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General Chemistry Exam 1 Study Guidance

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Q1. What is the atomic number of argon?

Background

Topic: Periodic Table

This question tests your ability to use the periodic table to identify the atomic number of a given element.

Key Terms:

  • Atomic number (Z): The number of protons in the nucleus of an atom, which defines the element.

  • Argon (Ar): A noble gas element found in Group 18 of the periodic table.

Step-by-Step Guidance

  1. Locate argon (Ar) on the periodic table. It is typically found in Group 18 (noble gases).

  2. Identify the atomic number listed above or below the symbol for argon. This number represents the number of protons in an argon atom.

  3. Compare the atomic number you find with the answer choices provided. Remember, atomic number is always a whole number, not a decimal or the atomic mass.

Try solving on your own before revealing the answer!

Final Answer: D. 18

Argon's atomic number is 18, which means it has 18 protons. The other numbers listed are either atomic masses or atomic numbers of other elements.

Q2. Which element has chemical behavior most similar to oxygen?

Background

Topic: Periodicity and Chemical Properties

This question tests your understanding of periodic trends and how elements in the same group have similar chemical properties.

Key Terms:

  • Periodicity: The recurring trends that are observed in the properties of elements across periods and groups of the periodic table.

  • Group: A vertical column in the periodic table. Elements in the same group often have similar chemical properties.

Step-by-Step Guidance

  1. Recall that elements in the same group (column) as oxygen will have similar valence electron configurations and thus similar chemical behavior.

  2. Find oxygen on the periodic table (Group 16).

  3. Look at the answer choices and identify which element is also in Group 16.

Try solving on your own before revealing the answer!

Final Answer: C. Sulfur

Sulfur is in the same group as oxygen (Group 16), so it has similar chemical properties.

Q3. Which set of elements contains only nonmetals?

Background

Topic: Classification of Elements

This question tests your ability to distinguish between metals, nonmetals, and metalloids on the periodic table.

Key Terms:

  • Nonmetals: Elements that are typically poor conductors of heat and electricity, and are found on the right side of the periodic table.

  • Metals: Elements that are good conductors and are found on the left and center of the periodic table.

Step-by-Step Guidance

  1. Review the periodic table and recall which elements are classified as nonmetals.

  2. For each answer choice, check if all listed elements are nonmetals.

  3. Eliminate any set that contains a metal or metalloid.

Try solving on your own before revealing the answer!

Final Answer: C. phosphorus, nitrogen, oxygen

All three are nonmetals. The other sets include at least one metal.

Q4. The observation that hydrogen and oxygen can react to form two compounds with different O:H mass ratios (8:1 and 16:1) is consistent with which law?

Background

Topic: Laws of Chemical Combination

This question tests your understanding of the fundamental laws that govern how elements combine to form compounds.

Key Terms:

  • Law of Multiple Proportions: When two elements form more than one compound, the ratios of the masses of the second element that combine with a fixed mass of the first element are ratios of small whole numbers.

  • Law of Definite Proportions: A chemical compound always contains exactly the same proportion of elements by mass.

  • Law of Conservation of Mass: Mass is neither created nor destroyed in a chemical reaction.

Step-by-Step Guidance

  1. Notice that two different compounds are formed from the same elements (hydrogen and oxygen).

  2. Compare the mass ratios of oxygen to hydrogen in the two compounds (8:1 and 16:1).

  3. Recall which law explains the existence of different compounds with the same elements but different mass ratios.

Try solving on your own before revealing the answer!

Final Answer: A. Law of Multiple Proportions

This law explains how elements can combine in different ratios to form different compounds.

Q5. Which drawing represents a product mixture consistent with the law of mass conservation? Briefly explain your reasoning.

Background

Topic: Conservation of Mass in Chemical Reactions

This question tests your ability to apply the law of conservation of mass to visual representations of chemical reactions.

Key Terms:

  • Law of Conservation of Mass: The total mass of reactants equals the total mass of products in a chemical reaction.

  • Atoms: The basic units of matter that are conserved during chemical reactions.

Step-by-Step Guidance

  1. Count the number of each type of atom in the starting mixture.

  2. For each product drawing, count the number of each type of atom present.

  3. Identify which drawing has the same number of each type of atom as the starting mixture.

Try solving on your own before revealing the answer!

Final Answer: Drawing 2

Only Drawing 2 preserves the number of each type of atom, consistent with the law of mass conservation.

Q6. Match each experiment to the property of the atom it determined.

Background

Topic: Atomic Structure and History

This question tests your knowledge of key experiments in atomic theory and what each revealed about atomic structure.

Key Terms:

  • Millikan Oil Drop Experiment: Determined the charge of the electron.

  • Rutherford Gold Foil Experiment: Revealed the structure of the atom (nucleus).

  • Radioactive Decay: Showed the existence of fundamental particles (alpha, beta, gamma).

Step-by-Step Guidance

  1. Recall what property each experiment is famous for discovering.

  2. Match the experiment to the property: (a) Radioactive decay, (b) Millikan, (c) Rutherford.

  3. Fill in the blanks with the correct property for each experiment.

Try solving on your own before revealing the answer!

Final Answer:

  • (a) building blocks

  • (b) charge of the electron

  • (c) structure of the atom

Each experiment revealed a different fundamental property of atomic structure.

Q7. Fill in the gaps in the isotope table for protons, neutrons, electrons, and mass number.

Background

Topic: Isotopes and Atomic Structure

This question tests your understanding of how to determine the number of protons, neutrons, and electrons in isotopes and ions.

Key Terms and Formulas:

  • Protons: Equal to the atomic number (Z).

  • Neutrons: Mass number (A) minus atomic number (Z):

  • Electrons: For neutral atoms, electrons = protons. For ions, adjust for charge.

  • Mass Number (A): Total number of protons and neutrons:

Step-by-Step Guidance

  1. For each isotope symbol, identify the atomic number (number of protons) from the periodic table.

  2. Calculate the number of neutrons by subtracting the atomic number from the mass number.

  3. For neutral atoms, set electrons equal to protons. For ions, adjust for the charge.

  4. Double-check that the sum of protons and neutrons equals the mass number.

Try solving on your own before revealing the answer!

Final Answer:

  • For : 15 protons, 16 neutrons, 15 electrons, mass number 31

  • For : 42 protons, 54 neutrons, 42 electrons, mass number 96

These values are determined using the relationships above.

Q8. Calculate the average atomic weight of an element with two isotopes: 28.00 amu (21.00%) and 30.00 amu (79.00%).

Background

Topic: Average Atomic Mass

This question tests your ability to calculate the weighted average atomic mass of an element based on isotope masses and abundances.

Key Formula:

Step-by-Step Guidance

  1. Convert the percent abundances to decimal fractions (e.g., 21.00% = 0.2100).

  2. Multiply each isotope's mass by its fractional abundance.

  3. Add the results to get the average atomic mass.

Try solving on your own before revealing the answer!

Final Answer: 29.58 amu

amu

This is the weighted average based on the given abundances.

Q9. Which copper isotope is more abundant: copper-63 or copper-65? Explain using the average atomic mass.

Background

Topic: Isotopic Abundance and Average Atomic Mass

This question tests your ability to interpret average atomic mass and relate it to the relative abundance of isotopes.

Key Terms:

  • Average atomic mass: The weighted average of all naturally occurring isotopes of an element.

  • Isotopic abundance: The relative amount of each isotope present in nature.

Step-by-Step Guidance

  1. Find the average atomic mass of copper on the periodic table (about 63.55 amu).

  2. Compare this value to the masses of copper-63 (62.93 amu) and copper-65 (64.93 amu).

  3. Recall that the average atomic mass will be closer to the mass of the more abundant isotope.

Try solving on your own before revealing the answer!

Final Answer: Copper-63 is more abundant

The average atomic mass (63.55 amu) is closer to 62.93 amu (copper-63) than to 64.93 amu (copper-65), indicating copper-63 is more abundant.

Q10a. If you have 0.65 moles of carbon, how many grams of carbon do you have?

Background

Topic: Mole-to-Mass Conversions

This question tests your ability to convert between moles and grams using molar mass.

Key Formula:

Step-by-Step Guidance

  1. Find the molar mass of carbon from the periodic table (about 12.011 g/mol).

  2. Multiply the number of moles (0.65) by the molar mass.

  3. Set up the calculation:

Try solving on your own before revealing the answer!

Final Answer: 7.81 g

g

This is the mass of 0.65 moles of carbon.

Q10b. If you have 0.65 moles of carbon, how many atoms of carbon do you have?

Background

Topic: Mole-to-Particle Conversions

This question tests your ability to convert between moles and number of atoms using Avogadro's number.

Key Formula:

Where atoms/mol

Step-by-Step Guidance

  1. Write down the number of moles (0.65 mol).

  2. Multiply by Avogadro's number:

  3. Set up the calculation, but do not compute the final value yet.

Try solving on your own before revealing the answer!

Final Answer: atoms

atoms

This is the number of carbon atoms in 0.65 moles.

Q11a. Which type of radiation (radio waves or visible light) has the shorter wavelength? Show your reasoning.

Background

Topic: Electromagnetic Spectrum

This question tests your understanding of the relationship between frequency and wavelength for electromagnetic radiation.

Key Formula:

Where is wavelength, is the speed of light, and is frequency.

Step-by-Step Guidance

  1. Recall that wavelength and frequency are inversely related: as frequency increases, wavelength decreases.

  2. Compare the frequencies given: radio waves ( s) and visible light ( s$^{-1}$).

  3. Determine which has the higher frequency and thus the shorter wavelength.

Try solving on your own before revealing the answer!

Final Answer: Visible light

Visible light has a much higher frequency, so it has a much shorter wavelength than radio waves.

Q11b. Which photon (radio waves or visible light) has the lower energy? Show your reasoning.

Background

Topic: Photon Energy

This question tests your understanding of the relationship between frequency and photon energy.

Key Formula:

Where is energy, is Planck's constant, and is frequency.

Step-by-Step Guidance

  1. Recall that photon energy is directly proportional to frequency.

  2. Compare the frequencies of radio waves and visible light.

  3. Determine which has the lower frequency and thus the lower energy per photon.

Try solving on your own before revealing the answer!

Final Answer: Radio waves

Radio waves have a much lower frequency, so their photons have lower energy than those of visible light.

Q12. According to the Balmer-Rydberg equation, which electron transition emits the greatest energy?

Background

Topic: Atomic Spectra and Electron Transitions

This question tests your understanding of how electron transitions between energy levels relate to the energy of emitted photons.

Key Formula:

Energy is proportional to , where .

Step-by-Step Guidance

  1. For each transition, identify the initial () and final () energy levels.

  2. Calculate for each transition.

  3. The transition with the largest value corresponds to the greatest energy emission.

Try solving on your own before revealing the answer!

Final Answer: B.

This transition has the largest energy difference and thus emits the greatest energy.

Q13. Why does the hydrogen emission spectrum consist of sharp lines?

Background

Topic: Atomic Emission Spectra

This question tests your understanding of quantized energy levels in atoms.

Key Terms:

  • Quantized energy levels: Electrons can only occupy specific energy levels in an atom.

  • Emission spectrum: The set of wavelengths emitted by excited electrons dropping to lower energy levels.

Step-by-Step Guidance

  1. Recall that sharp lines indicate discrete energy changes.

  2. Consider which answer choice explains why only certain wavelengths are emitted.

  3. Relate this to the concept of quantized energy levels in hydrogen.

Try solving on your own before revealing the answer!

Final Answer: B. The electron in hydrogen can only jump between specific energy levels

This quantization leads to sharp lines in the emission spectrum.

Q14. Use the de Broglie relationship to calculate the wavelength of a neutron traveling at m/s.

Background

Topic: Wave-Particle Duality

This question tests your ability to apply the de Broglie equation to calculate the wavelength of a particle.

Key Formula:

Where is Planck's constant, is mass, and is velocity.

Step-by-Step Guidance

  1. Write down the values: J·s, kg, m/s.

  2. Plug these values into the de Broglie equation:

  3. Set up the calculation, but do not compute the final value yet.

Try solving on your own before revealing the answer!

Final Answer: m

m

This is the de Broglie wavelength for the neutron at the given speed.

Q15. For each orbital, write the proper label and the value of the angular momentum quantum number (): and .

Background

Topic: Quantum Numbers and Atomic Orbitals

This question tests your understanding of quantum numbers and how they relate to orbital labels.

Key Terms:

  • Principal quantum number (): Indicates the energy level.

  • Angular momentum quantum number (): Determines the shape of the orbital ( for s, for p, for d, for f).

  • Orbital label: Combines and the letter corresponding to (e.g., 2s, 3p, 4d).

Step-by-Step Guidance

  1. For , determine which value and label are given (e.g., corresponds to d orbital, so label is 4d).

  2. For , corresponds to a p orbital, so label is 3p.

  3. Write the correct label and value for each case.

Try solving on your own before revealing the answer!

Final Answer:

  • n = 4: label = 4d,

  • n = 3: label = 3p,

These are the correct labels and quantum numbers for the given orbitals.

Q16. Given an electron configuration, determine the element, or vice versa.

Background

Topic: Electron Configurations

This question tests your ability to interpret and write electron configurations for elements.

Key Terms:

  • Electron configuration: The arrangement of electrons in an atom's orbitals.

  • Noble gas core: Abbreviated notation using the previous noble gas in brackets.

Step-by-Step Guidance

  1. For each configuration, count the total number of electrons to identify the element.

  2. For a given element, write its electron configuration, using noble gas notation if allowed.

  3. Double-check your answer using the periodic table.

Try solving on your own before revealing the answer!

Final Answer:

  • [Ar]4s2 = Ca (calcium)

  • 1s22s22p63s23p1 = Al (aluminum)

  • [Xe]6s24f145d7 = Ir (iridium, Z=77)

Each configuration matches the element with the corresponding number of electrons.

Q17. For each pair of atoms, which is larger: (a) germanium (Ge) vs. bromine (Br); (b) carbon (C) vs. tin (Sn)?

Background

Topic: Periodic Trends – Atomic Radius

This question tests your understanding of how atomic size changes across periods and down groups in the periodic table.

Key Concepts:

  • Atomic radius: Increases down a group, decreases across a period from left to right.

Step-by-Step Guidance

  1. For (a), compare Ge and Br, which are in the same period. Atomic radius decreases left to right, so the element further left is larger.

  2. For (b), compare C and Sn, which are in the same group. Atomic radius increases down a group, so the element lower in the group is larger.

  3. Circle the larger atom in each pair.

Try solving on your own before revealing the answer!

Final Answer:

  • (a) Germanium (Ge) is larger than bromine (Br).

  • (b) Tin (Sn) is larger than carbon (C).

These trends follow the rules for atomic radius across periods and down groups.

Q18. Why does an electron in a 3s orbital "feel" less of the nuclear charge than an electron in a 2s orbital?

Background

Topic: Effective Nuclear Charge and Electron Shielding

This question tests your understanding of how electron shielding affects the effective nuclear charge experienced by electrons in different orbitals.

Key Terms:

  • Effective nuclear charge (): The net positive charge experienced by an electron in a multi-electron atom.

  • Shielding: Inner electrons block some of the positive charge from the nucleus, reducing the attraction felt by outer electrons.

Step-by-Step Guidance

  1. Recall that electrons in higher energy levels (like 3s) are farther from the nucleus than those in lower levels (like 2s).

  2. Recognize that inner-shell electrons (e.g., 1s, 2s, 2p) shield outer electrons from the full nuclear charge.

  3. Explain how this shielding causes the 3s electron to experience a lower effective nuclear charge than the 2s electron.

Try answering in your own words before revealing the answer!

Final Answer:

The 3s electron is farther from the nucleus and has more inner-shell electrons between it and the nucleus. These inner electrons shield it from the full nuclear charge, so the 3s electron experiences a lower effective nuclear charge () than the 2s electron, which is closer to the nucleus and less shielded.

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