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WAEC 2016 CHEMISTRY QUESTIONS ANDANSWERS (Obj/Theory) NOW AVAILABLE FOR NOW.

WAEC 2016 CHEMISTRY QUESTIONS ANDANSWERS (Obj/Theory) NOW AVAILABLE FOR NOW.
4/18/2016
Chem-Theory-Answers no3d picture will be uploaded soon 3ai) structral isomerism is the existance of two or more compounds ( known as isomers) with the same molecular formula but different molecular structures 3ai) i) tertiary alkanol ii) secondary alkanol iii) primary alkanol 3bi) i) 2[CH3 CH2 COOH ] 2K(s)----> ii) [CH3 CH2 COO ] K + H2 i) 2[CH3 CH2 COOH ] +2 K(s) ------> 2[ CH3 CH2 COO ] K + H2 ii) CH3 CH2 COOH + C4 H9OH HEAT---->CH3 CH2 COO C4 H9 + H20 H2SO4 iii)CH3 CH2 CH2 CH20H + H^+ KMnO4------> CH3 CH2 CH2 CH2 - 0H EXCESS 3bii) i) potassium ethanoate,hydrogen ii) butyl propanoate and water iii) enthanal 3c) the percentage composition of the hydrocarbon = 100 H + C = 100 14.3 + C =100 C= 100 - 14.3 C= 85.7% ================================== 1a) nucleons is the collective name for two important sub-atomic particles:neutrons and protons 1b) Graham's law of diffusion states that at a constant temperature and presure,the rate of diffusion of a gas is inversely proportional to the square root of it density 1c) because when aluminum is exposed to moist air,a thin continuous coating of aluminum oxide is formed, which prevents futher attack of the aluminium by atmospheric oxygen and water or steam under normal conditions 1d) i) electron - affinity ii) electron - negativity iii) ionization - energy 1e) i) the molecule of real gases occupies space and there are forces of the attraction between them ii) real gases do liquefy when their temperature droops. 1f) i)used in seperation of different component of crude oil. ii)it is used for separating acetonic from water iii) it is used for obtaining different gases from air for industrial use. 1g) i)concentration of ions in electrolyte ii)the nature of the electrode iii)the position of ions in the electrochemical series 1h) i)substitution reaction ii)addition reaction 1i) i)carbon dioxide ii)water vapour iii) carbon monoxide 1j) i)centrifugation ii)sieving iii)evaporation to dryness ================================== 2a) i) isotopy ii) they have the same number of proton (atomic number) but different mass number iii) oxygen,carbon iv)1s^2,2s^2,2p^6,3s^2,3p^5 2bi) PHYSICAL PROPERTIES METAL i) high melting and boiling point ii) they are good conductor of heat and electricity NON-MATEL i)they have lower melting and boiling point ii)they are poor conductors of heat and electricity. CHEMICAL PROPERTIES METAL i) they form basic and ampheric oxides ii)the react by electron lose or donation of electron NON-METAL i)the form acidic oxides ii)the react by sharing and accepting electrons 2bii) i) Al2 O3(Aluminum oxide) ii) sodium hydride (NaH) iii)zinc trioxocarbonate(iv)ZnCo3 iv)silicon tetrachloride (SiCl4) 2ci) i) variable oxidation states ii) complex ion formation iii) they possess strong metallic bonding 2cii) 1s^2,2s^2, 2p^6, 3s^2, 3p^6, 3d^10, 4s^2 2ciii) Zinc is not considered as a typical transition element bacause it has only one oxidation state of +2, it metallic ion are not coloured and it is not used as a catalyst. 2d) Na2Co3 + MgCl2 ----> 2NaCl + MgCo3 1mole 1mole 2mole 1mole molar mass of Na2 Co3 = 2 * 23 + 12 * 1 + 16 * 3 = 46+12+48 = 106g 106g of Na2Co3 produced 84g of MgCo3 ; 106g = 84g xg= 3.36g 84 * x = 106 * 3.36 x = 106 * 3.36/84 x=356.16/84 x=4.24g 4.24g of Na2Co3 is needed to produce 3.36g of MgCo3 ================================== 5ai) Na2S2O3 ===> 2 + 2x + 6 = 0 2x = 6 - 2 = 4 X = 4/2 X = 2 The oxidation number of sulphur is 2 5aii) Rhombic & monoclinic 5aiii) - Both are tetravalent - both are allotrope of carbon 5bi) CO2 & Chloroflorocarbon 5bii) There is increase in sun radiation reaching the earths surface ie Global warming 5biii) ThunderStorm 5iv) I2KNo3. --------> 2KNO3 + O2 2AgNo3 --------->. 2Ag + 2No 5ci) Calcium chloride in a solution can give rise to crystal using filteration & evaporation to dryness. The sol is filtered into filtrate & residue b4 evaporation to dryness takes place. 5cii) - Because of presence of hydrogen bonding in NH3 - Because Iodine as higher molecular mass than chlorine 5di) Mol of Nacl = Mass / MM = 5.85/ 58.01 = 0.1mol From the equation 2mol of Nacl gives 2mol of HCL, 0.1mol of Nacl gives 0.1mol of Hcl Vol of Hcl = 0.

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