Chem 20 Review Answers by ghY0mL8

VIEWS: 68 PAGES: 11

									Chemistry 30                                    Chem. 20 Review Answers
1.         In the following chart you will find partial information about several
           atoms and ions. Fill in the missing information using your table of
           atomic weights. (You are to determine the atomic mass from the
           information in the question table, not from your atomic masses
           table).
     Name        Symbol     Atomic     Atomic        # of      # of          # of
                            Number      Mass       Protons   Neutrons     Electrons

calcium            Ca         20        40           20        20            20
atom

chromic           Cr3+        24        52           24        28            21
ion

chlorine           Cl         17        35           17        18            17
atom

oxide ion          O2-        8         16           8          8            10

calcium ion       Ca2+        20        40           20        20            18

nitride           N3-         7         40           7          7            10
ion


2.         Write the correct chemical (empirical) formulas for the following
           compounds using the information in your ion sheets.
           a)   beryllium chloride        f)    ferrous hydroxide
                       BeCl2                             Fe(OH)2
           b)   gallium iodide            g)    diphosphorous pentoxide
                        GaI3                               P2O5
           c)   calcium hydroxide         h)    lead (II) bicarbonate
                     Ca(OH)2                           Pb(HCO3)2
           d)   phosphoric acid           i)    ammonium phosphate
                       H3PO4                           (NH4)3PO4
           e)   ferric hydroxide
                      Fe(OH)3

3.         Calculate the numbers of moles of atoms present in the following
           samples:
           a)    3.6g C                  b)    54g Al
              # of moles =    sample mass
                               molar mass
           a)    moles C = 3.6g  12g/mol = 0.3 mol
           b)    moles Al = 54g  27g/mole = 2.0 mol

4.         Calculate the molecular weights (molar masses) of the following
           compounds:
            a)   HCl                          c)  ammonium dichromate
            b)   SrSO4                        d)  aluminum carbonate
                 a)     HCl    1 - H is 1 x 1.008    = 1.008
                               1 - Cl is 1 x 35.5   = 35.5
                                                      36.508


7ced7cdf-1e7e-46b1-9803-35d4c4121be4.docAugust 9, 2012                  Page 1 of 11
                           Chem. 20 Review Answers - pg. 2


      b)       SrSO4       1 - Sr is 1 x 87.6   = 87.6
                           1 - S is 1 x 32.1    = 32.1
                           4 - O is 4 x 16.00   = 64.00
                                                  183.7

      c)       (NH4)2Cr2O7        2 - N is 2 x 14.01      = 28.02
                                  8 - H is 8 x 1.008      =   8.064
                                  2 - Cr is 2 x 52.0      = 104.0
                                  7 - O is 7 x 16.00      = 112.00
                                                               252.084

      d)       Al2(CO3) 3         2 - Al is 2 x 27.0      = 54.0
                                  3 - C is 12.01          = 36.03
                                  9 - O is 9 x 16.00      = 144.00
                                                             234.03

5.   Calculate the numbers of moles of molecules present in the following
     samples:
     a)   4.4g CO2                 b)   34g AgNO3
     The general strategy is to divide sample mass by molar mass to
     determine numbers of moles. Before this can be done you
     must calculate the molar mass (molecular weight) of each
     compound as in #4.
     a)    # of moles = 4.4g  44g/mol = 0.1mol
     b)    # of moles = 34g  170g/mol = 0.2mol
6.   Calculate the mass in grams of one (1) water molecule. The chemical
     formula water, for those of you who forgot, is H2O.
     1 mole of anything contains Avogadro’s Number of particles
     which is 6.02 x 1023. The mass of 1 mole of water is 18.016g
     (using same procedure as in #4 & #5). From this we can write:
     6.02 x 1023 water molecules have a mass of 18.016g.
     1 water molecule has a mass of:
     18.016g  6.02 x 1023 = 2.99 x 1023g
7.   Calculate the number of oxygen atoms contained in 15g of calcium
     carbonate.
     The formula for calcium carbonate is CaCO3, which tells us that
     1 molecule of the compound contains 3 oxygen atoms. , #
     of CaCO3 molecules  3 should give us the number of oxygen
     atoms. The number of CaCO3 molecules should be the
     number of moles  Avogadro’s Number. Moles of CaCO3
     should be sample mass  molar mass.
     # of moles CaCO3 = 15g  100g/mol = 0.15mol
     The remaining two steps can be shown in one calculation:
     # of O atoms = 0.15mol  6.02 x 1023  3 = 2.71 x 1023 O atoms
8.   Balance the following equations:
     a) 1 C7H16        +    11   O2 7 CO2 + 8               H2O
     b)    4   Fe +    3    O2    2 Fe2O3

7ced7cdf-1e7e-46b1-9803-35d4c4121be4.docAugust 9, 2012                Page 2 of 11
                         Chem. 20 Review Answers - pg. 3


     c)   3     NO2 +   1   H2O     2    HNO3 +    1    NO
     d)   4     NaHCO3 +    2     HCl
                 1 Na2CO3 + 2 NaCl + 3 CO2 + 3 H2O
9.   Write a balanced equation for the complete combustion (burning) of
     propane (C3H8). Combustion means combining with oxygen gas
     (O2). The products of combustion are carbon dioxide and water. For
     the complete combustion of 66g of propane, calculate:
     a)    the mass of oxygen required
     b)    the mass of CO2 produced
     c)    the mass of H2O produced.
     C3H8       + 5 O2  3 CO2 + 4 H2O
     Assuming the coefficients represent moles of each substance,
     we can write the masses involved by determining molecular
     weights and multiplying them by the coefficients.
       C3H8    +     5 O2           3 CO2     +    4 H2O
          44g              160g             132g             72g
                          (5 x 32)         (3 x 44)        (4 x 18)
     a)       the mass of oxygen required
              This first method is assuming you can do grocery store
              arithmetic.
              44g C3H8 requires 160g O2
              1 g C3H8 requires 160g  44g O2
               66g C3H8 requires (160  44)  66 = 240g O2
              This could also be solved using ratios. Set up a ratio of
              mass C3H8 : mass O2 for the balanced equation. Make
              this equal to the ratio of the same two substances in our
              new situation.
                Balanced Equation:                 New Situation:
                       mass C3H8           =        mass C3H8
                        mass O2                         mass O2
                         44g               =             66g
                       160g                        mass O2
              When two ratios or fractions are equal to each other, the
              cross products are equal. Working these out we get:
                  mass O2  44g         =        160g  66g
                      mass O2           =        160g  66g
                                                          44g
                      mass O2          =                 240g
     b)       the mass of CO2 produced
              44g C3H8 rproduces 132g CO2
              1 g C3H8 produces 132g  44g CO2
               66g C3H8 produces (132  44)  66 = 198g CO2
              OR:
                Balanced Equation:            New Situation:

7ced7cdf-1e7e-46b1-9803-35d4c4121be4.docAugust 9, 2012            Page 3 of 11
                        Chem. 20 Review Answers - pg. 4


                      mass C3H8           =           mass C3H8
                      mass CO2                        mass CO2
                        44g               =             66g
                     132g                              mass CO2
                mass CO2  44g            =           132g  66g
                  mass CO2                =           132g  66g
                                                         44g
                    mass CO2           =                198g
      c)     the mass of H2O produced.
             44g C3H8 rproduces 72g H2O
             1 g C3H8 produces 72g  44g H2O
              66g C3H8 produces (72  44)  66 = 108g H2O
             OR:
               Balanced Equation:             New Situation:
                    mass C3H8          =       mass C3H8
                      mass H2O                        mass H2O

                        44g               =               66g
                        72g                           mass H2O

                 mass O2  44g            =           72g  66g

                      mass H2O            =           72g  66g
                                                          44g

                    mass H2O           =            108g
10.   Calculate the molar volume of any gas at 200C and 850mmHg
      pressure.
      The molar volume of any gas at S.T.P. is 22.4L. S.T.P. is standard
      temperature and pressure, which is 0C (273K)and 1
      atmosphere pressure (760mmHg). Our new conditions are
      200C (473K) and 850mmHg. The gas formula which deals
      with pressure, volume and temperature together is the
      combined gas law. Substituting         P1V1      =        P2V2
      our known values into this              T1                 T2
      relationship we get:
                P1V1            =            P2V2
                 T1                              T2
           760mmHg  22.4L                850mmHg  V2
            273K                        473K
      Once again, our cross products should be equal.
         760mmHg  22.4L        =    850mmHg  V2  273K
              473K
            760mmHg  22.4L          =                 V2
                 473K
            850mmHg  273K

7ced7cdf-1e7e-46b1-9803-35d4c4121be4.docAugust 9, 2012             Page 4 of 11
                       Chem. 20 Review Answers - pg. 5



                  34.7L                =             V2

11.   A gas from an unmarked cylinder was found to have a density of
      1.20g/L.at 25 C and one standard atmosphere (760mmHg) pressure.
      What must be the molecular weight of the gas?
      The density tells us the mass of 1L is 1.20g. The molecular
      weight would be the mass of however many litres there are in
      one mole at the stated conditions. We know molar volume is
      22.4L at S.T.P., so the first thing we do is determine the molar
      volume at 25C and 760mmHg. Since only the temperature is
      different from S.T.P., we use the formula:
            V1            =              V2
            T1                    T2
      Substituing the new temperature into the equation and solving
      for the new volume we get:
            V1          =         V2
           T1                              T2
         22.4L            =             V2
         273K                         298K
       V2  273K         =      22.4L  298K
           V2             =      22.4L  298K
                                     273K
           V2           =         24.45L
      Therefore, since 1 mole of gas at these conditions is 24.45L and
      1L of the gas has a mass of 1.20g, the molecular weight is:
       24.45     L     1.20 g = 29.34          g
                 mol               L              mol

12.   64g of oxygen gas and 64g of sulfur dioxide gas are placed into a
      closed container and found to exert a total pressure of 1,000mmHg.
      Calculate the partial pressure of each gas.
      The law of partial pressures states that each gas in a mixture
      exerts a partial pressure in proportion to the number of moles of
      gas present. We must first determine the moles of each gas
      present.
      Moles oxygen (O2) = 64g 32g  g/mol = 2 moles
      moles sulfur dioxide(SO2) = 64g  64g/mol = 1 mole
       there are 3 moles of gas in the container. Since oxygen
      accounts for 2/3 of the moles, then it also accounts for 2/3 of
      the pressure. Similarly, the sulfur dioxide accounts for 1/3 of the
      moles and 1/3 of the total pressure.
      Pressure O2 = 2/3  1,000mmHg = 666.7mmHg
      Pressure S O2 = 1/3  1,000mmHg = 333.3mmHg
13.   A steel cylinder is evacuated with a pump until the pressure gauge
      reads zero. Then, 48g of O2(g) is admitted into the container resulting
      in a pressure reading of 15 p.s.i. Next, 90g of gas X is added to the

7ced7cdf-1e7e-46b1-9803-35d4c4121be4.docAugust 9, 2012         Page 5 of 11
                      Chem. 20 Review Answers - pg. 6


      container and the pressure gauge reads 30 p.s.i. What must be the
      molecular weight of gas X?
         Since the pressure of the oxygen alone is 15psi and the
      pressure of the two together is 30psi, the pressure of the
      unknown alone would be 15psi. Since these measurements
      were taken in the same steel cylinder (equal volumes) at the
      same temperature, and each of them alone would exert the
      same pressure, this situation fits Avogadro’s Hypothesis nicely:
      “Equal volumes of different gases measured at the same
      temperature and pressure contain equal numbers of
      molecules”.  48g of oxygen contains the same number of
      moles as 90g of the unknown. Moles oxygen present is:
       48g  32g/mol = 1.5mol. If 1.5mol of the unknown has a
      mass of 90g, then 1 mol mas a mass of:
      90g  1.5mol = 60g/mol.
         We could also reason as follows: since the two gases have
      equal numbers of moles, we can say
      sample mass  molar mass for the two will be equal. We can
      then write:
             Oxygen                          Unknown:
          Sample mass           =          Sample mass
           Molar mass                       Molar mass
              48g               =              90g
            32g/mol                         molar mass
        molar mass  48         =             90  32
           molar mass           =         (90  32)  48
                                =            60g/mol

14.   A cylinder of propane is filled so that ¾ of the cylinder is liquid
      propane and the top ¼ is gaseous propane. The pressure in the tank
      is measured to be 500kPa. After some time enough propane is used
      so that the cylinder is ½ liquid and ½ gas. Assuming that the
      temperature has not changed, what would be the reading on the
      pressure gauge? Explain your answer.
      The pressure in the tank will still be 500kPa. We are measuring
      the vapor pressure, and as long as there is some liquid and
      some gas, the vaspor pressure will be the same. The only way
      we could change the pressure would be to change the
      temperature.
15.   If you were to dissolve 40g of calcium bromide into enough water to
      give 500mL of solution,
      a)    What would be the concentration (molarity) of the calcium
            bromide solution?
      Molarity is the moles of solute present in 1 litre of solution. It
      can be calculated as: moles solute  litres solution. The
      formula is:
          M = n  V, or          M =       n
                                           V

7ced7cdf-1e7e-46b1-9803-35d4c4121be4.docAugust 9, 2012       Page 6 of 11
                      Chem. 20 Review Answers - pg. 7


      Moles CaBr2 = 40g  200g/mol = 0.2mol
        M =          n       =    0.2mol      = 0.4M
                      V             0.5L
      b)     What would be the concentration of calcium and bromide ions
             in the solution?
      Calcium bromide dissolves in water according to the equation:
      CaBr2(s)  Ca2+(aq) = 2 Br(aq)
      Since 1 CaBr2 produces 1 Ca2+ and 2 Br , the concentration of
      the calcium ion is the same as that of the compound, while the
      concentration of the bromide ion is twice that of the
      compound.
       [Ca2+(aq)] =- 0.4M                [Br((aq)] = 0.8M
16.   How many grams of pure AgNO3 would be contained in 250mL of a
      0.5M AgNO3 solution?
      We need to know the number of moles of silver nitrate first.
      Rearranging the basic molarity formula we should get:
      n = M  V, or moles equals molarity times volume.
      Substituting our given values we get:
      n = M  V = 0.5mol/L  0.25L = 0.125mol
      mass AgNO3 = 0.125mol  169.91g/mol = 21.24g
17.   If you had only 10.3g of pure solid NaBr available and you wished to
      prepare a 0.25M solution of it, what would be the maximum volume
      of this solution that could be made?
      Rearranging the basic molarity formula we should get:
         V =       n           or      Volume =        moles
                   M                                  Molarity
      Before we can use this we need to know the moles of NaBr.
      Moles NaBr = 10.3g  102.9g/mol = 0.1mol
      Substituting this into our revised formula we get:
       V =       n       =       0.1mol         = 0.4L, or 400mL
                 M             0.25mol/L
18.   If you took 250mL of a 0.6M HCl solution and added 750mL of water
      to it, what would be the concentration of the final solution?
      Since adding water (or taking it out) does not alter the moles of
      HCl, we can say that moles HCl before the dilution is equal to
      moles of HCl after dilution. Since moles = molarity x volume,
      we can derive the formula:
      M1V1 = M2V2
      Substituting our values into this equality we get:
      0.6M  0.25L = M2  1.0L
      M2 = (0.6M  0.25L)  1.0L = 0.15M
19.   A student made the following three pairs of mixtures of aqueous
      ionic solutions and observed which combinations produced
      precipitates:
      i)     Ba(NO3)2 and Na2SO4 gave a white precipitate
      ii)    Ca(NO3)2 and NaBr gave no precipitate
      iii) CaCl2 and Na2CO3 gave a white precipitate.


7ced7cdf-1e7e-46b1-9803-35d4c4121be4.docAugust 9, 2012       Page 7 of 11
                       Chem. 20 Review Answers - pg. 8


      a)    For each combination that produced a precipitate, indicate
            what the two possibilities are for the solid product.
       Ba(NO3)2 has the Ba2+ and NO3 ions, Na2SO4 has Na+ and
         SO42. The new combinations of + and - ions would be
         NaNO3 and Ba SO4
       CaCl2 has the Ca2+ and Cl ions. The Na2CO3 has the Na+
         and the CO32 ions. The two new combinations are NaCl
         and CaCO3.
      b)    Determine which of the new combinations are precipitates, and
            which are soluble combinations. Explain your reasoning.
       Of the two possibilities from i) the NaNO3 can be eliminated
         since there was no reaction with set ii) which brought the
         sodium and nitrate ions together. The precipitate has to be
         Ba SO4.
       In set iii) the NaCl can be elininated since this is table salt,
         which as we all should know, is soluble in water. The
         precipitate has to be CaCO3.
      c)    Write the net ionic equation for the formation of the two
            precipitates.
       Ba2+(aq) + SO42(aq)  BaSO4(s)
       Ca2+(aq) + CO32(aq)  CaCO3(s)
      d)    Using the information presented so far, indicate for each of the
            following compounds whether you would expect it to be
            soluble or insoluble in water.
NaNO3                     Ba(NO3)2                  BaSO4
soluble, no ppt. In       soluble, since it is an insoluble, see 19b)
reaction ii) which        original solution.
brings sodium and
nitrate ions together.
CaBr2                     CaCO3                     CaSO4
soluble, for same         insoluble, see 19b)       can’t be determined
reason NaNO3                                        from given
                                                    information. Se
                                                    solubility chart,
                                                    which says it is
                                                    insoluble.

20.   A chemist mixed 18g of carbon, 6g of hydrogen gas and 160g of oxygen gas
      in a closed container. She then ignited it electrically so that the hydrogen
      united with oxygen to form water. The carbon then burned to form carbon
      dioxide. The reaction proceeded until one or more of the reactants was
      completely used up.
      a)     Write balanced equations to represent the two reactions happening..
                 2H2 + O2  2 H2O
                 C + O2  CO2
      b)    Which of the reactants would still be left over, and what mass of it
            would there be?
      The question indicates that the oxygen reacts with the
      hydrogen first, then any oxygen left over reacts with the
      carbon. We have to treat this as two separate reactions.
            2H2 + O2  2 H2O


7ced7cdf-1e7e-46b1-9803-35d4c4121be4.docAugust 9, 2012              Page 8 of 11
                             Chem. 20 Review Answers - pg. 9


            C + O2  CO2
        For the reaction of hydrogen, the equation says that 4g
     hydrogen needs 32g oxygen. Therefore, 6g hydrogen needs:
     (32  4)  6 = 48g oxygen.
        This can happen since we have 160g of oxygen gas.
     Following this reaction there will be:
     160g  48g = 112g of oxygen left.
        For the reaction of carbon, the equation says 12g carbon
     needs 32g oxygen. Therefore, 18g carbon needs:
     (32  12)  18 = 48g oxygen.
        This can also happen since there are 112g of oxygen left.
     Following this reaction there will be:
     112g  48g = 64g of oxygen left.
     c)      What mass of water would be formed?
     From the equation, 4g of hydrogen produces 36g water, so 6g
     hydrogen should make:
     (36  4)  6 = 54g water.
     d)      What volume of carbon dioxide gas would be formed at 30C and
             150kPa pressure?
     Steps to be taken:
     1) determine mass of carbon dioxide produced
     2) determine moles of carbon dioxide this is
     3) determine molar volume under the conditions specified
     4) multiply moles x molar volume.

     1) 12g carbon makes 44g carbon dioxide, therefore 18g
        carbon makes (44  12)  18 = 66g carbon dioxide
     2) moles carbon dioxide = 66g  44g/mol = 1.5mol
     3) molar volume at S.T.P. is 22.4L. This is 0C (273K) and
        101.325kPa. Our new conditions are 30C (303K) and
        150kPa. Using the combined gas law we get a new molar
        volume of:
              P1V1             =            P2V2
                  T1                                  T2
          101.325kPa  22.4L           =        150kPa  V2
                 273K                             303K
            V2           =          101.325kPa  22.4L 303K
                                           150kPa  273K

            V2           =         16.8L

     We could do a similar calculation using the value for molar
     volume a S.A.T.P., which is 24.8L at 25C and 100kPa.
            P1V1               =            P2V2
                  T1                                  T2
           100kPa  24.8L              =        150kPa  V2
                 298K                             303K


7ced7cdf-1e7e-46b1-9803-35d4c4121be4.docAugust 9, 2012          Page 9 of 11
                     Chem. 20 Review Answers - pg. 10


          V2        =          100kPa  24.8L 303K
                                    150kPa  298K

          V2        =       16.8L


      Since we have 1.5mol, the volume of gas would be:
      1.5mol  16.8L/mol = 25.2L
      e)    What would be the total mass of all materials in the container
            at the end of the reaction?
      The law of conservation of mass states that total mass of
      products will equal total mass of reactants. Total mass of
      reactants was:
          18g       +       6g         +   160g       = 184g
         carbon         hydrogen          oxygen
      f)    After the material had cooled down to its initial temperature,
            would the gas pressure in the container have changed? Which
            way? Explain your answer.
      To judge any change in presssure we must know total moles of
      gas before and after the reaction. If we rewrite our equation to
      show the states of reactants and products at room temperature
      we get:
            2 C(s) + 2H2(g) + 3O2(g)  2 H2O(l) + 2CO2(g)
      On the reactant side we had 6g  2g/mol = 3mol H2
       and 160g  32g/mol = 5mol O2 for a total of 8 moles of gas
      molecules.
      Assuming the water cools and condenses to the liquid state, for
      products we have 1.5mol CO2 (see part d).
      Before the reaction we had 8 moles of gas, after we have 1.5
      moles. The gas pressure in the container will go down.
21.   Calcualte the anount of heat necessary to raise the temperature of
      450g of water from 10C to 75C. The formula you need is:
      Q = mcT, where
            Q = quantity of heat
            m = mass in grams
            c = specific heat capacity ( in cal/gC)
            T = change in temperature
      The specific heat capacity of water is 1 cal/gC.
      Q = mcT, where
                  m = 450g
                  c = 1cal/gC
                  T = 75C  10C = 65C
      Q = 450g  1cal/gC  65C
      Q = 29,250cal or 29.25kcal
22.   Calculate the amount of heat released when 75g of steam at 100C
      condenses into water at 100C. The heat of vaporization of water is
      9.7kcal/mole. The formula you need is:
            Heat = Amount of substance  heat of fusion.
      The ‘amount of substance’can be expressed as a mass in grams, or as
      a number of moles. If you wish to work in grams, you must first

7ced7cdf-1e7e-46b1-9803-35d4c4121be4.docAugust 9, 2012      Page 10 of 11
                    Chem. 20 Review Answers - pg. 11


     determine the heat of vaporization in cal/g. If you wish to work in
     moles, you must first determine the moles of the substance in the
     sample.
     Working in moles:
     Moles water = 75g  18g/mol = 4.16mol
     Heat = 4.2mol  9.7kcal/mol = 40.352kcal or 40,352cal

     Working in grams:
     Heat of vaporization in cal/g:
     9,700cal/mol  18g/mol = 539cal/g
     Heat = 75g  539cal/g = 40,425cal

     These answers are a bit different, but if we round them off to
     three significant figures, we get 40.4kcal for both.




7ced7cdf-1e7e-46b1-9803-35d4c4121be4.docAugust 9, 2012      Page 11 of 11

								
To top