mastering chemistry chapter 2 homework answer key

The starting materials consist of one green sphere and two purple spheres. The products consist of two green spheres and two purple spheres. This violates Dalton’s postulate that that atoms are not created during a chemical change, but are merely redistributed.

This statement violates Dalton’s fourth postulate: In a given compound, the numbers of atoms of each type (and thus also the percentage) always have the same ratio.

Dalton originally thought that all atoms of a particular element had identical properties, including mass. Thus, the concept of isotopes, in which an element has different masses, was a violation of the original idea. To account for the existence of isotopes, the second postulate of his atomic theory was modified to state that atoms of the same element must have identical chemical properties.

Both are subatomic particles that reside in an atom’s nucleus. Both have approximately the same mass. Protons are positively charged, whereas neutrons are uncharged.

(a) The Rutherford atom has a small, positively charged nucleus, so most α particles will pass through empty space far from the nucleus and be undeflected. Those α particles that pass near the nucleus will be deflected from their paths due to positive-positive repulsion. The more directly toward the nucleus the α particles are headed, the larger the deflection angle will be. (b) Higher-energy α particles that pass near the nucleus will still undergo deflection, but the faster they travel, the less the expected angle of deflection. (c) If the nucleus is smaller, the positive charge is smaller and the expected deflections are smaller—both in terms of how closely the α particles pass by the nucleus undeflected and the angle of deflection. If the nucleus is larger, the positive charge is larger and the expected deflections are larger—more α particles will be deflected, and the deflection angles will be larger. (d) The paths followed by the α particles match the predictions from (a), (b), and (c).

(a) 133 Cs + ; (b) 127 I − ; (c) 31 P 3− ; (d) 57 Co 3+

(a) Carbon-12, 12 C; (b) This atom contains six protons and six neutrons. There are six electrons in a neutral 12 C atom. The net charge of such a neutral atom is zero, and the mass number is 12. (c) The preceding answers are correct. (d) The atom will be stable since C-12 is a stable isotope of carbon. (e) The preceding answer is correct. Other answers for this exercise are possible if a different element of isotope is chosen.

(a) Lithium-6 contains three protons, three neutrons, and three electrons. The isotope symbol is 6 Li or 3 6 Li . 3 6 Li . (b) 6 Li + or 3 6 Li + 3 6 Li +

(a) Iron, 26 protons, 24 electrons, and 32 neutrons; (b) iodine, 53 protons, 54 electrons, and 74 neutrons

(a) 3 protons, 3 electrons, 4 neutrons; (b) 52 protons, 52 electrons, 73 neutrons; (c) 47 protons, 47 electrons, 62 neutrons; (d) 7 protons, 7 electrons, 8 neutrons; (e) 15 protons, 15 electrons, 16 neutrons

Let us use neon as an example. Since there are three isotopes, there is no way to be sure to accurately predict the abundances to make the total of 20.18 amu average atomic mass. Let us guess that the abundances are 9% Ne-22, 91% Ne-20, and only a trace of Ne-21. The average mass would be 20.18 amu. Checking the nature’s mix of isotopes shows that the abundances are 90.48% Ne-20, 9.25% Ne-22, and 0.27% Ne-21, so our guessed amounts have to be slightly adjusted.

Turkey source: 20.3% (of 10.0129 amu isotope); US source: 19.1% (of 10.0129 amu isotope)

The symbol for the element oxygen, O, represents both the element and one atom of oxygen. A molecule of oxygen, O 2 , contains two oxygen atoms; the subscript 2 in the formula must be used to distinguish the diatomic molecule from two single oxygen atoms.

(a) molecular CO 2 , empirical CO 2 ; (b) molecular C 2 H 2 , empirical CH; (c) molecular C 2 H 4 , empirical CH 2 ; (d) molecular H 2 SO 4 , empirical H 2 SO 4

(a) C 4 H 5 N 2 O; (b) C 12 H 22 O 11 ; (c) HO; (d) CH 2 O; (e) C 3 H 4 O 3

(a) CH 2 O; (b) C 2 H 4 O

(a) ethanol

(b) methoxymethane, more commonly known as dimethyl ether

(c) These molecules have the same chemical composition (types and number of atoms) but different chemical structures. They are structural isomers.

(a) metal, inner transition metal; (b) nonmetal, representative element; (c) metal, representative element; (d) nonmetal, representative element; (e) metal, transition metal; (f) metal, inner transition metal; (g) metal, transition metal; (h) nonmetal, representative element; (i) nonmetal, representative element; (j) metal, representative element

(a) He; (b) Be; (c) Li; (d) O

(a) krypton, Kr; (b) calcium, Ca; (c) fluorine, F; (d) tellurium, Te

(a) 11 23 Na 11 23 Na ; (b) 54 129 Xe 54 129 Xe ; (c) 33 73 As 33 73 As ; (d) 88 226 Ra 88 226 Ra

Ionic: KCl, MgCl 2 ; Covalent: NCl 3 , ICl, PCl 5 , CCl 4

(a) covalent; (b) ionic, Ba 2+ , O 2− ; (c) ionic, NH 4 + , NH 4 + , CO 3 2− ; CO 3 2− ; (d) ionic, Sr 2+ , H 2 PO 4 − ; H 2 PO 4 − ; (e) covalent; (f) ionic, Na + , O 2−

(a) CaS; (b) (NH 4 ) 2 SO 4 ; (c) AlBr 3 ; (d) Na 2 HPO 4 ; (e) Mg 3 (PO 4 ) 2

(a) cesium chloride; (b) barium oxide; (c) potassium sulfide; (d) beryllium chloride; (e) hydrogen bromide; (f) aluminum fluoride

(a) RbBr; (b) MgSe; (c) Na 2 O; (d) CaCl 2 ; (e) HF; (f) GaP; (g) AlBr 3 ; (h) (NH 4 ) 2 SO 4

(a) ClO 2 ; (b) N 2 O 4 ; (c) K 3 P; (d) Ag 2 S; (e) AIF 3 ·3H 2 O; (f) SiO 2

(a) chromium(III) oxide; (b) iron(II) chloride; (c) chromium(VI) oxide; (d) titanium(IV) chloride; (e) cobalt(II) chloride hexahydrate; (f) molybdenum(IV) sulfide

(a) K 3 PO 4 ; (b) CuSO 4 ; (c) CaCl 2 ; (d) TiO 2 ; (e) NH 4 NO 3 ; (f) NaHSO 4

(a) manganese(IV) oxide; (b) mercury(I) chloride; (c) iron(III) nitrate; (d) titanium(IV) chloride; (e) copper(II) bromide

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12.E: Homework Chapter 12 Answers

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Chemical Species/Chemical Interactions:

1. a.) Cu (s)                Atomic

b.) H 2 O (s)         Molecular

c.) NaCl (s)          Ionic

d.) Fe (s)            Atomic

3. This mixture would be miscible because both molecules are polar.

5. A solution that mixes completely, without any particles left unmixed. Homogenous mixtures are most commonly thought of as a mixture behaving as a single substance.

Intermolecular Forces:

7.  a.) CH 4         Dispersion

b.) HF          H-Bond

c.) H 2 O        H-Bond

d.) CHCl 3    Dipole-Dipole

9. a.) CH 3 CH 2 OH   Dispersion, Dipole-Dipole, H-Bond

b.) CCl 4                Dispersion

c.) CHF 3               Dispersion, Dipole-Dipole

11. When a Hydrogen atom is bonded directly to a Nitrogen, Fluorine, or Oxygen atom.

13. CH 4 , CH 3 CHO 2 , HF

15. a.) CO Dispersion, Dipole-Dipole

b.) HBr Dispersion, Dipole-Dipole

c.) CCl 4 Dispersion

d.) KCl ion-ion

17. c.) H-Bond

19. a.) ammonia                 H-Bond

b.) silicon dioxide       Dispersion

c.) ethanol                     H-Bond

d.) acetic acid               H-Bond

21. b.) Dipole-Dipole

23. a.) CH 4

25. a.) I 2                    Dispersion

b.) F 2                   Dispersion

c.) SO 2                       Dipole-Dipole

d.) HCH 3 COO   H-Bond

Boiling/Melting Point:

27. The molecule with the higher boiling point will be the molecule with the strongest intermolecular force. If they all have the same intermolecular force, then you will need to determine which molecule is the longest.

29. *remember: Ionic bonding is stronger than Hydrogen bonding!*

NaCl > CH 3 CH 2 OH > CHCl 3 > Ar

31. At room temperature, the molecule with the stronger intermolecular force will be a liquid. CO 2 only possesses dispersion forces, while CH 3 OH possesses hydrogen-bonding forces. Therefore, CH 3 OH will be a liquid at room temperature.

33. NaCl will have the higher melting point because ionic bonding between sodium and chlorine is stronger than the dispersion forces in methane.

35. a.) NH 4 Cl          *Remember: NH 4 + and Cl - form an ionic bond*

37. c.) CH 4

39. b.) HCl

41. HI > HBr > HCl

43. Octane > Hexane > Butane

45. No, ethanol and propane will not have the same boiling point because we have to put the intermolecular forces into account when deciding boiling points—not just their molar mass. Because ethanol has hydrogen-bonding forces, ethanol will have the higher boiling point.

47. a.) propanol and methanol

b.) methane and pentane

c.) dihydrogen monoxide and dihydrogen monosulfide

49. Acetic acid would most likely be a liquid at room temperature because of the strong hydrogen-bonding forces.

51. CH 3 CH 2 COH will have the higher boiling point because of the strong hydrogen-bonding forces present in the molecule.

53. b.) CH 4

Cumulative/Challenge Problems:

55. No, a hydrogen-bond would not be able to form by connecting two acetone molecules together because the hydrogen atom would not be covalently bonded to the oxygen atom in the structure. For a hydrogen-bond to form, the hydrogen atom must be covalently bonded to either Nitrogen, Fluorine, or Oxygen atoms.

57. b.) vinegar

59. a.) acetone    Dipole-Dipole

b.) butane          Dispersion

c.) iso-butane   Dispersion

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