JAMB UTME - Chemistry - 2019

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

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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

How many electrons will be found in the nucleus of an atom with mass number 23 and 17 neutrons?

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Electrons are not found in the nucleus of an atom. The nucleus of an atom only contains protons and neutrons, while electrons are located outside the nucleus in the electron cloud. The mass number of an atom is equal to the sum of the number of protons and the number of neutrons in the nucleus. Therefore, if an atom has a mass number of 23 and 17 neutrons, then the number of protons in the nucleus can be calculated as: Protons = Mass number - Neutrons Protons = 23 - 17 Protons = 6 This means that the nucleus of the atom contains 6 protons. The number of electrons in a neutral atom is equal to the number of protons, so the atom also contains 6 electrons in the electron cloud surrounding the nucleus. In summary, the answer is that there are 6 protons and 6 electrons in the atom.

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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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

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

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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${}^{-3}$

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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The heat of formation of ethene, C${}_2$H${}_4$ is 50 kJmol${}^{-1}$, and that of ethane, C${}_2$H${}_6$ is -82kJmol${}^{-1}$. Calculate the heat evolved in the process:

C${}_2$H${}_4$ + H${}_2$ $\to$ C${}_2$H${}_6$

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The heat evolved in a chemical reaction can be calculated by subtracting the heat of formation of the reactants from the heat of formation of the products. In this case, the reactants are ethene (C2H4) and hydrogen (H2), and the product is ethane (C2H6). The heat of formation of ethene is 50 kJ/mol and that of hydrogen is 0 kJ/mol (because hydrogen is a reference element). The heat of formation of ethane is -82 kJ/mol. So, the heat evolved in the reaction is given by: Heat evolved = (Heat of formation of products) - (Heat of formation of reactants) = (-82 kJ/mol) - (50 kJ/mol + 0 kJ/mol) = -82 kJ/mol - 50 kJ/mol = -132 kJ/mol. Therefore, the heat evolved in the process is -132 kJ.

The IUPAC nomenclature of the compound

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

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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

How many alkoxyalkanes can be obtained from the molecular formula C${}_4$H${}_{10}$O?

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Alkoxyalkanes have a general formula of R-O-R', where R and R' are alkyl groups. From the given molecular formula C4H10O, we can see that there are four carbon atoms, so the longest possible alkyl group is butyl (C4H9-). To form alkoxyalkanes, we need to attach an oxygen atom to the alkyl group. This can be done in three ways - by attaching the oxygen to one of the terminal carbon atoms (forming a primary alcohol), by attaching it to one of the central carbon atoms (forming a secondary alcohol), or by attaching it to the carbonyl carbon atom (forming an ester). So, we can obtain a maximum of three alkoxyalkanes from the given molecular formula. However, we need to take into account that there are different isomers possible for each type of alcohol or ester, depending on which carbon atom the oxygen is attached to. Therefore, the correct answer is (at least) 3.

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

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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.

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

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Explanation:

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?

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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 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.

2-methylprop-1-ene is an isomer of

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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.

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.

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 = $\frac{k}{M^{\frac{1}{2}}}$

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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.

Which of the following alkaline metals react more quickly spontaneously with water?

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The alkaline earth metals (Be, Mg, Ca, Sr, Ba, and Ra) are the second most reactive metals in the periodic table, and, like the Group 1 metals, have increasing reactivity in the higher periods. Beryllium (Be) is the only alkaline earth metal that does not react with water or steam, even if metal is heated to red heat. Additionally, beryllium has a resistant outer oxide layer that lowers its reactivity at lower temperatures.

Magnesium shows insignificant reaction with water, but burns vigorously with steam or water vapor to produce white magnesium oxide and hydrogen gas:

Mg(s) + 2 H2O(l) ⟶ Mg(OH)2(s) + H2(g)

A metal reacting with cold water will produce metal hydroxide. However, if a metal reacts with steam, like magnesium, metal oxide is produced as a result of metal hydroxides splitting upon heating.

The hydroxides of calcium, strontium and barium are only slightly water-soluble but produce sufficient hydroxide ions to make the environment basic, giving a general equation of:

M(s) + 2 H2O(l) ⟶ M(OH)2(aq) + H2(g)

• If metals react with cold water, hydroxides are produced.
• Metals that react with steam form oxides.
• Hydrogen is always produced when a metal reacts with cold water or steam.

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

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

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

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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.

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${}^3$

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

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

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.

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

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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.

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

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

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Hydrogen bond is a sort of

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Hydrogen bond is a covalent intermolecular bond that exists between hydrogen and highly electronegative elements like nitrogen, oxygen and fluorine.

Which of the following does not support the fact that air is a mixture?

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The option that does not support the fact that air is a mixture is "the constituents of air are in a fixed proportion by mass". Air is a mixture of different gases, primarily nitrogen (78%) and oxygen (21%), with small amounts of other gases such as carbon dioxide, argon, and neon. The proportion of each gas in air is not fixed and can vary depending on the location and other factors. For example, the amount of carbon dioxide in air can increase in areas with high levels of pollution, while the proportion of oxygen can decrease at high altitudes. Therefore, the composition of air is not in a fixed proportion by mass. On the other hand, the fact that air cannot be represented with a chemical formula and its constituents can be separated by physical means support the fact that air is a mixture. A chemical formula represents a pure substance, and since air is a mixture of gases, it cannot be represented by a single formula. Air can be separated into its individual components through physical means such as distillation or filtration, which is a characteristic of mixtures.

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.

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

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.

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

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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.

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

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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.

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

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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 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.