mastering chemistry chapter 7 homework answers

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7.E: Chapter 7 Homework Answers

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Evidence of a Chemical Reaction

Writing and Balancing Chemical Equations

5. b) 

a) Zn(s) + 2HCl(aq) → ZnCl 2 (aq) + H 2 (g)

b) CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O(l)

c) Ca(OH) 2 (aq) + CO 2 (g) → CaCO 3 (s) + H 2 O(l)

d) 2Fe 2 O 3 (s) + 6Cl 2 (g) → 4FeCl 3 (aq) + 3O 2 (g)

a) N 2 O 3 (aq) + H 2 O(l) → 2HNO 2 (aq)

b) Xe(g) + 3F 2 (g) → XeF 6 (s)

c) NH 4 NO 3 (aq) → N 2 O(g) + 2H 2 O(l)

d) I 2 (s) + 6HNO 3 (aq) → 2HIO 3 + 6NO 2 (g) + 2H 2 (g)

11. C 6 H 24 (g) + 12O 2 (g) → 12H 2 O(l) + 6CO 2 (g)

13. C 6 H 12 O 6 (aq) → 6C(s) + 6H 2 O(l)

15. 2H 2 O(g) + 2Na(s) → 2NaOH(s) + H 2 (g)

a) 2C 4 H 10 (g) + 13O 2 (g) → 8CO 2 (g) + 10H 2 O(l)

b) 2MgS(s) + 3O 2 (g) → 2MgO(aq) + 2SO 2 (g)

c) 3P 4 S 3 (s) + 16KClO 3 (aq) → 6P 2 O 5 (aq) + 16KCl(aq) + 9SO 2 (g)

d) Fe 3 O 4 (aq) + 4CO(g) → 4CO 2 (g) + 3Fe(s)

a) Correct.

b) Incorrect. You would need a 2 coefficient before HNO 2 (aq).

c) Incorrect. Rather than having a 2 coefficient on F 2 (g) and Cl 2 (g), you need to put a 2 coefficient on both NaCl(aq) and NaF(g) in order for this equation to be balanced.

d) Correct.

a) H 2 O(l) → H 2 O(g)

b) H 2 O(l) → H 2 O(s)

c) H 2 O(g) → H 2 O (l)

Since all of the above reactions are only physical changes, they require no balancing because they are already balanced.

23. No. The water molecules will still attract both the negative and the positive ions in the substance, but the polyatomic ions will break into units where the ion is intact. NO 3 - ions would not break down to Nitrogen and Oxygen ions.

Monoatomic Ion: KCl dissociates into K + and Cl - ions

Polyatomic Ion: KNO 3 dissociates into K + and NO 3 - ions

27.        BaSO 4 (aq)       CaCO 3 (aq)                                                                                                       

29. a) strong electrolyte

a) CaCrO 4 (aq) + Cu(NO 3 ) 2 (aq) → Ca(NO 3 ) 2 (aq) + CuCrO 4 (s)

b) 2KBr(aq) + Hg 2 SO 4 (aq) → K 2 SO 4 (aq) + Hg 2 (Br) 2 (s)

c) 2LiNO 3 (aq) + 2(NH 4 ) 2 S(aq) → 2NH 4 NO 3 (aq) + Li 2 S(s)

d) Pb(NO 3 ) 2 (aq) + 2NaCl(aq) → 2NaNO 3 (aq) + PbCl 2 (s)

a) This reaction implies a phase change for both mixtures, which is incorrect.

Na 2 SO 4 (aq) + BaCl 2 (aq) → 2NaCl(aq) + BaSO 4 (s)

b) This reaction is not balanced. There should be a 2 coefficient before KI(aq) and KNO 3 (aq)

2KI(aq) + Pb(NO 3 ) 2 (aq) → 2KNO 3 (aq) + PbI 2 (s)

c) This reaction states that Ag 2 S is soluble when it is actually insoluble. It also states NaNO 3 is insoluble when it is in fact soluble.

2AgNO 3 (aq) + Na 2 S(aq) → Ag 2 S(s) + 2NaNO 3 (aq)

35. Beaker A would form a precipitate because the reaction would be:

2K 3 PO 4 (aq) + 3Ca(NO 3 ) 2 (aq) → 6KNO 3 (aq) + Ca 3 (PO 4 ) 2 (s).

Beaker B would form a precipitate because the reaction would be:

2NaOH(aq) + BaBr 2 (aq) → Ba(OH) 2 (s) + 2NaBr(aq).

Ionic and Net Ionic Equations

37. A single displacement reaction involves the replacement of an element in a compound.

KCl(aq) + Na(s) → NaCl(aq) + K(s)

A double displacement reaction involves the replacement of an element in each compound. You ultimately switch the anions (or the cations) present in the reactants to get your products.

KCl(aq) + AgNO 3 (aq) → AgCl(s) + KNO 3 (aq)

39. a) Total Ionic: 2Rb + (aq) + 2F - (aq) + Cu 2+ (aq) + SO 4 2- (aq) → 2Rb + (aq) + 2F - (aq) + Cu 2+ (aq) + SO 4 2- (aq)

Net Ionic: No Reaction

b) Total Ionic: Sr 2+ (aq) + 2Br - (aq) + 2K + (aq) + SO 4 2- (aq) → SrSO 4 (s) + 2K + (aq) + 2Br - (aq)

Net Ionic: Sr 2+ (aq) + SO 4 2- (aq) → SrSO 4 (s)

c) Total Ionic: Na + (aq) + OH - (aq) + H + (aq) + Cl - (aq) → Na + (aq) + Cl - (aq) + H 2 O(l)

Net Ionic:  H + (aq) +OH - (aq) → H 2 O(l)

d) Total Ionic: 2Bi(OH) 3 (s) + 6H + (aq) + 3SO 4 2- (aq) → 2Bi 3+ (aq) + 3SO 4 2- (aq) + 6H 2 O(l)

Net Ionic: 2Bi(OH) 3 (s) + 6H + (aq) → 2Bi 3+ (aq) + 6H 2 O(l)

Acid-Base Reactions

41. Neutralization, water, salt

43. a) 2HClO 3 (aq) + Zn(OH) 2 (s) → Zn(ClO 3 ) 2 (aq) + 2H 2 O(l)

b) H 2 SO 4 (aq) + Ba(OH) 2 (s) → BaSO 4 (s) + 2H 2 O(l)

c) HBr(aq) + LiOH(aq) → LiBr(aq) + H 2 O(l)

d) 2HF(aq) + Ca(OH) 2 (aq) → CaF 2 (s) + 2H 2 O(l)

Combustion Reactions

45. a) 2C 6 H 6 (l) + 15O 2 (g) → 6H 2 O(l) + 12CO 2 (g)

b) C 6 H 12 O 6 (aq) + 6O 2 (g) → 6H 2 O(l) + 6CO 2 (g)

c) C 2 H 5 OH(l) + 3O 2 (g) → 3H 2 O(l) + 2CO 2 (g)

Classifying Chemical Reactions

47. a) Double Displacement           b) Decomposition         c) Single Displacement       d) Synthesis

49. a) Decomposition                         b) Double Displacement                          c) Single Displacement

51. a) Decomposition                         b) Single Displacement                          c) Single Displacement

Cumulative Challenge Problems

a) Molecular: BaCl 2 (aq) + Na 2 SO 4 (aq) → 2NaCl(aq) + BaSO 4 (s)

Total Ionic: Ba 2+ (aq) + 2 Cl - (aq) + 2Na + (aq) + SO 4 2- (aq) → 2Na + (aq) + 2Cl - (aq) + BaSO 4 (s)

Net Ionic: Ba 2+ (aq) + SO 4 2- (aq) → BaSO 4 (s)

b) Molecular:  Na 2 S(aq) + 2LiF(aq) → No Reaction

c) Molecular: H 2 SO 4 (aq) + Pb(NO 3 ) 2 (aq) → PbSO 4 (s) + 2HNO 3 (aq)

Total Ionic: 2H + (aq) + SO 4 2- (aq) + Pb 2+ (aq) + 2NO 3 - (aq) → PbSO 4 (s) + 2H + (aq) + 2NO 3 - (aq)

Net Ionic: Pb 2+ (aq) + SO 4 2- (aq) → PbSO 4 (s)

55. a) Gas Evolution Reaction

Molecular: 2HClO 4 (aq) + CaCO 3 (s) → H 2 O(l) + CO 2 (g) + Ca(ClO 4 ) 2 (aq)

Sum of Coefficients: 6

b) Precipitation Reaction

Molecular: 2Fe(s) + 3Cu(NO 3 ) 2 (aq) → 3Cu(s) + 2Fe(NO 3 ) 3 (aq)

Sum of Coefficients: 10

c) Combustion Reaction

Molecular: C 9 H 20 (l) + 14O 2 (g) → 10H 2 O(l) + 9CO 2 (g)

Sum of Coefficients: 34

57. a) 2(CH 3 ) 2 CHOH(l) + 9O 2 (g) → 8H 2 O(l) + 6CO 2 (g)

b) C 10 H 8 (s) + 12O 2 (g) → 4H 2 O(l) + 10CO 2 (g)

c) 2CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 3 (l) + 37O 2 (g) → 26H 2 O(l) + 24CO 2 (g)

d) C 12 H 22 O 11 (aq) + 12O 2 (g) → 11H 2 O(l) + 12CO 2 (g)

The protons in the nucleus do not change during normal chemical reactions. Only the outer electrons move. Positive charges form when electrons are lost.

P, I, Cl, and O would form anions because they are nonmetals. Mg, In, Cs, Pb, and Co would form cations because they are metals.

(a) P 3– ; (b) Mg 2+ ; (c) Al 3+ ; (d) O 2– ; (e) Cl – ; (f) Cs +

(a) [Ar]4 s 2 3 d 10 4 p 6 ; (b) [Kr]4 d 10 5 s 2 5 p 6 (c) 1 s 2 (d) [Kr]4 d 10 ; (e) [He]2 s 2 2 p 6 ; (f) [Ar]3 d 10 ; (g) 1 s 2 (h) [He]2 s 2 2 p 6 (i) [Kr]4 d 10 5 s 2 (j) [Ar]3 d 7 (k) [Ar]3 d 6 , (l) [Ar]3 d 10 4 s 2

(a) 1 s 2 2 s 2 2 p 6 3 s 2 3 p 1 ; Al 3+ : 1 s 2 2 s 2 2 p 6 ; (b) 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 10 4 s 2 4 p 5 ; 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 10 4 s 2 4 p 6 ; (c) 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 10 4 s 2 4 p 6 5 s 2 ; Sr 2+ : 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 10 4 s 2 4 p 6 ; (d) 1 s 2 2 s 1 ; Li + : 1 s 2 ; (e) 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 10 4 s 2 4 p 3 ; 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 10 4 s 2 4 p 6 ; (f) 1 s 2 2 s 2 2 p 6 3 s 2 3 p 4 ; 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6

NaCl consists of discrete ions arranged in a crystal lattice, not covalently bonded molecules.

ionic: (b), (d), (e), (g), and (i); covalent: (a), (c), (f), (h), (j), and (k)

(a) Cl; (b) O; (c) O; (d) S; (e) N; (f) P; (g) N

(a) H, C, N, O, F; (b) H, I, Br, Cl, F; (c) H, P, S, O, F; (d) Na, Al, H, P, O; (e) Ba, H, As, N, O

N, O, F, and Cl

(a) HF; (b) CO; (c) OH; (d) PCl; (e) NH; (f) PO; (g) CN

(a) eight electrons:

; (b) eight electrons:

; (c) no electrons Be 2+ ; (d) eight electrons:

; (e) no electrons Ga 3+ ; (f) no electrons Li + ; (g) eight electrons:

In this case, the Lewis structure is inadequate to depict the fact that experimental studies have shown two unpaired electrons in each oxygen molecule . (b)

(a) SeF 6 :

; (b) XeF 4 :

; (c) SeCl 3 + : SeCl 3 + :

; (d) Cl 2 BBCl 2 :

Two valence electrons per Pb atom are transferred to Cl atoms; the resulting Pb 2+ ion has a 6 s 2 valence shell configuration. Two of the valence electrons in the HCl molecule are shared, and the other six are located on the Cl atom as lone pairs of electrons.

Each bond includes a sharing of electrons between atoms. Two electrons are shared in a single bond; four electrons are shared in a double bond; and six electrons are shared in a triple bond.

CO has the strongest carbon-oxygen bond because there is a triple bond joining C and O. CO 2 has double bonds.

(a) H: 0, Cl: 0; (b) C: 0, F: 0; (c) P: 0, Cl 0; (d) P: 0, F: 0

Cl in Cl 2 : 0; Cl in BeCl 2 : 0; Cl in ClF 5 : 0

The structure that gives zero formal charges is consistent with the actual structure:

(a) −114 kJ; (b) 30 kJ; (c) −1055 kJ

The greater bond energy is in the figure on the left. It is the more stable form.

HCl ( g ) ⟶ 1 2 H 2 ( g ) + 1 2 Cl 2 ( g ) Δ H 1 ° = −Δ H f [ HCl ( g ) ] ° 1 2 H 2 ( g ) ⟶ H ( g ) Δ H 2 ° = Δ H f [ H ( g ) ] ° 1 2 Cl 2 ( g ) ⟶ Cl ( g ) Δ H 3 ° = Δ H f [ Cl ( g ) ] ° ¯ HCl ( g ) ⟶ H ( g ) + Cl ( g ) Δ H 298 ° = Δ H 1 ° + Δ H 2 ° + Δ H 3 ° HCl ( g ) ⟶ 1 2 H 2 ( g ) + 1 2 Cl 2 ( g ) Δ H 1 ° = −Δ H f [ HCl ( g ) ] ° 1 2 H 2 ( g ) ⟶ H ( g ) Δ H 2 ° = Δ H f [ H ( g ) ] ° 1 2 Cl 2 ( g ) ⟶ Cl ( g ) Δ H 3 ° = Δ H f [ Cl ( g ) ] ° ¯ HCl ( g ) ⟶ H ( g ) + Cl ( g ) Δ H 298 ° = Δ H 1 ° + Δ H 2 ° + Δ H 3 ° D HCl = Δ H 298 ° = Δ H f [ HCl ( g ) ] ° + Δ H f [ H ( g ) ] ° + Δ H f [ Cl ( g ) ] ° = − ( −92.307 kJ ) + 217.97 kJ + 121.3 kJ = 431.6 kJ D HCl = Δ H 298 ° = Δ H f [ HCl ( g ) ] ° + Δ H f [ H ( g ) ] ° + Δ H f [ Cl ( g ) ] ° = − ( −92.307 kJ ) + 217.97 kJ + 121.3 kJ = 431.6 kJ

The S–F bond in SF 4 is stronger.

The C–C single bonds are longest.

(a) When two electrons are removed from the valence shell, the Ca radius loses the outermost energy level and reverts to the lower n = 3 level, which is much smaller in radius. (b) The +2 charge on calcium pulls the oxygen much closer compared with K, thereby increasing the lattice energy relative to a less charged ion. (c) Removal of the 4 s electron in Ca requires more energy than removal of the 4 s electron in K because of the stronger attraction of the nucleus and the extra energy required to break the pairing of the electrons. The second ionization energy for K requires that an electron be removed from a lower energy level, where the attraction is much stronger from the nucleus for the electron. In addition, energy is required to unpair two electrons in a full orbital. For Ca, the second ionization potential requires removing only a lone electron in the exposed outer energy level. (d) In Al, the removed electron is relatively unprotected and unpaired in a p orbital. The higher energy for Mg mainly reflects the unpairing of the 2 s electron.

4008 kJ/mol; both ions in MgO have twice the charge of the ions in LiF; the bond length is very similar and both have the same structure; a quadrupling of the energy is expected based on the equation for lattice energy

(a) Na 2 O; Na + has a smaller radius than K + ; (b) BaS; Ba has a larger charge than K; (c) BaS; Ba and S have larger charges; (d) BaS; S has a larger charge

The placement of the two sets of unpaired electrons in water forces the bonds to assume a tetrahedral arrangement, and the resulting HOH molecule is bent. The HBeH molecule (in which Be has only two electrons to bond with the two electrons from the hydrogens) must have the electron pairs as far from one another as possible and is therefore linear.

Space must be provided for each pair of electrons whether they are in a bond or are present as lone pairs. Electron-pair geometry considers the placement of all electrons. Molecular structure considers only the bonding-pair geometry.

As long as the polar bonds are compensated (for example. two identical atoms are found directly across the central atom from one another), the molecule can be nonpolar.

(a) Both the electron geometry and the molecular structure are octahedral. (b) Both the electron geometry and the molecular structure are trigonal bipyramid. (c) Both the electron geometry and the molecular structure are linear. (d) Both the electron geometry and the molecular structure are trigonal planar.

(a) electron-pair geometry: octahedral, molecular structure: square pyramidal; (b) electron-pair geometry: tetrahedral, molecular structure: bent; (c) electron-pair geometry: octahedral, molecular structure: square planar; (d) electron-pair geometry: tetrahedral, molecular structure: trigonal pyramidal; (e) electron-pair geometry: trigonal bypyramidal, molecular structure: seesaw; (f) electron-pair geometry: tetrahedral, molecular structure: bent (109°)

(a) electron-pair geometry: trigonal planar, molecular structure: bent (120°); (b) electron-pair geometry: linear, molecular structure: linear; (c) electron-pair geometry: trigonal planar, molecular structure: trigonal planar; (d) electron-pair geometry: tetrahedral, molecular structure: trigonal pyramidal; (e) electron-pair geometry: tetrahedral, molecular structure: tetrahedral; (f) electron-pair geometry: trigonal bipyramidal, molecular structure: seesaw; (g) electron-pair geometry: tetrahedral, molecular structure: trigonal pyramidal

All of these molecules and ions contain polar bonds. Only ClF 5 , ClO 2 − , ClO 2 − , PCl 3 , SeF 4 , and PH 2 − PH 2 − have dipole moments.

SeS 2 , CCl 2 F 2 , PCl 3 , and ClNO all have dipole moments.

(a) tetrahedral; (b) trigonal pyramidal; (c) bent (109°); (d) trigonal planar; (e) bent (109°); (f) bent (109°); (g) C H 3 CCH tetrahedral, CH 3 CC H linear; (h) tetrahedral; (i) H 2 C C CH 2 linear; H 2 C C C H 2 trigonal planar

; (d) CS 3 2− CS 3 2− includes three regions of electron density (all are bonds with no lone pairs); the shape is trigonal planar; CS 2 has only two regions of electron density (all bonds with no lone pairs); the shape is linear

The Lewis structure is made from three units, but the atoms must be rearranged:

The molecular dipole points away from the hydrogen atoms.

The structures are very similar. In the model mode, each electron group occupies the same amount of space, so the bond angle is shown as 109.5°. In the “real” mode, the lone pairs are larger, causing the hydrogens to be compressed. This leads to the smaller angle of 104.5°.

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• Book title: Chemistry
• Publication date: Mar 11, 2015
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• Book URL: https://openstax.org/books/chemistry/pages/1-introduction
• Section URL: https://openstax.org/books/chemistry/pages/chapter-7

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