SPTA Chemistry Department: Unit One NAT 5 Flash Card Answers
Answers for UNIT ONE NAT 5 Flash Cards
1. (a) rate increases (b) rate increases (c) rate increases (d) rate increases
2. Average rate = change in property / change in time
Where property = concentration, volume, mass i.e.
Average rate = ΔC/Δt OR = ΔV/Δt OR = Δm /Δt
3. At 2 hrs concentration = 20mg/l; at 6hrs = 10mg/l
Average rate = ΔC/Δt = 20 – 10 / 6-2 = 10/4 = 2.5 mg/l/hr
4. At 40s = 56cm3 ; at 60s = 73cm3
Average rate = ΔV/Δt = 73 – 56 / 60 – 40 = 6.45cm3 /s
5. (a) alkali metals (b) halogens (c) Noble gases
6. (a) very reactive metals (b) very reactive non-metals
(c) unreactive g non-metal gases
7. transition metals
8. (a) two atoms joined together by a covalent bond.
(b) Hydrogen (H2), Oxygen (O2), Nitrogen (N2), Fluorine (F2), Chlorine (Cl2),
Bromine (Br2), Iodine (I2)
9. An element contains only one type of atom.
10. (a) Only two: Bromine (non-metal) and mercury (metal)
(b) Only eleven: Hydrogen, nitrogen, oxygen, fluorine, chlorine and all of
group 8(0)
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SPTA Chemistry Department: Unit One NAT 5 Flash Card Answers
*
11.
*
.
nucleus
neutron
electron
in
energy level
*
proton
12.
Particle
Proton
Electron
Neutron
Mass (amu)
1
0
1
Charge
+
0
Position in Atom
In nucleus
Outside nucleus
Inside nucleus
13.(a) An atom contains equal numbers of positive protons and negative
electrons.
(b) Positive; as the nucleus contains + protons and neutral neutron making the
overall charge positive.
14. (a) The atomic number = the umber of protons
(b) In the data booklet in the periodic table.
15.(a) Mass number = number of protons + number of neutrons
(b) No, the mass number has to be calculated by adding together the
number of neutrons and protons in an individual isotope of an element.
16. (a) Calcium-40; p = 20; e= 20; n = 20;
40
(b) Carbon-13; p = 6; e = 6; n = 7;
13
(c) Chlorine-35; p = 17; e = 17; n = 18;
35
20Ca
6C
17Cl
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SPTA Chemistry Department: Unit One NAT 5 Flash Card Answers
17. The electronic configuration or arrangement shows how the electrons are
arranged in an atom.
18. (a) 2 (b) 8 (c) 8 (18)
19(a) sodium = 2,8,1
(b) Fluorine = 2,7
(c) Calcium = 2,8,8,2
20. They have the same number of OUTER electrons.
21. The OUTER energy level.
22. They have the same number of outer electrons or they are in the same
group in the periodic table.
23. (a) A and B (are in the same group)
(b) (i) D (noble gas)
(ii) E (group 7 – chlorine)
(iii) A or B (group 1 alkali metal)
24. An ion is a charged particle formed when an atom loses or gains
electrons.
25. (a) Metal atoms lose electrons to gain a stable electron arrangement.
(b) Metals form a positive ion.
26. (a) Non- metal atoms gain electrons to gain a stable electron
Arrangement.
(b) Non-metals form a negative ion.
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SPTA Chemistry Department: Unit One NAT 5 Flash Card Answers
27. (a)
23
+
11Na
= 1+ ion so the metal sodium has lost one electron.
P = 11
e = 10 (2,8)
n = 12
(b) 32 16S2- = 2- ion so the non-metal sulphur has gained two electrons.
P = 16
e = 18 (2,8,8)
n = 16
28. Isotopes are atoms of the same element which have the same atomic
number but a different mass number.
29. (a) The relative atomic mass is the average mass of all the isotopes in an
element. The mass is calculated relative to carbon-12.
(b) The relative atomic mass is found in the data booklet on p7.
30. It is an average.
31. There is more 11H isotopes or 11H is more abundant.
32.
79
35Br
and 8135Br are of equal abundance or in equal proportion.
33. A covalent bond is formed by the sharing of electrons.
34.
+
+
Positive nucleus
positive nucleus
Shared pair of electrons
A covalent bond is formed due to the attraction of the positive nuclei of
each atom for the shared pair of electrons in the overlap region.
35. Non-metals
36. Covalent discrete molecular and covalent network
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SPTA Chemistry Department: Unit One NAT 5 Flash Card Answers
37. (a) Any of group seven (except astatine) and hydrogen.
e.g. Cl—Cl or H—H
(b) Oxygen only: O = O
(c) Nitrogen only: N ≡ N
38. Chlorine is Cl—Cl and has an electron arrangement of 2,8,7
It is the third energy level which forms the covalent bonds.
OR
39. Oxygen is Cl—Cl and has an electron arrangement of 2,6
It is the 2nd energy level which forms the covalent bonds.
O
O
OR
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SPTA Chemistry Department: Unit One NAT 5 Flash Card Answers
40. Carbon has an electron arrangement of 2,4 and hydrogen is 1.
For Carbon it is the third energy level which forms the covalent bonds.
For hydrogen it is the first energy level which forms the covalent bonds.
OR
41. (a) H—Cl
42. (a)
(b) linear
O
(b) Planar or bent
H
43. (a)
H
44(a)
H
N
H
H
H
H
C
(c) Pyramidal
(d) Tetrahedral
H
H
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SPTA Chemistry Department: Unit One NAT 5 Flash Card Answers
45. (a) Low; as covalent bonds are NOT broken – only weak forces between
the molecules are broken.
(b) No, it is a covalent molecule so does not have any free moving
(delocalised) charged particles.
46. (a) High; as strong covalent bonds are broken on melting.
(b) No, it is a covalent network so does not have any free moving
(delocalised) charged particles.
47. Carbon dioxide (CO2) is a covalent molecule with a low melting
point/boiling point because only weak forces between the molecules are
broken, so CO2 exists as a gas.
Whereas silicon dioxide (SiO2) is a covalent network with a high melting
point/boiling point because strong covalent bonds between the atoms are
broken, so SiO2 exists as a solid.
48. (a)
(b) Yes, because it has delocalised electrons.
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SPTA Chemistry Department: Unit One NAT 5 Flash Card Answers
49.(a)
(b) No, it does not have any delocalised charged particles.
50. (a) Ionic bond is the electrostatic attraction between + and – ions .
(b) They are formed when by the complete transfer of electron(s) from the
metal to the non-metal forming + and – ions.
51. A simple diagram of an ionic lattice:
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SPTA Chemistry Department: Unit One NAT 5 Flash Card Answers
52. (a) High
(b) Not as a solid because the ions are not free to move; the ions are
trapped in the crystal lattice.
Conduct as a melt (liquid) or in solution (dissolved in water) because
the ions are free to move.
53. (a) B
(b) C
(c) A
54.
Discuss the following: (put only the first statement on your card, then whole
answer in revision jotter)
 What is its physical state?
(If a liquid or a gas then it must be a covalent molecule, as covalent
molecules have low melting points and boiling points because only weak
forces between the molecules are broken)
 Test the substance for electrical conductivity, as a solution and as a
solid if appropriate.
If the substance does not conduct in any state it is covalent. Covalent
substance do not have charged particles to carry the current.
If the substance conducts in solution (but not as a solid) it is ionic.
Ionic substance are made up of ions which are free to move in solution
but not as a solid. The free moving ions in solution can conduct
electricity.
55.(a) No
(b) The melting points do not compare ionic bonds with covalent bonds
because when a covalent molecule melts it is the weak forces between
the bonds which break NOT covalent bonds.
56. (a) Shows the ratio of atoms in the compound.
(b) Shows the ratio of ions in the compound.
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SPTA Chemistry Department: Unit One NAT 5 Flash Card Answers
57 (a) NH4NO3
(b) CuCl
(c) Al(OH)3
(d) Fe2(SO4)3
(e) NaHCO3
58. (a) (Ca2+)3(PO43-)2
(d) (Fe3+)2(SO4 2-)3
(b) (NH4+)2S2-
(c) Ba2+(OH-)2
(e) Mg2+(HSO4-)2
59.(a) Mole = formula mass expressed in grams.
(b)
m
n
GFM
60. (a) GFm of CaCl2 = 40 + (2 x35.5) = 111g
m = n x GFm = 2.5 x 111 = 277.5g
(b) GFm of Na2SO4 = (2x23) + 32 + (4 x 16) = 142g
m = n x GFm = 1.5 x 142 = 213g
(c) GFm of Mg3(PO4)2 = (24.5 x 3) + (31 x 2) + (16 x 8) = 263.5g
m = n x GFm = 5 x 263.5 = 1317.5g
61. (a) GFm of NaNO3 = 23 + 14 + (16 x 3) = 85g
n = m / GFm = 32 / 85 = 0.38 moles
(b) GFm of SiO2 = 28 + (16 x 2) = 60g
n = m / GFm = 100 / 28 = 3.6 moles
(c) GFm of Ca(OH)2 = 40 + (16 x 2) + (1x2) =74g
n = m / GFm = 250 / 74 = 3.4 moles
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SPTA Chemistry Department: Unit One NAT 5 Flash Card Answers
62.
(a) C5H12 + 8O2
5CO2 + 6H2O
(b) 2P + 3Cl2
2PCl3
(c) N2 + 3H2
2NH3
(d) C2H5OH + 3O2
2CO2 + 3H2O
63.
Mole statement: 1 mole C2H6
Change to mass:
Proportion sum:
2 moles CO2
30g
88g
60g
X
60 x 88 = 30 x X
X = 60 x 88 / 30
= 176g
64. Balanced equation: S + O2
SO2
Mole statement: 1 mole S
Change to mass:
Proportion sum:
1 mole SO2
32g
64g
32kg
64kg
25kg
X
25 x 64 = 32 x X
X = 25. x 64 / 32
= 50kg
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SPTA Chemistry Department: Unit One NAT 5 Flash Card Answers
65.
(a)(i) pH less than 7
(ii) pH greater than 7
(iii) pH = 7
(b) (i) red (pH= 1,2) to orange (pH = 3,4) to yellow (pH = 5,6)
(ii) blue/green (pH= 8,9) to purple (pH = 10-14)
(iii) green = pH =7
66. (a) Put the pH paper into sample and then compare colour to pH chart to get
accurate PH.
(b) Same as (a) only you would wet the pH paper first.
67. Sodium ethanoate has a lower pH than sodium hydroxide i.e. it is not as alkaline.
(or sodium hydroxide is a stronger alkali than sodium ethanoate.)
68. Hydrochloric acid has a lower pH than ethanoic acid i.e. hydrochloric acid is
more acidic. (or hydrochloric acid is a stronger acid)
69. Copper oxide is insoluble in water; the solution turned green because it was the
pH of the water which was measured not copper oxide.
70. (a) A base is a substance which reacts with an acid.
(b) metal oxide; metal hydroxide and metal carbonate
71(a) Acids: hydrochloric (HCl); Nitric (HNO3); Sulphuric (H2SO4)
(b) Alkalis: Ammonium hydroxide (NH4OH); Potasium hydroxide (KOH)
Sodium hydroxide (NaOH)
72. (a) Salt and water
(b) neutralisation
73. (a) sodium chloride
(b) ammonium sulphate
(c) potassium nitrate
74. (a) Calcium nitrate + water
(b) Calcium nitrate + water
(c) Calcium nitrate + water + carbon dioxide
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SPTA Chemistry Department: Unit One NAT 5 Flash Card Answers
75.(a) CuO
+
2HNO3
Cu(NO3)2
(b) 2NaOH + H2SO4
+
H2O
Na2SO4 + 2H2O
76 (a) hydrogen (H+) and hydroxide (OH-))
(b) H2 O(l)
H+(aq) + OH-(aq)
77. pH is a measure of the hydrogen ion, H+(aq) concentration.
78.(a) A neutral solution has a pH = 7 because
[H+] = [OH-]
i.e. the hydrogen ion concentration = hydroxide ion concentration.
(b) An acid solution has a pH less than 7 because
[H+] › [OH-]
i.e. the hydrogen ion concentration is greater than the hydroxide ion
concentration.
(c) An alkali solution has a pH greater than 7 because
[H+] ‹ [OH-]
i.e. the hydrogen ion concentration is less than the hydroxide ion
concentration
79. When acid is diluted the pH increases towards pH 7.
This is because the hydrogen ion concentration is decreasing.
80. When an alkali is diluted the pH decreases toward pH 7
This is because the hydroxide ion concentration is decreasing.
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SPTA Chemistry Department: Unit One NAT 5 Flash Card Answers
81. Soluble metal oxides form alkaline solutions because hydroxide ions are
produced which means [OH-] › [H+] giving a pH greater than 7.
82. Soluble non-metal oxides produce acidic solutions because H+ ions are
produced which means [H+] › [OH-] giving a pH less than 7.
83. (a) A spectator ion does not take part in the reaction.
(b) The spectator ion in a neutralisation reaction are the salt ions, so in
this example NaCl(aq) which contain the Na+ and the Cl- ion.
84.
H+(aq)
OH-(aq)
H2O(l)
85.
2H+(aq) +
O2-(s)
H2O(l)
86.
2H+(aq) +
CO32-(s)
H2O(l) + CO2(g)
+
87. The spectator ions in a precipitation reaction is the soluble salt in the
products, in this example, KNO3(aq) i.e. potassium ion, K+ and the
nitrate ion, NO3-(aq).
88. The spectator ions in a precipitation reaction is the soluble salt in the
products, in this example, Ca(NO3 )2(aq) i.e. calcium ion, Ca2+ and the
nitrate ion, NO3-(aq).
89. (a) Titration is the technique used to find out by experiment the
concentration of an acid or an alkali.
(b) The end point is a colour change which indicates the end of the
reaction.
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SPTA Chemistry Department: Unit One NAT 5 Flash Card Answers
90.(a) n = 3.5 moles
c = 2 moll-1
v=?
Use triangle:
v
n
c
v = n / c = 3.5 / 2 = 1.75 litres
(b) n = 1.25 moles v = ?
c = 0.01 moll-1
Use triangle:
n
v
c
v = n / c = 1.25 / 0.01 = 125 litres
91. (a) n = ?
c = 2 moll-1
Use triangle:
v = 100cm3 = 0.1 litres
v
n
c
n = v x c = 0.1 x 2 = 0.2 moles
(b) n = ?
Use triangle:
c = 0.1 moll-1
v = 250cm3 = 0.25 litres
v
n
c
n = v x c = 0.25 x 0.1 = 0.025 moles
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SPTA Chemistry Department: Unit One NAT 5 Flash Card Answers
92. (a) Work out moles first using n = v x c
n = v x c = 1 x 0.05 = 0.05 moles
Work out mass using m = n x GFM; GFM of NaOH = 23+16+1 = 40g
m = 0.05 x 40 = 2g.
(b) Work out moles first using n = v x c
n = v x c = 0.1 x 0.1 = 0.01 moles
Work out mass using m = n x GFM; GFM of KCl = 39+ 35.5 = 74.5g
m = 0.01 x 74.5 = 0.745g.
93.
NaOH: v = 50cm3 = 0.05 litres; c = 0.2 mol l-1
H2SO4: v = ?; c = 2 mol-1
Moles of NaOH = 0.05 x 0.2 = 0.01 moles
From balanced equation, NaOH:H2SO4 = 2:1
So moles of H2SO4 = 0.01 /2 = 0.005 moles
Use n = v x c triangle to calculate volume:
V = n / c = 0.005 / 2 = 0.0025 litres
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SPTA Chemistry Department: Unit One NAT 5 Flash Card Answers
94.
KOH: v = 10cm3 = 0.01 litres; c = 0.1 mol l-1
HNO3 : v = 25cm3 = 0.025 litres ; c = ?
Moles of KOH = 0.01 x 0.1 = 0.001 moles
From balanced equation, KOH:HNO3 = 1:1
So moles of HNO3 = 0.001
Use n = v x c triangle to calculate concentration:
C = n / v = 0.001 / 0.025 = 0.04 mol l-1.
95. Open Ended Question
Put this on a separate piece of paper to get full answer.
A metal oxide if soluble dissolves in water to produce hydroxide ions, (OH -)
which means [OH-] becomes greater than [H+] in solution. This gives a pH
greater than 7 i.e. alkaline. However, if the metal oxide is insoluble no extra
hydroxide ions are produced leaving water where [H+] = [OH-] which gives a
neutral solution, so pH = 7. So not all metal oxides produce an alkali, they
have to be soluble.
A non- metal oxide if soluble dissolves in water to produce hydrogen ions,
(H+) ) which means [H+] becomes greater than [OH-] in solution. This gives a
pH greater less 7 i.e. acidic. However, if the non-metal oxide is insoluble no
extra hydrogen ions are produced leaving water where [H+] = [OH-] which
gives a neutral solution, so pH = 7. So not all non-metal oxides produce an
acid they have to be soluble.
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SPTA Chemistry Department: Unit One NAT 5 Flash Card Answers
96. Open Ended Question
Put this on a separate piece of paper to get full answer.
An ionic compound is made up of a giant crystal lattice held together by the
electrostatic attraction of + ions for – ions.
A typical ionic compound is a solid at room temperature, has high melting
points and boiling points, conducts electricity in solution and as a melt but
does not conduct as a solid. So sodium chloride would be tested to see if it
conducts as a melt/ solution and Not as a solid. Would also look at its
physical state and see if it’s a solid.
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