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
1(d)(i) 0.001220 mol
1(d)(ii) 0.01220 mol
1(e) 1.85 g
1(f) 1.31 g1.31 g
1(g) 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.
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.
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)
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)(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.
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.
5(a) Ammonia is a colourless gas with a pungent odour.
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.
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.