Here are the links to the past year questions by year and variant:
2015
Oct/Nov 2015 - 11
Oct/Nov 2015 - 12
May/June 2015 - 11
May/June 2015 - 12
May/June 2015 - 13
2014
Oct/Nov 2014 - 11
Oct/Nov 2014 - 13
May/June 2014 - 11
Tuesday, 15 November 2016
Chemistry Solutions 9701/11/M/J/14
Ref: 9701/11/M/J/14 #14
What is the order of increasing melting point of the four chlorides shown?
CCl4 HCl MgCl2 PCl5
Solution
B
MgCl2 is an ionic compound with some covalent character. The other three compounds are covalent compounds. MgCl2 will have a significantly higher melting point than the rest due to its ionic bonding. The melting points of the other three compounds depend on the total number of electrons in the molecule. The order of increasing number of electrons is HCl - CCl4 - PCl5. The greater the number of electrons, the higher the melting point.
Ref: 9701/11/M/J/14 #15
When calcium is burnt in oxygen, what colour is the flame?
A. green
B. red
C. white
D. yellow
Solution
B. red
Flame colours of group 2 elements burnt in oxygen are as follows:
Mg - white
Ca - red
Sr - red
Ba - green
Ref: 9701/11/M/J/14 #28
Compound X has the molecular formula C4H10O2. X has an unbranched carbon chain and contains two OH groups.
On reaction with an excess of hot acidified, aqueous manganate(VII) ions, X is converted into a compound of molecular formula C4H6O4.
To which two carbon atoms in the chain of X are the two OH groups attached?
A. 1st and 2nd
B. 1st and 3rd
C. 1st and 4th
D. 2nd and 3rd
Solution
C
Upon realization that there is an increase of two oxygen atoms after oxidation, we can deduce that the new molecule contains two carboxylic acid groups. (It is not possible that the oxidation product is ketone or aldehyde because the number of oxygen atoms should remain the same.) In order to form ONLY carboxylic acid from alcohol through oxidation, the alcohol has to be a primary alcohol. So, the OH groups should be attached to the first and last carbon atoms.
Ref: 9701/11/M/J/14 #29
How many geometrical (cis-trans) isomers are there of hex-2,4-diene, CH3CH=CHCH=CHCH3?
A. none; hex-2,4-diene does not show geometric isomerism
B. 2
C. 3
D. 4
Solution
C. 3
This question is a special question. Usually, we use the formula of 2^n, n being the number of double bonds. In this case, however, if the first double bond has a cis arrangement, and the second double bond has a trans arrangement, it is actually the same as the first double bond having a trans arrangement, and the second double bond having a cis arrangement. (Please refer to diagram, they are actually the same molecule flipped.) So, the total number of isomers would be 2^2 - 1 = 4 - 1 = 3.
Ref: 9701/11/M/J/14 #31
Use of the Data Booklet is relevant to this question.
When the liquid N2F4 is heated, it decomposes into a single product, X.
Which statements are correct?
1. N-F bonds are broken during this decomposition.
2. The enthalpy change when N2F4 decomposes into X is approximately +160 kJ mol-1.
3. Molecules of X are non-linear.
Solution
C. 2 and 3 only
From the question, we need to deduce the following equation:
N2F4 -> 2NF2
From the equation, we can see that only N-N bonds are broken. So 1 is incorrect.
Based on Data Booklet, the bond energy of N-N bond is +160 kJ mol-1, so 2 is correct because that is the only bond broken in the reaction.
3 is correct because there are lone pair(s) on the NF2 molecule, so it has to be non-linear. (Please take note that we are not entirely sure of the nature of this ion/molecule as well, but the knowledge you will need to solve this question is that there are lone pairs on the nitrogen atom.)
What is the order of increasing melting point of the four chlorides shown?
CCl4 HCl MgCl2 PCl5
B
MgCl2 is an ionic compound with some covalent character. The other three compounds are covalent compounds. MgCl2 will have a significantly higher melting point than the rest due to its ionic bonding. The melting points of the other three compounds depend on the total number of electrons in the molecule. The order of increasing number of electrons is HCl - CCl4 - PCl5. The greater the number of electrons, the higher the melting point.
Ref: 9701/11/M/J/14 #15
When calcium is burnt in oxygen, what colour is the flame?
A. green
B. red
C. white
D. yellow
Solution
B. red
Flame colours of group 2 elements burnt in oxygen are as follows:
Mg - white
Ca - red
Sr - red
Ba - green
Ref: 9701/11/M/J/14 #28
Compound X has the molecular formula C4H10O2. X has an unbranched carbon chain and contains two OH groups.
On reaction with an excess of hot acidified, aqueous manganate(VII) ions, X is converted into a compound of molecular formula C4H6O4.
To which two carbon atoms in the chain of X are the two OH groups attached?
A. 1st and 2nd
B. 1st and 3rd
C. 1st and 4th
D. 2nd and 3rd
Solution
C
Upon realization that there is an increase of two oxygen atoms after oxidation, we can deduce that the new molecule contains two carboxylic acid groups. (It is not possible that the oxidation product is ketone or aldehyde because the number of oxygen atoms should remain the same.) In order to form ONLY carboxylic acid from alcohol through oxidation, the alcohol has to be a primary alcohol. So, the OH groups should be attached to the first and last carbon atoms.
Ref: 9701/11/M/J/14 #29
How many geometrical (cis-trans) isomers are there of hex-2,4-diene, CH3CH=CHCH=CHCH3?
A. none; hex-2,4-diene does not show geometric isomerism
B. 2
C. 3
D. 4
Solution
C. 3
This question is a special question. Usually, we use the formula of 2^n, n being the number of double bonds. In this case, however, if the first double bond has a cis arrangement, and the second double bond has a trans arrangement, it is actually the same as the first double bond having a trans arrangement, and the second double bond having a cis arrangement. (Please refer to diagram, they are actually the same molecule flipped.) So, the total number of isomers would be 2^2 - 1 = 4 - 1 = 3.
Ref: 9701/11/M/J/14 #31
Use of the Data Booklet is relevant to this question.
When the liquid N2F4 is heated, it decomposes into a single product, X.
Which statements are correct?
1. N-F bonds are broken during this decomposition.
2. The enthalpy change when N2F4 decomposes into X is approximately +160 kJ mol-1.
3. Molecules of X are non-linear.
Solution
C. 2 and 3 only
From the question, we need to deduce the following equation:
N2F4 -> 2NF2
From the equation, we can see that only N-N bonds are broken. So 1 is incorrect.
Based on Data Booklet, the bond energy of N-N bond is +160 kJ mol-1, so 2 is correct because that is the only bond broken in the reaction.
3 is correct because there are lone pair(s) on the NF2 molecule, so it has to be non-linear. (Please take note that we are not entirely sure of the nature of this ion/molecule as well, but the knowledge you will need to solve this question is that there are lone pairs on the nitrogen atom.)
Monday, 14 November 2016
Chemistry Solutions 9701/13/O/N/14
Ref: 9701/13/O/N/14 #8
Use of the Data Booklet is relevant to this question.
Ferrochrome is an alloy of iron and chromium. Ferrochrome can be dissolved in dilute sulfuric acid to produce a mixture of FeSO4 and Cr2(SO4)3. The FeSO4 reacts with K2Cr2O7 in acid solution according to the following reaction.
Use of the Data Booklet is relevant to this question.
Ferrochrome is an alloy of iron and chromium. Ferrochrome can be dissolved in dilute sulfuric acid to produce a mixture of FeSO4 and Cr2(SO4)3. The FeSO4 reacts with K2Cr2O7 in acid solution according to the following reaction.
14H+ +6Fe2+ +Cr2O72---->2Cr3+ +6Fe3+ +7H2O
When 1.00 g of ferrochrome is dissolved in dilute sulfuric acid, and the resulting solution titrated, 13.1cm3 of 0.100moldm-3 K2Cr2O7 is required for complete reaction.
What is the percentage by mass of Fe in the sample of ferrochrome?
A. 1.22
B. 4.39
C. 12.2
D. 43.9
Solution
D 43.
The number of moles of K2Cr2O7 required for the reaction is 13.1x0.1/1000=1.31 x 10^-3 mol. From the equation, 1 mol of K2Cr2O7 reacts with 6 mols of Fe. Hence the number of moles of Fe=7.86x10^-3. The mass of Fe present in 1.00 g of ferrochrome= 7.86x10^-3 x 55.8=0.4386g. Hence the percentage by mass of Fe in ferrochrome is 0.4386/1.00 x 100%=43.9%.
Ref: 9701/13/O/N/14 #14
X, Y and Z are compounds of three elements in Period 3. Their electrical conductivities are shown in the table.
What could be compounds X, Y and Z?
Solution
C
For this type of question, look to SiO2 for reference.
SiO2 cannot conduct electricity in any state. Z is very likely to be SiO2. Let's take a look at all the choices for Z.
NaF is ionic, so it can conduct electricity in both molten and aqueous state.
For now, A can be eliminated.
SiCl4 is a covalent compound, which hydrolyses in water to form H+ and Cl-, so it can conduct electricity when added to water, but does not conduct electricity in its molten state due to its covalent nature.
Referring to the properties of SiCl4 and all the choices left for X, we can eliminate D.
Al2O3 is an ionic compound with a lot of covalent character, so it should be able to conduct electricity in both molten and aqueous state.
Y cannot be Al2O3.
The best answer would have to be C.
Ref: 9701/13/O/N/14 #19
A student examines two semi-precious stones; one is agate, SiO2, and the other is calcite, CaCO3.
X, Y and Z are compounds of three elements in Period 3. Their electrical conductivities are shown in the table.
What could be compounds X, Y and Z?
C
For this type of question, look to SiO2 for reference.
SiO2 cannot conduct electricity in any state. Z is very likely to be SiO2. Let's take a look at all the choices for Z.
NaF is ionic, so it can conduct electricity in both molten and aqueous state.
For now, A can be eliminated.
SiCl4 is a covalent compound, which hydrolyses in water to form H+ and Cl-, so it can conduct electricity when added to water, but does not conduct electricity in its molten state due to its covalent nature.
Referring to the properties of SiCl4 and all the choices left for X, we can eliminate D.
Al2O3 is an ionic compound with a lot of covalent character, so it should be able to conduct electricity in both molten and aqueous state.
Y cannot be Al2O3.
The best answer would have to be C.
Ref: 9701/13/O/N/14 #19
A student examines two semi-precious stones; one is agate, SiO2, and the other is calcite, CaCO3.
How could they be distinguished?
A. Add a fixed amount of cold aqueous sodium hydroxide to each separately and measure any temperature change.
B. Heat each separately over a gentle Bunsen flame and note which one melts first.
C. Shake each separately with dilute hydrochloric acid and test any gas formed.
D. Shake each separately with distilled water and add a few drops of Universal indicator.
Solution
C Shake each separately with dilute hydrochloric acid and test any gas formed.
A is incorrect as SiO2 only reacts with hot, concentrated NaOH and CaCO3 does not react with NaOH at all. Hence there will be no reaction between cold NaOH with both compounds.
B is incorrect as SiO2 has a giant molecular structure and CaCO3 has a high thermal stability. Hence both will not melt over a gentle Bunsen flame.
C is correct as SiO2 will not react with HCl (it is an acidic oxide). CaCO3 will react with HCl to produce CO2 gas. Hence HCl can be used to distinguish between these two compounds.
D is incorrect as both SiO2 and CaCO3 are insoluble in water.
Ref: 9701/13/O/N/14 #20
In this question, structural isomerism and stereoisomerism should be considered.
How many isomeric aldehydes have the formula C5H10O?
A. 3
B. 4
C. 5
D. 6
Solution
C 5
Pentanal
2-methylbutanal (2)
3-methylbutanal
2,2-dimethylpropanal
* 2-methylbutanal has a chiral 2nd carbon atom, hence it displays optical isomerism.
Ref: 9701/13/O/N/14 #25
The compound known as 'gamma-linolenic acid' is found in the seeds of the evening primrose plant.
How many cis-trans isomers are there with this structural formula?
A. 3
B. 6
C. 8
D. 12
Solution
C 8
The number of cis-trans isomers can be determined using the formula 2^n, where n represents the number of alkene groups/C=C bonds in the molecule. Hence for this question it is 2^3=8.
Ref: 9701/13/O/N/14 #27
The compound 2-ethyl-3-methylbutanoic acid is used as a flavouring in some food.
Which compound will produce 2-ethyl-3-methylbutanoic acid when heated under reflux with sodium cyanide in ethanol, followed by acid hydrolysis of the reaction product?
Solution
B
When halogenoalkanes are heated under reflux with ethanolic sodium cyanide, there is an addition of a C atom across the chain when the halogen atom is substituted by CN. 2-ethyl-3-methylbutanoic acid has a total of 7 carbon atoms, hence the halogenoalkane used to synthesise it should have 6 C atoms. Option D can be eliminated as it has 7 C atoms.
Ref: 9701/13/O/N/14 #29
The diagram shows the structure of the naturally-occurring molecule cholesterol.
Student X stated that the seventeen carbon atoms in the four rings all lie in the same plane.
Student Y stated that this molecule displays cis-trans isomerism at the C=C double bond.
Which of the students are correct?
A. both X and Y
B. neither X nor Y
C. X only
D. Y only
Solution
B. neither X nor Y
Student X's statement is wrong because not all carbon atoms have sp2 hybridised orbitals (those which do possess trigonal planar structure). Some of the carbon atoms have sp3 hybridised orbitals, and so possess the tetrahedral structure, which means that not all carbon atoms lie in the same plane. (Concept of "same plane" = imagine a flat piece of paper where all the carbon atoms are lying on the piece of paper, none of them are above or below the piece of paper.)
Student Y's statement is wrong because the C=C double bond is part of a ring structure. Cis-trans isomerism cannot exist when both the carbon atoms in the double bond belong to a ring structure due to restriction of rotation.
Ref: 9701/13/O/N/14 #30
Methyl butanoate, C5H10O2, is an ester used in the food industry to give products the flavour of apples.
Including methyl butanoate, how many structural isomers are there of C5H10O2 that are esters?
A. 6
B. 8
C. 9
D. 10
Solution
C. 9
Please refer to the attached diagram for the 9 possible structural isomers.
Ref: 9701/13/O/N/14 #33
Which types of intermolecular forces can exist between adjacent urea molecules?
1 hydrogen bonding
2 permanent dipole-dipole forces
3 instantaneous dipole-induced dipole forces
Solution
A 1, 2 and 3
Hydrogen bonding exists when the hydrogen atoms, attached to the very electronegative N atom in one molecule, accepts lone pairs of electrons from the O or N atoms of other molecules.
The urea molecule is polar due to the polar C=O bond where there is the displacement of the bond pair of valence electrons closer to the more electronegative O atom. Hence the urea molecules can form permanent dipoles with one another.
Ref: 9701/13/O/N/14 #37
Which statements about the photochemical chlorination of ethane are correct?
1. A propagation step in the mechanism is C2H5 + Cl2 -> C2H5Cl + Cl
2. Butane is present in the products.
3. The initiation step is the heterolytic fission of chlorine.
Solution
B. 1 and 2 only
Statements 1 is one of the propagation steps in the mechanism.
Statement 2: Butane is formed according to the following equation: C2H5 + C2H5 -> C4H10.
Statement 3 is wrong because of the word "heterolytic". Fission of chlorine to form free radicals is homolytic. Heterolytic = to form nucleophiles and electrophiles.
Solution
C Shake each separately with dilute hydrochloric acid and test any gas formed.
A is incorrect as SiO2 only reacts with hot, concentrated NaOH and CaCO3 does not react with NaOH at all. Hence there will be no reaction between cold NaOH with both compounds.
B is incorrect as SiO2 has a giant molecular structure and CaCO3 has a high thermal stability. Hence both will not melt over a gentle Bunsen flame.
C is correct as SiO2 will not react with HCl (it is an acidic oxide). CaCO3 will react with HCl to produce CO2 gas. Hence HCl can be used to distinguish between these two compounds.
D is incorrect as both SiO2 and CaCO3 are insoluble in water.
Ref: 9701/13/O/N/14 #20
In this question, structural isomerism and stereoisomerism should be considered.
How many isomeric aldehydes have the formula C5H10O?
A. 3
B. 4
C. 5
D. 6
Solution
C 5
Pentanal
2-methylbutanal (2)
3-methylbutanal
2,2-dimethylpropanal
* 2-methylbutanal has a chiral 2nd carbon atom, hence it displays optical isomerism.
Ref: 9701/13/O/N/14 #25
The compound known as 'gamma-linolenic acid' is found in the seeds of the evening primrose plant.
How many cis-trans isomers are there with this structural formula?
A. 3
B. 6
C. 8
D. 12
Solution
C 8
The number of cis-trans isomers can be determined using the formula 2^n, where n represents the number of alkene groups/C=C bonds in the molecule. Hence for this question it is 2^3=8.
Ref: 9701/13/O/N/14 #27
The compound 2-ethyl-3-methylbutanoic acid is used as a flavouring in some food.
Which compound will produce 2-ethyl-3-methylbutanoic acid when heated under reflux with sodium cyanide in ethanol, followed by acid hydrolysis of the reaction product?
Solution
B
When halogenoalkanes are heated under reflux with ethanolic sodium cyanide, there is an addition of a C atom across the chain when the halogen atom is substituted by CN. 2-ethyl-3-methylbutanoic acid has a total of 7 carbon atoms, hence the halogenoalkane used to synthesise it should have 6 C atoms. Option D can be eliminated as it has 7 C atoms.
Ref: 9701/13/O/N/14 #29
The diagram shows the structure of the naturally-occurring molecule cholesterol.
Student X stated that the seventeen carbon atoms in the four rings all lie in the same plane.
Student Y stated that this molecule displays cis-trans isomerism at the C=C double bond.
Which of the students are correct?
A. both X and Y
B. neither X nor Y
C. X only
D. Y only
Solution
B. neither X nor Y
Student X's statement is wrong because not all carbon atoms have sp2 hybridised orbitals (those which do possess trigonal planar structure). Some of the carbon atoms have sp3 hybridised orbitals, and so possess the tetrahedral structure, which means that not all carbon atoms lie in the same plane. (Concept of "same plane" = imagine a flat piece of paper where all the carbon atoms are lying on the piece of paper, none of them are above or below the piece of paper.)
Student Y's statement is wrong because the C=C double bond is part of a ring structure. Cis-trans isomerism cannot exist when both the carbon atoms in the double bond belong to a ring structure due to restriction of rotation.
Ref: 9701/13/O/N/14 #30
Methyl butanoate, C5H10O2, is an ester used in the food industry to give products the flavour of apples.
Including methyl butanoate, how many structural isomers are there of C5H10O2 that are esters?
A. 6
B. 8
C. 9
D. 10
Solution
C. 9
Please refer to the attached diagram for the 9 possible structural isomers.
Ref: 9701/13/O/N/14 #33
Which types of intermolecular forces can exist between adjacent urea molecules?
1 hydrogen bonding
2 permanent dipole-dipole forces
3 instantaneous dipole-induced dipole forces
Solution
A 1, 2 and 3
Hydrogen bonding exists when the hydrogen atoms, attached to the very electronegative N atom in one molecule, accepts lone pairs of electrons from the O or N atoms of other molecules.
The urea molecule is polar due to the polar C=O bond where there is the displacement of the bond pair of valence electrons closer to the more electronegative O atom. Hence the urea molecules can form permanent dipoles with one another.
Ref: 9701/13/O/N/14 #37
Which statements about the photochemical chlorination of ethane are correct?
1. A propagation step in the mechanism is C2H5 + Cl2 -> C2H5Cl + Cl
2. Butane is present in the products.
3. The initiation step is the heterolytic fission of chlorine.
Solution
B. 1 and 2 only
Statements 1 is one of the propagation steps in the mechanism.
Statement 2: Butane is formed according to the following equation: C2H5 + C2H5 -> C4H10.
Statement 3 is wrong because of the word "heterolytic". Fission of chlorine to form free radicals is homolytic. Heterolytic = to form nucleophiles and electrophiles.
Sunday, 13 November 2016
Chemistry Solutions 9701/11/O/N/14
Ref: 9701/11/O/N/14 #4
Two glass vessels M and N are connected by a closed valve.
Two glass vessels M and N are connected by a closed valve.
M contains helium at 20 degree Celsius at a pressure of 1 x 10^5 Pa. N has been evacuated, and has three times the volume of M. In an experiment, the valve is opened and the temperature of the whole apparatus is raised to 100 degree Celsius.
What is the final pressure in the system?
A. 3.18 x 10^4 Pa
B. 4.24 x 10^4 Pa
C. 1.24 x 10^5 Pa
D. 5.09 x 10^5 Pa
Solution
A 3.18 x 10^4 Pa
Let the volume of M be V and the volume of N be 3 V. Initially, the gas helium is contained inside M only. When the valve is opened, the gas spreads to N as well, hence the total volume containing the gas increases from V to 4V. Pressure is inversely proportional to volume. Hence the new pressure in the system is calculated using the formula P1V1=P2V2.
(1 x 10^5)(V)=P2(4V)
By cancelling out V on both sides, the new pressure, P2 when the valve is opened is 25000 Pa. The temperature of the apparatus is raised from 20 to 100 degree Celsius. The new pressure is calculated from the formula P2/T1=P3/T2. Remember to convert degree Celsius to Kelvin by adding 273 for this equation.
25000/293=P3/373
Hence, P3 is 3.18 x 10^4 Pa. The answer is A.
Ref: 9701/11/O/N/14 #6
Aluminium carbide, Al4Cl3, reacts readily with aqueous sodium hydroxide. The two products of the reaction are NaAlO2 and a hydrocarbon. Water molecules are also involved as reactants.
What is the formula of the hydrocarbon?
A. CH4
B. C2H6
C. C3H6
D. C6H12
Solution
A CH4
To solve this question, work on finding the balanced equation for this reaction. The equation is
Al4C3+4NaOH+4H2O--->4NaAlO2+3CH4
So the answer is CH4.
Ref: 9701/11/O/N/14 #15
Use of the Data Booklet is relevant to this question.
A sample of potassium oxide, K2O is dissolved in 250 cm3 of distilled water. 25.0cm3 of this solution is titrated against sulfuric acid of concentration 2.00 moldm-3. 15.0cm3 of this sulfuric acid is needed for complete neutralisation.
Which mass of potassium oxide was originally dissolved in 250cm3 of distilled water?
A. 2.83g
B. 28.3g
C. 47.1g
D. 56.6g
Solution
B 28.3g
The balanced equation for this reaction is K2O+H2SO4---> K2SO4+H2O. The number of moles of sulfuric acid needed for neutralisation is 15x2/1000=0.03 mol. K2O and H2SO4 react in a 1:1 mole ratio, so 25 cm3 of the solution containing dissolved K2O contains 0.03 mol of K20. The number of moles in 250 cm3 of the solution 250/25x 0.03=0.3 mol. The Mr of K2O is 94.2. Hence the mass of K20 originally dissolved=94.2 x 0.3=28.26g.
Ref: 9701/11/O/N/14 #23
Considering only structural isomers, what is the number of alcohols of each type with the formula C5H12O?
Solution
C
Since the question is only asking for structural isomers, we do not need to consider stereoisomers.
Primary
Pentan-1-ol
2-methylbutan-1-ol
3-methylbutan-1-ol
2,2-dimethylpropan-1-ol
Secondary
Pentan-2-ol
Pentan-3-ol
3-methylbutan-2-ol
Tertiary
2-methylbutan-2-ol
Ref: 9701/11/O/N/14 #25
In the hydrolysis of bromoethane by aqueous sodium hydroxide, what is the nature of the attacking group and of the leaving group?
Solution
D
The reaction between bromoethane and sodium hydroxide is nucleophilic substitution. It begins with the nucleophilic attack of OH- from NaOH on the slightly positive carbon atom in the C-Br bond. The leaving group is the Br- ion, which is a species with a lone pair of electrons, thus both of the attacking and leaving groups are nucleophiles.
Ref: 9701/11/O/N/14 #30
B-carotene is responsible for the orange colour of carrots.
B-carotene is oxidised by hot, concentrated, acidified KMnO4.
When an individual molecule of B-carotene is oxidised in this way, many product molecules are formed.
How many of these product molecules contain a ketone functional group?
A. 4
B. 6
C. 9
D.11
Solution
B 6
We need to focus on the carbon atoms in double bonds which have two carbon atoms attached to each of them. In the molecule, there are 8 carbon atoms in double bonds which are able to form ketone functional groups once oxidised by KMnO4. When these double bonds are broken, 6 product molecules containing ketone functional groups will be produced. 4 of these contain one ketone group, whereas the other two (in boxes) contain 2 ketone groups each.
Ref: 9701/11/O/N/14 #32
Use of the Data Booklet is relevant to this question.
The bond energy of the Br-O bond is 235 kJmol-1.
Which reactions are exothermic?
1 OH+HBr---> H2+BrO
2 OH+HBr---> H2O+ Br
3 H+HBr--->H2+Br2
Solution
C 2 and 3 only
For reaction 1, an O-H bond and a H-Br bond are broken. A H-H bond and Br-O bond are formed. Bond breaking is always endothermic and bond formation is always exothermic. Hence the enthalpy change for the reaction is 460+366-436-235=+155kJmol-1. So the reaction is endothermic.
For reaction 2, a H-Br bond is broken and an O-H bond is formed. Enthalpy change= 366-460=-94kJmol-1. The reaction is exothermic.
For reaction 3, a H-Br bond is broken and a H-H bond is formed. Enthalpy change= 366-436= -70kJmol-1. The reaction is exothermic.
Ref: 9701/11/O/N/14 #36
Which statements about calcium oxide are correct?
1 It reacts with cold water
2 It is produced when calcium nitrate is heated
3 It can be reduced by heating with magnesium
Solution
B 1 and 2 only
Calcium oxide reacts with cold water to produce calcium hydroxide and is produced when calcium nitrate decomposes. Statement 3 is incorrect as calcium is more electropositive than magnesium and is a better reducing agent since it is lower down the group (greater electropositivity), hence calcium oxide cannot be reduced by magnesium.
Ref: 9701/11/O/N/14 #40
A reaction pathway diagram is shown.
Which reactions would have this profile?
1 (CH3)3CBr+NaOH--->(CH3)3COH+NaBr
2 CH3CH2Br + NaOH---> CH3CH2OH+NaBr
3 (CH3)3CCH2CH2Cl+2NH3---> (CH3)3CCH2CH2NH2+NH4Cl
Solution
D 1 only
The reaction pathway diagram shows that for a tertiary halogenoalkane which undergoes the SN1 mechanism for nucleophilic substitution reactions.
Only 1 is correct as (CH3)3CBr is a tertiary halogenoalkane. The halogenoalkanes in 2 and 3 are primary halogenoalkanes which undergo the SN2 mechanism for nucleophilic substitution.
Chemistry Solutions 9701/13/M/J/15
Ref: 9701/13/M/J/15 #6
Which statement can be explained by intermolecular hydrogen bonding?
A. Ethanol has a higher boiling point than propane.
B. Hydrogen chloride has a higher boiling point than silane, SiH4.
C. Hydrogen iodide forms an acidic solution when dissolved in water.
D. Propanone has a higher boiling point than propane.
Solution
A
Hydrogen bonds are formed when a H atom of a molecule, attached to a very electronegative atom (Fluorine, Oxygen, Nitrogen) accepts a lone pair from another very electronegative atom (FON) from another molecule. A is correct because ethanol molecules have OH groups and are able to form hydrogen bonds with one another, unlike propane which only has Van der Waal's forces between molecules.
B, C and D are incorrect because the compounds do not have any intermolecular hydrogen bonds.
If you have covered organic chemistry, it is worth noting that carboxylic acids and alcohols are able to form intermolecular hydrogen bonds whereas carbonyl compounds are unable to. This is because carboxylic acids and alcohols have -OH groups whereas carbonyl compounds do not.
Carboxylic acids have higher melting and boiling points than alcohols because they have the additional polar C=O groups.
Ref: 9701/13/M/J/15 #14
The compound (CH3)3NAlCl3 has a simple molecular structure. Which statement about (CH3)3NAlCl3 is correct?
A. (CH3)3NAlCl3 molecules attract each other by hydrogen bonds.
B. The Al atom has an incomplete valence shell of electrons.
C. The bonds around the Al atom are planar.
D. The molecules contain coordinate and covalent bonding.
Solution
D. The molecules contain coordinate and covalent bonding
A is incorrect because the H atoms in the molecule are not attached to very electronegative atoms (Fluorine, Oxygen, Nitrogen) hence it is not possible for the molecule to form hydrogen bonds.
B is incorrect. The Al atom originally has three valence electrons, two in the 3s orbital and 1 in the 3px orbital. After hybridisation, the Al atom has three sp2 hybrid orbitals and is able to form three polar covalent bonds with the three Cl atoms. The Al atom has an empty 3pz orbital, and it then accepts the lone pair of electrons from the nitrogen atom, forming a complete valence shell of electrons.
C is incorrect. Al is attached to three Cl atoms and 1 N atom, forming a tetrahedral structure, not planar.
D is correct as covalent bonding exists between the C and H atoms and a coordinate/dative bond is formed between the Al and N atom.
Ref: 9701/13/M/J/15 #25
Compound X, C2H12O, is oxidised by acidified sodium dichromate (VI) to compound Y.
Compound Y reacts with butan-2-ol in the presence of a little concentrated sulfuric acid to give liquid Z.
What could be the formula of Z?
A. CH3(CH2)3CO2CH(CH3)CH2CH3
B. CH3(CH2)3CO2(CH2)3CH3
C. CH3(CH2)2CO2CH(CH3)CH2CH3
D. (CH3)2CHCH2CO2C(CH3)3
Solution
A. CH3(CH2)3CO2CH(CH3)CH2CH3
Compound X is an alcohol since it has the general molecular formula CnH2n+2O. It is fully oxidised
by acidified sodium dichromate (VI) to pentanoic acid, compound Y. Compound Y then undergoes esterification with butan-2-ol to form CH3(CH2)3CO2CH(CH3)CH2CH3.
Ref: 9701/13/M/J/15 #26
Alkane X has molecular formula C4H10.
X reacts with Cl2(g) in the presence of sunlight to produce only two different monochloroalkanes, C4H9Cl. Both of these monochloroalkanes produce the same alkene Y, and no other organic products, when they are treated with hot ethanolic KOH.
What is produced when Y is treated with hot concentrated acidified KMnO4?
A. CO2 and CH3CH2CO2H
B. CO2 and CH3COCH3
C. HCO2H and CH3COCH3
D. CH3CO2H only
Solution
B CO2 and CH3COCH3
Alkane X is 2-methylpropane. It undergoes free radical substitution with Cl2 in sunlight to produce 1-chloro-2-methylpropane and 2-chloro-2-methylpropane. Both of these compounds, after undergoing dehydrohalogenation, produce the same alkene, 2-methylprop-1-ene.
Alkane X has to be a branched compound in order for the same alkene to be produced from two different monochloroalkanes. If alkane X were not branched, there would be two possible alkenes produced.
Which statement can be explained by intermolecular hydrogen bonding?
A. Ethanol has a higher boiling point than propane.
B. Hydrogen chloride has a higher boiling point than silane, SiH4.
C. Hydrogen iodide forms an acidic solution when dissolved in water.
D. Propanone has a higher boiling point than propane.
Solution
A
Hydrogen bonds are formed when a H atom of a molecule, attached to a very electronegative atom (Fluorine, Oxygen, Nitrogen) accepts a lone pair from another very electronegative atom (FON) from another molecule. A is correct because ethanol molecules have OH groups and are able to form hydrogen bonds with one another, unlike propane which only has Van der Waal's forces between molecules.
B, C and D are incorrect because the compounds do not have any intermolecular hydrogen bonds.
If you have covered organic chemistry, it is worth noting that carboxylic acids and alcohols are able to form intermolecular hydrogen bonds whereas carbonyl compounds are unable to. This is because carboxylic acids and alcohols have -OH groups whereas carbonyl compounds do not.
Carboxylic acids have higher melting and boiling points than alcohols because they have the additional polar C=O groups.
Ref: 9701/13/M/J/15 #14
The compound (CH3)3NAlCl3 has a simple molecular structure. Which statement about (CH3)3NAlCl3 is correct?
A. (CH3)3NAlCl3 molecules attract each other by hydrogen bonds.
B. The Al atom has an incomplete valence shell of electrons.
C. The bonds around the Al atom are planar.
D. The molecules contain coordinate and covalent bonding.
Solution
D. The molecules contain coordinate and covalent bonding
B is incorrect. The Al atom originally has three valence electrons, two in the 3s orbital and 1 in the 3px orbital. After hybridisation, the Al atom has three sp2 hybrid orbitals and is able to form three polar covalent bonds with the three Cl atoms. The Al atom has an empty 3pz orbital, and it then accepts the lone pair of electrons from the nitrogen atom, forming a complete valence shell of electrons.
C is incorrect. Al is attached to three Cl atoms and 1 N atom, forming a tetrahedral structure, not planar.
D is correct as covalent bonding exists between the C and H atoms and a coordinate/dative bond is formed between the Al and N atom.
Ref: 9701/13/M/J/15 #25
Compound X, C2H12O, is oxidised by acidified sodium dichromate (VI) to compound Y.
Compound Y reacts with butan-2-ol in the presence of a little concentrated sulfuric acid to give liquid Z.
What could be the formula of Z?
A. CH3(CH2)3CO2CH(CH3)CH2CH3
B. CH3(CH2)3CO2(CH2)3CH3
C. CH3(CH2)2CO2CH(CH3)CH2CH3
D. (CH3)2CHCH2CO2C(CH3)3
Solution
A. CH3(CH2)3CO2CH(CH3)CH2CH3
Compound X is an alcohol since it has the general molecular formula CnH2n+2O. It is fully oxidised
by acidified sodium dichromate (VI) to pentanoic acid, compound Y. Compound Y then undergoes esterification with butan-2-ol to form CH3(CH2)3CO2CH(CH3)CH2CH3.
Ref: 9701/13/M/J/15 #26
Alkane X has molecular formula C4H10.
X reacts with Cl2(g) in the presence of sunlight to produce only two different monochloroalkanes, C4H9Cl. Both of these monochloroalkanes produce the same alkene Y, and no other organic products, when they are treated with hot ethanolic KOH.
What is produced when Y is treated with hot concentrated acidified KMnO4?
A. CO2 and CH3CH2CO2H
B. CO2 and CH3COCH3
C. HCO2H and CH3COCH3
D. CH3CO2H only
Solution
B CO2 and CH3COCH3
Alkane X is 2-methylpropane. It undergoes free radical substitution with Cl2 in sunlight to produce 1-chloro-2-methylpropane and 2-chloro-2-methylpropane. Both of these compounds, after undergoing dehydrohalogenation, produce the same alkene, 2-methylprop-1-ene.
Alkane X has to be a branched compound in order for the same alkene to be produced from two different monochloroalkanes. If alkane X were not branched, there would be two possible alkenes produced.
The above diagram shows an example of an unbranched alkane X forming two possible monochloroalkanes and then two different alkenes. Two different alkenes would also be produced if the 2nd and 3rd hydrogen atoms were substituted by chlorine instead of the 1st and 2nd. Thus the compound must be branched to obtain the same alkene from the monochloroalkanes.
The alkene produced undergoes vigorous oxidation with hot concentrated acidified KMnO4 to produce CO2 and CH3COCH3. Thus the answer is B.
Ref: 9701/13/M/J/15 #34
Which physical properties are due to hydrogen bonding between water molecules?
1 Water has a higher boiling point than H2S
2 Ice floats on water
3 The H-O-H bond angle in water is approximately 104'
Solution
B 1 and 2 only
Hydrogen bonding between water molecules causes water to have a higher boiling point than H2S, which does not have intermolecular hydrogen bonding. Intermolecular hydrogen bonding leads to the solid state of water, ice, being less dense than its liquid state. Hence statements 1 and 2 are correct.
Statement 3 is incorrect as the bond angles in molecules or ions depend on the magnitude of electron pairs repulsion and not hydrogen bonding.
*Hydrogen bonding in molecules leads to higher viscosity, higher surface tension and greater solubility in water. They can also cause some molecules to dimerise (molecules join together to form one larger molecule called a dimer), so that the apparent Mr is greater than expected.
Ref: 9701/13/M/J/15 #35
Under atmospheric conditions, in which transformations is sulfur dioxide involved as either a reagent or a catalyst?
1 NO2 to NO
2 NO to NO2
3 CO to CO2
Solution
D 1 only
Nitrogen monoxide acts as a catalyst for the formation of SO3 from SO2. The reactions are as follow:
2NO+O2--->2NO2
NO2+SO2---> NO+SO3
Thus it can be seen that sulphur dioxide acts as a reagent in transformation 1.
Transformation 2 only involves atmospheric oxygen and not SO3. Transformation 3 does not occur under atmospheric conditions. Hence the answer is D.
Ref: 9701/13/M/J/15 #38
Which changes in bonding occur during the reaction of propanal and hydrogen cyanide?
1 A carbon-hydrogen bond is broken.
2 An oxygen-hydrogen bond is formed.
3 A carbon-carbon bond is formed.
Solution
A 1,2 and 3
Propanal is an aldehyde and it undergoes nucleophilic addition with hydrogen cyanide, HCN. This reaction involves cyanide ions as a catalyst and produces 2-hydroxybutanenitrile.
A carbon-carbon bond is formed due to the nucleophilic attack of the cyanide ions on the slightly positive carbon atom in the polar C=O bond in propanal. 3 is correct.
A carbon-hydrogen bond is broken between the H+ and CN- ions in HCN, after which the H+ ion accepts a lone pair of electrons from the negative O2- ion. Thus an oxygen-hydrogen bond is formed. 1 and 2 are correct.
Ref: 9701/13/M/J/15 #39
Which pairs of reagents will react together in a redox reaction?
1 CH3CHO+ Fehling's reagent
2 C2H4+ Br2 (aq)
3 CH4 + Cl2 (g)
Solution
A 1,2 and 3
1 is correct as CH3CHO is an aldehyde (ethanal), and will be oxidised by Fehling's solution to produce ethanoic acid. Fehling's solution is reduced by ethanal, so this is a redox reaction.
2 is correct as the oxidation state of each carbon atom in C2H4 is -2. With aqueous bromine, electrophilic addition occurs across the double bond and the oxidation number of carbon increases from -2 to -1 whereas the bromine atoms are each reduced from 0 to -1 respectively.
3 is correct as the oxidation number of carbon in CH4 was originally -4, and then increases to -2 when a H atom is substituted by a Cl atom through free radical substitution to form CH3Cl. The oxidation number of the chlorine atom decreases from 0 to -1, so this is a redox reaction.
From the diagram, it can be seen that a difference in electronegativity will cause C to have an oxidation state of -1 in C-H bond and +1 in a C-Br bond. The "net" oxidation state of the carbon atom in each compound is the addition of the oxidation numbers from each bond it forms.
Chemistry Solutions 9701/12/M/J/15
Ref: 9701/12/M/J/15 #2
The shell of a chicken's egg makes up 5% of the mass of an average egg. An average egg has a mass of 50g.
Assume the egg shell is pure calcium carbonate.
How many complete chicken's egg shells would be needed to neutralise 50cm3 of 2.0moldm-3 ethanoic acid?
A. 1
B. 2
C. 3
D. 4
Solution
B 2
Form the balanced equation for the reaction of ethanoic acid with calcium carbonate.
2CH3COOH+CaCO3-->(CH3COO)2Ca+CO2+H2O
For an average egg, the shell is 0.05 x 50=2.5g. The Mr of CaCO3 is 100.1. Hence a chicken's egg shell contains 2.5/100.1=0.025 mol of CaCO3. 50cm3 of 2.0moldm-3= 50 x 2/1000=0.1mol of ethanoic acid. From the balanced equation, 2 mols of ethanoic acid reacts with 1 mol of CaCO3. Hence, 0.05 mols of calcium carbonate would be required to react with the amount of ethanoic acid stated in the question.
Therefore, number of chicken's egg shells=0.-05/0.025= 2.
Ref: 9701/12/M/J/15 #9
When a sample of HI is warmed to a particular temperature the equilibrium below is established.
Ref: 9701/12/M/J/15 #10
Photochromic glass, used for sunglasses, darkens when exposed to bright light and becomes more transparent again when the light is less bright. The darkness of the glass is related to the concentration of silver atoms.
The following reactions are involved.
A. Cu+ and Cu2+ ions act as catalysts.
B. Cu+ ions act as an oxidising agent in reaction 2.
C. Reaction 3 increases the darkness of the glass.
D. Silver atoms are reduced in reaction 3.
Solution
A Cu+ and Cu2+ ions act as catalysts.
Statement 1 is correct as the Cu2+ and Cu+ ions are regenerated at the end of reactions 2 and 3 respectively, hence they act as catalysts.
Statement 2 is incorrect as the oxidation number of Cl changes from 0 to -1, hence the Cu+ ions act as a reducing agent.
Statement 3 is incorrect as silver atoms are converted to silver ions. Silver atoms increase the darkness of the glass whereas silver ions cause the glass to become more transparent.
Statement 4 is incorrect as the oxidation number of silver increases from 0 to +1, hence the silver atoms are oxidised.
Ref: 9701/12/M/J/15 #12
Use of the Data Booklet is relevant to this question.
When 3.00 g of an anhydrous nitrate of a Group II metal is decomposed, 1.53g of gas is produced. What is the nitrate compound?
A. beryllium nitrate
B. calcium nitrate
C. magnesium nitrate
D. strontium nitrate
Solution
D strontium nitrate
The decomposition of anhydrous Group II nitrates will produce the metal oxide, nitrogen dioxide gas and oxygen gas. The balanced equation for this process is
There are two types of gases produced. Hence it is not possible to find the number of moles of the gases produced with the information given. So we must calculate the amount of solid metal oxide produced, which is 3.00--1.53=1.47g.
From the balanced equation, it can be seen that 2 mols of the metal nitrate will produce 2 mols of the metal oxide. Hence it is a 1:1 mole ratio.
To solve this question, the Mr of each metal nitrate and metal oxide must be calculated. The answer would be the compound which has equal number of moles of metal nitrate and metal oxide for the masses given.
For this question, the number of moles of strontium nitrate= 3.00/211.6=0.01418 mol. The number of moles of strontium oxide=1.47/103.6=0.01419 mol. Thus the answer is D.
Ref: 9701/12/M/J/15 #30
The drug Sirolimus is used to treat patients after kidney transplants.
On reaction with hot aqueous sodium hydroxide, Sirolimus produces an equimolar mixture of two organic products.
What is the structural formula of the product with the lower relative molecular mass?
Solution
B
Ref: 9701/12/M/J/15 #33
Which statements are correct for all exothermic reactions?
1 H for the reaction is negative
2 On a reaction pathway diagram the products are shown lower than the reactants.
3 The reaction will happen spontaneously
Solution
B 1 and 2 only
Statement 3 is incorrect as the enthalpy change of a reaction does not determine its rate or whether it occurs. An example is the conversion of diamond to graphite which is an exothermic reaction, but occurs at an immeasurably slow rate due to the large activation energy for the reaction.
Statements 1 and 2 are true for all exothermic reactions, so the answer is B.
The shell of a chicken's egg makes up 5% of the mass of an average egg. An average egg has a mass of 50g.
Assume the egg shell is pure calcium carbonate.
How many complete chicken's egg shells would be needed to neutralise 50cm3 of 2.0moldm-3 ethanoic acid?
A. 1
B. 2
C. 3
D. 4
Solution
B 2
Form the balanced equation for the reaction of ethanoic acid with calcium carbonate.
2CH3COOH+CaCO3-->(CH3COO)2Ca+CO2+H2O
For an average egg, the shell is 0.05 x 50=2.5g. The Mr of CaCO3 is 100.1. Hence a chicken's egg shell contains 2.5/100.1=0.025 mol of CaCO3. 50cm3 of 2.0moldm-3= 50 x 2/1000=0.1mol of ethanoic acid. From the balanced equation, 2 mols of ethanoic acid reacts with 1 mol of CaCO3. Hence, 0.05 mols of calcium carbonate would be required to react with the amount of ethanoic acid stated in the question.
Therefore, number of chicken's egg shells=0.-05/0.025= 2.
Ref: 9701/12/M/J/15 #9
When a sample of HI is warmed to a particular temperature the equilibrium below is established.
2HI(g)--->< H2 (g)+ I2 (g)
At this temperature, it is found that the partial pressure of HI(g) is 28 times the partial pressure of H2 (g).
What is the value of Kp at this temperature?
A. 1.28 X 10-3
B. 0.035
C. 28
D. 784
Solution
A 1.28 x 10-3
Let the partial pressure of H2 be x. Since there are the same number of moles of H2 and I2, the partial pressure of I2 is x as well from the decomposition of HI. Hence the partial pressure of HI is 28x. Kp=x^2/(28x)^2= x^2/784x^2. By cancelling out x^2, Kp= 1/784= 1.28 x 10-3.
Photochromic glass, used for sunglasses, darkens when exposed to bright light and becomes more transparent again when the light is less bright. The darkness of the glass is related to the concentration of silver atoms.
The following reactions are involved.
reaction 1 Ag+ + Cl- --->< Ag+ Cl
reaction 2 Cu+ +Cl---> Cu2+ +Cl-
reaction 3 Cu2+ +Ag---> Cu+ + Ag+
Which statement about these reactions is correct?A. Cu+ and Cu2+ ions act as catalysts.
B. Cu+ ions act as an oxidising agent in reaction 2.
C. Reaction 3 increases the darkness of the glass.
D. Silver atoms are reduced in reaction 3.
Solution
A Cu+ and Cu2+ ions act as catalysts.
Statement 1 is correct as the Cu2+ and Cu+ ions are regenerated at the end of reactions 2 and 3 respectively, hence they act as catalysts.
Statement 2 is incorrect as the oxidation number of Cl changes from 0 to -1, hence the Cu+ ions act as a reducing agent.
Statement 3 is incorrect as silver atoms are converted to silver ions. Silver atoms increase the darkness of the glass whereas silver ions cause the glass to become more transparent.
Statement 4 is incorrect as the oxidation number of silver increases from 0 to +1, hence the silver atoms are oxidised.
Ref: 9701/12/M/J/15 #12
Use of the Data Booklet is relevant to this question.
When 3.00 g of an anhydrous nitrate of a Group II metal is decomposed, 1.53g of gas is produced. What is the nitrate compound?
A. beryllium nitrate
B. calcium nitrate
C. magnesium nitrate
D. strontium nitrate
Solution
D strontium nitrate
The decomposition of anhydrous Group II nitrates will produce the metal oxide, nitrogen dioxide gas and oxygen gas. The balanced equation for this process is
2X(NO3)2--->2XO+4NO2+O2
*X represents the Group II metalThere are two types of gases produced. Hence it is not possible to find the number of moles of the gases produced with the information given. So we must calculate the amount of solid metal oxide produced, which is 3.00--1.53=1.47g.
From the balanced equation, it can be seen that 2 mols of the metal nitrate will produce 2 mols of the metal oxide. Hence it is a 1:1 mole ratio.
To solve this question, the Mr of each metal nitrate and metal oxide must be calculated. The answer would be the compound which has equal number of moles of metal nitrate and metal oxide for the masses given.
For this question, the number of moles of strontium nitrate= 3.00/211.6=0.01418 mol. The number of moles of strontium oxide=1.47/103.6=0.01419 mol. Thus the answer is D.
Ref: 9701/12/M/J/15 #30
The drug Sirolimus is used to treat patients after kidney transplants.
On reaction with hot aqueous sodium hydroxide, Sirolimus produces an equimolar mixture of two organic products.
What is the structural formula of the product with the lower relative molecular mass?
Solution
B
At the top left corner, there is an ester bond and an amide bond. When heated with hot aqueous sodium hydroxide, the compound undergoes alkaline hydrolysis of amide and ester and the two bonds break, producing two compounds. The compound with the lower relative molecular mass contains the carboxylic part of the ester (COO- group), forming a salt with the sodium ion from NaOH, thus producing CO2Na. This compound also has a secondary amine group from the hydrolysis of the amide bond. So the answer is B.
Which statements are correct for all exothermic reactions?
1 H for the reaction is negative
2 On a reaction pathway diagram the products are shown lower than the reactants.
3 The reaction will happen spontaneously
Solution
B 1 and 2 only
Statement 3 is incorrect as the enthalpy change of a reaction does not determine its rate or whether it occurs. An example is the conversion of diamond to graphite which is an exothermic reaction, but occurs at an immeasurably slow rate due to the large activation energy for the reaction.
Statements 1 and 2 are true for all exothermic reactions, so the answer is B.
Saturday, 12 November 2016
Chemistry Solutions 9701/11/M/J/15
Ref: 9701/11/M/J/15 #3
Use of the Data Booklet is relevant to this question.
1.00 g of carbon is combusted in a limited supply of oxygen. 0.50 g of the carbon combusts to form CO2 and 0.50 g of the carbon combusts to form CO.
The resultant mixture of CO2 and CO is passed through excess NaOH (aq) and the remaining gas is then dried and collected.
What is the volume of the remaining gas? (All gas volumes are measured at 25 degree Celsius and 1 atmosphere pressure.)
A. 1 dm3
B. 1.5 dm3
C. 2 dm3
D. 3 dm3
Solution
A. 1 dm3
Form the balanced equation for the combustion of carbon. 4C+3O2--->2CO+2CO2
The RAM of carbon is 12. Hence 1g of carbon= 1/12=0.08333 mol. From the equation, 4 mols of carbon produce 2 mols of CO and 2 mols of CO2 respectively. Hence, 0.08333 mol of carbon will produce 0.041667 of CO and CO2 respectively.
NaOH absorbs acidic gases such as CO2 and SO2. Since there is limited amount of oxygen (not in excess), only CO remains once CO2 is absorbed by NaOH. At room temperature and pressure, 1 mol of gas occupies 24 dm3. Hence, 0.041667 mol of CO occupies 0.041667 x 24= 1 dm3.
Ref: 9701/11/M/J/15 #6
One mole of phosphorus(V) chloride, PCl5, is heated to 600 K in a sealed flask of volume 1 dm3. Equilibrium is established and measurements are taken.
Use of the Data Booklet is relevant to this question.
1.00 g of carbon is combusted in a limited supply of oxygen. 0.50 g of the carbon combusts to form CO2 and 0.50 g of the carbon combusts to form CO.
The resultant mixture of CO2 and CO is passed through excess NaOH (aq) and the remaining gas is then dried and collected.
What is the volume of the remaining gas? (All gas volumes are measured at 25 degree Celsius and 1 atmosphere pressure.)
A. 1 dm3
B. 1.5 dm3
C. 2 dm3
D. 3 dm3
Solution
A. 1 dm3
Form the balanced equation for the combustion of carbon. 4C+3O2--->2CO+2CO2
The RAM of carbon is 12. Hence 1g of carbon= 1/12=0.08333 mol. From the equation, 4 mols of carbon produce 2 mols of CO and 2 mols of CO2 respectively. Hence, 0.08333 mol of carbon will produce 0.041667 of CO and CO2 respectively.
NaOH absorbs acidic gases such as CO2 and SO2. Since there is limited amount of oxygen (not in excess), only CO remains once CO2 is absorbed by NaOH. At room temperature and pressure, 1 mol of gas occupies 24 dm3. Hence, 0.041667 mol of CO occupies 0.041667 x 24= 1 dm3.
Ref: 9701/11/M/J/15 #6
One mole of phosphorus(V) chloride, PCl5, is heated to 600 K in a sealed flask of volume 1 dm3. Equilibrium is established and measurements are taken.
PCl5(g)--->< PCl3(g)+Cl2(g)
The experiment is repeated with one mole of phosphorus (V) chloride heated to 600K in a sealed flask of volume 2dm3.
How will the measurements vary?
A. The equilibrium concentrations of PCl3 (g) and Cl2 (g) are higher in the second experiment.
B. The equilibrium concentration of PCl5 (g) is lower in the second experiment.
C. The equilibrium concentrations of all three gases are the same in both experiments.
D. The value of the equilibrium constant is higher in the second experiment.
Solution
B
Pressure is inversely proportional to volume. When the volume of the flask is increased from 1dm3 to 2dm3, the equilibrium pressure decreases and hence the equilibrium concentrations of PCl5, PCl3 and Cl2 are subsequently less than in the first experiment.
The equilibrium constant remains the same as the temperature remains the same. Hence D is incorrect.
Though there is an increase in the number of moles of PCl3 and Cl2 when pressure is decreased, the value of the equilibrium concentration is decreased due to an increase in volume from 1 dm3 to 2 dm3. Hence A is incorrect and the answer is B.
Ref: 9701/11/M/J/15 #10
The double bond between the two carbon atoms in an ethane molecule consists of one pi bond and one sigma bond.
Which orbitals overlap to form each of these bonds?
Solution
A. sp2-sp2 and p-p
In an ethene, C2H4 molecule, each carbon atom is surrounded by 2 hydrogen atoms and forms a double bond with the other carbon atom. The double bond consists of a sigma bond, and the pi bond which is formed after the sigma bond. Since each carbon atom is surrounded by 3 other atoms, they are both sp2 hybridised. Thus the sigma bond is formed between two sp2 hybrid orbitals. C and D are incorrect.
Pi bonds are only formed through the overlapping of p orbitals, hence B is incorrect. The answer is A.
* If you would like to increase your understanding on hybridisation, head over to Chemguide. Click on the bonding of methane and ethane.
Ref: 9701/11/M/J/15 #13
The three minerals below are obtained from mines around the world. Each one behaves as a mixture of two carbonate compounds. They can be used as fire retardants because they decompose in the heat, producing CO2. This gas smothers the fire.
Ref: 9701/11/M/J/15 #28
Ref: 9701/11/M/J/15 #10
The double bond between the two carbon atoms in an ethane molecule consists of one pi bond and one sigma bond.
Which orbitals overlap to form each of these bonds?
Solution
A. sp2-sp2 and p-p
In an ethene, C2H4 molecule, each carbon atom is surrounded by 2 hydrogen atoms and forms a double bond with the other carbon atom. The double bond consists of a sigma bond, and the pi bond which is formed after the sigma bond. Since each carbon atom is surrounded by 3 other atoms, they are both sp2 hybridised. Thus the sigma bond is formed between two sp2 hybrid orbitals. C and D are incorrect.
Pi bonds are only formed through the overlapping of p orbitals, hence B is incorrect. The answer is A.
* If you would like to increase your understanding on hybridisation, head over to Chemguide. Click on the bonding of methane and ethane.
Ref: 9701/11/M/J/15 #13
The three minerals below are obtained from mines around the world. Each one behaves as a mixture of two carbonate compounds. They can be used as fire retardants because they decompose in the heat, producing CO2. This gas smothers the fire.
barytocite BaCa(CO3)2
dolomite CaMg(CO3)2
huntite Mg3Ca(CO3)4
What is the order of effectiveness of fire retardant, from best to worst?
Solution
D
The effectiveness of the compounds as fire retardants is based on their ability to decompose under heat to produce CO2. Going down Group II, the stability of the metal carbonates increases, hence lower down the group, the more difficult and the more heat energy required for the metal carbonates to decompose. Barytocite, dolomite and huntite all have Ca. Barytocite has Ba, which is lower in the group than Mg (present in dolomite and huntite), hence barytocite decomposes least easily among all three, making it the worst fire retardant.
Although dolomite and huntite both have Ca and Mg, huntite has a higher proportion of Mg compared to dolomite. Thus huntite will decompose more under the same conditions, making it the best fire retardant.
Ref: 9701/11/M/J/15 #26
The presence of a halogen in an organic compound may be detected by warming the organic compound with aqueous silver nitrate.
Which compound would be the quickest to produce a precipitate?
Solution
C.
Warming the compounds would cause the bonds in the compounds to break, releasing the halide ions which then react with the silver ions in AgNO3 to form precipitates. The smaller the bond energy, the weaker the bond, the more quickly it breaks and the faster a precipitate is produced.
Bond energy depends on two factors: Bond length and Bond polarity. The greater the bond length, the weaker the bond. The less polar the bond, the weaker the bond.
Among F, Cl, Br and I, I has the largest atomic radius and the least electronegativity, hence the bond with I is the longest and least polar, thus the weakest bond. Compound C is the only compound with I in it, hence the bond containing I in compound C would be the first to break. Thus when heated, compound C produces a precipitate (yellow for AgI) most quickly.
Ref: 9701/11/M/J/15 #28
In 1869 Ladenburg suggested a structure for benzene, C6H6, in which one hydrogen atom is attached to each carbon atom.
A compound C6H4Cl2 could be formed with the same carbon skeleton as the Ladenburg structure.
How many structural isomers would this compound have?
How many structural isomers would this compound have?
A. 3
B. 4
C. 5
D. 6
Solution
A. 3
This question requires visualisation. The Ladenburg structure consists of two triangular faces and three rectangular faces.
The three structural isomers are as follows :
1. Two Cl atoms attaching to two adjacent carbon atoms on the triangular face
2. Two Cl atoms attaching to two adjacent carbon atoms on the rectangular face
3. Two Cl atoms attaching to two carbon atoms diagonally across the rectangular face
This 3-D model of the Ladenburg structure might help you to see it more clearly.
Ref: 9701/11/M/J/15 #29
The citrus flavour of lemons is due to the compound limonene, present in both the peel and the juice.
What is the mole ratio of carbon dioxide to water produced when limonene is completely burnt in oxygen?
Solution
B
When burnt in excess oxygen, hydrocarbons will produce CO2 (with the C atoms) and H2O (with the H atoms). By counting, you will obtain 10 carbon atoms and 16 hydrogen atoms. 10 mols of C atoms will produce 10 mols of CO2. 16 mols of H atoms will produce 8 mols of H2O (watch out for the '2' in the compound, hence 8 mols of H2O and not 16).
Thus the mole ratio of CO2 to H20 is 10:8=5:4.
Ref: 9701/11/M/J/15 #37
The diagram shows the structure of an alkene molecule.
Which statements about this molecule are correct?
1 All the carbon atoms are in the same plane
2 It has geometrical isomers
3 It is optically active
Which statements about this molecule are correct?
Solution
C 2 and 3 only
The carbon atoms can only be in the same plane if bonds are sp2-sp2 or sp3-sp2. They cannot be in the same plane if the bonds are sp3-sp3. Since there are sp3-sp3 bonds between carbon atoms (carbon atoms which form bonds with four other atoms), not all the carbon atoms are in the same plane. Hence statement 1 is incorrect.
The two C atoms in the double bond each have two different groups attached to them, hence they display cis-trans/ geometrical isomerism. Statement 2 is correct.
The compound has a chiral carbon atom, hence it is optically active. Statement 3 is correct.
Friday, 11 November 2016
Chemistry Solutions 9701/11/O/N/15
Ref: 9701/11/O/N/15 #13
Rat poison needs to be insoluble in rain water but soluble at the low pH of stomach contents. What is a suitable barium compound to use for rat poison?
A. barium carbonate
B. barium chloride
C. barium hydroxide
D. barium sulfate
Solution:
A. barium carbonate
Barium chloride and barium hydroxide are both soluble in rain water and also acids, so B and C are incorrect. Barium sulfate is insoluble in both water and acids, so D is incorrect as well. Barium carbonate is insoluble in water, but dissolves in acids to give off carbon dioxide.
BaCO3 + 2HCl -> BaCl2 + CO2 + H2O
Ref: 9701/11/O/N/15 #15
The melting points of the Period 3 elements sodium to aluminium are shown in the table.
Which factor explains the increase in melting points from sodium to aluminium?
A. the changes in first ionisation energy from sodium to aluminium
B. the increase in electronegativity from sodium to aluminium
C. the increase in the Ar of the elements from sodium to aluminium
D. the increase in the number of outer electrons in each atom from sodium to aluminium
Solution:
D. the increase in the number of outer electrons in each atom from sodium to aluminium
Strength of the metallic bonds depends on the number of delocalised electrons in the metal. The greater the number of delocalised electrons, the greater the strength of the metallic bonds, the higher the melting and boiling points. Take note that melting and boiling points only depend on metallic bonds for metals, intermolecular forces for non-metallic compounds (permanent dipoles, Van der Waals' forces/temporary induced dipoles, hydrogen bonds), ionic bonds for ionic compounds.
Melting and boiling points DO NOT depend on the strength of covalent bonds.
Ref: 9701/11/O/N/15 #19
Element X reforms a pollutant oxide Y. Y can be futher oxidised to Z. Two students made the following statements.
Student P "The molecule of Y contains lone pairs of electrons."
Student Q "The oxidation number of X increases by 1 from Y to Z."
X could be carbon or nitrogen or sulfur
Which student(s) made a correct statement?
A. P only
B. Q only
C. Both P and Q
D. neither P nor Q
Solution:
A. P only
-Nitrogen monoxide has 3 lone pairs , one from nitrogen , two from
oxygen. Hence student P is correct. If nitrogen was element X, it would form nitrogen monoxide/nitric oxide which would be oxidised to nitrogen dioxide. In this process, the oxidation number of nitrogen increases by 2, hence student Q is incorrect.
-If carbon was element X, it would form carbon monoxide which has lone pairs. Carbon monoxide is not oxidised in the atmosphere. Hence student P is correct and student Q is incorrect.
- If sulfur was element X, it would form sulfur dioxide which has 6 lone pairs. Hence P is correct. Sulfur dioxide would be oxidised to sulfur trioxide in the atmosphere.In this process, the oxidation number of sulfur increases by 2, hence student Q is incorrect.
Ref: 9701/11/O/N/15 #20
How many isomeric esters have the molecular formula C4H8O2?
A. 2
B. 3
C. 4
D. 5
Solution:
C. 4
Usually for isomeric esters, the first step is to list out the alcohol and carboxylic acid components, then consider the isomers for each component. In this case, only propanol has isomers, which are propan-1-ol, and propan-2-ol.
Methyl propanoate CH3CH2CO2CH3
Ethyl ethanoate CH3CO2CH2CH3
Propyl methanoate HCO2CH2CH2CH3
2-methyl-ethyl methanoate HCO2CH(CH3)CH3
Ref: 9701/11/O/N/15 #21
A new jet fuel has been produced that is a mixture of different structural isomers of compound Q.
Which skeletal formula represents a structural isomer of Q?
Solution:
C
Once you draw out the compounds and look carefully, you will see that A, D and Q are actually the same compound (notice that the parts circled orange are the same). A and D are laterally inverted. Hence A and D are incorrect. B has a total of 13 carbon atoms in its compound whereas Q has 14 carbon atoms. Hence B is incorrect. C is a structural isomer of Q, so C is the answer.
*note that the compounds are drawn for illustration purposes, it would be safer to draw the compound with CH3 branches tetrahedrally rather than V-shaped in structured questions.
Ref: 9701/11/O/N/15 #23
Compound Q contains 3 double bonds per molecule.
Which bond, X or Y, will be ruptured by hot, concentrated acidified KMnO4 and how many lone pairs of electrons are present in one molecule of Q ?
Solution:
B
Each oxygen atom has 2 lone pairs. Hence molecule Q has 6 lone pairs. Bond X will be broken when the alkene molecule undergoes vigorous oxidation with hot concentrated acidified KMnO4. Y is a ketone, therefore it cannot be oxidised.
Ref: 9701/11/O/N/15 #31
Which statements about orbitals in a krypton atom are correct ?
1. The 1s and 2s orbitals have the same energy as each other but different sizes.
2. The 3rd energy level ( n=3) has 3 subshells and 9 orbitals
3. the 3d subshell has 5 orbitals that have the same energy as each other in an isolated atom
Solution:
C 2 and 3
1 is incorrect because the 2s orbital is at a higher energy level than 1s orbital . The 2s orbital is larger than the 1s orbital.
2 is correct because the 3rd energy level consist of the 3s, 3p and 3d subshells. The 3s subshell has 1 orbital. The 3p subshell has 3 orbitals and the 3d subshell has 5 orbitals.
3 is correct because the 5 orbitals are degenerates and have the same energy level.
Ref: 9701/11/O/N/15 #34
Which statements correctly describe an effect of a rise in temperature on a gas phase reaction.
1. More particles now have energies greater than the activation energy .
2. The energy distribution profile changes with more particles having the most probable energy.
3. The activation energy of the reaction is decreased.
Solution:
D. 1 only
Statement 1 is correct because the peak of the Boltzmann Distribution curve is lowered and the whole graph shifts to the right. The activation energy remains the same, so more particles have energies more than or equal to the activation energy.
Statement 2 is wrong because most probable energy = energy possessed by the greatest number of molecules. (Refer to peaks in the attached diagram.) At higher temperature, the peak is lowered. Less particles have the most probable energy at a higher temperature.
Statement 3 is wrong because the activation energy of the reaction remains the same.
Ref: 9701/11/O/N/15 #38
Each of the compounds below is treated separately with excess NaBH4. The product of the reaction is then heated with excess concentrated H2SO4.
In each case, one ore more products are formed with molecular formula C7H10.
Which compounds give only one final product with the molecular formula C7H10?
Solution:
C. 2 and 3 only
1 produces two different compounds with the molecular formula C7H10, hence it is incorrect.
2 produces one compound only. The diagram shows the products being laterally inverted --- when flipped, they are actually the same compound.
3 produces one compound only, because a double bond cannot be formed between C1 and C2. This is because the H atom from C1 is eliminated along with the branched -OH group to form a double bond already.
Rat poison needs to be insoluble in rain water but soluble at the low pH of stomach contents. What is a suitable barium compound to use for rat poison?
A. barium carbonate
B. barium chloride
C. barium hydroxide
D. barium sulfate
Solution:
A. barium carbonate
Barium chloride and barium hydroxide are both soluble in rain water and also acids, so B and C are incorrect. Barium sulfate is insoluble in both water and acids, so D is incorrect as well. Barium carbonate is insoluble in water, but dissolves in acids to give off carbon dioxide.
BaCO3 + 2HCl -> BaCl2 + CO2 + H2O
Ref: 9701/11/O/N/15 #15
The melting points of the Period 3 elements sodium to aluminium are shown in the table.
A. the changes in first ionisation energy from sodium to aluminium
B. the increase in electronegativity from sodium to aluminium
C. the increase in the Ar of the elements from sodium to aluminium
D. the increase in the number of outer electrons in each atom from sodium to aluminium
Solution:
D. the increase in the number of outer electrons in each atom from sodium to aluminium
Strength of the metallic bonds depends on the number of delocalised electrons in the metal. The greater the number of delocalised electrons, the greater the strength of the metallic bonds, the higher the melting and boiling points. Take note that melting and boiling points only depend on metallic bonds for metals, intermolecular forces for non-metallic compounds (permanent dipoles, Van der Waals' forces/temporary induced dipoles, hydrogen bonds), ionic bonds for ionic compounds.
Melting and boiling points DO NOT depend on the strength of covalent bonds.
Ref: 9701/11/O/N/15 #19
Element X reforms a pollutant oxide Y. Y can be futher oxidised to Z. Two students made the following statements.
Student P "The molecule of Y contains lone pairs of electrons."
Student Q "The oxidation number of X increases by 1 from Y to Z."
X could be carbon or nitrogen or sulfur
Which student(s) made a correct statement?
A. P only
B. Q only
C. Both P and Q
D. neither P nor Q
Solution:
A. P only
-Nitrogen monoxide has 3 lone pairs , one from nitrogen , two from
oxygen. Hence student P is correct. If nitrogen was element X, it would form nitrogen monoxide/nitric oxide which would be oxidised to nitrogen dioxide. In this process, the oxidation number of nitrogen increases by 2, hence student Q is incorrect.
-If carbon was element X, it would form carbon monoxide which has lone pairs. Carbon monoxide is not oxidised in the atmosphere. Hence student P is correct and student Q is incorrect.
- If sulfur was element X, it would form sulfur dioxide which has 6 lone pairs. Hence P is correct. Sulfur dioxide would be oxidised to sulfur trioxide in the atmosphere.In this process, the oxidation number of sulfur increases by 2, hence student Q is incorrect.
Ref: 9701/11/O/N/15 #20
How many isomeric esters have the molecular formula C4H8O2?
A. 2
B. 3
C. 4
D. 5
Solution:
C. 4
Usually for isomeric esters, the first step is to list out the alcohol and carboxylic acid components, then consider the isomers for each component. In this case, only propanol has isomers, which are propan-1-ol, and propan-2-ol.
Methyl propanoate CH3CH2CO2CH3
Ethyl ethanoate CH3CO2CH2CH3
Propyl methanoate HCO2CH2CH2CH3
2-methyl-ethyl methanoate HCO2CH(CH3)CH3
Ref: 9701/11/O/N/15 #21
A new jet fuel has been produced that is a mixture of different structural isomers of compound Q.
Solution:
C
Once you draw out the compounds and look carefully, you will see that A, D and Q are actually the same compound (notice that the parts circled orange are the same). A and D are laterally inverted. Hence A and D are incorrect. B has a total of 13 carbon atoms in its compound whereas Q has 14 carbon atoms. Hence B is incorrect. C is a structural isomer of Q, so C is the answer.
Ref: 9701/11/O/N/15 #23
Compound Q contains 3 double bonds per molecule.
B
Each oxygen atom has 2 lone pairs. Hence molecule Q has 6 lone pairs. Bond X will be broken when the alkene molecule undergoes vigorous oxidation with hot concentrated acidified KMnO4. Y is a ketone, therefore it cannot be oxidised.
Ref: 9701/11/O/N/15 #31
Which statements about orbitals in a krypton atom are correct ?
1. The 1s and 2s orbitals have the same energy as each other but different sizes.
2. The 3rd energy level ( n=3) has 3 subshells and 9 orbitals
3. the 3d subshell has 5 orbitals that have the same energy as each other in an isolated atom
Solution:
C 2 and 3
1 is incorrect because the 2s orbital is at a higher energy level than 1s orbital . The 2s orbital is larger than the 1s orbital.
2 is correct because the 3rd energy level consist of the 3s, 3p and 3d subshells. The 3s subshell has 1 orbital. The 3p subshell has 3 orbitals and the 3d subshell has 5 orbitals.
3 is correct because the 5 orbitals are degenerates and have the same energy level.
Ref: 9701/11/O/N/15 #34
Which statements correctly describe an effect of a rise in temperature on a gas phase reaction.
1. More particles now have energies greater than the activation energy .
2. The energy distribution profile changes with more particles having the most probable energy.
3. The activation energy of the reaction is decreased.
Solution:
D. 1 only
Statement 1 is correct because the peak of the Boltzmann Distribution curve is lowered and the whole graph shifts to the right. The activation energy remains the same, so more particles have energies more than or equal to the activation energy.
Statement 2 is wrong because most probable energy = energy possessed by the greatest number of molecules. (Refer to peaks in the attached diagram.) At higher temperature, the peak is lowered. Less particles have the most probable energy at a higher temperature.
Statement 3 is wrong because the activation energy of the reaction remains the same.
Ref: 9701/11/O/N/15 #38
Each of the compounds below is treated separately with excess NaBH4. The product of the reaction is then heated with excess concentrated H2SO4.
In each case, one ore more products are formed with molecular formula C7H10.
Which compounds give only one final product with the molecular formula C7H10?
Solution:
C. 2 and 3 only
1 produces two different compounds with the molecular formula C7H10, hence it is incorrect.
2 produces one compound only. The diagram shows the products being laterally inverted --- when flipped, they are actually the same compound.
3 produces one compound only, because a double bond cannot be formed between C1 and C2. This is because the H atom from C1 is eliminated along with the branched -OH group to form a double bond already.
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