(O Level Chem) Macromolecules

1. We have seen in the chapter on Alkenes that ethene gas molecules CH₂=CH₂ can be made to undergo addition polymerization to form the addition polymer, poly(ethene) or polythene, [-CH₂-CH₂]_n where n = 2000 to 20000. There is a sharp increase in the relative molecular mass from that of ethene to that of poly(ethene). There is also a sharp increase in the melting and boiling points because the van der Waals forces between the long-chained poly(ethene) molecules require much more energy to break than the weaker van der Waals forces between the small ethene molecules.

2. Depending on the conditions used, we can produce poly(ethene) of different density. For example, to produce low density (0.91-0.94 g/cm³) poly(ethene) which is used in plastic carrier bags and cling films, a temperature of 2000 ºC and a pressure 2000 atm with a small amount of oxygen impurity are used. To produce high density (0.95-0.97 g/cm³) poly(ethene) which is used to make milk bottles, a temperature of 60 ºC, a pressure of a few atm and catalysts containing titanium compounds are used.

3. A polymer is a macromolecule or large molecule that is made up of many monomers chained up together. A monomer is a molecule that is able to bond in long chains. The monomer of poly(ethene), for example, is ethene. The linking up of monomers to form polymers is called polymerization. There are two ways of linking monomers, namely, addition polymerization and condensation polymerization.

4. Addition polymerization occurs due to the presence of C=C functional group in monomers. One example of addition polymers, other than poly(ethene), is polystyrene which is used to make food containers. The monomer of polystyrene is phenylethene or styrene which contains the C=C bond. Other examples include PVC (polyvinyl chloride or poly(chloroethene)) which is used to make water pipes, PTFE (poly(tetrafluoroethene)) which is used as the non-stick coating on cooking utensils, and PP (poly(propene)). Draw the structures of the monomers of PVC, PTFE and PP. All these monomers contain the C=C bond.

5. Addition polymers contain saturated C-C bonds. The strong C-C covalent bonds require a lot of energy to break making addition polymers generally unreactive. The advantage of unreactivity is that addition polymers are safe and durable. Unfortunately, addition polymers are non-biodegradable which means that they cannot be decomposed naturally by bacteria. If addition polymers are irresponsibly disposed of, they would stay for many years causing land and water pollution, and harming wildlife. Addition polymers are usually disposed of by burying in landfills, burning in incinerators and recycling. All these ways of disposing addition polymers have their disadvantages. Landfills take up precious land, incineration releases carbon dioxide and toxic gases and recycling is difficult and expensive.

6. In condensation polymerization, the monomers have two functional groups at both of its ends. The condensation polymers that we are learning are nylon and Terylene which are man-made fibres used in clothing, curtain materials, fishing line, parachutes and sleeping bags. Carbohydrates, proteins and fats are natural condensation macromolecules which are not in the syllabus.
   

7. There are two monomers of nylon, namely a dicarboxylic acid and a diamine. The dicarboxylic acid has one -COOH functional group on each of its two ends. The diamine has one -NH₂ on each of its two ends. The carboxylic acid and amine functional groups react to form an amide linkage -CONH₂- and a water molecule. Nylon is said to be an example of polyamides because of the amide linkages. Because a small molecule is formed as a by-product, the reaction is described as a condensation reaction. For your information, proteins also have amide linkages but their monomers, called amino acids, have one -COOH group on one end and one -NH₂ group on the other end.

8. Nylon is tough, lightweight and waterproof. Because nylon does not let water vapor through, nylon waterproof clothing traps sweat, making it unpleasant to wear. Is nylon biodegradable? Nylon takes 30 to 40 years to biodegrade but this is still less than 500 years for poly(ethene).

9. There are two monomers of Terylene. One of the monomers is also a dicarboxylic acid, as in nylon. The other monomer is a diol which contains two alcohol -OH groups on both of its ends. We have seen esterification in the chapter on Carboxylic Acids and Alcohols in which an alcohol and a carboxylic acid react to form an ester and a water molecule. Similarly, in the formation of Terylene, ester linkages -COO- and water molecules are formed. Terylene is said to be an example of polyesters because of the ester linkages.
   

(O Level Phy) Practical Electricity Teaching & Learning Notes

1. Power stations supply electricity for households, factories and office buildings. Fuel power stations burn fossil fuels (gas, oil, coal) while nuclear power stations make use of nuclear fission of uranium-235 to produce electrical energy. Fossil fuels and uranium-235 are non-renewable because they cannot be replaced. On the other hand, the hydroelectric, solar and wind power stations make use of renewable sources of energy. Hydroelectric power stations, for example, change kinetic energy of running water into electrical energy. Different solar power stations use different ways of changing solar energy to electrical energy. Wind power stations change kinetic energy of wind to electrical energy. One disadvantage of fuel power stations is that they produce air pollutants. Power stations that use renewable sources of energy do not produce air pollutants. 

2. Different power stations have different energy conversion efficiencies. Power plant efficiency is the ratio between the rate of useful electricity output and the rate of energy supplied by the source. For example, a hydroelectric power station may have an efficiency of 95% while an oil power station may have an efficiency of 45%. Why are their efficiencies so different? The hydroelectric power station uses running water to spin its turbine. The oil power station, on the other hand, uses steam to spin its turbine. Much energy has been lost during energy conversions from the oil to the steam and to the turbine. This makes the oil power station less efficient than the hydroelectric power station. Which power station, the hydroelectric station or the oil power station, do you think, would charge more per unit kWh of energy?

3. Electricity can be converted to other forms of energy by using different electrical components. These electrical components make use of different effects of electricity. A heating element uses the heating effect of electricity to convert electrical energy to heat. Heating elements can be found in electric kettles, electric irons and water heaters. A filament lamp uses the heating and lighting effects of electricity to convert electrical energy to light and heat. An electric motor uses the magnetic effect of electricity to convert electrical energy to mechanical energy. What electrical appliances make use of motors? 

4. The power of an electrical component is the rate of conversion of electrical energy into another form of energy. Its SI unit is the Watt which is equivalent to J/s. For example, an electric lamp that is rated as “60 W” converts 60 J of electrical energy to heat and light in 1 s. 

5. Using the definition of potential difference (V = W/Q), we can show that power (P = E/t) is the product of current and voltage i.e. P = IV. We can further derive P = I²R and P = V²/R by inserting the Ohm's Law relationship. Let's apply these formulas in this question. If a small heater operates at 12 V, 2 A, how much energy will the heater use if it is run for 5 minutes?

6. The SI unit of energy is the Joule. The commercial unit of energy, however, is the kilowatt hour (kWh). The kilowatt is the unit of power while the hour is the unit of time. Since energy = power x time, the kWh is a unit of energy. Knowing that 1 kW = 1000 J/s and 1 h = 3600 s, we can easily convert between J and kWh. Now, let's suppose you have an electric cooker that has a hotplate rated at 1500 W and an electric oven rated at 2000 W. During a day, the hotplate is switched on for a total of 1.0 h and the oven is switched on for 3.0 h hours. What is the cost of using the electric cooker during one day if the cost of electricity is 24 cents per kWh?

7. Electrical hazards include electric shocks and electric fires. Electric shocks are caused by damaged insulation and damp or wet conditions. Short circuits between the live and neutral wire occurs when there is damaged insulation. If you happen to touch the metal casing of an appliance that is “live” due to damaged insulation, you would get an electric shock. Damp conditions make it worst because water lowers the electrical resistance of our body from 500 000 Ω to 1000 Ω, making our body a better conductor of electricity. Damaged insulation can also cause electric fires. The large current that results from short circuits produce enough heat to start a fire. Another cause of electric fires is overloading. A power outlet that is overloaded tends to overheat because it draws in current greater than what the electrical wires can safely carry. Note that thinner wires produce heat faster (P = I²R) than thicker wires because thinner wires have greater resistances (R  = ρl/A).

8. The mains three-pin plug contains the brown Live wire, the yellow-green Earth wire and the blue Neutral wire. It also has a fuse that is connected to the Live wire which is at a high voltage, typically at 240 V. The neutral wire which is at 0 V completes the circuit and enables current to flow through the appliance. The Earth wire is the safety wire that conducts current to the earth in case the metal casing becomes “live”. While the Live and Neutral wire should never be in contact with the metal casing of appliances, the Earth wire is always connected to the metal casing. If the appliance has a double insulation, however, there would not be a need for earthing. That is why you see electric irons have a three-pin plug while the hair dryer which has a plastic casing has a two-pin plug.

9. Fuses and circuit breakers are safety devices used to protect appliances from damage caused by excessively large currents. Inside the fuse, there is a short wire that heats up and melts if current exceeds its fuse rating. Once the fuse melts, it has to be replaced. We should choose a fuse rating which is slightly larger than the current that normally flows through the appliance. Common fuse ratings include 3 A, 5 A and 13 A. Let's suppose you have an appliance that uses 3 A. What is the most suitable fuse rating and in which wire should the fuse be connected?

10. One common question is “How do the fuse and Earth wire work together as a safety precaution to prevent electric shocks when using electrical appliances?” If the metal casing becomes “live” due to damaged insulation, the low resistance Earth wire draws a very large current from the Live wire causing the fuse to melt and the circuit to be switched off.

11. Unlike the fuse, the circuit breaker, on the other hand, can just be reset and so does not need to be replaced. Inside the circuit breaker, there is an iron lever next to an electromagnet. The electromagnet is a coil of insulated wire that is wound around a soft iron core. The core magnetizes when a current flows through the coil of wire. If the current is large enough, the electromagnet will be strong enough to attract the iron lever. As the iron lever is pulled away from a contact, the circuit is broken. Note that fuses and circuit breakers are wired into the Live conductor that is at a high voltage. Now, design your own circuit breaker. You should include an  electromagnet, an iron lever that can rotate about a pivot and that makes or breaks a contact, and a spring that is attached to the lever.

12. The household electric circuit consists of the lighting circuit and a ring socket circuit that are both parallel circuits and are both connected to the electricity meter and consumer unit. The electricity meter records the electrical energy used in kWh. The consumer unit contains the main switch and some fuses and circuit breakers. The current in the lighting circuit is much lower than the current in the ring circuit. That is why the lighting circuit uses a lower fuse rating (5 A fuse) than the ring circuit (30 A fuse). That is also why thicker wires (2.5 mm²) are used in ring circuits than in lighting circuits (1.5 mm²). In the ring circuit, the Live wire, Neutral wire and Earth wire goes round the household in the loop without connection to one another. The Live and Neutral wires are connected only when an appliance is plugged to a socket and switched on. Now, draw a table to compare between the lighting circuit and the ring socket circuit. 

(O Level Chem) Alcohols and Carboxylic Acids Teaching & Learning Notes

1. We have seen in the chapter on alkenes that ethene C₂H₂ can be changed to ethanol C₂H₅OH by passing a mixture of ethene and steam over H₃PO₄ catalyst at 300 ºC and 70 atm. Ethanol is used as a solvent in perfumes and varnishes, and a useful fuel in many cars manufactured in the US. The ethanol that is present in alcoholic drinks such as wine and beer is made by fermentation of glucose.

2. Ethanol CH₃CH₂OH is the second member of a homologous series of alcohols which has a general formula of C_nH_2n+1OH. All alcohols contain the functional group -OH which is responsible for the chemical properties of alcohols. The first member of alcohols is methanol CH₃OH which contains only one carbon atom (n = 1). The third (n = 3) and fourth (n = 4) members of the alcohols are propanol and butanol respectively. When n is greater than 2, isomerism exists. Complete the following table and try to draw out all the full structural formulas of these alcohols including the isomers of propanol and butanol. 

Name	Number of carbon atoms, n	Molecular formula	Relative molecular mass	Structural formula
Methanol			32
Ethanol	2			CH3CH2OH
Propanol		C3H8O
Butanol				CH3CH2CH2CH2OH

3. Generally, down the series when the molecular mass of the alcohol increases, the boiling point increases. For example, methanol (n = 1) boils at 68 ºC, ethanol (n = 2) boils at 78 ºC and butanol (n = 4) boils at 118 ºC. Isomers however have different boiling points even though they have the same molecular formula. For example, propan-1-ol boils at 97.2 ºC but propan-2-ol boils at 82.3 ºC.

4. Alcohols, like alkanes and alkenes, are flammable. Ethanol, for example, undergoes complete combustion in sufficient supply of oxygen to produce carbon dioxide and water. Write down the balanced equation for the combustion of 1 mole of ethanol. Use the following information to calculate the enthalpy of combustion of 1 mole of ethanol. You should get -1020 kJ/mol. Because ethanol is runny and volatile and its combustion is so exothermic, ethanol can be used as  a fuel for motor vehicles. What is one advantage of using ethanol over fossil fuels to run motor vehicles?

Bond	Bond energy (kJ/mol)
H-C	413
C-C	348
C-O	743
O-H	463
O=O	498
C=O	743

5. Laboratory experiments sometimes use ethanol. For example, in Biology experiments, ethanol is used as a solvent to dissolve the green pigment, chlorophyll, in leaves. Care must be taken to keep ethanol from open Bunsen flames because ethanol is flammable. If you happen to spill ethanol on the laboratory bench and it catches fire, do not pour water because, water, being denser than ethanol would spread the ethanol over a larger area. Instead, you should use the fire blanket to put out the fire.

6. Besides undergoing combustion, alcohols can also be oxidized by atmospheric oxygen to form weak acids called carboxylic acids. This oxidation can also be achieved by other oxidizing agents such as acidified potassium dichromate or acidified potassium permanganate. For example, ethanol C₂H₅OH can be oxidized by acidified potassium dichromate to ethanoic acid CH₃COOH. In this redox reaction, the orange dichromate is changed to green Cr³⁺(aq). This reaction is used in breathalyzers to test for alcoholic breathe in drivers. Why is this a redox reaction? If potassium permanganate is used, what do you think would be observed?

7. Ethanoic acid is the second member of another homologous series called carboxylic acids which has the general formula, C_nH_2n+1COOH. The first member, methanoic acid HCOOH (n = 0), commonly called formic acid, boils at 101 ºC. Note that in this case, n for the first member is not 1. Ethanoic acid CH₃COOH (n = 1), commonly called acetic acid, boils at 118 ºC while propanoic acid CH₃CH₂COOH (n = 2) boils at 141 ºC. Therefore, like other homologous series, the boiling point of carboxylic acids increases as relative molecular mass increases. Try to draw out the full structural formulas of the first few members of the carboxylic acids.

8. The first few members of carboxylic acids are soluble in water. Down the series, however, the solubility of carboxylic acids decreases. From n = 9 onwards, the carboxylic acids become insoluble in water. The insoluble carboxylic acids which have long chain of carbon atoms are often referred to as fatty acids.

9. Fatty acids are said to be saturated if they contain only C-C single bonds, and are said to be unsaturated if they contain C=C double bonds. Do you still remember, from the topic on alkenes, how you can change unsaturated fatty acid to saturated fatty acid? What catalyst do you use?

10. The functional group of carboxylic acid is -COOH which is responsible for its acidic properties. Therefore, carboxylic acids can react with alkalis to form a salt and water, with metals to evolve hydrogen gas, and with carbonates to evolve carbon dioxide. Make sure you know how to describe the tests for hydrogen gas and carbon dioxide gas. Try to write out the chemical equations for the reaction between carboxylic acid with sodium hydroxide, with magnesium, and with calcium carbonate. Note that ethanoate CH₃COO^- has a charge of +1.

11. Carboxylic acids are, however, weak acids meaning that they dissociate partially to give H⁺(aq). In an aqueous solution of ethanoic acid, for example, only about 0.4% of the ethanoic acid molecules dissociate to become ethanoate ions CH₃COO^-(aq) and protons H⁺(aq). Vinegar, which contains about 1.0 mol dm^-3 of ethanoic acid, has a pH of 2.4. 

(12 and 13 are only for Pure Chemistry)

12. If we heat ethanoic acid and ethanol in the presence of sulfuric acid catalyst under reflux as shown below, we will get a sweet-smelling compound, called ethyl ethanoate CH₃COOC₂H₅, that floats on top of the reaction mixture. Ethyl ethanoate is a class of compounds called esters which contain the -COO- ester linkage. The reaction that forms esters is called esterification. At the same time, water is formed as a by-product. Since a small molecule is formed as a by-product, esterification is also described as a condensation reaction. Note that the reaction is reversible. To prevent the reaction from going backward reforming the reactants, concentrated sulfuric acid, which is a dehydrating agent, is used to absorb the water.

13. We can produce different esters by heating different alcohols and carboxylic acids using the same conditions. Try writing chemical equations for the formations of different esters. Esters tend to smell very pleasant and so they are used as perfumes and flavorings. They are also used as solvents such as in inks and adhesives as they can dissolve many organic compounds. 

Now, try to draw out the flow chart of reactions involving ethane, ethene, ethanol and ethanoic acid.

(O Level Chem) Alkanes and Alkenes Teaching & Learning Notes

1. Carbon is a Group IV and Period 2 element and so it has four valence electrons and two shells. Using its four valence electrons, a carbon atom is able to form a maximum of four covalent bonds. After forming four bonds, its shell is completely full and so the carbon atom can never form five or more bonds. Carbon is well known for its ability, called catenation, to form covalent bonds with other carbon atoms to form large structures such as diamond and graphite and long chains of organic compounds. Living things are made of carbohydrates, proteins and fats that are long-chain carbon compounds.

2. Fossil fuels include natural gas, crude oil and coal. They are called fossil fuels because they were made naturally from decomposing prehistoric plant and animal remains. Since living things are made of compounds of carbon, fossil fuels are made of compounds of carbon too. Coal is a black colored rock mainly made of carbon. Natural gas is a mixture of hydrocarbons consisting mainly of methane. Crude oil is a thick, often black, liquid consisting of various hydrocarbons of different relative molecular masses.

3. Hydrocarbons are covalent compounds. Between atoms in the hydrocarbon molecules, there are strong covalent bonds. Between molecules, however, there are weak van der Waals forces. The larger the hydrocarbon molecule, the greater the van der Waals forces which require more energy to overcome. That is why hydrocarbons with greater molecular masses have higher melting and boiling points than hydrocarbons with lower molecular masses. Now, arrange the compounds, butane C₄H₁₀, heptane C₇H₁₆, and methane CH₄, in order of increasing melting and boiling points.

4. Crude oil needs to be separated into fractions that are useful to us. Since it is a mixture of hydrocarbons with different boiling points, crude oil can be separated using fractional distillation. At the oil refinery, crude oil is heated to about 350 ºC to evaporate most of its compounds. The gases are pumped into the bottom of a fractionating tower about 10 m high. Hydrocarbons with higher boiling points will condense first and are collected at the lower parts of the tower. Hydrocarbons with lower boiling points will rise to the cooler top of the tower before they condense. The fractions obtained include petrol or gasoline (C₅-C₁₀, bp~120 ºC) used as a fuel in cars, naphtha (C₅-C₉, bp~70 ºC) used as feedstock for the chemical industry, kerosene or paraffin (C₁₀-C₁₆, bp~170ºC) for heating and cooking and for aircraft engines, diesel (C₁₄-C₂₀, bp~270 ºC) for diesel engines, lubricating oil (C₂₀-C₅₀) used as lubricants, polishes and waxes. The residue (>C₇₀), bitumen, would be used for making road surface.

5. A homologous series is a group of compounds with a general formula and similar chemical properties. The compounds in a homologous series show a gradual change in physical properties. For example their melting points and viscosity increase and flammability decreases as their molecular mass and size of molecules increase. The homologous series that we will study in O-level are alkanes, alkenes, alcohols and carboxylic acids. Alkanes and alkenes are hydrocarbons because they contain hydrogen and carbon only. Alkanes are also called saturated hydrocarbons because they only contain C-C single bonds, while alkenes are called unsaturated hydrocarbons because they contain C=C double bonds.

6. Alkanes is a homologous series of saturated hydrocarbons with a general formula of C_nH_2n+2. The first member of alkanes is methane CH₄ which has the lowest melting and boiling points. Methane is easily burned and so we say it is very flammable. The next three members are ethane C₂H₆, propane C₃H₈ and butane C₄H₁₀ which are also flammable gases. Pentane C₅H₁₂ and hexane C₆H₁₄ are flammable liquids. The greater the molecular mass of the liquid alkanes, the more viscous and the less flammable they are. The following diagram shows the full structural formulas of straight-chained alkanes up to butane. In drawing full structural formulas, all the bonds must be shown. Now, it is your turn to draw the straight-chained pentane and hexane. Check that all the structures of alkanes agree with the general general formula C_nH_2n+2.

7. Straight-chained alkanes are unbranched alkanes. From C₄ onwards, branched alkanes exist. This is an example of isomerism. Isomers are molecules with the same molecular formula but different structural formulas. Butane C₄H₁₀, for example, can only form two isomers, n-butane which is the unbranched butane, and isobutane which is the branched butane. Pentane C₅H₁₂can form three isomers, namely n-pentane, iso-pentane and neo-pentane. The more carbon atoms there are in the alkane, the more isomers it can form. Try drawing the isomers of hexane C₆H₁₄. How many isomers can hexane form?

8. Alkanes are generally less reactive than other classes of organic compounds because it requires a lot of energy to break their strong C-H and C-C covalent bonds. Alkanes can undergo two kinds of reactions, namely, combustion and free radical substitution. In plentiful supply of oxygen, methane burns to form carbon dioxide and water. If the supply of oxygen is limited, then incomplete combustion occurs and the poisonous carbon monoxide is produced. Use the following values to calculate the enthalpy of combustion of 1 mole of methane. The incomplete combustion of methane is less exothermic than its complete combustion. 

Bond	Bond energy (kJ/mol)
O=O	498
C-O	358
O-H	463
C=O	732
C-H	413

9. If methane and chlorine gases are mixed together in the dark at room temperature, there is no reaction. In the presence of ultraviolet, however, a chain reaction occurs. The ultraviolet has provided the activation energy for the reaction to occur by breaking the Cl-Cl bond to form the reactive chlorine atoms. This is the initiation stage. The Cl atom is called a free radical because it has an unpaired valence electron. A chlorine radical is written as Cl•. With the Cl• radical, the propagation to change methane CH₄ into methyl chloride CH₃Cl occurs. The Cl• is reformed to attack more CH₄ molecules. In effect, the Cl• radical acts as a catalyst to change CH₄ to CH₃Cl. The reaction stops or terminates when radicals combine together to form molecules. Write down the overall equation for the chain reaction by combining equations 2 and 3 together.

10. Alkenes are unsaturated hydrocarbons with the general formula C_nH_2n. The functional group of alkenes is C=C. Alkenes are called unsaturated because the carbon atoms are not bonded to the maximum number of 4 atoms. The first member of the alkenes has two, not one, carbon atoms and it is called ethene C₂H₂. Note the difference that the names of alkanes end with -ane while the names of alkenes end with -ene. Draw the structural formulas of unbranched but-1-ene, pent-1-ene and hex-1-ene. The number “1” in the names indicate that the C=C is at the first carbon atom. The melting and boiling points and viscosity increase down the series of alkenes, just like the trend in alkanes.

11. Like alkanes, alkenes undergo combustion to form carbon dioxide and water. Practice writing out the chemical equation for the complete combustion of ethene. Alkenes typically undergo a type of reactions called addition since two more atoms can be attached to the carbon atoms of each C=C bond. Although the C=C bond (bond energy = 610 kJ/mol) is stronger than the C-C bond (bond energy = 350 kJ/mol), alkenes are more reactive than alkanes. This is because one of the two bonds in C=C is easily broken.

12. When ethene and hydrogen are passed over nickel catalyst at 150 ºC, the hydrogen is added to the C=C and the saturated ethane is formed. This addition reaction is also called hydrogenation. Solid margarine is manufactured from liquid vegetable oils by hydrogenation too. Vegetable oils are polyunsaturated meaning that the chain of molecules contains several C=C bonds. 

13. When a mixture of gaseous ethene and steam is passed over a catalyst of phosphoric(V) acid at 300 ºC and 60-70 atm, ethanol C₂H₅OH is formed. This is the industrial manufacture of ethanol from ethene. In the chapter on alcohols, you will learn about fermentation which is another method of making ethanol. Note that ethanol can also be changed back to ethene by dehydration using concentrated sulfuric acid at 170 ºC.

14. How do we test for the presence of alkenes? In other words, how do you test for unsaturation in an organic compound? We use bromine. When ethene, for example, is blown into bromine that is  dissolved in an organic solvent such as tetrachloromethane under standard laboratory conditions with no ultraviolet, the reddish-brown liquid is decolorized. 1,2-dibromoethane, which is the addition product, is formed. Note that if aqueous bromine is used instead, the product is 2-bromo-ethanol.

15. How is ethene manufactured? Crude oil, unfortunately, contains too many large hydrocarbons which are useless because they are viscous and are not flammable. There is not enough small hydrocarbons to meet demand. Cracking is needed to break large hydrocarbon molecules to produce small hydrocarbons such as ethene. For example, C₁₅H₃₂ is vaporized and passed over a very fine zeolite catalyst powder at 500 ºC to form ethene, propene and octane C₈H₁₈. Try to write down the equation for this reaction which is an example of thermal decomposition. Cracking can also be done in the laboratory. 

16. The ethene and propene produced from the cracking of C₈H₁₈ are useful for making plastics. At high pressure of 10-80 atm and 70-300 ºC in the presence of an aluminum-based catalyst, the gaseous ethene is converted to white solid pellets of poly(ethene) which is our familiar plastic. The ethene molecules have joined together by addition reaction. Each poly(ethene) molecule can be made by 1000 to 1,000,000 ethene molecules and so there is a sharp increase in molecular mass from reactant to product. Poly(ethene) is an example of polymers which are very long molecules made up of repeating units called monomers. The formation of polymers by the addition of unsaturated monomers is called addition polymerization. Try to show the equation for the addition polymerization of propene. Use “n” to denote a very large number.

Draw a flow chart to show all the reactions of ethane and ethene that you have learned.

(O Level Chem) Atmosphere Teaching & Learning Notes

1. The atmosphere is a layer of air which is a mixture of several gases comprising mainly of nitrogen and oxygen. Suppose that in an experiment, a candle is burned in an enclosed 70.5 ml sample of dry air. The candle wax undergoes complete combustion with oxygen to give carbon dioxide and water. After the candle flame is extinguished, the air is shaken with sodium hydroxide solution to absorb the carbon dioxide produced. The volume of the air is found to be 55 ml. This shows that oxygen occupies approximately 21% in the air. In fact, the typical composition by volume of dry air is 78% nitrogen, 21% oxygen and 1% noble gases and carbon dioxide. Because the air is a mixture of gases of different boiling points, it can separated by fractional distillation after it is liquefied.

2. Oxygen exists as a diatomic gas, meaning that its molecules contains two atoms each. Oxygen is reactive and it is a strong oxidizing agent. It supports combustion. For example, oxygen oxidizes hydrocarbons such as methane to produce carbon dioxide and water. The burning of fuels is useful because it is exothermic. Oxygen is also responsible for the corrosion of metals such as the rusting of iron. In living things, oxygen is important for respiration which produces energy from food.

3. Nitrogen also exists as a diatomic gas but, unlike oxygen, it is chemically unreactive. This is because a large amount of energy is required to break its strong N≡N triple covalent bonds. Since nitrogen gas is unreactive, it is used in packing food and in providing an inert atmosphere to prevent oxidation of oil. Under extremely high temperature such as that during lightning and in combustion engines, however, nitrogen can undergo combustion with oxygen to form oxides of nitrogen. Note that nitrogen is important to living things to make proteins. The Haber Process is used to manufacture ammonia which is in turn used to make fertilizers. Plants need nitrogenous fertilizers to make proteins.

4. Only 0.03% of clean air comprises of carbon dioxide. This level of carbon dioxide is kept constant by the carbon cycle which includes photosynthesis, combustion, respiration and fermentation. Only during photosynthesis, carbon dioxide in the air is removed by plants and algae in the presence of sunlight to make food. Combustion, respiration and fermentation add carbon dioxide into the air. The rate of removing carbon dioxide from the air had been balanced by the rate of adding carbon dioxide into the air. We have, unfortunately, upset this balance as we increase the amount of carbon dioxide in the air. 

5. Polluted air contains air pollutants such as carbon monoxide, sulfur dioxide, oxides of nitrogen, methane, unburned hydrocarbons and ozone. Air pollutants are substances in the atmosphere that are harmful to living things and to the environment. Where do they come from and what consequences do they have?

6. Carbon monoxide, a colorless, odorless and poisonous gas comes from the incomplete combustion of fossil fuels in motor vehicles and fuel power stations. Carbon monoxide is poisonous because if it is inhaled, it combines with the haemoglobin in our red blood cells preventing haemoglobin from absorbing oxygen.

7. Sulfur dioxide is formed during the combustion of fuels such as coal which contain sulfur impurities. It is also given off by volcanic eruptions. Being an acidic gas, sulfur dioxide dissolves in rain water to form acid rain. Rain of pH less than 5.0 can kill plants and aquatic lives, and corrode limestone buildings. Sulfur dioxide gas is also a lung and eye irritant.

8. Oxides of nitrogen including nitrogen monoxide and nitrogen dioxide are formed when nitrogen is oxidized at high temperatures. Lightning activity and combustion engines provide high temperature for the oxidation of nitrogen to occur. Oxides of nitrogen damage the lungs while nitrogen dioxide causes acid rain.

9. Methane comes from natural gas, rice fields and cattle. The natural gas which contains mainly methane is a gaseous fossil fuel that is found together with crude oil. Cattle including cows, goats and sheep produce methane when bacteria in their stomachs digest food. Rice fields also give out methane that is produced from the decomposition of organic matter. Methane is one of the greenhouse gases that lead to global warming.   

10. Unburned hydrocarbons come from the incomplete combustion of hydrocarbon fuels in combustion engines, while ozone is created by the reaction between nitrogen oxides, carbon monoxide and unburned hydrocarbons in the presence of ultra-violet radiation from the Sun. Ozone and unburned hydrocarbons react together to form photochemical smog that is harmful to health.

11. Ozone O₃ is usually not present at sea level. It is, however, present as an ozone layer which is 10 km above us. This ozone layer absorbs ultraviolet radiation from the Sun and therefore protects us from this harmful radiation. Over exposure to ultraviolet radiation can cause skin cancer. Unfortunately, the ozone layer has been depleting creating a so called “ozone hole” allowing more ultraviolet radiation through. The reason for its depletion traces back to our use of aerosol sprays. We have been releasing compounds called chlorofluorocarbons (CFC) into the atmosphere when we use aerosol sprays. Over the years, these CFC compounds rise up to the ozone layer. At the ozone layer, the CFC molecules absorb ultraviolet radiation and release chlorine atoms which react with ozone.

12. We have mentioned that methane is a greenhouse gas which is responsible for global warming. Greenhouse gases are gases that trap heat from the Sun causing air temperature to increase. Another important greenhouse gas is carbon dioxide. Some consequences of global warming include melting of polar icebergs and ice caps, climatic change, floods and droughts.

13. How can we reduce the emission of atmospheric pollutants? All motor vehicles should be installed with catalytic converters which convert harmful gases to less harmful gases before they are released into the air. Catalytic converters make use of transition metals, rhodium, platinum and palladium, to speed up redox reactions. In the converters, oxides of nitrogen are reduced to nitrogen and oxygen, carbon monoxide is oxidized to carbon dioxide, and hydrocarbons are oxidized to carbon dioxide and water.

14. Fossil fuel power stations also reduce emissions of air pollutants such as sulfur dioxide. Flue-gas desulfurization refers to the technologies that remove sulfur dioxide from exhaust gases. Since sulfur dioxide is acidic, limestone, lime or any alkaline substance can be used to remove sulfur dioxide from the gases before they are released.

15. We have mentioned that acidic pollutants such as nitrogen dioxide and sulfur dioxide cause acid rain. Acid rain decreases the pH of soils and water bodies such as rivers and lakes and kills plants and aquatic lives. To neutralize acidic soil and water bodies, powdered limestone can be added. 

(O level Chem) Questions on reversible reactions

The knowledge on reversible reactions that includes Le Chatelier's Principle may not be in the current syllabus but I still write out the solutions for two Cambridge questions. The first question is on the Haber process while the second question is on steam reforming.

Le Chatelier's principle: When a reversible reaction at equilibrium is disturbed by a change in concentration, temperature, volume or pressure, the reaction shifts to minimize the effect of the disturbance.  

1. Ammonia is manufactured by the Haber Process. 

N₂(g) + 3H₂(g) = 2NH₃(g)

200 atmospheres

450 ºC

catalyst: iron

The forward reaction is exothermic.

(a) State one use of ammonia. [1]
(b) Describe how the reactants are obtained. [2]
(c) The percentage of ammonia in the equilibrium mixture varies with temperature and pressure.

1. Will increasing temperature increase or decrease the percentage of ammonia at equilibrium? Explain your answer. [2]
   
2. Will increasing pressure increase or decrease the percentage of ammonia at equilibrium? Explain your answer. [2]
   

(d) Catalysts do not alter the position of equilibrium. Explain why a catalyst is used in this process. [2]
[Solution]
(a) Ammonia is used to manufacture ammonium fertilizers. [1]
(b) Nitrogen is obtained from the fractional distillation of liquid air. [1]
Hydrogen is industrially obtained from the reaction between methane and steam. [1]
(c) (i) The forward reaction is exothermic and so the backward reaction is endothermic.
When temperature increases, the equilibrium shifts to the left so as to absorb the heat supplied. [1]
The percentage of ammonia decreases. [1]
(ii) There are 1 + 3 = 4 moles of gaseous reactants to 2 moles of gaseous product.
When the pressure increases, the equilibrium shifts to the right so as to result in lesser moles of gas. [1]
The percentage of ammonia increases. [1]
(d) The catalyst increases the rate of attainment of equilibrium. [1]

2. The hydrogen used in the Haber process is obtained by an industrial process called steam reforming. It's a reversible reaction between methane and steam.
   

CH₄(g) + H₂O(g) = CO(g) + 3H₂(g) ∆H = +210 kJ/mol

The reaction is carried out in the presence of nickel catalyst. The conditions used are 30 atmospheres pressure and a temperature of 750 ºC.

(a) If the temperature of the reaction mixture is increased to 900 ºC, explain what happens to the position of equilibrium. [2]
(b) If the pressure of the reaction mixture is increased to 50 atmospheres, explain in terms of collisions between reacting particles what happens to the speed of the forward reaction. [2]
(c) Explain the advantages of using a catalyst in this manufacture of hydrogen. [2]
(d) In the reaction, 210 kJ of heat energy is used to form 3.0 moles of hydrogen. Calculate how much heat energy is needed to make 1000 kg of hydrogen.
(e) Describe how hydrogen is used to manufacture margarine. [2]
[Solution]
(a) The forward reaction is endothermic while the backward reaction is exothermic.
When temperature increases, the equilibrium shifts to the right [1] so that heat is absorbed [1].
(b) When the pressure is increased, the number of molecules per unit volume increases. [1]
The frequency of collision increases and therefore the rate of reaction increases. [1]
(c) Using a catalyst, the rate of attainment of equilibrium increases. [1]
The heated catalyst supplies the heat for more hydrogen formed at equilibrium. [1]
(d) Mass of 3.0 moles of hydrogen H₂ = 3.0 mol x (1 + 1) g/mol = 6.0 g
To form 6.0 g of H₂, 210 kJ of heat is used.
To form 1000 kg of H₂, (210 kJ x 1000 000 g)/6.0 g [1] = 35000 000 kJ [1]
(e) Margarine is formed when liquid vegetable oils that are unsaturated react with hydrogen [1] at about 60 ºC in the presence of nickel catalyst [1].

(O level Phy) Comparing EM waves and Sound

Similarities

EM waves	Sound waves
They are both forms of energy.
They show wave properties through reflection, refraction and diffraction.
The formula speed = frequency x wavelength applies to them.

Differences

EM waves	Sound waves
They do not require a material medium to travel. They can propagate through vacuo.	They require a material medium to travel as they are mechanical waves. They cannot travel in vacuo.
In air, they travel at 3 x 108 m/s.	In air, they travel at 330 m/s.
They are transverse waves, meaning the vibration is perpendicular to the direction of travel.	They are longitudinal waves, meaning that the vibration is parallel to the direction of travel.
They consist of a vibrating electric field and a vibrating magnetic field coupled together. The electric field and magnetic field are vibrating perpendicular to each other.	They consist of a series of compressions (regions in which pressure is higher than normal) and rarefactions (regions in which pressure is lower than normal).

Attempt these questions before you refer to the answers that follow.

1. In outer space, we can see an explosion but cannot hear it. Why?
   
2. We can calculate the distance a storm is away from us by measuring the time between seeing a lightning and hearing the thunder. What other value is needed? What assumption is being made?
   
3. A tuning fork vibrates with a frequency of 423.5 Hz. How many compressions pass through a point in one second when the sound from the tuning fork passes by?
   
4. The audible range of frequency for humans is 20-20,000 Hz. Calculate the shortest wavelength and longest wavelength that can be heard by humans. Take the speed of sound be 330 m/s. 

Answers:

1. Light can travel through vacuo but sound cannot.
   
2. The speed of the sound in air is needed. Assumption:
   
3. 423.5 compressions pass through the point in one second.
   
4. Shortest wavelength = speed / highest frequency = 330 / 20000 = 0.0165 m. Longest wavelength = speed / lowest frequency = 330 / 20 = 16.5 m
   

(O level Chem) How to prepare salts

Given a salt, how do you decide what method to use? First of all, ask yourself whether it is soluble or insoluble.

• All Group I (sodium, potassium) compounds are soluble.
  
• All ammonium salts are soluble.
  
• All nitrates are soluble.
  
• Examples of insoluble salts: AgCl, PbCl₂, AgI, PbI₂, BaSO₄, CaSO₄.
  

Precipitation method: If it's insoluble, then the steps are simple. Mix two solutions in a beaker and the desired salt precipitates out. Then filter out the precipitate and dry it.

Worked Example: Describe, with the aid of balanced chemical and ionic equations, how you would prepare lead iodide.
[Solution]
Lead iodide PbI₂ is an insoluble salt. So, use the precipitation method.
Choose two solutions of soluble compounds that are readily available in the laboratory: lead(II) nitrate and potassium iodide (Remember: All nitrates are soluble. All potassium salts are soluble.)
Steps:

• Add colorless solution of lead(II) nitrate to colorless solution of potassium iodide in a beaker. A yellow precipitate of lead(II) iodide is formed.
  

Chemical equation: Pb(NO₃)₂(aq) + 2KI(aq) --> PbI₂(s) + 2KNO₃(aq)
Ionic equation: Pb²⁺(aq) + 2I^-(aq) --> PbI₂(s)

• Filter the suspension to obtain the residue of lead(II) iodide.
  
• Dry the lead(II) iodide salt under an infrared lamp.
  

Titration method: If it's a soluble salt of Group I metals or ammonium, use the titration method. The description of this method is pretty long. Basically, a titration between an acid and an alkali is performed using an indicator. This is to determine the volume of the acid or alkali needed to completely neutralize the other. The mixture is discarded and the step is repeated without the indicator. Now, you get the solution of the required salt. The salt is finally obtained from its solution by crystallization.

Worked Example: Describe how you would prepare rubidium sulfate.
[Solution]
Check the Periodic Table and you find that rubidium is in Group I. So Rb₂SO₄ is soluble. The acid part of the salt clearly comes from H₂SO₄ while the base part of the salt can come from rubidium hydroxide RbOH (similar to NaOH and KOH). The titration method should be used.
Steps:

• Place RbOH in a burette.
  
• Pipette 25.0 cm³ of H₂SO₄ into a conical flask.
  
• Add an indicator into the H₂SO₄.
  
• Run RbOH from the burette into the acid until the indicator just changes color.
  

2RbOH(aq) + H₂SO₄(aq) --> Rb₂SO₄(aq) + 2H₂O(l)

• Repeat the above steps without the indicator so that you obtain a colorless Rb₂SO₄ solution.
  
• Evaporate some of the water to obtain a saturated solution of Rb₂SO₄. Let the saturated solution to cool so that Rb₂SO₄ crystallizes out. Filter the crystals and dry them under an infrared lamp.
  

Now, salts of Group I metals and ammonium are not the only soluble salts. What if you are asked to prepare copper(II) sulfate or zinc chloride? You cannot simply suggest titrating Cu(OH)₂ or Zn(OH)₂ with H₂SO₄ or HCl because Cu(OH)₂ and Zn(OH)₂ are insoluble solids (precipitates) if you remember your Qualitative Tests for Metals. Here are other methods you can use.

• acid + insoluble base --> salt + water
  
• acid + insoluble carbonate --> salt + water + carbon dioxide
  
• acid + metal --> salt + hydrogen gas
  

All these three methods are similar in that the insoluble base / insoluble carbonate / metal are added in excess to the acid so as to ensure that all the acid has reacted. You then filter away the unreacted solid reagent so as to obtain the filtrate only. Finally, the salt is then obtained from the filtrate solution by crystallization.

• Most bases (e.g. CuO) are insoluble. Only oxides of Group I metals and some Group II metals are soluble.
  
• Most carbonates (e.g. CaCO₃, CuCO₃) are insoluble. Only ammonium carbonate, Group I carbonates are soluble.
  
• Copper does not react with dilute HCl and H₂SO₄. Do not suggest preparing CuSO₄ or CuCl₂ by reacting Cu with dilute acids.
  

Worked Example: Describe how you would prepare copper(II) sulfate.
[Solution]
The acid part of copper(II) sulfate is clearly from sulfuric acid. The base part can come from CuO or CuCO₃ but not from Cu because Cu does not react with dilute H₂SO₄.
Steps:

• Add excess CuO (a black solid powder) or CuCO₃ (a green solid powder) to a beaker of dilute H₂SO₄ and stir. The solid dissolves and the solution turns blue.
  

Either: CuO(s) + H₂SO₄(aq) --> CuSO₄(aq) + H₂O(l)
Or: CuCO₃(s) + H₂SO₄(aq) --> CuSO₄(aq) + H₂O(l) + CO₂(g)

• Filter the unreacted CuO or CuCO₃ to obtain the blue filtrate solution of CuSO₄.
  
• Evaporate some of the water to obtain a saturated solution of CuSO₄. Let the saturated solution to cool so that CuSO₄ crystallizes out. Filter the crystals and dry them under an infrared lamp.
  

(O Level Phy) Uses of EM waves

EM waves are waves that are made up of an electric field component and a magnetic field component oscillating at 90º to each other and propagating at 3 x 10⁸ m/s in vacuo.

EM waves have various uses depending on their properties.

Part of EM waves	Uses	Danger
Radio (longest wavelength, lowest frequency, lowest energy)	Used for FM and AM radio broadcasts Used for communication by sending information to and from satellites in space Reason: Radio waves are able to pass through the atmosphere. Also, having very long wavelengths, they are able to diffract or bend around buildings and hills so that places in the shadow of large buildings or hills can receive the radio waves. Note: The radio wave and MW frequencies used in commercial, scientific and military satellites are different so that they do not interfere with one another.
Microwave	Used for cooking Reason: The MW has a certain frequency that can be absorbed by water molecules. The water molecules then vibrate faster. This translates to heat and cooks the cells in food. Used for communication by sending information to and from satellites in space Reason: MW, especially the low frequency MW that is near the radio waves region, can travel through the Earth's atmosphere and through clouds.	May kill healthy cells Reason: Water in cells can be heated up by MW.
Infrared	Used in cameras to detect hot objects in the dark Reason: The surface of hot objects emit IR radiation faster than the surface of cold objects. Used in remote controls for TV and videos Used in optical light fibre	Can cause sunburn Reason: IR radiations are heat waves that can damage cells.
Visible from 400 nm (red) to 700 nm (violet)	Used in optical light fibre e.g. In endoscopes for seeing into human body How it works: Light travels through the glass fibre without emerging as they undergo total internal reflection on the glass-air boundary.	Light of high intensity such as lasers can damage the retina of the eye.
Ultraviolet	Used in sun beds to give people sun tans Reason: The skin absorbs UV rays causing melanin to undergo a chemical reaction. The skin then looks tanned.	Can cause skin cancer Reason: UV rays are ionizing.
X-way	To produce X-ray images of bones Reason: X-rays can pass through muscle tissue but cannot pass through bones.	Can cause cancer Reason: X-rays are ionizing.
Gamma (shortest wavelength, highest frequency, highest energy)	Used to kill cancer cells in chemotherapy Used to kill bacteria in foods and sterilize surgical equipment Reason: Gamma rays have high frequency and hence high energy. They are ionizing radiation, capable of knocking electron off from atoms and breaking chemical bonds in molecules such as DNA.	Can damage healthy cells Reason: Gamma rays are highly ionizing

Attempt these questions before you refer to the answers that follow.

1. What properties of EM radiations show that they are waves?
   
2. State two differences between EM waves and sound.
   
3. Ultraviolet, X-ray, infrared radiation, microwave radiation
   

(a) Arrange the EM radiations in order of increasing frequency.
(b) (i) What are ionizing radiations?
(ii) Which of the above four are ionizing radiations?
(iii) State one use for each of the four radiations.


Answers:

1. EM waves can undergo reflection, refraction and diffraction which are general properties of waves.
   
2. EM waves travel at 3 x 10⁸ m/s in air and sound travels at 330 m/s in air. / EM waves can travel through vacuo but sound needs a material medium to travel. / EM waves are transverse waves but sound waves are longitudinal waves. (any two)
   
3. (a) microwave (lowest frequency), infrared, ultraviolet, X-ray (highest frequency) (b) (i) Ionizing radiation is a radiation that is able to “knock off” electrons from atoms. (iii) Microwave – transmission of signals via satellites, infrared – in remote controls, ultraviolet – in sun bed, X-ray – to take X-ray images of bones in medicine
   

(Lower Sec Science) The path of the hamburger through the digestive system

How do you tackle this which may appear in a free-response-questions section?

Question: Describe what happens to a regular hamburger in the digestive system. [10]

You can either draw a diagram labeled with descriptions or write out as follow:

The hamburger consists of bread, meat, cheese, lettuce and tomato. The nutrients present are:

• Bread – carbohydrates
  
• Meat and cheese – Proteins and fats
  
• Lettuce and tomato – Fibre, water, mineral salts, vitamins
  

In the digestive system, the three classes of enzymes are:

• carbohydrase which breaks down carbohydrates
  
• protease which breaks down proteins
  
• lipase which breaks down fats
  

Digestion begins in the mouth where the hamburger is bitten and chewed. Salivary gland secretes the carbohydrase enzyme amylase to digest starch to sugars.

The bolus of food from the mouth is then moved to the stomach through the gullet (esophagus) by a wave-like action called peristalsis. No digestion occurs in the gullet.

In the stomach, the food is squashed by peristalsis. The stomach produces a gastric juice which contains hydrochloric acid and a protease enzyme. The acid kills bacteria and provide an optimum low pH condition (about pH = 2) for the protease to act on the proteins in the meat and cheese. The proteins are digested into small molecules called amino acids.

After about 3 hours, the thick mixture of food and gastric juice called chyme leaves the stomach and enters the duodenum of the small intestine. Pancreatic juice is secreted by the pancreas into the duodenum. The pancreatic juice is slightly alkaline and helps to neutralize the acidic chyme. It contains amylase, a protease (not the same protease at that in the stomach) and a lipase. The amylase continues to digest starch to sugars while the protease continues to digest proteins to amino acids.

Bile, a green liquid, is also produced by the liver and stored in the gall bladder. It is secreted into the duodenum. The bile acts as an emulsifier to separate the fat globules into smaller fat droplets. This makes it efficient for the lipase to digest the fats. The fats are digested to smaller molecules called fatty acids and glycerol.

The glucose, amino acids, fatty acids and glycerol are absorbed by the walls of the intestine into the blood stream by diffusion. The water, minerals and vitamins are absorbed by the walls of the large intestine. The undigested food which includes the fibre is stored in the rectum before it is passed out of the body through the anus.

Try these questions before you refer to the answers that follow.

1. Complete the following table.

Location	Enzyme	Role
	Salivary amylase	-
Stomach		- Breaks down proteins into amino acids
Small intestine (from pancreas)	Amylase Protease Lipase	- - -

2. At pH = 7, does the protease in the stomach able to work? Why?

3. (a) Why is the chyme that just leaves the stomach acidic?
(b) Can the enzymes in the pancreatic juice function if it is not alkaline? Why?

4. Without bile, is the lipase able to efficiently digest the fats? Why?

5. The small food molecules diffuse through the walls of the small intestine into the blood stream as the blood has a __________ (higher / lower) concentration of food molecules.


Answers:
1.

Location	Enzyme	Role
Mouth	Salivary amylase	Breaks down starch into sugars
Stomach	Protease	Breaks down proteins into amino acids
Small intestine (from pancreas)	Amylase Protease Lipase	- Breaks down starch into sugars - Breaks down proteins into amino acids - Breaks down fats into fatty acids and glycerol

2. pH 7 is not the optimum condition for the protease in the stomach to function properly. The protease can only function in acidic condition.
3. (a) The chyme is acidic because of the hydrochloric acid in the gastric juice.
(b) The enzymes in the pancreatic juice would not function properly in the acidic chyme. They function properly in neutral condition.
4. The fat globules present a small surface area for the lipase to digest the fats. Without bile to separate the fat globules into smaller droplets thereby increasing the surface area, it is inefficient for the lipase to act on the fats.

(O level Chem) Are reactions of acids redox reactions?

Acids and Alkalis is one of the easiest chapters to many students. The basic thing is that acids are substances that produce protons H⁺(aq) or hydroxonium ions H₃O⁺ when dissolved in water, not in alcohol or any other solvents. Do get the habit of writing “H⁺(aq)” instead of just “H⁺” because the proton is simply too unstable by itself. Because the concentration of H⁺(aq) in the acidic solutions is greater than the concentration of OH^-(aq), acidic solutions turn blue litmus to red, and yellow methyl orange to red. They have a pH (= -lg [H⁺]) of less than 7.

The properties of acids are due to their protons. 

H⁺(aq) ions are also responsible for the various properties of acids. They combine with OH^-(aq) in alkalis to produce water. This gives the general equation, acid + alkali --> salt + water, that's a neutralization. Its ionic equation is H⁺(aq) + OH^-(aq) --> H₂O(l). Note that the salt, which is soluble, don't appear in the ionic equation because they are spectator ions.

Acids can also be neutralized by carbonates. Carbonates are compounds that contain the ion CO₃^2-. Carbonates are either soluble such as Na₂CO₃ (Remember that all sodium compounds are soluble.) or insoluble such as CaCO₃ (in limestone) and CuCO₃. Effervescence is observed during the neutralization of an acid by a carbonate. The colorless gas gives a white precipitate with limewater indicating that it's carbon dioxide. The general equation is acid + carbonate --> salt water + carbon dioxide. The ionic equation is H⁺(aq) + CO₃^2-(aq) --> H₂O(l) + CO₂(g).

Acids also corrode metals producing a colorless and odorless gas that extinguishes a burning splint with a “pop” sound. The general equation for the reaction is acid + metal --> salt + hydrogen. If an acid reacts with zinc, then the ionic equation is Zn(s) + H⁺(aq) --> Zn²⁺(aq) + H₂(g). This is called a displacement reaction because the hydrogen is displaced by the metal. Remember, however, that dilute HCl and H₂SO₄ do not react with copper which is unreactive.

So, are the above three reactions redox reactions? To identify a redox reaction, we look for whether there are oxidation and reduction occurring in the reaction. The best way is to look for changes in oxidation states which are charges of atoms. When there's an increase in oxidation state, we say oxidation has occurred. When there's a decrease in oxidation state, we say reduction has occurred. A redox reaction has to have both oxidation and reduction occurring simultaneously. In the following, the oxidation states of atoms are already calculated for you. You've got to know how to calculate them. Only one of the reactions is a redox reaction. Which one? 

(A Level Phy) Pressure and root-mean-square speed

The relationship between pressure of gas and mean-square speed P = 1/3 × ρ<c²> where <c²> = mean of the square of speeds (= ∑c_n²/N) may not be in the current A Level Physics syllabus but I'll just leave this information here for future syllabuses. Questions on this would require the use of other formulas:

• Ideal gas law: PV = nRT = NkT where P = pressure in Pa, V = volume in m³, n = number of moles of gas, R = 8.31 J K^-1 mol^-1, T = temperature in K, k = 1.38 × 10^-23 J K^-1
  
• Number of moles n = mass (g) / molar mass (g mol^-1)
  
• Density of gas ρ = mass / volume V
  
• Mass of gas = N × m where N = number of molecules and m = mass of one molecule
  

Using the above equations, you should practice deriving this if it's in the syllabus:
P = 1/3 × ρ<c²>
P = 1/3 × Nm/V × <c²>
PV = 1/3 × Nm × <c²>
NkT = 1/3 × Nm × <c²>
<c²> = 3kT/m
√<c²> = √(3kT/m)

Know how to work from PV = NkT and P = 1/3 × ρ<c²> to PV = 1/3 × Nm × <c²> and to √<c²> = √(3kT/m)... 

(June 2012) (a) The kinetic theory of gases is based on some simplifying assumptions. The molecules of the gas are assumed to behave as hard elastic identical spheres. State the assumption about ideal gas molecules based on (i) the nature of their movement, [1] (ii) their volume. [1] (b) A cube of volume V contains N molecules of an ideal gas. Each molecule has a component cx of velocity normal to one side S of the cube.  The pressure p of the gas due to the component cx of velocity is pV = Nmcx2 where m is the mass of a molecule. Explain how the expression leads to the relation pV = 1/3 Nm<c2> where <c2> is the mean square speed of the molecules. [3] (c) The molecules of an ideal gas have a root-mean-square speed of 520 m s-1 at a temperature of 27 ºC. Calculate the rms speed of the molecules at a temperature of 100 ºC. [3]

(a) (i) The molecules have zero intermolecular forces on one another so do not affect the speeds at which they collide with the walls of the container. [1]
(ii) The molecules have zero volume such that volume of the gas V does not include the volume of the gas. [1]
(b) The mean of the sum of the square of the velocities in the x-direction is
<c_x²> = [c_1x² + c_2x² + c_3x² + ... + c_Nx²]/N [1]
This is one third of the mean of the square of the velocities in directions x-, y- and z-directions. [1]
<c_x²> = 1/3 x <c²> [1]
pV = Nm<c_x²> = 1/3 × Nm<c²>
(c) pV = 1/3 × Nm<c²>
Also, pV = NkT
NkT = 1/3 × Nm<c²>
<c²> = 3kT/m
√<c²> = √(3kT/m)
At T = 273 + 27 K = 300 K,
√<c²> = 520 m s^-1
√(3k x 300/m) = 520
m = 3(1.38 × 10^-23) × 300 K / 520² = 4.993 × 10^-26 kg [1]
At T = 273 + 100 = 373 K,
√<c²>
= √(3kT/m)
= √[3(1.38 × 10^-23)(373 K)/(4.993 × 10^-26)] [1]
= 556 m s^-1 [1]

(Nov 2011) The planet Mars may be considered to be an isolated sphere of diameter 6.79 × 106 m with its mass of 6.42 x 1023 kg concentrated at its centre. A rock of mass 1.40 kg rests on the surface of Mars. For this rock, (a) (i) determine its weight [3] (ii) show that its gravitational potential energy is -1.77 × 107 J. [2] (b) Use the information in (a)(ii) to determine the speed at which the rock must leave the surface of Mars so that it will escape the gravitational attraction of the planet. [3] (c) The mean translational kinetic energy of a molecule of an ideal gas is <Ek> = 3/2 × kT where T is the thermodynamic temperature of the gas and k is the Boltzmann constant. Determine the temperature at which the rms speed of hydrogen molecules is equal to the speed calculated in (b). A molecule of hydrogen has a mass of 2u. [2] State and explain one reason why hydrogen molecules may escape from Mars at temperatures below that calculated in (i). [2]

(a) (i) F = Gmm/r² [1]
F = (6.67 × 10^-11 N m² kg^-2)(6.42 x 10²³ kg)(1.40 kg)/(6.79 × 10⁶ m /2)² [1]
F = 1.3 N [1]
(ii) GPE = -GMm/r
= -(6.67 × 10^-11 N m² kg^-2)(6.42 x 10²³ kg)(1.40 kg)/(6.79 × 10⁶ m /2) [1]
= -1.77 × 10⁷ J [1]
(b) 1/2 mv² = -GPE [1]
v² = 2 × 1.77 × 10⁷ J / 1.40 kg [1]
v = 5000 m s^-1 [1]
(c) (i) <E_k> = 3/2 × kT
1/2 × m<v²> = 3/2 × kT
2u<v²> = 3kT
T = 2u<v²>/3k
T = 2(1.66 × 10^-27 kg)(5000 m s^-1)²/3(1.38 × 10^-23 J K^-1) [1]
T = 2000 K [1]
(d) The hydrogen molecules that are near the top of Mar's atmosphere experience low gravitational potential and thus have lower escape speed. [1] Lower temperatures are needed by the molecules to possess lower escape speed. [1]

(Jun 2008) (a) Explain qualitatively how molecular movement causes the pressure exerted by a gas. [3] (b) The density of neon gas at a temperature of 273 K and a pressure of 1.02 × 105 Pa is 0.900 kg m-3. Neon may be assumed to be an ideal gas. Calculate the root-mean-square (rms) speed of neon atoms at (i) 273 K, [3] (ii) 546 K [2] (c) The calculations in (b) are based on the density for neon being 0.900 kg m-3. Suggest the effect, if any, on the rms speed of changing the density at constant temperature. [2]

(a) Molecules are in constant motion. [1] They collide with the walls of container. [1] Their change in momentum causes an impact on the walls of container. [1] This gives rise to pressure.
(b) (i) At T = 273 K, P = 1/3 × ρ<c²>
1.02 × 10⁵ Pa = 1/3 × (0.900 kg m^-3)<c²> [1]
√<c²> = 583 m s^-1 [1]
(ii) PV = 1/3 Nm<c²>
NkT = 1/3 Nm<c²>
<c²> = 3kT/Nm
√<c²> = √(3kT/Nm)
The rms speed is directly proportional to √T.
<c₂²>/<c₁²> = T₂/T₁
<c₂²>/340000 = 546/273 [1]
√<c₂²> = 824 m s^-1 [1]
(c) P₁V₁ = P₂V₂
1/3 × ρ₁<c₁²>V₁ = 1/3 × ρ₂<c₂²>V₂
ρ₂/ρ₁ = <c₁²>V₁/<c₂²>V₂
m₂/V₂ ÷ m₁/V₁ = <c₁²>V₁/<c₂²>V₂ m₂V₁/m₁V₂ = <c₁²>V₁/<c₂²>V₂
m₂/m₁ = <c₁²>/<c₂²> 
A change in density due to a change in volume will not have a change in rms speed. [1] A change in density due to an increase in the amount of gas however will increase the rms speed. [1]


Try these questions before you check with the answers below.

1. (Nov 2012) An ideal gas has volume V and pressure p.
   

PV = 1/3 × Nm<c²>

where m is the mass of a molecule of the gas.
(a) State the meaning of the symbol (i) N [1] (ii) <c²> [1]
(b) A gas cylinder of volume 2.1 × 10⁴ cm³ contains helium-4 gas at pressure 6.1 × 10⁵ Pa and temperature 12 ºC. Helium-4 may be assumed to be an ideal gas.

1. Determine for the helium gas (1) the amount, in mole, [3] (2) the number of atoms [2]
   
2. Calculate the root-mean-square speed of the helium atoms. [3]

2. (OCR Jun 2005) The average translational kinetic energy per molecule of gas 1/2 × m<v²> is related to the absolute temperature T by 1/2 x m<v²> = 3/2 × kT where k, the Boltzmann constant, is 1.4 × 10^-23 J K^-1.
   

(a) Show that the root mean square speed √<v²> is proportional to 1/√m when T is kept constant. [1]
(b) The table shows the root mean square speed of molecules for three gases at 293 K.

Gas	Mass / 10-27 kg	Rms speed / m s-1
Helium	6.7	1400
Neon	33	600
Carbon dioxide	47	510

Propose and carry out an arithmetical test to decide whether the data in the table support the statement that the rms speed is proportional to 1/√m when T is kept constant.
(c) The distribution of molecular speeds in a sample of oxygen gas is as shown.
The area of each bar is proportional to the fraction of molecules having a speed within that range. The chart shows that the molecules have a range of speeds.
Use ideas from kinetic theory to explain why 
 

1. gas molecules have a range of speeds [1]
   
2. gas molecules are very unlikely to have speeds much greater than the most common speed. [1]
   

(d) The distribution of molecular speeds shows that the most probable speed is 400 m s^-1.
(i) Use √<c²> = 1.2 × most probable speed to calculate the mean square speed of the molecules in the sample of the gas. [2]
(ii) Use your answer to (d) (i) to estimate the temperature of the gas given the mass of the oxygen molecule is 5.3 × 10^-26 kg. [2]

3. (OCR Jun 2013) In 2005, the Huygens probe landed on Titan, one of Saturn's Moons. It recorded an atmosphere of nitrogen at a temperature of 94 K. Calculate the root-mean-square speed of a nitrogen molecule at this temperature given that the mass of a nitrogen molecule is 4.7 × 10^-26 kg and k = 1.4 × 10^-23 J K^-1. [3]

4. (OCR Jan 2011) The air in a football has a mass of 1.1 × 10^-2 kg.
   

(a) Show that the ball contains about 2 × 10²³ particles given the molar mass of air is 2.9 × 10^-2 kg mol^-1 and Avogadro's constant is 6.0 × 10²³ mol^-1. [1]
(b) Calculate the mean square speed <c²> of the particles in the football if the pressure of air in the ball is 1.7 × 10⁵ Pa and volume is 5.4 × 10^-3 m³. [2]
(c) The temperature of the air in the football increases during the game. Sketch a graph to show how the mean square speed of air particles in the football varies with absolute temperature T. Assume the air behaves as an ideal gas. [1]

5. (Jun 2004) The pressure p of an ideal gas is given by the expression p = 1/3 × Nm/V × <c²>.
   

(a) Explain the meaning of the symbol <c²>. [2]
(b) The ideal gas has a density of 2.4 kg m^-3 at a pressure of 2.0 × 10⁵ Pa and a temperature of 300 K.

1. Determine the root-mean-square (rms) speed of the gas atoms at 300 K. [3]
   
2. Calculate the temperature of the gas for the atoms to have an rms speed that is twice that calculated in (i). [3]
   

Answers: 1. (a) (i) N is the number of gas molecules. (ii) <c²> is the mean of the square of the speeds of molecules. (b) (i) (1) PV = nRT ==> n = PV/RT = (6.1 × 10⁵ Pa)(2.1 × 10⁴ × 10^-6 m³)/(8.31 J K^-1 mol^-1)(273 + 12 K) = 5.2 moles (2) Number of atoms = number of moles x Avogadro's constant = 5.2 mol × 6.02 × 10²³ mol^-1 = 3.2 × 10²⁴ (ii) <c²> = 3PV/(Nm) ==> √<c²> = √[3PV/(Nm)] = √[3(6.1 × 10⁵ Pa)(2.1 × 10⁴ × 10^-6 m³)/(3.2 × 10²⁴ mol)(4 x 1.66 x 10^-27 kg)] = 1300 m s^-1
2. (a) 1/2 x m<v²> = 3/2 × kT ==> <v²> = 3kT/m ==> √<v²> = √(3kT/m). Therefore, √<v²> is inversely proportional to √m. (b) √<v²> = √(3kT/m) ==> √<v²> √m = √(3kT). √<v²> √m should be constant if √<v²> is proportional to 1/√m. Helium: √[(6.7 × 10^-27 kg)(1400 m s^-1)] = 3.1 × 10^-12. Neon: √[(633 × 10^-27 kg)(600 m s^-1)] = 1.9 × 10^-11. Carbon dioxide: √[(47 × 10^-27 kg)(510 m s^-1)] = 4.9 × 10^-12. √(<v²> m) are different for each gas. (c) (i) Different molecules possess different kinetic energies. (ii) Gas molecules are not likely to possess kinetic energies much greater than their surrounding molecules. (c) (i) √<c²> = 1.2 × most probable speed = 1.2 × 400 m s^-1 = 480 m s^-1 ==> <c²> = 230 400 m² s^-2 (ii) 1/2 × m<v²> = 3/2 × kT ==> 1/2 × 5.3 × 10^-26 kg × 230 400 m² s^-2 = 3/2 x 1.38 x 10^-23 J K^-1 x T ==> T = 295 K
3. PV = 1/3 × Nm<c²> ==> NkT = 1/3 x Nm<c²> ==> √<c²> = √(3kT/m) = √[3(1.4 × 10^-23 J K^-1)(94 K)/(4.7 × 10^-26 kg)] = 289 m s^-1
4. (a) Number of molecules N = (1.1 × 10^-2 kg)(6.02 × 10²³ mol^-1)/(2.9 × 10^-2 kg mol^-1) = 2 × 10²³ (shown) (b) PV = 1/3 × Nm<c²> ==> <c²> = 3PV/Nm = 3(1.7 × 10⁵ Pa)(5.4 × 10^-3 m³)/(1.1 × 10^-2 kg) = 250 000 m² s^-2 (c) PV = 1/3 Nm<c²> ==> NkT = 1/3 Nm<c²> ==> <c²> = 3kT/m. Therefore, the graph of <c²> is directly proportional to T.
5. (a) <c²> is the mean of the square of the speeds of particles. <c²> = (c₁² + c₂² + c₃² + ... c_N²)/N (b) (i) P = 1/3 × ρ<c²> ==> 1.02 × 10⁵ Pa = 1/3 × (2.4 kg m^-3)<c²> ==> √<c²> = 360 m s^-1 (ii) PV = 1/3 Nm<c²> ==> NkT = 1/3 Nm<c²> ==> <c²> = 3kT/Nm ==> √<c²> = √(3kT/Nm). The rms speed is directly proportional to √T. √<c₂²>/√<c₁²> = √T₂/√T₁ ==> 2/1 = √T₂/√300 ==> √T₂ = 2 × √300 ==> T₂ = 1200 K