Chemistry 2 Midterm Review Answers. CHAPTER 1. 1. a. mixture ... 1 x 10-3 s. 7.
a. density = mass/volume density = 39.75 g x 1 ml = 1.59 g/cm3. 25 mL 1 cm3.
Chemistry 2 Midterm Review Answers CHAPTER 1 1. a. mixture, homogeneous b. mixture, heterogeneous (if particles in liquid) c. pure substance, element d. mixture, probably heterogeneous but would not be determined by visual inspection alone e. . pure substance, element f. pure substance, compound g. mixture, heterogeneous
2. A(s) B(s) + C(g) When carbon(s) is burned in excess oxygen, the two elements combine to form a gaseous compound, carbon dioxide. Clearly substance C is this compound. Since C is produced when A is heated in the absence of oxygen (from air), both the carbon and oxygen in C must have been present in A originally. A is, therefore, a compound composed of two or more elements chemically combined. Without more information on the chemical or physical properties of B, we cannot determine absolutely whether it is an element or a compound. However, few if any elements exist as white solids, so B is probably also a compound.
3. a. gas
b. solid
4. a. chemical e. chemical
5. a. mass
c. liquid
b. physical
b. density
e. energy
c. physical
c. volume
f. length
7. a. density = mass/volume
d. temperature
g. amount
6. a. 2.52 x 103 kg x 1 x 103 g = 2.52 x 106 g 1 kg c. 6.25 x 10-4 s x 1 ms 1 x 10-3 s
d. chemical
h. molar mass
b. 0.0023 mm x 1 x 10-3 m x 1 nm 1 mm 1 x 10-9 m
= 0.625 ms
density = 39.75 g x 1 ml 25 mL 1 cm3
= 1.59 g/cm3
b. 75.0 cm3 x 23.4 g/cm3 = 1.76 x 103 g (1.76 kg) c. 275 g x 1 cm3 1.74 g 8. a. 4 9. a. 1.44 x 105
b. 3
= 158 cm3 (158 mL)
c. 5 b. 9.75 x 10-2
d. 6 c. 8.90 x 105
e. 6 d. 6.76 x 104
= 2.3 x 103 nm
e. 3.40 x 104
f. -6.56
10. a. -2.3 x 103
b. [285.3 x 105 – 0.01200 x 105] x 2.8954 = 8.260 x 107
c. 3.4 x 104
d. 7.62 x 105
11. a. 76 mL
b. 50. nm
CHAPTER 2 12. a. 15 p, 17 n e. 53 p, 78 n
c. 6.88 x 10-4 s
b. 24 p, 27 n f. 81 p, 120 n
d. 1.55 g/L
c. 27 p, 33 n
d. 43 p, 56 n
13. SYMBOL
46
103
PROTONS
22
45
16
52
38
NEUTRONS
24
58
18
75
50
ELECTRONS
22
45
16
52
38
ATOMIC #
22
45
16
52
38
MASS #
46
103
34
127
88
Ti
Rh
14. a. lithium, metal b. scandium, metal d. ytterbium, metal e. manganese, metal g. tellurium, metalloid
34
S
127
Te
88
Sr
c. germanium, metalloid f. gold, metal
15. Average atomic mass = [23.98504 x 0.7870] + [24.98584 x 0.1013] + [25.98259 x 0.1117] = 24.31 amu Atomic weights are really average atomic masses, the sum of the mass of each naturally-occurring isotope of an element times is percent abundance. Each Mg atom will have the mass of one of the naturally-occurring isotopes, while the “atomic weight” (listed on the periodic table) is a weighted average value.
16. a. 13 protons, 13 neutrons c. 29 protons, 26 neutrons
b. 47 protons, 62 neutrons d. 9 protons, 10 neutrons
17. a. 6
b. 6
c. 12
18. a. SN
b. C7H15
c. C3H5O
d. P2O3
e. C3H5F4
f. SiO3 19. a. Al3+
b. Ca2+
d. I-
c. S2-
20. a. GaF3, gallium fluoride d. K2S, potassium sulfide
b. LiH, lithium hydride
21. Molecular (all nonmetal atoms): c a. scandium(III) oxide b. sodium iodide c. sulfur dichloride d. calcium nitrate e. iron(III) chloride f. lanthanum phosphide g. cobalt(II) carbonate h. ammonium sulfate
22. a. silver(I) sulfide d. strontium sulfite 23. a. Mg3N2 24. a. bromic acid e. H2SO3
e. Cs+ c. AlI3, aluminum iodide
Ionic: (metal and nonmetal) a, b, d, e, f, g, h
b. barium phosphate e. cobalt (II) bromide b. FeSO3
c. magnesium chlorate
c. Cr2(CO3)3
b. hydrobromic acid
c. phosphoric acid
d. HIO3
CHAPTER 3 25. a) P4 indicates that there are four phosphorus atoms bound together by chemical bonds into a single molecule, whereas 4P denotes four separate phosphorus atoms. b) This equation is inconsistent with the law of conservation of matter because the reactant side has a Mg2+ ion while the product side has a Ca2+ ion. Also, the reactant side has two more H atoms and one more O atom than the products side. 26. a. 2KNO3 2KNO2 + O2 c. NCl3 + 3H2O NH3 + 3HOCl 2KNO3
b. La2O3 + 3H2O 2La(OH)3 d. Mg3N2 + 8HCl 3MgCl2 + 2NH4Cl
27. a. CaC2(s) + 2H2O(l) Ca(OH)2(aq) + C2H2(g) c. Zn(s) + H2SO4(aq) H2(g) + ZnSO4(aq)
b. 2KClO3(s) 2KCl(s) + 3O2(g)
28. a. 2C3H6(g) + 9O2(g) 6CO2(g) + 6H2O(l) combustion b. NH4NO3(s) NO2(g) + 2H2O(l) decomposition c. N2(g) + 3H2(g) 2NH3(g) combination (aka synthesis)
29. a. 108.0 g/mol
b. 115.8 g/mol
e. 2AgNO3 + K2SO4 Ag2SO4 +
c. 158.1 g/mol
30. a. 46.0 g/mol; 69.6% O c. 238.0 g/mol; 60.5% O
b. 74.0 g/mol; 43.2% O
31. a. 162.28 g/mol b. 3.08 x 10-5 mol c. 1.86 x 1019 molecules d. 3.71 x 1019 S atoms 32. a. molar mass is 96.09 g/mol; 0.0116 mol ammonium (Calculate ammonium like you would calculate atoms, like you did in problem 31d.) b. molar mass is 150.82 g/mol; 6.61 g c. molar mass 180.15 g/mol; 80.5 g d. 284.7 g 33.
a. K2CO3
b. SnF4
c. NH2
34. C6H12 35. a. 4.0 mol HF
b. 3.15 g NaF
36. a. Al2S3(s) + 6H2O(l) 2Al(OH)3(s) + 3H2S(g) 37.
0.282 M
38.
0.834 L
39. a. 60.0 g
c. 0.229 g Na2SiO3 b. 10.9 g aluminum hydroxide
b. 27.0 g
40. 126 mL HCl 41. 51.5 g Fe 42. 44.4 g FeCl3 43. Theoretical yield is the maximum amount of product possible, as predicted by stoichiometry, assuming that the limiting reactant is converted entirely to product. Actual yield is the amount of product actually obtained in the lab, usually less than or equal to the theoretical yield. Percent yield is the ratio of actual yield to theoretical yield multiplied by 100.
44. a. NaHCO3 b. 0.524 g CO2 -3 c. 3.97 x 10 mol citric acid react so 1.24 x 10-3 mol remain Therefore, 0.238 g citric acid remain
45. 2.98 g CO2 formed 46. a. C6H6 is the limiting reactant and 60.3 g C6H5Br is the theoretical yield b. Percent yield is 94.0% CHAPTER 5 47. Kinetic energy is the energy of motion. Potential energy is stored energy or energy of position.
48. a) endothermic d) exothermic
b) exothermic
c) exothermic
49. The system is the solution. The surroundings are the flask, the stopper, and the area around the flask. 50. Heat is a form of energy that flows between two samples of matter because of their differences in temperature. Energy transfer as heat will occur spontaneously from an object at higher temperature to an object at a lower temperature. Heat flows between system and surroundings. 51. a) ΔE = -152 kJ (exothermic) c) ΔE = 14 kJ (endothermic)
b) ΔE=0.746 kJ (endothermic)
52. Specific heat capacity is the amount of energy needed to raise 1 g of the substance by 1°C. Water has a large value, meaning that it will absorb or release more energy than other substances without its temperature changing. Temperature changes will occur more slowly. 53. The PE diagram for an exothermic reaction has the following general shape:
54. A state function is only dependent upon the initial versus final states, not the pathway to reach the final state. A path function depends on, and will change, depending on the pathway from initial to final states. 55. a) ΔH is positive, endothermic b) 127 kJ absorbed c) 0.735 g H2 56. q = 9.04 x 105 J 57. q = 3.47 x 104 J 58. ΔH = -2486.3 kJ 59. Light has particle-like characteristics as it travels through space. It exhibits properties of both waves and particles. 60. c = λν 61. E=hv 62. a) inversely proportional c) directly proportional
b) inversely proportional
63. The AM station has a frequency of 1.440 x 106 Hz and an energy of 9.54 x 10-28 J.
The FM station has a frequency of 9.45 x 107 Hz and an energy of 6.26 x 10-26 J. 64. Electrons travel in orbits around the nucleus. Each energy level has a fixed energy meaning the energy is quantized. Electrons further from the nucleus have a greater energy than electrons closer to the nucleus. Excited electrons move from their ground state to an excited state. As they fall back down to the ground state, they emit a photon of light equal in energy to the difference between the ground state and excited state energies. 65. The orbital defines an area around the nucleus where the probability of finding the electron there is 90% at any given time, whereas the orbits of Bohr’s model defined the location of the electron at all times in its ground state. Different orbitals have different amount of energy like in Bohr’s model. 66. It is impossible to know both the position and velocity of an electron in an atom at any given point in time. 67. a. b. c. d. e.
Principle quantum number, n, whole # integers from 1 to 7, describes the main energy level and size of orbitals Angular momentum quantum number, l, values from 0 to n-1, describes the orbital shape and sublevels Magnetic quantum number, ml, integers from –l through +l, describes the orientation of the orbital shapes Spin quantum number, ms, +1/2 or -1/2, describes the direction spin of the two electrons in each orbital
68. No two electrons can have the same four quantum numbers. 69. a) l = 0, 1, 2, 3 b) ml= -2, -1, 0, 1, 2 70. Hydrogen only has two electrons which would both go in the 1s orbital. None of these sets of quantum numbers describes an electron in the 1s orbital. [“a” is 2p, “b” is incorrect as it would need ml of 0, “c” is 4d, and “d” does not exist] 71. a) For 2p: n=2, l=1, and ml = -1, 0, or +1; Each ml orbital contains two electrons with a spin of +1/2 and -1/2 b) For 5d: n=5, l=2, ml = -2, -1, 0, +1, or +2; Each ml orbital contains two electrons with a spin of +1/2 and -1/2
