NECO Chemistry Practical Questions and Answers2021/2022( Complete result)

 NECO Chemistry Practical Questions and Answers2021/2022( Complete result) 

 

 NECO Chemistry Practical 2021. numerous NECO campaigners find it delicate to answer the NECO Practical Chemistry Alternative A and B rightly due to not understanding the questions. I'll show you the stylish way to answer the 2021 NECO Chemistry Practical questions and make a good grade in your Chemistry examination. 

 

 It's material you understand the NECO Chemistry Practical timber scheme before trying any questions so as to present your answers in such a way that it'll look seductive to the monitor to earn you big ma


 


 Bearit in mind that every single procedure is awarded marks. So, insure you don't skip any step while reporting and presenting your practical results. 

 

 The Chemistry Practical question paper will correspond of two sections and general Chemistry questions videlicet 

 

 Quantitative Analysis 

 Qualitative Analysis 

 These involve titration and test of ions. You'll be anticipated to carry out an trial and report your conclusion and observation. What you'll see won't be far from the NECO Chemistry Specimen given to your academy. 

 

 Note Your titration endpoint will be given to you by your Chemistry schoolteacher. It isn't the same for every academy but within a range. 

 

 Don't use the endpoint then. It may vary from your center endpoint but you can follow the computation procedures. 

 



1.NECO Chemistry Specimen 2022

 2 NECO Practical Chemistry Questions and Answers( exhibit) 

 NECO Chemistry Specimen 2020 

 NECO Chemistry Specimen 2021 

 TITRATION 

 i) Adulterate sodium hydroxide result 

 ii) Adulterate ammonia result 

 iii) Adulterate hydrochloric acid 

 iv) Adulterate trioxonitrate( V) acid 

 v) Distilled water 

 vi) Red and blue litmus test paper 

 vii) Barium chloride result 

 viii) Phenolphthalein result 

 ix) Methyl orange 

 x) One boiling tube 

 xi) Five test tubes 

 xii) Source of heat 

 xiii) Wash bottle containing distilled water 

 xiv) Filtration apparatu 

 

 NECO Practical Chemistry Questions and Answers( exhibit) 

 Note The 2021 NECO Chemistry Practical answers will be posted then moment Wednesday, 14th July during the NECO Practical Chemistry test. Keep checking and reloading this runner to know when the answers are posted. Don't forget to reload this runner in order to see the answers. 

 

 NECO Practical Chemistry Answers 2022Loading.............

 

 moment’s Chemistry Practical Answers 

 

 

 

 1a) 

 

 Tabulate 

 

 Volume of pipette used = 25cm3. 

 

 Indicator used = Methyl orange 

 

 Under Titration, Rough, 1st, 2nd, 3rd, 

 

 Final burette reading;|25.00||23.40||23.50|23.60| 

 

 original burette reading;|0.00||0.00||0.00||0.00| 

 

 Volume of A used;|25.00||23.40||23.50||23.60| 

 

 1ai) 

 

 Average volume of Acid used = 23.4023.5023.60/ 3 = 70.5/ 3 = 23.50 cm3 

 

 1aii) 

 

 This is because it's a response between a weak acid and strong base 

 

 1aiii) 

 

 A channel should be used while transferring the acid into the burette. 

 

 1bi) 

 

 Conc of A in spook| dm ³ = ? 

 

 Conc of A in g| dm ³ = 12.75 g| dm ³ 

 

 Molar mass of A, NaHSO ⁴ = 23 1 32 64 

 

 = 120g| dm ³ 

 

 From. 

 

 Conc( spook/ dm ³) = conc( g/ dm ³) molar mass 

 

 Conc = 12.75/ 120 

 

 Conc = 0.106 spook/ dm ³ 

 

 1bii) 

 

 Con of B in spook/ dm ³ = ? 

 

 Ca = 0.106, Cb = ? Va = 23.50 cm ³, Vb = 25.00 cm ³ 

 

 na = 1, nb = 1 

 

 From; CaVa/ CbVb = na/ nb 

 

 Cb = CaVanb/ Vbna 

 

 Cb = 0.10623501/25.001 

 

 Cb = 2.5/ 25 

 

 Cb = 0.1 spook/ dm ³ 

 

 1biii) 

 

 Conc of B in g/ dm ³ = ? 

 

 Molar mass of B, NaOH = 23 16 1 = 40g/ spook 

 

 From; conc( g/ dm ³) = conc( spook/ dm ³) molar mass 

 

 = 0.140 

 

 = 4.0 g/ dm ³ 

 

 1biv) 

 

 Mass of swab formed = ? 

 

 Recall operative of NaHSO ⁴ = ? 

 

 From; operative = con( spook/ dm ³) * vol( cm ³) 1000 

 

 operative = 0.106 *23.50/ 1000 

 

 operative = 0.0025 spook of NaHSO ⁴ 

 

 By proportion 1mol of NaHSO ⁴ yield 1mol of Na ² SO ⁴0.0025 spook of NaHSO ⁴ will produce X spook of Na ² SO ⁴ X = 0.00251 = 0.0025 spook of NaSO ⁴ 

 

 Molar mass of Na ² SO ⁴ = ( 232) 32( 16 * 4) 

 = 46 32 64 

 

 = 142g/ spook 

 

 Recall; operative = operative/ molar mass 

 

 Mass = operative * molar mass 

 

 Mass = 0.0025 * 142 

 

 Mass = 0.36 g of Na ² SO ⁴ 

( 2a) 
 
 PLEASE PUT IN A TABULAR FORM) 
 TEST 
 
 C 5 cm ³ of distilled water 
 
 OBSERVATION 
 
 It dissolve fully 
 
 hindrance 
 
 C is a answerable swab 
 
 2bi) 
 
 TEST 
 
 result NaOH Heat 
 
 OBSERVATION 
 
 Effervescence occurs in which a colourless gas with pungent smell and turns red litmus test paper to blue is given off. 
 
 consequences 
 
 NH ³ gas from NH ⁴ is present 
 
 2bii) 
 
 TEST 
 
 Stirring rod of HCL gas given off 
 
 OBSERVATION 
 
 
 
 a gas which gives a pop sound is given off 
 
 consequences 
 
 H ² is present 
 
 2ci) 
 
 TEST 
 
 result C drops of BaCL ² 
 
 OBSERVATION 
 
 a white precipitate is formed 
 
 consequences 
 
 CO ²- 3, SO ²- 4, SO ²- 3, May be present. 
 
 2cii) 
 
 TEST 
 
 result in C( i) dil HCL 
 
 OBSERVATION 
 
 the white precipitate dissolve is dil HCL and effervescence occurs in which a colourless and odourless gas which turns blue litmus test red and lime water milky is given off 
 
 hindrance 
 
 CO ² gas from CO ² 3- verified 
 
 =================================================== 
 
 3a) 
 
 i) Ca ² 
 
 ii) Pb ² 
 
 3bi) 
 
 concentrated acid can be defined as an acid formed when a large volume of an acid dissolve in a small or little volume of water. 
 
 3bii) 
 
 A strong acid is defined as a type of acid that ionize fully in a result 
 
 3c) 
 
 Actuated watercolor is used as an adsorbent material 
 
 3d) 
 
 i) Sanguine – Brown 
 
 ii) I – red, II – brown. 
 
 — 
 
 NECO Chemistry Specimen 2020 
 CHEMISTRY PRACTICAL instance 
 
 Conditions 
 a) One burette( 50 cm ³) 
 b) One pippete( 20 cm ³/ 25 cm ³). still, all 
 campaigners in a center must use pipettes of the 
 same volume. 
 c) The usual outfit and reagents for 
 qualitative work including 
 i) Adulterate sodium hydroxide result 
 ii) Adulterate ammonia result 
 iii) Adulterate hydrochloric acid 
 iv) Barium chloride result 
 v) Distilled water 
 vi) Red and Blue litmus test paper 
 vii) Phenolphtalein 
 d) Methyl orange 
 e) One boiling tube 
 f) Five test tubes 
 g) Filtration outfit 
 h) Source of heat 
 Each seeker should be supplied with the 
 following Labelled An, Bn, Cn, where ‘ n ’ is the 
 seeker’s periodical number. 
 a) 150 cm ³ of chloride acid result in a bottle 
 labelled “ An ”. The acid result which should 
 be the same for all campaigners will contain 
 cm ³ of the concentrated hydrochloric acid 
 per dm ³ result. 
 b) 150 cm ³ of sodium hydroxide result in a 
 bottle labelled “ Bn ”. The result which should 
 be the same for all campaigners will contain 
 g of sodium hydroxide per dm ³ of result. 
 c) One spatulaful of bobby
 ( ii) 
 tetraoxosulphate( vi) swab in a instance bottle 
 labelled “ Cn ”. 
 
 The questions below are rigorously for practice not the 2021 Chemistry exhibit. 
 
 All your burette readings( initials and final) as well as the size of your pipette must be recorded but no account of experimented procedure is needed. All computations must be done in your folder. 
 
 A isO. 200 moldm3 of HCI. C is a result containing 14/ 3g of Na2CO3. x H2O in 500 cm3 of result.( a) Put A into the burette and titrate it against20.0 cm3 or25.0 cm3 portions of C using methyl orange as index. Repeat the titration to gain harmonious titre values. Tabulate your results and calculate the average volume of A used. The equation for the response is Na2CO3 x H2O 2HCL( aq) → 2NaCI( aq) CO2( g)( x 1) H2O( I). 
 b) From your results and the information handed. Calculate the( i) attention of C in moldm- 3;( ii) attention of C in gdm- 3;( iii) molar mass of Na2CO3, xH2O;( iv) the value of x in Na2CO3 xH2O.( H = 1.0; C = 12.0; O = 16.0; Na = 23.0) 
 
 Credit will be given for strict adherence to the instruction, for compliances precisely recorded and for accurate consequences. All tests, compliances and influences must be easily entered in the folder in essay at the same time they're made. 
 
 2. F is a admixture to two inorganic mariners. Carry out the following exercise onF. record your observation and identify any gas( s) evolved. State the conclusions you draw from the result of each test.( a) Put all of F in a teacup and add about 10cm3 of distilled water. Stir well and sludge. Keep the filtrate and the residue.( b)( i) To about 2cm3 of the filtrate, add NaOH( aq) in drops and also in redundant( ii) To another 2cm3 portion of the result, add a many drops of NHO3( aq) followed by many drops of AgNO3( aq).( d)( i) Put all the residue into a clean test- tube and add NHO3( aq) followed by many result from 2( d)( i) add NaOH( aq) in drops and also in excess. 

 3. State what would be observed if the following responses are carried out in the laboratory( i) methyl orange is dropped into a result of lime juice( ii) hydrogen sulphide gas is gurgled through iron( III) chloride result;( iii) sulphur( IV) oxide gas is gurgled into acidified result of KMnO4;( iv) ethanoic acid is added to a result of K2CO3. 
 
 CHEMISTRY PRACTICALS ANSWERS 
 
 Indicator = Methyl Orange 
 Volume of the base used = 25.00 cm3 
 
 Titration Rough Titre 1st Titre 2nd Titre 3rd Titre 
 Final Burette readings cm324.7024.8024.7024.90 
 original Burette reading cm30.000.000.000.00 
 Volume of acid used cm324.7024.8024.9024.90 
Average Titre = 1st 2nd 3rd/ 3 
 
 = 24.8024.7024.90/ 3 
 
 = 24.80 cm3 
 
 Alternately 2 accordant titres can be used to calculate average titre. 
 
 Equation of the response Na2CO3 XH2O 2HCI( aq) → 2NacI( aq) CO2( aq)( x 1) H2O( I) 
 
 CAVA/ CBVB = nA/ nB 
 
 CA = Molar attention of HCI( aq) in moldm3 
 
 VA = Volume of acid used in cm3 = 24.80 cm3 
 
 CB = Molar attention of Na2Cu3 xH2O in moldm3 
 
 nA = 2 
 
 nB = 1 
 
 b)( i) attention of C in moldm- 3 = ? From the equation of response 
 
 CAVA/ CBVB = nA/ nB 
 
 CA = 0.200 moldm3 VA = 24.80 cm3 
 
 CB = ? VB = 25.00 cm3 
 
 Negotiation of known values 
 
 x24.80/ CB x25.00 = 2/1 
 
 CB = 1 ×0.200 ×24.80/ 2 ×25.00 
 
 CB­ = 0.0992 moldm- 3 
 
 ii) attention of C in g dm- 3 = ? 
 
 500cm2 →14.3 g 
 
 1000cm3 →14.3/ 500 x 10002 g 
 
 = 28.6 g dm- 3 
 
 iii) Molar mass of Na2CO3 xH2O 
 
 Molar conc in moldm- 3 = conc in g dm- 3/ molar mass 
 
 thus Molar mass g spook- 1 = conc in g dm- 3/ molar conc in moldm- 3 
 
 = 28.6 g dm-3/0.0992 moldm- 3 
 
 = 288.3065 
 
 roughly 288 g spook- 1 
 
 iv) Value of x in Na2CO3 xH2O? 
 
 H = 1.0, C = 12.0, O = 16.0, Na = 23.0) 
 Na2CO3 xH2O = 288 
 
 2( 23) 12 3( 16) x( 2( I) 16) = 288 
 
 46 12 48 18x = 288 
 
 106 18x = 288 
18x = 288 
 
 18x = 182 
 
 x = 182/18 
 
 x = 10.11 
 
 x roughly 10 
 
 2 Test Observation Inference 
 a) F distil water stir sludge Effervescence( washing) occurs, colourless and odourless gas evolved after filtration blench blue/ colourless filtrate and green residue 
( b)( i) Filtrate NaOH( aq) in drops also in redundant White precipitate formed Precipitate undoable in redundant NaOH( aq) Ca2/ Pb2 Ca2 present 
 b)( ii) Filtrate NH3( aq) in drops also in redundant No visible response No visible response OR Ca2 present Ca2 verified 
 b)( i) Filtrate NaOH( aq) in drops also in redundant Blue glutinous precipitate formed Precipitate undoable in redundant Cu2 is present 
 b)( ii) Filtrate NH3( aq) in drops also in redundant Pale( light) blue glutinous precipitate formed Precipitate dissolves to form a deep blue result Cu2 Cu2 is verified 
 c) Filtrate NHO3( aq) AgNO3( aq) No visible response No gas involved White precipitate formed CuI is present 
 d)( i) Residue HNO3( aq) Efferveness occurs, colourless and odourless gas evolves that turn blue litmus test paper to red and turns lime water milky CO2( g) CO32-present 
( d)( ii) result from 2( d)( i) NaOH( aq) in drops also in redundant Blue precipitate formed Precipitate undoable in redundant NaOH( aq) Cu2 Cu2 present 
( ii) result from 2( d)( i) NaOH( aq) in drops also in redundant Blue precipitate formed Precipitate undoable in redundant NaOH( aq) Cu2 Cu2 present 
 3( a)( i) result of lime juice turns to pink or red.( ii) The FeCI3 result changes colour from brown to green and a unheroic deposit.( iii) The grandiloquent colour of the KMnO4 result turns colourless or decolourised.( iv) The result of K2CO3 reacts with the addition of ethanoic acid to evolve a colourless, odourless gas with washing of effervescence. 
 

 still, kindly let us know in the comment box, If you have any questions about the 2021 NECO Chemistry Practical questions and answers. 


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