GCSE Chemistry Higher Tier Paper 1 May 2019 Study Guide

General Examination Information

• Examination Body: GCSE CHEMISTRY Higher Tier Paper 1 H.
• Date and Time: Thursday 16 May 2019, Morning session.
• Duration: 1 hour 45 minutes.
• Total Marks: 100 marks.
• Materials Required: Ruler, scientific calculator, and the periodic table (enclosed).
• General Instructions:     - Use black ink or black ball-point pen.     - Answer all questions.     - Write calculations clearly to show the working process.     - Use calculators where appropriate.

Question 1: Development of the Periodic Table

• Historical Classification: In the 19th century, scientists classified elements by their atomic weights.
• Mendeleev's Periodic Table (1869):     - Mendeleev reversed the order of tellurium ($Te$, atomic weight $128$) and iodine ($I$, atomic weight $127$) because iodine has similar properties to other elements in its group (Group 7), ensuring elements with similar properties were in the same column.     - Asterisks (*): Mendeleev left spaces marked with asterisks for elements he predicted had not been discovered yet.     - Acceptance: His version became widely accepted because when the missing elements were later discovered, their properties closely matched Mendeleev’s predictions.
• Terminology and Modern Arrangement:     - The modern name for atomic weight is Relative atomic mass.     - In the modern periodic table, elements are arranged in order of atomic number.
• Group 7 (Halogens):     - Astatine ($At$): Positioned below Iodine ($I$).     - Prediction for Astatine:         - Formula: $At_2$ (diatomic molecule).         - State at room temperature: Solid (melting/boiling points increase down Group 7).
• Group 1 Reactions:     - Sodium and Chlorine Reaction: When sodium reacts with chlorine, an observer would see a bright yellow flame and the formation of a white smoke or white solid (sodium chloride).

Question 2: Acids, Alkalis, and Titration

• Aqueous Ions: All acids produce hydrogen ions ($H^+$) in aqueous solution.
• Neutralization Reaction:     - Word Equation: $\\text{calcium hydroxide} + \\text{hydrochloric acid} \\rightarrow \\text{calcium chloride} + \\text{water}$.
• Titration Apparatus (Figure 2):     - Apparatus A: Identified as a burette.     - Reading: Based on the diagram in Figure 2, the student must record the meniscus level for the volume in $cm^3$.
• Titration Procedure (Concentration Comparison):     - To find which sulfuric acid sample ($P$ or $Q$) is more concentrated:         1. Measure a fixed volume of sulfuric acid (e.g., $25.0\\,cm^3$) using a pipette and transfer to a conical flask.         2. Add a few drops of an indicator (e.g., phenolphthalein or methyl orange).         3. Fill a burette with sodium hydroxide solution of known concentration.         4. Perform the titration by adding sodium hydroxide until the indicator changes color (the end point).         5. Record the volume of sodium hydroxide used (the titre).         6. Repeat the experiment for both samples $P$ and $Q$.         7. The sample requiring the greater volume of sodium hydroxide is the more concentrated acid.

Question 3: Materials and Nanoparticles

• Carbon Nanotubes:     - Structure and bonding are similar to graphene.     - Electrical Conductivity: These conduct electricity because they have delocalized electrons that are free to move through the structure.
• Material Evaluation (Badminton Rackets):     - Properties Table (Table 1):         - Aluminium: Density: $2.7\\,g/cm^3$; Relative strength: $0.3$; Relative stiffness: $69$.         - Carbon Nanotube: Density: $1.5\\,g/cm^3$; Relative strength: $60$; Relative stiffness: $1000$.         - Wood: Density: $0.71\\,g/cm^3$; Relative strength: $0.1$; Relative stiffness: $10$.     - Evaluation: Carbon nanotubes are the best material for racket frames because they have the highest relative strength and stiffness while maintaining a low density (lighter than aluminium), though wood is the lightest but significantly weaker and less stiff.
• Nanoparticles (Zinc Oxide):     - Surface Area Calculation:         - A cube-shaped nanoparticle has sides of $82\\,nm$.         - Area of one face = $82\\,nm \\times 82\\,nm = 6724\\,nm^2$.         - Total surface area (6 faces) = $6 \\times 6724 = 40344\\,nm^2$.         - Standard Form: $4.0344 \\times 10^4\\,nm^2$.     - Suncream Economics: Using nanoparticles in suncreams costs less because they have a high surface area to volume ratio, meaning less of the material is required to provide the same level of coverage or protection compared to fine particles.

Question 4: Atomic Structure and Isotopes

• Subatomic Particles (Table 2):     - Neutron: Relative mass = $1$; Relative charge = $0$.     - Proton (implied by +1 charge): Relative mass = $1$; Relative charge = $+1$.
• Isotopes:     - Mass Number Definition: The sum of the number of protons and neutrons in the nucleus of an atom.     - Isotopic Difference: Isotopes of an element have different mass numbers because they contain different numbers of neutrons.
• Models of the Atom:     - Plum Pudding Model: Described the atom as a ball of positive charge with electrons embedded in it.     - Alpha Particle Scattering Experiment Change:         - Most alpha particles passed straight through gold foil, suggesting the atom is mostly empty space.         - Some particles were deflected or bounced back, suggesting a tiny, concentrated, positively charged nucleus at the center.         - This led to the nuclear model replacing the plum pudding model.

Question 5: Ammonia ($NH_3$)

• Bonding: Nitrogen (Group 5) shares 3 electrons with 3 Hydrogen atoms (Group 1) to form three covalent bonds, leaving one lone pair on the Nitrogen atom.
• Dot and Cross Limitations:     - It is a 2D representation and does not show the true 3D shape (trigonal pyramidal).     - It does not represent the relative sizes of the atoms or the lengths of the bonds.
• Physical Properties: Ammonia has a low boiling point because it is a small covalent molecule with weak intermolecular forces. Little energy is needed to overcome these forces to change state.
• Catalysts: Transition metal oxides like chromium oxide ($Cr_2O_3$) are most likely catalysts for ammonia reactions.
• Bond Energy Calculation:     - Equation: $4NH_3 + 3O_2 \\rightarrow 2N_2 + 6H_2O$.     - Bond Energies (Table 3):         - $N-H$: $391\\,kJ/mol$         - $O=O$: $498\\,kJ/mol$         - $N \\equiv N$: $945\\,kJ/mol$         - $O-H$: $464\\,kJ/mol$     - Energy to break bonds (Reactants):         - $12 \\times (N-H) = 12 \\times 391 = 4692\\,kJ$         - $3 \\times (O=O) = 3 \\times 498 = 1494\\,kJ$         - Total in = $6186\\,kJ$     - Energy released forming bonds (Products):         - $2 \\times (N \\equiv N) = 2 \\times 945 = 1890\\,kJ$         - $12 \\times (O-H) = 12 \\times 464 = 5568\\,kJ$         - Total out = $7458\\,kJ$     - Overall energy change: $6186 - 7458 = -1272\\,kJ$.
• Exothermic Nature: The reaction is exothermic because the energy released when making new bonds ($7458\\,kJ$) is greater than the energy taken in to break existing bonds ($6186\\,kJ$).

Question 6: Chemical Cells and Reactivity

• Cell Investigation:     - Control Variables: Volume of electrolyte, concentration of electrolyte, temperature of solution, distance between electrodes.
• Reactivity Order Based on Voltage (Table 4):     - Magnesium: $+2.71\\,V$ (Most reactive)     - Cobalt: $+0.62\\,V$     - Nickel: $+0.59\\,V$     - Tin: $+0.48\\,V$     - Copper: $0.00\\,V$     - Silver: $-0.46\\,V$ (Least reactive)
• Justification: The higher (more positive) the voltage when measured against a standard electrode (copper in this case), the more reactive the metal is.
• Greatest Voltage Pair: The pair with the largest difference in reactivity would be Magnesium and Silver (Silver has the most negative value, Magnesium the most positive).
• Hydrogen Fuel Cells: Converted trains are called "steam trains" because the only product of the reaction in a hydrogen fuel cell is water ($H_2O$), which is released as steam.

Question 7: Electrolysis

• Aluminium Production:     - Cryolite: Mixed with aluminium oxide to lower the melting point, reducing the energy/cost required for the process.     - Negative Electrode (Cathode): Aluminium ions ($Al^{3+}$) gain three electrons to form aluminium atoms (Reduction).     - Positive Electrode (Anode):         - Half-equation: $2O^{2-} \\rightarrow O_2 + 4e^-$.         - Replacement: The positive electrodes are made of carbon. The oxygen produced reacts with the carbon at high temperatures to form carbon dioxide gas, causing the electrodes to wear away.
• Mass Calculations:     - Equation: $2Al_2O_3 \\rightarrow 4Al + 3O_2$.     - $M_r$ of $Al_2O_3 = (2 \\times 27) + (3 \\times 16) = 102$.     - Moles of $Al_2O_3$ in $2000\\,kg$: $\frac{2,000,000\\,g}{102 \\,g/mol} \\approx 19607.8\\,moles$.     - Ratio is $2:3$, so moles of $O_2 = \\frac{3}{2} \\times 19607.8 = 29411.7\\,moles$.     - Mass of $O_2 = 29411.7 \\times 32 = 941,174\\,g \\approx 941\\,kg$.
• Sodium Chloride Electrolysis:     - Sodium metal cannot be produced from aqueous sodium chloride because hydrogen is produced at the negative electrode instead, as hydrogen is less reactive than sodium.
• Chlorine Gas Volume Calculation:     - Mass = $150\\,kg = 150,000\\,g$.     - Moles of $Cl_2 = \\frac{150,000}{71} \\approx 2112.7\\,moles$.     - Volume at room temperature = $2112.7 \\times 24.0\\,dm^3/mol = 50704\\,dm^3$.

Question 8: Oxides of Copper

• Reduction Experiment (Hydrogen):     - Step 8 (Repeat heating): Required to ensure that all the oxygen has been removed and the reaction is complete (attaining constant mass).     - Burning Excess Hydrogen: Hydrogen is highly flammable and explosive when mixed with air; burning it safely prevents buildup in the lab.
• Results Analysis (Table 5):     - Mass of oxide = $115.47 - 105.72 = 9.75\\,g$.     - Mass of copper produced = $114.38 - 105.72 = 8.66\\,g$.     - Mass of water produced (from Tube B) = $123.38 - 120.93 = 2.45\\,g$.
• Formula Determination:     - Using the teacher's second sample: $2.54\\,g$ Cu and $0.72\\,g$ water.     - Moles of $Cu = \\frac{2.54}{63.5} = 0.04\\,mol$.     - Moles of $H_2O = \\frac{0.72}{18} = 0.04\\,mol$.     - Reacting ratio of $Cu:H_2O$ is $1:1$. This matches Equation 2: $CuO + H_2 \\rightarrow Cu + H_2O$.

Question 9: Enthalpy and Concentration

• Experiment Design:     - A polystyrene cup is used instead of glass because it is a better insulator, minimizing heat loss to the surroundings for more accurate temperature change readings.
• Data Interpretation (Table 6 & Figure 11):     - Initial Temperature: 18.9\\,^\\circ C.     - Maximum Temperature: Reached when the volume of acid added neutralizes the alkali completely.     - Volume of acid for neutralization: Determined by the intersection of the two lines of best fit on the graph.
• Concentration Calculation:     - $15.5\\,cm^3$ of $0.500\\,mol/dm^3\\,H_2SO_4$ reacts with $25.0\\,cm^3\\,KOH$.     - Moles $H_2SO_4 = 0.500 \\times (15.5/1000) = 0.00775\\,mol$.     - Equation: $2KOH + H_2SO_4 \\rightarrow K_2SO_4 + 2H_2O$.     - Moles $KOH = 2 \\times 0.00775 = 0.0155\\,mol$.     - Concentration $KOH (mol/dm^3) = \\frac{0.0155}{0.025} = 0.62\\,mol/dm^3$.     - $M_r$ of $KOH = 39 + 16 + 1 = 56$.     - Concentration $KOH (g/dm^3) = 0.62 \\times 56 = 34.72\\,g/dm^3$.