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EKSIKA JOINT EVALUATION TEST CHEMISTRY MSCM PP2
1
i)
ii)
iii)
iv)
v)
vi)
vii)
2
D2O31
B/Potassium/k it has the biggest /largest atomic radius hence most reading loses
electrons /least ionization energy.
D has a larger/bigger atomic radius than G D has fewer number of protons ½
/lower 02 nuclear charge hence weaker nuclear attraction on its valence
A1
On the grid
2.8.8.21
Period 4 group II
Covalent bond. Element D has a higher ½ charge with smaller atomic radius
that allows it to share electrons with G.
F is more reactive than G.F has a smaller ½ atomic radius than G hence more
½ readily gains an electron.
a)
The enthalpy change that occurs when one mole of a substance is formed from its
constituent element under temperature of 298k and a pressure of 760mmHg.
b)
i)
-Enthalpy of combustion of hydrogen.
-Enthalpy of formation of water.
ii)
C2H6(g)+ 72 O2 ½
Labeling of both axis correctly ½
∆H=-1560KJ/mole ½
Energy
KJ 
2CO2(g)+3H 2O(l) ½
Reaction path/progress/course 
iii)
2CO2(g)+3H2O(l)→C2H6(g)+ 72 O2(g),∆H1=+1560 ½
2C(s)+2O2(g)→2CO2(g)∆H2=-394x2=-788 ½
3H2(g)+ 32 O2(g)→3H2O(l),∆H3=-286x3=-858 ½
2C(s)+3H2(g)→C2H6(g) ∆H=-86KJ/mole ½
iv)
i)
© 2015 Eksika Joint Evaluation Test
Heat change=mc∆+
 500 gx 4.2 Jxgk21.5 k
2mks
 45,150 J
233/2 Chemistry Marking Scheme
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If 86,000 J  30 g of C 2 H 6
ii)
c)
3
i)
 45,150 J 
45,150 Jx 30 g
86, 000 J
 15.75 g of C 2 H 6
lattice energy/Reserve lattice energy1
ii)
∆H3=∆H1+∆H2
=+2489+-2659 ½
=-170KJ/mole ½
a)
i)
ii)
iii)
Al3+/Aluminum/Aluminum (iii) ion1
Amphoteric property/Amphoterism1
Al(OH)3(s)+3H+(aq)→Al3+(aq)+3H2O(l)1
b)
i)
Step 2=Excess Carbon(iv) Oxide1
Step 4=Dilute hydrochloric acid1
ii)
Hydrated calcium sulphate is used in holdingbroken bones in
position/plaster of Paris
iii)
Ca(HCO3)2(aq)+Ca(OH)2(aq)→2CaCO3(s)+2H2O(l)1
iv)
To aqueous calcium chloride,add 1H2SO4/any soluble sulphate.Filter
1to get calcium sulphate as residue.Dry ½ the calcium sulphate
between filter papers.
c)
Fe(s)+2HCl(g)→FeCl2(s)+H2(g) ½
1800cm3
mole ratio
1800
No of moles of HCl  24, 000  0.075moles
mole ratio of Fe : HCl  1 : 2
No of moles of Fe 
0.075
2
 0.0375moles 03
Mass of iron reacted  0.0375 x56  2.1g
Mass of unreacted iron  5.0  2.1  2.9 g
4
a)
i)
Condensation polymerization1
ii)
1
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233/2 Chemistry Marking Scheme
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b)
i)
M
Q
-Propyl hydrogen1
-2-Chloropropane1
ii)
2CH3CH2CH2OH(l)+9O2(g)→6CO2(g)+8H2O(l)1
iii)
Ethanoic acid/CH3COOH1
1
iv)
c)
5
v)
Reagent
-hydrogen gas1
Conditions
-nickel catalyst and temperature tied 150-2500C do not
accept a range of temperature1
vi)
Put a spatula full of sodium hydrogen carbonate1/sodium carbonate to
test tubes of propanoic acid and propanal-01 separately. In propanoic acid
there is efference  ½ while in propan-1-01 there is no inference
i)
2-bromobutane1
ii)
2-chloropent-2-ene1
a)
S2│S . it has tQ value of 0.00v/ it is the standard reference electrode
b)
Q ½
c)
i)
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233/2 Chemistry Marking Scheme
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ii)
6
E.m.f = cell reduced – cell oxidated
=+0.34-(-2.90) ½
=+3.24v ½
d)
T(s)→T2+(aq)+2e-0.34
R2+(aq)+2e→R(s)-2.38
T(s)+R2+(aq)→T2+(aq)+R(s) tӨ=-2.72v1
There will be no ½ reaction between the solution of R2+ and the container of
T.Overal e.m.f is negative ½ (-2.72v)
a)
i)
X
Y
ii)
Finely divided iron catalyst1
iii)
Reducing property/Reduction1
iv)
4NH3(g)+5O2(g)→4NO(g)+6H2O(g)1
v)
Used as a nitrogenous fertilizer1
vi)
Insert ½ a glowing splint.The glowing slint will be rekindled/relit/re
ignited. ½
i)
Dehydration1
ii)
Solid B
Gas C
iii)
-Used as fuel1 any
-Used as a reducing agent of some metallic oxides
iv)
To absorb carbon(iv) oxide1
b)
© 2015 Eksika Joint Evaluation Test
-Nitrogen gas/N2(g) ½
-Hydrogen gas/H2(g) ½
-Copper ½
-Carbon(iv) oxide ½
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7
i)
Graph attached at the back 03
ii)
Correct reading from the graph
27.5mm ½ + 0.2
Shown on the graph ½ 01
a ) PbNO 3 2aq  2 NI aq  PbI 2  s  NaNO
iii)
mole ratio of Pb NO3  : NaI  1 : 2
3  aq 
b) No.of moles NaI  0.002 x 2  0.004moles
No of moles of Pb NO3 2 
8 x 0.25
1000
 0.002moles
iv)
Pb2+(aq)+2I-(aq)→PbI2(s)1
v)
Sodium nitrate solution │NaNO3(aq)1
vi)
Name a chemical reagent that can be added to the mixture to spread up the setting
of the precipitate1
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233/2 Chemistry Marking Scheme
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