Chemistry AS Revision Questions F332

Elements of the Sea

Calculations

1. Calculate the mass of propene required to produce 49.2g of bromopropane: C3H6+HBr→C3H7Br

Molecules involved C3H6: C3H7BrMolecular mass of each 42 : 122.9Divide to get one 42/122.9 : 1Multiply for mass you want 16.8: 49.2Answer 16.8g

2. 10.5 g of magnesium carbonate is heated strongly so that it fully decomposes. MgCO3(s)→MgO(s)+CO2(g)

a. Calculate the mass of magnesium oxide

Molecules involved MgCO3: MgOMolecular mass of each 84.3 : 40.3Divide to get one 1 : 40.3/84.3Multiply for mass you want 10.5: 5 Answer 5g

b. Calculate the volume of gas producedThe ratio is 1:1 so the same number of moles of carbon dioxide is produced as magnesium oxide.Number of moles = mass/molar mass = 5/40.3 =0.125 molesOne mole takes up 24dm3 at r.t.p.0.125 x 24 = 3dm3

3. A dry sample of barium sulphate was made as follows: BaCl2(aq)

+H2SO4(aq)→BaSO4(s)+2HCl(aq)

a. If 25cm3 of 1.0mol/dm3 barium chloride solution was used, how many moles of barium chloride were available?Number of moles = (concentration x volume)/1000 = (1 x 25)/1000 =0.025 moles

b. What is the theoretical yield of barium sulphate if the other reactants are in excess?Ratio of moles = 1:10.025 x (137.3 + 32.1 + 64) = 5.83g

c. If 5.20g of dry barium sulphate were recovered, what was the percentage yield of barium sulphate?Percentage yield= (actual/theoretical) x 100 = (5.2/5.83) x 100 = 89.2%

Titrations

1. Calculate the concentration of a solution of ethanoic acid, CH3COOH, if 25.4cm3 of it is neutralised by 14.6cm3 of 0.5M sodium hydroxide solution. CH3COOH+NaOH→CH3COONa+H2O

Molecules CH3COOH NaOHVolumes 25.4cm3 14.6cm3

Concentrations ? 0.5MMoles are in a 1:1 ratio in this reactionMoles of NaOH = (0.5 x 14.6)/1000 = 0.0073 = moles of ethanoic acidConcentration = (number of moles x 10000)/volume = (0.0073 x 1000)/ 25.4 =0.287M

2. You are supplied with 0.75g of calcium carbonate and a solution of 0.25M sulphuric acid. What volume of acid will be needed to neutralise the calcium carbonate? CaCO3+H2SO4→CaSO4+H2O+CO2

Moles of calcium carbonate= mass/molar mass = 0.75/100.1 0.0075 molesRatio =1:1 so 0.0075 moles of sulphuric acid.Volume= (1000 x moles)/concentration = (1000 x 0.0075)/0.25 = 30cm3

Electronic Structure

1. Potassium reacts with oxygen to form potassium oxide, K2O.a. Give the electron configuration of the potassium atom.

1s22s22p63s23p64s1

b. Using arrow in box notation, give the electron configuration of the oxygen atom.

c. Explain why it is the outer shell electrons, not those in the inner shells, which determine the chemistry the chemistry of potassium and oxygen.The outer shell electrons in potassium and oxygen can get close to the outer shells of neighbouring atoms so they can be transferred or shared. Inner shell electrons are held tightly to the nucleus and shielded from the electrons in other atoms.

2. This question concerns the electron configuration of atoms and ions.

a. What is the electron configuration of a manganese atom?1s22p22p63s23p63d54s2

b. Identify the element with the 4th shell configuration of 4s24p2

Manganese has 25 electrons, 5 more can fit in 3dAdd 2 in the 4p sub-shell so 25+5+2=32Atom with 32 electrons= Germanium

c. Using arrow in box notation, give the electronic configuration of the aluminium atom.

d. Suggest the identity of an atom, a positive ion and a negative ion with the configuration 1s22s22p63s23p6.Atom = Period 3, group 0 = ArgonPositive ion= one more proton = K+

Negative ion = one less proton = Cl-

Oxidation and Reduction

1. When hydrogen iodide gas is bubbled through warm concentrated sulphuric acid, hydrogen sulphide and iodine are produced.

a. Balance the equation below for the reactionH2SO4 (l) +8HI (g) →H2S (g) +4I2(s) +4H2O (l)

b. Calculate the oxidation state of sulphur in H2SO4 and H2S.H = +1 so +2 in totalO = -2 so -8 in total = -6Overall must = 0 so sulphur in H2SO4 =+6H = +1 so +2 in totalOverall must = 0 so sulphur in H2S = -2

c. In this reaction, which is the reducing agent? Give a reason.

Iodide because it was -1 in the reactants and 0 in the products. An increase in oxidation number means it is oxidised and is the reducing agent.

d. Write a half-equation to show the conversion of iodide, I-, into iodine, I2.2I- → I2 + 2e-

e. Write a half-equation to show the conversion of sulphuric acid into hydrogen sulphide.H2SO4 + 8H+ + 8e- → H2S + 4H2O

Electronegativity

1. Many covalent have a permanent dipole, due to differences in electronegativities.

a. Define Electronegativity.The ability for an atom to attract the electrons in a covalent bond

b. Draw the shapes and predict the overall polarity of the following molecules:

i. Br2

Non- polarii. H2O

More electronegative towards oxygen

iii. CCl4

iv. NH3

Br Br

Oδ-

Hδ+ Hδ+

c. Fluorine is the most electronegative element. NF3 is the same shape as NH3 yet it has no permanent dipole. Why is this?The lone pair of electrons enough to cancel out the dipole on NF3 as hydrogen is not electronegative; the lone pair is able to cause a dipole in the NH3 molecule.

2. Draw diagrams to show the shape of the covalently bonded molecules below. Indicate any permanent dipoles on your diagrams.

a. Boron(III)chloride (BCl3)

b. Nitrogen(III)chloride (NCl3)

Intermolecular Forces

1. The molecules in the table all have the molecular formula C5H12. Explain the differences in the strengths of the intermolecular forces.

Molecule Relative strength of intermolecular forces

Pentane Strong2-methylbutane Intermediate

2,2-dimethylpropane WeakPentane is a straight chained alkane so it has the greatest surface area. It also makes it easier for temporary dipoles to be induced on it which causes stronger intermolecular forces. As you go down the table there is more branching so there is less surface area so the intermolecular forces are weaker.

Ionisation Enthalpies

1. The graph shows how first ionisation enthalpy varies across Period 2.

a. Explain why there is a general increase in ionisation enthalpy across the period.As you go across the period the attraction between the outer shell electrons and the nucleus increases because the number of protons in the nucleus is increasing, giving more positive charge. The shielding is roughly constant because the shielding remains roughly the same.

b. Explain how the graph provides evidence for the electronic structure of these elementsFor Lithium and Beryllium the 2s sub-shell is filling. From Boron to Nitrogen the 2p sub-shell is half filling. This sub-shell is further out so it is easier to move electrons. From O to Ne, the electrons begin spin-pairing in the 2p sub-shell. There is extra repulsion between the electrons so they are easier to remove.

Group 7-The Halogens

1. a. Write an ionic equation for the reaction between iodide

solution and sodium astatide. (NaAt)I2 + 2At- → At2 + 2I-

b. For that equation, deduce which substance is oxidised.The sodium Astatide because its oxidation number increases from -1 to 2

2. The extraction of bromine from seawater can be represented by the following equation:Cl2 (aq) +2Br-

(aq) →Br2 (aq) +2Cl-(aq)

a. Explain why this is classed as a redox reactionBromine is oxidised chlorine is reduced

b. Identify the oxidising agent in this reactionChlorine

c. Using your knowledge of periodic trends, suggest which halogen(s) could be used to extract chlorine from a concentrated solution of chloride ions.Fluorine because it is more reactive so is able to displace the chlorine.

More about the Halogens

1. Chlorine is a very important industrial chemical.a. Explain why chlorine is normally produced on the same

site as where it is to be used.Chlorine is toxic and very reactive. It is difficult to store and expensive to transport in large quantities.

b. Name two important compounds made using chlorine, and give an example of how each is usedIt is used in PVC for insulating electric wires and it is used in bleach for water treatment.

c. Write the equations for the electrolysis of concentrated sodium chloride solution.2Cl- → Cl2 + 2e-

2H+ + 2e- → H2

2. Bromine can be extracted by the electrolysis of a concentrated solution of potassium bromide.

a. Write the equations for the reactions at the anode and the cathode.Anode: 2Br-

(aq) → Br2 (aq) + 2e-

Cathode : 2H+ + 2e- → H2

b. Bromine can be used to make the pesticide bromomethane. Suggest why bromomethane is likely to be made on the site where the bromine is produced.Bromine is a toxic substance and it is safer to transport it as a compound

The Chemical Industry

1. Ethanol is made by the addition of steam to ethane: C2H4(g)+H2OC2H5OH(g)

a. The atom economy of this reaction is 100%. Explain why reactions with high atom economies are used as often as possible in industry.Addition reactions produce no waste products so they make better use of the resources.

b. Suggest one reason why the percentage yield of this reaction is always less than 100%.Not all of the starting materials react.

c. Ethene is made by cracking oil or natural gas. Explain why this may affect the choice of location for a plant producing ethanol.

You would build it near an oil refinery so that it doesn’t have to be transported far.

2. Chemical A is used to make chemical B. Using 500Kg of chemical A, the reaction equation indicates that you should theoretically get 300Kg of chemical B. You actually get 250Kg.

a. Calculate the percentage yield of this reactionPercentage yield = (actual yield/theoretical) x 100 = (250/300) x 100 =83.3%

b. Chemical B is made from Chemical A in a continuous process. Give one advantage and one disadvantage of using a continuous process compared to using a batch process.There are less labour costs but the plant is more expensive to build

Halogenoalkanes

1. In 1956 the anaesthetic “Halothane” was developed by ICI. It has the chemical formula CF3CHBrCl. Give the chemical name for this structure and draw its displayed formula.2-bromo, 2-chloro, 1, 1, 1-trifluoroethane

2. A chemist has samples of three Halogenoalkanes in tubes labelled A, B and C. The boiling points of the three Halogenoalkanes are: Tube A=71⁰C, Tube B=46⁰C, Tube C=102⁰C. The Halogenoalkanes in the three tubes are 1-chloroprapane, 1-bromopropane and 1-iodopropane. Which halogenoalkane is in which tube? Explain your answer.

Tube A is 1-bromopropane, tube B is 1-chloropropane and tube C is 1-iodopropane. The more electrons the halogen has the stronger the instantaneous dipole induced dipole forces are and the more energy required to overcome them.

More about Halogenoalkanes

1. The equation for the reaction between water and 2-bromopropane is shown below:CH3CHBrCH3 + H2O → CH3CH(OH)CH3 + HBr

a. i. Name the type of reaction

Nucleophilic Substitutionii. Name the organic product

Propan-2-olb. Under the same conditions, 2-iodopropane was used in

place of 2-bromopane in the reaction above. What difference would you expect in the rate of reaction? Explain your answer.The reaction would happen faster because the C-I bond is weaker than the C-Br bond.

c. Draw the mechanism for the reaction between 2-bromopropane and aqueous potassium hydroxide.

The Atmosphere

Giant Structures

1. Look at the data below on the compound born nitride (BN).Melting point: 2700⁰CElectrical conductivity of liquid: zeroSolubility in water: zero

a. On the basis of this data suggest the bond type and structure in boron nitride. Explain your choices.It is a Giant covalent lattice. You can tell it is covalently bonded because it does not conduct electricity as it has no free electrons. It is giant as it has such a high melting point.

b. Boron(III) chloride has a boiling point of -107⁰C.i. Suggest the bond type and structure in

boron(III)chlorideSimple molecular, Covalent

ii. Explain why boron(III)chloride has a much lower melting point than boron nitride.Only weak instantaneous dipole induced dipole forces are holding the molecules together. These are weak so do not require much energy to overcome. This gives it a low melting point.

Reaction Rates

1. Nitrogen monoxide (NO) and ozone (O3) sometimes react to produce nitrogen dioxide (NO2) and oxygen (O2). A collision between the two molecules does not always lead to a reaction. Explain why.For a reaction to occur atoms and molecules must be in the right orientation or they will just bounce off of each other. They must also have enough activation energy.

2. Use collision theory to explain why the reaction between a solid and a liquid is generally faster than that between two solids.The particles in a liquid are able to move freely and collide with the solid particles. Particles in a solid only vibrate around fixed positions, so only those on the touching surface of the solid will be able to react.

3. This graph shows the rate that oxygen is evolved when hydrogen peroxide decomposes at 25⁰C.Which of the curves, X, Y or Z, shows the rate that oxygen is evolved when the same amount of hydrogen peroxide decomposes at 15⁰C? Explain your answer.Y. The same amount of hydrogen peroxide is decomposed, so the same amount of oxygen will be evolved therefore it cannot be a curve. Curve Y also shows it decomposing slower than at 25⁰C, which would be correct because reaction rates are slower at cooler temperatures.

More on Reaction Rates

1. Explain how homogeneous catalysts speed up chemical reactions.The homogeneous catalyst forms an intermediate compound. The activation energy required to form th intermediates and then the products is less than would be needed to make the products directly.

2. Enzymes are proteins that catalyse specific biological reactions. Draw a fully labelled enthalpy profile for an enzyme catalysed reaction and an uncatalysed reaction.

3. Homogeneous catalysts speed up a reaction by forming intermediate compounds. Explain why they are classed as catalysts, even though they are chemically altered during the reaction.The catalyst is reformed by the time the products form. It is chemically unchanged at the end of the process.

Reversible Reactions

1. Nitrogen and oxygen gasses were reacted together in a closed flask and allowed to reach equilibrium with the nitrogen monoxide formed. The forward reaction is endothermic.N2(g) + O2(g) 2NO(g)

a. State Le Chatelier’s principleIf a reaction at equilibrium is subject to a change in concentration, pressure or temperature, the equilibrium will shift to try and oppose the change.

b. Explain how the following changes would affect the position of the above reaction:

i. Pressure is increasedThere will be no change as there is the same number of molecules on each side of the equation

ii. Temperature is reduced

This removes heat which will favour the exothermic reaction so the equilibrium shifts to the left.

iii. Nitrogen monoxide is removed.Removing nitrogen monoxide reduces its concentration. They reaction will oppose this by shifting the equilibrium to the right to increase it again.

c. What would be the effect of a catalyst on the composition of the equilibrium mixture?There would be no affect on the equilibrium position because a catalyst speeds up both directions of the reaction equally.

The Atmosphere

1. The Earth absorbs radiation from the Sun. It also emits radiation.

a. What are the main types of radiation emitted by the Sun and the Earth?Visible light and infra red.

b. Name the type of radiation that can cause an increase in some covalent bonds’ vibrational energy. What types of molecules are not affected?Infrared. Molecules made up of identical atoms are not affected.

2. Calculate the energy absorbed when one molecule of HCl changes from its ground vibrational level to the next level, given that the frequency of radiation absorbed is 8.19x10¹³Hz.E = hf =6.63 x 10-34 x 8.19x10¹³= 5.43 x 10-20J

The Greenhouse Affect

1.a. Name the three greenhouse gasses

Water vapour, carbon dioxide and methaneb. Explain how greenhouse gasses keep the temperature in

the lower layers of the Earth’s atmosphere higher than it otherwise would be.The bonds absorb infrared radiation which increases their vibrational energy. Energy is transferred to other molecules by collision. This increases the average kinetic energy of the molecule which in turn increases the temperature.

2. The concentration of carbon dioxide in the Earth’s atmosphere has increased over the last 50 years.

a. Describe one piece of scientific evidence for the increaseOceans have become more acidic because carbonic acid has formed in them as they have dissolved more atmospheric carbon dioxide.

b. Describe two methods that chemists are developing as a way of reducing carbon dioxide emissions.Developing more renewable energy resources and by increasing photosynthesis.

Halogenoalkanes and CFC’s

1. 1-chloro-1,1-difluoroethane is an example of a hydrochlorofluoroalkane (HCFC).

a. Draw the displayed formula of this molecule

b. In the presence of strong sunlight, 1-chloro-1,1-difluoroethane undergoes homolytic fission to generate a chlorine radical and a molecular radical. Write an equation to represent this reaction.CH3CF2Cl → CH3CF2• + Cl•

2. CFC’s were invented in 1928. They were widely used in the 20th century.

a. Give three important uses of CFC’sCoolants in fridges

b. What useful properties do CFC’s have?Unreactive, non-flammable and non-toxic

c. Why was the use of CFC’s banned by the Montreal Protocol?Because they were destroying the ozone layer.

d. What alternatives to CFC’s have been used and what are their drawbacks?HFC’s and hydrocarbons are greenhouse gasses. HCFC’s still damage the ozone layer.

Ozone

1. The ‘ozone layer’ lies mostly between 15 and 30Km above the Earth’s surface.

a. Explain how ozone forms in this part of the atmosphereUV radiation from the Sun split oxygen molecules to form oxygen free radicals. These react with more oxygen molecules to form ozone.

b. What are the benefits to humans of the ozone layer?Ozone prevents most of the harmful UV radiation from the Sun from reaching the Earth’s surface.

c. How does the ozone layer absorb harmful radiation without being permanently destroyed?

The ozone molecules interact with the UV radiation to form an oxygen molecule and a oxygen free radical. The radical produced then goes and reacts with another oxygen molecule to make ozone again.

2.a. Explain why large amounts of ground-level ozone can be a

problemIt is toxic to humans.

b. Why are these problems most frequently encountered in heavily industrialised areas?Ground level ozone forms as a result of sunlight acting on mixtures of nitrogen dioxide and hydrocarbons which are emitted by power stations and vehicles.

The Polymer Revolution

Addition Reactions of Alkenes

1. Cyclohexane and hex-1-ene both have the molecular formula C6H12.

a. Hex-1-ene can be described as an unsaturated hydrocarbon. Explain what is meant by the term ‘unsaturated’ as applied to hydrocarbons.It contains one or more carbon-carbon double bond.

b. Draw the displayed formulae of cyclohexane and hex-1-ene.

c. Give details of a chemical test that would enable you to distinguish between samples of these substances. You

should include details of how your results should be interpreted.Add bromine water to samples of each substance. With cyclohexane the solution will remain orange. With hex-1-ene the solution will turn colourless.

2.a. There are two straight chain alkenes with the molecular

formula C5H10. Give the structural formulae and names of both isomers.CH3CH2CH2CH=CH2 and CH3CH2CH=CHCH3

b. Write an equation for the reaction of one of these alkenes with hydrogen gas.CH3CH2CH2CH=CH2 + H2 → CH3CH2CH2CH2CH3

3. Ethene can undergo a hydrolysis reaction with cold water in the presence of a catalyst.

a. Name a catalyst that can be used in the above reaction.Concentrated sulphuric acid

b. Describe the two stages in this reaction.The ethene reacts with the sulphuric acid to form

Alcohols and other Organic Compounds

1. The formula C4H9OH can represent the three alcohols shown. Name each alcohol and class it as primary, secondary or tertiary.

2. Alcohols A and B both have the molecular formula C3H8O. Alcohol A is a primary alcohol, alcohol B is a secondary alcohol.

a. State which of these alcohols can be oxidised to a carboxylic acid and name the product formed.

b. Describe the reagents and the conditions used for this oxidation.

3. A student wanted to produce an aldehyde from the alcohol propan-1-ol. He set up a reflux apparatus using acidified potassium dichromate(VI) as the oxidising agent.

a. Draw a labelled diagram of a reflux apparatus. Explain the purpose of a reflux apparatus

b. The student tested his product and found that he had not produced an aldehyde

i. What type of product had the student formed?ii. Write equations to show the two stage reaction. Use

[O] to represent the oxidising agentiii. What technique should the student have used and

why?c. The student also tried to oxidise 2-methylpropan-2-ol,

unsuccessfully.i. Draw the full structural formula for

2-methylpropan-2-ol.ii. Why is it not possible to oxidise

2-methylpropan-2-ol with potassium dichromate(VI)?

Polymers

1. Chloroethene CH2CHCl forms the polymer poly(Chloroethene), commonly known as PVC.

a. Write an equation to show the polymerisation of Chloroethene. Your equation should show clearly the structure of the repeating unit in poly(Chloroethene)

b. Poly(Chloroethene) is an example of a thermosoftening polymer. Explain, in terms of molecular structure, why thermosoftening can be remoulded.

2. The polymer poly(propene) can be made by an addition polymerisation reaction involving the monomer propene.

a. Drawi. The displayed formula of propene

ii. The repeat unit of poly(propene)b. A second polymer that can be

made by addition polymerisation has the repeat unit shown.Name the monomer that gives rise to this polymer.

E/Z Isomerism

1.a. Draw and name the E/Z isomers of pent-2-ene.b. Explain why alkenes can have E/Z isomerism but alkanes

cannot2. An alkane has four different groups attached, A, B, X and Y.

Which of the following is the E-isomer if A and X have priority?

Infrared Spectroscopy

1. A molecule with molecular mass of 74 produces the following IR spectrum

a. Which functional groups are responsible for the functional groups at A and B?

b. Suggest the molecular formula and name of this molecule. Explain your answer.