JAMB UTME - Chemistry - 2018

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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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.

(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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Beryllium and Aluminium have similar properties because they

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

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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.

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Coal is the fuel that is typically used to power steam engines. Coal is burned in a furnace to heat water and produce steam, which is then used to power a steam engine. The steam engine converts the energy from the steam into mechanical energy, which can be used to power machines or generate electricity. Coal is a fossil fuel that has been used for centuries as a source of energy, and it played a significant role in the industrial revolution, powering steam engines that were used to drive machines in factories and transport goods and people by train. Today, steam engines are less common as other forms of energy have taken their place, but they remain an important part of our history and technological development.

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

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What is the PH of 0.00 1 moldm${}^3$ solution of the sodium hydroxide

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In the shown experiment (Fig. 1) the litmus paper will initially

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Calculate the pH of 0.05 moldm${}^{?3}$ H${}_2$SO${}_4$

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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.

From the graph above, it may be concluded 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.

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

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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.

What volume of oxygen will remain after reacting 8cm of hydrogen gas with 20cm of oxygen gas

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

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Diamond is a bad conductor of electricity because its bonding electrons are used in

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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.

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.

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

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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.

If the molecular mass of tetraoxosulphate (VI) acid is 98, calculate its vapour density

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

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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.

The type of bonding in [Cu(NH${}_3$)${}^4$]${}^{2+}$ is

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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.

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

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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.

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.

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

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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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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.

The knowledge of half-life can be used to

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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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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."

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

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

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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).

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

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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.