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

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


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

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

results matching ""

    No results matching ""