Structure 1.1.1

1. The key principles of the particle model of matter are:
   • All matter is made up of particles
   • Particles have mass and volume
   • Particles are in constant motion
   • Particles attract each other through various forces
2. The particle model of matter explains changes in the state of matter by considering the energy of particles and their motion. As energy is added to a substance, the particles gain kinetic energy and move faster, causing the substance to change state from solid to liquid to gas.
3. Elements are the simplest form of matter and cannot be broken down into simpler substances. Compounds are formed when two or more elements chemically combine in a fixed ratio, resulting in unique properties. Mixtures are a combination of two or more substances that are not chemically bonded together and can be separated by physical means.
4. The particle model of matter aids in understanding the electrical and magnetic properties of materials by studying the behavior of charged particles (such as electrons and ions) and the interactions between these particles.
5. Solubility refers to the ability of a substance to dissolve in a solvent to form a homogeneous solution. Factors that can influence solubility include temperature, pressure, and the chemical nature of the substance itself.
6. The particle model of matter can be applied to the study of thermodynamics by providing a framework for understanding the relationship between the energy of particles and their temperature, as well as the behavior of gases at the molecular level.
7. Homogeneous mixtures, also known as solutions, have a uniform composition throughout, meaning the components are evenly distributed. Heterogeneous mixtures do not have a uniform composition throughout and have visible boundaries between the components.
8. The empirical formula is the simplest ratio of atoms in a molecule, derived from the molecular formula. To calculate the empirical formula, convert the masses or percentages of each element to moles and divide each mole value by the smallest mole value, resulting in the simplest whole number ratio of atoms.
9. The particle model of matter contributes to our understanding of the properties of biological molecules by examining the behavior of their constituent particles, such as atoms and molecules, and how these interactions determine the properties and behavior of biological molecules.
10. The relationship between the particle model of matter and the kinetic theory of gases lies in the explanation of the behavior of gas particles at the molecular level. The kinetic theory of gases is based on the particle model of matter and describes how gas particles move, collide, and transfer energy to each other, determining properties such as pressure, volume, and temperature.

Structure 1.1.2

1. The kinetic molecular theory is a scientific model that explains the physical properties of matter and how they change from one state to another by describing the behavior of tiny particles called molecules or atoms.
2. The kinetic molecular theory explains that all matter is made up of tiny particles that are in constant motion. In a solid, the particles are tightly packed and vibrate in fixed positions, which gives the solid its shape and rigidity. In a liquid, the particles are still close together but can move around more freely, which allows the liquid to flow and take the shape of its container. In a gas, the particles are far apart and move randomly in all directions, which makes the gas expand to fill its container.
3. The different states of matter are solids, liquids, and gases. Solids have a fixed shape and volume, liquids have a fixed volume but take the shape of their container, and gases have neither a fixed shape nor volume.
4. Changes of state occur when a substance gains or loses energy, causing its particles to move more rapidly or slowly. When a solid is heated, the particles gain energy and begin to vibrate more rapidly, eventually breaking free from their fixed positions and becoming a liquid. When a liquid is heated even further, the particles gain even more energy and begin to move around more rapidly, eventually breaking free of each other completely and becoming a gas.
5. The key assumptions of the kinetic molecular theory are that all matter is made up of tiny particles called molecules or atoms, the particles are in constant motion, the particles move independently of one another, and the temperature of a substance is related to the average kinetic energy of its particles.
6. Some limitations of the kinetic molecular theory are that it assumes the particles in a gas do not interact with each other, which is not always the case, and that the particles are point masses with no size or volume, which is also not entirely accurate.
7. Thermodynamics is the study of energy and its transformations, and the particle model of matter provides a framework for understanding the relationship between the energy of particles and their temperature.
8. When a substance is heated, the motion of its particles increases, causing them to vibrate more rapidly and with greater energy. When a substance is cooled, the opposite occurs, with particles losing energy and moving more slowly.
9. State symbols are used in chemical equations to indicate the physical state of reactants and products. The most common state symbols are (s) for solid, (l) for liquid, (g) for gas, and (aq) for aqueous (dissolved in water).
10. Sublimation is a physical change in which a substance transitions directly from the solid phase to the gas phase without passing through the intermediate liquid phase. This process typically occurs when a solid substance is heated under specific conditions of temperature and pressure, causing its molecules to gain enough energy to break free from their structured lattice and move freely as gas molecules. An example of sublimation is dry ice (solid carbon dioxide) transitioning directly into gas when exposed to room temperature and atmospheric pressure.

Structure 1.1.3

1. The Kelvin scale is the standard unit of temperature in the International System of Units (SI) and is based on the concept of absolute zero, making it an ideal temperature unit for scientific applications. It is preferred over other temperature units due to its unique properties, such as being an absolute temperature scale.
2. Absolute zero is the temperature at which all molecular motion ceases and is calculated to be -273.15°C, which is also known as 0 Kelvin. The Kelvin scale starts at absolute zero, and each Kelvin unit is equal to one Celsius degree.
3. The Celsius scale is based on the freezing and boiling points of water, while the Kelvin scale is based on the theoretical concept of absolute zero. This makes the Kelvin scale an ideal temperature unit for scientific applications, particularly in fields such as physics, chemistry, and engineering, where precise temperature measurements are essential.
4. The relationship between temperature, pressure, and volume is important in understanding the behavior of gases and other materials, particularly in thermodynamic calculations. Kelvin is the preferred unit of temperature in these experiments, as it is an absolute temperature scale that allows scientists to directly measure the amount of thermal energy in a system.
5. The Kelvin scale is used extensively in scientific experiments involving thermodynamic calculations, particularly in the study of the relationship between temperature, pressure, and volume.

Structure 1.2.1

Atomic Symbol	Atomic Number	Mass Number	Number of Protons	Number of Neutrons	Number of Electrons
Mg	12	24	12	12	12
Cl	17	35	17	18	17
Ti	22	48	22	26	22
Sr	38	88	38	50	38
Ag	47	108	47	61	47
Xe	54	131	54	77	54
Ba	56	138	56	82	56
La	57	139	57	82	57
Pt	78	195	78	117	78

1. The atomic number (Z) represents the number of protons in an atom’s nucleus.
2. To calculate the number of neutrons in an atom, subtract the atomic number (Z) from the mass number (A).
3. In a neutral atom, the number of electrons is equal to the number of protons.
4. In a neutral nitrogen atom (¹⁴N₇), there are 7 protons, 7 neutrons, and 7 electrons.
5. An oxide ion (O^2-) will have 10 electrons, as it has gained two electrons compared to a neutral oxygen atom.
6. The mass number (A) is 38 (20 neutrons + 18 protons), and the atomic number (Z) is 18 (equal to the number of protons or electrons in a neutral atom).
7. The number of electrons in an ion will be different from that in a neutral atom due to the gain or loss of electrons.
8. In a neutral aluminum atom (²⁷Al₁₃), there are 13 protons, 14 neutrons, and 13 electrons. The element is aluminum.
9. The nuclear symbol is ²⁴Mg₁₂, and it is an ion (specifically, Mg²⁺) because the number of electrons (10) is not equal to the number of protons (12).
10. In a Cl^– ion, there are 17 protons, 18 neutrons, and 18 electrons (one more electron than in a neutral chlorine atom).

Structure 1.2.2

1. An isotope is an atom of the same element with the same number of protons but a different number of neutrons. Isotopes of the same element differ in their neutron count and, consequently, in their mass.
2. The chemical properties of isotopes of the same element are similar because they have the same electron configuration and thus exhibit the same behavior in chemical reactions.
3. The relative atomic mass of an element can be a non-integer value because it is the weighted average mass of its naturally occurring isotopes, taking into account their relative abundances.
4. To calculate the relative atomic mass of an element, multiply each isotope’s mass by its relative abundance, and then sum up these values. The formula is: Relative atomic mass = Σ (isotope mass × isotope abundance).
5. The physical properties of isotopes of the same element differ due to their varying numbers of neutrons. Differences in physical properties may include mass, density, boiling and melting points, and rates of radioactive decay.
6. Using the given data for isotope A and isotope B, the relative atomic mass of the element can be calculated as follows: (15 × 0.40) + (16 × 0.60) = 6 + 9.6 = 15.6 u.
7. Understanding isotopes and their properties is important in chemistry because they play a significant role in various applications, such as analyzing isotopic patterns in mass spectrometry, understanding radioactive decay processes, and employing isotopes as tracers in chemical reactions and environmental studies.
8. Isotopes affect mass spectrometry results by producing distinct peaks in the mass spectrum corresponding to the masses of the different isotopes, allowing for the identification and determination of the isotopic composition of an element or compound.
9. Differences in stability and radioactivity among isotopes arise from variations in their nuclear structure. Some isotopes are stable, while others are radioactive and undergo decay processes, such as alpha decay, beta decay, or gamma radiation.
10. Isotopes can be used as tracers in chemical reactions and environmental studies by incorporating a specific isotope into a compound or system and then tracking its movement or distribution. This can provide valuable insights into reaction mechanisms, rates, and pathways, as well as help monitor and understand environmental processes such as pollution and nutrient cycling.

Structure 1.2.3

1. Mass spectrometry is a technique used to separate ions based on their mass-to-charge ratio. It involves ionizing a sample, accelerating the resulting ions, and then deflecting them using a magnetic field. The ions are detected by a detector, and a mass spectrum is generated that provides information about the identity and relative abundance of isotopes in the sample.
2. A mass spectrum provides information about the identity and relative abundance of isotopes in a sample. It can be used to identify isotopes based on the position of their peaks on the spectrum, which corresponds to their mass-to-charge ratio. The relative heights of the peaks can also be used to determine the relative abundance of isotopes in the sample.
3. Mass spectrometry can be used to determine the relative atomic masses of elements based on their isotopic composition. Isotopes have different masses and relative abundances, and mass spectrometry can be used to measure these values and calculate the relative atomic mass of an element.
4. The position of a peak on a mass spectrum corresponds to the mass-to-charge ratio of the ion. Heavier ions have a lower mass-to-charge ratio and are deflected less by the magnetic field, so they appear closer to the center of the spectrum. Lighter ions have a higher mass-to-charge ratio and are deflected more, so they appear farther from the center of the spectrum.
5. The isotopic composition of a sample can provide important information about its origin and history. For example, the isotopic composition of carbon in a sample can provide information about its age and origin. Mass spectrometry has practical applications in scientific research and analysis, such as in the fields of forensics, environmental science, and materials science. It can be used to identify unknown substances, analyze the purity of chemicals, and study the composition of complex mixtures.

Structure 1.3.1

1. Emission spectra are produced when atoms emit photons as their electrons transition from higher energy states to lower energy states. When an element is heated or subjected to an electric discharge, its electrons are excited to higher energy levels. As the electrons return to their lower energy states, they emit photons with energies specific to the energy level transitions.
2. In the context of the electromagnetic spectrum, color is dependent on the wavelength of light. Shorter wavelengths correspond to higher frequencies and higher energies, while longer wavelengths correspond to lower frequencies and lower energies. The relationship between wavelength (λ) and frequency (ν) is given by the equation c = λν, where c is the speed of light. Energy (E) is related to frequency by the equation E = hν, where h is Planck’s constant.
3. In the visible spectrum, the order of colors from highest to lowest energy is violet, blue, green, yellow, orange, and red. Violet light has the shortest wavelength, highest frequency, and highest energy, while red light has the longest wavelength, lowest frequency, and lowest energy.
4. A continuous spectrum is an unbroken band of colors that includes all the wavelengths within a specific range. It is produced when white light, which is a combination of all the colors in the visible range, is passed through a prism. In contrast, a line spectrum, also known as an atomic emission spectrum, consists of discrete lines of specific colors (wavelengths) corresponding to the unique energy levels of a particular element. It is produced when an element is heated or subjected to an electric discharge, causing its electrons to transition between energy levels and emit photons with specific energies.
5. The atomic emission spectrum reveals information about the unique energy levels of a specific element because each element has a distinct set of energy levels for its electrons. When the electrons transition between these energy levels, they emit photons with specific energies that correspond to the differences in energy levels. These emitted photons have specific wavelengths, which appear as distinct colored lines when the emitted light is passed through a spectrometer. The pattern of these lines serves as a “fingerprint” for the element, allowing scientists to identify the element based on its emission spectrum.

Structure 1.3.2

1. The hydrogen emission spectrum consists of a series of discrete lines that correspond to specific energy transitions within the hydrogen atom.
2. When a hydrogen atom is excited, its electron moves to a higher energy level. As the electron returns to lower energy levels, it emits photons with energies specific to the energy level transitions.
3. Energy transitions in the hydrogen atom are the changes in energy that occur when an electron moves from one energy level to another. These transitions are related to the emission spectrum because as the electron moves from a higher energy level to a lower one, it releases energy in the form of a photon with a specific wavelength.
4. The different series of lines in the hydrogen emission spectrum are the Lyman series, the Balmer series, and the Paschen series. The Lyman series involves transitions where the electron returns to the first energy level (n=1), the Balmer series involves transitions where the electron returns to the second energy level (n=2), and the Paschen series involves transitions where the electron returns to the third energy level (n=3).
5. The spectral lines in the hydrogen emission spectrum converge at higher energies because as the energy levels increase, the energy differences between adjacent levels decrease, causing the spectral lines in the series to converge at higher energies.
6. The Rydberg formula is 1/λ = R_H (1/n1² – 1/n2²), where λ is the wavelength of the emitted photon, R_H is the Rydberg constant for hydrogen, n1 is the lower energy level, and n2 is the higher energy level. This formula can be used to calculate the wavelengths of the spectral lines in the hydrogen emission spectrum for each series.
7. The Lyman series is significant because it involves transitions where the electron returns to the first energy level, which has the highest energy and corresponds to ultraviolet light. The Lyman series is the most energetic series of spectral lines in the hydrogen emission spectrum.
8. The Balmer series is significant because it involves transitions where the electron returns to the second energy level, which corresponds to visible light. The Balmer series is the most well-known series of spectral lines in the hydrogen emission spectrum.
9. The Paschen series is significant because it involves transitions where the electron returns to the third energy level, which corresponds to infrared light. The Paschen series is important for studying the infrared spectrum of hydrogen.
10. The line emission spectrum of hydrogen provides evidence for the existence of electrons in discrete energy levels because the spectral lines correspond to specific energy transitions within the hydrogen atom. This indicates that the electrons in the hydrogen atom can only exist at certain discrete energy levels, and not at any energy level in between.

Structure 1.3.3

1. The principal quantum number, represented by the symbol n, is an integer number given to each energy level in an atom. It represents the energy of the electron in that level.
2. The maximum number of electrons that an energy level can hold is calculated using the formula: Maximum number of electrons = 2n². This formula takes into account the number of subshells and orbitals in each energy level.
3. Each energy level in an atom contains one or more subshells, which are designated by the letters s, p, d, and f. Each subshell contains a certain number of orbitals, and each orbital can hold up to two electrons with opposite spins.
4. Based on the formula, the first energy level (n=1) can hold a maximum of 2 electrons, the second energy level (n=2) can hold a maximum of 8 electrons, the third energy level (n=3) can hold a maximum of 18 electrons, and the fourth energy level (n=4) can hold a maximum of 32 electrons.
5. The Aufbau principle is a rule that states that electrons fill the lowest energy levels and orbitals first before occupying higher energy levels and orbitals. This principle is important for understanding the distribution of electrons in energy levels, as it determines the order in which electrons occupy the subshells and orbitals within an energy level.

Structure 1.3.4

1. The main energy level is divided into sublevels, also known as subshells.
2. The subshells are denoted by the letters s, p, d, and f.
3. The s orbital has a spherical shape, while the p orbitals have a dumbbell-like shape with two lobes.
4. The s subshell has only one orbital.
5. The three p orbitals have different orientations in space, aligning along the x, y, and z axes.
6. The nodal plane is a plane where the probability of finding an electron is zero. In the p orbitals, there is a nodal plane passing through the nucleus.
7. The d orbitals appear in the third main energy level (n=3) and above.
8. There are five d orbitals in total: dxy, dxz, dyz, dx2-y2, and dz2.
9. The dx2-y2 orbital is square-shaped, while the double-lobe d orbitals have a cloverleaf pattern.
10. The dz2 orbital has a unique shape with two lobes along the z-axis and a torus (ring-shaped) region encircling the nucleus in the xy-plane. There is a nodal plane in the xy-plane that separates the two lobes along the z-axis.

Structure 1.3.5

1. The maximum number of electrons that can be present in a sublevel depends on the sublevel. The s sublevel can hold up to 2 electrons, the p sublevel can hold up to 6 electrons, the d sublevel can hold up to 10 electrons, and the f sublevel can hold up to 14 electrons.
2. The Pauli exclusion principle states that no two electrons in an atom can have the same set of quantum numbers, meaning that each orbital can hold a maximum of two electrons with opposite spins. This principle applies to electron configurations because it explains why orbitals are filled in a specific order, and why there can be a maximum of two electrons in each orbital.
3. Full electron configurations list all the orbitals in the order they are filled, while condensed electron configurations use the noble gas core as a shorthand notation. For example, the electron configuration of chlorine (Z = 17) can be written as [Ne]3s²3p⁵, where [Ne] represents the electron configuration of neon (Z = 10).
4. Chromium and copper have unusual electron configurations because they have increased stability with half-filled or fully filled d subshells. Chromium’s electron configuration is [Ar]4s¹3d⁵, while copper’s electron configuration is [Ar]4s¹3d¹⁰, where [Ar] represents the electron configuration of argon (Z = 18).
5. The Aufbau principle states that electrons fill orbitals in order of increasing energy, starting from the lowest energy orbital. This principle is used to determine electron configurations by following the order of orbital filling: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p.
6. A sublevel is a group of orbitals within a given energy level, while an orbital is a region of space where an electron is most likely to be found.
7. Degenerate orbitals are orbitals with the same energy. When filling degenerate orbitals in a sublevel, electrons occupy each orbital singly with parallel spins before any orbital is doubly occupied. This follows Hund’s rule.
8. Electron configurations can be represented using orbital diagrams, which visually represent the filling and relative energy of orbitals. In these diagrams, boxes represent orbitals, and arrows represent electrons. The direction of the arrow indicates the electron’s spin.
9. The electron configuration of chlorine is written as [Ne]3s²3p⁵ instead of 1s²2s²2p⁶3s²3p⁵ because the noble gas core is a shorthand notation that represents the electron configuration of the closest noble gas element (in this case, neon). Using this notation makes it easier to write out electron configurations for larger atoms.
10. The noble gas core in condensed electron configurations is significant because it represents the electron configuration of the closest noble gas element, which has a completely filled valence shell. This notation allows for a more concise way of representing electron configurations.

Structure 1.3.6

1. The limit of convergence at higher frequencies in an emission spectrum corresponds to the ionization energy, which is the energy required to remove an electron from the atom.
2. Across a period (left to right) in the periodic table, the first ionization energy generally increases due to the increasing effective nuclear charge experienced by the valence electrons.
3. Down a group (top to bottom) in the periodic table, the first ionization energy generally decreases because of the increasing atomic radius and the addition of electron shells, which result in increased shielding of the valence electrons from the nucleus.
4. Discontinuities in the first ionization energy trend across a period are caused by electron-electron repulsion in the p orbitals or when the electron being removed is from a new, higher energy sublevel.
5. The decrease in ionization energy as you move down a group in the periodic table is due to the increasing atomic radius and the addition of electron shells, which result in increased shielding of the valence electrons from the nucleus.
6. The energy of emitted photons is directly proportional to the frequency of electromagnetic radiation, as described by the equation E = hf, where E is the energy, h is the Planck constant, and f is the frequency.
7. To calculate the first ionization energy given the wavelength of the convergence limit, first use the equation c = λf to solve for the frequency (f), then use the equation E = hf to calculate the energy (E) in joules, and finally, convert the energy (E) to electron volts by dividing by the elementary charge (e = 1.602 x 10^-19 C).
8. Using the equation E = hf and the given frequency (4.0 x 10^-15 Hz), we calculate the energy (E) in joules. E = (6.626 x 10^-34 J s) * (4.0 x 10^-15 Hz) = 2.6504 x 10^-18 J. Converting to electron volts, we get 2.6504 x 10^-18 J / (1.602 x 10^-19 C) ≈ 16.54 eV.
9. First, use c = λf to find the frequency: (3.0 x 10⁸ m/s) / (1.5 x 10^-7) m) = 2.0 x 10¹⁵ Hz. Next, use E = hf to find the energy: (6.626 x 10^-34 J s) * (2.0 x 10¹⁵ Hz) = 1.3252 x 10^-18 J. Finally, convert to electron volts: 1.3252 x 10^-18 J / (1.602 x 10^-19 C) ≈ 8.27 eV.
10. Electron-electron repulsion in the p orbitals and the addition of a new, higher energy sublevel contribute to the discontinuities observed in the first ionization energy trend across a period because these factors lead to a reduction in the effective nuclear charge experienced by the valence electrons, resulting in a lower ionization energy than expected.

Structure 1.3.7

1. When plotting the first ionization energies for elements in Period 2, you will observe an overall increasing trend from left to right with two discontinuities between Be and B, and between N and O. The increasing trend can be attributed to the increase in effective nuclear charge experienced by the valence electrons as you move across the period. The discontinuity between Be and B occurs because B has one electron in the 2p orbital, which has a slightly higher energy and experiences greater electron-electron repulsion than Be’s fully-filled 2s orbital. The discontinuity between N and O occurs because in O, there is electron-electron repulsion due to the pairing of electrons in the 2p orbital, making it easier to remove one electron compared to N, which has one electron in each of the 2p orbitals.
2. When plotting the successive ionization energies for the unknown element X, you will observe a sharp increase in ionization energy between IE4 and IE5. This significant increase indicates that element X has four valence electrons, placing it in Group 14 (the carbon group) of the periodic table.
3. Upon plotting the first ionization energies for elements in Group 1 (alkali metals), you will observe a decreasing trend from Li to Fr. This trend can be explained by the increase in atomic radius and the addition of electron shells as you move down the group. The larger atomic radius and additional electron shells result in increased shielding of the valence electrons from the nucleus, making them easier to remove and thus requiring less energy.

Structure 1.4.1—The mole (mol) is the SI unit of amount of substance. One mole contains exactly the number of elementary entities given by the Avogadro constant.

1. The formula to convert the amount of substance (n) to the number of specified elementary entities using the Avogadro constant (N_A) is: Number of entities = n × N_A.
2. Number of carbon atoms = n × N_A = 2 moles × 6.022 × 10²³ mol⁻¹ ≈ 1.204 × 10²⁴ carbon atoms.
3. Number of water molecules = n × N_A = 5 moles × 6.022 × 10²³ mol⁻¹ ≈ 3.011 × 10²⁴ water molecules.
4. Number of sodium ions = Number of chloride ions = n × N_A = 3.5 moles × 6.022 × 10²³ mol⁻¹ ≈ 2.107 × 10²⁴ ions each.
5. Number of electrons = n × N_A = 1 mole × 6.022 × 10²³ mol⁻¹ ≈ 6.022 × 10²³ electrons.
6. Number of nitrogen molecules = n × N_A = 4 moles × 6.022 × 10²³ mol⁻¹ ≈ 2.408 × 10²⁴ nitrogen molecules.
7. Number of atoms of element X = n × N_A × 2 = 1.5 moles × 6.022 × 10²³ mol⁻¹ × 2 ≈ 1.803 × 10²⁴ atoms.
8. Number of calcium ions = n × N_A = 0.5 moles × 6.022 × 10²³ mol⁻¹ ≈ 3.011 × 10²³ calcium ions.
9. Number of specified groups of particles = n × N_A = 2.5 moles × 6.022 × 10²³ mol⁻¹ ≈ 1.505 × 10²⁴ groups.
10. Number of atoms of X = n × N_A = 3 moles × 6.022 × 10²³ mol⁻¹ ≈ 1.807 × 10²⁴ atoms of X. Number of atoms of Y = 2 × n × N_A = 2 × 3 moles × 6.022 × 10²³ mol⁻¹ ≈ 3.613 × 10²⁴ atoms of Y.

Structure 1.4.2—Masses of atoms are compared on a scale relative to 12C and are expressed as relative atomic mass Ar and relative formula mass Mr.

1. The relative atomic mass (A_r) represents the average mass of atoms of a specific element, taking into account the natural abundance of its isotopes. The relative formula mass (M_r) represents the sum of the relative atomic masses of all the atoms in a molecule or compound.
2. Atomic mass is defined in atomic mass units (amu) or unified atomic mass units (u), where 1 atomic mass unit is defined as one twelfth (1/12) of the mass of a carbon-12 (¹²C) isotope.
3. Carbon-12 (¹²C) is used as the reference standard for defining atomic mass units because it provides a stable, widely available, and universally recognized reference point for comparing the masses of different atoms.
4. The natural abundance of isotopes affects the calculation of atomic mass because it is a weighted average of the masses of an element’s isotopes, taking into account their proportions in nature. The more abundant an isotope is, the more it contributes to the overall atomic mass of the element.
5. Isotopes are atoms of the same element with the same number of protons but a different number of neutrons in their nucleus. They contribute to the atomic mass of an element because they have different masses, and their natural abundance affects the weighted average used to calculate the atomic mass.
6. To calculate the atomic mass of an element: a. Identify the isotopes of the element and their respective atomic masses. b. Determine the natural abundance of each isotope. c. Multiply the atomic mass of each isotope by its natural abundance (expressed as a decimal). d. Sum the products obtained in step 3 to get the atomic mass of the element.
7. The concept of relative formula mass is important in chemistry because it allows chemists to compare the masses of different molecules and perform stoichiometric calculations in chemical reactions.
8. To calculate the relative formula mass (M_r) of a compound: a. Identify the elements and the number of atoms of each element present in the molecule. b. Look up the relative atomic mass (A_r) of each element in the periodic table or a data booklet. c. Multiply the relative atomic mass (A_r) of each element by the number of atoms of that element in the molecule. d. Sum the products obtained in step 3 to get the relative molecular mass of the molecule.
9. You should use the values provided in the IBDP Chemistry data booklet when performing calculations involving relative atomic masses and relative formula masses to ensure consistency in calculations and answers.
10. To calculate the relative formula mass (M_r) of ammonia (NH₃): a. Nitrogen (N) has a relative atomic mass of 14.01, and hydrogen (H) has a relative atomic mass of 1.01. b. The formula for ammonia is NH₃, so there is one nitrogen atom and three hydrogen atoms. c. M_r (NH₃) = (1 × A_r of N) + (3 × A_r of H) = (1 × 14.01) + (3 × 1.01) = 14.01 + 3.03 = 17.04 Thus, the relative formula mass of ammonia is 17.04.

Structure 1.4.3a—Molar mass M has the units g mol^–1

1. The formula for calculating the number of moles in a substance is: n = m / M, where n is the number of moles, m is the mass of the substance, and M is the molar mass of the substance.
2. The mass of a substance can be determined using the number of moles by multiplying the number of moles by the molar mass: m = n × M.
3. The molar mass of a substance is the mass of one mole of the substance, and it is calculated by adding the atomic masses of all the atoms in the formula unit of the substance.
4. If you know the molar mass of a substance and its mass, you can find the number of moles by dividing the mass by the molar mass: n = m / M.
5. The number of moles of water in 50 g of H₂O is approximately 2.78 mol (using a molar mass of 18.02 g/mol for H₂O).
6. The number of moles of oxygen in 100 g of CO₂ is approximately 1.47 mol (using a molar mass of 44.01 g/mol for CO₂).
7. 2 mol of FeCl₃ contain 2 × 55.845 g/mol = 111.69 g of Fe.
8. 128 g of a substance with a molar mass of 32 g/mol would contain 4 mol (128 g ÷ 32 g/mol = 4 mol).
9. 4 g of CH₄ would contain 0.25 mol (4 g ÷ 16.04 g/mol = 0.25 mol, using a molar mass of 16.04 g/mol for CH₄).
10. The mass of 0.5 mol of NaCl is 29.23 g (using a molar mass of 58.44 g/mol for NaCl).
11. The limiting reactant is the one that is completely consumed in the reaction. To determine the limiting reactant, you need to compare the number of moles of each reactant to the stoichiometric coefficients in the balanced equation.
12. The combustion of 1 mol of C₃H₈ produces 3 mol of H₂O, so the number of moles of H₂O produced would be 3 mol (using the balanced equation: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O).
13. 100 g of CaCl₂ would contain 0.9 mol (100 g ÷ 110.98 g/mol = 0.9 mol, using a molar mass of 110.98 g/mol for CaCl₂).
14. 5 g of KMnO₄ would contain 0.031 mol of O (5 g ÷ 158.03 g/mol = 0.031 mol, using a molar mass of 158.03 g/mol for KMnO₄).
15. The reaction between Al and HCl produces H₂ gas according to the equation: 2Al + 6HCl → 2AlCl₃ + 3H₂. If 2 mol of Al react, then 3 mol of H₂ gas would be produced.

Structure 1.4.3b—Molar mass M has the units g mol^–1

1. The SI Unit system is the international standard for measuring physical quantities in science, engineering, and everyday life. It provides a consistent and standardized way of communicating measurements and ensures accurate and reliable measurement across different fields of study.
2. The seven base SI Units and their properties are:

• Meter (m) – Length
• Kilogram (kg) – Mass
• Second (s) – Time
• Ampere (A) – Electric Current
• Kelvin (K) – Thermodynamic Temperature
• Mole (mol) – Amount of Substance
• Candela (cd) – Luminous Intensity

3. Derived SI Units are formed by combining the base SI Units. Examples include square meters (m²) for area, cubic meters (m³) for volume, meters per second (m/s) for speed, newtons (N) for force, and joules (J) for energy.
4. Base SI Units represent fundamental physical quantities, while derived SI Units are formed by combining two or more base units. Base units cannot be expressed in terms of other units, while derived units can.
5. The kilogram is considered the base unit of mass in the SI system because it is an independent unit, unlike force, which depends on mass and acceleration. The kilogram is defined as the mass of the International Prototype of the Kilogram, a platinum-iridium cylinder kept at the International Bureau of Weights and Measures (BIPM) in France.
6. The meter is defined as the distance traveled by light in a vacuum during 1/299,792,458 of a second. It is related to other SI Units, such as the kilogram and the second, through physical constants and laws of nature, such as Newton’s laws of motion and the laws of electromagnetism.
7. The kelvin is used to measure temperature on a thermodynamic scale, which is based on the absolute zero point (0 K).
8. SI Units are used in everyday life to measure physical quantities, such as length (meter), mass (kilogram), time (second), and temperature (kelvin or degree Celsius). Examples include measuring the length of a room, the weight of a grocery item, or the temperature of a room.
9. SI Unit prefixes are used to indicate multiples or fractions of base units. Common prefixes include kilo (10³), mega (10⁶), giga (10⁹), milli (10^-3), micro (10^-6), and nano (10^-9).
10. The use of SI Units is important in international communication and collaboration among scientists and researchers because it provides a universal language for measuring physical quantities, regardless of geographical or cultural differences. It ensures that measurements are accurate, precise, and consistent across different fields of study, allowing for effective comparison and analysis of data.

Structure 1.4.4—The empirical formula of a compound gives the simplest ratio of atoms of each element present in that compound. The molecular formula gives the actual number of atoms of each element present in a molecule.

1. The empirical formula of the compound is C₂H₃O₂.
2. The molecular formula of the compound is C₆H₁₂O₆.
3. The empirical formula of the compound is N₂O₅.
4. The molecular formula of the compound is C₄H₈Cl₂.
5. The empirical formula of the compound is C₂H₅O.
6. The molecular formula of the compound is NH₃.
7. The molecular formula of the compound is C₉H₁₅BrO₃.
8. The empirical formula of the compound is Na₂SO₄.
9. The molecular formula of the compound is C₆H₁₂.
10. The empirical formula of the compound is NO₂.

Structure 1.4.5—The molar concentration is determined by the amount of solute and the volume of solution.

1. The standard unit of volume for molar concentration is cubic decimeters (dm³), also known as liters (L).
2. Molarity is a measure of the amount of a solute dissolved in a solution, expressed in moles per liter (mol/L) or mol/dm³.
3. To calculate the number of moles of solute in a solution, divide the mass of the solute by its molar mass.
4. The formula for calculating molarity is Molarity = Moles of solute / Volume of solution in liters.
5. The molarity of the solution would be 1.5 M (3 moles / 2 L = 1.5 M).
6. The volume of the solution would be 2 L (0.5 moles / 0.25 M = 2 L).
7. The concentration of the solution would be 0.5 M ([solute] = 25 g / 58.44 g/mol = 0.428 mol; 0.428 mol / 0.5 L = 0.856 M or 0.856 mol/L or mol/dm³).
8. To convert milliliters to cubic decimeters, divide by 1000.
9. Mol/dm³ and M are the same unit of measure for molarity, both representing moles of solute per liter (dm³) of solution.
10. To calculate the grams per cubic decimeter of a substance in a solution, divide the mass of the substance by the volume of the solution in cubic decimeters (dm³).

Structure 1.4.6

1. Avogadro’s law states that equal volumes of gases measured under the same conditions of temperature and pressure contain the same number of molecules. It means that the volume of a gas is directly proportional to the number of molecules it contains.
2. The Italian scientist Amedeo Avogadro is credited with first proposing Avogadro’s law in 1811.
3. Avogadro’s law is important in the study of chemistry because it provides a fundamental relationship between the volume and number of molecules of a gas. This relationship is used in the ideal gas law, molar volume calculations, and other important concepts in chemistry.
4. The ideal gas law cannot be derived directly from Avogadro’s law, but it is related to it. The ideal gas law incorporates Avogadro’s law, along with other gas laws, to describe the behavior of gases in terms of their volume, pressure, temperature, and number of molecules.
5. The kinetic molecular theory helps explain the behavior of gases in relation to Avogadro’s law by describing the motion and interactions of gas particles. According to this theory, gas particles move freely and independently of one another, and their collisions with each other and with the walls of a container cause pressure. Avogadro’s law can be explained by considering that gas particles occupy the same amount of space regardless of their identity or size.
6. Real-life gases may deviate from ideal gas behavior due to factors such as intermolecular forces or non-uniform conditions. These deviations can impact Avogadro’s law, as the volume and number of molecules of a gas may not be directly proportional under non-ideal conditions.
7. According to Avogadro’s law, the two gases should contain the same number of molecules since they have identical volumes and are measured at the same temperature and pressure conditions.
8. Avogadro’s law can be used to compare the number of molecules of a gas at different temperatures or pressures, as long as the volume of the gas is adjusted accordingly. The law assumes that the temperature and pressure are held constant, so any changes in these variables would require adjustments to the volume to maintain the same number of molecules.
9. Avogadro’s law relates to the concept of molar volume because it describes the relationship between the volume and number of molecules of a gas. Molar volume is the volume occupied by one mole of a substance, and for gases, this value is equal to the volume of one mole of gas under standard temperature and pressure conditions.
10. Avogadro’s number is the number of molecules in one mole of a substance, and it is significant in relation to Avogadro’s law because it allows us to calculate the number of molecules present in a gas sample. Avogadro’s number is approximately 6.02 x 10^23, and it is used in conjunction with the mass of a substance and its molar mass to determine the number of moles and molecules present.

Structure 1.5.3

1. P = 150 kPa, V = 5 dm³, T = 35°C
   • Convert: P = 150,000 Pa, V = 0.005 m³, T = 308.15 K
   • n = PV / RT ≈ (150,000 * 0.005) / (8.314 * 308.15) ≈ 0.031 mol
2. P = 200 kPa, n = 2 mol, T = 50°C
   • Convert: P = 200,000 Pa, T = 323.15 K
   • V = nRT / P ≈ (2 * 8.314 * 323.15) / 200,000 ≈ 0.026 m³ or 26 dm³
3. n = 0.5 mol, V = 3 dm³, T = 100°C
   • Convert: V = 0.003 m³, T = 373.15 K
   • P = nRT / V ≈ (0.5 * 8.314 * 373.15) / 0.003 ≈ 413,962 Pa or ≈ 414 kPa
4. P = 100 kPa, V = 8 dm³, T = 27°C
   • Convert: P = 100,000 Pa, V = 0.008 m³, T = 300.15 K
   • n = PV / RT ≈ (100,000 * 0.008) / (8.314 * 300.15) ≈ 0.322 mol
5. P = 300 kPa, n = 1.5 mol, T = 75°C
   • Convert: P = 300,000 Pa, T = 348.15 K
   • V = nRT / P ≈ (1.5 * 8.314 * 348.15) / 300,000 ≈ 0.026 m³ or 26 dm³
6. n = 0.25 mol, V = 2 dm³, T = 150°C
   • Convert: V = 0.002 m³, T = 423.15 K
   • P = nRT / V ≈ (0.25 * 8.314 * 423.15) / 0.002 ≈ 439,194 Pa or ≈ 439 kPa
7. P = 250 kPa, V = 12 dm³, T = 20°C
   • Convert: P = 250,000 Pa, V = 0.012 m³, T = 293.15 K
   • n = PV / RT ≈ (250,000 * 0.012) / (8.314 * 293.15) ≈ 0.123 mol
8. P = 120 kPa, n = 3 mol, T = 10°C
   • Convert: P = 120,000 Pa, T = 283.15 K
   • V = nRT / P ≈ (3 * 8.314 * 283.15) / 120,000 ≈ 0.058 m³ or 58 dm³
9. n = 0.75 mol, V = 4 dm³, T = 200°C (continued)
   • Convert: V = 0.004 m³, T = 473.15 K
   • P = nRT / V ≈ (0.75 * 8.314 * 473.15) / 0.004 ≈ 618,881 Pa or ≈ 619 kPa
10. P = 80 kPa, V = 6 dm³, T = 40°C
    • Convert: P = 80,000 Pa, V = 0.006 m³, T = 313.15 K
    • n = PV / RT ≈ (80,000 * 0.006) / (8.314 * 313.15) ≈ 0.196 mol

Structure 1.5.4

1. New volume (V₂):
   • Convert temperatures to Kelvin: T₁ = 298.15 K, T₂ = 323.15 K
   • V₂ = V₁ * (P₁ / P₂) * (T₂ / T₁) = 2 dm³ * (100 kPa / 200 kPa) * (323.15 K / 298.15 K) ≈ 1.08 dm³
2. New pressure (P₂):
   • Convert temperatures to Kelvin: T₁ = 293.15 K, T₂ = 313.15 K
   • P₂ = P₁ * (V₁ / V₂) * (T₂ / T₁) = 150 kPa * (5 dm³ / 10 dm³) * (313.15 K / 293.15 K) ≈ 79.65 kPa
3. New temperature (T₂):
   • Convert temperatures to Kelvin: T₁ = 303.15 K
   • T₂ = T₁ * (P₁ / P₂) * (V₂ / V₁) = 303.15 K * (300 kPa / 150 kPa) * (4 dm³ / 8 dm³) = 151.575 K
   • Convert back to Celsius: T₂ ≈ -121.6°C
4. Final temperature (T₂):
   • Convert temperatures to Kelvin: T₁ = 298.15 K
   • T₂ = T₁ * (P₂ / P₁) * (V₂ / V₁) = 298.15 K * (200 kPa / 100 kPa) * (6 dm³ / 3 dm³) = 596.3 K
   • Convert back to Celsius: T₂ ≈ 323.15°C
5. New volume (V₂):
   • Convert temperatures to Kelvin: T₁ = 323.15 K, T₂ = 293.15 K
   • V₂ = V₁ * (P₁ / P₂) * (T₂ / T₁) = 5 dm³ * (120 kPa / 60 kPa) * (293.15 K / 323.15 K) ≈ 8.97 dm³
6. New pressure (P₂):
   • Convert temperatures to Kelvin: T₁ = 303.15 K, T₂ = 283.15 K
   • P₂ = P₁ * (V₁ / V₂) * (T₂ / T₁) = 200 kPa * (10 dm³ / 5 dm³) * (283.15 K / 303.15 K) ≈ 186.52 kPa
7. New volume (V₂):
   • Convert temperatures to Kelvin: T₁ = 313.15 K, T₂ = 353.15 K
   • V₂ = V₁ * (P₁ / P₂) * (T₂ / T₁) = 7 dm³ * (180 kPa / 360 kPa) * (353.15 K / 313.15 K) ≈ 4.99 dm³
8. Final temperature (T₂):
   • Convert temperatures to Kelvin: T₁ = 323.15 K
   • T₂ = T₁ * (P₁ / P₂) * (V₂ / V₁) = 323.15 K * (250 kPa / 125 kPa) * (24 dm³ / 12 dm³) = 646.3 K
   • Convert back to Celsius: T₂ ≈ 373.15°C
9. New temperature (T₂):
   • Convert temperatures to Kelvin: T₁ = 308.15 K
   • T₂ = T₁ * (P₂ / P₁) = 308.15 K * (180 kPa / 90 kPa) = 616.3 K
   • Convert back to Celsius: T₂ ≈ 343.15°C
10. New volume (V₂): Convert temperatures to Kelvin: T₁ = 333.15 K, T₂ = 303.15 K, V₂ = V₁ * (P₁ / P₂) * (T₂ / T₁) = 6 dm³ * (300 kPa / 600 kPa) * (303.15 K / 333.15 K) ≈ 2.73 dm³