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MULTIPLE CHOICE QUESTIONS FOR CH

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					 MULTIPLE CHOICE QUESTIONS FOR
          CHAPTERS 5-7
1.   Naphthalene (C10H8), better known as mothballs,
     undergoes a combustion reaction with oxygen to
     form carbon dioxide and water. What is the
     coefficient in front of oxygen?
a.   12 b. 5           c. 7      d. 10
       Answer:    a

      Explanation: the balanced equation is
      C10H8 + 12O2       10CO2 + 4H2O
 MULTIPLE CHOICE QUESTIONS FOR
          CHAPTERS 5-7
2. In the chemical reaction between SiO2 and C to give
     SiC and CO, what is the coefficient in front of CO?
a. 1
b. 2
c. 3
d. 4

   Answer: b
   Explanation: the balanced equation is
   SiO2 + 3C        SiC + 2CO
 MULTIPLE CHOICE QUESTIONS FOR
          CHAPTERS 5-7
3. Which one of the following represents a
   decomposition reaction?
a. C2H4(g) + H2(g) → C2H6(g)
b. Fe(NO3)2(aq) + Na2S(aq) → FeS(s) +
   2NaNO3(aq)
c. Cr(s) + O2(g) → Cr2O3(s)
d. 2KClO3(s) → 2KCl(s) + 3O2(g)
  Answer:    d
 Comment: decomposition reactions are those where a
 single reactant gives more than one product
 MULTIPLE CHOICE QUESTIONS FOR
          CHAPTERS 5-7
4.   Aqueous solutions of potassium acetate and zinc
     sulfate are poured together and allowed to react.
     What is the identity of the precipitate, if one is
     produced?
a.   ZnSO4
b.   K2SO4
c.   K(SO4)2
d.   This reaction does not produce a precipitate
 Answer: d       Comment: all species are water-soluble
 MULTIPLE CHOICE QUESTIONS FOR
          CHAPTERS 5-7
5.     How many grams of iron are combined with 75.0
       grams of oxygen to make the iron(II) oxide
       compound?
     a. 21.5 g     b. 11.9 g   c. 262 g       d. 21.4 g
     Answer: c

     Explanation: balanced equation is
     2Fe + O2        2FeO . Moles of oxygen =
     75.0 g/32.0 g/mol = 2.34. Moles of iron = 2 x
     2.34 = 4.68
     Mass of iron = 4.68 mol x 55.8 g/mol = 261 g
 MULTIPLE CHOICE QUESTIONS FOR
          CHAPTERS 5-7
6.   You have already determined that there are 3.33 x
     10-6 moles in 2.92 x 10-4 grams of strontium. How
     many atoms of strontium are in this sample?
a.   2.01 x 1018 atoms
b.   1.76 x 1020 atoms            Answer: a
c.   5.68 x 10-21 atoms
d.   4.98 x 10-19 atoms

Explanation: There is Avogadro’s number of atoms in
1 mole, hence in 3.3 x 10-6 mol, there are 3.3 x 10-6 mol
x 6.022 x 1023 atoms/mol = 2.01 x 1018 atoms
 MULTIPLE CHOICE QUESTIONS FOR
          CHAPTERS 5-7
7.   What is the formula weight for iron(III) oxide?
a.   183.6 amu
b.   199.6 amu
                           Answer: c
c.   159.7 amu
d.   159.7 g/mol

Explanation: iron (III) oxide is Fe2O3. Formula weight
= 2 x 55.845 amu + 3 x 15.999 amu = 159.7 amu.
 MULTIPLE CHOICE QUESTIONS FOR
          CHAPTERS 5-7
8. The following mole ratio is used to convert from
    moles of barium to moles of oxygen, (6 mole O/1
    mole Ba). For which chemical compound is this
    ratio applicable?
a. Ba(NO3)2
b. BaO                 Answer: a
c. BaSO3
d. Ba(ClO4)2

 Comment: b is 1/1; c is 3/1 and d is 8/1
 MULTIPLE CHOICE QUESTIONS FOR
          CHAPTERS 5-7
9.   What can be said about the reaction of C5H12(l) with
     O2(g) to give CO(g) and H2O(l), if 13.3 g of C5H12 is
     allowed to react with 5.0 g of oxygen?
a.   More moles of CO produced than moles of water
b.   C5H12 is the limiting reactant
c.   it is a stoichiometric reaction      Answer: d
d.   Oxygen is the limiting reactant
 Explanation: balanced equation is 2C5H12 + 11O2
 10CO + 12H2O. The pentane:oxygen ratio is 2:11,
 hence by inspection, oxygen is the limiting
 reactant
 MULTIPLE CHOICE QUESTIONS FOR
          CHAPTERS 5-7
10. Given the balanced chemical equation, 2Mn(s) +
    3O2(g) → 2MnO3(s), how many molecules of O2
    are required to produce 15.0 g of MnO3?
a. 1.32 x 1023 molecules
b. 8.78 x 1022 molecules        Answer: a
c. 9.03 x 1024 molecules
d. 7.00 molecules
 Comment: 15 g of MnO3 = 15g/103 g/mol = 0.146
 mol. From stoichiometry, this requires 3/2 x
 0.146 mol of O2 = 0.219 mol = 0.219 mol x 6.022
 x 1023 molecules/mol = 1.32 x 1023 molecules
 MULTIPLE CHOICE QUESTIONS FOR
          CHAPTERS 5-7
11. The predicted amount of product produced, based
   on reaction stoichiometry, is known as
a. the actual yield.
b. the theoretical yield.
c. the percent yield.         Answer: b
d. the empirical yield.
 MULTIPLE CHOICE QUESTIONS FOR
          CHAPTERS 5-7
12. The reaction between nitrogen and oxygen gases
     requires the absorption of 43 kJ of energy per mole
     of nitrogen according to the following chemical
     equation, N2(g) + O2(g) → 2 NO(g). How much
     energy must be absorbed by the reaction in order
     to completely consume 4.06 g of oxygen?
a. 170 kJ       b. 5.5 kJ      c. 11 kJ      d. 0.0030 kJ
Answer: b
 Explanation: moles of nitrogen = moles of oxygen
 4.06 g = 4.06 g/ 32.0 g/mol = 0.127 mol of oxygen
 Heat absorbed = 43 kJ/mol x 0.127 mol = 5.5 kJ
       TRUE/FALSE QUESTIONS FOR
             CHAPTERS 5-7

1. (T/F) The net ionic equation contains only those
   ions that undergo a chemical change.
Answer: T. Comment: spectator ions are eliminated
2.   (T/F) When balancing chemical equations, it is
     sometimes acceptable to change a subscript in
     order to complete the balancing process.
      Answer: F. Comment: never do this
3.   (T/F) In all aqueous solutions, water is used as the
     solvent. Answer: T
4.   (T/F) For every mole of sodium phosphate there is
     one mole of sodium atoms.
     Answer: F. Comment: Na3PO4 requires 3 mol Na
          TRUE/FALSE QUESTIONS FOR
                CHAPTERS 5-7
   5. (T/F) The molecular formula for a compound is
       always presented in the smallest whole-numbered
       ratio.
 Answer: F. Comment: this describes the empirical formula
    6.(T/F) The mole is defined as the number of atoms
      in exactly 12 grams of matter.
  Answer: F. Comment: should be exactly 12 grams of 12C
    7.   (T/F) For the reaction between Fe(s) and HCl(aq) to
         give FeCl3(aq) and H2(g), two moles of HCl are
         consumed for every mole of H2 produced.
Answer: T. Comment: balanced equation is 2Fe(s) + 6HCl(aq)
   2FeCl3(aq) + 3H2(g)
       TRUE/FALSE QUESTIONS FOR
             CHAPTERS 5-7
8.    For the reaction between, Al(s) and H2SO4(aq) to
      give Al2(SO4)3(aq) and H2(g), for every mole of H2
      produced, one mole of H2SO4 is consumed.
     Answer: T. Comment: balanced equation is
     2Al(s) + 3H2SO4(aq)   Al2(SO4)3(aq) + 3H2(g)
9.    An exothermic reaction can be depicted by the
      following general reaction, A + B → C + D + DH.
      T
      Answer: T. Comment: a generally better
      depiction is A + B   C + D; DH (-ve)
         SHORT ANSWER QUESTIONS FOR
                CHAPTERS 5-7
    1. When aqueous solutions of silver nitrate and
       sodium iodide are allowed to react, the precipitate
       formed is silver iodide, AgI .
    2. An aqueous solution of potassium carbonate is
       allowed to react with an aqueous solution of barium
       chloride. Write the balanced chemical equation
       and indicate what ions are present in solution after
       the reaction has gone to completion and underline
       the precipitate if one is predicted to form
Answer:     K2CO3(aq) + BaCl2(aq) → 2KCl(aq) + BaCO3(s)
The ions remaining in solution are K+ and Cl-
  SHORT ANSWER QUESTIONS FOR
         CHAPTERS 5-7
3. Determine the empirical formula for a compound
   composed of 68.4% Cr and 32.6% O.
Answer: Cr2O3. Explanation: 68.4 g/51.996 g/mol =
1.3155 mol of Cr; 32.6/15.999 g/mol = 2.0376 mol
of O. Mole ratio O:Cr = 1.54:1 ~1.5:1 or 3:2
     SHORT ANSWER QUESTIONS FOR
            CHAPTERS 5-7
4.   Phenolphthalein, an organic dye used to
     indicate the endpoint in acid-base titrations,
     has a molar mass of 320.34 g/mol. It is
     composed of 74.99% C, 5.03% H, and 19.98% O.
     What is the molecular formula for
     phenolphthalein?        Answer: C20H16O4
                       C          H          O
     Moles           74.99       5.03         19.98
                      12.011     1.0079       15.999
                  = 6.2434       4.9905       1.2488
     Ratio               5           4           1
     Empirical formula = C5H4O, giving molar mass
     ~80
     g/mol . Hence molecular formula is C20H16O4
     SHORT ANSWER QUESTIONS FOR
            CHAPTERS 5-7
5.   During the process of a combustion reaction of a
                                  CO2
     carbon-containing compound, a complete
     combustion yields             and H2O whereas
     an CO
        incomplete combustion produces
     and H2O.
     SHORT ANSWER QUESTIONS FOR
            CHAPTERS 5-7
6.   One of the most notorious poisons in both fact and
     fiction is cyanide. The average lethal dose is only
     50 to 60 mg. An antidote that is available for
     cyanide poisoning is sodium thiosulfate. Sodium
     thiosulfate reacts with cyanide to produce sodium
     sulfite and thiocyanate according to the chemical
     equation
            CN- + Na2S2O3        SCN- + Na2SO3.

     However, very few victims of cyanide poisoning live
     long enough to be treated. If 55.8 mg of cyanide is
     accidentally ingested, how many grams of sodium
     thiosulfate must be consumed in order to react with
     all of the cyanide?
Answer:        339 mg of Na2S2O3 is required to react
  with all of the cyanide.
Explanation: 55.8 mg cyanide = 0.0558 g/ 26.018 g/mol
  = 2.14467 x 10-3 mol
From reaction stoichiometry, cyanide:thiosulfate is 1:1
Therefore 2.14467 x 10-3 mol of thio is needed =
  2.14467 x 10-3 mol x 158.109 g/mol = 339 mg (after
  changing from g to mg and rounding off).