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Ondo State Joint Exam
Chemistry Obj

01-10: C D A D D C D A A B
11-20: C D B C A D D D B D
21-30: B D C D D D C B C D
31-40: D B B B D D D B D C
41-50: C C A C B A C C B A
. .

>>>>THEORY PART<<<<<


1ai) Evaporation can be used to recover a solid solute from a solution. In this process, the solvent is usually sacrificed.
Solution—>solute + solvent
A water bath or a sand bath is used to bring about a steady rate of Evaporation. The solute required is left behind in the dish while the solvent escape into the air as vapour.
Evaporation is used in salt making industries.

1aii) Sublimation is the process when some solids are heated and they changed directly to the gaseous state without passing through the liquid state. Examples of such solids are iodine and ammonium chloride

1aiii) Chromatography is the separation technique use to separate colours as well as colourless complex substances. Examples are amino acids, ink, dye etc. Chromatography uses a solvent moving over a porous, adsorbents medium (e.g paper or gel) to separate a mixture of solutes.

1aiv) Fractional distillation is used to separate a mixture of two or more miscible liquids into it’s component parts or fraction. The fractional distillation over in ascending order of their boiling points, starting with the fraction with the lowest boiling point.

1av) Sieving is used to separate solid particles of different sizes. The mixture is placed on a sieve with a mesh of a particular size. Particles smaller than the mesh size of a the sieve will pass through the sieve while the bigger particles remain on the sieve.

1b) Mg + 2Hcl—> H2+ Mgcl2
(0.36gMg * 1molMg)/24gMg = 0.015molMg
Determines moles of 5MHcl
Convert 20cm³ to 20mL and then to 0.02L
Idm³= 1L
20* (1mL/1cm³)* (1L/1000mL) = 0.02LHcl
0.02* (5mol/1L)= 0.1molHcl
0.1molMg *(1molH2/1molMg) * 2= 0.2gH2
0.1molHcl* (1molH2/2molHcl)* (2gH2/1molH2)= 0.1gH2
The stoichiometry of the reaction is the, 0.1g of H2 were obtained when 0.36g of Mg was treated with 20cm³ 5M Hcl



2ai)a. Physical changes are easily reversible while chemical changes are not easily reversible
b. New substances are formed in physical changes while entirely new substances are always produced in chemical changes

2aii) a. It does not have a fixed boiling point
b. It’s constitutes can be easily separated or obtained Hy physical means
c. The constitutes of the air still retain their individual properties

2aiii)a. Zinc chloride
b. Sodium chloride
c. Ammonium chloride
d. Potassium trioxonitrate (v)
e. Calcium chloride


3i) a. high melting and boiling point
b. Solibiy

2.04g/12= 0.17
0.34g/1= 0.34
2.73g/16= 0.17
Divide the biggest one by the smallest one:
C 0.17/0.17, H 0.34/0.17, O 0.17/0.17
C=1, H=2, O=1
The empirical formula is CH2O

3iib) Relative molecular mass of the compound is 60
Atomic number of the elements; C=12, H=1, O=16
Divide the both side by 30
Since n=2, then the molecular formula is; (CH2O)2= C2H4O2
Therefore, the molecular formula is C2H4O2


4i) Charles law states that for a given mass of gas at constant pressure, the volume is directly proportional to it’s absolute temperature. Mathematically express as V1/T1= V2/T2 at constant pressure

4ii) V1=800cm³, V2= 940cm³, P1= 760mmHg, P2= 675mmHg, T1=273K, T2=?
Using the formula, (P1V1)/T1=(P2V2)/T2
Cross multiply
P1V1T2= P2V2T1
Substitute the following values

4iii) oxidation is a process involving a loss of electron(s). In other words, oxidation is the process by which an element donate or transfer it’s electron(s) to another element.
While reduction is a process involving a gain of electron(s). In other words, reduction is the process by which an element accept or gain an electron(s) from another element

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