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Chemistry (2017) QUIZ STUDY
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1.
The general formula of alkanones is
.
A. RCHO
B. R2CO
C. RCOOH
D. RCOOR
Explanation.
Alkanones also known as ketones have the general formula R2CO

2.
The constituent common to duralumin and alnico is
.
A. Co
B. Mn
C. Al
D. Mg
Explanation.
Constituents of duralumin are: Al, Cu, Mg, Mn.
Constituents of Alnico are: Al, Ni and Co

In 1909, the alloy of duralumin was discovered by Alfred Wilon consisting of 94% Al, 4% Cu, 1% Mg and 1% Mn(Manganese)
Alnico is an acronym referring to a family of iron alloys which in addition to iron are composed primarily of Al, Ni and Co.

3.
The shape of the S-orbital is
.
A. elliptical
B. spiral
C. circular
D. spherical
Explanation.
The shape of the S-orbital is spherical.

4.
Aluminium hydroxide is used in the dyeing industry as a
.
A. dye
B. dispersant
C. salt
D. mordant
Explanation.
Aluminium hydroxide is used in the dyeing industry as a mordant. It combines with a dye and thereby fixes it in a material.

5.
The tincture of iodine means iodine dissolved in
.
A. ethanol
B. bromine chloride
C. chlorine water
D. water
Explanation.
It is also called weak iodine solution. Tincture solutions are characterized by the presence of alcohol.

6.

Temporary hard water is formed when rain water containing dissolved carbon(IV) oxide flows over deposits of

.
A.

CaCO3

B.

Na2CO3

C.

Na2SO4

D.

CaSO4

Explanation.

Permanent hardness in water= Mg & Ca Sulphate.
Temporary hardness in water= Mg & Ca Carbonate.

Temporary hardness of water is caused by Magnesium and calcium hydrogencarbonate. It is formed when rainwater containing dissolved CO2 flows over deposit of CaCO3

i.e CaCO3 + CO2 + H2O → Ca(HCO3)2


7.

The acid anhydride that will produce weak acid in water is

.
A.

SO3

B.

NO2

C.

SO2

D.

CO2

Explanation.

H2CO3 is an example of a weak acid while H2SO4 and HNO3 are examples of a strong acid.

CO2 combines with water to give a weak trioxocarbonate (IV) acid.

CO2 + H2O → H2CO3


8.
The process that occurs when two equivalent forms of a compound are in equilibrium is
.
A. Isotopy
B. Resonance
C. Isomerism
D. Reforming
Explanation.
Resonance involves two forms of a compound.
Isomerism involves two or more forms of an element.
Reforming involves the rearrangement of molecule.

This deals with a two forms of a molecule where the chemical connectivity is the same but the electrons are distributed differently around the structure.

9.
In the laboratory preparation of ethyl ethanoate, the water present in the mixture is removed using a solution of
.
A. an hydrous CaCl4
B. concentrated NaCO4
C. dilute NaOH
D. concentrated H4SO4
Explanation.
This is used in the preparation of ethyl ethanoate to remove the water present since it serves as a dehydrated agent.

10.
The constituent of air necessary in the rusting process are
.
A. O2 and H2O
B. Ar and CO2
C. CO2 and H2O
D. O2 and CO2
Explanation.
The constituent of air includes O2, CO2, N2 and Noble gases. While for a rusting process to take place, the presence of O2, H2O and CO2 is important.

The constituent of air includes O2, CO2, N2 and Noble gases and for rusting process to take place.

The presence of O2 and CO2 as a constituent of air is indispensable

11.
For a general equation of the nature xP + yQ ? mR + nS, the expression for the equilibrium constant is
.
A. k [P]x [Q]y
B. \(\frac{[P]^x [Q]^y}{[R]^m [S]^n}\)
C. \(\frac{[R]^m [S]^n}{[P]^x [Q]^y}\)
D. \(\frac{m[R] n[S]}{x[P] y[Q] }\)
Explanation.
Expression for equilibrium constant

k = \(\frac{\text{concentration of product}}{\text{concentration of reactant}}\)

12.
A given mass of gas occupies 2dm3 at 300k. At what temperature will its volume be doubled, keeping the pressure constant?
.
A. 400k
B. 480k
C. 550k
D. 600k
Explanation.
At constant pressure connotes Charle's law

\(\frac{V_1}{T_1}\) = \(\frac{V_2}{T_2}\)

\(\frac{2dm^3}{300k}\) = 2 x \(\frac{2dm^3}{T_2}\)

T2= 600k

13.
The oxidation number of iodine in KIO3 is
.
A. 7
B. 3
C. 5
D. 6
Explanation.
If the rules are followed, it is non-negotiable that +5 is the suitable answer to the question.

14.
An isomer of C5H12 is
.
A. 2-ethyl butane
B. butane
C. 2-methyl butane
D. 2-methyl propane
Explanation.
C5H12 has 5 carbons acid 12 hydrogens.

2-methyl butane

No of carbon=5

No of hydrogen=12

15.

When few drops of concentrated trioxonitrate(V) acid is added to an unknown sample and wanned an intense yellow colouration is observed. The likely functional group present in the sample is

.
A.

NH- C - C=O

B.

CHO

C.

CO

D.

CNH2

Explanation.

Xanthopreitic test for the presence of protein, when conc nitric acid is added to the drop, an intense yellow colouration is observed.

It contains all the functional group of protein which includes the amino, alkanol and the carboxylic group. Adding few drops of conc HNO3 to a protein, gives an intense yellow colouration.

It is called Xanthopreitic test.


16.
A sample of orange juice is found to have a PH of 3.80. What is the concentration of the hydroxide ion in the juice?
.
A. 1.6 \(\times\) 10-4
B. 6.3 \(\times\) 10-11
C. 6.3 \(\times\) 10-4
D. 1.6 \(\times\) 10-11
Explanation.
PH = - Log[ H+ ]

PH + POH =14

POH = 14 - 3.8

POH = 10.2

POH = - Log[ OH- ]

10.2 = - Log[ OH- ]

10-10.2 = [ OH- ]

[ OH- ] = 6.3 x 10-11

17.
Incomplete oxidation of ethanol yields
.
A. CH3COOH
B. CH3COCH3
C. CH3CH2OCH2CH3
D. CH2CHO
Explanation.
C2H5OH ? CH3CHO ? CH3COOH
Ethanol.....oxidation Ethanal........Ethanoic Acid
Primary alcohol oxidises to aldehyde and later to carboxylic acid.
Secondary alcohol oxidises to ketones.
Ethanol is an example of primary alcohol and primary alcohol can be oxidised to aldehyde and carboxylic acid. Wherein, incomplete oxidation of primary alcohol yields aldehyde also known as alkaline while complete oxidation of primary alcohol yields carboxylic acid.

18.
The salt formed from a weak acid and a strong base hydrolyzes in water to form
.
A. A saturated solution
B. an acidic solution
C. a buffer solution
D. an alkaline solution
Explanation.
This is a solution formed from the hydrolyses of an alkali in water.

19.
The ideal gas laws and equations are true for all gases at
.
A. low pressures and lower temperatures
B. low temperatures and high pressures
C. high pressures and high temperatures
D. low pressure and high temperatures
Explanation.
At high temperature and low pressure, the vanderwaal equation is reduced to ideal gas equation.

i.e, [P + \(\frac{a}{v^2}\)] [v - b] = RT is reduced to PV = nRT

Generally, a gas behaves more like an ideal gas at higher temperature and lower pressure as the potential energy due to intermolecular forces, it becomes less significant compared with the particles kinetic energy and the size of the molecules, then it becomes less significant compared to the empty space between them.

20.

When ΔH is negative, a reaction is said to be

.
A.

endothermic

B.

exothermic

C.

reversible

D.

ionic

Explanation.

When ΔH is negative, heat is liberated to the surrounding and it connotes an exothermic reaction.
ΔH = -ve


21.

\(^{226}_{88}Ra\) → \(^x_{86}Rn\) + alpha particle

.
A.

226

B.

220

C.

227

D.

222

Explanation.

\(^{226}_{88}Ra\) → \(^x_{86}Rn\) + \(^4_{2}He\)

\(^4_{2}He\) = alpha particle

considering the summation of the mass number

226 = x + 4

x = 226 + 4

x = 222


22.
The shape of ammonia molecules is
.
A. trigonal planar
B. octahedral
C. square planar
D. tetrahedral
Explanation.
Ammonia = NH3

It has three bonds of hydrogen to the Nitrogen.

23.
The mass of silver deposited when a current of 10A is passed through a solution of silver salt for 4830s is - (Ag = 108 F = 96500(mol-1)
.
A. 54.0g
B. 27.0g
C. 13.5g
D. 108.0g
Explanation.
Recall that

m = \(\frac{MmIt}{96500n}\)

Mm =108, t = 4830s

I = 10A, n = 1

m = 108 x 10 x (\(\frac{4830}{96500}\)) x 1

m = 54.0g

Mass deposited = Mm x \(\frac{It}{96500n}\)

24.
Tin is unaffected by air at ordinary temperature due to its
.
A. Low melting point
B. Weak electropositive character
C. High boiling point
D. White lustrous appearance
Explanation.
Tin has a melting point of 232° which enables it to be unaffected by air at ordinary temperature coupled with the fact that it also helps in making a good metal for alloying.

Tin is relatively unaffected by both water and oxygen at room temperature due to its low melting point. It does not rust, corrode, or react in any other way. This explains one of its major uses: as a coating to protect other metals.

25.
Calculate the amount in moles of silver deposited when 9650C of electricty is passed through a solution of silver salt [= 96500 Cmol-1]
.
A. 0.05
B. 10.8
C. 10
D. 0.1
Explanation.
m = \(\frac{Mm \times Q}{96500n
}\)

where Q = IT

M = Mm x \(\frac{Q}{96500n}\)

where m = mass

Mm = Molar mass

Q = Quantity of electricity

n = number of change= +1

\(\frac{M}{Mm}\) = mole = \(\frac{mass}{Molarmass}\)

\(\frac{M}{Mm}\) = \(\frac{Q}{96500n}\)

= \(\frac{9650}{96500n}\) x 1

= \(\frac{1}{10}\) = 0.1mol

26.
The reaction of halogens in the presence of sunlight is an example of
.
A. oxidation reaction
B. addition reaction
C. hydrogenation reaction
D. substitution reaction
Explanation.
Alkanes undergoes substitution reaction and it is an example of halogenation substitution reaction.

Reaction of halogen with alkane in the presence of sunlight (ultraviolet) is termed halogenation. Halogenation is an example of substitution reaction. In addition, alkanes only undergo substitution reaction but not addition reaction.

27.
The enzyme used in the hydrolysis of starch to dextrin and maltose is
.
A. amylase
B. diastatse
C. invertase
D. zymase
Explanation.
This is any amylase or a mixture of amylases that converts starch to dextrin and maltose.

28.

The alkyl group is represented by the general formula

.
A.

CnH2n

B.

CnH2n - 2

C.

CnH2n + 1

D.

CnH2n + 2

Explanation.

Alkyl group has the general formula CnH2n + 1.

This formed when one hydrogen is removed from the alkane family. It has the general formula CnH2n + 1


29.
CH2S(g) + O2(g) ? 2Cn + SO2(g)
What is the change in the oxidation number of copper in the reaction?
.
A. 0 to + 2
B. 0 to + 1
C. C + 1 to 0
D. + 2 to + 1
Explanation.
In the reactant;

Cu2S

2 Cu - 2(1) = 0

2 Cu = 2

Cu = \(\frac{2}{2}\)

Cu = +1

In the product, Cu

Cu = O

The oxidation number of Cu in Cu2S and Cu respectively is +1 and 0 respectively

30. q-image

In the diagram above. X is

.
A.

SO3

B.

SO2

C.

S

D.

H2S

Explanation.
The setup represents the production of sulfur dioxide. And the cylinder marked X is SO2

31. q-image

The diagram above. Y is

.
A.

fused CaO

B.

H2O

C.

NaOH

D.

Concentrated H2SO4

Explanation.
In the preparation of sulphur dioxide by the action of dilute acids on sulphates and bisulphites. conc H2SO4 helps to release SO2 from the mixture.
The setup represents the production of sulphur dioxide

32.
Calculate the mass of copper deposited when a current of 0.5 ampere was passed through a solution of copper(II) chloride for 45 minutes in an electrolytic cell. [Cu = 64, F = 96500Cmol-1]
.
A. 0.300g
B. 0.250g
C. 0.2242g
D. 0.448g
Explanation.
M = \(\frac{\text{Molar mass x Quantity of Electricity}}{\text{96500 x no of charge}}\)

= \(\frac{MmIT}{96500n}\)

Copper II Chloride = CuCl2

CuCl2 ? Cu2+ + 2Cl2

Mass of compound deposited = \(\frac{\text{Molar mass x Quantity of Electricity}}{\text{96500 x no of charge}}\)

Q = IT

I = 0.5A

T = 45 x 60

T = 2700s

Q = 0.5 x 2700

= 1350c

Molarmass = 64gmol-1

no of charge = + 2

Mass = \(\frac{64 \times 1350}{96500 \times 2}\)

Mass = 0.448g

33.
C3H5 - COH3 =CH2
The IUPAC nomenclature of the structure above is
.
A. 3-methybut-3-ene
B. 2-methylbut-1-ene
C. 2-ethylprop-1-ene
D. 2-methylbut-2-ene
Explanation.
CH2CH2 - CCH3CH2
Start the numbering from the terminal carbon.

34.
The reddish-brown rust on ion roofing sheets consists of
.
A. Fe3 + (H2O)6
B. FeO.H2O
C. Fe2O3.3H2O
D. Fe3O4.2H22O
Explanation.
Iron [Fe] reacts with H2 in the presence of oxygen to form a rust.

4Fe + 3O2 ? 2 Fe2O2

Fe2O3 + H2O ? Fe2O3.H2O

35.

The densities of two gases, X and Y are 0.5gdm-3 and 2.0gdm-3 respectively. What is the rate of diffusion of X relative to Y?

.
A.

0.1

B.

0.5

C.

2

D.

4

Explanation.

The rate of dimension of a gas inversely proportional to the square root of its molecular mass or its density, which is Graham's Law of diffusion of gas.

R α \(\frac{1}{\sqrt{Mm}}\) or R α \(\frac{1}{\sqrt{D}}\)

Dx = 0.5gdm-3, Dy = 2gdm-3

R= \(\frac{K}{\sqrt{D}}\)

R\(\sqrt{D}\) = k

R1\(\sqrt{D_1}\) = R1\(\sqrt{D_2}\)

Rx\(\sqrt{D_x}\) = Ry\(\sqrt{D_y}\)

\(\frac{R_x}{R_y}\) = \(\frac{\sqrt{D_y}}{\sqrt{D_x}}\)

= \(\frac{\sqrt{2}}{\sqrt{0.5}}\)

= 2.0


36.
The carbon atoms on ethane are
.
A. sp2 hybridized
B. sp3 hybridized
C. sp4 hybridized
D. sp hybridized
Explanation.
Alkane's family are sp3 hybridized

37.

According to Charle's law, the volume of a gas becomes zero at

.
A.

100°c

B.

273°c

C.

373°c

D.

0°c

Explanation.

Where -273°c = 0K

i.e -273°c + 273 = 0K

At zero kelvin, the volume of a gas becomes zero.


38.
An oxide XO2 has a vapour density of 32. What is the atomic mass of X?
.
A. 20
B. 32
C. 14
D. 12
Explanation.
Molecular mass = vapour density X2

Mm of XO2 = x + 16(2) = x + 32

vapour density = 32

? x + 32 = 32 x 2

x + 32 = 64

x = 64 - 32

x = 32

? the relative molecular mass of X is 32
Relative molecular mass = vapour density x 2

39.
The gas that can be collected by downward displacement of air is
.
A. chlorine
B. sulphur (IV) oxide
C. carbon (IV) oxide
D. ammonia
Explanation.
Upward delivery works well for hydrogen and ammonia, which are both less densed than air. Sometimes, they are collected over water.

40.

In the laboratory preparation of trioxonitrate (V) acid the nitrogen(iv) oxide formed as a by-product is removed by

.
A.

further heating

B.

adding concentrated H2SO4

C.

cooling the acid solution with cold water

D.

bubbling air through the acid solution

Explanation.

Bubbling of air through the acid solution removes deposited oxides of nitrogen.

Nitric acid is prepared in the laboratory by heating a nitrate salt with the concentrated acid.
NaNO3 + H2SO4 → NaHSO4 + HNO3

Vapours of nitric acid are condensed to a brown liquid in a receiver cooled under cold water. "Dissolved oxides of nitrogen" e.g NO2 are removed by redistillation or blowing a current of carbondioxide or dry air through the warm acid


41.
A particle that contains a protons, 10 neutrons and 10 electrons is
.
A. positive ion
B. neutral atom of a metal
C. neutral atom of a non-metal
D. negative ion
Explanation.
protons = 9

neutrons = 10

electrons = 10

Electronic configuration = 2, 8

Ground state Electronic configuration=2, 7

It means that the atom has gained an electron thereby making it have a negative ion.

When an atom donates an electron, it becomes positively charged.

When an atom accepts an electron, it becomes negatively charged

42.

Ethene is prepared industrially by

.
A.

Reforming

B.

Polymerization

C.

Distillation

D.

Cracking

Explanation.

Ethene is produced from cracking which involves breaking up large hydrocarbon molecules into smaller and more useful bits. This is achieved by using high temperatures and pressures without a catalyst.

C15H32 → 2C2H4 + C3H6 + C8H14
......heat...ethene........propene.........octane


43.
Calculate the amount in moles of a gas which occupies 10.5 dm3 at 6 atm and 30oC [R = 082 atm dm3 K-1 mol-1]
.
A. 2.536
B. 1.623
C. 4.736
D. 0.394
Explanation.
For an ideal gas PV = nRT

Amount in moles = n

Volume v = 10.5dm3

Pressure P = 6atm

Temperature T = 30°C + 273 = 303k

R, Gas constant = 0.082 atmdm3k-1 mol

Recall from ideal gas equation

pv = nRT

n = \(\frac{RV}{RT}\)

n = \(\frac{6 \times 105}{0.082 \times 303}\)

n= 2.536mol

44.
If 100cm3 of oxygen pass through a porous plug is 50 seconds, the time taken for the same volume of hydrogen to pass through the same porous plug is? [O = 16, H = 1]
.
A. 10.0s
B. 12.5s
C. 17.7s
D. 32.0s
Explanation.
Rate of diffusion \(\frac{\alpha_1}{Density}\) or \(\frac{1}{\sqrt{Mm}}\)

Rate = \(\frac{1}{time}\)

\(\frac{1}{time}\) \(\frac{\alpha_1}{Density}\) or \(\frac{1}{\sqrt{Molarmass}}\)

Time \(\alpha \sqrt{Density}\) or \(\sqrt{Molarmass}\)

At constant volume of 100cm3

\(\frac{t_{o2}}{\sqrt{Mn_{o2}}}\) = \(\frac{t_{n2}}{\sqrt{1 \times 2}}\)

\(\frac{50}{4 \sqrt{2}}\) = \(\frac{t_{n2}}{\sqrt{2}}\)

tn2= 12.5s

45.
Due to the high reactivity of sodium, it is usually stored under
.
A. water
B. mercury
C. paraffin
D. phenol
Explanation.
Na is kept under kerosene (paraffin) to avoid reactivity with air.
Paraffin is also known as Kerosene.
Na(sodium) is kept in kerosene to prevent it from coming in contact with oxygen and moisture. If this happens, it will react with the moisture present in air and form sodium hydroxide.

46.
The furring of kettles is caused by the presence in water of
.
A. calcium hydrogentrioxocarbonate (IV)
B. calcium trioxocarbonate (IV)
C. calcium tetraoxosulphate (VI)
D. calcium hydroxide
Explanation.
Furring of kettles is caused by the temporary hardness in water. Temporary hardness in water is caused by calcium and magnesium trioxocarbonate (IV)
CaCO3 causes the furring of kettles

47.
Water for town supply is chlorinate to make it free from
.
A. bad colour
B. bacteria
C. temporary hardness
D. permanent hardness
Explanation.
Chlorine helps to keep water from germs and bacteria.

48.

Ca(OH)2(s) + 2NH4Cl(g) → CaCl2(s) + 2H2O2(l) + X. In the reaction above X is

.
A.

NO2

B.

NH3

C.

N2O

D.

NO2

Explanation.

Balancing the chemical equation.

Ca[OH]2 + 2NH4Cl → CaCl2 + 2H2O + 2NH3

X = NH3


49.

X(g) + 3Y(g) → 2z(g) H = +ve.

if the reaction above takes place at room temperature, the G will be

.
A.

negative

B.

zero

C.

positive

D.

indeterminate

Explanation.

\(\begin{array}{c|c}
\text{Enthalpy Change [ΔH]} & \text{Entropy Change [ΔS]} & \text{Gibbs free Energy[ΔG]} \\
\hline
\text{Positive} & \text{Positive} & \text{depends on T, may be + or -} \\ \hline \text{Negative} & \text{Positive} & \text{always negative} \\ \hline \text{Negative} & \text{Negative} & \text{depends on T, may be + or -} \\ \hline \text{Positive} & \text{Negative} & \text{always positive} \end{array}\)

ΔG= ΔH - TΔS.
To determine whether ΔG will be positive or negative, the value of ΔH(change in enthalpy) and ΔS (change in entropy) must be given. Likewise the temperature.


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