Question 2 (c)

(c) The tank is cooled to 250C, which is well below the boiling point of methanol. It is found that small amounts of H2(g) and CO(g) have dissolved in the liquid CH30H. Which of the two gases would you expect to be more soluble in methanol at 250C? Justify your answer. The only attractive forces between molecules of H2 and CH30H would be due to weak London dispersion forces (LDFs). In contrast, the LDFs are stronger between CO molecules and CH30H molecules because CO has more electrons than H . In addition CO is slightly polar; thus intermolecular dipole-dipole attractions can form between CO molecules and CH30H molecules. With stronger intermolecular interactions between molecules of CO and CH30H, CO would be expected to be more soluble in CH30H than 1--12. 1 point is earned for the correct answer and justification.

Question 3 (b)

In many organisms, glucose is oxidized to carbon dioxide and water, as represented by the following equation. C6H1206(s) + 6 02(g) 6 C02(g) + 6 H20(1) A 2.50 g sample of glucose and an excess of 02(g) were placed in a calorimeter. After the reaction was initiated and proceeded to completion, the total heat released by the reaction was calculated to be 39.0 kJ. (b) Calculate the value of AHO, in kJ mol-I, for the combustion of glucose. Im01CH O 6 12 6 2.50 g x 180.16 g C6H1206 Cb90kJ = 0.0139 mol C6H1206 1 point is earned for the correct answer. -(3,810 k.J mol-I 0.0139 mol

Question 5 (b)

(b) Describe the steps in a procedure to prepare 100.0 mL of 1.250 M NaOH solution using 5.000 M NaOH and equipment selected from the list below. Balance 50 mL buret Eyedropper 25 mL Erlenmeyer flask 100 mL Florence flask Drying oven 100 mL graduated cylinder 25 mL pipet Wash bottle of distilled H20 100 mL volumetric flask 100 mL beaker Crucible Pipet 25.00 mL of 5.000 M NaOH solution into the 100 mL volumetric flask. Fill the volumetric flask to the calibration line with distilled water, using an eyedropper for the last few drops is advised. Cap the volumetric flask and invert several times to ensure homogeneity. 1 point is earned for descriptions of any two of the three steps. An additional point is earned if all three steps are described.

  • Eyedropper

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Question 6 (a)

(ii) Estimate the numerical value of the bond angle in an ethyl methanoate molecule. Explain the basis of your estimate. The Cx is the central atom in a tetrahedral arrangment of bonding electron pairs; thus the angle would be approximately 109.50 1 point is earned for the correct angle with an appropriate explanation.

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In a tetrahedral molecular geometry, a central atom is located at the center with four substituents that are located at the comers of a tetrahedron. The bond angles are cos-l(-%) = 109.4712206...0 = 109.50 when all four substituents are the same, as in methane (CH4) as well as its heavier analogues. Tetrahedral molecular geometry - Wikipedia https://en.wikipedia.orgnvikinetrahedral\_molecular\_geometry

Question 6 (c)

(ii) Explain, in terms of processes occurring at the molecular level, why the pressure in the flask remained constant after 60. seconds. At the equilibrium vapor pressure, the rate of molecules passing from the liquid to the gas phase (vaporizing) equals the rate of gas phase molecules passing into the liquid phase (condensing). 1 point is earned for the correct explanation.

Phase changes Of matter solid @ 20" Encyclopædia Britannica, Inc gas melting freezng liquid

(iv) After 80. seconds, additional liquid ethyl methanoate is added to the container at 200C. Does the partial pressure of the ethyl methanoate vapor in the container increase, decrease, or stay the same? Explain. (Assume that the volume of the additional liquid ethyl methanoate in the container is negligible compared to the total volume of the container.) The artial ressure of the va or sta s the same because the equilibrium vapor pressure for 200C has already been reached. Because the temperature remains constant, the vapor pressure would remain unchanged. 1 point is earned for the correct answer with an explanation.

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