Unit 1 · Atomic Structure and PropertiesPractice · 原子结构与性质

The molar mass of $\ce{CaCO3}$ is 40.08 + 12.01 + 3 × 16.00 = 100.09 g/mol, and each formula unit contains 3 O atoms.

Significant figures: 5.00 has 3, 100.09 and 6.022 have more, and the subscript 3 is exact, so the answer has 3. Forgetting the factor of 3 gives the number of formula units, $3.01 \times 10^{22}$.

Estimate: $3.0 \times 10^{23}$ is about half of $6.022 \times 10^{23}$, so about 0.50 mol, and 0.50 mol × 26.98 g/mol ≈ 13 g (exactly, 13.4 g). 27 g is the mass of 1 mol of aluminum, and 81 g comes from forgetting to divide by $N_A$: 3.0 × 26.98.

Let x be the abundance of $\ce{^{71}Ga}$, so $\ce{^{69}Ga}$ is 1 − x. The average atomic mass is a weighted average:

So 1.999x = 0.797 and x = 0.399, or 39.9%. Check: the average, 69.723, is closer to 69, so $\ce{^{69}Ga}$ is the more common isotope and $\ce{^{71}Ga}$ must be under 50%.

Molar mass: 2 × 14.01 + 8 × 1.008 + 32.06 + 4 × 16.00 = 132.14 g/mol. One mole of $\ce{(NH4)2SO4}$ contains 2 mol of N, with a mass of 2 × 14.01 = 28.02 g.

The subscript 2 outside the parentheses applies to the whole $\ce{NH4+}$ group, so there are 2 N and 8 H. Counting only 1 N gives 10.6%.

All the C ends up in $\ce{CO2}$: n(C) = 0.440 ÷ 44.01 = 0.0100 mol. All the H ends up in $\ce{H2O}$, 2 H per molecule: n(H) = 2 × 0.180 ÷ 18.02 = 0.0200 mol. Find O by difference: m(O) = 0.300 − 0.0100 × 12.01 − 0.0200 × 1.008 = 0.160 g, so n(O) = 0.00999 mol. C : H : O = 1 : 2 : 1, so the empirical formula is $\ce{CH2O}$ (30.03 g/mol). 60 ÷ 30.03 ≈ 2, so the molecular formula is $\ce{C2H4O2}$. $\ce{CH2O}$ is only the empirical formula. $\ce{C3H8O}$ also has a molar mass near 60 g/mol (60.09), but its C : H : O ratio doesn’t fit the combustion data. Molar mass alone isn’t enough; you need it together with the composition.

Each $\ce{BaSO4}$ and each $\ce{BaCl2}$ contains exactly 1 Ba, and barium is conserved, so n($\ce{BaCl2}$) = n($\ce{BaSO4}$).

m($\ce{BaCl2}$) = 0.00300 mol × 208.23 g/mol = 0.625 g, and the mass percent is 0.625 ÷ 1.000 × 100% = 62.5%. The precipitate mass, 0.700 g, has 3 significant figures, so the answer has 3.

The calculation treats all the measured mass as $\ce{BaSO4}$. Leftover water makes the measured mass too large, so the calculated n($\ce{BaSO4}$), n($\ce{BaCl2}$), and mass percent all come out too large. For example, 0.020 g of leftover water changes the result from 62.5% to 64.2%. The fact in C is true (the barium in the sample really doesn’t change), but the calculated result depends on the measured mass, so that reason doesn’t lead to “unchanged.” The reasoning in B and D doesn’t hold.

When a transition metal forms a cation, it loses its highest-n electrons, the 4s electrons, before any 3d electrons. Ground-state configurations: $\ce{Fe}$: [Ar]4s²3d⁶; $\ce{Fe^2+}$: [Ar]3d⁶; $\ce{Mn+}$: [Ar]4s¹3d⁵; $\ce{Co^3+}$: [Ar]3d⁶; $\ce{Ni^2+}$: [Ar]3d⁸. The Fe atom has 3d⁶ too, but it also has 4s². Fe²⁺ has lost exactly its two 4s electrons, and Co³⁺ has lost its 4s² and one 3d electron.

From highest to lowest binding energy (left to right on the spectrum), the 5 peaks are 1s, 2s, 2p, 3s, and 3p, and each peak’s height is proportional to its number of electrons. The lowest-energy peak is 3p, next to the 3s peak with 2 electrons, so 3p holds 2.5 × 2 = 5 electrons: 1s²2s²2p⁶3s²3p⁵, which is Cl. The ratio is 1.5 for P, 2 for S, 3 for Ar.

Measured I₁: Mg 738 kJ/mol, Al 578 kJ/mol, so Mg’s is greater and A is wrong. Al is [Ne]3s²3p¹. The 3p electron it loses is higher in energy than a 3s electron and is partly shielded by the 3s electrons, so it feels a weaker net attraction and is easier to remove. B only says “filled is stable” without explaining in terms of the nucleus’s attraction for the electron, which earns no AP credit. C’s premise is false: Mg’s radius (141 pm) is larger than Al’s (121 pm).

Mg has only 2 valence electrons (3s²). Once both are gone, the third electron has to come from the inner 2p sublevel: it’s closer to the nucleus and much less shielded, so I₃ jumps, and I₃/I₂ = 7733 ÷ 1451 ≈ 5.3. Na’s big jump is from I₁ to I₂, Al’s from I₃ to I₄, and Si’s from I₄ to I₅; their I₃/I₂ ratios are about 1.5, 1.5, and 2.0.

Elements in the same group have the same number of valence electrons. Sr is in group 2 with 2 valence electrons, so X loses 2 electrons to form $\ce{X^2+}$. N has 5 valence electrons and gains 3 to form $\ce{N^3-}$, with the same configuration as Ne. The charges must add to zero: 3 × (+2) + 2 × (−3) = 0, so the formula is $\ce{X3N2}$ (for example, $\ce{Mg3N2}$).