistry
 

Revision for salt, electrochemistry, carbon compounds , thermochemistry

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    Revision for salt, electrochemistry, carbon compounds , thermochemistry Revision for salt, electrochemistry, carbon compounds , thermochemistry Presentation Transcript

    • SELARAS 2 FORM 5 2011
    • Question 1b) Write the formula of the following substances:• Hydrochloric acid – HCl• Sulphuric acid – H2SO4• Barium hydroxide – Ba(OH)2• Barium sulphate – BaSO 4
    • b) Hydrochloric acid is a strong acid. Why?HCl ionised completely in water toproduce high concentration ofhydrogen ion
    • c)Silver chloride and copper (II)sulphateare two salts that are required to beprepared in the labi) Which one is the soluble salt? Copper (II) sulphateii) State the name of acid needed toprepare copper (II) sulphate Sulphuric acid
    • iii) Silver chloride is prepared by the precipitation method.Write the chemicalEquation to prepare silver chloride saltAgNO3 + NaCl → AgCl + NaNO3
    • d) Calculate the following:i) What is the concentration of 50 cm3NaOH solution needed to neutralise100 cm3 0.5 moldm-3 H2SO4? 2 NaOH + H 2 SO4 → Na2 SO4 + 2 H 2O Where M 1V1 M 2V2 = M1 = molarity of acid V1 = volume of acid a b M2 = Molarity of alkali V2 = volume of alkali a = no. of mol of acid in the equatn b = no. of mol of alkali in the equatn
    • 2 NaOH + H 2 SO4 → Na2 SO4 + 2 H 2OM 1V1 M 2V2 Where = M1 = molarity of acid a b V1 = volume of acid0.5 x100 M 2 x50 M2 = Molarity of alkali = V2 = volume of alkali 1 2 a = no. of mol of acid in the equatnM 2 = 0.5 x100 x 2 / 50 b = no. of mol of alkali −3 in the equatnM NaOH = 2moldm
    • i) How much water is needed to form asolution of 1.0moldm-3 NaOH solutionfrom a solution of 50cm3 sodiumhydroxide 2.0moldm-3 Where M 1V1 = M 2V2 M1 = molarity of original 2(50) = 1(V2 ) solution V1 = volume of original V2 = 100 solution M2 = Molarity of diluted ∴VolumeH 2O = solution V2 = total volume of 100 − 50 = 50cm 3 diluted solution
    • 2.Diagram 2 shows the apparatus setup for two types of cells, P and Q
    • a) State one difference in energy change between cell P and cell Q:In cell P energy change is from electrical energy to chemical energyIn cell Q energy change is from chemical energy to electrical energy
    • b) What is the product at the anode incell P? Oxygen gasc) Write the half equation for thedischarge of cation in cell P 2+ − Cu + 2e → Cu
    • d) State one observation at electrodezinc in cell Q Zinc electrode becomes thinner
    • e) Diagram 2.2 shows the apparatusset-up for the electrolysis of sodiumchloride solution using carbonelectrodes
    • i) State all ions attracted towardselectrode P and Q Electrode P Electrode Q Na , H + + Cl , OH - -ii) State which ion discharge atelectrode Q. Explain whyChloride ion because theconcentration of Cl- is higher thanconcentration of OH-
    • iii) Describe one chemical test to identify gas XPlace burning wooden splinter intomouth of test tube, ‘pop’ sound willbe produced
    • Diagram 3 shows a series of reaction ofcompound P. Compound P andcompound Q are shown in condensedstructural formulae.
    • Compound Q CO2 + H2O CH3CH2 CH3 Combustion I IICompound P Compound R Steam andCH2=CHCH3 Phosphoric acid + Ethanoic acid III + concentrated sulphuric acid Compound S Diagram 3
    • (a) Compound P is propene . Statethe homologous series of compound P ALKENE
    • (b)Compound Q is used as a cooking gas.• i) State the molecular formula of compound Q C3H8ii) Write a chemical equation for thecombustion of compound Q to producecarbon dioxide and waterC3H8 + 5O2 3CO2 + 4H2O
    • (C) Compound P can be changed tocompound R which is an alcohol throughprocess II. • Name process II HYDRATION • State the general formula for alcohol CnH2n+1OH
    • (d) Explain briefly how compound Scan be prepared from compound R inlaboratory. In your answer include theappropriate reagents and condition forthe preparation.
    • • Add absolute propanol and glacial ethanoic acid into a beaker.• Add slowly concentrated H2SO4.• Heat under reflux for 6 hours• Distill the mixture to get the products.
    • e) An organic compound W has amolecular formula of C4 H8 . CompoundW has 3 isomers. Draw two structuralformulae of two isomers of W.
    • e) isomers of W.
    • 4) Natural rubber is a polymer, where themonomer is isoprene or 2- methylbuta-1,3-diene. Natural rubber is obtained from latex,which is a colloid made up of rubber particlesand water.Draw Structural formula of isoprene (1m)
    • Draw labelled diagram of latex particle in water (2m)
    • Explain why latex in the original form does not coagulate
    • However when latex is exposed to air or when acid isadded, coagulation of latex occur. Explain(3m) • Air contains bacteria that acts on latex to produce lactic acid
    • The properties of natural rubber can be improved by treating it with sulphur. Name this process (1m) VULCANISATION
    • State ONE use of the treated rubber above (1m)  TYRE  GLOVE  RUBBER HOSE
    • QUESTION 5 Initial temperature of both solutions = 29.5 o C Highest temp of the50 cm3 of 1.0 mol mixture = 34.5 o Cdm-3 Potassiumsulphate solution 50 cm3 of 1.0 mol dm-3 Barium Chloride solution
    • a. What is meant by the heat ofprecipitation of barium sulphate?• Heat change when 1 mole of barium sulphate is formed from barium ion and sulphate ion in aqueous solution 1m
    • b. Write an ionic equation for the reactionMethod:First, write the formula of theprecipitate on the right side of theequationSecond, write the ions of theprecipitate on the left side of equation• Ba 2+ + SO4 -2  BaSO4 1m
    • c. State one observation for theexperiment• White precipitate forms or the containers become warm 1m
    • d. Calculate i. The heat released in the experiment. [Specific heat capacity of water = 4.2 Jg-1oC-1, Density of water = 1.0 g cm-3]Heat release = mc =100x4.2x(34.5- 29.5)J = 2100J 2m
    • ii. The number of moles of bariumsulphate formed• Mole of BaSO4 = MV/1000 = 50 x 1.0 /1000 = 0.05 1m
    • iii. The heat of precipitation of barium sulphate0.05 mole BaSO4 produce heat 2100 J1 mole BaSO4 2100 / 0.05 J = 42000 J ∆ H = - 42 kJ mol-1
    • e.Draw the energy level diagram for the precipitation of barium sulphateEnergy Ba + SO 4 2+ -2 ∆ H = -42 kJ mol-1 BaSO 4 2m