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Georgia Unified Chemistry practice questions

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Question 1Structure, bonding and inorganic chemistry

An atom of the chromium isotope ⁵²Cr has an atomic number of 24. How many protons, neutrons, and electrons are present in a neutral atom of this isotope?

  1. 24 protons, 24 neutrons, and 28 electrons
  2. 24 protons, 28 neutrons, and 24 electrons Correct answer.
  3. 28 protons, 24 neutrons, and 24 electrons
  4. 24 protons, 52 neutrons, and 24 electrons

Explanation

The atomic number Z = 24 specifies that the neutral chromium atom has 24 protons and 24 electrons. The mass number A = 52 gives the total number of nucleons, so the number of neutrons is A - Z = 52 - 24 = 28.

Question 2Structure, bonding and inorganic chemistry

What is the ground-state electron configuration of the valence shell of a neutral phosphorus atom (Z = 15)?

  1. 2s² 2p³
  2. 3s² 3p¹
  3. 3s² 3p³ Correct answer.
  4. 3s¹ 3p⁴

Explanation

Phosphorus has 15 electrons with the complete electron configuration 1s² 2s² 2p⁶ 3s² 3p³. Its valence shell corresponds to the principal quantum number n = 3, which contains five electrons in the subshells 3s² 3p³.

Question 3Structure, bonding and inorganic chemistry

What is the ground-state electron configuration of the iron(III) cation, Fe³⁺ (atomic number of Fe = 26)?

  1. [Ar] 3d⁵ Correct answer.
  2. [Ar] 4s² 3d³
  3. [Ar] 4s¹ 3d⁴
  4. [Ar] 3d⁶

Explanation

Neutral iron has the ground-state configuration [Ar] 4s² 3d⁶. Transition metals lose electrons from the outermost 4s orbital first during ionization, so forming Fe³⁺ removes both 4s electrons and one 3d electron, yielding the half-filled, stable [Ar] 3d⁵ configuration.

Question 4Structure, bonding and inorganic chemistry

Which set of four quantum numbers (n, l, ml, ms) correctly describes the outermost valence electron of a potassium atom (Z = 19) in its ground state?

  1. n = 3, l = 0, ml = 0, ms = +1/2
  2. n = 4, l = 1, ml = 0, ms = +1/2
  3. n = 3, l = 2, ml = 0, ms = +1/2
  4. n = 4, l = 0, ml = 0, ms = +1/2 Correct answer.

Explanation

Potassium (Z = 19) has the configuration [Ar] 4s¹. The valence electron occupies the 4s orbital, which has principal quantum number n = 4 and orbital angular momentum quantum number l = 0. For l = 0, the magnetic quantum number ml must be 0, and the spin quantum number ms can be +1/2.

Question 5Structure, bonding and inorganic chemistry

How does the atomic radius vary across Period 3 of the periodic table from sodium (Na) to chlorine (Cl), and what is the primary cause?

  1. Increases, because additional electron shells are being added
  2. Decreases, because effective nuclear charge increases across the period Correct answer.
  3. Remains constant, because all valence electrons reside in the n = 3 shell
  4. Increases, because electron-electron repulsions expand the outer orbital

Explanation

Moving from left to right across Period 3, protons are added to the nucleus while electrons enter the same valence energy level (n = 3). The core electron shielding remains relatively constant, causing the effective nuclear charge (Zeff) to increase, which draws the valence electron cloud closer to the nucleus and decreases the atomic radius.

Question 6Structure, bonding and inorganic chemistry

Why is the first ionization energy of beryllium (Be, Z = 4) higher than that of boron (B, Z = 5), despite boron having a higher nuclear charge?

  1. Boron's outermost electron occupies a higher-energy 2p orbital that is shielded by the filled 2s subshell Correct answer.
  2. Beryllium has a completely filled 2p subshell providing enhanced electronic stability
  3. Boron has a significantly larger nuclear charge that destabilizes its valence electrons
  4. Beryllium has a higher electronegativity than boron, preventing electron loss

Explanation

Beryllium has the electronic configuration 1s² 2s² with a completely filled, stable 2s subshell. Boron has the configuration 1s² 2s² 2p¹; its single 2p electron is at a higher energy level and is shielded from the nucleus by both the 1s and 2s electrons, making it easier to remove.

Question 7Structure, bonding and inorganic chemistry

Naturally occurring chlorine consists of two stable isotopes: ³⁵Cl (isotopic mass = 34.97 u, natural abundance = 75.78%) and ³⁷Cl (isotopic mass = 36.97 u, natural abundance = 24.22%). What is the standard atomic mass of chlorine?

  1. 35.97 u
  2. 36.45 u
  3. 35.45 u Correct answer.
  4. 35.00 u

Explanation

The average atomic mass is calculated by taking the weighted sum of the isotopic masses: (0.7578 × 34.97 u) + (0.2422 × 36.97 u) = 26.500 u + 8.954 u = 35.45 u.

Question 8Structure, bonding and inorganic chemistry

Which of the following compounds contains both ionic and polar covalent bonds in its solid state?

  1. Calcium chloride (CaCl₂)
  2. Methane (CH₄)
  3. Potassium chloride (KCl)
  4. Sodium sulfate (Na₂SO₄) Correct answer.

Explanation

In sodium sulfate (Na₂SO₄), electrostatic ionic attractions exist between the sodium cations (Na⁺) and the polyatomic sulfate anions (SO₄²⁻). Within the sulfate polyatomic anion itself, the sulfur and oxygen atoms are held together by polar covalent bonds.

Question 9Structure, bonding and inorganic chemistry

How do ammonia (NH₃) and boron trifluoride (BF₃) compare in molecular geometry and overall molecular dipole moment?

  1. Both molecules are planar and nonpolar
  2. NH₃ is trigonal pyramidal and polar, while BF₃ is trigonal planar and nonpolar Correct answer.
  3. NH₃ is tetrahedral and nonpolar, while BF₃ is trigonal pyramidal and polar
  4. Both molecules are trigonal pyramidal and polar

Explanation

In NH₃, nitrogen has three single bonds and one lone pair (AX₃E), giving a trigonal pyramidal geometry where individual N–H bond dipoles do not cancel, producing a net dipole moment (polar). In BF₃, boron has three bonding pairs and zero lone pairs (AX₃), giving a symmetric trigonal planar geometry where the three B–F bond dipoles cancel completely, resulting in a nonpolar molecule.

Question 10Structure, bonding and inorganic chemistry

What is the hybridization of the central sulfur atom in sulfur hexafluoride (SF₆) and its corresponding molecular geometry?

  1. sp³d² hybridization with an octahedral molecular geometry Correct answer.
  2. sp³d hybridization with a trigonal bipyramidal geometry
  3. sp³ hybridization with a tetrahedral geometry
  4. dsp² hybridization with a square planar geometry

Explanation

In SF₆, the central sulfur atom forms six single covalent bonds with six fluorine atoms and has no non-bonding lone pairs. This arrangement corresponds to a steric number of 6, which requires six hybrid orbitals formed by sp³d² hybridization, resulting in an octahedral molecular geometry with 90° bond angles.

Question 11Structure, bonding and inorganic chemistry

Hydrogen fluoride (HF) has an unexpectedly high boiling point (19.5 °C) compared to hydrogen chloride (HCl, -85.1 °C), hydrogen bromide (HBr, -66.8 °C), and hydrogen iodide (HI, -35.4 °C). What primary intermolecular force accounts for this anomaly?

  1. Strong covalent network bonding in liquid HF
  2. Enhanced London dispersion forces due to high molecular polarizability
  3. Strong intermolecular hydrogen bonding between fluorine and hydrogen atoms Correct answer.
  4. Ion-dipole interactions arising from complete spontaneous ionization

Explanation

Fluorine is the most electronegative element with a small atomic radius, which polarizes the H–F bond intensely. The resulting large partial positive charge on hydrogen and high electron density on fluorine enable the formation of strong intermolecular hydrogen bonds, requiring substantial thermal energy to separate molecules into the gas phase.

Question 12Structure, bonding and inorganic chemistry

What mass of metallic iron is theoretically produced when 16.0 g of iron(III) oxide (Fe₂O₃, molar mass = 159.7 g/mol) is completely reduced by carbon monoxide according to the balanced equation: Fe₂O₃(s) + 3 CO(g) → 2 Fe(s) + 3 CO₂(g)?

  1. 5.59 g
  2. 8.00 g
  3. 16.0 g
  4. 11.2 g Correct answer.

Explanation

Moles of Fe₂O₃ reacted = 16.0 g / 159.7 g/mol = 0.1002 mol. From the stoichiometry, 1 mole of Fe₂O₃ produces 2 moles of Fe, yielding 0.1002 × 2 = 0.2004 mol of Fe. Using the molar mass of iron (55.85 g/mol), the theoretical mass produced is 0.2004 mol × 55.85 g/mol = 11.2 g.

Question 13Structure, bonding and inorganic chemistry

What volume of carbon dioxide gas measured at standard temperature and pressure (STP: 0 °C, 1 atm, molar volume = 22.4 L/mol) is released upon the complete thermal decomposition of 25.0 g of calcium carbonate (CaCO₃, molar mass = 100.1 g/mol)?

  1. 2.24 L
  2. 5.60 L Correct answer.
  3. 11.2 L
  4. 22.4 L

Explanation

The decomposition reaction is CaCO₃(s) → CaO(s) + CO₂(g). First calculate moles of CaCO₃: n = 25.0 g / 100.1 g/mol = 0.250 mol. Since the stoichiometric ratio is 1:1, 0.250 mol of CO₂ gas is formed. At STP, V = 0.250 mol × 22.4 L/mol = 5.60 L.

Question 14Structure, bonding and inorganic chemistry

How many total carbon dioxide molecules are present in a 4.40 g sample of pure carbon dioxide gas (CO₂, molar mass = 44.01 g/mol)?

  1. 6.02 × 10²² molecules Correct answer.
  2. 6.02 × 10²³ molecules
  3. 1.20 × 10²³ molecules
  4. 3.01 × 10²² molecules

Explanation

The number of moles in 4.40 g of CO₂ is n = 4.40 g / 44.01 g/mol = 0.100 mol. Multiplying by Avogadro's constant gives: N = 0.100 mol × 6.022 × 10²³ molecules/mol = 6.02 × 10²² molecules.

Question 15Structure, bonding and inorganic chemistry

A gaseous hydrocarbon consists of 85.63% carbon and 14.37% hydrogen by mass. If its experimental molar mass is 56.11 g/mol, what are its empirical and molecular formulas?

  1. Empirical formula CH₃; molecular formula C₂H₆
  2. Empirical formula CH; molecular formula C₄H₄
  3. Empirical formula CH₂; molecular formula C₄H₈ Correct answer.
  4. Empirical formula C₂H₅; molecular formula C₄H₁₀

Explanation

In 100 g of compound, moles of C = 85.63 g / 12.011 g/mol = 7.129 mol, and moles of H = 14.37 g / 1.008 g/mol = 14.256 mol. The molar ratio H:C is 14.256 / 7.129 = 2.00, giving the empirical formula CH₂ (empirical mass = 14.03 g/mol). The ratio of molar mass to empirical mass is 56.11 / 14.03 = 4, so the molecular formula is (CH₂)₄ = C₄H₈.

Question 16Structure, bonding and inorganic chemistry

A 5.40 g sample of powdered aluminum (molar mass = 27.0 g/mol) is placed into an aqueous solution containing 48.0 g of copper(II) sulfate (CuSO₄, molar mass = 159.6 g/mol) according to the equation: 2 Al(s) + 3 CuSO₄(aq) → Al₂(SO₄)₃(aq) + 3 Cu(s). What is the theoretical mass of copper metal precipitated?

  1. 12.7 g
  2. 6.36 g
  3. 25.4 g
  4. 19.1 g Correct answer.

Explanation

Calculate available moles: n(Al) = 5.40 g / 27.0 g/mol = 0.200 mol; n(CuSO₄) = 48.0 g / 159.6 g/mol = 0.3008 mol. The reaction requires 3 moles of CuSO₄ for every 2 moles of Al (ratio = 1.50). Here, 0.200 mol Al requires 0.300 mol CuSO₄, which matches the available 0.3008 mol CuSO₄. Thus, moles of Cu produced = (3/2) × 0.200 mol = 0.300 mol. Mass of Cu = 0.300 mol × 63.55 g/mol = 19.1 g.

Question 17Structure, bonding and inorganic chemistry

Why is the experimental bond angle in water (H₂O, 104.5°) smaller than the ideal tetrahedral angle observed in methane (CH₄, 109.5°)?

  1. The greater electronegativity of carbon relative to oxygen attracts bond pairs closer
  2. The two non-bonding lone pairs on oxygen exert greater electrostatic repulsion than bonding pairs Correct answer.
  3. Hydrogen bonds between adjacent water molecules physically compress the intramolecular angle
  4. Oxygen utilizes unhybridized p orbitals to form the covalent bonds with hydrogen

Explanation

According to VSEPR theory, lone pairs are held closer to the central nucleus and occupy a larger angular volume than bonding pairs. In water, the two lone pairs on oxygen exert stronger electron repulsion than the bonding pairs, compressing the H–O–H bond angle from the ideal tetrahedral 109.5° down to approximately 104.5°.

Question 18Structure, bonding and inorganic chemistry

Which of the following correctly ranks the isoelectronic chemical species S²⁻, Cl⁻, K⁺, and Ca²⁺ in order of decreasing ionic radius?

  1. S²⁻ > Cl⁻ > K⁺ > Ca²⁺ Correct answer.
  2. Ca²⁺ > K⁺ > Cl⁻ > S²⁻
  3. Cl⁻ > S²⁻ > Ca²⁺ > K⁺
  4. K⁺ > Ca²⁺ > S²⁻ > Cl⁻

Explanation

All four species are isoelectronic with 18 electrons (argon electron configuration). As the atomic number (nuclear charge Z) increases from sulfur (Z = 16) to chlorine (Z = 17), potassium (Z = 19), and calcium (Z = 20), the greater number of protons pulls the identical electron cloud more tightly toward the nucleus, decreasing ionic radius.

Question 19Structure, bonding and inorganic chemistry

A gaseous oxide of nitrogen has an experimental density of 2.054 g/L at standard temperature and pressure (0 °C and 1.00 atm, molar volume = 22.41 L/mol). What is the molar mass of the gas and its chemical identity?

  1. 30.0 g/mol; nitric oxide (NO)
  2. 44.0 g/mol; nitrous oxide (N₂O)
  3. 46.0 g/mol; nitrogen dioxide (NO₂) Correct answer.
  4. 92.0 g/mol; dinitrogen tetroxide (N₂O₄)

Explanation

The molar mass of an ideal gas at STP is calculated from its density using M = d × Vm. M = 2.054 g/L × 22.41 L/mol = 46.03 g/mol. The molar mass of nitrogen dioxide (NO₂) is 14.01 + 2(16.00) = 46.01 g/mol.

Question 20Structure, bonding and inorganic chemistry

Based on electronegativity differences (H = 2.20, C = 2.55, N = 3.04, O = 3.44, F = 3.98), which of the following single covalent bonds has the highest degree of ionic character (polarity)?

  1. C–H
  2. N–H
  3. O–H
  4. H–F Correct answer.

Explanation

Bond polarity is governed by the difference in Pauling electronegativity (ΔEN) between the bonded atoms. For H–F, ΔEN = 3.98 - 2.20 = 1.78, which is larger than O–H (1.24), N–H (0.84), or C–H (0.35), giving H–F the highest bond polarity and greatest dipole moment.

Question 21Structure, bonding and inorganic chemistry

What is the mass percentage of nitrogen in pure ammonium nitrate fertilizer (NH₄NO₃, molar mass = 80.04 g/mol; atomic mass of N = 14.01 g/mol)?

  1. 17.5%
  2. 35.0% Correct answer.
  3. 28.0%
  4. 42.5%

Explanation

One formula unit of NH₄NO₃ contains two nitrogen atoms: one in the ammonium cation and one in the nitrate anion. The total mass of nitrogen per mole is 2 × 14.01 g/mol = 28.02 g/mol. The mass percentage is (28.02 / 80.04) × 100% = 35.01%.

Question 22Structure, bonding and inorganic chemistry

Why is the experimental lattice energy of magnesium oxide (MgO, 3791 kJ/mol) dramatically higher than that of sodium chloride (NaCl, 786 kJ/mol)?

  1. The product of ionic charges in MgO (+2 and -2) is four times that in NaCl (+1 and -1), and the ions are smaller Correct answer.
  2. Magnesium forms covalent network coordinate bonds with oxygen within the crystal lattice
  3. Sodium has a higher ionization energy than magnesium, weakening the electrostatic field
  4. Chloride ions are smaller than oxide ions, increasing interionic repulsive forces

Explanation

Lattice energy is governed by Coulomb's law: U ∝ (q₁ × q₂) / r₀. In MgO, the ions carry charges of +2 and -2 (charge product = 4), whereas in NaCl the ions carry charges of +1 and -1 (charge product = 1). In addition, both Mg²⁺ and O²⁻ have smaller ionic radii than Na⁺ and Cl⁻, resulting in a shorter internuclear separation distance r₀.

Question 23Structure, bonding and inorganic chemistry

A 10.0 g piece of zinc metal (molar mass = 65.38 g/mol) is treated with 100.0 mL of 2.00 M hydrochloric acid (HCl). What volume of hydrogen gas at STP is generated upon completion of the reaction: Zn(s) + 2 HCl(aq) → ZnCl₂(aq) + H₂(g)?

  1. 3.43 L
  2. 4.48 L
  3. 2.24 L Correct answer.
  4. 1.12 L

Explanation

Calculate available moles: n(Zn) = 10.0 g / 65.38 g/mol = 0.153 mol; n(HCl) = 0.1000 L × 2.00 mol/L = 0.200 mol. Since the reaction requires 2 moles of HCl per mole of Zn, 0.200 mol of HCl can only react with 0.100 mol of Zn. Thus, HCl is the limiting reactant, and moles of H₂ produced = 0.200 mol / 2 = 0.100 mol. Volume at STP = 0.100 mol × 22.4 L/mol = 2.24 L.

Question 24Structure, bonding and inorganic chemistry

What is the ground-state electron configuration of the chromium(III) ion, Cr³⁺ (atomic number of Cr = 24)?

  1. [Ar] 4s² 3d¹
  2. [Ar] 4s¹ 3d²
  3. [Ar] 3d⁴
  4. [Ar] 3d³ Correct answer.

Explanation

Neutral chromium has the anomalous ground-state electron configuration [Ar] 4s¹ 3d⁵. Upon ionization to Cr³⁺, the outermost 4s electron is removed first, followed by two 3d electrons, leaving the configuration [Ar] 3d³ with three unpaired 3d electrons.

Question 25Structure, bonding and inorganic chemistry

What volume of concentrated 12.0 M hydrochloric acid (HCl) stock solution is required to prepare exactly 500.0 mL of a 0.600 M HCl analytical solution by dilution with distilled water?

  1. 15.0 mL
  2. 25.0 mL Correct answer.
  3. 50.0 mL
  4. 30.0 mL

Explanation

Using the dilution equation M₁V₁ = M₂V₂: V₁ = (M₂ × V₂) / M₁ = (0.600 mol/L × 500.0 mL) / 12.0 mol/L = 300.0 / 12.0 = 25.0 mL.

Question 26Physical chemistry, amounts and reactions

How does adding a chemical catalyst increase the rate of a reversible chemical reaction?

  1. It provides an alternative reaction mechanism with lower activation energy for both forward and reverse pathways Correct answer.
  2. It increases the average kinetic energy and temperature of the reacting particles
  3. It shifts the position of equilibrium toward the products by increasing the equilibrium constant Kc
  4. It makes the reaction enthalpy (ΔH) significantly more exothermic

Explanation

A catalyst functions by introducing an alternative reaction pathway with a lower activation energy (Ea). Because it lowers the energy barrier by the exact same amount for both the forward and reverse directions, both rates increase equally without altering the equilibrium position, Kc, or reaction enthalpy ΔH.

Question 27Physical chemistry, amounts and reactions

For the gaseous reaction 2 A + B → C, experimental kinetic studies show that doubling [A] while keeping [B] constant quadruples the initial rate, whereas doubling [B] while keeping [A] constant doubles the rate. What is the differential rate law and the overall reaction order?

  1. Rate = k[A][B]; overall order 2
  2. Rate = k[A][B]²; overall order 3
  3. Rate = k[A]²[B]; overall order 3 Correct answer.
  4. Rate = k[A]²[B]²; overall order 4

Explanation

When [A] doubles and the rate increases by 2² = 4, the reaction order with respect to A is 2. When [B] doubles and the rate increases by 2¹ = 2, the order with respect to B is 1. The rate law is Rate = k[A]²[B], and the overall reaction order is 2 + 1 = 3.

Question 28Physical chemistry, amounts and reactions

According to the empirical van 't Hoff rule, a reaction rate increases by a factor of 2 for every 10 °C temperature elevation. By what factor will the reaction rate increase if the temperature is raised from 20 °C to 60 °C?

  1. 4 times
  2. 8 times
  3. 12 times
  4. 16 times Correct answer.

Explanation

The temperature increase is ΔT = 60 °C - 20 °C = 40 °C. The number of 10 °C increments is n = 40 / 10 = 4. Applying van 't Hoff's temperature factor rule: Rate₂ / Rate₁ = γⁿ = 2⁴ = 16 times.

Question 29Physical chemistry, amounts and reactions

What is the correct equilibrium constant expression (Kc) for the thermal decomposition of solid calcium carbonate in a closed system: CaCO₃(s) ⇌ CaO(s) + CO₂(g)?

  1. Kc = [CaO][CO₂] / [CaCO₃]
  2. Kc = [CO₂] Correct answer.
  3. Kc = [CO₂] / [CaCO₃]
  4. Kc = 1 / [CO₂]

Explanation

In heterogeneous equilibria, pure solids and pure liquids have constant thermodynamic activities equal to 1. Because CaCO₃(s) and CaO(s) are pure solids, their terms are omitted from the equilibrium constant expression, leaving Kc = [CO₂].

Question 30Physical chemistry, amounts and reactions

Consider the Haber synthesis of ammonia: N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g), with ΔH = -92.2 kJ/mol. Which combination of changes will shift the equilibrium position to maximize the yield of ammonia?

  1. Increasing total pressure and decreasing temperature Correct answer.
  2. Decreasing total pressure and decreasing temperature
  3. Increasing total pressure and increasing temperature
  4. Decreasing total pressure and increasing temperature

Explanation

The forward reaction converts 4 moles of gas into 2 moles of gas; according to Le Chatelier's principle, increasing pressure shifts the system toward the side with fewer gas molecules (products). Because the reaction is exothermic (ΔH < 0), lowering the temperature shifts the equilibrium in the heat-producing forward direction, increasing ammonia yield.

Question 31Physical chemistry, amounts and reactions

In a 2.0 L rigid reaction vessel at 450 °C, 2.0 mol of H₂ and 2.0 mol of I₂ are allowed to reach equilibrium: H₂(g) + I₂(g) ⇌ 2 HI(g). If the equilibrium constant Kc is 64.0 at this temperature, what is the equilibrium concentration of HI?

  1. 0.80 mol/L
  2. 3.20 mol/L
  3. 1.60 mol/L Correct answer.
  4. 0.20 mol/L

Explanation

Initial concentrations are [H₂]₀ = [I₂]₀ = 2.0 mol / 2.0 L = 1.0 M. At equilibrium, [H₂] = [I₂] = 1.0 - x, and [HI] = 2x. Substitute into Kc: (2x)² / (1.0 - x)² = 64.0. Taking the square root of both sides: 2x / (1.0 - x) = 8.0. Solving gives 2x = 8.0 - 8.0x, so 10x = 8.0 and x = 0.80 M. Therefore, the equilibrium concentration of HI is [HI] = 2x = 2(0.80 M) = 1.60 M.

Question 32Physical chemistry, amounts and reactions

What are the algebraic signs of the standard enthalpy change (ΔH°) and standard entropy change (ΔS°) for the melting of ice at 0 °C: H₂O(s) → H₂O(l)?

  1. ΔH° < 0 and ΔS° < 0
  2. ΔH° < 0 and ΔS° > 0
  3. ΔH° > 0 and ΔS° < 0
  4. ΔH° > 0 and ΔS° > 0 Correct answer.

Explanation

Melting requires the absorption of heat to disrupt the hydrogen-bonded crystal lattice, so the process is endothermic and ΔH° > 0. Liquid water possesses greater positional and rotational disorder than the rigid ice lattice, so entropy increases and ΔS° > 0.

Question 33Physical chemistry, amounts and reactions

Given the following thermochemical equations: (1) C(graphite) + O₂(g) → CO₂(g), ΔH₁ = -393.5 kJ/mol (2) CO(g) + 1/2 O₂(g) → CO₂(g), ΔH₂ = -283.0 kJ/mol What is the standard enthalpy of formation of carbon monoxide: C(graphite) + 1/2 O₂(g) → CO(g)?

  1. -676.5 kJ/mol
  2. -110.5 kJ/mol Correct answer.
  3. +110.5 kJ/mol
  4. -283.0 kJ/mol

Explanation

According to Hess's Law, target equation = equation (1) - equation (2). Subtracting equation (2) from equation (1) yields C(graphite) + 1/2 O₂(g) → CO(g). Thus, ΔH = ΔH₁ - ΔH₂ = -393.5 kJ/mol - (-283.0 kJ/mol) = -393.5 + 283.0 = -110.5 kJ/mol.

Question 34Physical chemistry, amounts and reactions

How many grams of solid sodium hydroxide (NaOH) and distilled water are required to prepare 250.0 g of a 12.0% by mass aqueous NaOH solution?

  1. 30.0 g of NaOH and 220.0 g of water Correct answer.
  2. 12.0 g of NaOH and 238.0 g of water
  3. 24.0 g of NaOH and 226.0 g of water
  4. 30.0 g of NaOH and 250.0 g of water

Explanation

Mass fraction w = (mass of solute) / (mass of solution). Mass of NaOH = 250.0 g × 0.120 = 30.0 g. Mass of water required = total solution mass - solute mass = 250.0 g - 30.0 g = 220.0 g.

Question 35Physical chemistry, amounts and reactions

A student dissolves 9.80 g of pure sulfuric acid (H₂SO₄, molar mass = 98.08 g/mol) in distilled water to make a total solution volume of 500.0 mL. What is the molar concentration of the resulting solution?

  1. 0.100 M
  2. 0.050 M
  3. 0.200 M Correct answer.
  4. 0.400 M

Explanation

First calculate moles of H₂SO₄: n = 9.80 g / 98.08 g/mol = 0.0999 mol ≈ 0.100 mol. Convert volume to liters: V = 500.0 mL = 0.500 L. Molar concentration c = n / V = 0.100 mol / 0.500 L = 0.200 mol/L (0.200 M).

Question 36Physical chemistry, amounts and reactions

What is the pH of a 0.0050 M aqueous solution of barium hydroxide, Ba(OH)₂, at 25 °C assuming complete dissociation?

  1. 2.00
  2. 11.70
  3. 2.30
  4. 12.00 Correct answer.

Explanation

Barium hydroxide is a strong diprotic base that dissociates completely: Ba(OH)₂ → Ba²⁺ + 2 OH⁻. Therefore, [OH⁻] = 2 × 0.0050 M = 0.010 M = 1.0 × 10⁻² M. Calculate pOH = -log₁₀(1.0 × 10⁻²) = 2.00. At 25 °C, pH = 14.00 - pOH = 14.00 - 2.00 = 12.00.

Question 37Physical chemistry, amounts and reactions

A 0.100 M aqueous solution of acetic acid (CH₃COOH) has an acid dissociation constant Ka = 1.80 × 10⁻⁵ at 25 °C. What is the equilibrium concentration of hydronium ions [H₃O⁺] and the solution pH?

  1. [H₃O⁺] = 1.80 × 10⁻⁵ M, pH = 4.74
  2. [H₃O⁺] = 1.34 × 10⁻³ M, pH = 2.87 Correct answer.
  3. [H₃O⁺] = 4.24 × 10⁻³ M, pH = 2.37
  4. [H₃O⁺] = 1.00 × 10⁻¹ M, pH = 1.00

Explanation

For a weak monoprotic acid where percent ionization is less than 5%: [H₃O⁺] = √(Ka × C₀) = √(1.80 × 10⁻⁵ × 0.100) = √(1.80 × 10⁻⁶) = 1.34 × 10⁻³ M. The pH = -log₁₀(1.34 × 10⁻³) = 3 - 0.127 = 2.87.

Question 38Physical chemistry, amounts and reactions

Which of the following 0.10 M aqueous salt solutions will have a pH greater than 7.0 at 25 °C due to salt hydrolysis?

  1. Sodium carbonate (Na₂CO₃) Correct answer.
  2. Ammonium chloride (NH₄Cl)
  3. Sodium chloride (NaCl)
  4. Potassium nitrate (KNO₃)

Explanation

Sodium carbonate is the salt of a strong base (NaOH) and a weak diprotic acid (H₂CO₃). The carbonate anion undergoes basic hydrolysis: CO₃²⁻ + H₂O ⇌ HCO₃⁻ + OH⁻, generating hydroxide ions and producing a basic solution (pH > 7.0).

Question 39Physical chemistry, amounts and reactions

How does an acetic acid / sodium acetate buffer solution (CH₃COOH / CH₃COONa) resist a change in pH upon the addition of a small amount of hydrochloric acid (HCl)?

  1. The added H⁺ ions react with molecular acetic acid to form protonated water clusters
  2. The sodium cations precipitate the chloride ions, removing acid from the solution
  3. The added H⁺ ions are consumed by conjugate base acetate ions (CH₃COO⁻) to form weakly ionized CH₃COOH Correct answer.
  4. The buffer spontaneously evaporates excess chloride as hydrogen chloride gas

Explanation

A buffer contains significant concentrations of both a weak acid and its conjugate base. When strong acid (H⁺) is introduced, it is neutralized by the conjugate base: H⁺(aq) + CH₃COO⁻(aq) → CH₃COOH(aq). Because acetic acid is weak and poorly dissociated, the added hydronium ions are tied up, keeping the pH nearly unchanged.

Question 40Physical chemistry, amounts and reactions

What is the oxidation state of chromium in the dichromate polyatomic anion, Cr₂O₇²⁻?

  1. +3
  2. +4
  3. +5
  4. +6 Correct answer.

Explanation

Oxygen typically has an oxidation state of -2. For Cr₂O₇²⁻, the sum of all oxidation states must equal the ionic charge of -2: 2(Cr) + 7(-2) = -2. Thus, 2(Cr) - 14 = -2, which yields 2(Cr) = +12, or Cr = +6.

Question 41Physical chemistry, amounts and reactions

In the balanced oxidation-reduction equation in acidic aqueous solution: a MnO₄⁻ + b Fe²⁺ + c H⁺ → d Mn²⁺ + e Fe³⁺ + f H₂O, what is the stoichiometric coefficient ratio b : a (moles of Fe²⁺ per mole of MnO₄⁻)?

  1. 1 : 1
  2. 5 : 1 Correct answer.
  3. 3 : 1
  4. 2 : 5

Explanation

The reduction half-reaction is MnO₄⁻ + 8 H⁺ + 5 e⁻ → Mn²⁺ + 4 H₂O (Mn gains 5 electrons). The oxidation half-reaction is Fe²⁺ → Fe³⁺ + e⁻ (Fe loses 1 electron). To balance electron transfer, multiply the oxidation half-reaction by 5, giving MnO₄⁻ + 5 Fe²⁺ + 8 H⁺ → Mn²⁺ + 5 Fe³⁺ + 4 H₂O. The stoichiometric ratio is 5 moles of Fe²⁺ to 1 mole of MnO₄⁻.

Question 42Physical chemistry, amounts and reactions

Based on standard electrode potentials (E°(Mg²⁺/Mg) = -2.37 V, E°(Fe²⁺/Fe) = -0.44 V, E°(Cu²⁺/Cu) = +0.34 V, E°(Ag⁺/Ag) = +0.80 V), which metal will spontaneously displace copper from an aqueous copper(II) sulfate solution but CANNOT displace magnesium from magnesium sulfate?

  1. Iron (Fe) Correct answer.
  2. Silver (Ag)
  3. Gold (Au)
  4. Copper itself (Cu)

Explanation

A metal can reduce the cations of another metal if its standard reduction potential is more negative (it lies higher in the electrochemical activity series). Iron (E° = -0.44 V) is more active than copper (+0.34 V) and spontaneously reduces Cu²⁺ to metallic copper: Fe + Cu²⁺ → Fe²⁺ + Cu. However, iron is less active than magnesium (-2.37 V) and cannot displace Mg.

Question 43Physical chemistry, amounts and reactions

What is the standard cell potential (E°cell) for a galvanic Daniell cell operating under standard conditions: Zn(s) | Zn²⁺(1 M) || Cu²⁺(1 M) | Cu(s), given E°(Zn²⁺/Zn) = -0.76 V and E°(Cu²⁺/Cu) = +0.34 V?

  1. +0.42 V
  2. -1.10 V
  3. +1.10 V Correct answer.
  4. +0.98 V

Explanation

In this cell, copper undergoes reduction at the cathode and zinc undergoes oxidation at the anode. The standard cell potential is E°cell = E°cathode - E°anode = E°(Cu²⁺/Cu) - E°(Zn²⁺/Zn) = +0.34 V - (-0.76 V) = +0.34 V + 0.76 V = +1.10 V.

Question 44Physical chemistry, amounts and reactions

A steady direct electric current of 5.00 A is passed through an aqueous copper(II) sulfate (CuSO₄) electrolytic cell for 3860 seconds. What mass of metallic copper (molar mass = 63.55 g/mol; Faraday constant F = 96,500 C/mol) is deposited at the cathode?

  1. 12.7 g
  2. 3.18 g
  3. 1.93 g
  4. 6.36 g Correct answer.

Explanation

Calculate total electric charge passed: Q = I × t = 5.00 A × 3860 s = 19,300 C. Moles of electrons passed: n(e⁻) = Q / F = 19,300 C / 96,500 C/mol = 0.200 mol e⁻. Reduction of copper(II) requires 2 electrons: Cu²⁺ + 2 e⁻ → Cu. Moles of Cu deposited = 0.200 mol / 2 = 0.100 mol. Mass of Cu = 0.100 mol × 63.55 g/mol = 6.36 g.

Question 45Physical chemistry, amounts and reactions

What chemical products are evolved at the cathode and anode, respectively, during the commercial electrolysis of concentrated aqueous sodium chloride (brine) using inert electrodes?

  1. Sodium metal at the cathode and chlorine gas at the anode
  2. Hydrogen gas at the cathode and chlorine gas at the anode Correct answer.
  3. Hydrogen gas at the cathode and oxygen gas at the anode
  4. Sodium metal at the cathode and oxygen gas at the anode

Explanation

In aqueous solution, water is more easily reduced than sodium ions (E° for H₂O reduction is -0.83 V vs -2.71 V for Na⁺), evolving hydrogen gas (H₂) and generating OH⁻ at the cathode: 2 H₂O + 2 e⁻ → H₂ + 2 OH⁻. At the anode, chloride ions are oxidized to chlorine gas (Cl₂): 2 Cl⁻ → Cl₂ + 2 e⁻, due to overpotential effects compared to water oxidation.

Question 46Physical chemistry, amounts and reactions

The molar solubility of silver chloride (AgCl, molar mass = 143.3 g/mol) in pure water at 25 °C is 1.34 × 10⁻⁵ mol/L. What is the solubility product constant (Ksp) of AgCl at this temperature?

  1. 1.80 × 10⁻¹⁰ Correct answer.
  2. 1.34 × 10⁻⁵
  3. 2.68 × 10⁻¹⁰
  4. 3.60 × 10⁻¹⁵

Explanation

Silver chloride dissolves according to AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq). For each mole of AgCl dissolved, one mole of Ag⁺ and one mole of Cl⁻ enter solution, so [Ag⁺] = [Cl⁻] = s = 1.34 × 10⁻⁵ mol/L. The solubility product is Ksp = [Ag⁺][Cl⁻] = s² = (1.34 × 10⁻⁵)² = 1.80 × 10⁻¹⁰.

Question 47Physical chemistry, amounts and reactions

How does the addition of 0.10 M sodium chloride (NaCl) affect the molar solubility of silver chloride (AgCl) in water compared to its solubility in pure water?

  1. It increases solubility because Na⁺ ions form soluble coordination complexes with Ag⁺
  2. It has no effect because NaCl is a spectator salt
  3. It drastically decreases solubility because the high [Cl⁻] shifts the equilibrium to the left Correct answer.
  4. It increases solubility by decreasing the solution dielectric constant

Explanation

According to the common ion effect and Le Chatelier's principle, adding dissolved chloride ions from the fully soluble NaCl increases [Cl⁻]. To maintain the constant product [Ag⁺][Cl⁻] = Ksp, the dissolution equilibrium AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq) shifts strongly to the left, drastically suppressing the concentration of dissolved Ag⁺.

Question 48Physical chemistry, amounts and reactions

A 50.0 g piece of an unknown metal heated to 100.0 °C is placed into an insulated calorimeter containing 100.0 g of water at 20.0 °C (specific heat capacity of water = 4.184 J/(g·°C)). The final thermal equilibrium temperature is 22.0 °C. What is the specific heat capacity of the metal, assuming negligible heat loss?

  1. 0.450 J/(g·°C)
  2. 0.128 J/(g·°C)
  3. 0.897 J/(g·°C)
  4. 0.215 J/(g·°C) Correct answer.

Explanation

Heat gained by water: q_water = m_w × c_w × ΔT_w = 100.0 g × 4.184 J/(g·°C) × (22.0 - 20.0 °C) = 836.8 J. Heat lost by metal: q_metal = m_m × c_m × (T_initial - T_final) = 50.0 g × c_m × (100.0 - 22.0 °C) = 3900 g·°C × c_m. By conservation of energy, 3900 × c_m = 836.8 J, giving c_m = 836.8 / 3900 = 0.215 J/(g·°C).

Question 49Physical chemistry, amounts and reactions

When 100.0 mL of 1.00 M hydrochloric acid (HCl) is mixed with 100.0 mL of 1.00 M sodium hydroxide (NaOH) in a calorimeter, neutralization occurs: H⁺(aq) + OH⁻(aq) → H₂O(l), with ΔH° = -57.3 kJ/mol. How much heat is released into the solution?

  1. 11.5 kJ
  2. 5.73 kJ Correct answer.
  3. 57.3 kJ
  4. 2.87 kJ

Explanation

Calculate moles of reactants: n(HCl) = 0.100 L × 1.00 mol/L = 0.100 mol; n(NaOH) = 0.100 L × 1.00 mol/L = 0.100 mol. The stoichiometric reaction consumes 0.100 mol of H⁺ and 0.100 mol of OH⁻. Total heat released = n × |ΔH°| = 0.100 mol × 57.3 kJ/mol = 5.73 kJ.

Question 50Physical chemistry, amounts and reactions

According to collision theory and the Arrhenius equation, why does raising the temperature of a reaction mixture have such a pronounced effect on increasing the reaction rate?

  1. The fraction of colliding molecules with kinetic energy exceeding the activation energy increases exponentially Correct answer.
  2. The activation energy barrier (Ea) decreases significantly as temperature rises
  3. The frequency of collision between reactant molecules increases by more than tenfold per 10 °C
  4. The steric orientation requirement becomes irrelevant at higher kinetic energies

Explanation

According to the Maxwell-Boltzmann distribution and the Arrhenius relation k = A·exp(-Ea/RT), a modest increase in temperature slightly increases collision frequency (~2%), but exponentially increases the fraction of molecules with kinetic energies equal to or exceeding the activation energy threshold (Ea).