WYSE Academic Challenge 2004 Sectional Chemistry Solution Set

Transcription

1 WYSE Academic Challenge 2004 Sectional Chemistry Solution Set 1. Answer: d. Assume g of the compound. Thus, we have g of carbon, or 40.00/12.01 = 3.33 mol C. We have 6.71 g of hydrogen, or 6.71/1.008 = 6.66 mol H. We also have g of oxygen ( [ ]), or 53.29/16.00 = 3.33 mol O. The ratio of C:H:O = 3.33:6.66:3.33 = 1:2:1. Therefore, the empirical formula is CH 2 O. 2. Answer: e. Since both balloons are at the same temperature but contain different amounts of helium and are at different volumes, it is difficult to predict the exact pressure of each balloon. But, we can compare the pressure of each balloon to atmospheric pressure. Since the balloons are not expanding, we know that the helium pressures are not greater than atmospheric pressure. Since the balloons are not collapsing, we know that the helium pressures are not less than atmospheric pressure. Therefore, since the balloons are remaining the same size, they both must be equal to atmospheric pressure. 3. Answer: e. Atoms are conserved in a chemical reaction. The same number of each type of atom must be found among the reactants and products. The balanced chemical reaction is 3 NaNH 2 + NaNO 3 NaN NaOH + NH 3 No coefficient in front of a molecule represents 1. Therefore, the sum of the coefficients (from left to right) is = Answer: e. The balanced chemical equation is 3 Ba(NO 3 ) 2 (aq) + 2K 3 PO 4 (aq) 6 KNO 3 (aq) + Ba 3 (PO 4 ) 2 (s) We can use the solubility rules to conclude that potassium nitrate is soluble and barium phosphate is a solid. Initially, there are 25 mmol of barium nitrate (250.0mL x 0.10 mmol/ml) and 20 mmol of potassium phosphate (200.0mL x 0.10 mmol/ml). We need a 3:2 mole ratio of barium nitrate and potassium phosphate to react exactly. We have a 25:20 ratio, or 5:4 ratio, so the barium nitrate is the limiting reactant, and potassium phosphate is in excess. When the reaction is complete, solid barium phosphate, aqueous potassium nitrate, and aqueous potassium phosphate are present. The aqueous compounds consist of potassium, nitrate, and phosphate ions. 5. Answer: d. The valence electrons in chlorine, bromine, and iodine are in p-orbitals that are higher in energy than the 2s orbital, which is where the valence electron for lithium is found. 6. Answer: b. For trigonal bipyramid geometry, there would be too many lone pairs concentrated in one area to have a bent shape. 7. Answer: e. All four of the compounds are named correctly. The sulfide, oxide, and sulfate ions have charges of 2-. The iron ion has a charge of 2+, the lead ion has a charge of 4+, and the ammonium ion has a charge of 1+. Hexa is the correct prefix for Answer: d. All of these ions contain the same number of electrons, but the number of protons increase as the atomic number increases. Electrons experience a greater attraction as the positive charge on the nucleus increases. Therefore, as the number of protons increase, the ions become smaller.

2 9. Answer: a. Looking at the atomic numbers for sodium and chlorine, we know that sodium contains 11 protons and chlorine contains 17 protons. The sum of protons is 28. Na + means that an electron was lost, so the number of electrons is 10. Cl - means that an electron was gained, so the number of electrons is 18. The sum of electrons is Answer: c. Use the equation PV = nrt. We can assume moles of gas are constant. Since the volume increases by a factor of 2, the pressure must decrease by a factor of 2, so the P = 6.0 atm, assuming no temperature change. The temperature increases, however, although the Kelvin temperature does not double. So, the pressure is a bit greater than 6.0 atm (but not as high as 12.0 atm). The only reasonable answer is c. 11. Answer: c. We need a 2:7 mole ratio of ethane and oxygen to react exactly. Equal masses of ethane and oxygen are approximately equivalent to a 1:1 mole ratio, which means that oxygen is the limiting reactant (making a and b false). Having more mass of oxygen does not mean that ethane will be limiting (ex: If we have 5 g of oxygen and 4 g of ethane, oxygen is still the limiting reactant which makes d false.). But, in all cases, if we have more mass of ethane, then oxygen will always be the limiting reactant because the mole ratio of oxygen to ethane will never be greater than 7: Answer: a. The leaf is a semi-permeable membrane, which allows solvent but not solute molecules to pass through. Being a freshwater plant, the inside of the plant is relatively pure water (the concentration of stuff can be assumed to be less than the concentrated salt water solution. Therefore, the vapor pressure of the salt water solution is less than the vapor pressure of the solution in the plant cell. Therefore, the water will flow from the leaf to the salt solution. 13. Answer: a. CH 4, N 2, and He are all non-polar compounds. CO and H 2 O are both polar but water has hydrogen bonding, therefore it has the strongest intermolecular forces. 14. Answer: c. I, II, and V are oxidation-reduction reactions, in which there are a loss of electrons and a gain of electrons by different atoms. III and IV have no loss or gain of electrons by any of the atoms. 15. Answer: b. We start with the equation Rate = k[a] n [B] m and then solve for n and m using the data given. Rate = 4 = k[0.8] n [0.4] m (simplify by canceling) Rate = 1 = k[0.4] n [0.4] m Rate = 4 = [0.8] n / [0.4] n Rate = 4 = 2 n, therefore n = 2 (then solve for m) Rate = 16 = k[1.6] 2 [0.4] m (simplify by canceling) Rate = 4 = k[0.8] 2 [0.4] m Rate = 4 = 2.56[0.4] m / 0.64[0.4] m Rate = 1 = 1 n, therefore m = 1 Therefore, the rate law is Rate = k[a] 2 [B].

3 16. Answer: b. I is the only process accompanied by a chemical change because the given substances becomes new substances with different properties and different compositions, whereas II, III, and IV do not. The equation describing this process is HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) (process III consists of all ions that do not react) 17. Answer: b. III describes a neutral solution. The ph of a solution is temperature dependent (ph = 7.00 for a neutral solution only at 25 C). At the endpoint of a titration of a weak acid with a strong base, for example, the conjugates are in solution, and the conjugate of a weak acid acts as a base; thus, this solution is not neutral. Some salts also exhibit acid or base properties. 18. Answer: d. The relative masses of carbon and oxygen are 12:16 or 3:4, respectively. Thus, if we have 4 carbon atoms for every 3 oxygen atoms in a compound, the percent by mass values of carbon and oxygen will be the same. Or, we can determine the molar mass of each compound, and then determine the mass of oxygen and carbon in one mole of each compound. We find the percent by mass of oxygen and carbon by dividing each mass by the molar mass of each compound and multiplying by Answer: e. A general equilibrium expression is K = [C] l [D] m / [A] j [B] k for the equation ja + kb lc + md The solid compounds are not included in this equilibrium expression because the position of heterogeneous equilibria does not depend on the amounts of pure solids or liquids present (since their concentrations do not change). 20. Answer: a. A ketone consists of two alkyl groups connected to a carbonyl functional group. 21. Answer: a. We have a 1:1 mole ratio of A to C and a 3:4 mole ratio of B to C, thus the reaction is 4A + 3B 4C. Compound A is the limiting reactant because we need a 4:3 mole ratio of A to B and only a 2:3 ratio exists, therefore not enough A is present to react exactly with B. The maximum amount of C that is produced is (2 mol A)(4 mol C / 4 mol A) = 2.0 mol C 22. Answer: c. III and IV are the right answers. Process I will cause the piston to move away because the volume increases due to the temperature increase (but the pressure will remain constant). Process II the piston will move closer because the number of moles of helium will decrease causing the volume to decrease (the pressure will remain constant). Process III the piston is no longer pushing down on the helium because of gravity, therefore the same amount of helium gas will push the piston outward causing the pressure to decrease (and the volume to increase). Process IV less pressure is now pushing down on the piston from the outside, therefore the same amount of helium gas will push the piston upward causing the pressure to decrease (and the volume to increase).

4 23. Answer: e. The reaction is MCl(aq) + AgNO 3 (aq) AgCl(s) + MNO 3 (aq) The number of moles of AgCl is (0.0889g AgCl / g/mol) = x 10-4 mol AgCl. There is a 1:1 mole ratio of AgCl to MCl so we have x 10-4 mol of MCl also. Therefore, the molar mass of MCl is (0.1044g / x 10-4 mol) = g/mol. To find the molar mass of M, subtract the molar mass of Cl from g/mol. Molar Mass of M = g/mol g/mol = g/mol = Molar Mass of Cs 24. Answer: c. The second energy level contains both s and p orbitals. The third energy level contains s, p, and d orbitals. In general, as the energy levels increase, the types of orbitals increase. 25. Answer: a. If we assume that each sample contains x atoms, this represents 1 mole of zinc and 1 mole of aluminum. The mass of Zn is then g (its molar mass) and the mass of Al is g (its molar mass). The mass of Zn is (65.38 g / g) 2.42 times greater than the mass of Al. 26. Answer: c. By looking at the K b of each base, we can determine its strength the higher the K b, the stronger the base. When only the K a is given, we can determine the K b by K w = K a K b (K b = K w / K a ) where K w = 1.0 x Therefore, the K b for F - = 1.39 x and for CN - = 1.61 x The K b for Cl - is extremely small since its conjugate (HCl) is a very strong acid. Therefore, the order is CN -, F -, H 2 O, Cl Answer: d. There are a total of 22 electrons for [ICl 2 ] - when we add up the valence electrons for each element plus the extra electron (7+(2)7+1). When the Lewis structure is drawn, there are two bonding pairs of electrons and three lone pairs on the central I atom. 28. Answer: d. Once a system is at equilibrium, the concentration of reactant(s) and product(s) remain constant over time. The rate of the reverse reaction is equal to the rate of the forward reaction but their rates are not equal to zero (the concentrations never go to zero). Equilibrium is a highly dynamic situation in which the forward and reverse reactions continue to occur so new product molecules are formed even though it appears as though nothing is happening. 29. Answer: a. The energy required to remove a second electron from an ion is known as the second ionization energy. It takes more energy to remove the second electron because there is an increase in positive charge, which binds the electrons more firmly. 30. Answer: d. For the energy level n = 3, s, p, and d orbitals are present. The s orbital can hold two electrons, the p orbitals can hold six electrons, and the d orbitals can hold ten electrons. Thus, the sum is = 18 electrons.

5 31. Answer: b. Using HA as the monoprotic acid, the major species in solution before any reaction takes place are HA, OH -, Na +, and H 2 O. The equation is HA + OH - A - + H 2 O Titrating to the endpoint means that enough OH - has been added to react exactly with the H + from the monoprotic acid. Therefore, the number of moles of NaOH must equal the number of moles of HA. To determine the number of moles, we take ml x mmol/ml = mmol = mol NaOH = mol HA. To calculate the molar mass of HA we take g HA / mol HA = 64.7 g/mol. 32. Answer: e. In Dalton s atomic theory, he discusses how elements are made up of tiny particles called atoms, the atoms of a given element are identical and the atoms on different elements are different in some way(s), compounds are formed when atoms of different elements combine, and chemical reactions involve the reorganization of atoms. 33. Answer: d. A chiral carbon has four different substituents bonded to it. 34. Answer: a. When an electron is excited and moves up into a higher energy level, the process is endothermic because it requires energy. When an electron is excited and moves down into a lower energy level, the process is exothermic because energy is released and the atom is more stable. Furthermore, the atom that releases the most energy when changing energy levels is the most exothermic process. To determine this, we use the equation E = x J [(1/n f 2 ) - (1/n i 2 )]. For a, we get E = x J [(1/1 2 ) - (1/3 2 )] = x J. For b, we get E = x J [(1/2 2 ) - (1/4 2 )] = x J. 35. Answer: c. Use a shorthand table to solve this problem. N 2 (g) + 3H 2 (g) 2NH 3 (g) Initial 5.00M 4.00M 0.00M Change -x -3x +2x Equilibrium 5 x 4 3x 0 + 2x Since we know the concentration of H 2 at equilibrium is 1.00 M, we can determine the value of x to solve for the rest of the equilibrium concentrations. 4 3x = 1, so x = 1 M To solve for K, we use K = [NH 3 ] 2 / [N 2 ][H 2 ] 3 = [2(1)] 2 / [5-1][1] 3 = Answer: e. NaCl is an ionic compound because it is made of a metal and non-metal (and thus has the highest boiling point). SO 2 and NH 3 are polar compounds. C 6 H 14 and CH 4 are non-polar compounds, but C 6 H 14 is a larger molecule (with more electrons), thus it has stronger London Dispersion forces and a higher boiling point than CH 4.

6 37. Answer: b. The equation for percent ionization is % ionization = amount dissociated (mol/l) / initial concentration (mol/l) x 100. When we determine the amount each acid dissociates, we find that acetic acid is 1.34 % ionized. HC 2 H 3 O 2 H C 2 H 3 O 2 Initial 0.10M 0 M 0 M Change -x +x +x Equilibrium 0.10-x x x K a = [H + ][C 2 H 3 O 2 - ] / [HC 2 H 3 O 2 ] = [x][x] / [0.10-x] 1.8 x 10-5 = x 2 / 0.10 x 1.34 x 10-3 M % ionization = [1.34 x 10-3 M / 0.10 M] x 100 = 1.34% 38. Answer: a. We need a 2:1 mole ratio of A and B to react exactly. We only have a 1:1 mole ratio, therefore not enough A is present so this is the limiting reactant. (1 mol A)(1 mol A 2 B / 2 mol A) = 0.50 mol A 2 B 39. Answer: c. (23.0 g C 2 H 5 OH)(1 mol / g C 2 H 5 OH) = 0.50 mol C 2 H 5 OH (0.50 mol C 2 H 5 OH)( 1.37 x 10 3 kj) = x 10 2 kj 40. Answer: a. By looking at the K a of each acid, we can determine its strength the higher the K a, the stronger the acid (and the lower the ph). HCl has the highest K a because it s a strong acid, whereas the rest are weak acids and do not dissociate as well. Therefore, the order is HCN, HC 2 H 3 O 2, HF, HCl.