During the electrolysis of copper II sulphate between platinum electrodes, if litmus solution is added to the anode compartment

the litmus turns red and oxygen is evolved

the litmus turns blue but no gas is evolved

the litmus turns blue and oxygen is evolved

the litmus turns blue and hydrogen is evolved

the litmus turns red and oxygen is evolved

A sample of hard water contains some calcium sulphate and calcium hydrogen carbonate. The total hardness may therefore be removed by

adding excess calcium hydroxide

adding a calculated amount of calcium hydroxide

2KClO3(g) MNO3? 2KCl(s) + 3O2(g) The importance of the catalyst in the reaction above is that

the reaction takes place more rapidly at a lower temperature

heating may not be required before the reaction takes place

the reaction is controllable even at a high temperature

the reaction produces large quantity of oxygen

the reaction takes place more rapidly at a lower temperature

Beryllium and Aluminium have similar properties because they

are positioned diagonally to each other

The choice of method for extracting a metal from its ores depends on the

position of the metal in the electrochemical series

position of the metal in the periodic table

(I). 3CuO(s) + 2NH3(g) -----> 3Cu(s) + 3H2O(l) + N2(g) (II). 2NH3(g) + 3Cl2(g) -----> 6HCl(g) + N2(g) (III). 4NH3(g) + 3O2(g) -----> 6H2O(l) + N2(g) The reactions represented by the equations above demonstrate the

oxidizing properties of ammonia

The ionic radii of metals are usually

greater than their atomic radii

unaffected by the charge on the ion

Suitable reagents for the laboratory preparation nitrogen are

sodium dioxonitrate(III) and ammonium chloride

sodium trioxonitrate(V) and ammonium chloride

sodium chloride and ammonium trioxonitrate(V)

sodium chloride and ammonium di-ozonitrate(III)

The collision theory explains reaction rates in terms of

frequency of collision of the reactants

frequency of collision of the products

The conductivity of an acid solution depends on the

amount of ions present and their mobilities

Elements P, Q, R, S have 6, 11, 15, 17 electrons respectively, therefore,

Q will form an electrovalent bond with S

P will form an electrovalent bond with R

Q will form a covalent bond with S

R will form an electrovalent bond with S

Q will form an electrovalent bond with S

The solubility of the solids that dissolves in a given solvent with the liberation of heat will

increase with an increase in temperature

decrease with an increase in temperature

decrease with a decrease in temperature

not be affected by changes in temperature

The situation obtained when a perfect gas expands into a vacuum is

ΔH is zero and T Δ S is positive

ΔH is positive and TΔ S is zero

ΔH is positive and T Δ S is negative

ΔH is negative and TΔ S is zero

ΔH is zero and T Δ S is positive

In the laboratory preparation of oxygen, the gas cannot be collected by displacement of air because

the density of oxygen is greater than that of air

the density of air is nearly the same as that of oxygen

JAMB UTME - Chemistry - 2018 | Green Bridge CBT

Question 1 Report

Which of the following is an example of a chemical change?

dissolution of salt in water

rusting of iron

melting of ice

separating a mixture by distillation

Answer Details

Read lesson note on Acids, Bases And Salts (JAMB)

Acids, Bases And Salts

View Answer

Question 2 Report

What is the concentration of a solution containing 2g of NaOH in 100cm3 of solution? [Na = 23, O =16, H = 1]

0.40 moldm

^{- 3}

0.50 moldm

^{- 3}

0.05 moldm

^{- 3}

0.30 moldm

^{- 3}

Answer Details

The concentration of a solution containing 2g of NaOH in 100cm3 of solution is 0.40 moldm-3. This can be calculated by using the formula: molarity (M) = number of moles of solute / volume of solution (in liters) First, we need to calculate the number of moles of NaOH in the solution. The molar mass of NaOH is (23 + 16 + 1) = 40 g/mol. So, 2g of NaOH is equal to 2/40 = 0.05 moles. Next, we need to convert the volume of the solution from cm3 to liters. 1 cm3 = 0.001 liters, so 100 cm3 = 0.1 liters. Finally, we can calculate the molarity as follows: M = 0.05 moles / 0.1 liters = 0.5 mol/L = 0.50 moldm-3 So, the concentration of the solution is 0.50 moldm-3.

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

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

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

Solubility Of Substances (Ghana Only)

Kinetic Theory Of Matter And Gas Laws

Periodic Chemistry (Ghana Only)

View Answer

Question 3 Report

The derivative of benzene that can be used in making explosives is

aniline

xylene

phenol

toluene

Answer Details

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

Chemistry Of Carbon Compounds (Nigeria Only)

View Answer

Question 4 Report

Which of the following are mixtures?

I. Petroleum

II. Rubber latex

III. Vulcanizer's solution

IV. Carbon sulphide

I, II and III

I, II and IV

I and II only

I and IV

Answer Details

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

Separation Of Mixtures And Purification Of Chemical Substances

View Answer

Question 5 Report

The salt that reacts with dilute hydrochloric acid to produce a pungent smelling gas which decolourizes acidified purple potassium tetraoxomanganate (VII) solution is

_{2}

_{4}

_{2}

_{3}

_{2}

_{2}

_{3}

Answer Details

Read lesson note on Acids, Bases And Salts (JAMB)

Acids, Bases And Salts

View Answer

Question 6 Report

The boiling of fat and aqueous caustic soda is referred to as

hydrolysis

esterification

acidification

saponification

Answer Details

The boiling of fat and aqueous caustic soda is referred to as saponification. Saponification is the process of converting fat into soap through a reaction with an alkaline substance, such as caustic soda. The reaction results in the formation of soap (a salt of a fatty acid) and glycerol. This process is important in the manufacture of soap, as it allows the fat to be converted into a useful cleaning product.

Read lesson note on Acids, Bases And Salts (JAMB)

Acids, Bases And Salts

View Answer

Question 7 Report

To what volume must \(300\text{cm}^3\) of 0.60M sodium hydroxide solution be diluted to give a 0.40M solution?

450cm

^{3}

300cm

^{3}

200cm

^{3}

150cm

^{3}

Answer Details

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

Read lesson note on SOLUBILITY OF SUBSTANCES (Nigeria Only) (WAEC)

Kinetic Theory Of Matter And Gas Laws

SOLUBILITY OF SUBSTANCES (Nigeria Only)

View Answer

Question 8 Report

What mass of Cu would be produced by the cathodic reduction of \(\mathrm{Cu}^{2+}\) when 1.60A of current passes through a solution of \(\mathrm{CuSO}_{4}\) for 1 hour. (F=96500Cmol\(^{-1}\), Cu=64)

7.64g

3.82g

1.91g

0.96g

Answer Details

The reduction reaction that occurs at the cathode during the electrolysis of CuSO4" tabindex="0" class="mjx-chtml MathJax_CHTML" id="MathJax-Element-1-Frame">4, is: Cu2+" tabindex="0" class="mjx-chtml MathJax_CHTML" id="MathJax-Element-2-Frame">2+ + 2e- -> Cu(s) From this, we can see that each Cu2+ ion requires two electrons to be reduced to copper metal. Given the current (I = 1.60 A), time (t = 1 hour = 3600 s), and Faraday's constant (F = 96500 C/mol), we can calculate the total amount of charge that passes through the solution: Q = I*t = 1.60 A * 3600 s = 5760 C Using Faraday's law, we can relate the amount of charge that passes through the solution to the number of moles of electrons transferred during the reduction reaction: n = Q/F = 5760 C / 96500 C/mol = 0.0597 mol e- Since each Cu2+ ion requires 2 electrons to be reduced to copper metal, the number of moles of copper produced is half the number of moles of electrons transferred: mol Cu = 0.0597 mol e- / 2 = 0.0299 mol Cu Finally, we can convert the moles of copper produced to grams using the molar mass of copper: mass Cu = 0.0299 mol Cu * 64 g/mol = 1.91 g Therefore, the answer is 1.91 g of Cu produced. is correct.

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

Read lesson note on Chemical Bonds (Nigeria Only) (WAEC)

Read lesson note on Energy And Energy Changes (WAEC)

Kinetic Theory Of Matter And Gas Laws

Chemical Bonds (Nigeria Only)

Energy And Energy Changes

View Answer

Question 9 Report

In electrovalency, the oxidation number of the participating metal is always

decreasing

increasing

constant

neutral

Answer Details

View Answer

Question 10 Report

When air which contains the gases Oxygen, nitrogen, carbondioxide, water vapour and the rare gases, is passed through alkaline pyrogallol and then over quicklime, the only gases left are;

nitrogen and carbondioxide

the rare gases

nitrogen and oxygen

nitrogen and the rare gases

Answer Details

Read lesson note on Air (JAMB)

Air

View Answer

Question 11 Report

At what temperature is the solubility of potassium trioxonitrate(V ) equal to that of sodium trioxonitrate (V)?

^{o}

^{o}

^{o}

^{o}

Answer Details

Read lesson note on Solubility (JAMB)

Solubility

View Answer

Question 12 Report

Calculate the pH of \(0.05\ \mathrm{mol\,dm^{-3}}\ \mathrm{H_2SO_4}\)

1.30

1.12

1.00

0.30

Answer Details

To solve this problem, we need to use the formula for calculating the pH of a solution, which is: pH = -log[H+] where [H+] is the concentration of hydrogen ions in moles per liter. The given chemical equation is: H2SO4 + 2H2O → 2H3O+ + SO42- From this equation, we can see that one molecule of sulfuric acid (H2SO4) can donate two hydrogen ions (H+) to the solution, which means that the concentration of hydrogen ions is twice the concentration of sulfuric acid. Therefore, the concentration of hydrogen ions in this solution is: [H+] = 2 x 0.05 moldm^-3 = 0.1 moldm^-3 Now we can use the formula for pH: pH = -log[H+] pH = -log(0.1) pH = 1.00 Therefore, the pH of the solution is 1.00.

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

Periodic Chemistry (Ghana Only)

View Answer

Question 13 Report

Methanoic acid mixes with water in all proportions and has about the same boiling point as water. Which of the following methods would you adopt to obtain pure water from a mixture of Sand, water and methanoic acid?

Fractional distillation

Filtration followed by distillation

Neutralization with sodium hydroxide followed by distillation

Neutralization with sodium hydroxide followed by filtration

Answer Details

Read lesson note on Acids, Bases And Salts (Ghana Only) (WAEC)

Acids, Bases And Salts (Ghana Only)

View Answer

Question 14 Report

A basic postulate of the kinetic theory of gases is that the molecules of a gas move in straight lines between collisions. This implies that

collisions are perfectly elastic

forces of repulsion exis

forces of repulsion and attraction are in equilibrium

collisions are inelastic

Answer Details

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

Kinetic Theory Of Matter And Gas Laws

View Answer

Question 15 Report

To what temperature must a gas at 273k be heated in order to double both its volume and pressure?

298K

546K

819K

1092K

Answer Details

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

Kinetic Theory Of Matter And Gas Laws

View Answer

Question 16 Report

In the upper atmosphere, the ultra-violet light breaks off a free chlorine atom from chlorofluorocarbon molecule. The effect of this is that the free chlorine atom will

be very reactive and will attack ozone

be non-reactive and will not attack ozone

not change the level of ozone in the atmosphere

increase the level of ozone in the atmosphere

Answer Details

The free chlorine atom that breaks off from a chlorofluorocarbon molecule will be very reactive and will attack ozone in the upper atmosphere. Ozone is a molecule made up of three oxygen atoms, and when the free chlorine atom reacts with ozone, it breaks the ozone molecule into two separate oxygen molecules. This reaction reduces the amount of ozone in the atmosphere, which is known as ozone depletion. Over time, this can lead to a thinning of the ozone layer, which protects life on Earth from harmful ultraviolet radiation from the sun.

Read lesson note on Environmental Pollution (JAMB)

Environmental Pollution

View Answer

Question 17 Report

If 1 litre of 2.2M sulphuric acid is poured into a bucket containing 10 litres of water and the resulting solution mixed thoroughly, the resulting sulphuric acid concentration will be

2.2M

1.1M

0.22M

0.11M

Answer Details

When 1 liter of 2.2M sulphuric acid is added to 10 liters of water, the total volume of the resulting solution is 11 liters. To find the resulting concentration of sulphuric acid, we need to use the equation: M1V1 = M2V2 where M1 is the initial concentration, V1 is the initial volume, M2 is the final concentration, and V2 is the final volume. We can plug in the values we know: M1 = 2.2M (the initial concentration of the sulphuric acid) V1 = 1L (the initial volume of the sulphuric acid) M2 = ? (the final concentration we're trying to find) V2 = 11L (the final volume of the resulting solution) Solving for M2, we get: M2 = (M1 x V1) / V2 M2 = (2.2M x 1L) / 11L M2 = 0.2M Therefore, the resulting sulphuric acid concentration is 0.2M or 0.2 moles per liter. In summary, when 1 liter of 2.2M sulphuric acid is mixed with 10 liters of water, the resulting sulphuric acid concentration is diluted to 0.2M. This is because the total volume of the resulting solution is greater than the initial volume of the sulphuric acid, which leads to a decrease in concentration.

Read lesson note on Acids, Bases And Salts (Ghana Only) (WAEC)

Acids, Bases And Salts (Ghana Only)

View Answer

Question 18 Report

How many atoms are present in 6.0g of magnesium? [Mg = 24, N.A = \(6.02 \times 10^{23}\) mol]

1.20 x 10

^{22}

2.4 x 10

^{22}

1.51 x 10

^{22}

3.02 x 10

^{22}

Answer Details

Read lesson note on Atomic Structure And Bonding (JAMB)

Atomic Structure And Bonding

View Answer

Question 19 Report

An element X forms the following compounds with chlorine; \( \mathrm{XCl_4} \), \( \mathrm{XCl_3} \), \( \mathrm{XCl_2} \). This illustrates the

law of multiple proportions

law of chemical proportions

law of simple proportions

law of conservation of mass

Answer Details

The element X forming different compounds with chlorine (XCl4, XCl3, and XCl2) illustrates the law of multiple proportions. This law states that when two elements combine to form more than one compound, the ratio of the masses of one element that combine with a fixed mass of the other element is always a whole number ratio. In this case, the ratio of chlorine to X in the different compounds (XCl4, XCl3, and XCl2) is 4:1, 3:1, and 2:1, respectively, which are all whole number ratios.

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

Read lesson note on Acids, Bases And Salts (Ghana Only) (WAEC)

Kinetic Theory Of Matter And Gas Laws

Acids, Bases And Salts (Ghana Only)

View Answer

Question 20 Report

An aqueous solution of a metal salt, M. gives a white precipitate with NaOH which dissolves in excess NaOH. With aqueous ammonia, the solution of M also gives a white precipitate which dissolves in excess ammonia Therefore the cation in M is

Answer Details

Read lesson note on Chemical Bonds (Nigeria Only) (WAEC)

Chemical Bonds (Nigeria Only)

View Answer

Question 21 Report

The knowledge of half-life can be used to

Create an element

Detect an element

Split an element

Irradiate an element

Answer Details

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

Kinetic Theory Of Matter And Gas Laws

View Answer

Question 22 Report

Which of the following separation techniques can be employed in obtaining solvent from its solution?

Evaporation

Distillation

Filtration

Precipitation

Answer Details

The separation technique that can be employed in obtaining a solvent from its solution is evaporation. Evaporation is a process that involves heating a solution to vaporize the solvent, leaving behind the solute. The vaporized solvent can then be condensed and collected as a pure liquid. This technique is commonly used in industry and laboratory settings to recover solvents from solutions, as it is a simple and effective way to purify liquids. Distillation can also be used to separate a solvent from a solution, but it is a more complex process that involves boiling the solution and then condensing the vapors in a separate apparatus. Filtration and precipitation are not suitable for separating a solvent from a solution, as they are primarily used to separate solid particles from a liquid mixture.

Read lesson note on Solubility (JAMB)

Solubility

View Answer

Question 23 Report

The periodic classification is an arrangement of the elements

atomic weights

isotopic weights

molecular weights

atomic numbers

Answer Details

The periodic classification is an arrangement of the elements based on their atomic numbers. The periodic table is a chart that lists all the known chemical elements in order of increasing atomic number, arranged in rows and columns according to their electronic structure and chemical properties. The atomic number of an element is the number of protons in the nucleus of an atom of that element. Each element has a unique atomic number, which determines its position in the periodic table. The elements are arranged in rows called periods, and in columns called groups or families. Elements in the same group have similar properties because they have the same number of valence electrons, which are the electrons in the outermost shell of the atom. The periodic table is an incredibly useful tool for chemists because it allows them to predict the properties of elements based on their position in the table. For example, elements in the same group tend to form similar compounds, so if you know the properties of one element in a group, you can often predict the properties of the other elements in that group. In summary, the periodic classification is an arrangement of the elements based on their atomic numbers. The periodic table is a chart that organizes the elements into rows and columns based on their electronic structure and chemical properties, allowing scientists to make predictions about the behavior of the elements based on their position in the table.

Read lesson note on Atomic Structure And Bonding (JAMB)

Atomic Structure And Bonding

View Answer

Question 24 Report

A quantity of electricity liberates 3.6g of Silver from its salt. What mass of aluminium Will be liberated from its salt by the same quantity of electricity? [Al = 27, Ag = 108].

2.7g

1.2g

0.9g

0.3g

Answer Details

The amount of substance liberated at an electrode during electrolysis is directly proportional to the quantity of electricity passed through the solution. This is known as Faraday's laws of electrolysis. The key to solving this problem is to recognize that the same quantity of electricity is used to liberate both silver and aluminum from their respective salts. We can use the ratio of their molar masses to determine the mass of aluminum liberated. The molar mass of silver (Ag) is 108 g/mol, while the molar mass of aluminum (Al) is 27 g/mol. This means that it takes four times as many moles of aluminum to make the same mass as one mole of silver. Since the same quantity of electricity liberates 3.6g of silver from its salt, it will liberate four times as many moles of aluminum. Therefore, the mass of aluminum liberated is: (4 moles of Al) x (27 g/mol) = 108 g So, the mass of aluminum liberated is 0.108 g, or 0.1 g to one significant figure. Therefore, the answer is option D: 0.3g.

Read lesson note on Electrolysis (JAMB)

Electrolysis

View Answer

Question 25 Report

The reaction between an organic acid and an alcohol in the presence of an acid catalyst is known as;

saponification

dehydration

esterification

hydrolysis

Answer Details

The reaction between an organic acid and an alcohol in the presence of an acid catalyst is known as esterification. Esterification is the process of forming an ester, which is a type of organic compound, from an alcohol and an acid. The acid catalyst is used to speed up the reaction by providing a proton to the reaction mixture, which helps to form the ester. Esterification results in the loss of a water molecule from the reaction mixture, which makes the reaction a type of dehydration reaction. However, it is important to note that esterification is a specific type of dehydration reaction where the products are an ester and an alcohol. So, the answer is esterification.

Read lesson note on Acids, Bases And Salts (JAMB)

Acids, Bases And Salts

View Answer

Question 26 Report

The lUPAC name for

1-methyl pentane

3-methylbutane

2-methylbutane

1-dimethyl propane

Answer Details

View Answer

Question 27 Report

\(3\mathrm{H}_2(g) + \mathrm{N}_2 \rightleftharpoons 2\mathrm{NH}_3(g)\); H= -ve

In the reaction above, lowering of temperature will

have no effect on the equilibrium position

increase the rate of both the forward and reverse reactions equally

favour the forward reactio

favour the reverse reaction

Answer Details

Read lesson note on Chemical Kinetics And Equilibrium Systems (Ghana Only) (WAEC)

Chemical Kinetics And Equilibrium Systems (Ghana Only)

View Answer

Question 28 Report

The number of electrons in the valence shell of an element of atomic number 14 is?

Answer Details

The number of electrons in the valence shell of an element can be determined by using the periodic table and the electron configuration of the element. The valence shell is the outermost shell that contains electrons that are involved in chemical reactions. For an element with atomic number 14, which is silicon, the electron configuration is 1s2 2s2 2p6 3s2 3p2. The valence shell of silicon is the third shell, which contains 3s2 and 3p2 electrons. Therefore, the number of electrons in the valence shell of silicon is 4 electrons.

Read lesson note on Atomic Structure And Bonding (JAMB)

Atomic Structure And Bonding

View Answer

Question 29 Report

The refreshing and characteristic taste of soda water and other soft drinks is as a result of the presence of

carbon(IV)oxide

carbon(ll)oxide

soda

glucose

Answer Details

Read lesson note on Acids, Bases And Salts (Nigeria Only) (WAEC)

Acids, Bases And Salts (Nigeria Only)

View Answer

Question 30 Report

The hydrogen ion concentration of a sample of orange juice is 2.0 X 10\(^{-11}\)moldm\(^{-3}\). What is its p\(^{OH}\)? [log102 = 0.3010]

3.30

13.00

10.70

11.30

Answer Details

Read lesson note on Acids, Bases And Salts (Ghana Only) (WAEC)

Acids, Bases And Salts (Ghana Only)

View Answer

Question 31 Report

A correct electrochemical series can be obtained from Na, Ca, Al, Mg, Zn, Fe, Pb, H, Cu, Hg, Ag, Au by interchanging

Al and Mg

Zn and Fe

Zn and Pb

Pb and H

Answer Details

Read lesson note on Oxidation And Reduction (JAMB)

Oxidation And Reduction

View Answer

Question 32 Report

The figure above shows the electrolysis of molten sodium chloride. Z is the

anode where the Cl

^{-}

ions are oxidized

cathode where the Cl

^{-}

ions are reduced

anode where the Na

^{+}

ions are reduced

cathode where the Na

^{+}

ions are oxidized

Answer Details

The figure shows the electrolysis of molten sodium chloride. During electrolysis, an electric current is passed through a molten or dissolved ionic compound to separate the ions. The positive ions move towards the negative electrode (cathode) and the negative ions move towards the positive electrode (anode). In the figure, the electrode connected to the positive terminal of the battery is the anode and the electrode connected to the negative terminal is the cathode. At the anode, the negatively charged chloride ions (Cl-) lose electrons and are oxidized to form chlorine gas (Cl2). At the cathode, the positively charged sodium ions (Na+) gain electrons and are reduced to form liquid sodium metal (Na). Therefore, the answer is (a) anode where the Cl- ions are oxidized. Z is the anode in the figure.

Read lesson note on Electrolysis (JAMB)

Electrolysis

View Answer

Question 33 Report

The type of bonding in \(\left[\mathrm{Cu}(\mathrm{NH}_3)_4\right]^{2+}\) is

coordinate

electrovalent

metallic

covalent

Answer Details

The type of bonding in [Cu(NH3)4]2+ is coordinate bonding. Coordinate bonding (also known as dative covalent bonding) is a type of covalent bonding where one atom (in this case, the nitrogen atom in NH3) donates a pair of electrons to another atom or ion (in this case, the copper ion Cu2+). The donating atom is called the ligand, and the receiving atom or ion is called the central metal ion. In [Cu(NH3)4]2+, each ammonia molecule (NH3) donates a lone pair of electrons on the nitrogen atom to the copper ion, forming four coordinate bonds between the ligands and the central copper ion. The presence of coordinate bonds is indicated by the use of square brackets around the coordination compound, and the charge on the compound is indicated by the superscript outside the brackets. Therefore, the answer is option A: coordinate.

View Answer

Question 34 Report

In the preparation of oxygen by heating KCIO\(_2\), in the presence of MnO\(_2\) only moderate heat is needed because the catalyst acts by

lowering the pressure of the reaction

increasing the surface area of the reaction

increasing the rate of the reaction

lowering the energy barrier of the reaction

Answer Details

The presence of MnO2 acts as a catalyst in the reaction of KCIO2 to produce oxygen. A catalyst is a substance that increases the rate of a chemical reaction without being consumed in the reaction itself. MnO2 acts by lowering the energy barrier of the reaction, which means it reduces the amount of energy required for the reaction to take place. This makes it easier for the reaction to occur, and thus the reaction proceeds at a faster rate. As a result, only moderate heat is needed to provide the initial energy required for the reaction to start. Therefore, the correct answer is: lowering the energy barrier of the reaction.

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

Kinetic Theory Of Matter And Gas Laws

View Answer

Question 35 Report

Which of these alloys contains copper?

solder

steel

permallory

bronze

Answer Details

Bronze is the alloy that contains copper. Bronze is a metal alloy composed of copper and typically other elements such as tin, aluminum, silicon, or nickel. It is known for its strength, durability, and corrosion resistance. In fact, bronze is one of the earliest alloys created by humans, and it has been used for thousands of years to make tools, weapons, and decorative objects. Solder is an alloy of lead, tin, and sometimes other metals that is used to join metals together by melting the solder and allowing it to flow into the joint. Steel is an alloy of iron and carbon, and sometimes other elements like chromium, nickel, or manganese, that is known for its strength and durability. Permallory is a nickel-iron alloy with high magnetic permeability and low coercive force, which makes it useful in the production of electrical and electronic equipment. None of these alloys contain copper.

Read lesson note on Metals And Their Compounds (JAMB)

Metals And Their Compounds

View Answer

Question 36 Report

A given amount of gas occupies \(10.0\mathrm{dm}^{5}\) at 4atm and 273°C. The number of moles of the gas present is [Molar volume of gas at s.t.p = \(22.4\mathrm{dm}^{3}\)]

0.89 mol

1.90 mol

3.80 mol

5.70 mol

Answer Details

The ideal gas law is PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature. We can use this equation to solve for the number of moles of gas present. First, we need to convert the volume from dm5 to dm3, which is the same as liters (L). So, 10.0 dm5 is equal to 10.0/1000 = 0.01 dm3 or 0.01 L. Next, we need to convert the temperature from Celsius to Kelvin by adding 273 to get 546 K. Now we can plug in the values we have into the ideal gas law: 4 atm x 0.01 L = n x 0.0821 L·atm/K·mol x 546 K Simplifying, we get: 0.04 = n x 44.8 Solving for n, we get: n = 0.04/44.8 = 0.00089 mol Finally, we can compare this value to the molar volume of a gas at standard temperature and pressure (STP), which is 22.4 L/mol. To do this, we need to convert the volume of gas we have to STP conditions. Since the temperature is already at STP (273 K), we just need to adjust the pressure. Using the ideal gas law, we can solve for the volume at STP: 1 atm x V = 0.00089 mol x 0.0821 L·atm/K·mol x 273 K Simplifying, we get: V = 0.0224 L or 22.4 dm3 Therefore, the amount of gas present is equal to 0.00089 mol, which is less than 1 mol. So the answer is 0.89 mol.

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

Read lesson note on Stoichiometry And Chemical Reactions (WAEC)

Read lesson note on States Of Matter (WAEC)

Kinetic Theory Of Matter And Gas Laws

Stoichiometry And Chemical Reactions

States Of Matter

View Answer

Question 37 Report

The general formula of alkanones is

RCHO

_{2}

RCOOH

RCOOR

Answer Details

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

Chemistry Of Carbon Compounds (Nigeria Only)

View Answer

Question 38 Report

On which of the following is the solubility of a gaseous substance dependent?

I. Nature of solvent

II. Nature of solute

III. Temperature

IV. Pressure

I, II, III and IV

I and II only

II only

I, III and IV only

Answer Details

Read lesson note on Solubility (JAMB)

Solubility

View Answer

Question 39 Report

Calculate the percentage composition of oxygen in calcium trioxocarbonate(IV) [Ca=40, C=12, O=16]

Answer Details

To calculate the percentage composition of oxygen in calcium trioxocarbonate(IV), we first need to determine the molar mass of the compound. The compound has one calcium atom (Ca), one carbon atom (C), and three oxygen atoms (O). So, the molar mass of calcium trioxocarbonate(IV) can be calculated as follows: Molar mass = (1 × atomic mass of Ca) + (1 × atomic mass of C) + (3 × atomic mass of O) = (1 × 40) + (1 × 12) + (3 × 16) = 40 + 12 + 48 = 100 g/mol Next, we need to determine the mass of oxygen in one mole of calcium trioxocarbonate(IV). The compound has three oxygen atoms, each with an atomic mass of 16 g/mol. Therefore, the total mass of oxygen in one mole of the compound is: Mass of oxygen = 3 × 16 = 48 g/mol Finally, to determine the percentage composition of oxygen in calcium trioxocarbonate(IV), we divide the mass of oxygen by the molar mass of the compound and multiply by 100. Percentage of oxygen = (Mass of oxygen / Molar mass of compound) × 100 = (48 / 100) × 100 = 48% Therefore, the correct answer is 48, which represents the percentage composition of oxygen in calcium trioxocarbonate(IV).

Read lesson note on Chemical Combination (JAMB)

Chemical Combination

View Answer

Question 40 Report

The presence of ammonia gas in a desiccator can exclusively be removed by

phosphorus (V) oxide

fused calcium chloride

calcium chloride

concentrated tetraoxosulphate (VI) acid

Answer Details

Read lesson note on Oxidation And Reduction (JAMB)

Read lesson note on Acids, Bases And Salts (Ghana Only) (WAEC)

Oxidation And Reduction

Acids, Bases And Salts (Ghana Only)

View Answer






Press & Hold to Drag Around
Click Here to Close