JAMB UTME - Chemistry - 2019

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Question 3 Report

The IUPAC name of the compound $\mathrm{CF_3CHBrCl}$ is

2-chloro-2-bromo-1,1,1-trifluoroethane

2-bromo-2-chloro-1,1,1-trifluoroethane

1-bromo-1-chloro-2,2,2-trifluoroethane

1,1,1-trifluoro-2-bromo-2-chloroethane

Periodic Chemistry (Ghana Only)

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Question 4 Report

The molecular shape and bond angle of water are respectively

linear, 180°

bent, 109.5°

tetrahedral, 109.5°

bent, 105°

Answer Details

The shape of water molecule = Bent/ V- shaped
The bond angle of water = 104.5°/ 105°

Read lesson note on Atomic Structure And Bonding (JAMB)

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Question 5 Report

If the cost of electricity required to discharge 10g of an ion $X^{3+}$ is N20.00, how much would it cost to discharge 6g of ion $Y^{2+}$?

[1 faraday = 96,500C, atomic masses are X = 27, Y = 24]

N10.00

N6.00

N20.00

N9.00

Answer Details

X

^{3 +}

+ 3e

^{-}

\to

X
3F = 27g
xF = 10g

\frac{x}{3} = \frac{10}{27} ⟹ x = \frac{10}{9} F

\frac{10}{9}

F

\equiv

N20.00
1F is equivalent to x

\frac{1}{\frac{10}{9}} = \frac{x}{20}

\frac{9}{10} = \frac{x}{20} ⟹ x = N 18.00

1F is equivalent to N18.00.
Y

^{2 +}

+ 2e

^{-}

\to

Y
2F = 24g
xF = 6g
x =

\frac{6 \times 2}{24} = \frac{1}{2} F

1F = N18.00

\frac{1}{2}

F =

\frac{1}{2} \times N 18.00

= N9.00

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Question 6 Report

In the reaction between sodium hydroxide and tetraoxosulphate (VI) solutions, what volume of 0.5 molar sodium hydroxide would exactly neutralize 10cm$^{3}$ of 1.25 molar tetraoxosulphate (vi) acid?

25cm

^{3}

10cm

^{3}

20cm

^{3}

50cm

^{3}

Answer Details

Equation of reaction : 2NaOH + H

_{2}

SO

_{4}

→ Na

_{2}

SO

_{4}

+ 2H

_{2}

O

Concentration of a base, CB = 0.5M

Volume of acid, V $_{A}$ = 10cm $^{3}$

Concentration of an acid, C $_{A}$ = 1.25M

Volume of base, V $_{B}$ = ?

Recall:
$\frac{C_{A} V_{A}}{C_{B} V_{B}} = \frac{n_{A}}{n_{B}}$ ... (1)
N.B: From the equation,
$\frac{n_{A}}{n_{B}} = \frac{1}{2}$

From (1)
$\frac{1.25 \times 10}{0.5 \times V_{B}} = \frac{1}{2}$
$\frac{12.5}{0.5 V_{B}} = \frac{1}{2}$
25 = 0.5V $_{B}$

VB = 50.0 cm $^{3}$

Read lesson note on Stoichiometry And Chemical Reactions (WAEC)

Stoichiometry And Chemical Reactions

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Question 7 Report

When the end alkyl groups of ethyl ethanoate are interchanged, the compound formed is

propylmethanoate

ethylpropanoate

methyl ethanoate

methyl propanoate

Answer Details

The compound formed when the end alkyl groups of ethyl ethanoate are interchanged is ethyl propanoate. This is because ethyl ethanoate consists of two parts: the "ethyl" group and the "ethanoate" group. The ethyl group is a two-carbon chain, and the ethanoate group is a combination of a one-carbon chain and a carbonyl group (C=O) that is also attached to an oxygen atom. When the end alkyl groups are interchanged, the "ethyl" group is moved from the second carbon to the first carbon of the ethanoate group, and the "propanoate" group is formed. The "propanoate" group consists of a three-carbon chain and the carbonyl group. Therefore, the resulting compound is ethyl propanoate, which has a chemical formula of CH3CH2COOCH2CH3. This compound is commonly used as a flavoring agent and has a fruity odor reminiscent of pears.

Read lesson note on Chemistry Of Carbon Compounds (Nigeria Only) (WAEC)

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Question 8 Report

Which of the following pollutants will lead to the depletion of ozone layer?

chlorofluorocarbon

carbon (ii) oxide

hydrogen sulphide

sulphur (iv) oxide

Answer Details

The pollutant that leads to the depletion of the ozone layer is chlorofluorocarbon (CFCs). CFCs are man-made chemicals that were widely used in the past as refrigerants, solvents, and propellants. When CFCs are released into the atmosphere, they rise into the stratosphere, where they come into contact with ozone molecules. The chlorine atoms in CFCs react with ozone, breaking apart the ozone molecules and causing a reduction in the overall amount of ozone in the stratosphere. This process continues until all of the ozone-depleting chlorine atoms have been depleted. The resulting decrease in ozone in the stratosphere leads to an increase in the amount of harmful ultraviolet radiation that reaches the Earth's surface, which can have negative impacts on human health and the environment.

Read lesson note on Environmental Pollution (JAMB)

Environmental Pollution

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Question 9 Report

When chlorine water is exposed to bright sunlight, the following products are formed

HCl

_{(a q)}

and Cl

_{2}

HOCl

_{(a q)}

and Cl

_{2}

HCl

_{(a q)}

and O

_{2}

HOCl

_{(a q)}

and O

_{2}

Chemistry Of Carbon Compounds (Ghana Only)

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Question 10 Report

$ \mathrm{SO_3} $ is not directly dissolved in water in the industrial preparation of $ \mathrm{H_2SO_4} $ by the contact process because

SO

_{3}

is an acidic gas

Acid is usually added to water and not water to acid

The reaction between SO

_{3}

and water is violently exothermic

SO

_{3}

does not dissolve readily in water

Acids, Bases And Salts

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Question 11 Report

Which of the following is a set of neutral oxides?

N

_{2}

O, H

_{2}

O, CO

N

_{2}

O, SO

_{2}

, CO

N

_{2}

O, Na

_{2}

O, ZnO

N

_{2}

O, SO

_{3}

, NO

Oxidation And Reduction

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Question 12 Report

Which of the following conditions will most enhance the spontaneity of a reaction?

Δ

H is negative and greater than T

Δ

S

Δ

H is negative and

Δ

S = 0

Δ

H is positive and less than T

Δ

S

Δ

H is positive and equal to T

Δ

S

Answer Details

The condition that will most enhance the spontaneity of a reaction is when ΔH is negative (i.e., the reaction releases heat) and ΔS is positive (i.e., the reaction increases the disorder or randomness of the system). This is because a negative ΔH indicates that the reaction releases energy, which is favorable for a spontaneous reaction, while a positive ΔS indicates that the system becomes more disordered, which is also favorable for spontaneous reactions. Among the given options, the first condition of a negative and greater ΔH than ΔS is the best option for enhancing the spontaneity of a reaction. The other options have either a positive ΔH or a zero ΔS, which is not favorable for spontaneous reactions.

Read lesson note on Kinetic Theory Of Matter And Gas Laws (JAMB)

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Question 13 Report

The emission of two successive beta particles from the nucleus $ {}^{32}_{15}\mathrm{P} $ will produce

_{14}^{30}

Si

_{13}^{32}

Al

_{17}^{32}

Cl

_{16}^{32}

S

Answer Details

Explanation:

_{15}^{32}

P

\to

2

_{- 1}^{0}

e +

_{b}^{a}

X
32 = 2(0) + a
a = 32
15 = 2(-1) + b
b = 15 + 2
b = 17

∴

_{b}^{a}

X =

_{17}^{32}

Cl

Read lesson note on Structure Of The Atom (WAEC)

Structure Of The Atom

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Question 14 Report

Which of the following properties increases from left to right along the period but decreases down the group in the Periodic Table?

I. Atomic Number ii. Ionization energy iii. Metallic character iv. Electron affinity

ii and iii only

ii and iv only

iii and iv only

i and ii only

Answer Details

Ionization energy and electron affinity increase across a period, and decrease down a group.

Read lesson note on Periodic Chemistry (Nigeria Only) (WAEC)

Periodic Chemistry (Nigeria Only)

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Question 15 Report

When ammonia and hydrogen ion bond together to form ammonium ion, the bond formed is called

electrovalent bond

co-ordinate bond

ionic bond

covalent bond

Answer Details

When ammonia and hydrogen ion go into bonding, they form ammonium ion by combining with a dative/coordinate covalent bond.

Read lesson note on Oxidation And Reduction (JAMB)

Read lesson note on Atomic Structure And Bonding (JAMB)

Oxidation And Reduction

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Question 16 Report

Which of the following does not support the fact that air is a mixture?

the constituents of air are in a fixed proportion by mass

it cannot be represented with a chemical formula

the constituents of air can be separated by physical by physical means

none of the above

Answer Details

The option that does not support the fact that air is a mixture is "the constituents of air are in a fixed proportion by mass". Air is a mixture of different gases, primarily nitrogen (78%) and oxygen (21%), with small amounts of other gases such as carbon dioxide, argon, and neon. The proportion of each gas in air is not fixed and can vary depending on the location and other factors. For example, the amount of carbon dioxide in air can increase in areas with high levels of pollution, while the proportion of oxygen can decrease at high altitudes. Therefore, the composition of air is not in a fixed proportion by mass. On the other hand, the fact that air cannot be represented with a chemical formula and its constituents can be separated by physical means support the fact that air is a mixture. A chemical formula represents a pure substance, and since air is a mixture of gases, it cannot be represented by a single formula. Air can be separated into its individual components through physical means such as distillation or filtration, which is a characteristic of mixtures.

Read lesson note on Air (JAMB)

Air

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Question 17 Report

A solution X, on mixing with $ \mathrm{AgNO_3} $ solution gives a white precipitate soluble in aqueous $ \mathrm{NH_3} $, a solution Y, when also added to X, also gives a white precipitate which is soluble when heated solutions X and Y respectively contain

Al

^{3 +}

, Cl

^{-}

Hg

^{+}

, SO

_{4}^{2 -}

Pb

^{2 +}

, Cl

^{-}

Ag

^{+}

, SO

_{4}^{2 -}

Chemistry Of Carbon Compounds (Ghana Only)

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Question 18 Report

The shapes of water, ammonia, carbon (iv) oxide and methane are respectively

linear, tetrahedral, angular and pyramidal

angular, pyramidal, linear and tetrahedral

pyramidal, linear, tetrahedral and angular

tetrahedral, angular, pyramidal and linear

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Question 19 Report

$200\text{cm}^3$ of $0.50\text{mol}/\text{dm}^3$ solution of calcium hydrogen trioxocarbonate (IV) is heated. The maximum weight of solid precipitated is

10g

25g

20g

15g

Answer Details

To solve this problem, we need to use the concept of stoichiometry and the solubility product constant (Ksp) of calcium hydrogen trioxocarbonate (IV). First, we need to write the balanced equation for the reaction that occurs when the solution of calcium hydrogen trioxocarbonate (IV) is heated: Ca(HCO3)2(s) → CaCO3(s) + H2O(g) + CO2(g) From the balanced equation, we can see that 1 mole of calcium hydrogen trioxocarbonate (IV) produces 1 mole of calcium carbonate. Therefore, we need to determine the number of moles of calcium hydrogen trioxocarbonate (IV) in the solution: Number of moles = concentration x volume Number of moles = 0.50 mol/dm³ x 0.2 dm³ Number of moles = 0.1 mol Since 1 mole of calcium hydrogen trioxocarbonate (IV) produces 1 mole of calcium carbonate, the number of moles of calcium carbonate produced will also be 0.1 mol. Next, we need to use the solubility product constant (Ksp) of calcium carbonate to determine the maximum amount of solid that can be precipitated: Ksp = [Ca²⁺][CO3²⁻] Ksp = 3.3 x 10⁻⁹ (at 25°C) At the maximum amount of solid precipitated, all the calcium carbonate formed will have precipitated, and the concentration of calcium ions and carbonate ions will be equal. Therefore, we can assume that the concentration of calcium ions and carbonate ions is both x. Substituting into the Ksp expression: Ksp = x² 3.3 x 10⁻⁹ = x² x = 5.74 x 10⁻⁵ mol/dm³ The mass of calcium carbonate precipitated can now be calculated: Mass = number of moles x molar mass Mass = 0.1 mol x 100.1 g/mol Mass = 10.01 g Therefore, the maximum weight of solid precipitated is approximately 10 g. Note that this calculation assumes that all the calcium carbonate precipitated as a solid, which may not always be the case in a real-world experiment. Additionally, this calculation does not take into account any losses due to filtration or other experimental errors.

Read lesson note on Kinetic Theory Of Matter And Gas Laws (JAMB)

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Question 20 Report

The velocity, V of a gas is related to its mass, M by (k = proportionality constant)

V =

\frac{k}{M}

V =

\frac{k}{M^{\frac{1}{2}}}

V =

k M^{2}

V =

(\frac{k}{M})^{\frac{1}{2}}

Answer Details

Recall:
V =

\sqrt{\frac{3 R T}{M}}

∴ V \propto \frac{1}{\sqrt{M}}

V = \frac{k}{\sqrt{M}}

V =

\frac{k}{M^{\frac{1}{2}}}

Read lesson note on Kinetic Theory Of Matter And Gas Laws (JAMB)

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Question 21 Report

Elements in the periodic table are arranged in the order of their

atomic numbers

metallic character

mass number

relative atomic mass

Answer Details

Elements in the periodic table are arranged in the order of their atomic numbers. The atomic number of an element is the number of protons in the nucleus of an atom of that element. The elements are arranged in order of increasing atomic number from left to right and from top to bottom in the periodic table. The elements in each row, also known as a period, have the same number of electron shells, while the elements in each column, also known as a group or family, have the same number of valence electrons. This arrangement makes it possible to predict the chemical and physical properties of an element based on its position in the periodic table. Therefore, the correct answer is: - atomic numbers

Read lesson note on Atomic Structure And Bonding (JAMB)

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Question 22 Report

At 27°C, 58.5g of sodium chloride is present in 250cm$^3$ of a solution. The solubility of sodium chloride at this temperature is?

(molar mass of sodium chloride = 111.0gmol$^{-1}$)

2.0 moldm

^{- 3}

0.25 moldm

^{- 3}

1.0 moldm

^{- 3}

0.5 moldm

^{- 3}

Answer Details

Given the Mass of the salt = 58.5g
Volume = 250 cm

^{3}

= 0.25 dm

^{3}

Mass concentration =

\frac{M a s s}{V o l u m e}

=

\frac{58.5}{0.25}

= 234 gdm

^{- 3}

Solubility (in moldm

^{- 3}

=

\frac{234}{111}

= 2.11 moldm

^{- 3}

≊

2.0 moldm

^{- 3}

Read lesson note on Solubility Of Substances (Ghana Only) (WAEC)

Solubility Of Substances (Ghana Only)

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Question 23 Report

The IUPAC nomenclature of the compound

$\mathrm{H_3C - CH(CH_3) - CH(CH_3) - CH_2 - CH_3}$

3,4- dimethylhexane

2,3- dimethyl pentane

2- ethyl hexane

2,3- dimethylhexane

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Question 24 Report

Which of the following statements about catalyst is false?

biological catalysts are called enzymes

catalyst do not appear in the balanced equation

catalyst reduce the activation energy for a reaction

catalysts do not alter the mechanism of the reaction and never appear in the rate low

Answer Details

The false statement about catalysts is: "catalysts do not alter the mechanism of the reaction and never appear in the rate law." Catalysts are substances that speed up chemical reactions without being consumed in the process. They achieve this by reducing the activation energy needed for the reaction to occur. Enzymes are a type of biological catalysts. In a chemical reaction, a catalyst is not consumed and does not appear in the overall balanced equation. However, catalysts can alter the mechanism of a reaction by providing an alternative pathway with a lower activation energy. This alternative pathway can have a different rate-determining step, which means that the presence of the catalyst can change the rate law of the reaction. Therefore, the statement that catalysts do not alter the mechanism of the reaction and never appear in the rate law is false.

Read lesson note on Chemical Kinetics And Equilibrium System (Nigeria Only) (WAEC)

Chemical Kinetics And Equilibrium System (Nigeria Only)

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Question 25 Report

Which process(es) is/are involved in the turning of starch iodide paper blue-black by chlorine gas?

chlorine attacks the starch to give the blue-black color

chlorine attacks the iodide ion to give the blue-black color

chlorine oxidizes the iodide ion to produce iodine which attacks the starch to give the blue-black color

iodine attacks the starch to give the blue-black color

Answer Details

The process involved in the turning of starch iodide paper blue-black by chlorine gas is option number 3: chlorine oxidizes the iodide ion to produce iodine which attacks the starch to give the blue-black color. When chlorine gas comes in contact with iodide ions on the starch iodide paper, it oxidizes the iodide ion to form iodine. The iodine that is produced in this reaction is then able to react with the starch present on the paper to form a blue-black complex. This blue-black complex is formed due to the arrangement of the starch molecules and the iodine atoms in a way that causes them to absorb light at a specific wavelength, giving the blue-black color. Therefore, the blue-black color that is observed on the starch iodide paper is due to the reaction between iodine and starch, which is made possible by the oxidation of iodide ions by chlorine gas.

Read lesson note on Kinetic Theory Of Matter And Gas Laws (JAMB)

Read lesson note on Energy And Energy Changes (WAEC)

Energy And Energy Changes

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Question 26 Report

Which of the following will give a precipitate with an aqueous solution of copper (I) chloride?

But-2-ene

But-1-ene

But-2-yne

But-1-yne

Rates Of Chemical Reaction

Chemical Bonds (Ghana Only)

Redox Reactions

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Question 27 Report

A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen. If the molar mass of the compound is 180. Find the molecular formula.

[H = 1, C = 12, O = 16]

C

_{3}

H

_{6}

O

_{3}

C

_{6}

H

_{6}

O

_{3}

C

_{6}

H

_{12}

O

_{6}

CH

_{2}

O

Answer Details

The molecular formula of a compound is determined by the number of atoms of each element present in the molecule. To find the molecular formula, we need to determine the number of atoms of each element in the compound. First, we convert the percent composition to grams. For example, 40.0% carbon means 40.0 g of carbon per 100 g of compound. Then we divide the number of grams of each element by the molar mass of each element. For example, 40.0 g of carbon divided by the molar mass of carbon (12 g/mol) gives us 3.33 mol of carbon. Next, we convert the number of moles of each element to the number of atoms by multiplying the number of moles by Avogadro's number (6.022 x 10^23 atoms/mol). Finally, we balance the numbers of atoms of each element by dividing them by the smallest number of atoms of all the elements and rounding to the nearest whole number. In this case, the smallest number of atoms is 2, which is the number of hydrogen atoms. So, we divide the number of atoms of carbon and oxygen by 2 to balance the numbers of atoms of all the elements. Therefore, the molecular formula of the compound is C6H12O6.

Read lesson note on Chemistry Of Carbon Compounds (Nigeria Only) (WAEC)

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Question 28 Report

What technique is suitable for separating a binary solution of potassium chloride and potassium trioxochlorate (V)?

Fractional crystallization

Fractional distillation

Filtration

Evaporation

Answer Details

Fractional crystallization is the most suitable technique for separating a binary solution of potassium chloride and potassium trioxochlorate (V). This is because fractional crystallization is a process that separates a mixture of substances based on their solubility in a solvent at a particular temperature. In this case, potassium chloride and potassium trioxochlorate (V) have different solubilities in a solvent such as water at different temperatures. By carefully controlling the temperature, the solubility of each compound can be selectively increased or decreased, allowing them to be separated by crystallization. The less soluble compound will form crystals first and can be separated from the more soluble compound, which remains in the solution. Therefore, fractional crystallization can be used to separate potassium chloride and potassium trioxochlorate (V) in a binary solution.

Read lesson note on Separation Of Mixtures And Purification Of Chemical Substances (JAMB)

Separation Of Mixtures And Purification Of Chemical Substances

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Question 29 Report

Which of the following metals is the most essential in the regulation of blood volume, blood pressure and osmotic equilibrium?

Zinc

Manganese

Sodium

Iron

Answer Details

The metal that is most essential in the regulation of blood volume, blood pressure, and osmotic equilibrium is sodium. Sodium is a key electrolyte that helps maintain the balance of fluids in the body, including blood volume and blood pressure. Sodium ions are positively charged and are attracted to negatively charged ions, such as chloride (Cl-) and bicarbonate (HCO3-), which together help regulate the pH of the blood. Sodium is also essential for maintaining osmotic equilibrium, which refers to the balance of solutes between cells and the extracellular fluid. Osmotic equilibrium is critical for proper cellular function and is regulated by the movement of water and electrolytes, including sodium, in and out of cells. While the other metals listed (zinc, manganese, and iron) are important for various functions in the body, such as enzyme activity and oxygen transport, they are not directly involved in regulating blood volume, blood pressure, and osmotic equilibrium in the same way that sodium is. Therefore, the answer is not options 1, 2, or 4, and the correct answer is: sodium.

Read lesson note on Metals And Their Compounds (JAMB)

Metals And Their Compounds

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Question 30 Report

Which of the following alkaline metals react more quickly spontaneously with water?

Calcium

Beryllium

Magnesium

Strontium

Answer Details

The alkaline earth metals (Be, Mg, Ca, Sr, Ba, and Ra) are the second most reactive metals in the periodic table, and, like the Group 1 metals, have increasing reactivity in the higher periods. Beryllium (Be) is the only alkaline earth metal that does not react with water or steam, even if metal is heated to red heat. Additionally, beryllium has a resistant outer oxide layer that lowers its reactivity at lower temperatures.

Magnesium shows insignificant reaction with water, but burns vigorously with steam or water vapor to produce white magnesium oxide and hydrogen gas:

Mg(s) + 2 H2O(l) ⟶ Mg(OH)2(s) + H2(g)

A metal reacting with cold water will produce metal hydroxide. However, if a metal reacts with steam, like magnesium, metal oxide is produced as a result of metal hydroxides splitting upon heating.

The hydroxides of calcium, strontium and barium are only slightly water-soluble but produce sufficient hydroxide ions to make the environment basic, giving a general equation of:

M(s) + 2 H2O(l) ⟶ M(OH)2(aq) + H2(g)

Order of reactivity	Metal	Reactions with water or steam
most reactive	potassium (K)	very vigorous reaction with cold water
↑	sodium (Na)	vigorous reaction with cold water
↓	calcium (Ca)	less vigorous reaction with cold water
least reactive	magnesium (Mg)	slow reaction with cold water, vigorous with steam

If metals react with cold water, hydroxides are produced.
Metals that react with steam form oxides.
Hydrogen is always produced when a metal reacts with cold water or steam.

Read lesson note on Metals And Their Compounds (JAMB)

Read lesson note on Organic Compounds (JAMB)

Metals And Their Compounds

Organic Compounds

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Question 31 Report

The reaction $2H_2S + SO_2 \rightarrow 3S + 2H_2O$ is

not a redox reaction because there is no oxidant in the reaction

not a redox reaction because there is no reductant in the reaction

a redox reaction in which H

_{2}

S is the oxidant and SO

_{2}

is the reductant

a redox reaction in which SO

_{2}

is the oxidant and H

_{2}

S is the reductant

View Answer

Question 32 Report

Methane is prepared in the laboratory by heating a mixture of sodium ethanoate with soda lime. The chemical constituent(s) of soda lime is/are

Ca(OH)

_{2}

and NaOH

Na

_{2}

CO

_{3}

NaCl

NaOH

Answer Details

The chemical constituent of soda lime used to prepare methane in the laboratory is Ca(OH)2 (calcium hydroxide) and NaOH (sodium hydroxide). Soda lime is a mixture of these two compounds. When sodium ethanoate (NaC2H3O2) is heated with soda lime, it undergoes a reaction known as the Kolbe's reaction, which produces methane gas (CH4) as one of the products. The reaction can be represented as follows: 2NaC2H3O2 + 2Ca(OH)2 → 2CH4 + 2NaOH + 2CaCO3 In this reaction, the sodium ethanoate reacts with the calcium hydroxide to form calcium acetate (Ca(C2H3O2)2) and sodium hydroxide. The calcium acetate then decomposes to produce methane gas and calcium carbonate (CaCO3), which is a solid precipitate. Therefore, the chemical constituents of soda lime used to prepare methane in the laboratory are calcium hydroxide (Ca(OH)2) and sodium hydroxide (NaOH).

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Chemical Combination

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Question 33 Report

Which of the following statements does not show Rutherford's account of Nuclear Theory? An atom contains a region

which contains protons and neutrons

which is positively charged

which is massive and can cause deflection of a few projectiles

which is very large and in which close to 98% of projectiles pass undeflected

Answer Details

Rutherford's account of Nuclear theory does not include the fact that atoms contain a massive region and cause deflection of from projectiles.

Read lesson note on Atomic Structure And Bonding (JAMB)

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Question 34 Report

Which of the following sets of operation will completely separate a mixture of sodium chloride, sand and iodine?

addition of water, filtration, evaporation to dryness and sublimation

filtration, evaporation to dryness, addition of water and sublimation

sublimation, filtration, evaporation and addition of water

sublimation, addition of water, filtration and evaporation to dryness

Answer Details

The set of operations that will completely separate a mixture of sodium chloride, sand, and iodine is: - filtration, to separate the sand and iodine from the sodium chloride - evaporation to dryness, to concentrate the sodium chloride solution and remove any remaining water - sublimation, to separate the iodine as a solid from the remaining sodium chloride By using these operations, you can separate each component of the mixture into separate, pure forms. The order of the operations is important because each step must be done in a way that effectively separates the components and does not interfere with subsequent steps.

Read lesson note on Separation Of Mixtures And Purification Of Chemical Substances (JAMB)

Separation Of Mixtures And Purification Of Chemical Substances

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Question 35 Report

Which of the following represents the kind of bonding present in ammonium chloride?

ionic only

covalent only

ionic and dative covalent

dative covalent

Answer Details

Ammonium chloride contains both ionic and covalent bonds. In ammonium chloride, the ammonium ion (NH4+) is positively charged and the chloride ion (Cl-) is negatively charged. These ions are held together by ionic bonds, which are formed between positively and negatively charged ions. However, the bond between the hydrogen atom in the ammonium ion and the nitrogen atom in the ammonium ion is also a covalent bond. This type of covalent bond is known as a dative covalent bond, or a coordinate covalent bond, because the electron pair being shared is supplied by one atom only (the nitrogen atom in this case). So, the kind of bonding present in ammonium chloride is both ionic and dative covalent. In simple terms, ammonium chloride contains both ionic bonds between its positive and negative ions, and a dative covalent bond between the hydrogen atom and nitrogen atom within the ammonium ion.

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Question 36 Report

By what amount must the temperature of 200cm$^{3}$ of Nitrogen at 27°C be increased to double the pressure if the final volume is 150cm$^{3}$ (Assume ideality)

150°C

177°C

75°C

120°C

Answer Details

Using the ideal gas law and equation:

\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}

\frac{P_{1} \times 200 {cm}^{3}}{300 K} = \frac{2 P \times 150 {cm}^{3}}{T_{2}}

Cross multiply:

T_{2} = \frac{300 \times 150 \times 2 P}{200 \times P}

= 450 K

or

177^{\circ} C

Don't forget to convert to

^{\circ} C

Read lesson note on Kinetic Theory Of Matter And Gas Laws (JAMB)