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Chemistry
UTME / JAMB 2015
Matter & Separation Techniques
Solution
When steam is passed over red hot carbon the substances produced are
A
hydrogen and trioxocarbonate (IV) acid
B
hydrogen, oxygen and carbon (IV) oxide
C
hydrogen and carbon (ii) oxide
D
hydrogen and carbon (IV) oxide
medium Difficulty
Verified Correct Answer: Option C
Chemical Principles & Mechanism
Hydrogen and carbon (ii) oxide will be formed when red steam is passed ever red-hot carbon. Option A,B,D are not produced through this process
Source: Official Exam / Past Question Archive
Chemistry
WAEC / WASSCE 2015
Matter & Separation Techniques
Solution
All your burette readings (initials and final) as well as the size of your pipette must be recorded but no account experimental procedure is required. All calculations must be done in your answer booklet.
A is 0.100 mol dm\(^{-3}\) HNO\(_3\). B is a solution containing 2.50 g of a mixture of Na\(_2\)CO\(_3\) and Na\(_2\)SO\(_4\) in 250 cm\(^3\) of solution.
(a) Put into the burette and titrate it against 20.0 cm\(^3\) or 25.0cm\(^3\) portions of B using methyl orange as indicator. Repeat the exercise to obtain consistent titre values. Tabulate your readings and calculate the average volume of acid A used. The equation of reaction is Na\(_2\)CO\(_{3(aq)}\) + 2HNO\(_{3(aq)}\) \(\to\) NaNO\(_{3(aq)}\) + H\(_2\)O\(_{(l)}\) + CO\(_{2(g)}\)
(b) From your results and the information provided, calculate the:
(i) concentration of B in moldm\(^{-3}\);
(ii) concentration of Na\(_2\)CO\(_3\) in B in gdm\(^{-3}\);
(iii) percentage of Na\(_2\)CO\(_3\), in the mixture. [Na\(_2\)CO\(_3\) = 106]
A
Right leading to the production of more SO3(g)
B
Right leading to the production of more SO2(g)
C
Left leading to the production of more SO2(g)
D
Left leading to the production of more SO3(g)
medium Difficulty
Verified Correct Answer: Option A
Chemical Principles & Mechanism
(a)
Burette reading
Rough
1st titration
2nd titration
Final Burret Reading (cm\(^3\))
23.50
45.90
33.60
Initial Burret Reading (cm\(^3\))
01.10
23.50
11.30
Volume of acid used (cm\(^3\))
22.40
22.40
22.30
Average titre = \(\frac{22.40 + 22.40 + 22.30}{3}\) = 22.37 cm\(^3\)
(b)(i) Conc. of B in mol dm\(^{-3}\)
\(\frac{C_AV_A}{C_BV_B}\) = \(\frac{2}{I}\)
C\(_B\) = \(\frac{C_AV_A}{2VB}\)
= \(\frac{0.100 \times 22.37}{2 \times 25}\)
= 0.0447 mol dm\(^{-3}\)
(ii) Conc. of Na\(_2\)CO\(_3\) in Bg dm\(^{-3}\)
Conc. of Na\(_2\)CO\(_3\) = 0.0447 x Molar Mass of B
= 0.0447 x 106
= 4.74 g dm\(^{-3}\)
(iii) 250 cm\(^3\) of B contained 2.50 g of Na\(_2\)CO\(_3\)
1000 cm\(^3\) of B = \(\frac{2.50 \times 1000}{250}\)
= 10.0g
mass conc of B = 10 g dm\(^{-3}\)
% Na\(_2\)CO\(_3\) = \(\frac{4.74}{10} \times 100\)
= 47.4%
Source: Official Exam / Past Question Archive
Chemistry
WAEC / WASSCE 2015
Matter & Separation Techniques
Diagram
Solution
(a) Consider the following diagram ABOVE.
(i) Give the name of the experiment.
(ii) What does the experiment demonstrate?
(iii) Name one gas that could be used in place of HCl gas.
(iv) What colour could be observed in the flask during the spray of water?
(v) Could the gases used in the experiment be collected over water?
(vi) Explain briefly your answer in (a)(v).
(b) A solid substance U when strongly heated decomposes to give a white solid V and carbon (IV) oxide. When water is added to V, W is produced W can be used to test for carbon (IV) Oxide. Identify U, V and W.
Attached Figure / Diagram
medium Difficulty
Verified Correct Answer: Option
Chemical Principles & Mechanism
(a)(i) Fountain experiment
(ii) To demonstrate high solubility of hydrogen chloride gas
(iii) Ammonia
(iv) No colour/colourless
(v) No
(vi) The gases are highly soluble in water
(b) U - CaO\(_3\)/Calcium trioxocarbonate (IV)
V - CaO/Calcium oxide
W - Ca(OH)\(_3\)/Calcium hydroxide/Lime water
Source: Myschool Past Questions Archive (Record #59987)
Chemistry
WAEC / WASSCE 2015
Acids, Bases, and Salts
Solution
Credit will be given for strict adherence to the instructions, for observations precisely recorded and for accurate inferences. All tests, observations, and inferences must be clearly entered in your answer book, in ink, at the time they are made.
C and D are inorganic salts. E is a solution of an organic compound. Carry out the following exercises on C, D and E. Record your observations and identify any gas(es) evolved. State the conclusion you draw from the result of each test.
(a) To all of C, add about 2 cm\(^3\) of distilled water in a boiling tube. Shake to dissolve and test the solution with litmus paper.
(b)i) Test E with litmus paper.
(ii) Add all of E to the solution obtained in (a).
(c)(i) Add about 5 cm\(^3\) of distilled water to D in a test tube and shake thoroughly. Divide the resulting solution into two portions.
(ii) To the first portion, add few drops of BaCl\(_{2(aq)}\) followed by excess dil. HCl.
(iii) To the second portion, add HNO\(_{3(aq)}\) in drops and then in excess.
medium Difficulty
Verified Correct Answer: Option
Chemical Principles & Mechanism
TEST
Observation
Inference
(a) C\(_{(s)}\) + H\(_2\)O + litmus paper
Red litmus paper turned blue
Alkaline/Basic solution
(b)(i) E\(_{(aq)}\) + litmus paper
Blue Litmus paper turned red
Acidic solution
(ii) E\(_{(aq)}\) + C\(_{(aq)}\)
Effervescence occurred
Colourless and odourless gas which turned water milky was evolved
CO\(_2\) gas from HCO\(_3^-\)/CO\(_3^2\)
(c)(i) D\(_{(s)}\) + H\(_2\)O
D dissolved to give a colourless solution
D is a soluble salt
(ii) D\(_{(aq)}\) + BaCl\(_{2(aq)}\)
+ dil HCl(\(_{excess}\))
White precipitate
Precipitate insoluble
SO\(^{2-}\) or SO\(_3^{2-}\) or CO\(_3^2\)
or SO\(_4^{2-}\)
SO\(_4^{2-}\) present
(iii) D\(_{(aq)}\) + NH\(_{3(aq)}\) in drops
then in excess
White gelatinous precipitate
Precipitate dissolved
Zn\(^{2+-}\) or AL\(^{3+}\)
Zn\(^{2+-}\) confirmed
Source: Myschool Past Questions Archive (Record #59988)
Chemistry
WAEC / WASSCE 2015
Matter & Separation Techniques
Solution
All your burette readings (initials and final) as well as the size of your pipette must be recorded but no account experimental procedure is required. All calculations must be done in your answer booklet.
A is 0.100 mol dm\(^{-3}\) HNO\(_3\). B is a solution containing 2.50 g of a mixture of Na\(_2\)CO\(_3\) and Na\(_2\)SO\(_4\) in 250 cm\(^3\) of solution.
(a) Put into the burette and titrate it against 20.0 cm\(^3\) or 25.0cm\(^3\) portions of B using methyl orange as indicator. Repeat the exercise to obtain consistent titre values. Tabulate your readings and calculate the average volume of acid A used. The equation of reaction is Na\(_2\)CO\(_{3(aq)}\) + 2HNO\(_{3(aq)}\) \(\to\) NaNO\(_{3(aq)}\) + H\(_2\)O\(_{(l)}\) + CO\(_{2(g)}\)
(b) From your results and the information provided, calculate the:
(i) concentration of B in moldm\(^{-3}\);
(ii) concentration of Na\(_2\)CO\(_3\) in B in gdm\(^{-3}\);
(iii) percentage of Na\(_2\)CO\(_3\), in the mixture. [Na\(_2\)CO\(_3\) = 106]
medium Difficulty
Verified Correct Answer: Option
Chemical Principles & Mechanism
(a)
Burette reading
Rough
1st titration
2nd titration
Final Burret Reading (cm\(^3\))
23.50
45.90
33.60
Initial Burret Reading (cm\(^3\))
01.10
23.50
11.30
Volume of acid used (cm\(^3\))
22.40
22.40
22.30
Average titre = \(\frac{22.40 + 22.40 + 22.30}{3}\) = 22.37 cm\(^3\)
(b)(i) Conc. of B in mol dm\(^{-3}\)
\(\frac{C_AV_A}{C_BV_B}\) = \(\frac{2}{I}\)
C\(_B\) = \(\frac{C_AV_A}{2VB}\)
= \(\frac{0.100 \times 22.37}{2 \times 25}\)
= 0.0447 mol dm\(^{-3}\)
(ii) Conc. of Na\(_2\)CO\(_3\) in Bg dm\(^{-3}\)
Conc. of Na\(_2\)CO\(_3\) = 0.0447 x Molar Mass of B
= 0.0447 x 106
= 4.74 g dm\(^{-3}\)
(iii) 250 cm\(^3\) of B contained 2.50 g of Na\(_2\)CO\(_3\)
1000 cm\(^3\) of B = \(\frac{2.50 \times 1000}{250}\)
= 10.0g
mass conc of B = 10 g dm\(^{-3}\)
% Na\(_2\)CO\(_3\) = \(\frac{4.74}{10} \times 100\)
= 47.4%
Source: Myschool Past Questions Archive (Record #59993)
Chemistry
WAEC / WASSCE 2014
Basic Chemistry
Solution
An atom W has 17 electrons and 18 neutrons. Which of the following representations of the atom is correct?
A
1718W
B
1817W
C
3517W
D
3518W
medium Difficulty
Verified Correct Answer: Option C
Chemical Principles & Mechanism
The superscript is the atomic mass = 35
(The mass number also called atomic mass number or nucleon number, is the total number of protons and neutrons)
The subscript is the atomic or proton number = 17
The number of electrons in a neutral atom is equal to the number of protons.
Source: WAEC / JAMB / NECO Classroom Archive (Record #37828)
Chemistry
UTME / JAMB 2014
Acids, Bases, and Salts
Solution
The colour of litmus in a neutral medium is
A
purple
B
pink
C
yellow
D
orange
medium Difficulty
Verified Correct Answer: Option A
Chemical Principles & Mechanism
The colour of a neutral litmus solution is purple.
Litmus solution is made up of purple dye which is extracted from Lichens.
It is a natural Indicator and it is used to indicate the presence of an acid or a base.
A litmus paper will turn red in the presence of an acid and it turns blue in the presence of a base.
Source: WAEC / JAMB / NECO Classroom Archive (Record #30701)
Chemistry
UTME / JAMB 2014
Periodic Table & Periodicity
Solution
Beryllium and aluminum have similar properties because they?
A
are both metals
B
belong to the same group
C
belong to the same period
D
are positioned diagonally to each other
medium Difficulty
Verified Correct Answer: Option D
Chemical Principles & Mechanism
At the beginning of periods 2 and 3 of the Periodic Table, there are several cases where an element at the top of one group has some similarities with an element in the next group.
For example, boron is a non-metal with some properties rather like silicon. Unlike the rest of Group 2, beryllium has some properties resembling aluminium. And lithium has some properties which differ from the other elements in Group 1, and in some ways resembles magnesium.
There is said to be a diagonal relationship between these elements.
There are several reasons for this, but each depends on the way atomic properties like electronegativity vary around the Periodic Table.
Source: WAEC / JAMB / NECO Classroom Archive (Record #30681)
Chemistry
WAEC / WASSCE 2014
Periodic Table & Periodicity
Solution
Which of the following transition metals is not attracted to external magnetic field?
A
Cu
B
Fe
C
Ti
D
Zn
medium Difficulty
Verified Correct Answer: Option A
Chemical Principles & Mechanism
The period 4 transition metals are scandium (Sc), titanium (Ti), vanadium (V), chromium (Cr), manganese (Mn), iron (Fe), cobalt (Co), nickel (Ni), copper (Cu), and zinc (Zn).
Diamagnetism results from the lack of unpaired electrons. Diamagnetic substances are substances that are repelled by magnetic fields.
The examples of diamagnetic materials are copper, gold, antimony, silver, lead and hydrogen.
Source: WAEC / JAMB / NECO Classroom Archive (Record #37827)
Chemistry
WAEC / WASSCE 2014
Transition Metals and Their Compounds
Solution
The electron configuration of \(_{26}Fe^{3+}\) is?
A
[Ar]4s23d6
B
[AR]4s23d3
C
[AR]4s13d4
D
[AR]4s03d5
medium Difficulty
Verified Correct Answer: Option D
Chemical Principles & Mechanism
The neutral Iron atom has the electric configuration: 1s2 2s2 2p6 3s2 3p6 4s2 3d6.to write the electronic config. of Fe3+ we have to subtract 3 electrons from the outermost shell which is 4s and 3d orbitals. Thus we obtain the electronic configuration for Fe3+ as:
1s2 2s2 2p6 3s2 3p6 4s0 3d5
OR
[AR] 4s0 3d5.
Source: WAEC / JAMB / NECO Classroom Archive (Record #37826)
Chemistry
WAEC / WASSCE 2014
Crude Oil
Solution
The liquid hydrocarbon likely to be found in the fraction of crude oil used for domestic cooking is?
A
C3H8
B
C 5H 12
C
C13H28
D
C20H42
medium Difficulty
Verified Correct Answer: Option A
Chemical Principles & Mechanism
The gas used in gas bottles is LPG – liquefied petroleum gas – which are either propane or butane, flammable hydrocarbons used as fuel for home heating, cooking, hot water and LPG cars and vehicles.
Source: WAEC / JAMB / NECO Classroom Archive (Record #37880)
Chemistry
UTME / JAMB 2014
Matter & Separation Techniques
A mixture is different from a compound because
A
the properties of a compound are those of its individual constituents while those of a mixture differ from its constituents
B
a mixture is always homogeneous while a compound is not
C
the constituent of a compound are chemically bound together while those of a mixture are not
D
a mixture can be represented by a chemical formula while a compound cannot
medium Difficulty
Verified Correct Answer: Option C
Chemical Principles & Mechanism
No expanded working has been attached for this question yet. The verified answer is Option C.
Source: Myschool Past Questions Archive (Record #30675)
Chemistry
UTME / JAMB 2014
Stoichiometry & Mole Concept
Solution
What is the percentage of sulphur in sulphur (IV) oxide?
A
66%
B
25%
C
40%
D
50%
medium Difficulty
Verified Correct Answer: Option D
Chemical Principles & Mechanism
molar mass of SO 2=64 :.%S=32/64 x 100 = 50%
Source: Myschool Past Questions Archive (Record #30676)
Chemistry
UTME / JAMB 2014
Stoichiometry & Mole Concept
Solution
A gas X diffuses twice as fast as gas Y. if the relative molecular mass of X is 32, calculate the relative molecular mass of Y.
A
128
B
8
C
16
D
64
medium Difficulty
Verified Correct Answer: Option A
Chemical Principles & Mechanism
\(\frac{RY}{RX}\) = \(\frac{\sqrt{My}}{\sqrt{My}}\) 1/2 = \(\frac{\sqrt{32}}{\sqrt{y}}\) \(\frac{32}{y}\) = 1/4
Y = \(\frac{32x4}{1}\) = 128
Source: Myschool Past Questions Archive (Record #30677)
Chemistry
UTME / JAMB 2014
Matter & Separation Techniques
Solution
200 cm3 of a gas at 25ºC exerts a pressure of 700 mmHg. Calculate its pressure if its volume increase 350 cm3 at 75ºC.
A
342.53 mmHg
B
1430.54 mmHg
C
467.11 mmHg
D
400.00 mmHg
medium Difficulty
Verified Correct Answer: Option C
Chemical Principles & Mechanism
\(\frac{p1V 1}{T1}\) = \(\frac{P2V2}{T2}\) \(\frac{700x200}{298}\) = \(\frac{P2X 350}{348}\) P2 = \(\frac{700x200x348}{298x350}\) = 467.11mmHg
Source: Myschool Past Questions Archive (Record #30678)