Chemisty Chapter 4 and 5
One of the radioactive isotopes used in medical treatment or analysis is chromium-51. The half-life of chromium-51 is 28 days. How many days is/are required for the activity of a sample of chromium-51 to fall to 12.5 percent of its original value?
.Chromium-51 has a half-life of 28 days. It requires three half-lives, or 3 × 28 days = 84 days to fall to 12.5% of its initial value.
What is the energy (in J) of a mole of photons that have a wavelength of 701 nm? (h = 6.626 × 10⁻³⁴ J • s and c = 3.00 × 10⁸ m/s)
.The energy (E) of a photon is equal to the product of Planck's constant (h) and the speed of light (c) divided by the wavelength (λ): E=hc/λ First, convert the wavelength from nm to m, and then plug it into this equation. λ=701 nm×(1 m/10^9nm)=7.01×10^−7 m E=hcE=(6.626×10^−34 J⋅s)(3.00×10^8ms)/7.0110^−7 m=2.83×10^−19 J This is the energy of one photon; to determine the energy of a mole of photons with this wavelength, multiply the energy by Avogadro's number: Emole=Ephoton⋅NA=(2.83×10^−19 J)(6.022×10^23 mol−1)=1.70×10^5Jmol
A photon has a frequency of 8.6 × 10⁸ Hz. What is the energy of this photon?
.The relationship between energy and frequency is shown.E=hν Plug in the given values for frequency and Planck's constant and solve for energy. E=(6.626×10−34J⋅s)(8.6×10^81/8)
If n = 2, what is the maximum allowed value for ℓ?
1. According to the rules governing acceptable quantum numbers, the value for ℓ can only be as high as n-1. If n = 2, the maximum value for ℓ is 1.
What is the energy, in J, of light that must be absorbed by a hydrogen atom to transition an electron from n = 3 to n = 6? Submit an answer to three signficant figures.
1.82 × 10⁻¹⁹ J ΔEi→f=(2.18×10−18 J)(1n2f−1n2i)ΔE3→6=(2.18×10−18 J)(162−132)=−1.82×10−19 J∣ΔE3→6∣=1.82×10−19 J
The ground state electron configuration of an Al atom is
1s²2s²2p⁶3s²3p¹. The element Al is a main-group element on the right-hand side of the periodic table (meaning it is in the p block) and the third row, so its electron configuration will end with 3p. The electron configuration of Al is 1s²2s²2p⁶3s²3p¹.
The ground state electron configuration of a Se atom is
1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁴ The element Se is a main-group element on the right-hand side of the periodic table (meaning it is in the p block) and the fourth row, so its electron configuration will end with 4p. The electron configuration of Se is 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁴.
Which of the following is the electron configuration of a neutral K atom?
1s²2s²2p⁶3s²3p⁶4s¹ The element K is in the first column of the periodic table (meaning it is in the s block) and the fourth row, so its electron configuration will end with 4s¹. The electron configuration of K is 1s²2s²2p⁶3s²3p⁶4s¹.
Calculate the atomic mass of the magnesium-27 nucleus in amu (to three decimal places) assuming that the mass of a nucleon is 1.008 amu and the binding energy per nucleon is 1.326×10⁻¹² J/nucleon. (1 kg = 6.022×10²⁶ amu)
26.976 amu
What nuclide undergoes electron capture to produce ¹⁰⁸Pd?
A nuclide undergoing electron capture to produce ¹⁰⁸Pd, atomic number 46, would have a proton converted to a neutron, therefore an atomic number of 47 but still a mass number of 108, which is ¹⁰⁸Ag
If light has a lot of energy, it will have:
A small wavelength
A ground state atom of Mg could not have any electrons with which of the following configurations?
A) n = 3, ℓ = 1, mℓ = 0, ms = +½ A ground state Mg atom has the electronic structure 1s²2s²2p⁶3s². The highest-energy orbital in this atom is 3s, which represents the quantum numbers n = 3, ℓ = 0. Of the options given, the set of quantum numbers with n = 3, ℓ = 1 is higher in energy than the highest occupied atomic orbital, so a ground state atom of Mg could not have any electrons with the set of quantum numbers n = 3, ℓ = 1, mℓ = 0, ms = +½.
Which of the following sets of quantum numbers can describe a 4s electron?
A) n = 4, ℓ = 0, mℓ = 0 Atomic orbitals are described by their principal quantum number (n) followed by a letter that corresponds to the value for the angular momentum quantum number (ℓ). For a 4s electron, the number 4 represents n, whereas s represents ℓ. ℓ corresponds to the following values: s = 0, p = 1, d = 2, f = 3; this means that in this orbital, ℓ = 0. The value for mℓ must be between -ℓ and ℓ, so if ℓ = 0, mℓ must be 0 as well.
What is the most likely decay for the Mn-57 nucleus?
A) β⁻ decay The isotope Mn-57 has a neutron/proton ratio of 32/25 = 1.28, which is higher than the stable ratio of about 1.2 for a nucleus of this size. β⁻ decay, converting the ratio to 31/26 = 1.19m is the only choice that will lower the neutron/proton ratio.
Of the following, which sublevel is filled last?
According to the Aufbau principle, the order in which atomic orbitals are filled is as follows: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, ... Of the options given, the 4d orbital is at the highest energy, and therefore it will be filled with electrons last.
Which element has the electron configuration 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p²?
B) Ge. This electron configuration ends with 4p, meaning it is an element in the p block (a main group element on the right-hand side of the periodic table) in the fourth row. The 2 in 4p² indicates that it is the second element in this row, so the element must be Ge.
Which of the following electron configurations is incorrect?
B) Mn [Ar]4s²4d⁵. Because elements in the d block have the quantum number n one less than their row in the periodic table, the elements in the fourth row and in the d block will end in 3d. Therefore, the element Mn will not have the configuration [Ar]4s²4d⁵, but rather [Ar]4s²3d⁵.
For each of the following ground state electron configurations, determine what is incorrect. N: 1s²2s¹2p³
B) The configuration is missing an electron in the 2s orbital. The correct ground state configuration for nitrogen is 1s²2s²2p³ with 2 electrons in the 2s-orbital
Which of the following requires extremely high temperatures?
B) fusion reactions only.Nuclear fusion reactions require extremely high temperatures to overcome the repulsive forces of positive nuclei.
A ground state atom of As could not have any electrons with which of the following configurations?
B) n = 4, ℓ = 2, mℓ = 0, ms = +½ ground state As atom has the electronic structure 1s²2s²2p⁶3s²3p⁶4s²3dⁱ⁰4p³. The highest-energy orbital in this atom is 4p, which represents the quantum numbers n = 4, ℓ = 1. Of the options given, the set of quantum numbers with n = 4, ℓ = 2 is higher in energy than the highest occupied atomic orbital, so a ground state atom of As could not have any electrons with the set of quantum numbers n = 4, ℓ = 2, mℓ = 0, ms = +½.
The equation below is best described as ___________. ²³⁵₉₂U → ²³⁵₉₃Np + ⁰₋₁e
Beta decay is the emission on an electron from the nucleus. A neutron becomes and proton and emits the electron (beta particle). Therefore, this equation represents the beta decay of ²³⁵₉₂U.
Negatively charged particles of radiation emitted from the decay of radioactive substances are known as
Beta particles are negatively charged particles of radiation emitted from the decay of radioactive substances.
For each of the following ground state electron configurations, determine what is incorrect. Cr: 1s²2s²2p⁶3s²3p⁶4s²3d⁴
C) Cr is an exception and has only 1 electron is the 4s orbital. The correct ground state configuration for Cr is 1s²2s²2p⁶3s²3p⁶4s¹3d⁵. Cr is an exception and has only 1 electron is the 4s orbital.
Which of these terms describes the following process? ²³⁹₉₄Pu → ²³⁵₉₂U + ⁴₂He²⁺
C) Exothermic This process is a radioactive decay process in which a nucleus splits into two parts, and energy is released. Therefore all three terms describe it.
For each of the following ground state electron configurations, determine what is incorrect. Si: 1s²2s²2p⁶3s²3p³
C) The configuration has too many electrons in the 3p orbital. The correct ground state configuration for silicon is 1s²2s²2p⁶3s²3p² with 2 electrons in the 3p-orbital.
Which of the following sets of quantum numbers can describe a 3p electron?
C) n = 3, ℓ = 1, mℓ = 1 Atomic orbitals are described by their principal quantum number (n) followed by a letter that corresponds to the value for the angular momentum quantum number (ℓ). For a 3p electron, the number 3 represents n, whereas s represents ℓ. ℓ corresponds to the following values: s = 0, p = 1, d = 2, f = 3; this means that in this orbital, ℓ = 1. The value for mℓ must be between -ℓ and ℓ, so if ℓ = 1, mℓ must be -1, 0, or 1.
What is the product of alpha emission from uranium-238?
C) ²³⁴Th. Alpha emission from uranium 238 lowers the mass number by 4 from 238 to 234, and the atomic number by 2, from 92 to 90, so the product is ²³⁴Th.
Which element has the electron configuration 1s²2s²2p⁶3s²3p⁶4s¹3d⁵?
Cr. This electron configuration ends with 3d, meaning it is an element in the d block. Because elements in the d block have the quantum number n one less than their row in the periodic table, this element will be in the fourth row. While the 5 in 3d⁵ normally indicates that this electron configuration corresponds to the fifth entry in that row, the 4s orbital is only half-filled (with 1 electron instead of 2); if the 4s orbital were filled, it would take an electron from the 3d orbital, so 4s¹3d⁵ would change to 4s²3d⁴. This would correspond to the fourth element in that row, which is Cr. The reason the electron configuration of Cr ends in 4s¹3d⁵ instead of 4s²3d⁴ is that having two half-filled orbitals is more stable than having one partially filled (but not exactly half-filled) and one completely filled.
What is the purpose of boron or cadmium rods inserted into nuclear reactors?
D) They absorb excess neutrons to keep the chain reactions in the reactor under control. The cadmium or boron rods inserted in nuclear reactors are there to absorb excess neutrons and thereby keep the chain reaction under control.
What is the noble gas electron configuration of Ca?
D) [Ar]4s² The element Ca is in the second column of the periodic table (meaning it is in the s block) and the fourth row, so its electron configuration will end with 4s². The last noble gas filled prior to Ca is Ar, so its electron configuration will be [Ar]4s².
Which electron configuration violates Hund's rule?
D) [He]2s²2px²2py² Hund's rule states that electrons are added to the atomic orbitals with the same energy levels (degenerate orbitals) in such a way that each orbital is occupied by a single electron with the same spin before any orbital can be occupied by two electrons. Because there are three 2p orbitals, each of the individual 2p orbitals (2px, 2py, 2pz) needs to have a single electron before any of them can have two electrons. Therefore, the electron configuration [He]2s²2px²2py², which has two electrons in the 2px and 2py orbitals, but zero electrons in the 2pz orbital, is in violation of Hund's rule.
Because electron energy levels are quantized, which of the following is true? I) Atomic emission spectra are a series of discrete lines. II) Atomic emission spectra are continuous. III) Electrons may only occupy a discrete set of energy levels. IV) Electrons may occupy a continuous range of energy levels.
E) I and III
Rubidium-87 is radioactive, decaying by the emission of a single beta particle. Write the nuclear equation for this decay.
E) ⁸⁷Rb → ⁸⁷Sr + ⁰₋₁e Rubidium-87 decays by emission of a beta particle, as shown by the following nuclear equation.8737Rb→0−1e+8738Sr
Determine the energy of a photon with a wavelength of 364 nm.
E=hc/λ Plug in the given values for wavelength (converted from nm to m) and Planck's constant and the known speed of light. Then, solve for energy. E=(6.626×10^−34J⋅s)(3.00×10^8ms)/3.64×10^−7m E=5.46×10^−19J
What element is designated by the orbital diagram below?
F. This orbital diagram contains a total of 9 electrons (1s²2s²2p⁵). Since the question is asking for an element and not an ion, the element must have the atomic number 9, meaning that this element must be F.
Bombarding nuclei with neutrons is characteristic of
Fission Reactions only. Bombarding nuclei with neutrons is characteristic of fission reactions only. The extra neutrons lead to unstable nuclei that undergo the fission process.
Control rods are characteristic components of
Fission reactors only. Control rods are used to absorb neutrons to control the rate of the nuclear reaction. Neutron bombardment is characteristic of fission reactors. Therefore, control rods are an important component of fission reactors.
Which type of radiation has the greatest penetrating ability?
Gamma rays consist of electromagnetic radiation, which is more penetrating than the particles making up alpha and beta rays.
What type of radiation can be used to kill bacteria in food?
High energy gamma rays, often from Co-60, can be effectively used to kill bacteria in food because they are very penetrating and can pass through most materials except for lead.
A student draws the orbital diagram below for the 3p electrons in an S atom. What, if anything, is incorrect about the drawing?
Hund's rule states that electrons are added to the atomic orbitals with the same energy levels (degenerate orbitals) in such a way that each orbital is occupied by a single electron with the same spin (either +½ or -½) before any orbital can be occupied by two electrons. In this student's diagram, the two orbitals that contain a single electron have electrons with opposite spins (one is up and one is down), which is in violation of Hund's rule.
What is the noble gas core in the electron configuration for Sb?
Kr. The noble gas core for a given element is the last noble gas filled in the periodic table prior to the given element. Here, the last noble gas prior to Sb is Kr, so that is its noble gas core.
Which types of processes are likely when the neutron-to-proton ratio in a nucleus is too large? I. α decay II. β⁻ decay III. positron production IV. electron capture
Only β⁻ decay (D) II only) will reduce the neutron/proton ratio by producing a proton from a neutron. Positron production and electron capture will convert a proton to a neutron, and α decay, which removes two protons and two neutrons, will cause an increase in the ratio when there are more neutrons than protons.
What type of light on the electromagnetic spectrum has the lowest energy per photon?
Radio Waves
Who developed the equation that allowed the energy of the electron to be described quantum mechanically?
Schrödinger developed the equation that describes the energy of an electron quantum mechanically. This equation now bears his name (the Schrödinger equation).
The process that powers stars is
The energy in the stars is produced by nuclear fusion.
When two He atoms undergo fusion in a star, the resulting element is:
The fusion of two He atoms, atomic number 2, will form a nucleus with an atomic number of 4, which is Be.
What fundamental force is responsible for holding together the subatomic particles of an atomic nucleus?
The strong nuclear force acts on a subatomic scale, and is responsible for the attractions between protons and neutrons in the atomic nucleus.
Which of the following has the greatest mass?
The α particle contains two neutrons and two protons, so it is the most massive particle on the list.
What isotope is commonly used to determine thyroid activity and treat diseases of the thyroid?
Thyroid hormone contains iodine, so the isotope commonly used to determine thyroid activity and treat diseases of the thyroid is ¹³¹I.
Light wave A has a greater frequency than light wave B. Which has a greater wavelength?
Wave B. All light travels at the same speed (c), and since speed is equal to the product of wavelength and frequency, a higher frequency results in a shorter wavelength. Because wave A has a greater frequency than wave B, wave B must have the longer wavelength.
What nuclide is formed when Th-230 undergoes alpha emission?
When Th-230, atomic number 90, undergoes alpha emission, the nuclide formed will have an atomic number of 88 and a mass number of 236, which is Ra-226.
Which of the following occurs during positron emission?
When a positron is emitted, a positively charged proton is converted to a neutral neutron, so the charge on the nucleus, which is the atomic number, decreases.
Energy of a photon is ______ proportional to frequency, and _______ proportional to wavelength.
directly, inversely
In an investigation of the electronic absorption spectrum of a particular element, it is found that a photon having λ = 500 nm provides just enough energy to promote an electron from n = 2 to n = 3. From this information, we can deduce
the difference between the energies of the n = 2 and n = 3 levels. If we know that a photon that has a wavelength of 500 nm provides just enough energy to promote an electron from n=2 to n=3, we know the difference between the energies of the n=2 and n=3 levels.
A significant difficulty in using fusion as an energy source is
the extreme conditions required to initiate and sustain fusion reactions. The major difficulty in using nuclear fusion for an energy source is the extreme conditions required to initiate and sustain fusion reactions. The starting nuclides are relatively easy to obtain and are not terribly expensive, and the products are not highly radioactive needing special disposal procedures.
Heavy nuclides with too few neutrons to be in the band of stability are most likely to decay by what mode?
α-Particle production removes two protons and two neutrons from the nucleus. When there are more neutrons than protons in the nucleus, as there are in the heavy elements, removing two of each increases the neutron/proton ratio, thereby moving the nucleus more toward the band of stability.
Which of the following is true concerning ψ²?
ψ² describes the probability of finding an electron in space.
Which of the following symbols represents a beta particle?
⁰₋₁e is the symbol for a beta particle.
