Question 1 (c)

Beaker 1 о. м нс1 ВеаКет 2 О. 100 М ВеаКет З О. 100 М

(ii) The contents of beaker 2 are poured into beaker 3 and the resulting solution is stirred. Assume that volumes are additive. Calculate the pH of the resulting solution. In the resulting solution, \[NH3\] = \[NH'\] , Ka = 5.6 X 10-10 \[NH4+\] Thus \[H30+\] = 5.6 x 10-10; pH = -log(5.6 x 10-10) 1 point is earned for noting that the solution is a buffer with \[NH3\] = \[NH'\]. 1 point is eamed for the correct pH. = 9.25

  • Henderson–Hasselbalch equation

    ![[VHI + = Hd (t) [VHI + = [VHI [-VI[+HI ](./media/image120.png)

Question 1 (d)

(ii) Calculate the final \[NH4+\] in the resulting solution at 250C. moles = (volume)(molarity) moles H30+ in sol 1 . = = 0.00250 mol moles NH3 in 2 = = 0.00250 mol moles NH4+ in sol 3 . = = 0.00250 mol When the solutions are mixed, the H30+ and NH3 react to form NH + resulting in a total of 0.00500 mol NH 4 +. The final volume is the sum (25.0+25.0+25.0) = 75.0mL. The final concentration of NH + = (0.00500 mol/O.0750L) = 0.0667 M. 1 point is earned for the correct calculation of moles of NH 4. 1 point is eamed for the correct calculation of the final volume and concentration.

Question 2 (a)

  • Your answers should have the same decimal places as the equipment can read

    6 MHN03. moles before dilution (i) Calculate the volume, in mL, of 16 MHN03 that the student should use for preparing 50. mL of 114%. = MV = moles after dilution 1 point is earned for the correct volume. = (6 M(50.mL) Vi = 19 mL or 20 mL (to one significant figure)

  • Procedure for preparing solutions

    (ii) Briefly list the steps of an appropriate and safe procedure for preparing the 50. mL of 6 MHN03. Only materials selected from those provided to the student (listed above) may be used. Wear safe o les and rubber loves. Then measure 19 mL of 16 O using a 100 mL graduated cylinder. Measure 31 mL of distille H O using a 100 mL graduated cylinder. Transfer the water to a 100 mL beaker. Add the acid to the water with stirring. 1 point is earned for properly measuring the volume of 16 MHN03 and preparing a 6 MHN03 acid solution. 1 point is earned for wearing protective gear and for adding acid to water.

  • Volumetric flask vs. Graduated cylinder

    (iii) Explain why it is not necessary to use a volumetric flask (calibrated to 50.00 mL ±0.05 mL) to perform the dilution. The graduated cylinders provide sufficient precision in volume measurement to provide two significant 1 point is earned for an acceptable explanation. figures, making the use of the volumetric flask unnecessary.

Question 3 (a)

  • Notice the coefficient of the reactants and products in calculating standard enthalpy change

  • Standard enthalpy of formation of pure elements in their standard states are assigned zero

    2 H2(g) + 02(g) 2 H20(1) (a) Calculate the standard enthalpy change, AHO for the reaction represented by the equation above. 298 (The molar enthalpy of formation, AH AH0298 = - \[2(0) + 1(0)\] ; , for H20(1) is -285.8 kJ mol-I at 298 K.) = -571.6kJ mol-I 1 point is earned for the correct answer.

Question 5 (b)

(b) On the basis of the diagram you completed in part (a), do all six atoms in the N2H4 molecule lie in the same plane? Explain. No, they do not. The molecular geometry surrounding both 1 point is earned for a correct nitrogen atoms is trigonal pyramidal. Therefore the molecule answer with a valid explanation. as a whole cannot have all the atoms in the same plane.

  • N2H4 molecular geometry

    ydrog Nitrogeni o Nitrogen ydrog

  • Trigonal pyramidal

    C:\\8E425445\\852EE352-60A8-43EB-A33B-8042A84EBC3A\_files\\image128.png

Question 5 (c)

(c) The normal boiling point of N2H4 is 1140C, whereas the normal boiling point of C2H6 is —890C. Explain, in terms of the intermolecular forces present in each liquid, why the boiling point of N2H4 is so much higher than that of C2H6. N2H4 is a polar molecule with London dispersion forces, forces, and hydrogen bonding between molecules, whereas C2H6 is nonpolar and only has London dispersion forces between molecules. It takes more energy to overcome the stronger IMFs in 1 point is earned for correct reference to the two different types of IMFs. 1 point is earned for a valid explanation based on the relative strengths of the IMFs. hydrazine, resulting in a higher boiling point.

Ion-dipole H bond Dipole-dipole Ion—induced dipole Dipole—induced dipole Dispersion (London) Nonbonding (Intermolecular) 5-25 3-15 H—CI •a—a 2-10 Ion charge— dipole charge Polar bond to H— dipole charge (high EN of N, Dipole charges Ion charge— polarizable e— cloud Dipole charge— polarizable e— cloud Polarizable e— clouds 40-600 10-40 0.05-40

Question 6 (b)

(b) The flask is then heated to 450C, and the pressure in the flask increases. In terms of kinetic molecular theory, provide TWO reasons that the pressure in the flask is greater at 450C than at 350C. There are three possible reasons based on kinetic molecular theory. • At the higher temperature there are more, ethanol molecules in the gas phase, so there will be more collisions with the flask walls, resulting in a greater pressure. 1 point is earned for each • At the higher temperature the molecules will be moving faster on correct reason up to a maximum average, thus colliding with the flask walls more frequently, of 2 points. resulting in a greater pressure. • Because the molecules are moving faster on average, their collisions with the flask walls will exert a greater force, resulting in a greater pressure.

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