AP Unit 1 Solutions

1. Mass = 0.125 mol × 101.96 g mol⁻¹ = 12.745 g, reported as 12.7 g.
2. Formula-unit count = 0.125 × 6.022 × 10²³ = 7.5275 × 10²². Each formula unit represents three O atoms, so the count is 2.25825 × 10²³, reported as 2.26 × 10²³ O atoms. The subscript 3 is an exact counting factor; the measured 0.125 mol limits these results to three significant figures.
3. The average mass represented by one formula unit is 101.96 u. Multiplying by 1.66054 × 10⁻²⁴ g u⁻¹ gives about 1.6931 × 10⁻²² g per formula unit. Then multiplying by 6.022 × 10²³ formula units mol⁻¹ gives about 101.96 g mol⁻¹; with the rounded conversion constants, report this intermediate verification as approximately 102.0 g mol⁻¹. One mole therefore has a mass of 101.96 g using the supplied molar mass.
4. The numerical correspondence is valid at AP precision because the mass of 1 u in grams multiplied by the number of entities per mole is approximately 1 g mol⁻¹. The units are different: u expresses a particle-scale mass, while g mol⁻¹ expresses mass per mole. An ionic formula unit describes composition in a lattice, not an isolated molecule.

1. For each elemental sample, N = (m/M)N_A.
2. Take the ratio: the same m and N_A cancel, leaving 39.95/20.18.

1. Let x be the atom fraction at 10.0 u. Then 10.0x + 11.0(1 − x) = 10.80.
2. Rearrange to 11.0 − x = 10.80, so x = 0.20; the other fraction is 0.80.
3. For the stated ideal singly charged model, peak signals follow isotope counts.

1. The relative atom fractions are 0.79, 0.10 and 0.11 because the count sum is 100.
2. Average = 24(0.79) + 25(0.10) + 26(0.11) = 24.32 u.
3. The 24 u population dominates, so a center-peak midpoint argument ignores the evidence.

1. Choose a notional 100 g sample: 40.0 g C, 6.71 g H and 53.3 g O. A pure compound preserves these ratios at any scale.
2. Moles are 3.3306, 6.6567 and 3.33125. Divide by the smallest: about 1:2:1.
3. Use the simplest integer ratio.

1. Both samples have the same Mg:O mass ratio, 0.500/0.329 ≈ 1.52, consistent with definite proportions for MgO.
2. Convert one sample to moles using Mg = 24.31 and O = 16.00 g mol⁻¹: the ratio is approximately 1:1.
3. The evidence is consistent with MgO but does not independently exclude every mixture that happens to have the same bulk ratio. Ask whether additional independently obtained subsamples retain the same elemental composition, or whether a separation reveals more than one substance.

1. Let x be grams NaCl; KCl mass is 10.00 − x.
2. Set the Cl mass balance: 0.6066x + 0.4755(10.00 − x) = 5.542.
3. Solve x = (5.542 − 4.755)/(0.6066 − 0.4755) = 6.003 g. Subtract for KCl.

1. The measured oxygen fraction is inconsistent with the predicted composition if measurements and the proposed formula are reliable.
2. An exact purity cannot be identified from the oxygen deficit alone because an impurity may itself contain oxygen.
3. If the impurity is established to be oxygen-free, m_X = measured O mass / oxygen fraction in pure X.

The created 3p-box models are P: [↑] [↑] [↑] and P³⁻: [↑↓] [↑↓] [↑↓], with all three boxes labeled 3p. Parallel singly occupied arrows and opposite-spin pairs are required; equivalent spin-direction conventions are acceptable.

1. Neutral P has 15 electrons: 1s²2s²2p⁶3s²3p³. The 3− ion gains three electrons: 1s²2s²2p⁶3s²3p⁶.
2. Protons have charge +1 and neutrons charge 0; both are in the nucleus. Electrons have charge −1 and occupy the electron distribution outside the nucleus, described by orbitals rather than fixed planet-like paths.
3. ³¹P³⁻ has 15 protons, 31 − 15 = 16 neutrons and 15 + 3 = 18 electrons. The nucleus has charge +15, while the electrons contribute −18; the net charge is −3.
4. Electron gain changes only the electron count in this ordinary ion-formation model. Proton number, neutron number and mass number remain unchanged; adding protons would change the element.

1. Use F ∝ |q₁q₂|/r². The charge product grows by 2 and the distance-squared factor grows by 4.
2. The force ratio is 2/4 = 1/2.
3. Ionization energy is a change in energy involving electron removal; a real atom also involves shielding, electron repulsions and quantum distributions, not one fixed pairwise separation.

1. The submitted configuration has the right total electron count but removes electrons from the wrong subshell.
2. Transition-metal ion formation removes 4s electrons before 3d electrons, giving [Ar]3d⁶.
3. Neutral-atom filling patterns are not a universal sequence for ion formation because orbital energies and attractions change with occupancy and charge.

1. Three peaks correspond to 1s, 2s and 2p in this early-element pattern.
2. The proposed 2s⁴ occupancy violates the two-electron s-subshell capacity.
3. The consistent arrangement is 1s²2s²2p⁴, eight electrons in a neutral atom.

1. Both species have 10 electrons in 1s²2s²2p⁶, accounting for identical ideal populations.
2. Mg²⁺ has 12 protons; Ne has 10. Similar occupied shells give broadly similar shielding structure.
3. The stronger nuclear attraction in Mg²⁺ requires more removal energy.

1. Ca’s valence electrons are in n = 4 rather than n = 3, farther from the nucleus and more shielded, so they are generally easier to remove.
2. Na⁺ and Mg²⁺ both have 10 electrons; Mg²⁺ has the greater nuclear charge.
3. Stronger attraction contracts Mg²⁺ more.

1. The largest jump follows the third removal: IE₄ reaches core electrons, suggesting three valence electrons.
2. The third-period group-13 candidate is Al.
3. Across Mg to Al, nuclear charge increases while electrons enter the same principal shell and core shielding changes little; stronger effective attraction generally decreases radius.

1. The −200 kJ mol⁻¹ value is more favorable energetically under the stated convention: it releases energy, whereas +50 requires energy input. Electronegativity describes attraction to shared electrons in a bond; electron-addition enthalpy describes a gaseous atom gaining an electron. They are different processes.
2. Predict Cl as more favorable than S. Its larger nuclear charge attracts the incoming electron more strongly while the added electron enters the same principal shell and core shielding is comparable. This is a qualitative prediction, not an exact value or a universal rule for every neighboring pair.
3. The Cl value is more negative than the F value. An incoming electron encounters greater repulsion in F’s compact 2p region than in Cl’s larger 3p region; that repulsion offsets part of F’s attraction advantage. Considering attraction alone would miss this exception.
4. No. The first-addition data do not rank electronegativity. F remains more electronegative in the usual bonded-atom scale; the properties describe different electronic situations.

1. K and Na have one outer valence electron and commonly lose one, giving K⁺ and Na⁺. Ca and Mg have two and commonly lose two, giving Ca²⁺ and Mg²⁺. In this monatomic-ion model, O gains two electrons to complete its outer shell, giving O²⁻, and F gains one, giving F⁻. These are typical main-group predictions, not a claim that isolated electron transfer alone determines a reaction’s energy.
2. Two K⁺ ions give +2 and one O²⁻ gives −2: K₂O. The analogous Na charge gives Na₂O. One Ca²⁺ and two F⁻ give +2 − 2 = 0: CaF₂; MgF₂ follows the same balance.
3. Transition metals can have several charges, and oxygen can occur in different ions. Potassium superoxide KO₂, for example, contains O₂⁻, not the stipulated monatomic oxide ion. Such a different valid species does not invalidate charge balance within the stated model.

1. Both Na and K have one outer ns electron, but K’s is in a higher shell with greater average distance and core shielding. Its first ionization energy is lower, so electron removal is easier despite its greater proton count.
2. A reaction’s overall outcome depends on both partners and conditions, not only isolated-atom ionization energy.
3. In H₂, each positive nucleus attracts the shared negative valence-electron density. The attractions involve both nuclei; assigning an electron to its original atom misses the interaction that accompanies bonding.
4. The positive nuclei repel each other, and electrons also repel each other. At a stable bonded separation the total energy is lower than for separated atoms. At very short separation repulsive effects become important and the energy rises, so the nuclei do not simply collapse together. This qualitative model explains the competition but does not calculate an exact bond length.