2017/2018 Neco Gce Chemistry Obj And Theory Now Out

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
2diii)
AgBr
Mgcl2

Homogenous mixture is a mixture that is composed in equal amount of proportions .. eg CaCo3

Heterogenous is when it’s unequal eg floodwater

Molar mass = 2×VD
=2×68.75
=137.5g
Xcl3=137.5
X+3(35.5)=137.5
X+106.5=137.5
X=137.5-106.5
X=31g
The atomic mass of x is 31g

(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

5a
i) increasing the surface area
By adding catalyst to the reaction

5iii)
If 100g of Caco3—–1mole of co2(22.4dm3)
4g——X
X=4*22.4/100
X=0.896dm^3

:.0.896dm^3 of CO2 was produced by 4g of Caco3

17. Normal salt…. Na2co3
Acid salt… NaHSO4
Complex salt…Ca3[Co(Cl)6]2

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