CHEM 210 EXAMS 1 to 8 2026 2027
GRADE A+ DETAILED CORRECT
VERIFIED ANSWERS WITH
RATIONALES | INSTANT DOWNLOAD
CHEM 210: General Chemistry / Organic Chemistry
Foundations
Exam Prep:
1. A sample of an unknown pure element is found to consist of
two naturally occurring isotopes. Isotope A has a mass of
62.929 amu and a relative abundance of 69.15%, while Isotope
B has a mass of 64.928 amu and a relative abundance of
30.85%. What is the identity of this element based on its
calculated average atomic mass?
A. Zinc (Zn)
B. Copper (Cu)
C. Nickel (Ni)
D. Gallium (Ga)
Rationale: The average atomic mass is calculated using the weighted
average formula: \(\text{Average Mass} = (62.929 \times 0.6915) +
(64.928 \times 0.3085) = 43.515 + 20.030 = 63.545\text{ amu}\).
Looking at the periodic table, the element with an atomic mass closest
to 63.55 amu is Copper (Cu).
Correct Answer: B
2. An organic chemistry student is evaluating the structural
isomers of a hydrocarbon with the molecular formula
\(C_{5}H_{12}\). Which structural representation
corresponds to the isomer with the lowest boiling point due to
highly spherical branching?
A. n-pentane
B. Isopentane (2-methylbutane)
,C. Neopentane (2,2-dimethylpropane)
D. Cyclopentane
Rationale: Branching decreases the surface area of a molecule, which
minimizes the strength of London dispersion forces acting between
adjacent molecules. Neopentane is highly branched and compact
(spherical), giving it the weakest intermolecular interactions and the
lowest boiling point among the \(C_{5}H_{12}\) structural isomers.
Correct Answer: C
3. What is the correct systematic IUPAC name for a
coordination compound with the molecular formula
\([Co(NH_3)_5Cl]Cl_2\)?
A. Cobalt pentammine chloride dichloride
B. Pentamminechlorocobalt(III) chloride
C. Pentamminedichlorocobalt(II) chloride
D. Chloropentammocobaltate(III) chloride
Rationale: According to IUPAC nomenclature rules for coordination
complexes, ligands are listed alphabetically before the central metal
atom. Ammine (\(NH_{3}\)) comes before chloro (\(Cl^{-}\)). The
oxidation state of the metal is denoted by a Roman numeral in
parentheses. Here, Cobalt must balance three negative charges (one
coordinated chloride and two external counter-ion chlorides), giving it
an oxidation state of +3.
Correct Answer: B
4. A gas sample occupies a volume of 2.50 Liters at a pressure
of 740.0 Torr and a temperature of 25.0°C. If the gas is
compressed to a volume of 1.20 Liters while the temperature is
simultaneously elevated to 50.0°C, what is the final pressure
of the gas sample?
A. 336 Torr
B. 1432 Torr
C. 1671 Torr
D. 1280 Torr
Rationale: Using the combined gas law, \((P_1 \times V_1) / T_1 =
(P_2 \times V_2) / T_2\), where temperatures must be converted to
Kelvin (\(T_1 = 298.15\text{ K}\), \(T_2 = 323.15\text{ K}\)). Solving
,for \(P_{2}\): \(P_2 = (740.0 \times 2.50 \times 323.15) / (1.20 \times
298.15) = 597827..78 = 1671\text{ Torr}\).
Correct Answer: C
5. Which molecule possesses a molecular geometry that is
classified as see-saw based on Valence Shell Electron Pair
Repulsion (VSEPR) theory?
A. \(SF_{6}\)
B. \(CH_{4}\)
C. \(SF_{4}\)
D. \(XeF_{4}\)
Rationale: Sulfur tetrafluoride (\(SF_{4}\)) has 34 valence electrons,
yielding 5 electron domains around the central sulfur atom (4 bonding
pairs and 1 lone pair). This creates a trigonal bipyramidal electron-
pair geometry. When one equatorial position is occupied by a lone pair,
the resulting molecular shape is classified as see-saw.
Correct Answer: C
6. A chemist mixes 50.0 mL of 0.200 M \(HCl\) with 50.0 mL
of 0.200 M \(NaOH\) inside a coffee-cup calorimeter. The
temperature of the aqueous mixture rises from 22.00°C to
23.35°C. Assuming the solution has a specific heat capacity of
4.184 J/g·°C and a density of 1.00 g/mL, calculate the enthalpy
of neutralization (\(\Delta H_{rxn}\)) in kJ/mol of \(H_{2}O\)
formed.
A. \(-28.2\text{ kJ/mol}\)
B. \(-41.8\text{ kJ/mol}\)
C. \(-56.5\text{ kJ/mol}\)
D. \(-84.1\text{ kJ/mol}\)
Rationale: Total mass = 100.0 g. Heat absorbed by solution \(q = m
\cdot c \cdot \Delta T = 100.0\text{ g} \times 4.184\text{
J/g}\cdot^\circ\text{C} \times (23.35 - 22.00)^\circ\text{C} =
564.84\text{ J} = 0.56484\text{ kJ}\). Moles of \(H_{2}O\) produced =
\(0.0500\text{ L} \times 0.200\text{ M} = 0.0100\text{ mol}\).
Enthalpy \(\Delta H = -q / \text{mol} = -0.56484\text{ kJ} /
0.0100\text{ mol} = -56.5\text{ kJ/mol}\).
Correct Answer: C
, 7. Which set of quantum numbers (\(n, l, m_l, m_s\)) is
physically impossible for an electron residing within a multi-
electron atom?
A. 3, 2, \(-1\), \(+1/2\)
B. 4, 0, 0, \(-1/2\)
C. 2, 2, 1, \(+1/2\)
D. 5, 3, \(-2\), \(-1/2\)
Rationale: The angular momentum quantum number (\(l\)) is strictly
limited to integer values from 0 up to \((n - 1)\). If the principal
quantum number \(n = 2\), the maximum allowable value for \(l\) is 1
(corresponding to a 2p subshell). A value of \(l = 2\) when \(n = 2\) is
physically impossible.
Correct Answer: C
8. According to Molecular Orbital (MO) theory, what is the
calculated bond order and magnetic behavior of the
homonuclear diatomic oxygen ion (\(O_{2}^{+}\))?
A. Bond order = 2.0, Diamagnetic
B. Bond order = 2.5, Paramagnetic
C. Bond order = 1.5, Paramagnetic
D. Bond order = 3.0, Diamagnetic
Rationale: Neutral \(O_{2}\) has 16 valence electrons. \(O_{2}^{+}\)
has 15 valence electrons. Filling the molecular orbitals in order yields
10 bonding electrons and 5 antibonding electrons (as one electron is
removed from an antibonding \(\pi ^{*}\) orbital). \(\text{Bond
Order} = (10 - 5) / 2 = 2.5\). Because it contains an unpaired electron in
its antibonding \(\pi ^{*}\) system, it is paramagnetic.
Correct Answer: B
9. What is the hybridization state of the central nitrogen atom
in the chemical structure of the nitrate ion (\(NO_{3}^{-}\))?
A. \(sp\)
B. \(sp^{2}\)
C. \(sp^{3}\)
D. \(sp^{3}d\)
GRADE A+ DETAILED CORRECT
VERIFIED ANSWERS WITH
RATIONALES | INSTANT DOWNLOAD
CHEM 210: General Chemistry / Organic Chemistry
Foundations
Exam Prep:
1. A sample of an unknown pure element is found to consist of
two naturally occurring isotopes. Isotope A has a mass of
62.929 amu and a relative abundance of 69.15%, while Isotope
B has a mass of 64.928 amu and a relative abundance of
30.85%. What is the identity of this element based on its
calculated average atomic mass?
A. Zinc (Zn)
B. Copper (Cu)
C. Nickel (Ni)
D. Gallium (Ga)
Rationale: The average atomic mass is calculated using the weighted
average formula: \(\text{Average Mass} = (62.929 \times 0.6915) +
(64.928 \times 0.3085) = 43.515 + 20.030 = 63.545\text{ amu}\).
Looking at the periodic table, the element with an atomic mass closest
to 63.55 amu is Copper (Cu).
Correct Answer: B
2. An organic chemistry student is evaluating the structural
isomers of a hydrocarbon with the molecular formula
\(C_{5}H_{12}\). Which structural representation
corresponds to the isomer with the lowest boiling point due to
highly spherical branching?
A. n-pentane
B. Isopentane (2-methylbutane)
,C. Neopentane (2,2-dimethylpropane)
D. Cyclopentane
Rationale: Branching decreases the surface area of a molecule, which
minimizes the strength of London dispersion forces acting between
adjacent molecules. Neopentane is highly branched and compact
(spherical), giving it the weakest intermolecular interactions and the
lowest boiling point among the \(C_{5}H_{12}\) structural isomers.
Correct Answer: C
3. What is the correct systematic IUPAC name for a
coordination compound with the molecular formula
\([Co(NH_3)_5Cl]Cl_2\)?
A. Cobalt pentammine chloride dichloride
B. Pentamminechlorocobalt(III) chloride
C. Pentamminedichlorocobalt(II) chloride
D. Chloropentammocobaltate(III) chloride
Rationale: According to IUPAC nomenclature rules for coordination
complexes, ligands are listed alphabetically before the central metal
atom. Ammine (\(NH_{3}\)) comes before chloro (\(Cl^{-}\)). The
oxidation state of the metal is denoted by a Roman numeral in
parentheses. Here, Cobalt must balance three negative charges (one
coordinated chloride and two external counter-ion chlorides), giving it
an oxidation state of +3.
Correct Answer: B
4. A gas sample occupies a volume of 2.50 Liters at a pressure
of 740.0 Torr and a temperature of 25.0°C. If the gas is
compressed to a volume of 1.20 Liters while the temperature is
simultaneously elevated to 50.0°C, what is the final pressure
of the gas sample?
A. 336 Torr
B. 1432 Torr
C. 1671 Torr
D. 1280 Torr
Rationale: Using the combined gas law, \((P_1 \times V_1) / T_1 =
(P_2 \times V_2) / T_2\), where temperatures must be converted to
Kelvin (\(T_1 = 298.15\text{ K}\), \(T_2 = 323.15\text{ K}\)). Solving
,for \(P_{2}\): \(P_2 = (740.0 \times 2.50 \times 323.15) / (1.20 \times
298.15) = 597827..78 = 1671\text{ Torr}\).
Correct Answer: C
5. Which molecule possesses a molecular geometry that is
classified as see-saw based on Valence Shell Electron Pair
Repulsion (VSEPR) theory?
A. \(SF_{6}\)
B. \(CH_{4}\)
C. \(SF_{4}\)
D. \(XeF_{4}\)
Rationale: Sulfur tetrafluoride (\(SF_{4}\)) has 34 valence electrons,
yielding 5 electron domains around the central sulfur atom (4 bonding
pairs and 1 lone pair). This creates a trigonal bipyramidal electron-
pair geometry. When one equatorial position is occupied by a lone pair,
the resulting molecular shape is classified as see-saw.
Correct Answer: C
6. A chemist mixes 50.0 mL of 0.200 M \(HCl\) with 50.0 mL
of 0.200 M \(NaOH\) inside a coffee-cup calorimeter. The
temperature of the aqueous mixture rises from 22.00°C to
23.35°C. Assuming the solution has a specific heat capacity of
4.184 J/g·°C and a density of 1.00 g/mL, calculate the enthalpy
of neutralization (\(\Delta H_{rxn}\)) in kJ/mol of \(H_{2}O\)
formed.
A. \(-28.2\text{ kJ/mol}\)
B. \(-41.8\text{ kJ/mol}\)
C. \(-56.5\text{ kJ/mol}\)
D. \(-84.1\text{ kJ/mol}\)
Rationale: Total mass = 100.0 g. Heat absorbed by solution \(q = m
\cdot c \cdot \Delta T = 100.0\text{ g} \times 4.184\text{
J/g}\cdot^\circ\text{C} \times (23.35 - 22.00)^\circ\text{C} =
564.84\text{ J} = 0.56484\text{ kJ}\). Moles of \(H_{2}O\) produced =
\(0.0500\text{ L} \times 0.200\text{ M} = 0.0100\text{ mol}\).
Enthalpy \(\Delta H = -q / \text{mol} = -0.56484\text{ kJ} /
0.0100\text{ mol} = -56.5\text{ kJ/mol}\).
Correct Answer: C
, 7. Which set of quantum numbers (\(n, l, m_l, m_s\)) is
physically impossible for an electron residing within a multi-
electron atom?
A. 3, 2, \(-1\), \(+1/2\)
B. 4, 0, 0, \(-1/2\)
C. 2, 2, 1, \(+1/2\)
D. 5, 3, \(-2\), \(-1/2\)
Rationale: The angular momentum quantum number (\(l\)) is strictly
limited to integer values from 0 up to \((n - 1)\). If the principal
quantum number \(n = 2\), the maximum allowable value for \(l\) is 1
(corresponding to a 2p subshell). A value of \(l = 2\) when \(n = 2\) is
physically impossible.
Correct Answer: C
8. According to Molecular Orbital (MO) theory, what is the
calculated bond order and magnetic behavior of the
homonuclear diatomic oxygen ion (\(O_{2}^{+}\))?
A. Bond order = 2.0, Diamagnetic
B. Bond order = 2.5, Paramagnetic
C. Bond order = 1.5, Paramagnetic
D. Bond order = 3.0, Diamagnetic
Rationale: Neutral \(O_{2}\) has 16 valence electrons. \(O_{2}^{+}\)
has 15 valence electrons. Filling the molecular orbitals in order yields
10 bonding electrons and 5 antibonding electrons (as one electron is
removed from an antibonding \(\pi ^{*}\) orbital). \(\text{Bond
Order} = (10 - 5) / 2 = 2.5\). Because it contains an unpaired electron in
its antibonding \(\pi ^{*}\) system, it is paramagnetic.
Correct Answer: B
9. What is the hybridization state of the central nitrogen atom
in the chemical structure of the nitrate ion (\(NO_{3}^{-}\))?
A. \(sp\)
B. \(sp^{2}\)
C. \(sp^{3}\)
D. \(sp^{3}d\)
