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The questions of Chemistry for Life Sciences - Assignment Example

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Chemistry for Life Sciences Keep your sentences short and simple to understand. Correctly answer all questions please (100%). Do not directly copy sentences from Wikipedia as we check with Turnitin software. Answer all questions in Section A and Section B in the spaces provided…
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The questions of Chemistry for Life Sciences
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This paper consists of 6 questions on 6 pages. Section A 1 (a) Balance the equation, given below, for the reaction in which H2Se is formed from Al2Se3 Al2Se3 + 6 H2O > ---------------> 2 Al (OH) 3 + 3 H2Se (1 mark) 1 (b) Describe how the covalent bonds are formed in an H2Se molecule. Selenium has 6 valence electrons. Two of these electrons are involved in the formation of covalent bonds with two other hydrogen atoms, the other valence electrons remain as lone pairs. An unpaired electron of each hydrogen atom pairs with a valence electron of Selenium, forming two covalent bonds.

H : Se : H (2 marks) 1 (c) The hydrogen halides contain polar covalent bonds. 1 (c) (i) State what is meant by the term polar as it applies to a covalent bond. The covalent bonds in hydrogen halides are polar in nature, because halides are more electronegative than hydrogen. The shared electron pair is pulled towards the electronegative halide, leading to unequal sharing of electron pair. This causes charge separation, making the hydrogen more electropositive and the halide more electronegative (dipole). (1 mark) 1(c) (ii) Explain why the H–Cl bond is more polar than the H–I bond.

The electronegativity of Chlorine is 3.16 on the Linus Pauling scale, while that of Iodine is 2.66. Since the electronegativity of chlorine is more than iodine, the pull on the shared electron pair in H-Cl is more than in H-I. . Experiment Initial [P] / mol dm–3 Initial [Q] / mol dm–3 Initial rate / mol dm–3 s–1 1 1.2 ? 10–3 2.0 ? 10–3 1.8 ? 10–5 2 6.75 2.0 ? 10–3 2.7 ? 10–5 3 0.60 ? 10–3 6.0 ? 10–3 21.6? 10–9 4 1.8 ? 10–3 5.49? 10–2 0.30 ? 10–5 (3 marks) 2 (a) (ii) Using the data from Experiment 1, calculate a value for the rate constant, k, and state its units.

Calculation.Rate constant, k, for experiment 1 is calculated as follows: rate = k[P][Q]2 1.8 ? 10–5 mol dm–3 s–1= k [1.2 ? 10–3 mol dm–3] [2.0 ? 10–3 mol dm–3]2 Therefore, k= =0.375?104 mol-2 dm-6 s-1= 3.75?103 mol-2 dm-6 s-1 The rate constant, k, is calculated as 3.75?103 mol-2 dm-6 s-1 Units: mol-2 dm-6 s-1 (3 marks) 2 (b) The decomposition of compound R is a zero order reaction. On the axes below sketch a line to show the relationship between the initial rate of reaction of R and the initial concentration of R at constant temperature.

Initial rate Initial [R] (1 mark) 3. Define the following terms used in reaction kinetics. (a) Overall order of reaction The overall order of reaction is defined as the sum of the order of reaction of individual reactants, or sum of the powers of the concentrations of the individual reactants. If rate=k[A]x [B]y, then Overall order of reaction= x+y (1mark) (b) Rate constant Rate constant is defined as the coefficient of the ratio between rate of reaction and the product of the concentration of the reactants.

k= rate/[A]x [B]y (1mark) (Total 2 marks) 4. The equilibrium reaction 3Fe(s) + 4H2O(g) - Fe3O4(s) + 4H2(g) was at one time used to make hydrogen; the tri-iron tetroxide was returned to the blast furnace to be converted back to iron. 

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