JAMB UTME - Chemistry - 2018

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

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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.

2KClO3(g) MNO3? 2KCl(s) + 3O2(g)

The importance of the catalyst in the reaction above is that

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

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During the electrolysis of copper II sulphate between platinum electrodes, if litmus solution is added to the anode compartment, the litmus will turn red and oxygen gas will be evolved. This is because during electrolysis, the positively charged copper ions (Cu2+) in the copper II sulphate solution are attracted to the negative cathode electrode, where they gain electrons and are reduced to form solid copper. At the same time, the negatively charged sulphate ions (SO42-) are attracted to the positive anode electrode, where they lose electrons and are oxidized to form oxygen gas and water. The litmus added to the anode compartment turns red because of the formation of oxygen gas, which is a highly reactive oxidizing agent that can react with the litmus to cause it to turn red. No hydrogen gas is evolved because hydrogen is produced at the cathode, which is in a separate compartment from the anode where the litmus is added.

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

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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.

Which of the following are mixtures? 

I. Petroleum 

II. Rubber latex 

III. Vulcanizer's solution 

IV. Carbon sulphide

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The Consecutive members of an alkane homologous series differ by

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The consecutive members of an alkane homologous series differ by a CH2 unit. This means that each successive member of the alkane series has one more CH2 unit than the previous member. For example, consider the simplest alkane, methane (CH4). The next member of the series is ethane (C2H6), which differs from methane by one CH2 unit. The next member after that is propane (C3H8), which differs from ethane by another CH2 unit. This pattern continues for all members of the alkane homologous series. The reason for this is that each carbon atom in the alkane chain must be bonded to four other atoms, which are usually hydrogen atoms. This means that each carbon atom in the chain can only bond to one other carbon atom. Therefore, the length of the alkane chain can only increase by adding CH2 units to the end of the chain. In summary, the consecutive members of an alkane homologous series differ by a CH2 unit because this is the only way to add length to the alkane chain while maintaining the required number of bonds for each carbon atom in the chain.

H${}_2$ S${}_{\left(g\right)}$ + Cl${}_{2\left(g\right)}$ → 2HCl${}_{\left(g\right)}$ + S${}_{\left(g\right)}$ In the reaction above, the substance that is reduced is

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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

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2SO${}_2$ ${}_{\left(g\right)}$+ O${}_2$ ${}_{\left(g\right)}$ ↔ 2SO${}_3$${}_{\left(g\right)}$ ΔH = -395.7kJmol${}^{-1}$

In the equation, an increase in temperature will shift the equilibrium position to the

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Which of the following produces relatively few ions in solution?

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The correct answer is AI(OH)3. When ionic compounds dissolve in water, they dissociate into their constituent ions, producing charged particles in solution. The more ions a compound produces, the more conductive it is in solution. AI(OH)3, also known as aluminum hydroxide, produces relatively few ions in solution because it is a weak base. When AI(OH)3 dissolves in water, it releases a small amount of Al3+ and OH- ions. In contrast, NaOH, KOH, and Ca(OH)2 are strong bases that dissociate more completely in water and produce more ions in solution. NaOH and KOH produce one hydroxide ion for every sodium or potassium ion, while Ca(OH)2 produces two hydroxide ions for every calcium ion. Therefore, of the options listed, AI(OH)3 produces relatively few ions in solution.

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

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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.

The periodic classification is an arrangement of the elements

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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.

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

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Which of the following metals cannot replace hydrogen from water or steam?

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Which of the following pairs of substances will react further with oxygen to form a higher oxide?

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

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The choice of method for extracting a metal from its ores depends on the position of the metal in the electrochemical series. The electrochemical series is a list of metals arranged in order of their ability to gain or lose electrons. The metals at the top of the series (such as sodium and potassium) are very reactive and will readily lose electrons, while those at the bottom (such as gold and platinum) are less reactive and less likely to lose electrons. The position of a metal in the electrochemical series determines the method of extraction that should be used. For example, metals at the top of the series are usually extracted by electrolysis, which involves passing an electric current through a molten compound of the metal. This process is necessary because the metals at the top of the series are very reactive and are strongly bonded to other elements in their ores. On the other hand, metals at the bottom of the series are usually extracted by reduction with carbon or hydrogen. This is because these metals are less reactive and can be separated from their ores by reacting them with a reducing agent that can take away the oxygen and other impurities. Therefore, the position of the metal in the electrochemical series is a crucial factor in determining the method of extraction that should be used to extract it from its ores.

Aluminium does not react with either dilute or concentrated trioxonitrate (V) acid because

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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?

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What mass of Cu would be produced by the cathodic reduction of Cu${}^{2+}$  when 1.60A of current passes through a solution of CuSO${}_4$ for 1 hour. (F=96500Cmol${}^{-1}$ , Cu=64)

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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.

3H${}_{2\left(g\right)}$+ N2 ⇔ 2NH${}_{3\left(g\right)}$ ; H= -ve

In the reaction above, lowering of temperature will

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The lUPAC name for

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

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The collision theory explains reaction rates in terms of

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The collision theory explains reaction rates in terms of the frequency of collision of the reactants. In other words, the theory suggests that for a chemical reaction to occur, the reactant particles must collide with sufficient energy and with the correct orientation. The frequency of these collisions is an important factor in determining the rate of the reaction. The more frequently the reactant particles collide, the more likely it is that they will react and form products. Therefore, increasing the frequency of collisions between reactant particles can increase the rate of a chemical reaction. The size of the reactants or the products does not play a significant role in the collision theory.

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].

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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.

The Sulphide which is insoluble in dilute hydrochloric acid is

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The sulphide which is insoluble in dilute hydrochloric acid is Copper Sulphide (CuS). When metal sulphides react with hydrochloric acid, they undergo an acid-base reaction to produce hydrogen sulphide gas and the corresponding metal chloride. For example, when Iron Sulphide (FeS) reacts with hydrochloric acid, it forms hydrogen sulphide gas (H2S) and iron chloride (FeCl2) as follows: FeS + 2HCl → H2S + FeCl2 However, Copper Sulphide (CuS) does not react with dilute hydrochloric acid, as it is insoluble in this acid. This is due to the fact that CuS is a much less reactive metal sulphide compared to FeS and ZnS, and therefore it does not undergo an acid-base reaction with dilute hydrochloric acid. In summary, CuS is the sulphide which is insoluble in dilute hydrochloric acid due to its low reactivity with acids.

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

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Charles' law states that the volume of a gas is directly proportional to its temperature, provided that the pressure remains constant. This means that as the temperature of a gas increases, its volume also increases. However, it is important to note that this law only applies to ideal gases, which are theoretical gases that perfectly follow the laws of thermodynamics. According to Charles' law, the volume of a gas becomes zero at absolute zero, which is approximately -273°C. At this temperature, the gas particles would have no kinetic energy and would be in their lowest energy state. The volume of a real gas would not actually become zero at absolute zero because the gas particles would have some residual intermolecular interactions that would prevent them from completely collapsing to a single point.

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]

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

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

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The general formula of alkanones is

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According to the Kinetic Theory an increase in temperature causes the kinetic energy of particles to

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The kinetic energy of particles increases with an increase in temperature. In the Kinetic Theory, temperature is related to the average kinetic energy of the particles in a substance. The higher the temperature, the faster the particles move, and the more energy they have. Think of it like this: if you throw a ball, it will have more energy and travel farther if you throw it harder. Similarly, if you heat up a substance, its particles will move faster and have more energy. So, the answer is that an increase in temperature causes the kinetic energy of particles to increase.

In the preparation of oxygen by heating KCIO, in the presence of MnO${}_2$ only moderate heat is needed because the catalyst acts by ${}_2$

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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.

(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

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Which of the following statements is correct about the periodic table?

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An element X forms the following compounds with chlorine; XCl${}_4$, XCl${}_3$, XCl${}_2$. This illustrates the

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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.

The radio isotope used in industrial radiography for the rapid checking of faults in welds and casting is?

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In the reaction between sodium hydroxide and sulphuric acid solutions, what volume of 0.5 molar sodium hydroxide would exactly neutralise 10cm3 of 1.25 molar sulphuric acid?

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

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

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The knowledge of half-life can be used to

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