JAMB UTME - Chemistry - 2019

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

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

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

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

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

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The two ions responsible for hardness in water are

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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).

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

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

A synthetic rubber is obtained from the polymerization of

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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.

The IUPAC name of the compound CF${}_3$CHBrCl is

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

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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

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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

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

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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

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

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Na${}_2$CO${}_3$ + 2HCl $\to$ 2NaCl + H${}_2$O + CO${}_2$

The indicator most suitable for this reaction should have a pH equal to

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Methyl orange is the best indicator for the reaction with range 3.1 - 4.4.

Methane is prepared in the laboratory by heating a mixture of sodium ethanoate with soda lime. The chemical constituent(s) of soda lime is/are

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The chemical constituent of soda lime used to prepare methane in the laboratory is Ca(OH)2 (calcium hydroxide) and NaOH (sodium hydroxide). Soda lime is a mixture of these two compounds. When sodium ethanoate (NaC2H3O2) is heated with soda lime, it undergoes a reaction known as the Kolbe's reaction, which produces methane gas (CH4) as one of the products. The reaction can be represented as follows: 2NaC2H3O2 + 2Ca(OH)2 → 2CH4 + 2NaOH + 2CaCO3 In this reaction, the sodium ethanoate reacts with the calcium hydroxide to form calcium acetate (Ca(C2H3O2)2) and sodium hydroxide. The calcium acetate then decomposes to produce methane gas and calcium carbonate (CaCO3), which is a solid precipitate. Therefore, the chemical constituents of soda lime used to prepare methane in the laboratory are calcium hydroxide (Ca(OH)2) and sodium hydroxide (NaOH).

Consider the reaction: A + 2B${}_{\left(g\right)}\rightleftharpoons$ 2C + D${}_{\left(g\right)}$ ($\Delta$H = +ve)

What will be the effect of decrease in temperature on the reaction?

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The effect of a decrease in temperature on the reaction will be that the rate of the backward reaction will increase. In a chemical reaction, the rate of the forward and backward reactions are determined by the activation energy required for each step and the temperature of the system. When the temperature is decreased, the rate of the reaction decreases, and the rate of the backward reaction increases. This shift in the rate of the backward reaction means that there will be a shift in the position of the equilibrium of the reaction. As the rate of the backward reaction increases, the concentration of the reactants will increase and the concentration of the products will decrease, leading to a decrease in the overall yield of the products. In this reaction, as ΔH (the change in enthalpy) is positive, which means that the reaction is endothermic. Endothermic reactions absorb heat from the surroundings to proceed, so a decrease in temperature will lead to a decrease in the rate of the forward reaction and an increase in the rate of the backward reaction. This shift in the rate of the backward reaction will shift the position of the equilibrium of the reaction to the left, leading to an increase in the concentration of the reactants and a decrease in the concentration of the products.

For the general equation of the nature

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

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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 of the following statements about catalyst is false?

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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.

The IUPAC nomenclature of the compound

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

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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

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

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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.

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${}_2$

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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 of the following could not be alkane?

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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.

The emission of two successive beta particles from the nucleus ${}_{15}^{32}P$ will produce

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Explanation:

Elements X and Y have electronic configurations 1s${}^2$2s${}^2$2p${}^4$ and 1s${}^2$2s${}^2$2p${}^6$3s${}^2$3p${}^1$ respectively. When they combine, the formula of the compound formed is

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Which process(es) is/are involved in the turning of starch iodide paper blue-black by chlorine gas?

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The process involved in the turning of starch iodide paper blue-black by chlorine gas is option number 3: chlorine oxidizes the iodide ion to produce iodine which attacks the starch to give the blue-black color. When chlorine gas comes in contact with iodide ions on the starch iodide paper, it oxidizes the iodide ion to form iodine. The iodine that is produced in this reaction is then able to react with the starch present on the paper to form a blue-black complex. This blue-black complex is formed due to the arrangement of the starch molecules and the iodine atoms in a way that causes them to absorb light at a specific wavelength, giving the blue-black color. Therefore, the blue-black color that is observed on the starch iodide paper is due to the reaction between iodine and starch, which is made possible by the oxidation of iodide ions by chlorine gas.

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

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

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]

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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

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${}^{-1}$)

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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${}^{-3}$

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?

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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${}^3$

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

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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.

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

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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 pollutants will lead to the depletion of ozone layer?

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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.

If the cost of electricity required to discharge 10g of an ion X${}^{3+}$ is N20.00, how much would it cost to discharge 6g of ion Y${}^{2+}$?

[1 faraday = 96,500C, atomic masses are X = 27, Y = 24]

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X${}^{3+}$ + 3e${}^{-}$ $\to$ X
3F = 27g
xF = 10g
$\frac{x}{3}=\frac{10}{27}⟹x=\frac{10}{9}F$
$\frac{10}{9}$F $\equiv$ N20.00
1F is equivalent to x
$\frac{1}{\frac{10}{9}}=\frac{x}{20}$
$\frac{9}{10}=\frac{x}{20}⟹x=N18.00$
1F is equivalent to N18.00.
Y${}^{2+}$ + 2e${}^{-}$ $\to$ Y
2F = 24g
xF = 6g
x = $\frac{6\times 2}{24}=\frac{1}{2}F$
1F = N18.00
$\frac{1}{2}$F = $\frac{1}{2}\times N18.00$
= N9.00

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

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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

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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.

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

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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 reaction $2H_{2}S+S{O}_2\to 3S+2H_{2}O$ is

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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]

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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 following are isoelectronic ions except

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Two or more ions are said to be isoelectronic if they have the same electronic structure and the same number of valence electrons.
Na${}^{+}$ = 10 electrons = 2, 8
Mg${}^{2+}$ = 10 electrons = 2,8
O${}^{2-}$ = 10 electrons = 2,8
Si${}^{2+}$ = 12 electrons = 2,8,2
$⟹$ Si${}^{2+}$ is not isoelectronic with the rest.

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

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

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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.

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]

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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.