7.E: Chapter 7 Homework Answers
- Page ID
- 202166
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Evidence of a Chemical Reaction
1. a) b)
3. c)
Writing and Balancing Chemical Equations
5. b)
7.
a) Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
b) CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
c) Ca(OH)2(aq) + CO2(g) → CaCO3(s) + H2O(l)
d) 2Fe2O3(s) + 6Cl2(g) → 4FeCl3(aq) + 3O2(g)
9.
a) N2O3(aq) + H2O(l) → 2HNO2(aq)
b) Xe(g) + 3F2(g) → XeF6(s)
c) NH4NO3(aq) → N2O(g) + 2H2O(l)
d) I2(s) + 6HNO3(aq) → 2HIO3 + 6NO2(g) + 2H2(g)
11. C6H24(g) + 12O2(g) → 12H2O(l) + 6CO2(g)
13. C6H12O6 (aq) → 6C(s) + 6H2O(l)
15. 2H2O(g) + 2Na(s) → 2NaOH(s) + H2(g)
17.
a) 2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(l)
b) 2MgS(s) + 3O2(g) → 2MgO(aq) + 2SO2(g)
c) 3P4S3(s) + 16KClO3(aq) → 6P2O5(aq) + 16KCl(aq) + 9SO2(g)
d) Fe3O4(aq) + 4CO(g) → 4CO2(g) + 3Fe(s)
19.
a) Correct.
b) Incorrect. You would need a 2 coefficient before HNO2(aq).
c) Incorrect. Rather than having a 2 coefficient on F2(g) and Cl2(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.
21.
a) H2O(l) → H2O(g)
b) H2O(l) → H2O(s)
c) H2O(g) → H2O (l)
Since all of the above reactions are only physical changes, they require no balancing because they are already balanced.
Solubility
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. NO3- ions would not break down to Nitrogen and Oxygen ions.
Example:
Monoatomic Ion: KCl dissociates into K+ and Cl- ions
Polyatomic Ion: KNO3 dissociates into K+ and NO3- ions
25.
c) Na+
d) Mg2+
27. BaSO4(aq) CaCO3(aq)
29. a) strong electrolyte
31.
a) CaCrO4(aq) + Cu(NO3)2 (aq) → Ca(NO3)2 (aq) + CuCrO4(s)
b) 2KBr(aq) + Hg2SO4(aq) → K2SO4(aq) + Hg2(Br)2(s)
c) 2LiNO3(aq) + 2(NH4)2S(aq) → 2NH4NO3(aq) + Li2S(s)
d) Pb(NO3)2(aq) + 2NaCl(aq) → 2NaNO3(aq) + PbCl2(s)
33.
a) This reaction implies a phase change for both mixtures, which is incorrect.
Na2SO4(aq) + BaCl2(aq) → 2NaCl(aq) + BaSO4(s)
b) This reaction is not balanced. There should be a 2 coefficient before KI(aq) and KNO3(aq)
2KI(aq) + Pb(NO3)2(aq) → 2KNO3(aq) + PbI2(s)
c) This reaction states that Ag2S is soluble when it is actually insoluble. It also states NaNO3 is insoluble when it is in fact soluble.
2AgNO3(aq) + Na2S(aq) → Ag2S(s) + 2NaNO3(aq)
35. Beaker A would form a precipitate because the reaction would be:
2K3PO4(aq) + 3Ca(NO3)2(aq) → 6KNO3(aq) + Ca3(PO4)2(s).
Beaker B would form a precipitate because the reaction would be:
2NaOH(aq) + BaBr2(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) + AgNO3(aq) → AgCl(s) + KNO3(aq)
39. a) Total Ionic: 2Rb+(aq) + 2F-(aq) + Cu2+(aq) + SO42-(aq) → 2Rb+(aq) + 2F-(aq) + Cu2+(aq) + SO42-(aq)
Net Ionic: No Reaction
b) Total Ionic: Sr2+(aq) + 2Br-(aq) + 2K+(aq) + SO42-(aq) → SrSO4(s) + 2K+(aq) + 2Br-(aq)
Net Ionic: Sr2+(aq) + SO42-(aq) → SrSO4(s)
c) Total Ionic: Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) → Na+(aq) + Cl-(aq) + H2O(l)
Net Ionic: H+(aq) +OH-(aq) → H2O(l)
d) Total Ionic: 2Bi(OH)3(s) + 6H+(aq) + 3SO42-(aq) → 2Bi3+(aq) + 3SO42-(aq) + 6H2O(l)
Net Ionic: 2Bi(OH)3(s) + 6H+(aq) → 2Bi3+(aq) + 6H2O(l)
Acid-Base Reactions
41. Neutralization, water, salt
43. a) 2HClO3(aq) + Zn(OH)2(s) → Zn(ClO3)2(aq) + 2H2O(l)
b) H2SO4(aq) + Ba(OH)2(s) → BaSO4(s) + 2H2O(l)
c) HBr(aq) + LiOH(aq) → LiBr(aq) + H2O(l)
d) 2HF(aq) + Ca(OH)2(aq) → CaF2(s) + 2H2O(l)
Combustion Reactions
45. a) 2C6H6(l) + 15O2(g) → 6H2O(l) + 12CO2(g)
b) C6H12O6(aq) + 6O2(g) → 6H2O(l) + 6CO2(g)
c) C2H5OH(l) + 3O2(g) → 3H2O(l) + 2CO2(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
53.
a) Molecular: BaCl2(aq) + Na2SO4(aq) → 2NaCl(aq) + BaSO4(s)
Total Ionic: Ba2+(aq) + 2 Cl-(aq) + 2Na+(aq) + SO42-(aq) → 2Na+(aq) + 2Cl-(aq) + BaSO4(s)
Net Ionic: Ba2+(aq) + SO42- (aq) → BaSO4(s)
b) Molecular: Na2S(aq) + 2LiF(aq) → No Reaction
c) Molecular: H2SO4(aq) + Pb(NO3)2(aq) → PbSO4(s) + 2HNO3(aq)
Total Ionic: 2H+(aq) + SO42-(aq) + Pb2+(aq) + 2NO3-(aq) → PbSO4(s) + 2H+(aq) + 2NO3-(aq)
Net Ionic: Pb2+(aq) + SO42-(aq) → PbSO4(s)
55. a) Gas Evolution Reaction
Molecular: 2HClO4(aq) + CaCO3(s) → H2O(l) + CO2(g) + Ca(ClO4)2(aq)
Sum of Coefficients: 6
b) Precipitation Reaction
Molecular: 2Fe(s) + 3Cu(NO3)2(aq) → 3Cu(s) + 2Fe(NO3)3(aq)
Sum of Coefficients: 10
c) Combustion Reaction
Molecular: C9H20(l) + 14O2(g) → 10H2O(l) + 9CO2(g)
Sum of Coefficients: 34
57. a) 2(CH3)2CHOH(l) + 9O2(g) → 8H2O(l) + 6CO2(g)
b) C10H8(s) + 12O2(g) → 4H2O(l) + 10CO2(g)
c) 2CH3CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH3(l) + 37O2(g) → 26H2O(l) + 24CO2(g)
d) C12H22O11(aq) + 12O2(g) → 11H2O(l) + 12CO2(g)