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 ∝
==================
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
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
WAEC CHEMISTRY PRACTICAL ANSWERS BY WWW.SOLUTIONWAP.NET
===============
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
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
============+++++=======
2)
Tabulate
Test
(a ) ( i ) Fn + H 2O , then
filter
Observation
White residue and blue
filtrate was observed
Inference
Fn is a mixture of
soluble and insoluble
salts
Test
(ii ) Filtrate + NaOH(aq ) in
drops , then in excess
Observation
A blue gelatinous
precipitate which is
insoluble in excess
NaOH(aq ) was formed
Inference
Cu 2 + present
Test
(iii) Filtrate +NH 3 (aq ) in
drops , then in excess
Observation
A pale blue gelatinous
precipitate was
formed . The precipitate
dissolves or is soluble
in excess NH 3 (aq ) to give
a deep blue solution
Inference
Cu 2 + confirmed
Test
=======================
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=======================
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