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Neco Chemistry Practal Answers

chemistry practical answers

Neco chemistry answers 1st titre: Vol of acid used in cube : 28.00³ Initial reading : 0.00 Final reading: 28.00³ 2nd titre: Vol of acid used in cube : 30.00³ Initial reading : 0.00 Final reading: 30.00³ 3rd titre: Vol of acid used in cube : 26.00³ Initial reading : 0.00 Final reading: 26.00³ Average vol of acid used=28.00³ 30.00 26.00 =84 84/3= 28.00 .: the average titre value is 28.00 *note your school will give a value. Your value should not be more or less than /- 0.2 to what is being given* (1a) Titre value: Under 1: 20.70, 0.00, 20.70 Under2: 20.70, 0.00, 20.70 Under3: 20.70, 0.00, 20.70 Under Initial reading: 0.00, 0.00, 20.70, 20.70 Under volume of Acid used: 20.70, 20.70, 20.70 Average value= (20.70 20.70= 20.70)/3 VA=(63.10)/3= 20.70cm^3 (1bi) Given CA=? CB=0.10mol^-3 VA=20.70 cm^3 VB=25.0CM^3 NA=1 NB=2 H2C2O4 2NAOH=> Na2C2O4 2H20 CAVA/CBVB= NA/NB 0/0.10 *25.00cm^3= 1/2 CA= 2.5/2*20.70 CA2.50/41.40 =0.603mol dm^-3 (1bii) CA=mass concentration/molar mass H2C2O4=0 0 0) gmol 0 4 40)g/mol 2 24 48g/mol =72g/mol 0.0603= mass contration/72 =4.348g/dm^3 (1biii) mass of pure= 4.35gdm^-3 mass of impure=6.30g/dm^-3 % purity= mass of pure/ mass of impure * 100% =4.35/6.30 * 100% =4.35/6.30 % % purity= 69.01% (1biv) methyl orange is not used because it is a titration between weak acid and strong base, the end point will not coincide with the PH which the indicator exhibits of colour change ========================= =========== (2ai) OBservation: salt C dissolves in distilled water inference: Salt C is a soluble salt (2bi) observation: Rusty brown precipitate is formed in excess interference: Fe(3 ) is present (2bii) OBSERVATION: formation of blood-red precipitate Inference: Fe(3 ) is confirmed (2ci) Observation: formation of white precipitate which dissolves on heating Inference: SO32, SO42 are likely to present (2cii) Observation: A pungent gas is evolved, the gas turns moist blue litmus paper to red and turns acidified K2Cr2O7 from orange to green colour Inference: An acidic gas is evolved i.e SO2 gas is evolved from SO32 (2ciii) Observation: white precipitate is formed Inference: SO42, SO32 are likely present (2iv) observation: the white precipitate is soluble in dil HCL Inference: SO32 present. ========================= =============== (3ai) using lime water (3aii) by determining its boiling point and melting point (3aiii)by direct evaporation (3aiv)by sieving (3bi)addition of water, to be followed by filtration and then crystallization (3bii)in evaporation, the salt are always stable to heat. =================== COMPLETED =================== (1bii) CA=mass concentration/molar mass H2C2O4=0 0 0) gmol 0 4 40)g/mol 2 24 48g/mol =72g/mol 0.0603= mass contration/72 =4.348g/dm^3 (1biii) mass of pure= 4.35gdm^-3 mass of impure=6.30g/dm^-3 % purity= mass of pure/ mass of impure * 100% =4.35/6.30 * 100% =4.35/6.30 % % purity= 69.01% (1biv) methyl orange is not used because it is a titration between weak acid and strong base, the end point will not coincide with the PH which the indicator exhibits of colour change

Created at 2016-05-31
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