2015/2016 WAEC Chemistry 3(Practical Alt A) Real Questions And Answers

Note: ^ MEANS RAISE TO POWER / means division or divide * means multiplication ================== (2) Test | Observation | Inference == 1. Sample C water 5cm^3 of C NaoH in drop and in excess | soluble white geletenous precipitate in drop. White soluble or dissolve (in excess) | soluble salt. Zn^2 or Al^3 present. ====== 2. 5cm^3 of C NH4OH in drop. In excess | white geletenous precipitate in drop. White geletenous ppt dissolve (in excess) | Zn^2 or Al^3 present. Zn^2 confirmed (in excess). ====== 3. Solution of sample C dilute HN3 Bacl 2(AQ) HCL(AQ) | No visible reaction. White ppt form. White ppt insoluble in dilute Hcl | SO3^2- or SO4^2- may be present. SO4^2- confirmed.========== 4. Sample D dilute Hcl(aq) | A colourless and odourless gas evolve with a effervescence. The gas turn like water milky. The gas turn blue litmus to red | The gas in CO2 from CO3^2- or Hco3^- =================== PLS IN NUMBER 1, TAKE NOTE OF UR SCHOOL END POINT OR TITRE VALUE. EACH SCHOOL HAS A DIFFERENT TITRE VALUE FROM UR CHEMISTRY LAB. OUR END POINT HERE IS (1) TABULATE: Burrette Readings(cm3),1st Reading | 2nd Reading | 3rd reading| Final burette Reading|23.70|45.00|26.00| Initial burette Reading|2.00|23.50|4.30| Volume of acid used|21.50|20.70|20.90| Average titre value = (21.70 21.50 21.70)cm^3 /3 =(64.90)/3 = 21.70cm3 ============= (1) A IS 0.100 mol 2.50g of Na2CO3 and Na2SO4 in 250cm^3 Na2CO3 2HNO3->2NaNO3 H2O CO2 Conc of B in mol/dm^3 Conc of Na2CO3 in the mixture Na2CO3 =106 VA=21.7cm^3 CA=0.100 VB=25cm^3 CB-=? NA=2 NB=1 CAVA/CBVB=NA/NB 0.1*21.7/CB*25=2/1 CB=(0.1*21.7*1)/(25*2) =0.0434M ii) molar concentration =mass conc/molar mass mass conc=molar conc*molar mass =0.0434*106 =4.69g/dm^3 iii) 2.50g of Na2CO3 Na2SO4->250cm^3 Xg of Na2CO3 Na2SO4->1000cm^3 Xg=2.50*1000/250 Xg=10g % of Na2CO3=4.6/10*100 =46% 3 ai) fountain experiment aii) to show that HCL is extremely soluble in water aiii) Ammonia gas (NH3) aiv) The water turn red immediately av) No avi) because of its solubility in water 3b) u=>(heated) v CO2 CaCo3(s) =>(heated) Cao(s) Co2(g) v H20 => w Cao(s) H20(l) => Ca (OH)2(aq) U = CaCa(3), V = Cao, W = Ca (OH)2 ========== Keep Refreshing ==========