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Solutions 7

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  • These are solutions to select problems from homework set #7.


    a) 2K(s) + Br2(l) → 2KBr(s)

    b) 4Li(s) + O2(g) → 2Li2O(s)

    c) Na2CO3(s) → Na2O(s) + CO2(g)


    a) MgCl2(s) + Ca(s) → CaCl2(s) + Mg(s)

    b) Ca(s) + Cl2(g) → CaCl2(s)

    c) Ca(OH)2(s) → CaO(s) + H2O(l)


    Diamonds and graphites are allotropes of carbon. Diamonds are carbons bonded in a crystalline, tetrahedral structure while graphites are sheets of hexagonal carbon.


    a) Ga4C3(s) + 12H2O(l) → 3CH4(g) + 4Ga(OH)3(s)

    b) 3GeO2(s) + 4B(s) →2B2O3(s) + 3Ge(s)


    a) PbO(s) + 2HCl(aq) → PbCl2(s) + H2O(l)

    b) SiCO3(s) → SiO(s) + CO2(s)

    c) SnO(s) + C(s) →Sn(s) + CO(g)


    Write the chemical equations to represent the following: (a) reaction with sodium and chlorine gas (b) formation of cesium superoxide (c) combustion of sodium to form sodium peroxide.

    a) 2Na(s) + Cl2 (g) --> 2NaCl(aq)

    b) Cs(s) + O2(g) --> CsO2(s)

    c) 2Na(s) + O2(g) --> Na2O2 (s)


    Write the chemical reactions for the following: (a) reduction of BeCl2 to Be metal with Mg as a reducing agent (b) reduction of Sr metal with Br2

    (c) the formation of slaked lime from quicklime and water

    a) BeCl2 (s) + Mg(s) --> Be (s) + MgCl2(s)

    b) Sr (s) + Br2 (l) --> SrBr2(s)

    c) CaO (s) + H2O --> Ca(OH)2 (s)


    Do diamonds really last forever? Justify your answer.

    No, they do not. Under standard conditions, they will transform into graphite (the most stable allotrope of carbon). However, they will do so very slowly. Nevertheless, at elevated temperatures diamond will combust to form carbon dioxide gas.


    Complete the chemical equations:

    a) 3SiO2 (s) + 4Al (s) --> 2Al2O3 (s) + 3Si (s)

    c) Al4C3 (s) + 12H2O (l) --> 3CH4 (g) +4Al(OH)3 (s)


    Salts like NaH2PO4 are called acid salts because their anions undergo further ionization. What is the pH of 250 mL of water solution containing 10 g NaH2PO4?

    Molar mass of NaH2PO4 = 119.977 g/mol

    (10 g NaH2PO4)(1 mol NaH2PO4/119.977 g NaH2PO4)/(0.25L) = 0.333 M

    H2PO4- + H2O <---> HPO42- + H3O+

    I 0.333 M - 0 0

    C –x +x +x

    E 0.333 M – x x x

    x2/(0.333-x) = Ka = 6.3 * 10-8
    x= 1.45 * 10-4 = [H3O+]

    pH= -log[H3O+]= -log[1.45 * 10-4]= 3.84

    For more info, see Additional Resources.


    1. Write chemical equations to represent the following:

    1. Formation of hydrogen peroxide (H2O2)
    2. Reaction of potassium metal with fluorine gas
    3. Combustion of sodium to form sodium superoxide


    1. 2H2O(l) + O2(g) 2H2O2(s)
    2. 2K(s) + F2(g) 2KF(s)
    3. Na(s) + O2(g) NaO2(s)


    17. Write the following reactions

    1. Reduction of CaCl2 to Ca metal with Be as the reducing agent.
    2. The reaction of F2(l) and calcium metal.
    3. Neutralization of acetic acid by sodium hydroxide.


    1. CaCl2(s) + Be (s) Ca(s) + BeCl2(s)
    2. F2(l) + Ca(s) CaF2(s)
    3. NaOH(s) + CH3COOH(s) → NaCH3COO(s) + H2O(l)


    49. For each of the following, write plausible chemical equations

    a) dissolving of lead (IV) oxide in HCl
    Solution: PbO2(s) + 4HCl(aq) PbCl2(aq)
    b) heating of lead (II) sulfide
    Solution: 2PbS(s) + 3O2(g) + heat 2PbO(s) + 2SO2(g)
    c) reduction of lead (II) oxide by carbon
    Solution: 2PbO(s) + C(s) + heat 2Pb + CO2(g)
    d) reduction of Cl2(g) by Zn
    Solution: Zn(s) + Cl2(g) ZnCl2(aq)


    User VSPER theory to predict the probable geometric structures of: a) XeO2, b) XeF4, and c) XeF6.
    a) This is linear.
    b) This is square planar.
    c) This is octahedral.