Question 1
1(a) A
standard solution is one whose concentration is accurately known.
1(b) Mass of sample = 3.16 g
1(c)(i) Final
volumes:
Titration
1: 25.60 cm3
Titration
2: 26.30 cm3
Titration
3: 24.50 cm3
1(c)(ii) Volume
used:
Titration
1: 24.60 cm3
Titration
2: 24.25 cm3
Titration
3: 24.35 cm3
1(c)(iii) According
to the wording in this question, the student is required to use values from all three titrations to calculate the average volume used. This however, is not what should actually be done
in the lab. Correct treatment of
this data says that you should ignore Titration 1 and find the average of volumes used in Titrations 2 and 3. Both versions are shown here. The one in which all
three values were used, is
in red and the other in green. Please ask your teacher at school whether they subscribe to the literal interpretation of this question (in red) or
not.
Average
volume used = 24.40 cm3
Average
volume
used = 24.30
cm3
1(c)(iv) 0.000244 mol
0.000243
mol
1(d)(i) 0.001220
mol
0.001215 mol
1(d)(ii) 0.01220
mol
0.01215
mol
1(e) 1.85 g
1.85 g
1(f) 1.31 g
1.31 g
1(g) 0.07278
mol
0.07278 mol
1(h) 5.97 which
is 6 to the nearest whole number.
5.97
which is 6 to the nearest whole number.
1(i) The solution goes from pale green to very pale pink/purple.
1(j) The
unreacted potassium permanganate present show that the end point has been reached because its colour does not
disappear.
1(k)(i) Dirty
green precipitate is formed.
Precipitate is in insoluble in excess aqueous NaOH.
1(k)(ii) Precipitate
slowly changes colour from dirty green to dirty brown/rust brown.
1(k)(iii) White
precipitate formed.
Precipitate is insoluble in dilute
nitric acid.
Question 2
2(a)(i) Electrolysis
is defined as the chemical decomposition of an electrolyte by the passage of an electric current.
2(a)(ii) Electroplating
is the deposition of a metal on the cathode of an electrolytic cell, via the passage of an electric current,
if the right combination of electrolyte
and electrodes (or electrode system) is used.
2(a)(iii) The
extraction of reactive metals e.g. Na and Al from their ores/compounds.
Extraction of active non-metals
e.g. the halogens, like Cl2 from NaCl(aq) using the diaphragm cell.
Anodizing
of aluminium.
2(b)(i)
The
electrodes should be made of graphite.
2(b)(ii) Equation
at anode: 2Br-(l) → Br2(g) + 2e-
Equation
at cathode: Pb2+(l) + 2e- → Pb(l)
Question 3
3(a)(i) Compound
A is an alkane.
3(a)(ii) Compound
A can be used as a fuel, i.e. cooking gas/LPG.
3(a)(iii) One
necessary condition is UV light.
3(a)(iv)
3(a)(v) A
glass rod when dipped in (concentrated) aqueous ammonia and placed in the gaseous by-product, produces dense white
fumes.
3(b)(i)
A polymer is a substance with (macro) molecules built up from many smaller repeating units, which are connected
by covalent bonds: -P-P-P-P-P- or – (P)n -.
3(b)(ii) Type
of polymerization: Addition polymerization.
Name
of polymer: Polypropene
Use
of polymer: In the manufacture of bottles and containers.
3(c)(i) The
functional group present is the ester linkage (group).
3(c)(ii) Type
of polymerization: Condensation polymerization.
Use
of polymer: Manufacture of automobile bodies.
3(c)(iii) Water
is the by-product formed in this polymerization reaction.
Question 4
4(a) An isotope is a
species of an element having a different mass number, but the same atomic number as another species of
the same element, due to a difference in
the number of neutrons in the nuclei.
For carbon-12, number of protons = 6, and mass
number, i.e. number of (protons
+ neutrons) = 12 therefore number of neutrons = (12-6) = 6.
For carbon-13, number of protons = 6, and mass
number, i.e. number of (protons +
neutrons) = 13 therefore number of neutrons = (13-6) = 7.
So carbon-12 and carbon-13 have the same number of protons (6)
and different numbers of neutrons (6 and
7 respectively). They are therefore isotopes of carbon.
4(b) Radiocarbon dating uses carbon-14.
Treatment of cancer uses cobalt-60
Tracing of blood flow and locating of
obstructions in the circulatory system uses
sodium-24.
4(c)(i) Period number = number of occupied electron
shells. Group number = number of outer
electrons. W is in period 2, group 7 and X is in period 3, group 2. X is a metal and W is
a non-metal. The bonding between them is ionic
bonding. W forms W- and X
forms X2+, so the
formula of the compound formed is XW2.
4(c)(ii) The compound will dissolve in water. This is
because it consists of positive and
negative ions whose charges result in ion-dipole attractions with water, a polar solvent. In reality, this compound, MgF2 is one of a small number of insoluble fluorides so the correct answer is
that it is insoluble in water, but a CSEC
student normally does not know how to explain why a compound is insoluble. I can only guess what would be done
in a case where this question was
answered correctly in terms of lattice and hydration enthalpies.
Question 5
5(a) Ammonia is a
colourless gas with a pungent odour.
5(b)(i)
5(b)(ii) 2NH4Cl(s) + Ca(OH)2(s) → 2NH3(g) + CaCl2(aq) + 2H2O(l)
5(b)(iii) Ammonia gas is a weak base and will react, in
a neutralization reaction , with sulfuric
acid. Hence calcium oxide (CaO(s)) can be
used as the drying agent since CaO is basic
and will not react with/neutralize the ammonia gas.
5(c) Moist red litmus when placed in contact with
ammonia gas turns blue.
5(d) Three harmful effects of excessive nitrates in
the environment are:
1. Eutrophication of water bodies results from
high nitrate content.
2. Nitrates in drinking
water is poisonous to humans, especially babies.
3. Nitrosamines, which have been found in
foods are carcinogenic and are thought to orignate from
nitrates.
Question 6
6(a) 1. Water is a very
good solvent as it dissolves both ionic and covalent compounds. Since
chemical reactions occur much more readily when in solution, water is important
for metabolic processes such as digestion and respiration. The ability of water
to dissolve oxygen is necessary for the survival of aquatic life.
2. Water has a high specific heat capacity and
can absorb a large amount of heat without a large change in temperature. This
helps living organisms, which contain approximately 70% water, to maintain a
relatively constant body temperature.
6(b)(i) Ca2+(aq) + Na2CO3(aq) → CaCO3(s) + 2Na+(aq)
6(b)(ii) Boiling can remove temporary hardness. The
soluble hydrogencarbonates decompose to form the insoluble carbonates. Boiling
therefore softens the water by removing the Ca2+ and Mg2+ ions
that were in solution:
M(HCO3)2(aq) → MCO3(s) + H2O(l) + CO2(g)
Where M is Ca and/or
Mg.
6(c) Hard water is water that does not
readily form a lather with soap.
Soft water is water
that readily forms a lather with soap.
To distinguish between hard and soft water
samples, the following method can be used. Equal volumes of both samples are
added to identical test tubes and the test tubes labelled. 1 cm3 of soap solution is
added to both water samples and shaken for 30 seconds. The height of lather
produced by each sample is measured. The water sample giving the greater height
of lather is the soft water. The hard water may either give a lower height of
lather, or no lather at all depending on the degree of hardness.