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Solutions to Final Practice

# Solutions to Final Practice - 1 The general"formula for...

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1. The general “formula” for hydrogen bonding is B: …… H–A . The dotted line between B and H is the hydrogen bond. The B atom has at least one lone pair. Atom A is N, O, or F; and atom B is N, O, or F. HF can hydrogen bond with itself as follows: F: …… H–F. NH 3 can hydrogen bond with itself as follows: N: …… H–N. H 3 COCH 3 cannot hydrogen bond with itself; the oxygen atom in H 3 COCH 3 is covalently bonded to carbon atoms, and not to a hydrogen atom. H 3 COCH 3 can hydrogen bond with water as follows: O: …… H–O. HF and NH 3 can hydrogen bond with water. 2. 0.0817 / 2.0592045 = 0.040 (two significant figures in 0.0817; five in 2.0592045) 3. 76.57 g C 6.376 mol C 6.43 g H 6.38 mol H 17.00 g O 1.063 mol O C 6.376 H 6.38 O 1.063 C 6 H 6 O 4. 121.757 = (0.57300)(120.904) + (0.42700)(x) x = 122.90 5. The mass of one atom of X, m * X , is twice the mass of one atom of Y, m * Y . In X 2 Y 5 , the two atoms of X have the same mass as four atoms of Y. The ratio of the mass of Y to the mass of X in X 2 Y 5 is (5 m * Y / 2 m * X ) = (5 m * Y / 2 (2 m * Y )) = (5 m * Y / 4 m * Y ) = 1.25. 6. 1.000 mol Al 2 (SO 4 ) 3 (17 moles of atoms / mol Al 2 (SO 4 ) 3 ) (6.022 x 10 23 atoms / mol) = 1.024 x 10 25 atoms 7. (65.0 g H 2 SO 4 / 100 g solution)(1.55 g solution / mL solution) (1000 mL / L)(1 mol H 2 SO 4 / 98.076 g) = 10.27 mol / L (1 L / 10.27 mol)(2.00 mol) = 0.195 L = 195 mL 8. 3 Mg + N 2 Mg 3 N 2 35.00 g Mg 1.440 mol Mg 15.00 g N 2 0.5353 mol N 2 (1.440 / 0.5353) = (2.691 / 1) < 3 / 1 Mg is the limiting reactant. 1.440 mol Mg(1 mol N 2 / 3 mol Mg) = 0.4800 mol N 2 react 0.5353 mol N 2 – 0.4800 mol N 2 react = 0.0553 mol N 2 remain 1.55 g N 2 remain. Note: 13.45 g N 2 react. Note: 48.45 g Mg 3 N 2 are produced; this is not 50.00 g. 9. PV = nRT d = m/V PM = dRT P 2 M = d 2 RT 2 P 1 M = d 1 RT 1 P 2 / P 1 = (d 2 / d 1 ) (T 2 / T 1 ) d 2 = (P 2 / P 1 ) d 1 (T 1 / T 2 ) d 2 = (1.5)(1.76 g / L)( 283 K / 293 K) = 2.55 g / L 10. partial pressure of species i in the mixture = (mole fraction of i )(total pressure) Mole fraction is an intensive quantity, so any value for the mass of each species present can be chosen. Suppose that we choose the mass to be 1.000 gram for each of the species. Mole fraction of He = moles of He / total moles. But n = m / M, and m = 1.000 g, so mole fraction of He = (1.000 / 4.003) moles / total moles, where total moles = (1.000 / 4.003) + (1.000 / 28.02) + (1.000 / 32.00) = 0.3168 moles. Thus, mole fraction of He = 0.2498 / 0.3168 = 0.7887, and P He = (0.7887)(5.00 atm) = 3.94 atm.

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11. In N 2 O 5 , the oxidation number for N is +5, which is the same as the group number (5A) for N. The oxidation number cannot be higher than the group number; therefore, N 2 O 5 cannot be oxidized further. In NO 2 , the oxidation number for N is +4; and in HNO 2 , oxidation number for N is +3.
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