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1.1 Moles and Molar Mass

1.1

There are approximately 0.714 moles of O₃ present in the sample of 4.3 × 10²⁴ molecules of O₃.

Detailed explanation here.

1.2

There are approximately 2.108 × 10²⁵ atoms of Si present in the 35.0-mole sample.

Detailed explanation here.

There are approximately 1.75 × 10⁻³ formula units of Na₂SO₄ present in 0.248 g of Na₂SO₄.

Detailed explanation here.

1.3

1.2 Mass Spectrscopy

1.2.1

PART A. The element is most likely Bromine (Br).

Detailed explanation here.

PART B

For Bromine, the atomic number (which corresponds to the number of protons) is 35.

  • Mass Number 79:

    • Protons: 35

    • Neutrons: 44

  • Mass Number 81:

    • Protons: 35

    • Neutrons: 46

Detailed explanation here.

PART C

Mass number = 79 is more abundant in nature (50.7%).

Detailed explanation here.

1.3 Elemental Compoisition

1.3.1

a) Potassium bromide (KBr) is made up of potassium ions (K⁺) and bromide ions (Br⁻).

b) Carbon dioxide (CO₂) is made up of molecules.

c) Diamond (carbon) is made up of carbon atoms arranged in a crystal lattice.

d) Magnesium oxide (MgO) is made up of magnesium ions (Mg²⁺) and oxide ions (O²⁻).

Detailed explanation here.

1.3.2 & 1.3.3

Part A:

For the first sample: 52.87% Al and 47.13 % S

For the second sample: 52.89% Al and 47.11% S

Detailed explanation here.

Part B: The ratios of the masses of sulfur to aluminum are very close, indicating that the Law of Definite Proportions is observed.

Detailed explanation here.

Part C: The simplest whole-number ratio of Al to S is approximately 2:3, so the empirical formula is Al₂S₃.

Detailed explanation here.

1.4 Composition of a Mixture

PART A. The balanced equation is: 3Mg(s) + N2(g) → Mg3N2(s)

Detailed explanation here.

PART B. The mass of nitrogen in the sample is 14.70 g.

Detailed explanation here.

PART C. The mass of magnesium nitride in the mixture is 14.70 g.

Detailed explanation here.

PART D. The mass percent of magnesium nitride in the mixture is approximately 17.19%.

Detailed explanation here.

1.4.2

The sum all the percentages should add up to 100% from the sample. When adding up the given percentages in the problem, it equals 100 — which does not include the percentage for Cl (which definitely makes up some part of the composition since Cl is present in NaCl). Due to this, the percentages that are given are inaccurate.

Detailed explanation here.

1.4.3

Final Mass Percentages:

  • NaF: 28.47%

  • Na₂SO₄: 7.08%

  • NaNO₃: 36.97%

These percentages sum up to approximately 100%, accounting for rounding errors.

Detailed explanation here.

1.5 Atomic Structure

1.5.1

PART A: Protons: 92, Neutrons: 146, Electrons: 92

Detailed explanation here.

PART B: The charge of the uranium ion after losing two electrons is U²⁺.

Detailed explanation here.

PART C: The uranium-238 atom transforms into a thorium-234 atom​ after alpha decay.

Detailed explanation here.

1.5.2

PART A: The force between these two ions is attractive because opposite charges attract each other. This force pulls the Na+ ion and the Cl− ion towards each other.

Detailed explanation here.

PART B: Thus, if the distance between the ions is halved, the force quadruples (increases by a factor of 4).

Detailed explanation here.

1.5.3

The ground state electron configuration for an atom of tin (Sn) is:

  • Full configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p²

  • Condensed configuration: [Kr] 5s² 4d¹⁰ 5p²

Detailed explanation here.

1.5.4

The correct answer is (A): Binding Energy of 1s Electrons in Li: 6.26 MJ/mol, Be: 11.5 MJ/mol; Justification: Li atoms have a smaller nuclear charge than Be atoms.

Detailed explanation here.

1.6 Photoelectron Spectrscopy

The 2p electrons have a binding energy that is slightly lower than the 2s electrons.

  • Location: The 2p peak should be placed at a binding energy to the RIGHT of the 2s peak, which is around 2-3 MJ/mol.

  • Relative Height: The 2p peak should be taller than the 2s peak because fluorine has 5 electrons in the 2p sublevel. It should reach the fifth line from the bottom of the graph. (We know that each line represents one electron in this spectrum because 1s suborbital has a maximum of 2 electrons and it is at the second line from the bottom.)

Detailed explanation here.

1.7 Periodic Trends

1.7.2

PART A. Order of Ionization Energy (Highest to Lowest): Cl > Ar > S > Mg

Detailed explanation here.

PART B. Largest Atomic Radius: Mg

Detailed explanation here.

PART C. Highest Electron Affinity: Cl

Detailed explanation here.

PART D. Order of Electronegativity (Increasing): Mg < S < Ar < Cl

Detailed explanation here.

1.7.1

The best answer is (C) Ar, because of its completely filled valence shell. (D) is also technically correct, but (C) offers the most straightforward and commonly cited reason related to ionization energy.

Detailed explanation here.

1.8 Valence Electrons and Ionic Compounds

1.8.1

Pair 1: Sodium (Na) and Chlorine (Cl)

(a) Likely to form a bond.

(b) Type of bond: Ionic bond.

  • Reasoning:

    • Sodium has a strong tendency to lose its one valence electron to achieve a stable Na+ ion configuration (like neon, with a full 2s2 2p6 shell).

    • Chlorine has a strong tendency to gain one electron to complete its valence shell and form a Cl− ion (achieving the stable configuration of argon).

    • The transfer of the electron from Na to Cl results in the formation of oppositely charged ions (Na+ and Cl−), which are held together by strong electrostatic forces, forming an ionic bond.

(c) Not applicable since they form a bond.

Detailed explanation here.

Pair 2: Nitrogen (N) and Oxygen (O)

(a) Likely to form a bond.

(b) Type of bond: Covalent bond.

  • Reasoning:

    • Both nitrogen and oxygen are nonmetals with high electronegativities. Instead of transferring electrons, they share electrons to achieve full valence shells.

    • In a molecule like NO (nitric oxide) or NO₂ (nitrogen dioxide), nitrogen and oxygen share electrons to form covalent bonds, with each atom achieving a more stable electron configuration.

    • For instance, in NO₂, nitrogen shares electrons with two oxygen atoms, resulting in covalent bonds.

(c) Not applicable since they form a bond.

Detailed explanation here.

Pair 3: Helium (He) and Neon (Ne)

(a) Not likely to form a bond.

(b) Reasoning:

  • Helium (He) has the electron configuration 1s2 and is in Group 18 (noble gases) with a full valence shell.

  • Neon (Ne) has the electron configuration 2s2 2p6, also in Group 18, with a full valence shell.

  • Both helium and neon have complete valence shells, making them extremely stable and chemically inert. They have no tendency to gain, lose, or share electrons.

(c) Explanation:

  • Helium and Neon: Since both elements have filled valence shells, they do not need to form bonds to achieve stability. Their inert nature means they are highly unlikely to form any chemical bonds with each other or with other elements.

Detailed explanation here.

1.8.2

The most likely formula for the compound formed between potassium and element X would be KX.

Detailed explanation here.

1.8.3

PART A: Element X has 3 valence electrons.

Detailed explanation here.

PART B: The empirical formula for the compound between element X and oxygen is X₂O₃.

Detailed explanation here.

PART C: The symbol of the element in period 2 with the same number of valence electrons as element X is B.

Detailed explanation here.