Updated Dec.14/04 -- by Dr. Bird updated some typos: #33, 45, 63
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NUMERICAL ANSWERS TO ASSIGNED TUTORIAL PROBLEM SETS FOR CHEM205
FROM KOTZ & TREICHEL'S CHEMISTRY & CHEMICAL REACTIVITY
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NOTE: none of the answers from Ch.10 have been verified
NOTE: "^" means "raised to the power of..." to indicate scientific notation...
NOTE: "*" means "multiplied by"
NOTE: I cannot represent superscripts/subscripts here -- so...for polyatomic ions,
I have left a space between the ion's formula and its charge
NOTE: These answers have been taken from the instructors answer manual. In problems where
an intermediate result is calculated, the rounded intermediate result is
is used in the next step. THIS IS NOT CORRECT PROCEDURE, and can lead to rounding errors.
ALWAYS carry forward at least one extra figure to the next step.
NOTE: If the original question was broken into parts (a), (b), etc this is shown
in the Q# column. When a questiomn or part of a question involves several steps giving
intermediate answers, or implicit parts, this is indicated in the comments column by (i), (ii), etc.
Ch. Q# Answer Units SFs Comments
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12 5 (i) The pressure of a mixture of gases is the sum of the
partial pressures of the gases in the mixture.
0.21 none 2 (ii) X(O2).
1.6x10^2 mm Hg (torr) 2 (iii) from X(O2).
12 6 (i) See p 490.
(ii) Boyle's law states that pressure is proportional to
1/V when n and T are constant. If the temperature is
constant, the average impact force of molecules of a
given mass on the walls of the container must be constant.
If n is constant, while the volume of the container is
made smaller, the number of collisions with the container
walls per second must increase. This means the pressure
decreases.
12 13 Convert them all to atm to compare them:
Then: 0.256 atm < 363 mm Hg < 0.523 bar < 363 kPa
12 17 4.6 mL 2 Use P1/T1 = P2/T2
12 23 1.2x10^7 L 2 Use P1*V1/T1 = P2*V2/T2
12 25 18 L O2 2 (i)Use stoichiometric ratio 7/2
25 17 L H2O 2 (i) Use stoichiometric ratio 6/2
12 31 0.22 mol He 2 Use P*V = n*R*T
12 33 0.307 atm 3 then use d = (P*M)/(R*T)
0.929 g/L
12 39 0.856 g/L 3 (i) Use d = m/V
0.436 atm 3 (ii) Use 1 stm = 760 mm Hg
44.1 g/mol 3 (iii) Use M = d*R*T/P
12 43 8.3x10^-4 mol C8H18 2 (i) Use 114 g of C8H18 = 1 mol
0.0075 mol H2O 2 (ii) Use stoiciometric ratio 18/2
0.039 atm 2 (iii) Use n*R*T/V
0.010 mol O2 2 (iv) Use stoichiometric ration 25/2
0.053 atm 2 (v) Use P = n*R*T/V
12 45 1.01 atm 3 (i) Use 1 atm = 760 mm Hg
0.371 mol CO2 3 (ii) Use n = (P*V)/(R*T)
52.7 g KO2 3 (iii) Use stoiciometric ratio 4/2 and 1 mo KO2 = 71.10 g
12 47 92.5 % N2 3 (i) Use %N2 = 100% - %H2S - %CO2
43 atm 2 (ii) use Partial pressure = total pressure * %gas/100
2.1 atm 2
1.4 atm 2
12 49a 0.517 mol He 3 (i) Use n = (P*V)/(R*T)
2.07 g He 3 (ii) Use 1 mol = 4.003 g
49b 0.48 atm 2 Use P = n*R*T/V
49c 0.52 atm 2 Use P(total) = P(O2) + P(He)
49di 0.48 none 2 Use X(He) = P(He)/P(total)
49dii 0.52 none 2
12 51 The molar mass of Ar (40 g/mol) is greater than the
molar mass of N2 (28 g/mol). Therefore, for samples
with equal mass there are more moles of N2 present than
moles of Ar.
51a True. More moles, so more molecules.
51b False. Pressure is directly related to the number of
moles present. The pressure is therefore greater
in the nitrogen flask.
51c False. The molecules of the gas with the smaller molar
mass will have greater velocity.
51d True. The nitrogen molecules have greater velocity than
the argon molecules, and there are more molecules of
nitrogen present, so they will collide more
frequently with the walls of the flask.
12 55 Molar speed: OSCl2 < Cl2O < Cl2 < SO2
Molar mass: 119 > 87 > 71 > 64 g/mol
12 57 3.16 none 3 Use Graham's law of effusion
12 63 628. L 3 (i) Use V = pi*r^2*l
1.14 atm 3 (ii) Use 760 mm Hg = 1 atm
29.2 mol CO2 3 (iii) Use n = (P*V)/(R*T)
1.29x10^3 g 3 (iv) Use 1 mol CO2 = 44.01 g
12 69 58. deg C 2 Use T = (P*V)/(n*R), and deg C = K - 273.2
12 71 0.0801 mol Si 3 (i) Use 1 mol Si = 28.09 g
0.206 mol CH3Cl 3 (ii) n = (P*V)/(R*T) (Watch the units!)
(iii) Check for limiting reagent. It is Si.
10.3 g (CH3)2SiCl2 3 (iv) Use stoichuiometric ratio 1/1, and
1 mole (CH3)SiCl2 = 129.1 g
0.369 atm 3 (v) Use P = n*R*T/V
12 75 0.7868 mol S 4 (i) Use 1 mol S = 32.066 g
3.935 mol F 4 (ii) Use 1 mol F = 18.998 g
(iii) ratio moles F/S = 5/1 therefore
empirical formula is SF5.
254 g/mol 3 (iv) Use M = d*R*T/P
(v) Use ratio obsd mol. mass./emp. mol. mass. = 2:1,
so formula is S2F10.
12 77a 0.045 mol H2 2 Use 1 mol occupies 22.414 L at STP.
77b 0.045 mol Ar 2 See (a)
77c 0.041 mol H2 2 Use n = (P*V)/(R*T)
77d 0.05 mol Ar 1 See (c) (Note: Sig. figs. ambiguous for 900 mm Hg -
it has been taken as taken as just 1 sig. fig.)
The number of Molecules is proportional to the number of
moles, so sample (d) contains the greatest number of
molecules, and sample (c) contains the smallest number.
Argon has the greatest molar mass, so sample (b) contains
the largest mass of gas (1.8 g).
12 79a 29 g/mol 2 Use M = d*R*T/P (Watch the units!)
79b 0.18 none 2 (i) Use X(O2) + X(N2) = 1, and
total moles = X(O2)*mol.mass.O2 + X(N2)*mol.mass.N2
solving first for X(O2)
0.82 none 2 (ii) then solving for X(N2)
12 81 2.x10^-9 mol 1 (i) Use 1 mol P = 34 g
5.^-8 atm 1 (ii) Use P = n*R*T/V
12 85 0.0413 mol C2H2 3 (i) Theoretical yield from CaC2.
0.314 mol C2H2 3 (ii) Actual yield: use n = (P*V)/(R*T)
76.0 % 3 (iii) (actual/theo.)*100%
12 91a True.
91b False. Nitrogen has a smaller molecular mass
than O2, so an equal mass of N2 contains more
moles and more molecules than O2.
12 93 At constant pressure, the number of moles of a gas is inversely proportional to the temperature. Therefore
flask B(at a lower temperature) contains more moles
(and more molecules) of oxygen.
12 95a More significant.
95b More significant.
95c Less significant.
12 99 0.00362 mol CO2 3 (i) Use n = (P*V)/(R*T) (Watch the units!)
0.102 g CO 3 (ii) Use 1 mol CO = 28.01 g
0.040 g Fe 2 (iii) By difference.
7.3x10^-4 mol Fe 2 (iv) Use 1 mol Fe = 55.8 g
(v) Ratio mol CO/Fe = 5/1, empirical formula is Fe(CO)5
12 103a 0.00189 mol B2H6 3 (i) Use 1 mol B2H6 = 37.83 g
0.0160 atm 3 (ii) Use P = n*R*T/V
103b 0.0320 atm 3 (i) Use stoichiometric ratio = 2/1
0.0480 atm 3 (ii) Add the partial pressures
12 105a 19 valence e-.
105b The electron pair geometry of Cl is determied by
2 Cl-O bonds and 2 lone pairs as tetrahedral.
105c Hybridization: sp3; molecular geometry: bent.
105d The angel is larger in ozone where the central
O is sp2 hybridized.
105e 0.172 mol NaClO2 3 (i) Use 1 mol NaClO2 = 90.44 g
0.0828 mol Cl2 3 (ii) Use n = (P*V)/(R*T)
(iii) Determine limiting reagent. It is Cl2.
11.2 g ClO2 3 (iv) Use stoichiometric ratio = 2/1, and
1 mol ClO2 = 67.45 g