Chemistry 3 (Practical) Answers.
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
1000cm3ofCcontained0.002375x1000mol
2x25
=0.0475mol
concentration of C in moldm-3 =0.0475moldm-3
1bii)
Molar mass of Bing mol-1:
Molar mass of Na2CO3.yH2O=mass concentration of Bingdm-3
molar concentration of Binmoldm-3
=13.6gdm-3
0.0475moldm-3
=286gmol-1
1biii)
Molar mass of Na2CO3 =[(2×23) 12 (16×3)]=106gmol-1
Mass of anhydrous Na2CO3=106x0.0475gdm-3
=5.035gdm-3
Mass of water=13.6-5.035gdm-3
=8.565gdm-3
Mass of Na2CO3 =Molar mass of Na2CO3
Mass of water y×Molar mass of water
5.035=106
8.565 18y
y =106x8.565
5.035x18
=10
======================
(2)Tabulate
Test
(a)(i)Fn H2O,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
Cu2 present
Test
(iii)Filtrate NH3(aq)in
drops,then in excess
Observation
A pale blue gelatinous
precipitate was
formed. The precipitate
dissolves or is soluble
in excess NH3(aq)to give
a deep blue solution
Inference
Cu2 confirmed
Test
(iv)Filtrate dil.HNO3
AgNO3(aq)
Observation
No visible reaction
White precipitate
formed
Inference
Cl-present
Test
NH3(aq)in excess
Observation
Precipitatedissolvedin
excessNH3(aq)
Inference
Cl-confirmed
2bi)First portion of residue NaOH(aq) in drops, then in excess
Observation
White powdery precipitate which is insoluble in excess NaOH(aq)
Inference
Ca2 present
Test
2bii)Second portion of residue dil.HCl
Observation
Effervescence/bubbles; colourless, odourless gas evolved.
Gas turns lime water milky and turns damp blue litmus paper red.
Inference
Gas is CO2
CO3^2- or HCO3- present
======================
3i)
lime juice is acidic in nature and the colour of methyl orange in acidic medium is red
3ii)
Iron(iii)chloride will be reduced to iro(ii) with yellow deposit of sulphur
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
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DE CAMPUS===
UNDER MAINTENANCE
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