JAMB UTME - Chemistry - 2019

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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${}^3$

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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 ${}^{\circ }\text{C}$

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Détails de la réponse

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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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${}^{-3}$

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Potassium Dichromate (VI), when it is acidified, acts as an oxidizing agent. When this happens, the color changes from orange to green. This is because the orange color of the potassium dichromate is due to the presence of Cr(VI) ions, which are oxidized to Cr(III) ions. The green color that is produced is due to the formation of chromium(III) ions. In this reaction, the dichromate ions are being oxidized, which means that they are losing electrons, and the chromium ions are being reduced, which means that they are gaining electrons. The transfer of electrons causes the color change from orange to green.

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

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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.

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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.

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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.

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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.

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Increasing the temperature would increase the concentration of the product, P. The reaction rate, or the speed at which the reaction occurs, is influenced by temperature. An increase in temperature raises the kinetic energy of the reacting molecules, making it easier for them to collide and react. This leads to a higher rate of reaction and a higher concentration of the product, P. Adding a suitable catalyst can also increase the reaction rate, but it does not directly affect the concentration of the product. Increasing the concentration of P does not affect the reaction itself, but is a result of the reaction having taken place. Decreasing the temperature would slow down the reaction rate and reduce the concentration of the product.

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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.

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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.

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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

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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.

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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.

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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

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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.

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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.

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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.

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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.

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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.

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Ammonium chloride contains both ionic and covalent bonds. In ammonium chloride, the ammonium ion (NH4+) is positively charged and the chloride ion (Cl-) is negatively charged. These ions are held together by ionic bonds, which are formed between positively and negatively charged ions. However, the bond between the hydrogen atom in the ammonium ion and the nitrogen atom in the ammonium ion is also a covalent bond. This type of covalent bond is known as a dative covalent bond, or a coordinate covalent bond, because the electron pair being shared is supplied by one atom only (the nitrogen atom in this case). So, the kind of bonding present in ammonium chloride is both ionic and dative covalent. In simple terms, ammonium chloride contains both ionic bonds between its positive and negative ions, and a dative covalent bond between the hydrogen atom and nitrogen atom within the ammonium ion.

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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.

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

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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.

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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.

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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.

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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)

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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.

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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${}_4$.

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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.

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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.

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