JAMB UTME - Chemistry - 2019

Which of the following will give a precipitate with an aqueous solution of copper (I) chloride?

Detalles de la respuesta

Which of the following could not be alkane?

Detalles de la respuesta

An alkane is a type of hydrocarbon with only single bonds between the carbon atoms. It follows the general formula CnH2n+2, where "n" is the number of carbon atoms in the molecule. To determine whether a molecule is an alkane or not, we can calculate its molecular formula and check if it fits the general formula of alkane. Out of the given options, the third one (C7H14) cannot be an alkane. To see why, let's use the general formula of alkane, which is CnH2n+2. For C7H14 to be an alkane, it should have 2n+2 = 2(7) + 2 = 16 hydrogen atoms. However, C7H14 has only 14 hydrogen atoms, which means it does not follow the general formula of alkane. Therefore, C7H14 cannot be an alkane. The other options are as follows: - C4H10: This is butane, which is an alkane with four carbon atoms. - C5H12: This is pentane, which is an alkane with five carbon atoms. - C8H18: This is octane, which is an alkane with eight carbon atoms. In summary, the molecule C7H14 cannot be an alkane because it does not follow the general formula of alkane, while the other options are all examples of alkanes.

Which of the following properties increases from left to right along the period but decreases down the group in the Periodic Table?

I. Atomic Number   ii. Ionization energy  iii. Metallic character  iv. Electron affinity

Detalles de la respuesta

Ionization energy and electron affinity increase across a period, and decrease down a group.

200cm${}^3$ of 0.50mol/dm${}^3$ solution of calcium hydrogen trioxocarbonate (IV) is heated. The maximum weight of solid precipitated is

Detalles de la respuesta

To solve this problem, we need to use the concept of stoichiometry and the solubility product constant (Ksp) of calcium hydrogen trioxocarbonate (IV). First, we need to write the balanced equation for the reaction that occurs when the solution of calcium hydrogen trioxocarbonate (IV) is heated: Ca(HCO3)2(s) → CaCO3(s) + H2O(g) + CO2(g) From the balanced equation, we can see that 1 mole of calcium hydrogen trioxocarbonate (IV) produces 1 mole of calcium carbonate. Therefore, we need to determine the number of moles of calcium hydrogen trioxocarbonate (IV) in the solution: Number of moles = concentration x volume Number of moles = 0.50 mol/dm³ x 0.2 dm³ Number of moles = 0.1 mol Since 1 mole of calcium hydrogen trioxocarbonate (IV) produces 1 mole of calcium carbonate, the number of moles of calcium carbonate produced will also be 0.1 mol. Next, we need to use the solubility product constant (Ksp) of calcium carbonate to determine the maximum amount of solid that can be precipitated: Ksp = [Ca²⁺][CO3²⁻] Ksp = 3.3 x 10⁻⁹ (at 25°C) At the maximum amount of solid precipitated, all the calcium carbonate formed will have precipitated, and the concentration of calcium ions and carbonate ions will be equal. Therefore, we can assume that the concentration of calcium ions and carbonate ions is both x. Substituting into the Ksp expression: Ksp = x² 3.3 x 10⁻⁹ = x² x = 5.74 x 10⁻⁵ mol/dm³ The mass of calcium carbonate precipitated can now be calculated: Mass = number of moles x molar mass Mass = 0.1 mol x 100.1 g/mol Mass = 10.01 g Therefore, the maximum weight of solid precipitated is approximately 10 g. Note that this calculation assumes that all the calcium carbonate precipitated as a solid, which may not always be the case in a real-world experiment. Additionally, this calculation does not take into account any losses due to filtration or other experimental errors.

Consider the reaction

A${}_{\left(s\right)}$ + 2B${}_{\left(g\right)}$ → 2C${}_{\left(aq\right)}$ + D${}_{\left(g\right)}$

What will be the effect of a decrease in pressure on the reaction?

Detalles de la respuesta

Given: The equation below

A${}_{\left(s\right)}$ + 2B${}_{\left(g\right)}$ → 2C${}_{\left(aq\right)}$ + D${}_{\left(g\right)}$

Since we have a higher number of moles of gaseous species on the LHS, i.e 2, a decrease in pressure will favor the forward reaction.

The shapes of water, ammonia, carbon (iv) oxide and methane are respectively

Detalles de la respuesta

Sulphur exists in six forms in the solid state. This property is known as

Detalles de la respuesta

The property of sulfur existing in six different forms in the solid-state is known as allotropy. Allotropy is a phenomenon where an element can exist in multiple forms, called allotropes, that have different physical and chemical properties but are composed of the same atoms. These different forms arise due to differences in the arrangement of atoms or molecules within the substance. In the case of sulfur, it can exist in multiple solid-state allotropes, including rhombic, monoclinic, and plastic sulfur, among others. Each of these allotropes has a different crystal structure, melting point, and other physical and chemical properties, even though they are all composed of sulfur atoms. Allotropy is a common phenomenon observed in many elements, including carbon, oxygen, and phosphorus, among others.

The hybridization in the compound $C{H}_3-C{H}_2-C\equiv H$ is

Detalles de la respuesta

The hybridization in a and b is sp${}^3$ hybridization while in c and d is sp hybridization.

Hydrogen diffused through a porous plug

Detalles de la respuesta

Hydrogen gas (H2) diffuses faster than oxygen gas (O2) through a porous plug. This is because the rate of diffusion of a gas through a porous plug is inversely proportional to the square root of its molar mass. Since the molar mass of hydrogen (2 g/mol) is much smaller than that of oxygen (32 g/mol), the rate of diffusion of hydrogen through a porous plug is much faster than that of oxygen. To be more specific, the ratio of the diffusion rates of two gases through a porous plug is given by the equation: Rate of diffusion of gas A / Rate of diffusion of gas B = √(Molar mass of gas B / Molar mass of gas A) Using the molar masses of hydrogen and oxygen, we get: Rate of diffusion of hydrogen / Rate of diffusion of oxygen = √(32 g/mol / 2 g/mol) = √16 = 4 Therefore, hydrogen diffuses through a porous plug four times as fast as oxygen. Thus, the correct answer is: four times as fast as oxygen.

Hydrocarbons which will react with Tollen's reagent conform to the general formula

Detalles de la respuesta

A cell shorthand notation can be written as A / A${}^{+}$ // B${}^{2+}$ /B. The double slash in the notation represents the 

Detalles de la respuesta

The double slash in the cell shorthand notation represents the salt bridge. A salt bridge is a component of an electrochemical cell that connects the two half-cells and allows the flow of ions between them. It consists of an inert electrolyte solution (usually a salt) that is placed between the two half-cells. The purpose of the salt bridge is to maintain electrical neutrality in each half-cell by allowing the flow of ions to balance the charge buildup in the half-cells. In the cell shorthand notation, the double slash "//" represents the salt bridge that connects the two half-cells of the electrochemical cell. The first half-cell is represented on the left-hand side of the slash and the second half-cell is represented on the right-hand side of the slash. The anode (where oxidation occurs) is represented on the left side, and the cathode (where reduction occurs) is represented on the right side. Therefore, the correct answer is option number 3: salt bridge.

A synthetic rubber is obtained from the polymerization of

Detalles de la respuesta

A synthetic rubber is obtained from the polymerization of isoprene. Isoprene is a type of hydrocarbon that can be polymerized, or chemically joined together, to form long chains. This process is called polymerization, and the resulting material is called a polymer. When isoprene is polymerized, it forms a synthetic rubber, which is a type of polymer that is used in a wide range of products, including tires, hoses, and adhesives. Synthetic rubber offers several advantages over natural rubber, including improved durability and resistance to heat, ozone, and chemicals.

Elements in the periodic table are arranged in the order of their

Detalles de la respuesta

Elements in the periodic table are arranged in the order of their atomic numbers. The atomic number of an element is the number of protons in the nucleus of an atom of that element. The elements are arranged in order of increasing atomic number from left to right and from top to bottom in the periodic table. The elements in each row, also known as a period, have the same number of electron shells, while the elements in each column, also known as a group or family, have the same number of valence electrons. This arrangement makes it possible to predict the chemical and physical properties of an element based on its position in the periodic table. Therefore, the correct answer is: - atomic numbers

Burning magnesium ribbon in air removes which of the following

(i) oxygen (ii) nitrogen (iii) argon and (iv) carbon(iv)oxide?

Detalles de la respuesta

Burning magnesium ribbon in air will remove oxygen (option i) from the air, but not nitrogen (option ii), argon (option iii), or carbon dioxide (option iv). When magnesium burns, it reacts with oxygen in the air to form magnesium oxide. The reaction can be represented by the following equation: 2Mg(s) + O2(g) → 2MgO(s) The magnesium in the ribbon combines with oxygen in the air to form solid magnesium oxide. This reaction is exothermic, which means that it releases heat and light energy. So, when magnesium ribbon is burned in air, it consumes the oxygen in the air to form magnesium oxide. However, nitrogen, argon, and carbon dioxide are not chemically reactive with magnesium, and therefore are not removed from the air by the burning of magnesium ribbon. In summary, the correct option is (i) only - burning magnesium ribbon in air removes oxygen only.

The oxidation state(s) of nitrogen in ammonium nitrite is/are

Detalles de la respuesta

Ammonium nitrite = NH${}_4$NO${}_2$
NH${}_4^{+}$: Let the oxidation number of Nitrogen = x
x + 4 = 1 $⟹$ x = 1 - 4
x = -3
NO${}_2^{-}$: x - 4 = -1
x = -1 + 4 $⟹$ x = +3.
The oxidation numbers for Nitrogen in Ammonium Nitrite = -3, +3.

Detalles de la respuesta

Fractional crystallization is the most suitable technique for separating a binary solution of potassium chloride and potassium trioxochlorate (V). This is because fractional crystallization is a process that separates a mixture of substances based on their solubility in a solvent at a particular temperature. In this case, potassium chloride and potassium trioxochlorate (V) have different solubilities in a solvent such as water at different temperatures. By carefully controlling the temperature, the solubility of each compound can be selectively increased or decreased, allowing them to be separated by crystallization. The less soluble compound will form crystals first and can be separated from the more soluble compound, which remains in the solution. Therefore, fractional crystallization can be used to separate potassium chloride and potassium trioxochlorate (V) in a binary solution.

A radioactive nucleus has a half-life of 20 years, starting with 100,000 particles, how many particles will be left exactly at the end of 40 years

Detalles de la respuesta

The half-life of a radioactive nucleus is the time it takes for half of its particles to decay. This means that after 20 years, 100,000 particles will become 50,000 particles. After 40 years, we can find the number of particles remaining by counting the number of half-lives that have passed. Since 40 years is double the half-life of 20 years, this means that two half-lives have passed, so the number of particles will be halved twice. Starting with 100,000 particles: - After 1 half-life (20 years), there will be 50,000 particles remaining. - After 2 half-lives (40 years), there will be 25,000 particles remaining. So, exactly at the end of 40 years, there will be 25,000 particles remaining.

What volume of 0.100M sodium trioxonitrate (V) solution contains 5g of solute.

[Na = 23, N = 14, O = 16]

Detalles de la respuesta

To calculate the volume of a solution, we need to use the formula: moles of solute = concentration x volume First, let's find the number of moles of sodium trioxonitrate (V) in 5g of the solute. The molar mass of NaNO3 is: Na = 23 N = 14 3 x O = 3 x 16 = 48 Molar mass = 23 + 14 + 48 = 85 g/mol The number of moles of NaNO3 in 5g is: moles = mass / molar mass = 5 / 85 = 0.0588 moles Now, we can use the formula above to find the volume of the solution: moles of solute = concentration x volume volume = moles of solute / concentration volume = 0.0588 moles / 0.100 M volume = 0.588 litres Therefore, the correct answer is 0.588 litres of 0.100M sodium trioxonitrate (V) solution contains 5g of solute.

When ammonia and hydrogen ion bond together to form ammonium ion, the bond formed is called

Detalles de la respuesta

When ammonia and hydrogen ion go into bonding, they form ammonium ion by combining with a dative/coordinate covalent bond.

When chlorine water is exposed to bright sunlight, the following products are formed

Detalles de la respuesta

Which quantum divides shells into orbitals?

Detalles de la respuesta

The quantum that divides shells into orbitals is the "Azimuthal" quantum number, also known as the "angular momentum" quantum number. The azimuthal quantum number determines the shape of an electron's orbital, which is a region in space where there is a high probability of finding an electron. It describes the angular momentum of an electron in an atom and the number of subshells within a given shell. Each subshell is associated with a specific shape, and can hold a certain number of electrons. The azimuthal quantum number is represented by the letter "l" and can have integer values ranging from 0 to (n-1), where "n" is the principal quantum number. Each value of "l" corresponds to a different subshell shape: - l = 0 corresponds to an "s" subshell, which is spherical in shape. - l = 1 corresponds to a "p" subshell, which has a dumbbell shape with two lobes. - l = 2 corresponds to a "d" subshell, which has a more complex shape with four lobes and a doughnut-like ring. - l = 3 corresponds to an "f" subshell, which has an even more complex shape with eight lobes. The number of orbitals within a subshell is equal to 2l+1. For example, a "p" subshell (l = 1) has three orbitals (2l+1 = 3), which are labeled as "px", "py", and "pz". In summary, the azimuthal quantum number determines the shape of the electron's orbital and the number of subshells within a given shell, and it is represented by the letter "l".

A solution X, on mixing with AgNO${}_3$ solution gives a white precipitate soluble in aqueous NH${}_3$, a solution Y, when also added to X, also gives a white precipitate which is soluble when heated solutions X and Y respectively contain

Detalles de la respuesta

For the general equation of the nature

XP + yQ ⇌ mR + nS, the expression for the equilibrium constant is

Detalles de la respuesta

The expression for the equilibrium constant for the general equation XP + yQ ⇌ mR + nS is: Kc = [R]m[S]n / [P]x[Q]y where Kc is the equilibrium constant, [R] and [S] are the concentrations of the products, and [P] and [Q] are the concentrations of the reactants, all raised to the stoichiometric coefficients (m, n, x, y) in the balanced equation. This equation is known as the equilibrium constant expression and it represents the ratio of the concentrations of the products and reactants at equilibrium for a particular chemical reaction. The equilibrium constant is a measure of how far a reaction proceeds towards completion, with a larger value indicating a greater extent of reaction. The equilibrium constant expression is derived from the law of mass action, which states that the rate of a chemical reaction is proportional to the product of the concentrations of the reactants raised to their stoichiometric coefficients. At equilibrium, the rates of the forward and reverse reactions are equal, and the equilibrium constant expression represents the ratio of the rate constants for these two reactions. Therefore, the correct expression for the equilibrium constant for the general equation XP + yQ ⇌ mR + nS is Kc = [R]m[S]n / [P]x[Q]y.

Which important nitrogen-containing compound is produced in Haber's process?

Detalles de la respuesta

The important nitrogen-containing compound that is produced in Haber's process is NH3, which is also known as ammonia. Haber's process is a chemical process used to produce ammonia by reacting nitrogen gas (N2) and hydrogen gas (H2) under high pressure and temperature in the presence of an iron catalyst. The reaction between nitrogen and hydrogen produces ammonia as the main product, along with some nitrogen and hydrogen gases that do not react. NH3 is an important compound that is widely used in industry for the production of fertilizers, plastics, and other chemical products. It is also used as a cleaning agent, a refrigerant, and a fuel for engines. In addition, NH3 is an essential compound for life, as it is a key component of amino acids, which are the building blocks of proteins.

In the reaction between sodium hydroxide and tetraoxosulphate (VI) solutions, what volume of 0.5 molar sodium hydroxide would exactly neutralize 10cm${}^3$ of 1.25 molar tetraoxosulphate (vi) acid?

Detalles de la respuesta

Equation of reaction : 2NaOH + H${}_2$SO${}_4$ → Na${}_2$SO${}_4$ + 2H${}_2$O

Concentration of a base, CB = 0.5M

Volume of acid, V${}_A$ = 10cm${}^3$

Concentration of an acid, C${}_A$ = 1.25M

Volume of base, V${}_B$ = ?

Recall:
$\frac{C_A{V}_A}{C_B{V}_B}=\frac{n_A}{n_B}$ ... (1)
N.B: From the equation,
$\frac{n_A}{n_B}=\frac{1}{2}$

From (1)
$\frac{1.25\times 10}{0.5\times V_B}=\frac{1}{2}$
$\frac{12.5}{0.5V_B}=\frac{1}{2}$
25 = 0.5V${}_B$

VB = 50.0 cm${}^3$

A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen. If the molar mass of the compound is 180. Find the molecular formula.

[H = 1, C = 12, O = 16]

Detalles de la respuesta

The molecular formula of a compound is determined by the number of atoms of each element present in the molecule. To find the molecular formula, we need to determine the number of atoms of each element in the compound. First, we convert the percent composition to grams. For example, 40.0% carbon means 40.0 g of carbon per 100 g of compound. Then we divide the number of grams of each element by the molar mass of each element. For example, 40.0 g of carbon divided by the molar mass of carbon (12 g/mol) gives us 3.33 mol of carbon. Next, we convert the number of moles of each element to the number of atoms by multiplying the number of moles by Avogadro's number (6.022 x 10^23 atoms/mol). Finally, we balance the numbers of atoms of each element by dividing them by the smallest number of atoms of all the elements and rounding to the nearest whole number. In this case, the smallest number of atoms is 2, which is the number of hydrogen atoms. So, we divide the number of atoms of carbon and oxygen by 2 to balance the numbers of atoms of all the elements. Therefore, the molecular formula of the compound is C6H12O6.

The velocity, V of a gas is related to its mass, M by (k = proportionality constant)

Detalles de la respuesta

Recall:
V = $\sqrt{\frac{3RT}{M}}$
$\therefore V\propto \frac{1}{\sqrt{M}}$
$V=\frac{k}{\sqrt{M}}$
V = $\frac{k}{M^{\frac{1}{2}}}$

At 27°C, 58.5g of sodium chloride is present in 250cm${}^3$ of a solution. The solubility of sodium chloride at this temperature is?

(molar mass of sodium chloride = 111.0gmol${}^{-1}$)

Detalles de la respuesta

Given the Mass of the salt = 58.5g
Volume = 250 cm${}^3$ = 0.25 dm${}^3$
Mass concentration = $\frac{Mass}{Volume}$
= $\frac{58.5}{0.25}$ = 234 gdm${}^{-3}$
Solubility (in moldm${}^{-3}$ = $\frac{234}{111}$
= 2.11 moldm${}^{-3}$
$\approxeq$ 2.0 moldm${}^{-3}$

Which of the following pollutants will lead to the depletion of ozone layer?

Detalles de la respuesta

The pollutant that leads to the depletion of the ozone layer is chlorofluorocarbon (CFCs). CFCs are man-made chemicals that were widely used in the past as refrigerants, solvents, and propellants. When CFCs are released into the atmosphere, they rise into the stratosphere, where they come into contact with ozone molecules. The chlorine atoms in CFCs react with ozone, breaking apart the ozone molecules and causing a reduction in the overall amount of ozone in the stratosphere. This process continues until all of the ozone-depleting chlorine atoms have been depleted. The resulting decrease in ozone in the stratosphere leads to an increase in the amount of harmful ultraviolet radiation that reaches the Earth's surface, which can have negative impacts on human health and the environment.

A secondary alkanol can be oxidized to give an

Detalles de la respuesta

A secondary alkanol is an alcohol with two carbon atoms attached to the carbon bearing the hydroxyl group (-OH). Secondary alkanols can be oxidized by a strong oxidizing agent, such as potassium dichromate (K2Cr2O7), to give an alkanone. During the oxidation process, the oxygen atom from the oxidizing agent replaces the hydroxyl group of the secondary alkanol to form a carbonyl group (C=O) in the alkanone. Since alkanones contain a carbonyl group, they are also known as ketones. Therefore, the answer to the question is alkanone, as secondary alkanols can be oxidized to form ketones.

Which of the following conditions will most enhance the spontaneity of a reaction?

Detalles de la respuesta

The condition that will most enhance the spontaneity of a reaction is when ΔH is negative (i.e., the reaction releases heat) and ΔS is positive (i.e., the reaction increases the disorder or randomness of the system). This is because a negative ΔH indicates that the reaction releases energy, which is favorable for a spontaneous reaction, while a positive ΔS indicates that the system becomes more disordered, which is also favorable for spontaneous reactions. Among the given options, the first condition of a negative and greater ΔH than ΔS is the best option for enhancing the spontaneity of a reaction. The other options have either a positive ΔH or a zero ΔS, which is not favorable for spontaneous reactions.

Which of the following gases contains the least number of atoms at s.t.p?

Detalles de la respuesta

At standard temperature and pressure (s.t.p), all gases have the same number of atoms or molecules. What changes between them is the volume they occupy, and this is dependent on their molecular mass and the number of moles. Comparing the number of moles between the gases listed above, 7 moles of argon will contain the most number of atoms, followed by 4 moles of chlorine, then 3 moles of ozone, and finally 1 mole of butane would contain the least number of atoms. In summary, the number of atoms in a gas sample depends on the number of moles, but at s.t.p, the volume occupied by each gas depends on its molecular mass and the number of moles.

Which of the following describes the chemical property of acids?

Detalles de la respuesta

The molecular shape and bond angle of water are respectively

Detalles de la respuesta

The shape of water molecule = Bent/ V- shaped
The bond angle of water = 104.5°/ 105°

The electronic configuration of element Z is 1s${}^2$ 2s${}^2$ 2p${}^6$ 3s${}^2$ 3p${}^1$. What is the formula of the compound formed between Z and tetraoxosulphate (VI) ion.

Detalles de la respuesta

Z = 1s${}^2$ 2s${}^2$ 2p${}^6$ 3s${}^2$ 3p${}^1$
$?$ We have Z${}^{3+}$ and SO${}_4^{2?}$
The reaction : Z${}^{3+}$ + SO${}_4^{2?}$ $?$ Z${}_2$(SO${}_4$)${}_3$.

Which of the following statements does not show Rutherford's account of Nuclear Theory? An atom contains a region

Detalles de la respuesta

Rutherford's account of Nuclear theory does not include the fact that atoms contain a massive region and cause deflection of from projectiles.

Which of the following metals is the most essential in the regulation of blood volume, blood pressure and osmotic equilibrium?

Detalles de la respuesta

The metal that is most essential in the regulation of blood volume, blood pressure, and osmotic equilibrium is sodium. Sodium is a key electrolyte that helps maintain the balance of fluids in the body, including blood volume and blood pressure. Sodium ions are positively charged and are attracted to negatively charged ions, such as chloride (Cl-) and bicarbonate (HCO3-), which together help regulate the pH of the blood. Sodium is also essential for maintaining osmotic equilibrium, which refers to the balance of solutes between cells and the extracellular fluid. Osmotic equilibrium is critical for proper cellular function and is regulated by the movement of water and electrolytes, including sodium, in and out of cells. While the other metals listed (zinc, manganese, and iron) are important for various functions in the body, such as enzyme activity and oxygen transport, they are not directly involved in regulating blood volume, blood pressure, and osmotic equilibrium in the same way that sodium is. Therefore, the answer is not options 1, 2, or 4, and the correct answer is: sodium.

Which of the following factors will speed up the rate of evolution of carbon (iv) oxide in the reaction below?

2HCl + CaCO${}_3$ → CaCl${}_2$ + H${}_2$O + CO${}_2$

Detalles de la respuesta

The following factors increase a reaction rate

- Increase in concentration of reactants

- Increase in temperature

- Addition of catalyst

- Increase in the surface area of reactant(s)

SO${}_3$ is not directly dissolved in water in the industrial preparation of H${}_2$SO${}_4$ by the contact process because

Detalles de la respuesta

The IUPAC name for CH${}_3$CH${}_2$COOCH${}_2$CH${}_3$ is

Detalles de la respuesta

The IUPAC name for the given molecule is ethyl propanoate. To arrive at the IUPAC name, we first identify the longest continuous chain of carbon atoms, which in this case is a 4-carbon chain (propane). We then identify and name the substituent groups attached to this chain, which are a methyl group (CH3) attached to the second carbon atom and an ethoxy group (OC2H5) attached to the third carbon atom. The ethoxy group is named as an ethyl group, and the entire molecule is named as ethyl propanoate, following the standard IUPAC naming conventions for esters.