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chemistry practical answers

1)

Tabulate

Burette reading | Final burette reading

( cm^3 )|Initial burette reading ( cm^3)|

Volume of acid used ( cm^3)|

Rough - 24. 10, 0.00 , 24.10

First- 23 .80, 0.00 , 23. 80

Second - 23. 75, 0. 00, 23 .70

Third - 23.75 , 0. 00, 23 .75

Average volume of A used = 23 .80 + 23.70 +

23 .75cm ^3/3

=23 .75 cm^3

1bi)

CAVA /CcVc =2 /1

Cc =CAVA /2VC

=0 .100 *23 .75Moldm ^-3 /2*25 .00

=0 .0475 moldm ^- 3

amount of A used = 0.100 x

VA/1000 =0. 100* 23. 75/1000 = 0.00237

2moles Of A = 1mole of C

0. 002375 mol of A = 0.002375 mol/2

100 cm^3 of C contain 0. 00237 *100 mol/2 *25

=0

1000 cm3ofCcontained 0.002375 x 1000 mol

2x 25

=0 .0475 mol

concentration of C in moldm - 3

=0 .0475 moldm - 3

1bii )

Molar mass of Bing mol- 1:

Molar mass of Na 2CO 3. yH2O = mass

concentration of Bingdm- 3

molar concentration of Binmoldm - 3

=13 .6gdm - 3

0. 0475 moldm - 3

=286 gmol -1

1biii )

Molar mass of Na 2CO 3 =

[( 2×23) + 12+( 16 ×3)]=106 gmol- 1

Mass of anhydrous

Na 2CO 3= 106x 0 .0475 gdm -3

=5 .035 gdm -3

Mass of water =13. 6- 5.035 gdm -3

=8 .565 gdm -3

Mass of Na2CO 3 =Molar mass of Na 2CO 3

Mass of water y × Molar mass of water

5. 035= 106

8. 565 18y

y =106 x 8.565

5. 035x 18

=10

: ===============

NOTE THAT ^ MEANS Raise to

power.

Pls Draw Your Table As Usual And

Input The Following:-

Volume of pipette=25.00cm^3

indicator used- Methyl orange

colour change at end point-yellow to

orange/purple

Note Use Your School End Point.

Tabulate

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

1)

Tabulate

Burette reading|Final burette reading

(cm^3)|Initial burette reading (cm^3)|

Volume of acid used(cm^3)|

Rough- 24.10,0.00,24.10

First- 23.80, 0.00, 23.80

Second- 23.75, 0.00, 23.70

Third- 23.75, 0.00, 23.75

Average volume of A used = 23.80 +

23.70

+ 23.75cm^3/3

=23.75cm^3

1bi)

CAVA/CcVc=2/1

Cc=CAVA/2VC

=0.100*23.75Moldm^-3/2*25.00

=0.0475moldm^-3

amount of A used = 0.100x

VA/1000=0.100*23.75/1000 =0.00237

2moles Of A = 1mole of C

0.002375mol of A = 0.002375mol/2

100cm^3 of C contain

0.00237*100mol/2*25 =0

 (a)(i)Fn+H2O,then

filter

  

White residue and blue

filtrate was observed

Fn is a mixture of

soluble and insoluble

salts

(ii)Filtrate+NaOH(aq)in

drops,then in excess

A Blue gelatinous

precipitate which is

insoluble in excess

NaOH(aq)was formed

Cu2+present

(iii)Filtrate+NH3(aq)in

drops,then in excess

A pale blue gelatinous

precipitate was

formed.The precipitate

dissolves or is soluble

in excess NH3(aq)to give

a deep blue solution

Cu2+confirmed

(iv)Filtrate+dil.HNO3

+AgNO3(aq)

+NH3(aq)in excess

No Visible Reaction

White Precipitate

formed

Precipitate dissolved in

excess NH3(aq)

Cl-present

Cl-confirmed

(b)(i)First Portion Of

residue+NaOH(aq)in

drops,then in excess

White Powdery

precipitate which is

insoluble in excess

NaOH(aq)

Ca2+present

(ii)Second Portion Of

residue+dil.HCl

Effervescence/bubbles;

colourless,odourless

gas evolved.Gasturns

lime water milky and

turns damp blue litmus

paper red.

GasisCO2

CO3

2-orHCO3

-present

 NOTE:

Tita means θ

^ means Raise to power

Pie means π

 / (means) division or divide

2 whole no 3/4 means 2¾

* means

multiplication (×)

Sqr root means √

Proportional means ∝


(3i)
The colour of methyl orange will change from orange to yellow

(3ii)
Iron(III)chloride will be reduced from brown iron to green iron
Fe^3+(aq)->Fe^2+(aq)

(3iii)
The color of KM2O4 is decolorized because SO2(g) acts as a reducing agent.

(3iv)
Addition of ethanoic acid to KCO3 results to the liberation of a colorless and odorless gas CO2 which turns lime water milky