OCR Chemical Ideas Answer PDF File : For Checking Only

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SECTION 1
Section 1.1
M 1 a 2.0 b 5.3 c 1.3 d 10.0 e 5.0 b The relative number of moles of iodine and oxygen.
f 50.2 c To change the relative number of moles into the ratio
v 2 a 144 b Neodymium
of moles of oxygen relative to 1 mole of iodine.
d In order to produce a ratio involving whole numbers
3 e I2O5, I4O10, I6O15, etc.
?
Mass of Amount of Number of
sample/g sample/mol atoms f The molar mass is needed.
­ ­ 6.02 ¥ 1023 7 a H2O b CO c CS2 d CH4 e Fe2O3
­ 2.00 ­
56.0 ­ 6.02 ¥ 1023 f CuO g CaO h SO2 i MgH2
80.0 ­ 12.04 ¥ 1023 8 a 92.3 b 7.7 c CH
63.5 0.50 ­
v
9 a SiH4 b CO c CO2 d MgO e C2H6O
4 a 1 b 0.5 c 0.25 d 0.1 e 0.25
f CaCO3 g HClO3 h NaHCO3
f 0.5 g 0.25 h 0.1 i 2 j 5
10 a CH2 b P2O3 c AlCl3 d BH3 e C4H5
Atoms of copper are approximately twice as heavy as
f C3H4 g CH2O h C12H22O11
atoms of sulphur. Thus the same mass contains only half
as many moles of copper as it does of sulphur. 11 a H2O2 b CO c C2H2 d C6H6 e C6H12
5 Black copper(II) oxide (CuO) contains equal numbers of 12 a 2 b 11 c 2 d 10 e 2
copper and oxygen particles (Cu2+ and O2­ ions). Red 13 a 30 b 78 c 130 d 100 e 158
copper(I) oxide (Cu2O) contains twice as many copper f 242 g 132
particles as oxygen particles (Cu+ and O2­ ions).
14 a 2 b4 c 10 d 0.02 e 5
6 a The mass of the sample is needed to be sure that f 1 ¥ 106
iodine and oxygen are the only elements in the
compound.
Section 1.2
1 a 2Mg + O2 Æ 2MgO 2 a 2Ca + O2 Æ 2CaO
b 2H2 + O2 Æ 2H2O b Ca + 2H2O Æ Ca(OH)2 + H2
c 2Fe + 3Cl2 Æ 2FeCl3 c C + CO2 Æ 2CO
d CaO + 2HNO3 Æ Ca(NO3)2 + H2O d N2 + 3H2 Æ 2NH3
e CaCO3 + 2HCl Æ CaCl2 + CO2 + H2O e C3H8 + 5O2 Æ 3CO2 + 4H2O
f H2SO4 + 2NaOH Æ Na2SO4 + 2H2O
3 a Zn(s) + H2SO4(aq) Æ ZnSO4(aq) + H2(g)
g 2HCl + Ca(OH)2 Æ CaCl2 + 2H2O
b Mg(s) + 2HCl(aq) Æ MgCl2(aq) + H2(g)
h 2Na + 2H2O Æ 2NaOH + H2
c MgCO3(s) Æ MgO(s) + CO2(g)
i CH4 + 2O2 Æ CO2 + 2H2O
d 2C2H6(g) + 7O2(g) Æ 4CO2(g) + 6H2O(l)
j 2CH3OH + 3O2 Æ 2CO2 + 4H2O
e BaO(s) + 2HCl(aq) Æ BaCl2(aq) + H2O(l)
Section 1.3
1 a All the magnesium reacts. 6 a 56 tonnes
b So that we know the number of moles of each b S + O2 Æ SO2
substance involved in the reaction. c 64 g
c Mass of 1 mole of magnesium oxide. d 64 tonnes
d Because 2 moles of magnesium oxide are produced. e 2 tonnes
e To find the mass of magnesium oxide produced from f 112 tonnes
1 g of magnesium.
7 a 217 tonnes
f 80/48 would be multiplied by 50 rather than by 6.
b 0.58 tonnes
2 20 g
8 a Fe2O3 + 3CO Æ 2Fe + 3CO2
3 a 2.8 g b 3.1 g c 2.5 g b 160 g
c 1.43 g
4 3667 g (3.667 kg)
d 1.43 tonnes
5 a C8H18 + 12.5O2 Æ 8CO2 + 9H2O e 2.86 tonnes
b 175 kg f 11.9 tonnes
c 154 kg
162

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SECTION 2
Section 1.4
M 1 The particles in a gas are much further apart than in a 3 a H2(g) + Cl2(g) Æ 2HCl(g)
liquid or solid. In a gas, therefore, the volume of the b Volumes of hydrogen and chlorine are the same.
v particles is a very small part of the total volume and does
not significantly affect it. In a liquid or solid the particles
Volume of hydrogen chloride is twice the volume of
hydrogen or chlorine.…read more

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SECTION 2
Section 2.…read more

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SECTION 3
c 2+ X 2­ 5 a In a normal covalent bond, each atom supplies a single
M
X
Mg X O X electron to make up the pair of electrons involved in
XX
the bond. In a dative covalent bond one atom supplies
2+ XX ­ XX ­ both electrons.…read more

Page 5

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SECTION 3
Section 3.2
M 1 a Li(g) because it has an extra electron shell. 4 a Na+
b Li+(aq) because the ion is surrounded by water b Na+Cl­
v molecules.
c Cl­(g) because it has an extra electron.
c Na+Cl­
d Cl­(aq) because the ion is surrounded by water 5 a i 11, 12, 13, 15, 16, 17
? molecules.…read more

Page 6

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SECTION 3
6 a XX ­
M
X X
H N H About 109°
XX
b XX +
v
X X
H N H About 120°
?
Section 3.…read more

Page 7

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SECTION 3
Section 3.5
M 1 F F 4 aF H
C C cis-1,2-difluoroethene C C
v H H F H
F H b No
? C C trans-1,2-difluoroethene
c There must be the same two atoms or groups of atoms
on each C atom attached to the double bond.
H F Geometric isomers are not possible if two groups on
v
one side of the double bond are the same.…read more

Page 8

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SECTION 4
6 a H H H H b H3C
M H3C C C H3C C C CH CH3
H H H3C
C *C C CH3 C *C C CH3
v H C H C
H H
C H C H C
H
C
? H H H H H O
limonene carvone c No: the carbon which was chiral is
now bonded in the same way in both
v
directions around the ring.
v
Section 4.…read more

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SECTION 4
c DH1 + DH3 = DH2 13 a CH3CHO(l) +2 O2(g) Æ 2CO2(g) + 2H2O(l)
M DH1 = DH2 ­ DH3
= 4(­393) kJ mol­1 + 5(­286) kJ mol­1
b
DH1
CH3CHO(l) + 2 O2(g) 2CO2(g) + 2H2O(l)
­ (­2877) kJ mol­1
v = ­125 kJ mol­1
DH2 DH3
12 a Enthalpy change of combustion of methane.
? b Enthalpy change of formation of methane.…read more

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SECTION 4
Section 4.4
M 1 The entropies increase for the first four alkanes as the 3 Students' answers should be based on the following
molecules become heavier and composed of more atoms deductions.
v (the number of energy levels increases with the number
of atoms). Pentane is a liquid and so has a lower entropy
DSsys
/J K­1 mol­1
DSsurr
/J K­1 mol­1
Explanation
than butane.…read more

Comments

Adeel


is this for A level or GCSE?

Adeel


Do you have the exam without the answers aswell? I would like to do it

Tim Birkinshaw

There are at least two mistakes in this booklet; the answers for questions 6 & 7 in Section 10.3 are missing as are all the answers for Section 10.4 (2 questions).

The Sections labelled 10.4 and 10.5 should be 10.5 and 10.6 respectively.

Anyone got the correct answers for the missing questions or know where to find them?

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