June 2024 Chemistry Regents — Answer Key & Worked Solutions

Every question from the June 2024 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 2024 Chemistry New York State Regents examination, published by NYSED. Worked solutions are written by RegentsLab.

Question 1 · 2 pts

Which description of the atom is based on the results of the gold foil experiment in the early 1900s?

(A) Atoms are small, dense, indivisible spheres.
(B) Atoms are composed of protons, electrons, and neutrons.
(C) Atoms have small, dense, positively charged nuclei. ✓
(D) Atoms have electrons with wavelike properties.

Worked solution

$\text{Rutherford's gold foil experiment: most alpha particles passed through, some deflected sharply}$

$\text{This showed a small, dense, positively charged nucleus at the atom's center}$

Answer: Atoms have small, dense, positively charged nuclei.

Rutherford's gold foil experiment showed that most of an atom's mass and positive charge is concentrated in a tiny, dense nucleus.

Question 2 · 2 pts

According to the electron cloud model, which term is defined as the most probable location of an electron in an atom?

(A) configuration
(B) nucleus
(C) orbital ✓
(D) spectrum

Worked solution

$\text{The electron cloud (wave-mechanical) model describes regions of probability}$

$\text{An orbital is the region of space where an electron is most likely to be found}$

Answer: orbital

An orbital represents the region of space around the nucleus where there is the highest probability of finding an electron.

Question 3 · 2 pts

Which change occurs when an electron returns from a higher energy state to a lower energy state?

(A) An ionic compound is formed, and energy is emitted.
(B) An ionic compound is formed, and energy is absorbed.
(C) A specific amount of energy is absorbed.
(D) A specific amount of energy is emitted. ✓

Worked solution

$\text{Electrons absorb energy to move to a higher energy state (excited state)}$

$\text{Electrons emit a specific (quantized) amount of energy when returning to a lower state}$

Answer: A specific amount of energy is emitted.

Dropping to a lower energy level releases a fixed quantum of energy, often as light, corresponding to the exact energy difference between states.

Question 4 · 2 pts

Which phrase describes the different isotopes of an element?

(A) same number of electrons and a different number of protons
(B) same number of protons and a different number of electrons
(C) same number of protons and a different number of neutrons ✓
(D) same number of neutrons and a different number of protons

Worked solution

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

$\text{Isotopes differ in mass number, meaning they differ in number of neutrons}$

Answer: same number of protons and a different number of neutrons

Isotopes are atoms of the same element (same proton count) that have different numbers of neutrons, giving them different mass numbers.

Question 5 · 2 pts

Which statement describes a chemical property of sodium?

(A) It is silver in color.
(B) It has a density of 0.97 g/cm$^3$ at room temperature.
(C) It has atoms with an atomic radius of 160 pm.
(D) It is highly reactive with water. ✓

Worked solution

$\text{Chemical properties describe how a substance reacts/changes composition}$

$\text{Color, density, and atomic radius are physical properties; reactivity with water is chemical}$

Answer: It is highly reactive with water.

A chemical property describes a substance's tendency to undergo a chemical change, such as sodium reacting vigorously with water; the others are physical properties.

Question 6 · 2 pts

Oxygen can exist as diatomic oxygen gas, $O_2(g)$, or ozone, $O_3(g)$. These two forms of oxygen have

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

Worked solution

$O_2 \text{ and } O_3 \text{ are allotropes of oxygen}$

$\text{Different numbers of atoms per molecule give different structures and different physical/chemical properties}$

Answer: different molecular structures and different properties

O2 and O3 are allotropes with different numbers of atoms bonded together, resulting in different molecular structures and different properties.

Question 7 · 2 pts

Which formula contains a polyatomic ion?

(A) KCN ✓
(B) K$_2$S
(C) AlBr$_3$
(D) Al$_2$O$_3$

Worked solution

$\text{KCN} = K^+ + CN^- \text{ (cyanide, a polyatomic ion)}$

$K_2S, AlBr_3, Al_2O_3 \text{ are all made of monatomic ions only}$

Answer: KCN

KCN is composed of the K+ ion and the polyatomic cyanide ion CN-; the other choices contain only monatomic ions.

Question 8 · 2 pts

What information about C$_4$H$_{10}$ can be determined from its structural formula, but not determined from its molecular formula?

(A) physical state
(B) molar mass
(C) ratio of different atoms
(D) arrangement of the atoms ✓

Worked solution

$\text{Molecular formula } C_4H_{10} \text{ gives only atom counts (ratios) and hence molar mass}$

$\text{Structural formula additionally shows how atoms are bonded/arranged (e.g., butane vs isobutane)}$

Answer: arrangement of the atoms

A molecular formula only tells the number of each type of atom, while a structural formula also shows how those atoms are connected and arranged, distinguishing isomers.

Question 9 · 2 pts

Which quantity is conserved in all chemical reactions?

(A) charge ✓
(B) density
(C) moles
(D) volume

Worked solution

$\text{In balanced chemical reactions, mass and charge are conserved}$

$\text{Moles, density, and volume of substances can change during a reaction}$

Answer: charge

Charge (like mass) is always conserved in chemical reactions, while quantities such as moles, density, and volume of the reacting substances can change.

Question 10 · 2 pts

What do the coefficients in a balanced chemical equation indicate about the substances in the equation?

(A) mass ratios
(B) mole ratios ✓
(C) empirical formulas
(D) molecular formulas

Worked solution

$\text{Coefficients in a balanced equation represent relative numbers of moles (or molecules)}$

$\text{They do not directly give mass ratios, empirical, or molecular formulas}$

Answer: mole ratios

Coefficients in a balanced equation represent the mole (particle) ratios in which reactants and products react and form.

Question 11 · 2 pts

At STP, which substance has metallic bonding?

(A) argon
(B) barium chloride
(C) sulfur dioxide
(D) titanium ✓

Worked solution

$\text{Argon: atomic solid, no bonding; BaCl}_2\text{: ionic bonding; SO}_2\text{: covalent (molecular)}$

$\text{Titanium is a metal, so it exhibits metallic bonding}$

Answer: titanium

Titanium is a metallic element, so its atoms are held together by metallic bonding, unlike argon (no bonding), barium chloride (ionic), or sulfur dioxide (covalent).

Question 12 · 2 pts

How many pairs of electrons are shared between the nitrogen atoms in a molecule of N$_2$?

(A) 1
(B) 2
(C) 3 ✓
(D) 4

Worked solution

$N_2 \text{ has a triple covalent bond}$

$\text{A triple bond consists of 3 shared pairs of electrons}$

Answer: 3

The nitrogen molecule N2 is held together by a triple covalent bond, which consists of three shared pairs of electrons.

Question 13 · 2 pts

Which formula represents a molecule with an asymmetrical distribution of charge?

(A) Cl$_2$
(B) CO$_2$
(C) CH$_4$
(D) H$_2$O ✓

Worked solution

$\text{Cl}_2\text{: nonpolar (identical atoms); CO}_2 \text{ and CH}_4\text{: symmetrical, nonpolar overall}$

$\text{H}_2\text{O has a bent shape with unequal charge distribution, making it polar (asymmetrical)}$

Answer: H$_2$O

Water's bent molecular geometry gives it an uneven (asymmetrical) distribution of charge, making it a polar molecule, unlike the symmetrical, nonpolar Cl2, CO2, and CH4.

Question 14 · 2 pts

Given the equation representing a reaction: $I_2 \rightarrow I + I$ What occurs during this reaction?

(A) Energy is released and a bond is formed.
(B) Energy is released and a bond is broken.
(C) Energy is absorbed and a bond is broken. ✓
(D) Energy is absorbed and a bond is formed.

Worked solution

$I_2 \rightarrow I + I \text{ shows one molecule breaking into two atoms}$

$\text{Breaking a bond always requires (absorbs) energy}$

Answer: Energy is absorbed and a bond is broken.

The single covalent bond in I2 is being broken to form two separate iodine atoms, and breaking a bond always requires energy input.

Question 15 · 2 pts

The electronegativity difference between the atoms in a molecule of HF can be used to determine the

(A) energy of the molecule
(B) functional group of molecule
(C) polarity of the bond in the molecule ✓
(D) volume of the atoms in the molecule

Worked solution

$\text{Electronegativity difference between bonded atoms determines bond type/polarity}$

$\text{A large difference (as in HF) indicates a polar covalent bond}$

Answer: polarity of the bond in the molecule

The difference in electronegativity between the two bonded atoms is used to classify and determine how polar a covalent bond is.

Question 16 · 2 pts

Which sample of matter is classified as a mixture?

(A) NaCl(s)
(B) CH$_3$OH($\ell$)
(C) SO$_2$(g)
(D) KNO$_3$(aq) ✓

Worked solution

$\text{NaCl(s), CH}_3\text{OH}(\ell), \text{SO}_2(g) \text{ are each pure substances (single compound)}$

$\text{KNO}_3(aq) \text{ is KNO}_3 \text{ dissolved in water, i.e., a solution (a mixture)}$

Answer: KNO$_3$(aq)

The (aq) notation indicates KNO3 dissolved in water, which is a homogeneous mixture (solution), while the other choices are single pure compounds.

Question 17 · 2 pts

At STP, which property of tungsten remains the same for all samples of tungsten?

(A) density ✓
(B) mass
(C) surface area
(D) thermal energy

Worked solution

$\text{Density is an intensive property, independent of sample size}$

$\text{Mass, surface area, and thermal energy are extensive properties that depend on sample size}$

Answer: density

Density is an intensive physical property that stays constant for a pure substance at given conditions regardless of how much of the sample you have, unlike mass, surface area, or thermal energy.

Question 18 · 2 pts

An element is composed of atoms that must have

(A) the same atomic mass
(B) the same atomic number ✓
(C) a different number of protons
(D) a different number of electrons

Worked solution

$\text{An element is defined by its number of protons (atomic number)}$

$\text{Atoms of the same element can differ in atomic mass (isotopes) but must share the same atomic number}$

Answer: the same atomic number

All atoms of a given element share the same atomic number (number of protons), which is the defining characteristic of an element.

Question 19 · 2 pts

Compared to a 2.0 M aqueous solution of KI at 1.0 atm, water at 1.0 atm has a

(A) lower boiling point and a lower freezing point
(B) lower boiling point and a higher freezing point ✓
(C) higher boiling point and a lower freezing point
(D) higher boiling point and a higher freezing point

Worked solution

$\text{Dissolved solute in the KI solution raises its boiling point and lowers its freezing point (colligative properties)}$

$\text{So pure water, compared to the solution, has a lower boiling point and a higher freezing point}$

Answer: lower boiling point and a higher freezing point

Adding solute to water raises the boiling point and lowers the freezing point of the solution, so pure water has a lower boiling point and higher freezing point than the KI solution.

Question 20 · 2 pts

Which statement describes the particles of an ideal gas, based on the kinetic molecular theory?

(A) The particles have attractive forces between them.
(B) The particles move in random, constant, straight-line motion. ✓
(C) The particles collide, increasing the total energy of the system.
(D) The particles are separated by distances that are small compared to their size.

Worked solution

$\text{Kinetic molecular theory assumes ideal gas particles have negligible attractive forces}$

$\text{Particles move randomly in straight lines until collisions, with elastic collisions conserving total energy, and are far apart relative to their size}$

Answer: The particles move in random, constant, straight-line motion.

Kinetic molecular theory describes ideal gas particles as moving randomly in constant straight-line motion between elastic collisions, with negligible attractive forces and large distances relative to particle size.

Question 21 · 2 pts

Chemical reactions are most likely to occur when reacting particles collide with the

(A) proper energy and proper orientation ✓
(B) proper first ionization energy and proper molecular symmetry
(C) same mass and the same number of electrons
(D) same number of valence electrons and the same number of electron shells

Worked solution

$\text{Collision theory: reactions require sufficient activation energy}$

$\text{Collisions must also occur with the correct orientation for a reaction to occur}$

Answer: proper energy and proper orientation

According to collision theory, an effective (reaction-producing) collision requires both sufficient energy (activation energy) and the correct geometric orientation of colliding particles.

Question 22 · 2 pts

At STP, a 2.0-liter sample of nitrogen gas and a 2.0-liter sample of oxygen gas have equal

(A) atomic masses
(B) numbers of molecules ✓
(C) densities
(D) boiling points

Worked solution

$\text{Avogadro's principle: equal volumes of gases at the same temperature and pressure contain equal numbers of molecules}$

$\text{Atomic masses, densities, and boiling points of N}_2 \text{ and O}_2 \text{ differ since they are different substances}$

Answer: numbers of molecules

By Avogadro's principle, equal volumes of any gas at the same temperature and pressure (STP) contain equal numbers of molecules, regardless of the identity of the gas.

Question 23 · 2 pts

Based on Table $H$, which compound has the weakest intermolecular forces at 75°C?

From Reference Table H (Vapor Pressure of Four Liquids): at every temperature shown, the vapor pressure curves rank, from highest to lowest, propanone > ethanol > water > ethanoic acid. Reading the curves at 75 degrees C gives approximately: | Liquid | Vapor pressure at 75 C (kPa, approx.) | |---|---| | propanone | about 185 (highest) | | ethanol | about 89 | | water | about 39 | | ethanoic acid | about 24 (lowest) | Higher vapor pressure at a given temperature indicates weaker intermolecular forces.

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

Worked solution

$\text{Table H graphs vapor pressure vs. temperature for four liquids.}$

$\text{At any fixed temperature, the liquid with the highest vapor pressure evaporates most easily.}$

$\text{Higher vapor pressure} \Rightarrow \text{weaker intermolecular attractive forces holding molecules in the liquid.}$

$\text{At } 75^{\circ}C,\ \text{propanone has the highest vapor pressure of the four substances shown.}$

Answer: propanone

On Table H, the curve for propanone lies highest at 75 degrees C, meaning it has the greatest vapor pressure and therefore the weakest intermolecular forces of the four liquids.

Question 24 · 2 pts

Which term identifies a force of attraction that exists between molecules of water?

(A) covalent bonding
(B) hydrogen bonding ✓
(C) ionic bonding
(D) metallic bonding

Worked solution

$\text{Covalent and ionic bonding are intramolecular forces (within a molecule/compound), not between molecules.}$

$\text{Water molecules are polar, with H bonded to the highly electronegative O.}$

$\text{This produces strong dipole attractions between molecules called hydrogen bonds.}$

Answer: hydrogen bonding

Hydrogen bonding is the strong intermolecular attraction between the partially positive H of one water molecule and the partially negative O of an adjacent water molecule.

Question 25 · 2 pts

Which process results in an increase in disorder?

(A) the condensation of water vapor
(B) the freezing of liquid ethanol
(C) the sublimation of solid iodine ✓
(D) the solidification of melted sodium chloride

Worked solution

$\text{Disorder (entropy) increases when a substance moves to a less ordered state.}$

$\text{Order of increasing entropy: solid} < \text{liquid} < \text{gas}.$

$\text{Condensation, freezing, and solidification all go toward a more ordered state (entropy decreases).}$

$\text{Sublimation converts a solid directly to a gas, the most disordered state, so entropy increases.}$

Answer: the sublimation of solid iodine

Sublimation takes a substance from the highly ordered solid phase directly to the highly disordered gas phase, so it is the only listed process that increases disorder.

Question 26 · 2 pts

Systems in nature tend to undergo changes that result in

(A) an increase in both energy and entropy
(B) a decrease in both energy and entropy
(C) an increase in energy and a decrease in entropy
(D) a decrease in energy and an increase in entropy ✓

Worked solution

$\text{Natural systems favor lower potential energy (more stable) states.}$

$\text{Natural systems also favor higher entropy (greater disorder) states.}$

$\text{Together, spontaneous changes tend toward decreasing energy and increasing entropy.}$

Answer: a decrease in energy and an increase in entropy

Spontaneous natural processes tend to minimize energy while maximizing disorder, matching a decrease in energy paired with an increase in entropy.

Question 27 · 2 pts

What is the general formula for the homologous series that includes ethyne?

(A) $C_nH_n$
(B) $C_nH_{2n}$
(C) $C_nH_{2n+2}$
(D) $C_nH_{2n-2}$ ✓

Worked solution

$\text{Ethyne, } C_2H_2,\ \text{is a member of the alkyne series (contains a carbon-carbon triple bond).}$

$\text{The general formula for alkynes is } C_nH_{2n-2}.$

$\text{Check: for } n=2,\ C_2H_{2(2)-2}=C_2H_2,\ \text{matching ethyne.}$

Answer: $C_nH_{2n-2}$

Ethyne (acetylene) is the simplest alkyne, and all alkynes follow the general formula CnH2n-2.

Question 28 · 2 pts

Given the equation representing a reaction: $2KI + F_2 \rightarrow 2KF + I_2$ When 2.0 moles of electrons are gained by fluorine, how many moles of electrons are lost by iodine?

(A) 1.0 mol
(B) 2.0 mol ✓
(C) 3.0 mol
(D) 4.0 mol

Worked solution

$\text{In any redox reaction, electrons lost by one species must equal electrons gained by the other.}$

$\text{Fluorine gains 2.0 mol of electrons (given).}$

$\text{Therefore iodine must lose 2.0 mol of electrons.}$

Answer: 2.0 mol

By conservation of charge in a redox reaction, the total moles of electrons lost must equal the total moles gained, so iodine loses 2.0 mol of electrons.

Question 29 · 2 pts

Which compound is an Arrhenius base?

(A) $H_2SO_4$
(B) NaCl
(C) $Ca(OH)_2$ ✓
(D) $KNO_3$

Worked solution

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

$Ca(OH)_2(s) \rightarrow Ca^{2+}(aq) + 2OH^-(aq)$

$\text{H}_2SO_4 \text{ is an acid, and NaCl and } KNO_3 \text{ are neutral salts.}$

Answer: $Ca(OH)_2$

Calcium hydroxide dissociates in water to release hydroxide ions, the defining characteristic of an Arrhenius base.

Question 30 · 2 pts

Which statement describes a benefit of fission?

(A) Fission produces nuclear power. ✓
(B) Fission increases the possibility of nuclear accidents.
(C) Fission increases the possibility of biological exposure.
(D) Fission produces waste products that need long-term storage.

Worked solution

$\text{Benefits of fission are positive outcomes; the other three choices describe risks/drawbacks.}$

$\text{Producing usable electrical/nuclear power is a clear benefit of nuclear fission.}$

Answer: Fission produces nuclear power.

Generating nuclear power is a benefit, while accidents, biological exposure, and long-term radioactive waste storage are all risks or drawbacks of fission.

Question 31 · 2 pts

The numbers of protons, neutrons, and electrons in each of four different ions are shown in the table below. Which ion has the greatest mass?

(A) A
(B) E ✓
(C) G
(D) Z

Worked solution

$\text{Mass number} = \text{protons} + \text{neutrons (electrons contribute negligible mass).}$

$A:\ 49+60=109$

$E:\ 50+70=120$

$G:\ 50+69=119$

$Z:\ 51+59=110$

$\text{The largest sum, 120, belongs to ion E.}$

Answer: E

Mass is determined by protons plus neutrons; ion E has the highest total (120), giving it the greatest mass.

Question 32 · 2 pts

The mass of a sample of nickel is determined to be 20.40 grams. How many significant figures are used to express this mass?

(A) 1
(B) 2
(C) 3
(D) 4 ✓

Worked solution

$\text{In } 20.40,\ \text{all nonzero digits (2, 4) are significant.}$

$\text{The zero between nonzero digits (the 0 after the 2) is significant.}$

$\text{A trailing zero after a decimal point is significant.}$

$\text{Total significant figures: } 2,0,4,0 \Rightarrow 4.$

Answer: 4

In 20.40, every digit including both zeros counts as significant because one zero is sandwiched between nonzero digits and the other is a trailing zero after a decimal point, giving 4 significant figures.

Question 33 · 2 pts

What is a chemical name for the compound $PbO_2$?

(A) lead(I) oxide
(B) lead(II) oxide
(C) lead(III) oxide
(D) lead(IV) oxide ✓

Worked solution

$\text{Oxide ion is } O^{2-}.\ \text{With two oxide ions, total negative charge} = 2(-2) = -4.$

$\text{For the compound to be neutral, lead must have a charge of } +4.$

$\text{This is named lead(IV) oxide using the Stock system.}$

Answer: lead(IV) oxide

Since PbO2 contains two O2- ions (total charge -4), the lead ion must be Pb4+, named lead(IV) oxide.

Question 34 · 2 pts

Which formula is the empirical formula for ethane, $C_2H_6$?

(A) CH
(B) $CH_3$ ✓
(C) $C_2H_6$
(D) $C_4H_{12}$

Worked solution

$\text{Divide the subscripts of } C_2H_6 \text{ by their greatest common factor, 2.}$

$\frac{2}{2}=1,\quad \frac{6}{2}=3$

$\text{Empirical formula: } CH_3.$

Answer: $CH_3$

Dividing both subscripts in C2H6 by their greatest common divisor (2) gives the simplest whole-number ratio, CH3.

Question 35 · 2 pts

Given a balanced equation representing a reaction: $2CO(g) + O_2(g) \rightarrow 2CO_2(g) + \text{energy}$ Which mass of $O_2(g)$ reacts completely with 5.6 grams of $CO(g)$ to produce 8.8 grams of $CO_2(g)$?

(A) 1.6 g
(B) 2.8 g
(C) 3.2 g ✓
(D) 14.4 g

Worked solution

$\text{By conservation of mass: mass of reactants = mass of products.}$

$5.6\,\text{g CO} + x\,\text{g } O_2 = 8.8\,\text{g } CO_2$

$x = 8.8 - 5.6 = 3.2\,\text{g}$

Answer: 3.2 g

Using conservation of mass, the mass of oxygen gas needed equals the mass of product minus the mass of the given reactant, 8.8 g minus 5.6 g, which is 3.2 g.

Question 36 · 2 pts

Which numerical setup can be used to calculate the formula mass of $C_9H_{11}NO_2$?

(A) $9\text{ u} + 11\text{ u} + 1\text{ u} + 2\text{ u}$
(B) $12\text{ u} + 1\text{ u} + 14\text{ u} + 16\text{ u}$
(C) $9(6\text{ u}) + 11(1\text{ u}) + 1(7\text{ u}) + 2(8\text{ u})$
(D) $9(12\text{ u}) + 11(1\text{ u}) + 1(14\text{ u}) + 2(16\text{ u})$ ✓

Worked solution

$\text{Formula mass} = \sum (\text{number of atoms}) \times (\text{atomic mass}).}$

$\text{Atomic masses: C}=12\text{ u},\ H=1\text{ u},\ N=14\text{ u},\ O=16\text{ u}.$

$C_9H_{11}NO_2:\ 9(12\text{ u}) + 11(1\text{ u}) + 1(14\text{ u}) + 2(16\text{ u})$

Answer: $9(12\text{ u}) + 11(1\text{ u}) + 1(14\text{ u}) + 2(16\text{ u})$

Formula mass requires multiplying each element's atomic mass (C=12, H=1, N=14, O=16) by its subscript and summing, which matches choice 4.

Question 37 · 2 pts

Based on Table $F$, which compound is most soluble in water?

(A) calcium bromide ✓
(B) lead(II) sulfate
(C) silver carbonate
(D) tin(II) hydroxide

Worked solution

$\text{Table F lists solubility guidelines for common ions.}$

$\text{Bromides are generally soluble (calcium bromide is soluble).}$

$\text{Sulfates of Pb are exceptions and are insoluble; most carbonates are insoluble (silver carbonate); most hydroxides are insoluble (tin(II) hydroxide).}$

Answer: calcium bromide

According to the solubility guidelines in Table F, calcium bromide is soluble, while lead(II) sulfate, silver carbonate, and tin(II) hydroxide are all classified as insoluble or only slightly soluble.

Question 38 · 2 pts

Based on Table $G$, which combination of solute and solvent will form a saturated solution when thoroughly mixed at 20.°C?

From Reference Table G (Solubility Curves at Standard Pressure), solubility in grams of solute per 100. g of H2O at 20. degrees C, read from the curves: | Solute | Solubility at 20 C (g per 100. g H2O, approx.) | |---|---| | KClO3 | about 10 | | KCl | about 34 | | NaCl | about 36 | | KI | about 144 | A solution is saturated (or holds undissolved excess) when the amount of solute added equals or exceeds the solubility value.

(A) 15 grams $KClO_3$ in 100. grams $H_2O$ ✓
(B) 20. grams KCl in 100. grams $H_2O$
(C) 30. grams NaCl in 100. grams $H_2O$
(D) 35 grams KI in 100. grams $H_2O$

Worked solution

$\text{A solution is saturated when the amount of solute added equals or exceeds the solubility value read from the Table G curve at that temperature.}$

$\text{At } 20^{\circ}C,\ \text{the Table G curve for } KClO_3 \text{ lies at only about 8 g per 100 g water, so the 15 g sample cannot all dissolve.}$

$\text{The excess undissolved } KClO_3 \text{ leaves a saturated solution; the other amounts are below their curve values, so they dissolve fully and stay unsaturated.}$

Answer: 15 grams $KClO_3$ in 100. grams $H_2O$

At 20 degrees C, the solubility of KClO3 on Table G is only about 8 g per 100 g of water, so the 15 g sample exceeds it and cannot all dissolve, giving a saturated solution with excess solid. The KCl, NaCl, and KI amounts are all below their Table G values, so they dissolve completely and remain unsaturated.

Question 39 · 2 pts

A solution has a mass of 2000. grams and contains 0.050 gram of dissolved solute. What is the concentration in parts per million of this solution?

(A) 5.0 ppm
(B) 25 ppm ✓
(C) 50. ppm
(D) 100. ppm

Worked solution

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

$ppm = \dfrac{0.050\text{ g}}{2000.\text{ g}} \times 10^6$

$ppm = 0.000025 \times 10^6 = 25\text{ ppm}$

Answer: 25 ppm

Dividing the mass of solute by the mass of solution and multiplying by one million gives 25 parts per million.

Question 40 · 2 pts

A person with a body temperature of 37°C holds a cup of hot water in their hand. The temperature of the cup is 44°C. The air temperature is 22°C. Which phrase describes a heat transfer that occurs in this system?

(A) from the air to the cup
(B) from the hand to the cup
(C) from the air to the hand
(D) from the cup to the hand ✓

Worked solution

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

$\text{Cup} = 44^{\circ}C > \text{hand} = 37^{\circ}C > \text{air} = 22^{\circ}C.$

$\text{Since the cup is hotter than the hand, heat flows from the cup to the hand.}$

Answer: from the cup to the hand

Because the cup (44 degrees C) is warmer than the hand (37 degrees C), heat naturally transfers from the higher-temperature cup to the lower-temperature hand.

Question 41 · 2 pts

What is the amount of heat absorbed to completely melt 26.2 grams of $H_2O(s)$ at 0°C?

(A) $8.75 \times 10^3$ J ✓
(B) $3.66 \times 10^4$ J
(C) $5.92 \times 10^4$ J
(D) $2.48 \times 10^5$ J

Worked solution

$q = m H_f$

$\text{From the Reference Tables, } H_f \text{ (heat of fusion) of water} = 334\ \text{J/g}$

$q = 26.2\text{ g} \times 334\text{ J/g} = 8752.8\text{ J} \approx 8.75 \times 10^3\text{ J}$

Answer: $8.75 \times 10^3$ J

Using q = mHf with water's heat of fusion of 334 J/g, melting 26.2 g of ice requires about 8.75 times 10 to the 3 joules.

Question 42 · 2 pts

The graph below shows a compound being cooled, beginning in the liquid phase. Cooling Curve for a Compound: the y-axis is Temperature (°C) ranging from 10 to 90, and the x-axis is Time (min) ranging from 0 to 8. The line starts at about 78°C at time 0 and decreases steadily to 50.°C at time 2 min. The temperature then remains constant (flat) at 50.°C from time 2 min to time 6 min. After 6 min, the temperature decreases steadily again, reaching about 12°C at time 8 min. Which statement describes the average kinetic energy and the potential energy between minute 2 and minute 6?

(A) Average kinetic energy and potential energy both decrease.
(B) Average kinetic energy and potential energy both remain constant.
(C) Average kinetic energy remains constant and potential energy increases.
(D) Average kinetic energy remains constant and potential energy decreases. ✓

Worked solution

$\text{Between minute 2 and 6, temperature stays constant at } 50^{\circ}C \text{ (flat plateau), indicating a phase change.}$

$\text{Average kinetic energy depends only on temperature, so it remains constant during the phase change.}$

$\text{During this cooling phase change (liquid to solid), particles move closer together and attractions strengthen, so potential energy decreases.}$

Answer: Average kinetic energy remains constant and potential energy decreases.

The flat plateau on a cooling curve represents a phase change at constant temperature, so average kinetic energy stays the same while potential energy decreases as the substance solidifies.

Question 43 · 2 pts

Which formula represents a saturated hydrocarbon?

(A) Structure (1): all single C-C and C-H bonds, atoms are only carbon and hydrogen ✓
(B) Structure (2): contains a carbon-carbon double bond
(C) Structure (3): all single bonds but contains chlorine atoms
(D) Structure (4): contains an oxygen atom (an ether)

Worked solution

$\text{A saturated hydrocarbon contains only carbon and hydrogen atoms, joined entirely by single bonds.}$

$\text{Structure (2) has a C=C double bond, so it is unsaturated (an alkene).}$

$\text{Structure (3) contains Cl atoms, so it is a halogenated compound, not a hydrocarbon.}$

$\text{Structure (4) contains an oxygen atom, so it is an ether, not a hydrocarbon.}$

$\text{Structure (1) has only C-C and C-H single bonds and only C, H atoms, so it is a saturated hydrocarbon (ethane).}$

Answer: Structure (1)

Only structure (1), with exclusively single bonds and only carbon and hydrogen atoms, meets the definition of a saturated hydrocarbon.

Question 44 · 2 pts

Given an equation representing a reaction: (structural formula of a two-carbon molecule with H, H on each carbon, i.e. ethane) $+ Br_2 \rightarrow$ (structural formula of the same two-carbon chain with one H replaced by Br on one carbon) $+ HBr$ Which type of organic reaction is represented by this equation?

(A) addition
(B) polymerization
(C) fermentation
(D) substitution ✓

Worked solution

$\text{The reactant hydrocarbon has only single bonds (saturated), so no double bond is available for addition.}$

$\text{A hydrogen atom on the hydrocarbon is replaced by a bromine atom, and the displaced hydrogen combines with the remaining bromine to form HBr.}$

$\text{Replacing one atom (H) with another (Br) on a saturated compound is a substitution reaction.}$

Answer: substitution

Because the starting hydrocarbon is already saturated, bromine cannot add across a double bond; instead a hydrogen atom is substituted by a bromine atom, identifying this as a substitution reaction.

Question 45 · 2 pts

In which compound does sulfur have an oxidation state of +4?

(A) K$_2$S
(B) CaS
(C) SO$_2$ ✓
(D) SO$_3$

Worked solution

$K_2S:\ S=-2$

$CaS:\ S=-2$

$SO_2:\ O=-2\times2=-4,\ S=+4$

$SO_3:\ O=-2\times3=-6,\ S=+6$

Answer: SO$_2$

In SO2 the two oxygen atoms contribute a total charge of -4, so sulfur must be +4 to make the neutral molecule balance.

Question 46 · 2 pts

Which solution is the best conductor of electricity?

(A) 0.0010 M CaCl$_2$(aq)
(B) 0.010 M CaCl$_2$(aq)
(C) 0.10 M CaCl$_2$(aq)
(D) 1.0 M CaCl$_2$(aq) ✓

Worked solution

$\text{Conductivity increases with ion concentration}$

$1.0\ M\ CaCl_2 \text{ has the greatest concentration of dissolved ions}$

Answer: 1.0 M CaCl$_2$(aq)

Electrical conductivity of an electrolyte solution increases as the concentration of dissolved ions increases, so the most concentrated CaCl2 solution conducts best.

Question 47 · 2 pts

Given the incomplete equation representing a neutralization reaction: $Sr(OH)_2 + 2HNO_3 \rightarrow X + 2H_2O$ What is the formula of the missing product X?

(A) SrO
(B) SrO$_2$
(C) Sr(NO$_2$)$_2$
(D) Sr(NO$_3$)$_2$ ✓

Worked solution

$\text{acid} + \text{base} \rightarrow \text{salt} + \text{water}$

$Sr^{2+} + 2NO_3^- \rightarrow Sr(NO_3)_2$

Answer: Sr(NO$_3$)$_2$

Neutralizing the base Sr(OH)2 with the acid HNO3 produces the salt strontium nitrate, Sr(NO3)2, and water.

Question 48 · 2 pts

Given the equation representing a reversible reaction: $NH_3(aq) + H_2O(\ell) \rightleftharpoons NH_4^+(aq) + OH^-(aq)$ The water molecule acts as an acid because it

(A) accepts a hydrogen ion
(B) accepts a hydroxide ion
(C) donates a hydrogen ion ✓
(D) donates a hydroxide ion

Worked solution

$\text{A Bronsted-Lowry acid is a hydrogen-ion (proton) donor}$

$H_2O \rightarrow OH^- \text{ (loses an } H^+ \text{ to } NH_3\text{)}$

Answer: donates a hydrogen ion

Water acts as a Bronsted-Lowry acid here because it donates a hydrogen ion to ammonia, becoming OH- while NH3 becomes NH4+.

Question 49 · 2 pts

Which nuclear equation represents beta decay?

(A) $^{81}_{37}Rb + ^{0}_{-1}e \rightarrow ^{81}_{36}Kr$
(B) $^{13}_{4}Be \rightarrow ^{12}_{4}Be + ^{1}_{0}n$
(C) $^{14}_{7}N + ^{1}_{0}n \rightarrow ^{14}_{6}C + ^{1}_{1}H$
(D) $^{14}_{6}C \rightarrow ^{14}_{7}N + ^{0}_{-1}e$ ✓

Worked solution

$\text{Beta decay: a neutron converts to a proton, emitting a beta particle (electron)}$

$\text{Mass number unchanged (14), atomic number increases by 1 (6}\rightarrow\text{7), and an } ^{0}_{-1}e \text{ is emitted}$

Answer: $^{14}_{6}C \rightarrow ^{14}_{7}N + ^{0}_{-1}e$

Choice (4) shows carbon-14 decaying into nitrogen-14 with emission of an electron, the defining signature of beta decay.

Question 50 · 2 pts

Given the equations A and B: Equation A: $^{2}_{1}H + ^{3}_{1}H \rightarrow ^{4}_{2}He + ^{1}_{0}n$ Equation B: $^{235}_{92}U + ^{1}_{0}n \rightarrow ^{90}_{38}Sr + ^{143}_{54}Xe + 3^{1}_{0}n$ Which type of nuclear reaction do these equations represent?

(A) Both A and B are fission.
(B) Both A and B are fusion.
(C) A is fission and B is fusion.
(D) A is fusion and B is fission. ✓

Worked solution

$\text{Equation A: two small nuclei combine into a larger nucleus} \Rightarrow \text{fusion}$

$\text{Equation B: one large nucleus splits into two smaller nuclei} \Rightarrow \text{fission}$

Answer: A is fusion and B is fission.

A combines two light hydrogen isotopes into helium (fusion), while B splits a uranium-235 nucleus into two lighter nuclei (fission).

Question 51 · 1 pt · constructed response

Compounds that contain the element boron, such as boric acid and sodium borate, produce a green flame when a sample of one of these compounds is placed in a laboratory burner flame. If this flame is viewed through a spectroscope, a bright-line spectrum is seen. State how the bright-line spectra of known elements can be used to identify an unknown element whose bright-line spectrum was viewed through a spectroscope.

Model answer: Compare the pattern (wavelengths/positions) of lines in the unknown element's bright-line spectrum to the bright-line spectra of known elements; since each element produces its own unique set of spectral lines, a match with a known element's spectrum identifies the unknown element.

Worked solution

$\text{Match the line pattern of the unknown spectrum to a reference (known) spectrum}$

Answer: Compare the pattern (wavelengths/positions) of lines in the unknown element's bright-line spectrum to the bright-line spectra of known elements; since each element produces its own unique set of spectral lines, a match with a known element's spectrum identifies the unknown element.

Every element has a unique bright-line spectrum, so comparing an unknown spectrum to known reference spectra identifies the element that matches.

Question 52 · 1 pt · constructed response

State, in terms of electrons in shells, why the electron configuration 2-2-1 is an excited state electron configuration of a boron atom.

Model answer: In the ground state, electrons fill the lowest available shells first; because the second shell can hold up to 8 electrons and is not yet full (only 2 electrons in it) while an electron is present in the third shell, this configuration shows an electron occupying a higher-energy shell before the lower shell was filled, which is characteristic of an excited state.

Worked solution

$\text{Ground state boron: } 2\text{-}3 \text{ (2 in first shell, 3 in second shell)}$

$2\text{-}2\text{-}1 \text{ places an electron in shell 3 while shell 2 (max 8) is not full}$

Answer: In the ground state, electrons fill the lowest available shells first; because the second shell can hold up to 8 electrons and is not yet full (only 2 electrons in it) while an electron is present in the third shell, this configuration shows an electron occupying a higher-energy shell before the lower shell was filled, which is characteristic of an excited state.

An electron occupies the third shell even though the second shell, which can hold up to 8 electrons, still has room, so this is an excited-state configuration.

Question 53 · 1 pt · constructed response

Draw a Lewis electron-dot diagram for an atom of silicon in the ground state.

Model answer: The symbol Si surrounded by four single (unpaired) electron dots, one placed on each of the four sides (top, bottom, left, right) of the symbol, representing silicon's 4 valence electrons.

Worked solution

$\text{Si is in Group 14} \Rightarrow 4 \text{ valence electrons}$

$\text{Place one dot on each of the 4 sides of the symbol (Hund's rule, unpaired)}$

Answer: The symbol Si surrounded by four single (unpaired) electron dots, one placed on each of the four sides (top, bottom, left, right) of the symbol, representing silicon's 4 valence electrons.

Silicon, in Group 14, has 4 valence electrons, which in the ground state are drawn as one unpaired dot on each side of the element symbol.

Question 54 · 1 pt · constructed response

Compare the energy of an electron in the first shell of a silicon atom to the energy of an electron in the third shell of the same atom.

Model answer: An electron in the first shell of a silicon atom has less energy than an electron in the third shell of the same atom.

Worked solution

$\text{First shell electrons are closer to the nucleus} \Rightarrow \text{lower energy}$

$\text{Third shell electrons are farther from the nucleus} \Rightarrow \text{higher energy}$

Answer: An electron in the first shell of a silicon atom has less energy than an electron in the third shell of the same atom.

Electrons closer to the nucleus (first shell) are held more tightly and have lower energy than electrons farther away (third shell), which have higher energy.

Question 55 · 1 pt · constructed response

Silicon is an element in Group 14 on the Periodic Table. The atomic mass and natural abundance for the three naturally occurring isotopes of silicon are shown in the table below. Show a numerical setup for calculating the atomic mass of the element silicon.

Model answer: $(27.98\,u)(0.9222) + (28.98\,u)(0.0469) + (29.97\,u)(0.0309) \approx 28.09\,u$

Worked solution

$(27.98)(0.9222) + (28.98)(0.0469) + (29.97)(0.0309)$

$\approx 25.80 + 1.359 + 0.926 \approx 28.09\,u$

Answer: $(27.98\,u)(0.9222) + (28.98\,u)(0.0469) + (29.97\,u)(0.0309) \approx 28.09\,u$

The atomic mass of an element is the weighted average of its isotopes' masses, each multiplied by its fractional natural abundance.

Question 56 · 1 pt · constructed response

The unbalanced equation below shows a reaction between nitrogen monoxide gas and oxygen gas to produce nitrogen dioxide gas. $NO(g) + O_2(g) \rightarrow NO_2(g)$ State, in terms of reactants and product, why this reaction is a synthesis reaction.

Model answer: This reaction is a synthesis reaction because two reactants (NO and O2) combine to form a single product (NO2).

Worked solution

$\text{Synthesis: two (or more) reactants} \rightarrow \text{one product}$

Answer: This reaction is a synthesis reaction because two reactants (NO and O2) combine to form a single product (NO2).

A synthesis reaction is defined by two substances combining into a single compound, exactly as NO and O2 combine to form NO2.

Question 57 · 1 pt · constructed response

Balance the equation in your answer booklet for the reaction, using the smallest whole-number coefficients. $NO(g) + O_2(g) \rightarrow NO_2(g)$

Model answer: $2NO(g) + O_2(g) \rightarrow 2NO_2(g)$

Worked solution

$2NO(g) + O_2(g) \rightarrow 2NO_2(g)$

$\text{Check: N: }2=2,\ O: 2+2=4=4$

Answer: $2NO(g) + O_2(g) \rightarrow 2NO_2(g)$

Placing a coefficient of 2 in front of NO and NO2 balances both nitrogen (2 = 2) and oxygen (2+2 = 4) atoms.

Question 58 · 1 pt · constructed response

Determine the percent composition by mass of oxygen in the product, NO$_2$ (gram-formula mass = 46 g/mol).

Model answer: $\%O = \dfrac{2(16\,u)}{46\,g/mol}\times100\% \approx 69.6\%$

Worked solution

$\%O = \frac{2(16)}{46}\times100$

$= \frac{32}{46}\times100 \approx 69.6\%$

Answer: $\%O = \dfrac{2(16\,u)}{46\,g/mol}\times100\% \approx 69.6\%$

Percent composition of oxygen equals the total mass of oxygen atoms (2 x 16 u) divided by the gram-formula mass (46 g/mol), times 100%.

Question 59 · 1 pt · constructed response

The ionic radii of four Group 17 elements are given in the table below. Identify the noble gas that has atoms in the ground state with the same electron configuration as a Br$^-$ ion in the ground state.

Model answer: Krypton (Kr)

Worked solution

$\text{Br has 35 protons}$

$Br^- \text{ has } 35+1 = 36 \text{ electrons, same as neutral Kr (Z=36)}$

Answer: Krypton (Kr)

A Br- ion has gained one electron, giving it 36 electrons, the same electron configuration as neutral krypton (atomic number 36).

Question 60 · 1 pt · constructed response

The ionic radii of four Group 17 elements are given in the table below. State, in terms of electrons, why the ionic radius of a Group 17 element is larger than the atomic radius for the same element.

Model answer: When a Group 17 atom gains an electron to form its negative ion, the number of electrons increases while the number of protons stays the same, increasing electron-electron repulsion in the outer shell; this causes the electron cloud to spread out/expand, making the ionic radius larger than the atomic radius.

Worked solution

$\text{Atom} + e^- \rightarrow \text{ion (one extra electron, same protons)}$

$\text{More electron-electron repulsion} \Rightarrow \text{electron cloud expands}$

Answer: When a Group 17 atom gains an electron to form its negative ion, the number of electrons increases while the number of protons stays the same, increasing electron-electron repulsion in the outer shell; this causes the electron cloud to spread out/expand, making the ionic radius larger than the atomic radius.

Adding an extra electron to form the anion increases repulsion among the electrons without adding protons to pull them in, so the ion is larger than the neutral atom.

Question 61 · 1 pt · constructed response

The diagram below represents two sealed, rigid containers and the conditions of pressure, volume, and temperature for each container. Container A is filled with N$_2$(g) and container B is filled with O$_2$(g). Container A: N$_2$(g), P = 101.3 kPa, V = 750. mL, T = 298 K Container B: O$_2$(g), P = 101.3 kPa, V = 750. mL, T = 298 K Explain, in terms of the strength of intermolecular forces, why the gas in container B at standard pressure has a higher boiling point than the gas in container A at standard pressure.

Model answer: O2 molecules have stronger intermolecular (London dispersion) forces than N2 molecules because O2 has a greater molar mass/more electrons, so more energy (a higher temperature) must be added to overcome these attractive forces, giving O2 a higher boiling point than N2.

Worked solution

$\text{O}_2 \text{ has more electrons/greater mass than N}_2 \Rightarrow \text{stronger London dispersion forces}$

$\text{Stronger IMFs require more energy (higher temperature) to boil}$

Answer: O2 molecules have stronger intermolecular (London dispersion) forces than N2 molecules because O2 has a greater molar mass/more electrons, so more energy (a higher temperature) must be added to overcome these attractive forces, giving O2 a higher boiling point than N2.

Because O2 molecules are larger and have stronger London dispersion forces than N2 molecules, more energy is needed to separate them into the gas phase, giving O2 the higher boiling point.

Question 62 · 1 pt · constructed response

The diagram below represents two sealed, rigid containers and the conditions of pressure, volume, and temperature for each container. Container A is filled with N$_2$(g) and container B is filled with O$_2$(g). Container A: N$_2$(g), P = 101.3 kPa, V = 750. mL, T = 298 K Container B: O$_2$(g), P = 101.3 kPa, V = 750. mL, T = 298 K In the space in your answer booklet, show a numerical setup for calculating the pressure if the gas in container B is cooled to standard temperature.

Model answer: $P_2 = \dfrac{(101.3\text{ kPa})(273\text{ K})}{298\text{ K}} \approx 92.7\text{ kPa}$

Worked solution

$\frac{P_1}{T_1} = \frac{P_2}{T_2} \text{ (constant volume)}$

$P_2 = \frac{(101.3\text{ kPa})(273\text{ K})}{298\text{ K}}$

Answer: $P_2 = \dfrac{(101.3\text{ kPa})(273\text{ K})}{298\text{ K}} \approx 92.7\text{ kPa}$

Since volume is constant, Gay-Lussac's law (P/T = constant) is used with standard temperature 273 K to solve for the new pressure.

Question 63 · 1 pt · constructed response

The table below shows the formulas of three organic compounds, and a chemical name for two of the compounds. Three Organic Compounds Compound 1: propyl butanoate, formula is an ester: H-C-C-C-C(=O)-O-C-C-C-H chain (propyl butanoate structure) Compound 2: (unnamed), formula: H-C-C-C(=O)-OH chain with H's on carbons 1 and 2 Compound 3: 2-propanol, formula: H-C-C(OH)-C-H chain (2-propanol structure) Identify the element in these three compounds that allows them to be classified as organic compounds.

Model answer: Carbon

Worked solution

$\text{Organic compounds are defined by the presence of carbon (bonded to hydrogen)}$

Answer: Carbon

All organic compounds contain the element carbon, which is present in all three structures shown.

Question 64 · 1 pt · constructed response

The table below shows the formulas of three organic compounds, and a chemical name for two of the compounds. Three Organic Compounds Compound 1: propyl butanoate, formula is an ester: H-C-C-C-C(=O)-O-C-C-C-H chain (propyl butanoate structure) Compound 2: (unnamed), formula: H-C-C-C(=O)-OH chain with H's on carbons 1 and 2 Compound 3: 2-propanol, formula: H-C-C(OH)-C-H chain (2-propanol structure) Identify the class of organic compound to which compound 1 belongs.

Model answer: Ester

Worked solution

$\text{Propyl butanoate contains a } -C(=O)-O- \text{ linkage} \Rightarrow \text{ester}$

Answer: Ester

Compound 1, propyl butanoate, contains the carbon-oxygen-carbon ester linkage (-COO-) characteristic of esters, formed from an acid and an alcohol.

Question 65 · 1 pt · constructed response

The table below shows the formulas of three organic compounds, and a chemical name for two of the compounds. Three Organic Compounds Compound 1: propyl butanoate, formula is an ester: H-C-C-C-C(=O)-O-C-C-C-H chain (propyl butanoate structure) Compound 2: (unnamed), formula: H-C-C-C(=O)-OH chain, three carbons ending in a carboxylic acid group, with H's on carbons 1 and 2 Compound 3: 2-propanol, formula: H-C-C(OH)-C-H chain (2-propanol structure) Write a chemical name for compound 2.

Model answer: Propanoic acid

Worked solution

$\text{3-carbon chain ending in -COOH} \Rightarrow \text{propanoic acid}$

Answer: Propanoic acid

Compound 2 is a three-carbon chain terminating in a carboxylic acid (-COOH) group, which is named propanoic acid.

Question 66 · 1 pt · constructed response

The Periodic Table of the Elements has changed over the past 150 years. During this time, many elements have been added as they were discovered. The elements 113, 115, 117, and 118 in Period 7 were named in 2016. Their official names have been approved by the International Union of Pure and Applied Chemistry (IUPAC). The table below lists the names, symbols, and atomic numbers for these four elements. Identify the element in this table that is located on the Periodic Table of the Elements in the group classified as noble gases.

Model answer: Oganesson (Og), atomic number 118

Worked solution

$\text{Element 118 is in Group 18, the noble gases}$

Answer: Oganesson (Og), atomic number 118

Oganesson, with atomic number 118, is located in Group 18 of the Periodic Table, the group classified as the noble gases.

Question 67 · 1 pt · constructed response

Explain, in terms of valence electrons, why the elements nihonium and tennessine would be expected to differ in their chemical properties.

Model answer: Nihonium and tennessine are in different groups on the Periodic Table, so they have different numbers of valence electrons, which causes them to have different chemical properties.

Worked solution

$\text{Nihonium (Nh, Z=113) is in Group 13} \Rightarrow 3 \text{ valence electrons}$

$\text{Tennessine (Ts, Z=117) is in Group 17} \Rightarrow 7 \text{ valence electrons}$

$\text{Different numbers of valence electrons} \Rightarrow \text{different chemical properties}$

Answer: Nihonium and tennessine are in different groups on the Periodic Table, so they have different numbers of valence electrons, which causes them to have different chemical properties.

Chemical properties of an element are determined largely by its number of valence electrons, which corresponds to its group number on the Periodic Table; since nihonium and tennessine are in different groups, they have different valence electron counts and thus different chemical behavior.

Question 68 · 1 pt · constructed response

Describe the general trend in first ionization energy values as the five elements above moscovium in Group 15 on the Periodic Table of the Elements are considered in order of increasing atomic number.

Model answer: First ionization energy generally decreases as atomic number increases down Group 15 (N, P, As, Sb, Bi).

Worked solution

$\text{Elements above Mc (Z=115) in Group 15: N, P, As, Sb, Bi}$

$\text{Moving down a group, atomic radius increases}$

$\text{Outer valence electrons are farther from the nucleus and more shielded} \Rightarrow \text{easier to remove}$

$\text{Therefore first ionization energy generally decreases with increasing atomic number}$

Answer: First ionization energy generally decreases as atomic number increases down Group 15 (N, P, As, Sb, Bi).

Down any group, increasing atomic radius and electron shielding make the outermost electron easier to remove, so first ionization energy generally decreases as atomic number increases.

Question 69 · 1 pt · constructed response

State, in terms of protons, why tennessine is listed before oganesson in Period 7 on the Periodic Table of the Elements.

Model answer: Tennessine has fewer protons (117) than oganesson (118); elements are arranged in order of increasing number of protons (atomic number).

Worked solution

$\text{Tennessine: } Z=117, \quad \text{Oganesson: } Z=118$

$\text{Periodic Table is arranged by increasing atomic number (number of protons)}$

$117 < 118 \Rightarrow \text{tennessine is listed first}$

Answer: Tennessine has fewer protons (117) than oganesson (118); elements are arranged in order of increasing number of protons (atomic number).

The Periodic Table lists elements in order of increasing atomic number, i.e., increasing number of protons, and tennessine has one fewer proton than oganesson.

Question 70 · 1 pt · constructed response

Compare the rate of the forward reaction to the rate of the reverse reaction for the equilibrium system at equilibrium represented by equation 1. Equation 1: $\text{CH}_4(g) + \text{H}_2\text{O}(g) + 210\text{ kJ} \rightleftharpoons \text{CO}(g) + 3\text{H}_2(g)$

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

Worked solution

$\text{At equilibrium, macroscopic properties are constant}$

$\text{rate}_{forward} = \text{rate}_{reverse}$

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

By definition of chemical equilibrium, the forward and reverse reactions occur at equal rates so concentrations remain constant.

Question 71 · 1 pt · constructed response

State how the equilibrium shifts when the temperature of the equilibrium system at equilibrium represented by equation 1 is increased. Equation 1: $\text{CH}_4(g) + \text{H}_2\text{O}(g) + 210\text{ kJ} \rightleftharpoons \text{CO}(g) + 3\text{H}_2(g)$

Model answer: The equilibrium shifts to the right (toward the products), favoring the forward reaction.

Worked solution

$\text{Energy (210 kJ) is a reactant} \Rightarrow \text{forward reaction is endothermic}$

$\text{Increasing temperature adds heat, favoring the endothermic direction}$

$\text{Equilibrium shifts toward the products (right), increasing forward reaction rate/yield}$

Answer: The equilibrium shifts to the right (toward the products), favoring the forward reaction.

Since the forward reaction absorbs energy (endothermic), adding heat by raising the temperature shifts the equilibrium to the right, favoring formation of more CO and H2.

Question 72 · 1 pt · constructed response

State, in terms of activation energy, why the catalyst increases the rate of the forward reaction represented in equation 1.

Model answer: The catalyst provides an alternate reaction pathway that has a lower activation energy for the forward reaction, so more particles have enough energy to react, increasing the rate.

Worked solution

$\text{Catalyst provides an alternate pathway with lower activation energy } (E_a)$

$\text{Lower } E_a \Rightarrow \text{more collisions have sufficient energy to react}$

$\Rightarrow \text{rate of forward reaction increases}$

Answer: The catalyst provides an alternate reaction pathway that has a lower activation energy for the forward reaction, so more particles have enough energy to react, increasing the rate.

A catalyst lowers the activation energy needed for the forward reaction, so a greater fraction of colliding particles have enough energy to react, speeding up the reaction.

Question 73 · 1 pt · constructed response

On the labeled axes in your answer booklet, draw a potential energy diagram for the forward reaction represented in equation 2. Equation 2: $\text{CO}(g) + \text{H}_2\text{O}(g) \rightleftharpoons \text{CO}_2(g) + \text{H}_2(g) + \text{energy}$

Model answer: A curve starting at the reactants' potential energy (CO + H2O), rising to a peak (the activated complex, representing the activation energy of the forward reaction), then descending to a final potential energy for the products (CO2 + H2) that is lower than the starting reactant energy, showing that the forward reaction is exothermic (net energy released).

Worked solution

$\text{Since energy is a product, the forward reaction is exothermic}$

$PE_{reactants} > PE_{products}$

$\text{Curve rises from reactants to a peak (activated complex), then falls to products, ending lower than the start}$

Answer: A curve starting at the reactants' potential energy (CO + H2O), rising to a peak (the activated complex, representing the activation energy of the forward reaction), then descending to a final potential energy for the products (CO2 + H2) that is lower than the starting reactant energy, showing that the forward reaction is exothermic (net energy released).

Because equation 2 releases energy in the forward direction, the potential energy diagram must show the products at a lower energy level than the reactants, with a hump above the reactants representing the activation energy.

Question 74 · 1 pt · constructed response

State the form of energy that is converted to electrical energy during the operation of this voltaic cell.

Model answer: Chemical energy (chemical potential energy) is converted to electrical energy.

Worked solution

$\text{Voltaic cells convert chemical energy released by a spontaneous redox reaction into electrical energy}$

Answer: Chemical energy (chemical potential energy) is converted to electrical energy.

In a voltaic cell, the spontaneous redox reaction between Fe(s) and Cu2+(aq) releases chemical energy, which is converted into electrical energy that powers the voltmeter/circuit.

Question 75 · 1 pt · constructed response

Identify one metal from Table J that is more easily oxidized than copper and less easily oxidized than iron.

Excerpt from Reference Table J, Activity Series of Metals (most easily oxidized at the top, least easily oxidized at the bottom): | Metals (most active to least active) | |---| | ... | | Zn | | Cr | | Fe | | Co | | Ni | | Sn | | Pb | | H2 | | Cu | | Ag | | Au |

Model answer: Lead (Pb) is an acceptable answer (other acceptable answers from Table J include cadmium, cobalt, nickel, or tin).

Worked solution

$\text{In Table J activity series, order (most to least easily oxidized) includes: } Fe > Cd > Co > Ni > Sn > Pb > (H_2) > Cu$

$\text{Any metal listed between Fe and Cu satisfies the condition}$

Answer: Lead (Pb) is an acceptable answer (other acceptable answers from Table J include cadmium, cobalt, nickel, or tin).

Table J's activity series lists metals from most to least easily oxidized; any metal appearing between iron and copper on that list (such as lead, tin, nickel, cobalt, or cadmium) is more easily oxidized than copper but less easily oxidized than iron.

Question 76 · 1 pt · constructed response

State the purpose of the salt bridge in completing the circuit in this cell.

Model answer: The salt bridge allows ions to flow between the two half-cells to maintain electrical neutrality (balance the charge) in each half-cell, completing the circuit.

Worked solution

$\text{As oxidation/reduction proceed, charge builds up in each half-cell}$

$\text{Salt bridge allows ion migration between half-cells to balance/neutralize this charge}$

$\Rightarrow \text{completes the circuit, allowing current to continue flowing}$

Answer: The salt bridge allows ions to flow between the two half-cells to maintain electrical neutrality (balance the charge) in each half-cell, completing the circuit.

The salt bridge permits the flow of ions between the two half-cells, which maintains electrical neutrality in each half-cell and completes the internal circuit of the voltaic cell.

Question 77 · 1 pt · constructed response

Write a balanced equation for the half-reaction that occurs in the copper half-cell.

A copper electrode in Cu(NO3)2(aq) is connected by a wire and salt bridge to an iron electrode in Fe(NO3)2(aq), forming a voltaic cell. Copper is less active than iron, so reduction occurs at the copper electrode.

Model answer: $\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)$

Worked solution

$\text{Overall equation: } Fe(s) + Cu^{2+}(aq) \rightarrow Fe^{2+}(aq) + Cu(s)$

$\text{Cu}^{2+} \text{ gains electrons (is reduced) at the copper electrode}$

$Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$

Answer: $\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)$

In the overall reaction, copper(II) ions gain two electrons to form solid copper at the copper electrode, which is the reduction half-reaction occurring in the copper half-cell.

Question 78 · 1 pt · constructed response

State the color of the indicator litmus in a sample of the original NaOH(aq) solution.

Model answer: Blue

Worked solution

$\text{Original NaOH(aq) has pH} = 12.00, \text{ which is basic}$

$\text{Litmus turns blue in basic solutions}$

Answer: Blue

Since the NaOH solution has a pH of 12.00 (basic), litmus indicator turns blue in this solution.

Question 79 · 1 pt · constructed response

Determine the pH value of a solution that has a hydronium ion concentration 100 times greater than the original NaOH(aq) solution.

The original NaOH(aq) solution has a pH of 12.00.

Model answer: pH = 10.00

Worked solution

$\text{Original solution: pH} = 12.00 \Rightarrow [H_3O^+] = 10^{-12} \text{ M}$

$\text{New } [H_3O^+] = 100 \times 10^{-12} = 10^{-10} \text{ M}$

$pH = -\log(10^{-10}) = 10.00$

Answer: pH = 10.00

Increasing the hydronium ion concentration by a factor of 100 (10^2) lowers the pH by 2 units, so the new pH is 12.00 minus 2, or 10.00.

Question 80 · 1 pt · constructed response

Identify the positive ion in the original HCl(aq) solution.

Model answer: Hydronium ion, $H_3O^+$ (or hydrogen ion, $H^+$).

Worked solution

$HCl(aq) \rightarrow H^+(aq) + Cl^-(aq)$

$\text{In aqueous solution, } H^+ \text{ associates with water to form } H_3O^+$

Answer: Hydronium ion, $H_3O^+$ (or hydrogen ion, $H^+$).

HCl is an Arrhenius acid that ionizes in water to produce hydronium ions (H3O+) as the positive ion and chloride ions as the negative ion.

Question 81 · 1 pt · constructed response

Determine the molarity of the HCl(aq) solution using the titration data.

Titration data: a 10.0 mL sample of the HCl(aq) is exactly neutralized by 20.0 mL of 0.010 M NaOH(aq).

Model answer: 0.020 M

Worked solution

$M_{acid} \times V_{acid} = M_{base} \times V_{base}$

$M_{HCl} \times 10.0\text{ mL} = 0.010\text{ M} \times 20.0\text{ mL}$

$M_{HCl} = \frac{0.010 \times 20.0}{10.0} = 0.020\text{ M}$

Answer: 0.020 M

Using the titration formula MaVa = MbVb with the given volumes of acid and base and the known base molarity, the HCl molarity is calculated to be 0.020 M.

Question 82 · 1 pt · constructed response

The half-life of Am-241 is 432.7 years. Determine the fraction of Am-241 remaining from an original sample after 865.4 years.

Model answer: 1/4 (0.25) of the original sample remains.

Worked solution

$\text{Number of half-lives} = \frac{865.4 \text{ yr}}{432.7 \text{ yr}} = 2.0$

$\text{Fraction remaining} = \left(\frac{1}{2}\right)^{2.0} = \frac{1}{4}$

Answer: 1/4 (0.25) of the original sample remains.

Dividing the elapsed time by the half-life of Am-241 gives exactly 2 half-lives, so the fraction of the original sample remaining is (1/2)^2 = 1/4.

Question 83 · 1 pt · constructed response

Compare the penetrating power of the alpha particles to the penetrating power of the gamma radiation.

Model answer: Alpha particles have much less penetrating power than gamma radiation.

Worked solution

$\text{Alpha particles: large mass, } +2 \text{ charge} \Rightarrow \text{low penetrating power (stopped by paper/skin)}$

$\text{Gamma radiation: high-energy electromagnetic radiation} \Rightarrow \text{very high penetrating power (requires thick lead/concrete to stop)}$

Answer: Alpha particles have much less penetrating power than gamma radiation.

Alpha particles are relatively large and charged, so they penetrate matter poorly, while gamma radiation is high-energy electromagnetic radiation with much greater penetrating power.

Question 84 · 1 pt · constructed response

Complete the nuclear equation for the alpha decay of Pu-238 by writing a notation for the missing nuclide.

Model answer: $^{234}_{92}\text{U}$ (uranium-234)

Worked solution

$^{238}_{94}Pu \rightarrow ^{234}_{92}X + ^{4}_{2}He$

$\text{Mass number: } 238 - 4 = 234$

$\text{Atomic number: } 94 - 2 = 92 \Rightarrow \text{element with } Z=92 \text{ is uranium (U)}$

Answer: $^{234}_{92}\text{U}$ (uranium-234)

In alpha decay, the parent nuclide loses a helium-4 nucleus, reducing its mass number by 4 and atomic number by 2; for Pu-238 this yields U-234 as the missing nuclide.

Question 85 · 1 pt · constructed response

Compare the energy released by the radioactive decay of 1.0 kilogram of Pu-238 in this nuclear power source to the energy released by 1.0 kilogram of chemical reactants in a battery.

Model answer: The radioactive decay of 1.0 kg of Pu-238 releases far more (much greater) energy than the chemical reaction of 1.0 kg of battery reactants.

Worked solution

$\text{Nuclear reactions release energy from changes in the nucleus (mass-energy conversion), on the order of millions of times greater per unit mass than chemical reactions}$

$\text{Chemical reactions (e.g., in a battery) release energy only from rearrangement of electrons/bonds, a much smaller amount}$

$\Rightarrow \text{Nuclear decay of 1.0 kg Pu-238 releases much more energy than 1.0 kg of chemical battery reactants}$

Answer: The radioactive decay of 1.0 kg of Pu-238 releases far more (much greater) energy than the chemical reaction of 1.0 kg of battery reactants.

Nuclear decay releases energy from changes within the nucleus itself, which is orders of magnitude greater per given mass than the energy released by chemical reactions, which involve only electron/bond rearrangements.

June 2024 Chemistry Regents — Answer Key & Worked Solutions

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