JAMB UTME - Chemistry - 2018
Tambaya 1 Rahoto
The Sulphide which is insoluble in dilute hydrochloric acid is
Bayanin Amsa
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.
Tambaya 2 Rahoto
A substance that is used as a ripening agent for fruits is
Bayanin Amsa
The substance that is commonly used as a ripening agent for fruits is ethene. Ethene, also known as ethylene, is a natural plant hormone that is produced by fruits, especially during the ripening process. It is a colorless gas that can be easily synthesized and used as a ripening agent for fruits. When fruits are exposed to ethene, it triggers a series of biochemical reactions that accelerate the natural ripening process. This can help fruits to ripen faster and more uniformly, which is important for commercial purposes where fruits need to be sold quickly. The use of ethene as a ripening agent is regulated by food safety agencies, as excessive exposure to ethene can cause over-ripening and spoilage of fruits. However, when used in appropriate concentrations, ethene is a safe and effective way to promote the ripening of fruits.
Tambaya 4 Rahoto
Calculate the pH of \(0.05\ \mathrm{mol\,dm^{-3}}\ \mathrm{H_2SO_4}\)
Bayanin Amsa
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.
Tambaya 5 Rahoto
The refreshing and characteristic taste of soda water and other soft drinks is as a result of the presence of
Tambaya 6 Rahoto
The presence of ammonia gas in a desiccator can exclusively be removed by
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Tambaya 7 Rahoto
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?
Bayanin Amsa
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.
Tambaya 8 Rahoto
The type of bonding in \(\left[\mathrm{Cu}(\mathrm{NH}_3)_4\right]^{2+}\) is
Bayanin Amsa
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.
Tambaya 9 Rahoto
What is the concentration of a solution containing 2g of NaOH in 100cm3 of solution? [Na = 23, O =16, H = 1]
Bayanin Amsa
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.
Tambaya 10 Rahoto
To what volume must \(300\text{cm}^3\) of 0.60M sodium hydroxide solution be diluted to give a 0.40M solution?
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Tambaya 11 Rahoto
The derivative of benzene that can be used in making explosives is
Tambaya 12 Rahoto
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)
Bayanin Amsa
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.
Tambaya 13 Rahoto
Which of the following pairs of substances will react further with oxygen to form a higher oxide?
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Tambaya 14 Rahoto
The constituent common to duralumin and alnico is
Bayanin Amsa
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.
Tambaya 15 Rahoto
Which of the following is an example of a chemical change?
Tambaya 16 Rahoto
At what temperature is the solubility of potassium trioxonitrate(V ) equal to that of sodium trioxonitrate (V)?
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Tambaya 17 Rahoto
Diamond is a bad conductor of electricity because its bonding electrons are used in
Bayanin Amsa
Diamond is a bad conductor of electricity because of its unique structure and bonding. The carbon atoms in diamond form a covalent network, where each carbon atom is bonded to four other carbon atoms. These bonds are strong and hold the atoms in a rigid three-dimensional structure called a crystal lattice. In a covalent bond, atoms share electrons to form a stable compound. In diamond, each carbon atom shares its valence electrons with four neighboring carbon atoms, forming a very strong covalent bond. All the valence electrons in the crystal lattice are used in covalent bond formation, which means there are no free or mobile electrons to carry an electric current. In other words, the electrons are tightly held in the covalent bonds, making it difficult for them to move around the crystal lattice and conduct electricity. In contrast, metals conduct electricity well because they have delocalized or free electrons that can move through the lattice of positively charged ions. So, diamond, being a covalent network solid, does not have free electrons that can carry an electric current, which is why it is a bad conductor of electricity.
Tambaya 18 Rahoto
The general formula of alkanones is
Tambaya 19 Rahoto
The end products of burning a candle in the atmosphere are water and
Tambaya 20 Rahoto
The ionic radii of metals are usually
Bayanin Amsa
The ionic radii of metals are usually smaller than their atomic radii. The size of an atom is determined by the distance between the nucleus and the outermost electrons, which is known as the atomic radius. When a metal atom loses one or more electrons to form a positive ion (or cation), the resulting ion has a smaller size than the original atom. This is because the positive charge of the ion attracts the remaining electrons closer to the nucleus, making the ion smaller in size. So, when a metal forms a cation, its ionic radius is typically smaller than its atomic radius. This is a general trend in the periodic table, although there are some exceptions.
Tambaya 21 Rahoto
A correct electrochemical series can be obtained from Na, Ca, Al, Mg, Zn, Fe, Pb, H, Cu, Hg, Ag, Au by interchanging
Tambaya 22 Rahoto
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]
Tambaya 23 Rahoto
\(3\mathrm{H}_2(g) + \mathrm{N}_2 \rightleftharpoons 2\mathrm{NH}_3(g)\); H= -ve
In the reaction above, lowering of temperature will
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Tambaya 24 Rahoto
What is the PH of 0.00 1 moldm\(^{3}\) solution of the sodium hydroxide
Tambaya 25 Rahoto
Which of the following statements is correct about the periodic table?
Tambaya 26 Rahoto
The choice of method for extracting a metal from its ores depends on the
Bayanin Amsa
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.
Tambaya 27 Rahoto
ME + nF -----> pG + qH
In the equation shown, the equilibrium constant is given by?
Bayanin Amsa
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.
Tambaya 28 Rahoto
Suitable reagents for the laboratory preparation nitrogen are
Tambaya 29 Rahoto
The collision theory explains reaction rates in terms of
Bayanin Amsa
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.
Tambaya 30 Rahoto
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;
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Tambaya 31 Rahoto
The elements in the periodic table are listed in order of increasing
Tambaya 32 Rahoto
(I). \(3\mathrm{CuO}(s) + 2\mathrm{NH}_3(g) \longrightarrow 3\mathrm{Cu}(s) + 3\mathrm{H}_2\mathrm{O}(l) + \mathrm{N}_2(g)\)
(II). \(2\mathrm{NH}_3(g) + 3\mathrm{Cl}_2(g) \longrightarrow 6\mathrm{HCl}(g) + \mathrm{N}_2(g)\)
(III). \(4\mathrm{NH}_3(g) + 3\mathrm{O}_2(g) \longrightarrow 6\mathrm{H}_2\mathrm{O}(l) + \mathrm{N}_2(g)\)
The reactions represented by the equations above demonstrate the
Tambaya 33 Rahoto
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
Bayanin Amsa
Tambaya 34 Rahoto
If the molecular mass of tetraoxosulphate (VI) acid is 98, calculate its vapour density
Tambaya 35 Rahoto
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].
Bayanin Amsa
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.
Tambaya 36 Rahoto
The conductivity of an acid solution depends on the
Bayanin Amsa
The conductivity of an acid solution depends on the amount of ions present and their mobilities. When an acid dissolves in water, it forms ions that can carry an electric charge. These ions are what allows the solution to conduct electricity. The more ions there are in the solution, the better it can conduct electricity. However, not all ions have the same mobility or ability to move around in the solution. Ions with a higher mobility can move more easily through the solution, leading to a higher conductivity. Therefore, the conductivity of an acid solution is determined by both the amount of ions present and their mobilities. Other factors such as temperature can also affect conductivity, but the primary factors are the amount and mobility of ions.
Tambaya 37 Rahoto
The solubility of the solids that dissolves in a given solvent with the liberation of heat will
Bayanin Amsa
The solubility of solids in a given solvent is the amount of solid that can dissolve in the solvent to form a solution. When a solid dissolves in a solvent, it releases heat. The solubility of the solid in the solvent can be affected by changes in temperature. Generally, when the temperature of a solution increases, the solubility of the solid in the solvent increases as well. This is because the increased heat energy makes it easier for the solid particles to separate and dissolve in the solvent. As a result, the solubility of the solid in the solvent will increase with an increase in temperature. On the other hand, if the temperature decreases, the solubility of the solid in the solvent decreases. This is because the decreased heat energy makes it harder for the solid particles to separate and dissolve in the solvent. As a result, the solubility of the solid in the solvent will decrease with a decrease in temperature. In summary, the solubility of solids in a given solvent will generally increase with an increase in temperature and decrease with a decrease in temperature.
Tambaya 38 Rahoto
The reaction between an organic acid and an alcohol in the presence of an acid catalyst is known as;
Bayanin Amsa
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.
Tambaya 39 Rahoto
Which of the following are mixtures?
I. Petroleum
II. Rubber latex
III. Vulcanizer's solution
IV. Carbon sulphide
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Tambaya 40 Rahoto
The Consecutive members of an alkane homologous series differ by
Bayanin Amsa
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.
