June 2023 Chemistry Regents — Answer Key & Worked Solutions

Every question from the June 2023 Chemistry Regents exam, with the correct answer and a full step-by-step solution. Check your work, learn the method, then convert your raw score.

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These are real questions from the June 2023 Chemistry New York State Regents examination, published by NYSED. Worked solutions are written by RegentsLab.

Question 1 · 2 pts

Which phrase describes the nucleus of any atom?

(A) has an overall positive charge ✓
(B) has an overall negative charge
(C) contains negative electrons
(D) contains positive electrons

Worked solution

$\text{Nucleus} = \text{protons (+)} + \text{neutrons (0)}$

$\text{Net charge} > 0$

Answer: has an overall positive charge

The nucleus contains protons (positive) and neutrons (neutral), so it always carries an overall positive charge; electrons reside outside the nucleus.

Question 2 · 2 pts

Which two particles each have a mass of approximately one atomic mass unit?

(A) an electron and a proton
(B) an electron and a positron
(C) a neutron and a proton ✓
(D) a neutron and a positron

Worked solution

$m_{proton} \approx 1\,u, \quad m_{neutron} \approx 1\,u$

$m_{electron}, m_{positron} \approx 0.00055\,u$

Answer: a neutron and a proton

Protons and neutrons each have a mass of about 1 atomic mass unit, while electrons and positrons have negligible mass in comparison.

Question 3 · 2 pts

The wave-mechanical model of the atom describes the location of electrons

(A) as loosely packed in the nucleus of an atom
(B) as densely packed in the nucleus of an atom
(C) in circular paths around the nucleus
(D) in orbitals outside the nucleus ✓

Worked solution

$\text{Wave-mechanical model} \Rightarrow \text{probability regions (orbitals)}$

Answer: in orbitals outside the nucleus

Unlike the Bohr model's fixed circular paths, the wave-mechanical model places electrons in orbitals, regions of highest probability, outside the nucleus.

Question 4 · 2 pts

When a ground state electron in an atom moves to an excited state, the electron

(A) absorbs energy as it moves to a higher energy state ✓
(B) absorbs energy as it moves to a lower energy state
(C) releases energy as it moves to a higher energy state
(D) releases energy as it moves to a lower energy state

Worked solution

$E_{ground} + \Delta E_{absorbed} = E_{excited}$

Answer: absorbs energy as it moves to a higher energy state

Moving from ground to excited state requires the electron to gain (absorb) energy as it rises to a higher energy level.

Question 5 · 2 pts

Which statement describes a chemical property of iron?

(A) Iron is malleable.
(B) Iron conducts electricity.
(C) Iron reacts with nitric acid. ✓
(D) Iron has a high melting point.

Worked solution

$\text{Chemical property} \Rightarrow \text{describes reactivity/how substance changes composition}$

Answer: Iron reacts with nitric acid.

Malleability, conductivity, and melting point are physical properties; reacting with nitric acid describes a chemical change/property, since a new substance forms.

Question 6 · 2 pts

Diamond and graphite are two forms of solid carbon. These two forms of carbon have

(A) different crystal structures and different properties ✓
(B) different crystal structures and the same properties
(C) the same crystal structure and different properties
(D) the same crystal structure and the same properties

Worked solution

$\text{Diamond: tetrahedral network} \ne \text{Graphite: layered hexagonal sheets}$

Answer: different crystal structures and different properties

Diamond and graphite are allotropes of carbon with different crystal structures (tetrahedral vs. layered), which give them different physical properties such as hardness and conductivity.

Question 7 · 2 pts

Which substance can be broken down by a chemical change?

(A) cobalt
(B) ethane ✓
(C) krypton
(D) manganese

Worked solution

$\text{Cobalt, krypton, manganese: elements (cannot be decomposed)}$

$\text{Ethane (C}_2\text{H}_6\text{): compound (can be decomposed)}$

Answer: ethane

Cobalt, krypton, and manganese are elements and cannot be chemically decomposed further, whereas ethane is a compound made of carbon and hydrogen that can be broken down chemically.

Question 8 · 2 pts

Based on Table I, which equation represents conservation of mass and energy? $CH_4(g) + O_2(g) + 890.4\ kJ \rightarrow CO_2(g) + H_2O(\ell)$ $CH_4(g) + O_2(g) \rightarrow CO_2(g) + H_2O(\ell)$ $CH_4(g) + 2O_2(g) + 890.4\ kJ \rightarrow CO_2(g) + 2H_2O(\ell)$ $CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(\ell) + 890.4\ kJ$

(A) $CH_4(g) + O_2(g) + 890.4\ kJ \rightarrow CO_2(g) + H_2O(\ell)$
(B) $CH_4(g) + O_2(g) \rightarrow CO_2(g) + H_2O(\ell)$
(C) $CH_4(g) + 2O_2(g) + 890.4\ kJ \rightarrow CO_2(g) + 2H_2O(\ell)$
(D) $CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(\ell) + 890.4\ kJ$ ✓

Worked solution

$\text{Balance atoms: } CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$

$\text{Combustion is exothermic} \Rightarrow \text{energy released as product}$

Answer: $CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(\ell) + 890.4\ kJ$

The equation must be atom-balanced (needs 2O2 and 2H2O) and since methane combustion releases energy, the 890.4 kJ belongs on the product side, matching Table I data.

Question 9 · 2 pts

At STP, which property can be used to differentiate a 10.-gram sample of NaCl(s) from a 10.-gram sample of NaNO$_3$(s)?

(A) mass of the sample
(B) temperature of the sample
(C) solubility in water ✓
(D) phase at STP

Worked solution

$\text{Mass and temperature are equal/identical given conditions}$

$\text{Both are solids at STP} \Rightarrow \text{phase does not differ}$

$\text{Solubility values differ between NaCl and NaNO}_3$

Answer: solubility in water

Since both samples have the same mass, temperature, and phase (solid) at STP, only solubility in water, a property unique to each substance, can distinguish them.

Question 10 · 2 pts

What is the number of electrons shared between the two atoms in an O$_2$ molecule?

(A) 6
(B) 2
(C) 3
(D) 4 ✓

Worked solution

$O_2: \; O=O \; \text{(double bond)}$

$\text{Double bond} = 2 \text{ shared pairs} = 4 \text{ electrons}$

Answer: 4

An oxygen molecule has a double covalent bond between the two O atoms, consisting of two shared pairs, or 4 shared electrons.

Question 11 · 2 pts

Which changes in both charge and radius occur when an atom loses an electron?

(A) A negative ion is formed with a smaller radius than the atom.
(B) A negative ion is formed with a larger radius than the atom.
(C) A positive ion is formed with a smaller radius than the atom. ✓
(D) A positive ion is formed with a larger radius than the atom.

Worked solution

$\text{Atom loses electron(s)} \Rightarrow \text{net positive charge (cation)}$

$\text{Fewer electrons} \Rightarrow \text{less electron-electron repulsion} \Rightarrow \text{smaller radius}$

Answer: A positive ion is formed with a smaller radius than the atom.

Losing an electron produces a positively charged ion, and with fewer electrons the electron cloud contracts, making the ionic radius smaller than the original atom.

Question 12 · 2 pts

Which statement describes what occurs when two iodine atoms react to produce an iodine molecule?

(A) A bond forms and energy is absorbed.
(B) A bond forms and energy is released. ✓
(C) A bond breaks and energy is absorbed.
(D) A bond breaks and energy is released.

Worked solution

$I + I \rightarrow I_2 + \text{energy}$

Answer: A bond forms and energy is released.

Forming a covalent bond between two atoms is an exothermic process, so a bond forms and energy is released as the more stable I2 molecule is produced.

Question 13 · 2 pts

Which process can be used to separate a mixture of two liquids having different boiling points?

(A) deposition
(B) distillation ✓
(C) filtration
(D) sublimation

Worked solution

$\text{Heat mixture} \Rightarrow \text{lower-boiling liquid vaporizes first, then condenses separately}$

Answer: distillation

Distillation separates liquids based on differences in boiling point, since the more volatile liquid vaporizes first and can be collected separately.

Question 14 · 2 pts

Which statement describes a solution of sodium chloride in water?

(A) The mixture is heterogeneous, the solute is NaCl and the solvent is H$_2$O.
(B) The mixture is heterogeneous, the solute is H$_2$O and the solvent is NaCl.
(C) The mixture is homogeneous, the solute is NaCl and the solvent is H$_2$O. ✓
(D) The mixture is homogeneous, the solute is H$_2$O and the solvent is NaCl.

Worked solution

$\text{Solution} = \text{homogeneous mixture}$

$\text{Present in smaller quantity} = \text{solute (NaCl)}, \quad \text{larger quantity} = \text{solvent (H}_2\text{O)}$

Answer: The mixture is homogeneous, the solute is NaCl and the solvent is H$_2$O.

A salt water solution is a uniform, homogeneous mixture where sodium chloride is the solute dissolved in water, the solvent.

Question 15 · 2 pts

At STP, which property would be the same for 1.0 liter of helium and 1.0 liter of argon?

(A) boiling point
(B) density
(C) mass
(D) number of atoms ✓

Worked solution

$\text{Avogadro's principle: equal volumes of gases at STP} \Rightarrow \text{equal number of particles}$

$\text{He and Ar are monatomic} \Rightarrow \text{particles} = \text{atoms}$

Answer: number of atoms

By Avogadro's principle, equal volumes of any gas at STP contain the same number of moles (particles); since He and Ar are monatomic, this means the same number of atoms, while mass and density differ due to different molar masses.

Question 16 · 2 pts

The melting of an ice cube is an example of an

(A) endothermic, chemical change
(B) endothermic, physical change ✓
(C) exothermic, chemical change
(D) exothermic, physical change

Worked solution

$H_2O(s) + \text{heat} \rightarrow H_2O(\ell)$

Answer: endothermic, physical change

Melting is a phase change (physical, no new substance forms) that requires absorbing heat energy, making it endothermic.

Question 17 · 2 pts

Which statement explains the low boiling point of hydrogen, H$_2$, at standard pressure?

(A) Hydrogen has strong covalent bonds.
(B) Hydrogen has weak covalent bonds.
(C) Hydrogen has strong intermolecular forces.
(D) Hydrogen has weak intermolecular forces. ✓

Worked solution

$\text{Boiling requires overcoming intermolecular forces, not covalent bonds}$

Answer: Hydrogen has weak intermolecular forces.

Boiling point is determined by the strength of intermolecular (between-molecule) forces, not the intramolecular covalent bonds; H2 has very weak intermolecular forces, giving it a low boiling point.

Question 18 · 2 pts

In chemical reactions, which term is defined as the difference between the potential energy of the products and the potential energy of the reactants?

(A) heat of fusion
(B) heat of reaction ✓
(C) thermal conductivity
(D) electrical conductivity

Worked solution

$\Delta H = PE_{products} - PE_{reactants}$

Answer: heat of reaction

The heat of reaction is defined as the difference between the potential energy of the products and reactants in a chemical reaction.

Question 19 · 2 pts

Which phrase describes what happens to the reaction pathway and activation energy of a reaction to which a catalyst is added?

(A) the same pathway with the same activation energy
(B) the same pathway with a lower activation energy
(C) a different pathway with the same activation energy
(D) a different pathway with a lower activation energy ✓

Worked solution

$\text{Catalyst} \Rightarrow \text{new reaction pathway with lower } E_a$

Answer: a different pathway with a lower activation energy

A catalyst provides an alternate reaction pathway that has a lower activation energy, speeding up the reaction without being consumed.

Question 20 · 2 pts

An atom of which element is bonded to the carbon atom in the amide functional group?

(A) iodine
(B) nitrogen ✓
(C) phosphorus
(D) sulfur

Worked solution

$\text{Amide group: } -C(=O)-N<$

Answer: nitrogen

The amide functional group consists of a carbonyl carbon bonded to a nitrogen atom, so nitrogen is the element bonded to the carbon.

Question 21 · 2 pts

Which statement describes the two isomers of butane?

(A) They have the same molecular formula but different structural formulas. ✓
(B) They have the same molecular formula and the same structural formula.
(C) They have different molecular formulas and different structural formulas.
(D) They have different molecular formulas but the same structural formula.

Worked solution

$\text{n-butane and isobutane: both } C_4H_{10}$

$\text{Different arrangement of atoms} \Rightarrow \text{different structural formulas}$

Answer: They have the same molecular formula but different structural formulas.

Isomers by definition share the same molecular formula (C4H10 for butane) but differ in the arrangement of atoms, giving different structural formulas.

Question 22 · 2 pts

Which term represents an organic reaction that produces soap?

(A) esterification
(B) fermentation
(C) saponification ✓
(D) solidification

Worked solution

$\text{Fat/oil (ester)} + \text{base} \rightarrow \text{soap} + \text{glycerol}$

Answer: saponification

Saponification is the reaction of a fat or oil with a base to produce soap and glycerol.

Question 23 · 2 pts

In which part of an electrochemical cell does reduction occur?

(A) anode
(B) cathode ✓
(C) wire
(D) voltmeter

Worked solution

$\text{Reduction} = \text{gain of electrons}$

$\text{By convention, reduction always occurs at the cathode; oxidation occurs at the anode.}$

Answer: cathode

In any electrochemical cell (voltaic or electrolytic), reduction (gain of electrons) always takes place at the cathode, while oxidation occurs at the anode.

Question 24 · 2 pts

Which energy change occurs in an operating voltaic cell?

(A) chemical energy to electrical energy ✓
(B) chemical energy to nuclear energy
(C) electrical energy to chemical energy
(D) electrical energy to nuclear energy

Worked solution

$\text{Voltaic cell: spontaneous redox reaction produces electric current}$

$\text{Chemical energy (stored in bonds/reactants)} \rightarrow \text{electrical energy}$

Answer: chemical energy to electrical energy

A voltaic (galvanic) cell uses a spontaneous chemical redox reaction to generate an electric current, converting chemical energy into electrical energy.

Question 25 · 2 pts

Which substance is an Arrhenius base?

(A) $HNO_3$
(B) $KNO_3$
(C) $LiOH$ ✓
(D) $CH_3COOH$

Worked solution

$\text{An Arrhenius base is a substance that yields } OH^- \text{ as the only negative ion in aqueous solution}$

$LiOH(aq) \rightarrow Li^+(aq) + OH^-(aq)$

Answer: $LiOH$

LiOH dissociates in water to release hydroxide ions, OH-, which is the defining property of an Arrhenius base; HNO3 and CH3COOH are Arrhenius acids, and KNO3 is a salt.

Question 26 · 2 pts

Which statement represents neutralization?

(A) An Arrhenius acid and an Arrhenius base react to produce water and a salt. ✓
(B) An Arrhenius acid and a salt react to produce water and an Arrhenius base.
(C) Water and an Arrhenius acid react to produce an Arrhenius base and a salt.
(D) Water and a salt react to produce an Arrhenius base and an Arrhenius acid.

Worked solution

$\text{Acid} + \text{Base} \rightarrow \text{Salt} + H_2O$

Answer: An Arrhenius acid and an Arrhenius base react to produce water and a salt.

Neutralization is defined as the reaction of an Arrhenius acid with an Arrhenius base to produce water and a salt.

Question 27 · 2 pts

A tenfold increase in hydronium ion concentration is represented by

(A) a decrease of one unit of pH ✓
(B) a decrease of 10 units of pH
(C) an increase of one unit of pH
(D) an increase of 10 units of pH

Worked solution

$pH = -\log[H_3O^+]$

$\text{If } [H_3O^+] \text{ increases by a factor of 10, } pH \text{ decreases by exactly 1 unit}$

Answer: a decrease of one unit of pH

Because pH is a base-10 logarithmic scale, every tenfold increase in hydronium ion concentration corresponds to a decrease of exactly one pH unit.

Question 28 · 2 pts

Based on Table N, which particle is emitted by the radioactive decay of francium-220?

(A) an alpha particle ✓
(B) a beta particle
(C) a positron
(D) a neutron

Worked solution

$^{220}_{87}Fr \rightarrow ^{216}_{85}At + ^{4}_{2}He$

$\text{Table N lists francium-220 as an alpha emitter}$

Answer: an alpha particle

Reference Table N (Selected Radioisotopes) lists francium-220 as decaying by alpha emission, releasing an alpha particle (a helium-4 nucleus).

Question 29 · 2 pts

Which type of reaction releases the greatest amount of energy per kilogram of reactant?

(A) acid-base reaction
(B) fission reaction ✓
(C) organic reaction
(D) redox reaction

Worked solution

$\text{Nuclear reactions (fission) involve mass-energy conversion } E=mc^2$

$\text{This releases far more energy per kilogram than chemical reactions (acid-base, organic, redox)}$

Answer: fission reaction

Nuclear fission reactions convert a measurable amount of mass directly into energy, releasing vastly more energy per kilogram of reactant than any ordinary chemical reaction.

Question 30 · 2 pts

Which risk is related to the radioactive isotopes used to generate electricity?

(A) depletion of fossil fuels
(B) depletion of atmospheric ozone
(C) exposure to acid rain
(D) exposure to nuclear emissions ✓

Worked solution

$\text{Radioactive isotopes in nuclear power plants emit ionizing radiation and produce radioactive waste}$

Answer: exposure to nuclear emissions

The principal risk associated with using radioactive isotopes for electricity generation is human exposure to nuclear (radioactive) emissions from the fuel, waste, or an accident, not fossil fuel depletion, ozone loss, or acid rain, which are linked to fossil fuel combustion.

Question 31 · 2 pts

Which Lewis electron-dot diagram represents an atom of nitrogen in the ground state?

(A) Diagram (1): N symbol with one pair of dots (two dots) above it only, 2 dots total
(B) Diagram (2): N symbol with one pair of dots (a lone pair) on one side and three single, unpaired dots placed individually on the remaining three sides ✓
(C) Diagram (3): N symbol with three pairs of dots and one single dot (7 dots total)
(D) Diagram (4): N symbol surrounded by four pairs of dots (8 dots total, a full octet)

Worked solution

$\text{Ground-state nitrogen: } 1s^2\,2s^2\,2p^3$

$\text{Valence electrons} = 5 \;(2s^2 + 2p^3)$

$\text{Lewis structure: one lone pair (2 dots) plus three unpaired single dots} = 5 \text{ total dots}$

Answer: Diagram (2)

Ground-state nitrogen has 5 valence electrons: one pair is already filled (2s^2) and the three 2p electrons remain unpaired, so the correct Lewis dot diagram shows one paired lone pair and three single unpaired dots around the symbol N.

Question 32 · 2 pts

The atomic masses and natural abundances of the two naturally occurring isotopes of silver are shown in the table below. Which numerical setup can be used to calculate the atomic mass of silver?

(A) $(106.905\ u)(51.8) + (108.905\ u)(48.2)$
(B) $(106.905\ u)(51.8\%) + (108.905\ u)(48.2\%)$ ✓
(C) $(106.905\ u)(48.2) + (108.905\ u)(51.8)$
(D) $(106.905\ u)(48.2\%) + (108.905\ u)(51.8\%)$

Worked solution

$\text{Atomic mass} = \sum(\text{isotope mass} \times \text{fractional abundance})$

$= (106.905\,u)(0.518) + (108.905\,u)(0.482)$

Answer: $(106.905\ u)(51.8\%) + (108.905\ u)(48.2\%)$

The weighted average atomic mass is found by multiplying each isotope's mass by its own correct percent (fractional) natural abundance and summing the products, matching Ag-107 with 51.8% and Ag-109 with 48.2%.

Question 33 · 2 pts

A potassium atom has a mass number of 37. What is the number of neutrons in this atom?

(A) 15
(B) 18 ✓
(C) 22
(D) 37

Worked solution

$\text{Atomic number of K} = 19 \;(\text{protons})$

$\text{neutrons} = \text{mass number} - \text{protons} = 37 - 19 = 18$

Answer: 18

Potassium always has 19 protons; subtracting this from the given mass number of 37 leaves 18 neutrons.

Question 34 · 2 pts

At room temperature, a student determines the density of a sample of nickel to be 9.79 g/cm^3. Based on Table S, what is the student's percent error for the density of nickel?

(A) 0.091%
(B) 0.10%
(C) 9.1%
(D) 10.% ✓

Worked solution

$\text{Accepted density of Ni (Table S)} = 8.90\ g/cm^3$

$\%\text{error} = \frac{|\text{measured} - \text{accepted}|}{\text{accepted}} \times 100$

$= \frac{|9.79 - 8.90|}{8.90}\times 100 = \frac{0.89}{8.90}\times 100 = 10.\%$

Answer: 10.%

Using the accepted density of nickel from Table S (8.90 g/cm^3), the percent error between the measured 9.79 g/cm^3 and the accepted value works out to 10.%.

Question 35 · 2 pts

Compared to the metals in Period 2, the nonmetals in Period 2 have

(A) lower first ionization energies and lower electronegativity values
(B) lower first ionization energies and higher electronegativity values
(C) higher first ionization energies and lower electronegativity values
(D) higher first ionization energies and higher electronegativity values ✓

Worked solution

$\text{Across a period, ionization energy and electronegativity both generally increase left to right}$

$\text{Nonmetals are farther right than metals in the same period, so both values are higher for the nonmetals}$

Answer: higher first ionization energies and higher electronegativity values

Both first ionization energy and electronegativity generally increase across a period from metals toward nonmetals, so Period 2 nonmetals have higher values for both properties than Period 2 metals.

Question 36 · 2 pts

Which formula represents calcium hydride?

(A) $CaH$
(B) $CaH_2$ ✓
(C) $CaOH$
(D) $Ca(OH)_2$

Worked solution

$Ca^{2+} \text{ and } H^- \text{ combine in a 1:2 ratio to balance charge}$

$CaH_2$

Answer: $CaH_2$

Calcium forms a 2+ ion and hydride is H-, so two hydride ions are needed to balance one calcium ion's charge, giving the formula CaH2.

Question 37 · 2 pts

What is the number of moles in a 78.8-gram sample of $MgCO_3$ (gram-formula mass = 84.3 g/mol)?

(A) 0.949 mol
(B) 0.935 mol ✓
(C) 0.843 mol
(D) 1.070 mol

Worked solution

$n = \frac{\text{mass}}{\text{gram-formula mass}} = \frac{78.8\ g}{84.3\ g/mol}$

$n \approx 0.935\ mol$

Answer: 0.935 mol

Dividing the given mass of MgCO3 by its gram-formula mass yields the number of moles present, approximately 0.935 mol.

Question 38 · 2 pts

Given the equation representing a reaction: $F_2(g) + 2KCl(aq) \rightarrow 2KF(aq) + Cl_2(g)$ Which type of chemical reaction is represented by the equation?

(A) synthesis
(B) decomposition
(C) single replacement ✓
(D) double replacement

Worked solution

$F_2 \text{ (a free element) displaces Cl from KCl, forming KF and free } Cl_2$

$\text{One element replacing another element within a compound is a single replacement reaction}$

Answer: single replacement

A free halogen element, F2, displaces chlorine from the compound KCl, producing a new compound (KF) and a different free element (Cl2), which is the defining pattern of a single replacement reaction.

Question 39 · 2 pts

Based on Table H, which compound has the strongest intermolecular forces at 60 kPa?

(A) ethanoic acid ✓
(B) ethanol
(C) propanone
(D) water

Worked solution

$\text{Table H's vapor pressure curves show that, at any given pressure, ethanoic acid requires the highest temperature to vaporize}$

$\text{The higher the temperature needed to reach a given vapor pressure, the stronger the intermolecular forces holding the liquid together}$

Answer: ethanoic acid

Reference Table H's vapor pressure curves show ethanoic acid reaching 60 kPa vapor pressure at a higher temperature than ethanol, propanone, or water, indicating it has the strongest intermolecular forces (hydrogen bonding plus dimer formation) of the four.

Question 40 · 2 pts

Which particle diagram represents a sample of xenon at STP? Key: open circle = atom of xenon

(A) Diagram (1): a box containing about 9 single circles spread apart with open space between them, unevenly but individually distributed, no touching or pairing ✓
(B) Diagram (2): a box containing circles arranged in tightly packed, touching rows filling most of the box
(C) Diagram (3): a box containing about 9 circles, several of which are touching/paired together in twos
(D) Diagram (4): a box containing circles arranged as touching pairs in tightly packed rows

Worked solution

$\text{Xenon is a monatomic noble gas: individual atoms, not bonded together}$

$\text{At STP, a gas sample has widely, evenly spaced particles filling the container with much empty space}$

Answer: Diagram (1)

Xenon exists as separate, unbonded individual atoms, and as a gas at STP its particles are spread apart with large amounts of empty space between them, matching the diagram showing individual, well-separated circles rather than paired or tightly packed particles.

Question 41 · 2 pts

Based on Table G, which compound is less soluble in water as the temperature increases from 0°C to 100°C?

(A) $KNO_3$
(B) $NH_3$ ✓
(C) $KClO_3$
(D) $NH_4Cl$

Worked solution

$\text{On Table G's solubility curves, most ionic solids (KNO}_3\text{, KClO}_3\text{, NH}_4\text{Cl) increase in solubility as temperature rises}$

$NH_3 \text{ is a gas, and gas solubility in water decreases as temperature increases}$

Answer: $NH_3$

Ammonia, NH3, is a gas, and the solubility of gases in water decreases as temperature increases, unlike the three ionic solids listed whose solubility curves in Table G rise with increasing temperature.

Question 42 · 2 pts

How many moles of KI are needed to make 0.50 L of a 0.20 M aqueous solution?

(A) 0.10 mol ✓
(B) 0.25 mol
(C) 0.40 mol
(D) 0.70 mol

Worked solution

$M = \frac{mol}{L} \Rightarrow mol = M \times L$

$mol = (0.20\ M)(0.50\ L) = 0.10\ mol$

Answer: 0.10 mol

Multiplying molarity by volume in liters gives the number of moles of solute needed: 0.20 mol/L times 0.50 L equals 0.10 mol.

Question 43 · 2 pts

A solution is prepared using 9.80 grams of $NaHCO_3$ in enough water to make 1500. grams of total solution. What is the concentration of the solution expressed in parts per million?

(A) $6.49 \times 10^{-3}$ ppm
(B) $6.53 \times 10^{-3}$ ppm
(C) $6.49 \times 10^{3}$ ppm
(D) $6.53 \times 10^{3}$ ppm ✓

Worked solution

$ppm = \frac{\text{mass of solute}}{\text{mass of solution}} \times 10^6$

$ppm = \frac{9.80\ g}{1500.\ g}\times 10^6 \approx 6.53\times 10^3\ ppm$

Answer: $6.53 \times 10^{3}$ ppm

Parts per million is calculated as the mass of solute divided by the total mass of solution, multiplied by one million, giving approximately 6.53 x 10^3 ppm.

Question 44 · 2 pts

A potential energy diagram for a chemical reaction is given below. Each interval on the potential energy axis represents 100. kilojoules of potential energy. What can be concluded from the diagram?

(A) The reaction is endothermic, and the heat of reaction is $-200.$ kJ.
(B) The reaction is endothermic, and the heat of reaction is $+200.$ kJ. ✓
(C) The reaction is exothermic, and the heat of reaction is $-200.$ kJ.
(D) The reaction is exothermic, and the heat of reaction is $+200.$ kJ.

Worked solution

$\Delta H = PE_{products} - PE_{reactants}$

$\text{Product plateau is 2 intervals above the reactant level} = 2\times100.\ kJ = +200.\ kJ$

$\Delta H > 0 \Rightarrow \text{reaction is endothermic}$

Answer: The reaction is endothermic, and the heat of reaction is $+200.$ kJ.

Since the final product energy level sits two 100.-kJ intervals above the starting reactant level, the products have higher potential energy than the reactants, making the reaction endothermic with a heat of reaction of +200. kJ.

Question 45 · 2 pts

Given the formula representing a compound: $$\begin{array}{ccccccccccc} & H & & H & & H & & H & & H & \\ & | & & | & & | & & | & & | & \\ H-&C&-&C&-&C&-&C&-&C&-H \\ & | & & | & & | & & | & & | & \\ & Br & & Br & & H & & H & & H & \end{array}$$ What is a chemical name of this compound?

(A) 1,1-dibromopentane
(B) 2,2-dibromopentane
(C) 1,2-dibromopentane ✓
(D) 4,5-dibromopentane

Worked solution

$\text{Number the 5-carbon chain from the end nearer the substituents.}$

$\text{C1: bonded to H, H, Br; C2: bonded to H, Br; C3-C5: CH}_2\text{-CH}_2\text{-CH}_3$

$\text{Br atoms are located on carbon 1 and carbon 2 of the pentane chain}$

Answer: 1,2-dibromopentane

The parent chain has 5 carbons (pentane), and bromine atoms are substituted on carbons 1 and 2, giving the name 1,2-dibromopentane.

Question 46 · 2 pts

The diagram and equation below represent an electrochemical cell. A battery is connected by wires to two electrodes immersed in a container of molten NaCl. In this electrolytic cell the anode is the electrode connected to the positive terminal of the battery and the cathode is connected to the negative terminal, and the other electrode is labeled Cathode. The solution contains $Cl^-$ and $Na^+$ ions. $2NaCl(\ell) \rightarrow 2Na(\ell) + Cl_2(g)$ Which process is represented by this diagram?

(A) chromatography
(B) distillation
(C) electrolysis ✓
(D) polymerization

Worked solution

$\text{A battery supplies electrical energy to drive a non-spontaneous reaction}$

$2NaCl(\ell) \rightarrow 2Na(\ell) + Cl_2(g)$

$\text{Using electrical energy to force a chemical reaction to occur is electrolysis}$

Answer: electrolysis

The presence of a battery driving the decomposition of molten NaCl into its elements is the defining feature of electrolysis.

Question 47 · 2 pts

Which aqueous solution has the greatest ability to conduct an electric current?

(A) 0.10 M NaCl(aq) ✓
(B) 0.010 M NaCl(aq)
(C) 0.10 M $C_6H_{12}O_6$ (aq)
(D) 0.010 M $C_6H_{12}O_6$ (aq)

Worked solution

$\text{NaCl is an ionic compound that dissociates into mobile ions in solution}$

$\text{Glucose (}C_6H_{12}O_6\text{) is a molecular compound that does not ionize}$

$\text{Higher ion concentration means greater conductivity, so 0.10 M NaCl conducts best}$

Answer: 0.10 M NaCl(aq)

NaCl fully dissociates into Na+ and Cl- ions, and the higher 0.10 M concentration provides more charge carriers than the more dilute NaCl or either glucose solution (glucose does not ionize at all).

Question 48 · 2 pts

What fraction of an original sample of $^{131}I$ remains unchanged after 24.063 days?

(A) 1/8 ✓
(B) 1/2
(C) 1/3
(D) 1/4

Worked solution

$\text{Half-life of I-131 (Table N)} = 8.021\text{ days}$

$\text{Number of half-lives} = \frac{24.063\text{ days}}{8.021\text{ days}} = 3$

$\text{Fraction remaining} = \left(\frac{1}{2}\right)^3 = \frac{1}{8}$

Answer: 1/8

Since 24.063 days equals exactly 3 half-lives of I-131 (8.021 days each), the fraction remaining is (1/2)^3 = 1/8.

Question 49 · 2 pts

Given the equation representing a reaction: $_1^2H + _1^3H \rightarrow \ _2^4He + \ _0^1n + \text{energy}$ Which type of reaction is represented by the equation?

(A) nuclear fission
(B) nuclear fusion ✓
(C) combustion
(D) substitution

Worked solution

$\text{Two light nuclei (deuterium and tritium) combine to form a heavier nucleus (helium)}$

$\text{The combining of small nuclei into a larger nucleus is nuclear fusion}$

Answer: nuclear fusion

Deuterium and tritium nuclei combine to form a helium nucleus and a neutron, which is the joining of light nuclei characteristic of nuclear fusion.

Question 50 · 2 pts

Given the nuclear equation and isotopic masses: $_3^6Li + \ _1^2H \rightarrow 2(_2^4He) + \text{energy}$ $6.0151\text{ u} \quad 2.0140\text{ u} \quad 2(4.0026\text{ u})$ What is the amount of mass converted to energy as a result of the reaction between the two reactant nuclei?

(A) 0.0239 u ✓
(B) 4.0265 u
(C) 8.0052 u
(D) 16.0343 u

Worked solution

$\text{Mass of reactants} = 6.0151\text{ u} + 2.0140\text{ u} = 8.0291\text{ u}$

$\text{Mass of products} = 2(4.0026\text{ u}) = 8.0052\text{ u}$

$\Delta m = 8.0291\text{ u} - 8.0052\text{ u} = 0.0239\text{ u}$

Answer: 0.0239 u

The mass converted to energy equals the total mass of the reactants minus the total mass of the products, which is 8.0291 u - 8.0052 u = 0.0239 u.

Question 51 · 1 pt · constructed response

Base your answer on the information below and on your knowledge of chemistry. The element technetium, Tc, has several isotopes. The bright-line spectrum of technetium has been observed in the spectra of some stars. Compare the energy of an electron in the first shell of a technetium atom to the energy of an electron in the third shell of the same atom.

Model answer: An electron in the first shell has less energy than an electron in the third shell.

Worked solution

$\text{Electron shells farther from the nucleus have higher energy}$

$\text{Shell 1 is closer to the nucleus than shell 3}$

Answer: An electron in the first shell has less energy than an electron in the third shell.

Electrons in shells closer to the nucleus (lower principal quantum number) have lower energy, so an electron in the first shell has less energy than one in the third shell.

Question 52 · 1 pt · constructed response

State, in terms of protons and neutrons, why the various nuclides of technetium are isotopes of each other.

Model answer: All nuclides of technetium have the same number of protons but different numbers of neutrons.

Worked solution

$\text{Isotopes of an element have the same atomic number (number of protons)}$

$\text{Isotopes differ in mass number due to a different number of neutrons}$

Answer: All nuclides of technetium have the same number of protons but different numbers of neutrons.

Isotopes are atoms of the same element, so all technetium nuclides have the same number of protons, but they have varying numbers of neutrons, giving different mass numbers.

Question 53 · 1 pt · constructed response

Base your answer on the information below and on your knowledge of chemistry. A sample of a gas in a sealed, rigid cylinder with a movable piston has a volume of 0.250 liter at STP. Show a numerical setup for calculating the volume of this sample of gas at 298 K and 1.00 atm.

Model answer: V2 = (0.250 L x 298 K) / 273 K = 0.273 L

Worked solution

$\text{STP: } P_1 = 1.00\text{ atm}, T_1 = 273\text{ K}, V_1 = 0.250\text{ L}$

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \Rightarrow V_2 = \frac{(1.00\text{ atm})(0.250\text{ L})(298\text{ K})}{(273\text{ K})(1.00\text{ atm})}$

$V_2 = \frac{(0.250\text{ L})(298\text{ K})}{273\text{ K}} \approx 0.273\text{ L}$

Answer: V2 = (0.250 L x 298 K) / 273 K = 0.273 L

Since pressure is unchanged (1.00 atm at both STP and the new condition), the combined gas law reduces to a simple ratio of temperatures multiplying the original volume.

Question 54 · 1 pt · constructed response

State a change in pressure that will cause the gas in the cylinder to behave more like an ideal gas.

Model answer: Decreasing the pressure on the gas will cause it to behave more like an ideal gas.

Worked solution

$\text{Real gases deviate from ideal behavior most at high pressure, where particle volume and attractive forces matter}$

$\text{Lowering the pressure increases the average distance between gas particles}$

Answer: Decreasing the pressure on the gas will cause it to behave more like an ideal gas.

At lower pressures, gas particles are farther apart, so their volume and intermolecular attractions become negligible, making the gas behave more ideally.

Question 55 · 1 pt · constructed response

Base your answers on the information below and on your knowledge of chemistry. During a laboratory activity, a student heats a beaker containing 120.0 grams of water, as shown in a diagram of a ring stand setup with a candle heating a beaker of water (thermometer, wire gauze, iron ring, glass square shown). Data for Heating Water: Mass of 120.0 mL of water: 120.0 g Temperature of water before heating: 23.0 degrees C Temperature of water after heating 20.0 min: 86.0 degrees C State the direction of heat flow between the candle flame and the beaker of water during the time the candle is lit.

Model answer: Heat flows from the candle flame to the beaker of water.

Worked solution

$\text{Heat flows spontaneously from a higher-temperature object to a lower-temperature object}$

$\text{The candle flame is at a higher temperature than the water in the beaker}$

Answer: Heat flows from the candle flame to the beaker of water.

Because the flame is hotter than the water, thermal energy flows from the flame to the beaker of water, raising the water's temperature.

Question 56 · 1 pt · constructed response

Show a numerical setup for calculating the amount of heat, in joules, absorbed by the water in the beaker as a result of the burning candle.

Model answer: q = (120.0 g)(4.18 J/(g x degC))(86.0 degC - 23.0 degC) = 31,600.8 J

Worked solution

$q = mC\Delta T$

$q = (120.0\text{ g})(4.18\ \text{J/g}^{\circ}\text{C})(86.0^{\circ}\text{C} - 23.0^{\circ}\text{C})$

$q = (120.0)(4.18)(63.0) = 31600.8\text{ J}$

Answer: q = (120.0 g)(4.18 J/(g x degC))(86.0 degC - 23.0 degC) = 31,600.8 J

Using q = mCdeltaT with the specific heat of water (4.18 J/g degC), the mass of 120.0 g, and the temperature change of 63.0 degC gives the heat absorbed.

Question 57 · 1 pt · constructed response

State how the molecular motion of the water molecules in the beaker changes as the temperature increases.

Model answer: As temperature increases, the molecular motion of the water molecules increases (the molecules move faster).

Worked solution

$\text{Temperature is a measure of average kinetic energy of particles}$

$\text{As temperature increases, average kinetic energy increases, so molecules move faster}$

Answer: As temperature increases, the molecular motion of the water molecules increases (the molecules move faster).

Since temperature reflects the average kinetic energy of particles, increasing temperature means the water molecules move with greater speed and increased motion.

Question 58 · 1 pt · constructed response

Base your answers on the information below and on your knowledge of chemistry. Nitrogen dioxide gas, $NO_2(g)$, can reach equilibrium with dinitrogen tetroxide gas, $N_2O_4(g)$, in a closed system. At 1.0 atmosphere, the boiling point of $N_2O_4$ is 21 degrees C. The equation below represents this system. $2NO_2(g) \rightleftharpoons N_2O_4(g) + \text{energy}$ Compare the rate of the forward reaction to the rate of the reverse reaction when the system is at equilibrium.

Model answer: At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction.

Worked solution

$\text{At chemical equilibrium, the forward and reverse reaction rates are equal}$

Answer: At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction.

By definition, dynamic chemical equilibrium is reached when the forward and reverse reaction rates become equal, even though the reaction continues in both directions.

Question 59 · 1 pt · constructed response

State how the equilibrium shifts when the pressure on the equilibrium system is increased at constant temperature.

Model answer: The equilibrium shifts toward the forward reaction (toward the production of N2O4), the side with fewer moles of gas.

Worked solution

$2NO_2(g) \rightleftharpoons N_2O_4(g) + \text{energy}$

$\text{Increasing pressure shifts equilibrium toward the side with fewer moles of gas}$

$\text{2 moles of gas (NO}_2\text{) versus 1 mole of gas (N}_2O_4\text{): equilibrium shifts right, toward N}_2O_4$

Answer: The equilibrium shifts toward the forward reaction (toward the production of N2O4), the side with fewer moles of gas.

By Le Chatelier's principle, increasing the pressure shifts the equilibrium toward the side with fewer gas moles, which is the N2O4 (forward) side.

Question 60 · 1 pt · constructed response

Compare the entropy of a sample of dinitrogen tetroxide gas at 25 degrees C and 1.0 atmosphere to the entropy of the same sample of dinitrogen tetroxide liquid at 15 degrees C and 1.0 atmosphere.

Model answer: The entropy of the gaseous sample of N2O4 is greater than the entropy of the liquid sample of N2O4.

Worked solution

$\text{Entropy (disorder) increases in the order: solid < liquid < gas}$

$\text{The gas sample has greater molecular disorder than the liquid sample}$

Answer: The entropy of the gaseous sample of N2O4 is greater than the entropy of the liquid sample of N2O4.

Gas particles are more disordered and have greater freedom of motion than particles in a liquid, so the gaseous N2O4 sample has higher entropy than the liquid sample.

Question 61 · 1 pt · constructed response

Base your answers on the information below and on your knowledge of chemistry. When solid copper is placed in an aqueous silver nitrate solution, a reaction occurs, as represented by the equation below. $Cu(s) + 2AgNO_3(aq) \rightarrow 2Ag(s) + Cu(NO_3)_2(aq)$ State the change in oxidation state of copper in this reaction.

Model answer: The oxidation state of copper increases from 0 to +2.

Worked solution

$\text{In Cu(s), copper has an oxidation state of 0}$

$\text{In Cu(NO}_3)_2, \text{copper has an oxidation state of } +2$

$\text{Copper's oxidation state changes from 0 to +2 (it is oxidized, losing electrons)}$

Answer: The oxidation state of copper increases from 0 to +2.

Elemental copper (oxidation state 0) is oxidized to Cu2+ in copper(II) nitrate, so its oxidation state increases from 0 to +2 as it loses two electrons.

Question 62 · 1 pt · constructed response

Based on Table J, state why Cu(s) reacts spontaneously with $Ag^+(aq)$.

Model answer: Copper is above silver in Table J (the activity series), meaning copper is a more active metal than silver and will spontaneously lose electrons to (reduce) silver ions.

Worked solution

$\text{Table J lists metals in order of decreasing activity (tendency to lose electrons)}$

$\text{Cu is located above Ag in Table J, so Cu is more reactive/more easily oxidized than Ag}$

$\text{A more active metal will spontaneously reduce the ions of a less active metal}$

Answer: Copper is above silver in Table J (the activity series), meaning copper is a more active metal than silver and will spontaneously lose electrons to (reduce) silver ions.

Since copper appears higher (more active) than silver on the Table J activity series, copper atoms will spontaneously give up electrons to silver ions, making the reaction spontaneous.

Question 63 · 1 pt · constructed response

Write a balanced half-reaction equation to represent the reduction of the silver ions to silver atoms.

Model answer: Ag+(aq) + e- -> Ag(s)

Worked solution

$\text{Reduction is the gain of electrons}$

$Ag^{+}(aq) + e^{-} \rightarrow Ag(s)$

Answer: Ag+(aq) + e- -> Ag(s)

The reduction half-reaction shows each silver ion gaining one electron to become a neutral silver atom, which balances both charge and mass.

Question 64 · 1 pt · constructed response

Base your answers on the information below and on your knowledge of chemistry. During a laboratory activity, 15.0 mL of hydrochloric acid, HCl(aq), is exactly neutralized by 18.2 mL of 0.11 M sodium hydroxide, NaOH(aq). During the laboratory activity, appropriate safety equipment is used and safety procedures are followed. Write the name of the laboratory procedure used in this activity.

Model answer: Titration

Worked solution

$\text{A measured volume of base of known concentration is used to exactly neutralize an acid}$

$\text{This procedure is called titration}$

Answer: Titration

Using a solution of known concentration to exactly neutralize a solution of unknown concentration to determine that unknown concentration is titration.

Question 65 · 1 pt · constructed response

Show a numerical setup for calculating the molarity of the HCl(aq) solution.

Model answer: M(HCl) = (0.0182 L x 0.11 mol/L) / 0.0150 L = 0.1335 M ≈ 0.13 M

Worked solution

$\text{mol NaOH} = (0.0182\text{ L})(0.11\text{ mol/L}) = 0.0020\text{ mol}$

$\text{HCl + NaOH} \rightarrow \text{NaCl + H}_2O \text{ (1:1 mole ratio), so mol HCl} = 0.0020\text{ mol}$

$M_{HCl} = \frac{(0.0182\text{ L})(0.11\text{ M})}{0.0150\text{ L}} \approx 0.13\text{ M}$

Answer: M(HCl) = (0.0182 L x 0.11 mol/L) / 0.0150 L = 0.1335 M ≈ 0.13 M

Since HCl and NaOH react in a 1:1 mole ratio, the moles of NaOH used (volume times molarity) equal the moles of HCl present, which is then divided by the volume of HCl to get its molarity.

Question 66 · 1 pt · constructed response

Base your answers on the information below and on your knowledge of chemistry. Several elements are considered endangered because there is a risk of these elements becoming unavailable for commercial uses in the next 100 years. Helium, zinc, gallium, indium, and tellurium are included in the list of these endangered elements. Identify the three endangered elements listed in the passage that are classified as metals.

Model answer: Zinc, gallium, and indium

Worked solution

$\text{Helium is a nonmetal (noble gas); tellurium is a metalloid}$

$\text{Zinc, gallium, and indium are classified as metals on the Periodic Table}$

Answer: Zinc, gallium, and indium

Of the five elements listed, helium is a noble gas nonmetal and tellurium is a metalloid, leaving zinc, gallium, and indium as the three classified as metals.

Question 67 · 1 pt · constructed response

Explain, in terms of electrons, why gallium and indium have similar chemical properties.

Model answer: Gallium and indium have similar chemical properties because they each have the same number of valence electrons (3 valence electrons), since both elements are in Group 13 of the Periodic Table.

Worked solution

$\text{Ga valence configuration: } 4s^{2}4p^{1}$ 3 valence electrons

$\text{In valence configuration: } 5s^{2}5p^{1}$ 3 valence electrons

Answer: Gallium and indium have similar chemical properties because they each have the same number of valence electrons (3 valence electrons), since both elements are in Group 13 of the Periodic Table.

Elements in the same group have the same number of valence electrons, which largely determines chemical behavior; Ga and In are both Group 13 elements with 3 valence electrons.

Question 68 · 1 pt · constructed response

Compare the density of a sample of helium at STP with the density of a sample of tellurium at STP.

Model answer: The density of helium at STP is much less than (far lower than) the density of tellurium at STP.

Worked solution

$\text{He(g) at STP: particles far apart} \Rightarrow \text{very low density}$

$\text{Te(s) at STP: particles closely packed} \Rightarrow \text{high density}$

Answer: The density of helium at STP is much less than (far lower than) the density of tellurium at STP.

Helium exists as a gas at STP with widely spaced, fast-moving particles, giving it a very low density, while tellurium is a solid at STP with tightly packed particles, giving it a much higher density.

Question 69 · 1 pt · constructed response

Explain, in terms of electron shells, why the atomic radius of an atom of indium is greater than the atomic radius of an atom of gallium when both atoms are in the ground state.

Model answer: An atom of indium has one more occupied electron shell than an atom of gallium (indium's outermost electrons are in the 5th shell, gallium's in the 4th shell), so indium's electron cloud extends farther from the nucleus, making its atomic radius larger.

Worked solution

$\text{Ga: outermost (valence) electrons in shell } n=4$

$\text{In: outermost (valence) electrons in shell } n=5$

Answer: An atom of indium has one more occupied electron shell than an atom of gallium (indium's outermost electrons are in the 5th shell, gallium's in the 4th shell), so indium's electron cloud extends farther from the nucleus, making its atomic radius larger.

Because indium is one period below gallium in Group 13, it has an additional occupied electron shell, increasing its atomic radius compared to gallium.

Question 70 · 1 pt · constructed response

Determine the percent composition by mass of nitrogen in HNO$_3$ (gram-formula mass = 63.0 g/mol).

Model answer: 22.2%

Worked solution

$\%N = \dfrac{\text{mass of N}}{\text{gram-formula mass of HNO}_3}\times 100$

$\%N = \dfrac{14.0\ \text{g/mol}}{63.0\ \text{g/mol}}\times 100$

$\%N = 22.2\%$

Answer: 22.2%

Divide the mass contributed by nitrogen (14.0 g/mol) by the total gram-formula mass of HNO3 (63.0 g/mol) and multiply by 100 to get percent composition.

Question 71 · 1 pt · constructed response

Balance equation 1 in your answer booklet, using the smallest whole-number coefficients: NH$_3$(g) + O$_2$(g) $\rightarrow$ NO(g) + H$_2$O(g) + heat.

Model answer: 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g) + heat

Worked solution

$\text{Balance N: } 4\ \text{NH}_3 \rightarrow 4\ \text{NO}$

$\text{Balance H: } 4\times3=12\ \text{H atoms} \Rightarrow 6\ \text{H}_2\text{O}$

$\text{Balance O: } 4(1)+6(1)=10\ \text{O atoms} \Rightarrow 5\ \text{O}_2$

Answer: 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g) + heat

Balancing nitrogen, hydrogen, and then oxygen atoms in turn gives the smallest whole-number coefficients 4, 5, 4, 6.

Question 72 · 1 pt · constructed response

Show a numerical setup for calculating the gram-formula mass of the NO$_2$(g) produced in equation 2.

Model answer: 46.0 g/mol

Worked solution

$14.0\ \text{g/mol} + 2(16.0\ \text{g/mol})$

$= 46.0\ \text{g/mol}$

Answer: 46.0 g/mol

Add the atomic mass of nitrogen to twice the atomic mass of oxygen to obtain the gram-formula mass of NO2.

Question 73 · 1 pt · constructed response

Determine the number of moles of oxygen required to completely react with 4.0 moles of NO(g) in equation 2.

Model answer: 2.0 mol O2

Worked solution

$2\text{NO(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{NO}_2\text{(g)}$

$4.0\ \text{mol NO}\times\dfrac{1\ \text{mol O}_2}{2\ \text{mol NO}}$

$=2.0\ \text{mol O}_2$

Answer: 2.0 mol O2

Using the 2:1 mole ratio of NO to O2 from the balanced equation, 4.0 mol NO requires 2.0 mol O2.

Question 74 · 1 pt · constructed response

State the molecular polarity for each of the reactants in the equation: 6CO$_2$ + 6H$_2$O + light energy $\rightarrow$ C$_6$H$_{12}$O$_6$ + 6O$_2$.

Model answer: CO2 is a nonpolar molecule; H2O is a polar molecule.

Worked solution

$\text{CO}_2\text{: linear, symmetric bond dipoles cancel} \Rightarrow \text{nonpolar}$

$\text{H}_2\text{O: bent shape, bond dipoles do not cancel} \Rightarrow \text{polar}$

Answer: CO2 is a nonpolar molecule; H2O is a polar molecule.

CO2's linear symmetric shape causes its bond dipoles to cancel out (nonpolar), while water's bent shape leaves a net dipole (polar).

Question 75 · 1 pt · constructed response

In the space in your answer booklet, draw a Lewis electron-dot diagram for a molecule of water.

Model answer: H-O-H, with the oxygen atom having two bonding pairs (one shared pair with each hydrogen) and two lone (nonbonding) pairs of electrons.

Worked solution

$\text{O has 6 valence electrons; each H contributes 1 electron to a shared pair}$

$\text{Resulting structure: H:O:H with 2 lone pairs on O}$

Answer: H-O-H, with the oxygen atom having two bonding pairs (one shared pair with each hydrogen) and two lone (nonbonding) pairs of electrons.

Oxygen forms one single covalent bond with each hydrogen atom and retains two lone pairs, giving the correct Lewis structure for water.

Question 76 · 1 pt · constructed response

Based on Table S, compare the strength of attraction of a carbon atom for electrons with the strength of attraction of an oxygen atom for the electrons in a bond between them.

Model answer: The oxygen atom has a greater attraction for the shared electrons than the carbon atom (oxygen has a higher electronegativity than carbon).

Worked solution

$\text{Electronegativity(O)} \approx 3.4$

$\text{Electronegativity(C)} \approx 2.5$

$3.4 > 2.5 \Rightarrow \text{O attracts bonding electrons more strongly}$

Answer: The oxygen atom has a greater attraction for the shared electrons than the carbon atom (oxygen has a higher electronegativity than carbon).

Table S lists electronegativity values; oxygen's value is higher than carbon's, so oxygen attracts the shared bonding electrons more strongly.

Question 77 · 1 pt · constructed response

State, in terms of element classification, why CO$_2$ is a molecular compound.

Model answer: CO2 is a molecular compound because it is composed of two nonmetals (carbon and oxygen) bonded together, and nonmetal-nonmetal bonds are covalent, forming molecular compounds.

Worked solution

$\text{C = nonmetal, O = nonmetal}$

$\text{nonmetal + nonmetal} \rightarrow \text{covalent (molecular) bonding}$

Answer: CO2 is a molecular compound because it is composed of two nonmetals (carbon and oxygen) bonded together, and nonmetal-nonmetal bonds are covalent, forming molecular compounds.

Bonds between two nonmetal elements are covalent, and covalently bonded substances are classified as molecular compounds.

Question 78 · 1 pt · constructed response

Write the empirical formula for glucose (C$_6$H$_{12}$O$_6$).

Model answer: CH2O

Worked solution

$C_6H_{12}O_6 \div 6$

$= CH_2O$

Answer: CH2O

Dividing all subscripts in the molecular formula of glucose by their greatest common factor, 6, gives the empirical formula CH2O.

Question 79 · 1 pt · constructed response

State the class of organic compound to which the product in the equation belongs (ethene + H$_2$O $\xrightarrow{\text{catalyst}}$ product shown as an H-C-C-H chain with an -OH group).

Model answer: Alcohol

Worked solution

$\text{Product contains an -OH group bonded to a saturated carbon chain}$

Answer: Alcohol

The product, ethanol, contains a hydroxyl (-OH) functional group attached to a carbon chain, which classifies it as an alcohol.

Question 80 · 1 pt · constructed response

State, in terms of carbon-to-carbon bonds, why the hydrocarbon in the equation (ethene) is unsaturated.

Model answer: Ethene is unsaturated because it contains a carbon-to-carbon double bond (rather than only single bonds), meaning it does not have the maximum possible number of hydrogen atoms bonded to the carbons.

Worked solution

$\text{Ethene structure: } H_2C=CH_2$

$\text{Presence of C=C double bond} \Rightarrow \text{unsaturated}$

Answer: Ethene is unsaturated because it contains a carbon-to-carbon double bond (rather than only single bonds), meaning it does not have the maximum possible number of hydrogen atoms bonded to the carbons.

A hydrocarbon is unsaturated when it contains at least one carbon-to-carbon double or triple bond; ethene has a C=C double bond.

Question 81 · 1 pt · constructed response

Identify the element in the product of the reaction that allows it to be classified as an organic compound.

Model answer: Carbon

Worked solution

$\text{Organic compounds are defined as compounds containing carbon (bonded to hydrogen)}$

Answer: Carbon

Organic compounds are, by definition, carbon-based compounds; the presence of carbon in ethanol is what makes it an organic compound.

Question 82 · 1 pt · constructed response

Based on the table (aqueous ammonia pH 11.9, black coffee pH 4.9, lemon juice pH 2.1, vinegar pH 3.3, distilled water pH 7.0), identify the liquid tested by the student that is most acidic.

Model answer: Lemon juice

Worked solution

$\text{Lowest pH value} = 2.1 \text{ (lemon juice)}$

Answer: Lemon juice

The lower the pH value, the more acidic the solution; lemon juice has the lowest measured pH (2.1) of the liquids tested.

Question 83 · 1 pt · constructed response

State the color of bromcresol green after the indicator is added to a sample of lemon juice.

Model answer: Yellow

Worked solution

$\text{Bromcresol green: yellow below pH} \approx 3.8,\ \text{blue above} \approx 5.4$

$\text{Lemon juice pH} = 2.1 \Rightarrow \text{yellow}$

Answer: Yellow

Bromcresol green is yellow in acidic solutions (below about pH 3.8); since lemon juice has a pH of 2.1, the indicator turns yellow.

Question 84 · 1 pt · constructed response

Complete the equation in your answer booklet by writing the formula of the missing product in the reaction of aqueous potassium hydroxide with the vinegar, acetic acid.

The partial equation to complete: CH3COOH(aq) + KOH(aq) -> H2O(l) + __________

Model answer: KOH(aq) + CH3COOH(aq) → CH3COOK(aq) + H2O(ℓ); the missing product formula is H2O (water), with CH3COOK(aq) as the salt product.

Worked solution

$KOH(aq) + CH_3COOH(aq) \rightarrow CH_3COOK(aq) + H_2O(\ell)$

Answer: KOH(aq) + CH3COOH(aq) → CH3COOK(aq) + H2O(ℓ); the missing product formula is H2O (water), with CH3COOK(aq) as the salt product.

This is a neutralization reaction between a base (KOH) and an acid (acetic acid) producing a salt (potassium acetate) and water; the missing product to be filled in the given equation is water, H2O.

Question 85 · 1 pt · constructed response

Based on the pH value in the table, compare the concentration of hydronium ions to the concentration of hydroxide ions in the distilled water.

Model answer: The concentration of hydronium ions is equal to the concentration of hydroxide ions (both are 1 x 10^-7 M) because distilled water has a neutral pH of 7.0.

Worked solution

$pH = 7.0 \Rightarrow [H_3O^+] = 1\times10^{-7}\ M$

$[OH^-] = 1\times10^{-7}\ M$

$[H_3O^+] = [OH^-]$

Answer: The concentration of hydronium ions is equal to the concentration of hydroxide ions (both are 1 x 10^-7 M) because distilled water has a neutral pH of 7.0.

A pH of 7.0 indicates a neutral solution in which the hydronium ion concentration equals the hydroxide ion concentration, both at 1 x 10^-7 M.

June 2023 Chemistry Regents — Answer Key & Worked Solutions

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