Gr. 10 Physical Science based upon Andries Oliviers' Textbook with input from DocScientia

1. Particles that substances are
made up of

2. Atoms
Basic building blocks
of matter.
Element
Mono-atomic elements
Metals – giant structures
Diatomic molecules
Diatomic molecular elements
Polyatomic molecular elements

3. Compounds
Two or more atoms of different elements:
Molecular compound
Covalent network structures
Ionic compounds
Metallic structure
Fixed proportions
Electrically neutral

4. Investigating elements and
compounds

5. Hydrogen gas

6. Carbon dioxide gas

7. Dehydration of copper sulfate

8. Dehydration of copper sulfate

9. Electrolysis of water

10. Representation of elements and compounds
Molecular formula of a compound
H2O – actual number and type of atoms
in the molecule.
Empirical formula of a compound
NaCl – the simplest formula.
Strucutural formula of a compound
O = C = O – shows arrangement between
atoms in a compound.

11. Diagrams of elements and compounds

12. Covalent structures
Electrons shared.
Non-metals.
Form molecules.

13. Covalent molecular solids
Solids = molecules
that are bonded in a
molecular lattice.
Low boiling and
melting points.
Often sublimates.

14. Covalent network structure
When bonds extend throughout the
structure of the substance to form a giant
covalent network.
These solids are very stable and have high
melting and boiling points.
Diamond and graphite are two allotropes of
carbon. Allotropes are when the same
element occur in different crystal forms.

15. Diamond
Colourless transparent crystal.
Hard – strong covalent bonds (4)
Very high M.P. and B.P.
Cannot conduct electricity – no free e- or
ions.

16. Graphite
Soft/slippery – form layers held together
by weak intermolecular forces.
Conduct electricity due to one free e-.

17. Silica compound
Every silicon atom bonds covalently to four
oxygen atoms.

18. Boron compounds
BN share properties similar to graphite and
diamond, depending on its molecular
structure.

19. Uses of diamond, graphite and
silica
SUBSTANCE PROPERTIES USES
DIAMOND Hardest known substance In tools to cut and drill
Very high M.P. and B.P. Precious stones
Does not conduct Jewellery
electricity
GRAPHITE Soft and slippery Lubricants for engines
and locks
High M.P. and B.P. Lead of pencils
Conducts electricity Electrodes and
connectors in generators
SILICA Hard, used to sand On sand paper
objects
High M.P. And B.P. Make glass and lenses
Does not conduct In bricks to line furnace
electricity ovens.

20. Properties of covalent
compounds
Low melting and boiling points due to weak
intermolecular forces.
Covalent network structures have very high M.P.
and B.P due to the strong covalent bonds between
atoms.
Covalent structures are poor conductors of
electricity and heat – they are insulators because
they have no free electrons.

21. Ionic structures
Electrons transferred
Metal and non-metals
Forms crystal lattice

22. Complete the ion and metal
stuff

23. Chemical bond
Ionic
Covalent
Strong electrostatic forces between ions
Forces between atoms inside of
are broken. Requires large amounts of
molecules are broken
energy to break.
Rearrange atoms/molecules. Rearrange ions.
Energy is released
New bond, new New bond, new
substance substance
Only reversible in certain circumstances.
www.docscientia.co.za

24. Physical and chemical change

25. Physical change
Composition does not change.
Properties – appearance, form and state changes.
A physical change is a change where no new
chemical compounds are formed. The state may
change, but the identity does not.

26. Physical
Arrangement
Reversibility
Conservation of mass, atoms and molecules
Energy involved

27. Melting Evaporation
Absorbs energy Absorbs energy
Releases energy Releases energy
Solidification
Sublimation
Arrangement of particles
www.docscientia.co.za

28. Physical
Reversibility
Generally easily reversed

29. Physical
Conservation of mass, atoms, and molecules
Matter cannot created or destroyed, only changed.
Therefore mass stays the same after a physical
change, as does the amount of atoms and
molecules.

30. Physical changes
Energy involved
Physical changes can be caused by small
energy changes.
Only intermolecular forces break during physical
changes.

31. Chemical changes
Arrangement of the particles
Atoms, molecules or ions regroup.
Decomposition reaction
Synthesis reaction

32. Chemical changes
Reversible
A few are reversible, but most are difficult to
reverse.

33. Chemical changes
Conservation of mass, atoms and molecules
The quantity of a specific subsance can
change, but total amount of atoms stay the same.

34. Chemical changes
Energy involved
Absorbed or released during chemical reactions.
Much more energy involved.
Intramolecular forces have to be broken and
re-formed.

35. Decompostion and synthesis
reactions

36. Decompostion reactions
AB → A + B

37. Decompostion reactions

38. Synthesis reactions
A + B → AB

39. Synthesis reactions

40. Energy changes in chemical
reactions
Total energy absorbed to break bonds are less
than the energy required to form new bonds.
Heat is given off to the environment.
EXOTHERMIC REACTION

41. Energy changes in chemical
reactions

42. Energy changes in chemical
reactions
Total energy absorbed to break bonds are more
than the energy released to form new bonds.
Heat is absorbed from the environment.
ENDOTHERMIC REACTION

43. Total energy Total energy released
Energy transferred
absorbed to break when new bonds form
bonds
Product: warmer –=exothermic reaction -
colder – endothermic reaction
Chemical reaction
Exothermic Endothermic
reaction reaction
To calculate the 'energy required', you need the following:
• Amount of Ep that is necessary to break bonds, as well as
• Amount of Ep that is released after bonds form.
Energy values shown as kJ⋅mol-1.
www.docscientia.co.za

44. Conservation of atoms and
mass

45. Law of conservation of mass
Laviosier: Matter cannot be created or destroyed.
Reactant mass = Product mass

46.
During a chemical reaction or a physical change the sum of the reactants are
equal to the sum of the mass of the products.
Atoms in reactants = Atoms in products
Amount of atoms (reactants) = Amounts of atoms
(products)
Mass before = Mass after
Consider the following balanced equation:
→
2SO2
2(32 + 16 + 16)
+ O2
+ (16 + 16)
2SO3
→ 2(32 + 16 + 16 + 16)
Ar 128 + 32 → 160
Voor: 160 = Na: 160
www.docscientia.co.za

47. Law of conservation of atoms but non-
conservation of molecule in a chemical reaction
During a chemical reaction the number of atoms of
each element stays intact.

48. Law of constant composition
Balanced equation:
Everyting in front of the arrow, is still to be found at the back of the
arrow.
Left side has the same amount of matter as the right side.
A specific chemical reaction always has the same elements in the same ratios.
Any amount:
• Carbon dioxide: carbon and oxygen contains 1 C for every 2 O's.
• Ammonia: nitrogen and hydrogen contains 1 N for every 3 H’s.
• Dinitrogen tetraoxide: nitrogen and oxygen contains 2 N's for
every 4 O's.
Multiple compounds are sometimes possible:
For example: : H2O en H2O2
CO en CO2
www.docscientia.co.za

49. A specific amount of particles of any gas occupies the
same volume at a fixed temperature and pressure.
2SO2(g) + O2(g) → 2SO3(g)
2SO2 + O2 → 2SO3
2 particles SO2 + 1 particle O2 → 2 particle SO3
2 volume-units SO2 + 1 volume-units → 2 volume-units
O2 SO3
2 dm3 SO2 + 1 dm3 O2 → 2 dm3 SO3
2 cm3 SO2 + 1 cm3 O2 → 2 cm3 SO3
6 cm3 SO2 + 3 cm3 O2 → 6 cm3 SO3
www.docscientia.co.za

50. Balancing of equations

51. Balancing of equations

52. Calculating mass
So, we have learned that:
Mass stays the same before and after a
chemical reaction.
The amount of atoms on the left hand side is
the same as the amount of atoms on the
right hand side – they are simply arranged
differently.
All of this led us to balance our chemical
equations.

53. Calculating mass
The balanced equations show us that:
Mass stays the same on the left ans right.
The volumes of gas reactants are the same as
the balanced equations.
All of these facts can help us calculate the
actual mass of the reactants.

54. Calculating mass
So, we can say that:
Mass of reactants and products can be
calculated.
Volume of gas reactions can be calculated.
Actual mass of reactants or products can be
calculated.

55. Calculating mass – proving the
law of conservation of mass
Step 1: Balanced equation
Pb(NO3)2(aq) + 2NaI(ag) → PbI2 + 2NaNO3(aq)
Step 2: Relative mass
REACTANTS
Mr[Pb(NO3)2] = 207 + 2(14 + (16·3)) = 331
Mr(NaI) = 23 + 127 = 150
PRODUCTS
Mr(PbI2) = 207 + (2·127) = 461
Mr(NaNO3) = 23 + 14 + (3·16) = 85

56. Calculating mass – proving the
law of conservation of mass
Step 3: Determine total mass before and after
Pb(NO3)2(aq) + 2NaI(ag) → PbI2 + 2NaNO3(aq)
331 + 150 → 461 + 85
631 → 631

57. Calculating mass – Using the law of
conservation of mass to determine the actual mass of substances.
QUESTION:
Determine the mass of oxygen that is released
When 29,4 g potassium chlorate is heated and
Completely decomposed into potassium and
Oxygen.
Step 1: What is asked, and what is given?
29,4 g KClO3 → ? g O2

58. Calculating mass – Using the law of
conservation of mass to determine the actual mass of substances.
Step 2: Balanced equation
2KClO3 → 2KCl + 3O2
Step 3: Relative mass of substances in step 1
2[39 + 35,5 + (3·16) → 3(2·16)
245 → 96

59. Calculating mass – Using the law of
conservation of mass to determine the actual mass of substances.
Step 4: Use ratios to caculate the mass of O2
245 g KClO3 : 96 g O2
245 g/245 KClO3 : 96 g/245 O2
Which is the same as...
1 g KClO3 : 96/245 g O2
So, we can say that
29,4 g KClO3 : (29,4·96)/245 g O2
= 11,52 g O2
A mass of 11,52 g O2 is formed.

60. Calculating mass – Using the law of
conservation of mass to determine the actual mass of subst ances.
QUESTION:
Determine what mass of sodium will fully react with
chlorine to form 25 g of table salt.

61. Calculating mass – Using the law of
conservation of mass to determine the actual mass of subst ances.
BALANCED EQUATION:
2Na + Cl2 → 2NaCl
WHAT DO WE HAVE?
Na - ? 25 g NaCl
RELATIVE MOLAR MASS
Na – 46 : NaCl – 117

62. Calculating mass – Using the law of
conservation of mass to determine the actual mass of subst ances.
Na – 46 : NaCl – 117
46 g Na : 117 g NaCl
46/117 g Na : 1 g NaCl
25·46/117 g Na : 25 g NaCl
9,83 g Na will yield 25 g NaCl.

63. Calculating volume
Calculate the volume of nitrogen oxide that forms,
should 3 dm3 nitrogen react with oxygen.
Step 1: Balanced equation
N2 + O2 → 2NO
Step 2: substitute particles for volume units
1 dm3 N2 + 1 dm3 O2 → 2 dm3 NO

64. Calculating volume
Calculate the volume of nitrogen oxide that forms,
should 3 dm3 nitrogen react with oxygen.
Step 1: Balanced equation
N2 + O2 → 2NO
Step 2: substitute particles for volume units
1 dm3 N2 + 1 dm3 O2 → 2 dm3 NO