Look inside the OCR A-Level Chemistry B (Salters) guide (H433)

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OCR · A-Level · H433
Chemistry B
Active Recall Guide
3,430 questions
Chemistry BContents
Contents
13 topics, 219 subtopics
  1. Development of Practical Skills in Chemistry166
  2. Elements of Life267
  3. Developing Fuels364
  4. Elements from the Sea357
  5. The Ozone Story355
  6. What's in a Medicine?235
  7. The Chemical Industry250
  8. Polymers and Life376
  9. Oceans282
  10. Developing Metals304
  11. Colour by Design312
  12. Chemical Literacy41
  13. and 1 more
ii
Elements of LifeQuestions
Elements of Life
The mole, relative masses and empirical formulae
  1. What does the atomic number of an element tell you?
  2. Define the mass number of an atom.
  3. What are isotopes?
  4. State the value of the Avogadro constant with its unit, and use it to find the number of atoms in 0.500 mol of iron.
  5. Two definitions of the mole are accepted in chemistry examinations — what are they?
  6. Define relative isotopic mass.
  7. Define relative atomic mass.
  8. When would you use the term relative formula mass rather than relative molecular mass?
  9. Calculate the amount, in moles, in 5.00 g of calcium carbonate, CaCO₃ (Aᵣ: Ca 40.1, C 12.0, O 16.0).
12
Elements of LifeAnswers
Answers
The mole, relative masses and empirical formulae
  1. It is the number of protons in the nucleus of each atom of that element, and it is what fixes the element's identity and its place in the periodic table.
  2. The mass number is the total number of protons and neutrons in the nucleus of the atom.
  3. Isotopes are atoms of the same element that have the same number of protons but different numbers of neutrons, so they have the same atomic number but different mass numbers.
  4. The Avogadro constant is 6.02 × 10²³ mol⁻¹, so 0.500 mol of iron contains 0.500 × 6.02 × 10²³ = 3.01 × 10²³ atoms.
  5. The classical definition is the amount of substance containing as many particles as there are atoms in exactly 12 g of carbon-12; the revised definition is the amount of substance containing exactly 6.02214076 × 10²³ specified particles, normally rounded to 6.02 × 10²³.
  6. It is the mass of one atom of a particular isotope compared with one twelfth of the mass of one atom of carbon-12.
  7. It is the weighted mean mass of an atom of an element, taking account of the abundance of each isotope, compared with one twelfth of the mass of one atom of carbon-12.
  8. Relative molecular mass is used for substances made of discrete molecules, whereas relative formula mass is used for giant structures such as ionic compounds, where no separate molecule exists and the formula only gives the ratio of ions.
  9. M(CaCO₃) = 100.1 g mol⁻¹, so n = 5.00 ÷ 100.1 = 0.0500 mol.
13
What's in a Medicine?Questions
What's in a Medicine?
Primary, secondary and tertiary alcohols
  1. On what basis are alcohols classified as primary, secondary or tertiary?
  2. Define a primary alcohol.
  3. Define a secondary alcohol.
  4. Define a tertiary alcohol.
  5. How many hydrogen atoms are attached to the carbon bearing the –OH group in a secondary alcohol?
  6. How many hydrogen atoms are attached to the –OH carbon of a tertiary alcohol?
  7. Classify butan-1-ol.
  8. Classify butan-2-ol.
  9. Classify 2-methylpropan-2-ol.
34
What's in a Medicine?Answers
Answers
Primary, secondary and tertiary alcohols
  1. On the number of carbon atoms bonded directly to the carbon atom that carries the –OH group.
  2. A primary alcohol has the –OH group on a carbon atom that is bonded to at most one other carbon atom.
  3. A secondary alcohol has the –OH group on a carbon atom bonded to two other carbon atoms.
  4. A tertiary alcohol has the –OH group on a carbon atom bonded to three other carbon atoms.
  5. One hydrogen atom is attached to that carbon.
  6. None, because all three remaining bonds go to carbon atoms.
  7. Butan-1-ol is a primary alcohol.
  8. Butan-2-ol is a secondary alcohol.
  9. 2-methylpropan-2-ol is a tertiary alcohol.
35
Developing MetalsQuestions
Developing Metals
Manganate(VII) titrations and their calculations
  1. What is the oxidising agent in a manganate(VII) titration, and what is the oxidation state of the manganese in it?
  2. Why does a manganate(VII) titration need no added indicator?
  3. Describe what you see at the end point when MnO₄⁻ is run into acidified iron(II) solution.
  4. Which acid is used to acidify a manganate(VII) titration, and why is hydrochloric acid avoided?
  5. Why is nitric acid also unsuitable for acidifying a manganate(VII) titration?
  6. State the ionic equation for the reaction of manganate(VII) with iron(II) ions in acid solution.
  7. How many moles of iron(II) ions react with one mole of manganate(VII) ions?
  8. A 25.0 cm³ portion of an iron(II) solution, acidified with dilute sulfuric acid, needed 24.60 cm³ of 0.0200 mol dm⁻³ potassium manganate(VII). Calculate the concentration of the iron(II) solution.
  9. A 1.50 g sample of iron ore was dissolved in excess dilute sulfuric acid, all the iron reduced to Fe²⁺, and the solution made up to 250.0 cm³. A 25.0 cm³ portion needed 21.51 cm³ of 0.0100 mol dm⁻³ manganate(VII). Calculate the percentage of iron by mass in the ore.
56
Developing MetalsAnswers
Answers
Manganate(VII) titrations and their calculations
  1. The manganate(VII) ion MnO₄⁻ is the oxidising agent, and the manganese in it has oxidation state +7.
  2. MnO₄⁻ is intensely purple while Mn²⁺ is almost colourless, so the reagent indicates its own end point.
  3. The flask stays colourless or very pale until one drop gives a permanent faint pink colour.
  4. Dilute sulfuric acid is used, because manganate(VII) would oxidise the chloride ions of hydrochloric acid and the titre would be too large.
  5. Nitric acid is itself an oxidising agent, so it would oxidise some of the reducing agent being measured and give a titre that is too small.
  6. The equation is MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O.
  7. Five moles of Fe²⁺ react with every one mole of MnO₄⁻.
  8. The manganate(VII) used is 4.92 × 10⁻⁴ mol, so the Fe²⁺ is five times this, 2.46 × 10⁻³ mol, giving a concentration of 0.0984 mol dm⁻³.
  9. The portion contains 1.076 × 10⁻³ mol Fe²⁺, so the whole sample contains 1.076 × 10⁻² mol, a mass of 0.600 g, which is 40.0% of the ore by mass.
57
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