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Title: HIGHER CHEMISTRY REVISION.


1
HIGHER CHEMISTRY REVISION.
Unit 1- The Mole
1. The balanced equation for the
decomposition of hydrogen peroxide into water and
oxygen is 2H2O2 (l) ?
2H2O(l) O2 (g)
24000 cm3 of gas 1 mol of O2 So 40 cm3
0.00167 mol 1 mole of O2 weighs 32g So 0.00167
mol weighs 0.0534g
Using information from the above graph, calculate
the mass of hydrogen peroxide used in the
reaction, assuming all the hydrogen peroxide
decomposed. (Take the molar volume of oxygen to
be 24 litres mol-1)
2
2. . A student electrolysed dilute sulphuric
acid using the apparatus shown in order to
estimate the volume of one mole of hydrogen gas.
(a) Q I x t 0.5 x 14 x 60
420C 2H 2e - ? H2 2 x
96500C ? 1 mol 193000C ? 1 mol
420C gives 52 cm3 So 193000C gives
23 895 cm3 or 23.895
litres (b) Introduce a variable resistor so
that current can be kept constant.
Current 0.5 A Time 14 minutes.
Volume of hydrogen collected 52 cm3. (a)
Calculate the molar volume of hydrogen gas. (b)
What change could be made to the apparatus to
reduce a possible significant source of
error?
3
3. The symbol for the Avogadro Constant is L.
Identify the two true statements.
A 64.2 g of sulphur contains approximately L
atoms. B 16.0 g of oxygen contains
approximately L molecules. C 6.0 g of water
contains approximately L atoms. D 1.0g of
hydrogen contains approximately L protons.
E 2.0 litres of 0.50 mol l-1 sulphuric acid
contains approximately L
hydrogen atoms. F 1.0 litres of 1.0 mol l -1
barium hydroxide solution contains
approximately L hydroxide atoms.
C and D
4. A student heated a compound which gave off
carbon dioxide gas and water vapour.
The volume of carbon dioxide collected was 240
cm3. Calculate the number of molecules in
this volume. (Take the molar volume of carbon
dioxide to be 24 litres mol-1.)
24 litres 24000 cm3. 24000 cm3 contains 6.02 x
1023 CO2 molecules. So 240 cm3 contains
240/24000 x 6.02 x 1023 6.02 x 1021 CO2
molecules.
4
A student added 0.20 g of silver nitrate, AgNO3,
to 25 cm3 of water. This solution was then added
to 20 cm3 of 0.0010 mol l-1 hydrochloric acid as
shown in the diagram The equation for the
reaction which occurs is AgNO3 (aq)
HCl(aq) ? AgCl(s) HNO3 (aq) (a) Name the
type of reaction which takes place. (b) Show by
calculation which reactant is in excess.
5.
  • Precipitation.
  • No of moles of HCl C x V(litres)
    20/1000 x 0.001
  • 2 x 10-5 mol
  • No of moles of AgNO3 mass/gfm
    0.2/169.9
  • 1.2 x 10-3 mol
  • So the silver nitrate is in excess.

5
6. When sodium hydrogencarbonate is heated to
112oC it decomposes and the gas carbon
dioxide is given off 2NaHCO3(s) ?
Na2CO3(s) CO2(g) H2O(g) The
following apparatus can be used to measure the
volume of carbon dioxide produced in the
reaction. (a) Why is
an oil bath used and not a water bath?
(b) (i) Calculate the theoretical volume of
carbon dioxide produced by the complete
decomposition of 1.68 g of sodium
hydrogencarbonate. Take the
molar volume of carbon dioxide to be 23 litres
mol-1) (ii) Assuming that all of
the sodium hydrogen carbonate is decomposed,
suggest why the volume of
carbon dioxide collected in the measuring
cylinder would be less than
the theoretical value.
(a) Water boils at 100oC and so could not
raise the temperature to 112oC. (b) (i)
From equation 2mol of NaHCO3
gives 1 mol of CO2. 168 g NaHCO3 ? 23
litresCO2. So 1.68 g ? 0.23
litres b) (ii) Some of the carbon dioxide
dissolves in water.
6
7. The concentration of a solution of sodium
thiosulphate can be found by reaction with
iodine. The iodine is produced by
electrolysis of an iodide solution using the
apparatus shown. The
current is noted and the time when the indicator
detects the end-point of the
reaction is recorded. (a) Iodine
is produced from the iodide solution according to
the following equation 2I-(aq) ? I2(aq)
2e- Calculate the number
of moles of iodine generated during the
electrolysis given the
following results. Current 0.010
A Time 1 min 37 s
Q I x t 0.01 x 97 0.97C 2I-(aq)
? I2(aq) 2e- 1 mol 2F 1 mol
2 x 96500 C 1 mol 193000C So
0.97 C ? 0.97/193000 x 1 5 x 10-6 mol
7
7. (b) The iodine produced reacts with the
thiosulphate ions according to the
equation I2(aq) 2S2O32-(aq) ?
2I-(aq) S4O62-(aq)
iodine thiosulphate ions At the
end-point of the reaction, excess iodine is
detected by the indicator. (i) Name the
indicator which could be used to detect the
excess iodine present at the
end-point. (ii) In a second experiment it
was found that 1.2 x 10-5 mol of iodine reacted
with 3.0 cm3 of the sodium
thiosulphate solution. Use this
information to calculate the concentration of the
sodium thiosulphate solution in mol
l-1.
8
8. Aluminium is manufactured in cells by the
electrolysis of aluminium oxide dissolved
in molten cryolite. What mass of aluminium
is produced each hour, if the current passing
through the liquid is 180 000 A?
Al3(l) 3e- ? Al(l) 3 F ?
1 mol 3 x 96500C ? 27 g 289500C ?
27 g Q I x t 180 000 x 60 x 60 648
000 000 C 289500C ? 27 g So 648 000 000 C
? 648 000 000/289500 x 27
60 435 g 60.435
kg of Al
9
9. Calcite is a very pure form of calcium
carbonate which reacts with nitric acid as
follows CaCO3(s) 2HNO3 (aq) ?
Ca(NO3) 2(aq) H2O(l) CO2(g) A 2.14
g piece of calcite was added to 50.0 cm3 of 0.200
mol l-1 nitric acid in a beaker. (a) Calculate
the mass of calcite, in grams, left
unreacted. (b) Describe what could be done to
check the result obtained in (a)
  • No. of moles of calcite mass/gfm
    2.14/100 0.0214 mol.
  • No. of moles of acid C x V(litres)
    0.2 x 50/1000 0.01
  • From equation 0.0214 mol of calcite
    reacts with 0.0418 mol of acid.
  • So all the acid is used up
  • 0.01 mol of acid reacts with 0.005 mol
    of calcite 0.005 x 100 0.5 g
  • Mass of calcite left over 2.14 0.5g
    1.64 g.
  • (b) Filter off any unreacted calcite, dry and
    then weigh it.

10
  • 10. Diphosphine, P2H4, is a hydride of
    phosphorus. All of the covalent bonds
  • in diphosphine molecules are non-polar
    because the elements present have
  • the same electronegativity.
  • What is meant by the termelectronegativity?
  • (b) The balanced equation for the complete
    combustion of diphosphine is
  • 2P2H4(g) 7O2 (g) ? P4O10(s)
    4H2O(l)
  • What volume of oxygen would be required for
    the complete combustion o
  • 10 cm3 of diphosphine?
  • (c) Calculate the volume occupied by 0.330 g of
    diphosphine.
  • (Take the molar volume to be 24.0 litres
    mol-1.)
  • (a) Electronegativity is a measure of the
    attraction of an atom
  • for electrons it shares with other atoms.
  • (b) 2P2H4(g) 7O2 (g) ? P4O10(s)
    4H2O(l)
  • 2 mol 7 mol
  • 2 vol 7 vol
  • 10cm3 35cm3
  • 1 mole of P2H4 weighs 66g
  • 66g occupies 24.0 litres
  • So 0.330g occupies 0.33/66 x 24 0.12
    litres

11
11. In 1996, the scientists Robert Curl, Harold
Kroto and Richard Smalley won the Nobel
Prize in Chemistry for their contribution to the
discovery of new forms of carbon called
fullerines. (a) In what way does the
structure of fullerines differ from the other
forms of carbon, diamond and
graphite? (b) One form of fullerine,
C60, forms a superconducting crystalline compound
with potassium. Its
formula can be represented as K3C60. A
sample of this compound was found to contain 2.88
g of carbon. (i) Calculate the number of
moles of fullerine used to make this compound.
(ii) Calculate the mass of potassium, in
grams, in the sample.
  • (a) Fullerines are covalent molecular solids
    diamond and graphite are covalen
  • network solids.
  • 1 mole of K3C60 contains 60 x 12 720g of
    carbon.
  • So 2.88g of carbon 2.88/720
    0.004 mol of fullerine.
  • 1 mole of K3C60 contains 3 x 39 g of potassium.
  • So 0.004 moles contains 3 x 39 x 0.004 g
  • 0.468 g of potassium

12
12. Ionisation energies provide information
about the structure of atoms. (a) Write the
equation, showing state symbols, for the first
ionisation energy of sodium. (b)
Calculate the number of electrons lost when one
mole of boron atoms is converted into
one mole of boron atoms with a charge of 3.
  • Na(g) ? Na(g) e-
  • B(g) ? B3(g) 3e-
  • 1 mol 3
    mol
  • 3 x 6.02 x
    1023
  • 1.806 x 1024 electrons
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