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*TAKE NOTE* For question 1 Use your school's endpoint and use it wherever you see our endpoint (13.20cm³) ...and calculate with it. Use the answer you got in 1b(i) to solve in 1b(ii) i.e we got 0.0396 in 1bi after calculating with our average titre (endpoint) value 0.0396, as u can see, we then used the 0.0396 in 1bii which we got from our calculation in 1bi. *NB:* If you use our endpoint and your school teacher submits a different one with a wide margin to what u used, you will either fail or lose some mark in this particular question. Your school teacher should have in one way given you the average titre value(endpoint) or you might have gotten it in the lab from your past experiment. If your school wants to use mine, then the chemistry teacher must submit endpoint similar to ours without much difference...eg 13.10, 13.20, 13.30, etc

CHEMISTRY- PRACTICAL-ANSWERS (1) Tabulate Titration; | rough | 1 | 2 | 3 | Final; | 20.50 | 14.00 |14.00 | 14.00 | Initial; | 0.00 | 0.00 | 0.00 | 0.00 | Volume of A used; | 20.50 | 14.00 |14.00 | 14.00 | Volume of pipette used =25.00cm³ (1a) Average volume of A used = 14.00+14.00+14.00/3 =14.00cm³ (Va) 2HNO³ + X²CO³ ---> 2XNO³ + H²O + CO² (1bi) Conc - of B in molldm³ = ? Given; Ca= 0.15molldm³ Vb= 25.00cm³ Va= 14.00cm³ Cb= ?? Na= 2 Nb= 1 Using; CaVa/CbVb=Na/Nb Cb= CaVaNb/VbNa Cb= 0.15*14.00*1/25.00*2 Cb= 2.1/50 Cb= 0.042molldm³. (1bii) Molar mass of B =? Using; Conc(molldm³)= Conc(gldm³)/molar mass 0.042/1 * 4.20/m.m m.m=4.20/0.042 m.m=100glmol (1biii) Value of x in x²(i)³=? (2x)+12 + (16*3)= 100 2x+12+48=100 2x+60=100 2x=100-60=40 2x/2=40/2 x=20g/mol (1ci) (i)I ensure that the readings are taken at the lower meniscus (ii)I ensure that my saliva did not enter the pipette while taking the base (1cii) Pink ============================== (2) Tabulate (2ai) Under Observation; An effervescence occurs in which a colorless and odourless gas which burns blue litmus red is evolved. Under Inference; CO² gas from CO3^²is suspected (2aii) Under Observation; The gas turns lime water milky. Under Inference; CO² gas is confirmed. (2b) Under Observation; Y dissolves completely in water Under Inference; Soluble salt (2bi) Under Observation; White gelatinous PPT is formed. Under Inference; Al³+, Zn²+, Pb²+, present (2bii) Under Observation; The white gelatinous PPT dissolve in excess NaOH Under Inference; Al³+, Zn²+, Pb²+, present (2biii) Under Observation; White gelatinous PPT formed. Under Inference; Al³+, Zn²+, Pb²+, present (2biv) Under Observation; The white gelatinous PPT remain insoluble in excess NH³ Under Inference; Al³+, Pb²+, present. ============================== (3ai) It increases the boiling point of water (3aii) It requires heat to occur. (3bi) This is because it is deliquescent. (3bii) I. Calcium oxide II. Anhydrous copper(ii) tetraoxosulphate(vi) (3ci) Hydrogen Chloride (3cii) I. To separate a mixture of immiscible liquids II. To collect gases. ============================== _*SOLVED BY EAGLE EYE CEO THE BEST EXAMS ASSISTANCE*_ _*WHATSAPP US @+2348023343546*_

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*ONDO BECE SOLUTION* (2a) moves along a straight line Linear motion means that an object moves along a straight line. (2b) {PICK ANY 2} • Wheels of a Moving Vehicle. • Fan Blades. • Helicopter Rotor Blades. • Spinning Top. • Ferris Wheel. • Gears. • Clock Ticking (2ci) Adhesive, also known as glue, cement, mucilage, or paste, is any non-metallic substance applied to one or both surfaces of two separate items that binds them together and resists their separation. (2cii) An abrasive is a material, often a mineral, that is used to shape or finish a workpiece through rubbing which leads to part of the workpiece being worn away by friction. (2d) Alloy is a metallic substance composed of two or more elements, as either a compound or a solution. _*SOLVED BY EAGLE EYE CEO THE BEST EXAMS ASSISTANCE*_ _*WHATSAPP US @+2348023343546*_ ➖

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