Waec Gce Free Chemistry 3 (Alt. to Practical Work) Answers Now Available

CHEMISTRY PRACTICAL ANSWERS 1a) i) TABULATE burette reading/cm^3 _ 1 __ 2 ___ 3 final reading__________ 39.5 _ 29.80 _ 19.40 initial reading ____ 19.55 _ 10.50 __ 0.0 volume of E used ___ 19.70 _ 19.30 __ 19.40 ii) Na2CO3(aq)^2HNO3(aq) — 2NaNO3(aq) + CO2(g) + H2O(c) average volume of E used = 19.70+19.30+19.40/3 =19.47cm^3 1bi) From the information given, nD=1, nE=2, CD=?, CE=0.0905mol/dm^3 VD=20.0cm^3, VE=19.35cm^3 Using: CDVD/CEVE=nD/nE CD=CEVEnD/VDnE =(0.0905*19.35*1)/(20.0*2) =1.751175/40 =0.04377 CD=0.0438mol/dm^3 1bii) Since 0.04377mol is in 1000cm^3 xmol will be 350cm^3 x=0.04377*35.0/1000 =1.5322/1000 =0.00153mol/dm^3 at 25C Hence the solubility is 0.00153mol/dm^3 1biii) From the stoichiometry 1 mole of Na2CO3 produces 1 mole of CO2 0.00153mol of Na2CO3 produces 0.00153mole of CO2 mass=molar mass*number of mole molar mass Na2CO3=23*2+12+16*3 =46+12+48 =106g/mol mass of CO2=12+16*2 =12+32 =44g/mol 106g Na2CO3 produces 44g of CO2 0.0153*106 will produce y y=0.153*106*44/106 =0.153*44 y=6.732g of CO2 Hence mass of CO2 produced is 6.732g ========= 1]Conc.of B in moldm^3 Molar con.=mass conc÷molar mass =2.52g÷54 =0.046moldm^3 Conc. of A in moldm^3 Ca=0.046Ã?25Ã?2 ______ 250Ã?1 =0.0092moldm^3 (2A) i) INFERENCE: Soluble salts (2A) ii) OBSERVATION: White precipitate (2A) iii) OBSERVATION: Insoluble in excess (2A) iv) OBSERVATION: insoluble excess INFERENCE: Na+ present (2b) i) INFERENCE: Zn^2+, Al^3+, Pb^2+ present (2b) ii) INFERENCE: Al^3+, Pb^2+ present (2B) iii) OBSERVATION: black precipitate is formed and precipitate dissolve (3A) i) Measuring cylinder (3A) ii) Kipp's Appararatus (3A) iii) Separating funnel method (3B) i. It turnsred litmus paper to blue ii. colours remain unchanged iii. It turns blue litmus paper red (3C) i. AgNo3/NH3 ii. NaHCO3 iii. I2/NaoH (3D) i. carbon(iv)oxide, carbon(ii)oxide ii. Aluminium(III)oxide, Iron(III)oxide keep reloading