Chemistry 3 (Practical Alternative A) 09.30am-11.30am
Answer..... 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
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(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 Precipitate dissolved in excessNH3(aq)
Inference Cl-confirmed
2b)
(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
===========================
3i) Lime juice is cacitic in nature and the color of methyl orange in active medium is Red
3ii) Iron(iii)chloride will be reduced to iron(ii) with the yellow deposit of Sulphur
3iv)Addition of ethanoic acid to k2lo3 results to the liberation of a colourless or odourless gas co2 Which turns lime water milky
======================
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Answer..... 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
================================
(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 Precipitate dissolved in excessNH3(aq)
Inference Cl-confirmed
2b)
(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
===========================
3i) Lime juice is cacitic in nature and the color of methyl orange in active medium is Red
3ii) Iron(iii)chloride will be reduced to iron(ii) with the yellow deposit of Sulphur
3iv)Addition of ethanoic acid to k2lo3 results to the liberation of a colourless or odourless gas co2 Which turns lime water milky
======================
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