NECO Chemistry Practical Questions and Answers 2021/2022 (Complete Solution)

NECO Chemistry Practical 2021. Many NECO candidates find it difficult to answer the NECO Practical Chemistry Alternative A and B correctly due to not understanding the questions. I will show you the best way to answer the 2021 NECO Chemistry Practical questions and make a good grade in your Chemistry examination.

NECO Chemistry Practical Questions and Answers 2021/2022 (Complete Solution)

It is relevant you comprehend the NECO Chemistry Practical making plan prior to attempting any question in order to present your answers so that it will look appealing to the inspector to acquire you enormous Marks.

Remember it that each and every single procedure is granted marks. In this way, Be sure you don’t skip any progression while revealing and introducing your reasonable results.

The Chemistry Practical question paper will comprise of two segments and general Chemistry questions to be specific:

1. Quantitative Analysis

2. Qualitative Analysis

These include titration and trial of particles. You will be relied upon to complete a test and report your deduction and perception. What you will see won’t be a long way from the NECO Chemistry Specimen given to your school.

 

Note: Your titration endpoint will be given to you by your Chemistry instructor. It isn’t something similar for each school yet within a range.

Try not to use the endpoint here. It might fluctuate from your middle endpoint yet you can follow the computation methodology.

See Also:

NECO Chemistry Specimen 2021

TITRATION
(i) Dilute sodium hydroxide solution
(ii) Dilute ammonia solution
(iii) Dilute hydrochloric acid
(iv) Dilute trioxonitrate(V)acid
(v) Distilled water
(vi) Red and blue litmus paper
(vii) Barium chloride solution
(viii) Phenolphthalein solution
(ix) Methyl orange
(x) One boiling tube
(xi) Five test tubes
(xii) Source of heat
(xiii) Wash bottle containing distilled water
(xiv) Filtration apparatus

NECO Practical Chemistry Questions and Answers (Expo)

Note: The 2021 NECO Chemistry Practical answers will be posted here today Wednesday, 14th July during the NECO Practical Chemistry exam. Keep checking and reloading this page to know when the answers are posted. Do not forget to reload this page in order to see the answers.

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Today’s Chemistry Practical Answers:

(1a)

Tabulate

Volume of pipette used=25cm3.

Marker used= Methyl orange

Under Titration, Rough, first, second, third,

Last burette perusing; |25.00| |23.40| |23.50| 23.60|

Starting burette perusing; |0.00| |0.00| |0.00| |0.00|

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

(1ai)

Normal volume of Acid used=23.40+23.50+23.60/3 =70.5/3=23.50cm3

(1aii)

This is on the grounds that it is a response between a feeble corrosive and solid base

(1aiii)

A pipe ought to be used while moving the corrosive into the burette.

(1bi)

Conc of An in mol|dm³ =?

Conc of An in g|dm³ = 12.75g|dm³

Molar mass of A, NaHSO⁴ = 23+1+32+64

=120g|dm³

From.

Conc(mol/dm³) = conc(g/dm³)/molar mass

Conc =12.75/120

Conc =0.106mol/dm³

(1bii)

Con of B in mol/dm³ =?

Ca =0.106, Cb =? Va =23.50cm³, Vb =25.00cm³

na=1 , nb=1

From; CaVa/CbVb=na/nb

Cb=CaVanb/Vbna

Cb=0.10623501/25.001

Cb=2.5/25

Cb=0.1mol/dm³

(1biii)

Conc of B in g/dm³=?

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

From; conc(g/dm³)=conc(mol/dm³)molar mass

=0.140

=4.0g/dm³

(1biv)

Mass of salt framed =?

Review mole of NaHSO⁴=?

From; mole =con(mol/dm³)*vol(cm³)/1000

Mole=0.106*23.50/1000

Mole=0.0025mol of NaHSO⁴

By extent 1mol of NaHSO⁴ produce 1mol of Na²SO⁴ 0.0025mol of NaHSO⁴ will create X mol of Na²SO⁴ X=0.00251=0.0025mol of NaSO⁴

Molar mass of Na²SO⁴=(232) + 32 + (16*4)

=46+32+64

=142g/mol

Review; mole =mole/molar mass

Mass =mole * molar mass

Mass =0.0025*142

Mass =0.36g of Na²SO⁴

===================================================

(2a)

[If it’s not too much trouble, PUT IN A TABULAR FORM]
TEST

C + 5cm³ of refined water

Perception

It break up totally

Obstruction

C is a dissolvable salt

(2bi)

TEST

Arrangement + NaOH + Heat

Perception

Fizz happens in which a dull gas with sharp smell and becomes red litmus paper to blue is radiated.

Inductions

NH³ gas from NH⁴+ is available

(2bii)

TEST

Mixing pole of HCL + gas radiated

Perception

a gas which gives a pop solid is emitted

Surmisings

H² is available

(2ci)

TEST

Arrangement C + drops of BaCL²

Perception

a white accelerate is framed

Deductions

CO²^-3, SO²^-4, SO²^-3, May be available.

(2cii)

TEST

Arrangement in C(i) + dil HCL

Perception

the white accelerate break up is dil HCL and fizz happens in which a boring and unscented gas which becomes blue litmus red and lime water smooth is radiated

Obstruction

CO² gas from CO²^3-affirmed

===================================================

(3a)

(i)Ca²+

(ii)Pb²+

(3bi)

concentrated corrosive can be characterized as a corrosive framed when a huge amount of a corrosive break up in a little or little volume of water.

(3bii)

A solid corrosive is characterized as a kind of corrosive that ionize totally in an answer

(3c)

Actuated charcoal is utilized as an adsorbent material

(3d)

(i)Reddish – Brown

(ii) I – red, II – brown.

—————-

NECO Chemistry Specimen 2020

CHEMISTRY PRACTICAL SPECIMEN:

Prerequisites:
(a) One burette (50cm³)
(b) One pippete (20cm³/25cm³). Nonetheless, all
competitors in a middle should utilize pipettes of the
same volume.
(c) The standard mechanical assembly and reagents for
subjective work including:
(I) Dilute sodium hydroxide arrangement
(ii) Dilute smelling salts arrangement
(iii) Dilute hydrochloric corrosive
(iv) Barium chloride arrangement
(v) Distilled water
(vi) Red and Blue litmus paper
(vii) Phenolphtalein
(d) Methyl orange
(e) One bubbling cylinder
(f) Five test tubes
(g) Filtration device
(h) Source of hotness
Every up-and-comer ought to be provided with the
following: Labeled An, Bn, Cn, where ‘n’ is the
applicant’s chronic number.
(a) 150cm³ of chloride corrosive arrangement in a container
named “An”. The corrosive arrangement which ought to
be no different for all up-and-comers will contain
3.4cm³ of the concentrated hydrochloric corrosive
per dm³ arrangement.
(b) 150cm³ of sodium hydroxide arrangement in a
bottle marked “Bn”. The arrangement which ought to
be no different for all applicants will contain
4.0g of sodium hydroxide per dm³ of arrangement.
(c) One spatulaful of copper (ii)
tetraoxosulphate (vi) salt in an example bottle
named “Cn”.

The inquiries underneath are rigorously for training not the 2021 Chemistry exhibition.

All your burette readings (initials and last) as well as the size of your pipette should be recorded yet no record of tested methodology is required. All computations should be done in your booklet.

An is O.200 moldm3 of HCI. C is an answer containing 14/3g of Na2CO3. x H2O in 500 cm3 of arrangement. (a) Put An into the burette and titrate it against 20.0 cm3 or 25.0 cm3 parts of C involving methyl orange as marker. Rehash the titration to acquire Consistent titre values. Organize your outcomes and work out the normal volume of A utilized. The condition for the response is: Na2CO3 x H2O + 2HCL(aq) → 2NaCI(aq) + CO2(g) + (x+1) H2O(I).
(b) From your outcomes and the data gave. Work out the: (I) convergence of C in moldm-3; (ii) grouping of C in gdm-3; (iii) molar mass of Na2CO3, xH2O; (iv) the worth of x in Na2CO3 xH2O. [H = 1.0; C=12.0; O = 16.0; Na =23.0]

Credit will be given for severe adherence to the guidance, for perceptions unequivocally recorded and for exact deductions. All tests, perceptions and impacts should be plainly placed in the booklet in ink simultaneously they are made.

2. F is a combination to two inorganic salts. Do the accompanying activity on F. record your perception and distinguish any gas(s) advanced. Express the determinations you make from the aftereffect of each test. (a) Put all of F in a measuring utencil and add around 10cm3 of refined water. Mix well and channel. Keep the filtrate and the buildup. (b)(i) To around 2cm3 of the filtrate, add NaOH(aq) in drops and afterward in overabundance (ii) To one more 2cm3 part of the arrangement, add a couple of drops of NHO3(aq)followed by couple of drops of AgNO3(aq). (d)(i) Put all the buildup into a spotless test-cylinder and add NHO3(aq) trailed by barely any arrangement from 2(d)(i) add NaOH(aq) in drops and afterward in abundance.

3. State what might be noticed in the event that the accompanying responses are done in the research facility: (I) methyl orange is dropped into an answer of lime juice (ii) hydrogen sulfide gas is risen through iron (III) chloride arrangement; (iii) sulfur (IV) oxide gas is risen into fermented arrangement of KMnO4; (iv) ethanoic corrosive is added to an answer of K2CO3.

CHEMISTRY PRACTICALS ANSWERS

  1. Indicator                                 =          Methyl Orange

Volume of the base used        =          25.00cm3

Titration Rough Titre 1st Titre 2nd Titre 3rd Titre
Final Burette readings cm3 24.70 24.80 24.70 24.90
Initial Burette reading cm3 0.00 0.00 0.00 0.00
Volume of acid used cm3 24.70 24.80 24.90 24.90

Average Titre              =          1st + 2nd + 3rd/3

=          24.80 + 24.70 + 24.90/3

=          24.80cm3

On the other hand 2 concordant titres can be utilized to work out normal titre.

Condition of the response: Na2CO3 XH2O + 2HCI(aq) → 2NacI(aq) + CO2(aq) + (x+1)H2O(I)

CAVA/CBVB = nA/nB

CA = Molar centralization of HCI(aq) in moldm3

VA = Volume of corrosive utilized in cm3 = 24.80cm3

CB = Molar centralization of Na2Cu3 xH2O in moldm3

nA = 2

nB = 1

(b)(i) centralization of C in moldm-3 = ? From the situation of response:

CAVA/CBVB = nA/nB

CA = 0.200 moldm3 VA = 24.80cm3

CB = ? VB = 25.00cm3

Replacement of known values

0.200 x 24.80/CB x 25.00 = 2/1

CB = 1×0.200×24.80/2×25.00

CB­ = 0.0992 moldm-3

(ii) Concentration 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

Thusly: Molar mass g mol-1 = conc in g dm-3/molar conc in moldm-3

= 28.6 g dm-3/0.0992 moldm-3

= 288.3065

Around 288 g mol-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 around 10

2 Test Observation Inference
(a) F + distil water + stir + filter Effervescence (bubbling) occurs, colourless and odourless gas evolved after filtration pale blue/colourless filtrate and green residue
(b)(i) Filtrate + NaOH(aq) in drops then in excess White precipitate formed   Precipitate insoluble in excess NaOH(aq) Ca2+/Pb2+   Ca2+ present
(b)(ii) Filtrate + NH3(aq) in drops then in excess No visible reaction   No visible reaction OR Ca2+ present   Ca2+ confirmed
(b)(i) Filtrate + NaOH(aq) in drops then in excess Blue gelatinous precipitate formed   Precipitate insoluble in excess     Cu2+  is present
(b)(ii) Filtrate + NH3(aq) in drops then in excess Pale (light) blue gelatinous precipitate formed   Precipitate dissolves to form a deep blue solution Cu2+     Cu2+  is confirmed
(c) Filtrate + NHO3(aq) + AgNO3(aq) No visible reaction 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 paper to red and turns lime water milky CO2(g) CO32- present
(d)(ii) Solution from 2(d)(i) + NaOH(aq) in drops then in excess Blue precipitate formed   Precipitate insoluble in excess NaOH(aq) Cu2+    Cu2+   present

3 (a)(i) Solution of lime juice turns to pink or red. (ii) The FeCI3 arrangement changes tone from brown to green and a yellow store. (iii) The purple shade of the KMnO4 arrangement turns dull or decolourised. (iv) The arrangement of K2CO3 responds with the expansion of ethanoic corrosive to advance a dreary, scentless gas with rising of fizz.

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