Your FREE ACS General Chemistry Exam Practice Test 2026 – 240+ Q&A

Correct answer: n = 2 to n = 1

A hydrogen atom emits a photon when its electron drops to a lower level, and the photon's energy is proportional to 1/n_f^2 - 1/n_i^2, where n_f is the final level and n_i the starting level. For n = 2 to n = 1 that difference is 1 - 1/4 = 0.75, the largest of the four. The other drops are bigger jumps in n but end on higher levels, where the levels crowd closer together: n = 7 to n = 2 gives 1/4 - 1/49, about 0.23; n = 9 to n = 3 gives 1/9 - 1/81, about 0.099; and n = 6 to n = 4 gives 1/16 - 1/36, about 0.035. A larger change in n does not mean a larger energy change.

Correct answer: 2

An orbital can hold at most 2 electrons, and those two must have opposite spins. This is the Pauli exclusion principle: no two electrons in an atom can share all four quantum numbers, and within one orbital only the spin is left to differ. 6 is the capacity of a whole p sublevel, which has three orbitals. 14 is the capacity of an f sublevel, which has seven orbitals. 18 is the capacity of the entire third shell (2n^2 with n = 3). Each of those counts the electrons in a group of orbitals, not in a single orbital.

Correct answer: From 2s to 2p

From 2s to 2p absorbs the lowest-energy photon: in a multielectron atom the 2s and 2p sublevels belong to the same shell and lie close together in energy. From 2s to 3p starts in the same place but must also climb into the next shell, which costs much more energy. From 1s to 2p starts from the tightly held 1s level, far below 2s, so it needs more energy than any jump out of 2s. From 1s to 3p combines both, the deepest starting level and the higher finishing level, so it is the largest gap of the four.

Correct answer: [Ar]

Neutral calcium (Z = 20) is [Ar] 4s^2. A 2+ ion has lost two electrons, and they come from the outermost 4s orbital, so Ca^2+ keeps 18 electrons arranged exactly like argon: [Ar]. [Ar] 4s^2 still counts all 20 electrons, so it describes neutral calcium before ionization. [Ne] keeps only 10 electrons; it uses the neon core from one period too early and is really the configuration of Mg^2+, calcium's lighter group 2 neighbor. [Ne] 3s^2 makes both mistakes at once: it is the configuration of a neutral magnesium atom.

Correct answer: Their nuclei contain different numbers of neutrons.

Isotopes are atoms of the same element whose nuclei hold the same number of protons but different numbers of neutrons, so the correct statement is "Their nuclei contain different numbers of neutrons."; carbon-12 has 6 neutrons and carbon-14 has 8, while both have 6 protons. "Their atoms are charged forms of the same element." confuses isotopes with ions: an ion has gained or lost electrons, but every isotope of an element has the same number of protons, and a neutral atom of any isotope has just as many electrons, so its net charge is zero. "Their nuclei contain different numbers of protons." describes atoms of different elements: the number of protons is the atomic number, so a different proton count makes a different element, not another isotope of the same one. "Their atoms are radioactive forms of the same element." mistakes a property of some isotopes for the definition: carbon-12 and carbon-13 are stable, and only some isotopes, such as carbon-14, are radioactive.

Correct answer: Azimuthal quantum number (l)

The azimuthal quantum number (l), also called the angular momentum quantum number, sets the shape of an orbital: l = 0 gives a spherical s orbital, l = 1 a dumbbell-shaped p orbital, and l = 2 the d orbitals. The principal quantum number (n) sets the orbital's size and main energy level, not its shape. The magnetic quantum number (m_l) sets how an orbital of a given shape is oriented in space. The spin quantum number (m_s) describes the electron's own spin, +1/2 or -1/2, and says nothing about the orbital.

Correct answer: Electrons fill the orbitals of lowest energy first.

The Aufbau ("building up") principle is the rule "Electrons fill the orbitals of lowest energy first.": each added electron goes into the lowest-energy orbital still available. "Electrons complete the orbitals of each shell first." describes shell-by-shell filling, which is not what happens: 4s fills before 3d because 4s is lower in energy, even though it belongs to a higher shell. "Electrons pair up in the orbitals of equal energy first." contradicts Hund's rule, which says electrons occupy orbitals of equal energy singly before any pairing. "Electrons enter the p orbitals of each shell first." reverses the order within a shell: the s sublevel is lower in energy than the p sublevel, so s fills before p.

Correct answer: 9

The third energy level has three sublevels: 3s with 1 orbital, 3p with 3, and 3d with 5, so it holds 1 + 3 + 5 = 9 orbitals (n^2 with n = 3). 4 leaves out the 3d sublevel and counts only 3s and 3p; 3d fills after 4s, but it still belongs to the third level. 16 adds a 3f sublevel of 7 orbitals, but no such sublevel exists: l can only go up to n - 1 = 2, so f orbitals first appear in the fourth level. 18 counts electrons instead of orbitals (2 per orbital, or 2n^2).

Correct answer: 1s^2 2s^2 2p^5 3s^1

An excited state has an electron in a higher orbital while a lower one is left partly empty, and it must still respect orbital capacities. 1s^2 2s^2 2p^5 3s^1 has neon's ten electrons, but one has been promoted from 2p to 3s, leaving 2p one short, so it is a possible excited state. 1s^2 2s^2 2p^6 3s^1 is the ground state of sodium: every lower sublevel is full before 3s begins. 1s^2 2s^2 2p^5 3s^3 and 1s^2 2s^2 2p^6 3s^3 are not possible at all, because an s sublevel holds at most two electrons.

Correct answer: 4s^2 3d^10 4p^5

Bromine (Z = 35) has the ground-state configuration [Ar] 4s^2 3d^10 4p^5, so the part written after the argon core is 4s^2 3d^10 4p^5: 2 + 10 + 5 = 17 electrons, which with argon's 18 makes 35. 4s^2 4d^10 4p^5 puts the ten d electrons in 4d, but the d sublevel that fills across period 4 is 3d; 4d does not begin filling until period 5. 4s^2 3d^10 4p^6 has 18 electrons beyond argon, which is krypton, the noble gas after bromine. 4s^2 4d^10 4p^6 makes both of those errors.

Correct answer: Neon

First ionization energy rises across Period 2 as nuclear charge grows while electrons stay in the same shell, and neon, with its filled 2s^2 2p^6 shell, has the highest value (about 2081 kJ/mol). Fluorine is the strongest pull among the reactive elements (about 1681 kJ/mol) but still falls short of the noble gas that follows it. Oxygen (about 1314 kJ/mol) is lower still, since its first paired 2p electron is repelled by its partner and is easier to remove. Carbon (about 1086 kJ/mol), in the middle of the period, has a smaller nuclear charge holding its outer electrons and is lower again.

Correct answer: Sodium

Sodium is in Group 1, with a single 3s electron outside a neon core, so it readily loses one electron to form Na+, a +1 oxidation state. Calcium is in Group 2 and loses both 4s electrons to form Ca^2+. Gallium is in Group 13 and is normally +3, as in Ga2O3. Zinc gives up both of its 4s electrons and is found almost only as Zn^2+, so none of these forms a +1 state as readily as sodium.

Correct answer: -2

Group 16 elements have six valence electrons (ns^2 np^4), so gaining two electrons completes an octet, and -2 is their most common oxidation state, as in oxides (O^2-) and sulfides (S^2-). '-3' is the typical state one group to the left, in Group 15, where five valence electrons need three more, as in nitrides (N^3-) and ammonia (NH3). '0' is the state of the uncombined elements, such as O2 and S8, not of these elements in compounds. '-6' counts all six valence electrons as if each atom gained six more, but an atom with six valence electrons needs only two to complete its octet.

Correct answer: Manganese

Manganese can use all seven of its 4s and 3d electrons and reaches +7 in permanganate, MnO4-, the highest state of these metals. Chromium tops out one step lower, at +6 in chromate and dichromate. Zirconium has only four valence electrons and stops at +4, as in ZrO2. Palladium is found mainly as +2 and at most +4, as in K2PdCl6, so it falls well short of manganese.

Correct answer: Argon

Across the third period the nuclear charge climbs from 11 to 18 while the added electrons enter the same n = 3 shell, so each electron is pulled in more tightly and atomic radius shrinks; argon, at the end of the period, is the smallest. Magnesium, only one step into the period, is among its largest atoms, so choosing it runs against the trend. Silicon and phosphorus sit toward the middle of the period and are smaller than magnesium, but with fewer protons than argon (14 and 15 against 18) both are still larger.

Correct answer: Group 18

Group 18, the noble gases (helium, neon, argon, krypton, xenon and radon), is the only group whose members are all gases at room temperature. Group 17 begins with two gases, fluorine and chlorine, but bromine is a liquid and iodine a solid. Group 16 begins with oxygen gas, but sulfur, selenium and tellurium are solids. Group 15 begins with nitrogen gas, but phosphorus, arsenic, antimony and bismuth are solids.

Correct answer: Fluorine

Fluorine has the highest electronegativity of any element (3.98 on the Pauling scale): it is small, with a high effective nuclear charge acting on bonding electrons that sit close to its nucleus. Oxygen is second (3.44) but has one less proton in the same shell. Hydrogen (2.20) forms polar bonds with most elements and is often thought to pull electrons strongly, but with a single electron shell and one proton its pull is far weaker than fluorine's. Nitrogen (3.04) is lower again, two protons short of fluorine in the same shell.

Correct answer: 1

Group 1 elements (the alkali metals) have a single ns^1 valence electron, which they lose easily, so in compounds they are +1, as in NaCl and K2O. '-1' reverses the sign, as if the metal gained an electron; that is hydrogen's state in metal hydrides such as NaH, not the typical state of the alkali metals. '2' is the typical state of the Group 2 alkaline earth metals, as in Mg^2+ and CaCl2, which have two valence electrons to lose; an alkali metal has only one. '-7' imagines the atom gaining seven electrons to complete an octet, which would cost far more energy than losing the one electron it has.

Correct answer: s^2 p^1

Group 13 elements (boron, aluminum, gallium) have three valence electrons, arranged as s^2 p^1 in the outer shell; aluminum, for example, is [Ne] 3s^2 3p^1. 's^2 p^2' is the Group 14 pattern of carbon and silicon, one electron too many. 's^2 d^1' is the Group 3 pattern of scandium (4s^2 3d^1), a mix-up between Group 13 and Group 3 that the older IIIA and IIIB labels invite. 's^2 d^2' is the Group 4 pattern of titanium, which makes both slips at once: a d electron in place of p and one electron too many.

Correct answer: Group 15 (the nitrogen group)

This element is in Group 15 (the nitrogen group). In [Ar] 3d^10 4s^2 4p^3 the outermost shell (n = 4) holds 4s^2 4p^3, five valence electrons, so this p-block element (arsenic) sits in Group 15; the filled 3d^10 subshell belongs to the core and is not counted. Group 13 (the boron group) counts only the three 4p electrons and leaves out the two 4s electrons, which are valence electrons too. Group 5 (the vanadium group) puts the five valence electrons into the transition-metal column, the old VA versus VB mix-up. Group 2 (the beryllium group) counts only the two 4s electrons.

Correct answer: One electron in each 2p orbital, the three spins parallel

Hund's rule puts one electron in each 2p orbital, the three spins parallel, before any 2p orbital takes a second electron, so ground-state nitrogen has three unpaired electrons with parallel spins (↑ ↑ ↑). One electron in each 2p orbital, the middle spin reversed (↑ ↓ ↑) fills the orbitals singly but breaks the parallel-spin part of the rule. Two paired electrons in one 2p orbital, the third alone (↑↓ ↑ _) pairs electrons while an empty 2p orbital is still available, which Hund's rule rules out for the ground state. Two spin-up electrons in one 2p orbital, the third alone also breaks the Pauli exclusion principle, because two electrons sharing one orbital must have opposite spins.

Correct answer: A chlorine ion with a +1 charge

The species is a chlorine ion with a +1 charge. The proton count (17) fixes the element as chlorine whatever the electron count, and the net charge is protons minus electrons: 17 - 16 = +1, so this is the cation Cl^+. A chloride anion with charge -1 subtracts in the wrong order (electrons minus protons) and lands on the familiar chloride ion. A sulfur anion with a -1 charge names the element from its 16 electrons and also reverses the subtraction. A bromine ion with charge +1 gets the charge right but adds the protons and neutrons (17 + 18 = 35) and reads that mass number as the atomic number, which is bromine's.

Correct answer: n = 2, l = 2, m_l = 0, m_s = +1/2

The forbidden set is n = 2, l = 2, m_l = 0, m_s = +1/2. The angular momentum quantum number l can take only the values 0 through n - 1, so the n = 2 shell has l = 0 or 1 (the 2s and 2p subshells) and no l = 2 (d) subshell. The other three sets obey every rule (l less than n, m_l from -l to +l, m_s = +1/2 or -1/2): n = 3, l = 1, m_l = 0, m_s = -1/2 describes a 3p electron; n = 4, l = 3, m_l = 0, m_s = +1/2 describes a 4f electron, with the largest l that n = 4 allows; and n = 1, l = 0, m_l = 0, m_s = -1/2 describes a 1s electron.

Correct answer: The weighted average of the natural isotope masses

The atomic weight is the weighted average of the natural isotope masses. Natural chlorine is about 76% chlorine-35 (34.97 u) and 24% chlorine-37 (36.97 u), and weighting each mass by its abundance gives about 35.45 u, a value no single atom has. The unweighted average of the isotope mass numbers counts 35 and 37 equally and gives 36, not 35.45. The loss of mass as binding energy in each isotope is real, but it is already built into each isotope's mass: chlorine-35 has a mass of 34.97 u, within about 0.03 u of a whole number, and no such effect can produce 35.45, a value that lies between the masses of the two isotopes. The mean of repeated readings taken on one isotope would cluster around that isotope's own mass; measurement scatter does not create the 0.45 offset.

Correct answer: Effective nuclear charge increases across the period

The primary reason is that effective nuclear charge increases across the period. Each step to the right adds a proton while the new electron enters the same shell, and electrons in the same shell shield one another poorly, so the net positive charge felt by an incoming electron grows and more energy is released when it is captured. Covalent atomic radius increases across the period has the trend backward: radius shrinks from left to right because of that rising effective nuclear charge. Core-electron shielding decreases across the period is false, since the inner shells stay the same within a period and shield about equally. Outer electron repulsion decreases across the period is also false: each step adds another valence electron, so repulsion within the outer shell grows, which works against electron affinity rather than explaining it.

Correct answer: sp^3d^2

A square planar molecule such as XeF4 has four bonding pairs and two lone pairs on the central atom: six electron domains in an octahedral arrangement, with the lone pairs opposite each other. Six domains need six hybrid orbitals, built from one s, three p and two d orbitals: sp^3d^2. sp^3d gives only five hybrid orbitals, the set for a steric number of 5 (trigonal bipyramidal). sp^3 gives four, which fits the four bonded atoms but leaves no room for the two lone pairs. sp^2 is what the word "planar" suggests, but it gives only three hybrid orbitals, the set for a trigonal planar atom with a steric number of 3.

Correct answer: SO2

Resonance structures exist when a molecule's electrons can be arranged in two or more valid Lewis structures that differ only in where the electrons sit. In SO2, the S=O double bond can be drawn to either oxygen, giving two equivalent structures, and the real molecule is a blend with two identical S-O bonds. OF2 has only O-F single bonds and fixed lone pairs, so there is one Lewis structure. IF5 has five I-F single bonds and one lone pair on iodine; there is no double bond to move. H2S also has just two single bonds, so it has no resonance structures.

Correct answer: s-p

Each C-H sigma bond in methane forms where a hydrogen 1s orbital overlaps head-on with an orbital on carbon. Carbon's bonding orbitals are sp^3 hybrids, three-quarters p in character, so the overlap is classed as s-p (more precisely, s with sp^3). s-s would mean carbon bonds through its 2s orbital alone, which could not give four equivalent bonds at 109.5 degrees. s-d and p-d both require d orbitals, but neither carbon nor hydrogen has d orbitals in its valence shell.

Correct answer: Sigma and delocalized pi bonding

Each carbon in benzene is sp^2 hybridized: sigma bonds join it to its two carbon neighbors and to a hydrogen, and the six leftover p orbitals overlap all around the ring, so the six pi electrons are shared by all six carbons. That is sigma and delocalized pi bonding, and it is why every C-C bond in benzene has the same length, between that of a single and a double bond. Sigma and localized pi bonding describes fixed, alternating double bonds (a single Kekulé structure), which would give two different bond lengths. Ionic and polar covalent bonding is wrong because carbon and hydrogen have similar electronegativities, so benzene's bonds are essentially nonpolar and no ions form. Metallic and covalent bonding misreads delocalization: benzene's pi electrons spread over one ring of six atoms, not through a metal lattice.

Correct answer: Nitrogen

Nitrogen exists as N2, in which the two atoms share three pairs of electrons in a triple bond (bond order 3), one of the strongest bonds known at about 945 kJ/mol. Oxygen is also diatomic in its stable form, but O2 is held by a double bond (bond order 2), not a triple bond. Boron is a tempting choice because each atom has three valence electrons, but boron is a network solid in its stable form, and even the B2 molecule seen in the gas phase has a bond order of only 1. Beryllium is a metal, and a Be2 pair has a bond order of 0 by molecular orbital theory, so it forms no stable diatomic molecule at all.

Correct answer: +1

Nitrogen has 5 valence electrons; in NH4+ it has no lone-pair electrons and 8 bonding electrons in four N-H bonds, so FC = 5 - 0 - (1/2)(8) = 5 - 4 = +1, which accounts for the ion's overall +1 charge. '-3' comes from subtracting all 8 bonding electrons instead of half of them (5 - 0 - 8). '+5' leaves the bonding electrons out of the formula entirely and reports nitrogen's valence electron count. '+4' takes the number of bonds as the charge.

Correct answer: The contributors share nuclear positions and differ in electron placement, and the actual molecule is a hybrid of them

Resonance structures differ only in where electrons (lone pairs and pi bonds) are drawn: the contributors share nuclear positions and differ in electron placement, and the actual molecule is a hybrid of them, one structure with delocalized electrons. That is why the nitrate ion has three equal N-O bonds rather than one double and two single bonds. 'The contributors are separate molecules, and the actual molecule is one that flips back and forth between them rapidly' describes an equilibrium between real species, which resonance is not. 'The contributors differ in electron placement, and the actual molecule is simply the single most stable one among them' is wrong because the hybrid is a weighted blend; the most stable contributor counts the most, but the molecule is not that one structure. 'The contributors are separate isomers with their atoms moved around, and the actual molecule is a mere mixture of them' describes isomers, since moving atoms makes a different compound, not a resonance form.

Correct answer: A sigma bond forms by head-on overlap along the bond axis, while a pi bond forms by sideways overlap off the axis

A sigma bond forms by head-on overlap along the bond axis, while a pi bond forms by sideways overlap off the axis: sigma electron density lies directly between the two nuclei, and pi density sits in two lobes above and below the bond axis. 'A sigma bond forms by parallel p orbitals overlapping off the bond axis, while a pi bond forms by head-on contact' swaps the two descriptions. 'A sigma bond forms from the overlap of two s orbitals, while a pi bond forms from the overlap of two p orbitals' is too narrow, because sigma bonds also form when p orbitals or hybrid orbitals overlap end to end, as in F2 or the C-H bonds of methane. 'A sigma bond forms as the weaker second bond between two nuclei, while a pi bond forms as the stronger first bond' reverses the order: the first bond between two atoms is always a sigma bond, and the weaker pi bonds are added in double and triple bonds.

Correct answer: 1 sigma and 2 pi

A triple bond is 1 sigma and 2 pi. The first bond between two atoms is always a sigma bond, formed by head-on overlap along the axis between the nuclei; every additional bond in a double or triple bond is a pi bond, formed by side-by-side overlap of parallel p orbitals. The triple bond in N2 therefore has one sigma bond and two pi bonds, three bonds in all and a bond order of 3. 2 sigma and 1 pi swaps the two counts, but a triple bond has only one sigma bond, the first bond between the two atoms, and its second and third bonds are both pi bonds. 0 sigma and 3 pi treats all three bonds as pi bonds, but the first bond between two atoms is the sigma bond, and pi bonds are only the second and third bonds added to it. 3 sigma and 0 pi treats all three bonds as sigma bonds, but only one head-on overlap fits along the axis between two nuclei, so the second and third bonds must be pi bonds.

Correct answer: Add up the total valence electrons, make the least electronegative atom except hydrogen central, join the atoms with single bonds, then complete octets with lone pairs, adding multiple bonds if needed

The standard sequence is: Add up the total valence electrons, make the least electronegative atom except hydrogen central, join the atoms with single bonds, then complete octets with lone pairs, adding multiple bonds if needed. The electron count comes first because it fixes how many electrons the finished structure may show (adjusted for any charge). The central atom is the least electronegative one because it shares electrons with the most neighbors, and hydrogen is never central because it can form only one bond. Leftover electrons go on as lone pairs, and a lone pair is turned into a double or triple bond only when an atom is still short of an octet. Assign each atom its formal charge, make the atom having the greatest electronegativity central, join the atoms with single bonds, then complete octets with lone pairs, adding multiple bonds if needed is wrong on two counts: formal charges can only be worked out after a structure exists, and the most electronegative atoms (such as O, F and Cl) normally sit on the outside, not in the middle. Assign each atom its formal charge, make the least electronegative atom except hydrogen central, join the atoms with single bonds, then create as many multiple bonds as possible, ignoring octet limits skips the electron count and piles on extra bonds; multiple bonds are added only as needed to complete octets. Add up the total valence electrons, make the atom having the greatest electronegativity central, join the atoms with single bonds, then create as many multiple bonds as possible, ignoring octet limits starts correctly but then picks the wrong central atom and ignores the octet rule.

Correct answer: Hydrogen atoms normally take terminal positions and hold a maximum of 2 electrons

Hydrogen atoms normally take terminal positions and hold a maximum of 2 electrons. Hydrogen has only a 1s orbital, so it is complete with a duet of 2 electrons, which means it forms just one bond and sits on the outside of a Lewis structure. Hydrogen atoms normally occupy central positions and share a total of 4 electrons applies the least-electronegative-atom-in-the-center rule to hydrogen, but a central atom needs at least two bonds and hydrogen can form only one; the rule always excludes hydrogen. Hydrogen atoms frequently form double bonds and share a total of 4 electrons is wrong for the same reason: a double bond would put 4 electrons around hydrogen. Hydrogen atoms satisfy the octet rule and hold a complete set of 8 electrons stretches the octet rule to hydrogen, the best-known exception to it; the octet rule also has other exceptions, such as boron with 6 electrons and expanded octets on sulfur and phosphorus.

Correct answer: 2

The bond order of O2 is 2. Bond order = (bonding electrons - antibonding electrons)/2 = (8 - 4)/2 = 2, which matches the O=O double bond; the same molecular orbital picture also leaves two unpaired electrons, which is why O2 is paramagnetic. 4 comes from stopping at 8 - 4 and forgetting to divide by 2. 6 comes from adding the antibonding electrons instead of subtracting them: (8 + 4)/2. 1.5 is the bond order of the superoxide ion, O2-, whose extra electron goes into an antibonding orbital, (8 - 5)/2; neutral O2 has only 4 antibonding electrons.

Correct answer: 1.5

Each carbon-carbon bond in benzene has a bond order of 1.5. Resonance spreads benzene's three pi bonds evenly over all six ring bonds, so each bond is one sigma bond plus half a pi bond: (6 sigma + 3 pi)/6 links = 9/6 = 1.5, the average of a single bond and a double bond, which is why all six bonds have the same length. 1.3 borrows the carbonate-ion average of 4/3, which applies when one double bond is shared over three positions, not when three double bonds are shared over six. 1.25 also counts the six C-H bonds and the six C-H links, (6 + 3 + 6)/12, although only the carbon-carbon bonds are being averaged. 0.5 counts only the pi bonds, 3 pi over 6 links, and leaves out the sigma bond that every ring link also has.

Correct answer: sp^3

The carbon in methane is sp^3 hybridized. Mixing one 2s and three 2p orbitals gives four equivalent sp^3 hybrid orbitals that point to the corners of a tetrahedron, 109.5 degrees apart, one for each C-H bond (a steric number of 4 with no lone pairs). sp^2 gives only three hybrids, 120 degrees apart in a plane, as on the carbons of ethene. dsp^2 also gives four hybrids, but they point to the corners of a square, the flat picture a Lewis drawing of CH4 suggests; it uses a d orbital and describes square planar metal complexes, not carbon. sp^3d^2 gives six octahedral hybrids and likewise needs d orbitals, which a second-period atom such as carbon does not have.

Correct answer: sp^3

A steric number of 4 corresponds to sp^3 hybridization. Four regions of electron density need four hybrid orbitals, made from one s and three p orbitals, and they point to the corners of a tetrahedron. Lone pairs count as regions, so water (two bonds, two lone pairs) and ammonia (three bonds, one lone pair) are both sp^3 even though their molecular shapes are bent and trigonal pyramidal. The number of hybrid orbitals always equals the steric number: sp^3d gives five hybrids, for a steric number of 5, and sp^3d^3 gives seven. dsp^2 does give four hybrids, but they point to the corners of a square and use a d orbital; that set describes square planar metal complexes, while four regions of electron density around a main-group atom take a tetrahedral, sp^3 arrangement.

Correct answer: sp

Each carbon in ethyne is sp hybridized. A carbon in a triple bond has a steric number of 2 (one sigma bond to hydrogen and one sigma bond to the other carbon), so it mixes one s and one p orbital into two sp hybrids 180 degrees apart; its two leftover p orbitals form the two pi bonds, and the molecule is linear. sp^3 comes from counting four bonds around each carbon (three to the other carbon and one to hydrogen), but the two pi bonds use unhybridized p orbitals and do not count as regions of electron density. sp^2 is the hybridization of the carbons in ethene, where a double bond leaves three regions of electron density 120 degrees apart; a triple-bonded carbon has only two. sp^3d^2 gives six octahedral hybrids and needs d orbitals that carbon does not have.

Correct answer: As bond order increases from single to triple, bond energy rises and bond length falls

As bond order increases from single to triple, bond energy rises and bond length falls. The energies in the question climb from 348 to 614 to 839 kJ/mol as the bond order goes from 1 to 2 to 3, and each extra shared pair pulls the two nuclei closer together (roughly 154, 134 and 120 pm). As bond order increases from single to triple, bond energy falls and bond length rises is the classic reversal, the idea that a triple bond is the weakest and longest of the three, which the given energies rule out. As bond order increases from single to triple, bond energy rises and bond length rises gets the energy right but assumes that more electrons make a bigger bond; extra bonding pairs pull the atoms closer, so the stronger bond is also the shorter one. As bond order increases from single to triple, bond energy falls and bond length falls has the length trend right but contradicts the energies given.

Correct answer: About 0.4 to 1.7, so one atom pulls the pair harder

A polar covalent bond has an electronegativity difference of about 0.4 to 1.7, so one atom pulls the pair harder: the shared electrons sit closer to the more electronegative atom, which carries a partial negative charge while its partner carries a partial positive charge (the cutoffs are guidelines, not sharp borders). About 0.0 to 0.4, so both nuclei pull about equally describes a nonpolar covalent bond. About 2.0 to 3.3, so the electrons are transferred describes an ionic bond. About 0.0 to 1.7, so like atoms also pull unequally lumps nonpolar bonds in with polar ones, yet two identical atoms (a difference of 0.0) share their electrons equally.

Correct answer: Bent, about 104.5 degrees, as two lone pairs close the angle

Water is bent, about 104.5 degrees, as two lone pairs close the angle: oxygen's four electron domains are arranged roughly tetrahedrally, the shape is named from the atoms alone, and lone pairs repel bonding pairs more strongly than bonding pairs repel each other, so the H-O-H angle closes from 109.5 to about 104.5 degrees. Tetrahedral, about 109.5 degrees, as four domains set the angle gives the electron-domain geometry rather than the molecular shape, and it leaves out the lone-pair compression. Angular, about 115 degrees, as bonding pairs open the angle reverses the repulsion order: lone pairs repel most strongly, so the angle closes below 109.5 degrees instead of opening above it. Angular, about 106.7 degrees, as one lone pair closes the angle counts the squeeze of only one lone pair, the situation in a molecule with three bonds and one lone pair; water's second lone pair closes the angle further.

Correct answer: A single atom supplies two bonding electrons

In a coordinate covalent bond, a single atom supplies two bonding electrons: a donor with a lone pair shares it with an acceptor that has an empty orbital, as when ammonia's lone pair bonds to H+ to form NH4+. Once formed, it behaves like any other covalent bond. Each atom gives one electron to the shared pair describes an ordinary covalent bond, which is the contrast the question asks about. Electrons shift fully to one atom, forming ions describes ionic bonding, where electrons are transferred rather than shared. Each atom shares two pairs, like a double bond confuses where the electrons come from with how many pairs are shared; a coordinate bond is a single shared pair.

Correct answer: Its linear shape lets two equal bond dipoles cancel

Carbon dioxide is nonpolar because its linear shape lets two equal bond dipoles cancel: each C=O dipole points from carbon toward oxygen, and with the two bonds 180 degrees apart the dipoles are equal in size and opposite in direction, so their vector sum is zero. Polarity needs hydrogen, which carbon dioxide lacks: that claim confuses polarity with hydrogen bonding, and many molecules with no hydrogen, such as SO2 and COCl2, are polar. Lone pairs on carbon cancel the pull of each oxygen is wrong because the carbon atom in CO2 has no lone pairs. Both dipoles are too weak to create net polarity is also wrong: the C=O bond dipoles are large, and they cancel only because of the geometry.

Correct answer: Six electrons, which leaves the octet unfilled

Boron is surrounded by six electrons, which leaves the octet unfilled: it brings three valence electrons and forms three single B-F bonds, so six electrons surround it, and this electron deficiency makes BF3 a Lewis acid that readily accepts an electron pair. Eight electrons, as a lone pair completes its octet assumes boron also carries a lone pair, but all three of its valence electrons are used in the bonds. Ten electrons, as empty d orbitals expand its shell describes an expanded octet, which needs an element in period 3 or beyond; boron is in period 2. Three electrons, which sit one in each B-F bond counts only boron's own valence electrons; each B-F bond is a shared pair, with one electron from boron and one from fluorine, so the three bonds place six electrons around boron.

Correct answer: Larger charges on the ions and a shorter distance between them

Lattice energy grows with larger charges on the ions and a shorter distance between them: by Coulomb's law the attraction is proportional to the product of the ionic charges divided by the distance between the ion centers, which is why MgO (small 2+ and 2- ions) has a far larger lattice energy than NaCl. Smaller charges on the ions and a longer distance between them reverses both factors. Heavier ions, which hold more electrons for the attraction treats mass and electron count as the drivers; heavier ions are usually larger, which lowers the lattice energy. Bigger ions, which offer more surface area for the attraction gets the size effect backward, because larger ions sit farther apart and attract each other less strongly.

Correct answer: Higher bond orders pull the two nuclei closer

For a given pair of atoms, higher bond orders pull the two nuclei closer: more shared electron pairs between the nuclei hold them together more tightly, so a triple bond is shorter than a double bond, which is shorter than a single bond (C≡C about 120 pm, C=C about 134 pm, C-C about 154 pm). Higher bond orders force the two nuclei apart has the trend reversed. Bond polarity, not order, sets the length cannot explain the trend, because the polarity barely changes for a fixed pair of atoms while the length drops sharply from single to triple bond. The heavier atom sets the bond length alone cannot apply here, because for a fixed pair of atoms the atoms stay the same and the bond order is what changes the length.

Correct answer: It sits in period 3, with empty d orbitals for extra electrons

The expanded octet is possible because it sits in period 3, with empty d orbitals for extra electrons: in the traditional description, phosphorus's valence shell (n = 3) includes an empty 3d set, so it can hold ten electrons in five P-Cl bonds, while second-period atoms such as nitrogen have no d orbitals in their valence shell. It sits in group 15, with its s and p electrons for five bonds fails because nitrogen is also in group 15 and never forms NCl5; one s and three p orbitals can hold only eight electrons. It is a nonmetal more electronegative than each chlorine atom is false, since chlorine (3.16) is more electronegative than phosphorus (2.19). It is a compact atom, so five chlorines pack closely around it gets size backward: phosphorus is larger than nitrogen, and that extra room helps it bond to five atoms.

Correct answer: Seesaw

A steric number of 5 means five electron domains arranged as a trigonal bipyramid. The single lone pair takes an equatorial position, where it has only two neighbors at 90 degrees, and the four bonded atoms form a seesaw shape, as in SF4. Trigonal bipyramidal is the electron-domain geometry, which counts the lone pair as one of the five domains; the molecular geometry names only the arrangement of the atoms, and that is the seesaw. Tetrahedral counts only the four bonded atoms and ignores the lone pair, but four atoms around a five-domain center cannot spread out tetrahedrally. Trigonal pyramidal puts the lone pair in an axial position, where it would have three neighbors at 90 degrees, a higher-repulsion arrangement than the equatorial one.

Correct answer: 90 degrees

In a T-shaped molecule such as ClF3, the five electron domains form a trigonal bipyramid and the two lone pairs take equatorial positions. The three bonded atoms then sit in the two axial positions and one equatorial position, so each axial bond is at 90 degrees to the equatorial bond (slightly less in real molecules, because the lone pairs push the bonds together; the two axial bonds are nearly 180 degrees apart). 104.5 degrees is the angle in water, which also has two lone pairs, but on a tetrahedral framework of four domains, not five. 107 degrees is the angle in ammonia, a trigonal pyramidal molecule with one lone pair and four domains. 72 degrees would come from spreading five domains evenly around a flat pentagon, which is not how five domains arrange in three dimensions.

Correct answer: Carbon dioxide (CO2)

A molecule with polar bonds is nonpolar only if its shape makes the bond dipoles cancel. Carbon dioxide (CO2) is linear (O=C=O), and its two identical C=O dipoles point in opposite directions, so they cancel and the molecule has no net dipole. Sulfur dioxide (SO2) is bent because of the lone pair on sulfur, so its S-O dipoles do not cancel. Hydrogen bromide (HBr) has a single polar H-Br bond and nothing to cancel it. Hydrogen cyanide (HCN) is linear, but its two bonds are different (H-C and C≡N) and their dipoles point the same way along the axis, so they add instead of canceling.

Correct answer: Withdraws electrons by resonance

The nitro group pulls electron density out of a benzene ring. Its N=O pi system conjugates with the ring, and resonance structures put positive charge on the ring carbons, especially at the ortho and para positions, so it withdraws electrons by resonance (its electronegative nitrogen and oxygens also withdraw electrons by induction). Donates electrons by resonance describes groups with a lone pair next to the ring, such as -OH and -NH2, not the nitro group. Donates electrons by hyperconjugation describes alkyl groups, whose C-H sigma bonds feed electron density into the ring; the nitro group has no such bonds. Withdraws electrons by hyperconjugation is not how the nitro group acts either: its withdrawal comes from resonance and induction, not from hyperconjugation.

Correct answer: Tetrahedral

Tetrahedral is correct. VSEPR theory spreads four electron domains as far apart as possible, toward the corners of a tetrahedron with 109.5-degree angles, and when all four domains are bonding pairs the molecular shape is tetrahedral too, as in methane (CH4). Seesaw is the shape for four bonding pairs plus one lone pair, five domains in all, as in SF4. Trigonal pyramidal is the shape for three bonding pairs and one lone pair, as in ammonia. Square pyramidal is the shape for five bonding pairs and one lone pair, as in BrF5. Each of these needs a lone pair, and this central atom has none.

Correct answer: Trigonal planar, 120 degrees

Trigonal planar, 120 degrees is correct. Three electron domains with no lone pairs spread out in a flat triangle around the central atom, so the electron-domain geometry is trigonal planar and the bond angles are 120 degrees, as in BF3. Trigonal pyramidal, 107 degrees is a molecular shape rather than an electron-domain geometry: it is the shape of a center with four domains, one of them a lone pair, as in NH3. Tetrahedral, 109.5 degrees is the arrangement for four electron domains, as if every central atom had four. Square pyramidal, 90 degrees comes from six electron domains, one of them a lone pair, as in BrF5, not from three.

Correct answer: Linear

Linear is correct. The carbon in CO2 has two electron domains, one for each C=O double bond (a double bond counts as one domain), and two domains get as far apart as possible on opposite sides of the carbon, 180 degrees apart. Bent is the classic mix-up with water, but the oxygen in water carries two lone pairs while the carbon in CO2 has none. Tetrahedral comes from counting each C=O double bond as two separate bonding domains, four in all; a multiple bond occupies a single domain, so the carbon has only two. Square planar makes the same miscount and then lays the four bonds flat, as a Lewis drawing looks on paper; even four bonding domains would be tetrahedral, and a square planar shape needs six domains, two of them lone pairs.

Correct answer: 104.5 degrees, because two lone pairs on oxygen push both bonds closer together

104.5 degrees, because two lone pairs on oxygen push both bonds closer together, is correct. Oxygen in water has four electron domains (two bonding pairs and two lone pairs), so its electron geometry is tetrahedral, but lone pairs repel more strongly than bonding pairs and close the H-O-H angle from the ideal 109.5 degrees to about 104.5 degrees. 104.5 degrees, because one lone pair on oxygen squeezes both bonds a bit closer, has the right angle for the wrong reason: the oxygen in water carries two lone pairs, not one. 106.7 degrees, because two lone pairs on oxygen push both bonds closer together, gives the right reason but ammonia's angle; two lone pairs squeeze the bonds more than one does, so water's angle is smaller than ammonia's. 106.7 degrees, because one lone pair on oxygen squeezes both bonds a bit closer, describes ammonia (about 107 degrees), whose nitrogen has only one lone pair; the oxygen in water has two.

Correct answer: How many bonding domains and how many lone pairs it has

How many bonding domains and how many lone pairs it has are the two things VSEPR starts from. From the Lewis structure you count the electron domains around the central atom (a single, double or triple bond each counts as one bonding domain) and sort them into bonding domains and lone pairs; the total sets the electron-domain geometry, and the number of lone pairs decides which molecular shape results. How many neighboring atoms and how many pi bonds it has leaves out the lone pairs, which change the shape (water and carbon dioxide both have two neighboring atoms but different shapes), and pi bonds do not add domains, because a double or triple bond still counts as one. How many polar bonds and how many nonpolar bonds it has describes polarity, which follows from the shape rather than setting it. How many protons and how many unpaired electrons it has identifies the element but does not fix how the electron domains are arranged.

Correct answer: Trigonal pyramidal

Trigonal pyramidal is correct. The four electron domains take a tetrahedral arrangement, but a molecular shape is named for the positions of the atoms only, so with a lone pair at one corner the three bonded atoms and the central atom form a three-sided pyramid, as in ammonia (NH3), with bond angles near 107 degrees. Tetrahedral is the electron-domain geometry, which matches the molecular shape only when all four domains are bonding pairs. Trigonal bipyramidal is the arrangement for five electron domains, not four. Square planar is the shape of four bonded atoms around a center with six electron domains, two of them lone pairs, as in XeF4.

Correct answer: Electron-domain geometry includes lone pairs, and molecular geometry excludes them

Electron-domain geometry includes lone pairs, and molecular geometry excludes them. The electron-domain geometry is set by every electron domain around the central atom, bonding pairs and lone pairs alike, while the molecular geometry describes only where the bonded atoms are. Water shows the difference: its electron-domain geometry is tetrahedral but its molecular geometry is bent, because two of the four positions are lone pairs rather than atoms. Electron-domain geometry excludes lone pairs, and molecular geometry includes them swaps the two definitions. Electron-domain geometry includes lone pairs, and molecular geometry includes them would make the two geometries identical for every molecule, so water would have a tetrahedral shape. Electron-domain geometry excludes lone pairs, and molecular geometry excludes them would also make the two identical, and neither would account for the lone pairs that push the bonded atoms closer together.

Correct answer: Toward the more electronegative atom

Toward the more electronegative atom is correct. The more electronegative atom pulls the shared electrons closer and carries a partial negative charge, leaving the other atom partially positive, and in general chemistry the bond dipole is drawn as an arrow running from the partially positive atom to the partially negative one. Toward the atom with more total mass ties the direction to mass, but mass does not decide it: in an S-O bond the arrow points to the lighter oxygen atom. Toward the atom taking up more space ties it to size, but in a C-O bond the arrow points to oxygen, the smaller atom. Toward the more centrally placed atom ties it to position, but in CF4 every C-F dipole points away from the central carbon, toward the outer fluorine atoms.

Correct answer: Debye

The debye is the unit chemists use. A dipole moment is the size of the separated charge multiplied by the distance between the charges, and molecular values are conveniently expressed in debyes (1 D is about 3.34 × 10^-30 coulomb-meters); water, for example, has a dipole moment of about 1.85 D. The coulomb measures charge alone and the angstrom measures distance alone; a dipole moment combines the two, so neither unit by itself can express it. The volt measures electric potential, not a separation of charge.

Correct answer: Symmetric geometry that lets the bond dipoles cancel out

Symmetric geometry that lets the bond dipoles cancel out is correct. Each polar bond has a dipole, but when identical bonds point in balanced directions, such as the three B-F bonds of trigonal planar BF3 set 120 degrees apart or the two opposed C=O bonds of linear CO2, the bond dipoles add up to zero and the molecule is nonpolar overall. Lone pairs that offset the central atom's bond dipoles gets the effect of lone pairs backward: lone pairs usually break the symmetry and create a net dipole, as in water and ammonia. Resonance that spreads the bond dipoles over the molecule does not remove a net dipole; ozone has resonance and is still polar. Rapid electron motion that averages the bond dipoles away confuses a permanent bond dipole with the fleeting dipoles behind dispersion forces; the electronegativity difference keeps each bond polar all the time.

Correct answer: Check whether the bonds are polar, then whether the shape lets bond dipoles cancel

Check whether the bonds are polar, then whether the shape lets bond dipoles cancel is the standard two-step test. A molecule can be polar only if it has polar bonds, and even then it is nonpolar when its shape arranges the bond dipoles symmetrically so that they cancel: water is polar because its bent shape leaves a net dipole, while CO2 is nonpolar because its linear shape cancels its two bond dipoles. Check whether oxygen atoms are present, then whether the molecule mixes with water confuses polarity with composition and solubility; CO2 contains oxygen and dissolves in water, yet it is nonpolar. Check whether lone pairs are present, then whether the octet rule is satisfied never looks at the bonds; XeF4 has lone pairs on xenon but is nonpolar, and the octet rule says nothing about polarity. Check whether the bond energies are very high, then whether the substance is ionic tests bond strength and bond type, neither of which decides whether the whole molecule has a net dipole.

Correct answer: Carbon tetrachloride (CCl4)

Carbon tetrachloride (CCl4) is correct. Each C-Cl bond is polar, but the four identical bonds point to the corners of a regular tetrahedron, so their dipoles cancel and the molecule is nonpolar. Nitrogen trichloride (NCl3) has a lone pair on nitrogen that makes it trigonal pyramidal, so its bond dipoles do not cancel. Dichloromethane (CH2Cl2) is tetrahedral, but its two C-H bonds and two C-Cl bonds are not identical, so their dipoles do not cancel and the molecule is polar. Hydrogen sulfide (H2S) is bent like water, so its two bond dipoles add up to a net dipole.

Correct answer: The bent shape keeps the bond dipoles uncanceled, so it is polar

The bent shape keeps the bond dipoles uncanceled, so it is polar. In a linear AB2 molecule such as CO2 the two bond dipoles point in exactly opposite directions and cancel. Two lone pairs on the central atom bend the molecule, as in water, so the bond dipoles no longer oppose each other directly and they add up to a net dipole moment. The bent shape makes the bond dipoles balance, so it is nonpolar reverses this: only a straight, symmetric arrangement cancels them. The bent shape lets the lone pairs cancel out, so it is nonpolar misreads the geometry; both lone pairs sit on the side of the central atom away from the bonded atoms, so they add to the polarity rather than canceling. The bent shape makes the bonds ionic, so it is polar is wrong about the bonding: bending the molecule does not turn its covalent bonds into ionic ones, lone pairs stay as unshared (nonbonding) electron pairs, the molecule remains neutral and covalent, and its polarity comes from the partial charges set up by the polar bonds.

Correct answer: Octahedral, with 90-degree angles in three square planes

SF6 is octahedral, with 90-degree angles in three square planes: six bonding domains and no lone pairs point to the six corners of an octahedron, so each S-F bond is 90 degrees from four neighbors and 180 degrees from the bond opposite it, and the fluorine atoms lie in three mutually perpendicular square planes. Hexagonal planar, with 60-degree angles around a flat ring would crowd all six bonds into one plane only 60 degrees apart, far closer than VSEPR allows. Trigonal prismatic, with 60-degree angles in two triangles is a real six-coordinate shape, but it holds the bonding pairs closer together than the octahedron does, so VSEPR does not predict it for SF6. Seesaw, with 90-degree and 120-degree angles on its axes is the shape of SF4, which has four bonds and a lone pair, not six bonds.

Correct answer: T-shaped, with both lone pairs taking equatorial sites

The molecule is T-shaped, with both lone pairs taking equatorial sites: a steric number of 5 gives a trigonal bipyramidal electron arrangement, and the two lone pairs take equatorial sites, where they have fewer neighbors at 90 degrees, which leaves the three bonded atoms in a T, as in ClF3. Trigonal planar, with both lone pairs placed axially is what axial lone pairs would give, but each axial lone pair would sit at 90 degrees to three bonding pairs, a higher-energy arrangement. Trigonal pyramidal, with the lone pairs below the atoms describes a central atom with four domains and one lone pair, such as NH3. Bent, with the two lone pairs taking the place of bonds borrows the shape of water, yet here three atoms are bonded to the central atom.

Correct answer: It becomes more polar, with larger partial charges

As the electronegativity difference grows, the bond changes in one direction: it becomes more polar, with larger partial charges. The more electronegative atom pulls the shared pair closer, so the partial negative charge on that atom and the partial positive charge on its partner both grow. It becomes less polar, with smaller partial charges reverses the relationship. Its polarity stays fixed, then the bond turns ionic treats polarity as an on-off switch, but it grows steadily with the electronegativity difference, and the polar-covalent to ionic border is a gradual one. The less electronegative atom ends up more negative puts the negative end on the wrong atom; the shared pair shifts toward the more electronegative atom.

Correct answer: Trigonal pyramidal, about 107 degrees, as a lone pair closes the angle

Ammonia is trigonal pyramidal, about 107 degrees, as a lone pair closes the angle: nitrogen's four electron domains are arranged roughly tetrahedrally, the shape is named from the three hydrogen atoms, and the lone pair repels the bonding pairs more strongly than they repel each other, narrowing the H-N-H angle from 109.5 to about 107 degrees. Tetrahedral, about 109.5 degrees, as its four electron domains spread out gives the electron-domain geometry rather than the molecular shape. Trigonal pyramidal, about 109.5 degrees, as a lone pair repels like a bond pair names the shape correctly but keeps the ideal tetrahedral angle, as if the lone pair repelled no more strongly than the bonding pairs; in fact it repels more strongly, so the angle closes to about 107 degrees. Tetrahedral, about 107 degrees, as its four electron domains define it has the right angle but names the shape from all four domains, counting the lone pair as if it were an atom.

Correct answer: It donates electrons to the ring, activating it

Through resonance the methoxy oxygen acts as a donor: it donates electrons to the ring, activating it. A lone pair on the oxygen overlaps with the ring's pi system and raises the electron density, especially at the ortho and para positions, so -OCH3 is an activating, ortho/para-directing group. It acts by hyperconjugation, as a methyl group does: that borrows the mechanism of an alkyl group, but the methyl in -OCH3 is bonded to the oxygen rather than to the ring, so no C-H hyperconjugation reaches the ring; the oxygen donates through its lone pair, a much stronger resonance effect. It pulls density away by induction, deactivating it names oxygen's real but weaker inductive pull; resonance donation outweighs it, so the ring ends up activated. It moves density to its meta sites, activating them misplaces the effect, because resonance adds density at the ortho and para positions rather than the meta positions.

Correct answer: Its equal bond dipoles cancel by facing opposite ways

The linear shape makes CO2 nonpolar because its equal bond dipoles cancel by facing opposite ways: the two C=O dipoles are equal in size, and with an O=C=O angle of 180 degrees they point in exactly opposite directions, so their vector sum is zero; a linear molecule with unequal bond dipoles, such as OCS, is still polar. Its quick tumbling in the gas averages the dipoles out is wrong because polarity belongs to the structure of each molecule; rotation does not remove a dipole, and polar gases such as water vapor keep theirs. Its gas molecules sit too far apart to show polarity confuses the spacing between molecules with the charge distribution inside one molecule. Lone pairs on its oxygens offset the pull of each bond is not the reason either: those lone pairs are part of the same symmetric charge distribution, and the cancellation comes from the linear geometry.

Correct answer: Structure determines function: shape guides interactions

The relationship is the principle that structure determines function: shape guides interactions. Two molecules with the same atoms but different three-dimensional shapes differ in polarity, in how they fit binding sites, and in their intermolecular contacts, so their properties differ. Function determines structure: atoms shift to fit a role reverses cause and effect, since a molecule does not rearrange itself to perform a task. Like dissolves like: matching polarity lets liquids mix is a solubility guideline, not the general link between shape and behavior. VSEPR theory: electron pairs keep bonded atoms far apart predicts the shape of a molecule but does not explain how that shape governs its properties.

Correct answer: Trigonal bipyramidal, with 120- and 90-degree angles

PCl5 is trigonal bipyramidal, with 120- and 90-degree angles: five bonding domains and no lone pairs place three chlorine atoms in an equatorial triangle 120 degrees apart and two axial chlorine atoms at 90 degrees to that plane (and 180 degrees from each other). Square pyramidal, with 90-degree and 180-degree angles is the shape for six domains when one is a lone pair, as in BrF5. Pentagonal planar, with 72- and 144-degree bond angles would crowd all five bonds into one plane, much closer together than the trigonal bipyramid allows. Seesaw, with 120-degree and 180-degree ideal angles is the shape for five domains when one of them is a lone pair, as in SF4, and phosphorus in PCl5 has no lone pair.

Correct answer: Hydrogen bonding

Water molecules form hydrogen bonds: a hydrogen attached to a small, highly electronegative oxygen is strongly attracted to a lone pair on a neighboring oxygen. H2S cannot do this to any real extent, because sulfur is larger and far less electronegative, so water boils at 100 °C while H2S boils near -60 °C. That difference comes from hydrogen bonding. O-H covalent bonding holds the atoms together inside each water molecule; boiling separates whole molecules from each other and breaks no O-H bonds. London dispersion forces act in both substances and are actually larger for H2S, which has more electrons, so they cannot explain why water boils so much higher. Ion-dipole forces need ions, and pure water and H2S are molecular substances with essentially no ions present.

Correct answer: Low compressibility

Low compressibility is correct, because it is the property gases lack: gas particles are far apart, so a gas can be squeezed into a much smaller volume, while liquids and solids are nearly incompressible. Negligible cohesion does describe gases, whose particles attract one another so weakly that they spread out to fill any container instead of clumping together as a liquid does. Rapid diffusibility describes them too, since widely spaced gas particles move past one another with little hindrance and mix quickly. Total miscibility is also typical of gases: any two gases mix in all proportions, which is not true of many pairs of liquids.

Correct answer: Cooling the solution

Cooling the solution is correct: gas molecules with less kinetic energy are less able to escape from the liquid, so more gas stays dissolved at a lower temperature; cold water holds more dissolved oxygen than warm water. Heating the liquid does the opposite, which is why a warm soda goes flat faster; the rule that solids dissolve better when hot does not hold for gases. Agitating the mixture only changes how fast equilibrium is reached, and shaking helps dissolved gas escape rather than adding more. Reducing the pressure lowers solubility, because by Henry's law the amount of dissolved gas is proportional to the partial pressure of that gas above the liquid.

Correct answer: Color of the solute

Color of the solute is correct: a solute's color comes from which wavelengths of light it absorbs and has no bearing on how fast its particles break away from the surface and spread into the solvent. Grain size of the solute does affect the rate, since smaller grains expose more surface area to the solvent. Agitation of the solvent affects it as well, because stirring carries dissolved solute away from the surface and brings fresh solvent into contact with it. Heating of the solvent speeds dissolving too, because faster solvent particles strike the surface more often and with more energy.

Correct answer: Ideal solutions

Ideal solutions is correct: Raoult's law, P(solvent) = X(solvent) × P°(solvent), holds exactly only when the attractions between unlike molecules match those between like molecules, which is what defines an ideal solution. Real solutions deviate from it, positively or negatively, because those attractions differ. Ionic solutions deviate strongly, since ions attract water far more than water attracts itself and each formula unit breaks into several particles. Polar solutions are not a category the law is written for; mixtures of polar substances often deviate from it, because hydrogen bonding and dipole attractions make unlike molecules attract each other more or less strongly than like ones.

Correct answer: The boiling point rises

The boiling point rises is correct: a non-volatile solute lowers the solvent's vapor pressure, so the solution must be heated to a higher temperature before its vapor pressure reaches the outside pressure; this is boiling point elevation, ΔTb = Kb × m. The boiling point falls mixes this up with freezing point depression, the companion colligative effect. The boiling point is unchanged assumes that a solute which never evaporates cannot matter, but it still dilutes the solvent at the surface. The boiling point is uncertain treats the outcome as depending on which solute is used; for any non-volatile solute the boiling point goes up, by an amount set by the concentration of dissolved particles.

Correct answer: They move faster and are farther apart

They move faster and are farther apart is correct: gas molecules fly in straight lines between collisions and are separated by distances many times their own size, while liquid molecules stay in contact and slide past one another. They move slower and are widely spaced gets the speed backward. They move faster and are closer together gets the spacing backward, since a liquid, not a gas, is the closely packed fluid. They move slower and are packed together describes a liquid compared with a gas, the reverse of what is asked.

Correct answer: It is decreased by the dissolved particles

It is decreased by the dissolved particles is correct: the solute particles dilute the solvent, so fewer solvent molecules escape from the surface each second and the vapor pressure falls below that of the pure solvent; by Raoult's law it equals the solvent's mole fraction times the pure solvent's vapor pressure. It is increased by the dissolved particles reverses the effect, as if the solute added vapor of its own, but a non-volatile solute contributes essentially none. It is unchanged by the amount of solute assumes that a solute which never evaporates cannot affect the vapor pressure; the solute does not evaporate, but it still lowers the solvent's mole fraction at the surface, so the more solute is dissolved, the lower the vapor pressure. It is set by the amount of solution treats vapor pressure as an extensive property. It is intensive: a cup and a tank of the same solution at the same temperature have the same vapor pressure.

Correct answer: The highest temperature and pressure at which liquid and vapor can coexist

The highest temperature and pressure at which liquid and vapor can coexist is correct: the critical point is the upper end of the liquid–vapor curve on a phase diagram. Beyond it, liquid and gas are no longer separate phases and the substance is a supercritical fluid, so above the critical temperature no amount of compression produces a distinct liquid. The temperature and pressure at which the solid, liquid and vapor can coexist is the triple point, at the opposite, lower end of the liquid–vapor curve. The boiling temperature of a substance under standard pressure is the normal boiling point, one ordinary point on the liquid–vapor curve, well below the critical point. The melting temperature of a solid under standard pressure is the normal melting point, which lies on the solid–liquid line rather than at the end of the liquid–vapor curve.

Correct answer: Ion-dipole > hydrogen bonding > dipole-dipole > London dispersion

Ion-dipole > hydrogen bonding > dipole-dipole > London dispersion is the order for molecules of similar size. Ion-dipole forces act between the full charge of an ion and a polar molecule, as when NaCl dissolves in water, so they are the strongest; hydrogen bonding is the strongest attraction between neutral molecules; ordinary dipole-dipole forces are weaker; and London dispersion forces, present in every substance, are the weakest when sizes are comparable. Hydrogen bonding > ion-dipole > dipole-dipole > London dispersion and Hydrogen bonding > dipole-dipole > ion-dipole > London dispersion both put hydrogen bonding at the top, stretching 'the strongest force between neutral molecules' to cover ions as well; a hydrogen bond is an attraction between partial charges, while an ion carries a full charge, and the second ordering also ranks ion-dipole forces below ordinary dipole-dipole forces. Dipole-dipole > hydrogen bonding > ion-dipole > London dispersion ranks ordinary dipole-dipole forces above hydrogen bonds, which are the strongest kind of dipole-dipole attraction, and puts ion-dipole forces below both.

Correct answer: Nitrogen, oxygen, or fluorine

Nitrogen, oxygen, or fluorine is correct. These three atoms are both small and highly electronegative, so an N-H, O-H or F-H bond leaves the hydrogen strongly positive and nearly bare, able to attract a lone pair on an N, O or F atom of a neighboring molecule. Fluorine, chlorine, or sulfur includes fluorine, but chlorine and sulfur are larger atoms whose charge is too spread out, so HCl and H2S show only ordinary dipole-dipole attractions. Oxygen, chlorine, or carbon includes oxygen, but a C-H bond is barely polar, which is why CH4 cannot hydrogen bond. Nitrogen, sulfur, or carbon includes nitrogen, but sulfur and carbon fail for the same reasons, so every option except Nitrogen, oxygen, or fluorine includes at least one atom that does not form hydrogen bonds.

Correct answer: I2 has many more electrons, so its London dispersion forces are stronger

I2 has many more electrons, so its London dispersion forces are stronger, and that is why it is a solid. Both halogens are nonpolar, so London dispersion forces are the only attractions between their molecules, and those forces grow with the number of electrons and the size of the electron cloud: an I2 molecule has 106 electrons against 18 for F2, and its larger, more polarizable cloud holds the molecules together strongly enough to make a solid at room temperature. I2 has much stronger I-I bonds, so its crystal resists melting and boiling confuses the bond inside each molecule with the attractions between molecules; melting and boiling do not break the I-I bond, which is in fact slightly weaker than the F-F bond. F2 is much lighter, so its molecules are too fast to settle into a solid credits mass alone; at the same temperature both substances have the same average kinetic energy, and what decides the phase is how strongly the molecules attract one another. F2 is far more electronegative, so its molecules repel each other as a gas misapplies electronegativity, which matters only between unlike atoms; the two identical F atoms share their electrons equally, and F2 molecules do not repel one another.

Correct answer: Dipole-dipole attractions arising from its polar C=O bond

Dipole-dipole attractions arising from its polar C=O bond explain the high boiling point. The C=O bond in acetone (CH3COCH3) is strongly polar, so each molecule has a permanent dipole, and the partially positive carbon end of one molecule attracts the partially negative oxygen end of the next. Hydrogen bonds arising from its partly negative oxygen atom would need a hydrogen attached to N, O or F on the neighboring molecule; all of acetone's hydrogens are bonded to carbon, so pure acetone cannot hydrogen bond with itself. London dispersion forces arising from its C-H and C-C bonds are present, but they are about the same as in nonpolar molecules of similar mass, so they cannot explain the difference. Covalent bonding arising from its shared electron pairs holds the atoms within each molecule together; boiling separates whole molecules from one another and breaks no covalent bonds.

Correct answer: Lower vapor pressure and higher viscosity

Lower vapor pressure and higher viscosity both follow from stronger intermolecular forces. Strong attractions hold molecules in the liquid, so fewer escape into the vapor (low vapor pressure), and they make the molecules resist sliding past one another, so the liquid flows slowly (high viscosity). Higher vapor pressure and higher viscosity gets the viscosity right but the vapor pressure backward; molecules that are held more tightly escape less, not more. Lower vapor pressure and lower viscosity gets the vapor pressure right but the viscosity backward; stronger attractions make a liquid flow more slowly, not more easily. Higher vapor pressure and lower boiling point describes the liquid with the weaker intermolecular forces; stronger attractions lower the vapor pressure and raise the boiling point.

Correct answer: Hydrogen bonds draw the surface molecules inward

Water's surface tension is high because hydrogen bonds draw the surface molecules inward: each surface molecule has hydrogen-bonded neighbors beside and below it but none above, so the net pull is inward, and the strength of the hydrogen bonds makes that pull unusually large. Covalent bonds clamp the surface molecules inward confuses the O-H bonds inside each molecule with the attractions between molecules. Ionic forces hold the surface molecules together is wrong because liquid water is molecular; only a tiny fraction of it is ionized. High density packs the surface molecules together does not explain it either, since denser liquids such as dichloromethane have a much lower surface tension than water.

Correct answer: Adhesion to the glass and cohesion of the water

Capillary rise depends on adhesion to the glass and cohesion of the water: water molecules are attracted to the polar glass surface (adhesion) and creep up the walls, while hydrogen bonding among the water molecules (cohesion) pulls the rest of the column up behind them; the narrower the tube, the higher the column rises. Adhesion of the water to the walls, without cohesion leaves out the cohesive forces that lift the bulk of the liquid; without cohesion there would be no surface tension to pull the column up. Air pressure and suction raising the water up a tube describes drinking through a straw, not capillary action, which happens in a tube open to the air at both ends. Evaporation at the top pulling the water up the tube is not needed, because capillary rise also occurs in a closed, saturated tube where no water evaporates.

Correct answer: Its larger electron cloud gives stronger dispersion forces

Xenon boils highest because its larger electron cloud gives stronger dispersion forces: all four noble gases are nonpolar single atoms held together only by London dispersion forces, and xenon's large electron cloud is the most easily distorted (the most polarizable), so its temporary dipoles are the strongest. Its heavier atoms move too slowly to escape into the gas treats mass as the cause, but at the same temperature every gas has the same average kinetic energy; what holds atoms in the liquid is the attraction between them. Its atoms couple through weak covalent bonds in the liquid is false, because liquid xenon contains no Xe-Xe bonds. Its electron cloud sits closer to its nucleus than in neon gets the size backward: xenon's outer electrons are much farther from the nucleus, which is exactly why they are so easily distorted.

Correct answer: Its linear shape gives dispersion forces more surface area

n-Pentane boils higher because its linear shape gives dispersion forces more surface area: the extended chains can lie side by side along their whole length, giving a large contact area, while compact, nearly spherical neopentane molecules touch over a much smaller area, so their dispersion forces are weaker. Its carbon-carbon bonds are stronger and must break to boil is wrong because boiling separates whole molecules and breaks no covalent bonds. Its C-H bonds let its molecules hydrogen bond to one another is false, since C-H bonds are not polar enough for hydrogen bonding. Neopentane's branches let its molecules move faster when hot also fails: at the same temperature both compounds have the same average kinetic energy, and their molar masses are identical.

Correct answer: Iodine and the solvent are both nonpolar, so like dissolves like

The principle is like dissolves like. Iodine and the solvent are both nonpolar, so like dissolves like: nonpolar I2 molecules and nonpolar CCl4 molecules attract each other through dispersion forces of similar strength, so iodine mixes in readily, whereas iodine cannot break into water's strong hydrogen-bonding network. Iodine and the solvent are both polar, so polar binds with polar misclassifies both substances: I2 has no polar bond, and the four polar C-Cl bonds of CCl4 cancel in its tetrahedral shape. Iodine reacts with carbon tetrachloride, so a new compound forms confuses dissolving with a chemical reaction; the iodine is recovered unchanged when the solvent evaporates. Iodine is a heavy solid, so it sinks into carbon tetrachloride mistakes sinking for dissolving; dense solid iodine also sinks in water without dissolving much.

Correct answer: Ion-dipole attractions form between the ions and the polar water molecules

Ion-dipole attractions form between the ions and the polar water molecules. Each Na+ ion attracts the partially negative oxygen ends of nearby water molecules and each Cl- ion attracts their partially positive hydrogen ends; building these hydration shells gives off energy, the hydration energy. Hydrogen bonds develop between the sodium ions and the chloride ions is wrong because neither ion carries a hydrogen atom, and the two ions attract each other ionically, not by hydrogen bonding. Ionic bonds break down between the sodium ions and the chloride ions is the classic bond-breaking error: pulling the ions out of the crystal lattice absorbs energy, so it cannot be the source of the energy released. Electrons transfer directly between the ions and the polar water molecules describes a redox change, but dissolving moves no electrons; each ion keeps its charge throughout.

Correct answer: Ethanol has hydrogen bonds, which the ether lacks

Ethanol has hydrogen bonds, which the ether lacks: one hydrogen in ethanol is bonded directly to oxygen (its O-H group), so ethanol molecules hydrogen bond to one another, while every hydrogen in dimethyl ether is bonded to carbon, so ether molecules cannot hydrogen bond to each other. That extra attraction is why ethanol boils near 78 °C and the ether near -24 °C even though the two share a formula and a molar mass. Ethanol has ionic attraction, which the ether lacks is wrong because the O-H in ethanol is a covalent group, not a hydroxide ion; both compounds are molecular. Ethanol has a polar group, which the ether lacks is wrong because the bent C-O-C group of the ether is also polar and gives it dipole-dipole forces of its own. Ethanol has a straight chain, which the ether lacks is wrong because both molecules are unbranched three-atom chains, C-C-O and C-O-C; the name dimethyl does not mean branched.

Correct answer: Solubility increases in direct proportion to the partial pressure

Solubility increases in direct proportion to the partial pressure. Henry's law states that the concentration of a dissolved gas equals a constant times the partial pressure of that gas above the liquid, so doubling the pressure doubles the solubility; that is why a sealed bottle of soda holds far more CO2 than an open one. Solubility declines in inverse proportion to the partial pressure reverses the relationship, as if dissolved gas behaved like gas volume in Boyle's law. Solubility stays constant, as the temperature alone can change it is wrong because temperature is not the only factor: pressure changes gas solubility too, which is why bubbles appear the moment a bottle is opened. Solubility rises, then stays constant as the liquid saturates misreads saturation, because the saturation concentration is itself what rises with the partial pressure; there is no fixed ceiling.

Correct answer: Hinders crystal formation, so it takes lower temperatures to freeze

Hinders crystal formation, so it takes lower temperatures to freeze. Dissolved solute particles get in the way of water molecules settling into the ordered ice lattice, so the solution must be cooled below 0 °C before ice forms; the size of the drop depends on how many particles are dissolved, not on what they are, which is what makes the effect colligative. Gives off heat as it dissolves, so the water stays too warm to freeze confuses a one-time heat of mixing with an equilibrium property: any heat released soon escapes, yet the lower freezing point remains. Raises the vapor pressure of water, so ice crystals form less readily has the direction wrong, because a nonvolatile solute lowers the vapor pressure of the solvent. Reacts in solution, so a compound with a lower melting point appears is wrong because ethylene glycol does not react with water; it simply dissolves.

Correct answer: They are weak nonpolar dispersion forces, allowing sublimation at ordinary pressure

They are weak nonpolar dispersion forces, allowing sublimation at ordinary pressure. CO2 is linear, so its two C=O bond dipoles cancel and the only attractions between molecules are weak London dispersion forces; at 1 atm the solid passes straight into the gas at -78 °C, because the triple point of CO2 lies near 5.1 atm, above ordinary pressure. They are strong polar dipole-dipole forces, preventing melting at ordinary pressure is wrong: each C=O bond is polar but the molecule is not, and strong attractions would not let the solid vaporize at such a low temperature. They are covalent bonds within each molecule, breaking apart at ordinary pressure is wrong because sublimation separates whole CO2 molecules and leaves their C=O bonds intact. They are hydrogen bonds through its oxygen atoms, holding firm at ordinary pressure is wrong because hydrogen bonding needs a hydrogen atom bonded to N, O, or F, and CO2 contains no hydrogen at all.

Correct answer: Extensive hydrogen bonding among its sugar molecules resists flow

Extensive hydrogen bonding among its sugar molecules resists flow. Glucose and fructose each carry several O-H groups, so the dissolved sugars form a dense network of hydrogen bonds with one another and with water, and that network strongly resists molecules sliding past one another. Dispersion attraction among its nonpolar sugar molecules resists flow is wrong because sugars are highly polar; dispersion forces are present but are not what makes honey thick. Strong covalent bonding among its linked sugar molecules resists flow is wrong because the sugars in honey are separate molecules, not covalently joined to one another, and a flowing liquid never breaks covalent bonds. Strong ion-ion attraction among its sugar molecules resists flow is wrong because sugars are molecular nonelectrolytes that dissolve without forming ions.

Correct answer: 216 grams

The reaction is 2Na + Cl2 → 2NaCl, so each mole of sodium gives one mole of sodium chloride. 85 g of Na is 85/22.99 = 3.70 mol, which forms 3.70 mol of NaCl; at 58.44 g/mol that is 3.70 × 58.44 = 216 grams. 108 grams halves the mole ratio, as if 2 mol of Na gave only 1 mol of NaCl. 131 grams is the mass of Cl2 that reacts (1.85 mol × 70.90 g/mol), not the mass of product. 33 grams turns the molar-mass ratio upside down, multiplying 85 g by 22.99/58.44 instead of 58.44/22.99.

Correct answer: 120 grams

Copper(II) sulfate is CuSO4, with molar mass 63.55 + 32.07 + 4(16.00) = 159.6 g/mol, so 0.75 mol has a mass of 0.75 × 159.6 = 120 grams. 187 grams uses the blue pentahydrate, CuSO4·5H2O (249.7 g/mol), but the reaction calls for copper(II) sulfate itself. 167 grams reads the (II) as two copper atoms and uses Cu2SO4 (223.2 g/mol); the Roman numeral gives copper's charge, 2+, not a subscript. 213 grams divides 159.6 by 0.75 instead of multiplying.

Correct answer: 15 moles

The balanced equation is C3H8 + 5O2 → 3CO2 + 4H2O, so each mole of propane gives 3 moles of CO2, and 5.0 mol of propane gives 5.0 × 3 = 15 moles of CO2. 3 moles is just the coefficient of CO2, the amount from one mole of propane, not from 5.0 mol. 25 moles multiplies by the coefficient of oxygen (5) instead of carbon dioxide (3). 1.7 moles inverts the mole ratio: it divides 5.0 by 3 instead of multiplying, as if 3 mol of propane gave only 1 mol of CO2.

Correct answer: Al

The balanced equation is 4Al + 3O2 → 2Al2O3. 3.0 g of Al is 3.0/26.98 = 0.111 mol, and 4.0 g of O2 is 4.0/32.00 = 0.125 mol. Reacting all of the aluminum needs 0.111 × 3/4 = 0.083 mol of O2, less than the 0.125 mol present, so oxygen is left over and Al runs out first: Al is the limiting reagent. O2 would be limiting only if it ran out first; choosing it usually comes from inverting the 4:3 ratio (0.111 × 4/3 = 0.148 mol of O2 "needed"). Al2O3 is the product, and a product cannot limit a reaction. Neither would require the two reactants to be in exactly the 4:3 mole ratio, and they are not.

Correct answer: 12.0 grams

The balanced equation is 4Fe + 3O2 → 2Fe2O3. 8.4 g of iron is 8.4/55.85 = 0.150 mol, which gives 0.150 × 2/4 = 0.0752 mol of Fe2O3; at 159.7 g/mol that is 12.0 grams. 10.8 grams assumes the oxide is FeO (71.85 g/mol, one mole per mole of iron), the wrong formula. 3.6 grams is only the mass of oxygen that combines with the iron (12.0 - 8.4), not the mass of the oxide. 4.2 grams multiplies the 0.0752 mol of Fe2O3 by iron's molar mass (55.85 g/mol) instead of the oxide's.

Correct answer: 154.2 L

The balanced equation is 2C4H10 + 13O2 → 8CO2 + 10H2O, so each mole of butane gives 8/2 = 4 mol of CO2. 100 g of butane is 100/58.12 = 1.7206 mol, which gives 4 × 1.7206 = 6.882 mol of CO2; at STP that occupies 6.882 × 22.4 L/mol = 154.2 L. 38.5 L skips the 4:1 mole ratio and counts one mole of CO2 per mole of butane. 89.6 L treats the 100 g of butane as if it were 1 mol, skipping the molar mass. 192.7 L uses the water ratio, 10/2 = 5 mol per mole of butane, instead of the CO2 ratio.

Correct answer: 22.5 grams

The balanced equation is 2H2 + O2 → 2H2O. 5.0 g of H2 is 5.0/2.016 = 2.48 mol, and 20.0 g of O2 is 20.0/32.00 = 0.625 mol. Using all the hydrogen would need 1.24 mol of O2, but only 0.625 mol is present, so O2 is the limiting reagent. It forms 2 × 0.625 = 1.25 mol of water, and 1.25 × 18.02 = 22.5 grams. 44.7 grams treats hydrogen as the limiting reagent (2.48 mol of water), which would need more oxygen than is there. 25.0 grams simply adds the two starting masses; mass is conserved only for the material that reacts, and about 2.5 g of hydrogen is left over. 2.5 grams is that leftover hydrogen, not the water formed.

Correct answer: 8.5 grams

8.5 grams is correct. In N2 + 3H2 → 2NH3, 7.0 g of N2 is 7.0/28.0 = 0.25 mol and 3.0 g of H2 is 3.0/2.0 = 1.5 mol. Using up 0.25 mol of N2 takes only 0.75 mol of H2, so nitrogen is the limiting reactant, and 0.25 mol of N2 gives 2 × 0.25 = 0.50 mol of NH3, or 0.50 × 17.0 = 8.5 g. 17.0 grams treats hydrogen as the limiting reactant (1.5 mol H2 × 2/3 = 1.0 mol NH3), but hydrogen is in excess. 7.5 grams finds the right 0.50 mol of NH3 but takes the molar mass of ammonia as 15.0 g/mol, counting one hydrogen atom instead of three (14.0 + 1.0), so 0.50 × 15.0 = 7.5 g. 3.8 grams ignores the 2 in 2NH3, taking only 0.25 mol of NH3, and also uses 15.0 g/mol for ammonia: 0.25 × 15.0 ≈ 3.8 g.

Correct answer: 22 grams

22 grams is correct. CaCO3 → CaO + CO2 has a 1:1 mole ratio between calcium carbonate and carbon dioxide, so 50 g ÷ 100 g/mol = 0.50 mol of CaCO3 releases 0.50 mol of CO2, and 0.50 mol × 44 g/mol = 22 g. 44 grams is the mass of a whole mole of CO2; it skips converting the 50 g of calcium carbonate to 0.50 mol. 28 grams is the mass of the other product, calcium oxide (0.50 mol × 56 g/mol). 14 grams takes the molar mass of CO2 as 28 g/mol, counting only one oxygen atom (12 + 16), so 0.50 mol × 28 g/mol = 14 g.

Correct answer: Cathode

Reduction takes place at the cathode, the electrode where electrons arriving through the external wire are taken up by ions from the solution (for example, Cu^2+ + 2e- → Cu). The anode is where oxidation happens, the opposite half-reaction, so 'Anode' swaps the two electrodes. 'Solution' names the electrolyte that supplies the reacting ions, but the electron transfer itself happens on the electrode surface, not out in the bulk liquid. 'Membrane' names the porous barrier (or salt bridge) that only lets ions drift between the half-cells to keep each side electrically neutral; no electrons are gained or lost there.

Correct answer: A more positive potential means a stronger oxidizing agent

A standard reduction potential measures how strongly a species pulls electrons toward itself, so a more positive potential means a stronger oxidizing agent: the species is more easily reduced (F2, at +2.87 V, is the strongest common oxidant). That same pull makes the species harder to oxidize, so 'A more positive potential means an easier loss of electrons' runs the relationship backward. Standard potentials are intensive, so the claim that doubling a half-reaction doubles its standard potential is false: the coefficients change but E° does not. The claim that a catalyst raises the standard potential of a half-reaction is also false, because a catalyst only changes how fast equilibrium is reached and leaves every thermodynamic quantity, E° included, unchanged.

Correct answer: H2 → 2H+ + 2e-

When the standard hydrogen electrode is the anode, hydrogen gas is oxidized: H2 → 2H+ + 2e-, with each hydrogen going from 0 to +1 and two electrons released so the charge balances (0 on each side). '2H+ + 2e- → H2' is the same half-reaction run as a reduction, which is what happens when the hydrogen electrode serves as the cathode, so it is not the oxidation. 'H2 → 2H + 2e-' drops the positive charge on the hydrogen ions, leaving the equation unbalanced in charge (0 on the left, -2 on the right). '2H + 2e- → H2' makes both slips at once: it is written as a reduction and its hydrogen carries no charge.

Correct answer: 6

Potassium is +1 and oxygen is -2, and the oxidation states in neutral K2Cr2O7 must add up to zero: 2(+1) + 2x + 7(-2) = 0, so 2x = 12 and each chromium is +6. '5' treats potassium as +2 by reading its subscript as a charge, so 2(+2) + 2x - 14 = 0 gives 2x = 10. '8' comes from moving the potassium term across the equals sign without changing its sign, which gives 2x = 14 + 2 = 16. '2' reads the subscript in Cr2 as if it were the oxidation state; a subscript only counts atoms.

Correct answer: Mn

Manganese is the element that is reduced: it is +4 in the reactant MnO2 (each oxygen is -2) and 0 in the product Mn, so each manganese gains four electrons. 'Al' is the species oxidized: aluminum metal goes from 0 to +3 and loses electrons, so it is the reducing agent, not the species reduced. 'Al^3+' is the form aluminum takes in the product Al2O3 after that oxidation; it is produced by losing electrons, and nothing in this reaction adds electrons to it. 'O^2-', the oxide ion, stays at an oxidation state of -2 on both sides of the equation, in MnO2 and in Al2O3, so oxygen is neither oxidized nor reduced here.

Correct answer: 2

Each copper atom loses two electrons (Cu → Cu^2+ + 2e-) and each silver ion gains one (Ag+ + e- → Ag), so the silver half-reaction is doubled to balance: Cu + 2Ag+ → Cu^2+ + 2Ag, and 2 electrons pass from one copper atom to two silver ions. '4' counts each transferred electron twice, once as lost by copper and once as gained by silver, but the two electrons copper gives up are the same two that the silver ions take. '3' adds the electrons in the two half-reactions before balancing them (2 + 1). '6' adds up the coefficients of the balanced equation (1 + 2 + 1 + 2), which counts particles, not electrons.

Correct answer: 5

The coefficient on O2 is 5. Balance carbon first (3 C atoms need 3 CO2), then hydrogen (8 H atoms need 4 H2O), then oxygen: the products contain 3 × 2 + 4 × 1 = 10 O atoms, which is 5 O2 molecules, so the balanced equation is C3H8 + 5O2 → 3CO2 + 4H2O. 10 is the number of oxygen atoms, not O2 molecules; each O2 supplies two atoms. 7 forgets the second oxygen in each CO2: counting 3 + 4 = 7 oxygen atoms calls for 3.5 O2, which becomes 7 when every coefficient is doubled to clear the fraction. 2 copies the subscript in O2 as if it were the coefficient; a subscript is part of the formula and is never changed or read as a coefficient.

Correct answer: 2.0 M

The molarity is 2.0 M. Molarity is moles of solute per liter of solution: 0.50 mol ÷ 0.250 L = 2.0 mol/L. 0.0020 M comes from dividing by 250 without first converting milliliters to liters. 4.0 M counts the Na+ and Cl- ions separately, but the molarity of an NaCl solution counts formula units of NaCl, not the ions they release. 0.0040 M makes both mistakes, doubling for the ions and skipping the unit conversion.

Correct answer: 80%

The percent yield is 80%. Percent yield = (actual yield ÷ theoretical yield) × 100 = (27.2 g ÷ 34.0 g) × 100 = 80%. 20% is the percent lost, (34.0 - 27.2) ÷ 34.0 × 100, not the percent of the theoretical yield that was obtained. 6.8% reports the 6.8 g difference between the two masses as if it were a percentage. 93% reads the 6.8 g lost as 6.8% lost and subtracts that from 100% (100 - 6.8 = 93.2, about 93%); 6.8 g is 20% of the 34.0 g theoretical yield, not 6.8% of it.

Correct answer: Ag+(aq) + Cl-(aq) → AgCl(s)

Ag+(aq) + Cl-(aq) → AgCl(s) is the net ionic equation. Silver chloride is insoluble, so the white precipitate is AgCl, while Na+ and NO3- stay dissolved on both sides as spectator ions and are left out. Ag+(aq) + Na+(aq) + Cl-(aq) → AgCl(s) + Na+(aq) shows the right precipitate but still carries the spectator ion Na+ on both sides; a net ionic equation cancels every spectator ion. Na+(aq) + NO3-(aq) → NaNO3(s) shows sodium nitrate coming out of solution, but sodium salts and nitrates are soluble. Ag+(aq) + NO3-(aq) → AgNO3(s) shows silver nitrate precipitating, but nitrates are soluble; silver nitrate was one of the dissolved starting materials.

Correct answer: Cl2

Cl2 is the limiting reactant. Dividing each amount by its coefficient gives 0.20/2 = 0.10 for Al and 0.20/3 = 0.067 for Cl2, and the smaller value marks the reactant that runs out first; put another way, 0.20 mol Al would need 0.30 mol Cl2, but only 0.20 mol is present. Al is the excess reactant: the 0.20 mol of Cl2 uses up only about 0.13 mol of it. Neither, since equal moles of Al and Cl2 react completely ignores the 2:3 mole ratio; equal amounts would react completely only if the two coefficients were equal. Al2Cl6, the form aluminum chloride takes as a vapor, is a product, so it cannot limit the reaction.

Correct answer: +6

The oxidation number of sulfur in SO4^2- is +6. Each oxygen is -2, so four oxygens contribute -8, and the oxidation numbers must add up to the ion's charge of -2: S + (-8) = -2, so S = +6. +8 comes from setting the sum to zero, as if SO4 were a neutral molecule rather than an ion with a 2- charge. -2 gives the whole ion's charge to sulfur, as if the four oxygens contributed nothing; -2 is sulfur's oxidation number in sulfides such as H2S, not in sulfate. -6 has the right size but the wrong sign, from subtracting in the wrong order, -8 - (-2) = -6.

Correct answer: Cu^2+

Cu^2+ is the oxidizing agent. An oxidizing agent is the species that is reduced: Cu^2+ gains two electrons to become copper metal, and in doing so it takes electrons away from the zinc. Zn is the reducing agent, the species that loses electrons and is itself oxidized. Zn^2+ is what zinc becomes once it has been oxidized, and Cu is what the copper ion becomes once it has been reduced; both are products, not the species that takes electrons from zinc.

Correct answer: 25.0 mL

The stock volume is 25.0 mL. Dilution keeps the moles of HCl the same, so concentration times volume is equal before and after: 12.0 M × (stock volume) = 3.0 M × 100.0 mL, which gives a stock volume of (3.0 × 100.0) ÷ 12.0 = 25.0 mL. That 25.0 mL of stock is then diluted with water to a total of 100.0 mL. 75.0 mL is the volume of water to add, not the volume of stock. 33.3 mL comes from dividing 100.0 mL by 3.0 M and leaving the stock concentration out. 66.7 mL makes both mistakes: it leaves the stock concentration out (100.0 ÷ 3.0 = 33.3 mL) and then reports the water, 100.0 - 33.3 mL, instead of the stock.

Correct answer: 49.3 L

The hydrogen occupies 49.3 L. The 1:1 mole ratio of Zn to H2 gives 2.0 mol of H2, and V = nRT/P = (2.0 mol × 0.0821 L·atm/(mol·K) × 300 K) ÷ 1.00 atm = 49.3 L. 24.6 L uses 1.0 mol of H2, as if the 2 in front of HCl halved the hydrogen produced; each mole of Zn gives one mole of H2. 44.8 L is the volume of 2.0 mol at STP (22.4 L per mole), but this gas is at 300 K, not 273 K. 22.4 L is the molar volume at STP, the volume of just 1.0 mol at 273 K, so it makes both of those mistakes.

Correct answer: 0.20 mol

0.20 mol. The balanced equation 2KI + Pb(NO3)2 → PbI2 + 2KNO3 shows that 2 mol of KI produce 1 mol of PbI2, and KI limits the reaction because lead(II) nitrate is present in excess: 0.40 mol KI × (1 mol PbI2 / 2 mol KI) = 0.20 mol PbI2. 0.40 mol skips the 2:1 mole ratio and assumes one mole of precipitate per mole of KI. 0.10 mol divides by 2 for the coefficient of KI and again by 2 for the iodide subscript in PbI2, counting the same 2:1 relationship twice. 1.6 mol multiplies by 2 for the coefficient of KI and again by 2 for the iodide subscript in PbI2 instead of dividing.

Correct answer: ΔS < 0

ΔS < 0 is correct: joining several simpler molecules into one more complex molecule leaves fewer independent particles that can move and arrange themselves, so the system becomes more ordered and its entropy decreases. ΔS > 0 assumes that a bigger, more complicated molecule means more disorder, but fewer molecules have fewer possible arrangements. ΔS = 0 assumes entropy cannot change because the same atoms are present before and after. ΔS ≥ 0 applies the second law to the system alone; it is the entropy of the universe, not of the reacting system, that cannot decrease in a spontaneous change.

Correct answer: Temperature independent

Temperature independent is correct, because a reaction does not have to be spontaneous at every temperature to be spontaneous under standard conditions; many reactions are spontaneous only above or only below some temperature, as ΔG = ΔH − TΔS shows. Negative free energy change is required: ΔG < 0 is the test for spontaneity at constant temperature and pressure. Positive total entropy change is required as well, since the second law demands that the entropy of the system plus its surroundings increase in any spontaneous change. Product-favored equilibrium is required too, because ΔG° = −RT ln K is negative only when K is greater than 1.

Correct answer: Less than zero

Less than zero is correct: an exothermic reaction releases heat to the surroundings, and by the sign convention heat leaving the system makes its enthalpy change negative (ΔH < 0), whatever the states of the reactants and products. Greater than zero describes an endothermic reaction; it reverses the sign convention, or lets the gases being formed override the word exothermic. Equal to zero would mean no heat is exchanged at constant pressure, which contradicts a reaction that releases heat. Equal to the change in internal energy ignores the work of expansion: forming gases from solids makes the change in moles of gas positive, so ΔH is larger than ΔU, not equal to it.

Correct answer: Heat is absorbed, so ΔH is positive

Heat is absorbed, so ΔH is positive is correct: an endothermic reaction takes in heat from its surroundings, and by convention heat gained by the system makes the enthalpy change positive. Heat is released, so ΔH is negative describes an exothermic reaction instead. Heat is absorbed, so ΔH is negative has the direction of heat flow right but the sign convention backward. Heat is released, so ΔH is positive gets the direction of heat flow wrong and also breaks the convention, since heat leaving a system makes ΔH negative.

Correct answer: Negative

Negative is correct: with ΔS converted to kilojoules, ΔG = ΔH − TΔS = −150 kJ − (298 K)(−0.100 kJ/K) = −150 kJ + 29.8 kJ = −120.2 kJ. Positive is what results from leaving ΔS in joules while ΔH is in kilojoules: −150 − (298)(−100) = +29,650, a units error. Variable suggests the sign could go either way under these conditions, but at a fixed 298 K it is fixed; with ΔH and ΔS both negative, it would turn positive only above T = ΔH/ΔS = 1500 K. Unknown is wrong because ΔH, ΔS and T are all given, so the sign can be calculated.

Correct answer: Never spontaneous at any temperature

Never spontaneous at any temperature is correct: in ΔG = ΔH − TΔS, a positive ΔH contributes a positive term, and a negative ΔS makes −TΔS positive at every temperature above absolute zero, so ΔG stays positive and the reaction is never spontaneous; it can be made to occur only by continuously supplying energy from outside. Always spontaneous at any temperature describes the opposite case, a negative ΔH with a positive ΔS. Spontaneous only at high temperatures fits a positive ΔH with a positive ΔS, where a large TΔS eventually outweighs ΔH. Spontaneous only at low temperatures fits a negative ΔH with a negative ΔS, where ΔH dominates while T is small.

Correct answer: +25 kJ

+25 kJ is correct. By the first law, ΔU = q + w, with q positive for heat absorbed by the system and w negative for work the system does on its surroundings: ΔU = (+40 kJ) + (−15 kJ) = +25 kJ. +55 kJ treats the work done on the surroundings as energy gained, adding 15 kJ instead of subtracting it. -40 kJ counts the absorbed heat as heat released (q = −40 kJ) and leaves out the work entirely. -25 kJ reverses both signs, treating the heat as released and the work as done on the system; that is the energy change of the surroundings, not of the system.

Correct answer: The reaction is non-spontaneous.

A positive ΔG of +30 kJ/mol means the reaction as written cannot proceed on its own at 298 K: the reaction is non-spontaneous. Reading a positive ΔG as favorable reverses the sign convention and leads to the claim that the reaction is favored overall. In fact a positive standard free energy change corresponds to K less than 1, so reactants dominate at equilibrium. ΔG does not reveal ΔH, and a positive ΔG can come from an unfavorable entropy change, so nothing shows that the reaction is heat-absorbing. ΔG also says nothing about speed, so it cannot show that the reaction is inherently slow. How fast a reaction goes depends on its activation energy, not on the sign of ΔG.

Correct answer: The cell potential becomes larger

By the Nernst equation, E = E° − (RT/nF) ln Q, and the cathodic reactant sits in the denominator of Q. Raising its concentration lowers Q, so the cell potential becomes larger; Le Chatelier's principle points the same way, since more reactant pushes the cell reaction forward. 'The cell potential goes down' gets the direction of the change in Q backward. 'The standard cell potential rises' and 'The standard cell potential falls' both confuse E with E°: the standard potential is fixed by the 1 M, 1 bar reference state, so changing an actual concentration moves E but leaves E° where it was.

Correct answer: 2.71 V

Magnesium has the more negative reduction potential (Mg^2+/Mg = -2.37 V), so it is the anode, and copper (Cu^2+/Cu = +0.34 V) is the cathode: E°cell = E°cathode − E°anode = 0.34 − (−2.37) = 2.71 V. '-2.03 V' adds the two tabulated reduction potentials as listed, 0.34 + (−2.37), without reversing the magnesium half-reaction, which runs as an oxidation at the anode. '2.03 V' subtracts the smaller magnitude from the larger (2.37 − 0.34), ignoring the signs altogether. '0.34 V' reports only the copper half-cell, as if the magnesium electrode contributed nothing.

Correct answer: Concentration

The Nernst equation, E = E° − (RT/nF) ln Q, corrects the standard potential through the reaction quotient Q, which is built from the actual concentrations (or partial pressures) of reactants and products; concentration is what standard conditions fix at 1 M and what the Nernst equation lets vary. 'Conductivity' governs how easily current passes through the solution, not the equilibrium voltage, and it appears nowhere in the equation. 'Electrode area' and 'Electrode mass' change how much current or total charge a cell can supply, but neither enters E, which is an intensive quantity.

Correct answer: Exothermic; ΔH is negative

The correct answer is "Exothermic; ΔH is negative". The reaction gives heat to its surroundings (the solution and the thermometer), so the temperature reading rises while the enthalpy of the reacting system falls, which makes ΔH negative. "Endothermic; ΔH is positive" describes the opposite process, a reaction that absorbs heat and makes the thermometer reading drop. "Exothermic; ΔH is positive" names the process correctly but flips the sign convention, treating the heat given off as a gain for the system. "Endothermic; ΔH is negative" keeps the right sign but mislabels the process; an endothermic reaction always has a positive ΔH, so the two halves contradict each other.

Correct answer: -110.5 kJ

The enthalpy change is -110.5 kJ. By Hess's law, keep the first equation as written (ΔH = -393.5 kJ) and reverse the second, CO2 → CO + 1/2 O2, which changes its sign to ΔH = +283.0 kJ. Adding the two cancels CO2 and leaves C + 1/2 O2 → CO, so ΔH = -393.5 + 283.0 = -110.5 kJ. "-676.5 kJ" adds the second equation without reversing it, so its ΔH keeps the wrong sign (-393.5 - 283.0). "-252.0 kJ" reverses the second equation but also halves it to match the 1/2 O2, which is not needed because the CO coefficients already match (-393.5 + 141.5). "-221.0 kJ" doubles both equations to clear the 1/2 and then forgets to halve the result: 2(-393.5) + 2(283.0) = -221.0 kJ, which is the enthalpy change for forming 2 mol of CO.

Correct answer: 1.0 J/(g*C)

The specific heat capacity is 1.0 J/(g*C). Rearranging q = mcΔT gives c = q/(mΔT) = 1200 J / (50.0 g × 24.0 C) = 1200/1200 = 1.0 J/(g*C), where ΔT = 49.0 - 25.0 = 24.0 C. "0.49 J/(g*C)" uses the final temperature, 49.0 C, in place of the temperature change (1200 / (50.0 × 49.0)). "0.32 J/(g*C)" adds the two temperatures instead of subtracting them (1200 / (50.0 × 74.0)). "24 J/(g*C)" divides the heat by the mass only and leaves out the temperature change (1200 / 50.0).

Correct answer: 10,450 J

The water absorbs 10,450 J (about 10.45 kJ). Using q = mcΔT with m = 250.0 g, c = 4.18 J/(g*C) and ΔT = 30.0 - 20.0 = 10.0 C gives q = 250.0 × 4.18 × 10.0 = 10,450 J. "31,350 J" uses the final temperature, 30.0 C, in place of the temperature change (250.0 × 4.18 × 30.0). "598 J" divides by the specific heat instead of multiplying by it (250.0 × 10.0 / 4.18). "1,794 J" makes both slips, dividing by the specific heat and using the final temperature (250.0 × 30.0 / 4.18).

Correct answer: Zero, because the element provides the reference

The standard enthalpy of formation of an element in its most stable form is zero: "Zero, because the element provides the reference". ΔHf° is defined as the enthalpy change for making one mole of a substance from its elements in their standard states, so an element already in that state needs no formation step; it is itself the reference against which the formation enthalpies of compounds are measured, and its value is fixed at exactly zero by definition. "Negative, because the element releases bond energy" treats the element as if it were assembled from free atoms; a formation reaction starts from the elements themselves, so no bonds are made and no bond energy is released. "Variable, because the value depends on the element" treats elements like compounds, whose formation enthalpies differ from one substance to the next; every element in its reference form has the same defined value, zero. "Undefined, because the element is already formed" draws the wrong conclusion from a correct idea: because the element needs no formation step, its value is defined as zero rather than left undefined.

Correct answer: -890.3 kJ

The standard enthalpy of reaction is -890.3 kJ. Using ΔH° = (sum for products) - (sum for reactants): products = (-393.5) + 2(-285.8) = -965.1 kJ; reactants = (-74.8) + 2(0) = -74.8 kJ; so ΔH° = -965.1 - (-74.8) = -890.3 kJ. "-965.1 kJ" stops at the products total and never subtracts the reactants, as if CH4, like O2, had a formation enthalpy of zero. "-679.3 kJ" also drops the coefficient of 2 on water, so the products total only -393.5 + (-285.8) = -679.3 kJ, and again nothing is subtracted. "-754.1 kJ" drops the coefficient of 2 and then adds the reactant term instead of subtracting it (-679.3 - 74.8).

Correct answer: ΔG = +32.5 kJ; non-spontaneous

The answer is "ΔG = +32.5 kJ; non-spontaneous". Convert ΔS to kJ/K (250 J/K = 0.250 kJ/K) and substitute: ΔG = 120 - (350)(0.250) = 120 - 87.5 = +32.5 kJ. A positive ΔG means the forward reaction is non-spontaneous at 350 K; with both ΔH and ΔS positive, it becomes spontaneous only above T = ΔH/ΔS = 480 K. "ΔG = +32.5 kJ; spontaneous" has the right value but reads the sign backward, since a positive ΔG means non-spontaneous. "ΔG = +45.5 kJ; non-spontaneous" plugs in 298 K out of habit instead of the stated 350 K (120 - 74.5). "ΔG = +45.5 kJ; spontaneous" makes both mistakes.

Correct answer: ΔG is negative

A reaction is spontaneous in the forward direction when ΔG is negative, that is, less than zero, at constant temperature and pressure; the process can then proceed without a continuous input of outside energy. "ΔG is positive" describes a non-spontaneous forward reaction, for which the reverse reaction is favored. "ΔG is zero" describes a system at equilibrium, with no net change in either direction. "ΔG is below ΔH" only says that TΔS is positive, meaning the entropy of the system increases; ΔG can be below ΔH and still be positive, so it is not the test for spontaneity.

Correct answer: Negative, because the number of gas molecules drops

The answer is "Negative, because the number of gas molecules drops": four moles of gas (one N2 and three H2) become two moles of NH3 gas, so fewer gas particles share the energy and the number of available microstates drops. "Negative, because the loss of heat lowers disorder" has the right sign for the wrong reason; heat released by the reaction raises the entropy of the surroundings and does not set the sign of the system's entropy change. "Positive, because the size of product molecules grows" overlooks that halving the moles of gas outweighs the slightly more complex NH3 molecule. "Positive, because the breaking of bonds adds disorder" ignores that new N-H bonds form as the N2 and H2 bonds break; the net change in moles of gas decides the sign.

Correct answer: -183 kJ

-183 kJ. Using bond enthalpies, ΔH equals the energy of bonds broken minus the energy of bonds formed. Breaking one H-H bond (436 kJ) and one Cl-Cl bond (243 kJ) costs 679 kJ, and forming two H-Cl bonds releases 2 × 431 = 862 kJ, so ΔH = 679 - 862 = -183 kJ and the reaction is exothermic. +183 kJ subtracts the bonds broken from the bonds formed (862 - 679), which reverses the sign of ΔH. +1541 kJ adds the energy of the bonds broken to the energy of the bonds formed (679 + 862) instead of subtracting one from the other. -1541 kJ treats breaking bonds as if it released energy too, adding -679 kJ to -862 kJ; breaking a bond always absorbs energy, and only forming one releases it.

Correct answer: 6.01 kJ

6.01 kJ. The sample is 18.0 g ÷ 18.0 g/mol = 1.00 mol of ice, and melting at 0 °C absorbs the molar enthalpy of fusion for each mole: q = 1.00 mol × 6.01 kJ/mol = 6.01 kJ. 3.00 kJ divides the mass by the enthalpy of fusion (18.0 ÷ 6.01) instead of converting grams to moles and multiplying. 0.334 kJ divides the enthalpy of fusion by the molar mass (6.01 ÷ 18.0), which is the heat per gram, and stops before multiplying by the 18.0 g. 0.000 kJ applies q = mcΔT with ΔT = 0, but a phase change at constant temperature still absorbs heat; that latent heat is exactly what the enthalpy of fusion measures.

Correct answer: K is large, so products are favored

K is large, so products are favored. ΔG° = -RT ln K, so a large negative ΔG° makes ln K large and positive, which means K is much greater than 1 and the equilibrium mixture is mostly products. K is small, so reactants are favored drops the minus sign in that relationship, turning a negative ΔG° into a K below 1. K is large, so reactants are favored reads K upside down; K places products over reactants, so a large K means products dominate. K is small, so products are favored makes both of those slips at once.

Correct answer: It is spontaneous at all temperatures

It is spontaneous at all temperatures. With ΔH negative and ΔS positive, the -TΔS term is negative at every temperature above absolute zero, so ΔG = ΔH - TΔS stays negative; both the enthalpy and the entropy terms favor the reaction. It is spontaneous only at high temperatures describes a reaction with ΔH and ΔS both positive. It is spontaneous only at low temperatures describes ΔH and ΔS both negative. It is nonspontaneous at all temperatures describes ΔH positive with ΔS negative, where both terms oppose the reaction.

Correct answer: The total entropy of the universe increases

The total entropy of the universe increases. The second law states that ΔS(universe) = ΔS(system) + ΔS(surroundings) is greater than zero for every spontaneous process. The total entropy of the universe stays fixed treats entropy like energy: energy is conserved, but entropy is created in every spontaneous, irreversible change and stays constant only in an idealized reversible process. The entropy of the system by itself must rise is wrong because the system can lose entropy, as when water freezes below 0 °C, provided the surroundings gain more. The entropy of the surroundings must rise is wrong for the reverse reason: when ammonium nitrate dissolves spontaneously and endothermically, the surroundings lose entropy while the system gains more.

Correct answer: 0.33 s^-1

0.33 s^-1 is correct. The two-point Arrhenius equation gives ln(k2/k1) = (Ea/R)(1/T1 − 1/T2) = (75,000 J/mol ÷ 8.314 J/(mol·K)) × (1/300 K − 1/350 K) = 9021 K × 0.000476 K^-1 = 4.30, so k2/k1 = e^4.30 = 73 and k2 = (4.5 × 10^-3 s^-1)(73) = 0.33 s^-1. 0.14 s^-1 applies the rule of thumb that a rate doubles for every 10 K rise (2^5 = 32) instead of using the given activation energy. 0.0053 s^-1 scales k in direct proportion to the absolute temperature (350/300), which is not how rate constants respond. 0.0045 s^-1 enters Ea as 75 instead of 75,000 J/mol while R is in joules, so the exponent shrinks to 0.0043 and k barely changes.

Correct answer: 200 s

200 s is correct: for a second-order reaction t1/2 = 1/(k[A]0) = 1/((0.1 M^-1 s^-1)(0.05 M)) = 1/(0.005 s^-1) = 200 s. 6.9 s uses the first-order formula, t1/2 = 0.693/k, which ignores the starting concentration. 139 s carries the 0.693 from the first-order formula into the second-order one, 0.693/(k[A]0). 10 s drops the initial concentration and takes t1/2 = 1/k, as if the half-life were independent of [A]0, which is true only of first-order reactions.

Correct answer: Lowering the activation energy.

A catalyst speeds up a reaction by lowering the activation energy. It provides an alternative pathway with a smaller energy barrier, so a larger fraction of collisions have enough energy to react. It does not work by shifting the equilibrium mixture. A catalyst speeds the forward and reverse reactions equally, so the system only reaches the same equilibrium sooner. Nor does it act by raising the activation energy. A higher barrier would leave fewer collisions energetic enough to react and would slow the reaction down. It also cannot work by raising the equilibrium constant. K depends only on temperature and is the same with or without a catalyst.

Correct answer: The rate is cut to one-quarter

The rate is cut to one-quarter is correct: the rate law is second order in A, so halving [A] multiplies the rate by (1/2)^2 = 1/4. The rate is cut to one-half treats the reaction as first order in A and ignores the exponent of 2. The rate is multiplied by four uses the right exponent but the wrong direction; lowering the concentration of a reactant with a positive order slows the reaction. The rate is multiplied by eight reverses the direction and also uses the overall order of 3 (2 for A plus 1 for B) instead of the order in A.

Correct answer: The rate quadruples.

'The rate quadruples.' is correct: the rate depends on [B]^2, so doubling [B] multiplies the rate by 2^2 = 4, while [A] and k are unchanged. 'The rate doubles.' would hold only if the reaction were first order in B. 'The rate increases eightfold.' raises 2 to the overall order of 3 (first order in A plus second order in B) instead of to the order in B alone. 'The rate stays the same.' confuses the rate with the rate constant k, which does not change when a concentration changes.

Correct answer: A → B

A → B is correct: the slowest step in a mechanism limits the overall rate, the way the slowest stage of an assembly line limits its output, and the stem labels A → B as the slow step. B + C → D is the fast step; it can consume B only as quickly as the slow first step produces it, so it does not set the rate. A + C → D is the overall reaction, found by adding the two steps and canceling the intermediate B; it is not itself a step of the mechanism. Each step would be the answer only if the two steps ran at comparable rates, but here one step is slow and the other fast.

Correct answer: The half-life gets shorter

The half-life gets shorter is correct: for a first-order reaction t1/2 = 0.693/k, and by the Arrhenius equation a higher temperature increases k, especially with an activation energy as large as 125 kJ/mol, so the half-life falls. The half-life grows longer reverses the effect of temperature on k. The half-life cannot change overextends the fact that a first-order half-life does not depend on concentration, or borrows from radioactive decay, whose half-life is unaffected by temperature; a chemical first-order half-life still depends on k, which rises with temperature. The half-life depends on ΔH applies equilibrium reasoning to kinetics; whether a reaction is exothermic or endothermic, its rate constant rises with temperature when the activation energy is positive.

Correct answer: The redox reaction speeds up

Electrode reactions happen at the surface where the metal meets the solution, so doubling the electrode area gives more sites for electron transfer and the redox reaction speeds up; the cell can deliver a larger current. 'The redox reaction slows down' reverses that effect. The cell potential depends on which half-reactions are paired and on the concentrations, not on how big the electrodes are, so 'The cell potential is doubled' wrongly treats voltage as if it scaled with electrode size, and 'The cell potential drops' is wrong for the same reason in the other direction.

Correct answer: The minimum collision energy that molecules need to react

Activation energy is the minimum collision energy that molecules need to react: the height of the energy barrier between the reactants and the transition state, so only collisions carrying at least this much energy can form the activated complex. "The energy change that reactants undergo to form products" is the enthalpy change, ΔH, the energy difference between reactants and products, which is a separate quantity that does not set the rate. "The extra energy that a catalyst contributes to molecules" reverses the role of a catalyst, which lowers the barrier by offering another pathway rather than supplying energy. "The total bond energy that reactants need to break apart" confuses the barrier with the full bond energy of the reactants; the transition state forms while old bonds are only partly broken and new ones partly formed, so the barrier is far smaller.

Correct answer: k increases because the exponential factor approaches one

The correct statement is "k increases because the exponential factor approaches one". In k = A e^(-Ea/(RT)), a higher T makes Ea/(RT) smaller, so the exponent -Ea/(RT) becomes less negative and the factor e^(-Ea/(RT)) grows toward its upper limit of 1, which drives k up; A and Ea themselves are treated as constants. "k increases because heating reduces the activation energy" is a common misconception: Ea is a fixed property of the reaction pathway, and only a catalyst, not heating, provides a lower barrier. "k increases because the magnitude of the exponent grows larger" has the arithmetic backward; the size of Ea/(RT) shrinks as T rises, and a larger negative exponent would make k smaller. "k increases because the equilibrium constant also grows larger" confuses the rate constant k with the equilibrium constant K, which can rise or fall with temperature depending on ΔH and does not set the rate.

Correct answer: 3.7 × 10^-3 s^-1

The rate constant at 320 K is about 3.7 × 10^-3 s^-1. Ea/R = 50,000 J/mol ÷ 8.314 J/(mol·K) = 6014 K, and 1/310 - 1/320 = 1.008 × 10^-4 K^-1, so ln(k2/k1) = 6014 × 1.008 × 10^-4 = 0.606 and k2/k1 = e^0.606 = 1.83. Then k2 = 1.83 × 2.0 × 10^-3 s^-1 = 3.7 × 10^-3 s^-1; a 10 K rise nearly doubles k. "8.1 × 10^-3 s^-1" raises 10 rather than e to the power 0.606 (10^0.606 = 4.04), treating ln as log. "2.0 × 10^-3 s^-1" leaves Ea in kilojoules (50 instead of 50,000 J/mol), so ln(k2/k1) is only 0.0006 and k barely changes from its value at 310 K. "1.2 × 10^-3 s^-1" multiplies k1 by ln(k2/k1) = 0.606 instead of by e^0.606, forgetting to undo the natural logarithm.

Correct answer: Third order

The reaction is third order overall. The overall order is the sum of the exponents in the experimental rate law, rate = k[A][B]^2: 1 for A plus 2 for B gives 3, so the reaction is first order in A, second order in B and third order overall. "Second order" comes from adding the coefficients in the balanced equation (1 + 1) or from reading only the exponent on B; coefficients do not set the order, which is found by experiment. "First order" reports only the order in A. "Fourth order" mixes the two sources, adding the coefficient of B from the balanced equation to the rate-law exponents (1 + 2 + 1); only the exponents count.

Correct answer: Second order

The reaction is second order in X. Doubling [X] multiplies the rate by 4, so 2^n = 4 and n = 2; comparing initial rates while changing one concentration at a time is the standard way to find each exponent in a rate law. "First order" would only double the rate (2^1 = 2). "Fourth order" mistakes the factor of 4 for the order itself; a fourth-order dependence would raise the rate 16-fold (2^4). "Third order" would raise the rate 8-fold (2^3).

Correct answer: ln[A] versus time

A plot of ln[A] versus time gives a straight line for a first-order reaction, because ln[A] = ln[A]0 - kt has the form y = b + mx with slope -k. "1/[A] versus time" is linear for a second-order reaction, not a first-order one. "[A] versus time" is linear for a zero-order reaction, whose rate does not depend on concentration ([A] = [A]0 - kt). "1/[A]^2 versus time" is the straight-line plot for a third-order reaction. Testing which plot is straight is the standard way to find the order from concentration-time data, and for first order only the logarithmic plot is linear.

Correct answer: 30.0 s

The half-life is 30.0 s. For a first-order reaction t1/2 = ln 2 / k = 0.693/k, so 0.693 ÷ 0.0231 s^-1 = 30.0 s, and this value does not depend on the starting concentration. "13.0 s" uses log 2 = 0.301 instead of ln 2 = 0.693 (0.301 ÷ 0.0231). "62.5 s" puts 0.693 in the denominator alongside k, 1/(0.693 × 0.0231). "99.7 s" uses 2.303, the factor that converts natural logarithms to base-10 logarithms, in place of 0.693 (2.303 ÷ 0.0231).

Correct answer: It stays fixed whatever the starting concentration

For a first-order reaction the half-life is constant: "It stays fixed whatever the starting concentration". Because t1/2 = 0.693/k contains only the rate constant, each successive half-life takes the same time no matter how much reactant remains. "It varies inversely with the starting concentration" describes a second-order reaction, whose half-life is 1/(k[A]0). "It gets shorter as the reactant concentration falls" describes a zero-order reaction, whose half-life [A]0/(2k) shrinks as the reactant is used up. "It depends on k and on the reactant concentration" is true for zero- and second-order reactions, but the first-order half-life depends on k alone.

Correct answer: Colliding molecules need enough energy and a suitable orientation

Collision theory says a collision is effective only when "Colliding molecules need enough energy and a suitable orientation": they must carry at least the activation energy and strike with the reacting atoms lined up. Collisions that are too gentle or wrongly angled bounce apart unchanged, which is why only a small fraction lead to reaction. "Colliding molecules need enough energy and a catalyst present" adds a requirement that does not exist; a catalyst speeds many reactions, but collisions can react without one. "Colliding molecules need high pressure and a suitable orientation" wrongly requires high pressure and drops the energy condition; pressure changes how often molecules collide, not whether a given collision succeeds. "Colliding molecules need high pressure and a catalyst present" makes both mistakes and leaves out energy and orientation altogether.

Correct answer: It raises the share of collisions that exceed the activation energy

Raising the temperature shifts the Maxwell-Boltzmann distribution toward higher energies, so "It raises the share of collisions that exceed the activation energy": many more collisions are energetic enough to react, which is the main reason rates climb with temperature. "It chiefly raises the collision frequency that limits the rate" overstates a minor effect; collisions become only a few percent more frequent for a 10 K rise, far too little to explain rates that roughly double. "It improves the orientation that colliding molecules need to adopt" is wrong because the fraction of collisions with a suitable orientation does not depend on temperature. "It improves the yield of products that the final equilibrium allows" mixes kinetics with equilibrium; the equilibrium yield concerns how far a reaction goes, not how fast it gets there.

Correct answer: rate = k[A]^2

The predicted rate law is rate = k[A]^2. The overall rate is set by the slow, rate-determining step, A + A → C, which is bimolecular in A, so its rate law is second order in A. "rate = k[A]^2[B]" writes the rate law from the overall equation, 2A + B → D, and includes B, which reacts only in the fast step after the bottleneck. "rate = k[2A]" turns the two A particles into a coefficient inside the brackets instead of an exponent, which makes the rate first order in A. "rate = k[C][B]" writes the rate law of the fast second step instead of the slow first step; a fast step that follows the bottleneck cannot set the overall rate.

Correct answer: The slowest step, the one named the rate-determining step

The answer is "The slowest step, the one named the rate-determining step". Just as the narrowest section of a pipe limits the total flow, the slowest elementary step limits how fast products can form, so the observed rate law usually reflects that step. "The final step, the one forming the overall final product" assumes that position in the sequence matters, but the slow step can come first, last or in between. "The most favorable step, the one with the largest K value" confuses equilibrium with kinetics; how far a step lies toward products says nothing about how fast it goes. "The reversible step, the one that sets up the equilibrium" describes a fast pre-equilibrium in many mechanisms; such a step adjusts quickly and waits on the slow step that follows it.

Correct answer: It speeds the reaction by opening pathways with smaller activation energy

A catalyst works because "It speeds the reaction by opening pathways with smaller activation energy": with a smaller barrier, a larger fraction of collisions is effective at the same temperature, so both the forward and reverse reactions speed up. "It speeds the reaction by shifting the equilibrium to favor the products" is the classic misconception; because the forward and reverse rates rise by the same factor, the position of equilibrium and ΔH are unchanged, and equilibrium is simply reached sooner. "It speeds the reaction by supplying reactants with additional heat energy" is wrong because a catalyst adds no energy; it lowers the energy that is needed. "It speeds the reaction by being used up as the new products are formed" is wrong because a catalyst is regenerated at the end of the cycle and is not consumed.

Correct answer: 54 kJ/mol

The activation energy is about 54 kJ/mol. With k2/k1 = 2, ln 2 = 0.693 and 1/300 - 1/310 = 1.075 × 10^-4 K^-1, so Ea = R ln 2 ÷ (1/T1 - 1/T2) = 8.314 × 0.693 ÷ (1.075 × 10^-4) = 53,600 J/mol, or 54 kJ/mol, in line with the rule of thumb that many rates roughly double for each 10 K rise near room temperature. "6.4 kJ/mol" leaves R out of the calculation (0.693 ÷ 1.075 × 10^-4 = 6450 J/mol). "23 kJ/mol" uses log 2 = 0.301 in place of ln 2 (8.314 × 0.301 ÷ 1.075 × 10^-4 = 23,300 J/mol). "2.8 kJ/mol" makes both slips, using log 2 and leaving out R (0.301 ÷ 1.075 × 10^-4 = 2800 J/mol).

Correct answer: The rate stays constant whatever [A] is, until A is consumed

For a zero-order reaction, rate = k[A]^0 = k, so "The rate stays constant whatever [A] is, until A is consumed": changing [A] has no effect until the reactant is essentially used up. This often happens when a catalyst surface or an enzyme is saturated, so adding more reactant cannot speed the process. "The rate is directly proportional to [A], until A runs out" describes a first-order reaction. "The rate is zero, since [A] raised to the power zero is zero" misapplies the exponent; any nonzero quantity raised to the power zero equals 1, so the rate equals k, not zero. "The rate is fixed by the starting [A], since k depends on it" is wrong because k depends on temperature, not on concentration; the constant rate equals k whatever [A] was at the start.

Correct answer: L mol^-1 s^-1

L mol^-1 s^-1. For rate = k[A][B] the rate has units of mol L^-1 s^-1 and [A][B] has units of mol^2 L^-2, so k = rate / ([A][B]) has units of mol L^-1 s^-1 divided by mol^2 L^-2, which is L mol^-1 s^-1, the units of every second-order rate constant. mol^3 L^-3 s^-1 multiplies the rate by the two concentrations instead of dividing by them. L mol s^-1 subtracts the mol exponents the wrong way round (2 - 1 instead of 1 - 2), leaving mol on top. mol L^-1 s^-1 gives k the units of the rate itself, which are the units of a zero-order rate constant, not of a k multiplied by two concentrations.

Correct answer: rate = k[A]

rate = k[A]. Doubling [A] with [B] held constant doubles the rate, so the reaction is first order in A (2^1 = 2); doubling [B] with [A] held constant leaves the rate unchanged, so it is zero order in B (2^0 = 1) and [B] drops out of the rate law. rate = k[A]^2 mistakes the factor of 2 for the order; a second-order dependence would make the rate four times larger, since 2^2 = 4. rate = k[B] swaps the two experiments and attaches the dependence to the wrong reactant. rate = k[B]^2 makes both of those slips at once.

Correct answer: The reaction moves forward to create more products

The reaction moves forward to create more products is correct: Q < K means the mixture holds too little product relative to its equilibrium ratio, so reactants are converted to products, raising Q until it equals K. The reaction moves backward to form more reactants is the response to Q > K, when there is excess product. The reaction moves forward to use up all reactants overshoots: the net forward change stops once Q reaches K, which normally happens well before the reactants run out. The reaction moves backward to use up all products gets both the direction and the end point wrong.

Correct answer: Kc remains unchanged

Kc remains unchanged is correct: the equilibrium constant depends only on temperature, so adding reactant changes Q, not K, and the system then shifts toward products until the concentrations again satisfy the same Kc. Kc decreases twofold assumes that doubling a concentration in the denominator of the expression halves the constant itself. Kc is also doubled assumes the constant rises in proportion to the amount of reactant added. Kc drops fourfold goes further and squares the doubled concentration, but no change in concentration alters Kc at constant temperature.

Correct answer: 0.054

0.054 is correct. Forming 0.50 M of C uses x = 0.50/3 = 0.167 M of A and 2x = 0.333 M of B, leaving [A] = 0.833 M and [B] = 1.667 M, so Kc = [C]^3/([A][B]^2) = (0.50)^3/((0.833)(1.667)^2) = 0.125/2.31 = 0.054. 0.031 plugs in the initial concentrations of A and B, 0.125/((1.00)(2.00)^2), instead of their equilibrium values. 0.090 forgets to square [B], computing 0.125/((0.833)(1.667)) = 0.125/1.39. 0.36 leaves out all of the exponents, computing [C]/([A][B]) = 0.50/1.39.

Correct answer: Shrinking the container volume

Shrinking the container volume is correct: compressing the mixture raises the total pressure, and the equilibrium shifts toward the side with fewer moles of gas; one mole of N2O4 replaces two moles of NO2, so more N2O4 forms. Enlarging the container volume lowers the pressure and shifts the equilibrium toward the two moles of NO2. Adding neon at constant volume raises the total pressure but leaves the partial pressures of NO2 and N2O4 unchanged, so the equilibrium does not shift. Adding heat at constant volume favors the endothermic direction, and because forming N2O4 releases heat, warming the mixture turns N2O4 back into NO2 (the brown color deepens).

Correct answer: It is endothermic

It is endothermic is correct: by the van't Hoff equation, K rises with temperature only when the forward reaction absorbs heat, and going from K = 10 at 300 K to K = 25 at 350 K implies ΔH of about +16 kJ/mol. It is exothermic reverses that relationship; heating an exothermic reaction lowers K. It is thermoneutral cannot be right, because with ΔH = 0 the value of K would not change with temperature. It is endergonic confuses ΔH with ΔG; since K is greater than 1 at both temperatures, ΔG° = −RT ln K is negative and the reaction is exergonic.

Correct answer: If added at constant volume

If added at constant volume is correct: an inert gas added to a rigid container raises the total pressure but leaves the partial pressures of H2, I2 and HI unchanged, so Q still equals K and the equilibrium does not shift (and because both sides have two moles of gas, even a change in volume would not shift this reaction). If added with excess hydrogen, the extra H2 makes Q smaller than K and the equilibrium shifts toward HI. If added at higher temperatures, the temperature change alters K itself, so the equilibrium position moves. If added with excess iodine gas, the added reactant again drives the equilibrium toward HI.

Correct answer: Favors SO3 production

Favors SO3 production is correct: the forward reaction is exothermic, so heat behaves like a product; cooling removes heat, and the system partly replaces it by forming more SO3, which also means K increases. Favors SO3 breakdown is the response to heating an exothermic reaction, not to cooling it. Position is unchanged would hold only for a reaction with ΔH = 0; unlike adding a catalyst, a temperature change alters K itself. Position is uncertain is wrong because knowing that the reaction is exothermic is enough to predict the direction of the shift.

Correct answer: 1.33 M

1.33 M is correct. With extent x, [C] = [D] = x and [A] = [B] = 2.0 − x, so K = x^2/(2.0 − x)^2 = 4.0; taking the square root of both sides gives x/(2.0 − x) = 2, so x = 4.0 − 2x and x = 1.33 M. 0.67 M is 2.0 − 1.33, the concentration of A (or B) left at equilibrium, not of D. 1.00 M assumes the mixture ends with equal amounts of reactants and products, which would be true only if K were 1. 1.60 M skips the square root and solves x/(2.0 − x) = 4.0.

Correct answer: 0.6 atm

0.6 atm is correct. With x as the pressure of PCl5 that dissociates, Kp = x^2/(0.4 − x) = 0.25, so x^2 + 0.25x − 0.1 = 0 and x = 0.215 atm; the total pressure is (0.4 − x) + x + x = 0.4 + x = 0.615 atm, or 0.6 atm. 0.7 atm comes from the small-x shortcut: treating 0.4 − x as just 0.4 gives x = sqrt(0.25 × 0.4) = 0.32 atm and a total of about 0.72 atm, but x is not small compared with 0.4 atm, so the quadratic is needed. 0.4 atm assumes the total pressure stays at its starting value, but each PCl5 that dissociates becomes two gas molecules. 0.2 atm is x itself, the partial pressure of Cl2 (or of PCl3), not the total pressure.

Correct answer: Equilibrium favors the reactants

Equilibrium favors the reactants is correct: K is the ratio of the product terms to the reactant terms once equilibrium is reached, so a value below 1 means the reactant terms are larger and relatively little of the compound has decomposed. Equilibrium favors the products would require K greater than 1. Equilibrium is reached very slowly confuses the equilibrium constant with the rate; K tells how far a reaction goes, not how fast it gets there. Equilibrium is reached when heated confuses K with the enthalpy change or the activation energy; a small K does not mean the reaction needs heat, and it says nothing about whether the reaction absorbs or releases heat.

Correct answer: Decreases

Decreases is correct: the reaction is exothermic (ΔH = −30 kJ/mol), so by the van't Hoff equation raising the temperature lowers K. Equivalently, ΔG° = ΔH° − TΔS° = −30 kJ/mol + T(0.080 kJ/(mol·K)) becomes less negative as T rises, and K = e^(−ΔG°/RT) falls. Increases is the behavior of an endothermic reaction. Unchanged treats K as a fixed number, but an equilibrium constant changes with temperature even though it does not change with concentration or pressure. Uncertain is wrong because the sign of ΔH alone fixes the direction of the change.

Correct answer: 5.04

The pH is 5.04. The Henderson-Hasselbalch equation is pH = pKa + log([A-]/[HA]), where A- is acetate and HA is acetic acid: pH = 4.74 + log(0.20/0.10) = 4.74 + log 2 = 4.74 + 0.30 = 5.04. With more conjugate base than acid, the pH sits above the pKa. "4.44" inverts the ratio, using log([HA]/[A-]) = log 0.5 = -0.30. "5.43" uses the natural logarithm, ln 2 = 0.69, instead of log 2. "6.74" forgets to take the logarithm and adds the concentration ratio itself, 2, to the pKa (4.74 + 2).

Correct answer: Kc = [C] / ([A]^2 [B])

The correct expression is Kc = [C] / ([A]^2 [B]). The equilibrium constant puts product concentrations in the numerator and reactant concentrations in the denominator, each raised to the power of its coefficient in the balanced equation, so the 2 in front of A becomes the exponent on [A]. "Kc = [C] / (2[A] [B])" turns the coefficient into a multiplier instead of an exponent; coefficients become exponents, never factors. "Kc = (2[A] [B]) / [C]" makes the same multiplier error and also inverts the expression, putting reactants over products. "Kc = [C]^2 / ([A] [B])" attaches the coefficient to the wrong species; the 2 belongs to A, and C has a coefficient of 1.

Correct answer: Kp = Kc(RT)^-2

The correct relationship is Kp = Kc(RT)^-2. In general Kp = Kc(RT)^Δn, where Δn is the moles of gaseous product minus the moles of gaseous reactant; here Δn = 2 - (1 + 3) = -2. "Kp = Kc(RT)^2" takes Δn as reactants minus products (4 - 2), which flips its sign. "Kp = Kc(RT)^-4" leaves out the gaseous product and uses only the reactant moles (0 - 4). "Kp = Kc(RT)^4" makes both slips, subtracting the wrong way and leaving out the product (4 - 0). Kp equals Kc only when Δn = 0, which is not the case for ammonia synthesis.

Correct answer: Removing SO3 from the system

Removing SO3 from the system shifts the equilibrium toward the products: by Le Chatelier's principle, taking away a product makes the system react forward to replace it. "Removing SO2 from the system" takes away a reactant, so the equilibrium shifts back toward the reactants. "Adding heat to the system" raises the temperature; because the forward reaction is exothermic, heat acts as a product, so adding it drives the equilibrium toward the reactants. "Adding neon to the system" is the inert-gas trap: at constant volume it changes no partial pressure and causes no shift, and if the volume expands to keep the pressure constant, the equilibrium moves toward the side with more moles of gas, the reactants (3 moles versus 2).

Correct answer: 3.00

The pH is 3.00. HCl is a strong acid that ionizes completely, so [H+] = 0.0010 M = 1.0 × 10^-3 M, and pH = -log(1.0 × 10^-3) = 3.00. "11.00" subtracts 3.00 from 14, which gives the pH of a 0.0010 M strong base, not acid. "6.91" uses the natural logarithm, -ln(0.0010), instead of the base-10 logarithm. "4.00" counts the four decimal places in 0.0010 as the exponent, treating the concentration as 1 × 10^-4 M. A weak acid at the same concentration would have a higher pH because it ionizes only partly.

Correct answer: 5.6 × 10^-10

Ka for the ammonium ion is 5.6 × 10^-10. For a conjugate acid-base pair at 25 degrees C, Ka × Kb = Kw = 1.0 × 10^-14, so Ka = Kw/Kb = (1.0 × 10^-14)/(1.8 × 10^-5) = 0.56 × 10^-9 = 5.6 × 10^-10. Because ammonia is a weak base, its conjugate acid is correspondingly weak. "5.6 × 10^-9" moves the decimal point in 0.56 but forgets to lower the exponent by one. "5.6 × 10^-20" adds the exponents, -14 + (-5) = -19, instead of subtracting them, -14 - (-5) = -9, which gives 0.56 × 10^-19 = 5.6 × 10^-20. "5.6 × 10^-19" makes both slips: it adds the exponents and then writes 0.56 × 10^-19 as 5.6 × 10^-19 without lowering the exponent.