ADMINISTRATION AND LOCAL GOVERNMENT
BRIGHT EDUCATION DEVELOPMENT ACADEMY - KIVULE
FORM THREE MONTHLY TEST
CHEMISTRY-MARKING SCHEME
SECTION A (15 Marks)
1.
i |
ii |
iii |
iv |
v |
vi |
vii |
viii |
ix |
x |
A |
D |
A |
A |
C |
B |
B |
D |
A |
D |
2.
LIST A |
i |
ii |
iii |
iv |
v |
LIST B |
C |
B |
A |
E |
D |
SECTION B (70 Marks)
3(a)i. To Increase /speed up the rate of chemical reaction.
ii. The chemical reaction will be too slow.
(b) The gas(oxygen gas) is tested by introducing a wooden glowing splint in flame.
(C) i. Burning various materials(combustion)
ii. Breathing for living organisms eg. Plants and animals.
iii. Welding of metals.
4.(a)i. Mole, is the amount of substance, equal to the quantity containing as many elementary units as there are atoms in 12 g of carbon-12
(ii) Molar mass is the mass of one mole of a substance expressed in grams per mole (g/mol).
(b) Given;
Volume of oxygen gas=112dm³
Solution.
Chemical Equation for decomposition of Lead Nitrate(Pb(NO3)2)
2Pb(NO3)2(s) →2PbO(s) +4NO2(g) +O2(g)
Molar ratio for Nitrogen dioxide(4NO2(g)) and oxygen(O2(g)) =4:1
Hence,find the number of moles of oxygen when volume is 112dm³.
From,
1 mol of a gas at STP=22.4dm³
x =112dm³
x=112dm³x1 mol/22.4dm³
x=5 Moles
When is Nitrogen dioxide will be;
Number of moles =4x5 moles=20 moles.
Now,change those moles into volume.
From,
1 mol =22.4dm³ at STP.
20 Mol=x
x=22.4dm³x20=448dm³
Therefore, the volume of nitrogen dioxide is 448dm³
5.(a)(i)Dilution Factor is the factor that shows how times the volume of concentrated solution is diluted to obtain the diluted solution.
(ii) The dilution formula is;
c1v1 = c2v2
(b) Given;
Volume of concentrated acid(V1) =500cm³
Volume of diluted acid(V2) =?
Molarity of conc. Acid(C1) =3M
Molarity of Dil. Acid (C2) =0.5M
Solution.
From the dilution formula.
C1v1=C2v2
3Mx500cm³=0.5Mxv2
V2=3Mx500cm³/0.5M
V2=7,500cm³
The volume of Diluted acid will be 7,500cm³
6.(a)volume of a gas is the volume of one mole of the gas at STP.
(b) Given;
Mass of Hydrogen =0.5g
Solution.
Chemical Equation for water formed.
2H2(g) +O2)(g) →2H2O(l)
Molar ratio;
2:1:2
Mass of Hydrogen =0.5gx2=1g
NB:Molar ratio of hydrogen and water are equal(2:2)
Mass of water is 1g
7.(a)Given;
Moles of SO2(g) =0.5Moles
Solution.
From,
1 mol of a substance=6.022x1023
For SO2(g)
0.5x6.022x1023=3.011x1023
For Oxygen(O2(g))=2 atoms in SO2(g)
2x3.011x1023=6.022x1023
The number of moles of oxygen is 6.022x1023
(b) Given;
Mass of Aluminum sulphate=20g
Solution.
From,
n=N/LA, whereby:n=number of moles, N=Number of Ions, LA=Avogadro's constant=6.022x1023
But, n=mass/Molar mass
Molar mass of Aluminum sulphate(Al2(SO4)3
=27x2+32x3+16x4x3=342g/mol
n=20g/342g/mol
n=0.0585mol~0.1mol.
N=0.1molx6.022x1023
N=6.022x1022
8.(a)
(b)(i) HCl(aq) + NaOH(aq)→NaCl(aq) + H2O
Solution
(ii) AgNO3(aq) + HCl(aq)→AgCl(s) + HNO3(aq)
Solution
Split soluble ions:Ag+(aq)+NO-3(aq)+H+(aq) +Cl-(aq) →AgCl(s) +H+(aq) +NO-3(aq)
Remove spectator ions:
Ag+(aq)+NO-3(aq)+H+(aq)
+Cl-(aq) →AgCl(s)
+H+(aq)
+NO-3(aq)
Ionic Equation:
Ag+(aq)+Cl-(aq) →AgCl(s)
SECTION C:15 MARKS
9.(a) sample C, because the volume of soap used is higher.
(b)sample D, because the volume of soap is lower.
(c) i. Magnesium and calcium Hydrogen carbonate compounds.
(ii) Magnesium and calcium sulphate compounds
(d) i. Hard water causes limescale in water boiler, hot water pipes, kettles and other appliances
ii. It needs more soap than soft water.
iii. It leaves scummy deposition on clothing and in baths
iv.Destroy the quality of some special finishes on clothes
10. Advantages of hard water.
i. It tastes better due to dissolved compound
ii. It provide useful calcium for growth of bones and teeth
iii. Formation of limescale (form insulator) which prevent pipe(tap) from rust
iv.Formation of limescale(form insulator) which prevent poisonous metal salt in the water from water tap
v. Help in the formation of strong shells in some aquatic animals
Any four points
Disadvantages of hard water.
i. Hard water causes limescale in water boiler, hot water pipes, kettles and other appliances
ii. It need more soap than soft water
iii. It leaves scummy deposition on clothing and in baths.
iv.Destroy the quality of some special finishes on clothes.
v. It causes corrosion of materials.