NECO GCE CHEMISTRY EXPO ANSWERS & QUESTIONS 2017 (COMPLETE CHEMISTRY OBJ/THEORY RUNZ)
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_*CHEMISTRY OBJ*_
1-10: AAEBBDCAAC
11-20: ABBEBDDBCD
21-30: BABDCECEEC
31-40: CBDEBBBCBB
41-50: AECBBEDBDE
51-60: ACEBDCCEDB
====COMPLETED====
(1a)
In a tabular form
C; H; O
Mass/molar mass of C is 6.12g/12
Mass/molar mass of H is 0.85g/1
Mass/molar mass of O is 2.72/16
Under C: 0.51
Under H: 0.85
Under O: 0.17
C = 0.51/0.17
H = 0.85/0.17
O = 0.17/0.17
Empirical Formula is C3H5O
Molecular Formula is. ?
Molar mass is 134
[C3H5O]n = 134
[12 × 3 + 1×5 + 16 × 1]n = 134
[36 + 5 + 16]n = 134
57n = 134
n = 134/57
n = 2.35
n = 2
Molecular Formula [C3H5O]2
= C6H10O2
(1bi)
Isotopy - Atom of the same element are alike in every aspect and differ from atoms of all other element
35
17Cl and
37
18Cl
(1bii)
Sub atomic particle: All elements are made up of small indivisible particles called atoms
(1ci)
Diagram
(1cii)
I. 2C2H2 + 5O2 ---> 4CO2 + 2H2O
II. C2H2 + HCl ---> C2H3Cl
(1d)
Because they are in different states. Steam is in a gaseous state while water is in a liquid state.
========================================
(2a)
(i) Temporary hardness - Dissolved calcium hydrogencarbonate
(ii) Permanent hardness -Dissolved calcium sulfate.
(2bi)
(i) Temporary Hardness can be Removed By Boiling
(ii) Permanent hardness can be removed by use of washing soda (sodium carbonate); ion exchange; the use of polyphosphate water softeners.
(2bii)
-Advantages:
(i) Calcium ions in the water are good for children's teeth and bones.
(ii) The soothe it forms is good for ceramics purposes.
-Disadvantages:
(i) It is more difficult to form a lather with soap.
(ii) Scum may form in a reaction with soap, wasting the soap.
(2ci)
A saturated solution is a chemical solution containing the maximum concentration of a solute dissolved in the solvent.
(2c)
Mass of impure CaCo3 = 5.2g
Amount n; of CO2 formed = 0.05moles
CaCo3 + 2Hcl ---> Cacl2 + CO2 + H2O
1 mole of CaCo3 = 1 mole of Co2
100g of CaCo3 = 1 mole of Co2
Hence Xg of CaCo3 = 0.05mole CO2
X = 0.05 × 1/1 = 0.05mol of CaCo3
but n = m/M
Therefore; m = nM = 0.05 × 100
=5g of CaCo3
%purity of CaCo3
= mass of pure/mass of impure × 100/1
= 5/5.2 × 100/1 = 96.2%
(2ciii)
(i) Anthracite Coal
(ii) Bituminous coal
(iii) Subbituminous coal
(2civ)
Coal tar
(2di)
(i) Zinc is not a transition metal because it forms only Zn2+ ions with all the 3d electrons present.
(ii) Zn not a transition metal, because it forms only Sc3+ ions with n d-electrons
(2dii)
Fine chemicalChemical substances prepared to a very high degree of purity for use in research and industry.
(2diii)
(i) dyes
(ii) perfumes
========================================
(3ai)
Petroleum Gas and Petrol>Kerosene>Diesel>Asphalt
(3aii)
I. Asphalt
II. Kerosene
III. Petroleum Gas and petrol
Iv. Asphalt
(3bi)
I. Ethene
II. Propene
(3bii)
(i) Structural Isomerism
(ii) Stereo Isomerism
(3biii)
19R - 1s² 2s² 2p6 3s² 3p6 4s¹
10S - 1s² 2s² 2p6
24T - 1s² 2s² 2p6 3s² 3p6 4s² 3d4
(3biv)
Group 6
(3ci)
I. AgNO3 and BaCl2
II. Bacl2 and Na2SO4
III. Ca(NO3)2 and H2CO3
(3cii)
(i) Filtration
(ii) Flocculation
(iii) Disinfection
(3d)
Mass of NaCl
= mass of Nacl + mass of H2O
= (23 + 35.5) + 80g
= 58.5 + 80g = 138.5g
Hence
Since 39.8g at 100°C = 35.9g at 15°C
138.5g at 100°C = Xg at 15°C
X = 138.5 × 35.9/39.8g
= 124.9g at 15°C
Hence mass of Nacl precipitated
= 138.5 - 124.9g
= 13.6g of Nacl
========================================
(4a)
(i) Coke
(ii) Graphite
(iii) Diamond
(iv) Coal
(v) Charcoal
(4b)
Le chatelier's principle states that for a reversible reaction in equilibrium, if any of the factors(temperature, concentration, pressure) keeping the system in equilibrium is altered, the system will shift to adjust the change
(4bii)
I - It will shift from left to right
II - It will shift from left to right
III - It will favor the forward reaction by increasing the product
(4biii)
Kc = [NH]2
------
[N2][H2]3
(4c)
4H2O + 3Fe ---> Fe3O4 + 4H2
3 moles of Me = 4 moles of H2
3(56)g = 4(22.4dm³) of H2
Therefore; 3.03g of Fe = Xg of H2
X = 3.03 × 4 × 22.4dm³/3 × 56
= 1.62dm³ of H2
3(56)g of Fe = 3(56) + 4(16)g
Therefore; 3.03g of Fe = X
X = 3.03 × 232/168 × 1 = 4.18g of iron oxide
(4di)
I. Calcium Hydroxide Ca(OH)2
II. H2SO4
III. H2SO4
(4dii)
I. NaCl
II. NaHSO4
III. K3Fe(CN)6
_*CHEMISTRY OBJ*_
1-10: AAEBBDCAAC
11-20: ABBEBDDBCD
21-30: BABDCECEEC
31-40: CBDEBBBCBB
41-50: AECBBEDBDE
51-60: ACEBDCCEDB
====COMPLETED====
(1a)
In a tabular form
C; H; O
Mass/molar mass of C is 6.12g/12
Mass/molar mass of H is 0.85g/1
Mass/molar mass of O is 2.72/16
Under C: 0.51
Under H: 0.85
Under O: 0.17
C = 0.51/0.17
H = 0.85/0.17
O = 0.17/0.17
Empirical Formula is C3H5O
Molecular Formula is. ?
Molar mass is 134
[C3H5O]n = 134
[12 × 3 + 1×5 + 16 × 1]n = 134
[36 + 5 + 16]n = 134
57n = 134
n = 134/57
n = 2.35
n = 2
Molecular Formula [C3H5O]2
= C6H10O2
(1bi)
Isotopy - Atom of the same element are alike in every aspect and differ from atoms of all other element
35
17Cl and
37
18Cl
(1bii)
Sub atomic particle: All elements are made up of small indivisible particles called atoms
(1ci)
Diagram
(1cii)
I. 2C2H2 + 5O2 ---> 4CO2 + 2H2O
II. C2H2 + HCl ---> C2H3Cl
(1d)
Because they are in different states. Steam is in a gaseous state while water is in a liquid state.
========================================
(2a)
(i) Temporary hardness - Dissolved calcium hydrogencarbonate
(ii) Permanent hardness -Dissolved calcium sulfate.
(2bi)
(i) Temporary Hardness can be Removed By Boiling
(ii) Permanent hardness can be removed by use of washing soda (sodium carbonate); ion exchange; the use of polyphosphate water softeners.
(2bii)
-Advantages:
(i) Calcium ions in the water are good for children's teeth and bones.
(ii) The soothe it forms is good for ceramics purposes.
-Disadvantages:
(i) It is more difficult to form a lather with soap.
(ii) Scum may form in a reaction with soap, wasting the soap.
(2ci)
A saturated solution is a chemical solution containing the maximum concentration of a solute dissolved in the solvent.
(2c)
Mass of impure CaCo3 = 5.2g
Amount n; of CO2 formed = 0.05moles
CaCo3 + 2Hcl ---> Cacl2 + CO2 + H2O
1 mole of CaCo3 = 1 mole of Co2
100g of CaCo3 = 1 mole of Co2
Hence Xg of CaCo3 = 0.05mole CO2
X = 0.05 × 1/1 = 0.05mol of CaCo3
but n = m/M
Therefore; m = nM = 0.05 × 100
=5g of CaCo3
%purity of CaCo3
= mass of pure/mass of impure × 100/1
= 5/5.2 × 100/1 = 96.2%
(2ciii)
(i) Anthracite Coal
(ii) Bituminous coal
(iii) Subbituminous coal
(2civ)
Coal tar
(2di)
(i) Zinc is not a transition metal because it forms only Zn2+ ions with all the 3d electrons present.
(ii) Zn not a transition metal, because it forms only Sc3+ ions with n d-electrons
(2dii)
Fine chemicalChemical substances prepared to a very high degree of purity for use in research and industry.
(2diii)
(i) dyes
(ii) perfumes
========================================
(3ai)
Petroleum Gas and Petrol>Kerosene>Diesel>Asphalt
(3aii)
I. Asphalt
II. Kerosene
III. Petroleum Gas and petrol
Iv. Asphalt
(3bi)
I. Ethene
II. Propene
(3bii)
(i) Structural Isomerism
(ii) Stereo Isomerism
(3biii)
19R - 1s² 2s² 2p6 3s² 3p6 4s¹
10S - 1s² 2s² 2p6
24T - 1s² 2s² 2p6 3s² 3p6 4s² 3d4
(3biv)
Group 6
(3ci)
I. AgNO3 and BaCl2
II. Bacl2 and Na2SO4
III. Ca(NO3)2 and H2CO3
(3cii)
(i) Filtration
(ii) Flocculation
(iii) Disinfection
(3d)
Mass of NaCl
= mass of Nacl + mass of H2O
= (23 + 35.5) + 80g
= 58.5 + 80g = 138.5g
Hence
Since 39.8g at 100°C = 35.9g at 15°C
138.5g at 100°C = Xg at 15°C
X = 138.5 × 35.9/39.8g
= 124.9g at 15°C
Hence mass of Nacl precipitated
= 138.5 - 124.9g
= 13.6g of Nacl
========================================
(4a)
(i) Coke
(ii) Graphite
(iii) Diamond
(iv) Coal
(v) Charcoal
(4b)
Le chatelier's principle states that for a reversible reaction in equilibrium, if any of the factors(temperature, concentration, pressure) keeping the system in equilibrium is altered, the system will shift to adjust the change
(4bii)
I - It will shift from left to right
II - It will shift from left to right
III - It will favor the forward reaction by increasing the product
(4biii)
Kc = [NH]2
------
[N2][H2]3
(4c)
4H2O + 3Fe ---> Fe3O4 + 4H2
3 moles of Me = 4 moles of H2
3(56)g = 4(22.4dm³) of H2
Therefore; 3.03g of Fe = Xg of H2
X = 3.03 × 4 × 22.4dm³/3 × 56
= 1.62dm³ of H2
3(56)g of Fe = 3(56) + 4(16)g
Therefore; 3.03g of Fe = X
X = 3.03 × 232/168 × 1 = 4.18g of iron oxide
(4di)
I. Calcium Hydroxide Ca(OH)2
II. H2SO4
III. H2SO4
(4dii)
I. NaCl
II. NaHSO4
III. K3Fe(CN)6
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