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Check out what is available? Study the different examples then try the Quizzes!Doc Brown's Chemistry - GCSE/IGCSE/GCE (basic A level) Online Chemical Calculations

ANSWERS to Part 7 - the more advanced mole Q's

This page has the answers to the more advanced mole based questions from calculations section 7. for more advanced chemistry students.

On-line Quantitative Chemistry Calculations

Online practice exam chemistry CALCULATIONS and solved problems for KS4 Science GCSE/IGCSE CHEMISTRY and basic starter chemical calculations for A level AS/A2/IB * If you find spot a silly error please EMAIL query?comment or request for type of GCSE calculation?  ORIGINAL mole Q

Mole calculations introduction * Molar gas volume * Advanced Redox titration Q's * Non-redox titration Q's

Qa7.1 (a) f. mass Al2O3 = 102, 2 ÷ 102 x 3 x 6.02 x 1023 = 3.54 x 1022 oxide ions

(b) f. mass H2 = 2, 3 ÷ 2 x 6.02 x 1023 = 9.03 x 1023 molecules

(c) 1.2 ÷ 24000 x 6.02 x 1023 = 3.01 x 1019 molecules

(d) f. mass Cl2 = 71, 3 ÷ 71 x 6.02 x 1023 = 2.54 x 1022 molecules

(e) Neon exists as single atoms (Ar = 20), 10 ÷ 20 x 6.02 x 1023 = 3.01 x 1023 atoms

(f) 2Na + 2H2O ==> 2NaOH + H2,

1 mole sodium gives 0.5 moles hydrogen,

mole Na = 0.2 ÷ 23 = 0.008696, so mole H2 = 0.008696 ÷ 2 = 0.004348

so volume H2 = 0.004348 x 24000 = 104.3 cm3 or 0.104 dm3

(g) e.g. Mg + 2HCl ==> MgCl2 + H2

1 mole magnesium gives 1 mole hydrogen, mole Mg = 2 ÷ 24.3 = 0.0823

so mole H2 = 0.0823, so volume H2 = 0.0823 x 24 = 1.975 dm3

(h) both 1 mole of Na2CO3 or NaHCO3 will give 1 mole of CO2

(1) VCO2 = mol Na2CO3 x 24000 = 0.76 ÷ 106 x 24000 = 172cm3

(2) VCO2 = mol NaHCO3 x 24000 = 0.76÷ 84 x 24000 = 217cm3

(i) Zn + 2HCl ==> ZnCl2 + H2 , mole H2 = mole HCl ÷ 2

mol HCl = 50 ÷ 1000 x 0.2 = 0.01 mol

so mole H2 = 0.005, VH2 = 0.005 x 24000 = 120cm3

top sub-index(j) CaCO3 + 2HCl ==> CaCl2 + H2O + CO2

mole CO2 = mole HCl ÷ 2, mol HCl = 75 ÷ 1000 x 0.05 = 0.00375 mol

so mole CO2 = 0.001875, VCO2 = 0.001875 x 24000 = 45cm3

OTHER CALCULATION PAGES

  1. What is relative atomic mass?, relative isotopic mass and calculating relative atomic mass

  2. Calculating relative formula/molecular mass of a compound or element molecule

  3. Law of Conservation of Mass and simple reacting mass calculations

  4. Composition by percentage mass of elements in a compound

  5. Empirical formula and formula mass of a compound from reacting masses (easy start, not using moles)

  6. Reacting mass ratio calculations of reactants and products from equations (NOT using moles) and brief mention of actual percent % yield and theoretical yield, atom economy and formula mass determination

  7. Introducing moles: The connection between moles, mass and formula mass - the basis of reacting mole ratio calculations (relating reacting masses and formula mass)

  8. Using moles to calculate empirical formula and deduce molecular formula of a compound/molecule (starting with reacting masses or % composition)

  9. Moles and the molar volume of a gas, Avogadro's Law

  10. Reacting gas volume ratios, Avogadro's Law and Gay-Lussac's Law (ratio of gaseous reactants-products)

  11. Molarity, volumes and solution concentrations (and diagrams of apparatus)

  12. How to do volumetric titration calculations e.g. acid-alkali titrations (and diagrams of apparatus)

  13. Electrolysis products calculations (negative cathode and positive anode products)

  14. Other calculations e.g. % purity, % percentage & theoretical yield, volumetric titration apparatus, dilution of solutions (and diagrams of apparatus), water of crystallisation, quantity of reactants required, atom economy

  15. Energy transfers in physical/chemical changes, exothermic/endothermic reactions

  16. Gas calculations involving PVT relationships, Boyle's and Charles Laws

  17. Radioactivity & half-life calculations including dating materials

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