chemistry answer is now available

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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2a)
Tabulate
Test
(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
Precipitate dissolved in excess NH3(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
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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
2×25
=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=106×0.0475gdm-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 =106×8.565
5.035×18
=10