CHEM 115 Exam 4 Practice Questions

Question 1

• The distance between identical points on successive waves is defined as the wavelength.

Question 2

• All forms of electromagnetic radiation share the characteristic of being the transmission of energy in the form of waves.

Question 3

• When a solid is heated, it emits electromagnetic radiation known as blackbody radiation.

Question 4

• The photoelectric effect involves electrons being ejected from the surface of a metal exposed to light of a certain minimum frequency.

Question 5

• Amplitude is defined as the vertical distance from the midline of a wave to the top of the peak or the bottom of the trough.

Question 6

• To calculate the frequency of light with a wavelength of 360 nm, use the formula: $c = λν$, where $c$ is the speed of light ($3.00 × 10^8$ m/s), $λ$ is the wavelength, and $ν$ is the frequency.
• Convert nm to m: $360 nm = 360 × 10^{-9} m$
• Solve for $ν$: $ν = c / λ = (3.00 × 10^8 m/s) / (360 × 10^{-9} m) = 8.3 × 10^{14} s^{-1}$

Question 7

• The arrangement of electromagnetic radiation, starting with the shortest wavelength and increasing to the longest wavelength, is: gamma rays, ultraviolet, infrared, radio.

Question 8

• To find the energy of one photon of radio wave radiation with a frequency of $8.6 × 10^8 Hz$, use the formula: $E = hν$, where $h$ is Planck's constant ($6.63 × 10^{-34} J·s$).
• $E = (6.63 × 10^{-34} J·s) × (8.6 × 10^8 Hz) = 5.7 × 10^{-25} J$

Question 9

• The solar radiation spectrum peaks at a wavelength of approximately 500 nm. To calculate the energy of one photon, use the formula: $E = hc/λ$
• Where $h = 6.63 × 10^{-34} J·s$, $c = 3.00 × 10^8 m/s$, and convert $λ$ from nm to m: $500 nm = 500 × 10^{-9} m$
• $E = (6.63 × 10^{-34} J·s × 3.00 × 10^8 m/s) / (500 × 10^{-9} m) = 4 × 10^{-19} J$

Question 10

• To calculate the wavelength of light emitted by a hydrogen atom when its electron drops from $n = 7$ to $n = 4$, first calculate the energy change using the formula: $En = -2.18 × 10^{-18} J (1/n^2)$.
• $\Delta E = E<em>{final} - E</em>{initial} = -2.18 × 10^{-18} J (1/4^2 - 1/7^2) = -2.18 × 10^{-18} J (1/16 - 1/49) = -2.18 × 10^{-18} J (0.0625 - 0.0204) = -2.18 × 10^{-18} J (0.0421) = -9.18 × 10^{-20} J$
• Since energy is emitted, the value is negative. Use the absolute value $|\Delta E|$ to find the wavelength using $E = hc/λ$.
• Rearrange to solve for $λ$: $λ = hc/E = (6.63 × 10^{-34} J·s × 3.00 × 10^8 m/s) / (9.18 × 10^{-20} J) = 2.17 × 10^{-6} m$
• Convert meters to nanometers: $2.17 × 10^{-6} m = 2.17 × 10^3 nm$

Question 11

• To calculate the frequency of a photon absorbed during a transition from $n<em>1 = 2$ to $n</em>2 = 4$, use the Rydberg equation: $\frac{1}{λ} = R(\frac{1}{n<em>1^2} - \frac{1}{n</em>2^2})$, where $R = 1.096776 × 10^7 m^{-1}$.
• $\frac{1}{λ} = 1.096776 × 10^7 m^{-1} (\frac{1}{2^2} - \frac{1}{4^2}) = 1.096776 × 10^7 m^{-1} (\frac{1}{4} - \frac{1}{16}) = 1.096776 × 10^7 m^{-1} (0.25 - 0.0625) = 1.096776 × 10^7 m^{-1} (0.1875) = 2.056455 × 10^6 m^{-1}$
• Then use $c = λν$ to find the frequency $ν = c / λ$.
• Since we have $1/λ$, use $ν = c × (1/λ) = 3.00 × 10^8 m/s × 2.056455 × 10^6 m^{-1} = 6.17 × 10^{14} s^{-1}$.

Question 12

• The shape of an atomic orbital is associated with the angular momentum quantum number (l).

Question 13

• The set of quantum numbers that can correctly represent a 3p orbital is: $n = 3, l = 1, m_l = -1$

Question 14

• The set of quantum numbers that is not possible is: $n = 3, l = 0, m<em>l = 1, m</em>s = -1/2$. Because if $l = 0$, then $m_l$ must also be 0.

Question 15

• For the set of quantum numbers $n = 4, l = 3, m<em>l = –2, m</em>s = +1/2$, the maximum number of electrons is 1.

Question 16

• Electrons in an orbital with $l = 3$ are in an f orbital.

Question 17

• The element with the ground-state electron configuration $1s^22s^22p^63s^2$ is Magnesium (Mg).

Question 18

• The electronic structure $1s^22s^22p^63s^23p^64s^23d^8$ refers to the ground state of Nickel (Ni).

Question 19

• In a Lewis dot symbol, the dots represent valence electrons.

Question 20

• The Lewis dot symbol for argon has 8 dots around it.

Question 21

• The Lewis dot symbol for sodium has 1 dot around it.

Question 22

• The Lewis dot symbol for the chloride ion (Cl-) is

Question 23

• The Lewis dot symbol for the S2– ion is

Question 24

• The compound most likely to be ionic is KF.

Question 25

• The compound most likely to be ionic is SrBr2.

Question 26

• The compound most likely to be covalent is SF4.

Question 27

• The molecule with the largest dipole moment is HF.

Question 28

• 6 lone pairs of electrons need to be added to complete the Lewis structure shown in the implied image (the image is not present, so this answer is based on typical structures).

Question 29

• The Lewis structure for CS2 is

Question 30

• The Lewis structure for a chlorate ion, ClO3–, should show 3 single bonds, 0 double bonds, and 9 lone pairs.

Question 31

• Without an image, it is impossible to verify which Lewis structures are incorrect.

Question 32

• Without an image, it is impossible to verify which Lewis structures are incorrect.

Question 33

• In the resonance structure of sulfur dioxide, where the sulfur atom has one single bond and one double bond, the formal charge of the sulfur atom is +1.

Question 34

• The formal charge on phosphorus in a Lewis structure for the phosphate ion that satisfies the octet rule is +1.

Question 35

• In the Lewis structure of the iodate ion, IO3–, that satisfies the octet rule, the formal charge on the central iodine atom is +2.

Question 36

• In the Lewis structure for ClO3F, chlorine has a formal charge of +1 and an oxidation number of +7.

Question 37

• The number of resonance structures for the sulfur dioxide molecule that satisfy the octet rule is 2.

Question 38

• The best Lewis structure a resonance structure: I3– (C = central atom).

Question 39

• The central atom likely to violate the octet rule is XeO3.

Question 40

• Molecules with a central atom that does not follow the octet rule: (2) BCl3, (4) SF4.

Question 41

• The correct equation to calculate formal charge on an atom is: formal charge = valence electrons – lone pair electrons – 1/2 bonding electrons

Question 42

• For the reaction O(g) + O2(g) → O3(g)$, with$\,Delta H° = -106.3 kJ/mol, the average bond enthalpy in O3 is not directly calculable from the information given. This requires more complex calculations involving bond enthalpies of reactants and products.

Question 43

• Without the Bond energies table and the Thermodynamic properties of pure substances table, it is impossible to calculate:
  Part 1 of 2 Predict the enthalpy of reaction from the average bond enthalpies.
  Part 2 of 2 Calculate the enthalpy of reaction from the standard enthalpies of formation of the reactant and product molecules.

Question 44

• To find the standard enthalpy of formation of liquid octane, use the equation:
  \Delta H°{rxn} = \Sigma n \Delta H°f(products) - \Sigma n \Delta H°_f(reactants)
• -1.0940 × 10^4 kJ/mol = [16(-393.5) + 18(-285.8)] - [2(\Delta H°f(C8H_{18})) + 25(0)]
• -1.0940 × 10^4 = [-6296 - 5144.4] - [2(\Delta H°f(C8H_{18}))]
• -1.0940 × 10^4 = -11440.4 - 2(\Delta H°f(C8H_{18}))
• 2(\Delta H°f(C8H_{18})) = -11440.4 + 10940
• 2(\Delta H°f(C8H_{18})) = -500.4
• \Delta H°f(C8H_{18}) = -250.2 kJ/mol

Question 45

• To find the standard enthalpy of formation of solid glycine, use the equation:
  \Delta H°{rxn} = \Sigma n \Delta H°f(products) - \Sigma n \Delta H°_f(reactants)
• -3896 kJ/mol = [8(-393.5) + 10(-285.8) + 2(0)] - [4(\Delta H°f(C2H5O2N)) + 9(0)]
• -3896 = [-3148 - 2858] - [4(\Delta H°f(C2H5O2N))]
• -3896 = -6006 - 4(\Delta H°f(C2H5O2N))
• 4(\Delta H°f(C2H5O2N)) = -6006 + 3896
• 4(\Delta H°f(C2H5O2N)) = -2110
• \Delta H°f(C2H5O2N) = -527.5 kJ/mol

Question 46

• The equation that has a \,Delta H{rxn}$that is not equal to$\,Delta H°f$of the product is:$O2(g) + H2(g) → H2O2(g)

Question 47

• Using Hess's Law:
• Given:
  • 6C(graphite) + 3H2(g) → C6H6(l) \Delta H°{rxn} = +49.0 kJ/mol
  • C(graphite) + O2(g) → CO2(g) \Delta H°_{rxn} = -393.5 kJ/mol
  • H2(g) + \frac{1}{2}O2(g) → H2O(l) \Delta H°{rxn} = -285.8 kJ/mol
• Target:
  • C6H6(l) + \frac{15}{2} O2(g) → 6CO2(g) + 3H_2O(l)
• Reverse the first reaction and multiplying the second reaction by 6 and the third reaction by 3:
  • C6H6(l) → 6C(graphite) + 3H2(g) \Delta H°{rxn} = -49.0 kJ/mol
  • 6C(graphite) + 6O2(g) → 6CO2(g) \Delta H°_{rxn} = 6(-393.5) = -2361 kJ/mol
  • 3H2(g) + \frac{3}{2}O2(g) → 3H2O(l) \Delta H°{rxn} = 3(-285.8) = -857.4 kJ/mol
• Add the reactions to get the target and add the enthalpies:
  \Delta H°_{rxn} = -49.0 - 2361 - 857.4 = -3267.4 kJ/mol

Question 48

• Using Hess's Law:

  • 2C8H{18}(l) + 25O2(g) → 16CO2(g) + 18H2O(l) \Delta H°{rxn} = -11020. kJ/mol
  • 2CO(g) + O2(g) → 2CO2(g) \Delta H°_{rxn} = -566.0 kJ/mol

• Target Reaction:

  • 2C8H{18}(l) + 17O2(g) → 16CO(g) + 18H2O(l)

• Manipulate the equations:
  The first reaction is already in the correct orientation.
  Reverse and multiply the second reaction by 8:

  • 8CO2(g) → 8CO(g) + 4O2(g) \Delta H°_{rxn} = 8(566.0) = 4528 kJ/mol
  • However, since this new equation creates 8 CO2, we would need to double it to eliminate the CO2, leading to a total of 16 CO_2.
  • Thus we need to consider the number of moles of CO2 to eliminate by doubling the equation: 16CO2(g) → 16CO(g) + 8O2(g) \Delta H°{rxn} = 1.183404*10^{-25} (566.0) = 9056 kJ/mol

Thus

• \,Delta H°_{rxn} = -11020 + 9056 = -1964
• \,Delta H°_{rxn} = -1964 kJ/mol

Question 49

• Barium (Ba) would be expected to have properties similar to calcium because they are in the same group (Group 2, the alkaline earth metals).

Question 50

• Krypton (Kr) would be expected to have properties similar to argon because they are both noble gases (Group 18).

Question 51

• The general electron configuration for the outermost electrons of elements in the alkaline earth group is ns2.

Question 52

• A carbon atom has 4 valence electrons.

Question 53

• Atomic radius decreases moving from left to right across a period and increases from top to bottom.

Question 54

• Cesium (Cs) has the largest radius among the choices provided.

Question 55

• Chlorine (Cl) has the highest first ionization energy among the choices provided.

Question 56

• The equation that correctly represents the process relating to the ionization energy of X is: X(g) → X^+(g) + e^–

Question 57

• Chlorine (Cl) has the greatest electron affinity among the choices provided.

Question 58

• The equation that correctly represents the process involved in the electron affinity of X is: X(g) + e^– → X^–(g)$$

Question 59

• Fluorine (F) has the smallest atomic radius among the Group 17 elements.

Question 60

• Anions of Group 17 elements are isoelectronic with Group 18.