Neco 2017 Chemistry Practical Answers – June/July Expo
Neco 2017 Chemistry Practical Answers – June/July Expo
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2017 NECO CHEMISTRY PRACTICAL SOLUTION
NO1)
*Note:*
Candidates are strongly advised to check which size of pipette was used by their school.
We used 25cm³
-------------------------------
If your school used 20cm³, all you need do is use it in place of 25cm³ anywhere we used it and calculate accordingly
*NOTE:ALL MAKE USE OF YOUR SCHOOL ENDS POINT*
Volume of Pipette used 25.0cm^3
Titration | Rough | First | Second | Third
Final burette reading (cm^3) | 22.50 | 45.10 | 32.65 | 27.60
Intial burette reading (cm^3) | 0.00 | 22.50 | 10.00 | 5.00
Volume of A used (cm^3) | 22.50 | 22.60 | 22.65 | 22.70
1a) Average volume of A = 22.60 + 22.65 + 22.70/3
= 22.65cm^3
1bi) GIVEN: CB = 0.10moldm^-3
VB = 25cm^3
CA = ?
VA = 22.65cm^3
Equation of reaction:
H2C2O4(aq) + 2NaOH(aq) ====> Na2C2O4(aq) + 2H2O(l)
hence, nA = 1
Nb = 2
Using CAVA /CBVB = nA/nB
CA = CBVBnA/VanB
=0.10 × 25 × 1/22.65 × 2
CA = 0.055 moldm^-3
Concentration of A in moldm^-3 = 0.055
bii) Using:
gram concentration = molar concentration × molar mass
conc. Of A in gdm^-3 = 0.055 × [(1 × 2) + (12 × 2) + (16 × 4)]
= 0.055 × 90
= 4.95gdm^-3
======================
No2 solution
2a)
*TEST*
X + 5cm of distilled water, Divide the solution into 4 portion
*OBSERVATION*
Salt X is soluble in distilled water
*INFERENCE*
Salt X is soluble salt
=======================
2bi)
*Test*
-To the first portion, Add dilute NaOH in drop
-then in excess
*OBSERVATION*
-White precipitate solution is formed
-The white precipitate is soluble
*INFERENCE*
Zn^2+ Pb^2+ and AL^3+ are present
=====================
2bii)
*TEST*
-To the 2nd portion add dilute NH3 solution in drop
-then in excess
*OBSERVATION*
White gelatinous precipitate is formed which is insoluble in excesa
*INFERENCE*
Zn^2+,Pb^2+ and Al^3+ are present
======================
2biii)
*TEST*
To the 3rd portion, Add dilute HCL in drops
-then in excess
-heat the solution
*OBSERVATION*
White precipitate is formed which Dissolved on heating and re-appears on cooling
*INFERENCE*
Pb^2+ confirmed
======================
2ci)
*TEST*
Sample Y + concentrated HNO3
(Cii)
Heat the product formed in (ci)
*OBSERVATION*
Dirty white curd is formed
===========
It turn to yellow colouration on heating
*INFERENCE*
Protein is present
====================
2ciii)
*TEST*
Add concentrated NH3 to the product in (cii)
*OBSERVATION*
The product in (Cii) turns to orange colour
*INFERENCE*
Protein is present
=======================
No3
3ai)
i) SO2 change the colour of acidified K2Cr2O7 solution from Orange to green when it is passed through it by changing to Cr2(SO4)3
solution
ii) when zinc dust is added to CuSO4 solution, the solution turn from Blue to white
3aii)
i)Due to the formation of Cr2(SO4)3 solution
ii)Due to the formation of white ZnSO4 since zinc is higher in electrochemical series
3b)
DRAW and label a diagram for a set up that can be used to separate suspension of chalk dust in water
*COMPLETE*
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2017 NECO CHEMISTRY PRACTICAL SOLUTION
NO1)
*Note:*
Candidates are strongly advised to check which size of pipette was used by their school.
We used 25cm³
-------------------------------
If your school used 20cm³, all you need do is use it in place of 25cm³ anywhere we used it and calculate accordingly
*NOTE:ALL MAKE USE OF YOUR SCHOOL ENDS POINT*
Volume of Pipette used 25.0cm^3
Titration | Rough | First | Second | Third
Final burette reading (cm^3) | 22.50 | 45.10 | 32.65 | 27.60
Intial burette reading (cm^3) | 0.00 | 22.50 | 10.00 | 5.00
Volume of A used (cm^3) | 22.50 | 22.60 | 22.65 | 22.70
1a) Average volume of A = 22.60 + 22.65 + 22.70/3
= 22.65cm^3
1bi) GIVEN: CB = 0.10moldm^-3
VB = 25cm^3
CA = ?
VA = 22.65cm^3
Equation of reaction:
H2C2O4(aq) + 2NaOH(aq) ====> Na2C2O4(aq) + 2H2O(l)
hence, nA = 1
Nb = 2
Using CAVA /CBVB = nA/nB
CA = CBVBnA/VanB
=0.10 × 25 × 1/22.65 × 2
CA = 0.055 moldm^-3
Concentration of A in moldm^-3 = 0.055
bii) Using:
gram concentration = molar concentration × molar mass
conc. Of A in gdm^-3 = 0.055 × [(1 × 2) + (12 × 2) + (16 × 4)]
= 0.055 × 90
= 4.95gdm^-3
======================
No2 solution
2a)
*TEST*
X + 5cm of distilled water, Divide the solution into 4 portion
*OBSERVATION*
Salt X is soluble in distilled water
*INFERENCE*
Salt X is soluble salt
=======================
2bi)
*Test*
-To the first portion, Add dilute NaOH in drop
-then in excess
*OBSERVATION*
-White precipitate solution is formed
-The white precipitate is soluble
*INFERENCE*
Zn^2+ Pb^2+ and AL^3+ are present
=====================
2bii)
*TEST*
-To the 2nd portion add dilute NH3 solution in drop
-then in excess
*OBSERVATION*
White gelatinous precipitate is formed which is insoluble in excesa
*INFERENCE*
Zn^2+,Pb^2+ and Al^3+ are present
======================
2biii)
*TEST*
To the 3rd portion, Add dilute HCL in drops
-then in excess
-heat the solution
*OBSERVATION*
White precipitate is formed which Dissolved on heating and re-appears on cooling
*INFERENCE*
Pb^2+ confirmed
======================
2ci)
*TEST*
Sample Y + concentrated HNO3
(Cii)
Heat the product formed in (ci)
*OBSERVATION*
Dirty white curd is formed
===========
It turn to yellow colouration on heating
*INFERENCE*
Protein is present
====================
2ciii)
*TEST*
Add concentrated NH3 to the product in (cii)
*OBSERVATION*
The product in (Cii) turns to orange colour
*INFERENCE*
Protein is present
=======================
No3
3ai)
i) SO2 change the colour of acidified K2Cr2O7 solution from Orange to green when it is passed through it by changing to Cr2(SO4)3
solution
ii) when zinc dust is added to CuSO4 solution, the solution turn from Blue to white
3aii)
i)Due to the formation of Cr2(SO4)3 solution
ii)Due to the formation of white ZnSO4 since zinc is higher in electrochemical series
3b)
DRAW and label a diagram for a set up that can be used to separate suspension of chalk dust in water
*COMPLETE*
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We're working on the answers
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