JAMB UTME - Chemistry - 2018
Swali 1 Ripoti
ME + nF -----> pG + qH
In the equation shown, the equilibrium constant is given by?
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The equilibrium constant for a chemical reaction is a measure of the balance between the reactants and products of a reaction at a particular temperature. The equilibrium constant is given by the ratio of the product of the concentration of the products raised to their stoichiometric coefficients, to the product of the concentration of the reactants raised to their stoichiometric coefficients. In the equation ME + nF -> pG + qH, the correct expression for the equilibrium constant is [G]^p * [H]^q / [E]^m * [F]^n, represented by.
Swali 2 Ripoti
Elements P, Q, R, S have 6, 11, 15, 17 electrons respectively, therefore,
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Elements form bonds with other elements in order to attain a stable electron configuration, like the one found in noble gases. There are two types of bonds: covalent and ionic (also called electrovalent). In covalent bonds, two elements share electrons to attain a stable electron configuration. This type of bond is formed between two non-metal elements. In ionic bonds, one element donates electrons to another element, creating ions. This type of bond is formed between a metal and a non-metal element. Based on the information given, we can deduce the following: - P is a metal, as it has only 6 electrons. - Q is a non-Metal, as it has 11 electrons. - R is a metal, as it has 15 electrons. - S is a non-Metal, as it has 17 electrons. So, from this information, we can conclude that: - P will form an ionic bond with R, as P is a metal and R is a metal. - Q will form a covalent bond with S, as Q is a non-Metal and S is a non-Metal. Therefore, the correct answer is "Q will form a covalent bond with S."
Swali 3 Ripoti
When large hydrocarbon molecules are heated at high temperature in the presence of a catalyst to give smaller molecules, the process is known as
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The process of breaking down large hydrocarbon molecules into smaller molecules by heating them at high temperatures in the presence of a catalyst is known as cracking. This process is used to convert heavy, high-molecular-weight hydrocarbon molecules into lighter, more valuable products such as gasoline and diesel fuel. The high temperatures cause the large molecules to break apart into smaller ones, and the catalyst helps speed up the reaction. This process is important in the petrochemical industry, as it allows for the production of a wider range of useful products from crude oil.
Swali 5 Ripoti
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.
Swali 6 Ripoti
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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Swali 7 Ripoti
A given amount of gas occupies 10.0dm5 at 4atm and 273°C. The number of moles of the gas present is [Molar volume of gas at s.t.p = 22.4dm3
]
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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.
Swali 8 Ripoti
The reaction between an organic acid and an alcohol in the presence of an acid catalyst is known as;
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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.
Swali 9 Ripoti
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.
Swali 10 Ripoti
The constituent common to duralumin and alnico is
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The common constituent found in both duralumin and alnico is aluminum (Al). Duralumin is an alloy made up of aluminum, copper, manganese, and magnesium. It is known for its high strength and light weight, making it useful in various applications such as aerospace and construction. Alnico, on the other hand, is an alloy made of aluminum, nickel, cobalt, iron, and small amounts of other elements. It is used in the production of strong permanent magnets for various applications such as in motors, generators, and loudspeakers. So, even though duralumin and alnico have different properties and uses, they both contain the element aluminum.
Swali 11 Ripoti
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.
Swali 12 Ripoti
The presence of ammonia gas in a desiccator can exclusively be removed by
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Swali 13 Ripoti
What volume of oxygen will remain after reacting 8cm of hydrogen gas with 20cm of oxygen gas
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Swali 14 Ripoti
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.
Swali 15 Ripoti
H2 S(g) + Cl2(g) → 2HCl(g) + S(g) In the reaction above, the substance that is reduced is
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Swali 17 Ripoti
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.
Swali 18 Ripoti
An element used in the production of matches is
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The element used in the production of matches is sulphur. Matches are small sticks made of wood or cardboard with a chemical mixture at one end. This chemical mixture, called the match head, contains several compounds including sulphur. When the match is struck against a rough surface, the friction generates heat that ignites the sulphur in the match head, causing a flame. This flame then ignites the other compounds in the match head, which in turn ignites the wood or cardboard stick. Sulphur is an important component of the match head because it is highly flammable and burns easily. It also helps to ignite the other compounds in the match head. However, sulphur by itself is not a good fuel, which means that it cannot sustain a flame on its own. Therefore, it needs other combustible materials, such as potassium chlorate or phosphorus, to make the match head burn. Overall, sulphur plays a crucial role in the chemistry of matches and allows us to easily start fires for various purposes.
Swali 19 Ripoti
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.
Swali 20 Ripoti
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.
Swali 21 Ripoti
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
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Swali 22 Ripoti
The elements in the periodic table are listed in order of increasing
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Swali 23 Ripoti
If one of the following oxides is heated with hydrogen or carbon using a bunsen burner. it is not reduced to the metal, Which one is it?
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The oxide that cannot be reduced to the metal when heated with hydrogen or carbon using a Bunsen burner is magnesium oxide. Magnesium oxide is an ionic compound made up of positively charged magnesium ions and negatively charged oxygen ions. When heated with hydrogen or carbon, the oxygen ions are not easily removed from the compound. This is because the ionic bond between the magnesium and oxygen ions is very strong and requires a lot of energy to break. On the other hand, lead oxide, copper oxide, and tin oxide are all metal oxides and can be reduced to the metal by heating with hydrogen or carbon. This is because they have a weaker bond between the metal and oxygen ions, allowing the oxygen to be removed more easily when heated. In conclusion, magnesium oxide is the oxide that cannot be reduced to the metal when heated with hydrogen or carbon using a Bunsen burner.
Swali 24 Ripoti
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.
Swali 25 Ripoti
The end products of burning a candle in the atmosphere are water and
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Swali 26 Ripoti
Which of the following separation techniques can be employed in obtaining solvent from its solution?
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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.
Swali 27 Ripoti
Which of these alloys contains copper?
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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.
Swali 28 Ripoti
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.
Swali 29 Ripoti
How many atoms are present in 6.0g of magnesium? [Mg = 24, N.A = 6.02 x 10 23 mol]
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Swali 30 Ripoti
If the molecular mass of tetraoxosulphate (VI) acid is 98, calculate its vapour density
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Swali 31 Ripoti
Which of the following pairs of substances will react further with oxygen to form a higher oxide?
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Swali 32 Ripoti
The derivative of benzene that can be used in making explosives is
Swali 33 Ripoti
Which of these sources of water may likely contain the least concentration of Ca2+ and Mg2+ ?
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The source of water that is likely to contain the least concentration of Ca2+ and Mg2+ is tap water. Tap water is treated and processed before it is made available for consumption, which often involves removing minerals such as calcium and magnesium. Spring water and river water, on the other hand, are naturally occurring and generally contain higher levels of minerals. Sea water has the highest concentration of minerals, including Ca2+ and Mg2+.
Swali 34 Ripoti
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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Swali 35 Ripoti
3H2(g) + N2 ⇔ 2NH3(g) ; H= -ve
In the reaction above, lowering of temperature will
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Swali 36 Ripoti
The number of electrons in the valence shell of an element of atomic number 14 is?
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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.
Swali 38 Ripoti
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].
Maelezo ya Majibu
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.
Swali 39 Ripoti
2SO2 (g) + O2 (g) ↔ 2SO3 (g) ΔH = -395.7kJmol−1
In the equation, an increase in temperature will shift the equilibrium position to the
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Swali 40 Ripoti
Sieving is a technique used to separate mixtures containing solid particles of
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Sieving is a technique used to separate mixtures containing solid particles of different sizes. A sieve is a mesh or perforated screen that is used to separate particles based on their size. The mixture is poured onto the sieve, and the particles that are too large to pass through the holes are left on top, while the smaller particles fall through the holes and are collected below. This process allows for the separation of the different-sized particles, making it easier to purify or further process the mixture.
