Chem 1310 I Answers for Assignment VI Due September 27, 2004

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1 Chem 1310 I Answers for Assignment VI Due September 27, 2004 p. 233, #33 Increasing the volume by a factor of 3.25 lowers the pressure by a factor of Doubling the absolute temperature doubles the pressure. Thus, the new pressure is 3.38 atm (= 5.50 (2.00/3.25)). p. 233, #39 Substituting into PV = nrt gives a temperature of 262 K. Note that n = 29.8 g (1 mol/32.0 g). p. 233, #45 One mol of SF 6 weighs g. Substitute P= atm, n= one mol and T=288 K into PV = nrt to find that V =25.09 L. Thus, the density is 5.82 g/l (= g mol 1 /25.09 L mol 1 ). p. 234, #49 The formula mass of sodium chloride is g/mol. Thus, 2500 kg of NaCl contains mol NaCl. The theoretical yield of hydrogen chloride gas is mol NaCl. Under the conditions specified (550 C and torr), the gas would occupy L. (The volume is obtained by substituting P= atm, n= mol and T=823 K into PV = nrt). p. 234, #53 Under the conditions specified (35.3 C and torr), 4.48 L of gas contains mol (a reasonable number; one mol would occupy 22.4 L at STP, not far from our conditions and 4.48/22.4 = 0.200; of course, I got the more precise number by substituting into PV = nrt after converting to atm and K). Looking at the balanced equation reveals that mol Cl 2 is made from mol MnO 2 which represents a mass of g (molar mass of manganese(iv) oxide is g mol 1 ).

2 p. 236, #88 The Chemistry in Your Life essay is on p The reaction that occurs when the air bag inflates is: 2 NaN 3 2 Na + 3 N 2. n nitrogen = (1 atm)(70.0 L)/( L atm mol 1 K 1 )(298 K) = 2.86 mol N 2 n sodium azide = 2.86 mol N 2 (2 mol NaN 3 /3 mol N 2 ) = 1.91 mol NaN 3 g sodium azide = 1.91 mol NaN 3 (65.20 g NaN 3 /mol NaN 3 ) = 125 g NaN 3 p. 236, #90 Let 2y represent the number of particles originally present. After the change, y particles have reacted and their total number has been changed to y. Thus, P = atm (1.666y mol/2y mol) = atm p. 237, #93 The balanced equation for the reaction is: CS 2 (g) + 3 O 2 (g) CO 2 (g) + 2 SO 2 (g) Before the reaction begins, the pressure of the gas mixture is 3.00 atm, corresponding to mol of gas (use PV = nrt with V = 10.0 L and T = K). After the reaction is complete, the pressure of the gas mixture is 2.40 atm, corresponding to mol of gas (use PV = nrt with V = 10.0 L and T = K). We are told that the initial mixture contained carbon disulfide and excess oxygen, so we can construct an i-c-e table. CS 2 O 2 CS 2 SO 2 initial X 3x + y 0 0 change x 3x + x +2x end 0 y x 2x

3 p. 237, #93 (con t) Initially, 4x + y = mol Finally, 3x + y = mol Thus, x = mol CS 2 which is also the value of y (= mol of O 2 in excess). The molar mass of carbon disulfide is mol CS 2 (76.13 g CS 2 /mol CS 2 ) = 14.9 g CS 2 p. 237, #98 Use the Ideal Gas Law to find the chemical amount of oxygen gas produced. Substitute P = torr (1 atm/760 torr), V = 34.6 ml and T = K into PV = nrt, with R = ml atm mmol 1 K 1 to find that mmol of oxygen gas were produced. The balanced equation is 2 NaClO 3 (s) 2 NaCl (s) + 3 O 2 (g), meaning that mg of sodium chlorate reacted. Thus, the sample was 61.4% NaClO 3. p. 289, #3 Attractions between particles are always affected, to some degree, by van der Waals forces (also called dispersion forces or induced dipole interactions). a. In potassium fluoride, ion-ion forces will be far more important than van der Waals forces. b. In hydrogen iodide, dipole-dipole forces and van der Waals forces are probably of equal importance. c. In CH 3 OH (methanol), hydrogen bonding is more important than van der Waals forces. (Hydrogen bonding may be regarded as a special case of dipole-dipole interactions. Methanol is wood alcohol, a poisonous liquid. Ethane, C 2 H 6, has a molar mass comparable to that of methanol but ethane is non-polar and is a gas. (Statements about the state of the compound refer to conditions of room temperature and atmospheric pressure).

4 p. 289, #3 (con t) d. In radon, only van der Waals forces occur between atoms. e. In (di)nitrogen, only van der Waals forces occur between molecules. p. 289, #5 A sodium ion will be more strongly attracted to a bromide ion than to a hydrogen bromide molecule or a krypton atom. p. 289, #15 P acetylene = P wet gas P water vapor = torr Substitute in PV = nrt (P = torr (1 atm/760 torr), V = 1 L (exactly) and T = K). This yields a chemical amount of mol of acetylene in the 1 L volume visualized. The molar mass of acetylene, C 2 H 2 is Thus, the density of the wet gas is g/l. p. 289, #17 When the metal is in a high oxidation state, the bonding is more likely to be covalent than ionic. Waxes are typical covalent solids; they have melting points on the order of 100 C -200 C. Table salt (sodium chloride) is a typical ionic solid; its mp is 800 C. a. lowest mp Mn 2 O 7 < MnO 2 < Mn 3 O 4 highest mp Note: The oxidation state of manganese in Mn 3 O 4 is apparently +8/3. The compound is really an equimolar mixture of MnO and Mn 2 O 3. Thus, Mn 3 O 4 contains two Mn(III) for every Mn(II) for an average oxidation number of 2⅔. b. Potassium permanganate is an ionic compound that contains potassium ions and permanganate ions. The manganese is in the +7 state in the permanganate ion, a polyatomic ion in which oxygens are covalently bound to Mn(VII) ions.

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