BackGeneral Chemistry Practice Test #4 – Step-by-Step Study Guidance
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Q1. Apply Hess’s law to determine the enthalpy change (\( \Delta H \)) for the reaction: 2 C(s) + 2 H2O(g) \( \rightarrow \) CH4(g) + CO2(g), using the supplied equations and their \( \Delta H \) values.
Background
Topic: Thermochemistry – Hess’s Law
This question tests your ability to use Hess’s Law to determine the enthalpy change for a reaction by combining given chemical equations and their enthalpy changes.
Key Terms and Formulas:
Hess’s Law: The total enthalpy change for a reaction is the sum of the enthalpy changes for individual steps that lead to the overall reaction.
\( \Delta H \): Enthalpy change for a reaction.
Step-by-Step Guidance
Write out all the supplied equations and their \( \Delta H \) values. Label them for reference.
Compare the supplied equations to the target equation. Decide which equations need to be reversed or multiplied to match the reactants and products in the target equation.
Reverse or multiply equations as needed. Remember to change the sign of \( \Delta H \) if you reverse an equation, and multiply \( \Delta H \) by the same factor if you multiply the equation.
Add the modified equations together, canceling out any species that appear on both sides, to ensure you obtain the target equation.
Try solving on your own before revealing the answer!
Final Answer: -116 kJ
By appropriately reversing and combining the given equations, the overall enthalpy change for the target reaction is -116 kJ.
This is found by ensuring the sum of the modified equations matches the target reaction and summing their enthalpy changes.
Q2. Use the supplied enthalpy of formation data to determine \( \Delta H \) for the reaction: 3 Fe2O3(s) + CO(g) \( \rightarrow \) 2 Fe3O4(s) + CO2(g).
Background
Topic: Thermochemistry – Enthalpy of Formation
This question tests your ability to calculate the enthalpy change for a reaction using standard enthalpies of formation (\( \Delta H_f^\circ \)).
Key Terms and Formulas:
Standard Enthalpy of Formation (\( \Delta H_f^\circ \)): The enthalpy change when one mole of a compound is formed from its elements in their standard states.
Formula:
Step-by-Step Guidance
List the \( \Delta H_f^\circ \) values for each reactant and product, and note their coefficients from the balanced equation.
Multiply each \( \Delta H_f^\circ \) value by its respective coefficient.
Sum the total \( \Delta H_f^\circ \) for all products and for all reactants separately.
Set up the equation: .
Try solving on your own before revealing the answer!
Final Answer: -47.2 kJ
After plugging in the values and performing the calculation, the enthalpy change for the reaction is -47.2 kJ.
This is found by applying the enthalpy of formation formula to the given data.
Q3. The equation for the combustion of ethane (C2H6) is given. How much energy is released by the combustion of 30.00 g of ethane?
Background
Topic: Thermochemistry – Stoichiometry and Enthalpy
This question tests your ability to relate the mass of a reactant to the energy released using stoichiometry and the enthalpy change of reaction.
Key Terms and Formulas:
Molar Mass (MW): The mass of one mole of a substance (given as 30.08 g/mol for C2H6).
Stoichiometry: The calculation of reactants and products in chemical reactions.
\( \Delta H \): Enthalpy change for the reaction (given per 2 moles of C2H6).
Step-by-Step Guidance
Calculate the number of moles of ethane in 30.00 g using its molar mass.
Determine the energy released per mole of ethane by dividing the given \( \Delta H \) by the number of moles in the balanced equation.
Multiply the number of moles of ethane you have by the energy released per mole to find the total energy released.
Try solving on your own before revealing the answer!
Final Answer: -777.9 kJ
By converting grams to moles and using the stoichiometry of the reaction, the energy released is -777.9 kJ.
This matches the energy released for the combustion of 30.00 g of ethane.
Q4. Arrange the following spectral regions in order of increasing energy: infrared, microwave, ultraviolet, visible.
Background
Topic: Electromagnetic Spectrum
This question tests your understanding of the relative energies of different regions of the electromagnetic spectrum.
Key Terms and Concepts:
Energy and Frequency: Energy increases with frequency and decreases with wavelength.
Order (from lowest to highest energy): Microwave < Infrared < Visible < Ultraviolet
Step-by-Step Guidance
Recall the order of the electromagnetic spectrum from lowest to highest energy.
Arrange the given regions (microwave, infrared, visible, ultraviolet) accordingly.
Try solving on your own before revealing the answer!
Final Answer: Microwave < Infrared < Visible < Ultraviolet
This order reflects increasing frequency and energy across the electromagnetic spectrum.
Q5. Photons with relatively low energy are light with…
Background
Topic: Electromagnetic Radiation – Energy, Wavelength, and Frequency
This question tests your understanding of the relationship between photon energy, wavelength, and frequency.
Key Terms and Formulas:
Photon Energy:
Where:
= energy of a photon
= Planck’s constant
= frequency
= speed of light
= wavelength
Step-by-Step Guidance
Recall that energy is directly proportional to frequency and inversely proportional to wavelength.
Identify which combination (long/short wavelength, high/low frequency) corresponds to low energy.
Try solving on your own before revealing the answer!
Final Answer: long wavelength and low frequency
Low energy photons have long wavelengths and low frequencies.
Q6. What is the wavelength of light that has a frequency of Hz?
Background
Topic: Electromagnetic Radiation – Wavelength and Frequency
This question tests your ability to use the relationship between wavelength, frequency, and the speed of light.
Key Terms and Formulas:
Speed of Light Equation:
Where:
= speed of light ( m/s)
= wavelength (in meters)
= frequency (in Hz)
Step-by-Step Guidance
Write the equation and solve for .
Plug in the values for and .
Convert the resulting wavelength from meters to nanometers (1 m = nm) if needed.
Try solving on your own before revealing the answer!
Final Answer: 829 nm
Using , the wavelength is 829 nm.
Q7. Light is said to be quantized into packets of energy called…
Background
Topic: Quantum Theory of Light
This question tests your understanding of the concept of quantization of light energy.
Key Terms:
Photon: A quantum (packet) of electromagnetic energy.
Step-by-Step Guidance
Recall the term used for a single packet of light energy in quantum theory.
Try solving on your own before revealing the answer!
Final Answer: photons
Light energy is quantized in discrete packets called photons.
Q8. In a hydrogen atom, what is the wavelength of light emitted when an electron falls from the 4th to the 2nd energy level?
Background
Topic: Atomic Structure – Hydrogen Emission Spectrum
This question tests your ability to calculate the wavelength of light emitted during an electronic transition in hydrogen using the Rydberg equation.
Key Terms and Formulas:
Rydberg Equation:
= Rydberg constant ( m-1)
= lower energy level (here, 2)
= higher energy level (here, 4)
Step-by-Step Guidance
Identify and for the transition (from 4 to 2: , ).
Plug these values into the Rydberg equation.
Solve for (wavelength), and convert to nanometers if necessary.
Try solving on your own before revealing the answer!
Final Answer: 486.2 nm
This is the wavelength of visible light emitted for the n=4 to n=2 transition in hydrogen.
Q9. When the value of the quantum number n is small, what does this mean for the position and energy of the electron?
Background
Topic: Quantum Numbers and Atomic Structure
This question tests your understanding of the principal quantum number (n) and its relationship to electron energy and position.
Key Terms:
Principal Quantum Number (n): Indicates the main energy level or shell of an electron.
Smaller n = closer to nucleus, lower energy.
Step-by-Step Guidance
Recall what a small value of n means for the electron’s average distance from the nucleus.
Recall how energy changes with n.
Try solving on your own before revealing the answer!
Final Answer: It is closer to the nucleus and lower in energy.
Lower n means the electron is in a lower energy shell, closer to the nucleus.
Q10. What are the possible values of ml when l = 4?
Background
Topic: Quantum Numbers
This question tests your understanding of the magnetic quantum number (ml) and its possible values for a given angular momentum quantum number (l).
Key Terms:
Magnetic Quantum Number (ml): Can take integer values from -l to +l, including zero.
Step-by-Step Guidance
For l = 4, list all integer values from -4 to +4.
Try solving on your own before revealing the answer!
Final Answer: -4, -3, -2, -1, 0, +1, +2, +3, +4
ml ranges from -l to +l in integer steps.
Q11. Which of the following is a set of quantum numbers for a p-type orbital in the third principal energy level?
Background
Topic: Quantum Numbers and Atomic Orbitals
This question tests your ability to identify valid quantum numbers for a specific type of orbital and energy level.
Key Terms:
For a p-orbital: l = 1
For the third principal energy level: n = 3
ml can be -1, 0, or +1
Step-by-Step Guidance
Identify the correct values for n, l, and ml for a 3p orbital.
Check which option matches these values.
Try solving on your own before revealing the answer!
Final Answer: n = 3, l = 1, ml = +1
This set of quantum numbers describes a 3p orbital.
Q12. Which of the following best describes the photoelectric effect?
Background
Topic: Quantum Theory – Photoelectric Effect
This question tests your understanding of the photoelectric effect and its significance in quantum mechanics.
Key Terms:
Photoelectric Effect: The emission of electrons from a metal surface when light of sufficient energy shines on it.
Step-by-Step Guidance
Recall the definition of the photoelectric effect.
Identify which option matches this definition.
Try solving on your own before revealing the answer!
Final Answer: The emission of electrons from the surface of a metal when light shines on the metal.
This is the classic definition of the photoelectric effect.
Q13. Which scientist’s model of the atom had electrons traveling in fixed orbits around the nucleus?
Background
Topic: Atomic Models
This question tests your knowledge of historical atomic models and the scientists who proposed them.
Key Terms:
Bohr Model: Electrons travel in fixed orbits around the nucleus.
Step-by-Step Guidance
Recall which scientist proposed the fixed-orbit model of the atom.
Try solving on your own before revealing the answer!
Final Answer: Niels Bohr
Bohr’s model featured electrons in fixed orbits.
Q14. For a wave, the distance from the center line to a peak is called…
Background
Topic: Properties of Waves
This question tests your understanding of wave terminology.
Key Terms:
Amplitude: The maximum displacement from the center line to a peak (or trough).
Step-by-Step Guidance
Recall the definition of amplitude in wave mechanics.
Try solving on your own before revealing the answer!
Final Answer: Amplitude
Amplitude is the distance from the center line to a peak.
Q15. The Pauli Exclusion Principle states which of the following?
Background
Topic: Quantum Mechanics – Electron Configuration
This question tests your understanding of the Pauli Exclusion Principle and its implications for electron arrangements.
Key Terms:
Pauli Exclusion Principle: No two electrons in an atom can have the same set of four quantum numbers.
Step-by-Step Guidance
Recall the statement of the Pauli Exclusion Principle.
Identify which option matches this statement.
Try solving on your own before revealing the answer!
Final Answer: No two electrons in an atom can have the same set of four quantum numbers.
This is the correct statement of the Pauli Exclusion Principle.
Q16. When writing the shorthand notation for the electron configuration of chlorine, the symbol for which noble gas will be used?
Background
Topic: Electron Configuration – Noble Gas Notation
This question tests your understanding of shorthand (noble gas) notation for electron configurations.
Key Terms:
Noble Gas Notation: Uses the symbol of the previous noble gas to abbreviate the electron configuration.
Step-by-Step Guidance
Find chlorine (Cl) on the periodic table (atomic number 17).
Identify the noble gas that comes before chlorine.
Try solving on your own before revealing the answer!
Final Answer: Ne
Neon (Ne) is the noble gas preceding chlorine.
Q17. Which scientist proposed that if waves could have particle-like properties, then particles could have wavelike properties?
Background
Topic: Quantum Mechanics – Wave-Particle Duality
This question tests your knowledge of the scientist who proposed the wave-particle duality of matter.
Key Terms:
de Broglie Hypothesis: Particles such as electrons can exhibit wavelike properties.
Step-by-Step Guidance
Recall which scientist is associated with the wave-particle duality of matter.
Try solving on your own before revealing the answer!
Final Answer: Louis de Broglie
de Broglie proposed that particles have wavelike properties.
Q18. Which of the following types of light in the visible spectrum will have the highest frequency?
Background
Topic: Electromagnetic Spectrum – Visible Light
This question tests your understanding of the order of colors in the visible spectrum by frequency.
Key Terms:
In the visible spectrum: Red < Orange < Yellow < Green < Blue < Violet (increasing frequency)
Step-by-Step Guidance
Recall the order of visible light colors by increasing frequency.
Identify which color among the options has the highest frequency.
Try solving on your own before revealing the answer!
Final Answer: Blue
Blue light has a higher frequency than red, yellow, or green.
Q19. Interference between waves that causes them to cancel each other out is known as…
Background
Topic: Wave Interference
This question tests your understanding of constructive and destructive interference.
Key Terms:
Destructive Interference: Occurs when waves combine to produce a smaller amplitude (or cancel out).
Step-by-Step Guidance
Recall the term for wave interference that results in cancellation.
Try solving on your own before revealing the answer!
Final Answer: destructive interference
Destructive interference causes waves to cancel each other out.
Q20. Which type of orbital is represented with the quantum number l = 0?
Background
Topic: Quantum Numbers and Atomic Orbitals
This question tests your knowledge of the correspondence between quantum numbers and orbital types.
Key Terms:
l = 0: s orbital
l = 1: p orbital
l = 2: d orbital
l = 3: f orbital
Step-by-Step Guidance
Recall which orbital type corresponds to l = 0.
Try solving on your own before revealing the answer!
Final Answer: s
The s orbital is represented by l = 0.
Q21. Which atomic orbitals are present in the second principal energy level?
Background
Topic: Atomic Structure – Energy Levels and Orbitals
This question tests your understanding of which types of orbitals are available in a given principal energy level (n).
Key Terms:
n = 2: l can be 0 (s) or 1 (p)
Step-by-Step Guidance
For n = 2, list all possible values of l (0 and 1).
Identify the corresponding orbital types (s and p).
Try solving on your own before revealing the answer!
Final Answer: s, p
The second energy level contains s and p orbitals.
Q22. Which of the given elements will have the electron configuration 1s2 2s2 2p5 3s2 3p3?
Background
Topic: Electron Configuration
This question tests your ability to interpret electron configurations and match them to the correct element.
Key Terms:
Sum the superscripts to find the atomic number.
Step-by-Step Guidance
Add up all the electrons in the configuration to determine the atomic number.
Match the atomic number to the correct element on the periodic table.
Try solving on your own before revealing the answer!
Final Answer: Phosphorus
Phosphorus has the electron configuration shown.
Q23. How many electrons can be placed in the third principal energy level?
Background
Topic: Atomic Structure – Electron Capacity of Energy Levels
This question tests your understanding of how many electrons can fit in a given principal energy level (n).
Key Terms and Formulas:
Maximum electrons per energy level:
Step-by-Step Guidance
For n = 3, use the formula to calculate the maximum number of electrons.
Try solving on your own before revealing the answer!
Final Answer: 18
The third principal energy level can hold up to 18 electrons.