=========================
Complete NECO 2015 CHEMISTRY PRATICAL ANSWER NOW READY.
Q2
In tabular form
Tabulate test|observation|inference
1. Sample c water |dissolves to give a brown solution| c is a soluble salt.
2. solution of C litmus|turns damp blue litmus red|solution C is acidic.
3. C(aq) NaoH in drops and in excess,followed by KSCN|brown gel ppt,ppt insoluble,brown ppt turns red|Fe^3 present,Fe^3 ,Fe^3 confirmed.
4. C(aq) KsCN|blood colouration|Fe^3 confirmed.
5.C (aq) AgNo3|White ppt|Cl^-,co3^2-.
6.C(aq) HNO3|No visible reaction|Cl^- Present.
7. C(aq) NH3(aq)|white ppt dissolves to give a red brown ppt|Fe^3 confirmed.
Thursday 4th June
|
2021
|
Paper I: Practical – Chemistry
|
2hrs
|
10:00am – 12:00noon
|
Q2
In tabular form
Tabulate test|observation|inference
1. Sample c water |dissolves to give a brown solution| c is a soluble salt.
2. solution of C litmus|turns damp blue litmus red|solution C is acidic.
3. C(aq) NaoH in drops and in excess,followed by KSCN|brown gel ppt,ppt insoluble,brown ppt turns red|Fe^3 present,Fe^3 ,Fe^3 confirmed.
4. C(aq) KsCN|blood colouration|Fe^3 confirmed.
5.C (aq) AgNo3|White ppt|Cl^-,co3^2-.
6.C(aq) HNO3|No visible reaction|Cl^- Present.
7. C(aq) NH3(aq)|white ppt dissolves to give a red brown ppt|Fe^3 confirmed.
PLS IN NUMBER 2, MAKE SURE U DRAW A TABLE SHOWING: TEST | OBSERVATION | INFERENCE.
THEN PLACE THE ANSWERS WE SENT UNDER IT'S SUITING POSITION AS A CHEMISTRY STUDENT U ARE.
Goodluck!
1)
Rough : 30.00 , 0.00 , 30.00
Ist trial: 23.00 , 0.00 , 23.00
2nd trial: 23.10 , 0.00 , 23.10
3rd trial: 23.90 , 0.00 , 23.90
1) Average titre value = 23.00 23.10 23.90 / 3
= 23.00cm3
II) Vol of A used H2C2O4 = (1x2) (1x2) (16x4)
= 90g/mol
CaVa/CbVb x Ra/Rb
Ca = 0.10 x 23 x 1 / 23 x 2
= 0.0543mol/dm3
III) Percentage Purity of A
Molarity = Mass conc / molar mass
Molar mass of A = 90g/mol
= 0.0543 = Mass conc/ 90gmol
Mass conc = 0.0543 x 90 = 4.887g/mol
%purity = 4.887/6.3 x 100
= 77.57%
IV) Reason is that: Reaction between
weak acid and Strong base.
(2ai)
OBservation: salt C dissolves in distilled water
inference: Salt C is a soluble salt
(2bi)
observation:
Rusty brown precipitate is formed in excess
interference:
Fe(3+) is present
(2bii)
OBSERVATION: formation of blood-red precipitate
Inference: Fe(3+) is confirmed
(2ci)
Observation: formation of white precipitate which dissolves on heating
Inference: SO3(2-), SO4(2-) are likely to present
(2cii)
Observation:
A pungent gas is evolved, the gas turns moist blue litmus paper to red and turns acidified K2Cr2O7 from orange to green colour
Inference:
An acidic gas is evolved i.e SO2 gas is evolved from SO3(2-)
(2ciii)
Observation: white precipitate is formed
Inference: SO4(2-), SO3(2-) are likely present
(2iv)
observation: the white precipitate is soluble in dil HCL
Inference: SO3(2-) present.
(3ai)
using lime water
(3aii) by determining its boiling point and melting point
(3aiii)by direct evaporation
(3aiv)by sieving
(3bi)addition of water, to be followed by filtration and then crystallization
(3bii)in evaporation, the salt are always stable to heat.
(3c)Loss of water
=========================THEN PLACE THE ANSWERS WE SENT UNDER IT'S SUITING POSITION AS A CHEMISTRY STUDENT U ARE.
Goodluck!
1)
Rough : 30.00 , 0.00 , 30.00
Ist trial: 23.00 , 0.00 , 23.00
2nd trial: 23.10 , 0.00 , 23.10
3rd trial: 23.90 , 0.00 , 23.90
1) Average titre value = 23.00 23.10 23.90 / 3
= 23.00cm3
II) Vol of A used H2C2O4 = (1x2) (1x2) (16x4)
= 90g/mol
CaVa/CbVb x Ra/Rb
Ca = 0.10 x 23 x 1 / 23 x 2
= 0.0543mol/dm3
III) Percentage Purity of A
Molarity = Mass conc / molar mass
Molar mass of A = 90g/mol
= 0.0543 = Mass conc/ 90gmol
Mass conc = 0.0543 x 90 = 4.887g/mol
%purity = 4.887/6.3 x 100
= 77.57%
IV) Reason is that: Reaction between
weak acid and Strong base.
(2ai)
OBservation: salt C dissolves in distilled water
inference: Salt C is a soluble salt
(2bi)
observation:
Rusty brown precipitate is formed in excess
interference:
Fe(3+) is present
(2bii)
OBSERVATION: formation of blood-red precipitate
Inference: Fe(3+) is confirmed
(2ci)
Observation: formation of white precipitate which dissolves on heating
Inference: SO3(2-), SO4(2-) are likely to present
(2cii)
Observation:
A pungent gas is evolved, the gas turns moist blue litmus paper to red and turns acidified K2Cr2O7 from orange to green colour
Inference:
An acidic gas is evolved i.e SO2 gas is evolved from SO3(2-)
(2ciii)
Observation: white precipitate is formed
Inference: SO4(2-), SO3(2-) are likely present
(2iv)
observation: the white precipitate is soluble in dil HCL
Inference: SO3(2-) present.
(3ai)
using lime water
(3aii) by determining its boiling point and melting point
(3aiii)by direct evaporation
(3aiv)by sieving
(3bi)addition of water, to be followed by filtration and then crystallization
(3bii)in evaporation, the salt are always stable to heat.
(3c)Loss of water
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