Chemistry X Lab Review Answers
1) Candle Lab
a. observation, quantitative
b. observation, qualitative
c. interpretation
d. interpretation
e. observation, qualitative
2) Accuracy/Precision lab – see wall in classroom
9.23 ± 0.5 cm
13.6 ± 0.2 °C
0.126 ± 0.005 g
59.4 ± 0.5 mL
1.55 ± 0.05 mL
3) Expt. #7 – Mass Relations
a. (3.21 g. + 1.94 g.) / (1.99 g. _ 3.23 g. ) = 1.01 ~ 1 YES, mass is conserved.
b.
0.00999 Pb(NO3) 2 + 0.00999 K2CrO4  0.0197 KNO3 + 0.00999 PbCrO4
Pb(NO3) 2 + K2CrO4  2 KNO3 + PbCrO4
c. (1.99g/2.01 g.) x 100% = 98.5 %
d. PbCrO4
KNO3
i.
increase same
ii.
same
increase
iii.
decrease increase
iv.
same
increase
v.
same
decrease
vi.
decrease decrease
4) Expt. S-3 “Silver Tree Lab”
a. 0.0181 AgNO3+ 0.00913 Cu  0.00913 Cu(NO3)2 + 0.00187 Ag
2 AgNO3+ 1 Cu  1 Cu(NO3)2 + 2 Ag
b.
i. increase (Cu adds to Ag mass)
ii. decrease (Ag missing)
iii. same (affects Cu mass only)
iv. increase (water adds to Ag mass)
v. decrease (filter paper mass missing, makes Ag appear to have less mass)
5) Expt. #4 – Pressure vs. Volume Lab
a. inversely (PV = k)
b. ~1.52 books
6)
Expt. #5 – Masses of Equal Volumes of Gases
a. 43.8 g/mol (work below)
apparent mass oxygen = 0.12 g
apparent mass carbon dioxide = 0.54 g
mass air displaced (m= dV) m = (1.12 g/L)(0.902 L) = 1.01 g
true mass = apparent mass + mass of air displaced
true mass oxygen = 0.12 g + 1.01 g = 1.13 g.
true mass carbon dioxide = 0.54 g + 1.01 g
ratio carbon dioxide: oxygen 1.55 g / 1.13g = 1.37
molar mass carbon dioxide = 32 x 1.37 = 43.8 g/mol
b. (44.0 g. – 43.8 g.)44.0 g = 0.2/44.0 = 0.5%
c. Avogadro’s La says that equal volumes of gases contain equal numbers of molecules. The ratio of
carbon dioxide to oxygen gas would be meaningless unless they contained the same number of
molecules.
d. lower (less gas, thus lower mass)
7) Expt. #3 – Warming and Cooling Behavior
a. ~ 20 °C
b. lines B and D
c. lines A,C, E
d. i. lines A,C, and E would have smaller slopes (less steep, more spread out)
ii. lines B and D would be longer (take longer to melt and to boil)
8) Expt. #12 – Heat of Solidification of para-dichlorobenzene
a. 18 kJ/mol (work below)
28.73 g C6H4Cl2/ 147.0 g/mol = 0.1954 mol
∆T = 5.7 °C
q = mc∆T = (150. g)(4.18 J/g°C)(5.7 °C) = 3573.9 = 3600 J = 3.6 kJ
3.6 kJ/0.1954 mol = 18.4237 = 18 kJ/mol
b. C6H4Cl (l) <==> C6H4Cl (s) + 18 kJ
OR
C6H4Cl (l) <==> C6H4Cl (s) ; ∆H = -18 kJ
9) Expt. #9 – Heat of Combustion of a Candle
a. 3690 J/g (work below)
q = mc∆T = (100.g)(4.18 J/g°C)(21.9 °C) = 9154 = 9150 J
9150 J/2.48 g = 3689.5 = 3690 J/g
b. lower your results
10)
i. 96.15 g
ii. 25.0 °C
iii. 10.0 kJ
iv. 12.00 g
v. 10,000 J
vi. 833 J/g
vii. 15.0 kJ/mol
viii. 149%
11)
i. 5.37 g x (1mol/40.0 g) = 0.134 mol
ii. q = mc∆T = (200g)(4.18 J/g°C)(7.2°C) = 6019.2 = 6.0 x 10 3 J or 6.0 kJ
iii. 6.0 kJ/0.134 mol = 44.8 = 45 kJ/mol
iv. NaOH + HCl  H2O + NaCl + 45 kJ
or
NaOH+ HCl  H2O + NaCl; ΔH = - 45 kJ
12)
a.
b.
c.
d.
e.
darker (shift right)
no change
darker (shift right)
lighter (shift left)
no change (no shift - possibly lighter due to dilution)
13)
a. (ii) Ca(C18H33O2) 2
b. The precipitate must have the same mole ratio of ions Ca 2+ : oleate ion that existed in the testtube
with the maximum amount of precipitate formed.
c. n = MV = 0.100mol/L)(0.250 L) = 0.250 mol
0.250 mol (111.0 g/mol) = 2.77 g
14)
a. AgCl
b. AgNO3 and NaCl formed a ppt. which could either be AgCl or NaNO 3. NaNO3 is soluble in box c
(where Ba(NO3)2 and NaCl were mixed), thus AgCl must be the ppt. because NaNO3 is soluble.
c. molecular eqn: Ca(OH)2 (aq) + H2SO4 (aq)  2 H2O (l) + CaSO4 (s)
total ionic eqn: Ca2+ (aq) + 2 OH- (aq) + 2 H+ (aq) + SO42- (aq)  2 H2O (l) + CaSO4 (s)
net ionic eqn: Ca2+ (aq) + 2 OH- (aq) + 2 H+ (aq) + SO42- (aq)  2 H2O (l) + CaSO4 (s)
15)
a.
b.
c.
d.
Ca(OH)2 (s)  Ca2+ (aq) + 2 OH- (aq)
H2SO4  2 H+ (aq) + SO42- (aq)
2 OH- (aq) +H2SO3 (aq)  2 H2O (l) + SO32- (aq)
a dilute solution of an aqueous ionic compound, a dilute solution of a strong acid, and either a dilute
or concentration solution of a weak acid.
e.
ionic = NaCl, Al(OH)3
covalent = C6H12O6, candle wax (C30H62), H2SO4, HC2H3O2
i. NaCl, Al(OH)3, H2SO4, HC2H3O2 (ionic and acids)
ii. HC2H3O2 (a weak acid)
iii. C6H12O6, candle wax (C30H62) (molecular compounds that aren’t acids)
iv. NaCl, Al(OH)3,(conduct well – ionic compounds dissociate 100%) H2SO4, (conducts well –
strong acids ionize 100%) HC2H3O2 (conduct poorly – weak acids ionize partially typically
~5% or less)
v. the higher the concentration, the higher the conductivity ex. 1.0 M NaCl conducts better than
0.5M NaCl
f.
i. 0.100 M BaCl2 (twice as many Cl-‘s, thus more ions)
ii. 0.150 M NaCl (higher concentration, thus more ions)
iii. 0.100 M HCl (stronger acid, thus more ions)
iv. 0.100 M HC2H3O2 (acetic acid is a stronger weak acid than is HSO3-)
v. 0.100 M KOH (ionic has ions, molecular does not)
16)
a. M1V1 = M2V2 (0.350 M)(14.38 mL) = M2 (46.12 mL) M OH- = 0.109 M
b. n = MV = (0.250 mol/L)(0.01429 L) = 0.00357 mol
0.65 g/0.00357 mol = 180 g/mol
17)
a. Take 1 mL of 0.10 M HCl and dilute with water to a total volume of 10 mL. (0.10 M is 10x more
concentration, thus you need to dilute it by a factor of 10.)
b. M1V1 = M2V2 (1.00 M)(V1) = (0.0500 M (500.0 mL) V1 = 25 mL
Thus, take 25 mL of 1.00 M HCl and dilute with water to a total volume of 500.0 mL
c. pH = 3.00 ([H+ = 0.001 M because that’s where the indicator colors match.)
d. 9.9976 x 10 -4 = 1.0 x 10-3
Ka = [H+][A-]/[HA]
Ka = (10-1.85)(10-1.85)2/0.200
e. pH = 3.28
Ka = [H+][A-]/[HA]
2.7 x 10-6 =x2/0.100; x = [H+] = 5.1962 x 10-4 = 5.2 x 10-4 M
f. Ka = 0.0001 or 1.0 x 10 -4 (Ka = [H+] where we see the intermediate color