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Example of Chemistry Exams Involving Chemical Compounds (Coursework Sample)


chemistry test exam and it was required to type answers on microsoft word and draw diagram s


Chemistry Test
Name of Student
Name of Professor
Table of Contents TOC \o "1-3" \h \z \u SECTION 1- Numerical PAGEREF _Toc73091304 \h 21.solution PAGEREF _Toc73091305 \h 22. solution PAGEREF _Toc73091306 \h 3Section 2 – Theoretical PAGEREF _Toc73091307 \h 4Addition reaction of benzene compound PAGEREF _Toc73091308 \h 4Introduction PAGEREF _Toc73091309 \h 4Mechanism PAGEREF _Toc73091310 \h 4Uses of the compound obtained PAGEREF _Toc73091311 \h 5Bibliography PAGEREF _Toc73091312 \h 5
SECTION 1- Numerical
1. Solution
K2Cr2O7 solution cm3= 21.25 ml
Concentration of K2Cr2O7 solution: 0.02 mol/dm3 = 0.02 mol/L (M).
0.02M oxidize 25 ml of iron (ll) Complex solution.
Cr2O72-- + 6Fe (II) à 6Fe (III)
Mole ratio of
Cr2O72 : 6Fe (II)
So, 1 mole of Cr2O72—oxidizes 6 Fe2+
Therefore, 6 moles of Fe (NH4)2(SO4)2 are equal to 1 mole of K2Cr2O7.
Molarity of Fe (II)
= 6x molarity (Cr2O7) volume (Cr2O7)/ volume (Fe (II)
= 6*0.2*21.25 ml/25 ml
=1.102 M
When we dissolve 10g of iron (ll) complex to 200 ml solution we get the Molarity equals to 0.1202
No. of mole = 0.1202 M * 200/1000
=0.0204 moles
The mass of iron (ll) complex is 10 g therefore the molar mass will be,
2. solution
At first, we balance the equation, Iodine solution reacts with sodium thiosulphate as follows:
I2 (aq) + 2 S2 O32- (aq) → 2I- (aq) + S4 O6 2- (aq)
Using dilution law of concentration, we can say that:
V1 as I2 solution = 25 cm3,
V2 as S2 O32- solution = 17.60 cm3
And the molarity of S2 O32- solution = 0.1 mol dm-3
M1 is the concentration of iodine solution, using dilution law and chemical balance equation we can equate the law as:
2M1V1 = M2V2
=M1= M2V2/ 2 V1
0.1 mol dm-3x17.60 cm32x 25 cm3
M1=0.0352 mol dm-3
So, x = 0.0352 ml dm-3 x 254 g/mole
=8.9408 g dm -3
(y-x) = 8.94- 0.0352
= 8.9056
Section 2 – Theoretical
Addition reaction of benzene compound
Benzene is the simplest aromatic compound and it is the essential parent of various organic aromatic hydrocarbons compounds. The aromatic compounds are made up of benzene ring structures that contain bonds which are described using the resonance rotating structure model of valence bond theory and concept of the delocalization in the theory of molecular orbital (Handbook of heterocyclic chemistry 2010).
In this case, the benzene structure is made up of two substituent group that allows substitution reaction to take place to activate it. However, according to this reaction, the aromatic compound reacts to forms pyridine (Reactivity of Heterocyclic Compounds 2014). When the additional reaction of benzene compound reacts with tera methylamine group in the presence of sodium methoxide and methanol. Sodium methoxide and methanol are used to speed up the rate of the reaction and in the process, hydrogen is added by benzene at a higher pressure. The product that is deduced is cyclohexane that stabilizes the thermodynamic of benzene.
Uses of the compound obtained
Many organic chemical compounds are in cyclic form, these properties enable them to join in the ring and

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