January 2025 Chemistry Regents — Answer Key & Worked Solutions

Every question from the January 2025 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.

Practice Chemistry free Score calculator

These are real questions from the January 2025 Chemistry New York State Regents examination, published by NYSED. Worked solutions are written by RegentsLab.

Question 1 · 2 pts

The subatomic particles in the nucleus of an oxygen atom include

(A) electrons, only
(B) neutrons, only
(C) protons and neutrons ✓
(D) protons and electrons

Worked solution

$\text{Nucleus contains protons (+) and neutrons (neutral)}$

$\text{Electrons orbit outside the nucleus}$

Answer: protons and neutrons

The nucleus of an atom is composed of protons and neutrons; electrons occupy the space outside the nucleus.

Question 2 · 2 pts

Which phrase describes the charges and numbers of protons and electrons in an atom?

(A) have opposite charges and are equal in number ✓
(B) have opposite charges and are unequal in number
(C) have the same charge and are equal in number
(D) have the same charge and are unequal in number

Worked solution

$p^+ \text{ charge} = +1,\ e^- \text{ charge} = -1$

$\text{In a neutral atom, } \#p^+ = \#e^-$

Answer: have opposite charges and are equal in number

Protons are positive and electrons are negative, so they have opposite charges; a neutral atom has equal numbers of each.

Question 3 · 2 pts

In the wave-mechanical model, which term identifies a region in an atom of lithium that is the most probable location of electrons in the ground state?

(A) hard sphere
(B) polar bond
(C) orbital ✓
(D) spectrum

Worked solution

$\text{Wave-mechanical model: electrons exist in orbitals}$

$\text{An orbital is a region of highest probability of finding an electron}$

Answer: orbital

In the wave-mechanical model, an orbital is defined as the region around the nucleus where an electron is most likely to be found.

Question 4 · 2 pts

An atom of C-12 in the ground state and an atom of C-13 in the ground state are defined as isotopes of carbon because these atoms have

(A) the same number of neutrons
(B) the same number of electron shells
(C) a different number of neutrons ✓
(D) a different number of electron shells

Worked solution

$\text{Isotopes: same protons, different neutrons}$

$C\text{-}12: 6\ n^0,\quad C\text{-}13: 7\ n^0$

Answer: a different number of neutrons

Isotopes of the same element have the same number of protons but different numbers of neutrons, which is why C-12 and C-13 are isotopes of carbon.

Question 5 · 2 pts

At STP, which property can be used to differentiate between a 2.0-gram sample of oxygen, $O_2(g)$, and a 2.0-gram sample of helium, $He(g)$?

(A) mass
(B) temperature
(C) reaction with sodium ✓
(D) phase of each sample

Worked solution

$\text{Mass, temperature, and phase (both gases) are identical properties given}$

$O_2 \text{ reacts with Na to form } Na_2O; \ He \text{ is inert and does not react}$

Answer: reaction with sodium

Since both samples have the same mass and are gases at STP, only a chemical property, such as reactivity with sodium, distinguishes reactive oxygen from inert helium.

Question 6 · 2 pts

How do the molecular structures and properties of oxygen, $O_2$, and ozone, $O_3$, compare?

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

Worked solution

$O_2 \text{ is diatomic; } O_3 \text{ is triatomic (different structure)}$

$\text{Different structures} \Rightarrow \text{different physical/chemical properties}$

Answer: They have different molecular structures and different properties.

O2 and O3 are allotropes of oxygen with different numbers of atoms per molecule, giving them different structures and correspondingly different properties.

Question 7 · 2 pts

Which substance can be broken down by chemical means?

(A) aluminum
(B) ammonia ✓
(C) argon
(D) arsenic

Worked solution

$\text{Aluminum, argon, arsenic are elements (cannot be broken down chemically)}$

$\text{Ammonia, } NH_3, \text{ is a compound and can be decomposed into } N_2 \text{ and } H_2$

Answer: ammonia

Only compounds can be broken down into simpler substances by chemical means; ammonia is the only compound listed.

Question 8 · 2 pts

Which type of bonding is present in a sample of zinc at STP?

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

Worked solution

$\text{Zinc is a metal}$

$\text{Metals bond via a "sea" of delocalized valence electrons: metallic bonding}$

Answer: metallic bonding

Zinc is a metallic element, and metallic elements are held together by metallic bonding.

Question 9 · 2 pts

Which formula represents a nonpolar molecule?

(A) $CH_4$ ✓
(B) $HCl$
(C) $H_2O$
(D) $NH_3$

Worked solution

$CH_4 \text{ is symmetric (tetrahedral)} \Rightarrow \text{bond dipoles cancel}$

$HCl, H_2O, NH_3 \text{ have asymmetric shapes/unshared pairs} \Rightarrow \text{polar}$

Answer: $CH_4$

Methane's tetrahedral symmetry causes the individual bond dipoles to cancel, making the molecule nonpolar, unlike the other choices.

Question 10 · 2 pts

Given the equation representing a reaction: $F + F \rightarrow F_2$. What occurs during this reaction?

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

Worked solution

$\text{Forming a bond is exothermic}$

$F + F \rightarrow F_2 \text{ forms a new covalent bond, releasing energy}$

Answer: Energy is released as a bond is formed.

Bond formation always releases energy (exothermic), which is what occurs as two fluorine atoms combine to form F2.

Question 11 · 2 pts

Based on Table S, an atom of which element in Group 16 has the greatest attraction for electrons in a chemical bond?

(A) oxygen ✓
(B) sulfur
(C) selenium
(D) tellurium

Worked solution

$\text{Electronegativity decreases down a group}$

$\text{Oxygen is highest in Group 16, so it has the greatest electronegativity}$

Answer: oxygen

Electronegativity, the attraction for electrons in a bond, decreases going down a group, so oxygen (topmost in Group 16) has the greatest electronegativity.

Question 12 · 2 pts

A solution is made by completely dissolving a sample of sodium hydrogen carbonate in water. This solution is classified as a

(A) heterogeneous compound
(B) heterogeneous mixture
(C) homogeneous compound
(D) homogeneous mixture ✓

Worked solution

$\text{A solution has uniform composition throughout} \Rightarrow \text{homogeneous}$

$\text{A solution is a mixture (variable composition, physically combined), not a compound}$

Answer: homogeneous mixture

A solution formed by completely dissolving a solute in a solvent is uniform throughout (homogeneous) and is a mixture, not a compound.

Question 13 · 2 pts

In which sample of matter can the proportion of its components be varied?

(A) $Cl_2(g)$
(B) $Br_2(\ell)$
(C) $KCl(s)$
(D) $KBr(aq)$ ✓

Worked solution

$Cl_2, Br_2 \text{ are elements; } KCl(s) \text{ is a pure compound (fixed ratio)}$

$KBr(aq) \text{ is a mixture (solution)} \Rightarrow \text{variable proportions of KBr and water}$

Answer: $KBr(aq)$

Only mixtures, such as an aqueous solution like KBr(aq), have components whose proportions can vary; pure substances have fixed composition.

Question 14 · 2 pts

According to the kinetic molecular theory, the particles of an ideal gas have

(A) attractive forces between them
(B) random, constant, straight-line motion ✓
(C) a greater size than the distance between them
(D) collisions that increase the total energy of gas particles

Worked solution

$\text{KMT assumption: negligible volume, no attractive forces}$

$\text{Particles move in random, constant, straight-line motion between elastic collisions}$

Answer: random, constant, straight-line motion

The kinetic molecular theory assumes ideal gas particles move randomly, constantly, and in straight lines, with negligible volume and no attractive forces.

Question 15 · 2 pts

A chemical reaction is most likely to occur when reactant particles collide with the proper orientation and proper

(A) charge
(B) energy ✓
(C) mass
(D) volume

Worked solution

$\text{Effective collisions require proper orientation and sufficient (activation) energy}$

Answer: energy

Collision theory states that a reaction occurs when particles collide with correct orientation and with at least the activation energy required.

Question 16 · 2 pts

At STP, which sample contains the same number of molecules as 44 liters of $NO_2(g)$ at STP?

(A) 22 L of $CO_2(g)$
(B) 44 L of $O_2(g)$ ✓
(C) 66 L of $O_2(g)$
(D) 88 L of $O_2(g)$

Worked solution

$\text{Avogadro's principle: equal volumes of gas at same T, P contain equal numbers of molecules}$

$44\ L\ NO_2 \text{ has the same molecule count as } 44\ L\ O_2 \text{ at STP}$

Answer: 44 L of $O_2(g)$

By Avogadro's principle, equal volumes of any gas at the same temperature and pressure contain equal numbers of molecules, so 44 L of O2 matches 44 L of NO2.

Question 17 · 2 pts

Based on Table I, which chemical change is exothermic?

(A) $N_2(g) + O_2(g) \rightarrow 2NO(g)$
(B) $N_2(g) + 2O_2(g) \rightarrow 2NO_2(g)$
(C) $2C(s) + 3H_2(g) \rightarrow C_2H_6(g)$ ✓
(D) $2C(s) + 2H_2(g) \rightarrow C_2H_4(g)$

Worked solution

$\text{Table I lists } \Delta H \text{ for each reaction}$

$\text{Formation of } C_2H_6(g) \text{ from elements has a negative } \Delta H \text{ (exothermic)}$

Answer: $2C(s) + 3H_2(g) \rightarrow C_2H_6(g)$

Table I shows the formation of ethane, C2H6, from its elements releases energy (negative delta H), making it the exothermic reaction among the choices.

Question 18 · 2 pts

Which term identifies a factor that affects the rate of a chemical reaction?

(A) decay mode
(B) isotopic abundance
(C) nuclear stability
(D) surface area ✓

Worked solution

$\text{Reaction rate factors: concentration, temperature, surface area, catalyst, pressure}$

Answer: surface area

Surface area is one of the well-known factors that affects reaction rate; the other choices relate to nuclear chemistry, not reaction kinetics.

Question 19 · 2 pts

Which phrase represents the net amount of energy released or absorbed during a chemical reaction?

(A) the kinetic energy of the reactants plus the potential energy of the products
(B) the kinetic energy of the reactants minus the potential energy of the products
(C) the potential energy of the products plus the potential energy of the reactants
(D) the potential energy of the products minus the potential energy of the reactants ✓

Worked solution

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

Answer: the potential energy of the products minus the potential energy of the reactants

The net energy change (heat of reaction) equals the potential energy of the products minus the potential energy of the reactants.

Question 20 · 2 pts

A catalyst increases the rate of a chemical reaction by providing an alternate reaction pathway that has a

(A) higher activation energy
(B) lower activation energy ✓
(C) higher potential energy for the reactants
(D) lower potential energy for the reactants

Worked solution

$\text{Catalysts lower the activation energy } (E_a) \text{ pathway}$

$\text{Reactant/product potential energies (and } \Delta H\text{) are unchanged}$

Answer: lower activation energy

A catalyst speeds up a reaction by offering an alternate pathway with lower activation energy, without changing the potential energy of reactants or products.

Question 21 · 2 pts

Systems in nature tend to undergo changes toward

(A) lower energy and less disorder
(B) lower energy and more disorder ✓
(C) higher energy and less disorder
(D) higher energy and more disorder

Worked solution

$\text{Natural systems favor minimum potential energy (stability)}$

$\text{Natural systems favor maximum entropy (disorder)}$

Answer: lower energy and more disorder

Systems in nature spontaneously tend toward states of lower energy (greater stability) and greater disorder (higher entropy).

Question 22 · 2 pts

Atoms of which element bond with each other to form chains, rings, and networks?

(A) calcium
(B) carbon ✓
(C) chlorine
(D) cobalt

Worked solution

$\text{Carbon has 4 valence electrons and can form 4 covalent bonds}$

$\text{This allows carbon to self-bond into chains, rings, and networks}$

Answer: carbon

Carbon's ability to form four covalent bonds with other carbon atoms allows it to create extensive chains, rings, and network structures, the basis of organic chemistry.

Question 23 · 2 pts

A molecule of which compound contains an $-OH$ functional group?

(A) propanamide
(B) 1-propanamine
(C) propanal
(D) 1-propanol ✓

Worked solution

$\text{propanamide has } -CONH_2, \text{ 1-propanamine has } -NH_2, \text{ propanal has } -CHO$

$\text{1-propanol is an alcohol, containing the } -OH \text{ functional group}$

Alcohols are the organic compound class characterized by the hydroxyl (-OH) functional group, and 1-propanol is an alcohol.

Question 24 · 2 pts

Which statement explains why butane and methylpropane are isomers of each other?

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

Worked solution

$\text{Both butane and methylpropane have formula } C_4H_{10}$

$\text{They differ in structural arrangement (straight chain vs branched)}$

Isomers share the same molecular formula but have different structural arrangements of atoms.

Question 25 · 2 pts

Molecules of which two substances each contain a multiple covalent bond?

(A) ethene and ethane
(B) ethene and ethyne ✓
(C) methane and ethane
(D) methane and ethyne

Worked solution

$\text{Ethene, } C_2H_4, \text{ has a carbon-carbon double bond}$

$\text{Ethyne, } C_2H_2, \text{ has a carbon-carbon triple bond}$

$\text{Ethane and methane contain only single bonds}$

Ethene (double bond) and ethyne (triple bond) each contain a multiple covalent bond, unlike the saturated hydrocarbons ethane and methane.

Question 26 · 2 pts

Which term identifies a type of organic reaction?

(A) deposition
(B) substitution ✓
(C) solidification
(D) vaporization

Worked solution

$\text{Deposition, solidification, and vaporization are phase changes}$

$\text{Substitution is an organic reaction type where an atom or group replaces another on a hydrocarbon}$

Substitution is a classic organic reaction type; the other three choices are phase changes, not chemical reactions.

Question 27 · 2 pts

Which statement identifies where oxidation and reduction occur in an electrochemical cell?

(A) Both oxidation and reduction occur at the anode.
(B) Both oxidation and reduction occur at the cathode.
(C) Oxidation occurs at the anode and reduction occurs at the cathode. ✓
(D) Oxidation occurs at the cathode and reduction occurs at the anode.

Worked solution

$\text{By convention, oxidation (loss of electrons) occurs at the anode}$

$\text{Reduction (gain of electrons) occurs at the cathode}$

In any electrochemical cell (voltaic or electrolytic), oxidation always occurs at the anode and reduction always occurs at the cathode.

Question 28 · 2 pts

Which type of reaction yields a salt and water as the only products?

(A) combustion
(B) fermentation
(C) neutralization ✓
(D) polymerization

Worked solution

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

A neutralization reaction between an acid and a base produces a salt and water as the only products.

Question 29 · 2 pts

In which process are heavy nuclei split into lighter nuclei?

(A) nuclear fission ✓
(B) nuclear fusion
(C) single replacement
(D) double replacement

Worked solution

$\text{Fission: a heavy nucleus splits into two or more lighter nuclei}$

Nuclear fission is the process in which a heavy nucleus splits into lighter nuclei, releasing energy.

Question 30 · 2 pts

Which statement explains the large amount of heat produced during a fission reaction?

(A) Chemical bonds are broken.
(B) Chemical bonds are formed.
(C) Energy is converted to mass.
(D) Mass is converted to energy. ✓

Worked solution

$E = mc^2$

$\text{A small amount of mass is lost and converted into a very large amount of energy}$

Nuclear reactions release enormous energy because a small amount of mass is converted directly into energy, according to Einstein's mass-energy equivalence.

Question 31 · 2 pts

What is the approximate mass of an ion that has 12 protons, 13 neutrons, and 10 electrons?

(A) 22 u
(B) 23 u
(C) 25 u ✓
(D) 35 u

Worked solution

$\text{Mass number} = \text{protons} + \text{neutrons} = 12 + 13 = 25$

$\text{Electrons contribute negligible mass, so mass} \approx 25\text{ u}$

The mass of an atom or ion is approximately the sum of its protons and neutrons; the number of electrons (which determines charge) does not significantly affect mass.

Question 32 · 2 pts

Given the bright-line spectrum of element $X$ and the spectra of four mixtures: Which mixture contains element $X$?

(A) Mixture A
(B) Mixture B ✓
(C) Mixture C
(D) Mixture D

Worked solution

$\text{A mixture contains element } X \text{ if all of element } X\text{'s spectral lines appear at matching wavelengths within that mixture's spectrum}$

Mixture B is the one whose set of spectral lines includes every line present in element X's bright-line spectrum at the same wavelengths.

Question 33 · 2 pts

What is the number of valence electrons in an atom in the ground state with an electron configuration of 2-8-18-3?

(A) 18
(B) 2
(C) 3 ✓
(D) 31

Worked solution

$\text{Total electrons} = 2+8+18+3 = 31 \text{ (gallium)}$

$\text{Valence electrons are those in the outermost occupied shell} = 3$

Valence electrons are the electrons in the highest occupied energy level, which here is the last number in the configuration, 3.

Question 34 · 2 pts

Which statement describes the general trends in electronegativity and atomic radius as the first seven elements in Period 3 are considered in order of increasing atomic number?

(A) Electronegativity decreases and atomic radius decreases.
(B) Electronegativity decreases and atomic radius increases.
(C) Electronegativity increases and atomic radius decreases. ✓
(D) Electronegativity increases and atomic radius increases.

Worked solution

$\text{Across a period, increasing nuclear charge pulls electrons closer}$

$\text{Atomic radius decreases while electronegativity increases}$

Moving across a period, increasing nuclear charge draws the valence electrons closer, so atomic radius decreases while electronegativity increases.

Question 35 · 2 pts

What is a chemical name for $KClO_2$?

(A) potassium chlorate
(B) potassium chlorite ✓
(C) potassium hypochlorite
(D) potassium perchlorate

Worked solution

$ClO_2^- \text{ is the chlorite ion}$

$KClO_2 = \text{potassium chlorite}$

The polyatomic ion ClO2- is named chlorite, so KClO2 is potassium chlorite (compare to hypochlorite ClO-, chlorate ClO3-, perchlorate ClO4-).

Question 36 · 2 pts

What is the percent by mass of oxygen in $CH_3COOH$ (gram-formula mass = 60. g/mol)?

(A) 27%
(B) 32%
(C) 53% ✓
(D) 60.%

Worked solution

$\text{mass of O} = 2 \times 16 \text{ g/mol} = 32 \text{ g/mol}$

$\%O = \frac{32}{60} \times 100\% \approx 53\%$

Acetic acid contains two oxygen atoms (32 g/mol) out of a total gram-formula mass of 60 g/mol, giving about 53% oxygen by mass.

Question 37 · 2 pts

Given the formula for a compound: $H-C(=O)-H$ (formaldehyde, with H, H, and =O all bonded to the central C) How many electrons are shared between the carbon atom and the oxygen atom in a molecule of this compound?

(A) 1
(B) 2
(C) 3
(D) 4 ✓

Worked solution

$\text{The C=O bond is a double bond}$

$\text{A double bond consists of 2 shared electron pairs} = 4 \text{ electrons}$

The carbon-oxygen double bond shown in the structural formula consists of two shared pairs of electrons, totaling 4 shared electrons.

Question 38 · 2 pts

Which particle model diagram represents the arrangement of particles in a sample of neon at STP?

(A) Diagram (1) ✓
(B) Diagram (2)
(C) Diagram (3)
(D) Diagram (4)

Worked solution

$\text{Neon is a noble gas; at STP it exists as a gas}$

$\text{Gas particles are far apart and randomly arranged, filling the container}$

At STP, neon is a gas, so its particles should be widely and randomly spaced throughout the container, which matches diagram 1, not the closely packed arrangements of diagrams 3 and 4.

Question 39 · 2 pts

The volume of a 150.0-gram sample of copper at 298 K and 101.3 kPa is 16.7 $cm^3$. The mass of the copper sample is expressed to how many significant figures?

(A) 1
(B) 2
(C) 3
(D) 4 ✓

Worked solution

$150.0 \text{ g has a decimal point, so all digits including trailing zero are significant}$

$1,5,0,0 \Rightarrow 4 \text{ significant figures}$

Because 150.0 g contains a decimal point after trailing zeros, all four digits are significant.

Question 40 · 2 pts

Which statement describes the energy transfer that occurs when a sample of copper at 100.°C is placed in a sample of water at 20.°C?

(A) Thermal energy is transferred from the copper to the water. ✓
(B) Thermal energy is transferred from the water to the copper.
(C) Chemical energy is transferred from the copper to the water.
(D) Chemical energy is transferred from the water to the copper.

Worked solution

$\text{Heat flows spontaneously from a warmer object to a cooler object}$

$\text{Copper (100°C) is warmer than water (20°C), so thermal energy flows from copper to water}$

Thermal energy always flows spontaneously from a higher-temperature object to a lower-temperature object, so it transfers from the hot copper to the cooler water.

Question 41 · 2 pts

Which change will cause an open flask half-filled with water at 295 K and 1.0 atm to reach phase equilibrium?

(A) increase the pressure
(B) increase the temperature
(C) remove some of the water
(D) seal the flask with a stopper ✓

Worked solution

$\text{In an open flask, vapor escapes continuously, preventing equilibrium}$

$\text{Sealing the flask traps vapor so the rate of evaporation equals the rate of condensation}$

Phase equilibrium between liquid and vapor requires a closed system so vapor cannot escape; sealing the flask allows evaporation and condensation rates to equalize.

Question 42 · 2 pts

What is the amount of heat released when 1.0 mole of $NH_3(g)$ is formed from its elements at 101.3 kPa and 298 K?

(A) 45.9 kJ ✓
(B) 91.8 kJ
(C) 137.7 kJ
(D) 183.6 kJ

Worked solution

$\tfrac{1}{2}N_2(g) + \tfrac{3}{2}H_2(g) \rightarrow NH_3(g), \quad \Delta H = -45.9\text{ kJ (Table I)}$

$\text{For 1.0 mole of } NH_3 \text{ formed, } 45.9\text{ kJ of heat is released}$

According to Reference Table I, the formation of one mole of NH3(g) from its elements releases 45.9 kJ of heat.

Question 43 · 2 pts

Given the equation representing a phase change for carbon dioxide, $CO_2(s)$: $CO_2(s) + \text{energy} \rightarrow CO_2(g)$ Which statement describes this phase change in terms of energy and entropy?

(A) The phase change is endothermic and entropy increases. ✓
(B) The phase change is endothermic and entropy decreases.
(C) The phase change is exothermic and entropy increases.
(D) The phase change is exothermic and entropy decreases.

Worked solution

$\text{Energy is a reactant (absorbed), so the process is endothermic}$

$\text{Going from solid to gas increases disorder, so entropy increases}$

Since energy is absorbed (sublimation) and the substance changes from a highly ordered solid to a disordered gas, this phase change is endothermic with increasing entropy.

Question 44 · 2 pts

Which equation represents an oxidation half-reaction?

(A) $Br_2 \rightarrow 2Br^- + 2e^-$
(B) $Br_2 + 2e^- \rightarrow 2Br^-$
(C) $Mn^{2+} + 2e^- \rightarrow Mn$
(D) $Mn \rightarrow Mn^{2+} + 2e^-$ ✓

Worked solution

$\text{Oxidation is the loss of electrons, shown with electrons as a product}$

$Mn \rightarrow Mn^{2+} + 2e^-: \text{ oxidation state goes from } 0 \text{ to } +2, \text{ losing 2 electrons}$

Oxidation is the loss of electrons; in this equation manganese metal loses two electrons to form Mn2+, correctly showing electrons as a product.

Question 45 · 2 pts

Which equation represents a redox reaction?

(A) $C(s) + O_2(g) \rightarrow CO_2(g)$ ✓
(B) $CaO(s) + H_2O(\ell) \rightarrow Ca(OH)_2(s)$
(C) $MgCO_3(s) \rightarrow MgO(s) + CO_2(g)$
(D) $Cu(OH)_2(s) \rightarrow CuO(s) + H_2O(\ell)$

Worked solution

$C(s):\ 0 \rightarrow +4\ \text{in } CO_2$ carbon is oxidized, losing electrons

$O_2(g):\ 0 \rightarrow -2\ \text{in } CO_2$ oxygen is reduced, gaining electrons

$\text{The other three reactions show no change in oxidation number for any atom}$

Answer: $C(s) + O_2(g) \rightarrow CO_2(g)$

A redox reaction requires a change in oxidation number for at least one element. Only in the combustion of carbon do the oxidation states of C (0 to +4) and O (0 to -2) change; the other three are synthesis/decomposition reactions with no oxidation-state changes.

Question 46 · 2 pts

Based on Table J, which metal will react spontaneously with $Fe^{2+}(aq)$ ions?

(A) copper
(B) gold
(C) magnesium ✓
(D) silver

Worked solution

$\text{Table J ranks metals by activity (reducing strength)}$

$\text{A metal higher in the activity series will spontaneously reduce the ion of a metal lower in the series}$

$Mg \text{ is above } Fe \text{ on Table J, but Cu, Ag, and Au are below Fe}$

Answer: magnesium

On Table J, a free metal reacts spontaneously with the ion of a less active metal listed below it. Magnesium is above iron in the activity series, so it will spontaneously displace Fe2+ ions; copper, silver, and gold are all below iron and will not.

Question 47 · 2 pts

Based on Table F, a saturated solution of which compound is the best conductor of electricity?

(A) $CaCO_3$
(B) $BaSO_4$
(C) $Mg(OH)_2$
(D) $NaNO_3$ ✓

Worked solution

$\text{Table F (solubility guidelines) shows carbonates, sulfates of Ba, and hydroxides of Mg are generally insoluble/low solubility}$

$\text{Nitrate compounds are soluble in water}$

$\text{A saturated } NaNO_3 \text{ solution has a much higher concentration of dissolved ions than the other saturated solutions}$

Answer: $NaNO_3$

According to Table F's solubility guidelines, nearly all nitrate salts are soluble, while carbonates, most sulfates, and most hydroxides are of low solubility. A saturated NaNO3 solution therefore contains far more free-moving ions, making it the best electrical conductor of the four.

Question 48 · 2 pts

Given the equation representing a system at equilibrium: $HSO_3^-(aq) + H_2O(\ell) \rightleftharpoons SO_3^{2-}(aq) + H_3O^+(aq)$ According to one acid-base theory

(A) $HSO_3^-$ acts as an acid because it is an $H^+$ donor ✓
(B) $HSO_3^-$ acts as an acid because it is an $H^+$ acceptor
(C) $H_2O$ acts as an acid because it is an $H^+$ donor
(D) $H_2O$ acts as an acid because it is an $H^+$ acceptor

Worked solution

$HSO_3^-(aq) \rightarrow SO_3^{2-}(aq) + H^+$ HSO3- loses a hydrogen ion, becoming SO3(2-)

$H_2O(\ell) + H^+ \rightarrow H_3O^+(aq)$ water gains the hydrogen ion, acting as the base

$\text{By the Bronsted-Lowry theory, the species that donates } H^+ \text{ is the acid}$

Answer: $HSO_3^-$ acts as an acid because it is an $H^+$ donor

In the Bronsted-Lowry acid-base theory, the acid is the species that donates a hydrogen ion. HSO3- loses an H+ to become SO3(2-), so HSO3- is acting as the acid (H+ donor), while H2O accepts the H+ to become H3O+, acting as the base.

Question 49 · 2 pts

What is the pH value of an aqueous solution with a hydronium ion concentration 1000 times greater than an aqueous solution with a pH value of 5.0?

(A) 1.0
(B) 2.0 ✓
(C) 8.0
(D) 14.0

Worked solution

$pH = 5.0 \Rightarrow [H_3O^+] = 1\times10^{-5}\ M$

$[H_3O^+]_{new} = 1000 \times 1\times10^{-5} = 1\times10^{-2}\ M$

$pH = -\log(1\times10^{-2}) = 2.0$

Answer: 2.0

Each factor-of-10 increase in hydronium ion concentration decreases pH by 1 unit. A 1000-fold (10^3) increase in concentration lowers the pH by 3 units, from 5.0 to 2.0.

Question 50 · 2 pts

Which energy source is paired with a risk associated with its usage?

(A) Nuclear fission produces stable isotopes.
(B) Nuclear fission produces radioactive waste. ✓
(C) Voltaic cells produce stable isotopes.
(D) Voltaic cells produce radioactive waste.

Worked solution

$\text{Nuclear fission splits large, unstable nuclei, producing fission products that are typically radioactive}$

$\text{This radioactive waste is a known environmental/health risk of nuclear fission energy}$

$\text{Voltaic cells involve chemical redox reactions and do not involve nuclear processes or radioactive waste}$

Answer: Nuclear fission produces radioactive waste.

Nuclear fission reactions commonly produce unstable, radioactive fission products that must be safely stored, representing a genuine risk of this energy source. Voltaic cells are ordinary chemical (redox) energy sources and produce no radioactive material.

Question 51 · 1 pt · constructed response

State, in terms of electrons and energy states, how the light emitted by excited boron atoms is produced.

Model answer: Electrons in excited boron atoms fall from a higher (excited) energy state to a lower energy state, and the energy lost is released as light (a photon).

Worked solution

$\text{An electron absorbs energy and moves to a higher (excited) energy state}$

$\text{The electron is unstable in the excited state and falls back down to a lower energy state}$

$\text{As it falls, the electron releases the excess energy as a photon of light}$

Answer: Electrons in excited boron atoms fall from a higher (excited) energy state to a lower energy state, and the energy lost is released as light (a photon).

Line spectra are produced when excited electrons drop from higher to lower energy levels, emitting the energy difference as light of a specific wavelength.

Question 52 · 1 pt · constructed response

Base your answer to question 52 on the information below and on your knowledge of chemistry. The atomic mass and natural abundance for the two naturally occurring isotopes of copper are shown in the table below. Show a numerical setup for calculating the atomic mass of the element copper.

Model answer: $(62.93)(0.6915) + (64.93)(0.3085) = 63.55\ u$

Worked solution

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

$(62.93\ u)(0.6915) + (64.93\ u)(0.3085)$

$= 43.52 + 20.03 = 63.55\ u$

Answer: $(62.93)(0.6915) + (64.93)(0.3085) = 63.55\ u$

The atomic mass of an element is the weighted average of the masses of its naturally occurring isotopes, weighted by their percent (fractional) abundances.

Question 53 · 1 pt · constructed response

A technician determined the percent natural abundance of Cu-65 in a sample to be 31.47%. Determine the percent error for this percent natural abundance of Cu-65 compared to the accepted value shown in the table.

Model answer: 2.0% (percent error)

Worked solution

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

$= \frac{|31.47 - 30.85|}{30.85} \times 100$

$= \frac{0.62}{30.85} \times 100 \approx 2.0\%$

Answer: 2.0% (percent error)

Percent error compares the measured value to the accepted (table) value, dividing the absolute difference by the accepted value and multiplying by 100.

Question 54 · 1 pt · constructed response

Base your answer to question 54 on the information below and on your knowledge of chemistry. The elements beryllium, magnesium, calcium, strontium, and barium are located in Group 2 on the Periodic Table. These Group 2 elements have both similarities and differences in their physical and chemical properties. Based on the location of Group 2 elements on the Periodic Table, state the element classification of these elements.

Model answer: They are classified as (alkaline earth) metals.

Worked solution

$\text{Group 2 elements are located on the left side of the Periodic Table (metals region)}$

$\text{This group is specifically named the alkaline earth metals}$

Answer: They are classified as (alkaline earth) metals.

Elements in Group 2 of the Periodic Table are metals, specifically known as the alkaline earth metals.

Question 55 · 1 pt · constructed response

Explain, in terms of electron shells, why the atomic radius of a calcium atom in the ground state is greater than the atomic radius of a magnesium atom in the ground state.

Model answer: A calcium atom in the ground state has more occupied electron shells (energy levels) than a magnesium atom, so its outermost electrons are farther from the nucleus, giving it a larger atomic radius.

Worked solution

$\text{Ca ground state electron configuration has one more occupied shell than Mg}$

$\text{More shells means the valence electrons are, on average, farther from the nucleus}$

$\text{Therefore the atomic radius of Ca is greater than that of Mg}$

Answer: A calcium atom in the ground state has more occupied electron shells (energy levels) than a magnesium atom, so its outermost electrons are farther from the nucleus, giving it a larger atomic radius.

Atomic radius increases down a group because each successive element has an additional occupied electron shell, placing the outermost electrons farther from the nucleus. Calcium has more electron shells than magnesium, giving it the larger radius.

Question 56 · 1 pt · constructed response

State the general trend in first ionization energy for the Group 2 elements, beryllium through barium, as these elements are considered in order of increasing atomic number.

Model answer: First ionization energy generally decreases as atomic number increases from beryllium to barium.

Worked solution

$\text{Down a group, atomic radius increases and electron shielding increases}$

$\text{Valence electrons are held less tightly (farther from nucleus, more shielding)}$

$\text{Therefore first ionization energy decreases going down the group}$

Answer: First ionization energy generally decreases as atomic number increases from beryllium to barium.

As atomic number increases down Group 2, additional electron shells increase the distance and shielding between the nucleus and valence electrons, making them easier to remove, so first ionization energy decreases.

Question 57 · 1 pt · constructed response

Base your answers to questions 57 through 59 on the information below and on your knowledge of chemistry. Diagram 1 represents a sample of N2(g) and the conditions of pressure, volume, and temperature of the gas in a sealed, rigid cylinder with a movable piston. Diagram 2 represents a sample of O2(g) and the conditions of pressure, volume, and temperature of the gas in a sealed, rigid cylinder with a movable piston. Diagram 1: N2(g), P = 60.0 kPa, V = 1.8 L, T = 300. K Diagram 2: O2(g), P = 60.0 kPa, V = 1.8 L, T = 300. K Compare the average kinetic energy of the gas molecules in diagram 1 to the average kinetic energy of the gas molecules in diagram 2.

Model answer: The average kinetic energy of the gas molecules in diagram 1 is equal to the average kinetic energy of the gas molecules in diagram 2.

Worked solution

$\text{Average kinetic energy of a gas depends only on temperature (Kelvin)}$

$T_1 = T_2 = 300.\ K$

$\Rightarrow KE_{avg,1} = KE_{avg,2}$

Answer: The average kinetic energy of the gas molecules in diagram 1 is equal to the average kinetic energy of the gas molecules in diagram 2.

Average kinetic energy of gas particles is directly related to Kelvin temperature alone, regardless of the identity of the gas. Since both samples are at 300. K, their average kinetic energies are equal.

Question 58 · 1 pt · constructed response

Determine the volume of the O2(g) in diagram 2 when conditions are changed to STP.

Model answer: approximately 0.970 L

Worked solution

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

$V_2 = \frac{P_1V_1T_2}{T_1P_2} = \frac{(60.0\ kPa)(1.8\ L)(273\ K)}{(300.\ K)(101.3\ kPa)}$

$V_2 \approx 0.970\ L$

Answer: approximately 0.970 L

Using the combined gas law with STP defined as 101.3 kPa and 273 K, solving for the new volume gives approximately 0.970 L.

Question 59 · 1 pt · constructed response

Determine the pressure, in atmospheres, of the nitrogen gas in diagram 1.

Model answer: approximately 0.592 atm

Worked solution

$1\ atm = 101.3\ kPa$

$P = 60.0\ kPa \times \frac{1\ atm}{101.3\ kPa}$

$P \approx 0.592\ atm$

Answer: approximately 0.592 atm

Converting the given pressure from kPa to atm using the equivalence 101.3 kPa = 1 atm gives approximately 0.592 atm.

Question 60 · 1 pt · constructed response

Base your answers to questions 60 through 62 on the information below and on your knowledge of chemistry. The boiling point for a sample of a molecular compound is 80.°C at standard pressure. The heat of fusion, Hf, of this compound is 127 joules per gram. Compare the potential energy of the molecules in the sample before vaporization to the potential energy of the molecules after vaporization.

Model answer: The potential energy of the molecules after vaporization (gas) is greater than the potential energy of the molecules before vaporization (liquid).

Worked solution

$\text{Vaporization is an endothermic phase change (energy is absorbed)}$

$\text{During a phase change, absorbed energy increases potential energy while temperature (average KE) stays constant}$

$\text{Therefore PE(gas, after) > PE(liquid, before)}$

Answer: The potential energy of the molecules after vaporization (gas) is greater than the potential energy of the molecules before vaporization (liquid).

Because vaporization absorbs energy without changing temperature, that energy goes into increasing the potential energy (intermolecular spacing/attractions) of the particles, so the gas has greater potential energy than the liquid.

Question 61 · 1 pt · constructed response

Determine the amount of heat required to completely melt a 50.0-gram sample of this molecular compound at its melting point.

The boiling point for a sample of a molecular compound is 80.°C at standard pressure. The heat of fusion, Hf, of this compound is 127 joules per gram.

Model answer: 6350 J (6.35 kJ)

Worked solution

$q = m \times H_f$

$q = (50.0\ g)(127\ J/g)$

$q = 6350\ J = 6.35\ kJ$

Answer: 6350 J (6.35 kJ)

The heat needed to melt a sample equals its mass multiplied by the heat of fusion, giving 6350 J for 50.0 grams.

Question 62 · 1 pt · constructed response

Explain, in terms of intermolecular forces, why the boiling point of this molecular compound is lower than the boiling point of water at standard pressure.

The boiling point for a sample of a molecular compound is 80.°C at standard pressure. The heat of fusion, Hf, of this compound is 127 joules per gram.

Model answer: The intermolecular forces between the molecules of this compound are weaker than the intermolecular forces (hydrogen bonding) between water molecules, so less energy is needed to separate them, resulting in a lower boiling point.

Worked solution

$\text{Water molecules experience strong hydrogen bonding between molecules}$

$\text{This compound's molecules have weaker intermolecular forces than water}$

$\text{Weaker forces require less energy to overcome, giving a lower boiling point (80.°C vs 100.°C for water)}$

Answer: The intermolecular forces between the molecules of this compound are weaker than the intermolecular forces (hydrogen bonding) between water molecules, so less energy is needed to separate them, resulting in a lower boiling point.

Boiling point reflects the strength of intermolecular attractions that must be overcome. Water's strong hydrogen bonds require more energy (higher boiling point) than the weaker intermolecular forces in this compound.

Question 63 · 1 pt · constructed response

Base your answers to questions 63 through 65 on the information below and on your knowledge of chemistry. A NaOH(aq) solution and an acid-base indicator are used to determine the molarity of an HCl(aq) solution. A 15.0-milliliter sample of the HCl(aq) is exactly neutralized by 30.0 milliliters of 0.010 M NaOH(aq). Identify the positive ion in the sample of HCl(aq) solution.

Model answer: $H_3O^+$ (hydronium ion), also acceptable as $H^+$

Worked solution

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

$\text{In water, } H^+ \text{ is associated with water as } H_3O^+$

Answer: $H_3O^+$ (hydronium ion), also acceptable as $H^+$

HCl is an Arrhenius acid that ionizes in water to produce hydrogen ions (hydronium ions, H3O+) and chloride ions; the positive ion is H3O+ (or H+).

Question 64 · 1 pt · constructed response

Based on Table M, state the color of litmus when placed in a sample of the original HCl(aq) solution.

Model answer: Red

Worked solution

$HCl(aq) \text{ is an acid}$

$\text{Table M shows litmus is red in acidic solutions (pH below its range) and blue in basic solutions}$

Answer: Red

Litmus indicator turns red in the presence of an acid such as HCl(aq), per Table M.

Question 65 · 1 pt · constructed response

Using the titration data, determine the concentration of the HCl(aq) solution.

Model answer: 0.020 M

Worked solution

$M_aV_a = M_bV_b\ \text{(1:1 mole ratio for HCl and NaOH)}$

$M_a(15.0\ mL) = (0.010\ M)(30.0\ mL)$

$M_a = \frac{(0.010)(30.0)}{15.0} = 0.020\ M$

Answer: 0.020 M

Using the titration formula MaVa = MbVb for the 1:1 neutralization of HCl and NaOH, solving for the acid molarity gives 0.020 M.

Question 66 · 1 pt · constructed response

Base your answers to questions 66 through 68 on the information below and on your knowledge of chemistry. The table below shows three compounds that are used for medical purposes. Identify the two types of bonding in the compound used to make plaster casts.

Model answer: Ionic bonding and covalent bonding

Worked solution

$CaSO_4 \text{ is used to make plaster casts}$

$\text{Ionic bonding exists between the } Ca^{2+} \text{ cation and the } SO_4^{2-} \text{ polyatomic anion}$

$\text{Covalent bonding exists between the sulfur and oxygen atoms within the sulfate ion}$

Answer: Ionic bonding and covalent bonding

CaSO4 contains an ionic bond between the calcium cation and the sulfate polyatomic ion, and covalent bonds between the sulfur and oxygen atoms within the sulfate ion itself.

Question 67 · 1 pt · constructed response

Base your answers to questions 66 through 68 on the information below and on your knowledge of chemistry. The table below shows three compounds that are used for medical purposes. 67 Identify the noble gas that has atoms in the ground state with the same electron configuration as the positive ion in Mg(OH)$_2$ in the ground state.

Model answer: Neon (Ne)

Worked solution

$\text{Mg atom (ground state)}: 2\text{-}8\text{-}2$ Neutral magnesium electron configuration

$\text{Mg}^{2+}: 2\text{-}8$ The positive ion in Mg(OH)2 is Mg2+, formed by losing its 2 valence electrons

$\text{Ne (ground state)}: 2\text{-}8$ Neon's ground-state configuration matches Mg2+

Answer: Neon (Ne)

Mg(OH)2 contains the Mg2+ ion, which loses the two 3s electrons of neutral magnesium to leave a 2-8 configuration identical to that of neutral neon.

Question 68 · 1 pt · constructed response

Based on Table S, determine the electronegativity difference for the bond between zinc and oxygen in ZnO.

Model answer: 1.7

Worked solution

$EN_{O} = 3.4 \text{ (Table S)}$

$EN_{Zn} = 1.7 \text{ (Table S)}$

$\Delta EN = 3.4 - 1.7 = 1.7$

Answer: 1.7

Subtracting the Table S electronegativity value of zinc from that of oxygen gives the electronegativity difference for the Zn-O bond.

Question 69 · 1 pt · constructed response

Base your answers to questions 69 through 71 on the information below and on your knowledge of chemistry. During a laboratory activity, appropriate safety equipment is used and safety procedures are followed. In a beaker, a student dissolves 0.10 mole of potassium nitrate, KNO$_3$, in 100. grams of water at 25°C. The gram-formula mass of KNO$_3$ is 101 g/mol. 69 Compare the freezing point of water at standard pressure to the freezing point of the KNO$_3$(aq) solution at standard pressure.

Model answer: The freezing point of the KNO3(aq) solution is lower than the freezing point of pure water.

Worked solution

$\Delta T_f = K_f \cdot m$ Dissolved solute particles depress the freezing point (freezing point depression)

Answer: The freezing point of the KNO3(aq) solution is lower than the freezing point of pure water.

Adding a solute such as KNO3 to water lowers the freezing point of the resulting solution below 0°C, the freezing point of pure water at standard pressure.

Question 70 · 1 pt · constructed response

Based on Table G, classify, in terms of saturation, the type of KNO$_3$ solution made by the student.

During a laboratory activity, appropriate safety equipment is used and safety procedures are followed. In a beaker, a student dissolves 0.10 mole of potassium nitrate, KNO$_3$, in 100. grams of water at 25°C. The gram-formula mass of KNO$_3$ is 101 g/mol.

Model answer: Unsaturated

Worked solution

$\text{mass KNO}_3 = 0.10\ \text{mol} \times 101\ \text{g/mol} = 10.1\ \text{g}$ Grams of KNO3 dissolved in the 100. g of water

$\text{Table G solubility of KNO}_3 \text{ at } 25^\circ C \approx 38\ \text{g}/100\ \text{g H}_2O$

$10.1\ \text{g} < 38\ \text{g}$ Less solute dissolved than the maximum solubility at that temperature

Answer: Unsaturated

Since only 10.1 g of KNO3 is dissolved in 100 g of water at 25°C, which is well below the roughly 38 g/100 g solubility shown on Table G at that temperature, the solution is unsaturated.

Question 71 · 1 pt · constructed response

Describe a laboratory procedure the student can use to separate the original KNO$_3$ from the KNO$_3$(aq) solution.

Model answer: Heat the solution to evaporate the water, leaving the solid KNO3 behind.

Worked solution

$\text{Evaporation}$ Heating the solution drives off the water (a volatile solvent) as vapor

$\text{Solid KNO}_3 \text{ remains}$ The nonvolatile solute is left behind once all the water has evaporated

Answer: Heat the solution to evaporate the water, leaving the solid KNO3 behind.

Because KNO3 is a solid dissolved in water, gently heating (or otherwise evaporating) the solution boils away the water and leaves the solid KNO3 residue in the container.

Question 72 · 1 pt · constructed response

Base your answers to questions 72 through 74 on the information below and on your knowledge of chemistry. A major component of agricultural fertilizer is ammonium nitrate, NH$_4$NO$_3$. It is commercially produced by reacting ammonia gas with concentrated nitric acid, as shown in equation 1 below. Equation 1: NH$_3$ + HNO$_3$ $\rightarrow$ NH$_4$NO$_3$ + energy The ammonium nitrate can decompose to produce nitrogen dioxide gas, NO$_2$(g), which is a brown gas. When NO$_2$(g) is placed in a sealed, glass container, some NO$_2$ molecules react to produce colorless N$_2$O$_4$ gas. An equilibrium system forms in the glass container, as shown in equation 2 below. Equation 2: 2NO$_2$(g) $\rightleftharpoons$ N$_2$O$_4$(g) + energy (brown) (colorless) 72 State, in terms of reactants and products, why the chemical reaction represented by equation 1 is a synthesis reaction.

Model answer: Two reactants (NH3 and HNO3) combine to form a single product (NH4NO3).

Worked solution

$NH_3 + HNO_3 \rightarrow NH_4NO_3$ Two substances combine to form one substance

Answer: Two reactants (NH3 and HNO3) combine to form a single product (NH4NO3).

A synthesis reaction is defined by two or more simpler substances combining to form a single, more complex product, exactly as ammonia and nitric acid combine to form the single product ammonium nitrate.

Question 73 · 1 pt · constructed response

On the labeled axes in your answer booklet, draw a potential energy diagram for the reaction represented by equation 1.

Model answer: A curve starting at the reactants' potential energy level, rising through a peak (the activated complex, whose height above the reactants represents the activation energy), then falling to a products' potential energy level that is lower than the reactants' level, since equation 1 releases energy (exothermic).

Worked solution

$PE_{reactants} > PE_{products}$ Equation 1 shows energy as a product, so the reaction is exothermic

$PE_{activated\ complex} > PE_{reactants}$ The curve must rise to a peak higher than the reactants before descending, representing the activation energy

Answer: A curve starting at the reactants' potential energy level, rising through a peak (the activated complex, whose height above the reactants represents the activation energy), then falling to a products' potential energy level that is lower than the reactants' level, since equation 1 releases energy (exothermic).

Because equation 1 releases energy, the potential energy diagram should show the products at a lower energy level than the reactants, with a hump above the reactants representing the activated complex.

Question 74 · 1 pt · constructed response

Compare the rate of the forward reaction to the rate of the reverse reaction in the equilibrium system represented by equation 2.

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

Worked solution

$rate_{forward} = rate_{reverse}$ Definition of dynamic chemical equilibrium

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

In any system at chemical equilibrium, such as the NO2/N2O4 system in equation 2, the forward and reverse reactions occur at the same rate, even though both continue to occur.

Question 75 · 1 pt · constructed response

Base your answers to questions 75 through 77 on the information below and on your knowledge of chemistry. Pentane, hexane, heptane, and octane are components of gasoline. The structural formula below represents a straight-chain molecule of pentane. 75 State the number of hydrogen atoms in a molecule of hexane.

Model answer: 14 hydrogen atoms

Worked solution

$C_nH_{2n+2}$ General formula for a saturated straight-chain alkane

$n = 6 \Rightarrow H = 2(6)+2 = 14$

Answer: 14 hydrogen atoms

Hexane is the six-carbon alkane, C6H14, so applying the alkane formula CnH2n+2 gives 14 hydrogen atoms.

Question 76 · 1 pt · constructed response

State, in terms of carbon-carbon bonds, why octane is a saturated hydrocarbon.

Model answer: Octane is saturated because all of the carbon-carbon bonds in the molecule are single bonds.

Worked solution

$\text{All C-C bonds} = \text{single bonds}$ Definition of a saturated hydrocarbon

Answer: Octane is saturated because all of the carbon-carbon bonds in the molecule are single bonds.

A hydrocarbon is classified as saturated when every carbon-carbon bond in its molecule is a single covalent bond, which is the case for octane.

Question 77 · 1 pt · constructed response

Draw a structural formula for a straight-chain molecule of heptane.

Model answer: H3C-CH2-CH2-CH2-CH2-CH2-CH3 (a straight chain of seven carbon atoms, each carbon bonded to the correct number of hydrogen atoms so every carbon has four single bonds: H-C-C-C-C-C-C-C-H with hydrogens filling all remaining bonds, molecular formula C7H16).

Worked solution

$C_7H_{16}$ Heptane's molecular formula, following CnH2n+2 for n=7

$H-\underset{H}{\underset{|}{C}}-\underset{H}{\underset{|}{C}}-\underset{H}{\underset{|}{C}}-\underset{H}{\underset{|}{C}}-\underset{H}{\underset{|}{C}}-\underset{H}{\underset{|}{C}}-\underset{H}{\underset{|}{C}}-H$ Straight chain of seven carbons, each with two additional hydrogens (except the end carbons, which have three), all bonds single

Answer: H3C-CH2-CH2-CH2-CH2-CH2-CH3 (a straight chain of seven carbon atoms, each carbon bonded to the correct number of hydrogen atoms so every carbon has four single bonds: H-C-C-C-C-C-C-C-H with hydrogens filling all remaining bonds, molecular formula C7H16).

Following the same pattern shown for pentane, a straight-chain heptane structure has seven carbons joined by single bonds in a row, with each carbon bonded to enough hydrogen atoms to complete four total bonds.

Question 78 · 1 pt · constructed response

Base your answers to questions 78 through 80 on the information below and on your knowledge of chemistry. Chlorine gas and sodium metal are two valuable industrial elements that can be produced in an electrochemical cell by an electrolytic reaction. In the reaction, liquid sodium chloride is separated into the elements by an electric current. In the cell, a power source is connected by its positive terminal to the anode and by its negative terminal to the cathode; the cell contains molten NaCl(ℓ) with Na+ and Cl- ions between the two electrodes. The equation below represents this process. 2NaCl($\ell$) $\rightarrow$ 2Na($\ell$) + Cl$_2$(g) 78 State, in terms of energy, why the power source is required for the cell to operate.

Model answer: The power source supplies electrical energy needed to drive this nonspontaneous reaction (electrolysis).

Worked solution

$\Delta G > 0$ Electrolytic cell reactions are nonspontaneous and require an external energy source

Answer: The power source supplies electrical energy needed to drive this nonspontaneous reaction (electrolysis).

Because the decomposition of molten NaCl into Na and Cl2 is nonspontaneous, electrical energy from the power source must be supplied continuously to force the reaction to occur.

Question 79 · 1 pt · constructed response

State the number of moles of electrons lost when 4.0 moles of electrons are gained in this reaction.

Chlorine gas and sodium metal are two valuable industrial elements that can be produced in an electrochemical cell by an electrolytic reaction. In the reaction, liquid sodium chloride is separated into the elements by an electric current. In the cell, a power source is connected by its positive terminal to the anode and by its negative terminal to the cathode; the cell contains molten NaCl(ℓ) with Na+ and Cl- ions between the two electrodes. The equation below represents this process. 2NaCl($\ell$) $\rightarrow$ 2Na($\ell$) + Cl$_2$(g)

Model answer: 4.0 moles of electrons

Worked solution

$\text{mol e}^-_{lost} = \text{mol e}^-_{gained} = 4.0\ \text{mol}$ Conservation of charge: electrons lost in oxidation equal electrons gained in reduction

Answer: 4.0 moles of electrons

In any redox reaction the total number of electrons lost by the species oxidized must equal the total number of electrons gained by the species reduced, so 4.0 moles of electrons are lost.

Question 80 · 1 pt · constructed response

Determine the change in oxidation state that occurs for chlorine in this cell.

Model answer: The oxidation state of chlorine increases from -1 to 0, a change of +1.

Worked solution

$Cl^{-} \rightarrow Cl^0 + e^-$ Chlorine is oxidized at the anode

$(-1) \rightarrow (0)$ Oxidation state increases by one unit

Answer: The oxidation state of chlorine increases from -1 to 0, a change of +1.

Chloride ions (oxidation state -1) are oxidized to neutral chlorine gas (oxidation state 0) at the anode, so the oxidation state of chlorine increases by 1.

Question 81 · 1 pt · constructed response

Base your answers to questions 81 through 85 on the information below and on your knowledge of chemistry. The radioisotope U-238 changes to Pb-206 in a series of nuclear decays that can be used to determine the age of some rock layers. In one step of this nuclear decay series, U-234 produces Th-230 by alpha emission. The graph below represents the decay of a 40.-milligram sample of Th-230 (Mass of Th-230, in mg, on the y-axis from 0 to 40, versus Time, in years, on the x-axis marked at 1.0 x $10^5$, 2.0 x $10^5$, and 3.0 x $10^5$; the decay curve starts at 40 mg at t = 0 and decreases, passing near 20 mg before 1.0 x $10^5$ y and near 10 mg (marked by a horizontal reference line) between 1.0 x $10^5$ and 2.0 x $10^5$ y, continuing to decrease toward about 2-3 mg by 3.0 x $10^5$ y). 81 Using the graph, determine the half-life of Th-230.

Model answer: Approximately 7.5 x 10^4 years (75,000 years)

Worked solution

$40\ \text{mg} \rightarrow 20\ \text{mg}$ Reading the graph, the mass drops to half (20 mg) at about 0.75 x 10^5 years

$t_{1/2} \approx 7.5\times10^{4}\ \text{y}$

Answer: Approximately 7.5 x 10^4 years (75,000 years)

The half-life is the time for the sample's mass to drop to half its initial value; reading from the graph, the 40. mg sample falls to 20 mg (and 20 mg falls to 10 mg) after equal time intervals of about 7.5 x 10^4 years each.

Question 82 · 1 pt · constructed response

Determine the fraction of the Th-230 sample that remains unchanged after two half-lives.

Model answer: 1/4 (one-fourth) of the original sample remains

Worked solution

$\left(\frac{1}{2}\right)^2 = \frac{1}{4}$ After n half-lives, the fraction remaining is (1/2)^n

Answer: 1/4 (one-fourth) of the original sample remains

Each half-life reduces the remaining amount by half, so after two half-lives the fraction remaining is 1/2 x 1/2 = 1/4.

Question 83 · 1 pt · constructed response

Compare the penetrating power of a beta particle to the penetrating power of the alpha decay particle from a Th-230 atom.

Model answer: A beta particle has greater penetrating power than an alpha particle.

Worked solution

$\text{alpha}(\alpha): \text{low penetrating power, stopped by paper/skin}$

$\text{beta}(\beta): \text{greater penetrating power, stopped by e.g. aluminum foil/thin metal}$

Answer: A beta particle has greater penetrating power than an alpha particle.

Alpha particles are large and slow-moving, so they are stopped easily and have low penetrating power, while beta particles are smaller and faster, giving them significantly greater penetrating power than alpha particles.

Question 84 · 1 pt · constructed response

Complete the nuclear equation in your answer booklet for the decay of U-238 by writing a notation for the missing product.

Model answer: $^4_2$He (an alpha particle)

Worked solution

$^{238}_{92}U \rightarrow\ ^{234}_{90}Th + \ ^{4}_{2}He$ U-238 decays by alpha emission to Th-234, releasing an alpha particle

$\text{Missing product} = ^4_2He$ Mass number and atomic number balance: 238 = 234+4 and 92 = 90+2

Answer: $^4_2$He (an alpha particle)

U-238 decays by emitting an alpha particle (a helium-4 nucleus), producing Th-234; balancing mass numbers (238 = 234 + 4) and atomic numbers (92 = 90 + 2) identifies the missing product as an alpha particle, notation 4/2 He.

Question 85 · 1 pt · constructed response

Explain, in terms of elements, why the nuclear decay of U-234 to Th-230 is considered a transmutation.

Model answer: The decay changes an atom of one element (uranium) into an atom of a different element (thorium).

Worked solution

$U (Z=92) \rightarrow Th (Z=90)$ The atomic number, and therefore the identity of the element, changes

Answer: The decay changes an atom of one element (uranium) into an atom of a different element (thorium).

A transmutation is a nuclear change in which atoms of one element become atoms of a different element; since uranium (atomic number 92) decays into thorium (atomic number 90), the element identity changes, making this a transmutation.

January 2025 Chemistry Regents — Answer Key & Worked Solutions

More Chemistry Regents help