Paper 1 2017

Question 1

Which atom is not an isotope of iodine?
Neutrons Mass number
A 72 125
B 74 127
C 75 128
D 77 129

Answer 1

D

Question 2

What is the bonding between the ligands and the metal ion in [Fe(H2O)6]^2+?

Answer 2

Dative covalent

Question 3

What is the oxidation number of Mn in K2MnO4?

Answer 3

+6

Question 4

Which calcium compound contain the greatest percentage by mass of calcium?
A Calcium carbonate
B Calcium nitrate
C Calcium hyrdoxide
D Calcium sulfate

Answer 4

Calcium hydroxide

Question 5

0.0200 mol of calcium oxide is reacted completely with 2.00mol dm^-3 HCl. What is the volume, in cm^3, of 2.00 mol dm^-3 HCl required for this reaction?

Answer 5

20

Question 6

How many electrons are removed from 2.02 x 10^-2g of Ne(g) atoms to form Ne+(g) ions?

Answer 6

6.02 x 10^20

Question 7

Silicon can be made by heating silicon tetrachloride, SiCl4, with zinc.

SiCl4 + 2Zn –> Si + 2ZnCl2

8.50 g of SiCl4 is reacted with an excess of zinc. The percentage yield of silicon is 90%. What is the mass of silicon made?

Answer 7

1.26 g

Question 8

For pairs of solutions are mixed. Which pair of solutions forms a white precipitate?

A NH4Cl (aq) and NaOH (aq)
B KBr (aq) and AgNO3 (aq)
C FeCl3 (aq) and NH3 (aq)
D Cr2(SO4)3 (aq) and BaCl2 (aq)

Answer 8

D

Question 9

Enthalpy values are provided below.

H2(g) + I2(g) –> 2HI(g), △rH = -9kJ mol^-1

Bond Bond enthalpy/ kJ mol^-1
H-H +436
I-I +151

Answer 9

+298

Question 10

Aqueous Cr3+ ions are reacted with an excess of aqueous sodium hydroxide. Which product is formed?

Answer 10

Cr(OH)6^3-

Question 11

HA and HB are two strong monobasic acids. 25.0 cm^3 of 6.00 mol dm^-3 HA is mixed with 45 cm^3 or 3.0 mol dm^-3. What is the H+ (aq) concentration, in mol dm-3, in the resulting solution?

Answer 11

4.1

Question 12

A mixture of N2 and O2 gases has a total pressure of 1.42 atm. The mole fraction of N2 is 0.700. What is the partial pressure, in atm, of O2, in the mixture?

Answer 12

0.426

Question 13

A cell is constructed from the two redox systems below.

Cu2+(aq) +2e- –> Cu(s)
E = +0.34V

Ag+(aq) + e- –> Ag(s)
E = +0.80V

Which statement(s) is/are correct for the cell?
1. The cell potential is 1.14V.
2. The reaction at the copper electrode is Cu(s) Cu^ 2+ (aq)+2e^ -
3. The silver electrode increases in mass.

Answer 13

Only 2 and 3

Question 14

What are the electron configuration of a Cr atom and a Cu atom?

Answer 14

• Cr: 1s2 2s2 2p6 3s2 3p6 3d5 4s1
• Cu: 1s2 2s2 2p6 3s2 3p6 3d10 4s1

Question 15

Predict the name of the shape of, and H-N-H bond angle in, NH4+ and NH2-.

Answer 15

• NH4+: tetrahedral and 109.5
• NH2-: non-linear and 104.5

Question 16

Nitrogen, phosphorus and arsenic are in Group 15(5) of the periodic table. The boiling points of their hydrides are shown below.

Element Hydride Boiling point/°C
N NH3 -33
P PH3 -88
As AsH3 -55

i) Explain why the boiling point of PH3 is lower than the boiling point of NH3.

ii) Explain why the boiling point of PH3 is lower than the boiling point of AsH3.

Answer 16

i) NH3 has hydrogen bonding which requires more energy to overcome
ii) AsH3 (As) has more electrons than PH3 (P) and more London forces which require more energy required to overcome

Question 17

Barium chloride can be prepared from barium hydroxide in a neutralisation reaction.

Write the equation for this reaction. State symbols are not required.

Answer 17

Ba(OH)2 + 2HCl –> BaCl2 + 2H2O

Question 18

The reactivity of the Group 2 elements Mg-Ba increases down the group. Explain why.

Answer 18

• Atomic radius increases
• Nuclear attraction decreases
• Ionisation energy decreases

Question 19

On gentle heating, the compound KClO3 reacts as shown in the equation.

4KClO3 (s) –> KCl (s) + 3KClO4 (s)

This reaction is an example of disproportionation.

State what is meant by disproportionation and use oxidation numbers to show that disproportionation has taken place.

Answer 19

• Oxidation and reduction of the same element
• Cl is oxidised from +5 (in KClO3) to +7 (in KClO4)
• Cl is reduced from +5 (in KClO3) to –1 (in KCl)

Question 20

What is the systemic name for KClO4?

Answer 20

potassium chlorate (VII)

Question 21

Two changes are described below.
For each change:
- write an equation, including state symbols
- state and explain how the entropy changes

i) The reaction of aqueous barium nitrate with aqueous sodium sulfate.
ii) The change that accompanies the standard enthalpy change of atomisation of iodine.

Answer 21

i)
- Ba(NO3)2 (aq) + Na2SO4 (aq) –> BaSO4(s) + 2NaNO3 (aq)
- Entropy decreases and (BaSO4) has dispersal of energy

ii)
- ½ I2(s) –> I(g)
- Entropy increases and more dispersal of energy

Question 22

Iron can be extracted from its ore Fe3O4 using carbon. Several equilibria are involved including equilibrium 18.1, shown below.

Equilibrium 18.: Fe3O4 (s) + 4C (s) –> 3Fe (s) + 4CO (g)

ΔH = 676.4kJ mol^-1
ΔS = 703.1JK^-1 mol^-1

i) Why is equilibrium 18.1 a heterogeneous equilibrium?

Answer 22

Species have different states/ phases

Question 23

Iron can be extracted from its ore Fe3O4 using carbon. Several equilibria are involved including equilibrium 18.1, shown below.

Equilibrium 18.: Fe3O4 (s) + 4C (s) –> 3Fe (s) + 4CO (g)

ΔH = 676.4kJ mol^-1
ΔS = 703.1JK^-1 mol^-1

ii) Write the expression for Kp for equilibrium 18.1.

Answer 23

Kp = p(CO(g))4

Question 24

Iron can be extracted from its ore Fe3O4 using carbon. Several equilibria are involved including equilibrium 18.1, shown below.

Equilibrium 18.: Fe3O4 (s) + 4C (s) –> 3Fe (s) + 4CO (g)

ΔH = 676.4kJ mol^-1
ΔS = 703.1JK^-1 mol^-1

iii) The forward reaction in equilibrium 18.1 is only feasible at high temperatures.

• Show that the forward reaction is not feasible at 25 C
• Calculate the minimum temperature, in K, for the forward reaction to be feasible

Answer 24

• ∆G = ∆H – T∆S = 676.4 – (298  0.7031)
  = (+) 467 (kJ mol–1
• Non-feasible when ∆G > 0
• Minimum temperature = ∆H/∆S = 676.4/0.7031 = 962(.0) K

Question 25

Another equilibrium involved in the extraction of iron from Fe3O4 is shown below. Fe3O4 (s) +4CO (g) --> 3Fe (s) + 4CO2 (g) ΔH = -13.5kJmol^-1 Enthalpy changes of formation, ΔfH for Fe3O4 (s) and CO2 (g) are shown in the table. Compound ΔfH/kJ mol^-1 Fe3O4 (s) -1118.5 CO2 (g) -393.5 Calculate the enthalpy change of formation, ΔfH, for CO (g).

Answer 25

- –13.5 = (4 x –393.5) – (–1118.5 + 4 x ∆fH(CO)) - 4 x ∆fH(CO) = (4 x –393.5) – (–1118.5) + 13.5 = –442(.0) (kJ mol–1) - ∆fH(CO) = –442/4 = –110.5 (kJ mol–1)

Question 26

This question is about equilibrium reactions. Hydrogen gas is manufactured by the chemical industry using the reaction of methane and steam. This is a reversible reaction, shown in equilibrium 20.1 below. Equilibrium 20.1: CH4 (g) +H2O (g) --> 3H2 (g) + CO(g) H = 210kJmol^- 1 Explain, in terms of le Chatelier's principle, the conditions of pressure and temperature for a maximum yield of hydrogen from equilibrium 20.1, and explain why the operational conditions used by the chemical industry may be different.

Answer 26

- Low pressure and high temperature - Left-hand side/reactant side has fewer gaseous moles - Forward reaction is endothermic - Low pressure gives a slow rate or high temperature uses a large amount of energy

Question 27

A chemist investigates the equilibrium reaction between sulfur dioxide, oxygen, and sulfur trioxide, shown below. 2SO2 (g) + O2 (g) --> 2SO3 (g) - The chemist mixes together SO2 and O2 with a catalyst. - The chemist compresses the gas mixture to a volume of 400cm^3. - The mixture is heated to a constant temperature and is allowed to reach equilibrium without changing the total gas volume. The equilibrium mixture contains 0.0540 mol SO2 and 0.0270 mol O2. At the temperature used, the numerical value for Kc is 3.045 x 10^4 dm^3 mol^-1. i) Write the expression for Kc and the units of Kc for this equilibrium.

Answer 27

- Kc = [SO3]^2 / [SO2]^2 x [O2] - Units: dm3 mol–1

Question 28

A chemist investigates the equilibrium reaction between sulfur dioxide, oxygen, and sulfur trioxide, shown below. 2SO2 (g) + O2 (g) --> 2SO3 (g) - The chemist mixes together SO2 and O2 with a catalyst. - The chemist compresses the gas mixture to a volume of 400cm^3. - The mixture is heated to a constant temperature and is allowed to reach equilibrium without changing the total gas volume. The equilibrium mixture contains 0.0540 mol SO2 and 0.0270 mol O2. At the temperature used, the numerical value for Kc is 3.045 x 10^4 dm^3 mol^-1. ii) Determine the amount, in mol, of SO3 in the equilibrium mixture at this temperature. Give your final answer to an appropriate number of significant figures. Show all your working.

Answer 28

- SO2 = 0.135 (mol dm–3) - O2 = 0.0675 (mol dm–3) - [SO3] = √(Kc x [SO2]^2 x O2) = ((3.045 x 10^4) x 0.1352^2 x 0.0675) = 6.12039291 (mol dm–3) - n(SO3) = 6.12039291/2.5 = 2.45 (mol)

Question 29

This question is about the properties and reactions of ethanoic acid, CH3. Ethanoic acid is a weak acid with an acid dissociation constant, Ka of 1.75 x 10^- 5 mol dm^-3 at 25 C. A student uses a pH meter to measure the pH of a solution of CH3COOH at 25°C. The measured pH is 2.440. Calculate the concentration of ethanoic acid in the solution. Give your answer to three significant figures.

Answer 29

- [H+] = 10^–pH = 10^–2.440 = 3.63 x 10^–3 - [CH3COOH] = [H+]^2 / Ka = (3.63 x 10^–3)^2 / 1.75 x 10^–5 = 0.753 (mol dm^–3)

Question 30

The student plans to prepare a buffer solution that has a pH of 4.50. The buffer solution will contain ethanoic acid, CH3COOH and sodium ethanoate, CH3COONa. The student plans to add 9.08g CH3COONa to 250cm^3 of 0.8 mol dm^-3 CH3COOH. The student assumes that the volume of the solution does not change. When the student prepares the buffer solution, the volume of solution increases slightly. ii) Suggest whether the pH of the buffer solution would be the same, greater than, or less than your calculated value in (c) (i) Explain your reasoning.

Answer 30

- n(CH3COONa) = 9.08/82.0 = 0.111 - [CH3COO–] = 9.08/82.0 x 1000/250 = 0.443 (mol dm–3) - n(CH3COOH) = 0.800 x 250/1000 = 0.200 (mol) - [H+] = Ka x [CH3COOH] / [CH3COO–] = 1.75 x 10^–5 x 0.800/0.443 = 1.75 x 10^–5 = 3.16 x 10^-5 (mol dm–3) - pH = –log (3.16 x 10^–5) = 4.50

Question 31

A hydrated nickel(II) complex, A, is heated in a crucible to remove the water of crystallisation. The anhydrous complex B is formed. The results are shown below. Mass of crucible + hydrated complex A = 59.554g Mass of crucible + anhydrous complex B = 58.690 g Mass of crucible = 51.257g The anhydrous complex B is analysed and found to have a molar mass of 309.7g mol^-1 and to contain the following percentage composition by mass: Ni, 18.95%; C, 23.25%; N, 27.12%; H, 7.75%; Cl, 22.93%. The anhydrous complex B contains a cation C comprising Ni, C. N and H only. Cation C is six-coordinate, contains three molecules of the bidentate ligand D, and exists as optical isomers. Determine the formula of A, B, C and D. Show all your working.

Answer 31

- NiC6N6H24Cl2 - NiC6N6H24Cl2*2H2O - n(anhydrous salt) = 7.433/309.7 = 0.02400 (mol) - n(H2O) = 0.864/18.0 = 0.04800 (mol) There may be other methods - Formula of cation C: [NiC6N6H24]^2+ OR - Bidentate ligand D: H2NCH2CH2NH2