JAMB UTME - Chemistry - 2019

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${}^{-1}$)

Detalhes da Resposta

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 sets of operation will completely separate a mixture of sodium chloride, sand and iodine?

Detalhes da Resposta

The set of operations that will completely separate a mixture of sodium chloride, sand, and iodine is: - filtration, to separate the sand and iodine from the sodium chloride - evaporation to dryness, to concentrate the sodium chloride solution and remove any remaining water - sublimation, to separate the iodine as a solid from the remaining sodium chloride By using these operations, you can separate each component of the mixture into separate, pure forms. The order of the operations is important because each step must be done in a way that effectively separates the components and does not interfere with subsequent steps.

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.

Detalhes da Resposta

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${}_4$)${}_3$.

An element Z contains 80% of ${}_8^{16}$Z and 20% of ${}_8^{18}$Z. Its relative atomic mass is

Detalhes da Resposta

R.A.M of Z = $16\left(\frac{80}{100}\right)+18\left(\frac{20}{100}\right)$
= $12.8+3.6$
= 16.4

2-methylprop-1-ene is an isomer of

Detalhes da Resposta

2-methylprop-1-ene is an isomer of 3-methyl but-1-ene and 2-methyl but-1-ene. An isomer is a molecule that has the same molecular formula as another molecule, but a different arrangement of atoms. In this case, 2-methylprop-1-ene has the molecular formula C4H8, and so do 3-methyl but-1-ene and 2-methyl but-1-ene. The difference between these three molecules is in the arrangement of the carbon and hydrogen atoms. 2-methylprop-1-ene has a branched structure with a double bond between the first and second carbon atoms. 3-methyl but-1-ene is also a branched molecule, but the double bond is between the second and third carbon atoms. Similarly, 2-methyl but-1-ene has a double bond between the first and second carbon atoms, but it has a different branching pattern. On the other hand, pent-2-ene has five carbon atoms, so it has a different molecular formula than 2-methylprop-1-ene. Therefore, 2-methylprop-1-ene is an isomer of 3-methyl but-1-ene and 2-methyl but-1-ene, but not of pent-2-ene, because it has the same molecular formula and a different arrangement of atoms compared to the other two isomers.

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

Detalhes da Resposta

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.

The two ions responsible for hardness in water are

Detalhes da Resposta

The ions responsible for hardness in water are Ca2+ and/or Mg2+. Hardness in water refers to the presence of calcium and magnesium ions, which are commonly found in natural water sources such as rivers, lakes, and groundwater. These ions can react with soap to form insoluble compounds, reducing the effectiveness of soap and causing scaling in pipes and appliances. The hardness of water is often measured in terms of the concentration of calcium and magnesium ions, expressed as calcium carbonate equivalents (CaCO3).

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

Detalhes da Resposta

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.

The molecular shape and bond angle of water are respectively

Detalhes da Resposta

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

Which of the following reactions is an oxidation process?

Detalhes da Resposta

Which of the following is a set of neutral oxides?

Detalhes da Resposta

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]

Detalhes da Resposta

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 combustion of carbon(ii)oxide in oxygen can be represented by equation.

2CO + O${}_2$ ? 2CO${}_2$

Calculate the volume of the resulting mixture at the end of the reaction if 50cm${}_3$ of carbon(ii)oxide was exploded in 100cm${}_3$ of oxygen

Detalhes da Resposta

Which of the following could not be alkane?

Detalhes da Resposta

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.

What mass of magnesium would be obtained by passing a current of 2 amperes for 2 hours, through molten magnesium chloride?
[1 faraday = 96500C, Mg = 24]

Detalhes da Resposta

Current (I) = 2A; Time (t) = 2 hours = 7200 secs
Q = It
= 2 x 7200 = 14400C
1 F = 96500C
x = 14400C
x = $\frac{14400}{96500}$
= 0.15F
Mg${}^{2+}$ + 2e${}^{-}$ $\to$ Mg
2F $\to$ 24g
0.15F $\to$ x
2x = 24 x 0.15
x = $\frac{24\times 0.15}{2}$
= 1.8g

By what amount must the temperature of 200cm${}^3$ of Nitrogen at 27°C be increased to double the pressure if the final volume is 150cm${}^3$ (Assume ideality)

Detalhes da Resposta

Using the ideal gas law and equation:
$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$
$\frac{P_1\times 200{\text{cm}}^3}{300\text{K}}=\frac{2P\times 150{\text{cm}}^3}{T_2}$
Cross multiply:
$T_2=\frac{300\times 150\times 2P}{200\times P}$
$=450\text{K}$ or ${177}^{\circ }\text{C}$
Don't forget to convert to ${}^{\circ }\text{C}$

When the end alkyl groups of ethyl ethanoate are interchanged, the compound formed is

Detalhes da Resposta

The compound formed when the end alkyl groups of ethyl ethanoate are interchanged is ethyl propanoate. This is because ethyl ethanoate consists of two parts: the "ethyl" group and the "ethanoate" group. The ethyl group is a two-carbon chain, and the ethanoate group is a combination of a one-carbon chain and a carbonyl group (C=O) that is also attached to an oxygen atom. When the end alkyl groups are interchanged, the "ethyl" group is moved from the second carbon to the first carbon of the ethanoate group, and the "propanoate" group is formed. The "propanoate" group consists of a three-carbon chain and the carbonyl group. Therefore, the resulting compound is ethyl propanoate, which has a chemical formula of CH3CH2COOCH2CH3. This compound is commonly used as a flavoring agent and has a fruity odor reminiscent of pears.

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

Detalhes da Resposta

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

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

Detalhes da Resposta

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?

Detalhes da Resposta

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 IUPAC nomenclature of the compound

H${}_3$C - CH(CH${}_3$) - CH(CH${}_3$) - CH${}_2$ - CH${}_3$

Detalhes da Resposta

X is a substance which liberates CO${}_2$ on treatment with concentrated H${}_2$SO${}_4$. A warm solution of X can decolorize acidified KMnO${}_4$. X is

Detalhes da Resposta

It should be noted that for X to liberate CO${}_2$, X must be a carbonate or an oxalate. Since X decolorizes KMnO${}_4$, X must be an oxalate. 
Therefore, X is H${}_2$C${}_2$O${}_4$.

Which of the following statements about catalyst is false?

Detalhes da Resposta

The false statement about catalysts is: "catalysts do not alter the mechanism of the reaction and never appear in the rate law." Catalysts are substances that speed up chemical reactions without being consumed in the process. They achieve this by reducing the activation energy needed for the reaction to occur. Enzymes are a type of biological catalysts. In a chemical reaction, a catalyst is not consumed and does not appear in the overall balanced equation. However, catalysts can alter the mechanism of a reaction by providing an alternative pathway with a lower activation energy. This alternative pathway can have a different rate-determining step, which means that the presence of the catalyst can change the rate law of the reaction. Therefore, the statement that catalysts do not alter the mechanism of the reaction and never appear in the rate law is false.

Burning magnesium ribbon in air removes which of the following

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

Detalhes da Resposta

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.

Detalhes da Resposta

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.

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

Detalhes da Resposta

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

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

Detalhes da Resposta

The cost of discharging 6.0g of a divalent metal, X from its salt is ₦12.00. What is the cost of discharging 9.0g of a trivalent metal, Y from its salt under the same condition?

[X = 63, Y = 27, 1F = 96,500C]

Detalhes da Resposta

For X:  X${}^{2+}$ + 2e${}^{-}$ $\to$ X
2F = 63g
xF = 6g
x = $\frac{6\times 2}{63}=\frac{4}{21}F$
$\frac{4}{21}$F = N12.00
1F = $\frac{12}{\frac{4}{21}}$
= N63.00
1F is equivalent to N63.00.
For Y: Y${}^{3+}$ + 3e${}^{-}$ $\to$ Y
3F = 27g 
xF = 9g
x = $\frac{3\times 9}{27}$
= 1F
1F = N63.00

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

Detalhes da Resposta

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.

A certain hydrocarbon on complete combustion at s.t.p produced 89.6dm${}^3$ of CO${}_2$ and 54g of water. The hydrocarbon should be

Detalhes da Resposta

In the question above an Hydrocarbon combust to give CO2 and H20

Let Hydrocarbon be
CxHy + x+Y/4O2= xCO2 + Y/2H2O

Mass of C0=44g and H2O=18g
at STP vol= 22.4
Therefore, 1mole of CO2 contains 44g and 22.4dm³ at STP

1mole = 22.4dm³
xmole = 89.6dm³
Cross multiplying x=89.6/22.4 =4mole of CO2 produce

1mole of H2O = 18g
Xmole = 56g
Cross multiplying
X = 56/18 = 3mole of H20
Then....
CxHy + X + y/4O2 = 4CO2+ 3H2O

Balancing
C4H6 + 6O2 = 4CO2 + 3H2O

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${}_2$

Detalhes da Resposta

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)

Which two gases can be used for the demonstration of the fountain experiment?

Detalhes da Resposta

Two gases that can be used in the study of fountain experiment is ammonia gas and hydrogen chloride gas. The experiment introduces concepts like solubility and the gas laws at the entry level.

Which of the following pairs cannot be represented with a chemical formula?

Detalhes da Resposta

The pair that cannot be represented with a chemical formula is air and bronze. Air is a mixture of several gases, primarily nitrogen (N₂) and oxygen (O₂), with small amounts of other gases such as argon (Ar), carbon dioxide (CO₂), and neon (Ne). Since air is a mixture and not a pure substance, it cannot be represented by a chemical formula. Bronze, on the other hand, is an alloy composed mainly of copper (Cu) and tin (Sn) with small amounts of other metals. The composition of bronze can vary depending on the specific alloy, but it can be represented by a chemical formula such as CuSn. Sodium chloride (NaCl) is a compound composed of sodium (Na) and chlorine (Cl) in a fixed ratio of 1:1, and it can be represented by a chemical formula. Similarly, copper (Cu) and sodium chloride (NaCl) can each be represented by a chemical formula. Cu is an element, so its chemical formula is simply its symbol, while NaCl is a compound with a fixed ratio of sodium and chlorine atoms. Caustic soda (sodium hydroxide, NaOH) and washing soda (sodium carbonate, Na₂CO₃) are both compounds that can be represented by chemical formulas. NaOH consists of one sodium atom, one oxygen atom, and one hydrogen atom, while Na₂CO₃ consists of two sodium atoms, one carbon atom, and three oxygen atoms.

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

Detalhes da Resposta

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

Which quantum divides shells into orbitals?

Detalhes da Resposta

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".

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

Detalhes da Resposta

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.

The part of the total energy of a system that accounts for the useful work done in a system is known as

Detalhes da Resposta

The part of the total energy of a system that accounts for the useful work done in a system is known as "Gibbs free energy". Gibbs free energy is a thermodynamic property that represents the amount of energy that can be converted into useful work in a system. It takes into account both the energy of the system and the entropy, or disorder, of the system. In other words, Gibbs free energy is a measure of the energy available to do work, taking into account the energy that is unavailable due to entropy. In simple terms, if a system has a high Gibbs free energy, it has a lot of energy available to do work, and if a system has a low Gibbs free energy, it has little energy available to do work.

Hydrogen diffused through a porous plug

Detalhes da Resposta

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.

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

Detalhes da Resposta

Consider the equation below:

Cr${}_2$O${}_7^{2-}$ + 6Fe${}^{2+}$ + 14H${}^{+}$ $\to$ 2Cr${}^{3+}$ + 6Fe${}^{3+}$ + 7H${}_2$O.

The oxidation number of chromium changes from

Detalhes da Resposta

Cr${}_2$O${}_7^{2-}$ + 6Fe${}^{2+}$ + 14H${}^{+}$ $\to$ 2Cr${}^{3+}$ + 6Fe${}^{3+}$ + 7H${}_2$O
The oxidation of Cr in Cr${}_2$O${}_7^{2-}$ :
Let the oxidation of Cr = x; 
2x + (-2 x 7) = -2 $⟹$ 2x - 14 = -2
2x = 12 ; x = +6
Hence, the change in oxidation of Cr = +6 to +3