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Summary Chemistry BMAT, Section 2
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Full summary of all expected chemistry knowledge for section 2 of the BMAT.
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C1 Atomic Structure 1C2 The Periodic Table 3
C3 Chemical reactions, formulae and equations 4
C4 Quantitative chemistry 7
C5 Oxidation, reduction and redox 7
C6 Chemical bonding, structure and properties 9
C7 Group chemistry 11
C8 Separation techniques 12
C9 Acids, bases and salts 14
C10 Rates of reactions 15
C11 Energetics 16
C12 Electrolysis 18
C13 Carbon/Organic chemistry 18
C14 Metals 21
C15 Kinetic/Particle theory 23
C16 Chemical tests 24
C17 Air and Water 25
,C1 Atomic Structure
1
Atom contains 3 subatomic particles → protons and neutrons (nucleus = 20000 of atom size),
electrons
particle relative mass relative charge
proton 1 +1
neutron 1 0
electron negligible -1
Atom loses electrons → more protons than electrons → positively charged ion
Atom gains electrons → more electrons than protons → negatively charged ion
Mass number = protons + neutrons
Atomic number = amount of protons
/amount of electrons ( - charge)
To determine electron configuration → determine amount of electrons → fill shells with
maximum amount of electrons
shell maximum number of electrons
1st 2
2nd 8
1
, 3rd 8
Isotope = Atom has the same amount of protons ( same element) but different amount of
neutrons (different mass number)
Mass spectrum:
There are two peaks in the above mass spectrum of boron → two isotopes of boron with m/z
values of 10 and 11 (mass numbers are 10 and 11)
The ratio of the peaks for m/z 10:11 is 1:4 → four times as many atoms of boron with mass
11 10
of 11 as mass of 10 → 80% of the atoms are 5
B and 20% are 5
B
Relative atomic mass (Ar) = weighted mean of the mass numbers of the isotopes of an
element (the abundances of the various isotopes of the element are taken into account when
calculating the average)
35 37
For example, in a sample of chlorine atoms 75% are Cl and 25% are
17
Cl
17
75 25
Ar (Cl) = ( 100 x 35)+( 100 x 37)
3550
Ar (Cl) = 100
= 35.5
q r s
With data presented as percentages: a% of X, b% of X, c% of X, … the element
has a relative mass of:
(𝑎𝑎×𝑞𝑞)+(𝑏𝑏×𝑟𝑟)+(𝑐𝑐×𝑠𝑠)+...
Ar (X) =
100
With data presented on a mass spectrum, the relative numbers of the various isotopes are
obtained from values on the y-axis: a the value for qX, b the value for rX, c the value for sX,
... the expression becomes:
(𝑎𝑎×𝑞𝑞)+(𝑏𝑏×𝑟𝑟)+(𝑐𝑐×𝑠𝑠)+...
Ar (X) =
(𝑎+𝑏+𝑐+...)
2
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